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Compute genus via Riemann-Hurwitz and monodromy

  1. Jan 5, 2012 #1
    Hi guys,

    I new to Algebraic Geometry and was wondering if someone could help me with this problem:

    1. The problem statement, all variables and given/known data
    Given the algebraic curve [itex]w(z)[/itex] represented by [itex]w^5+w^2+z^2=0[/itex], show that the genus is one by employing the Riemann-Hurwitz formula.



    2. Relevant equations
    For the mapping [itex]f:X\to S[/itex], the Riemann-Hurwitz formula is given by:
    [tex]2g(X)-2=\text{deg}(f)\left[2g(S)-2\right)+\sum (e_w-1)[/tex]

    which I assume the expression [itex]f:X\to S[/itex] as applied to my problem can be written as [itex]w(z):\mathbb{C}\to S[/itex], that is, the function w(z) maps the (five-sheeted) complex plane to the Riemann surface given by [itex]S[/itex] and it is this surface that is topologically equivalent to a Riemann sphere with one handle thus having genus one. If that's not the correct way to state that, could someone correct that for me please?



    3. The attempt at a solution
    1. The problem statement, all variables and given/known data

    I believe I understand somewhat well, direct construction of the Riemann surface given by the expression [itex]w^2-p(z)=0[/itex] and the subsequent determination of its genus directly from the union of two Riemann spheres. However I'm not able to deduce that via Riemann-Hurwitz.

    But back to the problem above:

    I'm taking [itex]g(X)[/itex] to be the genus of the complex plane which is zero.

    The degree of [itex]f[/itex] I'm taking to be the highest degree of [itex]f(z,w)[/itex] in [itex]w[/itex] which is five.

    The sum is where I'm having the problem. This I belive can be computed from the monodromy group which I assume represents the ramifications around each branch-point which I take to mean the branching around each branch-point including infinity. Here's my analysis of that:

    The function has six finite branch points and the surface [itex]S[/itex] ramifies around each one as a two-valued branch and three single-valued branches. I think each monodromy group about these branch-points can be written in terms of [itex]\left\{(n,m)\right\}[/itex] with [itex]n[/itex] being the sheet number and [itex]m[/itex], the branch order as:

    [tex]((1,2),(3,1),(4,1),(5,1))[/tex]

    Maybe though that's not a valid way to write that. However, around the branch-point at infinity, the function ramifies into a single five-valued branch so the monodromy group is [itex]((1,5))[/itex]

    So my (incorrect) analysis of the sum would be [itex]\sum (e_w-1)=(2-1)+(1-1)+(1-1)+(1-1)[/itex] for each finite branch-point and [itex]\sum (e_w-1)=5-1[/itex] for the point at infinity. Plugging all this into the Riemann-Hurwitz formula I obtain:

    [tex]-2=5\left[2g-2\right]+10[/tex]

    and solving for g I obtain [itex]g=-1/5[/itex] which is obviously not correct.

    I don't know, maybe I'm not even qualified to even ask the question. Still I would like to see how it's solved at the very least if someone could help me with that and as I study more the subject, I'm sure it will come together eventually for me.
    Thanks,
    Jack
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
    Last edited: Jan 5, 2012
  2. jcsd
  3. Apr 12, 2012 #2
    Let [itex]f(z,w)=z^2+w^2+w^5[/itex]

    then it's much easier to compute it via the formula:

    [tex]p=\sum_{i=1}^M \frac{r_i-1}{2}-n+1[/tex]

    with [itex]n[/itex] the degree of the function and [itex]r_i[/itex] the ramifications over each critical point including infinity.

    The finite critical points are zeros to [itex]R(f,f_w)=0[/itex] where R is the resultant since the function has no poles. There are seven critical points, one of which is the origin. At the origin, the function splits into five single-cycles. At the other six, it ramifies into a 2-cycle and three single cycles. At infinity, it fully ramifies into a 5-cycle. So then we have:

    [tex]p=\sum_{i=1}^8 \frac{r_i-1}{2}-n+1=1/2(1+1+1+1+1+1+4)-5+1=1[/tex]
     
    Last edited: Apr 12, 2012
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