STRUCTURES AND COMPUTATIONS IN ANNULAR KHOVANOV
HOMOLOGY
by
CHAMP DAVIS
A DISSERTATION
Presented to the Department of Mathematics
and the Division of Graduate Studies of the University of Oregon
in partial fulfillment of the requirements
for the degree of
Doctor of Philosophy
June 2023
DISSERTATION APPROVAL PAGE
Student: Champ Davis
Title: Structures and Computations in Annular Khovanov Homology
This dissertation has been accepted and approved in partial fulfillment of the
requirements for the Doctor of Philosophy degree in the Department of Mathematics
by:
Robert Lipshitz Chair
Nicolas Addington Core Member
Boris Botvinnik Core Member
Daniel Dugger Core Member
Brittany Erickson Institutional Representative
and
Krista Chronister Vice Provost for Graduate Studies
Original approval signatures are on file with the University of Oregon Division of
Graduate Studies.
Degree awarded June 2023
ii
© 2023 Champ Davis
iii
DISSERTATION ABSTRACT
Champ Davis
Doctor of Philosophy
Department of Mathematics
June 2023
Title: Structures and Computations in Annular Khovanov Homology
Let L be a link in a thickened annulus. In [GLW18], Grigsby-Licata-Wehrli
showed that the annular Khovanov homology of L is equipped with an action of
sl2(∧), the exterior current algebra of the Lie algebra sl2. In this dissertation, we
upgrade this result to the setting of L∞-algebras and modules. That is, we show
that sl2(∧) is an L∞-algebra and that the annular Khovanov homology of L is an
L∞-module over sl2(∧). Up to L∞-quasi-isomorphism, this structure is invariant
under Reidemeister moves.
In proving the above result, we include explicit formulas to compute the
′
higher L∞-operations. Additionally, given a morphism I ∶ L → L of L∞-algebras,
we define a restriction of scalars operation in the setting of L∞-modules and prove
∗ ′
that it defines a functor I ∶ L-mod → L -mod. A more abstract approach to this
problem was recently given by Kraft-Schnitzer.
Finally, computer code was written to aid in the study of the above
L∞-module structure. We discuss various patterns that emerged from these
computations, most notably one relating the torsion in the annular Khovanov
homology groups and the location of the inner boundary of the annulus.
iv
CURRICULUM VITAE
NAME OF AUTHOR: Champ Davis
GRADUATE AND UNDERGRADUATE SCHOOLS ATTENDED:
University of Oregon, Eugene, OR
Boston College, Chestnut Hill, MA
DEGREES AWARDED:
Doctor of Philosophy, Mathematics, 2023, University of Oregon
Bachelor of Science, Mathematics, 2017, Boston College
AREAS OF SPECIAL INTEREST:
Low-dimensional Topology
PROFESSIONAL EXPERIENCE:
Graduate Employee, University of Oregon, 2017-2023
v
ACKNOWLEDGEMENTS
First and foremost, I would like to express my sincere gratitude to my
advisor, Robert Lipshitz, for his invaluable guidance and support throughout my
academic journey. This dissertation would not have been possible without the
countless conversations we had, and I am truly grateful for his mentorship.
I am also deeply grateful to John Baldwin, who introduced me to the subject
of knot theory and patiently taught me while I was an undergraduate student.
Additionally, I would like to extend my thanks to Dan Chambers, Maksym
Fedorchuk, Eli Grigsby, and Bill Keane for all of their encouragement.
To all of my friends and to Tom Heinonen, I am forever grateful for the
incredible memories we have shared over the past six years. Finally, I wish to thank
my family for their unconditional support. Thank you all.
vi
TABLE OF CONTENTS
Chapter Page
I. INTRODUCTION . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1
II. L∞-ALGEBRAS AND MODULES . . . . . . . . . . . . . . . . . . . . 4
2.1. Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4
2.2. Definitions and Examples . . . . . . . . . . . . . . . . . . . . . 6
2.3. Restriction of Scalars . . . . . . . . . . . . . . . . . . . . . . . 19
2.4. Chain Contractions . . . . . . . . . . . . . . . . . . . . . . . . 41
III. ANNULAR KHOVANOV HOMOLOGY . . . . . . . . . . . . . . . . . 53
3.1. Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53
3.2. The Lie algebras sl2, sl2(∧), and sl2(∧)dg . . . . . . . . . . . . 55
3.3. Annular Khovanov homology . . . . . . . . . . . . . . . . . . . 58
3.4. The L∞-algebra structure on sl2(∧) . . . . . . . . . . . . . . . 62
3.5. The L∞-module structure on CKh(L) . . . . . . . . . . . . . . 65
3.6. Reidemeister Moves . . . . . . . . . . . . . . . . . . . . . . . . 66
3.7. The L∞-module structure on AKh(L) . . . . . . . . . . . . . . 91
3.8. Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 92
vii
Chapter Page
IV. COMPUTATIONS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 95
4.1. Overview . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 95
4.2. Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 96
4.3. Observations . . . . . . . . . . . . . . . . . . . . . . . . . . . . 100
APPENDICES
A. RESTRICTION OF SCALARS: COMPOSITION . . . . . . . . . . . 102
B. RESTRICTION OF SCALARS: OBJECTS . . . . . . . . . . . . . . 106
C. RESTRICTION OF SCALARS: MORPHISMS . . . . . . . . . . . . 110
D. RESTRICTION OF SCALARS: FUNCTORIALITY . . . . . . . . . 114
E. TRANSFER OF STRUCTURE VIA CHAIN CONTRACTIONS . . 116
F. ANNULAR KNOT DIAGRAMS . . . . . . . . . . . . . . . . . . . . 121
G. HOMOLOGY CALCULATIONS . . . . . . . . . . . . . . . . . . . . 127
REFERENCES CITED . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 141
viii
LIST OF FIGURES
Figure Page
1 Two knot diagrams. Can the knot on the left be untangled to obtain the
knot on the right? . . . . . . . . . . . . . . . . . . . . . . . . . . . 1
2 A depiction of an (1, 1, 2, 3)-unshuffle in S7. Here σ = (124653)(7),
•
and σ (x1, x2, x3, x4, x5, x6, x7) = (x2, x4, x1, x6, x3, x5, x7). . . . . . . 8
3 A graphical depiction of the generalized Jacobi identity. This should be
•
interpreted as the sum of all compositions lj◦(li⊗Id)◦σ , applied to
the input x1⊗⋯⊗xn. That is, this picture represents ∑i+j=n+1 ∑σ lj◦
(li ⊗ Id) ◦ •σ (x1 ⊗⋯⊗ xn) = 0. . . . . . . . . . . . . . . . . . . . . 10
4 A graphical depiction of the L∞-module homomorphism relation. This should
• • •
be interpreted as ∑hj ◦ (li ⊗ Id) ◦ σ +∑hj ◦ δ ◦ (ki ⊗ Id) ◦ σ =
∑ ′kr ◦ (Id⊗hs) ◦ ( •τ ⊗ Id). . . . . . . . . . . . . . . . . . . . . . . 13
5 A graphical depiction of the composition of two L∞-module homomorphisms.
• •
This should be interpreted as (g ◦ f)n = ∑ gj ◦ δ ◦ (fi ⊗ Id) ◦ σ . . 14
4 The left-hand side represents first unshuffling n elements into two boxes
•
(with the module element by itself) via σ and then unshuffling these
• •
boxes further into r boxes and s boxes via φ and ψ , respectively. The
right-hand side represents first unshuffling n−1 elements into α boxes
• •
via τ and then unshuffling these α boxes via σ . . . . . . . . . . . 39
7 A graphical depiction of the map At. . . . . . . . . . . . . . . . . . . . 42
2
8 A diagram P (L) ⊂ S − {X,O} of an annular link L, where X and O
represent the inner and outer boundaries of the annulus, respectively. 58
9 The various ways the operations of merging and splitting along a crossing
(indicated by a dashed line) interact with a basepoint. The top illustrates
the case of two trivial circles merging into trivial circle (or a trivial circle
splitting into two trivial circles). The middle illustrates the case of a
trivial circle and a nontrivial circle merging into a nontrivial circle (or
a nontrivial circle splitting into a trivial circle and a nontrivial circle).
The bottom illustrates the case of nontrivial circles merging into a trivial
circle (or a trivial circle splitting into two nontrivial circles). . . . . 61
ix
Figure Page
10 The relevant complexes in the proof of RII invariance. A similar diagram
appears in [Bar02]. . . . . . . . . . . . . . . . . . . . . . . . . . . . 69
11 The transferred bracket on (C/C ′)/C ′′′. Here, the labeled edges represent
′
the application of that particular map. For example, k3(x1, x2,m) =
q ◦ k2(x1, H ◦ k2(x2, j(m))) + q ◦ k2(x2, H ◦ k2(x1, j(m))) . . . . . 71
12 The complexes involved in RIII invariance. We have suppressed the degree
shifts. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 74
′ ′
13 The complexes C and D . The w+ means that the trivial circle is labeled
w+. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 75
14 Each strand in the vertex 000 belongs to a circle. Denote these circles by
a, b, and c. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 79
15 This picture shows all possible configurations of the circles a, b, and c. 79
16 The first of four cases with a = b = c. In each case, the labeling of the
circle in 000 must be w−, which forces a labeling of w+ ⊗ w+ in 100. 80
17 The first of three cases with a = b ≠ c. The labeling of the circle in 000
must be w− ⊗ w−, which forces a labeling of w+ in 100. . . . . . . . 81
18 The first of three cases with a ≠ b = c. The labeling of the circle in 000
must be w− ⊗ w−, which forces a labeling of w− ⊗ w+ ⊗ w+ in 100. . 84
19 The first of four cases of a ≠ b ≠ c. The labeling of the circle in 000
must be w− ⊗ w− ⊗ w•, which forces a labeling of w+ ⊗ w• in 100. . 85
20 The relevant part of the RIII cube. If we start with an element in 001, i2 ∶
(C/C ′)/C ′′ → C/C ′ gives a sum of elements in 001 and 100. We then
act by x1, apply the homotopy T , act by x2, and then quotient back
to (C/C ′)/C ′′. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 86
21 The cube of resolutions for the left-handed trefoil knot with basepoint in
the center. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 94
22 A diagram for the knot 73. The arcs are labeled, as well as the possible locations
of the basepoint. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 96
x
Figure Page
23 The result of executing the program for the knot 73. Each row corresponds
to the annular Khovanov homology of 73 with respect to a particular
basepoint. The first row is the ordinary Khovanov homology. The second
row corresponds to basepoint 1 in Figure 22. The third row corresponds
to the homology computed with respect to basepoint 2 in Figure 22,
and so on. The various columns represent the various homological degrees. 97
24 A diagram for the Borromean rings, also known as the link L6a4. The possible
locations for the basepoint are labeled. . . . . . . . . . . . . . . . . 98
25 The result of executing the program for the link L6a4. Each row corresponds
to the annular Khovanov homology of L6a4 with respect to a particular
basepoint. The first row is the ordinary Khovanov homology. The second
row corresponds to basepoint 1 in Figure 24. The third row corresponds
to the homology computed with respect to basepoint 2 in Figure 24,
and so on. The various columns represent the various homological degrees. 98
26 The gradings of the generators in the annular Khovanov homology of the
Borromean rings with basepoint 1, as in Figure 24. The columns represent
the homological gradings and the rows represent the filtration-adjusted
quantum gradings, as described in [GLW18]. Each cell contains the k-
gradings of the generators in a particular homological grading and filtration-
adjusted quantum grading. . . . . . . . . . . . . . . . . . . . . . . . 99
27 The ordinary integral Khovanov homology of the Borromean rings. The
columns represent the homological gradings and the rows represent the
quantum gradings. Each cell contains the homology group present in
that particular homological grading and quantum grading. This data
was computed with Mathematica, using the KnotTheory package [Kno11]100
28 A graphical depiction of the L∞-module relation, as in [Dav22]. . . . . 116
29 31 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 121
30 41 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 121
31 51 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 121
32 52 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 121
33 61 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 122
34 62 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 122
35 63 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 122
xi
Figure Page
36 71 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 122
37 72 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 122
38 73 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 122
39 74 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 123
40 75 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 123
41 76 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 123
42 77 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 123
43 L2a1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 123
44 L4a1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 123
45 L5a1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 124
46 L6a1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 124
47 L6a2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 124
48 L6a3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 124
49 L6a4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 124
50 L6a5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 124
51 L6n1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 125
52 L7a1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 125
53 L7a2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 125
54 L7a3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 125
55 L7a4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 125
56 L7a5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 125
57 L7a6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 126
58 L7a7 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 126
59 L7n1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 126
60 L7n2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 126
xii
CHAPTER I
INTRODUCTION
Knots are familiar objects to almost everyone, and they have been widely
studied throughout history. Mathematically, they are defined as embeddings of the
circle, typically into three-dimensional Euclidean space. We can represent these
embeddings using two-dimensional diagrams if we keep track of the overstrand and
the understrand each time the knot crosses over itself; see Figure 1.
FIGURE 1. Two knot diagrams. Can the knot on the left be untangled to obtain
the knot on the right?
Note that for any particular embedding of a knot, we may move the strands
around to produce a different diagram for the same knot. Because one knot can
have many different embeddings and associated diagrams, it makes sense to study
knots up to isotopy, or, continuous deformation. Determining whether two knots
belong to the same isotopy class is already a challenging question, as illustrated by
Figure 1.
It is useful to associate auxiliary data to the knot that is invariant of the
knot’s isotopy class. In particular, if the auxiliary data is different for two knots,
then the knots themselves had to be different—there is no way to deform one into
1
the other. This data can take many forms. It could be a number or a polynomial;
it could be a group or a topological space. One of the most popular forms of data
is that of a chain complex, which is what we will be studying. We will explore the
Khovanov chain complex, constructed in [Kho00].
Our primary goal is to use the algebraic structure of the Khovanov complex
to understand topological properties of a particular knot or link. In fact, much
can be said just by examining the homology groups of the Khovanov complex. For
example, a knot is the unknot if and only if the (reduced) Khovanov homology has
rank one [KM11]. The Khovanov chain complex has also been used in the proofs
of significant topological results. For example, Rasmussen used it to give a purely
combinatorial proof of the Milnor conjecture [Ras10], which was first proved by
Kronheimer-Mrowka using gauge theory [KM93]. More recently, Piccirillo used
Khovanov homology to show that the Conway knot does not bound a smooth disk
in the 4-ball, a longstanding open question [Pic20].
If our knot is embedded into a thickened annulus, there is a refinement of
Khovanov homology, known as annular Khovanov homology. One of the main
benefits of annular Khovanov homology is that there is additional structure that
is not present in ordinary Khovanov homology. For example, annular Khovanov
homology is an sl2-representation; see [GLW18]. We will be studying the structure
of annular Khovanov homology. In particular, we will be understanding it in terms
of L∞-algebras and modules, structures first appearing in rational homotopy
theory, but recently seen in physics. We will review the background and theory
of L∞-algebras and L∞-modules in Chapter II. In Chapter III, we will review the
construction of Khovanov homology and annular Khovanov homology and show
that both the annular Khovanov chain complex and its homology are L∞-modules.
2
Finally, as the knots we study get large, the Khovanov chain complex
becomes increasingly complex. Computer computation becomes increasingly
necessary to guide our intuition and provide experimental data. In Chapter IV, we
will discuss various patterns that have emerged from computer computation while
studying the annular Khovanov chain complex.
3
CHAPTER II
L∞-ALGEBRAS AND MODULES
2.1. Introduction
The study of L∞-algebras, also known as strong homotopy Lie algebras
or sh-Lie algebras, can be traced back to rational homotopy theory and the
deformations of algebraic structures, where they first appeared in the form of Lie-
Massey operations [All77; Ret85; SS85]. Early applications centered around the
Quillen spectral sequence and rational Whitehead products, and there has been
continued interest in higher order Whitehead products recently; see [Bel+17].
There has also been much interest in L∞-algebras in physics, where Lie algebras
and their representations play a major role. In particular, L∞-algebra structures
have appeared in work on higher spin particles [BBD85], as well as in closed string
theory [WZ92; Zwi93]. Stasheff gives a nice overview in a recent survey article
[Sta19].
Attention has also been given to modules over L∞-algebras. The notion of
an L∞-module was introduced in [LM95], in which the correspondence between
Lie algebra representations and Lie modules was generalized to the L∞ setting.
Moreover, homomorphisms between L∞-modules were developed in [All14].
While it is possible to give a complex an L∞ structure by writing down
explicit formulas, another option is to use homological perturbation theory to
transfer an existing L∞ structure from a different complex. Information on how
to do so can be found in [Hue11; HS02; GLS91], where this idea is referred to as
the homological perturbation lemma, though sometimes it is referred to as the
4
homotopy transfer theorem, as in [LV12; Man10]. An approach using operads was
given in [Ber14], where explicit formulas are written down for the A∞ case. Explicit
formulas for the L∞ case can be found in [Mor22a].
Much of the literature deals with the transfer of L∞-algebra structures;
however, given a map between L∞-algebras, it is natural to want to use this map
to relate their respective categories of modules. In this chapter, we give one explicit
formula to do so, giving a proof of the following:
′ ′
Theorem. Suppose L,L are L∞-algebras over F2 and I ∶ L → L is a map of L∞-
∗ ′
algebras. Then there is an induced functor I ∶ L -mod → L -mod, called restriction
of scalars.
Given an L∞-module homomorphism f ∶ M → N , our definition will satisfy
( ∗ ∗I f)1 = f1. It follows that I preserves quasi-isomorphisms; that is, if M and
∗ ∗
N are quasi-isomorphic, then so too are I M and I N . We also observe that this
generalizes the analagous result in the Lie algebra setting:
′ ′
Corollary. If L and L are Lie algebras, and φ ∶ L → L is a Lie
∗
algebra homomorphism, φ is the usual restriction of scalars for Lie algebra
representations.
Because L∞ modules are defined in the graded setting, keeping track of
signs requires a great deal of care. We will ignore signs and work over F2. As
mentioned in [All14], A∞-modules and maps between them can be reinterpreted
in terms of differential comodules. The analagous reformulation in the L∞ case
is less-understood, but perhaps could facilitate the recording of signs. Moreover,
Kraft-Schnitzer recently gave a more abstract approach to the restriction of
scalars operation in [KS22]. We present an alternative interpretation, and we
5
emphasize that the explicit formulas developed here are of particular interest for
our applications. On the other hand, [KS22] might serve as a guide for how to deal
with signs in the future.
The outline of this chapter is as follows. In section 2, we review the definition
of an L∞-algebra and explain morphisms between them. We provide a similar
exposition for L∞-modules, and we describe how to compose morphisms between
∗
L∞-modules. In section 3, we describe I , the restriction of scalars functor. We
∗
define I on objects and morphisms, and then we prove that it is functorial.
In section 4, we define chain contractions to describe an additional way to
transfer a existing L∞-algebra or L∞-module structures. The appendix includes
supplementary graphics for the proofs presented in this chapter, which contain
somewhat complicated formulas.
2.2. Definitions and Examples
In this section, we review L∞-algebras and explain morphisms between them.
We start by introducting some notation that we will use throughout the rest of this
dissertation.
Definition 1. Let σ ∈ Sn be a permutation. If X is a set, then σ induces a map
• ∶ n n •σ X → X , defined by σ (x1, x2, . . . , xn) = (xσ(1), . . . , xσ(n)). If X is a vector
• ⊗n
space, σ induces a similarly-defined map on the n-fold tensor product σ ∶ X →
⊗n
X .
Definition 2. Fix non-negative integers i1, i2, . . . , ir, with i1 + i2 +⋯ + ir = n. A
permutation σ ∈ Sn is an (i1, i2, . . . , ir)-unshuffle if
σ(1) <⋯ < σ(i1)
6
σ(i1 + 1) <⋯ < σ(i1 + i2)
⋮
σ(i1 +⋯+ ir−1 + 1) <⋯ < σ(i1 +⋯+ ir)
We will denote the set of (i1, i2, . . . , ir)-unshuffles in Sn by S(i1, . . . , ir).
′
Definition 3. We will denote by S (i1, . . . , ir) the set of (i1, i2, . . . , ir)-unshuffles σ
in Sn satisfying i1 ≤ i2 ≤⋯ ≤ ir and σ(i1+⋯+il−1+1) < σ(i1+⋯+il+1) if il = il+1.
This second condition on σ says that the order is preserved when comparing the
′
first elements of blocks of the same size. Indeed, if σ is a (1, 2, 2, 3)-unshuffle in S8,
then i2 = i3 = 2, so the order must be preserved when comparing the first element
of the i2 block to the first element of the i3 block.
Definition 4. We will denote by S(i1, . . . , ir) the set of (i1, i2, . . . , ir)-unshuffles σ
′
in S (i1, . . . , ir) satisfying σ(1) = 1.
n
Definition 5. Let V be a graded vector space. For σ ∈ S and vi ∈ V , let (σ) ∶=
(σ, v1, . . . , vn) be the total Koszul sign of σ. To compute (σ), every time two
xy
elements of degrees x and y are transposed, we record a sign of (−1) , and (σ) is
the total product of such signs. Define χ(σ) ∶= (σ) sgn(σ) to be the product of the
Koszul sign and the sign of the permutation σ.
Remark. Let f ∶ A → B and g ∶ C → D be graded maps of graded algebras. We
will also follow the Kozsul sign convention of including a sign in the evaluation of
the map f ⊗ g. That is, for an element x⊗ y ∈ A⊗ C,
( )( ) = ( )∣x∣∣g∣f ⊗ g x⊗ y −1 f(x)⊗ g(y).
7
Example 1. Figure 2 is an example of a (1, 1, 2, 3)-unshuffle in S7. That is,
σ = ( •124653)(7), and we have drawn a picture describing σ . That is, xσ(1) = x2,
xσ(2) =
•
x4, and so on. The picture describes how σ permutes x1, . . . , x7.
x1 x2 x3 x4 x5 x6 x7
xσ(1) xσ(2) xσ(3) xσ(4) xσ(5) xσ(6) xσ(7)
FIGURE 2. A depiction of an (1, 1, 2, 3)-unshuffle in S7. Here σ = (124653)(7), and
•
σ (x1, x2, x3, x4, x5, x6, x7) = (x2, x4, x1, x6, x3, x5, x7).
In words, a (1, 1, 2, 3)-unshuffle places the numbers 1 through 7 into boxes of
size 1,1,2, and 3, where the order is preserved in each box. In this example, the
resulting boxes would be (2), (4), (1, 6), and (3, 5, 7).
Example 2. A special case of the above definition is if we only have two numbers
in our partition of n. In particular, σ ∈ Sn is a (p, n − p)-unshuffle if σ(k) <
σ(k + 1) whenever k ≠ p. In words, this permutation will place the numbers 1
through n into two boxes, where order is preserved in each. For brevity, we will
sometimes refer to a (p, n − p)-unshuffle as a p-unshuffle if n is clear.
Example 3. In S4, if we use the notation xyzw to denote the permutation
( 1 2 3 4x y z w ), then we can write down the 1, 2, and 3-unshuffles:
1-unshuffles: 1234, 2134, 3124, 4123
2-unshuffles: 1234, 1324, 1423, 2314, 2413, 3412
3-unshuffles: 1234, 1243, 1342, 2341
8
We can now state the definition of an L∞-algebra. We will include the general
definition involving signs, though in the theorems we prove, we will work over F2.
Definition 6. Let V be a graded vector space. An L∞-algebra structure on V is
⊗k
a collection of skew-symmetric multilinear maps {lk ∶ V → V } of degree k − 2.
That is, each lk is skew-symmetric in the sense that
•
lk ◦ σ (x1, x2, . . . , xk) = χ(σ)lk(x1, x2, . . . , xk)
for all σ ∈ Sk and xi ∈ V . These maps also must satsify the generalized Jacobi
identity:
∑ ∑ i(j−1) •χ(σ)(−1) lj ◦ (li ⊗ Id) ◦ σ = 0
i+j=n+1 σ
Here, i ≥ 1, j ≥ 1, n ≥ 1, and the inner summation is taken over all (i, n − i)-
unshuffles.
Remark. If we are working over characteristic two, then these maps are alternating
as well.
•
Remark. We could have also written the skew-symmetry condition as lk ◦ σ = lk
for σ ∈ Sk.
Remark. Another way to write the generalized Jacobi indentity is by using the
notation
∑ ∑ •lj ◦ (li ⊗ Id) ◦ σ = 0
i+j=n+1 σ
Remark. Figure 3 is a depiction of the generalized Jacobi identity.
9
x1 x2 ⋯ xn−1 xn
•
σ
⋯
l ⋯i
lj
FIGURE 3. A graphical depiction of the generalized Jacobi identity. This should
•
be interpreted as the sum of all compositions lj ◦ (li ⊗ Id) ◦ σ , applied to the input
x1⊗⋯⊗xn. That is, this picture represents ∑i+j=n+1 ∑σ lj ◦ (li⊗ Id)◦ •σ (x1⊗⋯⊗
xn) = 0.
Remark. This definition follows the chain complex convention. If instead our L∞-
algebra is a cochain complex, we require each lk to have degree 2 − k. There are
similar cohain complex conventions for the following definitions.
′ ′
Definition 7. Let (L, li) and (L , li) be L∞-algebras. An L∞-algebra
′
homomorphism from L to L is a sequence of skew-symmetric multilinear maps
{ ∶ ⊗n ′fn L → L } of degree n − 1 such that
∑ ∑ • ′ •1 ⋅ fj ◦ (lk ⊗ Id) ◦ σ + ∑ 2 ⋅ lr ◦ (fi ⊗⋯⊗ f1 i ) ◦ τ = 0r
j+k=n+1 σ∈S(k,n−k) ′τ∈S (i1,...,ir)
i1+...+ir=n
( − )+ r(r−1) r−1k j 1 1 +∑ i (r−s)
where 1 = χ(σ)(−1) and  = χ(τ)(−1) 2 s=1 s2 .
Example 4. The n = 2 morphism relation says that
−f1(l2(x1, x2)) + f2(l1(x1), x2) − ( )∣x1∣∣x2∣−1 f2(l1(x2), x1)
′
+ l1(f2( ′x1, x2)) + l2(f1(x1), f1(x2)) = 0
10
When ( ′ ′L, li) and (L , li) are L∞-algebras consisting of elements in degree 0 only,
′
the n = 2 morphism relation simplifies to f1(l2(x1, x2)) − l2(f1(x1), f1(x2)) = 0,
which is just a Lie algebra homomorphism: φ([x1, x2]) = [φ(x1), φ(x2)].
Definition 8. Let (L, lk) be an L∞-algebra. The data of an L∞-module over
L consists of a graded vector space M , together with skew-symmetric multilinear
{ ∶ ⊗n−1maps kn L ⊗M →M ∣ 1 ≤ n <∞} of degree n − 2 satisfying:
∑ ∑ • • •1 ⋅ kq ◦ (lp ⊗ Id) ◦ σ + ∑ ∑ 23 ⋅ kq ◦ δ ◦ (kp ⊗ Id) ◦ σ = 0
p+q=n+1 σ(n)=n p+q=n+1 σ(p)=n
p<n
p(q−1)
where 1 = 2 = χ(σ)(−1) and σ is a p-unshuffle in Sn. In the case of σ(p) = n,
•
we used the skew-symmetry of kq and introduced δ to permute the kp term past
⊗q−1
the remaining elements to ensure that kq ∶ L ⊗M →M . Explicitly,
•
kq( kp(x , . . . , x ), x , . . . , x ) =  ⋅ k (δ ( k (x , . . . , x ), x , . . . , x ))ÍÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒσÒÒÒÒÒÒÒ(ÒÒÒÒÒ1ÒÒÒÒÒÒ)ÒÒÒÒÒÒÒÒÒÒÒÑ ÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒσÒÒÒÒÒÒÒ(ÒÒÒÒÒpÒÒÒÒÒÒ)ÒÒÒÒÒÒÒ Ï σ(p+1) σ(n) 3 q Í ÒÒÒÒpÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒσÒÒÒÒÒÒÒ(ÒÒÒÒÒ1ÒÒÒÒÒÒ)ÒÒÒÒÒÒÒÒÒÒÒÑ ÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒσÒÒÒÒÒÒÒ(ÒÒÒÒÒpÒÒÒÒÒÒ)ÒÒÒÒÒÒÒ Ï σ(p+1) σ(n)
∈M ∈M
= 3 ⋅ kq(xσ(p+1), . . . , xσ(n), kp(x , . . . , x ) )ÍÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒσÒÒÒÒÒÒÒ(ÒÒÒÒÒ1ÒÒÒÒÒÒ)ÒÒÒÒÒÒÒÒÒÒÒÑ ÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒσÒÒÒÒÒÒÒ(ÒÒÒÒÒpÒÒÒÒÒÒ)ÒÒÒÒÒÒÒ Ï
∈M
p n
where 3 = χ(δ) = (− )q−1(− )(p+∑s=1 ∣xσ(s)∣)(∑s=p+1 ∣xσ(s)∣)1 1 .
