Section 3 (0P3U) · ScalingStacks

3 Bimodule homomorphisms

For a ring $A$ denote by $\mathcal{K}(A)$ the category of bounded complexes of $A$ -bimodules up to homotopies of complexes.The objects of $\mathcal{K}(A)$ are bounded complexes of $A$ -bimodules, the morphisms are morphisms of complexes of bimodules quotiented by homotopic to $0$ morphisms.

We say that a complex of bimodules $M\in\mathcal{K}(A)$ is invertible if there exists $N\in\mathcal{K}(A)$ such that $N\otimes_{A}M\cong A$ and $M\otimes_{A}N\cong A$ in $\mathcal{K}(A).$ Here $A$ denotes the complex $0\longrightarrow A\longrightarrow 0$ with $A$ in cohomological degree $0$ and the usual left and right multiplication action of $A$ on itself.

Let $Z(A)$ be the center of $A.$

0P3H

Proposition 1 If $M$ is invertible then

\mathrm{Hom}_{\mathcal{K}(A)}(M,M)\cong\mathrm{Hom}_{A\otimes A^{o}}(A,A)\cong Z(A),

Proof: The second isomorphism is obvious, since endomorphisms of $A$ as an $A$ -bimodule are multiplications by central elements.

Consider the following sequence of ring homomorphisms:

	$\displaystyle\mathrm{End}_{\mathcal{K}(A)}(M)\stackrel{{\scriptstyle f}}{{\longrightarrow}}\mathrm{End}_{\mathcal{K}(A)}(M\otimes_{A}N)\stackrel{{\scriptstyle g}}{{\longrightarrow}}\mathrm{End}_{\mathcal{K}(A)}((M\otimes_{A}N)\otimes_{A}M)\cong$
	$\displaystyle\mathrm{End}_{\mathcal{K}(A)}(M\otimes_{A}(N\otimes_{A}M))\cong\mathrm{End}_{\mathcal{K}(A)}(M\otimes_{A}A)\cong\mathrm{End}_{\mathcal{K}(A)}(M),$

where $f,$ respectively $g,$ is tensoring with the identity endomorphism of $N,$ respectively $M.$ The composition $gf$ is the identity, thus $f$ is injective. Multiplication on the left by central elements makes each of the above rings a $Z(A)$ -module, and $f$ and $g$ are $Z(A)$ -module homomorphisms. $f$ and $g$ take identity endomorphisms to identity endomorphisms, and $\mathrm{End}_{\mathcal{K}(A)}(M\otimes_{A}N)\cong\mathrm{End}_{\mathcal{K}(A)}(A)=Z(A).$ Therefore, $f$ is surjective, since $f(\mathrm{id})=\mathrm{id}$ generates $\mathrm{End}_{\mathcal{K}(A)}(M\otimes_{A}N)$ as a $Z(A)$ -module. Thus, $f$ and $g$ are isomorphisms. $\square$

If $A$ is graded, denote by $\mathcal{K}(A)$ the category of bounded complexes of graded $A$ -bimodules (with grading-preserving differential) up to homotopies. The morphisms are grading-preserving homomorphisms of complexes (modulo homotopies). If $M$ is an invertible complex in $\mathcal{K}(A)$ then $\mathrm{Hom}_{\mathcal{K}(A)}(M,M)\cong Z_{0}(A),$ the degree $0$ component of the center of $A.$ Furthermore, the group of automorphisms of $M$ in $\mathcal{K}(A)$ is isomorphic to $Z_{0}^{\ast}(A),$ the group of invertible elements in $Z_{0}(A).$

We now specialize to the rings $H^{n}.$

0P3I

Proposition 2 The only invertible central elements of degree $0$ in $H^{n}$ are $\pm 1:$

Z_{0}^{\ast}(H^{n})\cong\{\pm 1\}.

