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Algebra by Michael Artin Def 2.5.12

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Obviously $I$ and $-I$ are in the centre: $AI=IA,A(-I)=(-I)A$.

How exactly do I go about doing this?

I was thinking to solve for $j,f,g,h$ below

$$\begin{bmatrix} a & b\\ c & d \end{bmatrix} \begin{bmatrix} j & f\\ g & h \end{bmatrix} = \begin{bmatrix} j & f\\ g & h \end{bmatrix} \begin{bmatrix} a & b\\ c & d \end{bmatrix}, ad-bc=1=jh-fg.$$

So, I plug in b*g=c*f, a*f+b*h=b*j+d*f, c*j+d*g=a*g+c*h, j*h-f*g=1, a*d-b*c=1 in Wolfram Alpha or here or here (or here) to get a bunch of complicated solutions sets, some of which include the desired $\pm I$.

Ugh, where can I find a proof online?

Or if this is still folklore, how do I begin?

  • Do I for example take cases $c=0, c \ne 0$ and then solve for the centre in each case('s subcases)?

  • Another thing I thought was to suppose on the contrary that $f \ne 0$ and then arrive at a contradiction and then assume $f=0$ when supposing on the contrary that $g \ne 0$ and then assuming $f=g=0$ when supposing on the contrary that $j \ne \pm 1$ and then assuming $f=g=0, j = \pm 1$ when supposing on the contrary that $h \ne \pm 1$.

Not looking for a full solution, just a little nudge in the right direction. I've been lost in subcases of subcases of cases (like here) that I think I'm missing something elegant or simple. I guess I've been doing maths for quite awhile that I've forgotten how to do arithmetic.

To clarify, I am looking for a way to do this by systems or at least nothing high level like using facts like these About the Center of the Special Linear Group $SL(n,F)$ Note that this chapter is on homomorphisms. The reader just finished cyclic groups. The reader didn't reach Lagrange's Theorem, fields, rings or even isomorphisms.

BCLC
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2 Answers2

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A matrix $M$ is in the centre iff $MA=AM$ for all elements $A$ of $\text{SL}_2(\Bbb R)$. This means that $MA'=A'M$ for all $A'$ in the linear span of elements of $\text{SL}_2(\Bbb R)$. If you can show that all $2\times 2$ matrices can be such an $A'$ then $M$ will commute will all matrices, and so be a scalar matrix.

ADDED IN EDIT

For a slick proof, let $T=\pmatrix{1&1\\0&1}$ and $U=\pmatrix{1&0\\1&1}$. Then $T$, $U\in\text{SL}_2(\Bbb R)$ and the only matrices $M$ with $MT=TM$ and $MU=UM$ are the scalar matrices $M=aI$.

A follow-up exercise is to extend this to $\text{SL}_n(\Bbb R)$.

Angina Seng
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  • Thanks, Lord Shark the Unknown. 1. all $2\times 2$ matrices in $\text{SL}_2(\Bbb R)$? in $\text{GL}_2(\Bbb R)$? in $\mathbb R^{2 \times 2}$ ? 2. 'all $2\times 2$ matrices can be such an $A'$' --> ummmm, does this follow from $\text{SL}_2(\Bbb R)$'s being closed due to being a subgroup of $\text{GL}_2(\Bbb R)$? 3. How do you conclude scalar matrix? – BCLC Aug 31 '18 at 19:35
  • I had a feeling my issue was something logical. Thanks Lord Shark the Unknown! Lemme get this straight: Elements in the centre in the group must commute with all elements in the group including any careful choices of elements in the group. For this group, our careful choices are the matrices $T, U$ in your post: The elements of the centre in this group must commute with $T$ and also with $U$. From commuting with $T$, we deduce $j=h$ and $g=0$. From commuting with $U$, we deduce $j=h$ and $f=0$. Therefore, the centre satisfies all our deductions: $j=h, f=0, g=0$ and therefore centre is scalar? – BCLC Sep 01 '18 at 08:22
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    Exactly. Now, how are you getting on with $\text{SL}_n$? @BCLC – Angina Seng Sep 01 '18 at 08:51
  • Lord Shark the Unknown, thanks! I posted an answer, less elegant than yours. As for $n > 2$, I got the centre by modifying the proof here and understand where you got your $T$ and $U$. Thanks for the exercise! – BCLC Sep 01 '18 at 16:31
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We must show $centre(SL_2(\mathbb R)) = \{ \pm I \}$.

Pf: $\supseteq$ is easy. For $\subseteq$, we must show that for any $N=\begin{bmatrix} j & f\\ g & h \end{bmatrix}$ in $centre(SL_2(\mathbb R))$, we will have $N=\pm I$.

Let $N \in centre(SL_2(\mathbb R))$. By definition of $centre(SL_2(\mathbb R))$, we have that for any $M = \begin{bmatrix} a & b\\ c & d \end{bmatrix}$ in $SL_2(\mathbb R)$, $MN=NM$.

Observe that $MN=NM$ gives the following equations

$$bg=fc, c(j-h)=(a-d)g, (a-d)f=b(j-h) \tag{1}$$

For the fixed $N$, we must have that $MN=NM$ holds for any $M$ in $SL_2(\mathbb R)$, i.e. for any real numbers $a,b,c,d$ s.t. $ad-bc=1$. Therefore, by careful selection of matrix $M$, i.e. by careful selection of real numbers $a,b,c,d$, we can deduce $e,f,g,h$.

First careful selection: Let us choose $M$ s.t. $c=0$ and $b \ne 0$. For $c=0$, since $N \in centre(SL_2(\mathbb R))$, we must have that $(1)$ becomes:

$$bg=0, 0=(a-d)g, (a-d)f=b(j-h)$$

From the first two inequalities, we have that ($g=0$ or $b=0$) and ($g=0$ or $a=d$). Therefore, $g=0$ or ($a=d$ and $b=0$). Now $b \ne 0$ implies that we must have $g=0$.

Second careful selection: Similarly, let us choose $M$ s.t. $b=0$ and $c \ne 0$. Then $f=0$.

Third careful selection: Let us choose $M$ s.t. $b \ne 0$. From our deductions that $f=0=g$, we have that $(1)$ becomes:

$$0=0, c(j-h)=0, 0=b(j-h)$$

From the third equation, we have, $j=h$.

Hence, we have come up with three deductions as follows:

$$1. f=0$$

$$2. g=0$$

$$3. j=h$$

Hence, $N=\begin{bmatrix} j & 0\\ 0 & j \end{bmatrix}=jI$. Now, $1=\det(N) \implies j = \pm 1$.

Therefore, we have shown that if $N$ is in $centre(SL_2(\mathbb R))$, $N=\pm I$. QED

BCLC
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