Orientation of frames generated by consistent charts in a common point

Question

I have a smooth $k$-surface $S\subset \mathbb{R}^n$ and two charts $\varphi_1:I_t^n\to U_1\subset S$, $\varphi_2:I_\tau^n\to U_2\subset S$ with $U_1\cap U_2\neq \emptyset$ ($I^n$ is the unit open cube in $\mathbb{R}^n$). Picked a point $\underbrace{x_0}_{\in U_1\cap U_2}=\varphi_1(\underbrace{t_0}_{\in I_t^n})=\varphi_2(\underbrace{\tau_0}_{\in I_\tau^n})$ in such intersection, we know that if the two charts have positive transitions at $x_0$, then the induced two frames in $x_0$ by the two charts have the same orientation (and viceversa). Matematically this statement is written as:

$\text{sign}\det \left([\varphi_2^{-1}\circ \varphi_1]'(t_0)\right)=\text{sign}\det \left([\varphi_1^{-1}\circ \varphi_2]'(\tau_0)\right)>0\iff$ the two frames $F_{\varphi_1}=\{\mathbf{e}_1,...,\mathbf{e}_n\}$ and $F_{\varphi_2}=\{\mathbf{b}_1,...,\mathbf{b}_n\}$ induced by the two charts in $x_0$ are such that $\text{sign}\det \left(M_{F_{\varphi_1}\to F_{\varphi_2}}\right)=\text{sign}\det \left(M_{F_{\varphi_2}\to F_{\varphi_1}}\right)>0$.

I remind that:

1. $\left\{\begin{matrix} \mathbf{b}_1=a_{11}\mathbf{e}_1+...+a_{1n}\mathbf{e}_n \\ ...\\ \mathbf{b}_n=a_{n1}\mathbf{e}_1+...+a_{nn}\mathbf{e}_n \end{matrix}\right.\iff M_{F_{\varphi_1}\to F_{\varphi_2}}=\left(\begin{matrix}a_{11} & ...&a_{1n}\\ ...&...&...\\ a_{n1} & ...&a_{nn}\end{matrix}\right)$;
2. $M_{F_{\varphi_2}\to F_{\varphi_1}}=M_{F_{\varphi_1}\to F_{\varphi_2}}^{-1}$;
3. $\mathbf{e}_i=\varphi_1'(t_0)\cdot \left(\begin{matrix}0_1\\...\\1_i\\...\\0_n\end{matrix}\right)$, $i=1,...,n$;
4. $\mathbf{b}_i=\varphi_2'(\tau_0)\cdot \left(\begin{matrix}0_1\\...\\1_i\\...\\0_n\end{matrix}\right)$, $i=1,...,n$.

Now, my question...

I would prove also the following proposition for the tangent space at $S$ in $x_0$ (namely $TS_{x_0}$):

The two charts have positive transitions at $x_0\iff$ the induced two frames for $TS_{x_0}$ by the two charts have the same orientation.

My book (Zorich, Mathematical Analysis II, 1st ed., Page 173) says that it's true but I can't find a plausible reason to prove it.

Answer 1

(I believe you want $I^k$, not $I^n$ and all the indexes ranging from $1,\ldots, k$.)

Claim: The matrix $M_{F_{\varphi_1}\to F_{\varphi_2}}$ and the matrix $[\varphi_1^{-1}\circ \varphi_2]'(t_0)$ are transposes of each other.

