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In the text, it says:

Consider smooth function on a manifold $f: X \to \mathbb{R}$. If $f(x)$ is an extreme value, then $f$ cannot be a coordinate function near $x$, so $df_x$ must be zero.

I don't understand here - if $f(x)$ is an extreme value, why $f$ cannot be a coordinate function?

1LiterTears
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    Because a coordinate chart $\varphi$ maps a neighbourhood of $x$ diffeomorphically onto a neighbourhood of $\varphi(x)$. That means in each neighbourhood of $x$, each of the coordinates assumes larger and smaller values than $\varphi_k(x)$. – Daniel Fischer Jul 11 '13 at 22:44
  • So we assumed $X$ cannot be flat? But I thought a manifold can't be flat globally, but could be flat locally, like the bump function. – 1LiterTears Jul 11 '13 at 22:50
  • That has nothing to do with flatness (and of course a manifold can be globally flat, $\mathbb{R}^n$ is a manifold too). A coordinate chart for a $d$-dimensional manifold $X$ is by definition a homeomorphism of an open subset $U$ of $X$ with an open subset $V$ of $\mathbb{R}^d$. That is what is meant, not that $f$ cannot be a coordinate function of $\mathbb{R}^n$ when $X$ is a $d$-dimensional submanifold of $\mathbb{R}^n$ ($d < n$). – Daniel Fischer Jul 11 '13 at 22:55
  • Thanks @DanielFischer - Do you mean diffeomorphism calls for injectivity, which can not be achieved by a function with global extrema? However, if it is in this case, what about $x$ is just a point of reflection? Thanks! ;) – 1LiterTears Jul 12 '13 at 00:22
  • A diffeomorphism is, by definition, in particular bijective. I don't understand what "what about $x$ is just a point of reflection" should mean. Locally, a manifold looks like an open ball in $\mathbb{R}^d$, and a coordinate chart is such a local view. Hence a function that has a local extremum in $x$ cannot be a component of a coordinate chart around $x$. – Daniel Fischer Jul 12 '13 at 00:37
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    Just to make what I said clear - if $f(x)$ is an extrema, it implies that $f(x_1) = f(x_2)$ in a open ball with arbitrary small radius centered at $x$. Hence $f$ is not injective so can not be a coordinate function. And please ignore what I said related to reflection point, I made a mistake, since reflection point is not extrema is not an extrema. – 1LiterTears Jul 12 '13 at 01:19
  • Hope I got it right above - thanks @DanielFischer =) – 1LiterTears Jul 12 '13 at 04:50

1 Answers1

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Writing $f$ in terms of some coordinate system in a neighborhood of $f$, we can reformulate the question as: if $F=(f_1,\dots,f_n)$ is a diffeomorphism between open subsets of $\mathbb R^n$, can $f_1$ have a point of local extremum at $x$? The answer is no, for otherwise $\nabla f_1(x)$ would vanish, making $\det DF(x)=0$.

40 votes
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  • This really, really greatly clears the cloud in my mind. Thanks so much 40 votes! – 1LiterTears Jul 13 '13 at 01:15
  • Also, I couldn't really follow the logic there. I realize that $df_x$ is zero as a consequence of $f(x)$ being an extreme value. However, in the text, $df_x$ is zero seems to be a consequence of $f$ cannot be a coordinate function. Why? And how could this be true? Thank you~ =) – 1LiterTears Jul 13 '13 at 03:59
  • @Jellyfish I find the sentence you quoted badly written (unless it's misquoted). Can be deciphered as: "If $f(x)$ is an extreme value, then $f$ cannot be a coordinate function near $x$, for $df_x$ must be zero at an extremum". – 40 votes Jul 13 '13 at 04:18
  • Thank you very much for clearing that out for me, 40 votes. Yes I understand you very well. Thanks again. – 1LiterTears Jul 13 '13 at 04:38