Excuse me if this question is trivial or trivially false, or not at this sites level. Lets work over an algebraically closed field of characteristic $p>0$, say $k$. The action of the Frobenius on $H^r_{crys}(X/k)$ gives us a sequence of positive rational numbers $a_i$ by taking the order of the eigenvalues.

Now if $X$ has a good lift to $W(k)$, $Y$, say with torsion free $H^p(Y, \Omega_Y/W)$, I believe we can decompose the Frobenius morphism using Hodge decomposition to obtain an inequality $a_i\le r$.

My question is whether this is true in general. This may be trivial, but I am a bit new to Crystalline cohomology so I can't figure it out.

Excuse me if this question is trivial or trivially false, or not at this sites level. Lets work over an algebraically closed field of characteristic $p>0$, say $k$. The action of the Frobenius on $H^r_{crys}(X/W(k))$ gives us a sequence of positive rational numbers $a_i$ by taking the order of the eigenvalues.

Now if $X$ has a good lift to $W(k)$, $Y$, say with torsion free $H^p(Y, \Omega_Y/W)$, I believe we can decompose the Frobenius morphism using Hodge decomposition to obtain an inequality $a_i\le r$.

My question is whether this is true in general. This may be trivial, but I am a bit new to Crystalline cohomology so I can't figure it out.

Excuse me if this question is trivial or trivially false, or not at this sites level. Lets work over an algebraicly closed field of characteristic $p>0$, say $k$. The action of the Frobenius on $H^r_{crys}(X/k)$ gives us a sequeunce of positive rational numbers $a_i$ by taking the order of the eigenvalues.

Now if $X$ has a good lift to $W(k)$, $Y$, say with torsion free $H^p(Y, \Omega_Y/W)$, I believe we can decompose the Frobeinus morphism using Hodge decomposition to obtain an inequality $a_i\le r$.

My question is whether this is true in general. This may be trivial, but I am a bit new to Crystalline cohomology so I can't figure it out.

Excuse me if this question is trivial or trivially false, or not at this sites level. Lets work over an algebraically closed field of characteristic $p>0$, say $k$. The action of the Frobenius on $H^r_{crys}(X/k)$ gives us a sequence of positive rational numbers $a_i$ by taking the order of the eigenvalues.

Now if $X$ has a good lift to $W(k)$, $Y$, say with torsion free $H^p(Y, \Omega_Y/W)$, I believe we can decompose the Frobenius morphism using Hodge decomposition to obtain an inequality $a_i\le r$.

My question is whether this is true in general. This may be trivial, but I am a bit new to Crystalline cohomology so I can't figure it out.