As I understand it, quaternions are a type of object called a spinor. Spinors are objects that are negated under a full rotation and only return to their original state under two full rotations. But what does that mean in the case of quaternions?

A quaternion is a number $p=a+bi+cj+dk$ where $i^2=j^2=k^2=-1=ijk$. So it's just a $4$-d version of a complex number. And just like a complex number, we can write it in polar form as $p=|p|e^{\hat n\theta} = |p|(\cos(\theta)+\hat n\sin(\theta))$ where $\hat n$ is a unit imaginary vector that's not necessary $i$, $j$, or $k$ (but some combination of them).

So to rotate a quaternion around a full circle wouldn't you just add $2\pi$ to the angle of the polar representation? Then we'd just get $$p_{rot}=|p|e^{\hat n(\theta+2\pi)} = |p|(\cos(\theta+2\pi)+\hat n\sin(\theta+2\pi)) = |p|(\cos(\theta)+\hat n\sin(\theta)) = p$$

I guess I'm just not understanding what "under a full rotation" means because to me it seems like rotating a quaternion once about the origin (by adding $2\pi$ to its angle) just returns the quaternion to itself. So can someone explain this to me?

  • $\begingroup$ This is imprecise. The precise statement is that there is a group $SU(2) \cong Sp(1)$, one description of which is the quaternions of unit norm, which double covers the special orthogonal group $SO(3)$. One can attempt to lift a closed path corresponding to a full rotation in $SO(3)$ to $SU(2)$, and the result is no longer a closed path; it's now a path to $-1$. $\endgroup$ Commented Jan 16, 2016 at 16:57
  • $\begingroup$ Do you have a reference I can read through on this? $\endgroup$
    – user305623
    Commented Jan 16, 2016 at 17:08
  • $\begingroup$ Try, for example, qchu.wordpress.com/2011/02/12/su2-and-the-quaternions. $\endgroup$ Commented Jan 16, 2016 at 17:15
  • $\begingroup$ Incidentally, that "Sp" above stands for "symplectic," not "spinor." if I wanted to describe $SU(2)$ in spinor-related terms I would call it $Spin(3)$. $\endgroup$ Commented Jan 16, 2016 at 17:17
  • $\begingroup$ Cool. Thanks. :) $\endgroup$
    – user305623
    Commented Jan 16, 2016 at 17:17

1 Answer 1


Let $\mathbb{H}=\mathrm{span}\{1,\mathbf{i},\mathbf{j},\mathbf{k}\}$ be the quaterions. One may write $\mathrm{Im}(\mathbb{H})=\mathrm{span}\{\mathbf{i},\mathbf{j},\mathbf{k}\}$ to be the subspace of purely imaginary quaternions, and write $\mathrm{Sp}(1)$ for the set of unit quaternions, which forms a group under multiplication, and as a smooth manifold $\mathrm{Sp}(1)\simeq\mathbb{S}^3\subset\mathbb{R}^4$, i.e. just the three-dimensional sphere sitting inside four-dimensional space.

[If we have $\mathbb{C}$ act by multiplication from the right, then $\mathbb{H}$ is a "right" vector space over $\mathbb{C}$, and then left multiplication by unit quaternions amounts to right $\mathbb{C}$-linear transformations of the right $\mathbb{C}$-vector space $\mathbb{H}$, and thus every unit quaternion may be represented by a $2\times 2$ complex matrix. Writing $\mathbb{H}=\mathbb{C}\oplus\mathbb{C}\mathbf{j}$ and defining the obvious $\mathbb{C}$-valued dot product, it's clear that left multiplication by unit quaternions preserves this dot product, hence these $2\times 2$ complex matrices are unitary matrices. In fact, it is not too difficult to write down the form of an arbitrary unit quaternion and an arbitrary special unitary matrix to see that in fact $\mathrm{Sp}(1)\cong\mathrm{SU}(2)$.]

By the way, $\mathrm{Sp}$ stands for "symplectic" which is the Greek word for "complex." The word "complex" was already in use for complex numbers when Weyl coined the term in The Classical Groups.

Just to review, if $\mathbf{a},\mathbf{b}\in\mathrm{Im}(\mathbb{H})$ are pure imaginary quaternions, their product is $$\mathbf{ab}=-\mathbf{a}\cdot\mathbf{b}+\mathbf{a}\times\mathbf{b}. \tag{$\circ$}$$

The cyclic multiplication table for $\mathbf{i},\mathbf{j},\mathbf{k}$ follows from the cross product.

