# Quaternion–Spinor relationship?

I've known for some time about the rotation group action of the ('pure') quaternions on $\mathbf{R}^3$ by conjugation. I've recently encountered spinors and notice similarities in their definitions (for example, the use of half-angles for rotations).

Is the relationship that this suggested in my mind a real one, and if so what is its formal nature? Are the spaces isomorphic? If not, is there any relationship at all?

• “Rotation group action on the ('pure') quaternions”, you meant? In short, this unit quaternions’ action is the same thing as Spin(3) group on vectors, whereas (one) quaternion multiplication is the action on spinors from ℂ² (weight ½ representation). Nov 1, 2014 at 19:14

Some background in Lie theory first. To any Lie group $G$ one may associate its lie algebra $\mathfrak{g}$. Among all isomorphism classes of Lie groups $G$ with a given lie algebra $\mathfrak{g}$, there is a universal one. It is the universal covering group (up to isomorphism) of all of the others. That is, if $H$ is any other Lie group with lie algebra $\mathfrak{g}$, then there is a covering map $G\to H$, which means every Lie group $H$ having lie algebra $\mathfrak{g}$ is a quotient of $G$ by a discrete central subgroup $K$. In fact, covering space theory goes on to tell us that there is a canonical isomorphism $\pi_1(G/K,e)\cong K$.

Note the fundamental group $\pi_1(X,x)$ of a pointed space $(X,x)$ (i.e. a topological space $X$ with a distinguished point $x$) is the set of all homotopy classes of loops based at $x$ with concatenation of paths (which is well-defined on homotopy classes) as a group operation. In other words, $\pi_1(X,x)$ encodes the essentially different "ways of going around $X$" starting and ending at $x$. ("Homotopy" essentially means continuously varying a continuous map, in other words "wiggling" it around, or perturbing it.)

There is in fact a way of explicitly constructing the universal cover $G$ of a group $H$. First, one forms the path group comprised of paths in $H$ emanating from $\mathrm{id}$ (not ending anywhere in particular), then identifies paths which are equivalent by an endpoint-preserving homotopy. One may multiply these paths pointwise, which is well-defined on homotopy classes again. This is $G$. There is an obvious homomorphism $G\to H$ which just forgets the path and keeps its endpoint. To topologize $G$ correctly, simply take sufficiently small connected neighborhoods in $H$ and lift them to connected neighborhoods in $G$.

One may prove that $\pi_1(\mathrm{SO}(3),\mathrm{id})\cong\mathbb{Z}_2$. The image of any one-parameter subgroup, which is comprised of all rotations around a fixed axis, yields a representative of the nontrivial element of the fundamental group. (Of course $\mathrm{SO}(2)\simeq\mathbb{S}^1$ is topologically a circle, which has the bigger fundamental group $\pi_1(\mathbb{S}^1)\cong\mathbb{Z}$, each element encoding how many times one goes around the circle and in what direction.) These distinct loops are homotopy equivalent by simply moving the axis of rotation around.

Given any loop in $\mathrm{SO}(n)$ based at $\mathrm{id}$, the last column traces out a loop in $\mathbb{S}^{n-1}$, which is a simply connected space and thus the loop in the sphere is contractible. Such a contraction may be lifted up to $\mathrm{SO}(n)$ which pushes the loop into the subspace $\mathrm{SO}(n-1)$. Therefore, by induction, they all have fundamental groups $\pi_1(\mathrm{SO}(n))\cong\mathbb{Z}_2$ for $n>2$. One can also frame this argument in terms of the homotopy long exact sequence by using the orbit-stabilizer theorem, which says since $\mathrm{SO}(n)$ acts transitively on the sphere $\mathbb{S}^{n-1}$ with point-stabilizer $\mathrm{SO}(n-1)$, there is therefore a fiber bundle $\mathrm{SO}(n-1)\to\mathrm{SO}(n)\to\mathbb{S}^{n-1}$.

