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Let us work over $\mathbb{C}$. Let $G$ be a finite group, acting on $\mathbb{A}^1$ via a character, and let $H$ be the kernel of the action.

Assume that $\mathbb{A}^1$ is the coarse moduli space of the $[\mathbb{A}^1/G]$, and that $[\mathbb{G}_m/G$ maps onto $\mathbb{G}_m$. The way I picture the stack quotient $[\mathbb{A}^1/G$ is $\mathbb{A}^1$ with a $BG$ at the origin and $BH$ at every point different from the origin. In these terms, how should I think of the difference between $[\mathbb{G}_m/G]$ and $\mathbb{G}_m\times BH$? They ought to be different, but I keep making the mistake of confusing one with the other. Is it possible to write $[\mathbb{G}_m/G]$ as a quotient stack involving some twist of $H$?

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Often you hear the informal statement that a stack is like a scheme, except with a stabilizer group attached to each point. Your question shows why the intuition from this statement can be misleading. In a sense, you need to also remember the way these groups fit together. But if you try to make that statement precise you'll soon find yourself completely abandoning the attempt to define stacks as some sort of "extra data" over a coarse moduli space.

Anyway, here is an analogy. If you had defined a vector bundle over a space as just the data of a vector space $V_x$ attached to each point $x$ of the space, then you might expect any two vector bundles of the same rank over the same space to be isomorphic. This is because the definition is incomplete; if you don't specify how the different $V_x$'s fit together there can be no global twisting.

In your case, you should use the word "gerbe" (or $H$-gerbe in this case) rather than "bundle" - a gerbe is basically a bundle of stacks with fiber $BH$. So $[\mathbb G_m / G]$ is an $H$-gerbe over $\mathbb G_m$ (which is generally nontrivial), whereas $\mathbb G_m \times BH$ is the trivial $H$-gerbe over $\mathbb G_m$. The fact that the spaces are different is no weirder than that you can have two fiber bundles with same base and fiber but different total spaces.

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  • $\begingroup$ How can one easily prove that this particular gerbe $[\mathbb{G}_{\mathrm{m}}/G]$ is nontrivial? $\endgroup$ – Qfwfq Aug 7 '18 at 9:05
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One can see the difference by writing down the functor of points explicitly. For a test scheme $T$, \begin{align} (\mathbb G_m\times BH)(T) & =\{(f,p)\mid f:T\to \mathbb G_m, p:T'\to T\text{ $H$-torsor}\},\\ [\mathbb G_m/G](T) & =\{(q,g)\mid q:\tilde T\to T\text{ $G$-torsor, }g:\tilde T\to \mathbb G_m\text{ $G$-equivariant}\}\\ & =\{(f,p')\mid f:T\to \mathbb G_m,p':\tilde T\to T\times_{f,\mathbb G_m,\chi}\mathbb G_m\text{ $H$-torsor}\}, \end{align} where $\chi :\mathbb G_m\to \mathbb G_m$ is the quotient by $G/H$. Note that $H$-torsors $p$ and $p'$ have different targets! We can define a map $\mathbb G_m\times BH\to [\mathbb G_m/G]$ by pulling back $p$ from $T$ to $T\times_{f,\mathbb G_m,\chi}\mathbb G_m$, but it won't be invertible in general. One can see it by considering a single $G$-orbit on both sides. On the left, we get $pt/H=(G/G)/H$, and on the right $(G/H)/G$. But the two actions don't have to commute, so in general we get two non-isomorphic groupoids.

One can also see the difference by computing $K$-theory of both spaces after decomposing $\mathbb A^1$ as $\mathbb G_m\cup pt$.

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