Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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?
    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


    User Avatar
    Science Advisor
    Homework Helper

    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]
    Last edited: Jul 28, 2008
  6. Jul 28, 2008 #5
    Surprising because [tex]e^{1/sin(z)}[/tex] has essential singularities!
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook