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Manifolds

Let $\mathcal{M}$ be our manifold.
Let $n$ be the dimension of manifold $\mathcal{M}$.
Let $p \in \mathcal{M}$ be a point on manifold.
Let $V = T_p(\mathcal{M})$ be the tangent space on $\mathcal{M}$ at point $p$.
Let $V^* = T_p^*(\mathcal{M})$ be the dual tangent space on $\mathcal{M}$ at point $p$.

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Charts

TODO these are associated with charts, right? Should subsequent tensors be functions of points, or should they be functions of coordinates on charts (which themselves, by the chart, map to points?).
Since tensors are linear functions of the basis, and since the basis is in some way a function of the coordinate basis, and since the coordinate basis depends on the coordinates i.e. depends on the chart.

TODO charts ... atlases ... the domain and range functions can have different dimensions ... and if they have different dimensions then we implicitly get horizontal and vertical tangent spaces ... but I don't talk about tangent spaces for a while ...

For now lets only consider charts of equal domain and range dimension:
A manifold is represented by an atlas, which is a collection of charts.
Define a chart of the atlas, $\mathbf{u} : X \rightarrow \mathcal{M}, X \subseteq \mathbb{R}^n, \mathcal{M} \subseteq \mathbb{R}^n$
$\mathbf{u}(\mathbf{x}) = \mathbf{u}(x^1, ..., x^n) = \{u^1 (x^1, ..., x^n), ..., u^n(x^1, ..., x^n) \} $
For $\mathbf{x} = ( x^1, ..., x^n )$ a point in the domain of our chart.
Is this right? Can I say the manifold $\mathcal{M}$ is a subset of the $\mathbb{R}^n$ Cartesian space? Or is the collection-of-points of the manifold the subset of $\mathbb{R}^n$ Cartesian space?

Is the definition of a coordinate an input variable of a chart? Can I say this?

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Coordinate Basis

Let $\partial_a$ be the derivative operator of a chart coordinate applied to the chart, such that applying it to the chart $\mathbf{u}(x^1, ..., x^n)$ produces the coordinate vector basis:
$\partial_a = \partial_a \mathbf{u}(\mathbf{x}) = \frac{\partial}{\partial x^a} \mathbf{u}(\mathbf{x}) = \frac{\partial}{\partial x^a} \mathbf{u}(x^1, ..., x^n) = \left[ \begin{matrix} \frac{\partial u^1}{\partial x^a} & \cdots & \frac{\partial u^n}{\partial x^a} \end{matrix} \right]^T$.
Notice that the vector produced from $\partial_a$ is a vector on the tangent space of the manifold $V$.

Is $\partial_a$ a vector? Or is its application to the chart a vector? Or are the two synonymous?
Is it safe to say $\partial_a = \frac{\partial}{\partial x^a}$? Or is the difference that the $\partial_a$ implicitly has the chart $\mathbf{u}$ applied to its left? Or are the two synonymous and this is the same sort of implicit math as to why indexed tensors usually leave off the basis elements?
Such that every time we write a vector $\vec{v} = v^a \partial_a = \overset{n}{\underset{a=1}{\Sigma}} v^a \partial_a$ we are implying $\vec{v} = \overset{n}{\underset{a=1}{\Sigma}} v^a \partial_a (\mathbf{u}(\mathbf{x}))$?

Also notice that for now I'm enforcing that the chart domain and range is the same dimension. If we are dealing with a chart whose range is higher dimension than domain then we also now have the problem of defining a tangent space basis with more vectors than we have derivative operators. So long as the two dimensions are equal, we can define a tangent space basis from the coordinate vector operators. Provided they are not degenerate / provided they are linearly independent.

TODO put this later, after defining what I'm using?
For convenience, define $\partial = \overset{n}{\underset{a=1}{\otimes}} e^a \partial_a$.

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Non-Coordinate Basis

Let $e_a \in \{ e_1 ... e_n \}$ be a vector basis.
Let $\{ e_a \} = span(V)$.

