# PDE’s and Frobenius Theorem

I’ve started many blog posts on algebra/algebraic geometry, but they won’t get finished and posted for a little while. I’ve been studying for a test I have to take in a few weeks in differential geometry-esque things. So I’ll do a few posts on things that I think are usually considered pretty easy and obvious to most people, but are just things I never sat down and figured out. Hopefully this set of posts will help others who are confused as I recently was.

My first topic is about the Frobenius Theorem. I’ve posted about it before. Here’s the general idea of it: If ${M}$ is a smooth manifold and ${D}$ is a smooth distribution on it, then ${D}$ is involutive if and only if it is completely integrable (i.e. there is are local flat charts for the distribution).

What does this have to do with being able to solve partial differential equations? I’ve always heard that it does, but other than the symbol ${\displaystyle\frac{\partial}{\partial x}}$ appearing in the defining of a distribution or of the flat chart, I’ve never figured it out.

Let’s go through this with some examples. Are there any non-constant solutions ${f\in C^\infty (\mathbb{R}^3)}$ to the systems of equations: ${\displaystyle \frac{\partial f}{\partial x}-y\frac{\partial f}{\partial z}=0}$ and ${\displaystyle \frac{\partial f}{\partial y}+x\frac{\partial f}{\partial z}=0}$?

Until a few days ago, I would have never thought we could use the Frobenius Theorem to do this. Suppose ${f}$ were such a solution. Define the vector fields ${\displaystyle X=\frac{\partial}{\partial x}-y\frac{\partial}{\partial z}}$ and ${\displaystyle Y=\frac{\partial}{\partial y}+x\frac{\partial}{\partial z}}$ and define the distribution ${D_p=\text{span} \{X_p, Y_p\}}$.

Choose a regular value of ${f}$, say ${C}$ (one exists by say Sard’s Theorem). Then ${f=C}$ is a 2-dimensional submanifold ${M\subset \mathbb{R}^3}$, and since ${f}$ is a defining function ${T_pM=ker(Df_p)}$. But the very fact that ${f}$ satisfies, by assumption, ${X(f)=0}$ and ${Y(f)=0}$, we have ${T_pM=\text{span} \{X_p, Y_p\}}$. I.e. ${M}$ is an integral manifold for the distribution ${D}$. Thus ${D}$ must be involutive.

Just check now. ${\displaystyle [X,Y]=2\frac{\partial}{\partial z}}$, so in particular at the origin ${\displaystyle X_0=\frac{\partial}{\partial x}}$ and ${\displaystyle Y_0=\frac{\partial}{\partial y}}$ it is not in the span, and hence not involutive. Thus no such ${f}$ exists. This didn’t even use Frobenius.

Now let’s spice up the language and difficulty. Is it possible to find a function ${z=f(x,y)}$, ${C^\infty}$ in a neighborhood of ${(0,0)}$, such that ${f(0,0)=0}$ and ${\displaystyle df=(ye^{-(x+y)}-f)dx+(xe^{-(x+y)}-f)dy}$? Alright, the ${d}$ phrasing is just asking there is a local solution to the system ${\displaystyle \frac{\partial f}{\partial x}=ye^{-(x+y)}-f}$ and ${\displaystyle \frac{\partial f}{\partial y}=x^{-(x+y)}-f}$. Uh oh. The above method fails us now since it isn’t homogeneous.

Alright, so let’s extrapolate a little. We have a system of the form ${\displaystyle \frac{\partial f}{\partial x}=\alpha(x,y,f)}$ and ${\displaystyle \frac{\partial f}{\partial y}=\beta(x,y,f)}$. The claim is that necessary and sufficient conditions to have a local solution to this system is ${\displaystyle \frac{\partial \alpha}{\partial y}+\beta\frac{\partial \alpha}{\partial z}=\frac{\partial \beta}{\partial x}+\alpha \frac{\partial \beta}{\partial z}}$.

I won’t go through the details of the proof, but the main idea is not bad. Define the distribution spanned by ${\displaystyle X=\frac{\partial}{\partial x}+\alpha\frac{\partial}{\partial z}}$ and ${\displaystyle Y=\frac{\partial}{\partial y}+\beta\frac{\partial}{\partial z}}$.

