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

Riemann roch made easy

  1. Aug 14, 2005 #1

    mathwonk

    User Avatar
    Science Advisor
    Homework Helper
    2015 Award

    Summary of the classical proof by Riemann and Roch

    Let D = p1 + ....+pd be a divisor of distinct points on a compact connected Riemann surface X of genus g, and let L(D) be the space of meromorphic functions on X with at worst simple poles contained in the set {p1,...,pd}. For each point pj let µj be a meromorphic differential with exactly one pole, a double pole at pj, and hence residue zero at pj. Let w1,...,wg be basis for the holomorphic differentials on X. Then for every function f in L(D), the differential df belongs to the linear space V(D) of differentials with basis {µ1,...,µd,w1,...wg}. Indeed the subspace of differentials of form df in V(D) consist exactly of those differentials in V(D) with zero period, i.e. zero integral, around all loops {A1,...,Ag,B1,...,Bg} in a standard homology basis for X. Thus the period matrix defines a linear map
    V(D)--->C^2g whose kernel is isomorphic to L(D)/C. Since dimV(D) = d+g, the fundamental theorem of linear algebra implies
    d+g -(2g) = d-g <= dimL(D)/C <= d+g.

    Since the period map is injective on holomorphic differentials, in fact the kernel L(D)/C does not meet the g dimensional subspace of V(D) spanned by w1,...,wg. Hence we get a better upper bound, dimL(D)/C <= d,
    i.e. d-g <= dimL(D)-1 <= d. This is Riemann's part of the theorem.

    Roch then analyzed the period matrix defining the map V(D)--->C^(2g) to compute its cokernel. First of all he normalized the meromorphic differentials µ1,...,µd to have all A-periods equal to zero by subtracting suitable linear combinations of the wj, and defined W(D) to be the d dimensional space of meromorphic differentials with basis {µ1,...µd}. Then differentiation maps L(D) into W(D) and the image, isomorphic to L(D)/C, equals those differentials in W(D) whose B periods are also zero. Thus the B-period map S(D):W(D)--->C^g, again has kernel isomorphic to L(D)/C. Since dimW(D) = d, again we get d-g <= dimL(D)-1 <= d.

    Next Roch computes explicitly the rank of the B-period matrix S(D). For this he normalized also each holomorphic differentials wj to have all A-periods zero except over Aj where the period is 1. Then he observed that by residue calculus Riemann's bilinear relations as above show that the integral of µk over Bj equals -2<pi>i wj(pk). Hence Riemann's matrix S(D) =
    [ integral of µk over Bj ] of periods of meromorphic differentials, is proportional to Roch's matrix T(D) = [wj(pk)] of values of holomorphic differentials, (which has come to be called the "Brill Noether" matrix, apparently just because they displayed it in a larger format).

    It is then elementary that the rank of Roch's matrix equals g - dim(K(-D)), where K(-D) is the space of holomorphic differentials vanishing at every point p1,...,pd of D. I.e. then the kernel L(D)/C has dimension d - (g- dim(K(-D))) = d-g + dim(K(-D)), so we get dim(L(D)/C) = d-g + dim(K(-D)), i.e. dimL(D) = d+1-g + dim(K(-D)). This is the full classical Riemann Roch theorem.

    Having finally "understood" this after many years (I finjally read it in Riemann and Roch's papers), I wanted to share it somewhere. The existence of the differentials used above was the gap in Riemann's argument, and it is still tedious and difficult to prove this thoroughly for compact complex Riemann surfaces, but easier for plane curves.
     
    Last edited: Aug 14, 2005
  2. jcsd
  3. Aug 15, 2005 #2
    re

    what is genus?
     
  4. Aug 15, 2005 #3

    mathwonk

    User Avatar
    Science Advisor
    Homework Helper
    2015 Award

    all surfaces are spheres with handles attached, and the genus is the numebr of handles. there are 2g essentially diffwerent loops on a surface of genus g.

    i.e. to check whether a differential has a single valued iuntegral you have to check that the integral around all 2g of those loops is zero.
     
