I am working out example 2.4.13 in Lazarsfeld's "Positivity in Algebraic Goemetry I" and I got a conflicting answer with his. I am wondering if I made a mistake or if it's a typo in the book.

Let $\mathfrak{m} \subset \mathcal{O}$ be the ideal sheaf of a point in some projective space $\mathbf{P} = \mathbf{P}^n$, and $0 < c < 1$ a real number. We denote $\Gamma_\bullet(\mathbf{P}, \mathcal{O}(1) \otimes \mathfrak{m}^c )$ to be the graded linear series $$(\Gamma_\bullet(\mathbf{P}^b, \mathcal{O}(1) \otimes \mathfrak{m}^c ))_m := H^0(\mathbf{P}, \mathcal{O}(m) \otimes \mathfrak{m}^{\lceil cm\rceil }).$$ For a greaded linear series $|V_\bullet|$ we define its volume to be $\limsup_{m \to \infty} (n!\dim(V_m))/m^n$.

It is claimed in the book that the volume of $\Gamma_\bullet(\mathbf{P}, \mathcal{O}(1) \otimes \mathfrak{m}^c)$ is $1 - c$ but my calculation yielded $1 - c^n$. I am wondering where/if I went wrong.

Thanks! I'll write my calculation below.

We note that $\mathcal{O}(m) \otimes \mathfrak{m}^{\lceil cm\rceil }$ fits into the short exact sequence below, which remains exact when we take global sections.

$$0 \to \mathcal{O}(m) \otimes \mathfrak{m}^{\lceil cm\rceil } \to \mathcal{O}(m) \to \mathbf{C}[x_1, \dots, x_n]/((x_1, \dots, x_n)^{\lceil mc \rceil}) \to0$$ The sheaf on the right is a skyskraper sheaf at $x$, with local coordinates $x_1, \dots, x_n$. Now homogenizing polynomials allows us to identify $\mathbf{C}[x_1, \dots, x_n]/((x_1, \dots, x_n)^{\lceil mc \rceil}) $ with homogeneous polynomials in $\mathbf{C}[x_0, \dots, x_n]$ of degree exactly $\lceil mc \rceil - 1$. Hence, we get that $$h^0(\mathcal{O}(m) \otimes \mathfrak{m}^{\lceil mc \rceil}) = P(m) - P(\lceil mc \rceil - 1)$$ where $P$ is the hilbert polynomial of $\mathbf{P}^n$. Now, the growth of this with respect to $m$ is determined by the highest order terms, so $$\text{vol}(\Gamma_\bullet(\mathbf{P}^b, \mathcal{O}(1) \otimes \mathfrak{m}^c )) = \limsup_{m \to \infty} \frac{n!}{m^n} (m^n/n! - (\lceil mc \rceil - 1)^n/n!) = \limsup (1 - (\lceil mc \rceil - 1)^n/m^n).$$

But now, the last expression is a limit, and equal to $$1 - (\lim_{m \to \infty} \frac{\lceil mc \rceil - 1}{m})^n = 1 - c^n.$$

Edit: I wanted to clean up what was written above, and to include an additional simple explanation that confirms what we already know.

For rational $c$, global sections considerations allows us to identify the volume of $\Gamma_\bullet(\mathbf{P}, \mathcal{O}(1) \otimes \mathfrak{m}^c)$ with the volume of $H - cE$ in $N^1(\operatorname{Bl}_x\mathbf{P})_{\mathbf{R}}$, where $H$ is the pullback of a hyperplane. Since this is an ample $\mathbf{Q}$-divisor, its volume is $(H - cE)^n = 1 - c^n$. The continuity of volume and the intersection product allows us to extend this to any real $0 < c < 1$, which proves the result.


1 Answer 1


Let's double check a concrete example first to see what happens. The easiest one where we'll find a concrete difference between $1-c$ and $1-c^n$ is $n=2$ and $c=\frac12$. Let's take our point to be $[0:0:1]$, so $s\in H^0(\mathcal{O}(m)\otimes\mathfrak{m}^{\lceil cm\rceil})$ means that it is homogeneous of degree $m$, and if $x^py^qz^r$ is a monomial appearing in $s$ with nonzero coefficient, $p+q \geq \lceil cm \rceil$.

For $m=2k+1$ odd, we get $\lceil cm\rceil = k+1$, so the vector space we're looking at is $H^0(\mathcal{O}(2k+1)\otimes\mathfrak{m}^{k+1})$, which has a basis $x^py^qz^r$ with $p+q+r=2k+1$ and $p+q \geq k+1$ and therefore dimension $$\sum_{i=k+1}^{i=2k+1} i = \frac12 (k+1)(3k+2).$$ Then $$\limsup_{m \to \infty} (n!\dim(V_m))/m^n = \lim_{k\to\infty} \frac{(k+1)(3k+2)}{(2k+1)^2} = \frac34 = 1-c^2.$$ This is already enough to confirm your calculations, assuming I didn't make any mistakes.

Another (more conceptual) reason it has to be $1-c^n$ is that under the same choice of point $[0:\cdots:0:1]$, the standard basis monomials belong to $H^0(\mathcal{O}(m)\otimes\mathfrak{m}^{\lceil cm\rceil})$ iff the coefficients on $x_0,\cdots,x_{n-1}$ sum to at least $\lceil cm\rceil$, which means treating the exponents as lattice points in the $n$-simplex in $\Bbb R^{n+1}$ given by $t_i\geq 0$, $\sum t_i=m$, they're not in the similar $n$-simplex with side $m-\lceil cm\rceil$. The probability this happens is the same as the volume defined in your post, which clearly depends on $n$, because you can write it in terms of the volumes of these simplices which again depend on $n$. This is pretty much equivalent to what you wrote in your post, but perhaps restating it in these terms will be helpful too.

  • $\begingroup$ Your calculation makes sense to me; thanks! $\endgroup$
    – Daniel
    Sep 7, 2023 at 22:43

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