2.1 $\mathfrak{sl}_{2}$ and its finite-dimensional representations

We work over $\mathbb{C}$ throughout. Accordingly, we will denote the Lie algebras $\mathfrak{gl}_{k}(\mathbb{C})$ and $\mathfrak{sl}_{k}(\mathbb{C})$ by $\mathfrak{gl}_{k}$ and $\mathfrak{sl}_{k}$ , respectively.

We recall some elementary facts about the finite-dimensional representation theory of the Lie algebra $\mathfrak{sl}_{2}$ . The Lie algebra $\mathfrak{sl}_{2}$ has a $\mathbb{C}$ -vector space basis given by the set $\{e,f,h\}$

$$\begin{eqnarray}h=\left(\begin{array}{@{}cc@{}}1 & 0\\ 0 & -1\end{array}\right),\quad e=\left(\begin{array}{@{}cc@{}}0 & 1\\ 0 & 0\end{array}\right),\quad f=\left(\begin{array}{@{}cc@{}}0 & 0\\ 1 & 0\end{array}\right),\end{eqnarray}$$

with Lie brackets

(2.1) $$\begin{eqnarray}[e,f]=h,\quad [h,e]=2e,\quad [h,f]=-2f.\end{eqnarray}$$

As a $\mathbb{C}$ -vector space, any finite-dimensional representation, $U$ , of $\mathfrak{sl}_{2}$ decomposes into weight spaces, i.e., into eigenspaces for the action of $h$ . Explicitly,

$$\begin{eqnarray}U:=\bigoplus _{\unicode[STIX]{x1D706}\in \mathbb{Z}}U[\unicode[STIX]{x1D706}],\end{eqnarray}$$

where

$$\begin{eqnarray}U[\unicode[STIX]{x1D706}]:=\{v\in U\,\,\,\,hv=\unicode[STIX]{x1D706}v\}.\end{eqnarray}$$

The bracket relations tell us that the generators $e$ (respectively, $f$ ) act on the weight spaces as raising (respectively, lowering) operators $e_{\unicode[STIX]{x1D706}}:U[\unicode[STIX]{x1D706}]\rightarrow U[\unicode[STIX]{x1D706}+2]$ (respectively, $f_{\unicode[STIX]{x1D706}}:U[\unicode[STIX]{x1D706}]\rightarrow U[\unicode[STIX]{x1D706}-2]$ ).

Each finite-dimensional irreducible representation of $\mathfrak{sl}_{2}$ is determined by its highest weight, $N\in \mathbb{Z}^{{\geqslant}0}$ . Explicitly, for each $N\in \mathbb{Z}^{{\geqslant}0}$ , one constructs an $(N+1)$ -dimensional irreducible representation, $V_{(N)}$ , with

$$\begin{eqnarray}V_{(N)}:=\text{Span}_{\mathbb{C}}\{v,fv,\ldots ,f^{N}v\}\end{eqnarray}$$

and $hv=Nv$ (and, hence, $h(f^{i}(v))=(N-2i)f^{i}(v)$ ). All finite-dimensional $\mathfrak{sl}_{2}$ irreducible representations arise in this manner. The defining two-dimensional irreducible representation $V_{(1)}$ of $\mathfrak{sl}_{2}$ , which plays a central role in what follows, will simply be denoted $V$ .

2.2 The Lie superalgebra $\mathfrak{sl}_{2}(\wedge )$

A Lie superalgebra is a $\mathbb{Z}_{2}$ -graded vector space $V$ equipped with a bilinear Lie superbracket $[\cdot ,\cdot ]:V\times V\longrightarrow V$ which satisfies $\mathbb{Z}_{2}$ -graded versions of usual Lie algebra axioms:

• ∙ (Super skew symmetry) $[x,y]=(-1)^{|x||y|}[y,x]$ ;

• ∙ (Super Jacobi identity) $[x,[y,z]]=[[x,y],z]+(-1)^{|x||y|}[y,[x,z]]$ .

A $\mathbb{Z}$ -graded Lie superalgebra is a Lie superalgebra $V$ with a $\mathbb{Z}$ grading $V=\bigoplus _{n\in \mathbb{Z}}V(n)$ compatible with both the Lie superbracket, so that for $x\in V(n)$ and $y\in V(m)$ , $[x,y]\in V(n+m)$ , and with the $\mathbb{Z}_{2}$ grading on $V$ , so that the subspaces $V(2n)$ are in the degree-0 part of the $\mathbb{Z}_{2}$ grading, while the subspaces $V(2n+1)$ are in degree 1.

We now describe the exterior current algebra $\mathfrak{sl}_{2}(\wedge )$ , which is a $\mathbb{Z}$ -graded Lie superalgebra, by generators and relations. We have

$$\begin{eqnarray}\mathfrak{sl}_{2}(\wedge )\cong \mathfrak{sl}_{2}\oplus \mathfrak{sl}_{2},\end{eqnarray}$$

with the first summand in degree 0 and the second in degree 1 for the $\mathbb{Z}$ (and $\mathbb{Z}_{2}$ ) gradings. We fix the standard $\{e,f,h\}$ basis of $\mathfrak{sl}_{2}$ ; in order to distinguish the two distinct $\mathfrak{sl}_{2}$ summands in $\mathfrak{sl}_{2}(\wedge )$ from each other, we will write the standard basis of the degree-1 summand as $\{v_{2},v_{-2},v_{0}\}$ . In this basis, the adjoint action of $\mathfrak{sl}_{2}$ is

$$\begin{eqnarray}\displaystyle & \displaystyle e(v_{2})=0,\quad e(v_{0})=-2v_{2},\quad e(v_{-2})=v_{0}, & \displaystyle \nonumber\\ \displaystyle & \displaystyle f(v_{2})=-v_{0},\quad f(v_{0})=2v_{-2},\quad f(v_{-2})=0. & \displaystyle \nonumber\end{eqnarray}$$

Thus, in the basis $\{e,f,h,v_{2},v_{-2},v_{0}\}$ , the Lie superalgebra $\mathfrak{sl}_{2}(\wedge )$ has defining relations:

• ∙ $[e,f]=h$ ;

• ∙ $[h,e]=2e$ ;

• ∙ $[h,f]=-2f$ ;

• ∙ $[e,v_{2}]=0$ ;

• ∙ $[e,v_{0}]=-2v_{2}$ ;

• ∙ $[e,v_{-2}]=v_{0}=-[f,v_{2}]$ ;

• ∙ $[f,v_{0}]=2v_{-2}$ ;

• ∙ $[f,v_{-2}]=0$ ;

• ∙ $[h,v_{2}]=2v_{2}$ ;

• ∙ $[h,v_{0}]=0$ ;

• ∙ $[h,v_{-2}]=-2v_{-2}$ ;

• ∙ $[v_{i},v_{j}]=0$ for $i,j\in \{2,0,-2\}$ .

The $\mathbb{Z}$ and $\mathbb{Z}_{2}$ gradings on $\mathfrak{sl}_{2}(\wedge )$ induce $\mathbb{Z}$ and $\mathbb{Z}_{2}$ gradings on its enveloping algebra $U(\mathfrak{sl}_{2}(\wedge ))$ .

3.1 Sutured annular Khovanov homology and Lee homology

Let $A$ be a closed, oriented annulus and $I=[0,1]$ the closed, oriented unit interval. Via the identification

$$\begin{eqnarray}A\times I=\{(r,\unicode[STIX]{x1D703},z)\mid r\in [1,2],\unicode[STIX]{x1D703}\in [0,2\unicode[STIX]{x1D70B}),z\in [0,1]\}\subset (S^{3}=\mathbb{R}^{3}\cup \infty ),\end{eqnarray}$$

any link, $\mathbb{L}\subset A\times I$ , may naturally be viewed as a link in the complement of a standardly embedded unknot, $(U=z{-}\text{axis }\cup \infty )\subset S^{3}$ . Such an annular link $\mathbb{L}\subset A\times I$ admits a diagram, ${\mathcal{P}}(\mathbb{L})\subset A,$ obtained by projecting a generic isotopy class representative of $\mathbb{L}$ onto $A\times \{1/2\}$ , and from this diagram one can construct a triply graded chain complex, $\text{CKh}({\mathcal{P}}(\mathbb{L}))$ , using a version of Khovanov’s original construction [Reference KhovanovKho00] due to Asaeda, Przytycki and Sikora [Reference Asaeda, Przytycki and SikoraAPS04] and Roberts [Reference RobertsRob13] (see also [Reference Grigsby and WehrliGW10a]), briefly recalled here.

View ${\mathcal{P}}(\mathbb{L})\subset A$ instead as a diagram on $S^{2}-\{\mathbb{X},\mathbb{O}\}$ , where $\mathbb{X}$ (respectively, $\mathbb{O}$ ) are base points on $S^{2}$ corresponding to the inner (respectively, outer) boundary circles of $A$ . If we temporarily forget the data of $\mathbb{X}$ , we may view ${\mathcal{P}}(\mathbb{L})$ as a diagram on $\mathbb{R}^{2}=S^{2}-\{\mathbb{O}\}$ and form the ordinary bigraded Khovanov complex

$$\begin{eqnarray}\text{CKh}({\mathcal{P}}(\mathbb{L}))=\bigoplus _{(i,j)\in \mathbb{Z}^{2}}\text{CKh}^{i}({\mathcal{P}}(\mathbb{L});j),\end{eqnarray}$$

as described in [Reference KhovanovKho00].

Recall that the generators of $\text{CKh}({\mathcal{P}}(\mathbb{L}))$ correspond to oriented Kauffman states (cf. [Reference Grigsby and WehrliGW11, §4.2]). That is, in the language of [Reference Bar-NatanBar05], we identify a ‘ $v_{+}$ ’ (respectively, a ‘ $v_{-}$ ’) marking on a component of a Kauffman state with a counterclockwise (respectively, clockwise) orientation on that component. We now obtain a third grading on the complex by defining the ‘ $k$ ’ grading of a generator (up to an overall shift) to be the algebraic intersection number of the corresponding oriented Kauffman state with a fixed oriented arc $\unicode[STIX]{x1D6FE}$ from $\mathbb{X}$ to $\mathbb{O}$ that misses all crossings of ${\mathcal{P}}(\mathbb{L})$ . Note (see [Reference RobertsRob13, §2]) that component circles of a Kauffman state are either trivial (intersect the arc $\unicode[STIX]{x1D6FE}$ from $\mathbb{X}$ to $\mathbb{O}$ in an even number of points) or non-trivial (intersect $\unicode[STIX]{x1D6FE}$ in an odd number of points). Roberts proved (see [Reference RobertsRob13, Lemma 1]) that the Khovanov differential, $\unicode[STIX]{x2202}$ , is non-increasing in this extra grading. Decomposing $\unicode[STIX]{x2202}=\unicode[STIX]{x2202}_{0}+\unicode[STIX]{x2202}_{-}$ into its $k$ -grading-preserving and $k$ -grading-decreasing parts, we obtain a triply graded chain complex $(\text{CKh}({\mathcal{P}}(\mathbb{L})),\unicode[STIX]{x2202}_{0})$ whose homology,

$$\begin{eqnarray}\text{SKh}(\mathbb{L}):=\bigoplus _{(i,j,k)\in \mathbb{Z}^{3}}\text{SKh}^{i}(\mathbb{L};j,k),\end{eqnarray}$$

is an invariant of $\mathbb{L}\subset A\times I$ , called the sutured annular Khovanov homology of $\mathbb{L}$ . More can be said, as shown in the following lemma.

