Change of
differential forms,
differential operators,
and the chain rule

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Chain rule

There's only one real rule for how change of coordinates affects
differential forms and differential operators, and that's the chain
rule.

If $x \mapsto (y_1,\dots,y_n) \mapsto z$,
then
$$\frac{\partial z}{\partial x} = \sum \frac{\partial z}{\partial y_j}
 \frac{\partial y_j}{\partial x}$$
(more generally, replace $z$ by $z_k$ and $x$ by $x_i$)
If you multiply through by $\partial x$ and write
$\dy_j$ for $\partial y_j$ and $\dz$ for $\partial z$, you get

$$\dz = \sum \frac{\partial z}{\partial y_j} \dy_j$$
...which is the change of variables rule for differential forms.

If instead you think of $z$ as a variable function, we get

\begin{align*}
\frac{\partial}{\partial x} &= \sum \frac{\partial}{\partial y_j}
 \frac{\partial y_j}{\partial x}
\partial_x = \sum \frac{\partial y_j}{\partial x} \partial_{y_j}
\end{align*}
...which is the change of variables rule for differential operators.

It's easy to keep these straight by thinking of $\partial_x$
as ``an $x$ in the denominator'',
and then summing so as to cancel.
Symbolically,
$\partial_x = \frac{\partial}{\partial x}$.

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Abstractly

What's really happening is:
Given $f\from M \to N$ and $g\from N \to P$
(in coordinates, $\bR^m \to \bR^n \to \bR^p$)
we can -pull back- cotangent vectors from $P$ to $N$:
$g^*\from T^*P \to T^*N$
and we can -push forward- tangent vectors from $M$ to $N$:
$f_*\from T_*M \to T_*N$

Push-forward:
Thus we should properly write
\begin{align*}
f_* \partial_{x_i} &= \sum \frac{\partial y_j}{\partial x} \partial_{y_j}
    &= \sum \partial_{x_i} f_j \partial_{y_j}
\end{align*}

For push-forward, it's exactly the Jacobian:
$\partial{y_j}/\partial{x_i}$

Pull-back:
Thus we should properly write
\begin{align*}
g^* \dz_k &= \sum \frac{\partial z}{\partial y_j} \dy_j\\
          &= \sum \partial_{y_j} g_k \dy_j
\end{align*}
In other words,
pull-back is written in coordinates as
$$\partial z_1/\partial y_1 ... \partial z_p/\partial y_1
  ...                             ...
  \partial z_1/\partial y_n ... \partial z_p/\partial y_n$$

It's the -transpose- of the Jacobian:
$\partial{z_k}/\partial{y_j}^T$

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Recall
A key swerve to note is:
differential operators are -tangent vectors- (fields),
because given a tangent vector (field), you can take the -directional
derivative- in that direction.

Thus the coordinates on the tangent space are written
$\partial x_i$, while the coordinates on the cotangent space are
$\dx_i$ (this is a form that -detects- a change in $x_i$;
it is not the change itself)

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Conjugation
I don't really want to write this down;
you just have an inverse on one side.
[I'll write it down if it becomes useful,
 or when I write these up.]

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Conceptually

Conceptually, the picture is exactly the description of the chain
rule:
Given a function $x \mapsto (y_1,\dots,y_n)$,
the unit derivative in the $x$ direction (think of as unit speed)
maps to its velocity in all of the directions.
Writing this in coordinates,
\begin{align*}
\partial_x &\mapsto \left(\frac{\partial y_1}{\partial x}, \dots,
              \frac{\partial y_n}{\partial x}\right)\\
           &= \sum \frac{\partial y_j}{\partial x} \partial_{y_j}\\
\end{align*}
Like, duh.

Dually,
Given a function $(y_1,\dots,y_n) \mapsto z$,
$\dz$ measures ``the change in output''.
This is given by
$$\dz = \sum \frac{\partial z}{\partial y_j} \dy_j$$
For a given vector, the change in $z$ in that direction is the sum of
the contributions from the change in each of the $y_j$ directions.
Again, like duh.

Ah, but now look!
For $\dz$, the $\partial z$ is in the -numerator-,
while for $\partial_x$ the $\partial x$ is in the -denominator-!

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Connection to integration

A volume form is, in particular, a form,
and the induced action on exterior powers is just the determinant,
so the induced action on forms is the determinant of the appropriate
Jacobian.
(Note that transpose doesn't change determinant; I don't know if this
 matters here.)

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Which of these is -easier-?

Ah -- depends on perspective.
For differential -forms- it's nice in that an -equation- for
$z$ in terms of $y_j$ yields the equation for the form.
[This is because maps $. \to \bR^p$ are easy.]

For differential -operators-, you need to know equations
for -all- the $y_j$ in terms of -all- the $x_i$ just to compute where
$\partial_x$ goes.
[This is because maps $\bR^m \to .$ are messier.]
OTOH, this is just the Jacobian.

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Inverting a partial derivative

Given a diffeomorphism
$(x_1,\dots,x_n) \to (y_1,\dots,y_n)$,
the derivative one way is
$\partial y_j/\partial x_i$
while the other way it's
$\partial x_i/\partial y_j$.

These are necessarily inverse matrices.
Using the formula for the inverse matrix,
in terms of the classical adjoint,
we get
$\partial x_i/\partial y_j = (-1)^{i+j} \det J_{ji}/\det J$,
where $J$ is the Jacobian $\partial y_j/\partial x_i$
and $\det J_{ji}$ is the $(j,i)$-minor.

Now, if it's a diagonal matrix (the map is just dilation in each
axis), this reduces to
$\partial x_i/\partial y_j = (\partial y_j/\partial x_i)^{-1}$
but in general you have the whole mess.

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Concretely

- keep your tops and bottoms straight
- be careful of how a change of coordinates is given:
  + does it tell you how to
    -express the existing coordinates-
    in terms of new ones,
  + or does it tell you the -new ones-
    -in terms of the old ones-?
- in what terms is the Jacobian (and hence all factors) given?
  (hint: it's in terms of the independent variables,
   the bottom of the partials)
- natural is to ``substitute in''
- unnatural (shrike maps) is to ``recognize and eliminate''
  ...you're familiar with these from $u$-substitutions in calculus.

For instance, if $u=f(x)$, then $\du = f'(x) \dx$,
but $\partial_u = 1/f'(x) \partial_x$
...and $\partial_x = f'(x) \partial_u$,
and $\dx = 1/f'(x) \du$.

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Some examples

If
$u=x+y,
v=x-y$
then:
$x=(u+v)/2,
 y=(u-v)/2$
$\du = \dx + \dy$
$\dv = \dx - \dy$
$\dx = (\du + \dv)/2$
$\dy = (\du - \dv)/2$
$\partial_x = \partial_u + \partial_v$
$\partial_y = \partial_u - \partial_v$
$\partial_u = (\partial_u + \partial_v)/2$
$\partial_v = (\partial_u - \partial_v)/2$
...see how it acts oppositely on forms and operators?

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The algebra of forms and operators

We turn out to be interested in -exterior powers- of forms,
(which means they die after dimension $n$, and are pretty small)
and -symmetric powers- of differential operators
(which means they just keep growing, and act like polynomials)
(Clairut's theorem: $\partial_x \partial_y = \partial_y \partial_x$;
 note that this shows that the bracket is a (indeed the only,
 in dim 2 at least) obstruction to a set of vector fields being
 integrable, i.e., the vector fields in some coordinate system.).

For differential forms, this is an -algebra- over functions,
so $\dx f \dy = f \dx \dy$.
For differential -operators-, there's commutation relations!
