0

Let $M$ be a differentiable manifold and $f\colon M\to\mathbf{R}$ a differentiable function. If $\mathrm{d}f=0$, what can we say about $f$? Is it constant?

Motivation:

In Thermodynamics, we consider some "state space" $M$ and "state variables" $\newcommand{\dif}{\mathrm{d}}$ \begin{align} X_1,\ldots,X_n,\xi_1,\ldots,\xi_n\colon M&\to\mathbf{R} \end{align} such that the function $(X_1,\ldots,X_n)\colon M\to\mathbf{R}^n$ is a chart (in particular, injective).

The energy $E\colon M\to\mathbf{R}$ is then said to satisfy $$\dif E=\sum_{i=1}^n\xi_i\cdot\dif X_i$$ and I was wondering if $E$ is determined up to some constant by this.

Filippo
  • 3,536

2 Answers2

1

I expect it's piecewise constant. If the manifold is a pair of spheres, $f$ might take one value on one sphere and a different value on the other.

Empy2
  • 50,853
  • I understand. Do you think that $f$ can be shown to be constant under some additional assumptions regarding the topology, e.g. connectedness? – Filippo Sep 12 '21 at 07:35
  • Well, I guess that's exactly what your are implying with your answer :) – Filippo Sep 12 '21 at 08:10
  • To be honest, I don't have a grasp on the math details. If you pull back to $\mathbb R^n$ then df is still 0 so f is constant on each patch. I guess connectedness makes f constant – Empy2 Sep 12 '21 at 08:53
  • Allright, thanks for the reply. Btw, I added some motivation to my question. – Filippo Sep 12 '21 at 09:30
0

Firstly, let's consider a trivial manifold, i.e. a manifold covered by one single chart $X\colon M\to D\subset \mathbf{R}^n$. Then $f$ is obviously constant if and only if $$F:=f\circ X^{-1}\colon D\to\mathbf{R}$$ is constant. And if $D$ is connected, then the total derivative of $F$ is zero if and only if $F$ is constant.

The more general case can be solved by using the fact that each point has a neighbourhood which is a trivial manifold.

Filippo
  • 3,536