Gerbes for a cyclic group. (or maybe G_m too) - MathOverflow

Gerbes for a cyclic group. (or maybe G_m too)

2009-11-26T15:23:29Z

Let μ_n be the group scheme of n-th roots of unity. If X is a scheme and L is a line bundle on X, then I can construct a μ_n-gerbe Y over X by letting the S-points of Y be a S-point of X, a line bundle M on S and an isomorphism between the n-th tensor power of M and the pullback of L to S. Can anyone provide examples not of this form?

Commentary: It looks like I'm taking the image of an element of H¹(X,G_m) in H²(X,μ_n) under the long exact sequence associated to 0-->μ_n-->G_m-->G_m-->0. Thus examples of gerbes for the multiplicative group G_m will likely be relevant, so people providing such examples will also be appreciated.

Answer by David Ben-Zvi for Gerbes for a cyclic group. (or maybe G_m too)

2009-11-26T16:37:24Z

I think the answer is that you can take nth roots of a complex power of a line bundle - i.e. of a line bundle on a G_m gerbe (an example of a twisted sheaf), and this will account for preimages of elements in H^2(G_m) in your long exact sequence (see this related question).

As for interesting examples of $\mu_n$ gerbes, look at the moduli stack of stable $G$ bundles on an algebraic curve for $G$ with center $\mu_n$ (eg $SL_n$). It is a $\mu_n$ gerbe over the moduli space of bundles, whose nontriviality accounts for the lack of existence of a universal bundle. (It is described explicitly in many places - the ones that come to mind are King-Schofield arXiv:math/9907068, Beilinson-Drinfeld's Quantization of Hitchin Hamiltonians (Chapter 4) and Kapustin-Witten arXiv:hep-th/0604151 (Section 7).

Answer by Paul Johnson for Gerbes for a cyclic group. (or maybe G_m too)

2009-11-26T18:13:30Z

This probably deserves to be worked out in more detail, and with more sophistication, but quickly and easily...

The gerbes you get from your construction are all banded - so certainly any nonbanded gerbe would work, but this is kind of silly so I'll construct a banded example for you.

When thinking about gerbes I find it useful to think about the zero dimensional case: then we're really doing group theory, which feels more familiar.

In other words, let's think about BG: a line bundle over BG is just a one dimensional representation of G; a (banded) H-gerbe over BG is a (central) extension of G by H.

Let's take G=Z/2Z X Z/2Z, and H=Z/2Z. Which extensions do we get from your construction? Well, a representation will either be trivial, or have one element that acts by multiplication by (-1). In the group extension that we get, this representation should have a nontrivial square root. In the trivial representation case we can take just take the trivial extension G X H, and the representation where (0,0, 1) act by multiplication by -1. as our nontrivial square root. For the nontrivial representations, the extension will be (Z/4Z) X (Z/2Z), and the square root will be the representation where (1,0) acts by multiplication by i.

In particular, we notice that all the extensions we got in this manner were abelian groups. However, G has a central extension by H that isn't abelian: D_4, the dihedral group of order 8 - rotation by 180 degrees is central. So BD_4 is a banded H gerbe over BG that doesn't come from your construction.

One can then easily modify this example to get something a little more geometric: take your favorite space X with a free G action. Give it a D_4 action by having rotation by first mapping to D_4 to G and then acting on X, so rotation by 180 acts trivially. Then the global quotient [X/D_4] should be a banded Z/2Z gerbe over [X/(Z/2Z X Z/2Z)] that doesn't come from your construction.

Answer by Anatoly Preygel for Gerbes for a cyclic group. (or maybe G_m too)

2009-11-26T19:44:34Z

A bit of a response to your "Commentary":

As you point out, the failure of your construction to hit all $\mu_n$ -gerbes is governed by the exact sequence $H^1(X, \mathbb{G}_m) \to H^2(X, \mu_n) \to H^2(X,\mathbb{G}_m)[n] \to 0$ so answering your question is related to producing torsion $\mathbb{G}_m$ -torsors.

The question of doing so has been studied as part of the theory of Brauer groups:

Let $Br(X)$ ("Brauer group") denote the group of Azumaya algebras, which are a generalization of the central simple algebras over a field (that is the classical Brauer group).

Let $Br'(X)$ ("Cohomological Brauer group") denote the torsion part of $H^2(X, \mathbb{G}_m)$ .

In the case of $X = Spec k$, $k$ a field, the equivalence of these two groups is classical and they can be computed in various cases of number-theoretic interest (e.g., number fields/local fields/finite fields). In this case, $H^1(X, \mathbb{G}_m)=0$ by Hilbert's Theorem 90, and yet there are plenty of examples where $Br(X)$ is very much non-trivial. Grothendieck studied the relation between $Br(X)$ and $Br'(X)$ in general in Dix Exposes. The upshot is that there is an injective map $Br(X) \to Br'(X)$ and it is an isomorphism in reasonable cases (e.g., I think $X$ quasi-projective over a field). (See Dix Exp, or Ch. IV of Milne's "Etale Cohomology".)

I won't say more about the general picture, but I'll work out in detail the simple case of $X = Spec \mathbb{R}$. In this case, $Br(Spec \mathbb{R}) = \mathbb{Z}/2\mathbb{Z}$ generated by the class of the usual quaternions, viewed as a central simple algebra over $\mathbb{R}$. We can give a geometric description of the resulting $\mathbb{G}_m$-torsor:

Start with the smooth plane conic $C = Proj \mathbb{R}[x,y,z]/(x^2+y^2+z^2)$ . It's a smooth genus $0$ curve, but has no $\mathbb{R}$-points and so is not isomorphic to $\mathbb{P}^1_{\mathbb{R}}$ . However, after base-change to $\mathbb{C}$ it attains a point and so becomes isomorphic to $\mathbb{P}^1_{\mathbb{C}}$ ; such a Galois-twisted form of $\mathbb{P}^n_{\mathbb{C}}$ is known as a Brauer-Severi variety and the elements of the Brauer-group (of a field) can also be thought of as corresponding to them (the group structure is then a bit strange). Since $Aut(\mathbb{P}^n) = PGL_{n+1}$ , these correspond to $PGL_{n+1}$ -torsors and the relation to $\mathbb{G}_m$ -torsors is via the exact sequence for $PGL_{n+1}=GL_{n+1}/\mathbb{G}_m$ . So, a $T$-point of the corresponding $\mathbb{G}_m$ -torsor for a $\mathbb{R}$-scheme $T$ consists of the following data:

It is a rank $2$ vector bundle $V$ over $T$, together with an isomorphism of $T$-schemes $C_T \simeq \mathbb{P}(V)$ where $C_T = C \times_{Spec \mathbb{R}} T$ is the pullback of our genus $0$ curve to $T$, and $\mathbb{P}(V)$ is the associated projective space (here $\mathbb{P}^1$) bundle of our vector bundle $V$.

Why is this a $\mathbb{G}_m$ -gerbe? Well, $\mathbb{P}(V) \simeq \mathbb{P}(V')$ iff $V$ and $V'$ differ by tensoring by a line bundle. The gerbe is non-trivial since it has no $\mathbb{R}$-points, since $C$ itself is not isomorphic to projective space. It has $\mathbb{C}$-points because the base-change is isomorphic to projective space.