# Hodge Symmetry in characteristic Zero

In this post we prove that varieties over characteristic $0$ fields are all ‘Hodge symmetric’, meaning that $h^{p,q}=h^{q,p}$.

# Motivation

We call a smooth, projective, integral variety $X/K$ Hodge symmetric (or say that it satisfies Hodge symmetry), just in case $h^{p,q}=h^{q,p}$. Recall the definition of the Hodge numbers $h^{p,q}:= \dim_k H^p(X,\Omega^q_{X/k})$. The notion of Hodge symmetry is most likely first encountered in the context of complex manifolds $M$, where a similar form of Hodge symmetry holds:

$\dim_\mathbb{C} H^p(M,\Omega^q_\text{hol})=\dim_\mathbb{C} H^q(M,\Omega^p_\text{hol})$.

It turns out that this actually holds, more generally, over any field of characteristic $0$, and in fact, the complex geometry version is the deciding reason why it does hold. The key is that, by using general functoriality, and a healthy dose of GAGA, we can reduce this to the complex analytic case, a technique which should be widely applicable.

Remark: I would like to say that this is the only proof of ‘Hodge symmetry holds in characteristic $0$‘ (a widely made claim) that I can think of. That said, I have a sneaking suspicion that there is a more elementary way to attack this problem. If you can point out the silly argument that I am missing, please do!

# Reduction to complex case

Let’s rigorously state our theorem:

Theorem(Main): Let $K$ be a characteristic $0$ field. Then, any smooth, projective, integral variety $X/K$ satisfies Hodge symmetry.

The somewhat non-obvious part to the proof of this, is that we can reduce the proof to the case when $K=\mathbb{C}$. More specifically:

Theorem(Reduction to $\mathbb{C}$): The main theorem is true if and only if it holds for $K=\mathbb{C}$.

Lemma: Let $L$ and $K$ be two algebraically closed fields of the same characteristic, and the same uncountable cardinality. Then, $L\cong K$.

Proof: Let the prime subfields of $L$ and $K$ be $k$. Take transcendence bases $\{x_\alpha\}_{\alpha\in\mathcal{A}}$ for $L/k$ and $\{y_\beta\}_{\beta\in\mathcal{B}}$ for $K/k$. Note that since $L/k(\{x_\alpha\})$ is algebraic, and $L$ is algebraically closed, it must be the case that $L\cong \overline{k(\{x_\alpha\}}$. Similarly, $K\cong\overline{k(\{y_\beta\})}$. Now, since $k$ is countable, and $L$ and $K$ uncountable, a simple cardinality argument shows that $\mathcal{A}$ and $\mathcal{B}$ must be equipotent. So, $k(\{x_\alpha\})\cong k(\{y_\beta\})$, and thus

$L\cong\overline{k(\{x_\alpha\})}\cong\overline{k(\{y_\beta\})}=K$

as desired. $\blacksquare$

Proof of Reduction to $\mathbb{C}$: Let $K$ be an arbitrary characteristic $0$ field, and let $X/K$ be a smooth, projective, integral variety. Choose an embedding $i:X\hookrightarrow \mathbb{P}_K^n$. We know then that $X=V(f_1,\ldots,f_\ell)$ for some homogenous polynomials $f_1,\ldots,f_\ell\in K[x_0,\ldots,x_n]$. Suppose that

$\displaystyle f_j=\sum_I a_{I,j}x^I$

where $I$ is a tuple in $\mathbb{N}^{n+1}$, and $x^{(m_0,\ldots,m_n)}=x_0^{m_0}\cdots x_n^{m_n}$. Let $k=\mathbb{Q}(a_{I,j})\subseteq K$.

Now, note that if we let $t:X\to X_0$ be the obvious map, then $t^\ast \Omega^q_{X_0/k}=\Omega^q_{X/K}$. Moreover, it follows then that

$H^p(X,\Omega^q_{X/K})=H^p(X,\Omega_{X_0/K}^q)\otimes_k K$

and so, in particular, $h^{p,q}_X=h^{p,q}_{X_0}$ for all $(p,q)$. Thus, to prove that $h^{p,q}_X=h^{q,p}_X$, it suffices to prove such a statement for $X_0$.

Now, let $L=\overline{k(\{x_\alpha\})_{\alpha\in\mathbb{C}}}$. Note that since $k$ is countable, that $\# L=\#\mathbb{C}$, and so by the lemma, $L\cong\mathbb{C}$. Thus, we obtain an embedding of fields $k\hookrightarrow L\cong\mathbb{C}$. Let $X_1:= X_0\times_k \mathbb{C}$. Note that the same argument as before shows that $h^{p,q}_{X_0}=h^{p,q}_{X_1}$ for all $(p,q)$. Since $X_1$ is obviously smooth, projective, and integral, the claim follows. $\blacksquare$

# A meta-theorem

It’s clear that one could imitate the above proof if many situations. Most specifically, let us informally say that a collection of quasicoherent sheaves $\{\mathscr{F}_{X,K}\}$, as $K$ ranges over all characteristic $0$ fields, and $X$ ranges over all smooth integral varieties over $K$ (one could take any subcategory of varieties over $K$), is functorial if $t^\ast\mathscr{F}_{X,k}=\mathscr{F}_{X_K,K}$ where $k\subseteq K$ is a field extension, and $t:X_K\to X$ is the obvious map.

