The Biquaternion Unit Group as a Topological Group

Introduction

This article describes the group of units $\mathbb{B}^\times$ of the biquaternion algebra as a topological group: its polar decomposition, its retractions onto the compact subgroups, the structure of those subgroups, and its homotopy groups. It uses the algebra and the unitary elements of Biquaternion Algebra, the biquaternion norm and the invertibility criterion of Biquaternion Norm and Invertibility, and the ambient topology and the null cone of Biquaternion Topology. The exponential and the Lie-group correspondence are Biquaternion Lie Group and Exponential Structure, and the motions the group carries are Biquaternion Rotations and Lorentz Transformations.

Scope. This is the topological part of the Lie-theoretic block. The Lie algebra is in Biquaternion Lie Algebra, the topology is here, and the Lie-group theory is in Biquaternion Lie Group and Exponential Structure; the motions the group generates are geometry. Nothing here is a statement about the smooth structure, which belongs with the Lie group.

Conventions. The algebra is $\mathbb{B}=\mathbb{C}\otimes_{\mathbb{R}}\mathbb{H}$, with basis $e_0=1,e_1,e_2,e_3$ and central scalar imaginary $i$; a general element is $\tilde{Q}=\sum_{\mu=0}^{3}Q_\mu e_\mu$ with $Q_\mu\in\mathbb{C}$, written $\tilde{Q}=Q_0e_0+\mathbf{Q}$ with $\mathbf{Q}=\sum_{k=1}^{3}Q_ke_k$. The biquaternion norm is $N(\tilde{Q})=\tilde{Q}\tilde{Q}^{\natural}=\sum_\mu Q_\mu^2$, and $\tilde{Q}$ is a unit exactly when $N(\tilde{Q})\neq0$. The conjugations are ${}^{\natural}$ (quaternion), $\bar{\cdot}$ (complex), ${}^{*}={}^{\natural}\bar{\cdot}$ (Hermitian), and the Hermitian form is $\tilde{Q}\tilde{Q}^{*}$.

The Group of Units

$$ \mathbb{B}^\times=\{\tilde{Q}\in\mathbb{B} : N(\tilde{Q})\neq0\} $$

is a group under multiplication, because $N(\tilde{P}\tilde{Q})=N(\tilde{P})N(\tilde{Q})$ and $N(e_0)=1$: the product of two units has nonzero norm, and the inverse is $\tilde{Q}^{-1}=\tilde{Q}^{\natural}/N(\tilde{Q})$. It is an open dense subset of $\mathbb{B}$ and its boundary is the null cone (Biquaternion Topology, §The null cone as the boundary of the group of units).

Dimension. $\mathbb{B}^\times$ has complex dimension $4$ and real dimension $8$; the units are exactly the elements of nonzero norm.

The norm-one group. Because $N$ is multiplicative and $N(e_0)=1$, the level set

$$ \mathbb{B}^\times_1=\{\tilde{Q}\in\mathbb{B} : N(\tilde{Q})=1\} $$

is a subgroup, the norm-one group. It is a noncompact real $6$-manifold.

Two unit spheres. There are two candidate "unit spheres" in $\mathbb{B}$, and only one of them is a group. The Euclidean sphere $\|\tilde{Q}\|_E=1$ is a genuine sphere $S^7$ but is not a group, since $\|\cdot\|_E$ is not multiplicative and it contains zero divisors (Biquaternion Topology, §The Euclidean unit sphere). The level set $N(\tilde{Q})=1$ is a group but is neither Euclidean nor compact. The condition that makes a level set of a form on $\mathbb{B}$ a subgroup is $N=1$, not $\|\tilde{Q}\|_E=1$.

Physical reading. The invertible elements are the objects on which the motions act, and the norm-one slice is the slice in which the norm is normalised. The distinction between the two unit spheres is the distinction between the normalisation of a quantum state (Hermitian, $\|\cdot\|_E=1$) and the normalisation of a motion (biquaternion, $N=1$). The units are also what acts on the material four-position $\tilde{Q}=ict\,e_0+\mathbf{x}$: the boundary of $\mathbb{B}^\times$ is the null cone, whose intersection with that sector is the light cone $c^2t^2=\mathbf{x}^2$.

