4
$\begingroup$

I have just me this problem in my class on smooth manifolds from Lee's introduction to smooth manifolds, from the chapter on the tangent bundle stating the following:

Let $M, N$ be smooth manifolds, with $M$ being connected. Now we have a smooth map $ F : M \to N $ such that its push forward is the zero map. We are to show that the map $F$ is constant

I thought about it for a while, I figured maybe I should assume to get contradiction the map is not constant that would entail that in a connected neighborhood of M the coordinate representation of this map is non constant so due to smoothness some derivation on it is not zero, but how would I connect this with the push forward known to be the zero map? This is where I am stcuk. I thank all helpers on this

$\endgroup$
  • 2
    $\begingroup$ In coordinates, the pushforward is the Jacobian. $\endgroup$ – user98602 Jan 25 '16 at 7:46
  • $\begingroup$ @MikeMiller : thanks so if I understand correctly because it is the zero map its Jacobian is identically zero and because of connectivity the map must be constant, is this correct? $\endgroup$ – kroner Jan 25 '16 at 7:49
4
$\begingroup$

Let $p\in M$, let $f\in C^\infty\left(N\right)$, and let $X\in T_pM$. By assumption, $\left(F_*X\right)\left(f\right)=X\left(f\circ F\right)=0$. Let $\left(U,\varphi\right)$ be a smooth chart containing $p$. Then

$$X=\sum_iX^i\left.\frac{\partial}{\partial x^i}\right|_p=\sum_iX^i\left(\varphi^{-1}\right)_*\left.\frac{\partial}{\partial x^i}\right|_{\varphi\left(p\right)},$$

which implies that

$$\left(\sum_iX^i\left(\varphi^{-1}\right)_*\left.\frac{\partial}{\partial x^i}\right|_{\varphi\left(p\right)}\right)\left(f\circ F\right)=\sum_iX^i\left.\frac{\partial}{\partial x^i}\right|_{\varphi\left(p\right)}\left(f\circ F\circ\varphi^{-1}\right)=0,$$

which in turn, by basic calculus, implies that $F$ is constant on $U$, i.e., $F\left(x\right)=c$ for some $c\in N$ and every $x\in U$.

Since $M$ is connected, it is the case that $M$ is path connected. Let $q\in M$ and let $\gamma:\left[0,1\right]\to M$ be a path connecting $p$ and $q$. Since, as above, $F$ is constant on each smooth chart $\left(U_{\gamma\left(x\right)},\varphi_{\gamma\left(x\right)}\right)$ containing $\gamma\left(x\right)$ for every $x\in\left[0,1\right]$, it is the case that $F\equiv c$ on $M$ since $F\left(p\right)=c$ and $\gamma$ is continuous.

$\endgroup$
  • $\begingroup$ Could you tell me how to conclude that $F$ is constant? Because $f$ is not guaranteed to be some injective function, I cannot find a way to conclude that $F$ is constant... $\endgroup$ – Keith Jan 31 '18 at 4:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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