This is to understand better Example 3.9 on page 221 of Group actions on stacks and applications by M.Romagny.

For an action of an algebraic group (scheme) $G$ on an algebraic stack $\mathcal{M}$, the author defines a fixed point stack $\mathcal{M}^G$ as follows. Over a base scheme $T$, objects of $\mathcal{M}^G(T)$ are pairs

$$(x,\{\alpha^g\}_{g\in G(T)})$$

with $x\in\mathcal{M}(T)$ and $\{\alpha^g\}$ a system of (iso)morphisms in $\mathrm{mor}\mathcal{M}$ $$\alpha^g:x\to g.x$$

such that, for every $g,g'\in G(T)$, the following "non abelian group cohomology cocycle condition" holds


and morhisms in $\mathcal{M}^G(T)$ are the obvious ones.

Now, the (counter)example in question is to show that, in general, "taking the fixed point stack" does not commute with "taking the coarse moduli space" (when the latter exists). It works as follows.

Let $H=\{\pm1,\pm i, \pm j, \pm k \}$ be the $8$ element quaternion group and $1\to Z \to H \to G \to 1$ be the quotient by its center $Z=\{\pm 1\}\simeq \mathbb{Z}/2$. The abelian group $G\simeq \mathbb{Z}/2\times \mathbb{Z}/2$ acts by conjugation on $H$, and this induces an action $G \curvearrowright BH$ , where $BH=[*/H]$ is the classifying stack of $H$.

The point of the example is that $((BH)^{\mathrm{cms}})^G=*^G=*$, while already $(BH)^G=\textrm{Ø}$.

What is not clear to me is this latter equality $(BH)^G=\textrm{Ø}$.


1 Answer 1


I didn't check every detail, but I think it should work like this:

the point is that the existence of a $G$-invariant $H$-torsor (i.e. an object of $(BH)^G$) would give you a section $G\to H$, which does not exist (as Matthieu specifies in the paper).

To see that, assume you have your $H$-torsor $P\to T$, and pull it back to a geometric point of $T$ to get an $H$-torsor over an algebraically closed field $X\to Spec\; k$. Let's call $X^g$ the torsor (still over $Spec \; k$) that you obtain by twisting by $g\in G$. Then $X^g$ is just $X$ as a scheme, but the action of $H$ is defined by $h\cdot x= \phi(g)(h)\cdot x$, where $\phi(g)$ is the automorphism of $H$ given by $g$.

Now assume you have for every $g$ an isomorphism $\alpha_g\colon X\to X^g$ of $H$-torsors over $Spec \; k$, and they are compatible in the sense you specify. Fix any $x\in X$ (think about this as picking out $1\in H$), and define $f\colon G\to H$ as follows: for $g\in G$ consider $\alpha_g(x)\in X^g$, and define $f(g)$ to be the unique element of $H$ such that $\alpha_g(x)=f(g)\cdot x$.

This should define a section, and give a contradiction.


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