Suppose we take a unit vector $\mathbf{u}\in\mathrm{Im}(\mathbb{H})$. Define left and right multiplication maps

$$L_\mathbf{u}(x)=\mathbf{u}x, \quad R_\mathbf{u}(x)=x\mathbf{u}, \qquad x\in\mathbb{H}.$$

Observe these are both norm-preserving, hence inner product preserving. If we extend $\{\mathbf{u}\}$ to an ordered orthonormal basis $\{\mathbf{u},\mathbf{v},\mathbf{w}\}$ of $\mathrm{Im}(\mathbb{H})$ yielding the same orientation as $\{\mathbf{i},\mathbf{j},\mathbf{k}\}$, then because of $(\circ)$ they have he same cycling multiplication property. Define the pair of oriented 2D planes $\Pi=\mathrm{span}\{1,\mathbf{u}\}$ and $\Sigma=\mathrm{span}\{\mathbf{v},\mathbf{w}\}$. Observe $L_\mathbf{u}$ and $R_\mathbf{u}$ both restrict to a counter-clockwise right angle rotation on $\Pi$, but while $L_\mathbf{u}$ is still a counterclockwise right angle rotation on $\Sigma$ we find that $R_\mathbf{u}$ is a clockwise one.

In analogy with complex numbers, if we replace $\mathbf{u}$ by $\mathbf{t}=e^{\theta\mathbf{u}}=\cos(\theta)+\sin(\theta)\mathbf{u}$ we find that $L_\mathbf{t}$ and $R_\mathbf{t}$ are the same rotations by but angle $\theta$ instead of necessarily right angles. Therefore, $\varphi_\mathbf{t}=L_\mathbf{t}\circ R_{\mathbf{t}^{-1}}$ defined by $\varphi_\mathbf{t}(x)=\mathbf{t}x\mathbf{t}^{-1}$ will restrict to rotation of $\Sigma$ by an angle of $2\theta$ and the identity map on $\Pi$. In other words, if we restrict $\varphi_\mathbf{t}$ to $\mathrm{Im}(\mathbb{H})$, it is a rotation around the directed axis $\mathbf{u}$ by angle $2\theta$. This defines a map $\mathrm{Sp}(1)\to\mathrm{SO}(3)$, where $\mathrm{SO}(3)$ is the 3D rotation group (defined on 3D space $\mathrm{Im}(\mathbb{H})$), which is a smooth group homomorphism.

What is its kernel? One may readily check it is $\{\pm1\}$. Therefore, the map $\mathrm{Sp}(1)\to\mathrm{SO}(3)$, being onto with kernel of size two, is a $2$-to-$1$ map.

Let's step back for a second. A transformation of space, being a function, has only a before and after picture, no in between. But our intuition of a rotation is one of continuous motion. How do we bridge the gap? We may mentally picture a path in $\mathrm{SO}(3)$, i.e. a continuum of rotations (say starting at the identity), to be a continuous animation of a rotating figure. What if our path in $\mathrm{SO}(3)$ loops back to the identity? One such path is the act of rotating around a fixed axis by an angle which is a continuously varying parameter from $0$ to $2\pi$. This loop in $\mathrm{SO}(3)$ is in fact not contractible. As a matter of fact, since $\pi_1(\mathrm{SO}(3))=C_2$, this is essentially the only nontrivial kind of loop in $\mathrm{SO}(3)$ which is not contractible.

As one traces out such a loop in $\mathrm{SO}(3)$, starting from any rotation, there is a "lift" (or "pullback") of such a path from $\mathrm{SO}(3)$ to $\mathrm{Sp}(1)$. When the loop in $\mathrm{SO}(3)$ gets back to the rotation where it started at, the lifted path in $\mathrm{Sp}(1)$ has traversed to the opposite quaternion of where it started at.

This is all generalized in the relationship between covering space theory and Lie groups.

  • $\begingroup$ Why do you have $\mathbb{S}^3 \cong \mathbb{R}^4$ in the beginning? The are neither isomorphic as groups, nor as manifolds. I suppose you meant $\mathbb{S}^3 \subset \mathbb{R}^4$. $\endgroup$
    – lisyarus
    Commented Apr 17, 2016 at 7:41
  • $\begingroup$ @lisyarus Yes, I replaced the wrong symbol when I was revising. $\endgroup$
    – anon
    Commented Apr 17, 2016 at 8:09

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