**UPDATE - Feb. 9, 2017:** The original title of this post was
"The $\text{isometry}^+$ group of hyperbolic $n$-space as $\mathrm{PSL}\_2$ of a Clifford group."
The original question, which appears below,
did not receive answers, but I discovered in the meantime that the confusion arrises from conflicting notation in the literature.
So I've explained that in the form of an answer to my own question below, and I now pose the obvious follow up question: what is the correct definition of $\mathrm{PSL}\_2(C\_n)$?
I'll happily accept a convincing answer to that question
(I'm not going to accept my own answer to the original question).
Please see my "answer" for more detail.




---


In geometric algebra (which I've just become aware of),
there is a method of realizing the group of Möbius transformations
of hyperbolic $n$-space
using a $2\times2$
matrix representation of the Clifford group of the Clifford algebra $\mathscr{C}\_{n-2}(\mathbb{R})$
(of the quadratic space $\mathbb{R}^{n-2}$ with the negative-definite quadratic form).
The appeal of this is that it directly generalizes the well-studied
theory of modular forms on the hyperbolic plane and hyperbolic $3$-space,
but on the other hand,
there is a subtle aspect of the construction that is trivial on those
well-known examples.
I would like to see how this goes on the first couple of
non-trivial examples.


The first five $\mathscr{C}\_n(\mathbb{R})$ (that is, $n=0,\dots,4$)
are


$$\mathbb{R}\rightarrow \mathbb{C}\rightarrow \mathbb{H}\rightarrow
\mathbb{H}^2\rightarrow \mathrm{M}\_2(\mathbb{H}).$$


The Clifford group $C\_n$
of $\mathscr{C}\_n(\mathbb{R})$
is defined as follows.
Let $\alpha:\mathscr{C}\_n(\mathbb{R})\rightarrow\mathscr{C}\_n(\mathbb{R})$
be the involution induced by negation on $\mathbb{R}^n$.
Then $C\_n$
is the multiplicative group


$$\big\{c\in\mathscr{C}\_n(\mathbb{R})^\times\mid
\forall v\in\mathbb{R}^n: cv\alpha(c)^{-1}\in\mathbb{R}^n\big\}$$


(where the ${}^\times$ means take the invertible elements).
Since the first three
Clifford algebras are division algebras,
we have for $n=0,1,2$:
$\mathscr{C}\_n(\mathbb{R})=C\_n$.
But this is false for $n\geq3$.


I want to say the next two Clifford groups would be


$$C\_3=\big\{(q,r)\in\mathbb{H}^2\mid q,r\neq0\big\}\\
C\_4=\mathrm{GL}\_2(\mathbb{H}).$$


But one must be careful in how one defines $\mathrm{GL}\_2$
over a non-commutative algebra (discussed below).
Also
I am dodging the details of how to multiply elements of
$\mathbb{H}^2$
(resp. $\mathrm{M}\_2(\mathbb{H})$)
by elements of $\mathbb{R}^3$
(resp. $\mathbb{R}^4$)
using their identifications within $\mathscr{C}\_3(\mathbb{R})$
(resp. $\mathscr{C}\_4(\mathbb{R})$).
(I will probably get my hands dirty with these issues in the meantime
while this post goes out there.)


The next step is to define $\mathrm{PSL}\_2(C\_n)$,
which as I mentioned above,
is less obvious when $C\_n$
is non-commutative
(and this is part of what I don't understand).
If done correctly,
we get that $\mathrm{PSL}\_2(C\_n)$
acts on $\mathbb{R}^{n+1}\cup\{\infty\}$
by Möbius transformations,
and is in fact isomorphic to the Möbius group.
We can then include half the scalar axis of $\mathscr{C}\_n(\mathbb{R})$
and extend along it isometrically (as in the lower-dimensional examples),
to get $\mathrm{PSL}\_2(C\_n)\cong\mathrm{Iso}^+(\mathfrak{H}^{n+2})$,
which is awesome.


It's not hard to check this for $n=0,1$
because we know that
$\mathrm{PSL}\_2(\mathbb{R})\cong\mathrm{Iso}^+(\mathfrak{H}^{2})$
and
$\mathrm{PSL}\_2(\mathbb{C})\cong\mathrm{Iso}^+(\mathfrak{H}^{3})$,
and there is no ambiguity about the determinant.
But what happens next?


Here comes the problem.
I know (for different reasons)
that $\mathrm{Iso}^+(\mathfrak{H}^{5})\cong\mathrm{PSL}\_2(\mathbb{H})$
where this $\mathrm{PSL}\_2$
is defined using the Dieudonné
determinant,
and that $\mathrm{Iso}^+(\mathfrak{H}^{4})$
is a proper subgroup of $\mathrm{PSL}\_2(\mathbb{H})$.
But from the above we should have
$\mathrm{Iso}^+(\mathfrak{H}^{4})\cong
\mathrm{PSL}\_2(C\_2)\cong
\mathrm{PSL}\_2(\mathbb{H})$,
no?
Is there a lack of equivalence in how to define $\mathrm{PSL}\_2$?
Or is there something off in my set-up?



How should we define $\mathrm{PSL}_2$ of a Clifford group?
*I think made this a little too complicated, and I don't think there is a lot of interest in the topic in the first place. So let me just add something more conclusive for the sake of resolving the post.*


To think about how to extend the notation $\mathrm{PSL}$,
we should start by looking at what $\mathrm{GL}$
means.
The notation $\mathrm{GL}(V)$
where $V$
is a vector space over a field $F$,
can be thought of
as the group of invertible linear transformations on $V$.
But when we write $\mathrm{GL}\_n(F)$,
we mean, more explicitly,
the group of $n\times n$
invertible matrices with entries in $F$.
Of course, if $\mathrm{dim}\_F(V)=n$
then $\mathrm{GL}(V)\cong\mathrm{GL}\_n(F)$,
but to be very pedantic (which I mos. def. am being),
we shouldn't exactly write $\mathrm{GL}(V)=\mathrm{GL}\_n(F)$.


Now the notation $\mathrm{SL}(V)$
is very rarely used,
whereas $\mathrm{SL}\_n(F)$
is common.
This is because the $\mathrm{S}$
for "special"
means, explicitly,
that we are looking at the matrices with determinant $1$,
and a determinant is not exactly defined as a polynomial on $\mathrm{GL}(V)$
(though of course, it *could* be defined).


All the above applies equally well if we put a $\mathrm{P}$
in front
and talk about maps on the projective space $\mathrm{P}(V)$,
so does not really change anything as far as choosing notation.


In light of this,
my opinion is that $\mathrm{SL}\_n(A)$,
where $A$
is any structure that makes sense as matrix entries so that we have
multiplication and a concrete notion of determinant,
should be the subgroup of $\mathrm{GL}\_n(A)$
having determinant $1$,
with no additional conditions (as convenient as they might seem).
Then $\mathrm{PSL}\_2(A)=\mathrm{SL}\_2(A)/Z(\mathrm{SL}\_2(A))$.


This aligns with option 1 of my previous "answer." This way we can introduce new notation for the Vahlen matrices (I like $\mathcal{V}\_n$),
and still have $\mathrm{PSL}\_2(C\_n)$
as a separate thing.
The two happen to be the same in those first 2 example,
but thereafter it is useful in this context to have notation for both of those things separately,
since $\mathcal{V}\_n\leq\mathrm{PSL}\_2(C\_n)$,
and sometimes $\mathcal{V}\_n\leq\mathrm{PSL}\_2(C\_m)$
with $n\neq m$.