  5. Aug 16, 2005 #4
    re

    There is alot of interesting terminology which I don't really understand, could you recommend a book that will cover some background. Thank you.
     
  6. Aug 16, 2005 #5

    mathwonk

    User Avatar
    Science Advisor
    Homework Helper
    2015 Award

    well it assumes basic path integration for complex variables, so any complex book. say cartan, elementary theory of functions of one and several complex variables.
    i.e. riemann roch comes at the end of a course on complex variables.

    the statement at least, the proof is often given at the end of a second such course.
     
    Last edited: Aug 16, 2005
  7. Aug 17, 2005 #6

    selfAdjoint

    User Avatar
    Staff Emeritus
    Gold Member
    Dearly Missed

    May I ask a stupid question. What is a divisor in this sense? "Let D = p1 + ....+pd be a divisor of distinct points... ". The Riemann surface I am fine with, but this, right at the start threw me.
     
  8. Aug 17, 2005 #7

    mathwonk

    User Avatar
    Science Advisor
    Homework Helper
    2015 Award

    well riemann did not consider them, he just asked: "given a set of distinct points p1,...,pd, when does there exist a meromorphic function f with poles at worst at these points and at most simple poles there?"

    but since in the more general case, a pole can be of order two or three or more, one now defines a "divisor" as a formal linear combination of points with integer coefficients:

    such as D = n1p1 + n2p2+....+ndpd. then if the integers ni are > 0, a meromorphic function with pole divisor bounded by D would be a meromorphic function with poles at most at the points p1,..pd, and such that the pole at pi has order at most ni.

    So i was trying to compromise, and take riemann's case but in modern language, i.e. all ni = 1, thus succeeding only in confusing the issue.

    in general some of the coefficients of points of D could also be negativ, and then one has e.g. that L(D) consists of functions with poles at most of order ni at the points pi with positive coefficients ni > 0, and zeroes at least of order ni at the points pi with negative coefficients ni<0.

    i.e. to each function f there is an associated divisor div(f) = zeroes of f - poles of f, where each zero or pole is counetd according to its order.

    then L(D) is all functions such that div(f) + D >= 0 (plus the zero function, whose divisor is undefined).
     
    Last edited: Aug 17, 2005
  9. Aug 17, 2005 #8
    oh yea mathwonk, real simple. :)

    Real question: what significance did Riemann's work conclude?
     
  10. Aug 17, 2005 #9

    selfAdjoint

    User Avatar
    Staff Emeritus
    Gold Member
    Dearly Missed

    Thank you mathwonk, I did understand it. And I was glad to be introduced to this terminology. I would imagine, considering their concerns with Riemann surfaces, that string theorists must use it quite often.
     
  11. Aug 18, 2005 #10

    matt grime

    User Avatar
    Science Advisor
    Homework Helper

    lots of things, but in this case it was what zeroes and poles a holomorphic function may have on a surface. (thus it may allow us to show when two surfaces are not homeomorphic). a special case of riemann roch is the euler characteristic. algebraic geometers use it too as it gives a signed sum of dimensions of ext groups and stuff in (Serre?) duality theory.
     
  12. Aug 18, 2005 #11

    mathwonk

    User Avatar
    Science Advisor
    Homework Helper
    2015 Award

    e.g. the most recent abel prize (math analog of nobel prize) was given to atiyah and singer for their index theorem generalizing RR.
     