In dimension $n>2$, one may define $\mathrm{Spin}(n)$ to be the universal cover, or equivalently the double cover since $\pi_1(\mathrm{SO}(n),\mathrm{id})\cong\mathbb{Z}_2$, of the rotation group $\mathrm{SO}(n)$. The elements of $\mathrm{Spin}(n)$ don't really have a name that I'm aware of ("spinors" are something different, I'll get to them). I've decided to just call them "spins." Here's my interpretation of them:

Intuitively when we think of a rotation we picture a continuous motion. But an individual rotation, or indeed any particular transformation of space, only has a "before" and "after," no in between. In order to animate a rotation continuously, we need a continuous one-parameter family of rotations, in other words a path in $\mathrm{SO}(n)$. Since, by the explicit construction of the universal covering group, $\mathrm{Spin}(n)$'s elements can be interpreted as such paths in $\mathrm{SO}(n)$, we can think of spins as rotations of space but with a memory of how the rotation is performed.

By the universal property of the universal covering group, any $2$-to-$1$ group homomorphism $G\to\mathrm{SO}(n)$ (with $G$ connected) is equivalent to an isomorphism $G\cong\mathrm{Spin}(n)$. Therefore, to find spin groups in the wild, it suffices to find connected double covers of $\mathrm{SO}(n)$. In low dimensions, there are "accidental isomorphisms" between spin groups and classical matrix Lie groups (constructed from composition algebras $\mathbb{R},\mathbb{C},\mathbb{H},\mathbb{O}$). There's a lot written on it, but for some starting points I have some links in the next paragraph.

Over here I detail how the group of unit quaternions, called $\mathrm{Sp}(1)$, is a double cover of $\mathrm{SO}(3)$, and so $\mathrm{Spin}(3)\cong\mathrm{Sp}(1)$. Over here I explain the origin of the three spin groups $\mathrm{SO}(2)\cong\mathrm{Spin}(2)$, $\mathrm{SU}(2)\cong\mathrm{Spin}(3)$, and $\mathrm{Sp}(2)\cong\mathrm{Spin}(5)$ from projective geometry. (I haven't actually explained what $\mathrm{Spin}(2)$ is yet.) There is another question here which begins with a list of low-dimensional spin groups. If you allow spaces equipped with a nondegenerate bilinear form which is not positive-definite, and instead has mixed signature, then you get corresponding "rotation" groups $\mathrm{SO}(p,q)$ and spin groups $\mathrm{Spin}(p,q)$. In this generality, this answer on MO exhibits a complete roadmap of low-dimensional spin groups. Also see linked in the comments, Paul Garrett's vignette on spin groups. For generalization to octonions, see e.g. this paper, in particular $\mathrm{SU}(2,\mathbb{O})\cong\mathrm{Spin}(9)$. John Baez's The Octonions goes further and explains (among many things) the mixed signature spin groups $\mathrm{SL}(2,\mathbb{R})\cong\mathrm{Spin}(2,1)$, $\mathrm{SL}(2,\mathbb{C})\cong\mathrm{Spin}(3,1)$, $\mathrm{SL}(2,\mathbb{H})\cong\mathrm{Spin}(5,1)$ and octonionic one $\mathrm{SL}(2,\mathbb{O})\cong\mathbb{Spin}(9,1)$, which are related to spacetime in physics.

Note that the quaternions $\mathbb{H}$ are a two-dimensional complex vector space (with scalars multiplied from the right), and multiplying by a unit quaternion on the left is a special unitary map: this identification yields an isomorphism $\mathrm{Sp}(1)\cong\mathrm{SU}(2)$. Many people sometimes even call $\mathrm{SU}(2)$ the group of unit quaternions, but I'm not a fan of this practice. It's much the same as $\mathrm{U}(1)\cong\mathrm{SO}(2)$: the complex plane $\mathbb{C}$ as a real vector space is two-dimensional, and multiplying by a unit complex number is a special orthogonal map.

