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Kapranov and Voevodsky introduced the following 2-category $2{\rm Vect}$ of "2-vector spaces" over a field $K$:

  • 0-cells are natural numbers $\langle N\rangle,\langle M\rangle...$
  • 1-cells $\langle N\rangle\to \langle M\rangle $ are strictly-positive-integer valued matrices of size $M\times N$
  • 2-cells $A\to B$ are matrices whose entries are matrices, whose $(i,j)$ entry is a matrix having size $A_{ij}\times B_{ij}$.

So for example, this is a 2-cell $\langle 2\rangle\to \langle 2\rangle$ where all the entries in $\alpha$ belong to the field $K$.

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Since this is the best of possible worlds, this categorifies correctly (at least part of) the idea of a vector space as a "set of tuples made of scalars $a_i\in K$".

In fact, this is a way to make the following intuitive idea work:

just as the category of vector spaces is equivalent to the category of natural numbers, where $\hom(n,m)$ is the space of $m\times n$ $K$-valued matrices, so the 2-category of 2-vector spaces is equivalent to the 2-category of natural numbers, where $\hom(n,m)$ is the category of $K$-linear functors ${\rm Vect}^m \to {\rm Vect}^n$.

Now.

The 2-category of 2-vector spaces has two monoidal structures $\boxtimes$ and $\boxplus$ where

  • $\langle N\rangle\boxtimes \langle M\rangle = \langle N\cdot M\rangle$;
  • the integer-valued matrix $A\boxtimes B$ has in its $(ik,jl)$ entry the integer $A_{ik}B_{jl}$;
  • given 2-cells $\alpha,\beta$ the 2-cell $\alpha\boxtimes\beta$ is the entry-wise tensor product of matrices.

and $\boxplus$ (the monoidal sum) is defined similarly, so that $\langle N\rangle ⊞ \langle M\rangle = \langle N+M\rangle$, $A ⊞ B$ is the direct sum of matrices, and $\alpha ⊞ \beta$ is the entry-wise direct sum of matrices (in fact $2{\rm Vect}$ a rig category with respect to $\boxtimes$ and $\boxplus$).

Question: Is the $\boxtimes$-structure also closed?

Representing the category $2{\rm Vect}(\langle mn\rangle,\langle p\rangle)$ as the category of linear functors $\underbrace{{\rm Vect}^n \times\dots\times{\rm Vect}^n}_{m \text{ times}} \to {\rm Vect}^p$ does not seem to help, and before going deep down the computations I would like to know if there is a conceptual reason for this category to (not) be closed.

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  • $\begingroup$ Incidentally, there's an argument to be made that this is the wrong definition: for example, it fails to satisfy Galois descent. $\endgroup$ – Qiaochu Yuan Jun 25 '18 at 16:00
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    $\begingroup$ Finally an argument against Leibniz :-) $\endgroup$ – Fosco Jun 25 '18 at 17:29
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Conceptually, $2Vect$ should be viewed as a sub-2-category of the 2-category $Bim$ of $k$-algebras with bimodules between them. In fact, when $k$ is algebraically closed, $2Vect = Bim^{fin}$ is exactly the 2-category of finite $k$-algebras and bimodules between them, because every finite $k$-algebra is Morita equivalent to $\oplus$-sum of copies of $k$. And the inclusion $2Vect \to Bim$ sends $n \mapsto k^{\oplus n}$, landing in commutative algebras.

Note that $\boxtimes$ corresponds to $\otimes$ of algebras, which is the coproduct for the sub-2-category $CAlg$ of commutative algebras and homormophisms, so it is definitely not closed when restricted to $CAlg$ (a cocartesian monoidal structure is closed only in trivial cases). But it's a general fact that for any algebra $A$, the opposite algebra $A^\circ$ is dual to $A$. Having duals in particular implies that the monoidal structure is closed.

