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Let $X$ be a projective scheme over a field $k$. Let $\mathcal{O}(1)$ be an ample line bundle on $X$, then the Hilbert polynomial $P(E)$ is given by $m\mapsto\chi(E ⊗ O(m))$. The explicit polynomial form is given by the following result

$\textbf{Lemma}$: Let $E$ be a coherent sheaf of dimension $d$ and let $H_1 , . . . , H_d \in |\mathcal{O}(1)|$ be an $E$-regular sequence. Then $P(E, m) = \chi(E ⊗ \mathcal{O}(m)) =\Sigma_{i=0}^{d}\chi(E|_{\cap_{j\leq i }H_j}){m+i-1\choose i}$.

So this polynomial can be uniquely written in the form $\Sigma_{i=0}^d\alpha_i(E)\frac{m^i}{i!}$.

We then define the rank as $rk(E)=\frac{\alpha_d(E)}{\alpha_d(\mathcal{O}_X)}$. But for an integral scheme, the rank is defined to be the rank at the generic point. How are these two notions the same?

Similarly, the degree of $E$ is defined to be $deg(E)=\alpha_{d-1}(E)-rk(E).\alpha_{d-1}(\mathcal{O}_X)$. Again, for a projective variety, the degree is defined to be $c_1(E).H^{d-1}$. How do we show that these two are the same. Any help will be greatly appreciated!

gradstudent
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1 Answers1

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Use the Hirzebruch-Riemann-Roch Theorem. $$ \chi(O(m)) = \int \exp(mH) \cdot td(X) = m^dH^d/d! + m^{d-1}H^{d-1}td_1/(d-1)! + \cdots $$ so we see that $a_d(O) = H^d$ and $a_{d-1}(O) = H^{d-1}td_1$.

Next \begin{align*} \chi(E(m)) &= \int ch(E) \cdot \exp(mH) \cdot td(X) \\ &= rm^dH^d/d! + c_1(E)H^{d-1}m^{d-1}/(d-1)! + rm^{d-1}H^{d-1}td_1/(d-1)! + \cdots \end{align*} so we see that $a_d(E) = rH^d$ and $a_{d-1} = c_1(E)H^{d-1} + r\cdot td_1 H^{d-1}$.

If you take for granted that the rank as you defined is equal to the zeroth chern character, everything you want follows.

user347489
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paragon
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