Now, I haven't said what $\mathrm{Spin}(1)$ or $\mathrm{Spin}(2)$ is. They are not universal covers, in fact the first has two connected components and the second is only a double cover even though $\pi_1(\mathrm{SO}(2),\mathrm{id})\cong\mathbb{Z}$ (so the spin group is not a universal cover). What, then, is the definition of a spin group? The answer lies in Clifford algebras.

The Cartan-Dieudonne theorem implies that reflections generate $\mathrm{O}(n)$. (A nice inductive proof is given in Stillwell's Lie theory; I don't feel like reproducing it or searching for it on the internet at the moment.) Reflections correspond to $(n-1)$-dimensional hyperplanes which in turn correspond to pairs of unit vectors (orthogonal to said hyperplanes). Taking a cue from semidirect products, we could hope to construct a "multiplication" operation on vectors in which $uxu^{-1}$ would be reflection across $u^\perp$. The problem with this is that $uuu^{-1}$ would be $-u$, which doesn't make sense. The fix is to consider the negative of reflection across $u^\perp$, so $uxu^{-1}=-x$ when $x$ is orthogonal to $u$. More generally, writing $x=x_{\|}+x_\perp$ as parallel and perpendicular components with respect to $u$, we have $uxu^{-1}=x_{\|}-x_\perp$. If we stipulate $u^2=-1$ for all unit vectors $u$ (the alternative would be $u^2=1$, which gives a geometric algebra), this is equivalent to imposing the relation $uv+vu=-2\langle u,v\rangle$ for all $u,v\in V$.

Therefore, if $V$ is an inner product space, we define $\mathrm{Cliff}(V)$ to be the tensor algebra $TV$ modulo the relations $uv+vu=-2\langle u,v\rangle$. If $e_1,\cdots,e_n$ is an orthonormal basis, we could interpret $\mathrm{Cliff}(n)$ as the free associative algebra generated by anticommuting square roots of $-1$ labelled $e_1,\cdots,e_n$. As vector spaces, $\mathrm{Cliff}(V)\cong\Lambda V$ is isomorphic to the exterior algebera. Then $\mathrm{Pin}(V)$ is the group generated by unit vectors, and $\mathrm{Spin}(V)$ is the group generated by products of evenly-many unit vectors. As $V$ sits inside $\mathrm{Cliff}(V)$, one may conjugate vectors by pins or spins and they act by orthogonal transformations or rotations respectively, giving $2$-to-$1$ homomorphisms $\mathrm{Pin}(V)\to\mathrm{O}(V)$ and $\mathrm{Spin}(V)\to\mathrm{SO}(V)$.

On a lie algebra level, $\mathfrak{so}(n)=\mathfrak{spin}(n)$ are essentially the same. However if $n\equiv 3,7$ mod $8$ there are representations of $\mathfrak{so}(n)$ that do not lift to representations of $\mathrm{SO}(n)$, instead they lift to representations of $\mathrm{Spin}(n)$. These are called spin representations $\mathrm{Spin}(n)\to\mathrm{SO}(W)$, and the elements of the underlying vector spaces $W$ are valled spinors. The induced action on the projective space $\mathbb{P}(W)$ factors through a projective representation $\mathrm{SO}(n)\to\mathrm{Aut}(\mathbb{P}(W))$, so sometimes these are taken into consideration.

The spinors with $n=3$ and $n=7$ come from quaternions and octonions respectively. I'll explain the case with $n=7$ because the case for $n=3$ is basically the same. Let $\mathbb{O}$ be the octonions. While not associative, it is alternative, so in particular if $L_a(x)=ax$ denotes the left multiplication maps then $L_a^2=-\|a\|^2$ whenever $a\in\mathrm{Im}(\mathbb{O})$ is pure imaginary. As a result, there is an induced action of $\mathrm{Cliff}(\mathrm{Im}(\mathbb{O}))$ on $\mathbb{O}$, where $(a_1\cdot a_2\cdots\cdot a_k)x=a_1(a_2(\cdots(a_k x)\cdots))$. Restricting to spins, this gives a spin representation $\mathrm{Spin}(7)\hookrightarrow\mathrm{SO}(8)$.

