In this note, we will first prove the theorem of Riemann-Roch for vector bundles, and then give definitions and properties of Chow rings and Chern classes to understand the formulation of the Hirzebruch-Riemann-Roch's theorem. The Grothendieck-Riemann-Roch's theorem will be discussed in our next note.
1. Riemann-Roch's theorem for vector bundles. Let us begin with the classical Riemann-Roch's theorem. If $L$ is any line bundle on a smooth projective curve $X$, then we know that $h^0(L)-h^1(L)=\deg L + (1-g)$. We denote $\chi(X,L)$ the Euler characteristic of $X$ with respect to $L$, then $\chi(X, L)=\deg L +(1-g)$.
Now, if $E$ is any vector bundle on $X$ of rank $n$, then there exists a complete flag
$$0=E_0\subset E_1\subset...\subset E_{n-1}\subset E_n=E$$
such that $E_i$ is vector bundle on $X$ of rank $i$, and $L_i:=E_{i+1}/E_i$ is a line bundle. It can be seen as follows. If we take the dual bundle of $E$, say $E^\vee$, then it is sufficient for us to find a divisor $D$ such that $E^\vee(D):=E^\vee\otimes \mathscr{O}(D)$ has a non-zero section, because in this case we will have the non-zero sheaf map from $\mathscr{O}_X $ to $E^\vee(D)$ and therefore a non-zero sheaf map from $\mathscr{O}_X(-D)$ to $E^\vee$, and the image will be a line bundle. And if we have a subbundle $L$ of rank 1 of $E^\vee$. We can actually change $L$ by a line bundle $L'$ containing $L$ such that the quotient $E^\vee/L'$ is again a vector bundle (because if $F:=E^\vee/L$ has the torsion part $T$, we can just divide it, and take the kernel $L'$ of the composition: $E^\vee \to F \to F/T$, $L'$ now is a vector bundle, since it is a subbundle of $E^\vee$ and of rank 1 due to the snake lemma). By dualizing the exact sequence $0 \to L \to E^\vee \to F \to 0$, we will get $F^\vee$ as the subbundle of $E$ with the quotient $E/F^\vee$ is a line bundle. So, using induction, we will get the result.
Now, we want to find $D$ such that $E^\vee(D)$ has a non-zero section. From the short exact sequence for sky-scrapper sheaf $0\to \mathscr{O}\to \mathscr{O}(D)\to k_D\to 0$, we can tensor with $E^\vee$ and get $0 \to E^\vee \to E^\vee(D) \to E^\vee \otimes k_D \to 0$, because vector bundles are flat. This induces the long exact sequence:
$$0 \to H^0(X, E^\vee) \to H^0(X, E^\vee(D)) \to H^0(E^\vee \otimes k_D) \to H^1(E^\vee) ...$$
We now can choose $D$ such that $h^0(X, E^\vee(D))=\deg D.rank(E)>h^1(X,E^\vee(D))$ so that the last map above is not injective, and $h^0(X,E^\vee(D))>0$.
So, in particular, if we have the complete flag for $E$ as above, this yields the following short exact sequence
$$0\to E_i \to E_{i+1}\to E_{i+1}/E_i\to 0$$
This induces the following long exact sequence
$$0\to H^0(X, E_i)\to H^0(X, E_{i+1})\to H^0(X, L_i)\to H^1(X, E_i)\to H^1(X, E_{i+1})\to H^1(X,L_i)\to 0$$
Note that at this point we use the theorem: if $X$ is any projective variety of dimension $n$, and $F$ is any quasi-cohenrent sheaf on $X$ then $H^{m}(X,F)=0$, for all $m>n$. By using basic linear algebra, if we denote $\chi(X, E_i):=h^0(X,E_i)-h^1(X,E_i)$, then this follows
$$\chi(X,E_{i+1})=\chi(X, E_i)+\chi(X, L_i)=\chi(X,E_i)+\deg L_i+(1-g)$$
From this, by induction, we obtain
$$\chi(X,E)=\sum_{i=0}^{n-1}\deg L_i + n(1-g)=\sum_{i=0}^{n-1}\deg L_i + rank(E)(1-g)$$
One can see from this that the change of complete flag does not change the sum $\sum_{i=0}^{n-1}\deg L_i$. And we can define $\deg E= \sum_{i=0}^{n-1}\deg L_i$. So the Riemann-Roch's theorem for vector bundles can be stated
$$\chi(X, E)=\deg E+(1-g)rank(E)$$
As a natural question, one wants to compute the Euler's characteristic of any smooth projective variety, which is done by the formulation of the Hirzebruch-Riemann-Roch theorem.