  13. Aug 18, 2005 #12

    mathwonk

    User Avatar
    Science Advisor
    Homework Helper
    2015 Award

    here is the introduction to my notes on RR, from which the paragraph above was an extract:

    On Classical Riemann Roch and Hirzebruch's generalization


    0. Summary of contents. p. 2.

    I. RR for curves, statement, application to branched covers of P^1. p.8

    II. RR for surfaces, statement, applications to Betti numbers of surfaces in P^3, and line geometry on cubic surfaces. p.15

    III. Classical proof of RR for complex curves, (assuming existence and uniqueness of standard differentials of first and second kind). p.20

    IV. Modern proof of HRR for nodal plane curves and smooth surfaces in P^3, by induction using sheaf cohomology and elementary topology. p.37

    V. Hirzebruch’s formalism of Chern roots, Todd class, and the statement of the general HRR. Exercises. p.48


    Chapter 0. Summary of Contents

    Mittag Leffler problems
    We view the classical Riemann Roch theorem as a non planar version of the Mittag Leffler theorem, i.e. as a criterion for recognizing which configurations of principal parts can occur for a global meromorphic function on a compact Riemann surface. For example, the complex projective line P^1 = C U {?}, the one point compactification of the complex numbers, is a compact Riemann surface of genus zero. Any configuration of principal parts can occur for a global meromorphic function on P^1, and all such functions are just rational functions of z.
    On a compact Riemann surface of genus one,*i.e. a "torus", hence a quotient C/L of C by a lattice L, there is a residue condition that must be fulfilled. If f is a global meromorphic function on C/L and F is the L - periodic function on C obtained by composition C-->C/L, then the differential form F(z)dz must have residue sum zero in every period parallelogram for the lattice L. This necessary condition, properly stated, is also sufficient for existence of the function.
    In the case of surfaces of genus g, such as the one point compactification of the plane curve {y2 = f(x), where f is a square free complex polynomial of degree 2g+1} there are g corresponding residue conditions. Note this surface may be viewed as a branched cover of degree 2 over P^1 by projecting on the x coordinate (and ? is always a branch point). Riemann's point of view was in the reverse order, since he considered as the basic object of study, a compact, connected branched cover of the projective line, and then proved such a manifold is a plane algebraic curve, possibly acquiring singular points from the plane mapping.
    We discuss the classical proof of the theorem of Riemann and Roch, as well as the statement and some methods of proof for generalized versions in higher dimensions. We also present a few of the many applications of RR for curves and surfaces.

    Riemann’s approach
    The problem Riemann set himself was, given a distinct set of points on a surface of genus g branched over P^1, to calculate the dimension of the space of meromorphic functions with poles only in that that set, and at worst simple poles. It is now stated more generally and more formally using the concept of "divisors", which are essentially zeroes and poles with multiplicities. A "divisor" is a finite formal linear combination of points, with integer coefficients D = <sum>nipi, on a smooth projective curve X. Then every meromorphic function f has an associated divisor div(f) consisting of the zeroes of f minus the poles of f, each counted with its appropriate multiplicity. The degree of a divisor <sum>nipi, is the sum of the integer coefficients, and for every meromorphic function f on a compact Riemann surface, the basic fact is that the degree of div(f) is zero, i.e. the number of poles and zeroes is the same, counted properly. For example a complex polynomial of degree n, has n finite zeroes, but also has a pole of degree n at ?.
    If L(D) is the space of meromorphic functions whose divisor bounded below by -D, i.e. such that div(f) + D >= 0, in the sense that all coefficients of div(f) are*>= the corresponding coefficients of -D, then the RR problem is to compute the analytic invariant dimL(D).
    Riemann’s approach was to solve first the analogous but simpler Mittag Leffler problem for meromorphic differentials, and then apply the usual criterion for exactness of differentials to pass to the case of functions. He always assumed for argument's sake, the distinctness of his points, i.e. the simplicity of his poles, but was well aware of the general case and remarked that it was easily dealt with by taking additional derivatives.

