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Let $F:M \to N$ be a smooth map between manifolds, and suppose $M$ is connected and that

$$F_*: T_pM \to T_{F(p)}N$$

is zero for every $p \in M$. Then $F$ is the constant map.

Here is mt attempt:

Let $q \in N$ with $$F(p)=q.$$ I claim the set $$A=\{x \in M : F(x)=q\}$$ is both open and closed in $M$. Clearly it is nonempty as $p \in A$. And by connectedness of $M$, this would force $A=M$ thus $F$ is constant on $M$. For closed let $x_0 \in M$ and $\{x_n\} \subset A$ with $$\lim_{n \to \infty}x_n=x_0.$$ As $x_n \in A$ for every $n$, $F(x_n)=q$ and by continuity of $F$, we have that $$F(x_0)=q$$ forcing $x_0 \in A$ thus $A$ contains its limit points and is closed in $M$. To show $A \subset M$ is open, fix some $a \in A$ and let $\epsilon>0$ be given such that $$B_\epsilon(a) \subset M.$$ as $M$ is connected, it is path connected thus if $z \in B_\epsilon(a)$, then define $$g:[0,1] \to M$$ via $$g(t):=F(zt+(1-t)a)$$ so that $g(0)=F(a), g(1)=F(z).$ and $g$ is a path. Then $$g_*=F_*(zt+(1-t)a)(z-a)=0.$$ Thus $g_*(t)=0$ for $t \in [0,1]$ implying $g$ is constant thus $$F(z)=g(1)=g(0)=F(a).$$ I.e., $B_\epsilon(a) \subset A$ and $A \subset M$ is open thus is all of $M$ and $F$ is thus constant on $M$.

For an exmaple of when connecedness is dropped, then it fails, consider $$f: \Bbb{R} \setminus \{0\} \to \{-1,1\}$$ via $$f(x)=\begin{cases} 1 & x>0\\ -1 & x<0 \end{cases}.$$ This function has zero derivative but is non constant. Furthermore the domain is a disconnected manifold.

Added: I am a bit shaky on the showing $A$ is open in $M$ but certain my closed argument works just fine. If I am correct, an upvote or a "looks ok" comment will suffice, if I am incorrect somewhere point it out and give a hint as to how to fix it please, thanks in advance! Also I think my counterexample works just fine

Nikolaos Skout
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MyMathYourMath
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  • Typo: "Clearly it is nonempty as $F(p)\in A$". I think you mean $p\in A$. – user10354138 Oct 04 '23 at 23:51
  • sorry yes $p \in A$ as $F(p)=q$ @user10354138 – MyMathYourMath Oct 04 '23 at 23:52
  • Looks fine, but didn’t read in full detail. It really is better to distill the argument into its essentials. First, show for any locally constant map between topological space, if the domain is connected, then the map is constant. Then, show using the mean-value theorem that for a differentiable function from an open set in R^n to R^m, having zero derivative implies it is locally constant. Finally, upgrade to the manifold context: show $F_*=0$ implies for every chart $(U,\phi)$ on $M$ and $(V,\psi)$ on $N$, $\psi\circ F\circ\phi^{-1}$ has vanishing derivative, hence $F$ is locally constant – peek-a-boo Oct 04 '23 at 23:53
  • @peek-a-boo does my argument not work? – MyMathYourMath Oct 04 '23 at 23:54
  • I did say it looks fine – peek-a-boo Oct 04 '23 at 23:54
  • Showing the set of $x \in M$ such that $F(x)=q$ is open and closed in $M$. – MyMathYourMath Oct 04 '23 at 23:54
  • @peek-a-boo ahh ok youre just giving me an alternate method, i see, thanks!! – MyMathYourMath Oct 04 '23 at 23:54
  • no, not an alternate method. I’m trying to get you to see what constitutes a ‘better presentation’. It’s the same method/ideas. – peek-a-boo Oct 04 '23 at 23:56
  • Anyway, for closed you can just use Hausdorff property on $N$ so ${q}$ is closed, hence $F^{-1}(q)$ is closed. – user10354138 Oct 04 '23 at 23:56
  • @user10354138 true but my method works as well right? But youre saying since $F$ is continuous pull backs of closed are closed and in Hausdorff space singletons are closed. right? – MyMathYourMath Oct 04 '23 at 23:59
  • @ted shifrin any input? – MyMathYourMath Oct 05 '23 at 00:07
  • First, $p$ was a dummy variable, so you have to pick a particular $p\in M$. Next, I would assume by now that we know that the continuous preimage of a closed set is closed. But third, and most important, without going to a coordinate chart, I don’t even know what it means to talk about $B_r(a)\subset M$. – Ted Shifrin Oct 05 '23 at 00:21
  • @TedShifrin I meant to say $p\in M$ and you’re right as the manifold isn’t necessary the reals it wouldn’t make sense to say a ball is contained in $M$? Right? – MyMathYourMath Oct 05 '23 at 04:16
  • @TedShifrin Every manifold is metrisable. – M Khaled Bin-Lateef Oct 05 '23 at 21:47
  • @MoeKhaledBin-Lateef I’m well aware of that. But the OP is applying the chain rule in the setting of a literal straight line segment in a Euclidean ball. Oh, and $g$ is a path in $N$, not in $M$. – Ted Shifrin Oct 05 '23 at 21:52
  • @TedShifrin can my argument be modified using charts to show my $A$ is open in $M$? – MyMathYourMath Oct 05 '23 at 22:37
  • Yes, of course. – Ted Shifrin Oct 05 '23 at 23:47
  • @TedShifrin so you can use a path in  your chart, then compose with the chart map into M, then you have a path from a to one of its neighborhood point – MyMathYourMath Oct 06 '23 at 18:09
  • So like this let $g:[0,1] \to V$ be a path where $(V,\phi)$ is a chart on $N$, then define $\phi^{-1} \circ g:[0,1] \to \phi(V) \subset \Bbb{R}^n$ via $F(tz+(1-t)a)$ if $a \in A$ is fixed, and $\epsilon>0$ is given and $z \in B_\epsilon(a)$ is arbitrary – MyMathYourMath Oct 06 '23 at 18:25

