I ask a question on math stack exchange about Hodge conjecture and have not got any reply or comment.


so I decide to post this question on MO. Let's focus on smooth projective varieties defined over $\mathbb{Q}$. The Hodge conjecture could also be formulated as the Hodge realisation functor $\text{R}$ being 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 $\mathbb{Q}$. Its equivalence to the usual statement is explained carefully in the post


However a naive question has puzzled me. Suppose $\chi:(\mathbb{Z}/N\mathbb{Z})^\times \rightarrow \mathbb{Q}$ is a real non-trivial Dirichlet character, i.e. it takes values in $\{ -1,1\}$, 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 $h^0(\text{Spec}\,\mathbb{Q}(\zeta_N))$. Its construction could also be found in section 1.1 of the paper


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 $\mathbb{Q}(0)$ as its is one dimensional. 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 $\chi$, which is non-trivial. So the motive $\chi$ cannot be isomorphic to $\mathbb{Q}(0)$.

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


The Hodge conjecture implies that the functor $R\colon M_{num}(k,\mathbb{Q})→HS(\mathbb{Q})$ is fully faithful when $k$ is the algebraic closure of $\mathbb{Q}$, not $\mathbb{Q}$ itself, as your example illustrates.


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