Let $X\to \mathrm{Spec}\:\mathbb{Z}\_{(p)}$ be a smooth proper morphism with a geometrically connected generic fiber. Assume that the special fiber has an $\mathbb{F}\_p$-point.


Via the isomorphism $H^{\*}\_{\mathrm{cris}}(X\_{\mathbb{F}\_p}/\mathbb{Z}\_p)\approx H^{\*}\_{\mathrm{dR}}(X\_{\mathbb{Z}\_p})$ the absolute Frobenius of the special fiber induces a map $H^{\*}\_{\mathrm{dR}}(X\_{\mathbb{C}})\to H^\*\_{\mathrm{dR}}(X\_{\mathbb{C}})$. Is this map induced by an algebraic map $X\_{\mathbb{C}}\to X\_{\mathbb{C}}$?


This seems like a strong condition since if $X\_{\mathbb{C}}$ is a hyperbolic curve there is finitely many algebraic self-maps (any such map is constant or an automorphism) but the mapping class group is infinite.



Is the action of the absolute Frobenius on de Rham cohomology induced by an algebraic map?
The answer is no. For a generic elliptic curve E, every algebraic map $E\to E$ is, up to shift by elements of $E$, a multiplication map $[k]\colon E\to E$. In particular, it acts on $H^1\_{\mathrm{dR}}(E)$ as multiplication by $k$. But $\mathrm{Fr}$ acts on the 2-dimensional space $H^1(E)$ as a linear map with determinant $p$, and hence can not be scalar.


