# Centre of the special linear group $SL_2(\mathbb R)$ or $SL(2,\mathbb R)$

Algebra by Michael Artin Def 2.5.12 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.

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.

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 $\text{SL}_n(\Bbb R)$.