Explicit Map from $\mathbb{T}^{3}$ to $SU(2)$

Question

I'm looking for a smooth map from the 3-torus to $SU(2)$, i.e. some $f:\mathbb{T}^{3}\to SU(2)$, that is injective near identity, $f^{-1}(1)=\{p\}$. I know that the homotopy classes of maps from $\mathbb{T}^{3}\to S^{3}\simeq SU(2)$ are labeled by $\mathbb{Z}$, but I need to find an explicit example (I think of the $n=1$ class) for my work.

If there's an easy, explicit construction, I'm all for it. Below I've tabulated my attempt at first embedding $\mathbb{T}^{2}$ in $\mathbb{R}^{3}$ and projecting to the unit sphere, hoping that the same idea works with $\mathbb{T}^{3}$ in $\mathbb{R}^{4}$.

Because I can't think in four dimensions, my initial idea was to knock off one dimension from both spaces and try to find a map from a 2-torus to $S^{2}$ that satisfies this.

So I've got a Torus parameterized as: $$ x=(R+r\cos\theta)\cos\phi-R,\hspace{.5cm} y=(R+r\cos\theta)\sin\phi, \hspace{.5cm}z=r\sin\theta$$ Crucially, this torus has the origin inside of it, so that when I project to the unit sphere $$ (x, y, z)\mapsto \frac{(x, y, z)}{\sqrt{x^{2}+y^{2}+z^{2}}}$$ only one point ($\theta=0, \phi=0$) is mapped to $(1, 0, 0)$ on the unit sphere. So I can map $\mathbb{T}^{2}\to S^{2}$ by: $$(\theta, \phi) \mapsto \frac{(x, y, z)}{\sqrt{x^{2}+y^{2}+z^{2}}}$$

(1) I'm not sure that this is even a smooth map, even though it seems like it should be. The issue is that one ends up with a $\sqrt{...+\cos\theta}$ in the denominator, which contains arbitrarily high powers of $\cos\theta$. (Ultimately I'm using this for a physics application which needs a local formulation, and $\theta$ will be a momentum, so I need there to be a finite maximum power of $\cos\theta$, $\sin\theta$, $\cos\phi$, etc)

(2) If this lower dimensional case is smooth, can the same idea be extended to a map $\mathbb{T}^{3}\to SU(2)$? Can one parameterize $\mathbb{T}^{3}$ in a similar way to $\mathbb{T}^{2}$, then project to the unit sphere, and be done with it?

(3) Is there an altogether better way to construct a map from $\mathbb{T}^{3}\to SU(2)$ that avoids all of this messiness?

Answer 1

First of all, $SU(2)$ is diffeomorphic to $S^3$ (I can write down an explicit diffeomorphism if you like), so I will be constructing a map $T^3\to S^3$. I will identify $S^3$ with the 1-point compactification of $R^3$, $S^3=R^3\cup \{\infty\}$; this is done via a stereographic projection.

The idea is that $T^3$ is homeomorphic to the cube $Q=[-1,1]^3$ in 3-space with opposite faces identified via translations. Thus, I will construct a continuous map $f: Q\to S^3$ which is constant (equal to $\infty$) on the boundary and is a homeomorphism (to $R^3$) on the interior of $Q$.

The map $f$, of course, respects the identification of the opposite faces on the boundary of $Q$, hence, defines the required map $T^3\to S^3$.

Let $S$ denote the unit sphere in $R^4$ centered at $e_4=(0,0,0,1)$, it is given by the equation $$ x_1^2 +x_2^2 + x_3^2 + (1-x_4)^2=1. $$

I will construct $f$ as the composition of three explicit maps.

1. An explicit homeomorphism $h_1$ from $Q$ to the unit ball $B$ centered at the origin, see for instance here.

2. The homeomorphism $h_2$ from $B$ to the lower hemisphere $H$ in $S$: $$ h_2(x_1,x_2,x_3)= \left(x_1,x_2,x_3, 1- \sqrt{1- x_1^2 -x_2^2 - x_3^2}\right). $$ This map sends the unit sphere $$ x_1^2 +x_2^2 + x_3^2=1 $$ to the boundary of the hemisphere $H$, i.e. the intersection of the sphere $S$ with the hyperplane $x_4=1$.

3. The stereographic projection $h_3$ with the center at $e_4$ and defined on the slab $$ P= \{(x_1,x_2,x_3,x_4): 0\le x_4\le 1\}. $$ The map $h_3$ sends every point $p=(x_1,x_2,x_3,x_4)\in P$ satisfying $$ 0\le x_4< 1 $$ to the intersection point of the line through $p$ and $e_4$ with the hyperplane $x_4=0$, and sends points with $x_4=1$ to $\infty$. In particular, $h_3$ sends the boundary of the hemisphere $H$ to $\infty$. It sends $H$ minus its boundary sphere, homeomorphically to $R^3$

Now, $f= h_3\circ h_2 \circ h_1$.