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In $\mathbb{R}^3$, let $h$ be the height function of a torus standing vertically on the top of the table. A critical point of a function is those point where the differential of the function is a zero vector. But how do I show this $h$ has four critical points? I think I should need a differential structure on torus, and then I can do the composition of $h$ and the coordinate function, thus allowing me to take partial derivative in the usual sense. But, I only know that the torus is the product manifold of $S^1$ and $S^1$. I don't know how to use that knowledge to come up with a workable local coordinate chart on this specific problem.

Any help is greatly appreciated!

P.S. Why is the formula not properly displayed on my laptop? All I see is source code. I use Chrome.

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Maybe the thread Why is $S^1 \times S^1$ a torus? is helpful. – t.b. Sep 10 '12 at 20:07
PS: it seems GoDaddy is down atm., so this might explain why the formulas don't work for you (they look fine from where I am). – t.b. Sep 10 '12 at 20:10
Thanks, @t.b.! I will take a look. – henryforever14 Sep 10 '12 at 20:12
It is difficult to prove that any function $f:\ T\to{\mathbb R}$ satisfying some technical conditions has at least four critical points. But you are considering a particular function $h$ on a particular realization of $T$ in ${\mathbb R}^3$. Your example is set up in such a way that you actually can see the four critical points! Now translate the "obvious" into formulas! – Christian Blatter Sep 10 '12 at 20:31
up vote 0 down vote accepted

The height function can be expressed as a map of the form $h(\theta, \phi) = (R+r \cos \phi ) \cos \theta$, where $R> r$ , positive. Then, you should just be able to derive the four critical points from this (I think).

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Thank you. I wasn't aware of this form before. – henryforever14 Sep 10 '12 at 21:38

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