# In the quaternions, “any imaginary unit may be called i”

## Introduction

Suppose we are trying to prove that $\rm PSO_3\times PSO_3$ is isomorphic with $\rm PSO_4,$ and we catch on to the idea of using the quaternions to do so. We realize (as in Conway & Smith's On Quaternions and Octonions, whence the quotation) that we can encapsulate $\rm PSO_3$ as the set of all maps $x\mapsto\bar qxq$ for a unit quaternion $q$ operating on imaginary quaternions $x,$ and go on to trying to understand why $\rm PSO_4$ is the set of all maps $\pm(x\mapsto\bar lxr)$ for unit quaternions $l,r$ operating on quaternions $x.$

To show that all maps $\pm(x\mapsto\bar lxr)$ really are elements of $\rm PSO_4,$ we begin by showing this in the case $\bar l=\cos\theta+isin\theta,r=1.$ This is simple, since this operation rotates the plane spanned by $1$ and $i$ through an angle of $\theta,$ and rotates the plane spanned by $j$ and $k$ through an angle of $\theta$ at the same time.

But at this point, we really are done proving that all maps $\pm(x\mapsto\bar lxr)$ are elements of $\rm PSO_4,$ since, as in the book, "any imaginary unit may be called $i,$ and perpendicular one $j,$ and their product $k$" (although this was said at a different point), right-multiplication has the same geometric properties as left-multiplication, and the composition of any two elements of $\rm PSO_4$ is an element of $\rm PSO_4.$

## Idea

It makes all the intuitive sense in the world to me that "any imaginary unit may be called $i,$" and I can really visualize this geometrically. Furthermore, I could go back through a proof, change all the $i$'s to $u$'s for an arbitrary imaginary unit quaternion $u,$ etc. But suppose I had a collection of proofs written by someone who wasn't very careful, and she/he had used $i, j,$ and $k$ for simplicity and computed examples, stating at the end of each one that it generalizes to all quaternions. Suppose I pored through these proofs and discovered that about a third of them were careless to the point of being false, because of some lack of care in going back/forth between general ($u,v,w$) and specific ($i,j,k$) contexts. To formalize this imaginary formalization attempt, suppose I had a computer that understood category theory really well and wanted to scan these proofs in for it and get it to check whether this person's proofs really proved whatever facts from geometry they purported to prove.

## Question

In the specific example, the notion that any imaginary unit may be called $i$ and left-multiplication is like right-multiplication can be dealt with at a first approximation using the notion of automorphisms. There is a ring automorphism sending $u$ to $i$ and vice versa, and there is a group automorphism between the multiplicative group of the quaternions and its opposite group. But I wonder if it follows directly enough that geometric facts can be proved "by example."

Is there a category-theoretic context in which certain ring/group automorphisms are natural and in which their being natural is biconditional with their preserving geometric properties?

To explain why the notion of an automorphism by itself might not be enough, we can imagine $\mathbb{Q}[\sqrt{2}]$ acting on $\mathbb{R}$ by multiplication. Multiplication by $\sqrt{2}$ preserves order, but multiplication by $-\sqrt{2}$ reverses it, so in the context of orderings, it would not be accurate to say "any square root of 2 may be called $\sqrt{2}.$"

What I'm envisioning is a category $\mathcal{C}$ with an object $\mathbb{H},$ as well as a morphism for each automorphism of $\mathbb{H},$ an object $\mathbb{H}^\star,$ and a morphism for each group automorphism of $\mathbb{H}^\star,$ perhaps geometric objects or morphisms as well, and whenever we speak of "an instance of the quaternions" we are really speaking of a functor $\mathcal{C}\to\mathcal{C}$ at least one existing for each possibility of an imaginary unit being called $i,$ a perpendicular one $j,$ and their product $k.$ I know this doesn't work out as stated, because the identity morphism doesn't go to the identity morphism, but perhaps there's a way to fix that. Then an automorphism of the quaternions can be viewed as a natural transformation between functors $\mathcal{C}\to\mathcal{C},$ one preserving the hidden (say, setwise) structure of the quaternions, and another preserving their apparent ($i,j,k$) structure and for some reason, because it's natural, the geometry comes out alright.

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Do you have an use for your "vision"? If what you want is simply to know whether "all units are created equal", then you can simply prove that the automorphism group of the algebra acts transitively on the set of elements which square to $-1$. –  Mariano Suárez-Alvarez Jan 30 '11 at 22:34
Is that true that the 2 lines of Mariano's comments are equivalent to the 40+ lines of the question? Or did I miss something important? –  Mark Sapir Jan 30 '11 at 23:39
It's a standard (easy) fact that the quaternions that square to −1 form a 2-sphere in the purely imaginary quaternions. And automorphisms act transitively on that sphere... –  Qfwfq Jan 30 '11 at 23:39
Also, why are you using $PSO_3$, $PSO_4$ and not just $SO_3$, $SO_4$ ? –  Qfwfq Jan 31 '11 at 12:52
Dear unknowngoogle, it's not the case case that $SO(3) \times SO(3)$ is isomorphic to $SO(4)$; there is a central isogeny from the latter to the former, with kernel of order 2. (One has a succession of degree 2 central isogenies $SU(2) \times SU(2) \to SO(4) \to SO(3) \times SO(3)$. The latter isogeny induces the isomorphism $PSO(4) \to PSO(3) \times PSO(3) = SO(3) \times SO(3)$ which the OP is referring to.) –  Emerton Jan 31 '11 at 14:23

Just to elaborate on what is already in the comments, the algebra automorphisms of $\mathbb H$ act transitively on the set of pairs $(u,v)$ where $u$ and $v$ are imaginary quaternions of unit length that are orthogonal to one another.

