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I am hoping that someone can point out the error in the "proof" of the following "theorem":

Theorem: Let $k$ be a perfect field of characteristic $p>2$ and let $G$ be an ordinary $p$-divisible group over $W(k)$. Then the connected-etale sequence of G is split: $G\simeq G^m \times G^{et}$.

The "Theorem" is decidedly false. For example, extensions of $\mathbf{Q}_p/\mathbf{Z}_p$ by $\mathbf{G}_m[p^{\infty}]$ over $W(k)$ are classified by the abelian group $1+pW(k)$, so any nonzero element of this group gives an ordinary $p$-divisible group with non-split connected-etale sequence.

Proof: Let $G_0:=G\times_{W(k)} k$ be the special fiber of $G$. Since $k$ is perfect, $G_0= G_0^{m}\times G_0^{et}$. By Messing (LNM 264, Chap V. Theorem 1.6), there is an equivalence of categories between deformations of $G_0$ to $W(k)$ and free $W(k)$-submodules $L$ of $D(G_0)(W(k))$ lifting $\omega_{G_0}\hookrightarrow D(G_0)(k)$. This equivalence is induced by sending a lift $G'$ of $G_0$ to $\omega_{G'}$. Now $G' = G^{m}\times G^{et}$ and $G$ both lift $G_0$, and these lifts correspond to the submodules $\omega_{G'}$ and $\omega_G$ of $D(G_0)(W(k))$, respectively. But since $G$ is ordinary, so $G^0=G^{m}$, the pullback map $\omega_{G}\rightarrow \omega_{G^m}$ is an isomorphism. Via this isomorphism, the map $\omega_{G'}\rightarrow D(G_0)(W(k))$ coincides with the composite $$ \omega_{G'}\simeq \omega_{G^m}\times \omega_{G^{et}} = \omega_{G^m}\simeq \omega_G\rightarrow D(G_0)(W(k)) $$ and we conclude that $\omega_{G'}=\omega_G$ as submodules of $D(G_0)(W(k))$ lifting $\omega_{G_0}$. It then follows from Messing's Theorem above that $G\simeq G'$, as claimed.

I must be making some silly and obvious mistake...can you find it?

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I think you're implicitly asserting that the map $\omega_{G^m}\rightarrow D(G_0)(W(k))$ coming from $G'$ is the same as the composition $\omega_{G^m}\rightarrow \omega_G\rightarrow D(G_0)(W(k))$. –  Rebecca Bellovin Dec 25 '11 at 17:41
Or explicitly, as the case may be :), and this must be where the error lies. The formation of $$0\rightarrow \omega_G\rightarrow D(G)(W)\rightarrow Lie(G^t)\rightarrow 0$$ is functorial in $G$; applying this to the connected-etale sequence of $G$ shows that the map $\omega_{G^m}\rightarrow D(G^m)(W)$ coincides with the composition of $\omega_{G^m}\simeq \omega_G\rightarrow D(G)(W)$ with the projection to $D(G^m)(W)$, so I guess the problem is that the isomorphism $D(G)(W)\simeq D(G^m)(W)\times D(G^{et})(W)$ does not coincide with projection to $D(G^m)$ followed by inclusion (clearly). –  B. Cais Dec 25 '11 at 18:26
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