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mathwonk

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## Main Question or Discussion Point

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.

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.

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