Theorem 1.1. Let $X$ be any smooth projective variety, and $E$ is any line bundle on $X$. We define $\chi(X,E):=\sum_{i\ge 0}(-1)^ih^i(X,E)$. Then $\chi(X,E)=\int_Xch(E)Td(X)$.
The remaining of this note is devoted to understand the formulation of the theorem above. There are three main ingredients for this: Chow rings, Chern classes and Todd genus.
2. Chow's ring and proper morphism.
Let $X$ be a smooth projective variety of dimension $n$, $Z_k(X)$ is defined as a free abelian group generated by $[V]$, where $V$ is a $k$-dimensional subvariety of $X$. A sum $\sum_{V\in Z_k(X)}n_V[V]$ with $n_V\in \mathbb{Z}$, with finitely many $n_V\ne 0$ is rational equivalent to 0 if there exists finite number of subvarieties of $X$, say $V_i$, of dimension $k+1$ and $f_i\in k(V_i)^*$ such that $\sum_{V\in Z_k(X)}=\sum_i[div(f_i)]$. It can be seen that the sets of all divisors on $Z_k(X)$ that are rational equivalent to 0 forms a subgroup of $Z_k(X)$, and is denoted $B_k(X)$. The $k$-th Chow's group $A_k(X)$ is the quotient $Z_k(X)/B_k(X)$. We also denote $A^k(X)$ the Chow's group of codimension $k$, i.e. $A^k(X)=A_{n-k}(X)$.
If $V,W$ are subvarieties of $X$ in general positions of codimension $i,j$, resp., then the intersection $V\cap W$ is a subvarieties of $X$ of codimension $i+j$. Hence, one can see that $[V]\in A^i(X), [W]\in A^j(X)$ defines a class $[V\cap W]\in A^{i+j}(X)$. This can be extended linearly to the map $A^i(X)\times A^j(X)\to A^{i+j}(X)$. If we denote $A(X):=A^0(X)\oplus ... \oplus A^n(X)$, then $A(X)$ is a ring with the multiplication defined as intersection product.
If $f:X\to Y$ is any morphism between smooth projective varieties, then we know that $f$ is a proper morphism. This will induce the map $f_*: A(X)\to A(Y)$ defines as follows. If $V$ is a subvariety of $X$ of dimension $k$, then $W:=f(V)$ is a subvariety of $Y$, and $\dim W\le \dim V$. In the case $\dim V=\dim W$, then the induced map between function fields $f^*: k(W)\to k(V)$ satisfying $[k(V):f^*(k(W))]$ is finite. In this case, we define $f_*([V])=[k(V):f^*(k(W))][W]$. Otherwise, if $\dim W<\dim V$, we define $f_*([V])=0$. The map $f_*$ is called the proper push-forward map, and it is a ring homomorphism between the Chow's rings $A(X)$ and $A(Y)$.
In the case $Y=\{pt\}$ is just a point, then it can be seen that $A(Y)\cong \mathbb{Z}$. Hence, if $D:=\sum_{V\in Z_0(X)}n_V[V]\in A^n(X)$, i.e. $V$ is just a point, we have $f_*(D)=(\sum_{V}n_V)[pt]\in A(Y)$. And for any $D\in A^k(X)$, for $k<n$, we have $f_*(D)=0$. This gives us the well-defined map
$$\int_X: A(X)\to \mathbb{Z}$$
which sends $D=\sum_{V\in Z_0(X)}n_V[V]\in A^n(X)$ to $\sum_{V}n_V$.
3. Projective bundles and splitting principles. Let $E$ be a vector bundle of rank $r\ge 1$ over a variety $X$ with transition functions $\psi_{i,j}$ and trivializations $U_i\times \mathbb{A}^r$. Then we know that each fiber of a point $p$ on $X$ is of the form $\{p\}\times \mathbb{A}^r$, and $(p,x)$ is identified with $(p,\psi_{i,j}x)$ whenever $p\in U_i\cap U_j$.
