# Understanding the stack $B\mathbb{Z}$

Here, let $\mathbb{Z}$ be the group scheme whose functor of points is the constant functor which takes a connected affine scheme to the group $\mathbb{Z}$. I'm having a bit of trouble understanding the stack $B\mathbb{Z}$, in particular quasicoherent sheaves on it.

Here's how I want to approach $BG$ for $G$ say an affine algebraic group. One can take a smooth atlas $f: * \rightarrow BG$ and apply Barr-Beck to find that the category of coalgebras over the comonad $f^* f_*$ is equivalent to the category of quasicoherent sheaves on $BG$. One can use smooth base change ($f$ is smooth and thus flat, also separated and finite type, everything is noetherian I think) and see that this comonad is the same as $g_* g^*$ for $g: G \rightarrow *$. Then a coalgebra is a vector space $V$ (let's say everything is over a field of char 0) which is a comodule for $\mathcal{O}_G$ (ring of global functions). If $V^*$ is finite-dimensional then we can take symmetric powers and spectrum and get an algebraic action $G \times V^* \rightarrow V^*$ -- here we need that $G$ is affine. Thus a coherent sheaf on $BG$ is a vector space with a rational action of $G$. (Is the above correct? I understand there's also some argument involving Hopf duals, but I'm not sure there's a good analogue for the group ring for $G$ not finite.)

For $G$ not affine (e.g. $G = \mathbb{Z}$, or maybe $G$ an ind-scheme like $\hat{\mathbb{G}}_a$), do we have a way to understand (coherent sheaves on) these objects?

Edit: Extra question. In the above, I had to assume that $V$ as finite-dimensional, in order for a coaction $k[V] \rightarrow k[V] \otimes k[G]$ to give a rational/algebraic action. If $V$ is infinite-dimensional (i.e. quasicoherent sheaf instead of coherent sheaf), I'm also not really sure how to proceed.

• I would guess that quasicoherent sheaves on $B G$ for any discrete group $G$ is just $G$-modules. – Zhen Lin Dec 1 '14 at 8:46

If $p : G \to S$ is any smooth group scheme over some base $S$, then $BG:=B(G/S)$ is an algebraic stack over $S$ with a presentation $S \to BG$ (corresponding to the trivial $G$-bundle). (Actually we only need $G \to S$ to be flat, then $BG$ will be an algebraic stack in the sense of Goerss-Naumann.) A quasi-coherent sheaf on $BG$ thus consists of a quasi-coherent sheaf $M$ on $S$ together with an isomorphism $p_1^* M \cong p_2^* M$ on $S \times_{BG} S$ which satisfies the cocycle condition. Using the isomorphism $S \times_{BG} S \cong G$, we get an isomorphism $p^* M \cong p^* M$. This is the same as a $G$-action on $M$, which is a natural family of group homomorphisms $G(T) \to \mathrm{Aut}_{\mathcal{O}_T}(M_T)$ for $S$-schemes $T$. So we get $\mathsf{Qcoh}(BG)=\mathsf{Rep}_S(G)$.
If $\Gamma$ is a discrete group, we may look at its corresponding group scheme $G=\coprod_{g \in \Gamma} S$. Its functor of points is the sheaf associated to the constant presheaf with value $\Gamma$. Therefore, an action of $G$ on a quasi-coherent module $M$ on $S$ is just an action of $\Gamma$ in the usual sense, i.e. a group homomorphism $\Gamma \to \mathrm{Aut}_{\mathcal{O}_S}(M)$. For $\Gamma=\mathbb{Z}$ this action is just an automorphism of $M$.
Hence, quasi-coherent sheaves on $B (\coprod_\mathbb{Z} S)$ are just quasi-coherent sheaves on $S$ equipped with an automorphism of $S$.
• Ah, this makes sense. What I was missing is how to think of a $G$-action by functor of points. Thanks! – user148177 Dec 1 '14 at 11:10
• Perhaps someone can explain (I forgot how to do this) why the isomorphism $p_1^* M \cong p_2^* M$ corresponds to a $G$-action on $M$? – Martin Brandenburg Dec 1 '14 at 18:54
• I'd guess you can do it by functor of points definition for quasicoherent sheaves, i.e. a quasicoherent sheaf on $G$ is a functorial assignment to each affine $T$ and $\eta \in G(T)$ a $T$-module (morally, the pullback to $T$, you can say this more cleanly but I forget the terminology). So a map in $G(T)$ gives you an isomorphism of the pullback of $M$ to $T$, i.e. a $T$-automorphism of $M_T$, which is a $G$-action on $M$ – user148177 Dec 1 '14 at 20:44
• Yes, this is a very good explanation! And the cocycle condition will (hopefully) correspond to the property that $G(T) \to \mathrm{Aut}(M_T)$ is a group homomorphism. – Martin Brandenburg Dec 1 '14 at 21:28