We now come to some applications of the theorem of Hirzebruch-Riemann-Roch for abelian varieties.
1. Warm-up with Euler's exact sequence.
First, we recall the Euler's exact sequence (Chap VII, Gathmann's note).
$$0\to \mathscr{O}_{\mathbb{P}^n}\to \mathscr{O}_{\mathbb{P}^n}(1)^{\oplus n+1}\to T_{\mathbb{P}^n}\to 0$$
where $T_{\mathbb{P}^n}$ is the tangent sheaf of ${\mathbb{P}^n}$. If we denote $H:=\mathscr{O}_{\mathbb{P}^n}(1)$ the hyperplane divisor on ${\mathbb{P}^n}$, then this yields by axioms of Chern's characters that
$$c_t(T_{\mathbb{P}^n})=(1+H)^{n+1}$$
Example 41. $n=2$, then $c_t(T_{\mathbb{P}^2})=1+3Ht+3H^2t+t^3$, and hence, Chern characters of $T_{\mathbb{P}^2}$ is $c_1:=3H, c_2:=3H^2$. And this yields $Td({\mathbb{P}^2})=(1, \frac{1}{2}c_1, \frac{1}{12}(c_1^2+c_2))$. For trivial line bundle $\mathscr{O}_{\mathbb{P}^2}$, $ch(\mathscr{O}_{\mathbb{P}^2})=1$, HRR implies that
$$\chi(\mathbb{P}^2)=\deg_2 Td({\mathbb{P}^2})=\deg \frac{1}{12}(9H^2+3H^2)=1$$
Moreover, using the adjunction sequence, if $X$ is a smooth hypersurface of degree $d$ in ${\mathbb{P}^n}$, and $i: X\hookrightarrow {\mathbb{P}^n}$ the inclusion, then we have the short exact sequence
$$0\to T_X\to i^*T_{\mathbb{P}^n}\to N\to 0$$
And $N$ is called the normal bundle of $X$ in $\mathbb{P}^n$. In this case, it is a line bundle. This yields $c_t(T_{i^*\mathbb{P}^n})/c_t(N)=c_t(T_X)$. Because Chern characters commutes with the pull-back, we have $c_j(i^*\mathbb{P}^n)=i^*c_j(\mathbb{P}^n)$. Hence, if we denote $h:=i^*H$ the pullback of hyperplane divisor on ${\mathbb{P}^n}$, we have $c_t(T_{i^*{\mathbb{P}^n}})=(1+ht)^{n+1}$. And because $[X]$ can be consider as a divisor on ${\mathbb{P}^n}$, with $X\sim dH$ (because $A^k({\mathbb{P}^n})$ is generated by the class of $k$-dimensional linear subspace), we have $c_1(N)=i^*[X]=dH$. And $N$ is a line bundle implies that $c_n(N)$ vanishes, for $n>1$. This yields $c_t(N)=1+dHt$. Combining these things, we get
$$c_t(T_X)=(1+ht)^{n+1}/(1+dht)=(1+ht)^{n+1}(1-dht+d^2h^2t^2-...)$$
Remark 42. If $i$ is an embedding of codimension $d$, i.e. there exists $d$ divisors $D_1,...,D_d$ on $\mathbb{P}^n$ such that $D_1\cap D_2\cap...\cap D_d=X$, then the normal bundle is a vector bundle of rank $d$ in this case, and its Chern roots will be $i^*[D_i]$, which are $i^*d_iH$, for some hyperplane $H$ in $\mathbb{P}^n$.
Example 43. We will use our recent argument to compute the genus of a smooth projective plane curve of degree $d$. In this case, $c_t(T_X)=(1, \frac{1}{2}(3-d)h,...)$, where $h$ is a divisor on $X$ with $\deg h=d$. This yields by HRR that $\chi(X)=\frac{1}{2}d(3-d)$. Because $\chi(X)=1-g$, we obtain $g=\frac{1}{2}(d-2)(d-1)$.
