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I have a problem with Kuranishi's theorem in deformation theory. I'll try to formulate it in general terms, and then describe the particular situation.

Let $\pi : M \to S$ be a smooth fiber bundle - i.e. $M$ and $S$ are smooth manifolds, and $\pi$ is a surjective submersion. There is an associated surjective morphism of vector bundles $ \pi_* : T_M \to \pi^* T_S$. I want to find a lifting of $\pi^* T_S$ into $T_M$.

Suppose I can find a smooth map $f : S \to M$ which satisfies $\pi \circ f = id_S$. This induces an injective map $T_S \to f^*T_M$. Can I lift this to a map $\pi^* T_S \to T_M$? How about if some extra data is given, like a metric on $S$, or a family of metrics $g_s$ on the fibers $T_M |_{M_s}$ (where $s$ is a parameter in $S$)?

Basically I'm trying to use Kuranishi's theorem to get something like Siu's canonical lifts. In this situation $M$ is the product of a fixed smooth manifold and the space of its complex structures, and $S$ is a complex manifold (open ball, even). The map $\pi$ is the passing to the quotient by the action of the group of diffeomorphisms. If we fix a hermitian metric $h$ on $M_0$, then Kuranishi gives a map $f : S \to M$ which satisfies the above hypothesis. I'm told that Kuranishi should induce a lifting of $\pi^*T_S$ into $T_M$, but I can't seem to figure out how.

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I'm a bit confused. A map $\pi^* T_S\to T_M$ splitting the map $T_M\to\pi^*T_S$ is called an Ehresmann connection on $M\to S$. Ehresmann connections are sections of an affine $A$ bundle over $S$ (modeled by the vector bundle $\pi^* T_S^* \otimes K$, where $K$ is the kernel of $\pi^*$). Hence a $C^\infty$ Ehresmann connection always exists, and you can also extend a section of $A$ from a submanifold. Is $C^\infty$ not good enough? –  user8268 Apr 11 '11 at 19:38
    
Sure, $C^\infty$ is good enough, but there are lots of smooth splittings $T_M \to \pi^*T_S$ (any metric on $M$ gives one) and I want one induced by the map $f : S \to M$ somehow. Can we use $f$ to construct a specific Ehresmann connection? –  Gunnar Þór Magnússon Apr 11 '11 at 19:50

1 Answer 1

Perhaps now you no longer need an answer, and I am not sure to have really understood your question, but anyway I tried.

My notations
In this paragraph I briefly review my notations and definitions, so that you can easily check if and how they are in agree with yours.

Given a vector bundle $\pi:E\to P$, and a smooth map $f:M\to E$, with the pull-back of $\pi$ through $f$ I mean a vector bundle $f^\ast\pi:f^\ast E\to E$ together with a smooth bundle map $\pi^\ast f$ from $f^\ast\pi$ to $\pi$ over $f$, which solve the following universal mapping problem:

if $\rho:F\to N$ is a vector bundle, $g:N\to M$ is a smooth map and $\phi$ is a smooth bundle map from $\rho$ to $\pi$ over $f\circ g$ then there exists a unique smooth map $h:F\to f^\ast E$ with the property that $(f^\ast\pi)\circ h=g\circ\rho$ and $(\pi^\ast f)\circ h=\phi$.

The answer
When $f:M\to N$ is a smooth map, we construct $(f^\ast\tau_N,\tau_N^\ast f)$, the pullback through $f$ of $\tau_N$, the tangent bundle over $N$. From the commutative diagram $\tau_N\circ(Tf)= f\circ\tau_M$ and the previous universal property of vector bundle pullbacks we find that there exists a unique smooth map $f_\ast:TM\to f^\ast(TN)$ such that $$(\tau_N^\ast f)\circ f_\ast=Tf\ \mathrm{and}\ (f^\ast\tau_N)\circ f_\ast=\tau_M.$$

Analogously when $g:N\to M$ is another smooth map we make the same construction, and, combining commutative diagrams, it is easy to realize that $$(\tau_N^\ast f)\circ f_{\ast}\circ(\tau_M^\ast g)\circ g_\ast=T(f\circ g).$$

Now from this we conclude that:

if $f$ is a left inverse of $g$ then $(\tau_N^\ast f)\circ f_{\ast}\circ(\tau_M^\ast g)$ is a left inverse of $g_\ast$.

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