Connected Unitaries in $K_{1}(C(\mathbb{T}^{3}))$ Let $\mathbb{T}=\{\xi:|\xi|=1\}$ and set $A:=C(\mathbb{T}^{3})$. Then, $A$ is a unital commutative C$^{*}$-algebra. For any C$^{*}$-algebra $B$,  let $\mathcal{U}(B)$ denote the unitary elements in $B$. For each $n$, set $\mathcal{U}_{n}(A):=\mathcal{U}(M_{n}(A))$, and put $\mathcal{U}_{\infty}(A)=\bigcup_{n\geq 1}\mathcal{U}_{n}(A)$.  Then, the $K_{1}$ group of $A$ is defined to be 
$$
K_{1}(A):=\{[u]_{1}:u\in \mathcal{U}_{\infty}(A)\}.
$$
Here $[\ \cdot \ ]_{1}$ denotes the equivalence class of the relation $\sim_{1}$ defined on $\mathcal{U}_{\infty}(A)$ as follows: for $u\in\mathcal{U}_{n}(A)$ and $v\in\mathcal{U}_{m}(A)$, $u\sim_{1} v$ if and only if there is a $k\geq\max\{n,m\}$ such that $u\oplus 1_{k-n}\sim_{h} v\oplus 1_{k-m}$ in $\mathcal{U}_{k}(A)$. 
I know that $K_{1}(A)\cong \mathbb{Z}^{4}$ and that $\mathcal{U}(A)/\mathcal{U}_{0}(A)\cong \mathbb{Z}^{3}$, where $\mathcal{U}_{0}(A):=\{u\in\mathcal{U}:u\sim_{h} 1\}$. Therefore, there must be some $n\geq 2$ and some $u\in\mathcal{U}_{n}(A)$ such that for all $v\in\mathcal{U}(A)$, $[u]_{1}\not=[v]_{1}$.
However, in Rordam's C$^{*}$-algebra book, on page 201, he states that a more refined argument can show that $n=2$. I'm not sure why this is the case. I.e., 

I need to show that there is some $u\in\mathcal{U}_{2}(C(\mathbb{T}^{3}))$ such that for all $v\in\mathcal{U}(C(\mathbb{T}^{3}))$, $[u]_{1}\not=[v]_{1}$. 

 A: I don't know if there is another way to do this, but here is one proof. Recall that $K_1(A)$ is defined as the limit of the groups
$$
\pi_0(GL_1(A)) \to \pi_0(GL_2(A))\to \pi_0(GL_3(A))\to \ldots
$$
where the maps are induced by the natural map $GL_{n-1}(A)\to GL_n(A)$ given by
$$
a\mapsto \begin{pmatrix}
a & 0 \\
0 & 1
\end{pmatrix}
$$
The question you ask would be answered if we knew when this sequence of groups stabilized. ie. Does there exist $n\in \mathbb{N}$ such that
$$
K_1(A)\cong \pi_0(GL_{m-1}(A))\cong \pi_0(GL_m(A)) \quad\forall m\geq n\qquad (\dagger)
$$
This is answered by Rieffel in this paper. Specifically, he defines two numbers associated to a $C^{\ast}$-algebra $A$


*

*Connected stable rank csr(A)

*General stable rank gsr(A)


He shows (Theorem 2.9 in the paper) that if $n=\max\{csr(A),gsr(C(\mathbb{T},A))\}$, then $(\dagger)$ holds.
Now, if $A = C(\mathbb{T}^3)$, then one can show (See, for instance, this paper by Nica)


*

*$csr(A) = 3$

*$gsr(C(\mathbb{T},A)) = gsr(C(\mathbb{T}^4)) = 1$


Hence, $(\dagger)$ holds for $n=3$, so that
$$
K_1(C(\mathbb{T}^3)) \cong \pi_0(GL_2(C(\mathbb{T}^3))
$$
The result now follows from what you have mentioned.
