Recently I posted the following question on stack exchange, but it remained with no answer https://math.stackexchange.com/questions/1863658/symbol-of-differential-operator-and-change-of-coordinates I thought that maybe here I will find some help. Here is the discussion, cross posted from stack exchange:

"Some time ago I posted the question about the change of coordinates in differential operator. Here is the relevant discussion https://math.stackexchange.com/questions/1700651/symbol-of-differential-operator-transforms-like-a-cotangent-vector."

The answer which was given to this question explains why we can invariantly define the principal symbol of differential operator. However I would like to understand straightforward what happens with the principal symbol when we introduce new coordinates. Let me recall some formulas for transformation laws for tensors: if $(x^i)$ are old coordinates and $(y^i)$ are new coordinates then we have following formulas for canonical basis in tangent and cotangent spaces (everything takes place over manifold $M$): $$\frac{\partial}{\partial y^i}=\sum_{k=1}^n \frac{\partial x^k}{\partial y^i} \frac{\partial}{\partial x^k}, \qquad \frac{\partial}{\partial x^i}=\sum_{k=1}^n \frac{\partial y^k}{\partial x^i} \frac{\partial}{\partial y^k},$$ $$dy^i=\sum_{k=1}^n\frac{\partial y^i}{\partial x^k} dx^k, \qquad dx^i=\sum_{k=1}^n\frac{\partial x^i}{\partial y^k} dy^k $$ If we have vector field $X=\sum_{k=1}^nX^k \frac{\partial}{\partial x^k}=\sum_{k=1}^n Y^k \frac{\partial}{\partial y^k}$ and differential one-form $\omega=\sum_{k=1}^n \alpha_k dx^k=\sum_{k=1}^n\beta_k dy^k$ expressed in these two system of coordinates then we have formulas: $$Y_i=\sum_{k=1}^n\frac{\partial y^i}{\partial x^k}X^k, \qquad X^i=\sum_{k=1}^n\frac{\partial x^i}{\partial y^k}Y^k,$$ $$\beta_i=\sum_{k=1}^n\frac{\partial x^k}{\partial y^i}\alpha_k, \qquad \alpha_i=\sum_{k=1}^n\frac{\partial y^k}{\partial x^i}\beta_k.$$ For general tensors of type $(r,s)$ if we have in two different coordinate systems $$t=\sum_{i_1,...,i_r,j_1,...,j_s}t^{i_1,...,i_r}_{j_1,...,j_s}\frac{\partial}{\partial x^{i_1}} \otimes ... \otimes \frac{\partial}{\partial x^{i_r}}\otimes dx^{j_1} \otimes ... \otimes dx^{j_s}=\sum_{i_1,...,i_r,j_1,...,j_s}u^{i_1,...,i_r}_{j_1,...,j_s}\frac{\partial}{\partial y^{i_1}} \otimes ... \otimes \frac{\partial}{\partial y^{i_r}}\otimes dy^{j_1} \otimes ... \otimes dy^{j_s}$$ then we have $$\frac{\partial}{\partial y^{i_1}} \otimes ... \otimes \frac{\partial}{\partial y^{i_r}}\otimes dy^{j_1} \otimes ... \otimes dy^{j_s}=\sum_{k_1,...,k_r,l_1,...,l_s}\frac{\partial x^{k_1}}{\partial y^{i_1}} ... \frac{\partial x^{k_r}}{\partial y^{i_r}}\frac{\partial y^{j_1}}{\partial x^{l_1}} ... \frac{\partial y^{j_s}}{\partial x^{l_s}} \frac{\partial}{\partial x^{k_1}} \otimes ... \otimes \frac{\partial}{\partial x^{k_r}}\otimes dx^{l_1} \otimes ... \otimes dx^{l_s},$$

$$\frac{\partial}{\partial x^{i_1}} \otimes ... \otimes \frac{\partial}{\partial x^{i_r}}\otimes dx^{j_1} \otimes ... \otimes dx^{j_s}=\sum_{k_1,...,k_r,l_1,...,l_s}\frac{\partial y^{k_1}}{\partial x^{i_1}} ... \frac{\partial y^{k_r}}{\partial x^{i_r}}\frac{\partial x^{j_1}}{\partial y^{l_1}} ... \frac{\partial x^{j_s}}{\partial y^{l_s}} \frac{\partial}{\partial y^{k_1}} \otimes ... \otimes \frac{\partial}{\partial y^{k_r}}\otimes dy^{l_1} \otimes ... \otimes dy^{l_s}$$and also

