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I'm trying to understand the statement of this problem:

Let $x_1, \dots, x_N$ be the standard coordinate functions on $\mathbb{R}^N$, and let $X$ be a $k$-dimensional submanifold of $\mathbb{R}^N$. Prove that every point $x \in X$ has a neighborhood on which the restrictions of some $k$-coordinate functions $x_{i_1}, \dots, x_{i_k}$ form a local coordinate system.

I've been always confused by the terms "local coordinate system", "local coordinates", etc.

Guillemin and Pollack point out the following. If $\phi: U\rightarrow V$ is a parametrization around $x$ in a manifold $X$ (i.e., a diffeomorphism), then writing $\phi^{-1}$ in coordinates as $\phi^{-1}=(x_1,\dots, x_k)$ gives local coordinates $x_1,\dots, x_k$. So for example if $X=\mathbb R^2_{x > 0}$, consider this parametrization around $(2,2)$: $\phi: \mathbb R^2\rightarrow \mathbb R^2_{x > 0},\ (x,y)\mapsto (e^x,y)$. Its inverse is $\phi^{-1}: \mathbb R^2_{x > 0}\rightarrow \mathbb R^2,\ (r,s)\mapsto (\ln r, s)$, so $\phi^{-1}=(\ln r, s)$. Is it correct that in this case $\ln r, s$ is a local coordinate system around $(2,2)$ (and any other point) in $X=\mathbb R^2_{x > 0}$?

And so the statement of the problem is that for example in the case $k=2$, $\phi$ can be chosen in such a way that either $\phi^{-1}=(r,s)$ or $\phi^{-1}=(s,r)$ ?

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  • $\begingroup$ The example does not hold water, $(0,0)$ is not a point in $\mathbb{R}^2_{x > 0}$. $\endgroup$ – Lee Mosher Feb 1 '18 at 18:39
  • $\begingroup$ Initially I took $\mathbb R^2$ as $X$ but then switched to $\mathbb R^2_{x > 0}$, hence the mistake. I've replaced $(0,0)$ with $(2,2)$. $\endgroup$ – user500094 Feb 1 '18 at 19:16
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Whenever you read the words "local coordinate system" just think of a local diffeomorphism $\psi:U\to \mathbb{R}^k$ where $U$ is an open subset of your manifold $X$.

To answer your question, prove the following (as the ook suggests):

For any vector space $V\subset\mathbb{R}^n$, if $B=\lbrace e_1,\dots, e_n\rbrace$ is the standard basis, then there is a selection $\lbrace i_1,\dots,i_k\rbrace\subset \lbrace 1,\dots,n\rbrace$ so that the map $\pi:\mathbb{R}^n\to\mathbb{R}^k$ given by $$(x_1,\dots,x_k)\mapsto (x_{i_1},\dots,x_{i_k})$$ restricts to an isomorphism on $V$.

Once you've done so, then we can apply this locally at any point $x\in X$ so that $\pi|_{T_x(X)}$ is an isomorphism. Then letting $\psi=\pi|_X$ in some neighborhood $U$ we have that $\psi:U\to \mathbb{R}^k$ is a local diffeomorphism (why?).

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