If $A/k$ is a principally polarised ordinary abelian variety ($k$ a perfect field of characteristic $p$, we may assume it is finite for simplicity), we have a canonical lift $\hat{A}/W(k)$.
Now if I take a deformation $A\_{\epsilon}/k[\epsilon]$, does there still exist a canonical lift of this deformation to a deformation of (the generic fiber of) $\hat{A}$?


By the Kodaira-Spencer mapping deformations are essentially encoded by differentials of $A$, so the question boils down to whether differentials on $A$ lift canonically to differentials on $\hat{A}$. By Katz, Serre-Tate local moduli, Section 3, differentials on any lift $\tilde{A}$ correspond to points in $T\_p(A^\vee)(k)$, and his main theorem 3.7.1 describes the compatibility of this identification with the Kodaira-Spencer map. Is there a way to use this to lift differentials canonically on the canonical lift?



Canonical lift of the deformation of an ordinary abelian variety
No. The picture over a general base is this: let $A\_0\to S\_0$ be an ordinary abelian variety over a characteristic $p$ scheme $S\_0$, and let $S\_n$ ($n\geq 0$) be compatible flat liftings of $S\_0$ over $\mathbf{Z}/p^{n+1}$. Let $F\colon S\_0\to S\_0$ be the Frobenius. Then the pull-back $(F^n)^\* A\_0$ has a canonical lifting $A\_{n,\rm can}$ to $S\_n$.


Thus if $S\_0$ is perfect (in which case automatically $S\_n = W\_{n+1}(S\_0)$), then we can undo the Frobenius twist $(F^n)^\*$ above, and we obtain a compatible system of lifts $A\_n = ((F^{-n})^\* A\_0)\_{n, \rm can}$ to $S\_n$, i.e. a formal abelian variety over the formal scheme $S\_\infty = \varinjlim\_n\, S\_n$.


If you prefer, you can also say that $A\_0$ has a canonical lifting $\tilde A\_n$ to $W\_{n+1}(S\_0)$ for all $n$ (this is an equivalent point of view taken in a paper by Borger and Guerney). If $S\_n$ is as above, there is a canonical map $\phi\_n \colon S\_n\to W\_{n+1}(S\_0)$ such that $\phi\_n\circ i = j\circ F^n$ where $i\colon S\_0\hookrightarrow S\_n$ and $j\colon S\_0\to W\_{n+1}(S\_0)$ are the canonical closed immersions. Then $\phi\_n^\* \tilde A\_n \simeq A\_{n, \rm can}$ as abelian schemes over $S\_n$.


Unfortunately, in your case $S\_0 = \operatorname{Spec} k[\varepsilon]/(\varepsilon^2)$ the Frobenius factors through $\operatorname{Spec} k$, so the above gives you nothing. But it does give you something e.g. for $S\_0 = \operatorname{Spec} k[\![t]\!]$.


**EDIT.** Coming back to the heart of your question. Let $D=\operatorname{Def}\_{A\_0/W(k)}$ be the formal deformation space of $A\_0$ over $W(k)$. The essence of Serre-Tate theory is that $D$ carries a canonical lifting of Frobenius $F\colon D\to D$, which in particularly chosen multiplicative coordinates $q\_{ij}$ ($1\leq i,j\leq g$), i.e. $D\simeq \operatorname{Spf} W(k)[\![q\_{ij}-1]\!]$, has the particularly simple form $F^\*(q\_{ij})=q\_{ij}^p$. (These coordinates may only exist over a finite separable extension of $k$, so let us assume that $k$ is algebraically closed for simplicity.) The "canonical coordinates" $q\_{ij}$ can be recovered back from $F^\*$ once we fix a basis of the $p$-adic Tate module of $A\_0$, and then the Hodge $F$-crystal of the universal formal abelian variety over $D$ is explicitly described as in Katz's article or the accompanying article by Deligne and Illusie (with an appendix by Katz) "Cristaux ordinaires et coordonnees canoniques".


