TITLE: Understand the difference between two stacks
QUESTION [7 upvotes]: 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 $H$?

REPLY [8 votes]: 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 $\mathbb G_m\cup pt$.