One therefore obtains a spectral sequence converging to $\text{Kh}(\mathbb{L})$ whose $E^{1}$ page is $\text{SKh}(\mathbb{L})$ . Each page of this spectral sequence is an invariant of $\mathbb{L}\subset A\times I$ (cf. [Reference RobertsRob13]).

The reader is warned that the other spectral sequence associated to this bicomplex (whose $E^{1}$ page is the homology of $(\text{CKh}({\mathcal{P}}(\mathbb{L}),\unicode[STIX]{x2202}_{-}))$ is not an invariant of the annular link $\mathbb{L}$ ).

The chain complex $\text{CKh}({\mathcal{P}}(\mathbb{L}))$ also comes equipped with a natural involution $\unicode[STIX]{x1D6E9}$ , defined in the following lemma (cf. [Reference Auroux, Grigsby and WehrliAGW14, Proposition 7.2(3)]).

Lemma 2. Let $\mathbb{L}\subset (A\times I)\subset S^{3}$ be an annular link,

$$\begin{eqnarray}{\mathcal{P}}(\mathbb{L})\subset (S^{2}-\mathbb{O}-\mathbb{X})\subset (S^{2}-\mathbb{O})\sim \mathbb{R}^{2}\end{eqnarray}$$

a diagram for $\mathbb{L}$ , and

$$\begin{eqnarray}{\mathcal{P}}^{\prime }(\mathbb{L})\subset (S^{2}-\mathbb{X}-\mathbb{O})\subset (S^{2}-\mathbb{X})\sim \mathbb{R}^{2}\end{eqnarray}$$

the diagram obtained by exchanging the roles of $\mathbb{O}$ and $\mathbb{X}$ . The corresponding map

$$\begin{eqnarray}\unicode[STIX]{x1D6E9}:\text{CKh}({\mathcal{P}}(\mathbb{L}),\unicode[STIX]{x2202}_{0})\rightarrow \text{CKh}({\mathcal{P}}^{\prime }(\mathbb{L}),\unicode[STIX]{x2202}_{0})\end{eqnarray}$$

is a chain isomorphism inducing an isomorphism $\text{SKh}^{i,j^{\prime }}(L;k)\cong \text{SKh}^{i,j^{\prime }}(L;-k)$ for all $(i,j^{\prime },k)\in \mathbb{Z}^{3}$ .

Let $\unicode[STIX]{x1D6F7}$ denote Lee’s deformation of Khovanov’s differential, defined in [Reference LeeLee05, §4]. Explicitly, for an elementary merge cobordism, $\unicode[STIX]{x1D6F7}$ maps a $v_{-}\otimes v_{-}$ marking to a $v_{+}$ marking on the circles involved in the cobordism, and every other labeling is mapped to $0$ . Similarly, for an elementary split cobordism, $\unicode[STIX]{x1D6F7}$ maps a $v_{-}$ marking to $v_{+}\otimes v_{+}$ and every other labeling is mapped to $0$ . Lee proved that:

• ∙ $\unicode[STIX]{x1D6F7}^{2}=0$ ;

• ∙ $\unicode[STIX]{x2202}\unicode[STIX]{x1D6F7}+\unicode[STIX]{x1D6F7}\unicode[STIX]{x2202}=0$ .

As with the Khovanov differential above, we may write the Lee deformation as a sum

$$\begin{eqnarray}\unicode[STIX]{x1D6F7}=\unicode[STIX]{x1D6F7}_{0}+\unicode[STIX]{x1D6F7}_{+},\end{eqnarray}$$

where this time $\unicode[STIX]{x1D6F7}_{0}$ has $(i,j^{\prime },k)$ degree $(1,4,0)$ , while $\unicode[STIX]{x1D6F7}_{+}$ has degree $(1,2,2)$ .

One may check directly along the various split and merge maps in the cube that $\unicode[STIX]{x1D6E9}$ commutes with $\unicode[STIX]{x2202}_{0}$ and $\unicode[STIX]{x1D6F7}_{0}$ , and that conjugation by $\unicode[STIX]{x1D6E9}$ exchanges $\unicode[STIX]{x2202}_{-}$ and $\unicode[STIX]{x1D6F7}_{+}$ . This observation, together with the equations that arise by writing the $k$ -homogeneous components of the equations $\unicode[STIX]{x2202}^{2}=0$ , $(\unicode[STIX]{x1D6F7})^{2}=0$ , and $\unicode[STIX]{x2202}\unicode[STIX]{x1D6F7}+\unicode[STIX]{x1D6F7}\unicode[STIX]{x2202}=0$ , result in a number of relations involving $\unicode[STIX]{x2202}_{0},\unicode[STIX]{x2202}_{-},\unicode[STIX]{x1D6F7}_{0},\unicode[STIX]{x1D6F7}_{+}$ , and $\unicode[STIX]{x1D6E9}$ . We collect these relationships in the following.

4.1 Khovanov bracket for annular links

Let $\mathbb{L}\subset A\times I$ be an annular link and ${\mathcal{P}}(\mathbb{L})\subset A$ a diagram for $\mathbb{L}$ , as in the previous subsection. Following Bar-Natan [Reference Bar-NatanBar05, §§2 and 11], one can define an abstract chain complex

$$\begin{eqnarray}[{\mathcal{P}}(\mathbb{L})]=(\cdots \longrightarrow [{\mathcal{P}}(\mathbb{L})]^{i-1}\longrightarrow [{\mathcal{P}}(\mathbb{L})]^{i}\longrightarrow [{\mathcal{P}}(\mathbb{L})]^{i-1}\longrightarrow \cdots \,)\end{eqnarray}$$

by constructing a resolution cube for ${\mathcal{P}}(\mathbb{L})$ and then formally taking direct sums of resolutions that sit in the same ‘ $i$ ’ degree. The differential in this complex is defined in terms of (signed) saddle cobordisms associated to the edges of the resolution cube, and the resulting complex, $[{\mathcal{P}}(\mathbb{L})]$ , is viewed as an object in the category $\text{Kom}_{/h}(\text{Mat}({\mathcal{C}}ob_{/\ell }^{3}(A)))$ , defined below.

We will use the shorthand notation $\text{Kob}_{/h}(A):=\text{Kom}_{/h}(\text{Mat}({\mathcal{C}}ob_{/\ell }^{3}(A)))$ , and we will write $\text{Kob}_{/\pm h}(A)$ for the (non-additive) category obtained from $\text{Kob}_{/h}(A)$ by identifying each morphism with its negative. Bar-Natan proved in [Reference Bar-NatanBar05] that the homotopy type of the complex $[{\mathcal{P}}(\mathbb{L})]$ is invariant under Reidemeister moves, and thus the object $[{\mathcal{P}}(\mathbb{L})]\in \text{Kob}_{/h}(A)$ provides an invariant for the annular link $\mathbb{L}\subset A\times I$ when considered up to isomorphism in $\text{Kob}_{/h}(A)$ . We will see in Proposition 4 below that $[{\mathcal{P}}(\mathbb{L})]$ also has good functoriality properties.

4.2 $\text{CKh}$ as a TQFT valued in representations of $\mathfrak{sl}_{2}$

Let $\text{gRep}(\mathfrak{sl}_{2})$ denote the category of $\mathbb{Z}$ -graded representations of $\mathfrak{sl}_{2}$ . An object of $\text{gRep}(\mathfrak{sl}_{2})$ is a direct sum

$$\begin{eqnarray}Y=\bigoplus _{n\in \mathbb{Z}}Y(n),\end{eqnarray}$$

where each $Y(n)$ is a finite-dimensional representation of $\mathfrak{sl}_{2}$ . An object $Y\in \text{gRep}(\mathfrak{sl}_{2})$ is sometimes naturally regarded as a bigraded vector space, where the component gradings are the $\mathbb{Z}$ grading above and the $\mathfrak{sl}_{2}$ weight space grading. These component gradings will be referred to as the $j^{\prime }$ and $k$ gradings, respectively (in particular, $k$ grading means $\mathfrak{sl}_{2}$ -weight-space grading). For $m\in \mathbb{Z}$ , we will denote by $\{m\}$ the grading shift operator which acts on objects of $\text{gRep}(\mathfrak{sl}_{2})$ by raising the $j^{\prime }$ grading by $m$ . That is, if $Y$ is an object of $\text{gRep}(\mathfrak{sl}_{2})$ , then $Y\{m\}$ denotes the object with components $(Y\{m\})(n+m):=Y(n)$ .

We will define a $(1+1)$ -dimensional annular TQFT with values in $\operatorname{gRep}(\mathfrak{sl}_{2})$ (where by a $(1+1)$ -dimensional annular TQFT, we here mean any sufficiently nice functor from the category ${\mathcal{C}}ob^{3}(A)$ to a category of vector spaces, possibly equipped with extra structure). In order to define this annular TQFT, we will need to use three particular graded representations of $\mathfrak{sl}_{2}$ .

• ∙ Let

  $$\begin{eqnarray}V:=\text{Span}_{\mathbb{C}}\{v_{+},v_{-}\}\end{eqnarray}$$
  denote the two-dimensional defining representation of $\mathfrak{sl}_{2}$ . The bigrading on $V$ is $j^{\prime }(v_{\pm })=0$ and $k(v_{\pm })=\pm 1$ ; in particular, $v_{+}$ is a highest-weight vector and $v_{-}=f\cdot v_{+}$ a lowest-weight vector.

• ∙ Let

  $$\begin{eqnarray}V^{\ast }:=\text{Span}_{\mathbb{ C}}\{\overline{v}_{+},\overline{v}_{-}\}\end{eqnarray}$$
  denote the dual representation to $V$ , where $\overline{v}_{-}$ is the dual vector to $v_{+}$ and $\overline{v}_{+}$ is the dual vector to $v_{-}$ . The bigrading on $V^{\ast }$ is $j^{\prime }(\overline{v}_{\pm })=0$ and $k(\overline{v}_{\pm })=\pm 1$ .

• ∙ Let

  $$\begin{eqnarray}W:=\text{Span}_{\mathbb{C}}\{w_{+},w_{-}\}\end{eqnarray}$$
  be the trivial two-dimensional representation of $\mathfrak{sl}_{2}$ , graded with $j^{\prime }(w_{\pm })=\pm 1$ and $k(w_{\pm })=0$ .

Of course, the objects $V$ and $V^{\ast }$ are isomorphic in $\text{gRep}(\mathfrak{sl}_{2})$ . However, the matrices for the action of the standard basis $\{e,f,h\}$ with respect to the bases $\{v_{\pm }\}$ on $V$ and $\{\overline{v}_{\pm }\}$ on $V^{\ast }$ are different; explicitly, the actions on $V$ and $V^{\ast }$ are given by

$$\begin{eqnarray}e\cdot v_{-}=v_{+},\quad f\cdot v_{+}=v_{-},\quad e\cdot v_{+}=f\cdot v_{-}=0,\quad h\cdot v_{\pm }=\pm v_{\pm }\end{eqnarray}$$

and

$$\begin{eqnarray}e\cdot \overline{v}_{-}=-\overline{v}_{+},\quad f\cdot \overline{v}_{+}=-\overline{v}_{-},\quad e\cdot \overline{v}_{+}=f\cdot \overline{v}_{-}=0,\quad h\cdot \overline{v}_{\pm }=\pm \overline{v}_{\pm }.\end{eqnarray}$$

(Note also that Khovanov’s ‘ $j$ ’ grading used in [Reference RobertsRob13] is the sum of the ‘ $j^{\prime }$ ’ and the ‘ $k$ ’ gradings. See Remark 2.)

