Counting with weight

Let $G$ be a finite group acting on a nonempty finite set $X.$ The groupoid $[X/G]$ of this action is a category defined by the following data:

• objects: elements of $X$;
• morphism: $\mathrm{Hom}_{[X/G]}(x, y) = \{g \in G : gx = y\}.$

In this category, every morphism is invertible, so $[X/G]$ is indeed a groupoid. We have $$\mathrm{Aut}_{[X/G]}(x) = \mathrm{Stab}_{G}(x),$$ the stabilizer subgroup of $G$ at $x \in X.$ The set $X/G$ of orbits of this action is precisely the set of isomorpihsm classes of $[X/G].$ 

Given an isomorphism class $[x]$ of $[X/G]$ (i.e., an orbit of the given action), the orbit-stabilizer theorem translates to $$|[x]| = |G|/|\mathrm{Aut}_{[X/G]}(x)|.$$ This implies that $$\sum_{[x] \in X/G}\frac{1}{|\mathrm{Aut}_{[X/G]}(x)|} = \frac{|X|}{|G|}.$$ Now, say we have a subset $Y \subset X$ where the $G$-action restricts to. Then

$$\frac{\sum_{[y] \in Y/G}\frac{1}{|\mathrm{Aut}(y)|}}{\sum_{[x] \in X/G}\frac{1}{|\mathrm{Aut}(x)|}} = \frac{|Y|/|G|}{|X|/|G|} = \frac{|Y|}{|X|}.$$ Hence, we just obtained another expression of the probability that a random element of $X$ lies in $Y,$ when $X$ is given the uniform distribution.

Why do we bother? We could just write $|Y|/|X|$ instead of another fraction of some complicated summations. The first reason I can think of is quite philosophical. If we are just given categories $\mathcal{B} \subset \mathcal{A}$, where the latter is nonempty, with finitely many isomorphism classes that are not necessarily given as the action groupoids, how should we define the probability that a random object of $\mathcal{A}$ is in $\mathcal{B}$? One may say $|\mathrm{Ob}(\mathcal{B})|/|\mathrm{Ob}(\mathcal{A})|,$ but it is rarely the case that a category has finitely many objects. It would be more reasonable to define $$\mathrm{Prob}_{x \in \mathcal{A}}(x \in \mathcal{B}) := \frac{\sum_{[y] \in \mathcal{B}/\simeq}\frac{1}{|\mathrm{Aut}(y)|}}{\sum_{[x] \in \mathcal{A}/\simeq}\frac{1}{|\mathrm{Aut}(x)|}},$$ which is a "generalization" of the case of action groupoids. Perhaps, we should define the number of points on the category $\mathcal{A}$ as $$\| \mathcal{A} \| := \sum_{[x] \in \mathcal{A}/\simeq}\frac{1}{|\mathrm{Aut}(x)|}.$$ OK. Now I am going to speak about something that I am not quite familiar with. The sum over the isomorphism classes with the weights above also seems to be the definition of the number of points over a finite field for a Deligne-Mumford stack (e.g. van den Bogaard and Edixhoven). When we have a quotient stack, this seems to fit our discussion even more evidently, but I am quite daunted to check any formal "stacky" axioms, so let's think about an example.

Example. Consider the scheme $\mathrm{Mat}_{n} := \mathbb{A}^{n^{2}} = \mathrm{Spec}(\mathbb{F}_{q}[x_{ij}]_{1 \leq i, j \leq n})$ over a fixed finite field $\mathbb{F}_{q}.$ We chose the notation because we would like to think about $n \times n$ matrices $(x_{ij})_{i,j=1}^{n}$ in the back of our head. We can then think about the group scheme $$\mathrm{GL}_{n} := \mathrm{Mat}_{n} \smallsetminus V(\det),$$ which is an open subset of $\mathrm{Mat}_{n}.$ We have a scheme-theoretic action of $\mathrm{GL}_{n}$ on $\mathrm{Mat}_{n},$ given by the conjugation, and this gives a quotient stack $[\mathrm{Mat}_{n}/\mathrm{GL}_{n}].$ The set $[\mathrm{Mat}_{n}/\mathrm{GL}_{n}](\mathbb{F}_{q})$ of $\mathbb{F}_{q}$-points of this quotient stack is precisely the groupoid of the conjugate action of the group $\mathrm{GL}_{n}(\mathbb{F}_{q})$ of invertible matrices on the set $\mathrm{Mat}_{n}(\mathbb{F}_{q})$ of $n \times n$ matrices over $\mathbb{F}_{q},$ so in notation, we have $$[\mathrm{Mat}_{n}/\mathrm{GL}_{n}](\mathbb{F}_{q}) = [\mathrm{Mat}_{n}(\mathbb{F}_{q})/\mathrm{GL}_{n}(\mathbb{F}_{q})].$$ Moreover, note that the isomorphism classes of $\mathbb{F}_{q}[t]$-modules of $\mathbb{F}_{q}$-dimension $n$ are in one-to-one correspondence with the orbits of this conjugation action. The category $\mathrm{Mod}_{\mathbb{F}_{q}[t]}^{\dim_{\mathbb{F}_{q}} = n}$ of $\mathbb{F}_{q}[t]$-modules of $\mathbb{F}_{q}$-dimension $n$ surely has infinitely many objects, but it has only finitely many isomorphism classes, each of corresponds to an element of $\mathrm{Mat}_{n}(\mathbb{F}_{q})/\mathrm{GL}_{n}(\mathbb{F}_{q}).$ If we unravel the definition of probability we defined above, for any property $\mathscr{P}$ on the objects of the category $\mathrm{Mod}_{\mathbb{F}_{q}[t]}^{\dim_{\mathbb{F}_{q}} = n}$ that is invariant under its isomorphism classes, we have: $$\mathrm{Prob}_{M \in \mathrm{Mod}_{\mathbb{F}_{q}[t]}^{\dim_{\mathbb{F}_{q}} = n}}(M \text{ satisfies } \mathscr{P}) = \mathrm{Prob}_{A \in \mathrm{Mat}_{n}(\mathbb{F}_{q})}(A \text{ satisfies } \mathscr{P}),$$ where the right-hand side is given by the uniform distribution on $\mathrm{Mat}_{n}(\mathbb{F}_{q}).$

Remark. Another reason for taking the definition for the number of points with the weight $1/|\mathrm{Aut}|$ is due to Behrend's trace formula.