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Suppose that $D=\sum_{|\alpha| \leq k} A_{\alpha}(x)\frac{\partial^{\alpha}}{\partial x_1^{\alpha_1}...\partial x_n^{\alpha_n}}$ is a differential operator defined on vector valued functions on $\mathbb{R}^n$ (here $A_{\alpha}$ are matrices). Then one can form the expression $\sum_{|\alpha| = k} A_{\alpha}(x)\xi_1^{\alpha_1} \cdot ... \cdot \xi_n^{\alpha_n}$ which is called the principial symbol of $D$. This is fine while we deal with flat case, i.e. everything takes place on $\mathbb{R}^n$. But differential operators may be defined in the broader context of (say) compact manifolds and arbitrary vector bundle. Then the symbol is defined by the similiar procedure, but only locally. On the global level suddenly cotangent bundle somehow pops up: I've heard that this follows from the fact that the principial symbol transforms like a $(0,1)$ tensor more precisely that the variables $\xi$ transform like this. I'm a bit confuses what does it mean: I know transformation law for general tensors of type $(p,q)$ but here are formulas which involve higher order differential operators (not just first order as in standard vector fields). So

How one can show that the variables $\xi$ transform like a $(0,1)$ tensor? How it can be used to define symbol as a function from the cotangent bundle?

There is also more invariant way of defining symbol, I asked about this some time ago and I received the answer only to the half of my questions (the relevant discussion may be found here Symbol of the differential operator on vector bundles). So I would also like to ask

Why this invariant definition which can be found in this discussion produces locally the expression $\sum_{|\alpha| = k} A_{\alpha}(x)\xi_1^{\alpha_1} \cdot ... \cdot \xi_n^{\alpha_n}$

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Here's how I like to think about it. If we have a Riemannian structure, we can verify that a linear differential operator $L$ is coordinate-independent by writing it as

$$ Lf = \sum_{0\le i \le k} \langle A_i, \nabla^i f \rangle$$

where $A_i$ is a symmetric contravariant $i$-tensor, $\nabla^i$ is the iterated covariant derivative and $\langle,\rangle$ is the natural pairing of $(0,i)$-tensors with $(i,0)$-tensors. It then is completely natural to pair $A_k$ with covectors: for $\xi \in \Lambda^1(TM^*)$ we can form the coordinate-independent expression $$\langle A_k, \otimes^k \xi\rangle= A_k^\alpha\xi_{\alpha_1} \cdots \xi_{\alpha_k}.$$

The Riemannian structure is not necessary here - I just use it because it makes me feel better about higher derivatives. If you change the Riemannian structure then any change in $\nabla^k f$ will only involve lower derivatives of $f$ and Christoffel symbols, so the principal symbol will be untouched.

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  • $\begingroup$ Thank you for your answer. I struggled a little bit with the previous discussion and finally I see why the global definition of the symbol map (as explained in the linked discussion) locally gives just a substitution of $\frac{\partial^{|\alpha|}}{\partialx^{\alpha}$ by $i\xi^{\alpha}$. So as far as I understood, your argument ensures that the definition of symbol is "global". Nevertheless still I'm interested in seeing how does it transforms under the change of coordinates. In particular, the problem follows from the fact that I don't know formulas for transforming differential operators of $\endgroup$ – truebaran Mar 22 '16 at 0:50
  • $\begingroup$ ... higher orders. $\endgroup$ – truebaran Mar 22 '16 at 0:50

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