I don't know very much about differential geometry, so this may be trivial but I will try to make this question as clear as I can.

Suppose I have a $d$ dimensional smooth manifold $\mathcal{M}^d$, with a smooth maximal atlas $\mathcal{A}$. In the neighborhood of a point $m$ I have a chart $(U,\phi) \in \mathcal{A}$ that gives some local coordinates for my manifold, say $\phi(m) = (x^1(m), ..., x^d(m))$ and $\phi : U \subseteq \mathcal{M}^d \rightarrow \mathbb{R}^d$ is a homeomorphism. Suppose also that I have a smooth function defined on my manifold, $f: \mathcal{M}\rightarrow \mathbb{R}$.

My question is the following: Are there any conditions that guarantee that the chart $(V, \psi)$, with $m \in U \cap V$, $\psi = (x^1, ..., f)$ is also in $\mathcal{A}$?

In other words, how can I be sure that I can replace one of my coordinates with the smooth function $f$?


Take a look at Tu's Introduction to manifolds, Section 6.3 on the Inverse Function Theorem.

The way the IFT is written there is that some local functions $(f_1,\dots,f_n):U\to\mathbb R^n$ define a coordinate chart around $p\in U$ iff there's a chart $(x_1,\dots,x_n)$ around $p$ such that $$ \det\begin{bmatrix}\dfrac{\partial f_i}{\partial x^j}\end{bmatrix}\neq 0 $$

So we get a sufficient a condition: $(x_1,\dots,x_{n-1},f)$ defines a chart if $$ \det \begin{bmatrix} \dfrac{\partial x_1}{\partial x^1} &\cdots &\dfrac{\partial x_1}{\partial x^{n-1}}&\dfrac{\partial x_1}{\partial x^n}\\ \vdots & \ddots &\vdots & \vdots \\ \dfrac{\partial x_{n-1}}{\partial x^1} &\cdots &\dfrac{\partial x_{n-1}}{\partial x^{n-1}}&\dfrac{\partial x_{n-1}}{\partial x^n}\\ \dfrac{\partial f}{\partial x^1}&\cdots&\dfrac{\partial f}{\partial x^{n-1}}&\dfrac{\partial f}{\partial x^n} \end{bmatrix} = \det \begin{bmatrix} 1 & \cdots &0&0\\ \vdots & \ddots &\vdots & \vdots \\ 0 &\cdots &1&0\\ \dfrac{\partial f}{\partial x^1}&\cdots&\cdots&\dfrac{\partial f}{\partial x^n} \end{bmatrix} = \dfrac{\partial f}{\partial x^n} \neq 0. $$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.