Example 44. We will compute the Euler's characteristic of the line bundle $\mathscr{O}_{\mathbb{P}^n}(d)$ on $\mathbb{P}^n$. In this case
$$Td(T_X)=\frac{H^{n+1}}{(1-\exp(-H))^{n+1}}$$
And $ch(X)=\exp(dH)$. In particular, we want to compute the $n$-th coefficient of the power series
$$\frac{\exp(dH)H^{n+1}}{(1-\exp(-H))^{n+1}}$$
In terms of complex analysis, it is the residue at 0 of
$$\frac{\exp(dH)}{(1-\exp(-H))^{n+1}}dH$$
Let $x:=1-\exp(-H)$, then $\exp(H)=\frac{1}{1-x}$, and $\frac{dH}{dx}=\frac{1}{1-x}$. And it is sufficient for us to compute the residue at 0 of $\frac{(1-x)^{-d-1}}{x^{n+1}}$, and it is the $n$-th coefficient of $(1-x)^{-d-1}$, which is $(-1)^n{-d-1\choose n}={n+d\choose n}$. In short,
$$\chi(\mathbb{P}^n, \mathscr{O}_{\mathbb{P}^n}(d))={n+d\choose n}$$
Using the same method, we can also compute the Euler's characteristic of a smooth hypersurface in $\mathbb{P}^n$.
2. Embedding of abelian varieties. We will prove in this section that an abelian variety of dimension $g$ cannot be embedded into $\mathbb{P}^{2g-1}$, and it is embedded into $\mathbb{P}^{2g}$ iff it is an elliptic curve, or an abelian surface of degree 10 in $\mathbb{P}^4$. This shows, it is not easy to write explicitly equations that define an abelian variety in $\mathbb{P}^n$.
Now, let us denote $m$ the smallest integer such that $A$ can be embedded to $\mathbb{P}^m$. It is an embedding of codimension $m-g$. Hence, the normal bundle of $A$ in $\mathbb{P}^m$ is of rank $m-g$. Let us denote the Chern characters of $N$ as $c_1,...,c_{m-g}$. This yields by the adjunction sequence
$$0\to T_X\to i^*T_{\mathbb{P}^m}\to N\to 0$$
that $c_t(T_X)c_t(N)=c_t(i^*T_{ \mathbb{P}^m})$. For abelian varieties, the tangent sheaf is trivial, and hence
$$1+\sum_{i=1}^{m-g}c_it^i=c_t(N)=c_t(i^*T_{\mathbb{P}^m})=(1+ht)^{m+1}$$
From this, one can see for $n\ge m+1-g$, $h^n$ vanish. But when $n=g$, we have $\deg h^g$ is exactly the degree of $A$, which is never zero. Hence, $m+1-g\ge g+1$. This yields $m\ge 2g$.
Now, assume that $m=2g$. Look at the $g$-th coefficient in the last identity, we can see
$$c_g={2g+1\choose g}h^g$$
Taking the degree, we have $\deg c_g={2g+1\choose g}d$. We now make use of a following
Theorem 45 [A. Van de Ven - On the embedding of abelian varieties in Projective Spaces]. Let $X$ be a complete non-singular projective variety of dimension $2d$, and $Y$ is a non-singular projective subvariety of $X$ of dimension $d$. If $N$ is the normal bundle of $Y$ in $X$, and $c_g(N)$ the $g$-th Chern character of $N$, then $c_g(N)$ is equal to the self-intersection number of $Y$ on $X$.
As one can see from this, $\deg c_g$ is the self-intersection number of $A$ on $\mathbb{P}^m$. As we mentioned earlier, the $A^k(\mathbb{P}^m)$ is generated by $k$-th dimensional linear subspace of $\mathbb{P}^m$, we have $A\sim dH$ for some $g$-dimensional linear subspace $H$ on $\mathbb{P}^m$. This yields $\deg A.A=d^2=\deg c_g$. This follows $d={2g+1\choose g}$.