We will now define an additive functor

$$\begin{eqnarray}{\mathcal{F}}:{\mathcal{C}}ob_{/\ell }^{3}(A)\longrightarrow \text{gRep}(\mathfrak{sl}_{2}).\end{eqnarray}$$

4.2.1 ${\mathcal{F}}$ on objects

We refer to a compact, connected $1$ -manifold embedded in $A$ as a trivial circle if it represents 0 in $H_{1}(A,\mathbb{Z})$ , and refer to it as a non-trivial circle otherwise. Thus, any unoriented $1$ -manifold embedded in $A$ is a disjoint union of trivial circles and non-trivial circles. Let $C\in {\mathcal{C}}ob_{/\ell }^{3}(A)$ be an unoriented $1$ -manifold $C\subset A$ , with $\ell _{n}$ non-trivial circles and $\ell _{t}$ trivial circles. Enumerate the circles of $C$ as $C_{1},\ldots ,C_{\ell _{n}+\ell _{t}}$ , and regard $C$ as a submanifold of $S^{2}-\mathbb{X}-\mathbb{O}$ , where $\mathbb{X}$ (respectively, $\mathbb{O}$ ) is a base point on $S^{2}$ corresponding to the inner (respectively, outer) boundary of $A$ , as in § 3.1. For a non-trivial circle $C_{i}$ , we denote by $X(C_{i})\in \{0,\ldots ,\ell _{n}-1\}$ the number of non-trivial circles of $C$ which lie in the same component of $S^{2}-C_{i}$ as the base point $\mathbb{X}$ , and we define

$$\begin{eqnarray}\unicode[STIX]{x1D716}(C_{i}):=(-1)^{X(C_{i})}.\end{eqnarray}$$

We now set

$$\begin{eqnarray}{\mathcal{F}}(C):=\biggl(\bigotimes _{\unicode[STIX]{x1D716}(C_{i})=1}V\biggr)\otimes \bigg(\bigotimes _{\unicode[STIX]{x1D716}(C_{i})=-1}V^{\ast }\bigg)\otimes \bigg(\bigotimes _{s=1}^{\ell _{t}}W\bigg).\end{eqnarray}$$

Thus, non-trivial circles $C_{i}$ are assigned either $V$ or $V^{\ast }$ , depending on the sign $\unicode[STIX]{x1D716}(C_{i})$ , and trivial circles are assigned $W$ .

4.2.2 ${\mathcal{F}}$ on morphisms

To define ${\mathcal{F}}$ on morphisms, we use that morphisms of ${\mathcal{C}}ob_{/\ell }^{3}(A)$ are generated by elementary Morse cobordisms: cup cobordisms creating a trivial circle, cap cobordisms annihilating a trivial circle, and saddle cobordisms, which either merge two circles into one or split one circle into two. To saddle cobordisms, we now assign the merge/split maps defined by Roberts in [Reference RobertsRob13, §2]; to cup cobordisms, we assign the map $\unicode[STIX]{x1D704}:\mathbb{C}\rightarrow W$ given by $\unicode[STIX]{x1D704}(1):=w_{+}$ ; to cap cobordisms, we assign the map $\tilde{\unicode[STIX]{x1D716}}:W\rightarrow \mathbb{C}$ given by $\tilde{\unicode[STIX]{x1D716}}(w_{+}):=0$ and $\tilde{\unicode[STIX]{x1D716}}(w_{-}):=1$ .

There are two points about the above definition that require explanation. The first is that the above assignment to cups, caps, and saddles induces a well-defined linear map on any annular cobordism. To see this, note that $\unicode[STIX]{x1D704}$ and $\tilde{\unicode[STIX]{x1D716}}$ are precisely the unit and co-unit maps defined by Khovanov in [Reference KhovanovKho00]. Roberts further shows that his merge/split maps can be viewed as the degree-0 parts (with respect to the ‘ $k$ ’ filtration) of Khovanov’s multiplication/comultiplication maps. More precisely, let $\unicode[STIX]{x1D719}$ be a linear map between graded vector spaces which is filtered for the filtration induced by the grading (so that the matrix of $\unicode[STIX]{x1D719}$ is block upper-triangular when written in a basis consisting of graded-homogeneous vectors).Footnote ⁴ Let ${\mathcal{G}}(\unicode[STIX]{x1D719})$ be the degree-0 part of $\unicode[STIX]{x1D719}$ . Note that the degree-0 part of a filtered map $\unicode[STIX]{x1D719}$ between graded vector spaces $V$ and $W$ depends only on the map $\unicode[STIX]{x1D719}$ and on the gradings on $V$ and $W$ . In particular, if two filtered maps $\unicode[STIX]{x1D719}$ and $\unicode[STIX]{x1D713}$ between graded vector spaces $V$ and $W$ are equal to each other as linear maps, then so, too, are their degree-0 parts equal. Moreover, taking the degree-0 part of a filtered map is an operation which commutes with composition. The assignment $\unicode[STIX]{x1D719}\mapsto {\mathcal{G}}(\unicode[STIX]{x1D719})$ is thus a functor from filtered linear maps between graded vector spaces to linear maps between vector spaces. It now follows from what Roberts shows that ${\mathcal{F}}$ can be written as ${\mathcal{F}}={\mathcal{G}}\circ {\mathcal{F}}^{Kh}$ , where ${\mathcal{F}}^{Kh}$ is Khovanov’s $(1+1)$ -dimensional TQFT [Reference KhovanovKho00], viewed as a functor from ${\mathcal{C}}ob_{/\ell }^{3}(A)$ to the category of graded vector spaces and filtered linear maps between graded vector spaces (graded by the $k$ grading). Since the latter functor is well defined, it follows that ${\mathcal{F}}$ is well defined as well. In particular, this argument shows that the map ${\mathcal{F}}(S)$ associated to a cobordism $S$ is independent of how $S$ is decomposed into elementary cobordisms (cups, caps, and saddles).

The second point is to observe that the linear maps that ${\mathcal{F}}$ assigns to generating morphisms of ${\mathcal{C}}ob_{/\ell }^{3}(A)$ (cup, cap, and saddle cobordisms in $A\times I$ ) are maps of $\mathfrak{sl}_{2}$ -modules, so that the functor ${\mathcal{F}}$ can be regarded as taking values in $\text{gRep}(\mathfrak{sl}_{2})$ . For cup and cap cobordisms in $A\times I$ , this is obvious, because the ‘non-identity parts’ of these cobordisms only involve trivial circles, and hence the ‘non-identity parts’ of the associated linear maps only involve $W$ factors, on which the $\mathfrak{sl}_{2}$ action is trivial. For saddle cobordisms in $A\times I$ , we have the following lemma.

Lemma 4. With the above assignment of $\mathfrak{sl}_{2}$ -module structures to the vector spaces ${\mathcal{F}}(C)$ , each merge/split map (defined as in [Reference RobertsRob13, §2]):

$$\begin{eqnarray}\displaystyle W\otimes W & \longleftrightarrow & \displaystyle W,\nonumber\\ \displaystyle W\otimes V & \longleftrightarrow & \displaystyle V,\nonumber\\ \displaystyle W\otimes V^{\ast } & \longleftrightarrow & \displaystyle V^{\ast },\nonumber\\ \displaystyle V\otimes V^{\ast } & \longleftrightarrow & \displaystyle W\nonumber\end{eqnarray}$$

is an $\mathfrak{sl}_{2}$ -module map of $(j^{\prime },k)$ -bidegree $(-1,0)$ .

Proof. Since $W$ is a direct sum of trivial $\mathfrak{sl}_{2}$ -modules, there is nothing to check for line (1). Lines (2) and (3) have essentially the same proof. For example, in line (2) we have

$$\begin{eqnarray}{V^{\ast }}^{\otimes x}\otimes V^{\otimes y}\otimes (V\otimes W)\otimes W^{\otimes z}\xrightarrow[{}]{\unicode[STIX]{x1D7D9}\otimes \cdots \unicode[STIX]{x1D7D9}\otimes (\unicode[STIX]{x1D6F7})\otimes \unicode[STIX]{x1D7D9}}{V^{\ast }}^{\otimes x}\otimes V^{\otimes y}\otimes (V)\otimes W^{\otimes z},\end{eqnarray}$$

where $\unicode[STIX]{x1D6F7}$ is either the merge or split map, depending on the direction of the arrow, and

$$\begin{eqnarray}V\otimes W:=V\{-1\}\oplus V\{1\}\end{eqnarray}$$

is a direct sum of two irreducible graded representations of $\mathfrak{sl}_{2}$ .

If $C$ (respectively, $C^{\prime }$ ) is the non-trivial circle involved in the merge/split before (respectively, after) the merge/split, then $\unicode[STIX]{x1D716}(C)=\unicode[STIX]{x1D716}(C^{\prime })$ because the number of non-trivial circles in the same component of $S^{2}-C$ as $\mathbb{X}$ is unchanged by merging/splitting with a trivial circle.

It follows that the Roberts merge (respectively, split) map is precisely the canonical degree- $(0,0)$ projection of $\mathfrak{sl}_{2}$ representations:

$$\begin{eqnarray}V\{-1\}\oplus V\{1\}\longrightarrow V\{1\}\end{eqnarray}$$

(respectively, inclusion):

$$\begin{eqnarray}V\{0\}\longrightarrow V\{0\}\oplus V\{2\}.\end{eqnarray}$$

For line (4), we have

$$\begin{eqnarray}{V^{\ast }}^{\otimes x}\otimes V^{\otimes y}\otimes (V\otimes V^{\ast })\otimes W^{\otimes z}\xrightarrow[{}]{\unicode[STIX]{x1D7D9}\otimes \cdots \unicode[STIX]{x1D7D9}\otimes (\unicode[STIX]{x1D6F7})\otimes \unicode[STIX]{x1D7D9}}{V^{\ast }}^{\otimes x}\otimes V^{\otimes y}\otimes (W)\otimes W^{\otimes z},\end{eqnarray}$$

where $\unicode[STIX]{x1D6F7}$ again denotes the merge or split, depending on the direction of the arrow, and

$$\begin{eqnarray}V\otimes V^{\ast }:=V_{(0)}\oplus V_{(2)}\end{eqnarray}$$

is the decomposition into the irreducible trivial ( $V_{(0)}$ ) and adjoint ( $V_{(2)}$ ) $\mathfrak{sl}_{2}$ representations. Let $C_{1}$ and $C_{2}$ denote the two non-trivial circles involved in the merge; then, since $C_{1}$ is adjacent to $C_{2}$ , we have $-\unicode[STIX]{x1D716}(C_{1})=\unicode[STIX]{x1D716}(C_{2})$ . Moreover, with respect to the chosen bases of $V,V^{\ast }$ , we have

$$\begin{eqnarray}V_{(0)}=\text{Span}_{\mathbb{C}}\{v_{+}\otimes \overline{v}_{-}+v_{-}\otimes \overline{v}_{+}\}\subset V\otimes V^{\ast }.\end{eqnarray}$$

We conclude that the merge and split maps are non-zero scalar multiples of the composition of inclusion and projection maps

$$\begin{eqnarray}V_{(0)}\longleftrightarrow V\otimes V^{\ast }.\end{eqnarray}$$

We have shown that each vector space ${\mathcal{F}}(C)$ can be given the structure of a graded $\mathfrak{sl}_{2}$ -module, and that the maps associated to morphisms of ${\mathcal{C}}ob_{/\ell }^{3}(A)$ intertwine the $\mathfrak{sl}_{2}$ actions. Hence, ${\mathcal{F}}$ lifts to a functor with values in $\text{gRep}(\mathfrak{sl}_{2})$ , as desired.◻

Remark 4. In fact the functor ${\mathcal{F}}$ can take values in the category of graded representations of $\mathfrak{gl}_{2}$ , after declaring $V$ to be the defining two-dimensional representation of $\mathfrak{gl}_{2}$ , $V^{\ast }$ the linear dual, and $W$ the trivial two-dimensional representation. In some sense, the distinction between $V$ and $V^{\ast }$ in the construction is more natural when ${\mathcal{F}}$ takes values in $\text{gRep}(\mathfrak{gl}_{2})$ , since $V$ and $V^{\ast }$ are no longer isomorphic as $\mathfrak{gl}_{2}$ representations. On the other hand, the exterior current algebra which appears later in the paper is that of $\mathfrak{sl}_{2}$ , not $\mathfrak{gl}_{2}$ .

