What is the symbol of a differential operator? - MathOverflow

What is the symbol of a differential operator?

2009-10-30T22:12:06Z

I find Wikipedia's discussion of symbols of differential operators a bit impenetrable, and Google doesn't seem to turn up useful links, so I'm hoping someone can point me to a more pedantic discussion.

Background

I think I understand the basic idea on ℝⁿ, so for readers who know as little as I do, I will provide some ideas. Any differential operator on ℝⁿ is (uniquely) of the form Σ p_{i₁,...,i_k}(x) ∂^k/(∂x_i₁...∂x_{i_k}), where x₁,...,x_n are the canonical coordinate functions on ℝⁿ, the p_{i₁,...,i_k}(x) are smooth functions, and the sum ranges over (finitely many) possible indexes (of varying length). Then the symbol of such an operator is Σ p_{i₁,...,i_k}(x) ξ^i₁...ξ^i_k, where ξ¹,...,ξⁿ are new variables; the symbol is a polynomial in the variables {ξ¹,...,ξⁿ} with coefficients in the algebra of smooth functions on ℝⁿ.

Ok, great. So symbols are well-defined for ℝⁿ. But most spaces are not ℝⁿ — most spaces are formed by gluing together copies of (open sets in) ℝⁿ along smooth maps. So what happens to symbols under changes of coordinates? An affine change of coordinates is a map y_j(x) = a_j + Σ_i Y_jⁱx_i for some vector (a₁,...,a_n) and some invertible matrix Y. It's straightforward to describe how the differential operators change under such a transformation, and thus how their symbols transform. In fact, you can forget about the fact that indices range 1,...,n, and think of them as keeping track of tensor contraction; then everything transforms as tensors under affine coordinate changes, e.g. the variables ξⁱ transform as coordinates on the cotangent bundle.

On the other hand, consider the operator D = ∂²/∂x² on ℝ, with symbol ξ²; and consider the change of coordinates y = f(x). By the chain rule, the operator D transforms to (f'(y))² ∂²/∂y² + f''(y) ∂/∂y, with symbol (f'(y))² ψ² + f''(y) ψ. In particular, the symbol did not transform as a function on the cotangent space. Which is to say that I don't actually understand where the symbol of a differential operator lives in a coordinate-free way.

Why I care

One reason I care is because I'm interested in quantum mechanics. If the symbol of a differential operator on a space X were canonically a function on the cotangent space T*X, then the inverse of this Symbol map would determine a "quantization" of the functions on T*X, corresponding to the QP quantization of ℝⁿ.

But the main reason I was thinking about this is from Lie algebras. I'd like to understand the following proof of the PBW theorem:

Let L be a Lie algebra over ℝ or ℂ, G a group integrating the Lie algebra, UL the universal enveloping algebra of L and SL the symmetric algebra of the vector space L. Then UL is naturally the space of left-invariant differential operators on G, and SL is naturally the space of symbols of left-invariant differential operators on G. Thus the map Symbol defines a canonical vector-space (and in fact coalgebra) isomorphism UL → SL.

Answer by Qiaochu Yuan for What is the symbol of a differential operator?

2009-10-30T22:33:16Z

John Baez's week282 might help you out, although I can't say I really understand what's going on.

Answer by Ben Webster for What is the symbol of a differential operator?

2009-10-30T22:35:51Z

I think you have misunderstood the definition of "symbol." You should only take the term of highest order in the vector fields. Then the symbol is well defined. (EDIT: well, I guess I should have read Wikipedia first. I stick by my assertion that the symbol map one should consider is the leading order one).

More to the point, the symbol map isn't from differential operators to functions on the cotangent bundle, it's from the associated graded of differential operators for the order filtration to functions on the cotangent bundle. So, on operators of order less than n, you can do the operation you described to the highest order term, and you get something coordinate independent.

Answer by Ilya Nikokoshev for What is the symbol of a differential operator?

2009-10-30T22:55:37Z

The definition of symbol as presented in Wikipedia is not invariant — only the highest order terms. Some textbooks call those higher order terms symbols (Wikipedia suggests the name principal symbol), hence the Ben's answer, which refers to that definition.

