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Given a variety V and a locally free (coherent) sheaf $\mathcal{F}$ of rank 1 (equivalently a line bundle $L$), I can do a Cech cohomology on it. Then $H^0(V; \mathcal{F})$ are just global sections. Is there a similarly understandable meaning to elements of $H^1(V; \mathcal{F})$?

Thanks!

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Thanks to everyone! To summarize, a 1-cocycles are interesting because they can be used to define new sheaves which are locally isomorphic to $\mathcal{F}$; moreover any two 1-cocycles differing by a coboundary of a 0-chain are isomorphic. So naturally, we are led to consider 1-cocycles modulo coboundaries. One thing that still hounds me is the following. For curves, knowing $H^0(V; \mathcal{F})$ for a tangent bundle, for example, reveals the number of holomorphic vector fields (or differentials for cotangent bundle). Is there a special meaning for $H^1$ of those two bundles on curves? –  Paul Yuryev Feb 2 '10 at 1:13
    
See also this question: mathoverflow.net/questions/38966/… –  Peter Arndt Nov 20 '11 at 22:39
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2 Answers

up vote 14 down vote accepted

$H^1(V;\mathcal{F})$ is the space of bundles of affine spaces modeled on $\mathcal{F}$. An affine bundle $F$ modeled on $\mathcal{F}$ is a sheaf of sets that $\mathcal{F}$ acts freely on as a sheaf of abelian groups (i.e., there is a map of sheaves $F\times \mathcal{F}\to F$ which satisfies the usual associativity), and on a small enough neighborhood of any point, this action is regular (i.e., the action map on some point gives a bijection). You should think of this as a sheaf where you can take differences of sections and get a section of $\mathcal{F}$.

This matches up with what Anweshi said as follows: given such a thing, you can try to construct an isomorphism to $\mathcal{F}$. This means picking an open cover, and picking a section over each open subset and declaring that to be 0. The Cech 1-cochain you get is the difference between these two sections on any overlap, and if an isomorphism exists, the difference between the actual zero section and the candidate ones you picked is the Cech 0-chain whose boundary your 1-cochain is.

Another way of saying this is that a Cech 1-cycle is exactly the same sort of data as transition functions valued in your sheaf, so if you have anything that your sheaf acts on (again, as an abelian group), then you can use these transition functions to build a new sheaf; a homology between to 1-cycles (i.e. a 0-cycle whose boundary is their difference) is exactly the same thing as an isomorphism between two of these.

I'll note that there's nothing special about line bundles; this works for any sheaf of groups (even nonabelian ones). For example, if you take the sheaf of locally constant functions in a group, you will classify local systems for that group. If you take continuous functions into a group, you will get principal bundles for that group. If you take the sheaf $\mathrm{Aut}(\mathcal{O}_V^{\oplus n})$, you'll get rank $n$ locally free sheaves. A particularly famous instance of this is that line bundles are classified by $H^1(V;\mathcal{O}_V^*)$.

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Maybe you should also add a comment or two about the case $\mathcal{F} = \mathcal{O}^\ast$ ? –  Kevin H. Lin Jan 29 '10 at 22:59
    
A very picky point: The phrase "the action map on any point gives a bijection" is a little dangerous because under the usual interpretation of this phrase, it would be vacuously true if the bundle is empty at that point. (I.e. if the bundle is empty on the connected component containing the point.) But allowing that would give the wrong thing. So you should add that the fibers are nonempty. –  JBorger Jan 30 '10 at 8:55
    
Huh? If the fiber is empty it won't be in bijection with the group. –  Ben Webster Jan 30 '10 at 18:39
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I like the phrase "locally free heaves". Also, you mean $H^1(V;\mathcal O_V^*)$ for holomorphic line bundles. Blame Jim for inspiring the nitpicking ;) –  Kevin McGerty Jan 31 '10 at 18:48
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More generally, is there a similar interpretation for $H^2$, $H^3$, etc? I guess the relevant keyword will be "gerbe"... –  Kevin H. Lin Feb 1 '10 at 2:38
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$H^1$ is the first derived functor of the functor $H^0$ of global sections.

In the Cech cohomology construction, note that we look whether the local sections glue together to form global sections. On an affine space, this is indeed true. But, not so in general. We would like to address the obstruction using algebraic means. When things glue well, we have an exact sequence. In homological algebra, the question of exact sequences breaking down under a functor is addressed by the machinery of "derived functor".

So $H^1$ and the higher cohomology groups are in a sense the obstruction to local sections patching up to form global sections. Since $H^0$ behaves well on affine spaces, it in a sense measures the failure of affineness also. It captures geometry by seeing how the affine pieces glue together to form a projective variety, for instance. The dimensions in which geometry is interesting can be seen by at which dimension the derived functors are nontrivial.

This is my personal point of view to see how geometry is involved, based on derived functors.

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This response is completely geometric, based on Cech cohomology and gluing. This response has nothing at all to do with derived functors, besides one sentence on exact sequences which is vague and not relevant to the rest of the response. –  Marty Jan 29 '10 at 21:44
    
"sections not glueing properly" ==> Cech short exactness breaks down ==> enter derived functors ....... Anyway I have edited my post to reflect what you said. –  Anweshi Jan 29 '10 at 21:59
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Not really -- the "exactness breaking down" occurs when there's a short exact sequence of sheaves (not just one sheaf). Then, taking global sections of each term makes a left-exact sequence of abelian groups. This is where derived functors come into play, and a long exact sequence results. The comparison theorem between Cech cohomology and derived functor cohomology requires a bit of work -- it's a standard Grothendieck composite functor spectral sequence, I recall, with a bit of flabbiness. –  Marty Jan 29 '10 at 22:33
    
Is it so hard? It is there in Hartshorne, I think. –  Anweshi Jan 29 '10 at 22:47
    
@Marty. Yes, to introduce derived functors one must consider a long exact sequence of sheaves. I didn't mention this and was vague. Sorry. –  Anweshi Jan 29 '10 at 22:52
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