Example 5. The n = 1 module relation says that M is a chain complex with
differential k1:
k1(k1(m)) = 0
The n = 2 module relation says that the action satisfies the graded Leibniz rule:
−k2(l1(x1), x2) ( )∣x− −1 1∣k2(x1, k1(x2)) + k1(k2(x1, x2)) = 0
11
Using a different notation, we could also write
[ ∣x1∣− ∂x1, x2] − (−1) [x1, ∂x2] + ∂[x1, x2] = 0
to remind us of differential graded Lie algebras. For reference, the n = 3 module
relation is the following.
k3(l1(x1) )) ( )∣x1∣∣x ∣, x2, x3 − −1 2 k3( ( ∣x1∣+∣x ∣l 21 x2), x1, x3) + (−1) k3(x1, x2, k1(x3))
∣x ∣∣x ∣
+k2(l2(x1, x2), x3) + (−1) 1 2 k2(x2, k2(x1, x3)) − k2(x1, k2(x2, x3))
+k1(k3(x1, x2, x3)) = 0
Definition 9. Following [All14], let (L, li) be an L∞-algebra, and let (M,ki) and
( ′ ′ ′M ,ki) be L∞-modules over L. An L∞-module homomorphism from M to M
⊗(n−1) ′
is a collection of skew-symmetric multilinear maps {hn ∶ L ⊗ M → M } of
degree n − 1 satisfying:
∑ ∑ 1 ⋅ hj ◦ ( • • •li ⊗ Id) ◦ σ + ∑ ∑ 2 ⋅ hj ◦ δ ◦ (ki ⊗ Id) ◦ σ
i+j=n+1 σ(n)=n i+j=n+1 σ(i)=n
i<n
∑ ∑ ′ •+ 3 ⋅ kr ◦ (Id⊗hs) ◦ (τ ⊗ Id) = 0
r+s=n+1 τ
i(j−1)+1 ( − n−ss 1)(∑
where  =  = χ(σ)(−1) and  = χ(τ)(−1) t=1 xτ(t))1 2 3 , σ is an i-
unshuffle in Sn, and τ is an (n − s)-unshuffle in Sn−1. Similar to the definition of
L∞-module, we include the permutation δ to ensure the module element is in the
correct location.
Remark. Figure 4 is a depiction of the L∞-module homomorphism relation.
12
x1 x2 ⋯ xn−1 m x1 x2 ⋯ xn−1 m x1 x2 ⋯ xn−1 m
• •
σ σ •τ
⋯ ⋯ ⋯
l ⋯i + k ⋯i = ⋯ hs
′
hj hj kr
FIGURE 4. A graphical depiction of the L∞-module homomorphism relation. This
• • •
should be interpreted as ∑hj ◦ (li ⊗ Id) ◦ σ + ∑hj ◦ δ ◦ (ki ⊗ Id) ◦ σ =
∑ ′kr ◦ (Id⊗hs) ◦ ( •τ ⊗ Id).
Example 6. The n = 1 module homomorphism relation says that h1 is a chain
map: h1k1(m) = ′k1h1(m). Omitting signs, the n = 2 module homomorphism
relation says:
′ ′
h2(l1(x1),m) + h2(x1, k1(m)) + h1(k2(x1,m)) = k2(x1, h1(m)) + k1(h2(x1,m))
Definition 10. The identity map, IdM , of an L∞ module M is defined as follows.
(IdM)1 is the identity map of the underlying graded vector space M , and (IdM)r =
0 for r ≥ 2. It is straightforward to check that this satisfies the definition of an
L∞-module homomorphism.
Definition 11. Let L be an L∞-algebra, and let A,B, and C be L∞-modules over
f g
L. Given L∞-module homomorphisms A −→ B →− C, we define the composition g ◦ f
by
(g ◦ f)n = ∑ ∑ • •gj ◦ δ ◦ (fi ⊗ Id) ◦ σ
i+j=n+1 σ(i)=n
•
where σ is an i-unshuffle in Sn, and λ is the map that permutes the module
element to the final input.
13
x1 x2 ⋯ xn−1 m
•
σ
⋯
f ⋯i
gj
FIGURE 5. A graphical depiction of the composition of two L∞-module
• •
homomorphisms. This should be interpreted as (g ◦ f)n = ∑ gj ◦ δ ◦ (fi ⊗ Id) ◦ σ .
The following Lemma is perhaps well-known, but we do not know a reference
for it. Pictures representing each step in the proof are given in the appendix.
Lemma 1 (Composition). Let (L, li) be an L∞-algebra, and let A,B, and C be
L∞-modules over L, with module operations denoted by ai, bi, and ci, respectively.
f g
Given L∞-module homomorphisms A −→ B →− C, the composition g ◦ f is an L∞-
module homomorphism.
Proof. This follows from the fact that both f and g are L∞-module
homomorphisms. Below, we will apply the L∞-module homomorphism relation
for f , then we will apply the L∞-module homomorphism relation for g, and then we
will conclude the L∞-module homomorphism relation for g ◦ f .
Step 1. The relation that we need to show is
∑ ∑( • •g ◦ f)j ◦ (ai ⊗ Id) ◦ σ = ∑ ∑ cr ◦ (Id⊗(g ◦ f)s) ◦ τ
i+j=n+1 σ r+s=n+1 τ
where σ is an (i, n − i)-unshuffle and τ is an (n − s, s − 1)-unshuffle.
14
Step 2. Break the left-hand side into two parts, and replace (g ◦ f)j with its
definition
∑ ∑ ∑ ∑ • • • •gq ◦ δ ◦ (fp ⊗ Id) ◦ θ ◦ λ ◦ (ai ⊗ Id) ◦ σ
i+j=n+1 σ(i)=n p+q=j+1 θ(p)=j
+ ∑ ∑ ∑ ∑ • • •gq ◦ δ ◦ (fp ⊗ Id) ◦ θ ◦ (li ⊗ Id) ◦ σ
i+j=n+1 σ(n)=n p+q=j+1 θ(p)=j
i<n
•
where δ is the map that permutes the module element to the last input.
• •
Step 3. In the first sum, applying σ and θ results in a block of size i being
inputted to ai, a block of size p − 1 being inputted into fp, together with the output
of ai, and then a block of size j − p remaining elements (which will be inputted into
•
gq). An equivalent way to achieve this is to first apply a (p+ i− 1)-unshuffle η and
•
then an i-unshuffle ψ . If η(p + i − 1) = n and ψ(i) = p + i − 1, we again obtain a
block of size i being inputted into ai, then a block of size p − 1 being inputted into
fp, together with the output of ai, with j − p elements remaining.
In the second sum, we do the same thing, except the output of li can either
go into the first input of fp or the first input of gq, by the definition of unshuffle. So
we decompose the second sum to reflect these two cases.
∑ ∑ ∑ ∑ • ( ) • • •gq ◦ δ ◦ fp ⊗ Id ◦ λ ◦ (ai ⊗ Id) ◦ (ψ ⊗ Id) ◦ η
i+j=n+1 p+q=j+1 η∈S(p+i−1,j−p) ψ∈S(i,p−1)
η(p+i−1)=n ψ(i)=p+i−1
• • •
+ ∑ ∑ ∑ ∑ gq ◦ δ ◦ (fp ⊗ Id) ◦ (li ⊗ Id) ◦ (ψ ⊗ Id) ◦ η
i+j=n+1 p+q=j+1 η∈S(p+i−1,j−p) ψ∈S(i,p−1)
i<n p>1 η(p+i−1)=n ψ(i)=i
+ ∑ ∑ ∑ ∑ •gq ◦ δ ◦ (li ⊗ fp ⊗ Id) ( •◦ ψ ⊗ Id) •◦ η
i+j=n+1 p+q=j+1 η∈S(p+i,j−p−1) ψ∈S(i,p−1)
i<n η(p+i)=n ψ(i)=i
15
Step 4. Reindex over α = p + i.
n+1
∑ ∑ ∑ ∑ • • • •gn+2−α ◦ δ ◦ (fp ⊗ Id) ◦ λ ◦ (ai ⊗ Id) ◦ (ψ ⊗ Id) ◦ η
α=2 p+i=α η∈S(α−1,n−α+1) ψ∈S(i,α−1−i)
η(α−1)=n ψ(i)=α−1
n+1
+∑ ∑ ∑ ∑ •gn+2−α ◦ δ ◦ ( • •fp ⊗ Id) ◦ (li ⊗ Id) ◦ (ψ ⊗ Id) ◦ η
α=2 p+i=α η∈S(α−1,n−α+1) ψ∈S(i,α−1−i)
1<p,i<n η(α−1)=n ψ(i)=i
n+1
+∑ ∑ ∑ ∑ • • •gn+2−α ◦ δ ◦ (li ⊗ fp ⊗ Id) ◦ (ψ ⊗ Id) ◦ η
α=2 p+i=α η∈S(α,n−α) ψ∈S(i,p−1)
i<n η(α)=n ψ(i)=i
Step 5. Apply the module homomorphism relation for f in the first two sums. In
the third sum, change notation from i to t and from p to s.
n+1
∑ ∑ ∑ ∑ • • •gn+2−α ◦ δ ◦ (bt ⊗ Id) ◦ (Id⊗fs ⊗ Id) ◦ (τ ⊗ Id) ◦ η
α=2 t+s=α η∈S(α−1,n−α+1) τ∈S(t−1,s−1)
η(α−1)=n
n+1
∑ ∑ ∑ ∑ • • •+ gn+2−α ◦ δ ◦ (lt ⊗ fs ⊗ Id) ◦ (ψ ⊗ Id) ◦ η
α=2 t+s=α η∈S(α,n−α) ψ∈S(t,s−1)
t<n η(α)=n ψ(t)=t
Step 6. In the first sum, combine τ ∈ S(t − 1, s − 1) and η ∈ S(α − 1, n − α + 1)
into a single (t−1, s, n−α+1)-unshuffle, denoted by π. In the second sum, combine
16
ψ ∈ S(t, s − 1) and η ∈ S(α, n − α) into a single (t, s, n − α)-unshuffle, denoted by
π.
n+1
∑ ∑ ∑ • •gn+2−α ◦ δ ◦ (bt ⊗ Id) ◦ (Id⊗fs ⊗ Id) ◦ π
α=2 t+s=α π∈S(t−1,s,n−α+1)
π(α−1)=n
n+1
+∑ ∑ ∑ • •gn+2−α ◦ δ ◦ (li ⊗ fs ⊗ Id) ◦ π
α=2 t+s=α π∈S(t,s,n−α)
t<n π(α)=n
Step 7. In the first sum, π unshuffles the n elements into a block of size t − 1, a
block of size s, and a block of size n − α + 1. The block of size s is then inputted
into fs, and then the output of fs is then inputted into bt, as the module element,
with the block of size t − 1.
An equivalent way of achieving this is to apply an (n − s, s − 1)-unshuffle to
the (n − 1)-algebra elements, to form blocks of size (n − s) and s − 1, and then
input the s − 1 algebra elements into fs, with the module element. Then, apply an
•
t-unshuffle σ to these n − s + 1 elements. By requiring σ(t) = n − s + 1, we obtain
a block of size t − 1, plus a module element, that we input into bt. We can do an
analagous reformulation of the second sum.
n+1
∑ ∑ ∑ ∑ •gn+2−α ◦ δ ◦ (bt ⊗ Id) • •◦ σ ◦ (Id⊗fs) ◦ (φ ⊗ Id)
α=2 t+s=α φ∈S(n−s,s−1) σ∈S(t,n−s+1)
σ(t)=n−s+1
17
n+1
∑ ∑ ∑ ∑ • •+ gn+2−α ◦ (lt ⊗ Id) ◦ σ ◦ (Id⊗fs) ◦ (φ ⊗ Id)
α=2 t+s=α φ∈S(n−s,s−1) σ∈S(t,n−s+1)
t<n σ(n−s+1)=n−s+1
n+1 n n+1−s n
Step 8. Reindex, noting that ∑α=2 ∑t+s=α = ∑s=1 ∑t=1 = ∑s=1 ∑x+y=n+2−s.
n
∑ ∑ ∑ ∑ • • •gy ◦ δ ◦ (bx ⊗ Id) ◦ σ ◦ (Id⊗fs) ◦ (φ ⊗ Id)
s=1 x+y=n+2−s φ∈S(n−s,s−1) σ∈S(x,n−s+1)
σ(x)=n−s+1
n
∑ ∑ ∑ ∑ • •+ gy ◦ (lx ⊗ Id) ◦ σ ◦ (Id⊗fs) ◦ (φ ⊗ Id)
s=1 x+y=n+2−s φ∈S(n−s,s−1) σ∈S(x,n−s+1)
x<n σ(n−s+1)=n−s+1
Step 9. Apply the morphism relation for g.
n
∑ ∑ ∑ ∑ ( ) ( •cr ◦ Id⊗gq ◦ κ ⊗ Id) ◦ (Id⊗fs) ◦ ( •φ ⊗ Id)
s=1 r+q=n−s+2 φ∈S(n−s,s−1) κ∈S(r−1,q−1)
Step 10. Combine κ and φ into a single permutation π.
n
∑ ∑ ∑ ( •cr ◦ Id⊗gq) ◦ (Id⊗fs) ◦ (π ⊗ Id)
s=1 r+q=n−s+2 π∈S(r−1,q−1,s−1)
•
Step 11. Split π into τ and ψ. The map λ is needed to permute the module
element into the last input of gq.
18
n
∑ ∑ ∑ ∑ cr ◦ ( Id⊗[ • • •gq ◦ λ ◦ (fs ⊗ Id) ◦ ψ ]) ◦ (τ ⊗ Id)
s=1 r+q=n−s+2 τ∈S(r−1,n−r) ψ∈S(s,q−1)
Step 12. Change how we index over s, r, q.
n
∑ ∑ ∑ ∑ •cr ◦ ( Id⊗[gq ◦ λ ◦ (fs ⊗ Id) •◦ ψ ]) •◦ (τ ⊗ Id)
r=1 s+q=n+2−r τ∈S(r−1,n−r) ψ∈S(s,q−1)
Step 13. Use the definition of g ◦ f .
n
∑ ∑ cr ◦ (Id⊗(g ◦ f) •n+1−r) ◦ (τ ⊗ Id)
r=1 τ∈S(r−1,n−r)
Step 14. This is
∑ ∑ •cr ◦ (Id⊗(g ◦ f)s) ◦ (τ ⊗ Id)
r+s=n+1 τ∈S(r−1,s−1)
2.3. Restriction of Scalars
In this section, we define the restriction of scalars functor on objects, and
we prove that the result is an L∞-module. We then define the restriction of
scalars functor on morphisms, and we prove that the result is an L∞-module
homomorphism. Finally, we complete the proof of functoriality. The end of this
19
section contains a technical lemma that is applied several times throughout the
aforementioned proofs.
′ ′
Lemma 2 (Objects). Suppose I ∶ (L , l ) → (L, l) is a map of L∞-algebras. If
( ) ∗ ′ ′ ′ ⊗n−1M,k is an L-module, then I M ∶= (M,k ) is an L -module, where kn ∶ L ⊗
M →M is given by
n−1
′
kn = ∑ ∑ kr+1 ◦ (Ii ⊗⋯⊗ Ii ⊗ Id) ( •◦ τ ⊗ Id)1 r
r=1 ′τ∈S (i1,...,ir)
i1+...+ir=n−1
Proof. The idea of the proof is straightforward. We will first make a substitution
′
using the definition of k (steps 1-2). We will then use the L∞-algebra
′
homomorphism relation for I to exchange any I and l terms (steps 3-9). The terms
that remain will then cancel by applying the L∞-module relation for k (steps 10-
19). Pictures representing each step in the proof are given in the appendix.
′
Step 1. The L∞ relation for kn that we need to show is zero is:
∑ ∑ ′ ′ • ′ • ′ •kq ◦ (lp ⊗ Id) ◦ σ + ∑ ∑ kq ◦ δ ◦ (kp ⊗ Id) ◦ σ = 0
p+q=n+1 σ(n)=n p+q=n+1 σ(p)=n
p<n
′
Step 2. Focusing only on the first double sum for now, we substitute for kq using
its definition:
∑ ∑ ∑ • ′ •kr+1((Ii ⊗⋯⊗ Ii ⊗ Id) ◦ (τ ⊗ Id) ◦ (l1 r p ⊗ Id) ◦ σ )
p+q=n+1 σ(n)=n
< τ∈
′
S (i1,...,i )p n r1≤r≤q−1
i1+...+ir=q−1
20
′
Step 3. The goal now is to use the morphism relation to commute the lp and I
terms. To do so, we will break down this sum by the specific morphism relation
that we will apply (k = 1, . . . , n−1). In particular, this is determined by the sum of
′
p and the size of the block to which τ sends lp. We will denote the block containing
′
lp by il, and we will denote its size by s.
n−1
∑ ∑ ∑ (( ) ( • ) ( ′ ) •kr+1 Ii ⊗⋯⊗ Ii ⊗ Id ◦ τ ⊗ Id ◦ lp ⊗ Id ◦ σ )1 r
p=1 σ(n)=n ′τ∈S (i1,...,ir)
1≤r≤n−p
i1+...+ir=n−p
n−1 n−p
=∑ ∑ ∑ ∑ • ′ •kr+1((Ii ⊗⋯Ii ⊗⋯⊗ Ii ⊗ Id) ◦ (τ ⊗ Id) ◦ (lp ⊗ Id) ◦ σ )1 l r
p=1 σ(n)=n s=1 ∈ ′τ S (i1,...,ir)
1≤r≤n−p
i1+...+ir=n−p
il=s
We can now reindex over the sum of p and s (on the (p, s)-plane, this is
summing over the diagonal) to obtain
n−1
∑ ∑ ∑ ∑ (( • ′ •kr+1 Ii ⊗⋯⊗Ii ⊗⋯⊗Ii ⊗Id)◦(τ ⊗Id)◦(l ⊗Id)◦σ )1 l r p
k=1 p+s=k+1 σ(n)=n ′τ∈S (i1,...,ir)
1≤r≤n−p
i1+...+ir=n−p
il=s
′
Step 4. Here, we change τ to τ and introduce λ. Since τ is an unshuffle, we can
′
make two observations. First, τ sends lp to the first input of Ii . Second, in thel
partition i1 + . . . + ir = n − p, the block il is the first of its size (i.e. t < l
implies it < il), since the first elements of blocks of the same size are in order. This
′
information allows us to remove lp as an input to τ , and then put it back in the
correct spot after the remaining elements are permuted. That is, τ corresponds to
an (i1, . . . , il − 1, . . . , ir) ′ ′-unshuffle τ in Sn−p−1, and we will send lp to the first input
21
′
of Ii via a permutation λ after we apply τ . Special care is needed when s = 1, inl
′
which case τ ∈ S(0, i2, . . . , ir), and no element will go to the block of size 0.
′
Note: because τ ∈ S (i1, . . . , ir), we had conditions that i1 ≤ ⋯ ≤ ir and that
the order of the first elements among these blocks is preserved. In the rest of the
proof, we must remember these restrictions inherited from τ . We obtain,
n−1
∑ ∑ ∑ ∑
k=1 p+s=k+1 σ(n)=n ′τ ∈S(i1,i2,...,il−1,...,ir)
i1+...+ir=n−p
il=s
kr+1((Ii ⊗⋯⊗ Ii ⊗⋯⊗ Ii ⊗ Id) • ( ′• ′ •◦ λ ◦ Id⊗τ ⊗ Id) ◦ (lp ⊗ Id) ◦ σ )1 l r
′
Step 5. Combine σ and τ into ψ. Now we observe that applying a p-unshuffle and
′
then τ to the remaining inputs is equivalent to doing a (p, i1, . . . , ir)-unshuffle to
all of the inputs at once. We obtain
n−1
∑ ∑ ∑ (( ) • ( ′ ) •kr+1 Ii ⊗⋯⊗ Ii ⊗⋯⊗ Ii ◦ λ ◦ lp ⊗ Id ◦ ψ )1 l r
k=1 p+s=k+1 ψ∈S(p,i1,...,il−1,...,ir,1)
i1+...+ir=n−p
il=s
ψ(n)=n
Step 6. Change from ψ to µ, α, ω. Notice that a (p, i1, . . . , il − 1, . . . ir)-unshuffle is
the same as first doing a (p + il − 1)-unshuffle, and then doing a (p, il − 1)-unshuffle
on the (p+ il− 1)-block and an (i1, . . . , îl, . . . , ir)-unshuffle on the rest. Since we are
fixing il = s, note that p + il − 1 = k.
22
Afterwards, we need to apply a permutation ω to move the strands in the il
block back to their original position between the il−1 and il+1 blocks. That is, ω is
•
the block permuation so that applying ω to the blocks {1, il − 1, i1, . . . , îl, . . . , ir}
yields { • • ′1, i1, . . . , il − 1, . . . ir}. We apply λ after ω to move the lp term.
n−1
∑ ∑ ∑ ∑
k=1 p+s=k+1 µ∈S(k,i1,...,îl,...,ir,1) α∈S(p,k−p)
i1+⋯+ir=n−p
il=s
µ(n)=n
• • ′ • •
kr+1((Ii ⊗⋯⊗ Ii ⊗⋯⊗ Ii ⊗ Id) ◦ λ ◦ ω ◦ (lp ⊗ Id) ◦ (α ⊗ Id) ◦ µ )1 l r
Step 7. Since kr+1 is skew-symmetric, we can move the Ii term to the first input.l
n−1
∑ ∑ ∑ ∑
k=1 p+s=k+1 µ∈S(k,i1,...,îl,...,ir,1) α∈S(p,k−p)
i1+⋯+ir=n−p
il=s
µ(n)=n
′ • •
kr+1((Ii ⊗ Ii ⊗⋯⊗ Îl 1 i ⊗⋯⊗ Ii ⊗ Id) ◦ (lp ⊗ Id) ◦ (α ⊗ Id) ◦ µ )l r
Step 8. Rewrite the maps as
n−1
∑ ∑ ∑ ∑
k=1 p+s=k+1 µ∈S(k,i1,...,îl,...,ir,1) α∈S(p,k−p)
i1+⋯+ir=n−p
il=s
µ(n)=n
′ • •
kr+1([Ii ◦ (l ⊗ Id) ◦ α ]⊗ [(I ⊗⋯⊗ Î ⊗⋯⊗ I ⊗ Id)] ◦ µ )l p i1 il ir
23
′
Step 9. Apply the L∞-algebra homomorphism relation to the terms Ii ◦ (ll p ⊗ Id) ◦
•
α . Since we no longer are keeping track of p, we also use the fact that p+ s = k + 1
to rewrite the conditions for µ.
n−1
∑ ∑ ∑ ∑
k=1 1≤t≤k γ∈S ′(a1,...,a= t) µ∈S(k,i1,...,îl,...,ia +...+a k r,1)1 t
a ≥1 i1+⋯îl+⋯+ir=n−1−kr
µ(n)=n
kr+1([lt ◦ ( • •Ia ⊗⋯⊗ I1 a ) ◦ γ ]⊗ [(Ii ⊗⋯⊗ Î ⊗⋯⊗ I ⊗ Id)] ◦ µ )t 1 il ir
Step 10. Rewrite the maps as
n−1
∑ ∑ ∑ ∑
k=1 1≤t≤k γ∈S ′(a
= 1
,...,at) µ∈S(k,i1,...,îl,...,i ,1)a1+...+at k r
a ≥1 i1+⋯îl+⋯+ir=n−1−kr
µ(n)=n
• •
kr+1((lt ⊗ Id) ◦ (Ia ⊗⋯⊗ Ia ⊗ I ⊗⋯⊗ Î ⊗⋯⊗ I ) ◦ (γ ⊗ Id) ◦ µ )1 t i1 il ir
Step 11. We can combine µ and γ into one permutation η. Indeed, applying
µ and then an (a1, . . . , at)-unshuffle on the k-block is the same as applying an
(a1, . . . , at, i1, . . . , îl, . . . , ir, 1)-unshuffle all at once.
24
n−1
∑ ∑ kr+1(( •lt⊗Id)◦(Ia ⊗⋯⊗Ia ⊗Ii ⊗⋯⊗ Îi ⊗⋯⊗Ii ⊗Id)◦η )1 t 1 l r
k=1 η∈S(a1,...,at,i1,...,îl,...ir,1)
1≤t≤k
a1+...+at=k
i1+⋯+îl+⋯+ir=n−1−k
η(n)=n
Step 12. Since k = 1, . . . , n − 1, we can drop the sum over k from the notation
and just require that a1, . . . , at, i1, . . . , ir is a partition of n − 1, with a1 ≤ . . . ≤ at,
i1 ≤ . . . ≤ ir, and t ≥ 1 and r ≥ 1. If we fix η ∈ S(a1, . . . , at, i1, . . . , îl, . . . , ir), we
don’t have any relation between the two partitions a1 ≤ . . . ≤ at and i1 ≤ . . . ≤ ir.
That is, the sizes of the blocks are in order as part of their respective partitions,
but it might not be the case that a1, . . . , at, i1, . . . , ir is in increasing order as a
whole. However, from these two partitions, we can use an unshuffle to construct a
new partition where the sizes of the boxes are in order. Indeed, define σ so that
( −1σ )• arranges the a1, . . . , at, i1, . . . , ir in increasing order (to get a unique σ,
require that the order of the a’s is preserved, the order of the i’s is preserved,
and that, using η, the first elements of boxes of same size are in order). Then let
• −1
c1, . . . , cα ∶= (σ ) (a1, . . . , at, i1, . . . , ir). To summarize, what we have done is
define a new partition c1, . . . , cα of n − 1 so that cσ(1) = a1, . . . , cσ(t) = at, cσ(t+1) =
i1, . . . , cα = ir. Of course, since a1 ≤ . . . ≤ at, σ is a t-unshuffle. Moreover, we define
τ by requiring that the elements that η puts into the a1, . . . , at and i1, . . . , ir-boxes
are precisely those that τ puts into the cσ(1), . . . , cσ(t) and cσ(t+1), . . . , cα-boxes,
respectively. Finally, since α = t + r − 1, we relabeled kr+1 as kα+2−t. Note that
we can reverse this whole construction to obtain an inverse correspondence. This
process is similar to Lemma 4.
25
∑ ∑ kα+2−t ◦ ( • •lt ⊗ Id) ◦ σ ◦ (Ic ⊗⋯⊗ Ic ⊗ Id) ◦ (τ ⊗ Id)1 α
∈ ′τ S (c ,...,c ) σ∈S(t,α+1−t)1 α
c1+...+cα=n−1 σ(α+1)=α+1
1≤α≤n−1 1≤t≤α+1
Step 13. On the other hand, we now examine the second term in the original sum:
∑ ∑ ′ • ′ •kq ◦ δ ◦ (kp ⊗ Id) ◦ σ
p+q=n+1 σ(p)=n
′ ′ ′
Step 14. Use the definition of k to substitute for kp and kq. The cases p = 1 and
q = 1 require some care; they correspond to the cases r = 0 and s = 0, respectively.
