What do gerbes and complex powers of line bundles have to do with each other? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T21:20:07Z http://mathoverflow.net/feeds/question/1721 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/1721/what-do-gerbes-and-complex-powers-of-line-bundles-have-to-do-with-each-other What do gerbes and complex powers of line bundles have to do with each other? Ben Webster 2009-10-21T21:03:18Z 2010-07-20T16:57:06Z <p>We all know how to take integer tensor powers of line bundles. I claim that one should be able to also take fractional or even complex powers of line bundles. These might not be line bundles, but they have some geometric life. They have Chern classes, and one can twist differential operators by them. How should I think about these? What do they have to do with gerbes? </p> http://mathoverflow.net/questions/1721/what-do-gerbes-and-complex-powers-of-line-bundles-have-to-do-with-each-other/1728#1728 Answer by David Speyer for What do gerbes and complex powers of line bundles have to do with each other? David Speyer 2009-10-21T21:29:45Z 2009-10-21T21:29:45Z <p>Is this anything more than thinking about the Chern class of the line bundle in H^1(\Omega^1), which can of course be multiplied by any complex number? For example, if you have an elliptic curve E, a complex number alpha and two line bundles L and L' of the same degree, is there any difference between alpha x L and alpha x L'?</p> http://mathoverflow.net/questions/1721/what-do-gerbes-and-complex-powers-of-line-bundles-have-to-do-with-each-other/1735#1735 Answer by Johan for What do gerbes and complex powers of line bundles have to do with each other? Johan 2009-10-21T21:37:08Z 2009-10-21T21:37:08Z <p>Cadman defined a root stack for line bundles on a scheme, and variations on that theme such as Cartier divisors, in Cadman, Charles, Using stacks to impose tangency conditions on curves. Amer. J. Math. 129 (2007), no. 2, 405--427.<br /> Maybe this construction was known as folklore beforehand, in any case for a line bundle it gives you a flat gerbe, and might do the job for you for rational powers.</p> http://mathoverflow.net/questions/1721/what-do-gerbes-and-complex-powers-of-line-bundles-have-to-do-with-each-other/1737#1737 Answer by David Ben-Zvi for What do gerbes and complex powers of line bundles have to do with each other? David Ben-Zvi 2009-10-21T21:43:04Z 2010-07-13T17:31:38Z <p>Complex powers of line bundles are classes in $H^{1,1}$, or equivalently sheaves of twisted differential operators (TDO) (let's work in the complex topology). This maps to $H^2$ with $\mathbb{C}$ coefficients, or modding out by $\mathbb{Z}$-cohomology, to $H^2$ with $\mathbb{C}^\times$ coefficients. The latter classifies $\mathbb{C}^\times$ gerbes, ie gerbes with a flat connection (usual gerbes can be described by $H^2(X,\mathcal{O}^\times)$). Note that honest line bundles give the trivial gerbe.</p> <p>In fact the category of modules over a TDO only depends on the TDO up to tensoring with line bundles --- ie it only depends on the underlying gerbe, and can be described as ordinary $\mathcal{D}$-modules on the gerbe. Or if you prefer, regular holonomic modules over a TDO are the same as perverse sheaves on the underlying gerbe. This is explained eg in the encyclopedic Chapter 7 of Beilinson-Drinfeld's Quantization of Hitchin Hamiltonians document, or I think also in a paper of Kashiwara eg in the 3-volume Asterisque on singularities and rep theory (and maybe even his recent $\mathcal{D}$-modules book). B&amp;D talk in terms of crystalline $\mathcal{O}^\times$ gerbes rather than $\mathbb{C}^\times$ gerbes but the story is the same.</p> http://mathoverflow.net/questions/1721/what-do-gerbes-and-complex-powers-of-line-bundles-have-to-do-with-each-other/31726#31726 Answer by Michael Thaddeus for What do gerbes and complex powers of line bundles have to do with each other? Michael Thaddeus 2010-07-13T16:08:55Z 2010-07-20T16:57:06Z <p>To any homomorphism of Lie groups $\phi: G \to H$, and any principal $H$-bundle $P$ over a space $X$, you can associate what I like to call the "gerbe of liftings," namely the stack over $X$ whose objects over $f: Y \to X$ consist of a principal $G$-bundle $Q$ together with an isomorphism $Q_\phi \cong f^* P$. Here $Q_\phi$ denotes the associated $H$-bundle. This stack is nothing but $BG \times_{BH} X$. (Note that it carries a tautological $G$-bundle, pulled back from $BG$.) If $\phi$ is surjective with kernel $K$, this is a $K$-gerbe with trivial band. Morally speaking, its isomorphism class comes from the image in the connecting homomorphism $H^1(X,H) \to H^2(X,K)$, and this is literally true if $K$ is abelian. </p> <p>To relate this to the question, consider the case where $G = H = {\bf C}^\times$ and $\phi$ takes the $k$th power. Then $K = \mu_k$, a cyclic group of order $k$. The gerbe of liftings measures the failure of a given line bundle on $X$ to be a $k$th power. (For example, if it is not, then there will be no objects over $f =$ id.) It coincides with the root stack introduced by Cadman.