Automorphisms of Frobenius liftings and degeneration of the Hodge-de Rham spectral sequence

Question

I am still studying Deligne and Illusie's paper (https://eudml.org/doc/143480), and I am again stuck, this time on pages 262/263.

Assume $X\longrightarrow S$ is a smooth morphism of $\mathbb{F}_{p}$-schemes, then $\operatorname{Lif}(X,\tilde{S})$ is the gerbe of liftings to $\tilde{S}=S(\mathbb{Z}/p^{2})$, a morphism between two liftings $\tilde{U}'\longrightarrow\tilde{U}''$ (defined over an open $U\subseteq X$) being a morphism that is compatible with reduction mod $p$.

(1.) The authors make the claim that the sheaf of automorphisms of any given lifting of $U\subseteq X$ is the $\operatorname{Hom}$-sheaf $\operatorname{Hom}(\Omega^{1}_{U/S},\mathcal{O}_{U})$. Why is that the case?

(2.) While proving Theorem 3.5, the authors further claim that, if $a$ is any such automorphism, from the fact that $dF=0$ (where $F$ is the Frobenius on $X$) it follows that $\tilde{F}\circ a=\tilde{F}$ for any Lifting $\tilde{F}$ of $F$. Why?

I might not be seeing the wood for all the trees here, or maybe there is something that I just don't see. Can somebody explain this to me in detail?

Answer 1

(1) is elementary deformation theory: if $\alpha\colon \mathcal{O}_{\tilde U}\to \mathcal{O}_{\tilde U}$, then $\alpha-{\rm id}\colon \mathcal{O}_{\tilde U}\to \mathcal{O}_{\tilde U}$ vanishes modulo $p$ and hence there exists a unique $\delta\colon \mathcal{O}_U\to \mathcal{O}_U$ such that $$\alpha(f) = f + p\cdot \delta(f\text{ mod }p).$$ It is straightforward to check that $\delta$ is a derivation, and hence gives an element of ${\rm Hom}(\Omega^1_{U/S}, \mathcal{O}_U)$. Conversely, for every derivation $\delta\colon \mathcal{O}_U\to\mathcal{O}_U$, the above formula defines an automorphism $\alpha$.

(2) you can also verify directly. Say $\tilde A$ is a $\mathbf{Z}/p^2$-algebra, $A=\tilde A/p$, $\alpha\colon \tilde A\to \tilde A$ lifts the identity on $A$ and $\tilde F\colon \tilde A\to\tilde A$ lifts the Frobenius on $A$. Write $\alpha(f)=f+\delta(\overline{f})$ for some derivation $\delta\colon A\to p\tilde A$ (note that $p\tilde A$ is an $A$-module). Take $f\in \tilde A$, then $$\alpha(\tilde F (f)) = \tilde F(f) + \delta(\overline{f}{}^p)= \tilde F(f) + p\overline{f}{}^{p-1}\delta(f) = \tilde F(f). $$ Here we used $\delta(\overline{f}{}^p) = p\overline{f}{}^{p-1}\delta(f) = 0$, note that $\delta(f)$ is already an element of $p\tilde A$ (this is what Deligne--Illusie meant by $dF=0$!)