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Genus, differential forms, and algebraic geometry

  1. May 11, 2006 #1

    Hurkyl

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    I decided earlier this week that I was going to compute by hand the genus of an elliptic curve. I've had a miserable (but enlightening!) time!

    I eventually stumbled upon the trick in Shafaravich: I should be looking at the rational differential forms, and counting zeroes & poles of things.


    But I still feel like there should be a way to do it without resorting to the holomorphic stuff. (Though, I suppose I don't have enough intuition for algebraic geometry to have any right to think so. :biggrin:)


    This is what I had been doing:

    I decided to consider the complex projective elliptic curve E : y² z = x³ - x z².

    Let U be the open affine subset consisting of all of the points of the form (u : v : 1). This is the affine curve defined by v² = u³ - u.
    Let V be the open affine subset consisting of all of the points of the form (s : 1 : t). This is the affine curve defined by t = s³ - s t².

    Then {U, V} is an open cover of E. Let W be their intersection. On W, the change of variable relations are ut = s and vt = 1.

    (Does omega -- Ω -- show up right?)

    Ω, the differential forms on U, is simply the C-module generated by {du, dv}, satisfying the relation 2v dv = (3u² - 1) du.

    Ω[V] is the C[V]-module generated by {ds, dt}, satisfying the relation (1 + 2st) dt = (3s² - t²) ds

    And we have maps from both of these into the C[W]-module Ω[W]. The collection of all of the relevant relations is:

    v² = u³ - u
    t = s³ - s t²
    ut = s
    vt = 1
    2v dv = (3u² - 1) du
    (1 + 2st) dt = (3s² - t²) ds
    u dt + t du = ds
    v dt + t dv = 0


    and I was trying to find the intersection of the images of the two maps. (As C-vector spaces, I suppose) But I just couldn't figure out how to do it. After much work, I eventually stumbled across the global differential form... but I have absolutely no idea how I would go about proving that was the only one (up to a constant).

    So, I suppose my problem is that I just don't know how I would compute the intersection of the images of these two maps -- is this a tractable problem at all, in this case or in general?
     
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  3. May 11, 2006 #2

    mathwonk

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    what is your definition of the genus? if it is the topological one, then the compoutation is trivial, since genus is an integr it is unchanged under continuous defirmation, so deform your cubic to three lines in a triangle, then notice there is one hole in a triangle! so the genus is one.

    another approach is to project onto the x axis, sending (x,y) to x. and count branch points. i.e. you get a 2 to 1 cover which has only one preimage over certain points, in your example, over 0,1,-1, and infinity. the hurwitz formula says the genus of the double cover must be then one, since the genus of the x axis (a sphere) is zero. i.e. hurwitz just says that if you triangulate the double cover you get 2 edges over every edge and 2 faces over every face, but only one vertex over those 4 vertices, so v'-f'+e' = 2(v-f+e) - 4 = 2(2) - 4 = 0 = 2g-2,, so the genus is 1.

    if your definition is the vector dimension of the space of global holomorphic differentials, then you can easily construct one ala shafarevich, by considering dx/y. to prove there cannot be more than one, note that a holomorphic one form and its complex conjugate, give harmonic forms on the manifold, and if there are mnokre of these than the rank of the first homology group, then one would get a linear combination which is exact, in the sense of havng zero integral over every homology mcycle, hence defines a global harmonic function.

    but such a function is constant by the maximum principle.

    thus the topology always tells you there can never be more than g inmdependent differentials and the real problem is to construct them. constructing them via the rational forms approach in shafarevich is due to riemann, in his original paper.


    if your definition is the dimension of the cohomology group h^1(O), please consult my notes on the RRT on my webpage, p.41, where the formula is proved:

    1-g = 1 - (d-1)(d-2)/2, for a smooth curve of degree d. when d = 3, you get

    1-g = 0. or just notice this number g = (d-1)(d-2)/2 is the number of holes in a polygon with d sides, and again for a triangle we get one.

    or notice that you differential form has no zeroes, hence the "hopf theorem", due again to riemann in this case, says the euler characteristic of your surface is zero, = 2g-2, so g = 1.
     
