How to cook up an Artin motive from a positive-dimensional variety

Question

I am trying to make sense of the paper "Eigenvalues of Frobenius and Hodge Numbers" (Kisin--Lehrer). I have not succeeded after some hours of intent staring at the screen.

In the proof of Corollary 2.3, they say "the converse is precisely the content of [KW, 1.3]." How is that so? In Theorem 1.3 of [KW], we see that the associated graded of the Hodge filtration on the de Rham cohomology of our motive has to be concentrated in degree $0$. But the point (2) of Corollary 2.3 says the associated graded of the Hodge filtration is concentrated in $j$ (or $j-1$, I am bad with indices). No way that is $0$. Does Tate twist shift the Hodge filtration?

Answer 1

Tate twists do indeed shift the Hodge filtration. Since the de Rham realization of the Tate motive $\mathbb{Q}(1)$ is concentrated in the degree $-1$ part of the Hodge filtration, we have $F^n M(m)_{dR}=F^{n+m} M_{dR}$ for every motive $M$. So if condition (2) in Kisin-Lehrer holds, then $\mathrm{Gr}^0 H^{2j}_{dR}(X)(-j)=H^{2j}_{dR}(X)(-j)$, and Corollary 1.3 of Kisin-Wortmann applied to the motive $H^{2j}(X)(-j)$ implies the Galois action on $H^{2j}(X,\mathbb{Q}_p)(-j)$ factors through a finite quotient of the absolute Galois group. This means the eigenvalues of $\mathrm{Frob}_v$ on $H^{2j}(X,\mathbb{Q}_p)(-j)$ are all roots of unity. It follows that the eigenvalues of $\mathrm{Frob}_v$ on $H^{2j}(X,\mathbb{Q}_p)$ are all of the form $q_v^{j}$ times a root of unity, since $\mathrm{Frob}_v$ acts by multiplication by $q_v$ on $\mathbb{Q}_p(1)$.