# Clarify Hodge conjecture

For the purpose of this question, let's focus on the variety defined over $\mathbb{Q}$. The Hodge conjecture could also stated as the Hodge realisation functor $\text{R}$ is full-faithful, \begin{equation} \text{R}:\textbf{M}_{\text{num}}(\mathbb{Q},\mathbb{Q}) \rightarrow \textbf{HS}_{\mathbb{Q}} \end{equation} where $\textbf{M}_{\text{num}}(\mathbb{Q},\mathbb{Q})$ is the category of motives defined over numerical equivalence and $\textbf{HS}_{\mathbb{Q}}$ is the abelian category of pure Hodge structures over $\mathbb{Q}$. It equivalence to the usual Hodge conjecture is explained carefully in the following post.

Why is the Hodge conjecture equivalent to the assertion that $\mathcal{R}_{ \mathrm{Hodge} }$ is fully faithfull?

However a naive question has puzzled me. Suppose $\chi:(\mathbb{Z}/N\mathbb{Z}) \rightarrow \mathbb{C}$ is a real non-trivial Dirichlet character, then by decomposing the Artin motive $h^0(\text{Spec}\,\mathbb{Q}(\zeta_N))$ there is a pure motive $\chi$ which is a direct summand of $h^0(\text{Spec}\,\mathbb{Q}(\zeta_N))$. The construction could also be found in section 1.1 of the paper

https://arxiv.org/pdf/math/0101071.pdf

The Hodge realistion of the Artin motive $h^0(\text{Spec}\,\mathbb{Q}(\zeta_N))$ is just $\mathbb{Q}(0)^m$, where $m=[\mathbb{Q}(\zeta_N):\mathbb{Q}]$, so as a direct summand, the Hodge realisation of $\chi$ should be $\mathbb{Q}(0)$ (? not so sure about this). If the functor $\text{R}$ is full-faithful, this would imply the motive $\chi$ is isomorphic to the Tate motive $\mathbb{Q}(0)$.

However for a prime $\ell \neq N$, the $\ell$-adic representation of $\chi$ is just the Galois representation associated with the Dirichlet character, which is non-trivial. So the motive $\chi$ cannot be isomorphic to $\mathbb{Q}(0)$.

There must be something that I haven't understood, could some one point out the naive mistakes that I have committed?

The realization $\mathbf{M}(\mathbb{Q},\mathbb{Q})\to\mathbf{HS}(\mathbb{Q})$ can't be fully faithful because one has factorizations : $$\mathbf{M}(\mathbb{Q},\mathbb{Q})\to\mathbf{M}(K,\mathbb{Q})\to\mathbf{M}(\mathbb{C},\mathbb{Q})\to\mathbf{HS}(\mathbb{Q})$$ for any fields $\mathbb{Q}\subset K\subset\mathbb{C}$, but these extensions $\mathbf{M}(k,\mathbb{Q})\to\mathbf{M}(K,\mathbb{Q})$ are not full (except when both $k$ and $K$ are algebraically closed).

The true statement of Hodge conjecture is for any algebraically closed field $K$, $\mathbf{M}(K,\mathbb{Q})\to\mathbf{HS}(\mathbb{Q})$ is fully faithful. (This is equivalent to the statement for $K=\mathbb{C}$ only).

But you may have notice that sometimes the statement with $K=\mathbb{Q}$ is stated. In fact it has nothing to do with Hodge conjecture, nor with Hodge theory, since one expect these conjectures :

Conjecture 1 : The Betti realization : $\mathbf{M}(K,\mathbb{Q})\to \mathbb{Q}-$Vect is conservative for any field $K\subset\mathbb{C}$.

Conjecture 2 : The De Rham-Betti realization $\mathbf{M}(k,\mathbb{Q})\to (\mathbb{Q},k)-$Vect is fully faithful for any number field $k$ (and $\overline{\mathbb{Q}}$).

Note : these conjecture are known to be true for Artin motives...

So what is going on ? if $M$ is an Artin motive, its Betti (or Hodge) realization is just $\mathbb{Q}^n$, but $M$ is not necessarily isomorphic to $\mathbb{Q}(0)^n$. Is it a contradiction ?

No because the part of the isomorphism between Betti and de Rham is often hidden but this part is enough to get the Artin motive.

More precisely, let $M$ be an Artin motive (=finite dimensional $\mathbb{Q}$-representation of $\operatorname{Gal}(\overline{\mathbb{Q}},\mathbb{Q})$ which factor through a finite quotient). Its Betti realization is just the underlying vector space.

Its de Rham realization is the following. Assume that $M$ is the motive of $\operatorname{Spec}(k)$ for a number field $k$. Then this is $H^0_{dR}(\operatorname{Spec}(k))=\Omega^0_{k/\mathbb{Q}}=k$.

Finally, the de Rham-Betti comparison will give a morphism $$k\otimes_{\mathbb{Q}}\mathbb{C}\to \mathbb{C}^n$$ such that $x\otimes \lambda\mapsto (\lambda\sigma_1(x),\lambda\sigma_2(x),...,\lambda\sigma_n(x))$ where $\sigma_1,...,\sigma_n$ are the $n$ embedding $k\subset\mathbb{C}$. (I am skipping details here...)

Clearly this is enough to get back the representation of $\operatorname{Gal}(\overline{k}/k)$.

As a conclusion, Hodge and Betti don't see the Artin motive. But, if we remember the isomorphism with de Rham, you get it back.

• Thank you very much for your very clear answer! – Wenzhe Mar 24 '18 at 12:13
• From Levine's chapter mixed motives in K theory book, the Hodge realisation functor $R:NMM(k,\mathbb{Q}) \rightarrow MHS_{\mathbb{Q}}$ is conjectured to be fully faithful, which could be seen as a generalisation of Hodge conjecture, here $NMM(k,\mathbb{Q})$ is Nori's mixed motive, which is expected to be the conjectured abelian category of mixed motives if we assume some standard conjectures. So $NMM(k,\mathbb{Q})$ is expected to has the pure motives as its semi-simple objects. But Levine does not require $k$ to be algebraically closed, why? – Wenzhe Mar 24 '18 at 15:18
• I don't know why there is such a conjecture. But I know nothing about Nori motives. Did they really capture Artin motives ? (Since Nori motives are constructed using Betti's cohomology, they might not see them). – Roland Mar 24 '18 at 16:15
• I will check this, you are probably right. – Wenzhe Mar 24 '18 at 16:37
• Sorry for bothering again. I try to prove the statement that the functor $\textbf{M}(k,\mathbb{Q}) \rightarrow \textbf{M}(K,\mathbb{Q})$ is fully faithful when $k \hookrightarrow K$ both are algebraically closed, could you explain it a little more? – Wenzhe Mar 25 '18 at 7:41