How should they be defined?
As functions of points on the manifold: $e_a = e_a(x)$
Or as functions of chart coordinates:
$e_a = e_a \left( x^1, ..., x^n, \frac{\partial}{\partial x^1}, ..., \frac{\partial}{\partial x^n} \right)$ be a vector basis that spans $V$.
$e_a : X \times V \rightarrow V$ form a basis of the tangent vector space, where X was our chart domain containing elements $( x^1, ..., x^n)$.

Another thing to consider, the function of $e_a$ wrt $x^i$ is arbitrary, however the function of $e_a$ wrt $\partial_a$ should be linear, i.e. subject to $e_a( \alpha \partial_i + \beta \partial_j) = \alpha e_a (\partial_i) + \beta e_a (\partial_j)$.

How about defining the $\vec{v}$ vs $\underrightarrow{w}$?
$e_a = e_a(\vec{x}, \underrightarrow{\vec{\partial}})$
for $\underrightarrow{\vec{\partial}} = \{ \vec{\partial}_a \}$
... though while you call $\vec{\partial}_a$ a vector and an element in the set-of-vectors $\{ \vec{\partial}_a \}$, it might not be correct to use the $\underrightarrow{\vec{\partial}}$ especially if you reserve the $\underrightarrow{w}$ for one-forms.

Or specifying that my bold symbols represent sets-of-things, i.e. matrices?
$e_a = e_a(\mathbf{x}, \mathbf{\partial})$
Notice that the bold version of $\partial$ doesn't look any different.

Define our linear transform from the coordinate basis $\partial_\tilde{a}$ to the non-coordinate basis $e_a$ as:
$e_a = {e_a}^\tilde{a} \partial_\tilde{a}$.
I am accenting the indexes as $\tilde{a}$ just to help distinguish which indexes are associated with a coordinate basis and which are associated with a non-coordinate basis.

So the matrix $E$ formed by the elements $\{ {e_a}^\tilde{a} \}$ are a member of $GL(n, \mathbb{R})$,
where $GL(n, \mathbb{R})$ is the general linear group of linear transformations,
such that for $\mathbf{M} \in GL(n, \mathbb{R}), \mathbf{M} : \mathbb{R}^n \rightarrow \mathbb{R}^n$.

By the way, is a matrix a collection of coefficients, or is it a function mapping vectors/other matrices by linear transformation of its collection of coefficients, or is it both?
I think it is the function / linear transformation, and the coefficients are what parameterize the matrix.
So a matrix is just a tensor without the associated basis -- a matrix has an implied Cartesian basis.

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Differential-Form Coordinate Dual-Basis

Let $w \in V^* \otimes ... \otimes V^*$ be a k-form.
Let w as a function mapping k input elements in the vector space $V$ to the reals:
i.e. $w : \underbrace{V \otimes \dots \otimes V}_{\times k} \rightarrow \mathbb{R}$.
Let $dx^a$ be the one-form basis dual to $\frac{\partial}{\partial x^a}$ basis,
defined as $dx^a \in \{ dx^1 ... dx^n \}$, $dx^a : V \rightarrow \mathbb{R}$, $dx^a \left( \frac{\partial}{\partial x^b} \right) = \delta^a_b$

Notice this is very similar to the partial derivative of coordinate variable definition:
$dx^a \left( \frac{\partial}{\partial x^b} \right) = \frac{\partial}{\partial x^b} (x^a) = \delta^a_b$

Then we have fundamental theorem of calculus:
$\int_{x_L}^{x_R} dx = x_R - x_L$

If we let $x_L = 0$ then we can define $x = \int_{x_L}^x dx$,
Then we can say:
$\int_{x^b_L}^{x^b} \frac{\partial}{\partial x^a} dx^b = \frac{\partial}{\partial x^a} \int_{x^b_L}^{x^b} dx^b = \frac{\partial}{\partial x^a} x^b = \partial_a x^b = \delta^a_b$

Then there's the Dirac-delta vs Kronecker-delta:
$y = \int \delta(x,y) dx$, so $x^a = \int \delta(x^a, x^b) dx^b = \delta^a_b x^b$,
so $\delta^a_b = \int \delta(x^a, x^b) dx^b$.

