# How are rational algebraic Hodge classes of type $(p,p)$ defined?

• Definition ( Hodge classes ) :

For each integer $p \in \mathbb{N}$, let $H^{p,p} (X)$ denotes the subspace of $H^{2p} ( X ,\mathbb{C} )$ of type $(p,p)$.

The group of rational $(p,p)$- cycles : $H^{p,p} (X , \mathbb{Q} ) = H^{2p} ( X , \mathbb{Q} ) \cap H^{p,p} (X)$ is called the group of rational Hodge classes of type $(p,p)$.

• Definition :

An $r$ -cycle of an algebraic variety $X$ is a formal finite linear combination $\displaystyle \sum_{ i \in [1,h] } m_i Z_i$ of closed irreducible subvarieties $Z$ of dimension $r$ with integer coefficients $m_i$.

The group of $r$ -cycles is denoted by $\mathcal{Z}_r (X)$.

On a compact complex algebraic manifold, the class of closed irreducible subvarieties of codimension $p$ extends into a linear morphism : $\mathrm{cl}_{ \mathbb{Q} } \ : \ \mathcal{Z}_{p} (X) \otimes \mathbb{Q} \to H^{p,p} (X, \mathbb{Q} )$ defined by : $\mathrm{cl}_{ \mathbb{Q} } \big( \sum_{ i \in [1,h] } m_i Z_i \big) = \sum_{ i \in [1,h] } m_i \eta_{Z_{i}} \ , \ \forall m_i \in \mathbb{Q}$.

The elements of the image of $\mathrm{cl}_{ \mathbb{Q} }$ are called rational algebraic Hodge classes of type $(p,p)$.

• Hodge conjecture :

On a non-singular complex projective variety, any rational Hodge class of type $(p,p)$ is algebraic, i.e : in the image of $\mathrm{cl}_{ \mathbb{Q} }$.

• My questions are :

How to understand explicitly the nature of any element $[ \alpha ]$ of $H^{p,p} (X , \mathbb{Q} ) = H^{2p} ( X , \mathbb{Q} ) \cap H^{p,p} (X)$ the group of rational Hodge classes of type $(p,p)$ ?.

1) If $[ \alpha ] \in H^{p,p} (X , \mathbb{Q} ) = H^{2p} ( X , \mathbb{Q} ) \cap H^{p,p} (X)$, does it mean that $\alpha$ has the form : $\alpha = \sum \alpha_{j_{1}j_{2} ... j_{p}} dz_{j_{1}} \wedge ... \wedge z_{j_{p}} \wedge \overline{dz_{j_{1}}} \wedge ... \wedge \overline{dz_{j_{p}}}$ such that : $\alpha \in \mathcal{A}^{p,p} ( X , \mathbb{Q} )$ with $\alpha : X \to \bigwedge^{p,p} (TX)^*$ and $\forall x \in X \ : \ \alpha (x) \in \bigwedge^{p,p} (TX)^*$ an alternating $(p,p)$ - form on $(TX)^*$ ?

If this is true, $\alpha (x) = \sum \alpha_{j_{1}j_{2} ... j_{p}} (x) dz_{j_{1}} \wedge ... \wedge z_{j_{p}} \wedge \overline{dz_{j_{1}}} \wedge ... \wedge \overline{dz_{j_{p}}} \in \bigwedge^{p,p} (TX)^*$. Does it mean that : $\forall (v_1 , \dots , v_n) \in (TX)^* \times ... \times (TX)^* \ : \ \alpha (x) (v_1 , \dots , v_n ) \in \mathbb{C}$ or $\alpha (x) (v_1 , \dots , v_n ) \in \mathbb{Q}$ ? I think that : $\alpha (x) ( v_1 , \dots , v_n ) \in \mathbb{Q}$ because : $\alpha \in H^{2p} ( X , \mathbb{Q} ) \cap H^{p,p} (X)$ so $\alpha \in H^{2p} ( X , \mathbb{Q} )$, right ?

2) Otherwise, if : $\alpha = \sum \alpha_{j_{1}j_{2} ... j_{p}} dz_{j_{1}} \wedge ... \wedge z_{j_{p}} \wedge \overline{dz_{j_{1}}} \wedge ... \wedge \overline{dz_{j_{p}}} \in H^{2p} (X, \mathbb{Q} ) \cap H^{p,p} (X)$, have we $\alpha_{j_{1}j_{2} ... j_{p}} : X \to \mathbb{C} \ \$ or $\ \ \alpha_{j_{1}j_{2} ... j_{p}} : X \to \mathbb{Q}$ ?

