I am having trouble with a certain part of Sean Carroll's GR notes at the bottom of page 46 (electronic page 53). I thought I'd post my question here since the question is entirely mathematical.
Let's say $\omega = W_{\mu}dx^{\mu}$ is a one-form on a local coordinate chart $(U, (x^{\mu}))$. At the top of page 47 (electronic page 54), the text claims that the expression $\partial_{\mu}W_{\nu}$ does not transform as a tensor when changing coordinates. However, I am wondering what is wrong with my calculation below. (I apologize, but I changed the notation. The content should still be the same.)
Let's say $(V, (\overline{x}^{\mu}))$ is another local coordinate chart in which $\omega = \overline{W}_{\mu}d\overline{x}^{\mu}$.
Then
\begin{align*}
\frac{\partial \overline{W}_{\nu}}{\partial{\overline{x}^{\mu}}} &= \frac{\partial}{\partial\overline{x}^{\mu}} \Big( \frac{\partial x^{\kappa}}{\partial \overline{x}^{\nu}} W_{\kappa} \Big) \\[1.3ex]
&= \frac{\partial x^{\lambda}}{\partial \overline{x}^{\mu}} \frac{\partial}{\partial x^{\lambda}} \Big( \frac{\partial x^{\kappa}}{\partial \overline{x}^{\nu}} W_{\kappa} \Big) \\[1.3ex]
&= \frac{\partial x^{\lambda}}{\partial \overline{x}^{\mu}} \frac{\partial x^{\kappa}}{\partial \overline{x}^{\nu}} \frac{\partial W_{\kappa}}{\partial x^{\lambda}} + \frac{\partial x^{\lambda}}{\partial\overline{x}^{\mu}} \frac{\partial^{2} x^{\kappa}}{\partial x^{\lambda} \partial \overline{x}^{\nu}} W_{\kappa} .
\end{align*}
So far, I just rederived (2.26) from the text by the usual contravariance transformation rule and the product rule.
Now in the last expression, Carroll argues that the second term is what prevents from the usual tensor transformation rule to hold.
However, it seems like by applying the commutativity of partial derivatives, I can show that the second term is zero:
\begin{align*}
\frac{\partial x^{\lambda}}{\partial\overline{x}^{\mu}} \frac{\partial^{2} x^{\kappa}}{\partial x^{\lambda} \partial \overline{x}^{\nu}} W_{\kappa} = \frac{\partial x^{\lambda}}{\partial\overline{x}^{\mu}} \frac{\partial^{2} x^{\kappa}}{\partial \overline{x}^{\nu} \partial x^{\lambda}} W_{\kappa} = \frac{\partial x^{\lambda}}{\partial\overline{x}^{\mu}} \frac{\partial}{\partial \overline{x}^{\nu}} \left( \frac{\partial x^{\kappa}}{\partial x^{\lambda}} \right) W_{\kappa} = \frac{\partial x^{\lambda}}{\partial\overline{x}^{\mu}} \frac{\partial}{\partial \overline{x}^{\nu}} \left( \delta^{\kappa}_{\lambda} \right) W_{\kappa} = 0
\end{align*}
where in the last equality, we know the expression is zero because we are taking the derivative of the Kronecker delta symbol, which is constant with respect to the coordinates.
This is getting me profoundly confused at the moment. I'm sure that $\partial_{\mu}W_{\nu}$ must not follow the tensor transformation rules because covariant differentiation (which does follow the tensor rules) is explicitly a different kind of derivative than the one used here, but I don't see what is invalid in my calculation.
Edit: It seems as though my calculation can be used to show that any second derivative is zero, so my work is definitely wrong. However, I still don't understand what was the invalid step.
Edit 2: Okay, this is much more interesting than I expected. My understanding is that the mistake comes from a failure to make a distinction between partial derivatives and derivatives along vector fields. In the former, you cannot mix coordinate systems; in the latter you cannot commute operations.
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If you are working with partial derivatives, you cannot mix different coordinate systems, because to take partial derivatives you need to have a well-defined function with a well-defined tuple of arguments.
