[Math] Definition of a distribution and integral manifolds

differential-geometry

We are currently talking about the various forms of the Frobenius theorem in my differential geometry class in order to build up integrability. For one of the versions, we use distributions, and I'd like to get a few things straight.

  • Given a manifold $M$ of dimension $n$, I have that a distribution, $\mathcal{D}$, is a sub-bundle (subspace) of the tangent space $TM$ such that $\mathcal{D}\subset TM$ and $\mathcal{D}_p\subset T_pM$ for all $p\in M$. I was also told to think of the distribution as $\mathcal{D}=\text{span}\{X_1,\cdots,X_r\}$ where $X_1,\cdots,X_r$ are linearly independent vector fields in the tangent space of $M$.

  • We also distinguished a distribution from a smooth distribution by adding the requirement that vectors must vary smoothly from point to point.

  • Given two vector fields $X,Y \in \mathcal{D}$ with $X_p,Y_p\in \mathcal{D}_p$ if $[X,Y]\in \mathcal{D}$ then there exists an integral manifold which is a sub-manifold of the distribution; that is $\mathcal{D}$ is integrable. We also say that the distribution $\mathcal{D}$ is involutive.

Do you see any errors here? I'm fairly new to the material so any additional comments would be appreciated as well.

Thanks.

Best Answer

I think you're getting a handle on the basic idea, but there are a number of details in your statements that aren't quite correct.

Given a manifold $M$ of dimension $n$, I have that a distribution, $\mathcal{D}$, is a sub-bundle (subspace) of the tangent space $TM$ such that $\mathcal{D}\subset TM$ and $\mathcal{D}_p\subset T_pM$ for all $p\in M$.

Yes, a distribution is a subbundle of $TM$. But it's not really correct to say it's a subspace of $TM$. Of course, "subspace" can have different meanings, depending on context. It is true that $\mathcal D$ is a topological subspace of $TM$ if we endow it with the subspace topology, but then that's true of any subset of $TM$, or of any other topological space. On the other hand, it wouldn't make sense to say that $\mathcal D$ is a linear subspace of $TM$, because $TM$ is not a vector space.

The fact that it's a subbundle says everything that needs to be said -- by definition, this means that it's a subset of $TM$ that is an embedded submanifold, such that each fiber $\mathcal D_p$ is a linear subspace of $T_pM$, and such that $\mathcal D$ is a vector bundle in its own right when each fiber $\mathcal D_p$ is given the vector space structure that it inherits from $T_pM$. Implicitly, this also implies a few other things, such as that each fiber $\mathcal D_p$ is nonempty, and the fibers are all the same dimension.

I was also told to think of the distribution as $\mathcal{D}=\text{span}\{X_1,\cdots,X_r\}$ where $X_1,\cdots,X_r$ are linearly independent vector fields in the tangent space of $M$.

This is not quite true, but it is true locally -- Each point of $M$ has a neighborhood $U$ such that $TU\cap \mathcal D$ is locally the span of an $r$-tuple of linearly independent vector fields. More precisely, what this means is that there are $r$ vector fields $X_1,\dots,X_r$ on $U$ such that for each $p\in U$, $$ \mathcal D_p = \operatorname{span}(X_1|_p,\dots,X_r|_p). $$ The main thing wrong with your statement is that it might not be possible to find such vector fields globally on $M$.

We also distinguished a distribution from a smooth distribution by adding the requirement that vectors must vary smoothly from point to point.

"The requirement that vectors must vary smoothly from point to point" is a little too vague to be useful. What vectors? Here's a more precise statement of what it means for $\mathcal D$ to be a smooth distribution:

  • Each point of $M$ has a neighborhood $U$ on which there are smooth independent vector fields $X_1,\dots,X_r$ that span $\mathcal D\cap TU$ in the sense described above.

Given two vector fields $X,Y \in \mathcal{D}$ with $X_p,Y_p\in \mathcal{D}_p$ if $[X,Y]\in \mathcal{D}$ then there exists an integral manifold which is a sub-manifold of the distribution; that is $\mathcal{D}$ is integrable. We also say that the distribution $\mathcal{D}$ is involutive.

You've got some misplaced quantifiers here. It should say

  • We say $\mathcal D$ is involutive if it has the following property: for all smooth vector fields $X,Y$ on $M$ that satisfy $X_p,Y_p\in \mathcal D_p$ for each $p\in M$, we also have $[X,Y]_p\in \mathcal D_p$ for each $p\in M$.
  • If $\mathcal D$ is involutive, then it has the property that through every point of $M$ there is a integral manifold of $\mathcal D$, which is an $r$-dimensional submanifold $S\subseteq M$ whose tangent space at each point $q\in S$ is equal to $\mathcal D_q$. (This is called the Frobenius theorem.)

(It's not correct to say that an integral manifold is a "submanifold of the distribution," since the distribution is a subset of $TM$ while an integral manifold is a subset of $M$.)

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