Let $A$ be a finite dimensional central simple $k$-algebra. We proved in this theorem that $\text{Nrd}_A: A^{\times} \longrightarrow k^{\times}$ is a group homomorphism. The image of $\text{Nrd}_A$ is an abelian group because it lies in $k^{\times}.$ So if $a,b \in A^{\times},$ then $\text{Nrd}_A(aba^{-1}b^{-1})=\text{Nrd}_A(aa^{-1}) \text{Nrd}_A(bb^{-1})=1.$ Therefore $(A^{\times})',$ the commutator subgroup of $A^{\times},$ is contained in $\ker \text{Nrd}_A=\{a \in A: \ \text{Nrd}_A(a) = 1\}$ and so the following definition makes sense.

Definition. Let $A$ be a finite dimensional central simple algebra. The reduced Whitehead group of $A$ is the factor group $\text{SK}_1(A) = (\ker \text{Nrd}_A)/(A^{\times})'.$

Remark. Let $D$ be a finite dimensional central division $k$-algebra of degree $n.$ By the theorem in this post, for every $a \in D$ there exists $v \in (D^{\times})'$ such that $\text{Nrd}_D(a)=a^nv.$ So if $\text{Nrd}_D(a)=1,$ then $a^n \in (D^{\times})'.$ Therefore $g^n=1$ for all $g \in \text{SK}_1(D).$

Example 1. $\text{SK}_1(M_n(k))=\{1\}$ if $k$ is a field and either $n > 2$ or $n=2$ and $|k| > 3.$

Proof. Let $A=M_n(k).$ Then $A^{\times} = \text{GL}(n,k)$ and

$\ker \text{Nrd}_A = \{a \in M_n(k): \ \det(a) = 1 \}=\text{SL}(n,k).$

We proved in here that $(\text{GL}(n,k))' = \text{SL}(n,k).$ Thus $\text{SK}_1(A)=\{1\}. \Box$

Example 2. $\text{SK}_1(\mathbb{H})=\{1\},$ where $\mathbb{H}$ is the division algebra of quaternions over $\mathbb{R}.$

Proof. Let $z = \alpha + \beta i + \gamma j + \delta k \in \mathbb{H}$ and suppose that $\text{Nrd}_{\mathbb{H}}(z)=\alpha^2+ \beta^2 + \gamma^2 + \delta^2=1.$ We need to prove that $z$ is in the commutator subgroup of $\mathbb{H}^{\times}.$ We are going to prove a stronger result, i.e. $z = aba^{-1}b^{-1}$ for some $a,b \in \mathbb{H}^{\times}.$ If $z \in \mathbb{R},$ then $\beta = \gamma = \delta = 0$ and $z = \alpha = \pm 1.$ If $z = 1,$ then we can choose $a=b=1$ and if $z=-1,$ we can choose $a=i, \ b=j.$ So we may assume that $z \notin \mathbb{R}.$ We also have $z^2 - 2 \alpha z + 1 = 0$ (you may either directly check this or use this fact that every element of a finite dimensional central simple algebra satisfies its reduced characteristic polynomial).  Thus $(z-\alpha)^2 = \alpha^2 - 1.$ Note that since $\text{Nrd}_{\mathbb{H}}(z)=1$ and $z \notin \mathbb{R}$ we have $\alpha^2 < 1.$ So $1 - \alpha^2 = \alpha_0^2$ for some $0 \neq \alpha_0 \in \mathbb{R}.$ Let

$w = \alpha_0^{-1}(z - \alpha). \ \ \ \ \ \ \ \ \ (1)$

So, since the real part of $z$ is $\alpha,$ the real part of $w$ is zero and hence

$w^2 = -1, \ \ \overline{w} = -w. \ \ \ \ \ \ \ \ \ \ (2).$

Since the center of $\mathbb{H}$ is $\mathbb{R}$ and $w \notin \mathbb{R},$ there exists $c \in \mathbb{H}$ such that $u = cw-wc \neq 0.$ Therefore

$uw = -wu = \overline{w}u. \ \ \ \ \ \ \ \ \ \ (3)$

Now, by $(1),$ we have $z = \alpha + \alpha_0 w$ and $\alpha^2+\alpha_0^2=1.$ So we can write $\alpha = \cos \theta$ and $\alpha_0 = \sin \theta.$ Let $v = \cos(\theta/2) + \sin(\theta/2)w.$ Then $(2)$ gives us

$v^2=z, \ \ \ \overline{v}=\cos(\theta/2) + \sin(\theta/2)\overline{w} = \cos(\theta/2) - \sin(\theta/2)w=v^{-1}. \ \ \ \ \ \ \ \ (4)$

We also have $u v^{-1} = \overline{v^{-1}}u=vu,$ by $(3)$ and $(4).$ Thus, by $(4)$ again, $uv^{-1}u^{-1}v = v^2=z. \ \Box$

To be continued …