Proof: A degree $0$ element of $H^{n}$ has the form $v=\sum v_{a}e_{a}$ where $e_{a}$ is the minimal idempotent corresponding to the crossingless matching $a$ and $v_{a}\in\mathbb{Z}.$ For any $a,b$ choose $x\in\hskip 3.61371pt_{a}(H^{n})_{b},x\not=0.$ Then $vx=v_{a}x$ and $xv=v_{b}x.$ Therefore, if $v$ is central, $v_{a}=v_{b}$ for all $a,b,$ and $v=m\sum e_{a}=m,$ for some integer $m,$ so that $Z_{0}(H^{n})\cong\mathbb{Z}.$ The proposition follows. $\square$

Remark: We investigated the center of $H^{n}$ (and not just its degree $0$ component) in [8]. It turned out to be isomorphic to the cohomology ring of the $(n,n)$ Springer fiber.

0P3J

Corollary 1 If $M$ is an invertible complex of graded $H^{n}$ -bimodules, then $\mathrm{Id}$ and $-\mathrm{Id}$ are the only degree $0$ automorphisms of $M.$

We use notation $\mathcal{K}_{n}^{m}$ from [9] for the category of bounded complexes of geometric $(H^{m},H^{n})$ -bimodules up to chain homotopies. A bimodule is geometric if it is isomorphic to a finite direct sum of bimodules ${\mathcal{F}}(a)\{i\},$ for flat tangles $a$ and $i\in\mathbb{Z}$ (recall that $\{i\}$ denotes shift in the grading by $i$ ). Morphisms in $\mathcal{K}_{n}^{m}$ are grading-preserving homomorphisms of complexes up to chain homotopies.

From Corollary 1 we derive

0P3K

Corollary 2 If $f:M\to N$ is an isomorphism of invertible objects in $\mathcal{K}_{n}^{n}$ then the only other isomorphism of $M$ and $N$ is $-f.$

For now on we assume that the reader is familiar with the construction of [9, Section 2], which to a surface $S$ embedded in $\mathbb{R}^{3}$ and viewed as a cobordism between flat tangles $a$ and $b$ assigns a bimodule homomorphism ${\mathcal{F}}(S):{\mathcal{F}}(a)\longrightarrow{\mathcal{F}}(b).$

Standard cobordisms in $\mathbb{R}^{3}$ (the first is a birth move, the second and third are saddle points, the fourth is a death move)

$\displaystyle\mathrm{Vert}_{n-1}$	$\displaystyle\Longrightarrow$	$\displaystyle\cap_{i,n}\cup_{i,n-1},$
$\displaystyle\cup_{i,n-1}\cap_{i,n}$	$\displaystyle\Longrightarrow$	$\displaystyle\mathrm{Vert}_{n},$
$\displaystyle\mathrm{Vert}_{n}$	$\displaystyle\Longrightarrow$	$\displaystyle\cup_{i,n-1}\cap_{i,n},$
$\displaystyle\cap_{i,n}\cup_{i,n-1}$	$\displaystyle\Longrightarrow$	$\displaystyle\mathrm{Vert}_{n-1}$

induce grading-preserving bimodule homomorphisms

	$\displaystyle H^{n-1}\longrightarrow{\mathcal{F}}(\cap_{i,n})\otimes_{H^{n}}{\mathcal{F}}(\cup_{i,n-1})\{1\},$		(1)
	$\displaystyle{\mathcal{F}}(\cup_{i,n-1})\otimes_{H^{n-1}}{\mathcal{F}}(\cap_{i,n})\{1\}\longrightarrow H^{n},$		(2)
	$\displaystyle H^{n}\longrightarrow{\mathcal{F}}(\cup_{i,n-1})\otimes_{H^{n-1}}{\mathcal{F}}(\cap_{i,n})\{-1\},$		(3)
	$\displaystyle{\mathcal{F}}(\cap_{i,n})\otimes_{H^{n}}{\mathcal{F}}(\cup_{i,n-1})\{-1\}\longrightarrow H^{n-1},$		(4)

(we used that ${\mathcal{F}}(\mathrm{Vert}_{n})\cong H^{n},{\mathcal{F}}(\cup_{i,n-1}\cap_{i,n})\cong{\mathcal{F}}(\cup_{i,n-1})\otimes_{H^{n-1}}{\mathcal{F}}(\cap_{i,n}),$ etc.) Isotopies between compositions of these cobordisms translate into relations between homomorphisms. These relations imply that the functors of tensoring with ${\mathcal{F}}(\cap_{i,n})$ and ${\mathcal{F}}(\cup_{i,n-1})$ are biadjoint, up to grading shifts. Precisely, let $F_{\cup}$ be the functor of tensoring with ${\mathcal{F}}(\cup_{i,n-1})$ and $F_{\cap}$ the functor of tensoring with ${\mathcal{F}}(\cap_{i,n})$ (viewed as functors between categories of $H^{n}$ and $H^{n-1}$ -modules).