Then, there is an easy to generalize meta-theorem lurking the above reduction to $\mathbb{C}$. If we have two functorial collections of sheaves $\{\mathscr{F}_{X,K}\}$ and $\{\mathscr{G}_{X,K}\}$ we call a relation between $\mathscr{F}_{X,K}$ and $\mathscr{G}_{X,K}$ functorial if for all extensions of fields $k\subseteq K$ the relationship between $\mathscr{F}_{X,k}$ and $\mathscr{G}_{X,k}$ holds if and only if the relationship between $\mathscr{F}_{X_K,K}$ and $\mathscr{G}_{X_K,K}$ holds. Then, the meta-theorem should be as follows:

Meta-theorem: The relationship between $\mathscr{F}_{X,K}$ and $\mathscr{G}_{X,K}$ should hold for all $X$ and all $K$, if and only if it holds for all $X$ when $K=\mathbb{C}$.

For example, the collection of quasicoherent sheaves $\{\Omega^q_{X,K}\}$ and $\{\Omega^p_{X/K}\}$ is functorial, and the relationship that $\dim_K H^p(X,\Omega^q_{X/K})=\dim_K H^q(X,\Omega^p_{X/K})$ is functorial. So, the meta-theorem implies the proof of the reduction to $\mathbb{C}$ theorem. Of course, the proof of this meta-theorem is also just the same as in the proof of the reduction to $\mathbb{C}$ theorem.

This is a very powerful meta-theorem since if the $\{\mathscr{F}_{X,K}\}$ and $\{\mathscr{G}_{X,K}\}$ are coherent, then by reducing to $K=\mathbb{C}$, and applying the GAGA principle we can reduce these theorems to an analytic setting, where we might have more tools.

I think the above is, in some sense, just a disguised form of another well-known meta-theorem. Namely, that of the Lefschetz principle. This roughly says that all statements about varieties over algebraically characteristic $0$ fields $K$, which are first-order definable in the language of fields, hold for general $K$ if and only if they hold for $\mathbb{C}$. This is just a teased out statement that the theory $\mathsf{ACF}_0$ of algebraically closed fields of characteristic $0$ is complete.

This is further cemented by the fact that the proof that $\mathsf{ACF}_0$ is complete uses the following elementary result from model theory:

Theorem(Łoś–Vaught test): Let $\mathcal{T}$ be a (first-order) theory with no finite models, and which is $\kappa$-categorical for some uncountable cardinal $\kappa$. Then, $\mathcal{T}$ is complete.

Note then that the lemma above is just checking that $\mathsf{ACF}_0$ is $\kappa$-categorical for all uncountable $\kappa$! Since $\mathsf{ACF}_0$ obviously has no finite models, the Łoś–Vaught test tells us that $\mathsf{ACF}_0$ is complete.

Despite the obvious similarities between our meta-theorem and the Lefschetz principle, it’s not clear to me how they precisely relate.

Intuitively one would expect that any functorial relationship between two functorial collections of sheaves is first-order definable. This is because independence of the base field $K$ should make you think that the statement can be coded using only the basic operations of the fields (i.e. in the first order language of fields). If this were true, then our meta-theorem really would be a special case of the Lefschetz principle. But, I don’t see how to show that such functorial relationships are first-order in general.

# The complex case

So, now that we have the reduction to $\mathbb{C}$ out of the way, we can focus our attention entirely on the complex case.

Now, the following theorem is fairly elementary, and while a much, much, much easier special case, is still an instance of the above-cited GAGA principle.

Theorem(Complex Hodge Symmetry): Let $X/\mathbb{C}$ be a smooth, projective, integral variety. Then,

$H^p(X,\Omega^q_{X/\mathbb{C}})\cong H^p(X^\text{an},\Omega^q_\text{hol})$

where $X^\text{an}$ is the analytification of $X$, and $\Omega^q_\text{hol}$ denotes holomorphic $q$-forms.

So, we have reduced the case over $\mathbb{C}$ to showing that if $M$ is a compact Kahler manifold (which all analytifications of smooth projective integral varieties are), then

$\text{dim}_\mathbb{C} H^p(M,\Omega^q_\text{hol})=\dim_\mathbb{C} H^q(M,\Omega_M^p)$.

But, in case you haven’t noticed yet, our post has been a bit thematic so-far. Namely, this is the post of reductions, and after that, more reductions. Thus, you shouldn’t be surprised that we are not done with the reductions. Namely, we reduce this identity to a statement about Dolbeault cohomology.

