The Quaternion Unit Group as a Topological Group

Introduction

The quaternions form a division algebra, so the units of the quaternion product are exactly the nonzero elements and the punctured space is the group of units. The space is the article before it; the discrete subgroups of the sphere are the article after it.

The Group of Units

Proposition (the units are the nonzero elements). The group of units of the quaternion product is

$$ \boxed{\;\mathbb{H}^\times=\mathrm{GL}_1(\mathbb{H})=\{\tilde q:\tilde q\neq0\}\;} $$

of real dimension $4$, open and dense in the algebra; its centre is $\mathbb{R}^\times e_t\cong\mathbb{R}^\times$, of real dimension one.

Proof. The algebra is a division algebra, so every nonzero element is invertible, with inverse $\tilde q^{\natural}/\langle\tilde q\rangle_{\natural}$ (Norm and Invertibility in the Quaternion Algebra); the origin is a closed proper algebraic subset, so its complement is open and dense; an element commutes with all of $\mathbb{H}$ exactly when it is a real scalar multiple of $e_t$, and the central units are the nonzero ones.

Proposition (the subspace is a topological group). With the topology of $\mathbb{H}$ the multiplication $\mathbb{H}^\times\times\mathbb{H}^\times\to\mathbb{H}^\times$ and the inversion $\mathbb{H}^\times\to\mathbb{H}^\times$ are continuous, so $\mathbb{H}^\times$ is a topological group.

Proof. The product is polynomial in the coordinates, and the inverse has coordinates rational with denominator $\langle\tilde q\rangle_{\natural}\neq0$; both are continuous on the punctured space.

The Decomposition with the Sphere

Theorem (the units are a product). The map

$$ \boxed{\; \mathbb{H}^\times\longrightarrow\mathbb{R}_{>0}\times S^3,\qquad \tilde q\longmapsto\bigl(\lvert\tilde q\rvert,\ \tilde q/\lvert\tilde q\rvert\bigr) \;} ,$$

is an isomorphism of topological groups, the second factor being the group of unit quaternions and the first the multiplicative group of the positive reals.

Proof. The map is well defined and continuous; the inverse is $(t,\tilde u)\mapsto t\tilde u$, continuous, and satisfies $\lvert t\tilde u\rvert=t$ because $\lvert\tilde u\rvert=1$; multiplicativity of the norm gives $\lvert\tilde q_1\tilde q_2\rvert=\lvert\tilde q_1\rvert\lvert\tilde q_2\rvert$, so the map is a group homomorphism, and its kernel and image are as displayed.

Corollary (the units are connected and their topology is the sphere's). The group $\mathbb{H}^\times$ is connected, path-connected and not compact; it is homotopy equivalent to $S^3$, so it is simply connected, $\pi_1(\mathbb{H}^\times)=0$ and $\pi_3(\mathbb{H}^\times)\cong\mathbb{Z}$.

Proof. The product $\mathbb{R}_{>0}\times S^3$ is connected and path-connected because both factors are; the factor $\mathbb{R}_{>0}$ admits the retraction $t\mapsto1$ and is contractible, so the product deformation retracts onto $S^3$; the fundamental group of $S^3$ is trivial and its third homotopy group is infinite cyclic, and these are homotopy invariants. The group is not compact because $\mathbb{R}_{>0}$ is unbounded.

The Compact Subgroup of the Unit Quaternions

Proposition (the maximal compact subgroup). The unit quaternions

$$ \boxed{\; Sp(1)=S^3=\{\tilde u:\tilde u^{\natural}\tilde u=e_t\} \;} $$

are a compact Lie group of real dimension $3$, the maximal compact subgroup of $\mathbb{H}^\times$; the decomposition $\mathbb{H}^\times=\mathbb{R}_{>0}\cdot Sp(1)$ holds, and the centre of $Sp(1)$ is $\{\pm e_t\}$.

Proof. The set $S^3$ is closed and bounded in $\mathbb{H}$, hence compact; it is closed under the product and the inverse, so it is a subgroup; it is a differentiable submanifold of dimension three cut by $N=1$ with nonzero gradient, so it is a Lie group. Every unit $\tilde q$ is $\lvert\tilde q\rvert\cdot\tilde u$ with $\tilde u$ a unit quaternion, which is the displayed decomposition and shows maximality, a compact subgroup lying in $S^3$ by the norm bound. An element of $S^3$ commuting with all of $\mathbb{H}$ is a real scalar unit, that is $\pm e_t$.

