**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?

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