    Last edited: May 11, 2006
  4. May 11, 2006 #3

    Hurkyl

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    The definition I have at my disposal that I actually understand is the vector space dimension of the global regular differential forms.

    I also have the arithmetic genus defined in Hartshorne, but I don't understand the Hilbert polynomial at all, so that one's right out. :smile:


    I guess I should point out that I'm not specifically interested in complex elliptic curves -- I just figured that amongst all the possible choices of fields and varieties, that would be the simplest nontrivial case to look at.


    After seeing the idea, I think I can reproduce Shafarevich's approach to computing the genus of any hyperelliptic curve.

    I was sort of hoping there was a dumb brute force approach -- even if you would never do it that way in practice, sometimes it's instructive to see! I was trying to cobble something together with Gröbner bases and linear algebra, but I'm still relatively new to actually computing things with Gröbner bases and modules.


    But seeing different approaches is always good. How much of what you said can be formulated in the general case of an algebraic curve? Or an algebraic variety?
     
  5. May 12, 2006 #4

    matt grime

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    All of it. And it works in almost any characteristic. The Riemann-Hurwitz formula breaks down if the ramification indices are divisible by the characteristic of the field.

    I have come to the conclusion that geometry is actually relatively straight forward if we could just get the geometers to use plain English and stop blinding us with fancy terms.
     
  6. May 12, 2006 #5

    mathwonk

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    if you just want to construct a differential, just differentiate the equation,

    i.e. if your equation is y^2 = x^3 + x, then you get 2ydy = (3x^2+1)dx.

    then dx/y = 2dy/(3x^2+1). these two expressions show the differential is well defined. you can check holomorphicity by noting that the map defined by x ramifies exactly where y vanishes, so dx and y have the same zeroes.
     
  7. May 12, 2006 #6

    mathwonk

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    and note this differential has no zeroes or poles. since any two differentials can be divided to yield a function, if there were two of these then their quotient would yield a function with no zeroes or, poles hence constant.

    so the space of them is one dimensional.

    is that explicit enough?
     
    Last edited: May 12, 2006
  8. May 12, 2006 #7

    mathwonk

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    let me try to make this as detailed and explicit as possible.

    1) on a smooth projective curve, every non constant function has both zeroes and poles, and the same number of each, counted properly.

    (for a rational function f, p is a pole of f iff p is a zero of 1/f.)

    2) the quotient of two differentials (the one in the bottom being non zero) is a rational function.

    3) hence any smooth plane curve having a differential with no zeroes and no poles, has genus one, where genus is defiend as the vector dimension of the space of differentials with no poles.

    proof: if Z is a differential with neither zeroes nor poles, and W is any other differential with no poles, then W/Z is a rational function with no poles, hence no zeroes, hence constant. so W is a constant mkultiple of Z. thus Z is a basis of the space of differentials with no poles. mhence the genus is one.

    corollary: the plane cubic with equation y^2 z = x^3 - x z^2, has genus one.
    proof: we need only exhibit a differential with neither zeroes nor poles.
    we go affine by setting z = 1, getting equation y^2 = x^3 - x. now we take d of both sides, getting, 2y dy = [3x^2 - 1]dx.
    now we divide by, well you know, getting dx/y = 2dy/[3x^2-1]. we claim this is our desired differential.

    "clearly" the expression on the left shows there are poles at most where y = 0. but at such points the function x^3 - x, has a simple zero, hence its derivative 3x^2 - 1 has no zero. so the expression on the right shows there is no pole at these points either.

    so this is a differential with no poles. as to zeroes, the only zeroes of the expression on the elft must be where dy = 0, but the onlyt zeores of the expression on the right are where dx = 0, and since our curve is smooth, dx and dy do not have any common zeroes.

    thus this differential has no zeroes or poles, except possibly at infinity, and I leave it to you to check there.