The highest order terms are clearly most important for the properties of the differential equations, e.g. their positiveness allows to prove the existence of solutions (it's related to the fact that positively definite linear operators are invertible in linear algebra).

As for "Thus the map Symbol defines a canonical vector-space (and in fact coalgebra) isomorphism UL → SL.", this statement should be proved by induction order-by-order in a fixed coordinate system. It should be true in any coordinate system, but the homomorphism depends on it.

There is however, a canonical coordinate system given by the exp map, and this, I think (not sure here), is the canonical map referred to in the question.

As for quantum mechanics, while I have only some general knowledge, I think "inverse of this Symbol map would determine a "quantization" of the functions on T*X, corresponding to the QP quantization of ℝn" is true, but somewhat too optimistic. Yes, there are quantizations, but they are canonical to all terms only when you restrict yourself to linear change of coordinates (or when you do some additional constructions).

I risk being viewed as stepping on a slippery stone here, but my limited understanding is that you're actually hitting a fundamental question here — even some quantum field theory play nice with diffeomorphisms, but it's far from a simple exercise. Moreover, the theory that includes them as the degrees of freedom would be called quantum gravity and is the Holy Grail of high-energy physics rather then a theorem of Lie group geometry (though the latter is extensively used for the former).

Answer by S Kitchen for What is the symbol of a differential operator?

2009-10-31T02:44:00Z

The D-module course notes of Dragan Milicic contain a detailed construction of the symbol map -- they can be found on his webpage www.math.utah.edu/~milicic. There may be several versions linked there -- the 2007-2008 course should be most thorough. Start reading in Chapter 1, section 5. This goes through the construction of the filtration Ben mentions, then constructs the graded module and symbol map explicitly. Of course, this section only covers the (complex) affine case you already describe. The coordinate-free generalization for say smooth quasi-projective varieties over the complex numbers is done in Chapter 2, section 3. Basically, you look at the sheaf of differential operators on your variety, construct a degree filtration of that sheaf, then the corresponding graded sheaf is isomorphic to the direct image of the sheaf of regular functions on the cotangent bundle via the symbol map.

In the case G in your question is an algebraic group, the sheaf of differential operators on G are formed by localizing UL, and the pushforward of regular functions on the cotangent bundle is the localization of SL. Then UL and SL can be recovered by pulling these sheaves back to the identity element in G. I don't think the isomorphism the way you have stated it is true as-is. I think the content of any statement along these lines (as it relates to the proof of the PBW theorem) probably has to do with the construction of the filtration by top degree being well-defined.

Answer by David Bar Moshe for What is the symbol of a differential operator?

2009-10-31T14:51:45Z

I think that one shouldn't insist on the invariance of symbols. A symbol is just a realization of a differential operator on the cotangent bundle. If the symbol were inavariant under some transformation it would restrict the corresponding operator to some subset which may be less interesting. As a finite dimensional example, the subspace of linear transformations of a finite dimensional vector space invariant under the group of unitary transformations would be the dull subspace of multiples of the unit operator.

Answer by Rafe Mazzeo for What is the symbol of a differential operator?

2009-11-01T02:06:07Z

The principal symbol of a differential operator $\sum__ {|\alpha| \leq m} a__ \alpha(x) \partial__ x^\alpha$ is by definition the function $\sum__ {|\alpha| = m} a__ \alpha(x) (i\xi)^\alpha$ Here \alpha is a multi-index (so \partial__ x^\alpha denotes \alpha__ 1 derivatives with respect to x__ 1, etc.) At this point, the vector \xi = (\xi__ 1, \ldots, \xi__ n) is merely a formal variable. The power of this definition is that if one interprets (x,\xi) as variables in the cotangent bundle in the usual way -- i.e. x is any local coordinate chart, then \xi is the linear coordinate in each tangent space using the basis dx^1, \ldots, dx^n, then the principal symbol is an invariantly defined function on T^*X, where X is the manifold on which the operator is initially defined, which is homogeneous of degree m in the cotangent variables.

