What Frobenius acts on

Via the isomorphism $H^2_{\mathrm{crys}}(X_{\mathbf{F}_p}/\mathbf{Z}_p) \otimes \mathbf{Q}_p \simeq H^2_{\mathrm{dR}}(X/\mathbf{Q})$, we have

$$\operatorname{Frob}_p : H^2_{\mathrm{dR}}(X/\mathbf{Q}_p) \rightarrow H^2_{\mathrm{dR}}(X/\mathbf{Q}_p).$$