#### Problem

Prove that the torus $\mathbb{T}^3 = \mathbb{S}^1 \times \mathbb{S}^1 \times \mathbb{S}^1$ is a manifold.

#### Solution

First, we'll prove that $\mathbb{S}^1 = \mathbb{T}$ is a manifold. We'll show that $\mathbb{S}^1$ is the union of four compatible charts, namely $U_{1}, U_{2}, U_{3}$ and $U_{4}$ depicted above.

Let

$\Phi_{1}: U_{1}\rightarrow (-1,1)$ with $\Phi_{1}(x_{1},x_{2}) = x_{1}$ that maps $U_{1}$ bijectively onto the segment (-1,1) of the $x_{1}$ axis; thus, $U_{1}$ is a chart.

Similarly,

$\Phi_{2}: U_{2}\rightarrow (-1,1)$ with $\Phi_{2}(x_{1},x_{2}) = x_{1}$

$\Phi_{3}: U_{3}\rightarrow (-1,1)$ with $\Phi_{3}(x_{1},x_{2}) = x_{2}$

$\Phi_{4}: U_{4}\rightarrow (-1,1)$ with $\Phi_{4}(x_{1},x_{2}) = x_{2}$; therefore, $U_{2}, U_{3}$ and $U_{4}$ are charts as well.

Next we'll show that $U_{1}, U_{2}, U_{3}$ and $U_{4}$ are, in fact, compatible. This requires, for example, $U_{1}, U_{2}$ and $U_{1}, U_{3}$ to be compatible (the other cases are similar). The case of $U_{1}, U_{2}$ is trivial to prove because $U_{1} \cap U_{2} = \emptyset$. For the case of $U_{1}, U_{3}$ (the charts intersect on the second quadrant $V = U_{1} \cap U_{3}$), we need to show that $\Phi_{1} \circ \Phi_{3}^{-1}$ and $\Phi_{3} \circ \Phi_{1}^{-1}$ are smooth ($\in \mathbb{C}^{\infty}$).

First, we calculate an explicit formula for $\Phi_{1} \circ \Phi_{3}^{-1}$.

For $\Psi = \Phi_{1} \circ \Phi_{3}^{-1}: \Phi_{3}(V) \longrightarrow \Phi_{1}^{-1}(V)$ and for $t \in \Phi_{3}(V)$ we have $\Psi(t) = \Phi_{1} \circ \Phi_{3}^{-1}(t) = \Phi_{1}(\Phi_{3}^{-1}(t))$.

But $\Phi_{3}^{-1}: t \longmapsto (-\sqrt{1 - t^{2}}, t)$; hence, $\Psi(t) = \Phi_{1}(-\sqrt{1 - t^{2}}, t) = -\sqrt{1 - t^{2}}$, which is infinitely differentiable.

Therefore, $\Psi \in \mathbb{C}^{\infty}$.

Similarly, we calculate an explicit formula for $\Phi_{3} \circ \Phi_{1}^{-1}$.

For $\Upsilon = \Phi_{3} \circ \Phi_{1}^{-1}: \Phi_{1}(V) \longrightarrow \Phi_{3}^{-1}(V)$ and for $t \in \Phi_{1}(V)$ we have $\Upsilon(t) = \Phi_{3} \circ \Phi_{1}^{-1}(t) = \Phi_{3}(\Phi_{1}^{-1}(t))$.

But $\Phi_{1}^{-1}: t \longmapsto (t, \sqrt{1 - t^{2}})$; hence, $\Upsilon(t) = \Phi_{3}(t, \sqrt{1 - t^{2}}) = \sqrt{1 - t^{2}}$, which is, again, infinitely differentiable.

Therefore, $\Upsilon \in \mathbb{C}^{\infty}$.

So, $\mathbb{S}^1$ is a manifold.

Now that we've proved that $\mathbb{S}^1$ is a manifold, we'll show that $\mathbb{T}^3 = \mathbb{S}^1 \times \mathbb{S}^1 \times \mathbb{S}^1$ is also a manifold.

The torus $\mathbb{T}^3 = \mathbb{S}^1 \times \mathbb{S}^1 \times \mathbb{S}^1$ can be written as the union of $4^{3} = 64$ products of the form $A \times B \times C$ where $A, B, C \in \lbrace U_{1}, U_{2}, U_{3}, U_{4} \rbrace$ with $U_{1}, U_{2}, U_{3}$ and $U_{4}$ being the charts defined above, for example:

$U_{1} \times U_{1} \times U_{1} \subseteq \mathbb{T}^3$

$U_{1} \times U_{1} \times U_{2} \subseteq \mathbb{T}^3$

$U_{1} \times U_{2} \times U_{3} \subseteq \mathbb{T}^3$

etc

We'll now show that, for example, $U_{1} \times U_{2} \times U_{3}$ is a chart (the other 63 cases are similar).