Let $V^*$ be the space spanned by the one-form basis.
Or should I instead say that the one-form basis is defined as spanning $V^*$?

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One-Form Dual-Basis

Let $e^a \in \{ e^1, ..., e^n \}$, $e^a \in V^*$ be an element of the dual-basis of one-forms.
Let that dual-basis span the dual-tangent-space: $\{ e^a \} = span(V^*)$
Let $e^a = e^a(x^1, ..., x^n, dx^1, ..., dx^n)$.
$e^a : X \times V^* \rightarrow V^*$
$e^a$ is linear with regards to $dx^a$, such that $e^a (\alpha dx^i + \beta dx^j) = \alpha e^a (dx^i) + \beta e^a (dx^j)$.
such that $e^a(e_b) = \delta^a_b$

How do we specify that $e^a$ varies with point on manifold, is a linear combination of the one-form basis, and that it (like the one-form basis) maps vectors to reals? Ex:
$e^a : X \times V^* \rightarrow V^*$,
$V^* : V \rightarrow \mathbb{R}$,
Therefore $e^a : X \times V^* \times V \rightarrow \mathbb{R}$
$e^a(x^1, ..., x^n, dx^1, ..., dx^n, \partial_1, ..., \partial_n) \in \mathbb{R}$
$e^a(x^1, ..., x^n, dx^1, ..., dx^n) : V \rightarrow \mathbb{R}$
$e^a(x^1, ..., x^n, dx^1, ..., dx^n)(e_b) \in \mathbb{R}$

From here on out the manifold point (or chart coordinate?) of $e^a$ and the linear dependence on the one-form coordinate basis parameters will be hidden.
So $e^a(x^1, ..., x^n, dx^1, ..., dx^n)$ will simply be written $e^a$

Let $e^a = {e^a}_\tilde{a} dx^\tilde{a}$
For ${e^a}_\tilde{a}$ the coefficients of the linear transform from $dx^\tilde{a}$ to $e^a$.

The matrix $[ {e^a}_\tilde{a} ]$ is an element of the group: $[ {e^a}_\tilde{a} ] \in GL(n, \mathbb{R})$

Inverse components:
$[ {e_a}^\tilde{a} ] = [ {e^a}_\tilde{a} ]^{-1}$
Such that ${e_a}^\tilde{a} {e^b}_\tilde{a} = \delta_a^b$
And ${e_a}^\tilde{a} {e^a}_\tilde{b} = \delta^\tilde{a}_\tilde{b}$

If $e_a = \partial_a$ (such that ${e_a}^\tilde{a} = \delta_a^\tilde{a}$) and $e^a = dx^a$ (such that ${e^a}_\tilde{a} = \delta^a_\tilde{a}$) then we have $e^a(e_b) = dx^a (\frac{d}{dx^b}) = \frac{dx^a}{dx^b} = \delta^a_b$. The basis $\{e_a\}$ is called a coordinate basis, aka a holonomic basis.

Proving non-coordinate linear combination basis orthogonality:
$e^a (e_b)$ $= {e^a}_\tilde{a} dx^\tilde{a} ({e_b}^\tilde{b} \partial_\tilde{b})$
$= {e^a}_\tilde{a} {e_b}^\tilde{b} dx^\tilde{a} (\partial_\tilde{b})$
$= {e^a}_\tilde{a} {e_b}^\tilde{b} \delta^\tilde{a}_\tilde{b}$
$= {e^a}_\tilde{a} {e_b}^\tilde{a}$
$= \delta^a_b$

Basis linear transform determinant:
Let $e = det([{e^a}_\tilde{a}])$.
Let's associate $e > 0$ with right-handed basis and $e < 0$ with left-handed basis. (Can I do that? Or does handed-ness have to be a convention independent of the basis element ordering, in which case it would be proportional to the ordering, right?)
For simplicity I will assume $e > 0$ ... but should I?

