# Symbol of the differential operator on vector bundles

Suppose that we have a differential operator $D:C^{\infty}(\mathbb{R}^n) \to C^{\infty}(\mathbb{R}^n)$ of the form $(Df)(x)=\sum_{|\alpha| \leq k}a_{\alpha}(x)\frac{\partial^{|\alpha|}f}{\partial x_1^{\alpha_1} ... \partial x_n^{\alpha_n}}$ where we use standard multiindices notation. Here $a_{\alpha}$ are scalar valued functions. For such $D$ we can consider the expression coming from the highest order term in $D$, where we replace each partial derivative $\frac{\partial^{|\alpha}|}{\partial x_1^{\alpha_1} ... \partial x_n^{\alpha_n}}$ by $\xi^{\alpha}:=\xi_1^{\alpha_1}...\xi_n^{\alpha_n}$ where $\xi=(\xi_1,...,\xi_n) \in \mathbb{R}^n$ (note that some authors insert also factor $i$ in order to make fulfill some positivity conditions). This is rather standard and this construction may be generalized: the most general case is when everything takes place over some (say closed) manifold $M$ and two vector bundles $E,F$ over $M$. Then the symbol may be interpreted in the following way: let $T^*_0(M)$ be the cotangent bundle with the zero section deleted and $p:T^*_0M \to M$ be a canonical projection. One considers the pullback bundles $p^*(E),p^*(F)$ and symbol is defined as homomorphism $\sigma \in Hom(p^*(E),p^*(F))$ in the following way: for $(x,v) \in T^*_0(M)$ and $e \in E$ we pick $g \in C^{\infty}(M)$ and $s \in \Gamma^{\infty}(M,E)$ (a smooth section) such that $dg_x=v$ $g(x)=0$ and $s(x)=e$. Then we define $\sigma(D)(x,v)e=D(\frac{i^k}{k!}g^ks)(x)$. My question are then following:

Why this is well defined?

At least I can see that everything lands in the correct spaces and also all expressions makes sense: also it is clear for me that such choices are always possible. What is not clear for me, that this definition does not depend from the choices made.
Moreover I really would like to understand

How this definition relates to the most simple situation for example $M$ being euclidean space and $E,F$ be trivial bundles (but not necessarily of rank 1).

I would appreciate any explanation or references where this is explained in detail: I really would like to learn this material in detail.

To check that $D(g^k s)(x)$ is well-defined, we need to show that it's independent of the choices of the function $g$ and the section $s$, i.e., that $D(g^k s)(x)$ only depends on the value of $dg_x = v \in T_x^\ast M$ and $s(x) = e \in E_x$. (I will ignore the constant $\frac{i^k}{k!}$. It's there to normalize things.)

There may be a cleaner way to do this, but one way is to use a coordinate system on $M$ and local trivializations of $E$ and $F$ in a neighborhood of $x$. In case you're worried, this does not introduce a new question of well-definedness, since no mention of coordinates or trivializations was made in defining $D(g^k s)(x)$. (I hope this also answers your question about trivial bundles over Euclidean space, which come with canonical coordinates and trivializations. And that's really the most important case to understand for these local questions about vector bundles, which is one reason why most authors unfortunately gloss over some of these details.)

In coordinates, we may write $D$ as $\sum_{|\alpha| \leq k} A_\alpha \frac{\partial}{\partial x^\alpha}$, where each $A_\alpha$ is a bundle homomorphism $E \to F$. (With respect to our local trivializations, we can think of each $A_\alpha$ as an $(\operatorname{rank} E) \times (\operatorname{rank} F)$-matrix-valued function.

Let's think about applying the Leibniz rule to $\frac{\partial}{\partial x^\alpha} (g^k s)$. It's crucial that we chose $g$ such that $g(x) = 0$. It means that exactly one partial derivative needs to hit each of the $k$ factors of $g$ in $g^k s$, or else $\frac{\partial}{\partial x^\alpha} (g^k s)$ will vanish at $x$. We can conclude that:

1. $\frac{\partial}{\partial x^\alpha} (g^k s)(x)$ is zero unless $|\alpha| = k$ (i.e., the symbol only depends on the "highest derivatives" in $D$).

2. $\frac{\partial}{\partial x^\alpha} (g^k s)(x)$ only depends on the first partial derivatives of $g$ at $x$ (i.e., on $dg_x \in T_x^\ast M$) and on the value of $s(x) \in E_x$. (One derivative needs to act on each factor of $g$, leaving none to hit $s$, and no higher derivatives of $g$.)

The second fact is precisely the well-definedness that we were trying to check.

You may want to look at these two books by Liviu Nicolaescu: Lectures on the Geometry of Manifolds (see chapter 10) and Notes on the Atiyah-Singer index theorem (see chapter 2). I think the treatment of symbols is quite similar in both of them. The symbol is also explained (rather briefly) in, for example, Michael Taylor's PDE book (volume I, section 2.9) (your library may have free pdf's of this Springer book available).

Edit: since I mentioned Taylor's book, I will add that Taylor's definition of the symbol is the following, using the same notation as above ($dg_x = v$, $s(x) = v$): $$\sigma(D)(x, v) := \lim_{\lambda \to \infty} \lambda^{-k} e^{-i\lambda g} D \left( e^{i\lambda g} s \right)(x)$$ Checking that this is well-defined is essentially the same Leibniz rule argument as above. Note that here the assumption that $g(0) = 0$ is unnecessary.

• Thank you for your answer. If you don't mind, I would like to continue the discussion: so as far as I understood, in trivial (flat) case you can take $g$ to be affine function while $s$ to be a constant section. In general case this wil no longer works while you need to care about overlaps of maps. This approach (defining symbol as I explained) is obviously global, since it doesn't involve any charts, trivializations etc. However it is still by no means obvious for me, why in trivial case this definition yields the same result as "replacing each partial derivative by $\xi^{\alpha}$" (up to ... – truebaran Jul 25 '15 at 13:03
• ...some normalizing factors). Is it somehow possible to easily see this? – truebaran Jul 25 '15 at 13:04
• Even in the general case, you can still take $g$ to be an affine function near $x$ (with respect to whatever coordinate chart you like) and $s$ to be a "constant" local section (with respect to whatever local trivialization you like). The point is that only the values of $dg_x$ and $s(x)$ matter....I will comment on your last question when I have time later. – Phillip Andreae Jul 25 '15 at 14:16
• as the reply states, only the summands where each partial derivative hits one $g$ survive the leibniz rule (because of $g(x)=0$). they are all the same and there are $k!$ of them. and the first partial derivatives of $g$ evaluated at $x$ are the components of $v$, which in the flat version you called $\xi$. there's your replacement. – peter Oct 31 '17 at 7:43