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

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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?

  • 7
    $\begingroup$ Could you describe the axiom without using string diagrams as well, to help those that aren't familiar with this specific graphical language? $\endgroup$
    – Max New
    Nov 4, 2019 at 19:15
  • $\begingroup$ The formula is at the top of page 14 of Selinger's paper, arxiv.org/pdf/0908.3347. $\endgroup$
    – Neuromath
    Nov 5, 2019 at 18:14
  • $\begingroup$ Also now added in an edit. $\endgroup$
    – Neuromath
    Nov 8, 2019 at 12:29
  • $\begingroup$ Out of curiosity: Every symmetric monoidal category is spacial, right? $\endgroup$ Jan 10, 2020 at 21:03

3 Answers 3


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.

  • 2
    $\begingroup$ Objection: $S^1\vee S^2$ gives the easiest to understand counterexample. $\endgroup$ Nov 6, 2019 at 20:56
  • 2
    $\begingroup$ Ha! You speak the truth. $\endgroup$ Nov 6, 2019 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.


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