# Monoidal category that is not spacial

Selinger ("A survery of graphical languages for monoidal categories", 2009) defines a "spacial monoidal category" as one that satisifies the following string-diagram axiom

This diagram says that for every $$h : I \to I$$ and object $$A$$, $$\rho_A \circ (\text{id}_A \otimes h) \circ \rho^{-1}_A = \lambda_A \circ (h \otimes \text{id}_A) \circ \lambda_A^{-1},$$ where $$\lambda_A$$ and $$\rho_A$$ are the left and right unitors.

What is an example of a monoidal category that is not spacial by this definition?

• Could you describe the axiom without using string diagrams as well, to help those that aren't familiar with this specific graphical language? Nov 4 '19 at 19:15
• The formula is at the top of page 14 of Selinger's paper, arxiv.org/pdf/0908.3347. Nov 5 '19 at 18:14
• Also now added in an edit. Nov 8 '19 at 12:29
• Out of curiosity: Every symmetric monoidal category is spacial, right? Jan 10 '20 at 21:03

One of the simplest examples of a non-spacial category is $$\mathrm{End}(\mathrm{Vec}^{\oplus 2})$$, the category of $$2\times2$$ matrices with vector space coefficients. Working over a field $$\mathbb k$$, this is a multifusion category with four simple objects: $$\begin{pmatrix} \mathbb k & 0 \\ 0 & 0 \end{pmatrix}, \begin{pmatrix} 0 & \mathbb k \\ 0 & 0 \end{pmatrix}, \begin{pmatrix} 0 & 0 \\ \mathbb k & 0 \end{pmatrix}, \begin{pmatrix} 0 & 0 \\ 0 & \mathbb k \end{pmatrix}$$ and the tensor product is just matrix multiplication (with tensor product and direct sum of vector spaces, of course). The monoidal unit is not simple: $$\mathbb{1} = \begin{pmatrix} \mathbb k & 0 \\ 0 & 0 \end{pmatrix} \oplus \begin{pmatrix} 0 & 0 \\ 0 & \mathbb k \end{pmatrix}.$$ In particular, the endomorphisms of the monoidal unit is the commutative ring $$\mathbb k \oplus \mathbb k$$ (with componentwise addition and multiplication). Consider the endomorphism $$h = (\alpha,\beta) \in \mathrm{End}(\mathbb 1)$$ and object $$A = \begin{pmatrix} 0 & \mathbb k \\ 0 & 0 \end{pmatrix}.$$ Then one of your diagrams evaluates to $$\alpha \, \mathrm{id}_A$$ and the other one evaluates to $$\beta \, \mathrm{id}_A$$.

Remark: Write $$R = \mathrm{End}(\mathbb 1) = \mathbb k \oplus \mathbb k$$. Then it is an associative $$\mathbb k$$-algebra, and the multifusion category above is the monoidal category of $$R$$-$$R$$ bimodules. Categories of bimodules are typically not spacial except when the algebra has trivial (i.e. one-dimensional) centre. A fun example to think through is to consider $$\mathbb C$$ as an $$\mathbb R$$-algebra. Then the category of $$\mathbb C$$-$$\mathbb C$$ bimodules has two simple objects (Exercise: why?), and is not spacial (Exercise: why not?).

If your monoidal category is the fundamental 2-groupoid of a space, then this is exactly asking whether $$\pi_1$$ acts trivially on $$\pi_2$$ or not (or more precisely, that equality is saying that $$A \in \pi_1$$ acts trivially on $$h \in \pi_2$$). If I remember right, $$\mathbb{R}P^2$$ gives the easiest to understand counterexample.

• Objection: $S^1\vee S^2$ gives the easiest to understand counterexample. Nov 6 '19 at 20:56
• Ha! You speak the truth. Nov 6 '19 at 22:49

An example I learned about in Khovanov's "Heisenberg algebra and a graphical calculus", 2010, is the restriction and induction functors for the infinite chain $$S_0\subset S_1\subset S_2\subset\cdots$$ of symmetric groups. This is also explained in Likeng and Savage, "Embedding Deligne's category $$\operatorname{Rep}(S_t)$$ in the Heisenberg category," 2019.

Let $$\mathcal{S}=\prod_{m\in\mathbb{N}}\bigoplus_{n\in\mathbb{N}}\mathrm{Bim}{(S_n,S_m)}$$, where each $$\mathrm{Bim}(S_n,S_m)$$ is the category of $$(S_n,S_m)$$-modules over a field $$k$$. This has a monoidal structure given by tensor products of compatible bimodules, with the monoidal unit being $$I=\prod_{m}k[S_m]$$ with each $$k[S_m]$$ as an $$(S_m,S_m)$$-module. We may regard $$(S_n,S_m)$$-modules as objects of $$\mathcal{S}$$ by setting all non-$$m$$ indices to the $$0$$ bimodule.

The induction functor $$\mathrm{Ind}_{S_n}^{S_{n+1}}$$ can be given as an object $$\mathrm{Ind}=\prod_{n\geq 1} {}_n(k[S_n])_{n-1},$$ where the subscript notation means $$k[S_n]$$ is treated as an $$(S_n,S_{n-1})$$-module. This gives induction in the sense that, for each $$S_n$$-module $$M$$, the object $$\mathrm{Ind}\otimes M$$ is $$\operatorname{Ind}_{S_n}^{S_{n+1}}M$$. Similarly, restriction is given by $$\mathrm{Res}=\prod_{n\geq 1}{}_{n-1}(k[S_n])_n.$$

Just as induction and restriction are biadjoint functors, the $$\mathrm{Ind}$$ and $$\mathrm{Res}$$ objects are left and right duals to each other. Two of the four associated pairings and copairings are $$\mathrm{Ind}\otimes\mathrm{Res} \to I$$ and $$I\to \mathrm{Ind}\otimes\mathrm{Res}$$. Since $$\mathrm{Ind}\otimes\mathrm{Res} =\prod_{n\geq 1} {}_{n}(k[S_n]\otimes_{S_{n-1}}k[S_n])_n,$$ we may give their definitions in the form \begin{align*} \mathrm{Ind}\otimes\mathrm{Res} &\to I\\ (g\otimes h&\mapsto gh)_{n\geq 1}\\ \end{align*} and \begin{align*} I &\to \mathrm{Ind}\otimes\mathrm{Res}\\ (g& \mapsto \sum_{i=1}^n gg_i\otimes g_i^{-1})_{n\geq 1}, \end{align*} where $$g_1,\dots,g_n\in S_n$$ form a set of coset representatives for $$S_n/S_{n-1}$$.

The composition $$h:I\to \mathrm{Ind}\otimes\mathrm{Res} \to I$$ of these is $$h=(n\operatorname{id})_{n\geq 1}$$.

We can calculate \begin{align*} \operatorname{id}_{\mathrm{Ind}}\otimes h &= ((n-1)\operatorname{id})_{n\geq 1}\\ h\otimes \operatorname{id}_{\mathrm{Ind}} &= (n\operatorname{id})_{n\geq 1}, \end{align*} and therefore $$\mathcal{S}$$ is not spacial.

Graphically, this is that counter-clockwise loops cannot be dragged across an upward strand, imagining $$\mathrm{Ind}$$ as an up-arrow and $$\mathrm{Res}$$ as a down arrow.