I am a physicist trying to understand some of the mathematics used in condensed matter physics. I am getting stuck on what I think is probably a very simple mathematical question. In this paper, I found the statement that \begin{align} \text{Hilb}_{\mathbb{Z}_2} \cong \text{Rep}(\mathbb{Z}_2), \end{align} where $\text{Hilb}_{\mathbb{Z}_2}$ is the category of $\mathbb{Z}_2$-graded Hilbert spaces and $\text{Rep}(\mathbb{Z}_2)$ is the representation category of $\mathbb{Z}_2$. I believe this statement is true if we replace Hilbert spaces with vector spaces, i.e. $\text{Vec}_{\mathbb{Z}_2} \cong \text{Rep}(\mathbb{Z}_2)$, and I don't want to worry about the distinction between $\text{Hilb}_{\mathbb{Z}_2}$ and $\text{Vec}_{\mathbb{Z}_2}$, so I will state everything in terms of $\text{Vec}_{\mathbb{Z}_2}$. (Of course, please correct me if I am wrong about this.) Honestly, I don't really understand exactly what $\cong$ means here, but I interpret it as an isomorphism or equivalence between categories.

This statement is puzzling to me. I think that there should be two representations of $\mathbb{Z}_2$, so $\text{Rep}(\mathbb{Z}_2)$ should have two objects. But I think that I can have a distinct super vector space of any dimension $p|q$, so there should be infinitely many $\mathbb{Z}_2$-graded vector spaces. Thinking about it like this, I don't see how these two categories can be $\cong$.

I have read the answer to Representations of $\mathbb{Z}/2$ as super vector spaces? that shows how to associate representations of $\mathbb{Z}_2$ to $\mathbb{Z}_2$-graded vector spaces, and vice versa, but I am still confused. I think my confusion comes down to not really understanding the two categories involved, and perhaps to not understanding what $\cong$ means.

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    $\begingroup$ The original statement appears to be false as vector spaces are not necessarily Hilbert spaces and certainly not in a unique way. The way you have reformulated it in terms of vector spaces seems to be correct. Here the $\cong$ symbol could be taken to mean as $\mathbb{C}$-linear categories, or, more probably, as $\mathbb{C}$-linear tensor categories. In that case, one probably wants to restrict to finite dimensional spaces and one needs to specify the commutativity and associativity contraints on $\mathrm{Hilb}_{\mathbb{Z}_2}$. (The paper seems to suggest the choice does not matter.) $\endgroup$
    – Kapil
    Oct 23 '21 at 3:23

This isomorphism can be understood as an isomorphism of abelian symmetric monoidal categories. We also don't have to worry so much about the word "abelian", since all of these categories are semisimple.

Take the category of finite-dimensional vector spaces over a field. A simple object in this category (a nonzero object which has no subobjects other than 0 and itself) is a one-dimensional vector space, and all other vector spaces are isomorphic to direct sums of one-dimensional vector spaces. Vector spaces are isomorphic if and only if they have the same dimension, so we can label isomorphim classes in this category by natural numbers, say $\{[n] \mid n \in \mathbb{N}\}$, where $[n]$ refers to the class of $n$-dimensional vector spaces. On these isomorphism classes, direct sum goes across to addition $[n] \oplus [m] = [n + m]$, and the monoidal (tensor) product goes to multiplication $[n] \otimes [m] = [nm]$. In highbrow terms, we have categorified the natural numbers with addition and multiplication.

Note that the category of finite-dimensional representations is equivalent to the category of representations of the trivial group.

Now we move on to $\mathbb{Z}_2$-graded vector spaces, which some call super vector spaces. We can work out again that this category has two simple objects: a one-dimensional vector space in grade 0, and a 1-dimensional vector space in grade 1. We'll write $[p|q]$ with $p, q \in \mathbb{N}$ for the class of supervector spaces which have dimension $p$ in the $0$-graded part and dimension $q$ in the $1$-graded part. The $0$-graded and $1$-graded parts cannot really be distinguished by the abelian structure on the category (we know they are different, but we don't know which to call $0$ and which to call $1$) because $[p|q] \oplus [n|m] = [p+n|q+m]$. However the monoidal structure allows us to tell them apart, since we have $[1|0] \otimes [p|q] = [p|q]$ but $[0|1] \otimes [p|q] = [q|p]$.

Now take the category of finite-dimensional representations of the cyclic group $\mathbb{Z}_2$ over a field (of characteristic not equal to 2). We know that there are two simple objects, the trivial representation and the sign representation, and that all other objects are direct sums of these. Now you can check the tensor product rule to see that indeed this category "looks like" (at least on isomorphism classes) the category of super vector spaces, in both its additive and monoidal structure.


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