The Unapologetic Mathematician

Integrable Distributions Have Integral Submanifolds

Let’s say that we have a one-dimensional distribution on a manifold $M$. Around any point $p\in M$ we can find a patch $V$ and an everywhere-nonzero vector field $X$ on $V$ so that $X_q$ spans $\Delta_q$ for every $q\in V$. Then we know we can find a chart $(U,x)$ around $p$ such that $\frac{\partial}{\partial x^1}=X$ on $U$. Then the curve with coordinates $x^1=t$ and $x^i=0$ for all other $i\neq1$ is a one-dimensional integral submanifold of $\Delta$ through $p$.

Now, this doesn’t always work for every distribution $\Delta$. It turns out that the key ingredient is that all one-dimensional distributions are integrable; we can show that any integrable $k$-dimensional distribution has an integrable submanifold through any point $p$.

To give an even more detailed statement, let $\Delta$ be a $k$-dimensional integrable distribution on $M$. for every $p\in M$ there is a chart $(U,x)$ with $x(p)=0$, $x(U)=(-1,1)^n$, and so that for any $a_{k+1},\dots,a_n\in(-1,1)$ the set $\{q\in U\vert x^i(q)=a_i\}$ is an integral submanifold of $\Delta$. Further, any connected integral submanifold of $\Delta$ comes in this way.

Since this statement is purely local we can get away with working in some set of local coordinates to start with, although obviously not the one we’re trying to find. As such, we will assume that $M=\mathbb{R}^n$, that $p=0$, and that $\Delta_0$ is spanned by the $\left(D_i\right)_0$ for $1\leq i\leq k$. We’ll also let $\pi:\mathbb{R}^n\to\mathbb{R}^k$ be the projection onto the first $k$ components, so $\pi_*\vert_{\Delta_0}:\Delta_0\to\mathcal{T}_0\mathbb{R}^k$ is an isomorphism. By parallel translation, $\pi_*\vert_{\Delta_q}:\Delta_q\to\mathcal{T}_{\pi(q)}\mathbb{R}^k$ for all $q$ in a neighborhood $V$ of $0$.

Now, like we did yesterday we can use these isomorphisms to build $k$ vector fields $X_i$ on $V$ belonging to $\Delta$ that are $\pi$-related to the $D_i$ on $\mathbb{R}^k$. Then $\pi_*[X_i,X_j]=[D_i,D_j]=0$, but since $\Delta$ is integrable we know that $[X_i,X_j]\in\Delta$. Since $\pi_*$ is an isomorphism on $\Delta$, we conclude that $[X_1,X_j]=0$. Now we can find a coordinate patch $(U,x)$ around $0$ with $\frac{\partial}{\partial x^k}=X_k$ on $U$, just as in the one-dimensional case. It’s no loss of generality to tweak it until we have $x(U)=(-1,1)^n$. This gives us an integral submanifold through the origin.

But our assertion goes further! Let $f=\pi_2\circ x:U\to(-1,1)^{n-k}$, where $\pi_2$ is the complementary projection to $\pi$. This map has maximal rank everywhere, so we know that for each $a\in(-1,1)^{n-k}$ the preimage $f^{-1}(a)$ is an $n-(n-k)=k$-dimensional submanifold $N\subseteq M$. The tangent space to $N$ consists of exactly those vectors in the kernel of $f_*$, but since $x$ is a diffeomorphism these are exactly those vectors $v$ such that $x_*(v)$ is in the kernel of $\pi_{2*}$ — those $v\in\Delta$.

Conversely, if $N$ is a connected integral manifold of $\Delta$ contained in $U$. If $v\in\mathcal{T}_qN$, then $\iota_*v$ is in $\Delta_q$, which is spanned by the $\frac{\partial}{\partial x^i}$ for $1\leq i\leq k$. Thus $\iota_*v(x^{k+j})=0$. And so $\left((x^{k+j}\circ\iota)_*\right)_q=0$ for all $q\in N$. Since $N$ is connected, $x^{i+j}\circ\iota$ is constant, and thus $N$ comes from picking a value $a_{k+j}$ for each $1\leq j\leq n-k$.

June 30, 2011 - Posted by | Differential Topology, Topology