TITLE: What is the symbol of a differential operator?
QUESTION [69 upvotes]: 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 $\mathbb{R}^n$, so for readers who know as little as I do, I will provide some ideas.  Any differential operator on $\mathbb{R}^n$ is (uniquely) of the form $\sum p_{i_1,\dotsc,i_k}(x)\frac{\partial^k}{\partial x_{i_1}\dots\partial x_{i_k}}$, where $x_1,\dotsc,x_n$ are the canonical coordinate functions on $\mathbb{R}^n$, the $p_{i_1,\dotsc,i_k}(x)$ are smooth functions, and the sum ranges over (finitely many) possible indexes (of varying length).  Then the symbol of such an operator is $\sum p_{i_1,\dotsc,i_k}(x)\xi^{i_1}\dotso\xi^{i_k}$, where $\xi^1,\dotsc,\xi^n$ are new variables; the symbol is a polynomial in the variables $\{\xi^1,\dotsc,\xi^n\}$ with coefficients in the algebra of smooth functions on $\mathbb{R}^n$.
Ok, great.  So symbols are well-defined for $\mathbb{R}^n$.  But most spaces are not $\mathbb{R}^n$ — most spaces are formed by gluing together copies of (open sets in) $\mathbb{R}^n$ along smooth maps.  So what happens to symbols under changes of coordinates?  An affine change of coordinates is a map $y_j(x)=a_j+\sum_jY_j^ix_i$ for some vector $(a_1,\dotsc,a_n)$ 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 $1,\dotsc,n$, and think of them as keeping track of tensor contraction; then everything transforms as tensors under affine coordinate changes, e.g. the variables $\xi^i$ transform as coordinates on the cotangent bundle.
On the other hand, consider the operator $D = \frac{\partial^2}{\partial x^2}$ on $\mathbb{R}$, with symbol $\xi^2$; and consider the change of coordinates $y = f(x)$.  By the chain rule, the operator $D$ transforms to $(f'(y))^2\frac{\partial^2}{\partial y^2} + f''(y) \frac{\partial}{\partial y}$, with symbol $(f'(y))^2\psi^2 + f''(y)\psi$.  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 $T^\ast X$, then the inverse of this Symbol map would determine a "quantization" of the functions on $T^\ast X$, corresponding to the QP quantization of $\mathbb{R}^n$.
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 $\mathbb{R}$ or $\mathbb{C}$, $G$ a group integrating the Lie algebra, $\mathrm{U}L$ the universal enveloping algebra of $L$ and $\mathrm{S}L$ the symmetric algebra of the vector space $L$.  Then $\mathrm{U}L$ is naturally the space of left-invariant differential operators on $G$, and $\mathrm{S}L$ 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 $\mathrm{U}L\to\mathrm{S}L$.

REPLY [47 votes]: 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 $\mathbb R^n$, and one takes a function $\psi$ which is localised to a small neighbourhood of a point $x_0$, and whose Fourier transform is localised to a small neighbourhood of $\xi_0/\hbar$, for some frequency $\xi_0$ (or more geometrically, think of $(x_0,\xi_0)$ as an element of the cotangent bundle of $\mathbb R^n$).  Such functions exist when $\hbar$ is small, e.g. $\psi(x) = \eta( (x-x_0)/\epsilon ) e^{i \xi_0 \cdot (x-x_0) / \hbar}$ for some smooth cutoff $\eta$ and some small $\epsilon$ (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 $a(x_0,\xi_0)$ times the original wave packet.  This number $a(x_0,\xi_0)$ is the principal symbol of $a$ at $(x_0,\xi_0)$.  (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 $\xi_0\cdot(x-x_0)$ invariantly (up to lower order corrections) in the asymptotic limit when $x$ is close to $x_0$ and $(x_0,\xi_0)$ is in the cotangent bundle.  This gives a way to define the principal symbol on manifolds, which of course agrees with the standard definition.