Compute genus via Riemann-Hurwitz and monodromy

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In summary, the conversation discussed a problem in algebraic geometry involving an algebraic curve and its genus. The Riemann-Hurwitz formula was used to show that the genus of the curve is one. The conversation also discussed the attempt at solving the problem and a different method was suggested using the formula p=\sum_{i=1}^M \frac{r_i-1}{2}-n+1 with n as the degree of the function and r_i as the ramifications over each critical point including infinity. This method resulted in a genus of one.
  • #1
jackmell
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Hi guys,

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

Homework Statement


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.



Homework 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?



The Attempt at a Solution


Homework Statement



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 believe 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
 
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  • #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]
 
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1. What is the Riemann-Hurwitz formula?

The Riemann-Hurwitz formula is a mathematical formula used to compute the genus (or number of holes) of a complex algebraic curve. It relates the genus of a curve to the degrees of its branch points and the monodromy group of its cover.

2. How is the Riemann-Hurwitz formula used to compute genus?

The Riemann-Hurwitz formula states that the genus of a curve is equal to (d-1)(g-1) + sum of (m_i - 1), where d is the degree of the cover, g is the genus of the curve, and m_i is the order of the monodromy group at each branch point. By determining the degrees and orders of the cover and monodromy group, the genus can be computed using this formula.

3. What is monodromy in relation to the Riemann-Hurwitz formula?

Monodromy refers to the behavior of a curve when it is analytically continued around a branch point. In the Riemann-Hurwitz formula, the orders of the monodromy group at each branch point are used to compute the genus of the curve.

4. Can the Riemann-Hurwitz formula be used for any type of curve?

Yes, the Riemann-Hurwitz formula can be applied to any type of complex algebraic curve, as long as it has a well-defined cover and a monodromy group with finite order at each branch point.

5. Are there any limitations to using the Riemann-Hurwitz formula to compute genus?

While the Riemann-Hurwitz formula is a powerful tool in computing genus, it may not always be applicable in certain cases. For example, it cannot be used for curves with infinite branch points or when the monodromy group has infinite order at a branch point.

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