    If you consult shafarevich, in the chapter III on differential forms, either sections III.5.4-5 in the 1974 edition or sections III.6.4-5 of the 1994 edition, you will find the genus formula I gave above for all hypersurfaces. I.e. he explicitly computes the number of regular differentials on any smooth hypersurface.

    then he does it again on all plane hyperelliptic curves, which includes cubics, but by a method which works on all smooth plane curves.

    hows that? (the first three results I gave as facts are also proved in shafarevich.)
     
    Last edited: May 12, 2006
  9. May 12, 2006 #8

    mathwonk

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    remark: these results are not trivial, as genus >0 implkies non rational, and just proving non rational is quite hard. just try it by hand on your example without the theory of the genus, which came from geometry via riemann.

    the interesting and hard problem then is to decide rationality for example where the genus does not do it for you. e.g. a cubic threefold (hypersurface in P^4), has no global regualr differentials, but can still be shown to be non rational.

    this was a big spectacular result by clemens - griffiths in 1972, also by artin -mumford, and 2 russian mathematicians. one must find another, subtler invariant that vanishes on rational avrieties but not on these objects.

    the one used by clemens griffiths agin came from geometry, the intermediate jacobian, also computable as a cycle class group.

    these methods from geometry decide the purely algebraic question of whetehr every subfield of a purely transcendental field of rational functions, like k(X,Y,Z), must itself be pure transcendental, the "luroth problem".

    mumford also gave another proof (unpublished) using Prym varieties.

    the details were given by Beauville.

    A. Beauville, Les singularites du diviseur
     
    Last edited: May 12, 2006
  10. May 12, 2006 #9

    mathwonk

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    references on cubic threefolds

    mumford also gave another proof (unpublished) using Prym varieties.

    the details were given by Beauville.

    A. Beauville, Les singularites du diviseur theta de la Jacobienne intermediare de l'hypersurface cubique dans P^4, Algebraic threefolds (Proc. Varenna 1981), 190-208: Lecture Notes 947, Springer-Verlag, Berlin- Heidelberg- New York (1982).

    or in English:

    Smith, Varley, A Riemann singularities theorem for Prym theta
    divisors, with applications, Pacific Journal of Math, vol. 201, no.
    2, Dec 2001, 479-509.
     
  11. May 12, 2006 #10

    mathwonk

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    ok here is another calculation: suppose C is a smooth curve with a differential having exactly 2 zeroes and no poles. i claim C has genus 2.
    proof: i guess you have to construct another regular differential, hence also having 2 zeores. then in the 2 dimensional space spanned by these, you can find a non zero one with a zero at an given point. so if you are given another differential with no poles, it must have 2 zeroes, and you can find a linear combination of the first two with a zero at one of those zeroes. then the quotient of your differential by that linear combination has only one zero and one pole, hence defines a map from C to P^1 of degree one or less. but a curve with a non zero differential cannot have a map of degree one to P^1 since that would be an isomorphism, and P^1 has no non zero differentials.

    so our map is constant and hence our given differential must be a linear combination of the first two. so any smooth curve with a differential having 2 zeroes and no poles, has genus 2. (e.g. a plane quartic with one ordinary double point is the image of such a curve under a map identifying two points to the double point.)
     
    Last edited: May 12, 2006
  12. May 12, 2006 #11

    mathwonk

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    i am doing my best here to make this concrete (no cohomology!). is it helping?

    it was riemann by the way who showed how to write down differentials on plane curves using rational functions.
     
    Last edited: May 13, 2006
  13. May 13, 2006 #12

    mathwonk

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    by the way, the idea of the genus works to show non rationality for plane curves of degree 3 or mroe, surfaces in 3 space of degree 4 or more, threefolds of degree 5 or more, etc....

    but it is believed that actually smooth hypersurfaces of degree 4 or mroe are never rational, but no argument is known in high dimensions.
     
  14. May 13, 2006 #13

    mathwonk

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    helooo? helooo?
     
  15. May 13, 2006 #14

    Hurkyl

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    I'm still here. Just haven't worked out what to say yet.
     