Then use that assumption to see that ${[X,Y]=0}$ and hence the distribution is involutive and hence there is an integral manifold for the distribution by the Frobenius Theorem. If ${g}$ is a local defining function to that integral manifold, then we can hit that with the Implicit Function Theorem and get that ${z=f(x,y)}$ (the implicit function) is a local solution.

If we go back to that original problem, we can easily check that the sufficient condition is met and hence that local solution exists.

I had one other neat little problem, but it doesn’t really fit in here other than the fact that solutions to PDEs are involved.

# Everywhere Normal Vector Field

The title is maybe a little misleading, but here we go.

Not so much a standard problem, but a neat little result. Let $U\subset \mathbb{R}^3$. Let $X$ be a nowhere vanishing vector field on U. Then for each point $p\in U$ there is a surface passing through $p$ such that $X$ is normal to the surface if and only if $\langle X, curl(X)\rangle=0$.

This is nice since it ties back to the Frobenius posts. If $X=(X_1, X_2, X_3)$ in coordinates, then define $\omega=X_1dx+X_2dy +X_3dz$. Some texts use “musical notation,” which is amazingly effective once you get used to it. In which case, we would just say let $\omega=X^\flat$. Now $\omega$ is a smooth 1-form, so it defines a 2-dimensional distribution on $U$ in the standard way $D_p=ker\omega\big|_p$.

Now by definition, $T_pU=D_p\oplus X_p$, so our problem has been rephrased in Frobenius language to say D is integrable iff $\langle X, curl(X)\rangle=0$ since an integral manifold for $D$ through p will satisfy the surface condition.

Thus by the Frobenius Theorem, the problem is reduced to showing $D$ is involutive iff $\langle X, curl(X)\rangle=0$. But D is involutive iff given any two smooth sections Y and Z of D, $d\omega(Y, Z)=0$. In fancy notation, $d\omega=\beta(curl(X))$, in layman’s terms, if $curl(X)=(C_1, C_2, C_3)$, then $d\omega=C_3dx\wedge dy-C_2dx\wedge dz+C_1dy\wedge dz$ (maybe a sign is wrong there, it doesn’t really matter to finish the problem, but I didn’t actually work it out so don’t fully trust it).

So we now have D involutive iff $d\omega(Y, Z)=det(curl(X)|Y|Z)=0$. But this happens iff $curl(X)\in span(Y, Z)$ iff $curl(X)$ has trivial projection onto $X$, which is precisely $\langle X, curl(X) \rangle=0$.

# The Frobenius Theorem

What a joke to think that I could live without the internet.

Recall that the Frobenius Theorem is actually quite a concise statement given what we’ve defined and done. It just says that every involutive distribution is completely integrable.

Let D be involutive of dimension k and M an n-manifold and $p\in M$. We need a flat chart about p. We will work on a coordinate chart since it is a local property, and hence with loss of generality $M=U\subset \mathbb{R}^n$ we replace the manifold with an open subset of Euclidean space. Suppose $Y_1, \ldots , Y_k$ is a smooth local frame for D. We choose our coordinates such that $D_p$ is complementary to the span of $\displaystyle\left(\frac{\partial}{\partial x^{k+1}\big|_p}, \ldots , \frac{\partial}{\partial x^n}\big|_p\right)$.

Now let $\pi:\mathbb{R}^n\to\mathbb{R}^k$ be the projection onto the first k coordinates. Then we get a bundle morphism $d\pi:T\mathbb{R}^n\to T\mathbb{R}^k$. Explicitly this is just $\displaystyle d\pi\left(\sum_{i=1}^n v^i\frac{\partial}{\partial x^i}\big|_q\right)=\sum_{i=1}^k v^i\frac{\partial}{\partial x^i}\big|_{\pi(q)}$.

The morphism is smooth, since it is composition of the inclusion map from the distribution followed by $d\pi$. Thus, by definition, the component functions in any smooth frame are smooth. In particular, we have smooth frames $Y_1, \ldots, Y_k$ and $\frac{\partial}{\partial x^j}\big|_{\pi(q)}$. Thus the matrix entries of $d\pi\big|_{D_q}$ with respect to these frames are smooth functions of q.