The Retraction of $\mathbb{B}^\times$ onto Its Maximal Compact Subgroup

Every $\tilde{A}\in\mathbb{B}^\times$ has a unique polar decomposition $\tilde{A}=\tilde{U}\tilde{P}$, where $\tilde{U}$ is unitary ($\tilde{U}^{*}\tilde{U}=e_0$) and $\tilde{P}=(\tilde{A}^{*}\tilde{A})^{1/2}$ is Hermitian positive definite. For $t\in[0,1]$ put $\tilde{P}_t=(1-t)\tilde{P}+te_0$ and

$$ \tilde{H}(t,\tilde{A})=\tilde{U}\tilde{P}_t . $$

The eigenvalues of $\tilde{P}_t$ are $(1-t)\lambda+t$ with $\lambda>0$, hence positive, so $\tilde{P}_t$ is positive definite and $\tilde{H}(t,\tilde{A})\in\mathbb{B}^\times$; the map $\tilde{H}$ is continuous. Moreover $\tilde{H}(0,\tilde{A})=\tilde{A}$, $\tilde{H}(1,\tilde{A})=\tilde{U}$, and $\tilde{H}(t,\tilde{U})=\tilde{U}$ for unitary $\tilde{U}$. So the unitary biquaternions form a strong deformation retract of $\mathbb{B}^\times$, and the two are homotopy equivalent, whence $\pi_n(\mathbb{B}^\times)\cong\pi_n(\mathrm{U}(\mathbb{B}))$ for all $n$, writing $\mathrm{U}(\mathbb{B})$ for the group of unitary biquaternions. Every $\tilde{A}$ is joined to a unitary element, and $\mathrm{U}(\mathbb{B})$ is connected, so $\mathbb{B}^\times$ is connected. The retraction takes $\mathbb{B}^\times$ onto its maximal compact subgroup,

$$ \mathrm{U}(\mathbb{B})=\{\tilde{Q}\in\mathbb{B} : \tilde{Q}^{*}\tilde{Q}=e_0\} . $$

Physical reading. The deformation is the removal of the boost: the Hermitian positive-definite factor $\tilde{P}$ carries the boost and the unitary factor $\tilde{U}$ the rotation and the phase, so retracting $\tilde{P}$ to the identity leaves the rotation and the phase untouched. This is why the topology of the motion group is the topology of its maximal compact subgroup, and why the non-compact directions contribute no homotopy. In the $ict$ convention the removal is the removal of the time–space mixing: on the material four-position $\tilde{Q}=ict\,e_0+\mathbf{x}$ the factor $\tilde{P}$ is what carries the rapidity and moves $ict$ toward $x_k$, and setting $\tilde{P}=e_0$ sets the rapidity to zero and leaves a pure rotation of $\mathbf{x}$.

The Retraction of the Norm-One Group onto Its Maximal Compact Subgroup

Let $\tilde{A}\in\mathbb{B}^\times_1$ have polar decomposition $\tilde{A}=\tilde{U}\tilde{P}$. Then $1=N(\tilde{A})=N(\tilde{U})N(\tilde{P})$, with $N(\tilde{U})$ of modulus $1$ and $N(\tilde{P})$ a positive real, so $N(\tilde{U})=N(\tilde{P})=1$, that is $\tilde{U}$ lies in $S^3$. For $t\in[0,1]$ define

$$ \tilde{P}_t=\frac{(1-t)\tilde{P}+te_0}{N\big((1-t)\tilde{P}+te_0\big)^{1/2}} . $$

The denominator is a positive real number, so $\tilde{P}_t$ is Hermitian positive definite of norm $1$. The map $\tilde{H}(t,\tilde{A})=\tilde{U}\tilde{P}_t$ is continuous, lies in $\mathbb{B}^\times_1$ since $N(\tilde{U}\tilde{P}_t)=1$, and satisfies $\tilde{H}(0,\tilde{A})=\tilde{A}$, $\tilde{H}(1,\tilde{A})=\tilde{U}\in S^3$, and $\tilde{H}(t,\tilde{U})=\tilde{U}$ for unitary $\tilde{U}$. Hence $S^3$ is a strong deformation retract of $\mathbb{B}^\times_1$.

Thus $\mathbb{B}^\times_1\simeq S^3$: it is connected and simply connected with $\pi_3\cong\mathbb{Z}$ and the homotopy type of $S^3$, but is not homeomorphic to $S^3$, being a noncompact real $6$-manifold. The norm-one group is therefore simply connected, of homotopy type $S^3$.

Physical reading. On the norm-one slice the norm is normalised, and the slice retracts onto the unit quaternions, so the topology that remains is that of the unit quaternions. A closed loop in $\mathbb{B}^\times_1$ is contractible, so the norm-one slice carries no winding number; the winding appears only in the full group, through the phase.