So how does the Frobenius $F\colon D\to D$ give us canonical liftings? It is easy to check (see section 1 in Katz "Travaux de Dwork") that for every lifting of Frobenius $F$ on $R\simeq W(k)[\![x\_1, \ldots, x\_n]\!]$ there exists a unique homomorphism $\sigma\_F\colon R\to W(k)$ over $W(k)$ which commutes with the Frobenius lifts, i.e. $\sigma\_F F=F\_{W(k)}\sigma\_F$ where the second $F\_{W(k)}$ is the unique Frobenius lift $F\_{W(k)} = W(F\_k)\colon W(k)\to W(k)$ on the Witt vectors. If $\operatorname{Def}\_{A\_0/W(k)} \simeq \operatorname{Spf} R$ and $F$ is the Serre-Tate lift of Frobenius, then $\sigma\_F$ corresponds to an element $\tilde A\_{\rm can}$ of $\operatorname{Def}\_{A\_0/W(k)}(W(k))$, i.e. a (formal) lifting of $A\_0$ over $W(k)$.


What happens if we replace $W(k)$ with some $p$-adic $W(k)$-algebra $V$ (a quotient of $W(k)[\![x\_1, \ldots, x\_n]\!]$ for some $n$)? Do we get some canonical lift of $A\_0$ over $\operatorname{Spf} V$? For this to make sense, we must assume that we are given some deformation $A\_{V\_0}$ of $V\_0$ over $\operatorname{Spf} V\_0$ where $V\_0 = V/pV$, and that $V$ is endowed with a Frobenius lifting $F\_V\colon V\to V$. Thus the question is whether there exists a natural (possibly unique) $\sigma\colon R\to V$ with $\sigma F=F\_V\sigma$ lifting the $R\_0\to V\_0$ induced by $A\_{V\_0}$.


The answer to this is no. For example, let $V=W\_2(k)[\varepsilon]/(\varepsilon^2)$, so $V\_0 = k[\varepsilon]/(\varepsilon^2)$. Let $F\_V\colon V\to V$ be a lifting of Frobenius, so $F\_V(\varepsilon)=\varepsilon^p + p\beta = p\beta$ for some $\beta\in V\_0$. Let $R \simeq W(k)[\![q\_{ij}-1]\!]$ as in Serre--Tate theory, so $F(q\_{ij})=q\_{ij}^p$. Let $R\_0\to V\_0$ send $q\_{ij}$ to $\varepsilon + 1$ for all $i$ and $j$ (just so that the tangent direction does not lie in any of the hyperplanes/subtori $q\_{ij}=1$). Suppose that $\sigma\colon R\to V$ lifts $R\_0\to V\_0$, so $\sigma(q\_{ij}) = \varepsilon + 1 + pg\_{ij}$ for some $g\_{ij}\in V\_0$. The condition $\sigma F = F\_V\sigma$ means that $(\varepsilon+1+pg\_{ij})^p = p\beta + 1 + pg\_{ij}^p$. But the left hand side is $(1+pg\_{ij})^p + p\varepsilon(1+pg\_{ij})^{p-1} = 1 + p\varepsilon$. We get $\beta + g\_{ij}^p = \varepsilon$. Thus $\varepsilon - \beta$ has to be a $p$-th power in $V\_0$, which is a non-trivial condition on $\beta$. (Perhaps this can be simplified)


What is true though is this. We have a canonical homomorphism $W\_{n+1}(V\_0)\to V\_n$ ($V\_n=V/p^{n+1}$), taking $(x\_0, \ldots, x\_n)$ to $y\_0^{p^n} + py\_1^{p^{n-1}} + \cdots + p^ny\_n$, where $y\_i\in V\_n$ are lifts of $x\_i\in V\_0$. Moreover, the lifting of Frobenius $F$ on $R$ induces a homomorphism $t\_F \colon R\_n\to W\_{n+1}(R\_0)$ (called "Cartier's arrow" in Illusie "Complexe de de Rham-Witt...", chapitre 0). We can now form the composition $$ \tilde\sigma\_0\colon R\_n \to W\_{n+1}(R\_0)\xrightarrow{W\_{n+1}(\sigma\_0)} W\_{n+1}(V\_0)\to V\_n $$ where $\sigma\_0\colon R\_0\to V\_0$ is the given map. This map turns out to satisfy $\tilde\sigma\_0 F = F\_{V\_n}\tilde\sigma\_0$ for *every* lifting of Frobenius $F\_{V\_n}\colon V\_n\to V\_n$. However, modulo $p$ this map reduces to $F^n\sigma\_0 = \sigma\_0 F^n$. Therefore we obtain a lifting of $(F^n)^\*A\_{V\_0}$, not of $A\_{V\_0}$, over $V\_n$. This recovers what I said in the first paragraph.