We also get that at any given point p, this restricted map is bijective as a map to $D_p$. It is certainly surjective by construction, and when restricted $d\pi\big|_{D_p}$ is complementary to the kernel of the non-restricted bundle map by choice of coordinates. Thus it is bijective in an entire neighborhood of p. Thus we have a smooth inverse in this neighborhood $(d\pi\big|_{D_q})^{-1} : T_{\pi(q)}\mathbb{R}^k\to D_q$.

Define a frame for D on this neighborhood by $X_i\big|_q=(d\pi\big|_{D_q})^{-1}\frac{\partial}{\partial x^i}\big|_{\pi(q)}$. Note that if we can show that $[X_i, X_j]=0$ we will be done. This is because the frame will be a commuting frame, and the “canonical form” for a commuting frame is precisely our definition of a flat chart.

We have that $\frac{\partial}{\partial x^i}\big|_{\pi(q)}=\left(d\pi\big|_{D_q}\right)X_i\big|_q=d\pi_q (X_i\big|_q)$. i.e. we have that $X_i$ and $\frac{\partial}{\partial x^i}$ are $\pi$-related.

But now we have it by naturality of the Lie bracket, since $d\pi_q([X_i, X_j]_q)=[\frac{\partial}{\partial x^i}, \frac{\partial}{\partial x^j}]_{\pi(q)}=0$. And the involutivity of D tells us that $[X_i, X_j]$ is in D, and since $d\pi$ on the distribution is injective, $[X_i, X_j]_q=0$ for every point in the neighborhood.

# Distributions and Differential Forms

It’s time to put the development of differential forms to use. To me vector fields, and moreover, distributions are sort of messy and difficult when in the form they were presented in. It turns out we can define distributions in terms of differential forms.

Let’s intuitively figure out how we can do this. If we have a 1-form $\omega$, then $\ker \omega\Big|_p$ is going to be the span of vectors in the tangent space at p, such that $\omega(X_p)=0$. So intuitively this should be $n-1$-dimensional, since the span of the vector such that $\omega(X_p)=1$ will be the only non-vanishing direction. i.e. 1-forms will define codimension 1 subspaces, or k linearly independent 1-forms will define codimension k subspaces.

So they way we will define a k-distribution, is to choose $n-k$ linearly independent 1-forms $\omega^1, \ldots , \omega^{n-k}$ and define the k-dimensional distribution by $D_p=\ker \omega^1\Big|_p\cap \cdots \cap \ker \omega^{n-k}\Big|_p$. These are called the local defining forms for D.

Now we should still be a little careful here, since this has all been wishful thinking so far. What we really need is something of the form: Given any k-distribution $D\subset TM$, then D is smooth if and only if there exists smooth defining forms for D.

This turns out to be the case. The forward direction is just that in any neighborhood we can complete the defining forms to a smooth coframe $(\omega^1, \ldots , \omega^n)$. Then take the dual frame $(E_1, \ldots E_n)$. We see that in this neighborhood $D=span\{E_{n-k+1}, \ldots , E_n\}$, so by the local frame criterion of last post the distribution is smooth. This argument essentially reverses for the other direction (complete the frame and take the dual coframe).

Now we say that a p-form annihilates the distribution if $\omega(X_1, \ldots, X_p)=0$ whenever $X_1, \ldots, X_p$ are local sections of D.

It turns out that a p-form $\eta$ annihilates (locally/on some open set) a distribution iff it is of the form $\displaystyle\eta=\sum_{i=1}^{n-k}\omega^i\wedge \beta^i$ where the $\omega$‘s are the local defining forms and the $\beta$‘s are smooth (p-1)-forms.

That is just a calculation (maybe not easy), so I will skip its proof. The really interesting stuff is that I can now express the definitions from last post in the language of differential forms. For instance:

A distribution is involutive iff for any smooth 1-form $\eta$ that annihilates it on an open set, we also have that $d\eta$ annihilates it on that same set.