The Structure of the Maximal Compact Subgroup

The unit quaternions form $S^3$; $S^3$ is compact, connected and simply connected.

Every unitary biquaternion $\tilde{U}$ is a scalar multiple of a unit quaternion: if $N(\tilde{U})=z\in S^1$ and $\zeta^2=z$, then $\tilde{A}=\zeta^{-1}\tilde{U}$ has $N(\tilde{A})=1$ and $\tilde{U}=\zeta\tilde{A}$. Hence

$$ \mathrm{U}(\mathbb{B})=S^1\cdot S^3,\qquad S^1\cap S^3=\{\pm e_0\}, $$

and the multiplication map $S^1\times S^3\to\mathrm{U}(\mathbb{B})$ is a surjective homomorphism with kernel $\{(e_0,e_0),(-e_0,-e_0)\}\cong\mathbb{Z}/2$, so

$$ \mathrm{U}(\mathbb{B})\cong(S^1\times S^3)/\{\pm e_0\}, $$

with $\{\pm e_0\}$ acting diagonally. The biquaternion norm $N:\mathrm{U}(\mathbb{B})\to S^1$ is a principal $S^3$-bundle, each fibre being a coset of $S^3$, and it admits a section, so $\mathrm{U}(\mathbb{B})\cong S^1\times S^3$ as spaces. This is a homeomorphism, not an isomorphism of Lie groups: the map above is two-to-one, while the centre of $\mathrm{U}(\mathbb{B})$ is connected but that of $S^1\times S^3$ is not.

The central scalars form a maximal torus $T^2\cong S^1\times S^1\subset\mathrm{U}(\mathbb{B})$. It is not true that $\mathrm{U}(\mathbb{B})$ deformation retracts onto $T^2$: that would give $\pi_1(\mathrm{U}(\mathbb{B}))\cong\pi_1(T^2)$, but these are $\mathbb{Z}$ and $\mathbb{Z}^2$. Every element of $\mathrm{U}(\mathbb{B})$ does lie in some maximal torus, and the quotient is $\mathrm{U}(\mathbb{B})/T^2\cong P^1\cong S^2$, so $\mathrm{U}(\mathbb{B})$ is a fibre bundle over $S^2$ with fibre $T^2$; and $\mathrm{U}(\mathbb{B})$ is not homotopy equivalent to $T^2$, being homeomorphic to $S^1\times S^3$.

Physical reading. The two factors of the maximal compact subgroup are the two things a unitary element carries: the $S^3$ factor is the unit quaternions, and the $S^1$ factor is the central phase, the global phase of a quantum state, the phase of the scalar line $\mathbb{C}e_0$, which is also the line that carries the two times $ct'$ and $ict$. The identification $\mathrm{U}(\mathbb{B})\cong S^1\times S^3$ therefore says that the compact part of the group is a unit quaternion together with a phase, and the two are not independent as a group, only as a space: the quotient by $\{\pm e_0\}$ is what turns the unit quaternion into a rotation and not a spinor. The maximal torus is the pair of commuting factors, of the phase and of an axis, and the sphere $\mathrm{U}(\mathbb{B})/T^2\cong S^2$ is the sphere of axes.

Homotopy Groups and Generators

The retractions give $S^3$ for the unit quaternions, $\mathrm{U}(\mathbb{B})\cong S^1\times S^3$, $\mathbb{B}^\times_1\simeq S^3$, and $\mathbb{B}^\times\simeq\mathrm{U}(\mathbb{B})\simeq S^1\times S^3$. Hence

$$ \pi_1(S^3)=\pi_2(S^3)=0,\qquad \pi_3(S^3)\cong\mathbb{Z}, $$

$$ \pi_1(\mathrm{U}(\mathbb{B}))\cong\mathbb{Z},\qquad \pi_2(\mathrm{U}(\mathbb{B}))=0,\qquad \pi_3(\mathrm{U}(\mathbb{B}))\cong\mathbb{Z}, $$

and likewise $\pi_1(\mathbb{B}^\times_1)=0$, $\pi_1(\mathbb{B}^\times)\cong\mathbb{Z}$, with $\pi_2=0$ and $\pi_3\cong\mathbb{Z}$ for both.