3 Answers3

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If $F_{*}=0$ then its matrix representation is the zero matrix. Hence locally, the partial derivatives of $F$ vanish. Hence locally, $F$ is constant. Since the manifold is connected, $F$ is constant everywhere.

M Khaled Bin-Lateef
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Suppose that there are two points in the image, and pick a function $f:N\to\mathbb R$ that takes different values at those two points. The function $f\circ F$ is then not constant, but the chain rule tells you that its differential is zero.

Mariano Suárez-Álvarez
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  • Too vague.. I wanted to make my argument work. Am I on the right track somewhat? – MyMathYourMath Oct 05 '23 at 04:17
  • Your argument is just a rephrased version of the question. The question itself wants to show that if the differential is zero then the map is constant, and you just used it as a result? – PinkRabbit Oct 14 '24 at 17:51
  • @PinkRabbit, no, it is not, since the composition is a real valued function. You can use results of one variable calculus… – Mariano Suárez-Álvarez Oct 14 '24 at 17:53
  • You downvoted something that you did not understand? Great practice huh. – Mariano Suárez-Álvarez Oct 14 '24 at 17:56
  • @MarianoSuárez-Álvarez The composition is a real valued function but with domain $M$, this is not a leisure one-variable calcucus. If you want to have the result in calcucus, left-compose a smooth function $\mathbb{R} \to M$ that passes arbitrary two points is needed. And that is exactly the key to the question, and takes work to show, as in answers of this question. – PinkRabbit Oct 14 '24 at 18:04
  • Yes. The answer was written with people who did not want the whole solution spoon fed to them in mind, at a time when that was more or less the norm for people who’d gotten as far as, say, manifolds: you caught me! – Mariano Suárez-Álvarez Oct 14 '24 at 22:34
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I think you can generalize the "$A$ is open"'s argument to prove the entire statement. First of all lets fix an $x_0\in M$ and define $F(x_0)=q$, now we have only to prove that $F(x)=q\;\; \forall x\in M$.

By $M's$ connectedness there exists some smooth path $\phi(t)$ from $x_0$ to $x$ (i'd avoid using $tx_0+(1-t)x$ because we're not allow to treat points like vectors in a generic smooth manifold), now the following path on $N$ $$\alpha(t):=F(\phi(t)),\;\; \alpha'(t)=D_{\phi(t)}F (\phi'(t))\stackrel{F_*=0}{=} 0 \;\; \forall t, $$ has zero speed, so for the The Existence and Uniqueness Theorem for Solutions to ODEs, $\alpha(t)$ must be the constant path hence $$F(x)=F(\phi(1))=\alpha(1)\equiv \alpha(0)=F(\phi(0))=F(x_0)=q. $$

Bongo
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