This is fairly straightforward and quick to prove, but seems to me a much nicer way of going about involutivity. Suppose $D$ is involutive and $\eta$ annihilates D on $U$. Let $X, Y\subset D$ be smooth sections. We calculate that $d\eta(X, Y)=X(\eta(Y))-Y(\eta(X))-\eta([X,Y])$. But every term on the right is zero by either involutivity or that $\eta$ annihilates D.

Now suppose the hypothesis of the reverse direction. Let $X, Y\subset D$ be smooth local sections and D has the $\omega$ defining forms. Then the same formula form before shows that $\omega^i([X,Y])=X(\omega^i(Y))-Y(\omega^i(X))-d\omega^i(X, Y)=0$. i.e. $[X, Y]\in D$, and hence is involutive by the first definition.

Now granted I used some formula that magically did the work for me, but it is fairly standard and in both texts I have on smooth manifolds.

Using the above, we get an even better check for involutivity. TFAE:
1) D is involutive on U
2) $d\omega^1, \ldots , d\omega^{n-k}$ annihilate D
3) There are smooth 1-forms $\alpha_j^i$ such that $d\omega^i=\sum \omega^j\wedge\alpha_j^i$.

The proof is just applying the few things from this post that we’ve done.

# Distributions

We need to build up a lot of definitions now to properly state the Frobenius Theorem. The main definition will be a distribution on a manifold. Essentially, the theory of distributions is a way to generalize the notion of a vector field and the flow of a vector field.

A distribution is a choice of k-dimensional subspace $D_p\subset T_pM$ at each point of the manifold. Note that this is just a subbundle of the tangent bundle, so we have a nice notion of smoothness. In particular, just as we could check a local frame for smoothness of a vector field (i.e. a 1-dimensional distribution), we can check for smoothness of a distribution by checking if each point has a neighborhood on which there are smooth vector fields $X_1, \ldots , X_k$ such that $X_1\big|_q, \ldots , X_k\big|_q$ forms as basis for $D_q$ at each point of the open set.

The analogous thing for integral curves for vector fields will be what we call an integral manifold. If we think about the natural way to define this we would see that all we want is an immersed submanifold $N\subset M$ such that $T_pN=D_p \forall p\in N$. Thus in the one dimensional case, the immersed submanifold is just a curve on the manifold.

Unfortunately, it is not the case that integral manifolds exist for all distributions. Our goal is to figure out when they exist. This leads us to our next two definitions. A distribution is integrable if every point of the manifold is in some integral manifold for the distribution. A distribution is called involutive if for any pair of local sections of the distribution, the Lie bracket is also a local section. Note that a local section is really just a vector field where the vectors are chosen from the distribution rather than the whole tangent bundle.

Every integrable distribution is involutive. If $D\subset TM$ is involutive, then given any $p\in M$ and local sections $X, Y$ there is some integral manifold about p, say $N$. Since both $X, Y\in T_pN$, we have that $[X, Y]_p\in T_pN=D_p$, which is the definition of involutive.

This gives us an easy way to see that there are non-integrable distributions (recall, this is not going to happen for 1-distributions, i.e. vector fields, since every point has an integral curve). We don’t even need some weird manifold. Just take $\mathbb{R}^3$, and let the distribution be the span of two vector fields whose Lie bracket is not in the span. Thus something like $\displaystyle D=span\{X=\frac{\partial}{\partial x}+y\frac{\partial}{\partial z}, Y=\frac{\partial}{\partial y}\}$ will work, since $[X, Y]_0=-\frac{\partial}{\partial z}\notin D_0$.

I think we need only one more definition to be in a place to move on. A distribution is completely integrable if there exists a flat chart for the distribution in a neighborhood of every point. By this I mean that I can find a coordinate chart such that $\displaystyle D=span\{\frac{\partial}{\partial x^1}, \ldots , \frac{\partial}{\partial x^k}\}$. This is obviously the strongest condition.

So our definitions at this point satisfy completely integrable distributions are integrable, and integrable distributions are involutive. The utterly remarkable thing that the Frobenius Theorem says, is that all of these implications reverse, and so all of the definitions are actually equivalent! We’ll get there later, though.