# Invariant definition of the space of symbols on a vector bundle (pseudo-differential operators)

Normally, in the context of pseudo-differential operators, a symbol on a vector bundle $E$ is defined as a smooth function on $E$ which in each trivializing chart fulfills the usual symbol estimates \begin{equation} \sup_{x \in U, \xi \in \mathbb{R}^p} (1 + |\xi|^2)^{\frac{-m + |\beta|}{2}} \ |D^\alpha_x D^\beta_\xi a(x, \xi)| < \infty. \end{equation}

Does there exists a definition which does not directly rely on a local trivialization?

Note that classical symbols (symbols with an asymptotic expansion at infinity) can be interpreted as smooth functions on the (radial) compactification and thus for them there is a nice coordinate-free interpretation (I think this approach is publicized by Richard Melrose). Something along these lines would be perfect, but a definition based on additional data (like choosing a fiber metric and/or connection) is also ok.

• This is discussed in many places. See e.g. mathoverflow.net/questions/3477/… A possible global definition of a symbol uses jet bundles. Basically, a linear differential operator of order $k$ is a linear bundle map from the $k$-th jet prolongation of $J^kE$ into the target bundle. The symbol is then the associated map from $J^kE / J^{k-1}E \simeq \bigodot^k TM \times E$ to the target bundle. – Vít Tuček Aug 27 '14 at 20:11
• Thanks Vít Tuček for your comment. Maybe I should have made this clear in the question: I'm not interested in the symbol of a differential operator but of a pseudo-differential operator. These symbols are more generally defined via the above symbol estimates. Furthermore, pseudo-differential operators does not have such a nice interpretation in terms of jet bundles, see for example mathoverflow.net/questions/75976/symbol-of-pseudodiff-operator – Tobias Diez Aug 28 '14 at 20:24
• I see. In that case I am very interested in answers. :) You can edit your questions and in this case I would even consider adding "pseudodifferential" to the title. – Vít Tuček Aug 28 '14 at 20:57

I think the following should work:

Let $$M$$ be a compact manifold (just to be safe) and $$\pi :E \to M$$ a vector bundle. Since $$E$$ carries an action of $$\mathbb{R}^{\times}$$ there's an invariant notion of a function on $$\overset{\circ}{E}$$ ($$E$$ without the zero section) which is homogeneous of degree $$s$$ for every $$s \in \mathbb{R}$$ namely:

$$\{ f: \overset{\circ}{E} \to \mathbb{R} | f(\lambda v) = |\lambda|^s f(v) \}$$

Using this we can define what it means for a function to have growth of degree $$\le s$$. Namely you just take those functions $$f$$ on $$E$$ s.t. $$f = O(\psi)$$ for some homogeneous function $$\psi$$ of degree $$s$$ (where the $$O$$ notation is supposed to be interpreted fiberwise and not in the $$M$$-direction).

All that's left now is to take care of derivatives but the vertical subbundle $$VE \subset TE$$ is always globally well defined (as it is the kernel of the pushforward of tangent vectors). Moreover we can also consider vertical vector fields which are invariant w.r.t. the action of $$E$$ on itself by vector addition. That is

$$C^{\infty}(E,VE)^E = \{X \in C^{\infty}(E,VE) | \forall u \in C^{\infty}(M,E), t_{\pi^*u}^* X = X\}$$

Where $$t_{\pi^*u}^*$$ is the pullback along the tranlsation by $$\pi^*u$$. Call these vertical vector fields linear. In local coordinates these are the vector fields which are constant along the fibers.

Now we can say that a function $$f$$ on $$E$$ is of symbol class $$\le m$$ iff for every collection of $$r$$ linear vector fields the growth of the iterated derivative w.r.t. these vector fields is of degree $$\le m-r$$. This definition is obviously local on $$M$$ and it also recovers your definition in the case of the trivial vector bundle so it must coincide with it in the global case, i hope i'm not wrong...