0
$\begingroup$

Is there an example of a $n$ dimensional manifold $M$ and a natural number $k<n$ with a Lie subalgebra $L$ of $\chi^{\infty}(M)$ with the following property:

For every $x\in M$ the space $\{V_x \in T_x M\mid V\in L\} $ is a $k$ dimensional vector space $D_x$ but the distribution $D$ consisting of all $D_x,\;x\in M$ is not an integrable distribution.

In the other word we search for a non integrable distribution $D$ of a manifold and a Lie algebra $L$ of vector fields such that $L$ is $D$-ample where $D$- ample means that the evaluation $L_x$ of $L$ at every point $x$ is equal to $D_x$.

$\endgroup$

1 Answer 1

4
$\begingroup$

No, this a consequence of the Frobenius theorem. Let $x\in M$, you can find $X_1,..,X_k\in L$ which is a basis of $D_x$, for $Y,Z$ tangent to $D$ in a neighbourood $U$ of $x$, you can write $Y=f_1X_1+...+f_kX_k$ and $Y=g_1X_1+...+g_kX_k$, $[X,Y](y)\in D_y, y\in U$. You can apply Frobenius.

Let $f,g$ be functions and $U,V\in L$, $[fU,gV]=d(gV).(f(U)-d(fU).(gV)=$

$(fdg(U))V+fgdV.U-(gdf(V))U-fgdU(V)=(fdg(U))V-(gdf(V))U+fg[U,V]$, since $[U,V]\in L$, we deduce that $[fU,gV]$ is tangent to $D$.

https://en.wikipedia.org/wiki/Frobenius_theorem_(differential_topology)#Formulation_using_vector_fields

$\endgroup$
0

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .