Serre duality for Khovanov–Rozansky homology

Abstract

We prove that the full twist is a Serre functor in the homotopy category of type A Soergel bimodules. As a consequence, we relate the top and bottom Hochschild degrees in Khovanov–Rozansky homology, categorifying a theorem of Kálmán.

Appendix A: Semiorthogonal decompositions for coxeter groups

Let W be a Coxeter group with the set of reflections S. Let ${\mathfrak {h}}$ be a realization of W. Define $R=\mathbb {k}[{\mathfrak {h}}]$, and $B_s=R\otimes _{R^s}R$ for $s\in S$. The category ${\mathbb {S}}\mathrm {Bim}_W$ of Soergel bimodules is the smallest full subcategory of the category of $R-R$ bimodules containing R and all $B_s$ and closed under direct sums, direct summands, tensor products and grading shifts. For $W=S_n$ and ${\mathfrak {h}}=\mathbb {k}^n$ we recover the category ${\mathbb {S}}\mathrm {Bim}_n$ defined in Sect. 3.1.

Rouquier complexes can be defined in the homotopy category ${\mathcal {K}}^b({\mathbb {S}}\mathrm {Bim}_W)$ similarly to Sect. 3.4:

$$\begin{aligned} F_s=[B_s\rightarrow R],\ F_s^{-1}=[R\rightarrow B_s] \end{aligned}$$

In [26] Rouquier proved that they satisfy the relations in the braid group associated to W. Therefore for any $w\in W$ one can consider a Rouquier complex $F_w$ corresponding to the positive permutation braid associated to any reduced expression of w. It does not depend on the choice of a reduced expression up to homotopy equivalence.

We define triangulated subcategories $\mathcal {U}_{<w}$ and $\mathcal {U}_{\le w}$ of ${\mathcal {K}}^b({\mathbb {S}}\mathrm {Bim}_W)$ generated by the Rouquier complexes $F_v$ with $v<w$ (and $v\le w$) in Bruhat order.

For any $s\in S$, there exists a chain map $\psi _s:F_s\rightarrow F_s^{-1}$ such that

$$\begin{aligned} {\text {Cone}}[F_s\rightarrow F_s^{-1}]=[R\rightarrow R]. \end{aligned}$$

(A.1)

As an immediate corollary, we get the following:

Proposition A.1

For any $w\in W$ there is a chain map $\psi _w: F_w\rightarrow F_{w^{-1}}^{-1}$. The cone of $\psi _w$ is filtered by $F_{u^{-1}}^{-1}$ for $u<w$ in Bruhat order.

In [20] Libedinsky and Williamson proved a much stronger statement (conjectured by Rouquier in [27], p. 215 before Remark 4.12).

Theorem A.2

[20] If $w\ne v$ then ${\text {Hom}}(F_v,F^{-1}_{w^{-1}})=0$. If $w=v$ then ${\text {Hom}}(F_w,F^{-1}_{w^{-1}})=R$ is generated by the map $\psi _w$.

Corollary A.3

We have ${\text {Hom}}(F_w,R)=0$ for $w\ne 1$.

In type A this corollary is an easy consequence of Corollary 3.22. In fact, Theorem A.2 also can be deduced:

Proposition A.4

Theorem A.2 follows from Corollary A.3.

Proof

Assume that ${\text {Hom}}(F_w,R)=0$ for $w\ne 1$. Note that

$$\begin{aligned} {\text {Hom}}(F_v,F^{-1}_{w^{-1}}) \cong {\text {Hom}}(F_v F_{w^{-1}}, R). \end{aligned}$$

We induct on the number $\min (l(v), l(w))$. The base case follows from the assumption. Without loss of generality, we may assume $l(v)\le l(w)$. Let $v=v's$ for a simple reflection s and $l(v')=l(v)-1$. If $l(ws)>l(w)$ then $w\ne v$ and $ws\ne v'$, so we have

$$\begin{aligned} {\text {Hom}}(F_v F_{w^{-1}}, R) = {\text {Hom}}(F_{v'} F_{s w^{-1}}, R) = {\text {Hom}}(F_{v'} F_{(ws)^{-1}}, R) \cong 0, \end{aligned}$$

where the last equality follows from the induction hypothesis. So we assume $l(ws)<l(w)$. Let $w=w' s$. The map $\psi _s:F_s\rightarrow F_s^{-1}$ induces a map

$$\begin{aligned} {\text {Hom}}(F_{v'}, F_{{w'}^{-1}}^{-1})= & {} {\text {Hom}}(F_{v'} F_s^{-1}, F_{{w'}^{-1}}^{-1} F_s^{-1}) \rightarrow {\text {Hom}}(F_{v'} F_s, F_{{w'}^{-1}}^{-1} F_s^{-1}) \\= & {} {\text {Hom}}(F_v, F_{w^{-1}}^{-1}), \end{aligned}$$

whose cone is filtered by ${\text {Hom}}(F_{v'}, F_{w^{-1}}^{-1})$, which vanishes by the induction hypothesis since $l(v')<l(v)\le l(w)$. So we are reduced to the statement for the pair $v', w'$. $\square$

