Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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

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 $g_\ast$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.