MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top


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?

share|cite|improve this question
up vote 13 down vote accepted

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.

share|cite|improve this answer
@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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.