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.

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    $\begingroup$ 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. $\endgroup$ Aug 27, 2014 at 20:11
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    $\begingroup$ 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 $\endgroup$ Aug 28, 2014 at 20:24
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    $\begingroup$ 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. $\endgroup$ Aug 28, 2014 at 20:57

1 Answer 1


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...


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