If r = 0, then φ = Id, and if s = 0, then ψ = Id. We also disallow r and s to be zero
simultaneously.
∑ ∑ ∑ ∑ ( ) •ks+1 ◦ Ij ⊗⋯⊗ Ij ⊗ Id ◦ (ψ ⊗ Id) ◦ δ1 s
p+q=n+1 σ(p)=n ′ ′
≤ ≤ φ∈S (i1,...,ir) ψ∈S (j1,...,j )1 p n s0≤r≤p−1 0≤s≤n−p
i1+...+ir=p−1 j1+...+js=n−p
◦ ( •kr+1 ⊗ Id) ◦ (Ii ⊗⋯⊗ Ii ⊗ Id ) ◦ (φ ⊗ Id) •◦ σ1 r
Step 15. Commuting composition and tensor product, and replacing δ with an
′
analogous δ that ensures the module element is in the correct spot, we get
26
∑ ∑ ∑ ∑ ′•ks+1 ◦ δ ◦ (kr+1 ⊗ Id)
p+q=n+1 σ(p)=n
≤ ≤ φ∈
′
S (i1,...,ir) ψ∈ ′S (j1 p n 1,...,js)0≤r≤p−1 0≤s≤n−p
i1+...+ir=p−1 j1+...+js=n−p
( • • •◦ Ii ⊗⋯Ii ⊗ Id⊗Ij ⊗⋯⊗ I1 r 1 j ) ◦ (φ ⊗ Id⊗ψ ) ◦ σs
Step 16. Instead of summing over r and s separately, we can sum over the
diagonal α = r + s.
n
∑ ∑ ∑ ∑ ∑
p=1 σ(p)=n 1≤α≤n−1 ′ ′
+ = φ∈S (i ,...,i ) ψ∈S (j ,...,j )r s α 1 r 1 s
r,s≥0 i1+...+ir=p−1 j1+...+js=n−p
′•
ks+1 ◦ δ ◦ (kr+1 ⊗ Id) ( ) ( • • •◦ Ii ⊗⋯Ii ⊗ Id⊗Ij ⊗⋯⊗ Ij ◦ φ ⊗ Id⊗ψ ) ◦ σ1 r 1 s
Step 17. Apply Lemma 4, where r + 1 above corresponds to t below.
∑ ∑ ′•kα+2−t◦δ ◦( • •kt⊗ Id)◦(σ ⊗ Id)◦(Ic ⊗⋯⊗Ic ⊗ Id)◦(τ ⊗ Id)1 α
∈ ′τ S (c1,...,c ) σ∈S(t,α+1−t)α
c1+...+cα=n−1 σ(t)=α+1
1≤t≤α+1
Step 18. Summarizing what we’ve done so far, we’ve shown that the original sum
∑ ∑ ′ ′kq ◦ (lp ⊗ Id) •◦ σ
p+q=n+1 σ(n)=n
+
∑ ∑ ′ ( ′ •kq ◦ Id⊗kp) ◦ σ
p+q=n+1 σ(p)=n
27
can be rewritten as
∑ ∑ • •kα+2−t ◦ (lt ⊗ Id) ◦ σ ◦ (Ic ⊗⋯⊗ Ic ⊗ Id) ◦ (τ ⊗ Id)1 α
′
τ∈S (c1,...,cα) σ∈S(t,α+1−t)
c1+...+cα=n−1 σ(α+1)=α+1
1≤t≤α+1
+ ∑ ∑ ′•kα+2−t ◦ δ ◦ (kt ⊗ Id) •◦ σ ◦ ( •Ic ⊗⋯⊗ Ic ⊗ Id) ◦ (τ ⊗ Id)1 α
∈ ′τ S (c1,...,c ) σ∈S(t,α+1−t)α
c1+...+cα=n−1 σ(t)=α+1
1≤t≤α+1
Step 19. Letting F = (Ic ⊗⋯⊗ •Ic ⊗ Id) ◦ (τ ⊗ Id) and setting u = α+ 2− t, this1 α
becomes
∑ ∑ ∑ ( ′ ) •ku ◦ lt ⊗ Id ◦ σ ◦ F
τ∈ ′S (c ,...,c ) t+u=α+2 σ∈S(t,α+1−t)1 α
c1+...+cα=n−1 σ(α+1)=α+1
′• •
+ ∑ ∑ ∑ ku ◦ δ ◦ (kt ⊗ Id) ◦ σ ◦ F
τ∈ ′S (c1,...,c ) t+u=α+2 σ∈S(t,α+1−t)α
c1+...+cα=n−1 σ(t)=α+1
which cancel by the module relation.
′
Lemma 3 (Morphisms). Suppose L and L are L∞-algebras and M and N are L-
′
modules. Let I ∶ L → L be an L∞-algebra homomorphism, and let f ∶ M → N be
∗
an L-module homomorphism. Set (I f)1 = f1, and for n ≥ 2, define
( ∗ ) ′ ⊗n−1 ∗ ∗I f n ∶ (L ) ⊗ I M → I N
28
by
n−1
( ∗ •I f)n = ∑ ∑ fr+1 ◦ (Ii ⊗⋯⊗ Ii ⊗ Id) ◦ (τ ⊗ Id)1 r
r=1 ′τ∈S (i1,...,ir)
i1+...+ir=n−1
∗ ∗ ∗ ′
Then I f ∶ I M → I N is a homomorphism of L -modules.
Proof. We will start by examining the L∞-module homomorphism relation. After
∗ ′
replacing I f and mi with their definitions on the left-hand side (steps 1-4), we will
rearrange the sum (steps 5-6) and apply the L∞-algebra relation for I (step 7). We
then rewrite the terms (steps 8-9) and apply the module homomorphism relation
for f (step 10). We then show that the result is equal to the right-hand side (steps
11-16).
∗ ∗ ′ ′
Step 1. To start, we will denote the operations of M,N, I M, I N by m,n,m , n
∗ ′
respectively. To show that I f is a homomorphsism of L -modules, we must show
that it satisfies the L∞-module homomorphism relation
∑ ∑( ∗ ′ •I f)j ◦ (mi ⊗ Id) ◦ σ = ∑ ∑ ′ ∗ •nr ◦ (Id⊗(I f)s) ◦ τ
i+j=n+1 σ r+s=n+1 τ
where σ is an (i, n − i)-unshuffle and τ is an (n − s, s − 1)-unshuffle.
Step 2. Focusing only on the left-hand side, we break this sum up into two parts
∑ ∑ ( ∗ • ′ •I f)j ◦ λ ◦ (mi ⊗ Id) ◦ σ
i+j=n+1 σ(i)=n
+ ∑ ∑ ( ∗ ) ( ′ •I f j ◦ li ⊗ Id) ◦ σ
i+j=n+1 σ(n)=n
i<n
where we use skew-symmetry and introduce the permutation λ to insert the module
element in the correct spot.
29
∗
Step 3. Replace I f with its definition. Note that we’ve allowed r = 0 in the first
′ •
sum to include the case j = 1, which corresponds to f1 ◦ mn ◦ σ . If j is anything
but 1, r = 0 makes no contribution to the sum.
j−1
∑ ∑ ∑ ∑ • • ′ •fr+1 ◦ (Ii ⊗⋯⊗ Ii ⊗ Id) ◦ (τ ⊗ Id) ◦ λ ◦ (m ⊗ Id) ◦ σ1 r i
i+j=n+1 σ(i)=n r=0 τ∈ ′S (i1,...,ir)
i1+...+ir=j−1
j−1
+ ∑ ∑ ∑ ∑ fr+1 ◦ (Ii ⊗⋯⊗ Ii ⊗ Id) ◦ ( • ′ •τ ⊗ Id) ◦ (li ⊗ Id) ◦ σ1 r
i+j=n+1 σ(n)=n r=1 ′τ∈S (i1,...,i )
1≤ ri<n i1+...+ir=j−1
′
Step 4. Now focus on the first sum and replace mi with its definition. Similar
to the above, we’ve allowed for the case s = 0 to include the case i = 1, which
∗
corresponds to (I f)n ◦ •λ ◦ ( ′ •m1 ⊗ Id) ◦ σ . If i is anything but 1, s = 0 makes no
contribution to the sum.
j−1 i−1
∑ ∑ ∑ ∑ ∑ ∑ •fr+1 ◦ (Ii ⊗⋯⊗ Ii ⊗ Id) ◦ (τ ⊗ Id)1 r
i+j=n+1 σ(i)=n r=0 ∈ ′ ′τ S (i1,...,i ) s=0r ψ∈S (j1,...,ja)
i1+...+ir=j−1 j1+...+js=i−1
•
◦ λ ◦ [(ms+1 ◦ ( • •Ij ⊗⋯⊗ Ij ⊗ Id) ◦ ψ )⊗ Id] ◦ σ1 s
Step 5. Rewrite the sum by commuting composition and tensor product and
considering the diagonal α = r + s instead of r and s individually. Observe that
one of r and s can be 0, but not both at the same time.
30
∑ ∑ ∑ ∑ ∑ •fr+1 ◦ ω ◦ (ms+1 ⊗ Id)
i+j=n+1 σ(i)=n 1≤α≤n−1
+ = τ∈
′
S ( ′i1,...,ir s α r) ψ∈S (j1,...,js)
r,s≥0 i1+...+ir=j−1 j1+...+js=i−1
• • •
◦ (Ij ⊗⋯⊗ Ij ⊗ Id⊗I ⊗⋯⊗ I ) ◦ (ψ ⊗ Id⊗τ ) ◦ σ1 s i1 ir
Step 6. Apply Lemma 4 to obtain
∑ ∑ • •fα+2−t ◦ ω ◦ (mt ⊗ Id) ◦ θ ◦ (Ic ⊗⋯⊗ Ic ⊗ Id) ◦ ( •π ⊗ Id)1 α
π∈ ′S (c1,...,c θ∈S(t,α+1−t)α)
c1+...+cα=n−1 θ(t)=α+1
1≤t≤α+1
Step 7. Now, focusing on the l terms (the second sum in Step 3), our goal is to
apply the L∞-algebra relation for I. The steps we follow here are essentially the
same as in Lemma 2 (steps 3-12), and we direct the reader to them for details and
for diagrams. We start with
j−1
∑ ∑ ∑ ∑ • ′fr+1 ◦ (Ii ⊗⋯⊗ Ii ⊗ Id) ◦ (τ ⊗ Id) ◦ (l ⊗ Id) •◦ σ1 r i
i+j=n+1 σ(n)=n r=1 τ∈ ′S (i ,...,i )
1≤i< 1 rn i1+...+ir=j−1
′
Denote the block where li goes by Ii . Break down the sum by il = s.l
j−1 j−1
∑ ∑ ∑∑ ∑ fr+1 ◦ ( •Ii ⊗⋯⊗ Ii ⊗ Id) ◦ (τ ⊗ Id) ◦ ( ′ •li ⊗ Id) ◦ σ1 r
i+j=n+1 σ(n)=n r=1 s=1 τ∈ ′S (i1,...,i≤ < r)1 i n i1+...+ir=j−1
il=s
31
Remove j from the notation.
n−1 n−i n−i
∑ ∑ ∑∑ ∑ •fr+1 ◦ (Ii ⊗⋯⊗ Ii ⊗ Id) ◦ (τ ⊗ Id) ◦ ( ′ •li ⊗ Id) ◦ σ1 r
i=1 σ(n)=n r=1 s=1 τ∈ ′S (i1,...,ir)
i1+...+ir=n−i
il=s
Reindex over the sum of i and s.
n−1
∑ ∑ ∑ ∑ fr+1 ◦ (Ii ⊗⋯⊗ Ii ⊗ Id) ◦ ( •τ ⊗ Id) ′ •◦ (li ⊗ Id) ◦ σ1 r
k=1 i+s=k+1 σ(n)=n ∈ ′τ S (i1,...,ir)
i1+...+ir=n−i
il=s
• ′ ′
Use the map λ to permute li around τ and change τ to τ .
n−1
∑ ∑ ∑ ∑
k=1 i+s=k+1 σ(n)=n ′τ ∈S(i1,...,il−1,...,ir)
i1+...+ir=n−i
il=s
• ′• ′ •
fr+1 ◦ (Ii ⊗⋯⊗ Ii ⊗⋯⊗ Ii ⊗ Id) ◦ λ ◦ (Id⊗τ ⊗ Id) ◦ (li ⊗ Id) ◦ σ1 l r
′
Combine τ and σ into the permutation η.
n−1
∑ ∑ ∑ ∑
k=1 i+s=k+1 ρ∈S(i,il−1) η∈S(i+il−1,i1,...,îl,...,ir,1)
i1+...+ir=n−i
il=s
η(n)=n
fr+1 ◦ ( • • ′ • •Ii ⊗⋯⊗ Ii ⊗⋯⊗ Ii ⊗ Id) ◦ λ ◦ ω ◦ (li ⊗ Id) ◦ (ρ ⊗ Id) ◦ η1 l r
Use skew-symmetry of fr+1 to swap the order of the I’s
n−1
∑ ∑ ∑ ∑ fr+1◦( ′ • •Ii ⊗Ii ⊗⋯⊗Ii ⊗Id)◦(li⊗Id)◦(ρ ⊗Id)◦ηl 1 r
k=1 i+s=k+1 ρ∈S(i,il−1) η∈S(i+il−1,i1,...,îl,...,ir,1)
i1+...+ir=n−i
il=s
η(n)=n
32
Rewrite suggestively, noting that now the Ii is omitted from Ii ⊗⋯⊗ Il 1 i .r
n−1
∑ ∑ ∑ ∑ fr+1( (Ii ◦ ( ′li ⊗ Id) • •◦ ρ )⊗ (Ii ⊗⋯⊗ Ii )⊗ Id ) ◦ ηl 1 r
k=1 i+s=k+1 ρ∈S(i,il−1) η∈S(k,i1,...,îl,...,ir,1)
i1+...+ir=n−i
il=s
η(n)=n
Apply the morphism relations.
n−1
∑ ∑ ∑ ( ( ( ) • •fr+1 lz ◦ It ⊗⋯⊗ It ◦ γ )⊗(Ii ⊗⋯⊗ Ii )⊗Id )◦η1 z 1 r
k=1 ′γ∈S (t1,...,tz) η∈S(k,i1,...,îl,...,ir,1)
t1+...+tz=k i1+...+îl+...+ir=n−1−k
1≤z≤k η(n)=n
Combine γ and η into ψ.
n−1
∑ ∑ •fr+1((lz ⊗ Id) ◦ (It ⊗⋯⊗ It ⊗ Ii ⊗⋯⊗ Ii ⊗ Id) ◦ ψ )1 z 1 r
k=1 ψ∈S(t1,...,tz ,i1,...,îl,...,ir,1)
1≤z≤k
t1+...+tz=k
i1+...+îl+...+ir=n−1−k
ψ(n)=n
This is equivalent to
∑ ∑ fα+2−t ◦ (lt ⊗ Id) •◦ θ ◦ ( •Ic ⊗⋯⊗ Ic ⊗ Id) ◦ (π ⊗ Id)1 α
π∈ ′S (c ,...,c ) θ∈S(t,α+1−t)1 α
c1+...+cα=n−1 θ(α+1)=α+1
1≤α≤n−1 1≤t<α+1
Step 8. In total, combining this with Step 6, we have the sum
∑ ∑ •fα+2−t ◦ ω ◦ (mt ⊗ Id) • •◦ θ ◦ (Ic ⊗⋯⊗ Ic ⊗ Id) ◦ (π ⊗ Id)1 α
∈ ′π S (c ,...,c ) θ∈S(t,α+1−t)1 α
c1+...+cα=n−1 θ(t)=α+1
1≤t≤α+1
33
+ ∑ ∑ fα+2−t ◦ ( ) • ( ) ( •lt ⊗ Id ◦ θ ◦ Ic ⊗⋯⊗ Ic ⊗ Id ◦ π ⊗ Id)1 α
∈ ′π S (c ,...,c ) θ∈S(t,α+1−t)1 α
c1+...+cα=n−1 θ(α+1)=α+1
1≤t<α+1
Step 9. Change notation; change t to i and α + 2 − t to j.
∑ ∑ ∑ • ( ) • ( ) ( •fj ◦ ω ◦ mi ⊗ Id ◦ θ ◦ Ic ⊗⋯⊗ Ic ⊗ Id ◦ π ⊗ Id)1 α
π∈ ′S (c ,...,c ) i+j=α+2 θ∈S(i,α+1−i)1 α
c1+...+cα=n−1 θ(i)=α+1
+ ∑ ∑ ∑ • •fj ◦ (li ⊗ Id) ◦ θ ◦ (Ic ⊗⋯⊗ Ic ⊗ Id) ◦ (π ⊗ Id)1 α
∈ ′π S (c ,...,c ) i+j=α+2 θ∈S(i,α+1−i)1 α
c1+...+cα=n−1 i<α+1 θ(α+1)=α+1
Step 10. Applying the module homomorphism relation for f , we obtain
∑ ∑ ∑ ( •nr ◦ Id⊗fs) ◦ (ρ ⊗ Id) ◦ ( •Ic ⊗⋯⊗ I ⊗ Id) ◦ (π ⊗ Id)1 cα
π∈ ′S (c1,...,c ) r+s=α+2 ρ∈S(α−s,s)α
c1+...+cα=n−1
Step 11. It just remains to show that the sum above is equal to
∑ ∑ ′ ∗nr ◦ (Id⊗(I f)s) ◦ ( •τ ⊗ Id)
r+s=n+1 τ
∗
Therefore, use the definition of I f . Like usual, we start indexing at x = 0 to allow
for the f1 case.
∑ ∑ ∑ ′ [ •nr ◦ Id⊗(fx+1 ◦ [(Ii ⊗⋯⊗ Ii ◦ φ )⊗ Id])] •◦ (τ ⊗ Id)1 x
r+s=n+1 τ∈S(n−s,s) ′φ∈S (i1,...,ix)
i1+...+ix=s−1
0≤x≤s−1
34
′
Step 12. Now use the definition of n . Allow for y = 0 to deal with the n1 case.
s−1 n−s
∑ ∑ ∑ ∑ ∑ ∑ ny+1 ◦ (Ij ⊗⋯⊗ Ij ⊗ Id)◦1 y
r+s=n+1 τ∈S(n−s,s) x=0 φ∈S(i1,...,ix) y=0 γ∈S(j1,...,jy)
i1+...+ix=s−1 j1+...+jy=n−s−1
( •γ ⊗ Id) ◦ [ Id⊗( • •fx+1 ◦ (Ii ⊗⋯⊗ Ii ⊗ Id) ◦ (φ ⊗ Id))] ◦ (τ ⊗ Id)1 x
Step 13. Commute composition and tensor product to rewrite as
s−1 n−s
∑ ∑ ∑ ∑ ∑ ∑
r+s=n+1 τ∈S(n−s,s) x=0 φ∈ ′ ′S (i1,...,i ) y=0x γ∈S (j1,...,jy)
i1+...+ix=s−1 j1+...+jy=n−s−1
• • •
ny+1 ◦ (Id⊗fx+1) ◦ (Ij ⊗⋯⊗ Ij ⊗ Ii ⊗⋯⊗ Ii ⊗ Id) ◦ (γ ⊗ φ ⊗ Id) ◦ (τ ⊗ Id)1 y 1 x
Step 14. Reindex over the diagonal of α = x + y. Observe that one of x and y can
be 0, but not both at the same time.
∑ ∑ ∑ ∑ ∑
r+s=n+1 τ∈S(n−s,s) 1≤α≤n−1 ′ ′
+ = φ∈S (i1,...,ix) γ∈S (j1,...,jy)x y α
x,y≥0 i1+...+ix=s−1 j1+...+jy=n−s−1
ny+1 ◦ (Id⊗fx+1) ◦ (Ij ⊗⋯⊗ Ij ⊗ Ii ⊗⋯⊗ Ii ⊗ Id) • • •◦ (γ ⊗ φ ⊗ Id) ◦ (τ ⊗ Id)1 y 1 x
Step 15. Apply Lemma 4 to get
35
∑ ∑ • •nα+2−s ◦ (Id⊗fs) ◦ θ ◦ (Ic ⊗⋯⊗ Ic ⊗ Id) ◦ (π ⊗ Id)1 α
∈ ′π S (c θ∈S(α−s,s)1,...,cα)
c1+...+c =n−1 1≤s≤αα
1≤α≤n−1
Step 16. Rewriting this as
∑ ∑ ∑ • •nr ◦ (Id⊗fs) ◦ θ ◦ (Ic ⊗⋯⊗ I1 c ⊗ Id) ◦ (π ⊗ Id)α
∈ ′π S (c ,...,c ) r+s=α+2 θ∈S(α−s,s)1 α
c1+...+c =n−1 1≤s≤αα
shows that it is the same as the sum in Step 10, which completes the proof.
′ ′
Theorem 1 (Functoriality). Suppose I ∶ (L , l ) → (L, l) is a map of L∞-algebras.
∗
Then I ∶ ′L -mod→ L -mod is a functor.
Proof. Suppose we have L∞-modules M,N, and Q over L and L∞-module
f g ∗
homomorphisms M −→ N →− Q. We have defined I on objects and morphisms,
∗ ∗ ∗ ∗
so it remains to show that I (IdM) = IdI∗M and that I (g ◦ f) = I g ◦ I f . For the
∗
former, observe that (I (IdM))1 = (IdM)1, and for n ≥ 2,
n−1
( ∗ •I (IdM))n = ∑ ∑ (IdM)r+1 ◦ (Ii ⊗⋯⊗ I1 i ⊗ Id) ◦ (τ ⊗ Id)r
r=1 τ∈ ′S (i1,...,ir)
i1+...+ir=n−1
∗
But (IdM)r = 0 for r > 1, and so we conclude that (I (IdM))n = 0 for n ≥ 2. Hence
∗
I (IdM) = IdI∗M .
∗ ∗ ∗
In remains to show that I (g ◦ f) = I g ◦ I f . We will follow essentially the
same procedure as in Lemma 2, steps 13-17.
36
∗ ∗
Step 1. We start with the right-hand side, and replace [I g ◦ I f]n with its
definition
∑ ∑ ( ∗ ) • ∗ •I g j ◦ λ ◦ ((I f)i ⊗ Id) ◦ σ
i+j=n+1 σ(i)=n
∗ ∗
Step 2. Replace I g and I f with their definitions.
i−1 j−1
∑ ∑ ∑ ∑ ∑ ∑ [ •gs+1 ◦ (Ij ⊗⋯⊗ Ij ⊗ Id) ◦ (ψ ⊗ Id)]1 s
i+j=n+1 σ(i)=n r=0 φ∈ ′( ′S i1,...,i s=0r) ψ∈S (j1,...,js)
i1+...+ir=i−1 j1+...+js=j−1
•
◦ λ ◦ ([ • •fr+1 ◦ (Ii ⊗⋯⊗ Ii ⊗ Id) ◦ (φ ⊗ Id)]⊗ Id) ◦ σ1 r
Note that we include the cases r = 0 and s = 0 to include the cases f1 and g1,
respectively. In particular, r = 0 will contribute a nonzero term only when i = 1,
and s = 0 will only contribute a nonzero term when j = 1.
Step 3. Commute composition and tensor product to rewrite.
i−1 j−1
∑ ∑ ∑ ∑ ∑ ∑
i+j=n+1 σ(i)=n r=0 ∈ ′( ) s=0 ∈ ′φ S i1,...,ir ψ S (j1,...,js)
i1+...+ir=i−1 j1+...+js=j−1
′•
gs+1 ◦ λ ◦ ( • • •fr+1 ⊗ Id) ◦ (Ii ⊗⋯I ⊗ Id⊗I ⊗⋯⊗ I ) ◦ (φ ⊗ Id⊗ψ ) ◦ σ1 ir j1 js
′•
Here, λ is the map that permutes the module element into the last input of gs+1.
Step 4. By Lemma 4, we obtain
37
∑ ∑ ′• • •gα+1−t ◦ λ ◦ (ft+1 ⊗ Id) ◦ θ ◦ (Ic ⊗⋯⊗ Ic ⊗ Id) ◦ (τ ⊗ Id)1 α
∈ ′τ S (c ,...,c ) θ∈S(t+1,α−t)1 α
c1+...+cα=n−1 θ(t+1)=α+1
1≤α≤n−1 0≤t≤α
Step 5. Change notation; let p = t + 1 and q = α + 1 − t.
∑ ∑ ′• • •gq ◦ λ ◦ (fp ⊗ Id) ◦ θ ◦ (Ic ⊗⋯⊗ Ic ⊗ Id) ◦ (τ ⊗ Id)1 α
τ∈ ′S (c ,...,c ) θ∈S(p,α+1−p)1 α
c1+...+cα=n−1 θ(p)=α+1
1≤α≤n−1 1≤p≤α+1
Step 6. By the definition of g ◦ f , this is
∑ ( ) ( •g ◦ f α+1 ◦ Ic ⊗⋯⊗ I1 c ⊗ Id) ◦ (τ ⊗ Id)α
∈ ′τ S (c1,...,cα)
c1+...+cα=n−1
1≤α≤n−1
∗ ∗
Step 7. By the definition of I , this is precisely [I (g ◦ f)]n, as desired.
′ ′
Corollary 1. If L and L are Lie algebras, and φ ∶ L → L is a Lie
∗
algebra homomorphism, φ is the usual restriction of scalars for Lie algebra
representations.
′
Proof. Let ρ ∶ L → gl(M) be a Lie algebra representation. For x ∈ L and m ∈ M ,
the usual restriction of scalars for Lie algebra representations is given by x ⋅ m ∶=
′ ′ ′
φ(x) ⋅m. Indeed, ρ ∶ L → gl(M) defined by ρ (y) = ρ(φ(y)) is a homomorphism of
Lie algebras. Now, regarding φ as an L∞-algebra map with φi = 0 for i ≠ 1, because
38
there are also no higher operations on M as an L∞ L-module, the formulas given in
Lemma 2 for the induced operation simplify to give the usual restriction of scalars
operation described above.
We now prove the technical lemma that was used in the main results above.
In particular, this lemma gives two ways to interpret a particular composition of
unshuffles.
Lemma 4. For a fixed n,
n
∑ ∑ ∑ ∑ ∑ ( • • •Ii ⊗⋯Ii ⊗Id⊗Ij ⊗⋯⊗Ij )◦(φ ⊗Id⊗ψ )◦σ1 r 1 s
p=1 σ(p)=n 1≤α≤n−1 ∈ ′+ = φ S (i1,...,ir) ∈
′
ψ S (j1,...,js)r s α
r,s≥0 i1+...+ir=p−1 j1+...+js=n−p
is the same as
∑ ∑ • •θ ◦ (Ic ⊗⋯⊗ Ic ⊗ Id) ◦ (τ ⊗ Id).1 α
∈ ′τ S (c ,...,c ) θ∈S(r+1,α−r)1 α
c1+...+cα=n−1 θ(r+1)=α+1
0≤r≤α
x1 x2 ⋯ xn−1 m x1 x2 ⋯ xn−1 m
• •
σ τ
⋯ ⋯ ⋯ ⋯ ⋯
•
φ •ψ = Ic I1 c ⋯ I2 cα
⋯ ⋯ ⋯ ⋯
I ⋯ ⋯ •i I1 i Ir j I1 js σ
⋯ ⋯
FIGURE 4. The left-hand side represents first unshuffling n elements into two
•
boxes (with the module element by itself) via σ and then unshuffling these boxes
• •
further into r boxes and s boxes via φ and ψ , respectively. The right-hand side
•
represents first unshuffling n − 1 elements into α boxes via τ and then unshuffling
•
these α boxes via σ .