We now have the following.

Proposition 1. The sutured annular Khovanov complex can be obtained from the annular Khovanov bracket by applying the functor ${\mathcal{F}}$ :

$$\begin{eqnarray}(\text{CKh}({\mathcal{P}}(\mathbb{L})),\unicode[STIX]{x2202}_{0})\cong {\mathcal{F}}([{\mathcal{P}}(\mathbb{L})]).\end{eqnarray}$$

Proof. This follows immediately from the definition and properties of ${\mathcal{F}}$ and from the definitions of $[{\mathcal{P}}(\mathbb{L})]$ and $(\text{CKh}({\mathcal{P}}(\mathbb{L})),\unicode[STIX]{x2202}_{0})$ given in [Reference Bar-NatanBar05] and [Reference RobertsRob13], respectively.◻

The above proposition implies that the sutured annular Khovanov complex of ${\mathcal{P}}(\mathbb{L})$ can be viewed as a complex in the category $\text{gRep}(\mathfrak{sl}_{2})$ . We can refine this result by introducing the Schur algebra

$$\begin{eqnarray}S(2,n):=\operatorname{im}(\unicode[STIX]{x1D70C}_{n}),\end{eqnarray}$$

where $\unicode[STIX]{x1D70C}_{n}:U(\mathfrak{sl}_{2})\rightarrow \text{End}_{\mathbb{C}}(V_{(1)}^{\otimes n})$ denotes the usual representation of $U(\mathfrak{sl}_{2})$ on the $n$ th tensor power of the defining representation of $\mathfrak{sl}_{2}$ .

Proposition 2. If there exists an essential arc $\unicode[STIX]{x1D6FE}\subset A$ intersecting ${\mathcal{P}}(\mathbb{L})\subset A$ transversely in exactly $n$ points, none of which are crossings of ${\mathcal{P}}(\mathbb{L})$ , then the $\mathfrak{sl}_{2}$ action on $\text{CKh}({\mathcal{P}}(\mathbb{L}))$ factors through the Schur algebra $S(2,n)$ .

Proof. Suppose that there is an arc $\unicode[STIX]{x1D6FE}$ as in the proposition. Then the number $\ell _{n}$ of non-trivial circles in any given resolution $C$ of ${\mathcal{P}}(\mathbb{L})$ satisfies

$$\begin{eqnarray}0\leqslant \ell _{n}\leqslant n\quad \text{and}\quad \ell _{n}\equiv n\hspace{0.6em}({\rm mod}\hspace{0.2em}2)\end{eqnarray}$$

and hence the representation

$$\begin{eqnarray}V_{(1)}^{\otimes n}=V_{(1)}^{\otimes \ell _{n}}\otimes V_{(1)}^{\otimes (n-\ell _{n})}\end{eqnarray}$$

contains a copy of the representation $V_{(1)}^{\otimes \ell _{n}}$ because $V_{(1)}^{\otimes (n-\ell _{n})}$ contains a copy of the trivial representation. It therefore follows that elements of $\operatorname{ker}(\unicode[STIX]{x1D70C}_{n})\subset U(\mathfrak{sl}_{2})$ act trivially on $V_{(1)}^{\otimes \ell _{n}}$ .

Now, as abstract $\mathfrak{sl}_{2}$ representations, the spaces $V$ and $V^{\ast }$ used in the definition of $\text{CKh}({\mathcal{P}}(\mathbb{L}))$ satisfy $V\cong V^{\ast }\cong V_{(1)}$ ; thus, as an abstract $\mathfrak{sl}_{2}$ representation, ${\mathcal{F}}(C)\cong V^{\otimes \ell _{n}}\otimes \,W^{\otimes \ell _{t}}$ is isomorphic to a direct sum of $2^{\ell _{t}}$ copies of $V_{(1)}^{\otimes \ell _{n}}$ , and it follows that the $\mathfrak{sl}_{2}$ action on ${\mathcal{F}}(C)$ factors through $U(\mathfrak{sl}_{2})/\text{ker}(\unicode[STIX]{x1D70C}_{n})\cong \operatorname{im}(\unicode[STIX]{x1D70C}_{n})=S(2,n)$ .◻

Passing to homology, Propositions 1 and 2 immediately imply the following.

Proposition 3. We have $\text{SKh}^{i}({\mathcal{P}}(\mathbb{L}))\cong H_{i}({\mathcal{F}}([{\mathcal{P}}(\mathbb{L})]))\in \text{gRep}(\mathfrak{sl}_{2})$ for all $i$ , so that $\text{SKh}({\mathcal{P}}(\mathbb{L}))$ is a bigraded representation of $\mathfrak{sl}_{2}$ . The isomorphism type of this bigraded representation is an invariant of the isotopy class of the annular link $\mathbb{L}$ . The action of $\mathfrak{sl}_{2}$ on $\text{SKh}(\mathbb{L})$ factors through the Schur algebra $S(2,n)$ , where $n$ is the wrapping number of $\mathbb{L}$ .

Here the wrapping number of an annular link $\mathbb{L}\subset A\times I$ is defined as the smallest integer $n\geqslant 0$ such that there exists an arc $\unicode[STIX]{x1D6FE}\subset A$ as in Proposition 2, where the minimum is taken over all diagrams ${\mathcal{P}}(\mathbb{L})\subset A$ representing the annular link $\mathbb{L}$ .

The above proposition will be strengthened by enlarging the action of $\mathfrak{sl}_{2}$ to an action of $\mathfrak{sl}_{2}(\wedge )$ in Proposition 7, which implies Theorem 1.

5.1 Functoriality for annular link cobordisms

The $\mathfrak{sl}_{2}$ representation structure is functorial with respect to annular link cobordisms. To state this precisely, we must first introduce the following closely related topological categories.

We have a canonical functor ${\mathcal{L}}:{\mathcal{C}}omb^{4}(A)\rightarrow {\mathcal{C}}ob_{/i}^{4}(A)$ obtained by lifting each annular link diagram to a specific annular link in general position and each Carter–Saito movie to a specific annular link cobordism with Morse decomposition described by the movie. The following is an annular version of [Reference Bar-NatanBar05, Theorem 4].

Proposition 4. The assignment ${\mathcal{P}}(\mathbb{L})\mapsto [{\mathcal{P}}(\mathbb{L})]$ extends to a functor

$$\begin{eqnarray}[-]:{\mathcal{C}}omb^{4}(A)\longrightarrow \text{Kob}_{/h}(A).\end{eqnarray}$$

Up to signs, this functor factors through the category ${\mathcal{C}}ob_{/i}^{4}(A)$ via the canonical functor ${\mathcal{L}}:{\mathcal{C}}omb^{4}(A)\rightarrow {\mathcal{C}}ob_{/i}^{4}(A)$ . In particular, $[-]$ descends to a functor ${\mathcal{C}}ob_{/i}^{4}(A)\rightarrow \text{Kob}_{/\pm h}(A)$ .

5.2 The $\mathfrak{sl}_{2}$ action via marked points

Let ${\mathcal{P}}(\mathbb{L})\subset S^{2}-\mathbb{O}-\mathbb{X}$ be a diagram of a link $\mathbb{L}\subset A\times I\subset S^{3}$ and suppose that $p_{1},\ldots ,p_{n}\subset {\mathcal{P}}(\mathbb{L})$ is a collection of $n$ distinct marked points on ${\mathcal{P}}(\mathbb{L})$ in the complement of a neighborhood of the crossings. Temporarily forgetting the data of the base point $\mathbb{X}$ , recall [Reference KhovanovKho00, Reference KhovanovKho03, Reference Hedden and NiHN13] that we have an action of

$$\begin{eqnarray}{\mathcal{A}}_{n}:=\mathbb{C}[x_{1},\ldots ,x_{n}]/(x_{1}^{2},\ldots ,x_{n}^{2})\end{eqnarray}$$

on the chain complex $CKh({\mathcal{P}}(\mathbb{L}))$ defined as follows.

Let ${\mathcal{P}}^{\prime }(\mathbb{L})$ denote the diagram obtained from ${\mathcal{P}}(\mathbb{L})$ by placing, for each $i\in \{1,\ldots ,n\}$ , a tiny trivial circle $C_{i}$ in a region adjacent to $p_{i}$ . We then have

$$\begin{eqnarray}\text{CKh}({\mathcal{P}}^{\prime }(\mathbb{L}))\cong \text{CKh}({\mathcal{P}}(\mathbb{L}))\otimes {\mathcal{A}}_{n}\end{eqnarray}$$

along with a map

$$\begin{eqnarray}m:\text{CKh}({\mathcal{P}}(\mathbb{L}))\otimes {\mathcal{A}}_{n}\rightarrow \text{CKh}({\mathcal{P}}(\mathbb{L}))\end{eqnarray}$$

realized as the composition of the $n$ (commuting) multiplication maps associated to merging ${\mathcal{P}}(\mathbb{L})$ with $C_{1},\ldots ,C_{n}$ at $p_{1},\ldots ,p_{n}$ .

Proof. We verify statement (1) by showing that the chain-level map $\sum _{i=1}^{n}(-1)^{i-1}x_{i}$ corresponding to any arc $\unicode[STIX]{x1D6FE}$ from $\mathbb{X}$ to $\mathbb{O}$ agrees with the chain-level map $f$ described in § 4.2.

To see this, let ${\mathcal{P}}_{{\mathcal{I}}}(\mathbb{L})$ be a resolution of ${\mathcal{P}}(\mathbb{L})$ . Recall from § 4.2 that the action of the lowering operator $f$ on the vector space associated to ${\mathcal{P}}_{{\mathcal{I}}}(\mathbb{L})$ is the standard tensor product representation of the actions of $f$ on the vector spaces associated to each circle of the resolution considered separately.