Remark (the quotient by the centre). The centre $\{\pm e_t\}$ is a discrete normal subgroup, and the quotient $Sp(1)/\{\pm e_t\}$ is the group of the rotations of three-dimensional space; it is the two-to-one cover read as a group quotient, and it is Quaternion Rotations and Reflections.

The Exponential Map

Theorem (the exponential and its homeomorphism property). The series

$$ \exp(\tilde q)=\sum_{n\geq0}\frac{\tilde q^{n}}{n!} $$

converges everywhere, takes values in $\mathbb{H}^\times$, is continuous, and is surjective onto $\mathbb{H}^\times$; on the pure quaternions it is $\exp(\vec v)=\cos\lvert\vec v\rvert+\dfrac{\vec v}{\lvert\vec v\rvert}\sin\lvert\vec v\rvert$, a unit quaternion. Restricted to the open band $\{\tilde q:\lvert\vec q\rvert<\pi\}$ the map $\exp$ is a homeomorphism onto $\mathbb{H}^\times$ with the negative scalar ray removed.

Proof. The series converges by comparison with the real exponential, gives $\lvert\exp(\tilde q)\rvert=\exp(q_t)\neq0$, so the values are units; the pure formula is the splitting of the series into its even and odd parts. Writing $\tilde q=q_te_t+\vec v$, $\exp(\tilde q)=\exp(q_t)\exp(\vec v)$, and $\exp(\vec v)=\cos\lvert\vec v\rvert+\vec u\sin\lvert\vec v\rvert$ with $\vec u=\vec v/\lvert\vec v\rvert$; every unit quaternion is uniquely $(\cos\theta,\vec u\sin\theta)$ with $\theta\in[0,\pi)$ and a unit imaginary direction, except the value $\theta=\pi$, which is $-e_t$, so the unit part of the image misses exactly $-e_t$ and the positive factor misses nothing; the inverse assigns to a unit quaternion its angle $\theta\in[0,\pi)$ and direction. Continuity and openness of the inverse are the continuity of the arccosine and the direction, and the excluded set is the negative scalar ray $-\mathbb{R}_{>0}e_t$.

Corollary (the exponential generates the unit group). Every unit is $\exp(\tilde q)$ for some $\tilde q$, and every unit quaternion is $\exp(\vec v)$ for a pure $\vec v$ with $\lvert\vec v\rvert\in[0,\pi]$; the exponential is a local homeomorphism at the origin, with local inverse the quaternion logarithm of Quaternion Exponential and Lie Group Structure.

Proof. The surjectivity onto $\mathbb{H}^\times$ is the theorem; the unit statement is the restriction to the pure part, allowing the value $\lvert\vec v\rvert=\pi$ for $-e_t$; the local statement is the inverse function theorem applied on the band, where the differential at the origin is the identity.

Summary

The group of units of the quaternion algebra is the punctured space $\mathbb{H}^\times=\mathbb{H}\setminus\{0\}$, open, dense, of real dimension four, with centre $\mathbb{R}^\times e_t$. It is a topological group, and the map $\tilde q\mapsto(\lvert\tilde q\rvert,\tilde q/\lvert\tilde q\rvert)$ is an isomorphism onto $\mathbb{R}_{>0}\times S^3$; consequently it is connected, simply connected, homotopy equivalent to $S^3$ and not compact. The unit quaternions $Sp(1)=S^3$ form the compact maximal subgroup, of dimension three, with centre $\{\pm e_t\}$ and quotient the rotation group. The exponential is surjective onto the units, is a unit quaternion on the pure part, and is a homeomorphism from the band of pure norm less than $\pi$ onto the units with the negative scalar ray removed.

Summary of Notation

Further Reading

• Introduction to Topology on the Quaternions, for the space and the three objects
• Norm and Invertibility in the Quaternion Algebra, for the inverse formula
• Quaternion Orders and Finite Groups of Units, for the discrete subgroups of $S^3$
• Quaternion Exponential and Lie Group Structure, for the logarithm and the Lie structure