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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 metric on $L^k$ to define a Hodge star operator on sections of $L^k \otimes \Omega^{0,q}$. 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: $$\theta(L^k, q, t) = \sum e^{-\lambda^2 t} \dim{\LARGE (}\lambda\mathrm{-eigenspace \ of\ } C^{\infty}(L^k \otimes \Omega^{0,q}) {\LARGE )}.$$ 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 $e^{-\lambda^2 t}$ 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.

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.

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 $e^{-\lambda^2 t}$ 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 $\Theta(t) = a_{d} t^{-d/2}+a_{d-1} t^{-(d-1)/2} + \cdots + a_0 + O(t^{1/2})$. Plugging into $(\S)$ and comparing constant terms, one gets a formula for $h(k)$.

This computation is really complicated, involving integrals of products of chern classes of $M$ with the powers of the Fubini-Study form. You are only interested in leading behavior for $k$ large, and you want to make the transition to thinking about $\Delta$ rather than $\Delta_{L^k}$.

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 $e^{-\lambda^2 t}$ 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.

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 metric on $L^k$ to define a Hodge star operator on sections of $L^k \otimes \Omega^{0,q}$. 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: $$\theta(L^k, q, t) = \sum e^{-\lambda^2 t} \dim{\LARGE (}\lambda\mathrm{-eigenspace \ of\ } C^{\infty}(L^k \otimes \Omega^{0,q}) {\LARGE )}.$$ 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 $e^{-\lambda^2 t}$ 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 $\Theta(t) = a_{d} t^{-d/2}+a_{d-1} t^{-(d-1)/2} + \cdots + a_0 + O(t^{1/2})$. Plugging into $(\S)$ and comparing constant terms, one gets a formula for $h(k)$.

This computation is really complicated, involving integrals of products of chern classes of $M$ with the powers of the Fubini-Study form. You are only interested in leading behavior for $k$ large, and you want to make the transition to thinking about $\Delta$ rather than $\Delta_{L^k}$.

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 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.

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 metric on $L^k$ to define a Hodge star operator on sections of $L^k \otimes \Omega^{0,q}$. 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: $$\theta(L^k, q, t) = \sum e^{-\lambda^2 t} \dim{\LARGE (}\lambda\mathrm{-eigenspace \ of\ } C^{\infty}(L^k \otimes \Omega^{0,q}) {\LARGE )}.$$ 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 $e^{-\lambda^2 t}$ 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 $\Theta(t) = a_{d} t^{-d/2}+a_{d-1} t^{-(d-1)/2} + \cdots + a_0 + O(t^{1/2})$. Plugging into $(\S)$ and comparing constant terms, one gets a formula for $h(k)$.

This computation is really complicated, involving integrals of products of chern classes of $M$ with the powers of the Fubini-Study form. You are only interested in leading behavior for $k$ large, and you want to make the transition to thinking about $\Delta$ rather than $\Delta_{L^k}$.

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.