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Genus of aribtrary algebraic function

  1. Oct 13, 2012 #1

    You guys know of any (software) function which accepts as input an aribtrary algebraic function, then computes the genus as output?

    Here's an aribtrary one:

    [tex](-9+8 z^2-6 z^4)\text{}+(6 z-6 z^4)w+(-6+2 z^3)w^2+(1-7 z+5 z^5)w^3+(-7 z-8 z^2-3 z^3-7 z^5)w^4+(9-8 z-z^2-9 z^3+5 z^4)w^5=0[/tex]

    What's the genus? I don't know.

    I count 43 singular points and I suppose if I had to, I could manually (numerically) compute the ramification around each to determine the genus. Is this the only way to compute the genus for this function?


    I did that numerical computation, and using

    [tex]g=1/2 \sum (r-1)-n+1[/tex]

    arrived at a genus of 16. Anyone feel like checking this for me?

    Last edited: Oct 13, 2012
  2. jcsd
  3. Oct 13, 2012 #2


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    degree 9, means the genus if smooth is let's see, something like (1/2)(8)(7) = 28.

    so singular points reduce that genus. but there can only be at most 28 non separating singular points. thus if there are 43 singular points, it seems there must be hmmmm

    how many components? it seems there would be at least 16 components, which is impossible for a degree 9 plane curve.

    but anyway, i think the macaulay software can probably do this.

  4. Oct 14, 2012 #3
    Ok thanks for the software link mathwonk. I'm affraid I did not follow you above. Are you saying the genus of that function cannot be 16?

    I did find a Mathematica demonstration package on Monodromy groups which can be easily modifed to include functions like these. Seems to compute it the same way I suggested, that is numerically integrating around each singular point. It's stalls however on the function above but with other functions it can analyze, the genus it returns agrees with the genus I calculate with my code. For example, the function:

    [tex]-6-5 z^3-5 z^5+w \left(-1-6 z^2\right)+w^3 \left(3-5 z^3-2 z^4\right)+w^2 \left(-2-8 z-5 z^2-8 z^3+3 z^4-2 z^5\right)+w^4 \left(-9+5 z^5\right)+w^5 \left(9 z+2 z^2-2 z^3+5 z^5\right)=0[/tex]

    both my code and this demonstration return 16 as the genus.


    To be objective though, for the function:

    [tex]w^5 (3-7 z)+z+2 z^4-8 z^5+w^3 \left(3 z-5 z^3\right)+w^4 \left(7 z-z^2+8 z^3+5 z^4\right)+w \left(-2+7 z+8 z^4-z^5\right)+w^2 \left(-8-z+4 z^5\right)==0[/tex]

    The demonstration returns a genus of 31/2 which is obviously wrong but my code returns 15 so as you can see, I really don't have a good handle on this yet.

    I did however e-mail the Maculay2 author about this and asked if his software could compute these values.
    Last edited: Oct 14, 2012
  5. Oct 17, 2012 #4


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    i don't say it cannot be 16, but i do say that seems to contradict your statement that it has 43 singularities.
  6. Oct 18, 2012 #5
    Then I'm probably not understanding the concepts. I have several texts on the matter and they seem to use "singular point'' and "critical point'' interchangably. For example, Kirwan in ``Complex Algebraic Curves'' defines a singular point [itex](a,b)[/itex] as:

    [tex]f(a,b)=\frac{\partial f}{\partial z}(a,b)=\frac{\partial f}{\partial w}(a,b)=0[/tex]

    and when I apply that to the function, it doesn't have any singular points.

    However, Bliss in "Algebraic Functions'', defines a singular point as one in which one or both of the resultant [itex]R(f,f_w)[/itex] and the coefficient of the highest power of w is zero. And when I apply that definition, I get 42 finite points plus the point at infinity where the function also ramifies. Other texts define these points as critical points.

    Should I be using "critical points" for the points according to Bliss and only use singular point as in Kirwan?

    Here's an example from Walker:
    [tex]2z^4-3 z^2 w+w^2-2 w^3+w^4[/tex]

    That has two singular points (0,0), (0,1), using the definition above by Kirwan and Walker, but seven critical points according to the definition by Bliss.

    I'm having problems with all of this.
    Last edited: Oct 18, 2012
  7. Oct 18, 2012 #6


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    why do you say "the" point at infinity? isn't the lead coefficient of w^5 equal to zero at all 4 zeroes of the coefficient polynomial in z?

    bliss seems to defining ramification points of the projection map onto the z axis.

    rather than singular points. singular points are points where the curve does not look like a smooth manifold. as in kirwan and walker.

    to compiute the genus of a smooth curve via projection, one does need the number of ramifiication points, plus the riemann hurwitz formula.

    have you read my notes on the genus computation?


    walker also discusses the genus in terms of singular points in thm. III.7.5, page 85.

    you might use his formula to compute the genus of that degree 4 curve of his (with a tacnode and another singular point) that you cited, from his page 58.
    Last edited: Oct 18, 2012
  8. Oct 18, 2012 #7


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    first off, to understand this topic you must understand the implicit function theorem IFT.

    given a curve f(x,y) = 0, a point is on the curve if f = 0.

    Then the IFT says that the curve looks like a graph of a function y(x) of x near a point p on the curve, provided ∂f/∂y ≠ 0 at p.

    Recall that the graph of a smooth function y(x) looks like a horizontal smooth curve.

    the resultant R(f,∂f/dy) vanishes at points where both f and ∂f/∂y vanish, i.e. points of the curve f=0 where the IFT does not hold. Thus they are points where either the curve f=0 is not smooth (singular points), or points where the curve f=0 is instantaneously vertical (vertical tangent line).

    Note also that a point where ∂f/∂x ≠ 0 would satisfy IFT in the sense that the curve would look like a graph of a function x(y), hence would look like a smooth curve which is instantaneously vertical.

    thus a singular point, where both ∂f/∂x and ∂f/∂y = 0, is one where the curve does not look like a graph in any direction. thus at such a point the curve has a kink or crosses itself.

    i'm tired now. please read some of the many detailed explanations i have written.
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