To see this, I will include here some remarks on $\mathbb H$ and its automorphisms. Part of the OP's concern seems to be that it is not a priori automatic that metric concepts in $\mathbb H$ such as unit length or orthogonality (and hence the notion of being imaginary, since the imaginary quaternions are the orthogonal complement to $\mathbb R$ in $\mathbb H$) are preserved by Aut$(\mathbb H)$, and so one of my goals is to show that this concern is not necessary. Indeed, this geometry is intrinsic to the quaternions, as we will see. (This is not coincidence: Hamilton was led to his discovery by trying to algebraize the geometry of $\mathbb R^3$.)

Note first that imaginary quaternions are characterized by the condition that $\overline{u} = - u$, and thus for such quaternions, $|u|^2 = -u^2$. Thus if $u$ is imaginary, $u^2$ is a non-positive real number. Converesly, if $u^2$ is a negative real number, then one sees that $u$ is imaginary (exercise), and so the imaginary quaternions are also characterized by having non-positive real squares. In particular, the set of imaginary quaternions is preserved by Aut$(\mathbb H)$.

On imaginary quaternions, the inner product $u\overline{v} + v\overline{u}$ is simply $u v + u v$, and so is also preserved by Aut$(\mathbb H)$. In particular, metric concepts like "length one" and "orthogonal" are preserved by Aut$(\mathbb H)$.

If $u$ and $v$ are unit length orthogonal imaginary quaternions, we then have that $u^2 = v^2 = -1$ (unit length condition) and that $u v = - v u$ (orthogonality condition). Thus, from the defining relations of $\mathbb H$, we obtain an algebra map $\mathbb H \to \mathbb H$ that maps $i$ to $u$ and $j$ to $v$ (and then $k$ to $u v$). This map is non-zero (since $u$ and $v$ are non-zero, having unit length), and hence is necessarily injective ($\mathbb H$ is a division ring, hence has no non-trivial ideals), and thus in fact bijective (source and target are of the same dimension).

An automorphism of $\mathbb H$ is determined by its values on $i$ and $j$ (since they generate $\mathbb H$), and so the previous discussion shows that in fact Aut$(\mathbb H)$ is the same as the group or permutations of pairs $(u,v)$ of orthogonal pairs of unit vectors in the imaginary quaternions (also known as $\mathbb R^3$).

This group is well-known: it is precisely $SO(3)$. (If you like, $u$ and $v$ determine uniquely a mutually orthogonal vector --- their quaternionic product $u v$ --- which can be characterized geometrically in terms of $u$ and $v$ via the right hand rule; thus pairs $(u,v)$ are the same as positively oriented orthonormal bases of $\mathbb R^3$, permutations of which are precisely the group $SO(3)$.)

Incidentally, it is not coincidence that Aut$(\mathbb H) = SO(3)$.

Namely, there is a natural map $\mathbb H^{\times} \to$ Aut$(\mathbb H)$ (where $\mathbb H^{\times}$ means the non-zero --- equivalently invertible --- quaternions), given by mapping $q$ to the automorphism $x \mapsto q x q^{-1}$. The kernel of this map is precisely the centre, and so it induces an injection $\mathbb H^{\times}/\mathbb R^{\times} \hookrightarrow$ Aut$(\mathbb H)$.

Now the source of this map can be identified with the quotient of the unit quaternions (which form a copy of $SU(2)$) by $\pm 1$, and of course $SU(2)/\{\pm 1\} = SO(3)$. On the other hand, this injection is in fact a bijection (i.e. any automorphism of $\mathbb H$ is inner), by the Skolem--Noether theorem. This puts the description of Aut$(\mathbb H)$ obtained above into a more general perspective.

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Concerning your third paragraph: i think you can just say that 1) the automorphisms of $\mathbb{H}$ (as an $\mathbb{R}$-algebra) preserve $1$, hence they preserve $\mathbb{R}$, 2) the conjugation on $\mathbb{H}$ is an automorphism, 3) the scalar product on $\mathbb{H}$ is defined in terms of conjugation hence the orthogonality condition is invariant under automorphisms, 4) imaginary quaternions are just $\mathbb{R}^{\perp}$. Conclusion: $\mathrm{Aut}(\mathbb{H})$ preserves imaginary quaternions. –  Qfwfq Jan 31 '11 at 13:03
Dear unkonwngoogle, I'm probably being dense, but I don't follow your argument. For example, conjugation is not an automorphism, but an anti-automorphism (i.e. it switches the order of multiplication). And even given this, is it a priori clear that this anti-automorphism commutes with all automorphisms? This is true in the end, but I don't see how to prove it without making some argument (although it's likely that what I wrote is not the most elegant argument --- it's just what came to mind first!). Best wishes, Matthew –  Emerton Jan 31 '11 at 14:21
Dear Emerton, you're totally right! I've been too sloppy in it. –  Qfwfq Feb 1 '11 at 18:34