Gerbes for a cyclic group. (or maybe G_m too) - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T18:54:03Z http://mathoverflow.net/feeds/question/6891 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/6891/gerbes-for-a-cyclic-group-or-maybe-g-m-too Gerbes for a cyclic group. (or maybe G_m too) Peter McNamara 2009-11-26T15:23:29Z 2009-11-26T19:44:34Z <p>Let &mu;<sub>n</sub> 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 &mu;<sub>n</sub>-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?</p> <p>Commentary: It looks like I'm taking the image of an element of H<sup>1</sup>(X,G<sub>m</sub>) in H<sup>2</sup>(X,&mu;<sub>n</sub>) under the long exact sequence associated to 0-->&mu;<sub>n</sub>-->G<sub>m</sub>-->G<sub>m</sub>-->0. Thus examples of gerbes for the multiplicative group G<sub>m</sub> will likely be relevant, so people providing such examples will also be appreciated.</p> http://mathoverflow.net/questions/6891/gerbes-for-a-cyclic-group-or-maybe-g-m-too/6896#6896 Answer by David Ben-Zvi for Gerbes for a cyclic group. (or maybe G_m too) David Ben-Zvi 2009-11-26T16:37:24Z 2009-11-26T16:37:24Z <p>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 <a href="http://mathoverflow.net/questions/1721/what-do-gerbes-and-complex-powers-of-line-bundles-have-to-do-with-each-other/1737#1737" rel="nofollow">related question</a>).</p> <p>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).</p> http://mathoverflow.net/questions/6891/gerbes-for-a-cyclic-group-or-maybe-g-m-too/6906#6906 Answer by Paul Johnson for Gerbes for a cyclic group. (or maybe G_m too) Paul Johnson 2009-11-26T18:13:30Z 2009-11-26T18:13:30Z <p>This probably deserves to be worked out in more detail, and with more sophistication, but quickly and easily...</p> <p>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.</p> <p>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. </p> <p>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. </p> <p>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.</p> <p>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.</p> <p>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. </p> http://mathoverflow.net/questions/6891/gerbes-for-a-cyclic-group-or-maybe-g-m-too/6910#6910 Answer by Anatoly Preygel for Gerbes for a cyclic group. (or maybe G_m too) Anatoly Preygel 2009-11-26T19:44:34Z 2009-11-26T19:44:34Z <p>A bit of a response to your "Commentary": </p> <p>As you point out, the failure of your construction to hit all <code>$\mu_n$</code>-gerbes is governed by the exact sequence <code>$H^1(X, \mathbb{G}_m) \to H^2(X, \mu_n) \to H^2(X,\mathbb{G}_m)[n] \to 0$</code> so answering your question is related to producing torsion <code>$\mathbb{G}_m$</code>-torsors.</p> <p>The question of doing so has been studied as part of the theory of Brauer groups:</p> <p>Let $Br(X)$ ("Brauer group") denote the group of <em>Azumaya algebras</em>, which are a generalization of the central simple algebras over a field (that is the classical Brauer group).</p> <p>Let $Br'(X)$ ("Cohomological Brauer group") denote the <em>torsion part</em> of <code>$H^2(X, \mathbb{G}_m)$</code>.</p> <p>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, <code>$H^1(X, \mathbb{G}_m)=0$</code> 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 <code>$Br(X) \to Br'(X)$</code> 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".)</p> <p><hr /></p> <p>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, <code>$Br(Spec \mathbb{R}) = \mathbb{Z}/2\mathbb{Z}$</code> 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: </p> <p>Start with the smooth plane conic <code>$C = Proj \mathbb{R}[x,y,z]/(x^2+y^2+z^2)$</code>. It's a smooth genus $0$ curve, but has no $\mathbb{R}$-points and so is not isomorphic to <code>$\mathbb{P}^1_{\mathbb{R}}$</code>. However, after base-change to $\mathbb{C}$ it attains a point and so becomes isomorphic to <code>$\mathbb{P}^1_{\mathbb{C}}$</code>; such a Galois-twisted form of <code>$\mathbb{P}^n_{\mathbb{C}}$</code> 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 <code>$Aut(\mathbb{P}^n) = PGL_{n+1}$</code>, these correspond to <code>$PGL_{n+1}$</code>-torsors and the relation to <code>$\mathbb{G}_m$</code>-torsors is via the exact sequence for <code>$PGL_{n+1}=GL_{n+1}/\mathbb{G}_m$</code>. So, a $T$-point of the corresponding <code>$\mathbb{G}_m$</code>-torsor for a $\mathbb{R}$-scheme $T$ consists of the following data:</p> <p>It is a rank $2$ vector bundle $V$ over $T$, together with an isomorphism of $T$-schemes <code>$C_T \simeq \mathbb{P}(V)$</code> where <code>$C_T = C \times_{Spec \mathbb{R}} T$</code> 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 $V$. </p> <p>Why is this a <code>$\mathbb{G}_m$</code>-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.</p>