39
Proof. To see this, it is helpful to examine what the first sum does for a fixed p and
a fixed α. It unshuffles n elements into a box of size p − 1 and a box of size n − p,
with the module element in between. It then unshuffles the box of size p − 1 further
via φ into r smaller boxes and the box of size n − p further via ψ into s smaller
boxes.
So, if we iterate through α = r + s, this sum describes all possible ways of
unshuffling n elements into r boxes (which contain a total of p − 1 elements) and s
boxes (which contain a total number of n − p elements), with the module element
in between. Then, iterating through all possible p tells us that the sum describes
all ways of unshuffling n elements into r + s boxes, with the module element in
between. Note that the r boxes and the s boxes have to be of increasing size when
considered separately, but they need not be in order when considered all together
(e.g. some of the s boxes could be smaller than the last r box).
On the other hand, the second sum unshuffles the n − 1 algebra elements into
α boxes first (here, the boxes are all of increasing size), and then it picks out r of
these via an r-unshuffle θ in Sα. Since there was a module element between the
r boxes and s boxes in the first sum, we can view θ as an (r + 1)-unshuffle in Sα
where it puts the module element after the r boxes. So what we have done is the
same as before: unshuffle n elements into a group of r boxes, a module element,
and a group of s = α − r boxes, where the boxes are of increasing order when
considered separately (but not necessarily when considered all together), see Figure
4. An explicit correspondence between the two sums can be written down using
formulas.
40
2.4. Chain Contractions
It is possible to use an existing L∞-algebra or L∞-module to obtain a new
L∞-structure on a particular chain complex. In this section, we will use chain
contractions to transfer L∞-structures.
Definition 12. Let (A, dA) and (B, dB) be chain complexes. A chain
contraction from A onto B consists of two chain maps q ∶ A → B and i ∶ B → A
of degree 0, together with a homotopy K ∶ A → A of degree 1. That is, we have the
following diagram.
q
K A B
i
These maps q, i, and K must satisfy the following conditions:
q ◦ i = IdB and IdA−i ◦ q = K ◦ dA + dA ◦K
2
K = K ◦ i = q ◦K = 0
We will denote a chain contraction by (A,B, i, q,K).
Remark. If (A, dA) and (B, dB) are cochain complexes, we require ∣K∣ = −1.
′
If L is an L∞-algebra and L is a chain complex, formulas exist in the
′
literature for how to transfer the L∞-algebra structure from L to L , given a chain
′
contraction (L,L , i, q,K). Following [Mor22b, Theorem 1], the chain maps i and q
′ ′
can also be extended to L∞-algebra homomorphisms I ∶ L → L and Q ∶ L → L
′
such that Q ◦ I = IdL′ . The transferred L∞-algebra structure on L is unique up
′
to quasi-isomorphism, and the formula for the transferred bracket {lk} can be given
41
′ ⊗n
inductively as follows. Set Kθ1 = −i and define θn ∶ (L ) → L for n ≥ 2 by
n
•
θn(x1, . . . , xn) =∑ ∑ 1 ⋅ lk(Ii ⊗⋯⊗ Ii ) ◦ σ (x1 k 1, . . . , xn)
k=2 σ∈S(i1,...,ik)
i1+⋯+ik=n
i1≤⋯≤ik
where 1 is given by the Koszul sign convention. Then for all n ≥ 2, we define
′
ln = q ◦ θn and In = K ◦ θn.
We can also use chain contractions to transfer an L∞-module structure. We
will make use of this technique in the proof of the invariance of the sl2(∧)dg L∞-
module structure under Reidemeister moves in Chapter III.
Theorem 2. Let L be an L∞-algebra, and let M be an L∞-module over L. Given a
chain contraction
q
′
T M M
i
′
then M inherits the structure of an L∞-module over L, with transferred bracket
given by
′ •
kn ∶= ∑ q ◦ At ◦ (τ ⊗ i)
τ∈S(i1,...,it)
i1+⋯+it=n−1
∶ ⊗i1⊗⋯⊗ ⊗iwhere At L L
t⊗M →M is defined inductively as follows. Let A1 = ki1+1
and define At = A1 ◦
•
δ2 ◦ [(T ◦At−1)⊗ ] ◦ •Id δ1 , where i1, . . . , it are positive integers;
see Figure 7.
x1 xi x1 i1+1 xn−1 m
⋯ ⋯ ⋯
T T T
ki +1 ki +1 ⋯ k1 2 it+1
FIGURE 7. A graphical depiction of the map At.
42
Remark. The permutations δi in the definition of At above are required to ensure
⊗r
that the module element is the last input of each ki +1 ∶ L ⊗M → M . Explicitly,r
•
δi is the unique permutation so that δi shifts the module element to the required
position and preserves the order of the other elements. For example, in Figure 7, δ1
is the permuation
⎛1 ⋯ i1 i1 + 1 i1 + 2 ⋯ n ⎞
δ1 = ⎜⎜⎜ ⎟⎟⎟
⎝1 ⋯ i1 n i1 + 1 ⋯ n − 1⎠
Throughout the proof of Theorem 2, we will make use of similar permutations λi
to correctly place the module element while preserving the order of the remaining
elements. We will not write down these permutations explicitly, but they can be
readily determined by examining the figures in the appendix.
Remark. We remind the reader that we are ignoring signs in the above theorem
and that the result is proved over a field of characteristic two.
′
Proof. We must show that the above definition for kn satisfies the L∞-module
relation:
∑ ∑ ′ • ′ • ′kq ◦ (lp ⊗ Id) ◦ σ = ∑ ∑ kq ◦ λ ◦ (kp ⊗ Id) •◦ σ
p+q=n+1 σ(n)=n p+q=n+1 σ(p)=n
p<n
′ ′
The idea of the proof is as follows. Start by replacing the kq and kp terms
′
using the definition of kn. Next, apply the L∞-module relation for kn to the terms
involving lp on the left-hand side. After that, use fact that IdM −i ◦ q = k1T + Tk1
to replace terms on the right-hand side. Terms will then cancel in pairs. Graphical
representations of the formulas in this proof are provided in the appendix.
43
Step 1. Focusing on the left-hand side of the L∞-module relation, we can replace
′
kn using its definition to obtain the following sum.
∑ ∑ ∑ • •q ◦ At ◦ (τ ⊗ i) ◦ (lp ⊗ Id) ◦ σ
p+q=n+1 σ(n)=n τ∈S(i1,...,i< t
)
p n i1+⋯+it=n−p
Step 2. We can combine σ and τ into η and ψ. Since τ is an unshuffle, the lp term
will be the first element in some block, which we denote by il. Defining s = p+il−1,
we obtain the following sum.
n−1
∑ ∑ ∑ • •q ◦ At ◦ [Id⊗((lp ⊗ Id) ◦ ψ )⊗ Id] ◦ (η ⊗ i)
p=1 η∈S(i1,...,p+il−1,...,it) ψ∈S(p,s−p)
i1+⋯+it=n−p
1≤t≤n−p
1≤l≤t
Step 3. The goal now is to unpack the At terms using the definition of At in order
to apply the L∞-module relation. Because At only makes sense for t ≥ 1, we break
up the sum into several cases. In the first case, the lp term is in the first box. In
the second case, the lp term is somewhere in the middle, in which case we need at
least three boxes. In the third case, the lp term is in the last box. Note further that
the only way for there to be one box is if p = n − 1. We obtain the following sum.
n−2
∑ ∑ ∑ • •q ◦ At−1 ◦ λ3 ◦ (T ⊗ Id) ◦ [ [ki +1 ◦ (lp ⊗ Id) ◦ ψ ]⊗ Id ]l
p=1 η∈S(p+i1−1,i2,...,it) ψ∈S(p,s−p)
i1+⋯+it=n−p
2≤t≤n−p
l=1
• •
◦ λ1 ◦ (η ⊗ i)
44
n−3
∑ ∑ ∑ •+ q ◦ At−l ◦ λ3 ◦ (T ⊗ Id) •◦ [ [ki +1 ◦ (ll p ⊗ Id) ◦ ψ ]⊗ Id ]
p=1 η∈S(i1,...,p+il−1,...,it) ψ∈S(p,s−p)
i1+⋯+it=n−p
3≤t≤n−p
2≤l≤t−1
• • •
◦ λ2 ◦ [(T ◦ Al−1)⊗ Id] ◦ λ1 ◦ (η ⊗ i)
n−2
∑ ∑ ∑ [ [ ( • •+ q ◦ ki +1 ◦ lp ⊗ Id) ◦ ψ ]⊗ Id ] ◦ λl 2
p=1 η∈S(i1,...,it−1,p+it−1) ψ∈S(p,s−p)
i1+⋯+it=n−p
2≤t≤n−p
l=t
◦ [( • •T ◦ At−1)⊗ Id] ◦ λ1 ◦ (η ⊗ i)
s
∑ ∑ ∑ •+ q ◦ [ [ki +1 ◦ (lp ⊗ Id) ◦ ψ ]⊗ Id ] ◦ (Id⊗i)l
s=n−1 p=1 ψ∈S(p,s−p)
We now reindex over the size of s = p + il − 1.
n−2 s
∑∑ ∑ ∑ •q ◦ At−1 ◦ λ3 ◦ (T ⊗ Id) ◦ [ [ •ki +1 ◦ (lp ⊗ Id) ◦ ψ ]⊗ Id ]l
s=1 p=1 η∈S(s,i2,...,it) ψ∈S(p,s−p)
i1+⋯+it=n−p
2≤t≤n−p
l=1
•
◦ λ1 ◦ ( •η ⊗ i)
n−3 s
∑∑ ∑ ∑ •+ q ◦ At−l ◦ λ3 ◦ (T ⊗ Id) ◦ [ [ki ◦ ( •l ⊗ Id) ◦ ψ ]⊗ Id ]l+1 p
s=1 p=1 η∈S(i1,...,s,...,it) ψ∈S(p,s−p)
i1+⋯+it=n−p
3≤t≤n−p
2≤l≤t−1
• • •
◦ λ2 ◦ [(T ◦ Al−1)⊗ Id] ◦ λ1 ◦ (η ⊗ i)
n−2 s
+∑ ∑ ∑ ∑ q ◦ [ [ •ki +1 ◦ (lp ⊗ Id) ◦ ψ ]⊗ Id ]l
s=1 p=1 η∈S(i1,...,it−1,s) ψ∈S(p,s−p)
i1+⋯+it=n−p
2≤t≤n−p
l=t
•
◦ λ2 ◦ [(T ◦ At−1)⊗ Id] • •◦ λ1 ◦ (η ⊗ i)
s
∑ ∑ ∑ •+ q ◦ [ [ki +1 ◦ (lp ⊗ Id) ◦ ψ ]⊗ Id ] ◦ (Id⊗i)l
s=n−1 p=1 ψ∈S(p,s−p)
45
We can now apply the L∞-module relation.
n−2 s+1
∑∑ ∑ ∑ • ( ) [ [ ( ) •q ◦ At−l ◦ λ3 ◦ T ⊗ Id ◦ ks−p+2 ◦ kp ⊗ Id ◦ ψ ]⊗ Id ]
s=1 p=1 η∈S(s,i2,...,it) ψ∈S(p−1,s−p+1)
i1+⋯+it=n−p
2≤t≤n−p
l=1
• ( •◦ λ1 ◦ η ⊗ i)
n−3 s+1
+∑∑ ∑ ∑ • •q ◦ At−l ◦ λ3 ◦ (T ⊗ Id) ◦ [ [ks−p+2 ◦ (kp ⊗ Id) ◦ ψ ]⊗ Id ]
s=1 p=1 η∈S(i1,...,s,...,it) ψ∈S(p−1,s−p+1)
i1+⋯+it=n−p
3≤t≤n−p
2≤l≤t−1
•
◦ λ2 ◦ [( • •T ◦ Al−1)⊗ Id] ◦ λ1 ◦ (η ⊗ i)
n−2 s+1
+∑ ∑ ∑ ∑ •q ◦ λ3 ◦ (T ⊗ Id) ◦ [ [ •ks−p+2 ◦ (kp ⊗ Id) ◦ ψ ]⊗ Id ]
s=1 p=1 η∈S(i1,...,it−1,s) ψ∈S(p−1,s−p+1)
i1+⋯+it=n−p
2≤t≤n−p
l=t
•
◦ λ2 ◦ [(T ◦ At−1)⊗ Id] • •◦ λ1 ◦ (η ⊗ i)
s+1
•
+ ∑ ∑ ∑ q ◦ [ [ks−p+2 ◦ (kp ⊗ Id) ◦ ψ ]⊗ Id ] ◦ (Id⊗i)
s=n−1 p=1 ψ∈S(p−1,s−p+1)
Step 4. Combine ψ and η into κ, and reintroduce At into the notation, treating
the cases p = 1 and p = s + 1 separately. Indeed, we observe that for 1 < p < s + 1,
we may combine both the kp and ks−p+2 terms into an At term. Otherwise, we will
have a k1 term.
n−2
∑∑ ∑ • • •q ◦ At ◦ λ3 ◦ (k1 ⊗ Id) ◦ λ1 ◦ (κ ⊗ i)
s=1 p=1 κ∈S(s,i2,...,it)
i1+⋯+it=n−p
2≤t≤n−p
l=1
46
n−2 s
∑∑ ∑ • ( ) ( • •+ q ◦ At ◦ λ3 ◦ T ⊗ Id ◦ A1 ⊗ Id) ◦ λ1 ◦ (κ ⊗ i)
s=1 p=2 κ∈S(p−1,s−p+1,i2,...,it)
i1+⋯+it=n−p
2≤t≤n−p
l=1
n−2
∑ ∑ ∑ •+ q ◦ At−1 ◦ λ3 ◦ ((T ◦ k1)⊗ Id) ◦ (A1 ⊗ Id) • •◦ λ1 ◦ (κ ⊗ i)
s=1 p=s+1 κ∈S(s,i2,...,it)
i1+⋯+it=n−p
2≤t≤n−p
l=1
n−3
+∑∑ ∑ • • •q ◦ At−l+1 ◦ λ3 ◦ ((k1 ◦ T )⊗ Id) ◦ (Al−1 ⊗ Id) ◦ λ1 ◦ (κ ⊗ i)
s=1 p=1 κ∈S(i1,...,il−1,s,il+1,...,it)
i1+⋯+it=n−p
3≤t≤n−p
2≤l≤t
n−3 s
∑∑ ∑ • • •+ q ◦ At−l+1 ◦ λ3 ◦ Al ⊗ Id) ◦ λ1 ◦ (κ ⊗ i)
s=1 p=2 κ∈S(i1,...,il−1,p−1,s−p+1,il+1,...,it)
i1+⋯+it=n−p
3≤t≤n−p
2≤l≤t
n−3
∑ ∑ ∑ • • •+ q ◦ At−l ◦ λ3 ◦ ((T ◦ k1)⊗ Id) ◦ (Al ⊗ Id) ◦ λ1 ◦ (κ ⊗ i)
s=1 p=s+1 κ∈S(i1,...,il−1,s,il+1,...,it)
i1+⋯+it=n−p
3≤t≤n−p
2≤l≤t
n−2
+∑∑ ∑ • • •q ◦ A1 ◦ λ3 ◦ ((k1 ◦ T )⊗ Id) ◦ (At−1 ⊗ Id) ◦ λ1 ◦ (κ ⊗ i)
s=1 p=1 κ∈S(i1,...,it−1,s)
i1+⋯+it=n−p
2≤t≤n−p
l=t
n−2 s
∑∑ ∑ • • •+ q ◦ A1 ◦ λ3 ◦ (At ⊗ Id) ◦ λ1 ◦ (κ ⊗ i)
s=1 p=2 κ∈S(i1,...,it−1,p−1,s−p+1)
i1+⋯+it=n−p
2≤t≤n−p
l=t
n−2
+∑ ∑ ∑ • • •q ◦ k1 ◦ λ3 ◦ (At ⊗ Id) ◦ λ1 ◦ (κ ⊗ i)
s=1 p=s+1 κ∈S(i1,...,it−1,s)
i1+⋯+it=n−p
2≤t≤n−p
l=t
•
+ ∑ ∑ ∑ q ◦ ks+1 ◦ (k1 ⊗ Id) ◦ λ1 ◦ (Id⊗i)
s=n−1 p=1 κ=Id
s
∑ ∑ ∑ • •+ q ◦ ks−p+2 ◦ (kp ⊗ Id) ◦ λ1 ◦ (κ ⊗ i)
s=n−1 p=2 κ∈S(p−1,s−p+1)
47
+ ∑ ∑ ∑ • •q ◦ k1 ◦ (ks+1 ⊗ Id) ◦ λ1 ◦ (κ ⊗ i)
s=n−1 p=s+1 κ=Id
Step 5. We can combine the sums above. The first term is obtained by combining
terms 4 and 7 above. The second term is obtained by combining terms 3 and 6
above. The third term is obtained by combining terms 2, 5, 8, and 11 above. The
fourth term is obtained by combining terms 1 and 10. The last term is obtained by
combining terms 9 and 12 above.
n−2
∑ ∑ • • •q ◦ At−l+1 ◦ λ2 ◦ [(k1 ◦ T ◦ Al−1)⊗ Id ] ◦ λ1 ◦ (κ ⊗ i)
s=1 κ∈S(i1,...,il−1,s,il+1...,it)
2≤t≤n−1
2≤l≤t
i1+⋯+it=n−1
il=s
n−2
+∑ ∑ • • •q ◦ At−l ◦ λ2 ◦ [(T ◦ k1 ◦ Al)⊗ Id ] ◦ λ1 ◦ (κ ⊗ i)
s=1 κ∈S(i1,...,il−1,s,il+1...,it)
2≤t≤n−1−s
1≤l≤t−1
i1+⋯+it=n−1−s
il=0
n−1 s
∑ ∑ ∑ •+ q ◦ At−l+1 ◦ λ2 ◦ [(Id⊗Al)⊗ Id ] • ( •◦ λ1 ◦ κ ⊗ i)
s=1 p=2 κ∈S(i1,...,il−1,p−1,s−p+1,il+1,...,it)
1≤t≤n−p
1≤l≤t
i1+⋯+it=n−p
il=s+1−p
n−1
+∑ ∑ •q ◦ At ◦ λ3 ◦ ( • •k1 ⊗ Id) ◦ λ1 ◦ (κ ⊗ i)
s=1 κ∈S(s,i2,...,it)
1≤t≤n−1
l=1
i1+⋯+it=n−1
il=s
n−1
+∑ ∑ • • •q ◦ k1 ◦ λ3 ◦ (At ⊗ Id) ◦ λ1 ◦ (κ ⊗ i)
s=1 κ∈S(i1,...,it−1,s)
1≤t≤n−1−s
l=t
i1+⋯+it=n−1−s
il=0
48
Step 6. Change notation. In the first sum, let c1, . . . , cw be i1, . . . , il−1 and
d1, . . . , dx be s, il+1, . . . , it. The conditions t ≥ 2 and 2 ≤ l ≤ t imply that
w, x ≥ 1. Make similar changes to the other sums. In the second sum, let c1, . . . , cw
be i1, . . . , il−1, s and d1, . . . , dx be il+1, . . . , it, and in the third sum let c1, . . . , cw be
i1, . . . , il−1, p − 1 and d1, . . . , dx be s − p + 1, il+1, . . . , it.
∑ • [( ) ] • •q ◦ Ax ◦ λ2 ◦ k1 ◦ T ◦ Aw ⊗ Id ◦ λ1 ◦ (κ ⊗ i)
κ∈S(c1,...,cw,d1...,dx)
c1+⋯+cw+d1+⋯+dx=n−1
w,x≥1
• • •
+ ∑ q ◦ Ax ◦ λ2 ◦ [(T ◦ k1 ◦ Aw)⊗ Id ] ◦ λ1 ◦ (κ ⊗ i)
κ∈S(c1,...,cw,d1...,dx)
c1+⋯+cw+d1+⋯+dx=n−1
w,x≥1
• • •
+ ∑ q ◦ Ax ◦ λ2 ◦ [(Id⊗Aw)⊗ Id ] ◦ λ1 ◦ (κ ⊗ i)
κ∈S(c1,...,cw,d1...,dx)
c1+⋯+cw+d1+⋯+dx=n−1
w,x≥1
+ ∑ • ( ) • ( •q ◦ Ax ◦ λ3 ◦ k1 ⊗ Id ◦ λ1 ◦ κ ⊗ i)
κ∈S(d1...,dx)
d1+⋯+dx=n−1
x≥1
• • •
+ ∑ q ◦ k1 ◦ λ3 ◦ (Aw ⊗ Id) ◦ λ1 ◦ (κ ⊗ i)
κ∈S(c1,...,cw)
c1+⋯+cw=n−1
w≥1
′
Step 7. Focusing now on the right-hand side, we substitute for kn using its
definition. We consider the cases p = 1 and q = 1 separately, and use the fact
′
that k1 ◦ q = q ◦ k1 and k1 ◦ i = i ◦
′
k1, since i and q are chain maps.
n−1
∑ ∑ ∑ ∑ • • • •q ◦ As ◦ (β ⊗ i) ◦ λ ◦ [(q ◦ Ar ◦ (α ⊗ i))⊗ Id] ◦ σ
p=2 σ(p)=n α∈S(a1,...,ar) β∈S(b1,...,bs)
a1+⋯+ar=p−1 b1+⋯+bs=q−1
49
∑ ∑ ∑ ( • ) • •+ q ◦ As ◦ β ⊗ i ◦ λ ◦ [(k1 ◦ i)⊗ Id] ◦ σ
p=1 σ(p)=n β∈S(b1,...,bs)
b1+⋯+bs=n−1
+∑ ∑ ∑ • • •q ◦ k1 ◦ (β ⊗ i) ◦ λ ◦ [q ◦ Ar ◦ (α ⊗ i)⊗ Id] •◦ σ
p=n σ(p)=n α∈S(a1,...,ar)
a1+⋯+ar=n−1
Step 8. We can combine σ, α, and β into one unshuffle θ.
n−1
∑ ∑ • • •q ◦ As ◦ λ2 ◦ [(i ◦ q ◦ Ar)⊗ Id] ◦ λ1 ◦ (θ ⊗ i)
p=2 θ∈S(a1,...,ar,b1,...,bs)
a1+⋯+ar=p−1
b1+⋯+bs=q−1
+∑ ∑ • • •q ◦ As ◦ λ2 ◦ (k1 ⊗ Id) ◦ λ1 ◦ (θ ⊗ i)
p=1 θ∈S(b1,...,bs)
b1+⋯+bs=n−1
+∑ ∑ •q ◦ k1 ◦ λ2 ◦ (Ar ⊗ Id) • •◦ λ1 ◦ (θ ⊗ i)
p=n θ∈S(a1,...,ar)
a1+⋯+ar=n−1
Step 9. Use the fact that IdM −i ◦ q = k1 ◦ T + T ◦ k1.
n−1
∑ ∑ • • •q ◦ As ◦ λ2 ◦ [(k1 ◦ T ◦ Ar)⊗ Id] ◦ λ1 ◦ (θ ⊗ i)
p=2 θ∈S(a1,...,ar,b1,...,bs)
a1+⋯+ar=p−1
b1+⋯+bs=q−1
n−1
• • •
+∑ ∑ q ◦ As ◦ λ2 ◦ [(T ◦ k1 ◦ Ar)⊗ Id] ◦ λ1 ◦ (θ ⊗ i)
p=2 θ∈S(a1,...,ar,b1,...,bs)
a1+⋯+ar=p−1
b1+⋯+bs=q−1
n−1
+∑ ∑ •q ◦ As ◦ λ2 ◦ [( • •Id◦Ar)⊗ Id] ◦ λ1 ◦ (θ ⊗ i)
p=2 θ∈S(a1,...,ar,b1,...,bs)
a1+⋯+ar=p−1
b1+⋯+bs=q−1
• • •
+∑ ∑ q ◦ As ◦ λ2 ◦ (k1 ⊗ Id) ◦ λ1 ◦ (θ ⊗ i)
p=1 θ∈S(b1,...,bs)
b1+⋯+bs=n−1
50
∑ ∑ • • •+ q ◦ k1 ◦ λ2 ◦ (Ar ⊗ Id) ◦ λ1 ◦ (θ ⊗ i)
p=n θ∈S(a1,...,ar)
a1+⋯+ar=n−1
These terms are precisely the terms in Step 6, and so the terms cancel in
′
pairs. Hence L∞-module relation holds for kn.
′
Lemma 5. In the setting of Theorem 2, if M has the L∞-module structure induced
′
by a chain contraction, then the map i ∶ M → M can be extended to an L∞-module
homomorphism, where I1 = i, and for n ≥ 2, we define In by
•
In ∶= ∑ T ◦ At ◦ (τ ⊗ i)
τ∈S(i1,...,it)
i1+⋯+it=n−1
⊗i ⊗i
The map A ∶ L 1t ⊗⋯⊗ L
t ⊗M → M is defined as in the statement of Theorem
2; see Figure 7.
Proof. (Sketch). We will prove that I satisfies the L∞-module homomorphism
′
relation for n = 2. Indeed, for x ∈ L and m ∈M we must show that
I2(l1( ′ ′x),m) + I2(x, k1(m)) + I1(k2(x,m)) = k2(x1, I1(m)) + k1(I2(x,m)) (2.4.1)
′
Working on the left-hand side and substituting in the definitions of I and k , we get
′
T ◦ k2(l1(x), i(m)) + T ◦ k2(x1, i ◦ k1(m)) + i ◦ q ◦ k2(x, i(m)) (2.4.2)
51
Next, we use the fact that i is a chain map and that i ◦ q = T ◦ k1 + k1 ◦ T + IdM , to
see that (2) is equal to
T◦k2(l1(x), i(m))+T◦k2(x1, k1(i(m)))+T◦k1◦k2(x, i(m))+k1◦T◦k2(x, i(m))+k2(x, i(m))
(2.4.3)
Applying the L∞-module relation to the first two terms in (3), we obtain
T ◦k1 ◦k2(x, i(m))+T ◦k1 ◦k2(x, i(m))+k1 ◦T ◦k2(x, i(m))+k2(x, i(m)) (2.4.4)
Now, the first two terms cancel, and what remains is k2(x1, I1(m)) + k1(I2(x,m)),
as desired. The proof of the general case is similar to the proof of Theorem 2
52
CHAPTER III
ANNULAR KHOVANOV HOMOLOGY
3.1. Introduction
In [Kho00], Khovanov defined a bigraded homology group for oriented
3
links in S which is a categorification of the Jones polynomial. Following this,
for a compact, oriented surface Σ, Asaeda, Przytycki, and Sikora introduced a
generalization of Khovanov homology for links in Σ × [0, 1] that categorifies the
Kauffman skein module of Σ; see [APS04]. The case where Σ is an annulus is
known as annular Khovanov homology and has since garnered much attention.
For example, there have been various detection results that have been obtained
by exploiting the relationship of annular Khovanov homology with various Floer
theories. In [XZ19], Xie-Zhang use instanton Floer homology to show that annular
Khovanov homology detects both the unlink and the closure of the trivial braid.
They also show that it distinguishes braid closures from other links. More recently,
Binns-Martin showed that knot Floer homology detects various torus links, and
they used this to show that annular Khovanov homology detects certain braid
closures; see [BM20].
A key feature of annular Khovanov homology is that it is endowed with extra
structure not present in ordinary Khovanov homology. In [GLW18], Grigsby-
Licata-Wehrli show that the annular Khovanov homology of a link is both an
sl2-representation and an sl2(∧)-representation, where sl2(∧) is a Z-graded Lie
superalgebra related to sl2. This structure has been studied in several contexts.
In one direction, Quefflec-Rose generalized this to show that annular Khovanov-
53
Rozansky homology carries an sln-action; see [QR18]. In another direction,
Akhmechet-Krushkal-Willis have made progress towards lifting the sl2-action to
the stable homotopy refinement of the annular Khovanov homology; see [AKW22].