Notice that, by linear independence, $e = 0$ implies that $\{ e_a \}$ does not span V, and $e \ne 0$ implies that $\{ e_a \}$ spans V.

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Tensor

Tensors are objects invariant to change of basis.
Tensors exist within (p,q) tanganet and dual-tangent spaces:
$x = {x^{a_1 ... a_p}}_{b_1 ... b_q} e_{a_1} \otimes ... \otimes e_{a_p} \otimes e^{b_1} \otimes ... \otimes e^{b_q}$
Examples of tensors:
scalars: $\phi : X \rightarrow \mathbb{R}$ (TODO is the chart coordinates the input, or is the manifold points? If the chart-domain then I suppose I should introduce a chart-inverse?)
rank-1 contravariant tensors (vectors): $v = v^a e_a$, $v \in V$
rank-2 contravariant tensors: $v = v^{ab} e_a \otimes e_b$, $v \in V \otimes V$
rank-p contravariant tensors: $v = v^{a_1 ... a_p} e_{a_1} \otimes ... \otimes e_{a_p}$, $v \in \underbrace{V \otimes ... \otimes V}_{\times p}$
rank-1 covariant tensors (one-form): $w = w_a e^a$, $w \in V*$, $w \in V^*$, $w : V \rightarrow \mathbb{R}$
rank-2 covariant tensors: $w = w_{ab} e^a \otimes e^b$, $w \in V^* \otimes V^*$, $w : V \otimes V \rightarrow \mathbb{R}$
rank-p covariant tensors: $w = w_{a_1 ... a_p} e^{a_1} \otimes ... \otimes e^{a_p}$, $w \in \underbrace{V^* \otimes ... \otimes V^*}_{\times p}$, $w : \underbrace{V \otimes ... \otimes V}_{\times p} \rightarrow \mathbb{R}$

Using multi-index notation:
rank-p contravariant tensors: $v = v^A e_A$
rank-q covariant tensor: $w = w_A e^A$
rank-(p,q) tensor: $x = {x^A}_B {e_A}^B$
Notice I am shorthanding $e_A = \overset{p}{\underset{i=1}{\otimes}} e_{a_i}$, $e^A = \overset{q}{\underset{i=1}{\otimes}} e^{a_i}$, ${e_A}^B = \overset{p}{\underset{i=1}{\otimes}} e_{a_i} \overset{q}{\underset{i=1}{\otimes}} e^{b_i}$
Older texts use capital letters to denote groups of indexes, i.e. multi-index notation:
However some other texts use capital letters to denote non-coordinate diagonalized basis, so the use of capital indexes will be context-dependent. Maybe I'll use vector arrows over capital letters to denote multi-indexes: $\vec{I}$.
This is close to the Ricci calculus definition of sequential summation of multi-indexes using a vector underneath $\underset{\rightharpoondown}{I}$ to denote sequential summing, $|i_1 i_2 ... i_n|$. Why not just use the same notation as sequential summing of individual indexes, i.e. $|I| = |i_1 i_2 ... i_n|$?

Change of basis of a (p,q) tensor from non-coordinate basis $e_a$ to coordinate basis $e_\bar{a}$:
Assuming the non-coordinate basis is a linear combination of another coordinate basis: $e_a = {e_a}^\bar{a} e_\bar{a}$
${x^{a_1 ... a_p}}_{b_1 ... b_1} e_{a_1} \otimes ... \otimes e_{a_p} \otimes e^{b_1} \otimes ... \otimes e^{b_q}$
$= {x^{a_1 ... a_p}}_{b_1 ... b_1} ({e_{a_1}}^\bar{a_1} e_{\bar{a}_1}) \otimes ... \otimes ({e_{a_p}}^\bar{a_p} e_{\bar{a}_p}) \otimes ({e^{b_1}}_{\bar{b}_1} e^{\bar{b}_1}) \otimes ... \otimes ({e^{b_q}}_{\bar{b}_q} e^{\bar{b}_q})$
$= {e_{a_1}}^\bar{a_1} \cdot ... \cdot {e_{a_p}}^\bar{a_p} \cdot {e^{b_1}}_\bar{b_1} \cdot ... \cdot {e^{b_q}}_\bar{b_q} \cdot {x^{a_1 ... a_p}}_{b_1 ... b_1} e_{\bar{a}_1} \otimes ... \otimes e_{\bar{a}_p} \otimes e^{\bar{b}_1} \otimes ... \otimes e^{\bar{b}_q}$
$= {x^{\bar{a}_1 ... \bar{a}_p}}_{\bar{b}_1 ... \bar{b}_1} e_{\bar{a}_1} \otimes ... \otimes e_{\bar{a}_p} \otimes e^{\bar{b}_1} \otimes ... \otimes e^{\bar{b}_q}$
In multi-index notation: ${x^A}_B {e_A}^B = {x^\bar{A}}_\bar{B} {e_\bar{A}}^\bar{B}$

