# Volume of the graded linear series $\Gamma_\bullet(\mathbf{P}^n, \mathcal{O}(1) \otimes \mathfrak{m}^c)$. Is it $1 - c$ or $1 - c^n$?

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.

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.