In proving that there is an sl2-representation structure on the annular
Khovanov complex CKh(L), Grigsby-Licata-Wehrli showed that the boundary
maps of CKh(L) commute with the sl2-action, which shows that the sl2-action
holds at the chain level. In contrast, the sl2(∧)-action is well-defined on the
annular Khovanov homology AKh(L), but at the chain level, it only holds up to
homotopy. This observation suggests the existence of an L∞-module structure on
AKh(L). In this chapter, we exhibit sl2(∧) as an L∞-algebra and upgrade the
sl2(∧)-representation structure to that of an L∞-module. This module structure
is an invariant of the annular link at both the chain level and on homology. In
particular, we will prove the following theorem.
Theorem. Let L ⊂ A × I be an annular link. There is an L∞-module structure
on both CKh(L;Z/2Z) and AKh(L;Z/2Z) over the L∞-algebra sl2(∧). Up to L∞-
quasi-isomorphism, this module structure only depends on the isotopy class of L in
A × I.
The organization of this chapter is as follows. In section 2, we recall the
definitions of sl2, sl2(∧), and sl2(∧)dg and review some key results obtained by
Grigsby-Licata-Wehrli. In section 3, we provide a more detailed background of
annular Khovanov homology. In section 4, we explain how sl2(∧)dg and sl2(∧)
are L∞-algebras. In sections 5 and 6, we explain how CKh(L) and AKh(L) are
L∞-modules. In section 7, we prove the invariance of these structures under
Reidemeister moves. In section 8, we provide some examples showing this structure
is nontrivial.
54
Remark. The proof of the above theorem relies on several results about L∞-
modules. In particular, the proofs of Theorem 1, Theorem 2, and Lemma 5 are
given over Z/2Z. We expect these results to hold with signs, but tracking them
through their respective proofs is intricate. Outside of these three proofs, we will
include signs when appropriate. Working without signs affects the bracket relations
in sl2, sl2(∧), and sl2(∧)dg; see section 3.2. The absence of signs also affects the
higher operations involved in the sl2(∧) L∞-module structure on CKh(L); see
Theorem 4.
3.2. The Lie algebras sl2, sl2(∧), and sl2(∧)dg
In this section, we review the Lie algebras of interest. We first recall the
definition of sl2. Next, we define the Lie superalgebra sl2(∧), which will be our
main L∞-algebra of study. Finally, we define an auxiliary Lie superalgebra,
sl2(∧)dg, which is closely related to sl2(∧) and will help us prove several key
results.
3.2.1. The Lie algebra sl2
To fix notation, we will denote the standard basis for the Lie algebra sl2 by
{e, f, h}. Over Z, the Lie bracket relations are given by:
[e, f] = h, [e, h] = −2e, [f, h] = 2f.
55
3.2.2. The Lie superalgebra sl2(∧)
In [GLW18], Grigsby-Licata-Wehrli introduce a larger Lie algebra sl2(∧)
containing sl2 as a subalgebra. In fact, sl2(∧) has the structure of a Z-graded Lie
superalgebra.
Definition 13. A Lie superalgebra g is a Z/2Z-graded vector space geven ⊕ godd
equipped with a bilinear map [⋅, ⋅] ∶ g × g → g, called the super Lie bracket,
satsfying the following conditions:
• [ ] = −(− )∣x∣∣y∣x, y 1 [y, x]
• (−1)∣x∣∣y∣[x, [y, z]] + (− )∣y∣∣x∣1 [y, [z, x]] + (− )∣z∣∣y∣1 [z, [x, y]] = 0
The first condition is known as super skew-symmetry, and the second condition
is known as the super Jacobi identity. Here, x, y and z are homogeneous elements
with respect to the Z/2Z-grading. The notation ∣x∣ represents the degree of x, and
the degree of [x, y] is required to be the sum of the degrees of x and y, modulo 2.
These conditions should be thought of as analogs of the usual Lie algebra axioms,
but with gradings taken into consideration.
We now describe the exterior current algebra sl2(∧) by generators and
relations, as presented in [GLW18]. As vector spaces,
sl2(∧) ≅ sl2 ⊕ sl2,
where the first summand is in degree 0 and the second in degree 1 with respect
to both the Z-grading and the Z/2Z-grading. The Z/2Z-grading required for the
Lie superalgebra structure is the mod 2 reduction of the Z-grading. Denoting
56
the standard basis of the first sl2 summand by {e, f, h} and that of the degree 1
summand by {v2, v−2, v0}, the bracket relations for the Lie superalgebra sl2(∧) are
[e, f] = h [h, e] = 2e [f, v0] = 2v−2
[e, v2] = 0 [h, f] = −2f [f, v−2] = 0
[e, v0] = −2v2 [h, v0] = 0 [h, v2] = 2v2
[e, v−2] = v0 = −[f, v2] [h, v−2] = −2v−2 [vi, vj] = 0 for i, j ∈ {2, 0,−2}.
3.2.3. The Lie superalgebra sl2(∧)dg
Following [GLW18], we describe the Z-graded Lie superalgebra sl2(∧)dg. As a
Z-graded super vector space, the degree 0 generators are {e, f, h}, and the degree 1
generators are {v2, v−2, d,D}. The defining bracket relations are
[e, f] = h [e, v−2] = −[f, v2]; [d, y] = 0 for all y ∈ {e, f, h, v2, v−2};
[h, e] = 2e; [f, v−2] = 0; [D, y] = 0 for all y ∈ {e, f, h, v2, v−2};
[h, f] = −2f ; [h, v2] = 2v2; [d, d] = [D,D] = [v2, v2] = [v−2, v−2] = 0.
[e, v2] = 0; [h, v−2] = −2v−2; [v2, v−2] + [d,D] = 0.
The structure of sl2(∧)dg becomes more clear with the following two lemmas.
The first gives us a basis for sl2(∧)dg, and the second exhibits sl2(∧) as a direct
summand of the homology of sl2(∧)dg by regarding sl2(∧)dg as a chain complex
with differential given by the adjoint action of d. Both lemmas are proved in
[GLW18].
57
Lemma 6 ([GLW18]; Lemma 6). Let ṽ0 = [e, v−2] = −[f, v2], and let x =
[v2, v−2] = −[d,D] = 1[ṽ0, ṽ0]. Then the set {e, f, h, v2, v−2, ṽ0, d,D, x} forms a2
basis of sl2(∧)dg.
Lemma 7 ([GLW18]; Lemma 7). The homology of the chain complex (sl2(∧)dg,
[d, ⋅]) is isomorphic to the direct sum of sl2(∧) and the trivial Lie superalgebra.
That is, H(sl2(∧)dg, [d, ⋅]) ≅ sl2(∧)⊕ Z.
3.3. Annular Khovanov homology
In this section, we review the construction of annular Khovanov homology
and recall some of its structure. For other expositions; see [Rob13], [SZ18], and
[GLW18]. To start, let L ⊂ A × I be a link in the thickened annulus. The link L
admits a diagram P (L) ⊂ A by considering the projection A×I → A×{0}, and this
2
diagram can be regarded as sitting inside of S − {X,O}, where X is a basepoint
representing the inner boundary of A, and O is a basepoint representing the outer
boundary of A; see Figure 8.
X O
2
FIGURE 8. A diagram P (L) ⊂ S − {X,O} of an annular link L, where X and O
represent the inner and outer boundaries of the annulus, respectively.
If we ignore the basepoint X, we can form the ordinary Khovanov complex
CKh(P (L)). CKh(P (L)) is generated by oriented Kauffman states, where circles
are labeled either v+ or v−. CKh(P (L)) is also bigraded, where an element of
58
i,j
CKh (P (L)) is said to have homological grading i and quantum grading j.
Formulas for these gradings are given in [Zha18] and [Rob13].
The addition of the basepoint X endows CKh(P (L)) with a third grading
k, called the k-grading or the winding-number grading. For a fixed generator, the
associated Kauffman state is a collection of oriented circles, and the k-grading is
defined to be the algebraic intersection number of this collection of circles with
an oriented arc from X to O that misses all crossings of P (L). Another way to
compute the k-grading is to count the number of positively-labeled nontrivial
circles and subtract the number of negatively-labeled nontrivial circles, where a
nontrivial circle is a circle that separates X and O. In [Rob13], it is proved that the
Khovanov differential ∂ does not increase the k-grading, and so this gives rise to
a filtration on CKh(P (L)). The annular Khovanov homology AKh(P (L)) is the
homology of the associated graded object. Said differently, we can decompose the
Khovanov differential as ∂ = ∂0 + ∂−, where ∂0 and ∂− are the k-preserving and
k-decreasing parts of ∂, respectively. AKh(P (L)) is the homology of the triply-
graded chain complex (CKh(P (L)), ∂0). Moreover, up to isomorphism, the annular
Khovanov homology does not depend on the diagram P (L) representing L, so it
makes sense to write AKh(L).
It is instructive to see how the differential ∂0 of the annular Khovanov
complex differs from the usual Khovanov differential ∂. To do so, we need to
examine how the k-gradings of generators change under merge and split maps.
Denoting trivial circles by T’s and nontrivial circles by N’s, the three possibilities
are TT ↔ T, NT ↔ N, and NN ↔ T; see Figure 9.
59
The formula for the differential ∂0 depends on the types of circles involved,
and we list the explicit formulas for each case below. Recall that trivial circles are
labeled by either w+ or w− and nontrivial circles are labeled by either v+ or v−.
1. When two trivial circles merge into trivial circle, or when a trivial circle splits
into two trivial circles:
Merge Split
w+ ⊗w+ ↦ w+ w+ ↦ w+ ⊗w− +w− ⊗w+
w+ ⊗w− ↦ w− w− ↦ w− ⊗w−
w− ⊗w+ ↦ w−
w− ⊗w− ↦ 0
2. When a trivial circle and a nontrivial circle merge into a nontrivial circle, or
when a nontrivial circle splits into a trivial circle and a nontrivial circle:
Merge Split
w+ ⊗ v+ ↦ v+ v+ ↦ w− ⊗ v+
w+ ⊗ v− ↦ v− v− ↦ w− ⊗ v−
w− ⊗ v+ ↦ 0
w− ⊗ v− ↦ 0
3. When two nontrivial circles merge into a trivial circle, or when a trivial circle
splits into two nontrivial circles:
60
Merge Split
v+ ⊗ v+ ↦ 0 w+ ↦ v+ ⊗ v− + v− ⊗ v+
v+ ⊗ v− ↦ w− w− ↦ 0
v− ⊗ v+ ↦ w−
v− ⊗ v− ↦ 0
↔ or ↔
↔ or ↔
↔
FIGURE 9. The various ways the operations of merging and splitting along a
crossing (indicated by a dashed line) interact with a basepoint. The top illustrates
the case of two trivial circles merging into trivial circle (or a trivial circle splitting
into two trivial circles). The middle illustrates the case of a trivial circle and a
nontrivial circle merging into a nontrivial circle (or a nontrivial circle splitting into
a trivial circle and a nontrivial circle). The bottom illustrates the case of nontrivial
circles merging into a trivial circle (or a trivial circle splitting into two nontrivial
circles).
We end this section by briefly describing the sl2 representation structure on
AKh(L), referring the reader to [GLW18] for details. Fix a resolution of P (L).
Nontrivial circles, with respect to the basis {v+, v−}, are assigned the 2-dimensional
defining representation of sl2, defined by
⎛ 1 0 ⎞ ⎛ 0 1 ⎞ ⎛ 0 0 ⎞
h↦ ⎜⎜⎜ ⎟⎟⎟ , e↦ ⎜⎜⎜ ⎟⎟⎟ , f ↦ ⎜⎜⎜ ⎟⎟⎟ .
⎝ 0 −1 ⎠ ⎝ 0 0 ⎠ ⎝ 1 0 ⎠
61
Trivial circles are assigned the 2-dimensional trivial representation. The resolution
is then assigned the tensor product of these representations. We take the direct
sum of all of these representations to obtain the structure of an sl2-representation
on CKh(P (L)). This action descends to an action on the homology AKh(L), which
Grigsby-Licata-Wehrli then upgrade to an action of sl2(∧). They show that the
annular boundary maps commute with the sl2-action, implying that the sl2-action
holds at the chain level. In contrast, the sl2(∧)-action is well-defined on AKh(L),
but at the chain level, it only holds up to homotopy [GLW18]. This observation
leads us to consider this situation in terms of L∞-algebras and modules.
3.4. The L∞-algebra structure on sl2(∧)
Since sl2(∧)dg is a Lie superalgebra, it is an L∞-algebra with no higher
operations. In this section, we will use a cochain contraction to transfer this L∞-
algebra structure on sl2(∧)dg to sl2(∧), and then we will show that all higher
operations in the L∞-algebra structure on sl2(∧) vanish.
Lemma 8. There exist maps i and q so that the data
q
K sl2(∧)dg H(sl2(∧)dg) (∗)
i
satisfies the definition of a cochain contraction.
Proof. By Lemma 7, H(sl2(∧)dg) ≅ sl2(∧) ⊕ Z with basis {v2, v−2, v0, d, e, f, h}.
Writing out the basis elements of sl2(∧)dg and H(sl2(∧)dg), with their degrees
above them, we have
62
2 1 0
sl2(∧)dg x v2, v−2, ṽ0, d,D e, f, h
H(sl2(∧)dg) 0 v2, v−2, ṽ0, d e, f, h
The maps i and q are easy to define. Let i lift every element to its corresponding
element in sl2(∧)dg, and let q be the projection back down, sending x and D to 0.
Define the chain homotopy K to be 0 for every element except for x, in which case
we define K(x) = −D.
It is straightforward to check that i and q are chain maps. The differential in
sl2(∧) is 0, so i∂ = 0. Also, ∂i = 0, since the elements in the image of i are in the
kernel of [d, ⋅], which is the differential in sl2(∧)dg. On the other hand, ∂q = 0.
Also, q∂ = 0, since the only element in the image of [d, ⋅] is x, which is sent to 0 by
q. It is also straightforward to check that all of the chain contraction conditions are
satisfied.
Lemma 9. The Lie superalgebra H(sl2(∧)dg) inherits an L∞-algebra structure
induced by (∗), and this L∞-algebra structure has no higher operations.
Proof. Following section 2.4, the formulas for the transfered bracket tell us that
n−1
In = ∑ ∑ (σ) ⋅K ◦ l2 ◦ (Ij ⊗ In−j) ◦ σ•
j=1 σ∈S(j,n−j)
n−1
′
ln = ∑ ∑ (σ) ⋅ q ◦ l2 ◦ (Ij ⊗ In−j) ◦ σ•
j=1 σ∈S(j,n−j)
63
Recall that I1 = −i. For n = 2, the only unshuffle in S(1, 1) is the identity.
So, I2(x1, x2) = K(l2(I1(x1), I1(x2))). Since K(x) = −D and is 0 otherwise,
I2(v2, v−2) = K(x) = −D
I2(v−2, v2) = K(x) = −D
I2(ṽ0, ṽ0) = K(2x) = −2D
I2(x1, x2) = 0 otherwise
′
Moreover, l2(x1, x2) = q(l2(I1(x1), I1(x2))), and so to compute the bracket
of two elements in H(sl2(∧)dg), we lift them to sl2(∧)dg, take their bracket in
sl2(∧)dg, and then quotient back to H(sl2(∧)dg).
Now, let n ≥ 3. For all m ≥ 2, Im is in the image of K, and so
Im(x1, . . . , xm) = cD for some scalar c. But then for any 1 ≤ j ≤ n − 1 and
σ ∈ S(j, n− j), q◦ l2 ◦(Ij⊗In−j)◦σ• is 0, since either the l2 ◦(Ij⊗In−j)◦σ• term is
0, as [D, y] = 0 for all y ∈ {e, f, h, v2, v−2, ṽ0, D}, or q will send this term to 0 since
′
the only nonzero bracket involving D is [d,D] = −x, and q(x) = 0. Hence ln = 0 for
n ≥ 3, and so the Lie superalgebra H(sl2(∧)dg) has no higher operations.
Theorem 3. The Lie superalgebra sl2(∧) inherits an L∞-algebra structure as a
subalgebra of H(sl2(∧)dg), and this L∞-algebra structure has no higher operations.
Proof. The map H(sl2(∧)dg) → sl2(∧) that sends e, f, h, v2, v−2, ṽ0, d to
e, f, h, v2, v−2, v0, 0 is surjective with 1-dimensional kernel. So H(sl2(∧)dg) ≅
sl2(∧) ⊕ Z, where the Z summand is generated by the element d. But the bracket
of d with everything in the sl2(∧) summand is 0, so sl2(∧) is a direct sum not
only as a vector space, but also as an L∞-algebra. So sl2(∧) is an L∞-algebra as
a subalgebra of H(sl2(∧)dg).
64
3.5. The L∞-module structure on CKh(L)
Viewing sl2(∧)dg and sl2(∧) as L∞-algebras, in this section we will exhibit
CKh(P (L)) as an L∞-module over sl2(∧). Also, fix a diagram P (L) of the annular
link L. We will simplify notation and write CKh(L) and AKh(L) instead of
CKh(P (L)) and AKh(P (L)).
Theorem 4. Let L be an annular link and m ∈ CKh(L). Then CKh(L) is an L∞-
module over the L∞-algebra sl2(∧). One of the higher operations is given in terms
Lee
of the Lee differential: k3(v2, v−2,m) = ∂0 (m). In particular, the L∞-module
Lee
structure is nontrivial if ∂0 ∶ CKh(L)→ CKh(L) is nonzero.
Proof. To start, CKh(L) is an L∞-module over sl2(∧)dg, where the k2 operation
is given by the usual module action, and kn = 0 for n ≥ 3. The module actions
of elements of the basis {e, f, h, v2, v−2, ṽ0, d,D, x} are as follows. The actions of
Lee
e, f, h were described at the end of section 3.3, and v2, v−2, d, and D act by ∂+ ,
Lee
∂−, ∂0, and ∂0 , respectively; see [GLW18]. The actions of ṽ0 and x can then be
determined by the bracket relations.
Now, we have a cochain contraction from sl2(∧)dg onto its homology, so we
can transfer the sl2(∧)dg-module structure to obtain a new module structure over
H(sl2(∧)dg). We can then restrict this module structure to the copy of sl2(∧) that
sits inside of H(sl2(∧)dg).
To see that the induced module structure is nontrivial, recall the cochain
contraction from Lemma 8.
q
K sl2(∧)dg H(sl2(∧)dg)
i
65
Examining the restriction of scalars formulas from Theorem 1, we see that
′
k3(x1, x2,m) = k3(I1(x1), I1(x2),m) − k2(I2(x1, x2),m) = −k2(I2(x1, x2),m)
for x1, x2 ∈ H(sl2(∧)dg) and m ∈ CKh(L). Here, kn is the L∞-module
operation for sl2(∧)dg, and recall that kn = 0 for n ≥ 3. Since I2(v2, v−2) =
− LeeD and I2(v−2, v2) = −D, and since D acts by ∂0 , we conclude that
′
k3( Lee ′v2, v−2,m) = ∂0 (m) and k3(v−2, v2,m) = Lee∂0 (m), showing that we obtain
higher operations.
3.6. Reidemeister Moves
3.6.1. Invariance of the sl2(∧)dg-module structure
In this section, we follow Khovanov’s original proof that Khovanov homology
is invariant under Reidemeister moves; see [Kho00]. There, Khovanov constructs
quasi-isomorphisms between a given Khovanov complex and the complex obtained
after applying a particular Reidemeister move. Here, we upgrade these quasi-
isomorphisms to sl2(∧)dg L∞-module quasi-isomorphisms.
Theorem 5. The L∞-module structure on CKh(L) is invariant under Reidemeister
I.
Proof. Let J K and J K denote the annular chain complexes before and after
applying an RI move, respectively. Our goal is to construct a quasi-isomorphism
of L∞-modules {hn} ∶ J K → J K. Because the sl2(∧)dg-module structures on
these complexes have no higher operations, it suffices to give a quasi-isomorphism
h1 ∶ J K → J K that respects the module action, since we can then take hn = 0 for
66
n ≥ 2. To this end, let C be the complex
C m∶= J K = J K −→ J K{1}
and let C ′ be the subcomplex
C ′ ∶= J K mw −→ J K{1},+
where J Kw means that the extra circle is labeled w+. A straightforward check of+
′ ′
the actions of the basis elements {e, f, h, v−2, v2, ṽ0, d,D, x} on C shows that C is an
(∧) C ′ ′sl2 dg-submodule. Moreover, is acyclic, since we can write C as the mapping
cone of the isomorphism m.
⋯ J K dw J Kw ⋯+ +
C ′ = J K mw −→ J K{1} = ⊕ m+ ⊕
⋯ J K{1} d J K{1} ⋯
′
Therefore, the quotient complex C/C is the complex J K/w =0 → 0, and it is+
isomorphic to J K as chain complexes via the map z ⊗ w− ↦ z. To summarize,
we have constructed a chain map J K→ J K given by
y ⊗ w+ + z ⊗ w− + x↦ z ⊗ w− ↦ z
for y, z ∈ C(∗0) and x ∈ C(∗1) (we have labeled the crossing formed by
the Reidemeister I move last in the chain complex), and this map induces an
67
isomorphisms on homology
H(J K) = H(C) ≅ H(C/C ′) ≅ H(J K)
To complete the proof, we need to check that this composition respects
the sl2(∧)dg action. Certainly the first map does, as it is the quotient map of an
sl2(∧)dg-submodule. For the second map, if s ∈ sl2(∧)dg, mapping over and then
acting by s gives z ⊗ w− ↦ sz. On the other hand, acting first by s and then
mapping over gives s(z ⊗ w−) = sz ⊗ w− ↦ sz.
Theorem 6. The L∞-module structure on CKh(L) is invariant under Reidemeister
II.
Proof. There is a more direct way to prove RII invariance, but the method that
follows will be useful in proving RIII invariance. Consider the diagrams in Figure
10.
J K{ } m1 // J K{2}
OO OO J Kw {1}
m // J K{2}
OO + OO
∆ C ⊃ C ′
J K // J K{1} 0 // 0
J K{1}/ //w+=0 0 β //OO OO OO 0OO
= −1τ d⋆0∆
∆ ∆⊃ ))
J K d⋆0 // J K{1} dα ⋆0 // τβ
C/C ′ C ′′′
68
β //OO 0OO
β=τβ
))
0 // γ
(C/C ′)/C ′′′
FIGURE 10. The relevant complexes in the proof of RII invariance. A similar
diagram appears in [Bar02].
As complexes, the composition
J K q ′ p f= C→− C/C →− (C/C ′)/C ′′′ −→ J K
is a chain of quasi-isomorphisms; see [Bar02]. Our goal is to show that these
′′′
complexes are actually quasi-isomorphic as L∞-modules. Since C is not an L∞-
′
submodule, we do not immediately have an L∞-module structure on (C/C )/C ′′′.
′ ′
Our strategy then will be to give chain contractions from C to C/C and from C/C
to (C/C ′)/C ′′′ in order to equip these quotients with L∞-module structures. Doing
so will give us our desired L∞-module quasi-isomomorphisms. To this end, define
i ∶ C/C ′ → C to be the map
⎪⎪⎧
⎪⎪
⎪⎪z − −1m ∂C(z) if z is in the top left
⎪⎪⎪
⎪⎪⎪⎪⎪0 if z is in the top righti(z) = ⎪⎨
⎪⎪⎪⎪⎪⎪z if z is in the bottom left⎪⎪⎪⎪⎪⎪ −1⎪z −m ∂C(z) if z is in the bottom right⎩
69
−1
where the map m ∶ C → C is zero except on the top right vertex. There, it will
be the inverse to the isomorphism that merges a circle with the small circle labeled
w+.
′
Remark. The map i above takes an element z ∈ C/C and views it as an element
′
of C. The complex C has a preferred basis of Khovanov generators, and C/C has a
′
preferred basis consisting of basis elements of C not in C . So, before applying i, we
′
should apply a map i0 ∶ C/C → C as F2 vector spaces, but we will suppress this for
brevity.
Now, if K ∶ C → C is the map
⎧
⎪⎪⎪⎪⎪⎪0 if z is in the top left
⎪⎪⎪
⎪⎪⎪ −1⎪⎪m (z) if z is in the top rightK(z) = ⎪⎨⎪⎪⎪⎪⎪⎪0 if z is in the bottom left⎪
⎪⎪⎪⎪⎪⎪0 if z is in the bottom right⎩
the data
q
K C C/C ′
i
satisfies the requirements of a chain contraction, which we can use to transfer the
′
L∞-module structure from C to C/C . In particular, since i was a quasi-isomorphism
′
of chain complexes, we obtain a quasi-isomorphism of L∞-modules In ∶ C/C →
C ′, where C/C has the induced L∞-module structure. In fact, there are no higher
′
operations on C/C . After examining the formula for the induced operation, this
follows from the fact that C itself has no higher operations, that the image of K is
′ ′
in C , and that C is an L∞-submodule of C.
70
′ ′′′
Next, since every nonzero element of (C/C )/C is equivalent to some element
′
γ in the bottom right, we can define j(γ) ∶= γ, thought of as an element of C/C .
′ ′
Then if H ∶ C/C → C/C is the map
⎪⎧⎪⎪ −1
⎪⎪⎪∆ (z) if z is in the top left⎪
⎪⎪⎪
⎪⎪⎪⎪0 if z is in the top rightH(z) = ⎨⎪⎪⎪⎪⎪⎪
⎪0 if z is in the bottom left
⎪⎪⎪⎪⎪⎪⎪0 if z is in the bottom right⎩
the data
p
H C/C ′ (C/C ′)/C ′′′
j
also satisfies the requirements of a chain contraction. We obtain a quasi-
′ ′ ′′′
isomorphism of L∞-modules Jn ∶ C/C → C, where (C/C )/C has the induced
′ ′ ′′′
L∞-module structure from C/C . There are no higher operations on (C/C )/C as
′
well. To see this, note that because C/C has no higher operations, the induced
′ ′′′ ′ •
module operation on (C/C )/C is of the form kn = ∑ τ∈S(1,...,1) q ◦ An−1 ◦ (τ ⊗ i);
i1=⋯=in−1=1
see Figure 11.
x1 x2 x3 ⋯ xn−1 m
•
τ j
H H H
k k ⋯2 2 k2 q
′ ′′′
FIGURE 11. The transferred bracket on (C/C )/C . Here, the labeled edges
′
represent the application of that particular map. For example, k3(x1, x2,m) =
q ◦ k2(x1, H ◦ k2(x2, j(m))) + q ◦ k2(x2, H ◦ k2(x1, j(m)))
.
71
′
Since the image of j is concentrated in the bottom right corner of C/C ,
and H is zero everywhere except the top-left, it follows that all higher operations
vanish. As for the module operation k2, an element s ∈ sl2(∧)dg acts on γ ∈
(C/C ′)/C ′′′ by
s ⋅ γ = p(s ⋅ j(γ)) = p(q(s ⋅ (i ◦ j(γ)))) = ( −1p ◦ q)(s ⋅ γ − s ⋅m ∂C(γ))
−1
That is, we consider the difference s ⋅ γ − s ⋅m ∂C(γ) as an element of C, and then
′ ′′′
quotient twice. It remains to show that the degree shift map f ∶ (C/C )/C → J K
′ ′′′
respects this action, that is, f(s ⋅ γ) = s ⋅ f(γ) for s ∈ sl2(∧)dg and γ ∈ (C/C )/C .
We compute that
f( −1s ⋅ γ) − s ⋅ f(γ) = f((p ◦ q)(s ⋅ γ) − (p ◦ q)(s ⋅m ∂C(γ))) − s ⋅ γ
= −(p ◦ q)( −1s ⋅m (∂Cγ))
−1
Using the fact that any term m (∂Cγ) will be labeled by w+, the action of any
s ∈ sl2(∧)dg on this term will quotient to 0 under p ◦ q. In particular, we have
shown that the composition
J K q p f= C→− C/C ′ C ′′→− −→ J K
is a chain of L∞-quasi-isomorphisms, since J K has no higher operations.