</p> <p>It is tempting to regard the $k$th root gerbe $B$ of a line bundle $L$ as $L^{1/k}$. But it is a gerbe with structure group $\mu_k$, not a bundle with structure group ${\bf C}^\times$. Topologists should think of it as like an orbifold over $X$, but with orbifold structure spread out all over. How, then, is $B$ related to a $k$th root of $L$? As noted above, $B$ carries a tautological line bundle --- and its $k$th power is the pullback of $L$ from $X$! </p> <p>Moral: fractional line bundles on $X$ can be realized as bona fide line bundles on $\mu_k$-gerbes over $X$.</p> http://mathoverflow.net/questions/1721/what-do-gerbes-and-complex-powers-of-line-bundles-have-to-do-with-each-other/31734#31734 Answer by Ryan Reich for What do gerbes and complex powers of line bundles have to do with each other? Ryan Reich 2010-07-13T17:21:36Z 2010-07-13T17:21:36Z <p>If L is any line bundle on a space (scheme, whatever) X, A is any (additive) abelian group, and a an element of A, there is a natural construction of an A-gerbe $L^a$ as follows. By definition, $L^a$ should be a "sheaf of categories", or stack (not algebraic) on X, and here are its categories of sections. Identify L with its total space, which is a $\mathbb{G}_m$-bundle on X, and for any open set U in X, let $L^a(U)$ be the category of all A-torsors on $L|_U$ whose monodromy about each fiber of $L|_U \to U$ is a.</p> <p>One can check that this really is a gerbe: it is locally nonempty, since if L is trivial over U, you can write $L|_U = \mathbb{G}_m \times U$ and then pull back the unique A-torsor on $\mathbb{G}_m$ with monodromy a. It has a natural action of A-torsors on X, given by pulling up along the bundle map $L \to X$ and tensoring. And this action is free and transitive, since the difference of two a-monodromic torsors on $L|_U$ has trivial monodromy on each fiber and therefore descends to X.</p> <p>Why do I call this $L^a$? Suppose that $L = \mathcal{O}_X(D)$ for a divisor D, where for simplicity let's say that D is irreducible of degree n; then L gets a natural trivialization on $U = X \setminus D$ having a pole of order n along D. As shown above, this induces a trivialization $\phi$ of $L^a$ on U, and if we pick a small open set V intersecting D and such that D is actually defined by an equation f of degree n, then we get a second (noncanonical) trivialization $\psi$ of $L^a$ on V. You can check that the difference $\psi^{-1} \phi$, which is an automorphism of the trivial gerbe on $U \cap V$, is in fact described by the A-torsor $\mathcal{T} = f^{-1}(\mathcal{L}_a)$, where $f \colon U \cap V \to \mathbb{G}_m$ and $\mathcal{L}_a$ is the A-torsor of monodromy a. Since f has degree n, $\mathcal{T}$ has monodromy na about D. Thus, it is only reasonable to say that the natural trivialization $\phi$ has a pole of order na, which is consistent with the behavior of the trivialization of L itself on U, when raising to integer powers.</p> <p>What does this have to do with twisting of differential operators? Suppose we have some kind of sheaves (D-modules, locally constant sheaves, perverse sheaves; technically, they should form a stack admitting an action of A-torsors). On the one hand, one could mimic the above construction of $L^a$ to describe a-monodromic sheaves on L, and this is what is often called twisting. On the other hand, there is a natural way to directly twist sheaves by the gerbe $L^a$ without mentioning L at all (that is, you can twist by any A-gerbe). The procedure is as follows: a twisted sheaf is the assignment, to every open set U in X, of a collection of sheaves on U parameterized by the sections of $L^a(U)$, and compatible with tensoring by A-torsors. Of course, since if $L^a(U)$ is nonempty this is the same as giving just one sheaf, this is sort of overkill, but the choice of just one such sheaf is noncanonical whereas this description is canonical. These collections should be compatible with the restriction functors $L^a(U) \to L^a(V)$ when $V \subset U$. It is an exercise to reader to check that this is the same as the other definition of twisting :)</p> <p>Man, you asked the right question at the right time. My thesis is all about this stuff.</p>