8
$\begingroup$

I am currently trying to understand the local expression of a (pseudo)differential operator $ \int_{R^n} e^{(x - y)\cdot \xi} \sigma(x,\xi) \, d \xi $ on a manifold $M$ (compact and boundaryless, say), where $\sigma(x,\xi)$, as usual, is the symbol of the operator. (alternatively one may also write this as $ \int_{R^n} e^{(x - y)\cdot \xi} a(x,y,\xi) \, d \xi $ where $a(x,y,\xi)$ belongs to the same symbol class as $\sigma$.)

now the picture I have in my head is that the variable $(x,\xi)$ stands for a local coordinate in the cotangent bundle $T^*M$. but then, at the same time, the variables $(x,y)$ are local coordinates for some subset $U \times V$ of $M \times M$, so I fear I am mixing something up here?

Also, I was wondering whether the dot product $(x - y) \cdot \xi$ in the exponential term has anything to do with the pairing of elements from $T^*M$ with elements of the tangent bundle $TM$? some sources for example write the dot product in terms of a braket, $(x - y) \cdot \xi = \langle (x - y), \xi \rangle$ which makes this even more suggestive. but I struggle to make sense of this, I don't really know how to relate $(x - y)$ to the tangent bundle $TM$, making it very likely that this picture is wrong anyways.

Many thanks for your comments!

1 Answers 1

3

A $\Psi$DO $P$ on a compact manifold $M$ is (modulo smoothing operators) a finite sum $P = \sum_k P_k$, where each $P_k$ is in local coordinates a $\Psi$DO on $\mathbb{R}^n$ with compact support.

Each $P_k$ has a symbol $\sigma_k$ on $T^\ast \mathbb{R}^n$. Clutching them together with a fixed partition of unity gives a function on $T^\ast M$. In order to make this independent of the partition of unity, we have to quotient out the lower order symbols, and so we get the principal symbol $[\sigma_P]$ of $P$.

Now given a symbol $\sigma$ on $T^\ast M$ (not a principal symbol, but a concrete function on $T^\ast M$ satisfying the needed estimates in local charts), we can construct a $\Psi$DO $P$ on $M$ with the same principal symbol, i.e., $[\sigma_P] = [\sigma]$. This seems like what you want: Going from the symbol to the corresponding $\Psi$DO defined by it. But this construction is a little bit technical, since we first have to define the Fourier transform on a manifold. Details for this may be found in, e.g., Lawson, Michelsohn, Spin Geometry, Princeton University Press, 1989, at the end of Chapter III, ยง3.