Theorem 7. The L∞-module structure on CKh(L) is invariant under Reidemeister
III.
Proof. Step 1: Overview. For RIII, the situation is summarized in Figure 12.
We start by decomposing the complexes J K and J K into C and D (these are the
72
top left and top right cubes in Figure 12, respectively). We will then transfer the
L∞-module structures by quasi-isomorphisms q2 ◦ q1 and p2 ◦ p1 to the quotient
complexes (C/C ′)/ ′′C and (D/D′)/ ′′D (the bottom row) and show that these
quotients are L∞-quasi-isomorphic via an L∞-module map f .
Step 2: The structure on C/C ′ and D/D′. Analagous to the RII case, we
′ ′
have subcomplexes C ⊂ C and D ⊂ D, which are L∞-submodules; see Figure 13.
′ ′ ′ ′
Because C and D are submodules, the quotients C/C and D/D have no higher
operations as L∞-modules. Alternatively, this quotient structure agrees with the
one obtained by using cochain contractions
q1 p1
C C/C ′ ′H and K D D/D
i1 j1
to transfer the structure. Here, the maps i1 and i2 are
⎪⎧
⎪⎪⎪⎪z, if z ∈ 000, 001, 010, 100⎪⎪⎪
i ⎪1(z) = ⎪⎪⎨
−1
z −m (∂Cz), if z ∈ 011, 101, 110
⎪⎪⎪
⎪⎪⎪⎩⎪0, if z ∈ 111
and
⎧
⎪⎪⎪
⎪⎪z, if z ∈ 000, 001, 010, 100
⎪⎪
( ) ⎨⎪⎪j z = ⎪ − −11 ⎪z m (∂Dz), if z ∈ 011, 101, 110
⎪⎪⎪
⎪⎪
⎪⎪⎩0, if z ∈ 111
73
m
∆ m
∆
q1 i1 p1 j1
/ mw+=0 0 0
∆ m
∆ /w+=0
q2 i2 p2 j2
/ mw+=0 0 0
m
∆
∆
0 0 /w+=0
f
FIGURE 12. The complexes involved in RIII invariance. We have suppressed the
degree shifts.
74
m 0
w+ m
0 0 0 w+
0 0 0 0
0 0 0 0
′ ′
FIGURE 13. The complexes C and D . The w+ means that the trivial circle is
labeled w+.
The coordinates above refer to different corners of the cubes, i.e.
(101) (111)
(100) (110)
(001) (011)
(000) (010)
′ ′′ ′ ′′ ′
Step 3: The structure on (C/C )/C and (D/D )/D . To go from C/C to
(C/C ′)/C ′′ D/D′ (D/D′)/D′′and to , we identify elements in vertices 101 and 110
′
by imposing the relation β1 = τ1β1 in (C/C )/C ′′ and the relation β2 = τ2β2 in
(D/D′)/D′′, analagous to the RII case. Similar to before, we are not quotienting
′ ′
by a submodule, so we need to transfer the structure from C/C and D/D . To this
end, we define maps i2 and j2. Let
⎪⎧⎪
⎪⎪⎪z, if z ∈ 000, 010, 011, 110
( ) ⎪
⎪⎪
i2 z = ⎨⎪⎪ − −1⎪z ∆ (∂C/C′z), if z ∈ 001⎪⎪⎪
⎩⎪
⎪
⎪⎪0, if z ∈ 100, 111
75
Note that if z ∈ 101, then z is equivalent via τ1 to some element in 110. Also, let
⎧⎪⎪⎪⎪⎪z, if z ∈ 000, 001, 011, 101⎪⎪
j2(
⎪
z) = ⎪⎨⎪ − −1⎪z ∆ (∂⎪ D/D
′z), if z ∈ 010
⎪⎪
⎪⎪⎪⎪⎩0, if z ∈ 100, 111
where we again note that if z ∈ 110, then z is equivalent via τ2 to some element in
′ ′ ′ ′
101. Then, if we define T ∶ C/C → C/C and S ∶ D/D → D/D by
⎪⎪⎧ ⎧⎪ −1⎪∆ (z)
⎪
, if z ∈ 101 ⎪⎪⎪
−1
⎪ ⎪∆ (z), if z ∈ 110T (z) = ⎪⎪⎨ and S(z) = ⎨⎪ ⎪⎪⎪⎩⎪0, otherwise
⎪⎪⎪⎩0, otherwise
both
q2 p
C/C ′ (C/C ′)/C ′′ D/D′
2
T and S (D/D′)/D′′
i2 j2
satisfy the requirement of a cochain contraction. In particular, this allows us to
′ ′
transfer the L∞-module structures from C/C and D/D to their respective quotient
complexes.
′ ′′ ′ ′′
Step 4: (C/C )/C and (D/D )/D have no higher operations. The next
′ ′′ ′ ′′
goal is to show that there is no higher structure on (C/C )/C or (D/D )/D . We
′ ′′
will explain the case of (C/C )/C . The case of (D/D′)/D′′ is analagous. Indeed,
′
because C/C has no higher L∞-module operations, the transferred structure looks
like
′
kn( •x1, . . . , xn−1,m) = ∑ q ◦ An−1 ◦ (τ ⊗ i2)
τ∈S(1,...,1)
i1=⋯=in−1=1
76
•
See Figure 11. We will show that q ◦ An−1 ◦ (τ ⊗ i2) = 0 for any τ ∈ Sn−1. That is,
for n ≥ 3, it suffices to show that q ◦ An−1(x1, x2, . . . , xn−1, i2(m)) = 0 for any choice
of x1, x2, . . . , xn−1 ∈ sl2(∧)dg, where i1, . . . , in−1 = 1 in the definition of An−1.
Step 4.1: The case n > 3. We start with the case n > 3. Because T is only
nonzero on the vertex 101, for q ◦ An−1(x1, x2, . . . , xn−1, i2(m)) to be nonzero, it
must contain a nonzero composition
T xj T
/w+=0 /w+=0
101 100 101 100
Here, the map xj represents acting by the element xj ∈ sl2(∧)dg. We
will show that if xj is any element of the basis {e, f, h, v2, v−2, ṽ0, d,D, x}, then
this composition is zero. Indeed, xj cannot be e, f, h, since it must change the
homological degree by one to have nonzero image in vertex 101. Moreover, modulo
the relation w+ = 0, the actions of the elements v2, v−2, ṽ0, and D are all the zero
map. Finally, if xj = −x = [d,D], then the component that lies in the vertex 101 is
D101d10∗ +D10∗d100 + d101D10∗ + d10∗D100
where, for example, the notation D101 represents the component of D that remains
in vertex 101, and d10∗ represents the component of d obtained by acting along the
edge 100 → 101. Now we observe that the middle terms D10∗d100 and d101D10∗ are
both zero, because the relation w+ = 0 implies that D10∗ is the zero map. Also,
the terms D101d10∗ and d10∗D100 cancel, because d10∗ just appends a trivial circle
labeled w− to the resolution in vertex 100.
77
Therefore, we have reduced the possible nonzero q◦An−1(x1, x2, . . . , xn−1, i2(m))
to either the case of q ◦ A2(x1, x2, i2(m)) or q ◦ An−1(x1, x2, . . . , xn−1, i2(m)), where
x2 =⋯ = xn−2 = d.
Step 4.2: The case n = 3. We now examine the case n = 3. From the
formula for q ◦ A2(x1, x2, i2(m)), we need x1 ⋅ i2(m) to be in vertex 101. This
implies that m is either in the vertex 000 or the vertex 001. If m ∈ 000, then the
only possibility for x1 is x1 = x. But then
x ⋅m = −[d,D] = Lee Lee⋅m ∂0∂0 m + ∂0 ∂0m
Lee
Since the boundary map ∂∗01 is a split map, and w+ = 0 in vertex 101, ∂0 = 0
Lee
along this edge. So we only have a term ∂0∂0 m. Therefore, we need to focus on
the composition
/w+=0
∆ 101
Lee
∂ −10 ∂0 ∆
+
t1 ∆ /w+=0
000 001 101 100
t2
110
where t1, t2 can be either merge or split. Let a, b, c denote the circles to which the
three strands in vertex 000 belong; see Figure 14. Then we have four cases: either
a = b = c, a = b ≠ c, a ≠ b = c, or a ≠ b ≠ c.
78
b c
a
FIGURE 14. Each strand in the vertex 000 belongs to a circle. Denote these circles
by a, b, and c.
a = b = c
a = b ≠ c
a ≠ b = c
a ≠ b ≠ c
FIGURE 15. This picture shows all possible configurations of the circles a, b, and c.
We have not drawn the basepoint, which can be anywhere outside of the
dashed circles. We have also not drawn the possible other circles coming from the
other crossing resolutions.
Step 4.2.1: m ∈ 000 and a = b = c. If a = b = c, then in each case, t1 is a
Lee
split map and t2 is a merge map. Because ∂0 needs to be nonzero, we must label
our circle by w−. This forces a labeling of w+ ⊗ w+ in 100; see Figure 16.
79
/w+=0
∆ 101
Lee
∂ −10 ∂0 ∆
+
t1 ∆ /w+=0
000 001 101 100
t2
110
FIGURE 16. The first of four cases with a = b = c. In each case, the labeling of the
circle in 000 must be w−, which forces a labeling of w+ ⊗ w+ in 100.
The possibilities for x2 are {e, f, h, v2, v−2, ṽ0, d,D, x}. It cannot be e, f, h,
since x2 must change the homological be degree by one. Moreover, v2, v−2 and ṽ0
are each the 0 map, since we are only involving trivial circles. The labeling w+ ⊗w+
implies that D is the 0 map. Finally, the terms obtained from acting by either d or
′ ′′
x will cancel when we quotient to (C/C )/C . For example, if we act by d, then the
relation β1 = τ1β1 identifies the terms obtained by acting by d10∗ and d1∗0, and so
they will cancel. On the other hand, if we act by x, the terms we obtain in vertices
101 and 110 are
d10∗D + d D +D d +D d + d D + d D +D d +D dÍ ÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒ1ÒÒÒÒÒÒ0ÒÒÒÒÒÒ0ÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒ1ÒÒÒÒÒÒ0ÒÒÒÒÒ1ÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒ1ÒÒÒÒÒÒ0ÒÒÒÒÒ∗ÒÒÒÒÒÒÒÒÒÒÒÒÒÑÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒ1ÒÒÒÒÒÒ0ÒÒÒÒÒÒ∗ÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒ1ÒÒÒÒÒÒ0ÒÒÒÒÒÒ0ÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒ1ÒÒÒÒÒÒ0ÒÒÒÒÒÒ1ÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒ1ÒÒÒÒÒÒ0ÒÒÒÒÒÒ∗ÒÒÒÏ Í ÒÒÒÒ1ÒÒÒÒÒÒ∗ÒÒÒÒÒÒÒ0ÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒ1ÒÒÒÒÒÒ0ÒÒÒÒÒÒ0ÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒ1ÒÒÒÒÒÒ1ÒÒÒÒÒ0ÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒ1ÒÒÒÒÒÒ∗ÒÒÒÒÒÒÒ0ÒÒÒÒÒÒÒÒÒÒÒÑÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒ1ÒÒÒÒÒÒ∗ÒÒÒÒÒÒÒÒ0ÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒ1ÒÒÒÒÒÒ0ÒÒÒÒÒÒ0ÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒ1ÒÒÒÒÒÒ1ÒÒÒÒÒÒ0ÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒ1ÒÒÒÒÒÒ∗ÒÒÒÒÒÒÒ0ÒÒÏ
vertex 101 vertex 110
Now, the terms involving D10∗ and D1∗0 are zero, because w+ = 0 in vertex 101 and
both circles in vertex 100 are labeled by w+. We are left with
d10∗D100 +D101d10∗ + d1∗0D100 +D110d1∗0
80
Because of the w+ labelings in vertex 100, the only nonzero parts of D101 and D110
come from applying D amongst the other circles in the resolution. It follows that
D101d10∗ and D110d1∗0 will be identified when we quotient, and so they will cancel.
The d10∗D100 and d1∗0D100 terms will also cancel.
Step 4.2.2: m ∈ 000 and a = b ≠ c. If a = b ≠ c, then in each case, t1
is a split map and t2 is a split map. Again, we need to involve trivial circles for t1,
Lee
otherwise ∂0 = 0; see Figure 17.
/w+=0
∆ 101
Lee
∂0 ∂
−1
0 ∆
/ +t1 ∆ w+=0
000 001 101 100
t2
110
FIGURE 17. The first of three cases with a = b ≠ c. The labeling of the circle in
000 must be w− ⊗ w−, which forces a labeling of w+ in 100.
Since the circle in vertex 100 must be labeled by w+, by a similar argument
to the case of a = b = c, acting by v2, v−2, ṽ0, D are all 0, and the terms obtained
′ ′′
from acting by either d or x will cancel when we quotient to (C/C )/C . Seeing that
the terms will cancel in the quotient if we act by x in vertex 100 is slightly different
than before. To see this explicitly, we start as in the case of a = b = c by examining
81
the terms
d10∗D1 + d D +D d +D d + d D + d D +D d +D dÍ ÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒ0ÒÒÒÒÒÒ0ÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒ1ÒÒÒÒÒÒ0ÒÒÒÒÒ1ÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒ1ÒÒÒÒÒÒ0ÒÒÒÒÒ∗ÒÒÒÒÒÒÒÒÒÒÒÒÒÑÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒ1ÒÒÒÒÒÒ0ÒÒÒÒÒÒ∗ÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒ1ÒÒÒÒÒÒ0ÒÒÒÒÒÒ0ÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒ1ÒÒÒÒÒÒ0ÒÒÒÒÒÒ1ÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒ1ÒÒÒÒÒÒ0ÒÒÒÒÒÒ∗ÒÒÒÏ Í ÒÒÒÒ1ÒÒÒÒÒÒ∗ÒÒÒÒÒÒÒ0ÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒ1ÒÒÒÒÒÒ0ÒÒÒÒÒÒ0ÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒ1ÒÒÒÒÒÒ1ÒÒÒÒÒ0ÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒ1ÒÒÒÒÒÒ∗ÒÒÒÒÒÒÒ0ÒÒÒÒÒÒÒÒÒÒÒÑÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒ1ÒÒÒÒÒÒ∗ÒÒÒÒÒÒÒÒ0ÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒ1ÒÒÒÒÒÒ0ÒÒÒÒÒÒ0ÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒ1ÒÒÒÒÒÒ1ÒÒÒÒÒÒ0ÒÒÒÒÒÒÒÒÒÒÒÒÒÒÒ1ÒÒÒÒÒÒ∗ÒÒÒÒÒÒÒ0ÒÒÏ
vertex 101 vertex 110
Now, d10∗D100 and d1∗0D100 will cancel in the quotient. Also, D10∗ is the zero map
due to the relation w+ = 0 in vertex 101. The w+ label implies that d110D1∗0 is
zero. It remains to show that the terms
D101d10∗ +D1∗0d100 +D110d1∗0
cancel. Label the circles in 100 by c1 ⊗⋯⊗ cn ⊗ w+. The idea is to show that part
of D110d1∗0 will cancel with D101d10∗ (the part involving the ci themselves) and that
the rest will cancel with D1∗0d100 (the part involving the w+). Indeed, we may write
the D110d1∗0 term as
D110d1∗0 = D110(c1 ⊗⋯⊗ cn ⊗ w+ ⊗ w−) +D110(c1 ⊗⋯⊗ cn ⊗ w− ⊗ w+)
= c cD110(c1 ⊗⋯⊗ cn ⊗ w+ ⊗ w−) +D110(c1 ⊗⋯⊗ cn ⊗ w− ⊗ w+)
w
+D +
w
110(c1 ⊗⋯⊗ cn ⊗ w+ ⊗ w +−) +D110(c1 ⊗⋯⊗ cn ⊗ w− ⊗ w+)
w w
+D −110(c1 ⊗⋯⊗ cn ⊗ w+ ⊗ w−) +D −110(c1 ⊗⋯⊗ cn ⊗ w− ⊗ w+)
c
where D110 is the part of D110 that involves only crossings among the circles
w
c1, . . . , c
+
n, D is the part of D110 that involves only crossings with the circle
w
labeled w+, and D
− is the part of D110 that involves only crossings with the
circle labeled w−. By the definition of the Lee differential, the labels imply
82
w+ ( ) = wD110 c1 ⊗⋯⊗ c +n ⊗ w+ ⊗ w− D110(c1 ⊗⋯⊗ cn ⊗ w− ⊗ w+) = 0, and so
D110d1∗0 =
c
D110( cc1 ⊗⋯⊗ cn ⊗ w+ ⊗ w−) +D110(c1 ⊗⋯⊗ cn ⊗ w− ⊗ w+)
w
+D −
w
110(c1 ⊗⋯⊗ cn ⊗ w− ⊗ w+) +D −110(c1 ⊗⋯⊗ cn ⊗ w+ ⊗ w−)
On the other hand, D101d10∗ can be written as
c
D101(c1 ⊗⋯⊗ cn ⊗ w+ ⊗ w−) = D101(c1 ⊗⋯⊗ cn ⊗ w+ ⊗ w−)
w
+D101(c1 ⊗⋯⊗ cn ⊗ w+ ⊗ w−)
w
where D101 is the part of D101 involving a crossing with either the (outermost)
circle labeled w+ or the circle labeled w−. The w+ label together with the relation
w+ =
w
0 in vertex 101 implies that D101(c1 ⊗⋯⊗ cn ⊗w+ ⊗w−) = 0. In the quotient
(C/C ′)/C ′′ c, D101(c1 ⊗⋯⊗ cn ⊗ w+ ⊗ w−) is identified with
D110(c1 ⊗⋯⊗ cn ⊗ w+ ⊗ w−) +D110(c1 ⊗⋯⊗ cn ⊗ w− ⊗ w+)
Therefore, it remains to examine the D1∗0d100 term, which we may write as
c
D1∗0d100(c1 ⊗⋯⊗ cn ⊗ w+) = D1∗0d100(c1 ⊗⋯⊗ cn ⊗ w+)
w
+D1∗0d100(c1 ⊗⋯⊗ cn ⊗ w+)
c
Because of the w+ label, D1∗0d100(c1 ⊗⋯⊗ cn ⊗ w+) = 0, and so it remains to show
that
w
D1∗0d100(c1 ⊗⋯cn ⊗ w+)
83
and
w
D +
w
110(c1 ⊗⋯⊗ cn ⊗ w− ⊗ w −+) +D110(c1 ⊗⋯⊗ cn ⊗ w+ ⊗ w−)
′
cancel in (C/C )/C ′′ w. This is indeed the case since to compute D +110(c1 ⊗⋯ ⊗ cn ⊗
w−⊗w+), we need only consider crossings where either a circle ci labeled w− merges
w
with the w− or the circle labeled w− splits. The same is true to compute D
−
110(c1 ⊗
⋯ ⊗ wcn ⊗ w+ ⊗ w−). On the other hand, to compute D1∗0d100(c1 ⊗ ⋯ ⊗ cn ⊗
w+) we again have two cases. The first case consists of crossings where a circle ci
labeled w− merges with the w+. These terms will cancel with those from the first
case above. The second case consists of the crossings where a w+ splits to w− ⊗
w+ + w+ ⊗ w−. These terms will cancel with the second case above.
Step 4.2.3: m ∈ 000 and a ≠ b = c. We can now study the case a ≠ b = c. In
this scenario, t1 is a split map and t2 is a merge map; see Figure 18.
/w+=0
∆ 101
Lee
∂ −10 ∂0 ∆
/ +t1 ∆ w+=0
000 001 101 100
t2
110
FIGURE 18. The first of three cases with a ≠ b = c. The labeling of the circle in
000 must be w− ⊗ w−, which forces a labeling of w− ⊗ w+ ⊗ w+ in 100.
In each case, the labeling in 000 must be w− ⊗ w−, and this forces a labeling
of w− ⊗ w+ ⊗ w+ in 100 in each case. Again, v2, v−2, ṽ0, D are all 0, and a similar
84
argument shows that the terms obtained from acting by either d or x will cancel
when we quotient to (C/C ′)/C ′′.
Step 4.2.4: m ∈ 000 and a ≠ b ≠ c. Finally, if a ≠ b ≠ c, then t1 is a merge
map and t2 is a merge map; see Figure 19.
/w+=0
∆ 101
Lee
∂0 ∂
−1
0 ∆
+
t1 ∆ /w+=0
000 001 101 100
t2
110
FIGURE 19. The first of four cases of a ≠ b ≠ c. The labeling of the circle in 000
must be w− ⊗ w− ⊗ w•, which forces a labeling of w+ ⊗ w• in 100.
In each case, the labeling in 000 must be w− ⊗ w− ⊗ w•, where w• denotes
that the innermost circle can be labeled either w+ or w−. This forces a labeling of
w+ ⊗ w• in 100 in all cases. For the last time, we verify that v2, v−2, ṽ0, D are all 0,
and a similar argument shows that the terms obtained from acting by either d or x
′ ′′
will cancel when we quotient to (C/C )/C . To summarize, we have thus shown that
q ◦A2(x1, x2, i2(m)) = 0 for all m ∈ 000 and x1, x2 ∈ sl2(∧)dg, and we conclude that
′
k3(x1, x2,m) = 0 for all m ∈ 000 and x1, x2 ∈ sl2(∧)dg as well.
Step 4.2.5: m ∈ 001. We next examine q ◦ A2(x1, x2, i2(m)) = 0 in the case
m ∈ 001. The relevant composition in the RIII cube is given in Figure 20.
85
/w+=0x1
x2
∆
/ Tw+=0
x x1 2
∆
FIGURE 20. The relevant part of the RIII cube. If we start with an element in
001, i2 ∶ (C/C ′)/C ′′ → C/C ′ gives a sum of elements in 001 and 100. We then act by
′ ′′
x1, apply the homotopy T , act by x2, and then quotient back to (C/C )/C .
As before, the possibilities for x1 are {e, f, h, v2, v−2, ṽ0, d,D, x}. Because x1
needs to increase the homological degree of m, it cannot be e, f , or h. Since we are
= Leeworking modulo w+ 0, both ∂ and ∂− are the zero map, and so D, v2, v−2, and
ṽ0 are all the zero map. Moreover, x1 cannot be d, since the resolutions in 001 and
100 have the same label, which means that they will cancel when mapped to 101.
Similarly, the fact that both resolutions have the same label also implies that the
terms in x = −[d,D] will cancel. We conclude that q ◦ A2(x1, x2, i2(m)) = 0 for all
m ∈ ′001 and x1, x2 ∈ sl2(∧)dg, and so we have thus shown that k3(x1, x2,m) = 0 on
(C/C ′)/C ′′.
Step 4.2.6: Conclusion. From the above case analysis, the only possible
′
higher operation is kn for n > 3, which could include a nonzero term of
q ◦ An−1(x1, x2, . . . , xn−1, i2(m)) with x2 = ⋯ = xn−2 = d. But because d is just
the inverse to the chain homotopy T , this will cycle the module element back and
forth between vertices 101 and 100. In particular, q ◦An−1(x1, x2, . . . , xn−1, i2(m)) =
q ◦ A2(x1, xn−1, i2(m)), which we have already shown is zero. We conclude that
′ ′
kn = 0 on (C/C )/C ′′ > (C/C ′)/C ′′for n 3, and this completes the proof that has no
′ ′′ ′ ′′
higher operations. The symmetry between (C/C )/C and (D/D )/D implies that
(D/D′)/D′′ also has no higher operations.
86
Step 5: The cubes (C/C ′)/C ′′ ′and (D/D )/D′′ are quasi-isomorphic.
′ ′′ ′ ′′
It remains to construct the map f ∶ (C/C )/C → (D/D )/D and show that it
′ ′′
respects the (trivial) L∞-module structures. Indeed, in (C/C )/C , each β1 ∈ 101
is equivalent via τ1 to some γ1 ∈ 110. The map f will send an element in 110 to
′ ′′
itself, but as an element of 101 in (D/D )/D , and it will keep the bottom layer of
the cube fixed. This is an isomorphism on spaces, and Bar-Natan checks that this
′ ′′
map is a chain map; see [Bar02]. So, for s ∈ sl2(∧)dg and x ∈ (C/C )/C we need to
compare f(s ⋅ x) and s ⋅ f(x), where the module structure is s ⋅ x = q2(s ⋅ i2(x)) =
q2(q1(s ⋅ i1(i2(x))).
Step 5.1: The case s ∈ {e, f, h}. Suppose that s ∈ {e, f, h}. First we
examine the case where z is on the bottom face of the cube. If z is in 000 or 010,
then
q2(q1(s ⋅ i1(i2(z))) = q2(q1(s ⋅ z))
Note that we abuse notation and think of z as an element of C on the right-hand
side. If z is in 001, then
−1
q2(q1(s ⋅ i1(i2(z))) = q2(q1(s ⋅ i1(z −∆ (∂C/C′z))))
= −1q2(q1(s ⋅ z − s ⋅∆ (∂C/C′z)))
= q2(q1(s ⋅ z)) −1− q2q1(s ⋅∆ (∂C/C′z))
= q2(q1(s ⋅ z))
−1
because s ⋅∆ (∂C/C′z) is in 100, which quotients to 0. If z is in 011, then
q2(q1(s ⋅ i1(i2(z))) = q2(q1(s ⋅ i1(z)))
87
= q2( −1q1(s ⋅ (z −m ∂Cz)))
= −1q2(q1(s ⋅ z)) − q2q1(s ⋅m (∂Cz))
= q2(q1(s ⋅ z))
−1
because s ⋅m (∂Cz) is labeled w+, which quotients to 0. A similar argument shows
′ ′′
that s ⋅ z = p2(p1(s ⋅ z)), if z is thought of as an element of (D/D )/D . Since
f is the identity on the bottom face, it follows that s ⋅ f(z) = f(s ⋅ z) for z ∈
000, 010, 001, 100.
If z is on the top face, we need only consider the case z ∈ 110, since any
′ ′′
element in 101 is equivalent to some z ∈ 110. Then in (C/C )/C ,
s ⋅ z = q2(q1(s ⋅ i1(i2(z))) = q2(q1(s ⋅ i1(z)))
= ( ( ( −1q2 q1 s ⋅ z −m ∂Cz)))
= q2(q1( −1s ⋅ z)) − q2q1(s ⋅m (∂Cz))
= q2(q1(s ⋅ z))
−1
because s ⋅m (∂Cz) is labeled w+, which quotients to 0. On the other hand, if we
′ ′′
consider z as an element of 101 in (D/D )/D ,
s ⋅ z = p2(p1(s ⋅ j1(j2(z))) = p2(p1(s ⋅ j1(z)))
= −1p2(p1(s ⋅ (z −m ∂Dz)))
= −1p2(p1(s ⋅ z)) − p2p1(s ⋅m (∂Dz))
= p2(p1(s ⋅ z))
88
′ ′′
Since f identically maps elements in 110 in (C/C )/C to those in 101 in
(D/D′)/D′′, it follows that s ⋅ f(z) = f(s ⋅ z) on the top face.
Step 5.2: The case s ∈ {v2, v−2, ṽ0, d,D}. Suppose that s ∈
{v2, v−2, ṽ0, d,D}. We again start with the case that z is on the bottom face of
(C/C ′)/C ′′. The cases z ∈ 000 and z ∈ 011 are straightforward to check, since f
′ ′′
is the identity on the bottom face. If z ∈ 001, then in (C/C )/C ,
q2( −1q1(s ⋅ i1(i2(z))) = q2(q1(s ⋅ i1(z −∆ (∂C/C′z))))
= −1q2(q1(s ⋅ z − s ⋅∆ (∂C/C′z)))
= q2( −1q1(s ⋅ z)) − q2(q1(s ⋅∆ (∂C/C′z)))
= q2(q1(s0 ⋅ z + s∗01 ⋅ z + s0∗1 ⋅ z))
− q2(q1( −1 −1s10∗ ⋅∆ ∂C/C′z + s1∗0 ⋅∆ ∂C/C′z))
′ ′′
and in (D/D )/D ,
s ⋅ f(z) = p2(p1(s0 ⋅ f(z) + s∗01 ⋅ f(z) + s0∗1 ⋅ f(z)))
′ ′′
and we must show that f maps the former to the latter. Indeed, in (C/C )/C ,
the terms q2q1(s∗01 ⋅ z) ( ⋅ −1and q2q1 s10∗ ∆ (∂C/C′z)) will cancel. This is because
−1
∆ (∂C/C′z) has the same labeling as z, and both maps to 101 are split maps.
