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Trivial holomorphic first sheaf cohomology

  1. Jul 27, 2008 #1
    What's an example of a set with trivial holomorphic first sheaf cohomology? I was thinking of subsets of C, and trying to think what would satisfy this.
    For example, suppose you covered C by [tex]U_1=re^{i\theta}:r < 2[/tex] and [tex]U_2=re^{i\theta}:r> 1/2.[/tex] Then if [tex]f(z)=1/z[/tex], [tex]\oint_{|r|=1} f(z)\neq 0,[/tex] and he says the contour integral of a function in [tex]U_1\cap U_2[/tex] should be 0. So I don't think subdividing C this way is a good limiting open cover.
    But what would be a good limiting open cover, other than just C by itself? One should be able to refine an arbitrary open cover of C into one which has trivial holomorphic first sheaf cohomology, since C by itself does.
    Supposing you covered C by [tex]U_1=z: Re(z) > \pi/2[/tex] and [tex]U_2=z: Re(z) < \pi.[/tex] I don't see how that open cover would have trivial first sheaf cohomology either since [tex]\displaystyle e^{1/ sin(z)}[/tex] would be analytic in [tex]U_1\cap U_2[/tex] and I don't think [tex]e^{1/ sin(z)}[/tex] could be expressed as the sum of a function that's analytic in [tex]U_1[/tex] and a function that's analytic in [tex]U_2.[/tex] So would such an open cover need more refinement? Into what?
    What's a good book on it?
    Laura
     
    Last edited: Jul 27, 2008
  2. jcsd
  3. Jul 27, 2008 #2
    I'm not sure I follow precisely what is going on (cohomology of which sheaf in particular?), but I'd suggest that simply connectedness is something to think about - if you have C\{0} then there's that standard logarithm thing isn't there.
     
  4. Jul 27, 2008 #3

    mathwonk

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    I guess you mean H^1(O) = {0}. This is true of all connected open sets in C, but not in C^2, as I recall.

    the point is whether there exists a function holomorphic in the set which does not extend to a larger connected open set. this is true of all regions in C.

    another related problem is whether there is a divisor in that set which is not the zero set of a meromorphic function, but mittag leffler says there is not.
     
  5. Jul 28, 2008 #4
    I guess the Mittag-Leffler theorem does say you could express [tex]e^{1/sin(z)}[/tex] as the sum of two functions, one analytic in one half plane, one analytic in the other. It's pretty surprising.
    I guess what Penrose says about trivial cohomology meaning that the contour integral is 0 for a function that's a coboundary, doesn't apply to the complex plane.
    The idea of the sheaf cohomology group is that you have an open cover [tex]U_\alpha[/tex] of the manifold, and holomorphic functions [tex]f_{\alpha \beta}[/tex] defined on each intersection [tex]U_\alpha\cap U_\beta,[/tex], such that on triple intersections [tex]f_{\alpha \beta}+f_{\beta \gamma}+f_{\gamma \alpha}=0[/tex] (i.e. the [tex]f[/tex]'s are a cocycle), and a set of [tex]f[/tex]'s is considered equivalent to 0 iff [tex]f_{\alpha\beta}=g_\alpha-g_\beta[/tex], with [tex]g_\alpha[/tex] analytic on [tex]U_\alpha[/tex], [tex]g_\beta[/tex] analytic on [tex]U_\beta.[/tex]
    Laura
     
    Last edited: Jul 28, 2008
  6. Jul 28, 2008 #5
    Surprising because [tex]e^{1/sin(z)}[/tex] has essential singularities!
    [tex]Laura[/tex]
     
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