Now, let $L$ be any line bundle on $A$, the Riemann-Roch theorem for abelian variety reads
$$\chi(A,L)=\frac{1}{g!}\deg c_1(L)^g$$
Take $L:=h$, we have $\deg c_1(L)^g=\deg h^g = d = {2g+1\choose g}$. And because the Euler characteristic is an integer, we have $g!\mid {2g+1\choose g}$. This can happen only if $g=1,2$. When $g=1$, it is an elliptic curve, embedded into $\mathbb{P}^2$, and of degree 3. If $g=2$, we have $d=10$, and it is an abelian surface of degree 10, embedded into $\mathbb{P}^4$. Such an abelian surface exists, due to the result of Mumford.
Remark 46. It also follows from the Riemann-Roch's theorem that $\chi(A)=0$.
3. HRR for smooth surfaces in $\mathbb{P}^4$.
We will first visit HRR for smooth surfaces $X$. The tangent bundle of $X$ is of rank 2, with Chern characters $c_1(T_X), c_2(T_X)$. We have $c_1(T_X)=-c_1(\Omega_X)$, where $\Omega_X$ is the cotangent bundle of $X$. It is also of rank $2$. Recall that if $E$ is a vector bundle on $X$ of rank $n$, with Chern roots $\alpha_i$, then $\bigwedge^r E$ has Chern roots $\sum_{1\le i_1<...<i_r\le n}\alpha_{i_1}+...+\alpha_{i_r}$. From this, one can see $c_1(\Omega_X)=c_1(\bigwedge^2\Omega_X)=K$, where $K$ is the canonical bundle of $X$. This yields $c_1(T_X)=-K$. We can compute $c_2$ via the genus of $X$. The HRR for smooth surfaces reads
$$\chi(X)= \frac{1}{12}(K^2+c_2(T_X))$$
This yields $c_2(T_X)=12\chi(X)- K^2$. Now, if $X$ can be embedded into $\mathbb{P}^4$, we have by Theorem 45, the adjunction sequence yields
$$c_t(T_X)c_t(N)=(1+ht)^5=1+5ht+10h^2t^2+...$$
We know that $c_t(T_X)=1-Kt+c_2(T_X)t^2$ and $c_t(N)=1+c_1(N)+c_2(N)$. Equating both sides of the last identity, we get
$$c_1(N)=5h+K$, c_2(N)-Kc_1(N)+c_2(T_X)=10h^2$$
We know from Theorem 45 that $\deg c_2(N)=d^2$, $\deg h^2=d$, $c_2(T_X)=12\chi(X)-K^2$. So, taking the degree of the 2-nd graded part, we have
$$d^2 - 10d + 5hK - 2K^2 + 12\chi(X)=0$$
Now, for an abelian surface that can be embedded into $\mathbb{P}^4$, because the canonical sheaf is just trivial, and $\chi(X)=0$, by Remark 46, we have $d^2-10d=0$. This yields $d=10$. And we obtain a part of our result in the previous section.
4. An exercise in the book of Van der Geer and Ben Moonen.
We will prove in this exercise that an abelian variety $A$ of dimension $g$ cannot be embedded into $P:=\mathbb{P}_1^{2g-1}$. Assume that we have such an embedding $i: A\to P$, then the adjunction sequence yields
$$0\to T_A\to i^*T_P\to N\to 0$$
And in this case, again, we have $c_t(N)=c_t(i^*T_P)$. If we denote $pr_i: P\to \mathbb{P}^1$ the $i$-th projection, then it follows that $T_P=\oplus pr_i^*T_{\mathbb{P}_1}$, and because $c_t(P_1)=(1+H)^2=1+2Ht$, where $H$ is a point in $\mathbb{P}^1$, we have $c_t(T_P)=\prod_{i=1}^{2g-1}(1+2h_i)$, where $h_i$ is the pullback of $H_i$ via $pr_i$. Because $N$ is a vector bundle on $A$ of rank $g-1$, we have $c_g(N)=0$. And this yields the $g$-th part of the product $\prod_{i=1}^{2g-1}(1+2h_i)$ vanish, i.e. all the term of the form $h_{i_1}...h_{i_g}$ vanish, and there are ${2g-1\choose g}$ such terms.