Let $f_{123}: (U_{1} \times U_{2} \times U_{3}) \longrightarrow (-1, 1) \times (-1, 1) \times (-1, 1) \subseteq \mathbb{R}^3$

with $f_{123}((x_{1}, x_{2}), (y_{1}, y_{2}), (z_{1}, z_{2})) = (\Phi_{1}(x_{1}, x_{2}), \Phi_{2}(y_{1}, y_{2}), \Phi_{3}(z_{1}, z_{2}))$

$f_{123}$ is onto:

Given a point in $\mathbb{R}^3$, for example, $(k_{1}, k_{2}, k_{3})$, we can consider each of its coordinates separately.

So, the first coordinate of this point in $\mathbb{R}^3$, $k_{1}$, is mapped onto by $\Phi_{1}$ (since $\Phi_{1}: U_{1} \longrightarrow (-1, 1) \subseteq \mathbb{R}^1$) since $\Phi_{1}$ is onto. Therefore we can assume that there is a point $(\alpha_{1}, \alpha_{2})$ in $\mathbb{S}^1$ such that $\Phi_{1}: (\alpha_{1}, \alpha_{2}) \longmapsto k_{1}$. Similarly, since $\Phi_{2}$ is onto, we can assume that there is a point $(\beta_{1}, \beta_{2})$ in $\mathbb{S}^1$ such that $\Phi_{2}: (\beta_{1}, \beta_{2}) \longmapsto k_{2}$, and, finally, since $\Phi_{3}$ is onto, we can assume that there is a point $(\gamma_{1}, \gamma_{2})$ in $\mathbb{S}^1$ such that $\Phi_{3}: (\gamma_{1}, \gamma_{2}) \longmapsto k_{3}$.

Therefore, $f_{123}$ maps $((\alpha_{1}, \alpha_{2}), (\beta_{1}, \beta_{2}), (\gamma_{1}, \gamma_{2})) \longmapsto (k_{1}, k_{2}, k_{3})$ and, thus, $f_{123}$ is onto.

Furthermore, $f_{123}$ is 1-1:

Assume that $((\alpha_{1}, \alpha_{2}), (\beta_{1}, \beta_{2}), (\gamma_{1}, \gamma_{2})) \in S^3$ and $((\alpha_{1}', \alpha_{2}'), (\beta_{1}', \beta_{2}'), (\gamma_{1}', \gamma_{2}')) \in S^3$ are different. Then, at least one of the following three relations:

$(\alpha_{1}, \alpha_{2}) \neq (\alpha_{1}', \alpha_{2}')$

$(\beta_{1}, \beta_{2}) \neq (\beta_{1}', \beta_{2}')$

$(\gamma_{1}, \gamma_{2}) \neq (\gamma_{1}', \gamma_{2}')$

hold true.

But then, $\Phi_{1}(\alpha_{1}, \alpha_{2}) \neq \Phi_{1}(\alpha_{1}', \alpha_{2}')$ or $\Phi_{2}(\beta_{1}, \beta_{2}) \neq \Phi_{2}(\beta_{1}', \beta_{2}')$ or $\Phi_{3}(\gamma_{1}, \gamma_{2}) \neq \Phi_{3}(\gamma_{1}', \gamma_{2}')$.

Therefore, $(\Phi_{1}(\alpha_{1}, \alpha_{2}), \Phi_{2}(\beta_{1}, \beta_{2}), \Phi_{3}(\gamma_{1}, \gamma_{2}))$ and $(\Phi_{1}(\alpha_{1}', \alpha_{2}'), \Phi_{2}(\beta_{1}', \beta_{2}'), \Phi_{3}(\gamma_{1}', \gamma_{2}'))$ are different.

Hence, $f_{123}$ is 1-1.

To prove compatibility, take two such maps, for example

$f_{123}: U_{1} \times U_{2} \times U_{3} \longrightarrow (-1, 1) \times (-1, 1) \times (-1, 1) \subseteq \mathbb{R}^3$

and

$f_{112}: U_{1} \times U_{1} \times U_{2} \longrightarrow (-1, 1) \times (-1, 1) \times (-1, 1) \subseteq \mathbb{R}^3$

Then

$f_{123} \circ f_{112}^{-1} = f_{123}(f_{112}^{-1}) = f_{123}(\Phi_{1}^{-1}, \Phi_{2}^{-1}, \Phi_{3}^{-1}) = (\Phi_{1} \circ \Phi_{1}^{-1}, \Phi_{2} \circ \Phi_{1}^{-1}, \Phi_{3} \circ \Phi_{2}^{-1})$

and $f_{123} \circ f_{112}^{-1} \in \mathbb{C}^{\infty}$ because, as shown above, $\Phi_{1} \circ \Phi_{1}^{-1}, \Phi_{2} \circ \Phi_{1}^{-1}$ and $\Phi_{3} \circ \Phi_{2}^{-1}$ are smooth.

Therefore, the torus $\mathbb{T}^3 = \mathbb{S}^1 \times \mathbb{S}^1 \times \mathbb{S}^1$ is a manifold.

Scott agrees! (And was, in fact, convinced before reading all the way to the end…)