Now suppose that $C$ is a non-trivial circle of ${\mathcal{P}}_{{\mathcal{I}}}(\mathbb{L})$ . Any arc $\unicode[STIX]{x1D6FE}$ as above will then intersect $C$ in an odd number of points $p_{i_{1}},\ldots ,p_{i_{k}}$ according to their order of intersection with the oriented arc $\unicode[STIX]{x1D6FE}$ . As the actions of $x_{i_{1}},\ldots ,x_{i_{k}}$ on the vector space associated to ${\mathcal{P}}_{{\mathcal{I}}}(\mathbb{L})$ all agree, we then have

$$\begin{eqnarray}\mathop{\sum }_{j=1}^{k}(-1)^{i_{j}-1}x_{i_{j}}=(-1)^{i_{1}-1}x_{i_{1}}.\end{eqnarray}$$

Since $i_{1}$ differs, mod $2$ , from the number of non-trivial circles separating $C$ from $\mathbb{X}$ (i.e., using the terminology of § 4.2, we have $i_{1}-1\equiv \unicode[STIX]{x1D716}(C)\hspace{0.6em}{\rm mod}\hspace{0.2em}2$ ), the above agrees with the action of $f$ on the vector space associated to $C$ described in § 4.2. Similarly, if $C$ is a trivial circle of ${\mathcal{P}}_{{\mathcal{I}}}(\mathbb{L})$ , any arc $\unicode[STIX]{x1D6FE}$ will intersect $C$ in an even number of points $p_{i_{1}},\ldots ,p_{i_{k}}$ , so the action of $\sum _{j=1}^{k}(-1)^{i_{j}-1}x_{i_{j}}$ on $C$ is $0$ , agreeing with the action from § 4.2. This concludes the proof of statement (1), and statement (2) now follows from Lemma 5.◻

Remark 6. Recall that it is shown in [Reference Hedden and NiHN13, Proposition 2.2] that the chain-level action of ${\mathcal{A}}_{n}$ described above induces a well-defined action (modulo signs) of

$$\begin{eqnarray}{\mathcal{A}}_{\ell }:=\mathbb{C}[X_{1},\ldots ,X_{\ell }]/(X_{1}^{2},\ldots ,X_{\ell }^{2})\end{eqnarray}$$

on the ordinary Khovanov homology of an $\ell$ -component link (one need only ensure that each link component contains at least one marked point). In particular, $X_{i}$ denotes the map induced on the ordinary Khovanov homology of $\mathbb{L}$ by (any one of) the base point(s) marking the $i$ th component of $\mathbb{L}$ .

Recalling that there is a spectral sequence relating $\text{SKh}(\mathbb{L}\subset A\times I)$ to $\text{Kh}(\mathbb{L}\subset S^{3})$ , it is tempting to conclude that the map induced by the lowering operator $f$ on $\text{Kh}(\mathbb{L})$ agrees with $g=\sum _{i=1}^{n}\unicode[STIX]{x1D716}_{i}X_{i}$ for some choices $\unicode[STIX]{x1D716}_{i}\in \{\pm 1\}$ .

However, the $E^{\infty }$ page of the spectral sequence is the associated graded $\text{gr}\text{Kh}(\mathbb{L})$ of $\text{Kh}(\mathbb{L})$ with respect to the induced filtration. As a result, we can only conclude the weaker statement that the highest-degree terms of the maps agree. More precisely, noting that $f$ is a filtered map of degree $-2$ on the filtered complex described in Remark 1, we can regard $f$ as a filtration-preserving map

$$\begin{eqnarray}f:\text{CKh}(\mathbb{L})\rightarrow \text{CKh}(\mathbb{L})\{\{2\}\},\end{eqnarray}$$

where in the above $\{\{n\}\}$ is the operator that shifts $k$ gradings (and hence the induced filtration) up by $2$ . Let $f_{\infty }$ denote the map induced by $f$ on the $E^{\infty }$ page of the spectral sequence associated to the $k$ filtration, and decompose $g=g_{-2}+g_{-4}+\cdots \,$ into its $k$ -homogeneous terms with respect to the induced $k$ grading on the $E^{\infty }$ page. Then

$$\begin{eqnarray}f_{\infty }=g_{-2}.\end{eqnarray}$$

6.1 Chain complexes and Lie superalgebras

Let $(C^{\bullet },\unicode[STIX]{x2202})$ be a $\mathbb{Z}$ -graded chain complex. The $\mathbb{Z}$ grading $C^{\bullet }=\bigoplus _{i\in \mathbb{Z}}C^{i}$ induces a $\mathbb{Z}_{2}$ grading

$$\begin{eqnarray}C=C^{\text{even}}\oplus C^{\text{odd}},\end{eqnarray}$$

where

$$\begin{eqnarray}C^{\text{even}}=\bigoplus _{n\in \mathbb{Z}}C^{2n}\text{ and }C^{\text{odd}}=\bigoplus _{n\in \mathbb{Z}}C^{2n+1}\end{eqnarray}$$

and hence the structure of a super vector space. In fact, as we now explain, $C^{\bullet }$ is naturally a representation of its endomorphism space, which itself may be regarded as a Lie superalgebra.

Let $\text{End}(C^{\bullet })$ denote the hom complex of $C^{\bullet }$ , which is a $\mathbb{Z}$ -graded super vector space in its own right:

$$\begin{eqnarray}\text{End}(C^{\bullet })=\bigoplus _{n\in \mathbb{Z}}\text{End}^{n}(C^{\bullet }),\end{eqnarray}$$

$$\begin{eqnarray}\text{End}^{n}(C^{\bullet })=\{f:C^{\bullet }\rightarrow C^{\bullet +n}\}.\end{eqnarray}$$

Elements of $\text{End}^{n}(C^{\bullet })$ are linear maps of homological degree $n$ , and are not required to intertwine the differential $\unicode[STIX]{x2202}$ . In fact the differential $\unicode[STIX]{x2202}\in \text{End}(C^{\bullet })$ is itself a degree-one endomorphism of $C^{\bullet }$ .

We endow $\text{End}(C^{\bullet })$ with the structure of a Lie superalgebra by declaring, for $f\in \text{End}^{n}(C^{\bullet })$ and $g\in \text{End}^{m}(C^{\bullet })$ ,

$$\begin{eqnarray}[f,g]=fg-(-1)^{nm}gf.\end{eqnarray}$$

The superalgebra $\text{End}(C^{\bullet })$ is also a chain complex, with differential

$$\begin{eqnarray}{\mathcal{D}}:\text{End}(C^{\bullet })\longrightarrow \text{End}(C^{\bullet +1}),\quad {\mathcal{D}}(f)=[\unicode[STIX]{x2202},f].\end{eqnarray}$$

Let $H(\text{End}(C^{\bullet }))$ be the homology of $\text{End}(C^{\bullet })$ . Then there is a canonical morphism of Lie superalgebras

$$\begin{eqnarray}H(\text{End}(C^{\bullet }))\longrightarrow \text{End}(H(C^{\bullet })).\end{eqnarray}$$

Explicitly, if $\unicode[STIX]{x1D703}\in \text{End}(C^{\bullet })$ and $x\in C^{\bullet }$ are cycles representing homology classes $[\unicode[STIX]{x1D703}]$ and $[x]$ , respectively, then one makes the well-defined assignment

$$\begin{eqnarray}[\unicode[STIX]{x1D703}]([x]):=[\unicode[STIX]{x1D703}(x)].\end{eqnarray}$$

Note that the cycles in $(\text{End}^{n}(C^{\bullet }),{\mathcal{D}})$ are precisely the chain maps (or skew-chain maps, depending on the parity of $n$ ):Footnote ⁶

$$\begin{eqnarray}C^{\bullet }\rightarrow C^{\bullet }[n],\end{eqnarray}$$

and the boundaries are precisely those chain maps that are chain homotopic to $0$ . Informally, one views the image of $H(\text{End}(C^{\bullet }))$ under the canonical morphism above as the collection of (graded) chain maps on $C^{\bullet }$ , modulo homotopy.

6.2 The Lie superalgebra $\mathfrak{sl}_{2}(\wedge )_{dg}$

We now describe a $\mathbb{Z}$ -graded Lie superalgebra $\mathfrak{sl}_{2}(\wedge )_{dg}$ that is closely related (cf. Lemma 7) to the Lie superalgebra $\mathfrak{sl}_{2}(\wedge )$ defined in § 2.2.

The underlying $\mathbb{Z}$ -graded super vector space of $\mathfrak{sl}_{2}(\wedge )_{dg}$ is generated (as a Lie superalgebra) by $\{e,f,h,v_{2},v_{-2},d,D\}$ , with the generators $\{e,f,h\}$ in degree $0$ and $\{v_{2},v_{-2},d,D\}$ in degree $1$ . The $\mathbb{Z}_{2}$ grading on $\mathfrak{sl}_{2}(\wedge )$ is induced from the $\mathbb{Z}$ grading, and the defining super commutation relations are as follows:

• ∙ $[e,f]=h$ ;

• ∙ $[h,e]=2e$ ;

• ∙ $[h,f]=-2f$ ;

• ∙ $[e,v_{2}]=0$ ;

• ∙ $[e,v_{-2}]=-[f,v_{2}]$ ;

• ∙ $[f,v_{-2}]=0$ ;

• ∙ $[h,v_{2}]=2v_{2}$ ;

• ∙ $[h,v_{-2}]=-2v_{-2}$ ;

• ∙ $[d,y]=0$ for all $y\in \{e,f,h,v_{2},v_{-2}\}$ ;

• ∙ $[D,y]=0$ for all $y\in \{e,f,h,v_{2},v_{-2}\}$ ;

• ∙ $[d,d]=[D,D]=[v_{2},v_{2}]=[v_{-2},v_{-2}]=0$ ;

• ∙ $[v_{2},v_{-2}]+[d,D]=0$ .

Let $\tilde{v_{0}}=[e,v_{-2}]$ $x=[v_{2},v_{-2}]$ . One may check using the above relations and the super Jacobi identity that $\tilde{v_{0}}=-[f,v_{2}]$ and that $x=-[d,D]=\frac{1}{2}[\tilde{v}_{0},\tilde{v}_{0}]$ . Then we have the following.

It is clear from the above description that $\mathfrak{sl}_{2}(\wedge )_{dg}$ is closely related to $\mathfrak{sl}_{2}(\wedge )$ . Indeed, we may regard $\mathfrak{sl}_{2}(\wedge )_{dg}$ as a chain complex by declaring the adjoint action of $d$ to be the differential, as described in the previous subsection. We have the following.

Lemma 7. The homology of $\mathfrak{sl}_{2}(\wedge )_{dg}$ taken with respect to the differential $[d,\cdot ]$ is isomorphic to the direct sum of $\mathfrak{sl}_{2}(\wedge )$ and the trivial Lie super algebra:

$$\begin{eqnarray}H(\mathfrak{sl}_{2}(\wedge )_{dg},[d,\cdot ])\cong \mathfrak{sl}_{2}(\wedge )\oplus \mathbb{C}.\end{eqnarray}$$

Moreover, if $(C^{\bullet },d)$ is a $\mathbb{Z}$ -graded $\mathfrak{sl}_{2}(\wedge )_{dg}$ representation, regarded as a chain complex with differential given by the action of $d$ , then the canonical map

$$\begin{eqnarray}H(\mathfrak{sl}_{2}(\wedge )_{dg})\rightarrow \text{End}(H(C^{\bullet },d))\end{eqnarray}$$

factors through $\mathfrak{sl}_{2}(\wedge )$ .

6.3 The current algebra $\mathfrak{sl}_{2}(\wedge )$ and its action on $\text{SKh}(\mathbb{L})$

We are now ready to extend the action of $\mathfrak{sl}_{2}$ on $\text{CKh}(\mathbb{L})$ to an action of $\mathfrak{sl}_{2}(\wedge )_{dg}$ by defining

$$\begin{eqnarray}\unicode[STIX]{x1D6F7}:\mathfrak{sl}_{2}(\wedge )_{dg}\longrightarrow \text{End}(\text{CKh}(\mathbb{L}))\end{eqnarray}$$

via:

• ∙ $v_{2}\mapsto \unicode[STIX]{x1D6F7}_{+}$ ;

• ∙ $v_{-2}\mapsto \unicode[STIX]{x2202}_{-}$ ;

• ∙ $d\mapsto \unicode[STIX]{x2202}_{0}$ ;

• ∙ $D\mapsto \unicode[STIX]{x1D6F7}_{0}$ .

