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?

  • 1
    $\begingroup$ I have not thought everything through, but it seems that this holds only in low dimensions. If you tensor with $\mathbb{C}$, then the Clifford algebra $C^*$ is just the matrix algebra (or two copies of it). Hence the units form a group of type $A$. Same for $SL_2(C^*)$, which is a group of type $A$. This it cannot be of type $B$ or $D$ (except in low dimensions) $\endgroup$ Feb 4, 2017 at 12:13
  • $\begingroup$ It seems to me that your dimension indices are wrong. According to Wikipedia, the Clifford group for $(\mathbb{R}^n, -\|x\|^2)$ gives the Möbius transformations over $\mathbb{R}^n$. In your definition of Clifford group I think you have off by one error. Also, Clifford group is certainly not the same object as the Clifford algebra, so writing $PSL_2(C_2) \simeq PSL_2(\mathbb{H})$ seems suspicious. $\endgroup$ Feb 4, 2017 at 21:18
  • $\begingroup$ @Venkataramana I'm no expert but the literature suggest that Vahlen matrices do correspond to Möbius transformations. But there is some extra condition on the entries. $\endgroup$ Feb 4, 2017 at 21:20
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    $\begingroup$ @j0equ1nn If $C_0 = \mathbb{R}$ then it shouldn't act on elements of $\mathbb{R}^0$, should it? You are right except the Clifford group does not contain zero. But this minute detail shouldn't matter in these issues. $\endgroup$ Feb 5, 2017 at 22:48
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    $\begingroup$ @j0equ1nn Yeah, on a second look it seems that it is me who's off by one. Sorry. $\endgroup$ Feb 7, 2017 at 11:30

2 Answers 2


After reading a bunch, I've noticed there is much disagreement about how to define $\mathrm{PSL}_2(\mathbb{H})$.

The conditions one might use to define this are best expressed in terms of the Clifford algebra $\mathscr{C}_2$. But it is no more complicated to discuss $\mathrm{PSL}_2(C_n)$ in terms of $\mathscr{C}_n$, so I'll do that.

The involution $\alpha$ discussed in the post is the one induced by negating the algebra's generators, i.e. $e_\ell^*=-e_\ell$ (so that its action on a product $e_{\ell_1}\dots e_{\ell_m}$ of distinct $e_{\ell_k}$ depends only on the parity of $m$). The definition of the Clifford group $C_n$ in the post is actually redundant because the condition "$c\in\mathscr{C}_n(\mathbb{R})^\times$" is equivalent to the condition "$\forall v\in\mathbb{R}^n: cv\alpha(c)^{-1}\in\mathbb{R}^n$."

Now consider the group $\mathrm{GL}_2(C_n)$, which is the group of $2\times 2$ matrices with entries in $C_n$ that have both a left and right inverse. Define on this group the map $$\Delta:\mathrm{GL}_2(C_n)\rightarrow\mathscr{C}_n\smallsetminus\{0\},\quad \begin{pmatrix} a & b\\ c & d \end{pmatrix}\mapsto ad^*-bc^*.$$ When $n=0$ or $1$, this is equivalent to the usual determinant. When $n=2$, this is equivalent to the Dieudonné determinant.

  1. Some authors define $\mathrm{SL}_2(\mathbb{H})$ as the subgroup of $\mathrm{GL}_2(\mathbb{H})$ of matrices $m$ where $\Delta(m)=1$, then define $\mathrm{PSL}_2(\mathbb{H})$ as the quotient of that by $\{\pm I\}$. This group is isomorphic to $\mathrm{Isom}^+(\mathfrak{H}^5)$.

  2. Other authors define $\mathrm{SL}_2(\mathbb{H})$ as the subgroup of $\mathrm{GL}_2(\mathbb{H})$ of matrices $m=\begin{pmatrix} a & b\\ c & d \end{pmatrix}$ where $\Delta(m)=1$, and $ab^*,cd^*\in\mathbb{R}^n$, where $\mathbb{R}^n$ is identified with the span of $\{1,e_1,\dots,e_{n-1}\}$ over $\mathbb{R}$, within $\mathscr{C}_n$. (This still generalizes the determinant over $\mathbb{R}$ and $\mathbb{C}$ because the additional condition is trivial there.) They then define $\mathrm{PSL}_2(\mathbb{H})$ is the quotient of that by $\{\pm I\}$. This group is isomorphic to $\mathrm{Isom}^+(\mathfrak{H}^{4})$. More generally, if we use the same conditions to define $\mathrm{PSL}_2(C_n)$, it will be isomorphic to $\mathrm{Isom}^+(\mathfrak{H}^{n+2})$.

The generalization to $C_n$ is a point for the second definition. But on the other hand, the second definition makes it difficult to decide how to write the group defined in the first definition, which remains relevant (e.g. the first definition's $\mathrm{PSL}_2(\mathbb{H})$ is isomorphic to the second definition's $\mathrm{PSL}_2(\mathbb{H}^2)$).

This may answer my original question, but as a researcher working on a paper that utilizes these objects it puts me in an awkward position as far as which convention to adopt. I've been taking a tally among papers I'm reading that use these constructions, and so far definition 1 is winning but the score is close: 4 to 3. There is even a paper with each choice by the same author. Then there are 2 that use $\mathrm{PSL}_+(C_n)$ for the second definition and nothing for the first (one of those is by Ahlfors who remarks, right after introducing the notation, that it is inadequate). And there is one that uses $\mathrm{P}\mathbb{G}\mathrm{L}_2(C_n)$ for the first definition and nothing for the second.

So I'd like to do something unconventional for Mathoverflow, which is the following. Rather than accept my own answer, I'd like to pose the follow-up question: which definition is better? Then I will accept whatever answer makes a solid case for choosing one over the other. Some might object to the subjective nature of that question, but I feel that it should have a rigorous mathematical answer rooted in the motivation for how $\mathrm{PSL}_2$ is defined. Hopefully this is agreeable to the moderators.


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 $n\neq m$.

  • $\begingroup$ Anyone interested in this with a better answer, feel free to post, I'm totally open to accepting a different one. $\endgroup$
    – j0equ1nn
    Feb 13, 2017 at 23:10

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