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I am wondering if there is a general coordinate-independent way to define a Partial Differential Equation on a Smooth manifold.

It is definitely true that in each coordinate neighborhood you could define a function to satisfy a differential equation, but when you change coordinates then the differential equation will most likely have a different form.

For example, in $\mathbb{R}^3$ one says a function is harmonic if it satisfies

$ \frac{\partial^2}{\partial x^2} f + \frac{\partial}{\partial y^2} f + \frac{\partial}{\partial z^2} f = 0 $

but it is not true that in spherical coordinates a function is harmonic if

$ \frac{\partial^2}{\partial r^2} f + \frac{\partial}{\partial \theta^2} f + \frac{\partial}{\partial \phi^2} f = 0 $

The change of variables to spherical coordinates gives you a much different PDE. So basically, how can you define a differential operator which is coordinate independent on a smooth manifold, so that you can have some notion of a PDE on a manifold.

2 Answers 2

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You could view the concept of a "Jet bundle" to be precisely what you need to make partial differential equations coordinate-independent on manifolds, as points in jet bundles represent precisely the Taylor polynomials of functions in local coordinates. Take a look at Chapter 4 in Hirsch's "Differential Topology" textbook for details.

The underlying problem of course is that the tangent bundle of a manifold is not trivial in general so you have to do something new to make sense of partial differential equations.

Perhaps the simplest would be to view a vector field as a 1st order differential equation via the directional derivative operator. From there you can take mixed directional derivatives. This isn't ideal because you can't express "natural" differential equations like the Laplacian in this manner (for example, think of $S^2$ -- every vector field has a zero...). edit: as pointed out, on $S^2$ there is a formulation of the Laplacian in terms of iterated 1st order DE's. I suppose I should have phrased things as "even if you can express a PDE as iterated 1st order DEs on a manifold, it may not be desirable due to the breaking of symmetries brought on by bringing in vector fields."

The most direct approach is to think of partial differential equations as functions on jet bundles. But you can frequently express interesting differential equations in other manners -- having a connection on a manifold is one of the simplest ways of turning 2nd derivatives into tangential data. The relevant formulation is called an Ehresmann Connection. But there's all kinds of other structures on manifolds where PDEs can lurk. Differential forms give you a standard language -- exterior derivative and Hodge star operators on Riemann manifolds give you a language where you can express the Laplacian.

I think this is a very good question in that to a large extent the language you need to express a PDE on a manifold very much informs on what you actually know geometrically about the PDE. So it's a key step in getting insight into the nature of a PDE. Having a PDE at the jet bundle level basically means you know nothing about it, other than it's a PDE.