But if we look at ${2g-1\choose g}$ (with index $i_1,...,i_g$ as above) projections from $P$ to $P_g:=\mathbb{P}_1\times ... \times \mathbb{P}^1$ ($g$ times), and $A\to P_g$ the composition, we will have at least one of them is surjective, since $\dim A\ge g$, and that means the pullback of points on $P_g$ via this map is never empty. And this yields the corresponding product $h_{i_1}...h_{i_g}$ is non zero, because they are just the pullback of points in each $\mathbb{P}^1$. This is a contradiction. Hence, an abelian variety of dimension $g$ cannot be embedded into $\mathbb{P}_1^{2g-1}$.
The later question is also interesting, that is can an abelian variety be embedded into $\mathbb{P}_1^{2g}$? First, if we are able to prove that there exists an elliptic curve $E$ that can be embedded into $\mathbb{P}^1\times \mathbb{P}^1$, then $E\times E\times ...\times E$ ($g$-times) is an abelian variety of dimension $g$, and it can be embedded into $\mathbb{P}_1^{2g}$. This rest of this section is devoted for the proof. We refer to the work of Peter Bruin. In this paper, he shows the existence of non-singular curve of type $(a,b)$, for $a,b>0$ in $\mathbb{P}^1\times \mathbb{P}^1$, and the genus of this curve is $(a-1)(b-1)$. If we choose $a=b=2$, we get the curve of type $(2,2)$ in $\mathbb{P}^1\times \mathbb{P}^1$, and it is an elliptic curve. It is actually the Edwards curve, and it turns out to be useful in cryptography. Now, let us make a sketch of his proof.
First, it is known that if $Z:=\mathbb{P}^1\times \mathbb{P}^1$, then $Pic(Z)$ is generated by $L_1, L_2$, where $L_1$ is the pull-back $p_1^*\mathscr{O}_{\mathbb{P_1}}(1)$, and $L_2$ is the pullback $p_2^*\mathscr{O}_{\mathbb{P}_1}(1)$. This yields, if $L$ is a divisor on $Z$, then $L=aL_1+bL_2$, for some $a,b\in \mathbb{Z}$. We call this line bundle is of type $(a,b)$. From this decomposition, we get
$$L=aL_1+bL_2\cong p_1^*(\mathscr{O}_{\mathbb{P}_1}(a))\otimes p_2^*(\mathscr{O}_{\mathbb{P}_1}(b))$$
From this, any line bundle of type $a(,b)$ on $Z$ is isomorphic, and we denote them $\mathscr{O}_Z(a,b)$ Now, making use of Kunneth's formula, we have
$$H^n(Z,\mathscr{O}_Z(a,b))\cong \bigoplus_{p+q=n}H^p(\mathbb{P}^1, \mathscr{O}_{\mathbb{P}_1}(a))\otimes H^q(\mathbb{P}^1, \mathscr{O}_{\mathbb{P}_1}(b))$$
Using Riemann-Roch, when $a,b\ge 0$, because $\deg K_{\mathbb{P}_1}=-2$, the higher cohomology groups of $H^*(\mathbb{P}_1,\mathscr{O}_{\mathbb{P}_1}(a))$ vanish, hence $h^0(\mathbb{P}_1,\mathscr{O}_{\mathbb{P}_1}(a))=a + 1$, and similarly for $b$. This yields $H^i(Z,\mathscr{O}_Z(a,b))$ vanish for $i=1,2$, and $h^0(Z, \mathscr{O}_Z(a,b))=(a+1)(b+1)$. Similarly, if $a,b<0$, we get $H^i(Z, \mathscr{O}_Z)$ vanish for $i=0,1$, and $h^2(Z,\mathscr{O}_Z(a,b))=(a+1)(b+1)$.