I could talk about a lot more things: positive versus negative spinors, how they relate to the classification of Clifford algebras and Bott periodicity, what the connected components of $\mathrm{O}(p,q)$ are and what the "rotations" in $\mathrm{SO}(n,1)$ look like, the "naturality" of spin representations and how they correspond to binary cross products, and the triality symmetry of $\mathrm{Spin}(8)$. But it looks like I've already written too much at this point.

• What a heroic answer! Thank you - sorry it's taken me so long to give it its proper appreciation: I took a bit of a break from StackExchange and other similar web platforms. I'll work through your answer over the next few days - it seems richly detailed and I expect a bit of work spent understanding it to pay dividends. Thanks again.
– JCW
Oct 11, 2016 at 23:09
• +1000! This is a great answer - I learnt a lot. Sep 9, 2018 at 18:10
• I wish there is a dictionary that contains all isomorphism relations for G(p,q,K), where K=R,C,H,O and G=(S)O, (S)U, Sp. Sep 16, 2019 at 15:20
• Great answer! But now I am curious about the condition n=3,7 mod 8. Where does it come from? Jul 10, 2021 at 1:47

Yes, quaternions are related to spinous in three dimensions. You might first try the subsection of the spinor wiki talking about quaternions where it mentions

Thus the (real) spinors in three-dimensions are quaternions, and the action of an even-graded element on a spinor is given by ordinary quaternionic multiplication.

More generally, spinors of higher dimension are related to Clifford algebras. The quaternions can be considered a special case of a Clifford algebra.

• Thanks for the prompt response. I've already read that article; I didn't think it answered my question, as there it only discusses the action of quaternions on spinors. Is there a subtlety I'm missing?
– JCW
Oct 24, 2014 at 11:55
• @JCW Actually I had meant to include the link that led me to that page too. I added it also. Oct 24, 2014 at 12:32
• @JCW There are no doubt hundreds of web pages explaining the connection, but I know that many of them are written by physicists, and hence long, involved and example-centric. I bet there's a terse algebraic bundling, but unfortunately I haven't internalized spinors enough yet to say what it is. I used to think they were related to simple modules of Clifford algebras, but I never learned the details well. Oct 24, 2014 at 12:38
• Thanks for the addition. I'm not very familiar with the Clifford algebra connection, though I know the basics. I'll explore that further.
– JCW
Oct 24, 2014 at 12:56

The spin group is the double cover of the orthogonal group. In dimension three, the spinors are $SU(2)$, which is the double cover of $SO(3)$. It can be identified with the unit quaternions, and is homeomorphic to a three-sphere. ($SO(3)$ is homeomorphic to the three-dimensional real projective space.)

• Dear Neal: Quick question: Of course there are many important topological projective spaces, but I'm wondering about the phrase "topologically a projective space." It seems to suggest that the projectiveness of the space is characterized by its topology, which I don't think is what you meant. Or is there actually a way to characterize topological spaces as projective based on their topology? Regards Oct 24, 2014 at 12:40
• Thanks for this, it looks promising. I have a couple of questions: why is it that spinors are $SU(2)$? Certainly they transform (perhaps tautologously, 'like spinors') under matrix multiplication by $SU(2) \cong \mathrm{Spin}(3)$, but I don't quite see how they are $SU(2)$ themselves. Could you elaborate?
– JCW
Oct 24, 2014 at 12:51
• Dear @rschwieb: What I meant was something more like, "As a topological space, $SO(3)$ is a projective space" or "$SO(3)$ is homeomorphic to $\mathbb{R}P^3$." I do not know of a way to characterize topological spaces as projective based on their topologies.
– Neal
Oct 24, 2014 at 12:52
• @JCW, my understanding is this is an accident of low dimension: in general, the spin group is not isomorphic to the special unitary group. However, at the moment i do not have time to double check this.
– Neal
Oct 24, 2014 at 13:02
• @Neal OK: I know what you mean then. Thanks for the comment. Oct 24, 2014 at 13:33