The projective bundle $\mathbb{P}(X)$ is constructed by gluing $U_i\times \mathbb{P}^{r-1}$ and $U_j\times \mathbb{P}^{r-1}$ via $\psi_{i,j}$, i.e. for any $p\in U_i\cap U_j, x\in \mathbb{P}^{r-1}$, we can identify $(p,x)$ with $(p,\psi_{i,j}x)$. Now, the projection map $\pi: \mathbb{P}(E)\to X$ sends $(p,x)\mapsto p$ is a proper morphism. From this, $\pi^*: A(X)\to A(\mathbb{P}(E))$ is well-defined map and it is injective. Also, $\pi^*E$ is a vector bundle of rank $r$ on $\mathbb{P}(E)$, with transition functions is given by
$$(U_i\cap U_j)\times \mathbb{P}^{r-1}\times \mathbb{A}^{r}\to (U_i\cap U_j)\times \mathbb{P}^{r-1}\times \mathbb{A}^r$$
which sends $(p,x,y)\mapsto (p,\psi_{i,j}x,\psi_{i,j}y)$, where $(x_1:..:x_r)$ are coordinates of $x$, and $(y_1,...,y_r)$ are coordinates of $y$. Let us consider the subbundle of $\pi^*E$ of the form $L:=\{(p,(x_1:...:x_r), (\lambda x_1,...,\lambda x_r))\}$. It can be seen that $L$ is a line bundle on $\mathbb{P}(E)$, which is called tautological line bundle of $\mathbb{P}(E)$. We will prove that there exists a complete flag for $\pi^*E$, i.e. there exists a chain
$$0=E_0\subset E_1\subset ...\subset E_{r-1}\subset E_r=\pi^*E$$
where $E_i$ are vector bundles on $\mathbb{P}(E)$ of rank $i$, and the quotient $E_i/E_{i-1}$ are line bundles. By induction, if $rank(E)=1$, then there is nothing to prove. Otherwise, we let $L^\vee$ be the tautological line bundle of $\mathbb{P}(E^\vee)$, with the quotient $F^\vee:=\pi^*E^\vee/L^\vee$. This yields the following short exact sequence
$$0\to L^\vee \to \pi^*E^\vee \to F^\vee\to 0$$
This will induce the dual exact sequence
$$0\to F \to \pi^*E \to L\to 0$$
So, $F$ is a subbundle of $E$ of rank $r-1$ and the quotient $E/F$ is a line bundle. Applying the induction hypothesis to $F$, we will obtain the splitting chain for $\pi^*E$.
3. Chern classes and Todd genus. Let $F(X)$ be the free abelian group generated by the isomorphism class $[E]$ of vector bundles on $X$, and $Q(X)$ the subgroup of $F(X)$ generated by elements of the form $[E]-[E']-[E'']$, where $E, E',E''$ satisfy the short exact sequence
$$0\to E'\to E\to E'' \to 0$$
Then the Grothendieck's group of vector bundle is defined $K^0(X):=F(X)/Q(X)$. The tensor product with respect to $\mathscr{O}_X$ defines a ring structure on $K^0(X)$. The construction of Chern classes is the following
Theorem 3.1. For each vector bundle $E$ on $X$, there exists a elements $c_i(E)\in A^i(X)$, which is called the $i$-the Chern class with the following properties:
(C0) $c_0(E)=1$.
(C1) $c_1(E) = [E]\in A^1(X)$ if $rank(E)=1$.
(C2) For any morphism of smooth projective variety $f: X\to Y$, $f^*(c_i(E))=c_i(f^*E)$.
Let us denote the Chern's polynomial $c_t(E):=\sum_{i\ge 0}c_i(E)t^i$, then
(C3) If $0\to E'\to E\to E''\to 0$ is a short exact sequence of vector bundles on $X$, then $c_t(E)=c_t(E')c_t(E'')$.
(C4) $c_i(E)=0$ for all $i>rank(E)$.
