Thread: involutive subbundle and Lie bracket

1. involutive subbundle and Lie bracket

Hello,

i have a question about involutive subbundles.
If M is a manifold and E is a subbundle. We say E is involutive, if for each smooth vector fields X,Y of E, (that is X(p),Y(p) $\displaystyle \in E_p$) also the lie bracket [X,Y] is a smooth vector field to E.

I think that not any subbundle is involutive, but i don't know why?
Whats wrong with this proof?

Claim:
If X,Y are smooth vector fields to E, =>[X,Y] is also a vect. field:

Pf:
[X,Y](p)=$\displaystyle X_p (Y) - Y_p (X)$. since X,Y are vector fields and E_p is a linear subspace, therefore the sum above is also in E_p??

Where is the mistake? I don't see it.

Regards

2. Take $\displaystyle M=R^3$. Let $\displaystyle X=\frac{\partial}{\partial x}$, $\displaystyle Y=\frac{\partial}{\partial y}+x \frac{\partial}{\partial z}$.
Obviously X, Y are nowhere dependent, So they span a sub-space span{X,Y} in the tangent space of each point of M. Let $\displaystyle E=\cup_{p\in R^3} span\{X, Y\}$. E is a smooth sub-bundle.
Now you can get easily compute the result $\displaystyle [X,Y]=\frac{\partial}{\partial z}$, which cannot be expressed by linear combination of X and Y, thus doesn't belong to E.

3. Actually, according to Frobenius Theorem

In the vector field formulation, the theorem states that a subbundle of the tangent bundle of a manifold is integrable (or involutive) if and only if it arises from a regular foliation. In this context, the Frobenius theorem relates integrability to foliation;

So involutivity means the subbundle is locally the tangent bundle of a sub-manifold.