# Weyl's law on asymptotic of Laplacian vs Hilbert's theorem on degree of a projective variety

Hi,

Let $M$ be a compact Riemannian manifold of dimension $n$. Define the integer-valued function $N(k)$ to be the number of eigenvalues of the Laplacian on $M$ which are less than or equal to $k$. Weyl's law states that the function $N(k)$ has asymptotic: $$N(k) = C_n Vol(M) k^n + O(k^{n-1}),$$ for some explicit constant C_n depending only on n. Now suppose $M$ is a smooth complex projective subvariety of complex dimension d in ${\mathbb C}P^r$ equipped with the Fubini-Study metric from projective space. Hilbert's theorem states that the Hilbert function $H(k)$ of $M$ has asymptotic $$H(k) = (1/d!)Vol(M) k^d + O(k^{d-1}),$$ where $Vol(M)$ is the volume of $M$ with respect to Fubini-Study metric which in turn is equal to degree of M as a subvariety of projective space. Noting that $2d$ is real dimension of $M$, the two functions $H(k^2)$ and $N(k)$ have very similar asymptotic (given by dimension and volume).

Question: for smooth projective subvarieties of projective space with Fubini-Study metric, is there a connection between Hilbert function and number of eigenvalues of Laplacian? or the similarity in asymptotic is a coincidence?

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There should be a relationship like this, because the heat function proof of Hirzebruch-Riemmann-Roch uses a much more refined equality between hilbert functions and Laplacian spectra. Let me see if I can put this together.

Let $M$ be a $d$-dimensional smooth projective variety. Let $L^k$ be the line bundle on $M$ obtained by restricting $\mathcal{O}(k)$ from projective space. Equip $M$ with the Fubini-Study metric. We have the complex of vector spaces:

$$0 \to C^{\infty}(L^k) \overset{\bar{\partial}}{\longrightarrow} C^{\infty}(L^k \otimes \Omega^{0,1}) \overset{\bar{\partial}}{\longrightarrow} \cdots \overset{\bar{\partial}}{\longrightarrow} C^{\infty}(L^k \otimes \Omega^{0,d}) \to 0. \quad (\dagger)$$

Here $\Omega^{0,q}$ is the vector bundle of $(0,q)$-differential forms and $C^{\infty}(\mathrm{vector \ bundle})$ means $C^{\infty}$ sections of that vector bundle. We can use the two metrics on $L^k$ and on $X$ to define an inner products on $C^{\infty}(L^k \otimes \Omega^{0,q})$. Let $\bar{\partial}^{\ast}$ be the adjoint to $\bar{\partial}$ and let $\Delta_{L^k} = \bar{\partial} \bar{\partial}^{\ast} + \bar{\partial}^{\ast} \bar{\partial}$. This is not the ordinary Laplacian -- it acts on sections of $L^k$ tensored with differential forms rather than acting on differential forms. All of this is pretty standard : See Wells Differential Geometry on Complex Manifolds or Voisin's Complex Algebraic Geometry and Hodge Theory for background on this kind of construction.

Since $\bar{\partial}$ commutes with $\Delta_{L^k}$, the sequence $(\dagger)$ splits up into finite dimensional sequences for every eigenspace of $\Delta_{L^k}$. (There is some deep analysis with Sobolev spaces necessary to make this precise.) Let $(\dagger)_{\lambda}$ be the corresponding sequence of $\lambda$ eigensequences. Since $\Delta_{L^k}$ is positive semidefinite, you only get terms with $\lambda \geq 0$.

For $\lambda >0$, we have the homotopy $\mathrm{Id} = (1/\lambda) (\bar{\partial} \bar{\partial}^{\ast} + \bar{\partial}^{\ast} \bar{\partial})$, so $(\dagger)_{\lambda}$ is exact for $\lambda>0$. On $(\dagger)_0$, we have $\bar{\partial}=0$ so the sequence is trivial. For $k$ large, it turns out that only the first term has a nontrivial $0$-eigenspace. That eigenspace is the kernel of $\bar{\partial}$ acting on $C^{\infty}(L^k)$, which is to say, the holomorphic sections of $L^k$. So its dimension is the Hilbert function. More generally, for all $k$, the alternating sum of the dimensions of the $0$-eigenspaces is the Hilbert polynomial.

Gather up the eigenvalues into a generating function: $$\theta(L^k, q, t) = \sum e^{-\lambda t} \dim{\LARGE (}\lambda\mathrm{-eigenspace \ of\ } C^{\infty}(L^k \otimes \Omega^{0,q}) {\LARGE )}.$$ Then the above argument shows that $$h(k) = \sum_q (-1)^q \theta(L^k, q, t). \quad (\S)$$ Note that $(\S)$ holds for all $t$.

Now, in the height function proof, one engages in a detailed analysis of the asymptopics of $\theta(L^k, q, t)$ as $t \to 0$. When $t \to 0$, all of the $e^{-\lambda t}$ terms go to $1$, so $\theta$ blows up, and the rate at which it blows up depends on the growth rate of the eigenvalues of $\Delta_{L^k}$. One gets an asymptopic formula that looks like $\Theta(t) = a_{2d} t^{-d}+a_{2d-1} t^{-d+1/2} + \cdots + a_0 + O(t^{1/2})$. Plugging into $(\S)$ and comparing constant terms, one gets the following formula, which is Hirzebruch-Riemann-Roch slightly specialized to our setting: $$h(k) = \int_M \frac{ e^{k \omega} \prod \alpha_i}{\prod (1-e^{- \alpha_i})}.$$ Here $\omega$ is the Fubini-Study form, $\alpha_i$ are the Chern roots of $T^{\ast} M$ and we are using the convention that you ignore all terms not in top degree when integrating. You are interested only in the leading power of $k$. We have $e^{k \omega} = 1+k \omega + k \omega^2/2+ \cdots + k^d \omega^d/d!$, where the sum stops because we have reached top degree, so the leading term is $\frac{k^d}{d!} \int_M \omega^d = \frac{k^d}{d!} \mathrm{Vol}(M)$, which is your left hand side.

Is there a way to get the large $k$ behavior of the $\theta$'s without getting the exact result? I don't know. Also, you would want to switch from $\Delta_{L^k}$ to $\Delta$ at some point.

I haven't seen anyone work this out, but I do remember there is a lemma which relates $\Delta$ and $\Delta_{L^k}$ for $k$ large and that Griffiths and Harris use it in their proof of Serre vanishing. I covered this proof in my Hodge theory course, so I am a little embarrassed not to remember the details! It should be in the notes for April 12. Anyway, possibly you can figure out how to finish the proof from here.

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@david :You should look at Demailly's asymptotic Morse Inequalities which is a generalization of these ideas. –  Mohan Ramachandran Feb 18 '12 at 2:35
Sounds to me like Kiumars should look at Demailly's book :). But thanks for the reference! –  David Speyer Feb 18 '12 at 8:17
Thanks David and Mohan. –  Kiu Feb 19 '12 at 21:02