# Can exterior calculus be used to solve differential equations?

I know one can formulate partial differential equations in terms of exterior derivatives, etc but I have been wondering for a while now how one might be able to use that formalism to extrapolate information about the solutions.

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Are you interested in real or complex setting? The solution of the dbar problem on pseudoconvex domains is a nice complex example. –  user53153 Jan 14 '13 at 15:03
I would prefer to start with the real setting. –  Alex Eftimiades Jan 14 '13 at 15:05
de Rham cohomology sounds relevant to your interests. –  Gunnar Þór Magnússon Jan 14 '13 at 15:24
It probably is. I took a course on differential forms that used "baby Spivak" as the textbook a year or two ago and I have been fascinated ever since. However, I have had trouble teaching myself more on the subject. I have read about de Rham cohomology, but I know nothing about relevent computations. I am hoping to gain some intuition on a higher level than baby Spivak so I could more easily read texts on these subjects. –  Alex Eftimiades Jan 14 '13 at 15:32

I'll write a very quick motivation for de Rham cohomology.

Say we have some manifold $M$ (or open set $M \subset \mathbb R^n$ if you will) and that we want to solve the differential equation $$du = v$$ on $M$, where $v$ is a $k$-form (so $k = 1$ has us searching for a function $u$ that solves the equation, higher $k$ have us searching for $(k-1)$-forms).

If this equation has a solution, then we necessarily have $$dv = d^2 u = 0.$$ This gives a necessary condition for our problem to have a solution, but not a sufficient one, since there may be "obstructions" to finding a solution of $du = v$ even if $dv = 0$.

To get some sense of these obstructions, we form a quotient vector space $$H^k(M,\mathbb R) := \{ \text{k-forms v} \mid dv = 0 \} / \{du \mid \text{u is a (k-1)-form}\}.$$ This is the $k$-th de Rham cohomology group of the manifold $M$. If our problem $du = v$ is well-posed, that is if $d v = 0$ so it has a chance of having a solution, then $v$ defines an element $[v]$ in $H^k(M,\mathbb R)$. Our problem also has a solution if and only if $[v] = 0$ in this space. A remarkable and nontrivial fact is that these cohomology groups are finite-dimensional if the space $M$ is compact.

This may not seem very useful, since it is in general quite difficult to show that some specific $v$ defines a zero cohomology class on $M$. What makes this formalism useful is that in many cases one can use geometric properties of the manifold $M$ to prove that some or all of its nontrivial cohomology groups (those with $k > 0$) are zero, and thus that the differential equation that interests us has a solution.

These geometric properties come in two large categories; topological and differential-geometric. The first includes facts like that if $M$ is contractible, then all its cohomology groups are zero. The second is known as Hodge theory and rests on the Hodge isomorphism theorem, which says that if you pick a Riemannian metric on $M$, then the space of harmonic $k$-forms is isomorphic to $H^k(M,\mathbb R)$ (if $M$ is compact). Both can give valuable information about the cohomology groups and the geometry of $M$, and the second has extremely important connections to algebraic geometry.

In short, passing to cohomology replaces the problem of solving one particular differential equation with the problem of solving all differential equations on a manifold, which has connections to the geometry of the underlying space. Knowledge of the geometry can then perhaps help us to solve our original problem.

Since you asked for an explicit example of these calculations, we can consider the $1$-dimensional torus $S^1 = \mathbb R / \mathbb Z$. Actually I'll cheat and not do any explicit calculations at all, but leave them to you to figure out. The torus admits a flat Riemannian metric $g$, induced by the standard Euclidean metric on $\mathbb R$. Hodge theory will tell you that $$H^1(S^1, \mathbb R) \cong \mathbb R,$$ and that and explicit isomorphism between the two is $\mathbb R \to H^1(S^1,\mathbb R)$, $\lambda \mapsto \lambda dx$, where $dx$ is the constant $1$-form on $\mathbb R$ (see Hodge isomorphism theorem).

Now let $dV$ be the volume form of the Riemannian metric $g$, which we'll suppose to be of volume 1. I claim that we cannot solve the differential equation $$df = dV$$ on $S^1$. If we could, then Stokes' theorem would give us $$\int_{S^1} dV = \int_{S^1} df = \int_{\partial S^1} f = 0$$ because $S^1$ has no boundary. But by hypothesis $$\int_{S^1} dV = \mathrm{Vol}(S^1,g) = 1,$$ which gives a contradiction.

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I am trying to come up with a strategy to prove the Hodge isomorphism theorem. I strongly suspect I can get there from Hodge decomposition. Am I on the right track? –  Alex Eftimiades Jan 14 '13 at 18:09
(+1) Nice answer. (I don't understand everything because I'm beginner in this topic, but your answer is clear and give some intuition.) –  vesszabo Jan 14 '13 at 19:59
@Feynman: The Hodge decomposition theorem I know holds for compact Kahler manifolds and follows from Hodge isomorphism. The latter is proved in the first couple of sections of Chapter VI of tinyurl.com/2dcnycm [Demailly]. –  Gunnar Þór Magnússon Jan 15 '13 at 7:54