(We will update the new index of theorems, etc..., as in my new PDF file, link here).
Before coming to the theorem of Riemann-Roch for abelian varieties, let us take a look on the cohomology groups of Poincare's bundle and Mumford's bundle. The circle of ideas mentioned in our previous notes still works, since for Poincare's bundle, $P|_{A\times \{L\}}=L$ for all $L\in Pic^0(A)$, and the restriction is never trivial unless $L$ is trivial. This yields the support of the higher direct image of the push-forward is of dimension 0, and hence, the higher cohomology group of this sheaf vanish. The same holds for Mumford's line bundle $\Lambda(L)$ in the case $L$ is ample. And the two line bundles related via the pull-back, due to our construction.
1. Cohomology groups of Poincare and Mumford's line bundles.
Let $P$ be the Poincare's bundle on $A\times \hat{A}$, and $p_2': A\times \hat{A}\to \hat{A}$ the projection on the second coordinate, by our earlier argument, for any line bundle $L\in Pic^0(A)$, and $L$ is not trivial, $P|_L=L$ is also not trivial. Using our Proposition 30, the cohomology groups of $P|_L$ vanish. And by Theorem 31, $Supp(R^qp'_{2,*}P)$ is of dimension 0, because it just has at most one point. This implies $H^p( \hat{A}, R^qp'_{2,*}P)$ vanish for all $p>0$. Now, making use of the Leray spectral sequence, we have
$$E_2^{p,q}=H^p( \hat{A}, R^qp'_{2,*}P)\Rightarrow H^{p+q}(A\times \hat{A}, P)$$
When $p=0$, this yields $E_2^{0,q}=R^qp'_{2,*}P\Rightarrow H^{q}(A\times \hat{A}, P)$. Look at the arrows of page 2 in our spectral sequence, we can see $0=H^1( \hat{A}, R^{q-2}p'_{2,*}P)\to E_2^{p,q}\to 0$, and hence $E_2^{p,q}$ is actually stable. This yields $R^qp'_{2,*}P\cong H^q(A\times \hat{A}, P)$, for all $q$.
Now, if $q>g$, the dimension of $A$, $R^qp'_{2,*}P$ vanishes. This yields $H^q(A\times \hat{A}, P)$ also vanishes whenever $q>g$. Making use of Serre's duality, we have $H^{2g-q}(A\times \hat{A}, P)\cong H^q(A\times \hat{A}, P^{-1})$, because the dualizing sheaf is isomorphic to the canonical bundle, which is trivial in the case $A$ is abelian. So, for all $q\ne g$, we have $H^q(A\times \hat{A}, P)=0$. This is the easy first part of the following
Theorem 36. Let $A$ be an abelian variety of dimension $g$, and $P$ the Poincare's sheaf of $A$, then $H^q(A\times \hat{A}, P)=0$, for $q\ne g$, and $H^g(A\times \hat{A}, P)=i_0(k)$, where $i_0(k)$ is the skyscrapper sheaf at $0$.
Proof. Theorem 9.1 in the book of G. Van der geer and Ben Moonen.
Now, we remember that if $L$ is an ample line bundle on $A$, then $A\times A\xrightarrow{(id, \lambda_L)}A\times \hat{A}$ is an isogeny, and $(id,\lambda_L)^*P=\Lambda(L)$, where $\Lambda(L)$ is the Mumford's line bundle of $L$. Take a look on the following commutative diagram
(DIAGRAM)
where $A\times \hat{A}\to \hat{A}$ is a proper morphism, and hence, separated of finite type, and $\lambda_L$ is an isogeny, which is flat. These things combined allows us to use the flat base change [HAG-...], which states
$$\lambda_L^*R^ip'_{2,*}P=R^ip_{2,*}(id,\lambda_L)^*P=R^ip_{2,*}\Lambda(L)$$
And from Theorem 36, we have $R^ip_{2,*}\Lambda(L)=0$ if $i\ne g$, otherwise, its dimension is $\deg\lambda_L$. Because $\Lambda(L)|_{A\times \{a\}}=\lambda_L(a)$, which is trivial iff $a\in K(L)$, which is finite. Hence, using Theorem 31, $Supp(R^ip_{2,*}\Lambda(L))\subset K(L)$, which is of dimension $0$. This yields the higher cohomology groups of this sheaf vanish.
