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 - Posted by | Differential Topology, Topology

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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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

5. [...] say that we have a one-dimensional distribution on a manifold . Around any point we can find a patch and an everywhere-nonzero vector field on [...]

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6. [...] of the foliation. We also ask that the tangent spaces to the leaves define a -dimensional distribution on , which we say is “induced” by , and that any connected integral submanifold of [...]

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