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 …

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