One can see that Chern classes of vector bundles are determined uniquely by Chern's polynomial. Now, how can we construct Chern's polynomial of a given vector bundle on a smooth projective variety? This can be done by using the splitting principle and the properties of Chern classes above. Let $X$ be any smooth projective variety and $E$ is a vector bundle on $X$ of rank $r$, then it follows from Fulton [...] that there exists a smooth quasi-projective variety $X'$ with the map $\pi: X'\to X$ such that $\pi^*: A(X)\to A(X')$ is injective and $E':=\pi^*E$ splits (i.e. there exists a complete flag $0=E'_0\subset E'_1\subset ...\subset E'_r=E'$ such that $E'_i$ is a vector bundle of rank $i$, and $L_{i+1}:=E'_{i+1}/E'_i$ is a line bundle). Due to Section 3, we can think of $X'$ as $\mathbb{P}(E)$.
By using this, we have
$$\pi^*(c_t(E))=c_t(\pi^*E)=\prod_{i=1}^{r}c_t(L_i)=\prod_{i=1}^{r}(1+c_1(L_i)t)=\prod_{i=1}^r(1+\alpha_it)$$
where $\alpha_i:=c_1(L_i)$ is called Chern's roots of $E$. Because $\pi^*$ is injective, one can see $c_t(E)$ is uniquely determined by $\alpha_i$. Thus, we can write $c_t(E)=\prod_{i=1}^r(1+\alpha_it)$. And in particular, one can write: $c_1(E)=\sum_{i=1}^r\alpha_i$, $c_2(E)=\sum_{i< j}\alpha_i\alpha_j,...,c_r(E)=\prod_{i=1}^r\alpha_i$. We will use the splitting construction to compute the Chern roots of dual bundles.
Example 3.2. First, if a vector bundle $E$ admits the splitting principles
$$0=E_0\subset E_1\subset ... \subset E_{n-1}\subset E_r=E$$
where $E_i$ are vector bundles of rank $i$, and $L_i:=E_i/E_{i-1}$ are line bundles. Then $E^\vee$ admits the following chain
$$0=E_0^\vee \subset (E_n/E_{n-1})^\vee \subset ... \subset (E_n/E_1)^\vee\subset E_n^\vee = E^\vee$$
It is because if we have $0\to F\to E\to E/F\to 0$ is a short exact sequence of vector bundles then $0\to (E/F)^\vee \to E^\vee\to F ^\vee\to 0$ is the induced exact sequence for dual bundles. From this, we can regard $(E_n/E_{i})^\vee$ as a subbundle of $E_n^\vee$. And also, since $(E_n/E_{i-1})/(E_i/E_{i-1})\cong E_n/E_{i}$, we have
$$0\to E_i/E_{i-1}\to E_n/E_{i-1} \to E_n/E_i\to 0$$
is exact. This induces the following short exact sequence
$$0\to (E_n/E_i)^\vee \to (E_n/E_{i-1})^\vee\to (E_i/E_{i-1})^\vee\to 0$$
That means, we can consider $(E_n/E_i)^\vee$ the subbundle of $(E_n/E_{i-1})^\vee$. And hence, the chain for $E^\vee$ is well-defined. Also, one has $L_i^\vee=(E_i/E_{i-1})^\vee \cong (E_n/E_{i-1})^\vee/(E_n/E_i)^\vee$. And this will yield the chain for $E^\vee$ defined above also admits the splitting principle. If $L_i$ are the line bundles with respect to the splitting of $E$, then one can see $L_i^\vee$ are the line bundles with respect to the splitting of $E^\vee$. And hence, one has
$$c_t(E^\vee)=\prod_{i=1}^r (1+c_1(L_i^\vee))$$
But then, because $L_i\otimes_{\mathscr{O}_X} L_i^\vee \cong \mathscr{O}_X$, we have
$$1 + (c_1(L_i)+c_1(L_i^\vee))t=c_t(L_i\otimes_{\mathscr{O}_X} L_i^\vee)=c_t(\mathscr{O}_X)=1$$
This yields $c_1(L_i^\vee)=-c_1(L_i)$. From this, we have
$$c_t(E^\vee)=\prod_{i=1}^r(1-\alpha_i t)$$
We are now ready to define the Chern's character and Todd's genus for $E$.
