Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

This seems obvious, but I'm having trouble carrying through the details.

Suppose there is a smooth function $f$ with zero derivative on a manifold $M$ with $n$ connected components. Why is $f$ constant on each connected component?

Detailed answers are very much appreciated. Thanks!

share|cite|improve this question
What exactly do you mean by derivative here? – Mariano Suárez-Alvarez Jan 7 '12 at 5:58
I suggest assuming $n=1$. Do you know how to prove it when $M=\mathbb R^k$? – Jonas Meyer Jan 7 '12 at 6:00
@MarianoSuárez-Alvarez I guess that's part of my question. I only know how to prove this when $M=\mathbb{R}$, and I don't know enough differential geometry to define derivatives on manifolds. But wikipedia claims this is true. – Potato Jan 7 '12 at 6:15
But then you should probably pick a textbook dealing with the subject and learn that first! It is extraordinarily understandable that you be having problems with proving this if you do not know what derivatives are in this context. – Mariano Suárez-Alvarez Jan 7 '12 at 6:17
(The only way to detailedly answer this starting from what a derivative is to the claim you want to prove is to more or less write out an exposition of what a manifold is and what a smooth function on it is: this is not the best way to use this site) – Mariano Suárez-Alvarez Jan 7 '12 at 6:19
up vote 6 down vote accepted

Let $M$ be an $m$-manifold. We'll concentrate on a connected component of $M$, say $U$. Pick $p\in U$, let $V_p$ be a neighborhood of $p$ in $U$ admitting a local Euclidean chart, and let $\phi: D\subset\mathbb{R}^m\rightarrow V_p$ be a coordinate chart. If $f: M\rightarrow \mathbb{R}$ is a differentiable function on $M$, this really means that $f\circ \phi: \mathbb{R}^m\rightarrow \mathbb{R}$ is a differentiable function (in fact, this is the definition of a differentiable function on $M$). Now prove that $f\circ \phi$ is constant using standard calculus. So $f$ is constant on $V_p$. From the fact that $U$ is connected, conclude by standard topological arguments that $f$ is constant on $U$.

I would suggest picking up a book on differential geometry and topology.

share|cite|improve this answer
This is crystal clear. Thank you! – Potato Jan 7 '12 at 7:09
One wonders how you followed this along, being that it does not contain a definition of the derivative you mention in your question (in fact, it does not even mention the hypothesis! :D ) – Mariano Suárez-Alvarez Jan 7 '12 at 9:41
Is there something unclear about the differential of a map from $\mathbb{R}^m$ to $\mathbb{R}^n$? Presumably he does know how to differentiate multivariable maps. If not, then he should pick up a book on elementary calculus, before looking to geometry (my answer does make a reference to calculus, so he knows where to look :-D). – William Jan 7 '12 at 10:02
@MarianoSuárez-Alvarez What it means to be a differentiable function on a manifold is defined in the answer. – Potato Jan 7 '12 at 22:51
@WNY Do not worry. I am familiar with multivariable analysis, just not so much with differential geometry. – Potato Jan 7 '12 at 22:52

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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