Making use of Leray spectral sequence as above, we have $R^ip_{2,*}\Lambda(L)\cong H^i(A\times A, \Lambda(L))$, which is $0$ if $i\ne g$, and is of dimension $\deg \lambda_L$ if $i=g$. We have proved the following
Proposition 37. Let $L$ be an ample line bundle on an abelian variety $A$ of dimension $g$, and $\Lambda(L)$ the Mumford's line bundle on $A\times A$, then $H^i(A\times A, \Lambda(L))=0$ if $i\ne g$, and $h^g( \Lambda(L))=\deg \lambda_L$. Furthermore, $\chi(\Lambda(L))=(-1)^gh^g(\Lambda(L))=(-1)^g\deg\lambda_L$.
2. Another computation for $\chi(\Lambda(L))$.
We will compute $\chi(\Lambda(L))$ by another way. Note that since $\Lambda(L)=m^*L\otimes p_1^*L^{-1}\otimes p_2^*L^{-1}$, we can make use of projection formula to make it have a simpler form. The map $f: A\times A\to A$ is a morphism, this yields by the projection formula
$$R^np_{2,*}(\Lambda(L))\otimes L\cong R^np_{2,*}(\Lambda(L)\otimes p_2^*L)\cong R^np_{2,*}(m^*L\otimes p_1^*L^{-1})$$
Because $Supp(R^np_{2,*}\Lambda(L))\subset K(L)$, which is finite, and $L$ can be trivialized on $K(L)$, this yields
$$R^np_{2,*}(m^*L\otimes p_1^*L^{-1})\cong R^np_{2,*}(\Lambda(L))\otimes L\cong R^np_{2,*}\Lambda(L)$$
If we restrict $m^*L\otimes p_1^*L^{-1}$ on $A\times \{a\}$, we will get $\lambda_L(a)$, which is trivial on $Pic^0(A)$ iff $a\in K(L)$, and hence, the support of this sheaf is of dimension 0, and the higher cohomology groups then vanish. Making use of Leray spectral sequence, we have $H^n(A\times A, m^*L\otimes p_1^*L^{-1})\cong R^np_{2,*}(m^*L\otimes p_1^*L^{-1})$. And hence, by our previous results, $H^n(A\times A,\Lambda(L))\cong H^n(A\times A, m^*L\otimes p_1^*L^{-1})$.
Now, the map $(m,p_1): A\times A\to A\times A$ sending $(a,b)$ to $(a+b,a)$ is an isomorphism. If we pullback $p_1^*L\otimes p_2^*L^{-1}$ via this map, we obtain $m^*L\otimes p_1^*L^{-1}$. To see this, we use the see-saw principle, for any line bundle $M$ on $A\times A$, we have $(m,p_1)^*M|_{\{a\}\times A}=\tau_a^*(M|_{A\times \{a\}})$, hence $(m,p_1)^*(p_1^*L\otimes p_2^*L^{-1})|_{\{a\}\times A}=\tau_a^*L$. Also, $m^*L\otimes p_1^*L^{-1}|_{\{a\}\times A}=\tau_a^*L$, for all $a\in A$. Furthermore, $(m,p_1)^*M|_{A\times \{0\}}=M|_{\Delta}$, where $\Delta:=\{(a,a)\in A\times A\}$. From this, $(m,p_1)^*(p_1^*L\otimes p_2^*L^{-1})|_{\Delta}$ is trivial. Also, $m^*L\otimes p_1^{*}L^{-1}|_{A\times \{0\}}$ is also trivial. This yields by the see-saw principle that $(m,p_1)^*(p_1^*L\otimes p_2^*L^{-1})\cong m^*L\otimes p_1^*(L^{-1})$. And hence, their cohomology groups agree. This yields
$$H^n(A\times A,\Lambda(L))\cong H^n(A\times A, p_1^*L\otimes p_2^*L^{-1})\cong \bigoplus_{p+q=n}H^p(A,L)\otimes H^q(A,L^{-1})$$
by the Kunneth's formula. From this,
$$h^n(\Lambda(L))=\sum_{p+q=n}h^p(L)h^q(L^{-1})$$
Using the result of Proposition 37, $h^n(\Lambda(L))=0$ for all $n\ne g$ yields for all $p$, $h^p(L)=0$ or $h^{n-p}(L)=0$ for $n\ne g$. On the other hand, $h^g(\Lambda(L))=\sum_{p=0}^nh^p(L)h^{n-p}(L^{-1})$. We will prove that there exists a unique index $i\le g$ such that both $h^i(L)$ and $h^{g-i}(L^{-1})$ are not zero. The existence of the index $i$ is clear. Assume that there exists $j\ne i$ such that both $h^j(L)$ and $h^{g-j}(L^{-1})$ are not zero. If $i<j$, then $g-j+i<g$, and both $h^i(L)$ and $h^{g-j}(L^{-1})$ are not zero, this is a contradiction. The case $i>j$ is similar. By Serre's duality, for all $k$, we have $h^k(L)=h^{g-k}(L^{-1})$, and this yields there exists a unique $i\le g$ such that $h^i(L)\ne 0$. Such an $i$ is called the index of $L$, which is denoted $i(L)$. This yields $\chi(L)=(-1)^{i(L)}h^{i(L)}(L)$, and hence, $h^g(\Lambda(L))=h^{i(L)}(L)^2=\chi(L)^2$. And by Proposition 37, $\deg(\lambda_L)=\chi(L)^2$.
