The Unapologetic Mathematician

Mathematics for the interested outsider

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

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