Here is a more invariant way of defining it: fix (x__ 0,\xi__ 0) to be any point in T^*X and choose a function \phi(x) so that d\phi(x__ 0) = \xi__ 0. If L is the differential operator, then L( e^{i\lambda \phi}) is sum complicated sum of derivatives of \phi, multiplied together, but always with a common factor of e^{i\lambda \phi}. The `top order part' is the one which has a \lambda^m, and if we take only this, then its coefficient has only first derivatives of \phi (lower order powers of \lambda can be multiplied by higher derivatives of \phi). Hence if we take the limit as \lambda \to \infty of \lambda^{-m} L( e^{i\lambda \phi}) and evaluate at x = x__ 0, we get something which turns out to be exactly the principal symbol of L at the point (x__ 0, \xi__ 0).

There are many reasons the principal symbol is useful. There is indeed a `quantization map' which takes a principal symbol to any operator of the correct order which has this as its principal symbol. This is not well defined, but is if we mod out by operators of one order lower. Hence the comment in a previous reply about this being an isomorphism between filtered algebras.

In special situations, e.g. on a Riemannian manifold where one has preferred coordinate charts (Riemann normal coordinates), one can define a total symbol in an invariant fashion (albeit depending on the metric). There are also other ways to take the symbol, e.g. corresponding to the Weyl quantization, but that's another story.

In microlocal analysis, the symbol captures some very strong properties of the operator L. For example, L is called elliptic if and only if the symbol is invertible (whenever \xi \neq 0). We can even talk about the operator being elliptic in certain directions if the principal symbol is nonvanishing in an open cone (in the \xi variables) about those directions. Another interesting story is wave propagation: the characteristic set of the operator is the set of (x,\xi) where the principal symbol p(L) vanishes. If its differential (as a function on the cotangent bundle) is nonvanishing there, then the integral curves of the Hamiltonian flow associated to p(L), i.e. for the Hamiltonian vector field determined by p(L) using the standard symplectic structure on T^*X, `carries'' the singularities of solutions of Lu = 0. This is the generalization of the classical fact' that singularities of solutions of the wave equation propagate along light rays.

Answer by David Speyer for What is the symbol of a differential operator?

2009-11-01T03:19:53Z

The original questioner already knows this, but anyone else who is interested in this question should check out the conversation at John Baez's blog.

Answer by Terry Tao for What is the symbol of a differential operator?

2009-11-01T15:55:30Z

One way to understand the symbol of a differential operator (or more generally, a pseudodifferential operator) is to see what the operator does to "wave packets" - functions that are strongly localised in both space and frequency.

Suppose, for instance, that one is working in R^n, and one takes a function psi which is localised to a small neighbourhood of a point x0, and whose Fourier transform is localised to a small neighbourhood of xi0/hbar, for some frequency xi0 (or more geometrically, think of (x0,xi0) as an element of the cotangent bundle of R^n). Such functions exist when hbar is small, e.g. psi(x) = eta( (x-x0)/eps ) e^{i xi0 . (x-x0) / hbar} for some smooth cutoff eta and some small eps (but not as small as hbar).

Now apply a differential operator L of degree d to this wave packet. When one does so (using the chain rule and product rule as appropriate), one obtains a bunch of terms with different powers of 1/hbar attached to them, with the top order term being 1/hbar^d times some quantity a(x0,xi0) times the original wave packet. This number a(x0,xi0) is the principal symbol of a at (x0,xi0). (The lower order terms are related to the lower order components of the symbol, but the precise relationship is icky.)

Basically, when viewed in a wave packet basis, (pseudo)differential operators are diagonal to top order. (This is why one has a pseudodifferential calculus.) The diagonal coefficients are essentially the principal symbol of the operator. [While on this topic: Fourier integral operators (FIO) are essentially diagonal matrices times permutation matrices in the wave packet basis, so they have a symbol as well as a relation (the canonical relation of the FIO, which happens to be a Lagrangian submanifold of phase space).]

One can construct wave packets in arbitrary smooth manifolds, basically because they look flat at small scales, and one can define the inner product xi0 . (x-x0) invariantly (up to lower order corrections) in the asymptotic limit when x is close to x0 and (x0,xi0) is in the cotangent bundle. This gives a way to define the principal symbol on manifolds, which of course agrees with the standard definition.