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Exercise and Hint

In the picture (P106 Do Carmo, Riemannian Geometry), why the red line implies $K = const$. I have two questions about this.

  1. I have seen in another answers, which strictly follows the hint in the book. It says $X(K)=0, \forall X \in TpM $ implies $K=const$ in a neighborhood of $p$. Why does $K=const$ in a neighborhood of $p$? I think, from $X(K)=0, \forall X \in TpM $, we only can say that in any coordinate neighborhoods $U$ about $p$, all partial derivatives of $K$ at $p$ (not $U$) are zero. But how could we say that $K=const$ in a neighborhood of $p$?

  2. Starting from the blue arrow, I think, by the arbitrariness of $p$, we have for any $p\in M$, $X_p(K)=0, \forall X_p \in T_pM$. So the partial derivatives of $K$ are identically zeros on $M$, which means that $K=const$. But, in this way, the connectedness of $M$ would be redundant. So where did I miss?

Edit

Why does $(df)_p=0, \forall p \in M$ only implies $f$ is locally constant (why is $f$ not globally constant). The following is my thinking.

In Euclidean space, if all partial derivatives of $f:R^n \rightarrow R^m$ are identically zero on $R^n$, then $f$ is constant on $R^n$ (no need for connectedness). Back to this exercise, for the "locally constant", is it because the coordinate neighborhoods? More precisely, at any point $q \in M$, for a chart $\{\phi, U\}$ ($\phi : U \in M \rightarrow R^n$) about $q$, $df=0$ on $U$ implies $f \circ \phi^{-1}$ is constant on $\phi(U)$. This in turn means $f$ is constant on $U$. So, the "$f$ is locally constant" you mentioned is that $f$ is constant on every charts.

gsoldier
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  • This is really a question from vector calculus: Suppose you have a $C^1$-smooth function $f$ of $n$ variables such that on an open ball $B$ in ${\mathbb R}^n$, we have $\frac{\partial f}{\partial x_i}=0$ for ll $i$. Then $f$ is constant on $B$. This is something one typically learns in a vector calculus class, while learning the mean value theorem. – Moishe Kohan Feb 08 '23 at 21:31
  • Yes, you are right. But, Here (Question 1) all partial derivatives $\frac {\partial K}{\partial x} =0 $ at $p$, not in $U$. – gsoldier Feb 08 '23 at 22:09
  • That's a mistake in do Carmo. I found several other small mistakes. – Moishe Kohan Feb 08 '23 at 22:16
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    Even though it does not appear obvious, the proof that $df=0$ in $\Bbb R^n$ implies that $f$ is constant really uses the connectedness of $\Bbb R^n$: in the proof, you apply the $1$-dimensional Theorem along a path joining any two points. – Didier Feb 09 '23 at 13:27

1 Answers1

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If a smooth function on a connected manifold $M$, and $X$ is a vector field, then the condition $Xf = 0$ is equivalent to the fact that $f$ is constant on the integral curves of $X$. Indeed: $$ Xf = 0 \iff df(X)=0 \iff \forall \gamma \text{ integral curve of } X, (f\circ \gamma)' = 0. $$ Let $p$ and $q$ be any two distinct points and let $\gamma\colon [0,1]\to M$ be any smooth injective path jointing $p$ and $q$ (here comes the connectedness!). Consider $Y = \gamma'$ along $\gamma$, and extend it arbitrarily on all of $M$. Assume that $Xf=0$ for all $X$. Then $Yf = 0$, which implies that $f(p) = f(q)$. Finally, $f$ is constant.

Apply this to $f = K$, which is supposed to be independent of the tangent plane $\sigma \subset T_pM$, and therefore, is assumed to be a function on $M$.

Edit

Here is what I think do Carmo has in mind.

  1. For $p\in M$ and $X_p\in T_pM$, he shows that $X_p K = 0$.
  2. This being true for all $X_p\in T_pM$, one can conclude that $d_pK = 0$.
  3. This being true for all $p\in M$, one can conclude that $dK = 0$ on all of $M$.
  4. For a smooth function $f$ on a manifold $M$, $df = 0$ implies that $f$ is locally constant, and if $M$ is connected, then $f$ is constant. This implies that $K$ is constant.
Didier
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  • Yes, I understand your points. But in the textbook, it is that $X_p(K)=0, \forall X_p \in T_pM$, where $X_p$ is a tangent vector, not a field. My question is that why at a point $p$, $X_p(K)=0, \forall X_p \in T_pM$ implies $K=0$ around $p$ – gsoldier Feb 09 '23 at 09:59
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    @gsoldier I see. I think do Carmo takes the not-so-obvious following shortcut: $XK=0$ for all $X\in T_pM$. Hence, $d_pK = 0$. This being true for all $p\in M$, then $dK = 0$. Since $M$ is connected, $K$ is constant. – Didier Feb 09 '23 at 11:41
  • Thanks for your kind answering! I still have a tiny question. Why does $(df)_p=0, \forall p \in M$ only implies $f$ is locally constant (why is $f$ not globally constant). I think the mentioned "locally constant" is that $f$ is constant on every coordinate neighborhoods. I have write my thinking in the main question (see Edit). Please check if my logic is right, if you are available. – gsoldier Feb 09 '23 at 12:50
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    @gsoldier Locally constant = constant on each connected component! The proof is a straightforward application of the mean value inequality in a chart. For instance, the map $f\colon \Bbb R^* \to \Bbb R$ defined by $f(x) = -1$ if $x<0$ and $f(x) = 1$ if $x > 0$ has $f'=0$ but is not constant. It is however locally constant. – Didier Feb 09 '23 at 13:04
  • Thanks! I do totally forget the connectedness in the case of $R^n$. Thank you again!! – gsoldier Feb 09 '23 at 14:45