• What do you mean by $\mathcal{A}^{p,p}(X,\mathbb{Q})$ ? Anyway, I think what is crucial here is that there is two cohomology theory at hand : singular cohomology with rational coefficients $H^{2p}(X,\mathbb{Q})$ and de Rham cohomology together with its decomposition into $(p,q)$-forms. In particular the subspace $\dim_{\mathbb{Q}}H^{p,p}(X,\mathbb{Q})$ is in general smaller than $\dim_{\mathbb{C}}H^{p,p}(X,\mathbb{C})$, and there is no "De Rham cohomology with rational coefficients" – Roland Jan 23 '18 at 19:35
• For me : $\alpha \in \mathcal{A}^{p,p} ( X , \mathbb{Q} ) = \mathcal{A}^{2p} ( X , \mathbb{Q} ) \cap \mathcal{A}^{p,p} (X)$ means that $\forall x \in X , \ , \forall (v_1 , \dots , v_n ) \in (TX)^* \times \dots \times (TX)^*$ : $\alpha (x) (v_1 , \dots , v_n ) \in \mathbb{Q}$. But, i still don't know if it's true or not. I still don't understand all of that. – YoYo Jan 24 '18 at 1:14
• I don't know what is : $\mathcal{A}^{2p} (X, \mathbb{Q} )$ explicitely. All that i know is that $\mathcal{A}^{2p} (X, \mathbb{Q} )$ induces the groupes of singular cohomology over $\mathbb{Q}$ : $H^{2p} ( X , \mathbb{Q} ) = \ker \big( \mathcal{A}^{2p} (X , \mathbb{Q} ) \to \mathcal{A}^{2p+1} (X , \mathbb{Q} ) \big) / \mathrm{im} \big( \mathcal{A}^{2p} (X , \mathbb{Q} ) \to \mathcal{A}^{2p+1} (X , \mathbb{Q} ) \big)$. $\mathcal{A}^{p,p} (X)$ is the space of alternating $(p,p )$ - forms. – YoYo Jan 24 '18 at 1:27

There is two cohomology theories (singular and de Rham) and an isomorphism between them. Often the isomorphism is not written, but it may be misleading. For example, the intersection $H^{p,p}(X,\mathbb{C})\cap H^{2p}(X,\mathbb{Q})$ is an intersection of one group belonging to the de Rham theory and one group belonging to the singular theory.

Let's have a look at the simplest example of a projective algebraic variety : $\mathbb{P}^1$. The cohomology theories are :

• singular cohomology : $H^2(\mathbb{P}^1,\mathbb{Q})=\mathbb{Q}.u$ where $u$ is the class of a cochain such that $\langle u,[\mathbb{P}^1]\rangle = 1$.
• de Rham cohomology : $H^2_{dR}(\mathbb{P}^1,\mathbb{C})=\mathbb{C}.[\omega]$ with $\omega=\frac{1}{2i\pi}\frac{dz\wedge d\overline{z}}{(1+z\overline{z})^2}$ (check that this is well defined at $\infty$). I put the coefficient $\frac{1}{2i\pi}$ so that $\int_{\mathbb{P}^1}\omega=1$ (easy computation with polar coordinates).

Now, integration of differential forms gives a morphism : $$\Psi_{dR}:H_{dR}^n(X,\mathbb{C})\rightarrow H^n(X,\mathbb{Q})\otimes\mathbb{C}=Hom(H_i(X,\mathbb{Z}),\mathbb{C}) \text{ such that }[\omega]\mapsto (\gamma\mapsto \int_{\gamma}\omega)$$

With $X=\mathbb{P}^1$, this mean that our class (with the choice of the coefficient $\frac{1}{2i\pi}$) satisfies $\Psi_{dR}([\omega])=u$.

Now de Rham cohomology has a decomposition $H_{dR}^n(X,\mathbb{C})=\bigoplus_{p+q=n}H^{p,q}(X,\mathbb{C})$. The group $H^{p,p}(X,\mathbb{Q})$ is either $\Psi_{dR}(H^{p,p}(X,\mathbb{C}))\cap H^{2p}(X,\mathbb{Q})$ or $H^{p,p}(X,\mathbb{C})\cap\Psi_{dR}^{-1}(H^{p,p}(X,\mathbb{Q}))$ (depending on whether we want to see Hodge classes as singular cochain or differential forms).

Say we prefer differential forms. Then $H^{p,p}(\mathbb{P}^1,\mathbb{Q})=\mathbb{Q}.[\omega]$. But note that $\omega$ is not an element of $\mathcal{A}(X,\mathbb{Q})$ whatever this means : it has a factor $\frac{1}{2i\pi}$ so it is really not $\mathbb{Q}$-valued. (This is what I meant in my comment by : there is no de Rham cohomology with rational coefficients)

I want to add that I don't like the notation $H^{p,p}(X,\mathbb{Q})$ simply because we could be tempted to say that $H^{p,p}(X,\mathbb{Q})\otimes_{\mathbb{Q}}\mathbb{C}=H^{p,p}(X,\mathbb{C})=H^{p,p}(X)$. This is wrong : the group $H^{p,p}(X)$ is a subspace of $H^{2p}_{dR}(X,\mathbb{C})$ and the group $\Psi_{dR}^{-1}H^{2p}(X,\mathbb{Q})$ is a sub-$\mathbb{Q}$-vector space of $H^{2p}_{dR}(X,\mathbb{C})$. Their intersection might be smaller than expected : in fact it might be zero even if $H^{p,p}(X)\neq 0$.

• Thank you very much Roland. It's clear now. :-) – YoYo Jan 24 '18 at 13:10