If $T:\mathbb{R}^n\to\mathbb{R}^n$ is a linear bijection and $y=Tx$, then why is $T^t\frac{\partial}{\partial y}=\frac{\partial}{\partial x}$

Question

A material I am reading on differential operators on manifolds starts with motivating examples in the Euclidean space. In one such example it is stated that if $T:\mathbb{R}^n\to\mathbb{R}^n$ is a linear bijection and $y=Tx$, then $T^t\frac{\partial}{\partial y}=\frac{\partial}{\partial x}$ holds with $T^t$ being to transpose of $T$. Maybe it is standard notation in differential geometry to write $\frac{\partial}{\partial x}$ for $x\in\mathbb{R}^n$, I don't know. Unfortunately this is the first instance of the said notation in my materials so your guess is as good as mine. Having said this, my question is that assuming you can give a reasonable meaning to $\frac{\partial}{\partial x}$, do you happen to recall any similar looking result from differential geometry?

Answer 1

I think we can all agree that for all differentiable functions $f: \mathbb R^n \to \mathbb R$, we have

$$ \begin{bmatrix} T_{11} & \dots & T_{n1} \\ \vdots & \ddots & \vdots \\ T_{n1} & \dots & T_{nn}\end{bmatrix} \begin{bmatrix} \left. \frac{\partial f}{\partial y^1} \right|_{\mathbf y = T(\mathbf x)} \\ \vdots \\ \left. \frac{\partial f}{\partial y^n} \right|_{\mathbf y = T(\mathbf x)} \end{bmatrix} = \begin{bmatrix} \left. \frac{\partial f}{\partial x^1} \right|_{\mathbf x} \\ \vdots \\ \left. \frac{\partial f}{\partial x^n} \right|_{\mathbf x} \end{bmatrix} \qquad \qquad (\star) $$ for all $\mathbf \in \mathbb R^n$. This follows from the chain rule.

I would interpret the equation $T^t \frac{\partial }{\partial \mathbf y} = \frac{\partial}{\partial \mathbf x}$ as nothing more than shorthand for equation $(\star)$.

Answer 2

This is not an answer but a long comment. @KennyWong's answer is correct. However, I do not like the notation that your material is using. I use the following conventions:

1. Given a matrix $M$, the element in the $i$-th row and $j$-th column is denoted $M^i_j$. In particular, a superscript labels the row and a subscript labels the column.
2. Given a vector space $V$, a basis of $V$ is labeled using subscripts, $(e_1, \dots, e_n)$. It therefore can be written formally as a row matrix $$ E = \begin{bmatrix} e_1 & \cdots & e_n \end{bmatrix} $$
3. Any vector $v \in V$ can be written with respect to the basis as $$ v = e_ja^j, $$ where the coefficients $a_1, \dots, a_n$ are scalars and labeled using subscripts. Therefore, they form a column matrix $$ A = \begin{bmatrix} a^1 \\ \vdots \\ a^n \end{bmatrix}. $$
4. Therefore, if $v = e_ja^j$, then it can be written in terms of matrix multiplication as $$ v = EA. $$
5. Now let $V = \mathbb{R}^n$ and $L: \mathbb{R}^n \rightarrow \mathbb{R}^n$ be a linear map. If $$ E = \begin{bmatrix} e_1 & \cdots & e_n \end{bmatrix} $$ is the standard basis, then for each $1 \le i \le n$, there exist coefficients $T_i^1, \cdots, T_i^n$ such that $$ L(e_i) = e_jT^j_i. $$ In matrix notation, this is written $$ \begin{bmatrix} L(e_1) & \cdots & L(e_n)\end{bmatrix} = ET. $$
6. Therefore, if $x = e_jx^j$, then $$ L(x) = L(e_jx^j) = L(e_j)x^j = e_iT^i_jx^j = ETX, $$
7. The partial differential operators, whose indices are lower, can be written as row matrices \begin{align*} \frac{\partial}{\partial x} = \begin{bmatrix} \frac{\partial}{\partial x^1} & \cdots & \frac{\partial}{\partial x^n} \end{bmatrix}\\ \frac{\partial}{\partial y} = \begin{bmatrix} \frac{\partial}{\partial y^1} & \cdots & \frac{\partial}{\partial y^n} \end{bmatrix}. \end{align*}
8. If $y = L(x)$, i.e., $y^i = T^i_jx^j$, then by the chain rule, $$ \frac{\partial}{\partial x^j} = \frac{\partial}{\partial y^i}\frac{\partial y^i}{\partial x^j} = \frac{\partial}{\partial y^i}T^i_j. $$ In matrix form, this can be written as $$ \frac{\partial}{\partial x} = \frac{\partial}{\partial y}T. $$ It turns out that the matrix of partial derivatives $$ \frac{\partial}{\partial x} = \begin{bmatrix} \frac{\partial}{\partial x^1} & \cdots & \frac{\partial}{\partial x^n} \end{bmatrix} $$ behaves, in formal calculations, just like the basis $$ E = \begin{bmatrix} e_1 & \cdots & e_n \end{bmatrix}. $$
9. By the way, the assumption that $T$ is an isomorphism is not needed for either @KennyWong's answer or my comment.

The reason I like my notation is that you do not have to worry about when to use $T$ or $T^t$ or $T^{-1}$. The notation, which is based only standard conventions used for matrices and matrix multiplication, guides you to the right formulas.