One-form applied to a vector:
For one-form $w \in V^*$ and vector $v \in V$
$w(v) = w (v^b e_b)$ by expanding $v$.
$= v^b w(e_b)$ by linearity of one-form functions.
$= v^b w_a e^a (e_b)$ by expanding $w$ to its linear coefficients of the one-form basis.
$= v^b w_a \delta^a_b$ by orthogonality of $e^a(e_b)$.
$= v^a w_a$

Outer product: maps (p,q) tensors and (r,s) tensors to (p+r,q+s) tensors
outer product of vectors: $ x \otimes y = x^a e_a \otimes y^b e_b = x^a y^b e_a \otimes e_b $
outer product of one-forms: $ x \otimes y = x_a e^a \otimes y_b e^b = x_a y_b e^a \otimes e^b $

Inner product: maps vectors and vectors to reals.
$\langle a, b \rangle$
$= \langle a^u e_u, b^v e_v \rangle$
$= a^u b^v \langle e_u, e_v \rangle$
$= a^u b^v g_{uv}$ for $g_{uv} = \langle e_u, e_v \rangle$
This depends on the definition of the metric tensor, and more on that later.
TODO some like Misner Thorne Wheeler distinguish between vector inner product $a \cdot b$ and form inner product $\langle a , b \rangle$

P-vector inner product:
$\langle a, b \rangle$
$= a^U b_U$

Kronecker delta:
$\delta^a_b = 1$ for $a = b$, otherwise 0.
$\delta^a_b = \overset{a\downarrow b\rightarrow}{\left[\matrix{1&0&\dots&0 \\ 0&1&\dots&0 \\ \dots&\dots&\dots&\dots \\ 0&0&\dots&1 }\right]}$

Kronecker delta extension:
$\delta^{a_1 \dots a_p}_{b_1 \dots b_p} = p! \delta^{a_1}_{[b_1} \cdot ... \cdot \delta^{a_p}_{b_p]}$
$ = 0 $ if any indexes are duplicated
$ = +1 $ if $a_1 ... a_p$ is a positive permutation of $b_1 ... b_p$
$ = -1 $ if $a_1 ... a_p$ is a negative permutation of $b_1 ... b_p$
$ = 0 $ otherwise
In multi-index notation:
$= \delta^A_B$

Product of generalized Kronecker delta:
$\delta^{a_1 ... a_p b_1 ... b_q}_{c_1 ... ... c_{p+q}} \delta^{u_1 ... u_q}_{b_1 ... b_q} = q! \delta^{a_1 ... a_p u_1 ... u_q}_{c_1 ... ... c_{p+q}}$
But if you only sum increasing indexes:
$\delta^{a_1 ... a_p |b_1 ... b_q|}_{c_1 ... ... c_{p+q}} \delta^{u_1 ... u_q}_{|b_1 ... b_q|} = \delta^{a_1 ... a_p u_1 ... u_q}_{c_1 ... ... c_{p+q}}$

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Change-of-Coordinates

How do we represent a change-of-basis?

