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For a smooth projective variety $X$ over a field $k$ the Tate conjecture says that the cycle class maps $$CH^i(X)\otimes \mathbb{Q}_l \to H^{2i}(X_{\bar{k}},\mathbb{Q}_l(i))^{G_k}$$

are surjective. To my knowledge both the Chow groups and l-adic cohomology are representable in the motivic category by some ring spectra. Is there a way to interpret the cycle class map, and thus Tate's conjecture, in terms of these ring spectra? I haven't touched algebraic homotopy theory for a long time so this may be a naive question.

I will just add my attempt to understand this so far. In Mixed Weil Cohomologies, Theorem 1 there is a cycle class map from motivic cohomology to any "Mixed Weil Cohomology" $E$ given by $$H^q(X,Q(p)) \to H^q(X,E(p))$$ and from (2.3.24.3) it should arise from a cycle class map on spectra $cl: HQ \to \mathcal{E}$ when $\mathcal{E}$ represents $E$. We know that $H^{2n}(X,Q(n)) = CH^n(X)_Q$. Is $Q$ just an algebra which we can take to be $\mathbb{Q}_l$? In this case we would want $l$-adic cohomology to be a mixed Weil cohomology.

This is not exactly the case because $l$-adic cohomology is an inverse limit of étale cohomology tensored by $\mathbb{Q}_l$. I would be satisfied with using $H^{2i}(X,\mathbb{Z}/l\mathbb{Z}(i))$ instead. Alternatively, we could use Scholze and Bhatt's pro-étale site to define $l$-adic cohomology and then maybe it will be representable?

I'm just curious if we can use this to reduce Hodge conjecture-type questions to a question about a single morphism of spectra.

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    $\begingroup$ You seem to be eliding the difference between $H^{2i}(X, \mathbb{Q}_\ell(i))$ and $H^{2i}(X_{\overline{k}}, \mathbb{Q}_\ell(i))^{G_k}$, which are different things in general. $\endgroup$ Commented Mar 18, 2023 at 16:09
  • $\begingroup$ @DavidLoeffler Yes this seems like a problem. Is there a way around this, or maybe just take $k$ to be algebraically closed? $\endgroup$
    – TCiur
    Commented Mar 18, 2023 at 16:24
  • $\begingroup$ Tate's conjecture is formulated for fields which are finitely generated over their prime field, so that rules out the case of algebraically-closed $k$. $\endgroup$ Commented Mar 18, 2023 at 17:46

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