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!

  • $\begingroup$ Why do you expect these spaces to have different classes in the Grothendieck ring? $\endgroup$ Dec 14, 2016 at 17:39
  • $\begingroup$ 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. $\endgroup$ Dec 14, 2016 at 18:28
  • $\begingroup$ 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 $X$. $\endgroup$ Dec 14, 2016 at 20:01
  • 1
    $\begingroup$ @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. $\endgroup$ Dec 15, 2016 at 2:28

1 Answer 1


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.

  • $\begingroup$ I see, $X\to X/G$ being a $G$-torsor does not involve the action, that is what I was missing. Thanks Dan! $\endgroup$ Dec 15, 2016 at 22:54

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