$dx^u = \frac{\partial x^u}{\partial y^v} dy^v$
$\frac{\partial}{\partial y^v} = \frac{\partial x^u}{\partial y^v} \frac{\partial}{\partial x^u}$
$dy^v = \frac{\partial y^v}{\partial x^u} dx^u$
$\frac{\partial}{\partial x^u} = \frac{\partial y^v}{\partial x^u} \frac{\partial}{\partial y^v}$

Ex. spherical coordinates:
Let $\mathbf{y} = \{ x, y, z \}$ and $\mathbf{x} = \{ r, \theta, \phi \}$
$x = r sin \theta cos \phi$
$y = r sin \theta sin \phi$
$z = r cos \theta$
For $r \ge 0, \theta \in [ 0, \pi ], \phi \in [0, 2 \pi )$
Therefore our derivatives are, in the form of a Jacobian:
$ \frac{\partial y^u}{\partial x^v} = \left[ \begin{matrix} \frac{\partial x}{\partial r} & \frac{\partial x}{\partial \theta} & \frac{\partial x}{\partial \phi} \\ \frac{\partial y}{\partial r} & \frac{\partial y}{\partial \theta} & \frac{\partial y}{\partial \phi} \\ \frac{\partial z}{\partial r} & \frac{\partial z}{\partial \theta} & \frac{\partial z}{\partial \phi} \end{matrix} \right] = \left[ \begin{matrix} sin \theta cos \phi & r cos \theta cos \phi & -r sin \theta sin \phi \\ sin \theta sin \phi & r cos \theta sin \phi & r sin \theta cos \phi \\ cos \theta & -r sin \theta & 0 \end{matrix} \right] $

Likewise in terms of total-derivatives:
$dx = sin \theta cos \phi dr + r cos \theta cos \phi d\theta - r sin \theta sin \phi d\phi$
$dy = sin \theta sin \phi dr + r cos \theta sin \phi d\theta + r sin \theta cos \phi d\phi$
$dz = cos \theta dr - r sin \theta d\theta$
...in matrix form...
$ \left[ \begin{matrix} dx \\ dy \\ dz \end{matrix} \right] = \left[ \begin{matrix} sin \theta cos \phi & r cos \theta cos \phi & -r sin \theta sin \phi \\ sin \theta sin \phi & r cos \theta sin \phi & r sin \theta cos \phi \\ cos \theta & -r sin \theta & 0 \end{matrix} \right] \left[ \begin{matrix} dr \\ d\theta \\ d\phi \end{matrix} \right] $
...and its inverse...
$ \left[ \begin{matrix} dr \\ d\theta \\ d\phi \end{matrix} \right] = \left[ \begin{matrix} \sin \theta \cos \phi & \sin \theta \sin \phi & \cos \theta \\ \frac{1}{r} \cos \theta \cos \phi & \frac{1}{r} \sin \phi \cos \theta & -\frac{1}{r} \sin \theta \\ -\frac{\sin \phi}{r \sin \theta} & \frac{\cos \phi}{r \sin \theta} & 0 \end{matrix} \right] \left[ \begin{matrix} dx \\ dy \\ dz \end{matrix} \right] $

Now the partial description is in terms of contravariant objects. The total derivative description is in terms of covariant objects. Which is correct in relation to our change-of-basis of tensors, and how?

Now how about the four different ways we can represent a change-of-basis in our index notation.

Let's use a convention of contravariant-index-implies-up-down, covariant-index-implies-left-right. Notice that a contravariant index in the partial denominator is equivalent to a covariant index.

