Let $\mathcal X \to S$ be the local universal family of an elliptic curve, and let $E \to S$ be a vector bundle over $S$. Then we can form the fiber product $\mathcal Y = \mathcal X \times_S E$, which will be a smooth complex manifold as $\mathcal X$ and $S$ are smooth, and look at the tangent bundle of this fiber product.
Question: If the vector bundle $E$ is trivial, $E = S \times \mathbb C^r$, then does the tangent bundle of $Y$ split holomorphically as $T_{\mathcal Y} = T_{\mathcal X} \oplus \mathbb C^r$? Or equivalently, as $\mathbb C^r = T_{E/S}$, does $T_{\mathcal Y} = T_{\mathcal X} \oplus T_{E/S}$?
I'm really only interested in the situation where $E$ is the relative Hodge bundle $H^{1,1}(\mathcal X/S, \mathbb C)$, whose fibers are the groups $H^{1,1}(X_s, \mathbb C)$. This is a holomorphic line bundle for any family of Riemann surfaces over a smooth base, and it seems to be trivial for a family of elliptic curves because the (1,1) form $dz \wedge d\bar z$ on $\mathbb C$ is invariant under everything we do and should descend to give a generator of each fiber of the relative Hodge bundle.
I've got a feeling the splitting is true, but I haven't been able to show it so far. The thing is we can index the family $\mathcal X$ by a parameter $\tau$ in the upper half-plane. We can then write down the dual basis of the lattice which defines $X_\tau$ and get two 1-forms on $X_\tau$ which depend on $\tau$ and generate each $H^1(X_\tau,\mathbb C)$ as a $\mathbb R$-vector space. Then we can take the wedge product of those forms and multiply by a smooth function to obtain the holomorphic (1,1)-form above. We can now give any element in a fiber of the relative Hodge bundle as a scalar multiple of this (1,1)-form. But this (1,1)-form doesn't depend on $\tau$, but only on the scalar we multiply by. My head keeps telling me this means the tangent bundle splits, but like I said, I can't prove it so far.