Definition 3.2. Let $E$ be any vector bundle of rank $r$ on a smooth projective variety $X$, and $\alpha_1,...,\alpha_r$ is the Chern roots of $E$, then we define the Chern character of $E$ as
$$ch(E):=\sum_{i=1}^n\exp(\alpha_i)=r+c_1(E)+\frac{1}{2}(c_1(E)^2-2c_2(E))+...$$
Also, the Todd's genus of $E$ is defined as
$$Td(E):=\prod_{i=1}^r \frac{\alpha_i}{1-\exp^{-\alpha_i}}=1+\frac{1}{2}c_1(E)+\frac{1}{12}(c_1^2(E)+c_2(E))+...$$
From the definition, it can be seen that $ch(E), Td(E)$ are well-defined elements of $A(X)\otimes \mathbb{Q}=: A(X)_{\mathbb{Q}}$. We will denote $Td(X)$ is the Todd genus of the tangent sheaf $T_X$ of $X$.
Proposition 3.3. There exists a ring homomorphism $ch: K^0(X)\to A(X)_{\mathbb{Q}}$, which is uniquely determined by the following properties
(1) For any morphism of smooth projective varieties, it holds that $f^*\circ ch=ch\circ f^*$.
(2) For any line bundle $L$ with $c_1(L)=[D]$, it holds that
$$ch(L)=\sum_{i\ge 0}\frac{1}{i!}[D]^i$$
4. The Hirzebruch-Riemann-Roch theorem. We now look back the Hirzebruch-Riemann-Roch's theorem in the first section
Theorem 1.1. Let $X$ be any smooth projective variety, and $E$ is any line bundle on $X$. We define $\chi(X,E):=\sum_{i\ge 0}(-1)^ih^i(X,E)$. Then $\chi(X,E)=\int_Xch(E)td(X)$.
It can be understood as follows. We first assume that $\dim X=n$. It can be seen that
$$ch(E)=(r, c_1(E), \frac{1}{2}(c_1(E)^2-2c_2(E)),...)\in A(X)\otimes \mathbb{Q}$$
$$Td(X)=(1, \frac{1}{2}c_1(T_X), \frac{1}{12}(c_1^2(T_X)+c_2(T_X)),...)\in A(X)\otimes \mathbb{Q}$$
And the $0$-graded part of $ch(E)Td(X)$ is $r.1=r$, the $1$-st graded part of $ch(E)Td(X)$ is $\frac{1}{2}rc_1(T_X)+c_1(E)$ and so on...From this, $\int_X ch(E)Td(X)$ is just the $n$-th graded part of the product $ch(E)Td(X)$, which we know is an integer by our discussion at Section 2. So, the Hirzebruch-Riemann-Roch states that the Euler's characteristic of a smooth projective variety $X$ of dimension $n$ with respect to the lint bundle $E$ is just the $n$-the graded part of $ch(E)Td(X)$. We will use this to obtain the theorem of Riemann-Roch for vector bundles on smooth projective curves.
Due to the splitting principle for curves, if $0=E_0\subset E_1\subset...\subset E_r=E$, where $L_i:=E_i/E_{i-1}$ is a line bundle and $E_i$ is a vector bundle of rank $i$. Then it can be seen that $c_t(E)=\prod_{i=1}^r(1+c_1(L_i)r)$. This yields $ch(E)=(r, c_1(L_1)+...+c_1(L_r))=(r,c_1(E))$.
Also, $Td(X)=(1,\frac{1}{2}c_1(T_X))$. But we know that $T_X$ is the dual bundle of $K_X$-the canonical bundle of $X$. This yields by Example 3.2 that $c_1(T_X)=-c_1(K_X)$. Hence, $Td(X)=(1,-\frac{1}{2}c_1(K_X))$. And one has $ch(E)Td(X)=(r,c_1(E)-\frac{1}{2}rc_1(K_X))$. Due to the Hirzebruch-Riemann-Roch's theorem, we have
$$\chi(X,E)=\int_Xch(E)Td(X)=\deg c_1(E)-\frac{1}{2}r\deg c_1(K_X)=$$
$$=\sum_{i=1}^r\deg L_i+r(1-g)=\deg E+rank(E)(1-g)$$
And the theorem of Riemann-Roch for vector bundles on smooth projective curves now follows. In the next part, we will formulate the more general version of HRR, the Grothendieck-Riemann-Roch's theorem. By genius observation, Grothendieck realizes that HRR is just the case when the "implicit" smooth variety $Y$ is just a point, and in this case, the degree map, i.e. $\int_X$ just counts the highest part in the Chow's ring. He then develops a theory that collects all coefficients in the Chow's ring whenever we have a proper morphism between two smooth quasi-projective varieties.
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