$\DeclareMathOperator{I changed index to upper one}{but edit has to take al least 6 elements}$
Some time ago I posted the question about the change of coordinates in differential operator. Here is the relevant discussion Symbol of differential operator transforms like a cotangent vector
The answer which was given to this question explains why we can invariantly define the principial symbol of differential operator. However I would like to understand straightforward what happens with the principial 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 differenet 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 appriopriate 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 principial synmbol 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:
- 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 $\xi_k$?
- 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 remeber, when I once saw the computations for spherical Laplacian, it was not clear whether such method will work in general arbitrary example.
In your tensor transoformation formulas you gracefully use lower and upper indices, but when you start talking about linear differential operators (LDO) you write it strangely as $a_\alpha\frac{\partial^{|\alpha|}}{\partial x_1^{\alpha_1} ... \partial x_n^{\alpha_n}}.$
Notice that any vector field $X$ is a LDO of order 1 and locally can be described, as you wrote above, as $X^i\frac{\partial}{\partial x^i}.$ Hence LDOs are generalizations (almost compositions) of vector fields.
So let us use the convention to write LDOs in map as $a^\alpha\frac{\partial^{|\alpha|}}{\partial x_1^{\alpha_1} ... \partial x_n^{\alpha_n}}.$ Now it looks like vector field case, i.e $a^\alpha=X^i.$
Notice that we write $\frac{\partial^{|\alpha|}}{\partial x_1^{\alpha_1} ... \partial x_n^{\alpha_n}}$ with indices on bottom. However we still apply functions defined on $M$ to this operator. So it rather should be denoted as $$\frac{\partial^{|\alpha|}}{\partial (x^1)^{\alpha_1} ... \partial (x^n)^{\alpha_n}}$$ But convention is different, I guess due to lack of space above $x$'s.
For simplicity let's say that we have LDO $\mathcal{D}:C^\infty(M)\to C^\infty(M).$ By this constrain we get rid of matrices $a^\alpha$ and simply we have functions instead. Assume that $\mathcal{D}$ is of order $m.$ Hence locally $\mathcal{D}$ takes a form $$D=\sum_{\alpha\leqslant m}a^\alpha\frac{\partial^{|\alpha|}}{\partial x_1^{\alpha_1} ... \partial x_n^{\alpha_n}}$$
At this point you should complain that we still do not know how does $a^\alpha$'s transform.
However $a^\alpha$'s transformation rule depends on the order $|\alpha|.$ To make long story short: transformation of $k$ order part influences parts of $k,k-1,\dots, 1$ orders. Leibniz rule is guilty for such behavior.
For example consider 3rd order LDO $\mathcal{D},$ which in some map has only $3$ and $0$ order parts, i.e $\mathcal{D}$ locally takes form
$$Df=\sum_{i_i,i_2i_3}a^{i^1i^2i^3}\frac{\partial^3f}{\partial x^{i_1}\partial x^{i_2}\partial x^{i_3}}+fh.$$
Recall that $0$ part is just a multiplication by other function (in our case $h$).
Now we change coordinates from $(x^1,\dots,x^n)$ to $(y^1,\dots,y^n).$ We use transformation rules that you wrote above and we see that in new coordinates $\mathcal{D}$ takes form
$$Df=\sum_{i_i,i_2i_3}a^{i_1i_2i_3}\sum_{j_1}\frac{\partial y^{j_1}}{\partial x^{i_1}}\frac{\partial}{\partial y^{j_1}}\left(\sum_{j_2}\frac{\partial y^{j_2}}{\partial x^{i_2}}\frac{\partial}{\partial y^{j_2}}\left( \sum_{j_3}\frac{\partial y^{j_3}}{\partial x^{i_3}}\frac{\partial f}{\partial y^{j_3}} \right)\right)+fh.$$ As I mentioned now we will use only Leibniz rule. Hence $$Df=\sum_{j_1,j_2,j_3}\sum_{i_i,i_2i_3}a^{i_1i_2i_3}\frac{\partial y^{j_1}}{\partial x^{i_1}}\frac{\partial}{\partial y^{j_1}}\left(\frac{\partial y^{j_2}}{\partial x^{i_2}}\left(\overbrace{\frac{\partial}{\partial y^{j_2}}\left(\frac{\partial y^{j_3}}{\partial x^{i_3}}\right) \frac{\partial f}{\partial y^{j_3}}}^{(1)} +\overbrace{\frac{\partial y^{j_3}}{\partial x^{i_3}}\frac{\partial^{i_2i_3}f}{\partial y^{i_2}y^{i_3}}}^{(2)}\right)\right)+fh.$$ Again we have to use Leibniz rule, but now to 3 factors. But I guess you can see that $(1)$ would produce $2$nd and $1$st order parts and $(2)$ would produce $3$rd and $2$nd order parts.