−1 ′ ′′′
Furthermore, q2q1(s1∗0 ⋅ ∆ (∂C/C′z)) in (C/C )/C will be mapped via f to
⋅ ( ) −1s∗01 f z . This is because ∆ (∂C/C′z) has the same labeling as z and the maps
′
∂1∗0 in (C/C )/C ′′ and ∂∗01 in (D/D′)/D′′ are of the same type (i.e. they are either
both merge or both split), meaning s will act the same across these maps. The
case z ∈ 010 is analogous. Next, suppose that z is in the top face of the cube. If
89
z ∈ 110, then
s ⋅ z = q2(q1(s ⋅ i1(i2(z))) = q2(q1(s ⋅ i1(z)))
= q2(q1( −1s ⋅ (z −m ∂Cz)))
= q2(q1( −1s110 ⋅ z)) − q2q1(s101 ⋅m (∂Cz))
= q2(q1(s110 ⋅ z))
−1
where s110 ⋅ z is the part of s ⋅ z that remains in 110 and s101 ⋅m (∂Cz) is the part
−1
of s ⋅m (∂Cz) that remains in 101. But the latter quotients to 0, as it is labeled by
w+. On the other hand, if we consider z as an element of 101 in (D/D′)/D′′,
s ⋅ z = p2(p1(s ⋅ j1(j2(z))) = p2(p1(s ⋅ j1(z)))
= −1p2(p1(s ⋅ (z −m ∂Dz)))
= p2( −1p1(s101 ⋅ z)) − p2p1(s110 ⋅m (∂Dz))
= p2(p1(s101 ⋅ z))
−1
where s101 ⋅ z is the part of s ⋅ z that remains in 101 and s110 ⋅m (∂Dz) is the part
−1
of s ⋅m (∂Dz) that remains in 110. Similar to before, the latter quotients to 0, as
it is labeled by w+. Since f identically maps elements in 110 in (C/C ′)/C ′′ to those
′
in 101 in (D/D )/D′′, we conclude that s ⋅ f(z) = f(s ⋅ z) on the top face.
Step 5.3: The case s = x = −[d,D]. Finally, suppose that s = x = −[d,D].
′ ′′
For z ∈ (C/C )/C ,
f(s ⋅ z) = f((−dD−Dd) ⋅ z) = −df(D ⋅ z)−Df(d ⋅ z) = (−dD−Dd) ⋅f(z) = s ⋅f(z)
90
Step 5.4: Conclusion. To summarize, we have shown that for every element
s in a basis of sl2(∧) ′dg, f(s ⋅ z) = s ⋅ f(z). We conclude that f ∶ (C/C )/C ′′ →
(D/D′)/D′′ is an L∞-module quasi-isomorphism, and so up to quasi-isomorphism,
the L∞-module structure on CKh(L) is invariant under the Reidemeister III move.
3.6.2. Invariance of the sl2(∧)-module structure
Now that we have shown the invariance of the sl2(∧)dg L∞-module structure
on CKh(L) under Reidemeister moves, we can show that the sl2(∧) L∞-module
structure on CKh(L) is invariant as well.
Theorem 8. Up to L∞-quasi-isomorphism, the sl2(∧) L∞-module structure is
invariant under Reidemeister moves.
Proof. This follows from the fact that the sl2(∧) L∞-module structure on CKh(L)
was obtained from the sl2(∧)dg L∞-module structure by restricting scalars through
an L∞-algebra homomorphism I ∶ H(sl2(∧)dg) → sl2(∧)dg. In particular,
restricton of scalars preserves L∞-quasi-isomorphisms (see [Dav22]), so applying the
restriction of scalars functor to the quasi-isomorphisms constructed in the proof of
invariance for sl2(∧)dg yields quasi-isomorphisms of these complexes considered as
L∞-modules over H(sl2(∧)). Finally, the sl2(∧) L∞-module structure is invariant,
since sl2(∧) is an L∞-subalgebra of H(sl2(∧)dg).
3.7. The L∞-module structure on AKh(L)
In this section, we explain how the annular Khovanov homology AKh(L) has
an L∞-module structure that is invariant under Reidemeister moves.
91
Theorem 9. Let L be an annular link. There is an L∞-module structure on
AKh(L), invariant under Reidemeister moves. It is well-defined up to L∞-quasi-
isomorphism.
Proof. The situation can be summarized by the following diagram.
sl2(∧)dg CKh(L)
sl2(∧) AKh(L)
Theorem 8 proved that, up to L∞-quasi-isomorphism, the L∞-module
structure on CKh(L) over sl2(∧) is invariant under Reidemeister moves. By
Theorem 2, AKh(L) inherits an L∞-module structure over sl2(∧) via any choice of
chain contraction CKh(L) → AKh(L). By Lemma 5, AKh(L) is quasi-isomorphic
′
to CKh(L), so if L and L differ by Reidemeister moves, we have the following
diagram:
≅ ′
CKh(L) CKh(L )
≅ ≅
′
AKh(L) AKh(L )
This shows that AKh(L) and AKh( ′L ) are quasi-isomorphic as L∞-modules
over sl2(∧), and so this L∞-module structure is well-defined up to L∞-quasi-
isomorphism.
3.8. Examples
In this section, we explore the L∞-module structure of several knots and links.
3 Lee
Example 7. Let L be any link in S where ∂ is nonzero on Khovanov
homology. We may view L as an annular link by placing the basepoint away
92
from the link. If we denote the L∞-module operation on AKh(L) by kn,
Lee
∂0 will yield a corresponding nontrivial k3(v2, v−2,m) on AKh(L). Indeed,
the L∞-module structure on AKh(L) is induced from a cochain contraction
q
T CKh(L) ′AKh(L). If kn is the L∞-module operation on CKh(L), the
i
following equation gives a formula for k3(x1, x2,m).
′ ′ ′
k3(x1, x2,m) = k3(x1, x2,m)+q◦k2(x1, T ◦k2(x2, i(m)))+q◦k2(x2, T ◦k2(x1, i(m)))
′
The k2 operations vanish because all of the circles involved are trivial.
Example 8. In the above example, suppose we put an unknot U around the
Lee
basepoint. Let w ∈ AKh(L) be a generator on which ∂0 acts nontrivially. After
choosing a cochain contraction that respects CKh(U ⊔ L) = V ⊗ CKh(L), then in
AKh(U ⊔ L), the generators v± ⊗ w have both nontrivial k2 and k3 actions.
Example 9. The left-handed trefoil with the basepoint in the center is an example
of a knot K where AKh(K) has both nontrivial k2 and k3 operations; see Figure
21.
Lee
Indeed, in resolution 000, k3(v2, v−2, w− ⊗ w− ⊗ w−) = ∂0 (w− ⊗ w− ⊗ w−)
is nonzero in homology. Also, the usual module action of sl2(∧) acts nontrivially on
the generator v+ ⊗ v+ in resolution 111. Notice that the mirror (the right-handed
trefoil) does not have a nontrivial k3 operation in the lowest homological degree.
Example 10. The above example generalizes to any torus knot or link where
the basepoint is in the center. If every boundary map coming from the lowest
homological degree is a merge map, the resolution with each circle labeled w− will
have a nontrivial k3 operation, and the module will act nontrivially on a generator
in the highest homological degree.
93
X X
001 011
X X X X
000 010 101 111
X X
100 110
FIGURE 21. The cube of resolutions for the left-handed trefoil knot with basepoint
in the center.
The examples above illustrate that for an annular link L, AKh(L) can have
both nontrivial k2 and k3 operations. In the case where L is a split link (i.e., at
least one component is disjoint), it is further possible for a specific generator to
have both nontrivial k2 and k3 operations. On the other hand, we end this section
with a question regarding non-split links.
Question 1. Does there exist a non-split link L ⊂ A× I such that AKh(L) contains
a homology class on which the k2 and k3 operations of sl2(∧) are nontrivial? In
other words, for a non-split annular link L, can there exist m ∈ AKh(L) and
x, y1, y2 ∈ sl2(∧) such that k2(x,m) ≠ 0 and k3(y1, y2,m) ≠ 0?
94
CHAPTER IV
COMPUTATIONS
4.1. Overview
In trying to further understand the L∞-module structure on the annular
Khovanov homology of a given knot or link, implementing a computer program
to compute the annular Khovanov homology has been beneficial.
Given a knot or link diagram, we can record the diagram as follows. First,
number both the crossings and the arcs in the diagram. Each crossing then
corresponds to an array of four numbers. This array is obtained by listing the
surrounding arcs in clockwise order, starting with one of the arcs that is part of
the understrand. The collection of all of the crossing arrays is enough to determine
the knot or link diagram. The basepoint is recorded by drawing an arc from the
basepoint to infinity, documenting the arcs it passes through in a separate array.
This method of inputting knots has been used by others to compute ordinary
Khovanov homology; see, for example, the KnotTheory package in Mathematica
[Kno11].
Having inputted the knot or link, executing the program will generate the
annular Khovanov chain complex. The computation of the homology is performed
by importing these complexes into the computational algebra system, Sage. The
program can also compute the gradings of the generators that are outputted by
Sage during the homology computation. This computation requires some additional
user input, such as the number of positive and negative crossings.
95
4.2. Examples
This section presents a selection of data generated by the code. Data is
available for all knots with up to 8 crossings and all links with up to 7 crossings.
For knots with more than 11 crossings, the computational demands begin to render
the program impractical.
To illustrate the patterns that emerge, we have included two examples: the 73
knot and the Borromean rings. These examples demonstrate general patterns that
are observed in the data.
4.2.1. The knot 73
1 9
•1 •2
13 8 2
6
• 104
• •3 5 •6 3
7
14 412
•7 •8
5 11
FIGURE 22. A diagram for the knot 73. The arcs are labeled, as well as the
possible locations of the basepoint.
We can see how the code works by studying the knot 73, shown in Figure 22.
After labeling the crossings, the crossing array might be inputted as the following
code.
96
1 crossings =
[[9,2,8,1],[14,5,1,6],[6,13,7,14],[13,8,12,7],[2,9,3,10],
[10 ,3,11,4],[4,11,5 ,12]]
On the other hand, we can document the basepoint by drawing a line from
the basepoint to infinity and recording the arcs that this line intersects. For
example, to record basepoint 5, we might input the following code.
1 specialarcs = [4 ,11]
Executing the program produces the data in Figure 23.
0 1 2 3 4 5 6 7
Z2 Z Z2 × 2 3 2C2 Z × C2 Z × C2 Z × C2 × C2 Z Z × C2
Z2 Z2 Z4 Z4 Z6 Z4 Z2 Z2
Z2 Z2 Z4 Z4 Z6 Z4 Z2 Z2
Z3 Z4 Z7 Z8 Z9 Z6 × Z2C2 Z × C2
Z4 Z6 Z8 Z8 Z9 Z6 × C2 Z
2 Z × C2
Z3 Z4 Z5 Z3 × C2 Z
3 × 2C2 Z × C2 × C2 Z Z × C2
Z2 Z2 Z4 Z4 Z6 Z4 Z2 Z2
Z2 Z2 Z4 Z4 Z6 Z4 Z2 Z2
Z2 Z2 Z4 Z4 Z6 Z4 Z2 Z2
FIGURE 23. The result of executing the program for the knot 73. Each row
corresponds to the annular Khovanov homology of 73 with respect to a particular
basepoint. The first row is the ordinary Khovanov homology. The second row
corresponds to basepoint 1 in Figure 22. The third row corresponds to the
homology computed with respect to basepoint 2 in Figure 22, and so on. The
various columns represent the various homological degrees.
4.2.2. Borromean Rings
The program can also compute the annular Khovanov homology for links.
One particular example is the Borromean rings; see Figure 24.
97
•
• 21 •3
•5
•4 •6
•7
FIGURE 24. A diagram for the Borromean rings, also known as the link L6a4. The
possible locations for the basepoint are labeled.
The result of executing the code is presented in the table in Figure 26.
−3 −2 −1 0 1 2 3
Z Z3 × 2C2 Z × C2 × Z
8
C2 Z
2 Z3 × C2 × C2 Z × C2
Z2 Z6 Z4 Z8 Z4 Z6 Z2
Z Z3 × Z4 × 12 9 10 3C2 C2 × C2 Z Z Z Z
Z2 Z6 Z4 Z8 Z4 Z6 Z2
Z Z3 × Z4 × 12 9 10 3C2 C2 × C2 Z Z Z Z
Z2 Z6 Z6 Z12 Z6 Z6 Z2
Z Z3 × Z4 × 12 9 10 3C2 C2 × C2 Z Z Z Z
Z2 Z6 Z4 Z8 Z4 Z6 Z2
FIGURE 25. The result of executing the program for the link L6a4. Each
row corresponds to the annular Khovanov homology of L6a4 with respect to a
particular basepoint. The first row is the ordinary Khovanov homology. The second
row corresponds to basepoint 1 in Figure 24. The third row corresponds to the
homology computed with respect to basepoint 2 in Figure 24, and so on. The
various columns represent the various homological degrees.
As mentioned, it is possible to extract the gradings of the generators of the
above homology groups. For example, if we select basepoint 1, we obtain data in
Figure 26.
98
-3 -2 -1 0 1 2 3
1
6
-1
1,1,1
4
-1,-1,-1
1,1,1
2
-1,-1,-1
3
1,1,1,1,1
0
-1,-1,-1,-1,-1
-3
1,1,1
-2
-1,-1,-1
1,1,1
-4
-1,-1,-1
1
-6
-1
FIGURE 26. The gradings of the generators in the annular Khovanov homology
of the Borromean rings with basepoint 1, as in Figure 24. The columns represent
the homological gradings and the rows represent the filtration-adjusted quantum
gradings, as described in [GLW18]. Each cell contains the k-gradings of the
generators in a particular homological grading and filtration-adjusted quantum
grading.
99
−3 −2 −1 0 1 2 3
7 Z
Z25 Z2
2
3 Z⊕ Z2
1 Z4 Z2
−1 Z2 Z4
− Z Z23 2
− Z25 ⊕ Z2
−7 Z
FIGURE 27. The ordinary integral Khovanov homology of the Borromean
rings. The columns represent the homological gradings and the rows represent
the quantum gradings. Each cell contains the homology group present in that
particular homological grading and quantum grading. This data was computed
with Mathematica, using the KnotTheory package [Kno11]
.
It is perhaps instructive to compare this data to the ordinary Khovanov
homology of the Borromean rings; see Figure 27. For one, it gives a way to
understand the spectral sequence from annular Khovanov homology to ordinary
Khovanov homology.
4.3. Observations
These two examples highlight a general phenomenon. In particular, the
annular Khovanov homologies with torsion correspond to basepoints with respect
to which the knot or link has even winding number—that is, basepoints where we
can draw an arc from the basepoint to infinity intersecting the knot or link an even
number of times. Another observation that we can make is that it seems as if more
torsion is occuring the more crossings the region containing the basepoint abuts.
In the case of the first observation, we would ideally like to make the claim
that if a knot or link has an odd winding number around the basepoint, then this
100
implies that there is no torsion in the annular Khovanov homology. However, the
knot 819 is the only known counterexample to this claim. In particular, 819 has
winding number three around one of the possible basepoints, but the annular
Khovanov homology has 3-torsion. It is perhaps worth noting that 819 is not
alternating, and it also has 3-torsion in its odd Khovanov homology. Therefore, we
can amend the claim in several ways. The first way is to simply make a conjecture
about alternating knots, as 819 is non-alternating.
Conjecture 1. If a non-split alternating link has an odd winding number around
the basepoint, then its annular Khovanov homology has no torsion.
Alternatively, because 819 has winding number three around the basepoint in
the counterexample, it is also possible to make the following conjecture.
Conjecture 2. If a non-split link has winding number one around the basepoint,
then its annular Khovanov homology has no torsion.
Finally, because the torsion involved in the 819 counterexample is 3-torsion,
we could also conjecture the following.
Conjecture 3. If a non-split link has an odd winding number around the basepoint,
then its annular Khovanov homology has no 2-torsion.
A search for counterexamples has begun with connected sums of knots,
though not much is known at this point.
101
APPENDIX A
RESTRICTION OF SCALARS: COMPOSITION
This appendix contains graphical representations of the formulas presented in
the proof of Lemma 1.
x1 x2 ⋯ xn−1 m x1 x2 ⋯ xn−1 m x1 x2 ⋯ xn−1 m
• •
σ σ •τ
⋯ ⋯ ⋯
ai ⋯ + l ⋯ = ⋯i (g ◦ f)s
(g ◦ f) cj (g ◦ f)j r
Step 1
x1 x2 ⋯ xn−1 m x1 x2 ⋯ xn−1 m
• •
σ σ
⋯ ⋯
ai ⋯ l ⋯i
• •
θ + θ
⋯ ⋯
f ⋯ f ⋯p p
gq gq
Step 2
102
x1 x2 ⋯ xn−1 m x1 x2 ⋯ xn−1 m x1 x2 ⋯ xn−1 m
• • •
η η η
⋯ ⋯ ⋯
• • •
ψ ψ ψ
⋯ ⋯ ⋯
ai ⋯ ⋯ + li ⋯ ⋯ + li ⋯ ⋯
fp fp fp
gq gq gq
Step 3
x1 x2 ⋯ xn−1 m x1 x2 ⋯ xn−1 m
η• η•
⋯ ⋯
• •
τ ψ
⋯ ⋯
⋯ f ⋯ + lt ⋯ ⋯s
bt fs
gn+2−α gn+2−α
Step 5
103
x1 x2 ⋯ xn−1 m x1 x2 ⋯ xn−1 m
• •
π π
⋯ ⋯ ⋯ ⋯ ⋯ ⋯
fs + lt
bt fs
gn+2−α gn+2−α
Step 6
x1 x2 ⋯ xn−1 m x1 x2 ⋯ xn−1 m
φ• φ•
⋯ ⋯
⋯ f ⋯s fs
• •
σ σ
⋯ ⋯
b ⋯ ⋯x + lx
gy gy
Step 8
104
x1 x2 ⋯ xn−1 m
x1 x2 ⋯ xn−1 m
•
φ
⋯ •π
⋯ fs ⋯
⋯ fs
•
κ ⋯
⋯ gq
⋯ gq
cr
cr
Step 9 Step 10
x1 x2 ⋯ xn−1 m
•
τ x1 x2 ⋯ xn−1 m
⋯
• •
ψ τ
⋯ ⋯
⋯ f ⋯s ⋯ (g ◦ f)n+1−r
gq cr
cr
Step 11 Step 13
105
APPENDIX B
RESTRICTION OF SCALARS: OBJECTS
This appendix contains graphical representations of the formulas presented in
the proof of Lemma 2.
x1 x2 ⋯ xn−1 m x1 x2 ⋯ xn−1 m
• •
σ σ
⋯ ⋯
′
l ⋯ +
′ ⋯
p kp
′ ′
kq kq
Step 1
x1 x2 x3 ⋯ xn−1 m
x1 x2 x3 ⋯ xn−1 m
•
σ
•
σ ⋯
⋯ ′l ⋯
′ p
l ⋯p
′•
τ
•
τ
⋯ ⋯ ⋯ ⋯
I I ⋯ ⋯ ⋯ ⋯i1 i ⋯ Ii ⋯ I2 l ir
Ii Ii ⋯ Ii ⋯ I1 2 l ir
kr+1
kr+1
Step 2 Step 4
106
x1 x2 x3 xn−1 m
•
x1 x2 x3 ⋯ xn−1 m µ
⋯
•
• α
ψ
⋯ ⋯
′
l ′p lp il − 1 i1 i2 ⋯ ir
⋯ ⋯ ⋯ ⋯
Ii I ⋯ ⋯ I1 i2 Iil ir
k I I ⋯r+1 i I
⋯ I
1 i2 il ir
kr+1
Step 5 Step 6
x1 x2 x3 xn−1 m
•
µ
⋯
•
α
⋯
′
lp il − 1 i1 i2 ⋯ ir
Ii Ii Ii ⋯ Il 1 2 ir
kr+1
Step 7
107
x1 x2 x3 xn−1 m
x1 x2 x3 xn−1 m
•
µ
⋯ •η
•
γ
⋯ ⋯ ⋯ ⋯
⋯ ⋯ ⋯ ⋯
Ia ⋯ Ia Ii ⋯ I1 t 1 ir
Ia ⋯ Ia Ii ⋯ I1 t 1 ir
lt
lt
kr+1
kr+1
Step 10 Step 11
x1 x2 x3 xn−1 m
•
τ x1 x2 ⋯ xn−1 m
⋯ ⋯ ⋯
I I ⋯ I •c1 c2 cα σ
⋯
•
σ ′k ⋯p
⋯
l ⋯t ′kq
kα+2−t
Step 12 Step 13
108
x1 x2 x3 xn−1 m
x1 x2 ⋯ xn−1 m
•
τ
•
σ ⋯ ⋯ ⋯
⋯ ⋯ Ic I1 c ⋯ I2 cα
•
φ •ψ
⋯ ⋯ ⋯ ⋯
I ⋯
•
i Iir I ⋯1 j I1 j σs
⋯
kr+1 k ⋯t
ks+1
kα+2−t
Step 15 Step 17
x1 x2 x3 xn−1 m x1 x2 x3 xn−1 m
• •
τ τ
⋯ ⋯ ⋯ ⋯ ⋯ ⋯
Ic Ic ⋯ Ic Ic I ⋯ I1 2 α 1 c2 cα
• •
σ + σ
⋯ ⋯
l ⋯t k ⋯t
ku ku
Step 19
109
APPENDIX C
RESTRICTION OF SCALARS: MORPHISMS
This appendix contains graphical representations of the formulas presented in
the proof of Lemma 3.
x1 x2 ⋯ xn−1 m x1 x2 ⋯ xn−1 m x1 x2 ⋯ xn−1 m
• •
σ σ •τ
⋯ ⋯ ⋯
′ ′
m ⋯ + l ⋯ = ⋯i i (
∗
I f)
s
( ∗ ∗ ′I f) (I f)
j j nr
Step 1
x1 x2 ⋯ xn−1 m x1 x2 ⋯ xn−1 m
• •
σ σ
⋯ ⋯
′
m ⋯
′ ⋯
i li
•
τ + •τ
⋯ ⋯ ⋯ ⋯ ⋯ ⋯
Ii Ii ⋯ Ii I1 2 r i I ⋯ I1 i2 ir
fr+1 fr+1
Step 3
110
x1 x2 ⋯ xn−1 m
•
σ
x1 x2 ⋯ xn−1 m
⋯
•
ψ ⋯ •σ
⋯ ⋯ ⋯ ⋯
Ij ⋯ I1 js •ψ •τ
⋯ ⋯ ⋯ ⋯
ms+1 Ij ⋯ I I ⋯ I1 js i1 ir
ms+1
•
τ
⋯ ⋯ ⋯
Ii ⋯ I1 ir fr+1
fr+1
Step 4 Step 5
x1 x2 x3 xn−1 m x1 x2 ⋯ xn−1 m
•
π •σ
⋯ ⋯ ⋯ ⋯
Ic Ic ⋯ Ic ′1 2 α l ⋯i
•
θ •τ
⋯ ⋯ ⋯ ⋯
mt ⋯ Ii I ⋯ I1 i2 ir
f fr+1α+2−t
Step 6 Step 7
111
x1 x2 x3 xn−1 m x1 x2 x3 xn−1 m
• •
π π
⋯ ⋯ ⋯ ⋯ ⋯ ⋯
Ic I ⋯ I1 c2 c Ic Ic ⋯ Iα 1 2 cα
•
θ +
•
θ
⋯ ⋯
mt ⋯ l ⋯t
fj fj
Step 9
x1 x2 x3 xn−1 m
•
π x1 x2 ⋯ xn−1 m
⋯ ⋯ ⋯
Ic I •1 c ⋯ I2 cα τ
⋯
•
ρ ⋯ ( ∗I f)
s
⋯
⋯ fs ′
nr
nr
Step 10 Step 11
112
x1 x2 ⋯ xn−1 m
•
τ x1 x2 ⋯ xn−1 m
⋯
•
•
φ τ
⋯
⋯ ⋯ ⋯
⋯ I ⋯ I • •i1 ix γ φ
⋯ ⋯ ⋯ ⋯
fx+1 Ij ⋯ Ij I ⋯ I1 y i1 ix
•
γ fx+1
⋯ ⋯
Ij ⋯ Ij n1 y y+1
ny+1
Step 12 Step 13
x1 x2 x3 xn−1 m
•
π
⋯ ⋯ ⋯
Ic I ⋯ I1 c2 cα
•
θ
⋯
⋯ fs
nα+2−s
Step 15
113
APPENDIX D
RESTRICTION OF SCALARS: FUNCTORIALITY
This appendix contains graphical representations of the formulas presented in
the proof of Theorem 1.
x1 x2 ⋯ xn−1 m
•
σ
⋯
x x •1 2 ⋯ xn−1 m φ ⋯
⋯ ⋯
•
σ I ⋯i I1 ir
⋯
( ∗
fr+1
I f) ⋯
i
( ∗I g)
j •ψ
⋯ ⋯
Ij ⋯ I1 js
gs+1
Step 1 Step 2
114
x1 x2 x3 xn−1 m
x1 x2 ⋯ xn−1 m
•
τ
•
σ ⋯ ⋯ ⋯
⋯ ⋯
Ic Ic ⋯ I• 1 2 cα
φ •ψ
⋯ ⋯ ⋯ ⋯
⋯ I ⋯ •Ii1 ir Ij I1 j σs
⋯
fr+1 f ⋯p
gs+1
gq
Step 3 Step 5
x1 x2 x3 xn−1 m
•
τ
⋯ ⋯ ⋯
Ic I ⋯ I1 c2 cα
(g ◦ f)α+1
Step 6
115
APPENDIX E
TRANSFER OF STRUCTURE VIA CHAIN CONTRACTIONS
This appendix contains graphical representations of the formulas presented in
the proof of Theorem 2.
x1 x2 ⋯ xn−1 m x1 x2 ⋯ xn−1 m
• •
σ σ
⋯ ⋯
′ ′
l ⋯ + k ⋯p p
′ ′
kq kq
FIGURE 28. A graphical depiction of the L∞-module relation, as in [Dav22].
x1 x2 ⋯ xn−1 m
•
σ
⋯
lp ⋯
•
τ i
⋯ ⋯
At
q
Step 1. We start with the terms on the left-hand side of the L∞-module relation
′
and replace kq with its definition.
116
x1 x2 x3 xn−1 m
•
η i
⋯
•
ψ
⋯
i1 i ⋯ l ⋯2 p il − 1 it
At
q
Step 2. By the definition of unshuffle, the lp term in Step 1 goes to the first element
in one of the boxes of size i1, . . . , it determined by τ . This observation allows us to
combine σ and τ into an unshuffle η.
x1 x2 x3 ⋯ xn−1 m x1 x2 x3 ⋯ xn−1 m
• •
η i η i
⋯ ⋯ ⋯ ⋯ ⋯ ⋯
Al−1 • Al−1 •
ψ ψ
⋯ ⋯ ⋯ ⋯
T T
lp kp
ks−p+2 ks−p+2
T A Tt−l At−l
q q
Step 3. After unpacking the definition of At, the left-hand side in the above figure
represents the second term in the proof. The cases where l = 1, l = t, and p = n − 1
are not pictured here. We obtain the right-hand side after applying the L∞-module
relation.
