# Effect of tensor product on euler characteristic of line bundles

Suppose X is a curve.

Under sufficiently nice conditions we have that every line bundle on X corresponds to an equivalence class of divisors modulo principal divisors, with tensor product of bundles corresponding to addition of divisors.

Given a line bundle L on X, I will call the Euler Characteristic of L minus the Euler characteristic of the trivial line bundle the cohomological degree of L. By the first half of the Riemann Roch theorem, we have that this is equal to the degree of the equivalence class of divisors corresponding to L. Thus cohomological degree is a homomorphism from the Picard group of X to Z.

Is there a more direct proof of the fact that cohomological degree is a homomorphism, that does not go through divisors and the Riemann-Roch theorem? Hopefully this is just a matter of homological algebra.

Is there a similar, cohomological definition of degree that works in higher dimensions?

Thank you.

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## 2 Answers

I think there is a complete answer to your question in SGA 6 Exp X $\S$ 5 : for a quasi-compact scheme there is a natural isomorphism (first Chern class) ${\rm Pic} X \simeq Gr^1(X)$ where $Gr^1(X)$ is defined by the filtration of the Grothendieck group $K_.(X)$ of coherent sheaves on $X$ given by the dimension of supports. It is defined by sending the class of an invertible sheaf $\mathcal L$ to the class of $\mathcal O_X$ minus the class of $\mathcal L^{-1}$. For curves $Gr^1(X)=Filt^1(X)$, so that you have an homomorphism ${\rm Pic} X \simeq Gr^1(X)=Filt^1(X)\subset K_.(X)$, that you can compose with the Euler characteristic $K_.(X)\rightarrow \mathbb Z$. In fact the situation is even nicer for curves : you have a natural isomorphism $K_.(X)\simeq \mathbb Z \oplus {\rm Pic} X$ given by rank and determinant. Another classical reference is Yu. I. Manin
Lectures on the K-functor in algebraic geometry.

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I am not really sure I have understood your question. First of all I guess you suppose your curve $C$ is smooth, compact, connected and over the field of complex numbers, right?

In this case, consider the exponential short exact sequence $$0\to \mathbb Z\to\mathcal O_C\to\mathcal O_C^*\to 0.$$ A piece of derived long exact sequence in cohomology is thus $$\operatorname{Pic}(C)\simeq H^1(C,\mathcal O_C^*)\overset{c_1}\to H^2(C,\mathbb Z)\simeq\mathbb Z.$$ Now, $H^1(C,\mathcal O_C^*)$ is the group of isomorphism classes of line bundles over $C$, that is the Picard group of $C$, and $c_1$ is the connection homomorphism which assigns to an isomorphism class of line bundles its first Chern class in $H^2(C,\mathbb Z)$.

Since the curve is (smooth) compact and connected, the latter group is canonically isomorphic to $\mathbb Z$ (the isomorphism been given by integrating over $C$) and under this identification $c_1(\bullet)$ is nothing else than the degree homomorphism.

Hope this will answer to your question.

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