Proof: Consider the image of the vector $\left(\begin{matrix}0_1\\...\\1_i\\...\\0_k\end{matrix}\right)$ under $[\varphi_1^{-1}\circ \varphi_2]'$. The idea of the proof is that by chain rule, we can first map it by $\varphi_2'(\tau_0)$ (and it goes to $\mathbf{b}_i$), and then take the resulting vector

$\mathbf{b}_i=a_{i1}\mathbf{e}_1+...+a_{ik}\mathbf{e}_k$

and map it by $(\varphi_1^{-1})'(x_0)$ considered as a map from $TS_{x_0}$ to $TI^k_{\tau_0}$, which is the inverse of $(\varphi_1)'(t_0)$ (considered as map to $TS_{x_0}$), and so sends it to $\left(\begin{matrix}a_{i1}\\...\\a_{ik}\end{matrix}\right)$ (since it's linear and sends each $\mathbf{e}_j$ to $\left(\begin{matrix}0_1\\...\\1_j\\...\\0_k\end{matrix}\right)$).

Thus the matrix of $[\varphi_1^{-1}\circ \varphi_2]'$ has $i$th column equal to $\left(\begin{matrix}a_{i1}\\...\\a_{ik}\end{matrix}\right)$, proving the claim.

This is a rigorous argument once one knows the appropriate chain rule. To avoid this more advanced version of chain rule, one can argue as follows:

Extend each $\phi_1$ and $\phi_2$ to maps $\Phi_1$ and $\Phi_2$ from $I^n \to \mathbb{R}^n$ (using Proposition on page 162). Moreover, extend $(\mathbf{e_1},\ldots, \mathbf{e_k})$ to a basis or $\mathbb{R}^n$. Then we write $\Phi_1'(t_0)$ using the standard basis on $TI^k_{t_0}$ and this newly constructed basis on $\mathbb{R}^n$. We have that $\Phi_1'(t_0)$ is block upper-triangular, with a $k$ by $k$ upper left block being identity, corresponding to the fact that

$$[\Phi_1'(t_0)](\left(\begin{matrix}0_1\\...\\1_j\\...\\0_n\end{matrix}\right))=[\phi_1'(t_0)](\left(\begin{matrix}0_1\\...\\1_j\\...\\0_k\end{matrix}\right))=\mathbf{e}_j$$ for $j=1, \ldots, k$.

Similarly, $\Phi_2'(\tau_0)$ is block upper-triangular, with a $k$ by $k$ upper left block being $\left(M_{F_{\varphi_1}\to F_{\varphi_2}}\right)^T$, corresponding to

$$[\Phi_2'(\tau_0)](\left(\begin{matrix}0_1\\...\\1_j\\...\\0_n\end{matrix}\right))=[\phi_2'(\tau_0)](\left(\begin{matrix}0_1\\...\\1_j\\...\\0_k\end{matrix}\right))=\mathbf{b}_j=\sum a_{ji}\mathbf{e}_i$$ for $j=1, \ldots, k$.

Then, by chain rule applied to $[\Phi_1^{-1} \cdot \Phi_2]$ we have that the $n$ by $n$ matrix $[\Phi_1^{-1} \cdot \Phi_2 (\tau)]'$ is composition of two $n$ by $n$ matrices $ [\Phi_2'] (\tau_0)$ and $[\Phi_1^{-1}]'(x_0)$. Both of these are block-upper-triangular, with the $k$ by $k$ left upper block of $ [\Phi_2'] (\tau_0)$ equal to $\left(M_{F_{\varphi_1}\to F_{\varphi_2}}\right)^T$ and the $k$ by $k$ left upper block of $[\Phi_1^{-1}]'(x_0)$ equal to inverse of identity, i.e. identity. Then we conclude that the $k$ by $k$ left upper block of $[\Phi_1^{-1} \cdot \Phi_2]$, being product of these two $k$ by $k$ blocks, is also just $\left(M_{F_{\varphi_1}\to F_{\varphi_2}}\right)^T$.

But this block is of course just $[\varphi_1^{-1}\circ \varphi_2]'(\tau_0)$, since the map $[\Phi_1^{-1} \cdot \Phi_2]$ restricted to $I^k$ is $[\varphi_1^{-1}\circ \varphi_2]$.

This reproves the claim (using onluy chain rule for maps between open subsets of $\mathbb{R}^n$).

Now from the claim the result follows, since transpose matrices have the same determinant.