117
x1 x2 x3 ⋯ xn−1 m x1 x2 x3 ⋯ xn−1 m x1 x2 x3 ⋯ xn−1 m
• i • •κ κ i κ i
⋯ ⋯ ⋯ ⋯ ⋯ ⋯ ⋯
k1 kp ks+1
⋯ ⋯
ks+1 ks−p+2 k1
T A T A Tt−1 t−1 At−1
q q q
x1 x2 x3 ⋯ xn−1 m x1 x2 x3 ⋯ xn−1 m x1 x2 x3 ⋯ xn−1 m
• i •κ κ i
•
κ i
⋯ ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ ⋯
Al−1 Al−1 Al−1
⋯ ⋯ ⋯ ⋯
T T T
k1 kp ks+1
ks+1 ks−p+2 k1
T T TAt−l At−l At−l
q q q
x1 x2 x3 ⋯ xn−1 m x1 x2 x3 ⋯ xn−1 m x1 x2 x3 ⋯ xn−1 m
•
κ i
•
κ i
•
κ i
⋯ ⋯ ⋯
At−1 At−1 At−1
⋯ ⋯ ⋯ ⋯
T T T
k1 kp ks+1
ks+1 ks−p+2 k1
q q q
x1 x2 x3 ⋯ xn−1 m x1 x2 x3 ⋯ xn−1 m x1 x2 x3 ⋯ xn−1 m
•
κ i
• i •κ κ i
⋯ ⋯ ⋯ ⋯
k1 kp ks+1
⋯ ⋯
ks+1 ks−p+2 k1
q q q
Step 4. Combine the permuations ψ and η into κ. There are four terms in step
three, and each row in this figure represents one of those terms, where the cases
p = 1, 2 ≤ p ≤ s, and p = s + 1 are considered separately (pictured left to right). For
1 < p < s + 1, we may combine the kp and ks−p+2 operations into the At operation
to obtain the formulas in Step 4.
118
x1 x2 ⋯ xn−1 m x1 x2 ⋯ xn−1 m x1 x2 ⋯ xn−1 m
•
κ i • •κ i κ i
⋯ ⋯ ⋯ ⋯ ⋯ ⋯
Al−1 Al Al
T ⋯ + k1 ⋯ + ⋯
k1 T
⋯ ⋯ ⋯
At−1+1 At−l At−l+1
q q q
Step 5. We can combine some of the terms in Step 4. In the graphic for Step 4
above, label the terms in the first row by 1, 2, 3, the terms in the second row by
4, 5, 6, the terms in the third row by 7, 8, 9, and the terms in the last row by 10,
11, 12. Then terms 4 and 7 combine to give the first term above on the left. The
middle term is obtained by combining terms 3 and 6. The last term is obtained by
combining terms 2, 5, 8, and 11. Moreover, the terms 1 and 10 combine, and so too
do 9 and 12, but these two cases are not pictured here.
x1 x2 ⋯ xn−1 m x1 x2 ⋯ xn−1 m
•
σ x1 x• 2 ⋯
xn−1 m
⋯ σ
•
•
α i ⋯ ′ σ⋯
⋯ ⋯ ⋯ k1 ⋯
Ar •
α i
q ⋯ ⋯ ⋯
•
β i
• A
β i r⋯ ⋯ ⋯
⋯ ⋯ ⋯ q
A Ass ′
k1
q q
Step 7. Focusing now on the right-hand side of the original L∞-module relation, we
′
substitute for kn using its definition. On the left is the case 2 ≤ p ≤ n − 1, in the
center is the case p = 1, and on the right is the case p = n − 1. After using the fact
′ ′
that i ◦ k1 = k1 ◦ i and k1 ◦ q = q ◦ k1, we obtain the formulas in Step 7.
119
x1 x2 ⋯ xn−1 m
•
θ i
⋯ ⋯
Ar
q ⋯
i
⋯
As
q
Step 8. Combine σ, α, and β into one unshuffle θ. Drawn above is the case 2 ≤ p ≤
n − 1. The cases of p = 1 and p = n − 1 are omitted.
x1 x2 ⋯ xn−1 m x1 x2 ⋯ xn−1 m x1 x2 ⋯ xn−1 m
•
θ i • •θ i θ i
⋯ ⋯ ⋯ ⋯ ⋯ ⋯
Ar Ar Ar
T ⋯ + k1 ⋯ + ⋯
k1 T
⋯ ⋯ ⋯
As As As
q q q
Step 9. In Step 8, we can replace i ◦ q with IdM +k1 ◦ T + T ◦ k1. The result is
precisely what we had in Step 5. Again, the cases of p = 1 and p = n − 1 are not
included in this picture.
120
APPENDIX F
ANNULAR KNOT DIAGRAMS
This appendix contains knot diagrams for all knots and links with up to seven
crossings. The possible basepoints are labeled in the diagrams.
•1
•1
• •2 4
•2 •3 •4
• •3 5
FIGURE 29. 31 FIGURE 30. 41
•1 •1
•2
•2 •
• 45 •• 36
•6
•3 •5•4
FIGURE 31. 51 FIGURE 32. 52
121
•6 •1
•2 •1
•5 •7
•2 •4
• •53
•6
•4 •3 •7
FIGURE 33. 61 FIGURE 34. 62
•1
•1 •2
•7
•3
• • •4 8 •32 •5 •6
•6
•7 • •45
FIGURE 35. 63 FIGURE 36. 71
• •1 1 •2
• • 27 2
•4
• •3 5 •6•8
•6 •3
•7 •
• 85 •4
FIGURE 37. 72 FIGURE 38. 73
122
•1 •2
•3 •1 •2
• •54
• •3 5
• •• 4 66
• •8•7 8 •7
FIGURE 39. 74 FIGURE 40. 75
•1
•2
•1 •2 •3
•3
• •4 •54 •5
•7
•7 •6
•6
•8
•8
FIGURE 41. 76 FIGURE 42. 77
•2
•1 •3
• • • •2 1 5 3
•4
FIGURE 43. L2a1 FIGURE 44. L4a1
123
•2 •
• 31
•2
•5
• • • • •1 5 6 3 4
•6
• •4 7
FIGURE 45. L5a1 FIGURE 46. L6a1
• •21
•3
•1 •2
• •64
• • •• 35 6 7
•7
•
• 45
FIGURE 47. L6a2 FIGURE 48. L6a3
•1
•1 •2
•3 •2 •3
•5 •6
•4 •6 • •4 5
•7
•7
FIGURE 49. L6a4 FIGURE 50. L6a5
124
•2
•1 •2 •3 •1
•3
•5
•4 • •6 4 •5
•8
•6
•7 •7
FIGURE 51. L6n1 FIGURE 52. L7a1
•1
•2 •2
•3 •1
•6 •3
•5 •6 • •7 8
•4
•
• 8 •47
•5
FIGURE 53. L7a2 FIGURE 54. L7a3
•1 •3
•2 •2
•4
•1 •3
•8 • •7 8 •5
•4
•7
• •66 •5
FIGURE 55. L7a4 FIGURE 56. L7a5
125
• •21 •3
•1
•2
•7 •4
•6 • •7 • •3 88
•6 •5
•5 •4
FIGURE 57. L7a6 FIGURE 58. L7a7
•2
•2
• •1 1
• •68
• • •6 7 •5 8
• •33
•4 •7
•5 •4
FIGURE 59. L7n1 FIGURE 60. L7n2
126
APPENDIX G
HOMOLOGY CALCULATIONS
This appendix contains annular Khovanov homology data for all knots and
links with up to seven crossings. These results were generated by the computer
code described in Chapter IV. The first row contains the ordinary Khovanov
homology. The ith row thereafter contains the annular Khovanov homology
computed with respect to the ith basepoint, as labeled in Appendix F. Each
column represents a different homological grading.
Knots
31
Z Z × C2 0 Z × Z
Z × Z Z × Z 0 Z × Z
Z Z × C2 Z Z × Z × Z
Z × Z Z × Z 0 Z × Z
Z × Z Z × Z 0 Z × Z
41
Z Z × C2 Z × Z Z Z × C2
Z × Z Z × Z Z × Z Z × Z Z × Z
Z × Z × Z Z × Z × Z × Z Z × Z × Z × Z Z × Z Z × C2
Z × Z × Z Z × Z × Z × Z Z × Z × Z Z Z × C2
Z × Z Z × Z Z × Z Z × Z Z × Z
Z × Z Z × Z Z × Z Z × Z Z × Z
127
51
Z Z × C2 Z Z × C2 0 Z
2
Z2 Z2 Z2 Z2 0 Z2
Z2 Z2 Z2 Z2 Z20
Z2 Z2 Z2 Z2 20 Z
Z2 Z2 Z2 Z2 0 Z2
Z2 Z2 Z2 Z2 0 Z2
Z Z × C2 Z Z × C2 Z Z
3
52
Z Z × C2 Z Z
2 × 2C2 Z × C2 Z
Z2 Z2 Z2 Z4 Z2 Z2
Z2 Z2 Z2 Z4 Z2 Z2
Z3 Z4 Z4 Z5 Z2 × Z2C2
Z3 Z4 Z2 Z2 × C2 Z × C2 Z
2
Z2 Z2 Z2 Z4 Z2 Z2
Z2 Z2 Z2 Z4 Z2 Z2
61
0 Z Z × 2C2 Z Z Z × C2 0
2 2 2 2 2
0 Z Z Z Z Z 0
Z2 Z2 Z20 Z2 Z2 0
Z2 Z2 Z2 Z20 Z2 0
Z2 Z2 Z2 Z2 20 Z 0
Z3 Z4 Z40 Z2 Z × C2 0
Z2 Z2 Z2 Z2 Z20 0
Z3 Z40 Z3 Z Z × C2 0
128
62
Z Z2 × Z2C2 × C2 Z
2 × C2 Z
3 × C2 Z Z × C2
Z2 Z4 Z4 Z4 Z4 Z2 Z2
Z2 Z4 Z4 Z4 Z4 Z2 Z2
Z2 Z4 Z4 Z4 Z4 Z2 Z2
Z Z2 × Z3 × Z5C2 C2 ×
6 4 3
C2 Z Z Z
Z Z2 × C2 Z
2 × C2 Z
3 × C2 Z
5 × C2 Z
4 Z3
Z2 Z4 Z4 Z4 Z4 Z2 Z2
Z2 Z4 Z4 Z4 Z4 Z2 Z2
63
Z Z2 × 2 4 2 2C2 Z × C2 Z × C2 Z × C2 Z × C2 Z × C2
Z2 Z4 Z4 Z6 Z4 Z4 Z2
Z2 Z4 Z4 Z6 Z4 Z4 Z2
Z3 Z7 Z8 Z8 Z3 × 2C2 Z × C2 Z × C2
Z3 Z7 Z6 Z6 × Z3C2 × C2 Z
2 × C2 Z × C2
Z2 Z4 Z4 Z6 Z4 Z4 Z2
Z3 Z7 Z8 Z10 Z6 Z3 × C2 Z × C2
Z2 Z4 Z4 Z6 Z4 Z4 Z2
129
71
0 Z Z × C2 Z Z × C2 0 Z
2
0
0 Z2 Z2 Z2 Z2 Z20 0
Z2 Z20 Z2 Z2 0 Z2 0
Z2 Z2 Z2 Z2 20 0 Z 0
2 2 2 2 2
0 Z Z Z Z 0 Z 0
Z2 Z2 Z2 Z20 0 Z2 0
Z2 Z20 Z2 Z2 0 Z2 0
Z20 Z2 Z2 Z2 20 Z 0
0 Z Z × 3C2 Z Z × C2 Z Z 0
72
Z Z × Z Z2C2 × C2 Z
2 × 2C2 Z × C2 Z × C2 Z
2
Z2 Z2 Z2 Z4 Z4 Z4 Z2 Z2
Z2 Z2 Z2 Z4 Z4 Z4 Z2 Z2
Z2 Z2 Z2 Z4 Z4 Z4 Z2 Z2
Z2 Z2 Z2 Z4 Z4 Z4 Z2 Z2
Z2 Z2 Z2 Z4 Z4 Z4 Z2 Z2
Z3 Z4 Z4 Z5 Z5 × 5 2 2C2 Z Z × C2 Z
Z2 Z2 Z2 Z4 Z4 Z4 Z2 Z2
Z3 Z4 Z2 Z2 × Z2C2 × C2 Z
2 × 2C2 Z × C2 Z
130
73
Z2 Z Z2 × Z2 × Z3 2C2 C2 × C2 Z × C2 × C2 Z Z × C2
Z2 Z2 Z4 Z4 Z6 Z4 Z2 Z2
Z2 Z2 Z4 Z4 Z6 Z4 Z2 Z2
Z3 Z4 Z7 Z8 Z9 Z6 × C2 Z
2 Z × C2
Z4 Z6 Z8 Z8 Z9 Z6 × C2 Z
2 Z × C2
Z3 Z4 Z5 Z3 × Z3 × Z2C2 C2 × C2 × C2 Z Z × C2
Z2 Z2 Z4 Z4 Z6 Z4 Z2 Z2
Z2 Z2 Z4 Z4 Z6 Z4 Z2 Z2
Z2 Z2 Z4 Z4 Z6 Z4 Z2 Z2
74
Z2 Z2 Z3 × C2 ×
2
C2 Z × C2 Z
3 × Z2C2 × C2 × C2 Z Z × C2
Z2 Z4 Z6 Z4 Z6 Z4 Z2 Z2
Z2 Z4 Z6 Z4 Z6 Z4 Z2 Z2
Z2 Z2 Z3 × 3 6 5 4 3C2 × C2 Z × C2 Z × C2 Z × C2 Z Z
Z2 Z4 Z6 Z4 Z6 Z4 Z2 Z2
Z2 Z4 Z6 Z4 Z6 Z4 Z2 Z2
Z2 Z2 Z3 × 3 6C2 × C2 Z × C2 Z × Z
5
C2 ×
4 3
C2 Z Z
Z2 Z4 Z6 Z4 Z6 Z4 Z2 Z2
Z2 Z4 Z6 Z4 Z6 Z4 Z2 Z2
131
75
Z Z2 × 3C2 Z × C2 Z
3 × C2 ×
3 3 2
C2 Z × C2 Z × C2 × C2 Z × C2 Z
Z2 Z4 Z6 Z6 Z6 Z6 Z2 Z2
Z2 Z4 Z6 Z6 Z6 Z6 Z2 Z2
Z2 Z4 Z6 Z6 Z6 Z6 Z2 Z2
Z Z2 × Z4 × Z7 × × Z9 Z8 4 3C2 C2 C2 C2 × C2 Z Z
Z Z3 × Z7 × Z10 × 12 11 6 4C2 C2 C2 Z Z Z Z
Z Z2 × 3C2 Z × C2 Z
4 × 7 8 4 3C2 × C2 Z × C2 Z × C2 Z Z
Z2 Z4 Z6 Z6 Z6 Z6 Z2 Z2
Z2 Z4 Z6 Z6 Z6 Z6 Z2 Z2
76
Z Z2 × 3 4 3 4 2C2 Z × C2 Z × C2 × C2 Z × C2 × C2 Z × C2 Z × C2 Z × C2
Z2 Z4 Z6 Z8 Z6 Z6 Z4 Z2
Z2 Z4 Z6 Z8 Z6 Z6 Z4 Z2
Z Z2 × Z3C2 ×
5
C2 Z × C2 × Z
7 × × Z10 Z7 Z3C2 C2 C2
Z Z3 × 7C2 Z × Z
10
C2 × C2 Z
10 × 11 7 3C2 Z Z Z
Z2 Z4 Z6 Z8 Z6 Z6 Z4 Z2
Z2 Z4 Z6 Z8 Z6 Z6 Z4 Z2
Z Z2 × C2 Z
4 × 7C2 Z × × Z
7
C2 C2 × C2 Z
9 × Z7C2 Z
3
Z2 Z4 Z6 Z8 Z6 Z6 Z4 Z2
132
77
Z Z3 × 3C2 Z × C2 × C2 Z
5 × 4 3 2C2 Z × C2 × C2 Z × C2 × C2 Z × C2 Z × C2
Z2 Z6 Z6 Z8 Z8 Z6 Z4 Z2
Z2 Z6 Z6 Z8 Z8 Z6 Z4 Z2
Z Z3 × 4C2 Z × C2 × Z
9 × 10 9 7 3C2 C2 Z × C2 Z × C2 Z Z
Z Z3 × 4 8 11 11 7 3C2 Z × C2 × C2 Z × C2 Z × C2 × C2 Z Z Z
Z2 Z6 Z6 Z8 Z8 Z6 Z4 Z2
Z2 Z6 Z6 Z8 Z8 Z6 Z4 Z2
Z Z3 × C2 Z
4 × C2 × C2 Z
9 × 10C2 Z × Z
9
C2 ×
7 3
C2 Z Z
Z2 Z6 Z6 Z8 Z8 Z6 Z4 Z2
Links
L6a1
Z Z2 × 3 2 3 2C2 Z × C2 Z Z × C2 × C2 Z × C2 Z
Z2 Z4 Z4 Z4 Z6 Z2 Z2
Z3 Z7 Z8 Z6 Z6 × 2 2C2 Z × C2 Z
Z2 Z4 Z4 Z4 Z6 Z2 Z2
Z3 Z7 Z7 Z3 Z3 × C2 ×
2
C2 Z × C2 Z
Z3 Z7 Z8 Z6 Z6 × 2 2C2 Z × C2 Z
Z2 Z4 Z4 Z4 Z6 Z2 Z2
Z2 Z4 Z4 Z4 Z6 Z2 Z2
133
L6a2
Z2 Z Z2 × Z2C2 ×
2 2
C2 Z × C2 Z × C2 Z
Z2 Z2 Z4 Z4 Z4 Z2 Z2
Z2 Z2 Z4 Z4 Z4 Z2 Z2
Z3 Z4 Z5 Z3 × Z2C2 ×
2
C2 Z × C2 Z
Z2 Z2 Z4 Z4 Z4 Z2 Z2
Z2 Z2 Z4 Z4 Z4 Z2 Z2
Z4 Z6 Z8 Z8 Z6 Z2 × 2C2 Z
Z3 Z4 Z7 Z8 Z6 Z2 × Z2C2
L6a3
Z2 0 Z Z × 2C2 Z Z × C2 Z
Z2 Z2 Z2 Z2 Z2 Z20
Z2 0 Z2 Z2 Z2 Z2 Z2
Z2 Z2 Z20 Z2 Z2 Z2
Z2 Z2 Z2 Z20 Z2 Z2
Z2 Z2 Z2 Z2 Z2 Z20
Z2 0 Z2 Z2 Z2 Z2 Z2
Z3 Z Z Z × C2 Z Z × Z
2
C2
134
L6a4
Z Z3 × 2C2 Z × C2 × C2 Z
8 Z2 Z3 × C2 × C2 Z × C2
Z2 Z6 Z4 Z8 Z4 Z6 Z2
Z Z3 × 4C2 Z × × Z
12 Z9 Z10C2 C2 Z
3
Z2 Z6 Z4 Z8 Z4 Z6 Z2
Z Z3 × 4C2 Z × × Z
12 Z9 Z10C2 C2 Z
3
Z2 Z6 Z6 Z12 Z6 Z6 Z2
Z Z3 × Z4 × × Z12 Z9C2 C2 C2 Z
10 Z3
Z2 Z6 Z4 Z8 Z4 Z6 Z2
L6a5
Z2 Z2 Z3 × C2 × C2 Z × C2 Z
6 Z Z × C2
Z2 Z4 Z6 Z2 Z6 Z2 Z2
Z3 Z7 Z11 Z6 Z8 Z2 Z × C2
Z3 Z7 Z11 Z6 Z8 Z2 Z × C2
Z2 Z4 Z6 Z2 Z6 Z2 Z2
Z2 Z4 Z6 Z2 Z6 Z2 Z2
Z4 Z9 Z12 Z6 Z7 Z Z × C2
Z3 Z7 Z11 Z6 Z8 Z2 Z × C2
135
L6n1
Z2 0 Z Z × 6C2 Z 0 0
Z2 0 Z2 Z2 Z6 0 0
Z3 Z Z Z × C2 Z
6
0 0
Z2 Z2 Z20 Z6 0 0
Z3 Z Z Z × 6C2 Z 0 0
Z4 Z2 Z2 Z2 Z6 0 0
Z3 Z Z Z × C2 Z
6
0 0
Z2 Z20 Z2 Z6 0 0
L7a1
Z Z3 × C2 Z
4 × C2 × C2 Z
4 × C2 × C2 Z
7 × C2 × C2 Z
3 × Z3C2 × C2 × C2 Z × C2
Z2 Z6 Z8 Z8 Z10 Z6 Z6 Z2
Z2 Z6 Z8 Z8 Z10 Z6 Z6 Z2
Z Z3 × Z4 × × Z6 × × Z11C2 C2 C2 C2 C2 × C2 × C2 Z
10 × C2 Z
10 Z3
Z Z3 × Z5 × 9 15 12 10 3C2 C2 × C2 Z × C2 × C2 Z × C2 Z Z Z
Z2 Z6 Z8 Z10 Z14 Z8 Z6 Z2
Z Z3 × Z5 × × Z9 × × Z15 × Z12C2 C2 C2 C2 C2 C2 Z
10 Z3
Z2 Z6 Z8 Z8 Z10 Z6 Z6 Z2
Z2 Z6 Z8 Z8 Z10 Z6 Z6 Z2
136
L7a2
Z2 Z2 Z4 × C2 × Z
3
C2 ×
5
C2 × C2 Z × Z
3
C2 ×
2
C2 × C2 Z × C2 Z × C2
Z2 Z4 Z8 Z6 Z8 Z6 Z4 Z2
Z2 Z4 Z8 Z6 Z8 Z6 Z4 Z2
Z3 Z7 Z12 Z8 × 7 4 2C2 Z × C2 Z × C2 × C2 Z × C2 Z × C2
Z2 Z4 Z8 Z6 Z8 Z6 Z4 Z2
Z4 Z9 Z15 Z13 Z10 Z4 × C2 ×
2
C2 Z × C2 Z × C2
Z3 Z7 Z14 Z13 Z12 Z7 × 3C2 Z × C2 Z × C2
Z3 Z7 Z14 Z13 Z12 Z7 × 3C2 Z × C2 Z × C2
Z2 Z4 Z8 Z6 Z8 Z6 Z4 Z2
L7a3
Z Z2 × 3 3 2 5C2 Z × C2 Z × C2 × C2 Z × C2 Z × C2 Z Z × C2
Z2 Z4 Z6 Z6 Z4 Z6 Z2 Z2
Z2 Z4 Z6 Z6 Z4 Z6 Z2 Z2
Z Z2 × 3C2 Z × C2 Z
3 × 4C2 × C2 Z × C2 Z
8 × 4C2 Z Z
3
Z2 Z4 Z6 Z6 Z4 Z6 Z2 Z2
Z2 Z4 Z6 Z6 Z4 Z6 Z2 Z2
Z2 Z4 Z6 Z6 Z6 Z10 Z4 Z2
Z Z2 × 3C2 Z × C2 Z
4 × C2 × C2 Z
6 × 9 4 3C2 Z Z Z
Z2 Z4 Z6 Z6 Z4 Z6 Z2 Z2
137
L7a4
Z Z2 × 2 3 3 5C2 Z × C2 Z × C2 Z × C2 × C2 Z × C2 Z Z × C2
Z2 Z4 Z4 Z6 Z6 Z6 Z2 Z2
Z2 Z4 Z4 Z6 Z6 Z6 Z2 Z2
Z2 Z4 Z4 Z6 Z6 Z6 Z2 Z2
Z3 Z7 Z8 Z9 Z7 × 7C2 Z × C2 Z
2 Z × C2
Z2 Z4 Z4 Z6 Z6 Z6 Z2 Z2
Z3 Z7 Z8 Z9 Z7 × 7 2C2 Z × C2 Z Z × C2
Z2 Z4 Z4 Z6 Z6 Z6 Z2 Z2
Z3 Z7 Z6 Z4 × 3C2 Z × C2 × Z
5
C2 × C2 Z Z × C2
L7a5
Z Z2 × 4 3 5 2 2C2 Z × C2 Z × C2 Z × C2 × C2 Z × C2 Z × C2 Z × C2
Z2 Z4 Z6 Z6 Z8 Z4 Z4 Z2
Z Z2 × 5 7 11 8 7 3C2 Z × C2 Z × C2 Z × C2 Z Z Z
Z2 Z4 Z6 Z6 Z8 Z4 Z4 Z2
Z Z3 × C2 Z
7 × C2 Z
10 × Z13C2 Z
8 Z7 Z3
Z2 Z4 Z6 Z6 Z8 Z4 Z4 Z2
Z2 Z4 Z6 Z6 Z8 Z4 Z4 Z2
Z2 Z4 Z6 Z6 Z8 Z4 Z4 Z2
Z Z2 × Z4 × Z4 × Z7C2 C2 C2 × ×
6 7 3
C2 C2 Z × C2 Z Z
138
L7a6
Z Z2 × Z2 × Z4 × Z2 × Z3C2 C2 C2 C2 × C2 Z Z × C2
Z2 Z4 Z4 Z6 Z4 Z4 Z2 Z2
Z2 Z4 Z4 Z6 Z4 Z4 Z2 Z2
Z2 Z4 Z4 Z6 Z4 Z4 Z2 Z2
Z2 Z4 Z4 Z6 Z4 Z4 Z2 Z2
Z2 Z4 Z4 Z6 Z4 Z4 Z2 Z2
Z2 Z4 Z4 Z6 Z4 Z4 Z2 Z2
Z Z2 × Z2C2 ×
4
C2 Z × Z
3
C2 ×
5
C2 Z × C2 Z
4 Z3
Z Z2 × Z3 × Z6 × Z5 × Z6C2 C2 C2 C2 Z
4 Z3
L7a7
Z Z3 × Z3 × 7 3 4 2C2 C2 × C2 Z × C2 Z Z × C2 × C2 × C2 Z × C2 Z
Z2 Z6 Z6 Z8 Z6 Z8 Z2 Z2
Z3 Z10 Z12 Z14 Z8 Z7 × × Z2C2 C2 × C2 Z
2
Z2 Z6 Z6 Z8 Z6 Z8 Z2 Z2
Z3 Z10 Z12 Z14 Z8 Z7 × 2 2C2 × C2 Z × C2 Z
Z2 Z6 Z6 Z8 Z6 Z8 Z2 Z2
Z2 Z6 Z6 Z8 Z6 Z8 Z2 Z2
Z3 Z10 Z12 Z14 Z8 Z7 × 2 2C2 × C2 Z × C2 Z
Z3 Z10 Z12 Z12 Z4 Z4 × C2 × C2 × C2 Z ×
2
C2 Z
139
L7n1
Z2 0 Z Z × C2 Z
3 Z × C2 0 0
Z2 Z2 Z20 Z4 Z2 0 0
Z2 2 2 4 20 Z Z Z Z 0 0
Z2 0 Z2 Z2 Z4 Z2 0 0
Z2 Z Z4 Z4 Z7 Z4 0 0
Z2 Z2 Z2 Z4 Z20 0 0
Z2 Z Z4 Z4 Z7 Z4 0 0
Z2 0 Z2 Z4 Z6 Z2 × C2 × C2 0 0
Z3 Z Z Z × C2 Z
4 Z2 × C2 0 0
L7n2
4
0 0 Z Z Z2 × C2 Z × C2 Z Z × C2
Z4 Z2 Z4 Z2 Z2 Z20 0
Z4 Z2 Z4 Z2 Z20 0 Z2
Z4 Z2 Z4 Z2 Z2 Z20 0
Z4 Z20 0 Z4 Z2 Z2 Z2
7 7 8 5 2
0 0 Z Z Z Z Z Z × C2
0 0 Z6 Z5 Z5 Z2 × C2 Z Z × C2
5
0 0 Z Z2 Z2 × 2C2 Z × C2 Z
4 Z3
6 4 4 2 2 2
0 0 Z Z Z Z Z Z
140
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