Let $X=E\_1\times\cdots\times E\_n$, where $E\_i$ is the elliptic curve $E\_i=\mathbb{C}/(\mathbb{Z}+\mathbb{Z}\alpha\_i)$. In Grothendieck's ["The Hodge Conjecture is False for Trivial Reasons,"](https://webusers.imj-prg.fr/~leila.schneps/grothendieckcircle/HodgeConj.pdf) $X$ is stated to fail the original formulation of the Generalized Hodge Conjecture, for appropriate choices of $\alpha\_i$. Namely, there exist classes in $F^1H^n(X,\mathbb{C})\cap H^n(X,\mathbb{Q})$ that are not generated by images of Gysin maps $H^{n-2q}(Y,\mathbb{Q})\to H^{n}(X,\mathbb{Q})$, where $Y$ is a resolution of a codimension $q\ge1$ subscheme of $X$. (Here $F^1H^n(X,\mathbb{C})=H^{1,n-1}(X)\oplus\cdots\oplus H^{n,0}(X)$ is the first term in the Hodge filtration for $X$.)


He states without proof that the dimension of the subspace of $H^n(X,\mathbb{Q})$ generated by images of Gysin maps is $2^n-N$, where $N$ is the dimension of the $\mathbb{Q}$-vector space generated by distinct products of $\alpha\_i$. I am confused by this, because it seems that when the $\alpha\_i$ are generic, this number is 0. In particular, there are no non-zero Hodge classes in $H^2(E\_1\times E\_2,\mathbb{Q})$, but don't $E\_1\times 0$ and $0\times E\_2$ define nonzero Hodge classes?


Would appreciate it if someone could clear up my confusion and also provide a hint for how to prove the claim in the above paragraph.



Which rational cohomology classes on a product of elliptic curves come from subschemes?
This definitely looks like an error in Grothendieck's paper! I think I have figured out what he meant. To simplify notation, I will concentrate on the case where Grothendieck builds his example: $H^3$ of a product of $3$ elliptic curves. I prefer to think in homology for this purpose, so I'll be talking about $H\_3(E\_1 \times E\_2 \times E\_3)$. 


Put $X = E\_1 \times E\_2 \times E\_3$ where $E\_i = \mathbb{C}/(\mathbb{Z}+\mathbb{Z} \tau\_i)$. I'll write $z\_i$ for the coordinate on the universal cover of $E\_i$, so $dz\_i$ is the nonvanishing holomorphic $(1,0)$-form on $E\_i$. 


Let $V$ be the subspace of $H\_3(X, \mathbb{Q})$ spanned by the images of all $H\_3(Y, \mathbb{Q})$, as $Y$ ranges over $2$-dimensional subvarieties of $X$. Hodge's "general conjecture" is an attempted description of $V$. (The famous Hodge conjecture deals with $H\_{2 \dim Y}(Y)$, so we are talking about the span of the fundamental classes; this less famous one allows the subscript on homology to be independent of the dimension of $Y$.)


Specifically, Hodge conjectures that $\alpha \in H\_3(X, \mathbb{Q})$ lies in $V$ if and only if $\int\_{\alpha} \omega=0$ for any $\omega \in H^{(3,0)}(X)$. In fact, $H^{(3,0)}(X)$ is one dimensional, spanned by $d z\_1 \wedge d z\_2 \wedge dz\_3$, so this criterion says that $\int\_{\alpha} d z\_1 \wedge d z\_2 \wedge dz\_3=0$. Let's denote the space of $\alpha$ obeying this condition by $V'$. It is easy to see that $V \subseteq V'$, since $ d z\_1 \wedge d z\_2 \wedge dz\_3|\_Y=0$ for any algebraic surface $Y$.


Grothendieck observes that, for "trivial reasons", $V$ is even dimensional. However, he shows that $V'$ need not be. 