  16. May 14, 2006 #15

    mathwonk

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    well the post i stand by is #7. that is what I think may be what you want. and it has nothing fancy in it.

    i.e. to compute dimension of space of differentials as one, we need to produce one and then prove there are no others. the trick of dividing two differentials to get a rational function is the most direct way, and was also classical, going at least back to weyl and maybe riemann.

    i.e. looking at the differewntials as a vector space over the rational functions as well as over the complex numbers is a basic trick. in fact that is the underlying trick in the serre weil proof of "serre duality" (the roch part of riemann roch) via adeles.

    sorry i keep going off on a tangent. but i have spent 30 years contemplating these objects and I like them.
     
  17. May 14, 2006 #16

    Hurkyl

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    Oh bleh, my response got killed. :frown:


    I'm not opposed to seeing fancy techniques -- it's just nice to see the "simple" techniques applied too. It was easy to turn the problem into a "simple" one, but I was mildly irritated I couldn't solve the problem with "simple" techniques!


    Once I saw to use rational differentials, it was easy enough to compute the genus of the elliptic curve -- and would now be even easier using the tricks you've mentioned.


    Post #10 was very useful -- based on the elliptic curve case, I had been wondering if differential forms had the property that {number of zeroes} - {number of poles} = 0.

    Of course, I already knew that wasn't true, but forgot: I had worked out that if x is the coordinate on the line, that dx has a double pole at infinity, and no zeroes.


    And now that I realize that in the three cases: g=0,1,2, that {number of zeroes} - {number of poles} = 2g - 2, the Riemann-Roch theorem (at least a special case of it) finally makes sense! I had never quite fathomed the canonical class before -- it's amazing how quickly a definition can make sense when one starts wondering about the idea it defines!
     
  18. May 14, 2006 #17

    mathwonk

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    i am very glad you asked this question as it motivated me to give a simple proof of the result.

    i have more difficulty showing that a smooth plane quartic has genus 3, but again shafarevich's proof of the adjunction formula for hypersurfaces will do iit. i just do not remember how he proceeded and dont like to peek.

    the key point seems to be that the 4 zeroes of a regular differential are all collinear. but why?

    i.e. I recall his proof hinges on showing all regular differentials are given by multiplying dx/[
     
  19. May 14, 2006 #18

    mathwonk

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    post#10 hides another useful fact: i.e. it is not entirely trivial that a degree one map to a smooth curve is an isomorphism.

    e.g. the degree one map t goes to (t^2,t^3) is not an isomorphism from P^1 to the cuspidal curve x^3=y^2.
     
  20. May 14, 2006 #19

    Hurkyl

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    Oh, something I forgot to repost -- I (think I) understand the topological definition of genus for a dimension 2 manifold, but not in a way that carries over to the case of an algebraic variety!


    What, then, is a "smooth plane quartic"? I know elliptic curves can be written as quartic equations, and they certainly don't have genus 3!
     
    Last edited: May 14, 2006
  21. May 14, 2006 #20

    mathwonk

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    the degree of a (reduced)( plane curve of degree d equals (d-1)(d-2)/2 - a contribution coming from any singular points.

    a smooth plane quartic has genus 3 but there do exist plane quartics of form y^2 = x^4 + 1 say, which have genus 1 because they have a singularity at infinity that subtracts 2 from the genus.

    for more general algebraic varieties i was taking your preferred definition of genus as the number of differential n - forms where n is the dimension of the variety, so for surfaces it would be 2forms and for threefolds it would be 3 forms. etc...


    there are homology groups of course of such varieties, lets see, the first homology group has dimension twice that of the space of 1 forms, but for 3 folds say, the hodge decomposition of the 3 cycles yields forms of type (0,3), (1,2), (2,1) and (3,0) of which only the forms of type (3,0) give the genus.

    for special 3 folds with no (0,3) forms, such as smooth 3 folds in P^4, we have the space of (1,2) forms and the space of (2,1) forms reflect the topology of the 3 fold but there are no holomorphic 3 forms so the "genus" is zero.


    i have some notes on plane curves i will try to put on my webpage, with pictures of curves with singular points, showing how the singularity contributes to the genus.

    i have course notes on plane curves (a la walker), basic algebraic geometry ( al la shafarevich), and geometry of surfaces ( a la beauville).
     
    Last edited: May 14, 2006
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