Now, if we assume that there exists a smooth curve $Y$ of type $(a,b)$ on $Z$, i.e. if we denote homogeneous coordinate of $Z$ as $(x_0:x_1:y_0:y_1)$, then $Y$ is defined by homogeneous polynomial of the form $\sum_{i,j}c_{i,j}x_0^ix_1^{a-i}y_0^jy_1^{b-j}$. We call $Y$ is bi-degree of the type $(a,b)$. The closed immersion $Y\to Z$ induces the short exact sequence of sheaves
$$0\to \mathscr{O}_Z(-Y)\to \mathscr{O}_Z\to i_*\mathscr{O}_Y\to 0$$
And this induces the following long exact sequence
$$0\to H^0(Z, \mathscr{O}_Z(-Y))\to H^0(Z, \mathscr{O}_Z)\to H^0(Z,i_*\mathscr{O}_Y)\to $$
$$\to H^1(Z, \mathscr{O}_Z(-Y))\to H^1(Z, \mathscr{O}_Z)\to H^1(Z, i_*\mathscr{O}_Y)\to $$
$$\to H^2(Z, \mathscr{O}_Z(-Y))\to H^2(Z, \mathscr{O}_Z)\to H^2(Z, i_*\mathscr{O}_Y)\to 0$$
Now, if $Y$ is bi-degree of the type $a,b$, with $a,b>0$, we have by our earlier arguments
$$0\to H^0(Z,\mathscr{O}_Z(-a,-b))\to k\to H^0(Z, i_*\mathscr{O}_Y)\to H^1(Z, \mathscr{O}_Z(-a,-b))\to 0$$
$$H^1(Z, i_*\mathscr{O}_Y)\cong H^2(Z, \mathscr{O}_Z(-a,-b))$$
And this yields $H^0(Z, i_*\mathscr{O}_Y)=k, h^1(Z, i_*\mathscr{O}_Z)=(a-1)(b-1)$. However, since $i$ is a closed immersion, we have $H^*(Z, i_*\mathscr{O}_Y)\cong H^*(Y, \mathscr{O}_Y)$. And this implies $H^0(Y, \mathscr{O}_Y)=k$, i.e. $Y$ is connected, and $h^1(Y,\mathscr{O}_Y)=(a-1)(b-1)$, i.e. the genus of $Y$ is $(a-1)(b-1)$. So, if we choose $a=b=2$, $Y$ is of genus 1, and it is an elliptic curve!
And everything will be done if we point out the existence of smooth irreducible curve of type $(a,b)$ in $\mathbb{P}_1\times \mathbb{P}_1$. This can be done by Bertini's theorem and Serge's embedding.
Theorem 47 [Bertini's theorem]. Let $X$ be a non-singular variety on $\mathbb{P}^n$, then there exists a hyperplane $H$ in $\mathbb{P}^n$, not containing $X$, such that $H\cap X$ is a regular scheme.
Making use of this theorem, we can embed $Z$ into $\mathbb{P}^n$, where $n=ab+a+b$ as follows. First, we embedded $\mathbb{P}^1$ into $\mathbb{P}^a$ by sending $(x_0:x_1)$ to $(x_0^a:x_0^{a-1}x_1:...:x_1^a)$, and $\mathbb{P}^1$ into $\mathbb{P}^b$ by sending $(y_0:y_1)$ to $(x_0^b:x_0^{b-1}x_1...:x_1^b)$. And we then embed $\mathbb{P}^a\times \mathbb{P}^b$ into $\mathbb{P}^m$ via Serge's embedding. This sends $((s_0:...:s_a),(t_0:...:t_b))$ to $(...:s_it_j:...)$. Now using Bertini's theorem, let $H$ be the hyperplane in the theorem above, $H\cap \mathbb{P}_1\times \mathbb{P}_1$ is a curve in $\mathbb{P}^n$ defined by a bi-degree equation in terms of $x_0,x_1, y_0, y_1$ as $\sum_{i,j}c_{i,j}x_0^ix_1^{a-i}y_0^iy_1^{b-j}$. And due to the Bertini's theorem, it is regular, i.e. the local ring at every point is the regular local ring, this yields our curve is actually irreducible, and smooth.
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