8
$\begingroup$

I'm reading Tu's intro to manifolds. He defines a critical point of a smooth map $F:N\to M$ to be a point where the differential $F_{*,p}:T_pN\to T_{F(p)}M$ fails to be surjective. He then gives illustration of the $2$ torus embedded in three space, standing vertically with a height function $f(x,y,z)=z$ given. So in this example, the differential at a critical point would have to be the zero map, as it would otherwise be surjective onto $\mathbb{R}$.

enter image description here

So I should be able to see just from the picture that there are four critical points as shown. But I don't know why. Is it because the tangent plane at each of these points is flat (i.e. every point in the tangent plane is at the same height)? If so, why would that make the differential not onto? After this example, he explains that $p$ is critical if $\frac{\partial f(p)}{\partial x^1}=,...,=\frac{\partial f(p)}{\partial x^n}=0$ for some coordinate neighboorhood of $p$, but since the example comes before this theorem, I should be able to understand the example without it.

$\endgroup$
9
$\begingroup$

Your map $f: \mathbb{T}^2 \to \mathbb{R}$ is the restriction of the projection function $\pi_z : \mathbb{R}^3 \to \mathbb{R}$. Since this is a linear map, identifying $\mathbb{R}^3$ with its tangent space at each point and likewise for $\mathbb{R}$, the differential of $\pi_z$ is just $\pi_z$. In particular, its kernel is the $xy$ plane.

This helps us to figure out the differential of $f$. After all, $f$ is the composition of the inclusion map $\mathbb{T}^2 \hookrightarrow \mathbb{R}^3$ and $\pi_z$, so $df_p$ is the composition of $T_p \mathbb{T}^2 \hookrightarrow \mathbb{R}^3$ and $\pi_z$. This is the zero map exactly when the image of $T_p \mathbb{T}^2$ in $\mathbb{R}^3$ is contained in the $xy$ plane. This is the condition you should be looking for; the points you've identified are clearly the only points at which the tangent plane is the $xy$ plane, hence they're the critical points.

$\endgroup$
  • $\begingroup$ Alex, I'm having some doubts about your answer... How is the differential the zero map? I get $(001)$. What am I doing wrong? $\endgroup$ – An old man in the sea. Dec 18 '18 at 12:30
  • $\begingroup$ @Anoldmaninthesea. It appears that you're just taking the derivative of $\pi_z$. Of course, you need to take the derivative of $f$ instead. My answer describes how to compute $df_p$ for each $p$. Is there a part of it which isn't clear or which you disagree with? $\endgroup$ – Alex G. Mar 4 at 13:25

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.