We use Theorem A.2 to deduce a very important fact about Rouquier complexes which does not appear to be explicitly stated in the literature.

Theorem A.5

We have ${\text {Hom}}(F_w,F_v)=0$ unless $w\le v$ in Bruhat order.

Proof

By Proposition A.1 $F_v$ is homotopy equivalent to a complex filtered by $F_{u^{-1}}^{-1}$ with $u\le v$. Therefore ${\text {Hom}}(F_w,F_v)=0$ unless ${\text {Hom}}(F_w,F_{u^{-1}}^{-1})\ne 0$ for some $u\le v$. But by Theorem A.2 this is possible only if $u=w$, and hence $w\le v$. $\square$

Corollary A.6

For all w we have semiorthogonal decompositions $\mathcal {U}_{\le w}=\langle \mathcal {U}_{<w}, F_w\rangle$ and $\mathcal {U}_{\le w}=\langle F_{w^{-1}}^{-1}, \mathcal {U}_{<w}\rangle$.

Proof

By Proposition A.1 the category $\mathcal {U}_{\le w}$ is generated by $\mathcal {U}_{<w}$ and $F_w$, or, equivalently, by $\mathcal {U}_{<w}$ and $F_{w^{-1}}^{-1}$ (since ${\text {Cone}}[F_w\rightarrow F_{w^{-1}}^{-1}]\in \mathcal {U}_{<w}$). Now for all $u<w$ we have ${\text {Hom}}(F_w,F_u)=0$ by Theorem A.5 and ${\text {Hom}}(F_{u},F_{w^{-1}}^{-1})=0$ by Theorem A.2. $\square$

Corollary A.7

If W is a finite Coxeter group then for all $w\in W$ the inclusion $\mathcal {U}_{\le w}\hookrightarrow {\mathcal {K}}^b({\mathbb {S}}\mathrm {Bim}_W)$ has both left and right adjoints.

Proof

Fix an arbitrary total order $\prec$ on W refining the Bruhat order, let $w_0$ be the longest element in W. Then for all w we have a chain

$$\begin{aligned} w=w^{(1)}\prec w^{(2)}\prec \cdots \prec w^{(k)}=w_0. \end{aligned}$$

Similarly to Corollary A.6, the inclusions $\mathcal {U}_{\le w^{(i)}}\hookrightarrow \mathcal {U}_{\le w^{(i+1)}}$ have both left and right adjoints, and by combining these we get adjoints to the inclusion

$$\begin{aligned} \mathcal {U}_{\le w^{(i)}}\hookrightarrow \mathcal {U}_{\le w_0}={\mathcal {K}}^b({\mathbb {S}}\mathrm {Bim}_W). \end{aligned}$$

$\square$

If $W'$ is a parabolic subgroup of W, we can consider the category of Soergel bimodules ${\mathbb {S}}\mathrm {Bim}_{W'}$ associated to the same realization ${\mathfrak {h}}$.

Corollary A.8

Let W be a finite Coxeter group and $W'$ a parabolic subgroup. Then the inclusion ${\mathcal {K}}^b({\mathbb {S}}\mathrm {Bim}_{W'})\rightarrow {\mathcal {K}}^b({\mathbb {S}}\mathrm {Bim}_W)$ has both left and right adjoints.

Proof

We have ${\mathcal {K}}^b({\mathbb {S}}\mathrm {Bim}_{W'})=\mathcal {U}_{\le w}$ where w is the longest element of $W'$. $\square$

Note that in type A this gives an alternative construction of adjoints to inclusions of ${\mathbb {S}}\mathrm {Bim}_{n,1,\ldots ,1}$ in ${\mathbb {S}}\mathrm {Bim}_{m}$. However, it seems that the direct construction of adjoints in Sect. 5 is easier to work with than the induction on Bruhat graph as in Corollary A.7.