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Jul 27, 2022 at 11:34 comment added Avi Steiner @JasonStarr Do you mean the paper "Riemann-Roch type inequalities" by Koll'ar and Matsutaka? From taking a quick glance at the paper, it seems to only be giving inequalities about the dimensions of $H^0(X, \mathcal{L}^{\otimes n})$, hence not the Hilbert polynomial. Can you point me to which part of the paper you were referring to?
Jul 27, 2022 at 0:01 comment added Jason Starr Those two terms in the Hilbert polynomial (the highest degree term together with the next highest degree term) also arise in several invariants considered in algebraic geometry: the "alpha invariant" and the Futaki invariant, for instance.
Jul 26, 2022 at 22:35 comment added Avi Steiner @jasonstarr morally, the Hilbert polynomial thing is where this quantity came from
Jul 26, 2022 at 15:31 comment added Jason Starr @SamHopkins That is true, but usually those terms are only defined if the input weights are roots (they certainly could be defined for other weights, but they usually only come up for roots).
Jul 26, 2022 at 15:25 comment added Jason Starr It is related to Hilbert polynomial for the dimensions $H^0(X,\mathcal{L}^{\otimes n})$ as a polynomial in $n$. The top degree term in the polynomial is a homogeneous polynomial in $n$. If you divide through by this homogeneous polynomial, you get something like $1-(2\langle \rho,\mu\rangle/\langle \mu,\mu \rangle) (\text{dim}(X)/n)$ plus lower order terms. This follows from the Koll'ar -- Matsusaka version of "asymptotic Riemann-Roch".
Jul 26, 2022 at 15:24 comment added Sam Hopkins In the usual root system notation (following e.g. en.wikipedia.org/wiki/Root_system#Definition), where $(\cdot, \cdot)$ is the Killing form, the quantity you are talking about would usually be denoted $\langle \rho, \mu \rangle$. Unless I'm confusing something...
Jul 26, 2022 at 14:34 history asked Avi Steiner CC BY-SA 4.0