Proposition 6. The above assignment defines a homomorphism of Lie superalgebras

$$\begin{eqnarray}\unicode[STIX]{x1D6F7}:\mathfrak{sl}_{2}(\wedge )_{dg}\longrightarrow \text{End}(\text{CKh}(\mathbb{L})).\end{eqnarray}$$

Proof. The $\mathfrak{sl}_{2}$ relations involving only $\{e,f,h\}$ were established in the course of proving Proposition 3. The relations involving commutators between pairs of degree-one elements $\{v_{-2},v_{2},d,D\}$ follow immediately from the relations in Lemma 3.

The mixed relations involving commutators of one of $\{e,f,h\}$ with one of $\{v_{-2},v_{2},d,D\}$ follow from a straightforward case-by-case check along the three possible types of edges in the cube of resolutions.

For example, consider the relation $[e,v_{-2}]=[v_{2},f]$ along an edge in which two non-trivial circles merge to form a single trivial circle. Recalling that $v_{-2}$ is identified with $\unicode[STIX]{x2202}_{-}$ (respectively, $v_{2}$ is identified with $\unicode[STIX]{x1D6F7}_{+}$ ), one computes in this case that

$$\begin{eqnarray}\displaystyle (\ast )\otimes (v_{-}\otimes \overline{v}_{-}) & \longmapsto & \displaystyle 0,\nonumber\\ \displaystyle (\ast )\otimes (v_{+}\otimes \overline{v}_{+}) & \longmapsto & \displaystyle 0,\nonumber\\ \displaystyle (\ast )\otimes (v_{-}\otimes \overline{v}_{+}) & \longmapsto & \displaystyle -(\ast )\otimes w_{+},\nonumber\\ \displaystyle (\ast )\otimes (v_{+}\otimes \overline{v}_{-}) & \longmapsto & \displaystyle (\ast )\otimes w_{+},\nonumber\end{eqnarray}$$

under both the map $e\unicode[STIX]{x2202}_{-}-\unicode[STIX]{x2202}_{-}e$ and the map $\unicode[STIX]{x1D6F7}_{+}f-f\unicode[STIX]{x1D6F7}_{+}$ , where in the above we are denoting by $(\ast )$ a fixed marking on the circles in the resolution uninvolved in the merge cobordism.◻

Passing to homology and using Lemma 7, we arrive at the following proposition, which completes the proof of Theorem 1 from the introduction.

Proposition 7. The homomorphism of Lie superalgebras $\unicode[STIX]{x1D6F7}:\mathfrak{sl}_{2}(\wedge )_{dg}\longrightarrow \text{End}(\text{CKh}(\mathbb{L}))$ induces a homomorphism of Lie superalgebras

$$\begin{eqnarray}\unicode[STIX]{x1D6F9}:\mathfrak{sl}_{2}(\wedge )\longrightarrow \text{End}(\text{SKh}(\mathbb{L})).\end{eqnarray}$$

This action of $\mathfrak{sl}_{2}(\wedge )$ is functorial for annular link cobordisms.

Before proving this proposition, we prove a useful lemma involving the involution $\unicode[STIX]{x1D6E9}$ defined in Lemma 2.

Lemma 8. Let ${\mathcal{P}}(\mathbb{L}),{\mathcal{P}}(\mathbb{L}^{\prime })$ be annular link diagrams whose underlying links are connected by an annular link cobordism $F$ , let

$$\begin{eqnarray}m_{0}:\text{CKh}(\mathbb{L},\unicode[STIX]{x2202}_{0})\rightarrow \text{CKh}(\mathbb{L}^{\prime },\unicode[STIX]{x2202}_{0})\end{eqnarray}$$

be the chain map induced on the annular Khovanov chain complex by any decomposition of $F$ into elementary cobordisms and Reidemeister moves, and let $\unicode[STIX]{x1D6E9}$ be the involution on ${\mathcal{P}}(\mathbb{L}),{\mathcal{P}}(\mathbb{L}^{\prime })$ defined in Lemma 2. Then we have $\unicode[STIX]{x1D6E9}m_{0}=m_{0}\unicode[STIX]{x1D6E9}.$

Proof of Proposition 7.

Since the generator $d$ of $\mathfrak{sl}_{2}(\wedge )_{dg}$ is sent to the annular differential $\unicode[STIX]{x2202}_{0}$ , the homology of $\mathfrak{sl}_{2}(\wedge )_{dg}$ taken with respect to $d$ acts on $\text{SKh}(\mathbb{L})$ . By Lemma 7, the action of this homology factors through the current algebra $\mathfrak{sl}_{2}(\wedge )$ .

What remains is to show that any annular link cobordism commutes (up to homotopy) with the operators $v_{2},v_{-2}$ . To see this, let $m$ be the chain map induced on the ordinary Khovanov chain complex by an annular link cobordism. Write $m=m_{0}+m_{-}$ , where $m_{0}$ preserves the annular $k$ grading and $m_{-}$ is a linear combination of components of negative $k$ degree. The claim is that $m_{0}$ and $v_{-2}(=\unicode[STIX]{x2202}_{-})$ commute up to homotopy. Since $m$ is a chain map, Khovanov’s differential $\unicode[STIX]{x2202}=\unicode[STIX]{x2202}_{0}+\unicode[STIX]{x2202}_{-}$ commutes with $m$ and it follows that

$$\begin{eqnarray}m_{0}\unicode[STIX]{x2202}_{-}-\unicode[STIX]{x2202}_{-}m_{0}+m_{-}\unicode[STIX]{x2202}_{0}-\unicode[STIX]{x2202}_{0}m_{-}=0.\end{eqnarray}$$

Thus, $m_{-}$ provides a homotopy between $[m_{0},\unicode[STIX]{x2202}_{-}]$ and $0$ , as desired. The analogous statement for $v_{2}$ follows from the above, combined with the observations that $\unicode[STIX]{x1D6F7}_{+}=\unicode[STIX]{x1D6E9}\unicode[STIX]{x2202}_{-}\unicode[STIX]{x1D6E9}$ (Lemma 3) and that $\unicode[STIX]{x1D6E9}$ commutes with $m_{0}$ (Lemma 8).◻

The observant reader may now wonder whether the $\mathfrak{sl}_{2}(\wedge )_{dg}$ action on the sutured annular chain complex $\text{CKh}(\mathbb{L})$ gives rise to any interesting new actions on Khovanov or Lee homology. Sadly, the answer is no, as we see in Lemmas 9 and 10. In what follows, let ${\mathcal{L}}$ denote the two-dimensional abelian Lie superalgebra with a single degree- $0$ generator, $y_{0}$ , and a single degree- $1$ generator, $y_{1}$ .

Lemma 9. The homology $H(\mathfrak{sl}_{2}(\wedge )_{dg},[d+v_{-2},\cdot ])$ has a codimension-one direct summand isomorphic to ${\mathcal{L}}$ . Moreover, if $(C^{\bullet },d+v_{-2})$ is any $\mathbb{Z}$ -graded $\mathfrak{sl}_{2}(\wedge )_{dg}$ representation, regarded as a chain complex with differential $d+v_{-2}$ , then the canonical map

$$\begin{eqnarray}H(\mathfrak{sl}_{2}(\wedge )_{dg})\rightarrow \text{End}(H(C^{\bullet },d+v_{-2}))\end{eqnarray}$$

factors through ${\mathcal{L}}$ .

Proof. The set $\{[f],[v_{2}+D],[d]\}$ is a basis for $H(\mathfrak{sl}_{2}(\wedge )_{dg},[d+v_{-2},\cdot ])$ , and we calculate that all pairwise brackets are nullhomologous. The map

$$\begin{eqnarray}H(\mathfrak{sl}_{2}(\wedge )_{dg},[d+v_{-2},\cdot ])\rightarrow {\mathcal{L}}\end{eqnarray}$$

sending $[f]\mapsto y_{0}$ , $[v_{2}+D]\mapsto y_{1}$ , and $[d]\mapsto 0$ is a Lie superalgebra homomorphism, and $[d]=[d+v_{-2}]$ will act trivially on the homology of any $\mathfrak{sl}_{2}(\wedge )_{dg}$ representation.◻

In particular, the action of $\mathfrak{sl}_{2}(\wedge )_{dg}$ on the annular chain complex $\text{CKh}(\mathbb{L})$ induces two commuting endomorphisms of $\text{Kh}(\mathbb{L})$ , one of which can be described in terms of the standard action by base points on Khovanov homology (compare § 5.2) and the other of which can be described in terms of the Lee deformation.

Lemma 10. The homology $H(\mathfrak{sl}_{2}(\wedge )_{dg},[d+v_{-2}+D+v_{2},\cdot ])$ has a codimension-one direct summand isomorphic to ${\mathcal{L}}$ . Moreover, if $(C^{\bullet },d+v_{-2}+D+v_{2})$ is any $\mathbb{Z}$ -graded $\mathfrak{sl}_{2}(\wedge )_{dg}$ representation, regarded as a chain complex with differential $d+v_{-2}+D+v_{2}$ , then the canonical map

$$\begin{eqnarray}H(\mathfrak{sl}_{2}(\wedge )_{dg})\rightarrow \text{End}(H(C^{\bullet },d+v_{-2}+D+v_{2}))\end{eqnarray}$$

factors through ${\mathcal{L}}$ .

Thus, the action of $\mathfrak{sl}_{2}(\wedge )_{dg}$ on the annular chain complex $\text{CKh}(\mathbb{L})$ induces two commuting endomorphisms of the Lee homology of $\mathbb{L}$ . It is straightforward to verify (e.g., by using the canonical generators described in [Reference RasmussenRas10, §2.4]) that each Lee homology class associated to an orientation of $\mathbb{L}$ is an eigenvector for the endomorphism represented by $[e\,+\,f]$ . The endomorphism represented by $[d+v_{-2}]$ increases homological grading by $1$ and hence must act trivially, since Lee homology is supported in even homological gradings [Reference LeeLee05, Proposition 4.3].

7.1 Cobordism maps associated to tangles

Let $K$ be a smooth framed oriented knot in $A\times I$ and

$$\begin{eqnarray}\unicode[STIX]{x1D704}:S^{1}\times D^{2}\longrightarrow A\times I\end{eqnarray}$$

an embedding which sends the circles $S^{1}\times \{0\}$ and $S^{1}\times \{1\}$ respectively to $K$ and to a longitude of $K$ specifying the framing, where $D^{2}$ is the closed unit disk in $\mathbb{C}$ and $S^{1}:=\unicode[STIX]{x2202}D^{2}$ . Moreover, let $P_{n}\subset \operatorname{int}(D^{2})$ be a collection of $n$ evenly spaced points on the real axis. In this situation, the $n$ -cable of $K$ can be defined as follows.

Now suppose that $T$ is a smooth $(n,n)$ tangle, i.e., a smooth properly embedded $1$ -manifold $T\subset D^{2}\times I$ such that $\unicode[STIX]{x2202}T=P_{n}\times \unicode[STIX]{x2202}I$ . To $T$ , we can associate the ‘surface of revolution’ $S^{1}\times T\subset S^{1}\times D^{2}\times I$ .

Figure 1. Schematic depiction of a tangle $T\subset D^{2}\times I$ (left) and the associated surface of revolution $S^{1}\times T\subset S^{1}\times D^{2}\times I$ (right). The cobordism $K^{T}:K^{n}\rightarrow K^{n}$ is obtained from $S^{1}\times T$ by applying the embedding $\overline{\unicode[STIX]{x1D704}}:S^{1}\times D^{2}\times I\rightarrow A\times I\times I$ .

