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I was asked to prove that a standard torus(which means we don't consider those pathological cases where it intersects with itself, e.g horn torus) is diffeomorphic to $ \mathbb S^1\times \mathbb S^1$.

I was thinking if we could prove it this way: Since every point on the torus can be uniquely defined with a pair of angles $(\theta_1, \theta_2)$. Then we construct a diffeomorphism $\phi(\theta_1, \theta_2)=(\tilde{\theta}_1 ,\tilde{\theta}_2)$ which maps every point on the torus to every point on $\mathbb S^1 \times \mathbb S^1$. Since the map is definitely bijective and smooth with a smooth inverse. We're basically done...


Thanks a lot for everyone's help!

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Depending on what your definition of “standard torus” is, that seems like you have it right. (To me, “standard torus” is by definition something diffeomorphic to $\mathbb{S}^1\times\mathbb{S}^1$ after all.) – Harald Hanche-Olsen Jun 8 '13 at 13:05
Hi Prof. Hanche-Olsen! The definition I had was "the set of points in $\mathbf R^3$ at distance $b$ from the circle of radius $a$ in the $xy$ plane, where $0<b<a$". – Evariste Jun 8 '13 at 13:07
@HaraldHanche-Olsen Btw Prof. Hanche-Olsen, I know your name from reading your notes on the Buckingham Pi Theorem more than a year ago. The notes you wrote were wonderful! – Evariste Jun 8 '13 at 13:19
Okay, then I think you have it right. But to nail it, you might want to write up an explicit parametrization of your torus using two angles. (Good to hear you enjoyed my notes. Thanks!) – Harald Hanche-Olsen Jun 8 '13 at 13:19
@HaraldHanche-Olsen Thanks Prof. Hanche-Olsen! I'll go find the explicit parametrisation. Thanks for the help! – Evariste Jun 8 '13 at 13:21

For completeness: $$F(\theta,\phi) = (a\cos\theta,a\sin\theta,0)+(0,b\cos\phi,b\sin\phi)$$ is a bijection between $S^1\times S^1$ and the torus as a subset of $\mathbb R^3$. Things to check:

  1. $F$ is smooth
  2. $F$ is a bijection
  3. The inverse of $F$ is smooth.

Part 3 is harder than 1 and 2. One approach is to show that the the derivative matrix $DF$ has rank 2, which allows for the inverse function theorem to be used.

The fact that the standard torus is indeed a submanifold of $\mathbb R^3$ can be shown by considering its non-parametric equation $(\sqrt{x^2+y^2}-a)^2+z^2=b^2$ and using the fact that the gradient of the left-hand side is nonzero on the surface.

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Hi Thanks a lot for the help. Yeah I actually did the first two bits last night. But I got stuck on the third bit... And it seems impossible to me... – Evariste Jun 9 '13 at 4:31
Also I don't think we can use the inverse function theorem. Because we don't know that they are manifolds a priori. Of course we can assume that they're manifolds, but then there are little point to do this question, because they're both diffeomorphic to $\mathbf R^2$. – Evariste Jun 9 '13 at 4:33
@Evariste They cannot be diffeomorphic to $\mathbb R^2$, since they are compact. Maybe you meant "locally diffeomorphic". I added another suggestion to the question, on how to show that the standard torus is a submanifold of $\mathbb R^3$. – ˈjuː.zɚ79365 Jun 9 '13 at 5:07
The image of a one-to-one immersion of a compact manifold is a submanifold. So checking the derivative has rank $2$ nails this one shut! – Ted Shifrin Jun 9 '13 at 5:13
@user79365 Yes I meant locally diffeomorphic. Thanks for the suggestion. I'll think about it now. – Evariste Jun 9 '13 at 6:09

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