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Fix a hyperkähler manifold $X$ and an identification of $S^2$ with the hyperkähler sphere of $X$. Now consider the twistor space $T := S^2\times X$ equipped with the tautological complex structure. For each $x\in X$, we have a holomorphic map $u_x:S^2\to T$ defined by $u_x(\theta):=(\theta,x)$.

Question: Is every holomorphic map $u:S^2 \to T$ which satisfies ${\rm pr}_1\circ u={\rm id_{S^2}}$ of this form?

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@680: I edited the post to fix a math display problem. Please double-check to see no error was introduced. – Willie Wong Dec 14 '10 at 1:25
up vote 3 down vote accepted

I think preprint arXiv:1006.0440 of Jardim is Verbitsky will answer your question. In short the answer is no, since if $\dim_C X = n$ then the deformation space of sections has dimension $\dim H^0(S^2,N_{S^2/S^2\times X}) = \dim H^0(P^1,O_{P^1}(1)^n) = 2n$ which is twice the dimension of the space which you consider.

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