Since the formal Khovanov bracket of annular links is functorial with respect to smooth annular link cobordisms (by Proposition 4), the $T$ -cable cobordism of $K$ induces a chain map

$$\begin{eqnarray}[K^{T}]:[K^{n}]\longrightarrow [K^{n}],\end{eqnarray}$$

which is well defined up to sign and homotopy. Likewise, $K^{T}$ induces maps

$$\begin{eqnarray}\unicode[STIX]{x1D719}_{K^{T}}:\text{SKh}(K^{n})\longrightarrow \text{SKh}(K^{n})\quad \text{and}\quad \unicode[STIX]{x1D719}_{K^{T}}^{\prime }:\text{Kh}^{\prime }(K^{n})\longrightarrow \text{Kh}^{\prime }(K^{n}),\end{eqnarray}$$

where $\text{Kh}^{\prime }(K^{n})$ denotes the Lee homology of $K^{n}$ (viewed as a link in $\mathbb{R}^{3}$ ). Note that the latter maps are well defined up to sign.

7.2 Fixing the sign ambiguity

In the following, let $E_{i}$ , $\unicode[STIX]{x1D6F4}_{i}$ , and $\unicode[STIX]{x1D6F4}_{i}^{-1}$ denote $(n,n)$ tangles which represent respectively the generator $e_{i}$ of the Temperley–Lieb algebra $\text{TL}_{n}(a)$ , the generator $\unicode[STIX]{x1D70E}_{i}$ of the braid group $\mathfrak{B}_{n}$ , and the inverse of the generator $\unicode[STIX]{x1D70E}_{i}\in \mathfrak{B}_{n}$ (see Figure 2).

Figure 2. The tangles $E_{i}$ (left) and $\unicode[STIX]{x1D6F4}_{i}$ (right).

The goal of this subsection is to pin down the sign in the definition of the maps $[K^{T}]$ , $\unicode[STIX]{x1D719}_{K^{T}}$ , and $\unicode[STIX]{x1D719}_{K^{T}}^{\prime }$ for the case where $T$ is one of the above tangles. We will need the following theorem, due to Rasmussen.

Theorem 3 (Rasmussen [Reference RasmussenRas05, Proposition 3.2]).

The Lee homology group $\text{Kh}^{\prime }(L)$ has a canonical basis whose vectors correspond bijectively to possible orientations on $L$ . Furthermore, if $S:L\rightarrow L^{\prime }$ is a smooth link cobordism with no closed components, then the matrix entries of $S$ relative to the canonical bases of $\text{Kh}^{\prime }(L)$ and $\text{Kh}^{\prime }(L^{\prime })$ satisfy

$$\begin{eqnarray}(\unicode[STIX]{x1D719}_{S}^{\prime })_{o^{\prime }o}=2^{-\unicode[STIX]{x1D712}(S)}\left\{\begin{array}{@{}ll@{}}\unicode[STIX]{x1D716}_{o^{\prime }o}\quad & \text{if }o\cup o^{\prime }\text{extends over }S,\\ 0\quad & \text{else}\end{array}\right.\end{eqnarray}$$

for any two orientations $o$ and $o^{\prime }$ on $L$ and $L^{\prime }$ , where $\unicode[STIX]{x1D716}_{o^{\prime }o}\in \{\pm 1\}$ .

The above theorem implies that if $T=\unicode[STIX]{x1D6F4}_{i}$ or $T=\unicode[STIX]{x1D6F4}_{i}^{-1}$ , then

$$\begin{eqnarray}(\unicode[STIX]{x1D719}_{K^{T}}^{\prime })_{o_{p}o_{p}}\in \{\pm 1\},\end{eqnarray}$$

where $o_{p}$ denotes the parallel orientation of $K^{n}$ , i.e., the orientation for which all strands of $K^{n}$ are oriented parallel to the orientation of $K$ . Likewise, the theorem implies that if $K=E_{i}$ , then

$$\begin{eqnarray}(\unicode[STIX]{x1D719}_{K^{T}}^{\prime })_{o_{a}o_{a}}\in \{\pm 1\},\end{eqnarray}$$

where $o_{a}$ denotes the alternating orientation of $K^{n}$ , i.e., the orientation for which the strands of $K^{n}$ are alternatingly oriented parallel and antiparallel to $K$ , in such a way that the left-most strand of $K^{n}$ (corresponding to the left-most point of $P_{n}\subset \operatorname{int}(D^{2})\cap \mathbb{R}$ ) is oriented parallel to $K$ .

Now note that

$$\begin{eqnarray}\unicode[STIX]{x1D719}_{K^{T}}={\mathcal{F}}([K^{T}])\quad \text{and}\quad \unicode[STIX]{x1D719}_{K^{T}}^{\prime }={\mathcal{F}}^{\prime }([K^{T}]),\end{eqnarray}$$

where ${\mathcal{F}}$ is the functor defined in § 4 and ${\mathcal{F}}^{\prime }$ denotes Lee’s TQFT [Reference RasmussenRas10]. Since ${\mathcal{F}}$ and ${\mathcal{F}}^{\prime }$ are additive functors, it follows that any sign choice for $[K^{T}]$ induces corresponding sign choices for $\unicode[STIX]{x1D719}_{K^{T}}$ and $\unicode[STIX]{x1D719}_{K^{T}}^{\prime }$ , and we can therefore pin down the sign of $[K^{T}]$ (and hence of $\unicode[STIX]{x1D719}_{K^{T}}$ ) by imposing the following conventions.

Convention 1 immediately implies the following result.

Proof. Since the cobordism maps induced on the formal Khovanov bracket are isotopy invariants when considered up to sign and homotopy, it is clear that the maps $[K^{\unicode[STIX]{x1D6F4}_{i}}],[K^{\unicode[STIX]{x1D6F4}_{i}^{-1}}]\in \operatorname{End}_{{\mathcal{C}}}([K^{n}])$ satisfy the braid group relations up to possible signs. In order to determine these signs, it suffices to compute the actions of $[K^{\unicode[STIX]{x1D6F4}_{i}}][K^{\unicode[STIX]{x1D6F4}_{i+1}}][K^{\unicode[STIX]{x1D6F4}_{i}}]$ and $[K^{\unicode[STIX]{x1D6F4}_{i+1}}][K^{\unicode[STIX]{x1D6F4}_{i}}][K^{\unicode[STIX]{x1D6F4}_{i+1}}]$ on a single non-zero vector.

Now Rasmussen’s theorem together with Convention 1 implies that

$$\begin{eqnarray}(\unicode[STIX]{x1D719}_{K^{\unicode[STIX]{x1D6F4}_{i}}}^{\prime })_{oo_{p}}=(\unicode[STIX]{x1D719}_{K^{\unicode[STIX]{x1D6F4}_{i}^{-1}}}^{\prime })_{oo_{p}}=\left\{\begin{array}{@{}ll@{}}1,\quad & o=o_{p},\\ 0,\quad & o\neq o_{p},\end{array}\right.\end{eqnarray}$$

and so the canonical basis vector associated to the orientation $o_{p}$ is an eigenvector for $\unicode[STIX]{x1D719}_{K^{\unicode[STIX]{x1D6F4}_{i}}}^{\prime }$ and for $\unicode[STIX]{x1D719}_{K^{\unicode[STIX]{x1D6F4}_{i}^{-1}}}^{\prime }$ for the eigenvalue $1$ . Thus, when applied to this vector, $[K^{\unicode[STIX]{x1D6F4}_{i}}][K^{\unicode[STIX]{x1D6F4}_{i+1}}][K^{\unicode[STIX]{x1D6F4}_{i}}]$ and $[K^{\unicode[STIX]{x1D6F4}_{i+1}}][K^{\unicode[STIX]{x1D6F4}_{i}}][K^{\unicode[STIX]{x1D6F4}_{i+1}}]$ are equal. Consequently, it follows that the maps $[K^{\unicode[STIX]{x1D6F4}_{i}}]$ and $[K^{\unicode[STIX]{x1D6F4}_{i}^{-1}}]$ satisfy the braid group relations.◻

7.3 Temperley–Lieb algebra relations for cobordism maps

We will now show that the representation $\unicode[STIX]{x1D70C}$ described in Proposition 8 factors through the symmetric group $\mathfrak{S}_{n}$ . This will in turn imply that the corresponding braid group action on $\text{SKh}(K^{n})$ (given by $\unicode[STIX]{x1D70E}_{i}\mapsto \unicode[STIX]{x1D719}_{K^{\unicode[STIX]{x1D6F4}_{i}}}$ and $\unicode[STIX]{x1D70E}_{i}^{-1}\mapsto \unicode[STIX]{x1D719}_{K^{\unicode[STIX]{x1D6F4}_{i}^{-1}}}$ ) factors through $\mathfrak{S}_{n}$ , and thus the main statement of Theorem 2 will follow.

Specifically, we will prove the following proposition, which holds under the assumption of Conventions 1 and 2, and which shows that $[K^{\unicode[STIX]{x1D6F4}_{i}}]$ , $[K^{\unicode[STIX]{x1D6F4}_{i}^{-1}}]=[K^{\unicode[STIX]{x1D6F4}_{i}}]^{-1}$ , and $[K^{E_{i}}]$ satisfy the symmetric group relation $\unicode[STIX]{x1D70E}_{i}=\unicode[STIX]{x1D70E}_{i}^{-1}$ and the Kauffman bracket skein relations $\unicode[STIX]{x1D70E}_{i}=a+a^{-1}e_{i}$ and $\unicode[STIX]{x1D70E}_{i}^{-1}=a^{-1}+ae_{i}$ at $a=1$ .

Proposition 9. The endomorphisms $[K^{\unicode[STIX]{x1D6F4}_{i}}],[K^{\unicode[STIX]{x1D6F4}_{i}^{-1}}],[K^{E_{i}}]\in \operatorname{End}_{{\mathcal{C}}}([K^{n}])$ satisfy

$$\begin{eqnarray}[K^{\unicode[STIX]{x1D6F4}_{i}}]=\operatorname{id}_{[K^{n}]}+[K^{E_{i}}]=[K^{\unicode[STIX]{x1D6F4}_{i}^{-1}}].\end{eqnarray}$$

Instead of proving this proposition directly, we break it into two lemmas.

Lemma 11. There exist signs $\unicode[STIX]{x1D716}_{ij}\in \{\pm 1\}$ , $i=1,\ldots ,n-1$ , $j=1,\ldots ,4$ , such that

$$\begin{eqnarray}[K^{\unicode[STIX]{x1D6F4}_{i}}]=\unicode[STIX]{x1D716}_{i1}\operatorname{id}_{[K^{n}]}+\unicode[STIX]{x1D716}_{i2}[K^{E_{i}}]\quad \text{and}\quad [K^{\unicode[STIX]{x1D6F4}_{i}^{-1}}]=\unicode[STIX]{x1D716}_{i3}\operatorname{id}_{[K^{n}]}+\unicode[STIX]{x1D716}_{i4}[K^{E_{i}}].\end{eqnarray}$$

The proof of Lemma 11 will be deferred to § 7.4.

Proof of Lemma 12.