    The cohomological approach
    The modern approach to Riemann Roch for curves is to define analytic cohomology groups H^0(D), H^1(D), such that H^0(D) and L(D) are isomorphic, and*prove that "chi(X,D)" = dimH^0(D) - dimH^1(D) [the analytic euler characteristic] is a topological invariant. Then the problem has two stages, (i) (Hirzebruch Riemann Roch or HRR) prove chi(D) = d+1-g, where d is the degree of D and g is the topological genus of X; this is essentially Riemann's theorem. (ii) (Serre Duality or SD) prove that H^1(D) is isomorphic to the dual space L(K-D)*, where K is the divisor of a differential form on X. This is Roch's contribution to Riemann's theorem.
    Together these give the classical RRT: dimL(D) - dimL(K-D) = d+1-g.
    More precisely, if D is a positive divisor of distinct points, i.e. one with all coefficients equal to +1, Riemann essentially proved that L(D)/C is the kernel of a linear map from Cd -->Cg, hence d-g <= dimL(D)-1 <= d, and also that dimL(K) = g. Roch later proved the codimension of the image of Riemann's map was dimL(K-D), hence that dimL(D)-1= d-(g - dimL(K-D)), where by Riemann's computation L(K) = g, then dimL(K-D) <= g.

    The classical argument via residues
    We give an account below of the classical proof of RR for curves, assuming Riemann’s existence and uniqueness results for holomorphic and meromorphic differentials of second kind. The only other tool is classical residue calculus of differential forms, i.e. Stokes theorem. The proof in section III is the one originally given more briefly by Roch.

    A recursive computation of the arithmetic genus of curves
    After presenting Riemann's and Roch's arguments, we give a proof of stage (i) in the modern approach to the HRR for curves as follows: We use the simplest results of sheaf theory to prove that chi(D) - chi(O) = deg(D), by induction on deg(D). Then we prove chi(O) = 1-g, by immersing X in the plane as a curve of degree n with only nodes, and proving the formula for chi(O) by induction on n. The essential fact is that chi(O) depends only on the degree n, and can be defined for a reducible curve D+E with transverse components D, E , satisfying the inductive axiom chi(D+E) = chi(D)+chi(E)-chi(D.E), where chi(D.E) = deg(D.E). The formula 1-g has the same properties. This lets us go from knowing the equation chi(O) = 1-g for D and E separately, to deducing it for a smooth deformation of their union. The argument in section IV was inspired by the introduction of Fulton’s paper [Am.J.Math, 101(1979)].

    Most modern proofs of HRR in higher dimensions seem to be of this type: they establish axiomatic properties which characterize an invariant uniquely, then show that both chi(D) and an appropriate topological expression in the chern classes of D satisfy the axioms. The first such argument may be due to Washnitzer, although his published proof omitted part of the verification, later supplied by Fulton.

    Arithmetic genus of smooth projective surfaces in P^3
    To illustrate this method in higher dimensions we give a similar inductive argument in section IV for HRR on a smooth complex surface S which can be embedded in P^3. Here a divisor is a linear combination of irreducible curves on the surface. It is possible to define the intersection of a curve in D "with itself" obtaining at least an equivalence class of divisors on that curve, and a self intersectiion number, and then to do induction by appealing to the RRT on the curve.

    Sheaf theory reduces the calculation of the difference chi(D) -chi(O) to the known HRR formula on the lower dimensional variety D, as follows: chi(D)-chi(O) = chi(O(D)|D) = (D.D) + 1-g(D), where g(D) is the genus of the curve D. Then the adjunction formula relating the canonical divisors of S and D, gives 1-g(D) = -(1/2)[(K+D).D], hence chi(D)-chi(O) = (1/2)([D.(D-K)]. To get a formula for chi(D), it suffices to calculate chi(O) for the surface S.

    Knowing in advance the answer to be chi(O) = (1/12)(K^2 + e(S)), where K is a “canonical” divisor of a differential 2 form, K^2 is the self intersection number of K, and e is the topological euler characteristic, we can check the validity of this equation for chi(O) by induction on the degree of the embedded surface S in P^3. Again chi(O) depends only on deg(S) = n, and satisfies the same inductive rule as before for transverse unions, chi(Y+Z) = chi(Y)+chi(Z)-chi(Y.Z). Since chi(P2) = 1, we get chi(Sn,O) = chi(Sn-1,O) + 1 - chi(Cn-1,O) where Cn-1 is a smooth plane curve of degree n-1. This determines chi(Sn,O) uniquely.

