Chart Transformation in $T^*_x M$

Question

I am confused about a statement in my differential geometry script. It states: $(U,\varphi = (x_1,...,x_n))$ and $(U,\psi = (y_1,...,y_n))$ are charts of a smooth manifold, then $$(dy_i)_x=\sum^n_{\alpha=1}\frac{\partial(\psi \circ \varphi^{-1})_i}{\partial x_\alpha}(x_1,...,x_n)(dx_\alpha)_x$$ holds for the base in case of a change of charts.

I tried to recreate the formula with our definitions but im confused about the (x_1,...,x_n) in the formula, how do these get there and what does it state? My try was: \begin{align*} d(y_i)_x & =d(x_i\circ \psi^{-1} \circ \psi)_x\\ &= d(x_i\circ \psi^{-1})_{\psi (x)}\circ(\sum^n_{\alpha=1}d(y_\alpha)_x\cdot e_\alpha)\\ & =\sum^n_{\alpha=1} \frac{\partial(x_i\circ \psi^{-1})}{\partial y_\alpha} (\psi(x))dy_\alpha\\ & =\sum^n_{\alpha=1} \frac{\partial(\varphi\circ \psi^{-1})_i}{\partial y_\alpha} (\psi(x))dy_\alpha &(?)\\ \end{align*} But there is no $(x_1,...,x_n)$ and i have no idea how i could get it.

Answer 1

It is just the chain rule! Call $f = \psi \circ \varphi^{-1}$. If $\varphi: U \to \tilde{U}$ and $\psi : U \to \tilde{V}$ then $f : \tilde{U} \to \tilde{V}$ and

$$ {\rm d}y_i = \sum_j \frac{\partial f(x_1,\cdots, x_n)}{\partial x_j}{\rm d}x_j $$

where $x \in \tilde{U}$, $y \in \tilde{V}$ and $y = f(x)$

You see from this diagram that $f = \psi \circ \varphi^{-1}$ connects directly $\tilde{U}$ and $\tilde{V}$ without going through the manifold

Answer 2

The book is saying that $$(dy_i)_p=\sum^n_{\alpha=1}\frac{\partial(\psi \circ \varphi^{-1})_i}{\partial x_\alpha}(\varphi(p))(dx_\alpha)_p.$$

[To avoid confusion, I'm using the letter $p$ to denote the point on the manifold $M$. So $\varphi(p) \in \mathbb R^n$ is the representation of $p$ in $\{ x_\alpha\}$ coordinates, while $\psi(p) \in \mathbb R^n$ is the representation of $p$ in $\{ y_i \}$ coordinates, and $\psi \circ \varphi^{-1} : \mathbb R^n \to \mathbb R^n$ is the transition map from $\{ x_\alpha \}$ coordinates to $\{ y_i \}$ coordinates.]

Perhaps the best way to prove this equation is to let $(dy_i)_p$ act on the basis of a tangent vectors $\{ \frac{\partial}{\partial x_\alpha}\vert_p\} $ at $p$. Specifically, we want to show that

$$ (dy_i)_p \left( \frac{\partial }{\partial x_\alpha}\vert_p\right) = \frac{\partial(\psi \circ \varphi^{-1})_i}{\partial x_\alpha}(\varphi(p)) \ \ \ (\star)$$ This follows immediately from the definitions: for any function $f : M \to \mathbb R$, the definition of $df$ says \begin{align} (df)_p \left( \frac{\partial }{\partial x_\alpha}\vert_p\right) &= \frac{\partial \left( f \circ \varphi^{-1}\right) }{\partial x_\alpha} (\varphi(p)).\end{align}

Applying this with $y_i$, which is the $i$th component of $\psi$, we get $(\star)$, as required.