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Given a K3 surface $X$, the cup product defines a non-degenerate even unimodular structure on the lattice $H^2(X,\mathbb{Z})$. Inside this lattice we have the Neron-Severi group $\text{NS}(X)$, which is also a primitive lattice. The rank of $\text{NS}(X)$, denoted by $\rho(X)$, is called the Picard number of $X$. The orthogonal complement of $\text{NS}(X)$ is by definition the transcendental lattice \begin{equation} T(X):=\text{NS}(X)^\perp \subset H^2(X,\mathbb{Z}). \end{equation}

In the note "Arithmetic of K3 surfaces" by Matthias Schutt, the author says that

"If $X$ is defined over some number field, the lattices of algebraic and transcendental cycles give rise to Galois representations of dimension $\rho(X)$ resp. $22-\rho(X)$."

Does he mean that the Galois representation arise from the etale cohomology $H^2_{et}(X,\mathbb{Q}_\ell)$ splits into the direct sum of two sub-representations with dimension $\rho(X)$ (associated to the algebraic cycles) and $22-\rho(X)$ (associated to the transcendental cycles)?

I guess this statement might be true generally for algebraic surfaces. Could anyone explain it more carefully, and give a reference if possible?

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    $\begingroup$ I don't know a precise reference, but it is clear that the subgroup of $H^2(X_{\bar{k}},\mathbb{Z}_{\ell}(1))$ spanned by algebraic classes is invariant under Galois, hence so is its orthogonal. $\endgroup$
    – abx
    May 26, 2019 at 4:11
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    $\begingroup$ @abx But why the orthogonal complement corresponds to transcendental cycles, is there an intuitive explanation? $\endgroup$
    – Wenzhe
    May 26, 2019 at 6:17
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    $\begingroup$ This is by definition! What do you think a transcendental cycle is? $\endgroup$
    – abx
    May 26, 2019 at 6:21
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    $\begingroup$ I think the fact you are missing might be $H^2_{et}(X,\mathbb Q_\ell) = H^2(X, \mathbb Z) \otimes_{\mathbb Z} \mathbb Q_\ell$ (as vector spaces). $\endgroup$
    – Will Sawin
    May 26, 2019 at 11:49
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    $\begingroup$ Yes, this follows from the compatibility of the cup product with the Galois action and the compatibility of the cup product with the comparison of cohomology. $\endgroup$
    – Will Sawin
    May 26, 2019 at 20:09

1 Answer 1

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This is essentially the Tate conjecture for K3 surfaces which says that the image of the NS lattice tensored with $\mathbb{Q}_l(1)$ in the etale cohomology $H^2_{et}(X, \mathbb{Q}_l(1))$ under the cycle class map is an invariant subspace under the action of the Galois group on the etale cohomology.

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    $\begingroup$ It may be good to point out that the Tate conjecture is known for K3 surfaces. $\endgroup$
    – jmc
    May 30, 2019 at 19:54
  • $\begingroup$ Indeed, Tate conjecture is true for K3 surfaces over finite fields and over number fields. It is culmination of the work of N. Nygaard, A .Ogus, Andre, Tankev, Maulik. As a reference please see Hyuybrecht's book on K3 surfaces. $\endgroup$
    – shehryar sikander
    May 30, 2019 at 20:04
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    $\begingroup$ Minor point: It is also true over finitely generated subfields of $\mathbb{C}$, which is slightly more general than number fields. $\endgroup$
    – jmc
    May 31, 2019 at 6:45
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    $\begingroup$ I keep waiting for someone to say "no, this is not what the Tate conjecture says" but it looks like that's not going to happen. The comments of @abx and Will Sawin suffice to answer the question. $\endgroup$
    – Pound Sterling
    Jun 1, 2019 at 20:58
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    $\begingroup$ Thanks for your comments. MO would not allow me to delete the answer, which would be the easiest thing to do. I'll fix it over the next few days when Im free of other obligations. $\endgroup$
    – shehryar sikander
    Jun 7, 2019 at 18:14

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