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Let $E$ be a coherent sheaf on a scheme $X$. Let $d = \text{dim}X$ be the dimension of $E$.

Huybrechts and Lehn define the degree of $E$ to be: $$ \text{deg} E := \alpha_{d-1}(E) - \text{rk}(E)\cdot\alpha_{d-1}(\mathcal{O}_X) $$ where $\alpha_i$ is the $i$th coefficient of the Hilbert polynomial, and $\text{rk}(E) := \frac{\alpha_d(E)}{\alpha_d(\mathcal{O}_X)}$. Then they say that it follows from the Hirzebruch-Riemann-Roch formula that on a smooth projective variety this definition gives $\text{deg}(E) = c_1(E)\cdot H^{d-1}$ where $H$ is an ample divisor, and in particular that $\text{deg}(E) = \text{deg}(\text{det}(E))$. It is not clear to me why this follows from the HRR. Could you help me clear it up? Thanks.

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1 Answer 1

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The Hilbert polynomial of $E$ is some Euler characteristic of the sheaf with the tensor product of some power of the twisting sheaf $\mathcal{O}(1)$. But HRR just gives you this Euler characteristic in terms of the Chern classes.

See here Lemma 1.20. for a proof, if you get lost in the computations.

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