# What is the motivic class of a quotient stack?

The Grothendieck ring of complex varieties $K(Var_\mathbb C)$ is the free abelian group generated by isomorphism classes $[X]$ of $\mathbb C$-varieties, modulo the scissor relation $[X]=[Z]+[X\setminus Z]$ for every closed subvariety $Z\subset X$. The product is given by $[X]\cdot [Y]=[X\times_\mathbb CY]$. There is a similar construction for stacks, leading to a ring $$K(St_\mathbb C).$$ One has to restrict to stacks with affine stabilizers. This construction is treated for example in section 3 of this paper by Bridgeland. One can show that $K(St_\mathbb C)$ can be obtained from $K(Var_\mathbb C)$ by inverting the classes of all special groups, or equivalently all classes $[GL_d]$ for $d\geq 1$, or equivalently, the classes $\{\mathbb L,\mathbb L^i-1:i\geq 1\}$, where $\mathbb L=[\mathbb A^1]$.

I am missing some basic point because it seems to me that all classes $[G]$ (for $G$ special) are invertible in $K(St_\mathbb C)$ so that if I have a quotient stack $X/G$ its class will be $$[X/G]=[X]/[G]\in K(St_\mathbb C)$$ regardless of the action of $G$ on $X$, and this is confusing. Here are a couple of examples I would like to understand:

1. What is the class of $GL_2/GL_2$, where $GL_2$ acts on itself by conjugation? I would like this to be different from $[pt]=1$.
2. What is the class of weighted projective space $\mathbb P(a_0,\dots,a_{n-1})=(\mathbb C^{n}\setminus 0)/\mathbb C^*$? I would like this to be different from the class $[\mathbb P^{n-1}]=1+\mathbb L+\cdots+\mathbb L^{n-1}$.

Thank you!

• Why do you expect these spaces to have different classes in the Grothendieck ring? Dec 14, 2016 at 17:39
• Just because the spaces are very different, and the Grothendieck ring should be able to keep track of stabilizers, say. For $X/G$, I find it strange that the action plays no role, but maybe I am wrong. Dec 14, 2016 at 18:28
• To give a very simple toy model, if $X$ is a finite set and $G$ a finite group acting on it then the groupoid cardinality ("orbifold Euler characteristic," if you prefer) of the homotopy quotient $X/G$ is equal to $\frac{|X|}{|G|}$ regardless of the action of $G$ on $X$. Dec 14, 2016 at 20:01
• @AndreaRicolfi: I think it might be true that weighted projective space is equal to $1 + \mathbb L + \ldots + \mathbb L^n$ in the usual Grothendieck ring already, even without talking about stacks. One way to try to do this by hand is using the toric description of weighted projective space. Dec 15, 2016 at 2:28

Yes: if $G$ is a special group, then $[X/G] = [X]/[G]$ in the Grothendieck group of stacks. This is the analogue of the fact that if $X \to E$ is a $G$-torsor over an algebraic variety (and $G$ is still special) then $[X] = [E]\cdot [G]$ in the Grothendieck group of varieties. In the category of stacks, the map $X \to X/G$ is always a $G$-torsor, regardless of the action of $G$ on $X$. Does this help?
Qiaochu's toy example is very relevant but also somewhat orthogonal to this situation: finite groups are not special, and $[X/G] = [X]/[G]$ is almost never true in the Grothendieck group of stacks (or varieties, for that matter) if $G$ is finite. For example, $\mathbb G_m$ is a finite étale double cover of itself, but its class in the Grothendieck group is nonzero. A more interesting example is that the class of $BG$ is in many cases (but not always!) just the class of a point: see "A geometric invariant of a finite group" by Torsten Ekedahl.
• I see, $X\to X/G$ being a $G$-torsor does not involve the action, that is what I was missing. Thanks Dan! Dec 15, 2016 at 22:54