The Unapologetic Mathematician

Mathematics for the interested outsider


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


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 Posted by | Differential Topology, Topology | 7 Comments