Canonical lift of the deformation of an ordinary abelian variety 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?
 A: 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.
