# Orientation of frames generated by consistent charts in a common point

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.

(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.

• No, the matrix $[\varphi_1^{-1}\circ\varphi_2]'(t_0)$ has dimension $n\times n$ and the matrix $M_{F_{\varphi_1}\to F_{\varphi_2}}$ has dimension $k\times k$. In fact, the frame generated by a chart at a point is composed by $n$ linear independent vectors, but only $k$ of these $n$ are in (generates) $TS_{x_0}$. Sep 2 '20 at 8:41
• ...at least, this is what I have understood from what I have studied, of course I can be wrong... Sep 2 '20 at 8:49
• Usual definition of a chart says it is a homeomorphism to (or, sometimes, from) a piece of a surface/manifold. As a homeomorphism, it better map between things that have the same dimension, and you are working with a k-surface. I believe Zorich is using the same definition.
– Max
Sep 2 '20 at 12:41
• There could be differences in definitions, but any reasonable definition would hard to square away with what you have written in the question (i.e. $\varphi_1:I_t^n\to U_1\subset S$ etc.) without modifying something.
– Max
Sep 2 '20 at 12:43
• Zorich uses the following def. for smooth k-surfaces: A set $S\subset\mathbb{R}^n$ is a smooth k-surface if for each point $x\in S$ there exists a diffeomorphism $\varphi:I^n\to U(x)$ such that every point of $U(x)$ is described by $(t_1,...,t_k,0,...,0)$ under $\varphi$. In other words, the diffeomorphism is between points of $\mathbb{R}^n$ and the points of the surface have the property that they are the image of points in $I^n$ with the last n-k coordinates equals to 0. Sep 2 '20 at 13:33