Posts Tagged ‘general linear group’

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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We will keep the notation in here and here . For a group G and a,b \in G we let [a,b]=aba^{-1}b^{-1}. Recall that G', the commutator subgroup of G, is the subgroup generated by the set \{[a,b]: \ a,b \in G \}.

Remark 1. Using the second part of Problem 1, it is easy to show that E_{ij}(\alpha \beta)=[E_{ir}(\alpha), E_{rj}(\beta)].

We are now ready to prove that the commutator subgroup of the general linear group GL(n,k) is the special linear group SL(n,k) unless n=2 and k has at most 3 elements.

Problem. Prove that GL(n,k)'=SL(n,k) unless n=2 and |k| \leq 3.

Solution. Clearly for any A,B \in GL(n,k) we have \det (ABA^{-1}B^{-1})=1. So [A,B] \in SL(n,k) and hence GL(n,k)' \subseteq SL(n,k). In order to show that SL(n,k) \subseteq GL(n,k)', we only need to prove that GL(n,k)' contains all elementary matrices because, by Remark 2, SL(n,k) is generated by elementary matrices. So we consider two cases.

Case 1 . n \geq 3 : For any i \neq j choose r \notin \{i,j \}. Then, by the above remark, E_{ij}(\alpha)=[E_{ir}(\alpha), E_{rj}(1)].

Case 2 .  n=2 and |k| > 3 : The equation x(x^2-1)=0 has at most three solutions in the field k and so, since k has more than three elements, we can choose a non-zero element \gamma \in k such that \gamma^2 \neq 1. Thus \gamma^2 - 1 is invertible in k. Now given \alpha \in k, let \beta_1=\alpha (\gamma^2 - 1)^{-1} and \beta_2 = \alpha \gamma^2 (1-\gamma^2)^{-1}. Let

A = \begin{pmatrix} \gamma & 0 \\ 0 & \gamma^{-1} \end{pmatrix}.

A quick calculation shows that

E_{12}(\alpha)=[A, E_{12}(\beta_1)] and E_{21}(\alpha)=[A, E_{21}(\beta_2)]. \Box

Remark 2. In the solution of the above problem, we actually showed that every elementary matrix is in the form [a,b] for some a,b \in SL(n,k). Thus SL(n,k)' = SL(n,k).

We will keep our notations in part (1). We now generalize the result proved in Problem 2.

Problem. Let A \in M_n(k) with \det A = 1. Prove that A is a product of elementary matrices.

Solution. Suppose that A = (a_{ij}), i.e. the (i,j)-entry of A is a_{ij}. The first column of A cannot be all zero because then \det A = 0, which is not true. So if a_{11}=0, then choosing a_{i1} \neq 0, we will have the matrix E_{1i}(1)A whose (1,1)-entry is a_{i1} \neq 0. Then we can replace A by E_{1i}(1)A and so we may assume that a_{11} \neq 0. Now let B_1=\prod_{i=2}^n E_{i1}(-a_{11}^{-1}a_{i1}) and C_1= \prod_{j=2}^n E_{1j}(-a_{11}^{-1}a_{1j}). Then the first row and the first column entries of B_1AC_1 are all zero except for the (1,1)-entry which is a_{11}. We can now do the same with the second row and the second column and continue this process until we get matrices B_1, \cdots , B_{n-1} and C_1, \cdots , C_{n-1} each of which is a product of elementary matrices and

D=B_{n-1}B_{n-2} \cdots B_1 A C_1 C_2 \cdots C_{n-1}

is a diagonal matrix. Note that \det D=1 because \det B_i=\det C_i=1 for all i (see the first part of Problem 1) and \det A=1. Thus by Problem 2, D is a product of elementary matrices. Hence

A=B_1^{-1} \cdots B_{n-1}^{-1}D C_{n-1}^{-1} \cdots C_1^{-1}

is also a product of elementary matrices, by the second part of Problem 1. \Box

Definition 1. The general linear group of order n over a field k is

GL(n,k)=\{A \in M_n(k): \ \det A \neq 0 \}.

Definition 2. The special linear group of order n over a field k is

SL(n,k)=\{A \in M_n(k): \ \det A=1 \}.

Remark 1. It is clear that GL(n,k) is a group with matrix multiplication and that SL(n,k) is a subgroup of GL(n,k). If, as usual, we let k^{\times}:=k \setminus \{0\} be the multiplicative group of the field k, then we can define the group homomorphism f : GL(n,k) \longrightarrow k^{\times} by f(A)=\det A. Obviously f is onto and \ker f = SL(n,k). So SL(n,k) is a normal subgroup of GL(n,k) and GL(n,k)/SL(n,k) \cong k^{\times}.

Remark 2. By the first part of Problem 1, every elementary matrix is in SL(n,k). Thus any product of elementary matrices is also in SL(n,k). On the other hand, by the above problem, every element of SL(n,k) is a product of elementary matrices. So we get the following important result:

SL(n,k), as a group, is generated by all elementary matrices.