$$u^{i_1,...,i_r}_{j_1,...,j_s}=\sum_{k_1,...,k_r,l_1,...,l_s}\frac{\partial y^{i_1}}{\partial x^{k_1}} ... \frac{\partial y^{i_r}}{\partial x^{k_r}} \frac{\partial x^{l_1}}{\partial y^{j_1}} ... \frac{\partial x^{l_s}}{\partial y^{j_s}}t^{k_1,...,k_r}_{l_1,...,l_s},$$

$$t^{i_1,...,i_r}_{j_1,...,j_s}=\sum_{k_1,...,k_r,l_1,...,l_s}\frac{\partial x^{i_1}}{\partial y^{k_1}} ... \frac{\partial x^{i_r}}{\partial y^{k_r}} \frac{\partial y^{l_1}}{\partial x^{j_1}} ... \frac{\partial y^{l_s}}{\partial x^{j_s}}u^{k_1,...,k_r}_{l_1,...,l_s}.$$

However general differential operator contains higher order differentials for which I haven't seen the appropriate formulas. For general differential order $m$ operator of the form $D=\sum_{|\alpha| \leq m}a_{\alpha}(x)\frac{\partial^{|\alpha|}}{\partial x_1^{\alpha_1} ... \partial x_n^{\alpha_n}}$ (where $\alpha=(\alpha_1,...,\alpha_n)$ a multiindex and $a_{\alpha}(x)$ are matrices) the principal symbol is defined as the expression $a(x,\xi)=\sum_{|\alpha|=m}a_{\alpha}(x)\xi_1^{\alpha_1} ... \xi_n^{\alpha_n}$ where $\xi=(\xi_1,...,\xi_n)$ is a vector: what we do, we replace each $\frac{\partial}{\partial x_k}$ by $\xi_k$. So far we have some formal expression $a$ which takes a point $x$ of a manifold and vector $\xi \in \mathbb{R}^n$ and produces some matrix. And here are my questions:

  1. What transformation law the function $a$ must obey in order to correctly define a mapping from the cotangent bundle $T^*M$? Or maybe we look for some transformation laws for the variables $\xi_k$?
  2. How to show that such transformation law holds?

I repeat once again: I already understood how to define symbol globally as a map from cotangent space and that this definition gives exactly the local expression described above-however I would like to understand the reason for such definition which at the moment is pulled out of hat.Concerning the second question I would like to understand whether there are some general formulas for transformation laws for an arbitrary differential operators: as far as I remember, when I once saw the computations for spherical Laplacian, it was not clear whether such method will work in general arbitrary example."

I'm afraid that some of the users may consider this question as not research level. However before voting to close let me argue a little:
1. I tried to get an satisfying answer on stack exchange but I didn't, despite posting two closely related questions.
2. I didn't find this issue explained in the literature properly.
3. I hope that maybe someone would also like to see this explained.


closed as off-topic by Michael Renardy, Willie Wong, user21574, Stefan Kohl, Stefan Waldmann Jul 23 '16 at 6:13

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question does not appear to be about research level mathematics within the scope defined in the help center." – Michael Renardy, Willie Wong, Community, Stefan Waldmann
If this question can be reworded to fit the rules in the help center, please edit the question.

  • 4
    $\begingroup$ Transforming higher order derivatives between different coordinate systems is rather messy. However, you indicated you are interested only in the principal order terms. This is not complicated at all. $\endgroup$ – Michael Renardy Jul 22 '16 at 15:06
  • 2
    $\begingroup$ In particular, the principal part transforms tensorially. This follows from chain rule and product rule. $\endgroup$ – Willie Wong Jul 22 '16 at 15:41
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    $\begingroup$ I will put forward the hypothesis that your question didn't receive much attention because it is rather long winded, compared to the complexity of what you are asking. To your question, $D = (a,\partial) = a^\alpha \partial_\alpha$, with implicit sums over the multi-index $\alpha$. Elementary calculus says that there exists a matrix $M^\alpha_\beta \partial_\alpha = \frac{\partial^{|\beta|}}{\partial y^\beta} =: \partial'_\beta$, which is triangular and involves $\partial x/\partial y^\alpha$. Then $D = (a,M^{-1} \partial') = ((M^{-1})^T a, \partial') =: (a',\partial')$ defines $a'$. $\endgroup$ – Igor Khavkine Jul 22 '16 at 20:02