# Relative tangent bundle of the pullback of a vector bundle splits?

Let $S$ be a complex manifold, which for our purposes we can take to be a small ball in $\mathbb C^n$. Let $p : E \to S$ be a holomorphic vector bundle over $S$, and let $\pi : X \to S$ be a family of compact complex manifolds over $S$ (i.e. $X$ is a complex manifold, and $\pi$ is a holomorphic proper submersion).

We can pull the vector bundle $E$ back to $X$. This gives a vector bundle $\pi^*E \to X$, which we can regard as a fiber bundle $\pi^*E \to S$ over $S$.

Does the relative tangent bundle of the fiber bundle $\pi^*E \to S$ split holomorphically as the sum of the relative tangent bundles of $X \to S$ and $E \to S$? In symbols, do we have $T_{\pi^*E/S} = T_{X/S} \oplus T_{E/S}$ (the latter two bundles having been pulled back to $\pi^*E$)?

For a situation where such a thing appears naturally, take $E \to S$ to be the Hodge bundle of the fibers of $X$, so it has fibers $E_s = H^k(X_s,\mathbb C)$. Then $\pi^*E \to S$ has fibers $X_s \times H^k(X_s,\mathbb C)$.

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Yes, of course. Both E and X have holomorphic submersions to $S$,and so does their fiber product over $S$. The fiber of the latter over $s\in S$ is the product $E_s\times X_s$. The tangent space of this fiber splits canonically just as you suggest. – Tom Goodwillie Apr 26 '11 at 14:59
@Tom: Thanks for that. You wouldn't happen to know of a reference for this? (I imagine this is pretty basic stuff, so it should be in a textbook somewhere, right?) – Gunnar Þór Magnússon Apr 26 '11 at 15:32
See Hartshorne exercise II.8.3 for a more general statement. – mdeland Apr 26 '11 at 16:27