Review of tensors in index notation, as matrix-multiply using this convention:

$v = v^a e_a = \left[ \begin{matrix} e_1 ... e_n \end{matrix} \right] \left[ \begin{matrix} v^1 \\ \vdots \\ v^n \end{matrix} \right] $

$v = v_a e^a = \left[ \begin{matrix} v_1 ... v_n \end{matrix} \right] \left[ \begin{matrix} e^1 \\ \vdots \\ e^n \end{matrix} \right] $

Mind you since I am choosing contravariant for up-down, covariant for left-right indexes, combined with using partial-operators as my matrix elements, this does not imply an apply-rightward of the derivatives as other notation-abuses might do.
So that in context of our change-of-basis formula:

$dx^u = \frac{\partial x^u}{\partial y^v} dy^v$ becomes
$ \left[ \begin{matrix} dx^1 \\ ... \\ dx^n \end{matrix} \right] = \left[ \begin{matrix} \frac{\partial x^1}{\partial y^1} & ... & \frac{\partial x^1}{\partial y^n} \\ \vdots & \ddots & \vdots \\ \frac{\partial x^n}{\partial y^1} & ... & \frac{\partial x^n}{\partial y^n} \end{matrix} \right] \left[ \begin{matrix} dy^1 \\ ... \\ dy^n \end{matrix} \right] $

$\frac{\partial}{\partial y^v} = \frac{\partial x^u}{\partial y^v} \frac{\partial}{\partial x^u}$ becomes
$ \left[ \begin{matrix} \frac{\partial}{\partial y^1} & ... & \frac{\partial}{\partial y^n} \end{matrix} \right] = \left[ \begin{matrix} \frac{\partial}{\partial x^1} & ... & \frac{\partial}{\partial x^n} \end{matrix} \right] \left[ \begin{matrix} \frac{\partial x^1}{\partial y^1} & ... & \frac{\partial x^1}{\partial y^n} \\ \vdots & \ddots & \vdots \\ \frac{\partial x^n}{\partial y^1} & ... & \frac{\partial x^n}{\partial y^n} \end{matrix} \right] $

$dy^v = \frac{\partial y^v}{\partial x^u} dx^u$ becomes
$ \left[ \begin{matrix} dy^1 \\ ... \\ dy^n \end{matrix} \right] = \left[ \begin{matrix} \frac{\partial y^1}{\partial x^1} & ... & \frac{\partial y^1}{\partial x^n} \\ \vdots & \ddots & \vdots \\ \frac{\partial y^n}{\partial x^1} & ... & \frac{\partial y^n}{\partial x^n} \end{matrix} \right] \left[ \begin{matrix} dx^1 \\ ... \\ dx^n \end{matrix} \right] $

$\frac{\partial}{\partial x^u} = \frac{\partial y^v}{\partial x^u} \frac{\partial}{\partial y^v}$ becomes
$ \left[ \begin{matrix} \frac{\partial}{\partial x^1} & ... & \frac{\partial}{\partial x^n} \end{matrix} \right] = \left[ \begin{matrix} \frac{\partial}{\partial y^1} & ... & \frac{\partial}{\partial y^n} \end{matrix} \right] \left[ \begin{matrix} \frac{\partial y^1}{\partial x^1} & ... & \frac{\partial y^1}{\partial x^n} \\ \vdots & \ddots & \vdots \\ \frac{\partial y^n}{\partial x^1} & ... & \frac{\partial y^n}{\partial x^n} \end{matrix} \right] $

For our spherical coordinates example:

$dx^u = \frac{\partial x^u}{\partial y^v} dy^v$ becomes
$ \left[ \begin{matrix} dr \\ d\theta \\ d\phi \end{matrix} \right] = \left[ \begin{matrix} \sin \theta \cos \phi & \sin \theta \sin \phi & \cos \theta \\ \frac{1}{r} \cos \theta \cos \phi & \frac{1}{r} \sin \phi \cos \theta & -\frac{1}{r} \sin \theta \\ -\frac{\sin \phi}{r \sin \theta} & \frac{\cos \phi}{r \sin \theta} & 0 \end{matrix} \right] \left[ \begin{matrix} dx \\ dy \\ dz \end{matrix} \right] $

