5
$\begingroup$

Try to prove the following proposition.

$SL_n(\mathbb R)$ is a subgroup of kernel of $f\in$ Hom($GL_n(\mathbb R), A)$, where $A$ is an abelian group.

My first attempt is to show that $SL_n(\mathbb R)$ is in form of $ABA^{-1}B^{-1}$, then it's clear that $ABA^{-1}B^{-1}$ is in $ker(f)$. However, there exists an elementary row reduction matrix, i.e. $E_{ij}(m)$ adding m times $j$-th row to $i$-th row. Then $E_{ij}(m)\in SL_n(\mathbb R)$. Now, it is possible that $E_{ij}(m)$ is not in $ker(f)$.
Then I try to find any relation between $\mathbb R^{\times}$ and $A$, but, except knowing they are both abelian groups, I cannot figure out what else connects those two group.
Could you give me some hints?

$\endgroup$
  • 2
    $\begingroup$ ysharifi.wordpress.com/2011/01/29/… seems to be what you want? disclaimer - I haven't read the proof. See also Jacobson Basic Algebra I, 6.7, lemma 2: $\text{SL}_n(k)$ is its own commutator ($k$ a field), except if $n=2$ and the cardinality of $k\le 3$. I bet other standard algebra books (Lang?) must do this too. $\endgroup$ – peter a g Jun 6 '17 at 0:07
  • $\begingroup$ @peterag Thanks Peter. I will read the proof. $\endgroup$ – Hamio Jiang Jun 6 '17 at 1:21
  • 1
    $\begingroup$ It is important to understand that this question is equivalent to asking you to prove that ${\rm SL}_n({\mathbb R})$ is the commutators subgroup of ${\rm GL}_n({\mathbb R})$. And in fact this statement is true for all fields in place of ${\mathbb R}$, so there nothing specific about ${\mathbb R}$ involved. $\endgroup$ – Derek Holt Jun 6 '17 at 8:03
  • $\begingroup$ @peterag Yes Peter, this is the proof I want. The proof is very neat. I think I will post a proof by referring to the material. $\endgroup$ – Hamio Jiang Jun 6 '17 at 13:47
1
$\begingroup$

Commutator Subgp. of GL(n,k)

The main idea of the proof in the above link is to show the commutator subgroup of $GL_n(k)$ is $SL_n(k)$ unless $n=2$ and $|k|\le 3$, we need to show the commutator subgroup of $GL_n(k)$, denoted by $GL'_n(k)$, satisfies $GL'_n(k)\subseteq SL_n(k)$ and $SL_n(k)\subseteq GL'_n(k)$, then $GL'_n(k)=SL_n(k)$.

Notation: $[a,b]:=aba^{-1}b^{-1}$, called commutator. $k$ is a field.

Recall that $GL'_n(k):=\{[a,b]|\ a,b\in GL_n(k)\}$. Then it is clear that any $A\in GL'_n(k),\ A=[a,b]$ for some $a,b\in GL_n(k)$. Then $detA=det([a,b])=det(aba^{-1}b^{-1})=1$. Hence, for any element in $GL'_n(k)$, it is in $SL_n(k)$$\Rightarrow\ GL'_n(k)\subseteq SL_n(k)$.

On the other hand, notice $SL_n(k)$ is generated by all elementary matrices, say $E_{ij}(\alpha):= I+\alpha e_{ij}$ where $\alpha \in k$ and $1\le i \neq j \le n$. Then it suffices to show that any elementary matrix is in $GL'_n(k)$. Here are two cases:

Case 1: $n\ge 3$. Notice that $E_{ij}(\alpha \beta)=[E_{ir}(\alpha),E_{rj}(\beta)]$, since we could let $r\neq i\neq j$ in the case $n\ge 3$. Hence, for $n\ge 3$, $SL_n(k)\subseteq GL'_n(k)$.

Case 2: $n=2$ and $|k| \gt 3$. Then the equation $x(x^2-1)=0$ has at most three solutions in the field $k$, and since $|k|\gt 3$, we can pick another non-zero element such that $x^2\neq 1$, say $\gamma$. Thus $\gamma^2-1$ is invertible in $k$. Now given $\alpha \in k$, let $\beta_1=\alpha (\gamma^2-1)^{-1}$ and $\beta_2=\alpha \gamma (1-\gamma^2)^{-1}$. Let $$A=\begin{pmatrix} \gamma & \\ & \gamma^{-1}\\ \end{pmatrix}$$

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

Then from case 1 and 2, we can conclude that $SL_n(k)\subseteq GL'_n(k)$ if $n=2$ and $|k|\gt 3$, or $n\ge 3$.

Hence, $SL_n(k)=GL'_n(k)$ unless $n=2$ and $|k|\le 3$.

Here, $k=\mathbb R$, $|\mathbb R|=2^{\aleph_0}\gt 3$. Hence, $SL_n(\mathbb R)=GL'_n(\mathbb R)$. In other words, any element in $SL_n(\mathbb R)$ can be expressed by $[a,b]$for some $a,b\in GL_n(\mathbb R)$. Then $f([a,b])=f(aba^{-1}b^{-1})=f(a)f(a^{-1})f(b)f(b^{-1})=1$. Hence, for any element in $SL_n(\mathbb R)$, it is also in the kernel of homomorphism from $GL_n(\mathbb R)$ to any abelian group$\Rightarrow$ $SL_n(\mathbb R)$ is subgroup of $ker (f)$.

Q.E.D

| cite | improve this answer | |
$\endgroup$
  • 1
    $\begingroup$ Just a little, pedantic remark, in reference to your paragraph starting with "Recall that ..." If $G$ is a group, and $G'$ its commutator subgroup, then $G'$ is not usually the set $C$ of commutators of $G$; rather, it is the subgroup generated by $C$: $G' = \langle C \rangle $. $\endgroup$ – peter a g Jun 7 '17 at 12:42
  • $\begingroup$ @peterag Thanks. You revealed the hidden truth I ignored. I just keep the notation from the link you gave. For general linear groups, this notation is fine, but others may not work. $\endgroup$ – Hamio Jiang Jun 7 '17 at 13:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.