Extension of integrable distribution over a subset

2011-08-21T06:21:18Z

Let $M$ be a smooth manifold and $G_k(M)$ be the $k$-dimensional Grassmian bundle of $M$. Let $K\subset M$ be a compact subset and $E:K\to G_k(M)$ be a continuous distribution on $K$.

We say $E$ is integrable on $K$ if there exists a foliation $\mathcal{F}$ (or lamination, since it may only foliates a subset of $M$), such that $T_x\mathcal{F}(x)=E_x$.

My quesion is: if $(E,K)$ is integrable, will there exist an open neighborhood $U\supset K$ that admits an integrable extension $\widetilde{E}:U\to G_k(M)$?

For example $K$ is a hyperbolic invariant set of a diffeomorphism $f:M\to M$. It is known that there are stable and unstable foliations (manifolds) through $K$. I donot know if we can extend the foliations to an open neighborhood of $K$.

Answer by jvp for Extension of integrable distribution over a subset

2011-08-24T03:34:13Z

The answer in general is no.

If $K$ is a submanifold of $M$ then tangent bundle of $K$ defines an integrable distribution on $K$. To wit we are talking about the foliation with just one leaf: $K$.

If we can extend this foliation to a neighborhood of $K$ then the restriction of Bott's connection to $K$ induces a flat connection on the normal bundle of $K$. This imposes restrictions on $K$. For instance $K$ cannot be a projective curve in $\mathbb P^2(\mathbb C)$.

Recall that Bott's connection is the partial connection on the normal bundle of $\mathcal F $ defined as follows. If $T\mathcal F$ is the tangent bundle of a foliation and $N\mathcal F$ is its normal bundle then they fit into the exact sequence: $$ 0 \to T \mathcal F \to TM \stackrel{\pi}{\to} N\mathcal F \to 0 . $$ Now, if $v$ is a local section of $T\mathcal F$ and $w$ is a local section of $N\mathcal F$ then Bott's partial connection $\nabla : T\mathcal F \times N\mathcal F \to N \mathcal F$ is defined by the formula
$$ \nabla_{v}(w) = \pi ([v, \pi^{-1}(w)]) $$ where $\pi^{-1}(w)$ is an arbitrary lifting of $w$ to $TM$. The involutiviness of $T\mathcal F$ implies that $\nabla$ is well-defined. It is not hard to check that $\nabla^2=0$.

Extension of integrable distribution over a subset - MathOverflow

Extension of integrable distribution over a subset

Answer by jvp for Extension of integrable distribution over a subset