# The Unapologetic Mathematician

## Distributions

A vector field $v$ defines a one-dimensional subspace of $\mathcal{T}_pM$ at any point $p$ with $v_p\neq0$: the subspace spanned by $v_p\in\mathcal{T}_pM$. If $v$ is everywhere nonzero, then it defines a one-dimensional subspace of each tangent space. A distribution generalizes this sort of thing to higher dimensions.

To this end, we define a $k$-dimensional distribution $\Delta$ on an $n$-dimensional manifold $M$ to be a map $p\mapsto\Delta_p\subseteq\mathcal{T}_pM$, where $\Delta$ is a $k$-dimensional subspace of $\mathcal{T}_pM$. Further, we require that this map be “smooth”, in the sense that for any $q\in M$ there exists some neighborhood $U$ of $q$ and $k$ vector fields $X_1,\dots,X_k$ such that the vectors $X_i(r)\in\mathcal{T}_rM$ span $\Delta_r$ for each $r\in U$.

Notice here that the $X_k$ don’t have to work for the whole manifold $M$. Indeed, we will see that in many cases there are no everywhere-nonzero vector fields on a manifold $M$. But over a small patch $U$ we might more easily find $k$ vector fields that are linearly independent at each point, and thus define a smooth $k$-dimensional distribution over $U$. Then more general smooth distributions come from patching these sorts of smooth distributions together.

A vector field $X$ on $M$ “belongs to” a distribution $\Delta$ — which we write $X\in\Delta$ — if $X_p\in\Delta_p$ for all $p\in M$. We say that $\Delta$ is “integrable” if $[X,Y]\in\Delta$ for all $X$ and $Y$ belonging to $\Delta$.

Every one-dimensional manifold is integrable. To see this, we note that if $X$ and $Y$ belong to $\Delta$ then $X_p=f(p)Y_p$ for some constant $f(p)$, at least at those points $p\in M$ where $Y_p\neq0$. Thus we see that

$\displaystyle[X,Y]=[fY,Y]=f[Y,Y]-(Yf)Y=-(Yf)Y$

and so $[X,Y]$ is proportional to $Y$, and thus belongs to $\Delta$. To handle points where $Y_p=0$, we can put the scalar multiplier on the other side.

June 28, 2011