The Unapologetic Mathematician

Mathematics for the interested outsider

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.

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June 28, 2011 - Posted by | Differential Topology, Topology

7 Comments »

  1. I’m not understanding the last equation [fY,Y] = f[Y,Y] – (Y f)Y — could you elaborate?

    Thanks!

    Comment by Joe English | June 29, 2011 | Reply

  2. Well, let’s work it out:

    \displaystyle\begin{aligned}{}[fX,Y]g&=fXYg-Y(fX)g\\&=fXYg-Y(fXg)\\&=fXYg-\left((Yf)Xg+fYXg\right)\\&=f(XYg-YXg)-(Yf)Xg\\&=f[X,Y]g-(Yf)Xg\end{aligned}

    Now just use Y for both X and Y.

    Comment by John Armstrong | June 29, 2011 | Reply

    • OK, I get it now, thanks.

      I think my difficulty comes from confusion over when juxtaposition denotes multiplication in the ring, multiplication of the ring over the module, function application, function composition, or even something else. (I had similar difficulties with representation theory :-) In a term like “fXYg”, the empty spaces between the letters can mean different things depending on how it’s parenthesized; I still haven’t wrapped my head around whether all the different ways make sense and that (among the ones that do make sense — all of them?) they all mean the same thing.

      Comment by Joe English | June 30, 2011 | Reply

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    Pingback by Integral Submanifolds « The Unapologetic Mathematician | June 30, 2011 | Reply

  4. It’s possible to fully-parenthesize these sorts of expressions, but then they get confusing for a different reason. In general, I try to write things out so they associate to the right, because of function application, and use parentheses for particularly ambiguous setups.

    In fXYg, we could write Y(g), but we’ll conventionally drop the parens and write Yg for the application of the vector field to the function. There’s no way of “multiplying” vector fields, so XYg has to be interpreted as X(Yg). And it doesn’t matter whether we multiply X by f before applying it to Yg or after, since f(X(Yg)) and [fX](Yg) give the same result.

    Comment by John Armstrong | June 30, 2011 | Reply

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    Pingback by Integrable Distributions Have Integral Submanifolds « The Unapologetic Mathematician | June 30, 2011 | Reply

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