This might be a naive question. But since I haven't seen this in any reference, I'll try to ask it here. Let $T$ be a smooth scheme over the algebraically closed field $k$ of characteristic $p>0$ (we can assume that $T$ is affine). Let $\mathbb{T}$ be a smooth lift of $T$ over the ring $W_{2}(k)$, the ring of witt vectors of length 2 over k. Now, if $H_{\mathbb{T}}$ is an Fcrystal on $\mathbb{T}$ and $F_{1}$ and $F_{2}$ are two lifts of Frobenius $F_{T}$ on T to $\mathbb{T}$, I know that there must exist an isomorphism $X(F_{1}, F_{2})$ between $F_{1}^*H_{\mathbb{T}}$ and $F_{2}^*H_{\mathbb{T}}$ coming from the nconnection on $H_{\mathbb{T}}$. What is the explicit formula of this isomorphism?
1 Answer
This is parallel transport. Note that you only get an explicit formula for the isomorphism between $F_1^*H(\mathbb{T})$ and $F_2^*H(\mathbb{T})$. To get explicit formulas over other thickenings of $\mathbb{T}$, you would need lifts of both $\mathbb{T}$ and also of the Frobenius lifts. For convenience, set $H_1=H(\mathbb{T})$. We will assume that $\mathbb{T}$ has \'etale coordinates $x_1,\ldots,x_n$.
Since $H_{\mathbb{T}}$ is a crystal over $\mathbb{T}$, there is a flat connection $\nabla:H_1\to H_1\otimes\Omega^1_{\mathbb{T}/W_2(k)}$. Note that, for any $a\in\mathcal{O}_{\mathbb{T}}$, $\delta(a)=F_1(a)F_2(a)$ is in $p\mathcal{O}_{\mathbb{T}}$; in particular, $\delta(a)^2=0$. Now, for any $h\in H_1$, the isomorphism sends $F_1^*h$ to $F_2^*h+\sum_{i=1}^n\nabla(\partial_i)(m)\delta(x_i)$. This is just a Taylor expansion up to the first degree.

1$\begingroup$ Dear Keerthi. Thank you for your answer. what do you mean by $\nabla(\partial_i)(m)$ ? I think you really mean $\nabla(h)(\partial_i)$. Am I right? $\endgroup$– JackSep 29, 2011 at 13:30

$\begingroup$ Sorry, $m$ should be $h$; and I suppose $h$ and $\partial_i$ commute in this notation! $\endgroup$ Sep 29, 2011 at 13:46