Let us quickly recall the definition of Dolbeault cohomology on a complex manifold $M$. Let is fix $p$, and note that we have the following chain complex of sheaves

$\cdots\to \Omega^{p,q-1}_M\xrightarrow{\overline{\partial}}\Omega^{p,q}_M\xrightarrow{\overline{\partial}}\Omega^{p,q+1}_M\to\cdots$

Here $\Omega^{p,q}_M$ is the sheaf of $(p,q)$-forms on $M$ which, in local (complex) coordinates $(z_1,\ldots,z_n)$ are the $C^\infty(M)$-span of forms of the type

$dz_{i_1}\wedge\cdots\wedge dz_{i_p}\wedge d\overline{z_{j_1}}\wedge\cdots\wedge d\overline{z_{j_q}}$

The map $\overline{\partial}:\Omega^{p,q}\to\Omega^{p,q+1}$ is just the map defined in local coordinates by

$\displaystyle \overline{\partial}\left(\sum_{I,J}f_{I,J}dz^I\wedge d\overline{z}^J\right)=\sum_{I,J}\sum_{\ell=1}^n \frac{\partial f}{\partial\overline{z_{\ell}}}d\overline{z_j}\wedge dz^I\wedge d\overline{z}^J$

It’s easy to check that $\overline{\partial}^2=0$, and so we really do have a chain complex.

We then define the $(p,q)^\text{th}$ Dolbeault cohomology of $M$, denoted $H^{p,q}(M)$, to be the cohomology of

$\displaystyle \cdots\to \Omega^{p,q-1}_M(M)\xrightarrow{\overline{\partial}}\Omega^{p,q}_M(M)\xrightarrow{\overline{\partial}}\Omega^{p,q+1}_M(M)\to\cdots$

at the $q^\text{th}$ group.

Just as in the case of de Rham cohomology, we can relate the Dolbeault cohomology $H^{p,q}(M)$ to the Hodge groups $H^q(M,\Omega^p_\text{hol})$ (notice the switch in indexing!) by realizing that the Dolbeault sequence is a resolution of the sheaf $\Omega^p_\text{hol}$. More explictly

Theorem(Dolbeault’s theorem): For all pairs $(p,q)$, we have an isomorphism, and any complex manifold $M$

$H^{p,q}(M)\cong H^q(M,\Omega^p_\text{hol})$

Proof: Note that we have the Dolbeault sequence

$0\to \Omega^p_\text{hol}\to \Omega^{p,0}_M\xrightarrow{\overline{\partial}}\Omega^{p,1}_M\cdots$

is exact.

Before we justify this, note the difference between $\Omega^p_\text{hol}$ and $\Omega^{p,0}_M$. In local coordinates, both look something like

$f dz_{i_1}\wedge\cdots dz_{i_p}$

but in $\Omega^p_\text{hol}$, we have that $f$ must be holomorphic, whereas in $\Omega^{p,0}_M$, $f$ is just a smooth $\mathbb{C}$-valued function. The map $\Omega^p_\text{hol}\to\Omega^{p,0}_M$ is just inclusion.

Now, let’s prove exactness of this sequence. Clearly the first map is injective. The fact that $\Omega^p_\text{hol}$ is the kernel of $\overline{\partial}:\Omega^{p,0}_M\to \Omega^{p,1}_M$ reduces to the fact that a smooth function $f$ is holomorphic if and only if $\displaystyle \frac{\partial f}{\partial \overline{z_i}}=0$ for $i=1,\ldots,n$. The exactness at the other points reduces to the so-called Dolbeault lemma, which is an exercise in tedium, very much similar to the similar de Rham theorem.

Now, note that since each $\Omega^{p,q}_M$ is a $C^\infty_M$-module ($C^\infty_M$ being the sheaf of smooth $\mathbb{C}$-valued functions on $M$), it is fine, and so acyclic. Thus, since we can compute the cohomology of a sheaf from an acyclic resolution, we see that $H^q(X,\Omega^p_\text{hol})$ is the $q^\text{th}$-cohomology of the sequence

$0\to \Omega^{p,0}_M\to\Omega^{p,1}_M\to\cdots$

but, this is precisely the group $H^{p,q}(M)$. $\blacksquare$

So, why is Dolbeault’s theorem helpful in proving complex Hodge symmetry? Well, because then we may apply a well-known (but difficult!) theorem:

Theorem(Hodge symmetry for Kahler manifolds): Let $M$ be a compact Kahler manifold, then

$H^{p,q}(M)\cong H^{q,p}(M)$

The proof of this result requires too much background to discuss here (a rough sketch of the proof, and its importance are discussed in this post of mine). Roughly the idea is that one takes the Kahler metric on $M$, and then uses this to define the notion of harmonic $(p,q)$-forms with respect to this metric. The set of harmonic $(p,q)$-forms are conjugate to the group of harmonic $(q,p)$-forms, and so the groups are obviously isomorphic. One then checks, that $H^{p,q}(M)$ can be identified with the harmonic $(p,q)$-forms.