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  • $\begingroup$ I see the correspondence, but I don't get if the structure is closed or not, then... $\endgroup$ – Fosco Jun 25 '18 at 17:07
  • $\begingroup$ $Bim$ has duals for objects, given by taking the opposite algebra (exercise!). It follows that $Bim$ is monoidal closed. $2Vect$ is closed in $Bim$ under taking duals, so $2Vect$ also has duals. It follows that $2Vect$ is monoidal closed. $\endgroup$ – Tim Campion Jun 25 '18 at 17:09
  • $\begingroup$ Yes. I wasn't able to link this to the comment for $CAlg$, that isn't closed because it's cocartesian. $\endgroup$ – Fosco Jun 25 '18 at 17:20
  • $\begingroup$ Oh, that's actually irrelevant. I was still thinking as I was writing. $\endgroup$ – Tim Campion Jun 25 '18 at 17:21
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    $\begingroup$ Yes, it's the componentwise multiplication. The bimodule corresponding to an $n \times m$ matrix of vector spaces is the direct sum of the vector spaces, where the scalar $e_i \otimes e_j$ acts by projecting onto the $(i,j)$th summand. Here $e_i = (0,\dots, 0, 1, 0, \dots, 0) \in k^n$ with the 1 in the $i$th spot. This is enough to determine the full bimodule structure. $\endgroup$ – Tim Campion Jun 25 '18 at 18:23
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As Tim says, for a conceptual ("coordinate-independent") picture we should start with $\text{Bim}(k)$, the 2-category of $k$-algebras, $k$-bimodules between them, and $k$-bimodule homomorphisms, which we can think of as a reasonably well-behaved 2-category of $\text{Mod}(k)$-modules. By Eilenberg-Watts we can equivalently think of this 2-category as the 2-category whose objects are categories of the form $\text{Mod}(A)$, $A$ a $k$-algebra, and whose morphisms are $k$-linear cocontinuous functors.

This 2-category is closed monoidal with monoidal structure given by tensor product, which has a universal property: cocontinuous $k$-linear functors out of $\text{Mod}(A \otimes_k B)$ correspond to $k$-linear functors out of $\text{Mod}(A) \times \text{Mod}(B)$ which are cocontinuous in each variable. Every object is dualizable, with dual given by $A \mapsto A^{op}$, and hence the closed structure is given by $[A, B] = A^{op} \otimes_k B$. This reflects the fact that $\text{Mod}(A^{op} \otimes_k B)$ is precisely $(A, B)$-bimodules over $k$, which are precisely the morphisms $A \to B$ in this 2-category.

KV considers the sub-2-category of this consisting of $k$-algebras of the form $k^n$ and finitely generated projective $(k^n, k^m)$-bimodules over $k$, which are described explicitly by $n \times m$ matrices of finite-dimensional $k$-vector spaces. These morphisms are singled out by the property of also being dualizable (that is, they have adjoints; see this blog post), and these $k$-algebras are 2-dualizable in addition to being dualizable, but this does not exhaust all 2-dualizable objects. In general the 2-dualizable objects are the $k$-algebras $A$ which are finite-dimensional and separable over $k$ (and this definition, unlike the KV definition, should satisfy Galois descent); the algebras $k^n$ exhaust these iff $k$ is separably closed. These can be used to define 2-dimensional topological field theories via the cobordism hypothesis. For example, if we take $A$ to be a twisted group algebra for a finite group $G$ with twist defined by a $2$-cocycle $\alpha \in H^2(G, k^{\times})$, then the resulting TFT is 2d Dijkgraaf-Witten theory with gauge group $G$ twisted by $\alpha$.

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  • $\begingroup$ Thank you. Your answer makes me guess that dualization can't be seen at the level of 0-cells, as passing to the dual algebra of $\langle n\rangle \approx A\cong k^n$ keeps the number $n$ unchanged. So everything happens at the level of 1- and 2-cells (if this approach seems down to earth, you're right: I'm trying to make the two ends of the string -abstract and concrete- converge) $\endgroup$ – Fosco Jun 26 '18 at 16:48
  • $\begingroup$ @Fosco: right. if you think of 1-morphisms of 2-vector spaces as matrices of vector spaces, then dualization acts on 1-morphisms by taking the transpose of these matrices. $\endgroup$ – Qiaochu Yuan Jun 26 '18 at 23:23

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