In the case of a curve of genus g, $g$, this is the standard Riemann-Roch theorem, it says \chi(L^k) $\chi(L^k) = k \cdot deg(L) + 1 - gg$. In higher dimensions, this is a result of the more general Grothendieck-Riemann-Roch theorem, though in the case I am about to state, it is commonly called the Hirzebruch-Riemann-Roch theorem. In the case of a line bundle on a n $n$ dimensional projective variety, it says \chi(L^k) $\chi(L^k) = (exp(1 + k \cdot L) \cdot td(X))_ntd(X))_n$. Here td(X) $td(X)$ means the Todd class of the tangent bundle of X $X$ (a fixed cohomology class) and the subscript n $n$ means we take the degree n $n$ piece of the above expression. If you expand this out, you will exactly find a degree n $n$ polynomial in k $k$ (the Hilbert polynomial). A proof can be found, for example, in Fulton's book on Intersection Theory.
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In the case of a curve of genus g, this is the standard Riemann-Roch theorem, it says \chi(L^k) = k \cdot deg(L) + 1 - g. In higher dimensions, this is a result of the more general Grothendieck-Riemann-Roch theorem, though in the case I am about to state, it is commonly called the Hirzebruch-Riemann-Roch theorem. In the case of a line bundle on a n dimensional projective variety, it says \chi(L^k) = (exp(1 + k \cdot L) \cdot td(X))_n. Here td(X) means the Todd class of the tangent bundle of X (a fixed cohomology class) and the subscript n means we take the degree n piece of the above expression. If you expand this out, you will exactly find a degree n polynomial in k (the Hilbert polynomial). A proof can be found, for example, in Fulton's book on Intersection Theory. |
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