Picard lattice, over finite fields

$X_p/\mathbf{F}_q$ a K3 surface; $q=p^n$, $\ell\nmid q$.

$$P_2(t) := \det(t - \operatorname{Frob}_q \mid H^2_{\mathrm{et}}(X_p^{\mathrm{al}}, \mathbf{Q}_\ell)) \in \mathbf{Z}[t].$$

Tate conjecture

$$\operatorname{Pic}(X_p)_{\mathbf{Q}_\ell} = \ker\left( \operatorname{Frob}_q - q \cdot \operatorname{id} \mid H^2_{\mathrm{et}}(X_p^{\mathrm{al}}, \mathbf{Q}_\ell)\right)$$

Tate conjecture is known for K3 surfaces over finite fields.

Since $\operatorname{Frob}_q$ acts semisimply, we have:

$$\rho\bigl(X_{\mathbf{F}_{q^m}}\bigr) = \sum_{\zeta^m=1} \operatorname{ord}_{t=q\zeta} P_2(t).$$

$$\rho(X_p^{\mathrm{al}}) = \sum_{\zeta} \operatorname{ord}_{t=q\zeta} P_2(t),$$

where $\zeta$ runs over all roots of unity.

Note: $\rho(X_p^{\mathrm{al}}) \equiv 0 \bmod 2$

SOURCE: O:L1108-1142

Compressed from Lecture 1, where it was stated properly.

NOTATION: P_2(T) is the H^2 characteristic polynomial, the name Lecture 1 standardises on. A source slide that calls it chi(t) means the same object.

PANEL FLAG: the frame writes chi(t) = det(1 - t Frob | H^2), which is the reciprocal of the deck's P_2(T) = det(T - Frob | H^2), not literally the same polynomial. The frame's chi(t) is kept here because the zeta function line needs that normalisation; if the course is to use one name throughout, the author should say which.

Spoken: Reminder only.

Spoken: The Tate conjecture is a theorem for K3 surfaces over finite fields, so, with P_2(T) = det(T - Frob | H^2) as in Lecture 1,

Spoken: rho(X_p bar) = #{roots of P_2(pT) that are roots of unity, with multiplicity} = sum_i deg Phi_i,

Spoken: the sum of the degrees of the cyclotomic factors Phi_i of the normalised polynomial, not the number of them.

Spoken: Slide 14: four factors, rank 10.

TRANSCRIBED FROM: frobenious-dist/nyc-jnts.tex:1108-1142