3

I'm trying to prove that the $n$-torus $ T^n =\mathbb R^n / \mathbb Z ^n$ is diffeomorphic to the product of $n$ circles $ S^1 $. Thanks a lot for everyone's help!

ctst
  • 1,424
Mahsa Beizaei
  • 81
  • What is $n$-torus ? – HK Lee Oct 14 '16 at 08:21
  • 1
    @HKLee https://en.wikipedia.org/wiki/Torus#n-dimensional_torus I always thought of this as the definition of a $n$-torus. Or do you mean $T^n=\mathbb R^n/\mathbb Z^n$ (which is the same, just take the unit cube as representation and project each single coordinate ($i$) to the corresponding ($i$th) circle) – ctst Oct 14 '16 at 08:24
  • $ T^n $ is the product of $ n $ circles $ S^1 $. – Mahsa Beizaei Oct 14 '16 at 08:33
  • By wikipedia cited by ctst we know that $n$-torus is product of $n$ circles So I can not understand what Beizavi are asking – HK Lee Oct 14 '16 at 08:33
  • 3
    show that $T^n = \frac{{R^n }}{{Z^n }} $ is diffeomorphic to the product of n circles $ S^1 $. – Mahsa Beizaei Oct 14 '16 at 08:39

2 Answers2

2

You'll need a solid understanding of the theory of quotient spaces and quotient maps, for which I recommend Munkres Topology.

For your problem, consider the function $f : \mathbb{R}^n \to (S^1)^n$ defined by the formula $$f(t_1,...,t_n) = (e^{2\pi i t_1},...,e^{2\pi i t_n}) $$

You can check easily that $f$ is a quotient map.

You can also check easily that the set of point pre-images $$\{f^{-1}(p) \,\bigm|\, p \in (S^1)^n\} $$ is exactly the set $\mathbb{R}^n / \mathbb{Z}^n$ which means the set of cosets of $\mathbb{Z}^n$ in $\mathbb{R}^n$, or if you prefer the set of orbits of the additive action of $\mathbb{Z}^n$ on $\mathbb{R}^n$.

Once you've checked the above two things, you'll see that they exactly match the hypotheses of an important corollary in Munkres book. Applying that corollary, from its conclusion you get that $\mathbb{R}^n / \mathbb{Z}^n$, equipped with the quotient topology, is homeomorphic to $(S^1)^n$.

In order to go further and prove that you get a diffeomorphism, it will suffice to check that each restriction of the function $f$ to an open cube of the form $(s_1,s_1+1) \times ... \times (s_n,s_n+1)$ is a smooth bijection onto an open subset of $(S^1)^n$ having smooth inverse (the smoothness statements are evident from the given formula for $f$).

Lee Mosher
  • 120,280
0

As said in the comment above you can look at your $T^n$ as the unit cube in $\mathbb R^n$ with glued sides, in other words every element in $[(x_1,\dots, x_n)] \in T^n$ has a unique represantation $(y_1, \dots, y_n)\in [(x_1,\dots, x_n)]$ with $y_i \in [0,1)$. Hence the function: $[(x_1, \dots, x_n)] \mapsto (y_1, \dots, y_n)$ is a well definied function from $T^n$ to $\prod_{i=1}^n S^1$. You can easily check, that this is indeed a diffeomorphism.

ctst
  • 1,424