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If you are working with derivatives along vector fields (or derivations if we are talking about differential geometry), then commutativity of derivatives does not hold! This is not a new fact, because this failure of commutativity is commonly measured by the Lie bracket $[X, Y]$.
It seems that Carroll's equation can be justified by interpreting the derivatives as derivations along the appropriate directions, but in such a case we cannot commute the two derivatives like I did in my post.
Example. I thought an example would be very instructive here. Let $M = \mathbb{R}^{2}$, let $(x, y)$ be the standard coordinates, and let $(r, \theta)$ be the polar coordinates.
Let $f$ be the function given by $f(x, y) = x$.
(Here $f$ is just one of the coordinate functions, but really the reader can use any function he or she desires.)
On one hand, we have
\begin{align*}
\partial_{y}\overline{\partial}_{r}(f) &= \partial_{y}\overline{\partial}_{r}(r\cos\theta) = \partial_{y}(\cos\theta) = \partial_{y}\left( \frac{x}{\sqrt{x^{2}+y^{2}}} \right) = \frac{-xy}{(x^{2}+y^{2})^{3/2}}.
\end{align*}
On the other hand, we have
\begin{align*}
\overline{\partial}_{r}\partial_{y}(f) = \overline{\partial}_{r}(0) = 0.
\end{align*}
From this we see that
$$ [\partial_{y}, \overline{\partial}_{r}]\ne 0. $$
Indeed, I remember a theorem that said that a local frame $(X_{1}, \ldots, X_{n})$ can yield a local coordinate system about any desired point if and only if the vector fields commute with one another (the coordinate system may have to be defined on a strictly smaller set than that of the vector fields however).
Since $\partial_{y}$ and $\overline{\partial}_{r}$ do not commute, $y$ and $r$ as a pair cannot form a coordinate system on any region of the plane! Thus, there is no context in which the derivative wrt $y$ and the derivative wrt $r$ can commute!
Best Answer
There are some identifications/abuses of notation being used here. These have their uses, but this compuation makes a bit more sense when written out more explicitly.
I'll use $x^a$ (with Latin indices) and $y^\alpha$ (with Greek indices) to denote the two sets of coordinate functions, with the shorthands $x:=(x^1,\cdots,x^n)$ and $y:=(y^1,\cdots,y^n)$ for the full sets. Both are simply smooth functions on the manifold, but we may identify them with their representatives in the other chart, giving $x^a(y)$ and $y^\alpha(x)$, which are both fuctions $\mathbb{R}^n\to\mathbb{R}^n$. When one writes expressions like $\frac{\partial y^\alpha}{\partial x^a}$, this is being done implicitly. In this sense, terms like $\frac{\partial^2 x^a}{\partial x^b\partial y^\alpha}$ don't make sense, since we can identify $x^a$ with a function of $x$ or of $y$, but not both.
The local representatives of the $1$-form $\omega$ are functions of their respective coordianates, $W_a(x)$ and $W_\alpha(y)$. The computation can then be carried out with arguments included, starting with the transformation rule for $1$-forms (with $y$ is a variable and $x$ is a function of $y$). $$ W_\alpha(y)=\frac{\partial x^a}{\partial y^\alpha}(y)W_a(x(y)) $$ Differentiating, $$ \frac{\partial W_\alpha}{\partial y^\beta}(y)=\frac{\partial}{\partial y^\beta}\left(\frac{\partial x^a}{\partial y^\alpha}(y)W_a(x(y))\right) $$ applying product rule, $$ =\frac{\partial^2 x^a}{\partial y^\alpha\partial y^\beta}(y)W_a(x(y))+\frac{\partial x^a}{\partial y^\alpha}(y)\frac{\partial}{\partial y^\beta}\left(W_a((x(y))\right) $$ and chain rule, $$ =\frac{\partial^2 x^a}{\partial y^\alpha\partial y^\beta}(y)W_a(x(y))+\frac{\partial x^a}{\partial y^\alpha}(y)\frac{\partial W_a}{\partial x^b}(x(y))\frac{\partial x^b}{\partial y^\beta}(y) $$ we obtain the two terms without issue.