Combining things together, we get the following important
Theorem 38 (Vanishing Theorem). Let $L$ be an ample line bundle on an abelian variety $A$, then there exists a unique integer $i$ such that $h^i(L)\ne 0$, and $\deg(\lambda_L)=\chi(L)^2$.
Example 39. Again, we will prove the vanishing theorem for elliptic curves by direct computation. A divisor $D\in Div(E)$ is ample iff $\deg E>0$. This yields by R-R that $h^0(D)\ne 0$, and it is $\deg D$. Because the genus of $E$ is 1, it yields by Riemann-Roch again that $h^1(D)=0$, and for $n>1$, we have $h^n(D)=0$, due to the vanishing theorem of Grothendieck. From this, $\chi(D)=\deg D$. Now, $\lambda_D$ is the map $a\mapsto \tau_a^*D-D$. We can write $D=[P_1]+...+[P_n]$, and $\lambda_D(a)=[P_1-a]+...+[P_n-a]-[P_1]-...-[P_n]$. Via the isomorphism between $Pic^0(E)$ and $E$, $\lambda_D$ can be identified with the map $[n]$, the multiplication by $n$ map. And our previous result in Section 1 shows that $\deg(\lambda_D)=\deg [n]=n^2=\deg(D)^2=\chi(D)^2$.
3. Riemann-Roch's theorem.
We now turn to the Riemann-Roch's theorem for abelian varieties. It can be seen that for abelian varieties $A$ of dimension $g$, the canonical sheaf on $A$ is trivial, and the tangent bundle of $A$ is also trivial. This yields the $Td(A)=1$, where $Td(A)$ is the Todd's genus of the tangent sheaf of $A$. Let $L$ be any line bundle on $A$, this yields by the theorem of Hizerbruch-Riemann-Roch that
$$\chi(L)=\int_X ch(L)= \frac{\deg c_1(L)^g}{g!}$$
And this is the theorem of Riemann-Roch for abelian varieties. We can use it to deduce the following
Corollary 40. Let $f:A\to B$ be an isogeny of abelian varieties, then for any line bundle on $B$, $\chi(f^*L)=\deg f\chi(L)$.
Proof. Pull-back of a line bundle is a line bundle, so $\chi(f^*L)=\deg (c_1(f^*L)^g)/g!$. Using axioms (C2) in our previous post for Chern's classes, we have $f^*(c_1(L))=c_1(f^*(L))$. And use Proposition 10.2.6 (iv) (Gathmann's note), that states, for $f: X\to Y$ is a morphism, such that $f^*: A_*(X)\to A_*(Y)$ exists. Let $F$ be a vector bundle on $Y$, and $\alpha\in A_*(Y)$, we have
$$c_i(f^*F).f^*\alpha=f^*(c_i(F).\alpha)$$
Now, let $F\equiv L$, and $\alpha=c_1(L)$, we have
$$c_1(f^*L)^2=c_1(f^*L).c_1(f^*L)=c_1(f^*L)f^*(c_1(L))=f^*(c_1(L)^2)$$
Using induction, we get $c_1(f^*L^g)=f^*(c_1(L)^g)$. And by Riemann-Roch's theorem, it is sufficient to check $\deg f^*[P]=\deg f$. But it is clear, since if $f$ is separable, then the last identity holds. Otherwise, $f$ is a composition between a separable isogeny $h$ and a purely inseparable isogeny $g$. If we pull-back $P$ via $h$, we will get exactly $\deg(h)$ points, and for each $\deg h$ points, if we pull them back via $g$, we have exactly $\deg h$ points, with multiplicity $\deg g$ for each. And our desired identity follows from the fact that $\deg f=\deg g\deg h$.
(Q.E.D)
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