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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?](http://math.stackexchange.com/questions/103621/) is helpful.2012-09-10
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    PS: it seems [GoDaddy is down atm.](http://techcrunch.com/2012/09/10/godaddy-outage-takes-down-millions-of-sites/), so this might explain why the formulas don't work for you (they look fine from where I am).2012-09-10
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    Thanks, @t.b.! I will take a look.2012-09-10
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    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!2012-09-10

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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.2012-09-10