# What is the groupoid cardinality of the category of vector spaces over a finite field?

For any groupoid, it's groupoid cardinality is the sum of the reciprocals of the automorphism groups over the isomorphism classes. Let us consider the category of vector spaces over a finite field $$\mathbb F_q$$ with only invertible morphisms allowed.

Then, the size of the automorphism groups are $$g_n= \prod_{i=1}^n(q^n-q^{i-1}) = q^{n\choose 2}\prod_{i=1}^n(q^i-1)$$ so it's groupoid cardinality is the following infinite sum: $$\sum_{n \geq 1}\frac{1}{g_n}.$$

Is there a closed form expression for it? Note that as $$q\to 1$$ in an appropriate sense, the general linear group is supposed to converge to the symmetric group and the groupoid of vector spaces should converge to the groupoid of finite sets.

In this sense, the above infinite sum is an analog of the usual one for $$e$$ and can perhaps reasonably be called a q-analog of $$e$$.

Alternatively, in the case of abelian, (even semi-simple) categories, is there a better replacement for the groupoid cardinality that I should be using?

• Is the sum $q$-hypergeometric, i.e., the ratio $g_{n+1}/g_n$ is a rational function of $q^n$? If it is then you can apply the $q$-Gosper algorithm. Jun 17 '20 at 5:54

Upon substituting $$x=\frac{1}{q}$$ we obtain $$\sum_{n\geq 0}\frac{1}{|\mathrm{GL}_n(\mathbb F_q)|}=\sum_{n\geq 0}\frac{x^{n^2}}{(1-x)(1-x^2)\cdots(1-x^n)}$$ and this evaluates to the product $$\prod_{i\geq 1}\frac{1}{(1-x^{5i-4})(1-x^{5i-1})}$$ by the first Rogers-Ramanujan identity.
You can also interpret the second Rogers-Ramanujan identity as giving a groupoid cardinality for the category of affine spaces over $$\mathbb F_q$$. where the sum becomes $$\sum_{n\geq 0}\frac{1}{|\mathrm{AGL}_n(\mathbb F_q)|}$$. (Note that I wrote these sums to contain a term for $$n=0$$, corresponding to the zero dimensional vector/affine space.)