    Then the adjunction formula for surfaces in P^3 and a Lefschetz pencil on Sn give (1/12)(e(Sn) + K^2) = n(n^2 - 6n + 11)/6. Checking that this formula satisfies the same inductive properties as does chi(O), gives the result, and hence the HRR for the smooth hypersurface Sn.

    For stage (ii) of the modern proof of RRT for curves (the Serre duality result that H^1(D)* is isomorphic to L(K-D)), we sketch Serre’s argument using the algebraic version of the classical residue pairing. He lumps the relevant spaces H^1(D)* for all D, into one large union, Weil’s space of "adeles", which is infinite dimensional over the base field k, but one dimensional as a vector space over the field k(X) of meromorphic functions on the curve X. He then shows this union is isomorphic to the analogous one dimensional union of the spaces L(K-D) of meromorphic differentials for all D. It is then easy to check this isomorphism restricts for each D to the desired isomorphism H^1(D)* = L(K-D).

    Kodaira vanishing theorems
    In higher dimensions to go from HRR to a computation of dimL(D) even in special cases, requires more than Serre’s duality. I.e. Serre duality only transforms H^dim(X) into an H^0, and we want our formula entirely in terms of H^0's when possible. Hence we appeal to Kodaira vanishing which gives a sufficient criterion for the higher cohomology groups H^1(D),....,H^dim(X)(D) to vanish, and which holds for all sufficiently "positive" divisors. We state a modern version of this result apparently due to Rananujam, Mumford, Kawamata and Viehweg, but we do not give any proof.
    The proof of the original vanishing theorem in the book of Kodaira and Morrow involves lengthy calculations in differential geometry and analysis to establish an inequality of “Bochner type” for curvature operators on harmonic forms on complex manifolds from which the result is easily deduced by Hodge theory. A short account of a more recent proof by Kolla’r can be found at the end of volume 3 of the introductory series by Ueno on algebraic geometry and scheme theory, (AMS translations). The Kodaira vanishing theorem is available only in characteristic zero, i.e. it is false in positive characteristic.

    Computations using the Todd genus
    Since the arguments we gave for HRR in dimensions 1 and 2 required knowing the topological formula for chi(O) in advance, we recall finally the formalism of Hirzebruch which shows in principle how to write down such formulas in all dimensions in terms of “chern roots”. The problem of giving the formulas in terms of chern classes is thus reduced to the elementary but tedious problem of expressing a given symmetric function in terms of elementary symmetric functions. Finally we compute the invariants of a cubic threefold, noting they agree with those of P^3, and discuss briefly how Clemens and Griffiths showed nonetheless a smooth cubic threefold is not even birational to P^3. Grothendieck’s relative version of the HRR for a map of two varieties is not discussed here; for this see Borel-Serre, or Fulton.
     
  14. Aug 19, 2005 #13

    mathwonk

    User Avatar
    Science Advisor
    Homework Helper
    2015 Award

    the previous discussion is obviously not aimed at the average high school student, but i hope it has some value and interest say to graduate students in physics and mathematics. :bugeye:
     
  15. Aug 20, 2005 #14

    selfAdjoint

    User Avatar
    Staff Emeritus
    Gold Member
    Dearly Missed

    Mathwonk you should really put these notes online. In fact I think they would make an excellent contribution to the arxiv, perhaps in the history and overview section of the matn board. You and Euler!
     
  16. Aug 20, 2005 #15

    mathwonk

    User Avatar
    Science Advisor
    Homework Helper
    2015 Award

    there is an old version on my webpage at http://www.math.uga.edu/~roy/

    but when I wrote those I had not read riemann, i.e. I did not know what I was talking about, so there is a new version coming in the next few days.

    thank you for the vote of confidence. If there is some other place to put them I would be happy to do so.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Riemann roch made easy
  1. Easy but interesting. (Replies: 11)

  2. Easy question (Replies: 2)

  3. Riemann and Ricci (Replies: 3)

  4. Riemann surfaces (Replies: 2)

Loading...