By Kunneth, we have 
$$H\_3(X, \mathbb{Q}) = \bigoplus\_{i\_1+i\_2+i\_3=3} H\_{i\_1}(E\_1, \mathbb{Q}) \otimes H\_{i\_2}(E\_2, \mathbb{Q}) \otimes H\_{i\_3}(E\_3, \mathbb{Q}).$$
Every summand other than $H\_1 \otimes H\_1 \otimes H\_1$ is easily seen to be in $V$. So it is natural to consider the inclusion $V \cap (H\_1\otimes H\_1 \otimes H\_1) \subseteq V' \cap (H\_1\otimes H\_1 \otimes H\_1)$, which will be equality if and only if $V \subseteq V'$ is equality. I believe that Grothendieck, without bothering to say so, has switched to working with these subspaces in his dimension computation. Each of the other summands is two-dimensional, so this doesn't effect the parity of the dimension.


So, let's compute the dimension of $V' \cap (H\_1\otimes H\_1 \otimes H\_1)$. Let $\gamma^0\_i$ be the $1$-cycle $\mathbb{R}/\mathbb{Z}$ in $E\_i$, and let $\gamma^1\_i$ be the $1$-cycle $\mathbb{R} \tau\_i/\mathbb{Z} \tau\_i$. So $\gamma^0\_i$, $\gamma^1\_i$ is a basis of $H\_1(E, \mathbb{Q})$. So a basis for $H\_1 \otimes H\_1 \otimes H\_1$ is the $8$ products $\gamma\_1^{j\_1} \times \gamma\_2^{j\_2} \times \gamma\_3^{j\_3}$, where each $j\_i$ is $0$ or $1$. We have $\int\_{\gamma\_i^j} dz\_i = \tau\_i^j$, so
$$\int\_{\gamma\_1^{j\_1} \times \gamma\_2^{j\_2} \times \gamma\_3^{j\_3}} d z\_1 \wedge d z\_2 \wedge dz\_3 = \tau\_1^{j\_1} \tau\_2^{j\_2} \tau\_3^{j\_3}.$$
Putting 
$$\alpha = \sum\_{j\_1, j\_2, j\_3 =0}^1 c(j\_1, j\_2, j\_3)\ [\gamma\_1^{j\_1} \times \gamma\_2^{j\_2} \times \gamma\_3^{j\_3}] \in H\_3(X, \mathbb{Q}),$$
we see that $\alpha \in V'$ if and only if 
$$\sum\_{j\_1, j\_2, j\_3 =0}^1 c(j\_1, j\_2, j\_3) \ \tau\_1^{j\_1} \tau\_2^{j\_2} \tau\_3^{j\_3}=0.$$
Thus, $V' \cap (H\_1\otimes H\_1 \otimes H\_1)$ is identified with the $\mathbb{Q}$-vector space
$$\left\{ c \in \mathbb{Q}^{\{0,1\}^3} : \sum\_{j\_1, j\_2, j\_3 =0}^1 c(j\_1, j\_2, j\_3) \ \tau\_1^{j\_1} \tau\_2^{j\_2} \tau\_3^{j\_3}=0 \right\}.$$
We see that the dimension of $V' \cap (H\_1\otimes H\_1 \otimes H\_1)$ is $2^3-N$ where $N$ is the dimension of the $\mathbb{Q}$-vector space spanned by the monomials $\tau\_1^{j\_1} \tau\_2^{j\_2} \tau\_3^{j\_3}$. 


As Grothendieck points out, if $\tau\_1=\tau\_2=\tau\_3$ is a cubic irrational, these monomials span a $3$-dimensional vector space. As you say, for generic $\tau\_i$, the monomomials span an $8$ dimensional vector space, so $V' \cap (H\_1\otimes H\_1 \otimes H\_1)=0$. This reflects that the obvious classes, coming from $H\_3(E\_i \times E\_j)$ with $1 \leq i < j \leq 3$, are all in the other summands of $H\_3(X)$.


