# The Quaternions and $SO(4)$

I am interested in the map $\phi:S^3 \times S^3 \to GL_4(\mathbb{R})$ given as follows:

Let $(p,q) \in S^3 \times S^3$. We identify $p$ and $q$ as real quaternions with unit norms and define $\phi(p,q)$ to be the map sending $v \in \mathbb{H}$ to the product $pvq^{-1}$, where $\mathbb{R}^4$ and $\mathbb{H}$ are identified in the obvious way as $\mathbb{R}$-vector spaces. It isn't hard to show that $\phi$ is a group homomorphism.

I would like to prove the following in complete detail:

1) The image of $\phi$ is $SO(4)$.
2) The kernel of $\phi$ is $\{(1,1), (-1, -1)\}$

I proceed by noting that the maps $\phi(p,q)$ are clearly $\mathbb{R}$-linear, and since $p$ and $q$ have unit norm, these maps are also easily seen to be orthogonal. Thus, we have $\phi(S^3 \times S^3) \subseteq O(4)$. Now suppose $\phi(p,q)$ is the identity map on $\mathbb{R}^4$. That is, $pvq^{-1}=v\ \ \ \forall v \in \mathbb{H}$. Then for $v=1 \in \mathbb{H}$, we have $pq^{-1}=1$, and so $p=q$. By considering the equations $piq^{-1}=i$, $pjq^{-1}=j$, and $pkq^{-1}=k$, we conclude that $p=q \in \mathbb{R}$. Since $\mathbb{R} \cap S^3 = \{\pm1\}$, an easy compuation shows that the kernel of $\phi$ is $\{(1,1), (-1, -1)\} \cong \mathbb{Z}_2$. I am not very confident about the remainder of this argument.

The map $\phi$ is smooth since the entries of the matrices $\phi(p,q)$ are polynomials in the $\{1, i, j, k\}$-coordinates of $p$ and $q$ realized as quaternions. Since $S^3 \times S^3$ is connected, we know that $\phi(S^3 \times S^3)$ is contained in the identity component of $O(4)$. That is, $\phi(S^3 \times S^3) \subseteq SO(4)$ (We could have also checked the determinant explicitly, but this is a bit tedious). Since the kernel acts freely and properly on $S^3 \times S^3$, we see that the image of $\phi$ is a smooth six-dimensional manifold. We can count the dimension of $SO(4)$ by considering its Lie algebra of skew-symmetric matrices. It is then easy to see that the dimension of $SO(4)$ is six. It follows that the image is an open submanfiold of $SO(4)$.

Note also that $S^3 \times S^3$ is compact, and so is its image. Since $SO(4)$ is Hausdorff, it follows that the image of $\phi$ is also closed in $SO(4)$. Since $SO(4)$ is connected, it follows that the image of $\phi$ is all of $SO(4)$.

My questions are as follows:
1) Is there anything wrong with this argument? Even picky corrections are much appreciated.
2) Can you identify any arguments that, while correct, you would prefer to do another way?
3) What more can be said about this map? Is there a common sense reason that these left and right quaternion multiplication actions produce all of $SO(4)$? Can we rephrase this argument in another language which might be more natural? This question is a little bit open ended, but only because I am not sure how to phrase it. The best answer would give me some clue about where this map fits into the general picture, if possible. Perhaps some application of the covering map would be appropriate here.

Thanks in advance for any responses.

• If I'm not mistaken, your map is precisely the construction of the double cover $\operatorname{Spin}(4) \cong SU(2) \times SU(2)$ of $SO(4)$, where $\mathbb{H} \supset S^3 \cong SU(2)$ via the isomorphism $$(w,z) \mapsto \begin{pmatrix} w & -\overline{z}\\ z & \overline{w} \end{pmatrix}.$$ In any event, it's an extremely well-known construction in spin geometry. – Branimir Ćaćić Aug 4 '13 at 19:04
• Thank you, that is interesting. I haven't yet studied spin geometry, but this is just the sort of link I was looking for. – James Staff Aug 4 '13 at 21:33
• In fact, your construction really is precisely the $n=4$ case, in extremely mild disguise, of the general Clifford-algebraic construction of the double cover $\operatorname{Spin}(n)$ of $SO(n)$; see, for instance, the beginning of Lawson and Michelson's Spin Geometry or Friedrich's Dirac Operators in Riemannian Geometry. – Branimir Ćaćić Aug 4 '13 at 21:50
• Thanks, this is just the sort of thing I was grasping for in question three above. – James Staff Aug 4 '13 at 22:18
• This also appears as Proposition 8.27 in Porteous (1995). – Mike Sep 1 '18 at 23:53

One important thing that comes out of this argument is that $SO(4)$ almost factors as a direct product, namely its universal (double) cover Spin$(4)$ factors as the direct product of Spin$(3)$ (which is $S^3$, or equivalently $SU(2)$) with itself.
So regarding question (3), in some sense your are asking whether there is a way to see that $SO(4)$ should factor in this way. One way to see it is to consider the Dynkin diagram for the Lie algebra $\mathfrak{so}_4$; it is the Dynkin diagram $D_2$, which is the disjoint union of two points; this shows that $\mathfrak{so}_4$ factors as the direct product of two copies of $\mathfrak{so_3}$, which is the Lie algebra analogue of the direct product factorization of the spin groups.
For larger $n$ there is no such factoriation: the Lie algebra $\mathfrak{so}_n$ is simple if $n > 4$, and the associated Dynkin diagram is connected.
What comes next is not really an answer to your question, but may be of interest: the fact that $SO(4)$ (almost) factors, while other $SO(n)$ don't, is one of the reasons that $4$-dimensional geometry has such a different flavour (e.g. Donaldson's theory) compared to geometry in other dimensions.