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Proposition 3 $F_{\cup}\{1\}$ is left adjoint to $F_{\cap},$ and $F_{\cap}\{-1\}$ is left adjoint to $F_{\cup}.$

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Corollary 3 The only grading-preserving endomorphisms of bimodules ${\mathcal{F}}(\cap_{i,n})$ and ${\mathcal{F}}(\cup_{i,n-1})$ are multiplications by integers. The only grading-preserving automorphisms of bimodules ${\mathcal{F}}(\cap_{i,n})$ and ${\mathcal{F}}(\cup_{i,n-1})$ are $\mathrm{Id}$ and $-\mathrm{Id}.$ Moreover, these bimodules have no graded endomorphisms of negative degree.

Proof of corollary: From adjointness,

$\displaystyle\mathrm{Hom}_{(n,n-1)}({\mathcal{F}}(\cup_{i,n-1}),{\mathcal{F}}(\cup_{i,n-1}))$	$\displaystyle\cong$	$\displaystyle\mathrm{Hom}_{(n-1,n-1)}(H^{n-1},{\mathcal{F}}(\cap_{i,n}\cup_{i,n-1})\{1\})$
	$\displaystyle\cong$	$\displaystyle\mathrm{Hom}_{(n-1,n-1)}(H^{n-1},H^{n-1}\oplus H^{n-1}\{2\})$
	$\displaystyle\cong$	$\displaystyle\mathrm{Hom}_{(n-1,n-1)}(H^{n-1},H^{n-1})$
	$\displaystyle\cong$	$\displaystyle\mathbb{Z}.$

Subscripts of the form $(m,n)$ in the above formula mean that the homomorphisms considered are those of graded $(H^{m},H^{n})$ -bimodules. We used that

\mathrm{Hom}_{(n-1,n-1)}(H^{n-1},H^{n-1}\{k\})=0

for any positive $k,$ since the ring $H^{n-1}$ is $\mathbb{Z}_{+}$ -graded. Similar computations establish the result for ${\mathcal{F}}(\cap_{i,n})$ and the last claim of the corollary. $\square$

0P3N

Corollary 4 If $M$ is a tensor product of ${\mathcal{F}}(\cap_{i,n})$ and invertible complexes of bimodules, and if $f:M\to N$ is an isomorphism in $\mathcal{K}_{n}^{n-1},$ then the only other isomorphism from $M$ to $N$ is $-f.$ Same with $\cup_{i,n-1}$ instead of $\cap_{i,n}.$

More generally, let $b$ be a flat tangle without closed components (circles), with $k$ arcs connecting bottom endpoints, $l$ arcs connecting top endpoints, and some number of arcs connecting a top endpoint with a bottom endpoint. Let $W(b)$ be the reflection of $b$ about the $x$ -axis. Representing $b$ as a product of $U$ -turns and using Proposition 3 repeatedly we obtain

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Proposition 4 The functor of tensoring with the bimodule ${\mathcal{F}}(W(b))\{k-l\}$ is left adjoint to tensoring with ${\mathcal{F}}(b)$ and the functor of tensoring with ${\mathcal{F}}(W(b))\{l-k\}$ is right adjoint to tensoring with ${\mathcal{F}}(b).$

0P3Q

Corollary 5 If $b$ is a flat tangle without closed components then the only grading-preserving endomorphisms of the bimodule ${\mathcal{F}}(b)$ are multiplications by integers, the only grading-preserving automorphisms are $\mathrm{Id}$ and $-\mathrm{Id},$ and ${\mathcal{F}}(b)$ has no endomorphisms of negative degree. If $f:M\to{\mathcal{F}}(b)$ is an isomorphism in $\mathcal{K}_{n}^{m}$ then the only other isomorphism between $M$ and ${\mathcal{F}}(b)$ is $-f.$

Mikhail Khovanov

Original source: arXiv:math/0207264v1