$\frac{\partial}{\partial y^v} = \frac{\partial x^u}{\partial y^v} \frac{\partial}{\partial x^u}$ becomes
$ \left[ \begin{matrix} \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \end{matrix} \right] = \left[ \begin{matrix} \frac{\partial}{\partial r} & \frac{\partial}{\partial \theta} & \frac{\partial}{\partial \phi} \end{matrix} \right] \left[ \begin{matrix} \sin \theta \cos \phi & \sin \theta \sin \phi & \cos \theta \\ \frac{1}{r} \cos \theta \cos \phi & \frac{1}{r} \sin \phi \cos \theta & -\frac{1}{r} \sin \theta \\ -\frac{\sin \phi}{r \sin \theta} & \frac{\cos \phi}{r \sin \theta} & 0 \end{matrix} \right] $

$dy^v = \frac{\partial y^v}{\partial x^u} dx^u$ becomes
$ \left[ \begin{matrix} dx \\ dy \\ dz \end{matrix} \right] = \left[ \begin{matrix} sin \theta cos \phi & r cos \theta cos \phi & -r sin \theta sin \phi \\ sin \theta sin \phi & r cos \theta sin \phi & r sin \theta cos \phi \\ cos \theta & -r sin \theta & 0 \end{matrix} \right] \left[ \begin{matrix} dr \\ d\theta \\ d\phi \end{matrix} \right] $

$\frac{\partial}{\partial x^u} = \frac{\partial y^v}{\partial x^u} \frac{\partial}{\partial y^v}$ becomes
$ \left[ \begin{matrix} \frac{\partial}{\partial r} & \frac{\partial}{\partial \theta} & \frac{\partial}{\partial \phi} \end{matrix} \right] = \left[ \begin{matrix} \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \end{matrix} \right] \left[ \begin{matrix} sin \theta cos \phi & r cos \theta cos \phi & -r sin \theta sin \phi \\ sin \theta sin \phi & r cos \theta sin \phi & r sin \theta cos \phi \\ cos \theta & -r sin \theta & 0 \end{matrix} \right] $

Then of course instead of using row-vectors, we can use column-vectors and transposed-jacobians:

$\frac{\partial}{\partial y^v} = \frac{\partial x^u}{\partial y^v} \frac{\partial}{\partial x^u}$ becomes
$ \left[ \begin{matrix} \frac{\partial}{\partial x} \\ \frac{\partial}{\partial y} \\ \frac{\partial}{\partial z} \end{matrix} \right] = \left[ \begin{matrix} \sin \theta \cos \phi & \frac{1}{r} \cos \theta \cos \phi & -\frac{\sin \phi}{r \sin \theta} \\ \sin \theta \sin \phi & \frac{1}{r} \sin \phi \cos \theta & \frac{\cos \phi}{r \sin \theta} \\ \cos \theta & -\frac{1}{r} \sin \theta & 0 \end{matrix} \right] \left[ \begin{matrix} \frac{\partial}{\partial r} \\ \frac{\partial}{\partial \theta} \\ \frac{\partial}{\partial \phi} \end{matrix} \right] $

$\frac{\partial}{\partial x^u} = \frac{\partial y^v}{\partial x^u} \frac{\partial}{\partial y^v}$ becomes
$ \left[ \begin{matrix} \frac{\partial}{\partial r} \\ \frac{\partial}{\partial \theta} \\ \frac{\partial}{\partial \phi} \end{matrix} \right] = \left[ \begin{matrix} sin \theta cos \phi & sin \theta sin \phi & cos \theta \\ r cos \theta cos \phi & r cos \theta sin \phi & -r sin \theta \\ -r sin \theta sin \phi & r sin \theta cos \phi & 0 \end{matrix} \right] \left[ \begin{matrix} \frac{\partial}{\partial x} \\ \frac{\partial}{\partial y} \\ \frac{\partial}{\partial z} \end{matrix} \right] $

So in terms of unifying partial-vs-total-derivatives and tensor change-of-basis, of all of these my vote is for $dx^u = \frac{\partial x^u}{\partial y^v} dy^v$ looks the most natural.

A distant second is $\frac{\partial}{\partial x^u} = \frac{\partial y^v}{\partial x^u} \frac{\partial}{\partial y^v}$ but this uses row vectors and left-multiply rather than row-vectors and right-multiply.

The others seem to be byproducts of linear systems.

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