Meaning of “modulo the identity matrix”

Question

I’m reading about $PSL(2,R)$. Can someone explain the meaning of: the matrices in $SL(2,R)$ “modulo plus or minus the identity matrix.” I understand the first part, but what does the quoted part mean? thanks

I'm new to math stackex; trying to get the hang of it. I'm attempting to respond to all 4 responses to my initial 10/22 Q (paragraph 1 above). Thanks Somos, J.W., lulu, and Ben for extremely fast answers. (Some of this is over my head, so I'd be grateful for concrete examples whenever possible. Thanks.)

1. Is there a homomorphism from (2,) to (2,) with kernel +- the identity matrix?
2. Is this homomorphism the squaring function? If not, what's the homomorphism?
3. The 2x2 matrices ([3,1],[-7,-2]) and ([-3,-1],[7,2]) are both in (2,). But only one is in (2,), right? Which one and why?

Thanks very much for your help, Somos (is the this the right way to respond to you?)

I understand that (2,) can be viewed as a group of cosets (each coset containing a matrix and its negative) where the kernal {+I and -I} is the identity. Each coset, I suppose, would represent one linear fractional transformation. (The 2x2 matrices ([3,1],[-7,-2]) and ([-3,-1],[7,2]) represent the same transformation.)

However, can't (2,) also be viewed as a group of individual 2x2 matrices where I is the identity element? If so, which of the above 2x2 matrices is in (2,)?

Answer 1

You asked about the quote

the matrices in $SL(2,R)$ modulo plus or minus the identity matrix.

In the context of $\,PSL(2,R),\,$ a matrix $\,A={a\, b\choose c\,d}\,$ is mapped to the linear fractional transformation $\,z\mapsto \frac{az+b}{cz+d}\,$ which is a homomorphism of groups as in Wikipedia SL(2,R). The matrices $\,A\,$ and $\,-A\,$ both map to the same transformation (in general, any scalar multiple of $\,A\,$ maps to the same transformation). The identity matrix and its negative form a normal subgroup of $\,SL(2,R)\,$ whose quotient is $\,PSL(2,R)\,$ where, indeed, $\,A\,$ and $\,-A\,$ are identified in the quotient.

Of course, if $\,R=\mathbb{Z}\,$ then this is the only identification of matrices when passing to linear transformations. For other rings, there may be more identifications based on units in the ring $\,R\,$ as indicated in the Wikipedia article.

You asked:

2. Is this homomorphism the squaring function? If not, what's the homomorphism?

The "squaring" function is not a homomorphism.

3. The 2x2 matrices ([3,1],[-7,-2]) and ([-3,-1],[7,2]) are both in (2,). But only one is in (2,), right? Which one and why?

The elements of (2,) are cosets (or equivalence classes) and each coset consists of a matrix and its negative, so both are in the same coset of (2,).

Answer 2

Formally, in a quotient $G/H$, two elements $g,g'$ are in the same coset iff (def.) $$ab^{-1} \in H $$.

In our case, per this definition, for $G:=SL(2, R)$ ; $H:=\{ I,-I\}$,:

Two matrices $M, M' $are equivalent, aka , in the same coset iff : $$MM'= \pm Id$$ , i.e., if $MM'=Id$ or $MM'= -Id $

Multiplying each equation by $M^{-1}$, this comes down to $M, M'$ being in the same coset iff:

$M'=M^{-1}$ or $M'=-M^{-1}$ .

This means we collapse a matrix and its inverse to a single element in the coset.

I leave to you the exercise of trying this with $S_3$ the permutation group of order 3 , see what you get for $S_3/ \pm Id$