So in different coordinates we produced orders of $2$ and $1$ as well, but haven't touched $0$ order part.
It is sufficient to see what have happened to the order $3$ part. Keeping track of $3$rd order part in the above, we see that it takes form
$$\sum_{j_1,j_2,j_3}\sum_{i_i,i_2i_3}a^{i_1i_2i_3}\frac{\partial y^{j_1}}{\partial x^{i_1}}\frac{\partial y^{j_2}}{\partial x^{i_2}}\frac{\partial y^{j_3}}{\partial x^{i_3}}\frac{\partial^{i_1i_2i_3}f}{\partial y^{j_1}y^{j_2}y^{j_3}}.$$
Hence we get that $$a^{j_1j_2j_3}=\sum_{i_i,i_2i_3}a^{i_1i_2i_3}\frac{\partial y^{j_1}}{\partial x^{i_1}}\frac{\partial y^{j_2}}{\partial x^{i_2}}\frac{\partial y^{j_3}}{\partial x^{i_3}}$$
In order to make principal part $C^\infty$ linear we have to kill $\frac{\partial y^{j_1}}{\partial x^{i_1}}\frac{\partial y^{j_2}}{\partial x^{i_2}}\frac{\partial y^{j_3}}{\partial x^{i_3}}$ somehow. So for covector $\alpha$ we define $\sigma(x,\alpha):C^\infty(M) \to C^\infty(M)$ in $(x^1,\dots,x^n)$ coordinates as $$\sigma(x,\xi_i dx^i)f=\xi_{i_1}\xi_{i_2}\xi_{i_3}a^{i_1i_2i_3}f.$$
(From now I will skip sums and use Einstein notation) By your formula above we know that $\xi_j=\xi_i\frac{\partial x^i}{\partial y^j}$ hence principal symbol in $(y^1,\dots,y^n)$ coordinates takes form
$$\sigma(x,\xi_jdy^j)f=\xi_{j_1}\xi_{j_2}\xi_{j_3}a^{j_1j_2j_3}f=\xi_{i_1}\xi_{i_2}\xi_{i_3}\frac{\partial x^{i_1}}{\partial y^{j_1}}\frac{\partial x^{i_2}}{\partial y^{j_2}}\frac{\partial x^{i_3}}{\partial y^{j_3}}a^{i'_1i'_2i'_3}\frac{\partial y^{j_1}}{\partial x^{i'_1}}\frac{\partial y^{j_2}}{\partial x^{i'_2}}\frac{\partial y^{j_3}}{\partial x^{i'_3}}=\\ \xi_{i_1}\xi_{i_2}\xi_{i_3}a^{i_1i_2i_3}f=\sigma(x,\xi_i dx^i)f.$$
Hence $\sigma(x,\alpha)$ is well defined.
BTW. To see how general transformation rules are sick, let's think about LDO as a composition of vector fields $\mathcal{D} = X_1\circ\dots\circ X_m.$ Now try compute $X_1\circ\dots\circ gX_k\circ\dots\circ X_m.$ Notice that it will produce a lot of summands. While changing coordinates you have to multiply by such $g_k$ each $X_k.$ This is why it is hard.