Generators. The group $\pi_3(S^3)\cong\mathbb{Z}$ is generated by the class $[\operatorname{id}_{S^3}]$ of the identity map under $S^3=\{$unit quaternions$\}$, and $\pi_3(\mathrm{U}(\mathbb{B}))\cong\mathbb{Z}$ by the image of that class under the inclusion $S^3\hookrightarrow\mathrm{U}(\mathbb{B})$, which induces an isomorphism on $\pi_3$. The group $\pi_1(\mathrm{U}(\mathbb{B}))\cong\mathbb{Z}$ is generated by the central loop $\gamma(t)=e^{2\pi it}e_0$, $t\in[0,1]$, and $N_*:\pi_1(\mathrm{U}(\mathbb{B}))\to\pi_1(S^1)\cong\mathbb{Z}$ is an isomorphism, so a generator is a loop whose biquaternion norm winds once. The universal covers are

$$ \widetilde{\mathrm{U}(\mathbb{B})}\cong\widetilde{\mathbb{B}^\times}\cong\mathbb{R}\times S^3, $$

while $S^3$ and $\mathbb{B}^\times_1$ are their own universal covers. By Hurewicz, $H_1(\mathrm{U}(\mathbb{B}))\cong H_1(\mathbb{B}^\times)\cong\mathbb{Z}$ and $H_1(S^3)=H_1(\mathbb{B}^\times_1)=0$; and $\pi_n(\mathrm{U}(\mathbb{B}))\cong\pi_n(S^3)$ for $n\ge2$.

Physical reading. The factor $\pi_1(\mathbb{B}^\times)\cong\mathbb{Z}$ is the winding of the phase, the homotopy invariant that a closed loop of transformations can carry, and it is the topological home of the winding numbers the framework uses. The factor $\pi_3\cong\mathbb{Z}$ is the winding of the unit quaternions, the invariant of a Skyrme field, and it is generated by the identity map of the unit quaternions because a unit quaternion is a map from the spatial sphere to the group. The universal cover $\mathbb{R}\times S^3$ is the statement that unwinding the phase and unwinding the rotation are independent: the first gives the real line, the second gives the spin cover.

Summary

The group of units $\mathbb{B}^\times$ is the complement of the null cone, of real dimension $8$; it is connected, and it deformation retracts onto its maximal compact subgroup $\mathrm{U}(\mathbb{B})$, so it is homotopy equivalent to $S^1\times S^3$. The norm-one group $\mathbb{B}^\times_1$ is a noncompact real $6$-manifold of the homotopy type of $S^3$: connected, simply connected, with $\pi_3\cong\mathbb{Z}$. The compact group is $\mathrm{U}(\mathbb{B})\cong(S^1\times S^3)/\{\pm e_0\}$, homeomorphic to $S^1\times S^3$, with the $S^3$ factor the rotations and the $S^1$ factor the central phase. In homotopy, $\pi_1(\mathbb{B}^\times)\cong\mathbb{Z}$ generated by the winding of the phase, $\pi_3\cong\mathbb{Z}$ generated by the identity of the unit quaternions, $\pi_2=0$, and the universal cover of the full group is $\mathbb{R}\times S^3$.

Summary of Notation

Symbol	Meaning
$\mathbb{B}^\times=\{\tilde{Q} : N(\tilde{Q})\neq0\}$	Group of units; open and dense, boundary the null cone
$\mathbb{B}^\times_1=\{\tilde{Q} : N(\tilde{Q})=1\}$	Norm-one group; noncompact real $6$-manifold, $\simeq S^3$
$\mathrm{U}(\mathbb{B})=\{\tilde{Q} : \tilde{Q}^{*}\tilde{Q}=e_0\}$	Unitary biquaternions, the maximal compact subgroup
$\tilde{A}=\tilde{U}\tilde{P}$	Polar decomposition; $\tilde{U}$ unitary, $\tilde{P}$ Hermitian positive definite
$\mathrm{U}(\mathbb{B})\cong(S^1\times S^3)/\{\pm e_0\}$	Maximal compact subgroup as a group
$\mathrm{U}(\mathbb{B})\cong S^1\times S^3$	Homeomorphism, not a group isomorphism
$S^1$ factor	Central phase; the phase of the scalar line $\mathbb{C}e_0$ that carries $ct'+ict$
$\tilde{Q}=ict\,e_0+\mathbf{x}$	Material four-position; its light cone $c^2t^2=\mathbf{x}^2$ is the boundary of $\mathbb{B}^\times$
$S^3$ factor	Unit quaternions
$T^2\cong S^1\times S^1$	Maximal torus; not a deformation retract
$\pi_1(\mathbb{B}^\times)\cong\mathbb{Z}$	Winding of the phase
$\pi_3\cong\mathbb{Z}$	Winding of the unit quaternions; the Skyrme invariant
$\widetilde{\mathbb{B}^\times}\cong\mathbb{R}\times S^3$	Universal cover