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Fix a perfect field $k$. Fix a field $K$ of characteristic $0$.

A Weil cohomology induces a functor from the category of smooth projective geometrically connected $k$-schemes to the category of $\mathbb{Z}_{\geq 0}$-graded finite-dimensional commutative associative unital $K$-algebras.

For any two Weil cohomologies are the corresponding functors naturally equivalent?

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    $\begingroup$ This statement implies the conjecture, basically part of the standard conjectures, that homological equivalence defines the same relation on algebraic cycles regardless of the Weil cohomology theory used. Probably there is a reverse implication but maybe one needs some other standard conjectures as well. Regardless, no one knows how to prove this standard conjecture. $\endgroup$
    – Will Sawin
    Sep 27, 2020 at 18:50
  • $\begingroup$ @WillSawin is it clear that if the functors are equal on objects they are naturally equivalent? I think one could construct a counterexample by messing with where the arrows get sent. $\endgroup$
    – user164740
    Sep 27, 2020 at 19:35
  • $\begingroup$ @JoeT Indeed, one theoretically could do this. The algebraic cycles are relevant precisely because they tell us about arrows - they are the image of the fundamental class under some arrow. $\endgroup$
    – Will Sawin
    Sep 27, 2020 at 19:50
  • $\begingroup$ For $k=\mathbb{Q}$, one has Betti cohomology and de Rham cohomology, and there is a god-given isomorphism between them with complex complex coefficients. I would expect these two cohomology functors are not isomorphic for $K=\mathbb{Q}$, but I don't know how to prove they are not isomorphic. $\endgroup$
    – Julian Rosen
    Sep 28, 2020 at 13:59

1 Answer 1

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If $k$ is a number field, each embedding $\sigma:k\hookrightarrow\mathbb{C}$ determines a Weil cohomology theory $H^*_{B,\sigma}$ on smooth projective $k$-varieties given by taking the topological cohomology of the complex manifold obtained form $X$ via $\sigma$. In [1], Charles gives an example of $k$, $X$, and two complex embeddings $\sigma_1$, $\sigma_2$ such that $H^*_{B,\sigma_1}(X)$ and $H^*_{B,\sigma_2}(X)$ are not isomorphic as algebras with real coefficients. This gives a counterexample to your question with $K=\mathbb{R}$, since the functors $H^*_{B,\sigma_1}$ and $H^*_{B,\sigma_2}$ are not even pointwise-isomorphic.

[1] Charles, François, Conjugate varieties with distinct real cohomology algebras, J. Reine Angew. Math. 630, 125-139 (2009). ZBL1222.14122.

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