I'm doing this problem because I my email record has no one claiming it, and it's still empty here. I'm sorry if I've stolen your problem. -AndrewC

#### Problem

A *distribution* $D$ on a manifold $M$ assigns to each $x \in M$ a linear subspace of $T_x M$. A distribution $D$ is said to be *involutive* if $[X, Y] \in D$ for any vector fields $X, Y \in D$, where $[\cdot, \cdot]$ denotes the Jacobi-Lie bracket.

Let $D = \text{span} \{X, Y\}$ on $\mathbb{R}^3$, where $X = z \frac{\partial}{\partial x} + \frac{\partial}{\partial z}$ and $Y = \frac{\partial}{\partial y} + \frac{\partial}{\partial z}$. Is $D$ involutive? Give an example of a nonconstant involutive distribution on $\mathbb{R}^2$.

#### Solution

$D$ is not involutive. To show it's not involutive we need only find a particular pair of vector fields which are not closed under the the bracket. We'll pick the basis elements as our vector fields and show the bracket operation is not closed.

(1)This does not lie in the span of $X$ and $Y$, so $D$ is not involutive.

An example of a non-constant involutive distribution on $\mathbb{R}^2$ is the span of the vector field $X = x \frac{\partial}{\partial x}$. This is trivially involutive as the Jacobi-Lie bracket $[X,X] = 0$, and 0 is in every subspace of $T_xM$. To be more explicit, let $V$ and $U$ be any vector fields in the span of $X$. They can be written as $V = aX$ and $U = bX$; thus, $[U,V] = ab[X,X] = 0$ by bi-linearity and anti-symmetry of the bracket.

Scott agrees!

#### Discussion

*[kcrane]* Hey - either your definition of the Jacobi-Lie bracket or your evaluation of the bracket is backwards. (Check: $J(X)$ shouldn't be zero since $X$ isn't constant.)

His typo was in the definition of the Jacobi-Lie bracket; I corrected it.