By Lemma 11, we have

$$\begin{eqnarray}[K^{\unicode[STIX]{x1D6F4}_{i}}]=\unicode[STIX]{x1D716}_{i1}\operatorname{id}_{[K^{n}]}+\unicode[STIX]{x1D716}_{i2}[K^{E_{i}}]\end{eqnarray}$$

for $\unicode[STIX]{x1D716}_{i1},\unicode[STIX]{x1D716}_{i2}\in \{\pm 1\}$ and hence the corresponding maps in Lee homology satisfy

$$\begin{eqnarray}\unicode[STIX]{x1D719}_{K^{\unicode[STIX]{x1D6F4}_{i}}}^{\prime }=\unicode[STIX]{x1D716}_{i1}\operatorname{id}+\unicode[STIX]{x1D716}_{i2}\unicode[STIX]{x1D719}_{K^{E_{i}}}^{\prime }.\end{eqnarray}$$

By considering the matrices of the above maps relative to the basis of Theorem 3 and comparing the diagonal entries corresponding to the parallel orientation $o_{p}$ , we thus obtain

$$\begin{eqnarray}1=\unicode[STIX]{x1D716}_{i1}+0\end{eqnarray}$$

and hence $\unicode[STIX]{x1D716}_{i1}=1$ , where we have used Convention 1 and Theorem 3 to conclude that $(\unicode[STIX]{x1D719}_{K^{\unicode[STIX]{x1D6F4}_{i}}}^{\prime })_{o_{p}o_{p}}=1$ and $(\unicode[STIX]{x1D719}_{K^{E_{i}}}^{\prime })_{o_{p}o_{p}}=0$ . Similarly, by comparing the diagonal entries corresponding to the alternating orientation $o_{a}$ , we obtain

$$\begin{eqnarray}0=\unicode[STIX]{x1D716}_{i1}-\unicode[STIX]{x1D716}_{i2}=1-\unicode[STIX]{x1D716}_{i2}\end{eqnarray}$$

and hence $\unicode[STIX]{x1D716}_{i2}=1$ , where we have used Theorem 3 and Convention 2 to conclude that $(\unicode[STIX]{x1D719}_{K^{\unicode[STIX]{x1D6F4}_{i}}}^{\prime })_{o_{a}o_{a}}=0$ and $(\unicode[STIX]{x1D719}_{K^{E_{i}}}^{\prime })_{o_{a}o_{a}}=-1$ . The proof of $\unicode[STIX]{x1D716}_{i3}=\unicode[STIX]{x1D716}_{i4}=1$ is analogous.◻

We can now use Proposition 9 to prove the following proposition, which implies that the endomorphisms $[K^{E_{i}}]$ satisfy the Temperley–Lieb algebra relations:

1. (1) $e_{i}^{2}=-(a^{2}+a^{-2})e_{i}$ ;

2. (2) $e_{i}e_{i\pm 1}e_{i}=e_{i}$ ;

3. (3) $e_{i}e_{j}=e_{j}e_{i}$ whenever $|i-j|\geqslant 2$

at $a=1$ .

Proof. (1) By Proposition 9, we have $[K^{E_{i}}]=[K^{\unicode[STIX]{x1D6F4}_{i}}]-\operatorname{id}_{[K^{n}]}$ and hence

$$\begin{eqnarray}[K^{E_{i}}]^{2}=([K^{\unicode[STIX]{x1D6F4}_{i}}]-\operatorname{id}_{[K^{n}]})^{2}=2\operatorname{id}_{[K^{n}]}-2[K^{\unicode[STIX]{x1D6F4}_{i}}]=-2[K^{E_{i}}],\end{eqnarray}$$

where in the second equality we have used that $[K^{\unicode[STIX]{x1D6F4}_{i}}]$ squares to $\operatorname{id}_{[K^{n}]}$ because $[K^{\unicode[STIX]{x1D6F4}_{i}}]^{-1}=[K^{\unicode[STIX]{x1D6F4}_{i}^{-1}}]=[K^{\unicode[STIX]{x1D6F4}_{i}}]$ by Propositions 8 and 9.

(2) Since the chain maps induced by annular link cobordisms are isotopy invariants when considered up to sign and homotopy and since the tangles $E_{i}\circ E_{i\pm 1}\circ E_{i}$ and $E_{i}$ are isotopic, it is clear that (2) holds up to an overall sign. It is therefore clear that the induced maps in Lee homology satisfy

$$\begin{eqnarray}\unicode[STIX]{x1D719}_{K^{E_{i}}}^{\prime }\circ \unicode[STIX]{x1D719}_{K^{E_{i\pm 1}}}^{\prime }\circ \unicode[STIX]{x1D719}_{K^{E_{i}}}^{\prime }=\unicode[STIX]{x1D716}\unicode[STIX]{x1D719}_{K^{E_{i}}}^{\prime }\end{eqnarray}$$

for an $\unicode[STIX]{x1D716}\in \{\pm 1\}$ . To conclude that $\unicode[STIX]{x1D716}=1$ , we now use Proposition 9 to write the right-hand side of the above equation as

$$\begin{eqnarray}\unicode[STIX]{x1D716}(\unicode[STIX]{x1D719}_{K^{\unicode[STIX]{x1D6F4}_{i}}}^{\prime }-\operatorname{id})\end{eqnarray}$$

and the left-hand side as

$$\begin{eqnarray}(\unicode[STIX]{x1D719}_{K^{\unicode[STIX]{x1D6F4}_{i}}}^{\prime }-\operatorname{id})\circ (\unicode[STIX]{x1D719}_{K^{\unicode[STIX]{x1D6F4}_{i\pm 1}}}^{\prime }-\operatorname{id})\circ (\unicode[STIX]{x1D719}_{K^{\unicode[STIX]{x1D6F4}_{i}}}^{\prime }-\operatorname{id})=\unicode[STIX]{x1D719}_{K^{\unicode[STIX]{x1D6F4}_{i}}}^{\prime }\circ \unicode[STIX]{x1D719}_{K^{\unicode[STIX]{x1D6F4}_{i\pm 1}}}^{\prime }\circ \unicode[STIX]{x1D719}_{K^{\unicode[STIX]{x1D6F4}_{i}}}^{\prime }-2\operatorname{id}+\unicode[STIX]{x1D6FF},\end{eqnarray}$$

where $\unicode[STIX]{x1D6FF}:=-\unicode[STIX]{x1D719}_{K^{\unicode[STIX]{x1D6F4}_{i}}}^{\prime }\circ \unicode[STIX]{x1D719}_{K^{\unicode[STIX]{x1D6F4}_{i\pm 1}}}^{\prime }-\unicode[STIX]{x1D719}_{K^{\unicode[STIX]{x1D6F4}_{i\pm 1}}}^{\prime }\circ \unicode[STIX]{x1D719}_{K^{\unicode[STIX]{x1D6F4}_{i}}}^{\prime }+2\unicode[STIX]{x1D719}_{K^{\unicode[STIX]{x1D6F4}_{i}}}^{\prime }+\unicode[STIX]{x1D719}_{K^{\unicode[STIX]{x1D6F4}_{i\pm 1}}}^{\prime }$ . This yields

$$\begin{eqnarray}\unicode[STIX]{x1D719}_{K^{\unicode[STIX]{x1D6F4}_{i}}}^{\prime }\circ \unicode[STIX]{x1D719}_{K^{\unicode[STIX]{x1D6F4}_{i\pm 1}}}^{\prime }\circ \unicode[STIX]{x1D719}_{K^{\unicode[STIX]{x1D6F4}_{i}}}^{\prime }-2\operatorname{id}+\unicode[STIX]{x1D6FF}=\unicode[STIX]{x1D716}(\unicode[STIX]{x1D719}_{K^{\unicode[STIX]{x1D6F4}_{i}}}^{\prime }-\operatorname{id})\end{eqnarray}$$

and, by considering the matrices of the above maps relative to the basis of Theorem 3 and comparing the diagonal entries corresponding to the alternating orientation $o_{a}$ , we obtain the equation

$$\begin{eqnarray}(\unicode[STIX]{x1D719}_{K^{\unicode[STIX]{x1D6F4}_{i}}}^{\prime }\circ \unicode[STIX]{x1D719}_{K^{\unicode[STIX]{x1D6F4}_{i\pm 1}}}^{\prime }\circ \unicode[STIX]{x1D719}_{K^{\unicode[STIX]{x1D6F4}_{i}}}^{\prime }-2\operatorname{id}+\unicode[STIX]{x1D6FF})_{o_{a}o_{a}}=\unicode[STIX]{x1D716}(\unicode[STIX]{x1D719}_{K^{\unicode[STIX]{x1D6F4}_{i}}}^{\prime }-\operatorname{id})_{o_{a}o_{a}}.\end{eqnarray}$$

It is now straightforward to check that the alternating orientation $o_{a}$ is compatible with the cobordism $K^{\unicode[STIX]{x1D6F4}_{i}}\circ K^{\unicode[STIX]{x1D6F4}_{i\pm 1}}\circ K^{\unicode[STIX]{x1D6F4}_{i}}$ , but not with the cobordisms $K^{\unicode[STIX]{x1D6F4}_{i}}\circ K^{\unicode[STIX]{x1D6F4}_{i\pm 1}}$ , $K^{\unicode[STIX]{x1D6F4}_{i\pm 1}}\circ K^{\unicode[STIX]{x1D6F4}_{i}}$ , $K^{\unicode[STIX]{x1D6F4}_{i}}$ , and $K^{\unicode[STIX]{x1D6F4}_{i\pm 1}}$ . Indeed, by Remark 8, one can think of these cobordisms in terms of isotopies which permute the strands of $K^{n}$ ; and while the permutation corresponding to $K^{\unicode[STIX]{x1D6F4}_{i}}\circ K^{\unicode[STIX]{x1D6F4}_{i\pm 1}}\circ K^{\unicode[STIX]{x1D6F4}_{i}}$ takes the alternating orientation to itself, the permutations corresponding to the other cobordisms do not. By Theorem 3, we thus have

$$\begin{eqnarray}(\unicode[STIX]{x1D719}_{K^{\unicode[STIX]{x1D6F4}_{i}}}^{\prime }\circ \unicode[STIX]{x1D719}_{K^{\unicode[STIX]{x1D6F4}_{i\pm 1}}}^{\prime }\circ \unicode[STIX]{x1D719}_{K^{\unicode[STIX]{x1D6F4}_{i}}}^{\prime })_{o_{a}o_{a}}=\unicode[STIX]{x1D716}^{\prime }\quad \text{and}\quad (\unicode[STIX]{x1D6FF})_{o_{a}o_{a}}=(\unicode[STIX]{x1D719}_{K^{\unicode[STIX]{x1D6F4}_{i}}}^{\prime })_{o_{a}o_{a}}=0\end{eqnarray}$$

for a sign $\unicode[STIX]{x1D716}^{\prime }\in \{\pm 1\}$ and, inserting these expressions into the above equation, we get

$$\begin{eqnarray}\unicode[STIX]{x1D716}^{\prime }-2+0=\unicode[STIX]{x1D716}(0-1).\end{eqnarray}$$

However, since $\unicode[STIX]{x1D716},\unicode[STIX]{x1D716}^{\prime }\in \{\pm 1\}$ , the equation $\unicode[STIX]{x1D716}^{\prime }-2=-\unicode[STIX]{x1D716}$ can only hold if $\unicode[STIX]{x1D716}^{\prime }=\unicode[STIX]{x1D716}=1$ , and hence (2) follows.

(3) Relation (3) follows because $[K^{E_{i}}]$ can be written as $[K^{E_{i}}]=[K^{\unicode[STIX]{x1D6F4}_{i}}]-\operatorname{id}_{[K^{n}]}$ (by Proposition 9) and because the $[K^{\unicode[STIX]{x1D6F4}_{i}}]$ satisfy the braid group relations (by Proposition 8).◻