About the filtration of crystalline cohomology

Question

Suppose $K$ is an finite unramified extension of $\mathbb Q_p$ with residue field $k$, and let $Y$ be an proper smooth variety defined over $k$. We know if $Y$ admits a proper smooth lifting $X/W(k)$ we have the comparison theorem $H^i_{cris}(Y/W(k))\otimes K\simeq H_{dR}^i(X_K/K)$, and the filtration of the de Rham cohomology induces a filtration of crystalline cohomology. I wonder is this filtration (of crystalline cohomology) independent of the lifting $X$?

I think the answer is no since otherwise $H^i_{cris}(Y/W(k))\otimes K\simeq D_{cris}(H_{{e}t}^i(X))$ would be isomorphic for all lifting $X$ (as a filtered $\phi$-module) which means the Galois representations $H_{et}^i(X,\mathbb Q_p)$ are isomorphic for all $X$ and it seems impossible.

Answer 1

Let me upgrade my comment to an answer:

The answer is no, the filtration is not independent of the lifting. In fact the relationship between liftings of $Y$ and filtrations lifting the Hodge filtration on $H^{i}_{\mathrm{cris}}(Y/W(k))\otimes_{W(k)}k=H^{i}_{\mathrm{dR}}(Y/k)$ is often very interesting. The classic example, I think, is if your $Y$ is an abelian variety of dimension $g$. In that case the liftings of $Y$ are in bijection with liftings of the Hodge filtration $\mathrm{Fil}^{1}$ on $H^{1}_{\mathrm{dR}}(Y/k)$ to rank $g$ submodules of $H^{1}_{\mathrm{cris}}(Y/W(k))$ which are isotropic with respect to the cup product. A reference is W. Messing, The crystals associated to Barsotti-Tate groups: with applications to abelian schemes, Lecture Notes in Math., Vol. 264 Springer-Verlag, Berlin-New York, 1972.