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    Can't one write the Laplacian on the sphere as a combination of vector fields? I would have thought that one gets it by considering the action of the Casimir in $U(\mathfrak{so(3)})$ on it, or something :)2010-11-03
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    (I mean on the round sphere only, of course)2010-11-03
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    I suppose it depends on how complicated a formula you allow -- but if you're just taking directional derivatives you've got the problem that every vector field on $S^2$ has to have a zero. So there's no formula of the form $\Delta f = D_vD_wf$ for vector fields $v$ and $w$ on $S^2$.2010-11-03
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    @Ryan: that's not true strictly speaking, what Mariano mentions is true if you allow the standard set of three rotation vector fields: let $\omega_{x,y,z}$ be the vector field representing rotations around the three axes. Then $\triangle f = \omega_x\omega_x f + \omega_y\omega_y f + \omega_z\omega_z f$ up to a constant factor which you can absorb into $\omega$. I mean, even on $\mathbb{R}^2$ there cannot be two real vector fields such that $D_vD_w f = \triangle f$.2010-11-03
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    Perhaps the main issue one might have with expressing PDEs in terms of iterated 1st order DE's via vector fields is that because vector fields have to have zeros, any such expression kills the symmetry of $S^2$.2010-11-03
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    @willie: I take it you're not responding to my comment but to the answer above the comment.2010-11-03
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    Uh... the Casimir operator that Mariano was talking about is generated *from* the symmetry of the sphere... ; and yes, I am responding to your comment (sort of) in the context of Mariano's comment.2010-11-03
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    The vector fields have zeros, and those aren't invariant under the isometry group. It may be generated "from" the symmetry of $S^2$ but it's very much like how $\sin$ and $\cos$ are generated from the symmetry of $S^1$ -- they're not invariant under the symmetries of $S^1$.2010-11-03
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    I think the fact that you cannot *factor* a general differential operator (like, say, the Laplacian) into a simple iteration of vector fields is better explained not by the fact vector fields have to have zeroes on the sphere (I think that is a red herring, since the same fact is true for Laplacian on any manifold with dimension greater than 1), but by the fact that differential operators corresponding to iterated vector fields have very precise and restrictive characterstic variety (union of several flat planes in the tangent space), and so cannot include any operator whose2010-11-03
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    (cont'd) character variety is curved. This is also why the wave operator in two dimensions can be factored as two vector fields, but not in three or higher dimensions. As for the Laplacian: its characteristic set is empty, so it cannot be represented as (single) iteration of vector fields. This of course doesn't rule out representation by a sum of such iterations. (Which you won't want anyway, since such a representation is always possible in a coordinate neighborhood.)2010-11-03
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    I think we're straying from my point. The idea was bringing in vector fields destroys symmetry for pretty simple reasons. Zeros is one, but you lose isotropy symmetry for more tautological reasons.2010-11-03
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    At least in the algebro-geometric context this always happens, by the way: on a smooth affine variety, derivations and functions generate the algebra of differential operators.2010-11-03
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    Right, I suppose this is true on manifolds as well -- by partitions of unity it boils down to a local statement, and it's true locally basically by design.2010-11-03
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Ryan already gave a very good answer. I just want to give a few clarifications on how one should think about PDEs in general. Most importantly: you should not think a differential operator as defined by "a formula". What do I mean by that? In your question, you wrote that the Laplacian in Euclidean coordinates is $\frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2} + \frac{\partial^2}{\partial z^2}$, but in Spherical coordinates it is not $\frac{\partial^2}{\partial r^2} + \frac{\partial^2}{\partial \theta^2} + \frac{\partial^2}{\partial \phi^2}$. But why should one have thought that the two expressions are the same?

To illustrate, think of the function on the plane $f(x,y) = x^2 + y^2$. This represents the square of the distance of a point from the origin. In radial coordinates, the same function is written as $f(r,\theta) = r^2$. Not $r^2 + \theta^2$, which is a very different beast. Just like how the direct replacement $x \to r$ and $y\to \theta$ changes the function you are considering, the replacements $\frac{\partial}{\partial x} \to \frac{\partial}{\partial r}$ etc changes the operator you are considering.

Now: to think geometrically (on manifolds), a (real valued) function is an assignment of a number to each point on the manifold. Then in any coordinate system there is a representation of the function by a formula. Since we like to work with formulas, we can interpret the notion of a function as a "map". This map takes as its input a coordinate system, and outputs a formula that represents your function in that coordinate system. (If you are familiar with category theory, this is similar to how one can think in terms of arrows instead of dots.)

Similarly, a partial differential operator can be thought of as an object in itself. Our usual way of writing it as partial derivatives in some coordinate system with coefficients is just a convenient representation of the real object upstairs. Therefore, analogous to how a function is merely a "map" from coordinate system to the formulaic representation, you can think of a partial differential operator also as a "map" which takes as its input a coordinate system and outputs an expression which is a sum of partial derivatives with some coefficients.

The jet bundle formulation is just a sophisticated, rigorous way of formulating this idea. The main difficulty is making sure the intuition above is compatible with the change of variables formulas. Suppose we are given a coordinate system $A$, and an expression in terms of partial derivatives which we'll call $L_A$. The question becomes: does there exist some abstract partial differential operator $L$ such that its representation in the coordinate system $A$ is precisely $L_A$? And if there is such an operator, is the fact that, for some function $f$ (written in coordinate system $A$ as $f_A$), $L_A f_A = 0$ invariant under change of coordinate system? (That given any other coordinate system $B$, $L_Bf_B = 0$ also.) The tools of jet bundles allows you to make such a consistent definition/description of partial differential equations.

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    Wow, I got two fantastic answers. Thank you Willie Wong and Ryan Budney. I gave the "answer accepted" to Ryan only because he was the first to answer.2010-11-03