The Unitary Group of the Biquaternion Algebra

Introduction

The Hermitian form $\tilde{Q}\tilde{Q}^{*}$ singles out, inside the group of units of the biquaternion algebra, the elements that leave it invariant: the unitary biquaternions $\tilde{U}$ with $\tilde{U}^{*}\tilde{U}=e_0$. This is the maximal compact subgroup $U(\mathbb{B})\cong U(2)$ of $\mathbb{B}^{\times}$, the compact real form of the algebra, and it is the single object through which the whole topology of the group of units is read: the polar decomposition gives a strong deformation retraction of $\mathbb{B}^{\times}$ onto $U(\mathbb{B})$ and of the norm-one group onto its compact part $S^{3}$, and the homotopy groups, the generators and the universal cover follow from the resulting homotopy equivalence $\mathbb{B}^{\times}\simeq S^{1}\times S^{3}$.

This article collects the Hermitian half of the topology of the unit group: the unitary slice, its structure, the two retractions that the dagger supplies, and the homotopy invariants. The bilinear half — the units as the complement of the null cone, the centre, the distribution of the units among the three classes — is The Biquaternion Unit Group as a Topological Group and Biquaternion Norm and Invertibility, and the ambient Euclidean topology and the contractibility of the algebra are The Euclidean Topology of the Biquaternion Algebra.

Conventions. As in the sibling articles: $\mathbb{B}=\mathbb{C}\otimes_{\mathbb{R}}\mathbb{H}$, $\tilde{Q}=\sum_\mu Q_\mu e_\mu$, $\|\tilde{Q}\|_E=(\sum_\mu|Q_\mu|^{2})^{1/2}$, $N(\tilde{Q})=\sum_\mu Q_\mu^{2}$, dagger ${}^{*}={}^{\natural}\circ\bar{\cdot}$, and $\Phi:\mathbb{B}\to M_2(\mathbb{C})$ the matrix isomorphism of Biquaternion 2×2 Matrix Element Representation, which carries the dagger to the conjugate transpose.

The Unitary Biquaternions

Definition. The unitary biquaternions are

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

Theorem ($U(\mathbb{B})\cong U(2)$). $\Phi$ restricts to an isomorphism of groups $\Phi:U(\mathbb{B})\to U(2)$. In particular $U(\mathbb{B})$ is a compact Lie group of real dimension four, and its elements are exactly the biquaternions of the form

$$ \tilde{U}=A\,\tilde{q},\qquad A\in\mathbb{C},\ |A|=1,\quad \tilde{q}\in\mathbb{H},\ N(\tilde{q})=1, $$

that is, scalar multiples of unit quaternions by phases.

Proof. Since $\Phi$ is an algebra isomorphism and $\Phi(\tilde{Q}^{*})=\Phi(\tilde{Q})^{\dagger}$ (Biquaternion 2×2 Matrix Element Representation, §The Conjugations in Matrix Form), the equation $\tilde{U}^{*}\tilde{U}=e_0$ is equivalent to $\Phi(\tilde{U})^{\dagger}\Phi(\tilde{U})=I$, which defines $U(2)$; the restriction of an injective homomorphism is an injective homomorphism onto that group. For the normal form: the identity $N(\tilde{U})=\det\Phi(\tilde{U})$ makes $|N(\tilde{U})|=1$, so if $\zeta^{2}=N(\tilde{U})$ then $\tilde{A}=\zeta^{-1}\tilde{U}$ has $N(\tilde{A})=1$; a norm-one element is a unit quaternion (Biquaternion Norm and Invertibility, §The Relation to the Hermitian Decomposition), and $\tilde{U}=\zeta\tilde{A}$.

Corollary (the determinant and the norm coincide). On $U(\mathbb{B})$ the biquaternion norm is the determinant through $\Phi$, and $N:U(\mathbb{B})\to S^{1}$ is a surjective homomorphism with kernel $S^{3}$, the unit quaternions. Hence $N$ realises an isomorphism $U(\mathbb{B})/S^{3}\cong S^{1}$.

Proof. $N(\tilde{U}\tilde{V})=N(\tilde{U})N(\tilde{V})$ is the multiplicativity of the norm ($\det$ is multiplicative), $N(e_0)=1$, and $|N(\tilde{U})|=1$ by the theorem, so the image lies in $S^{1}$ and is a subgroup; it is all of $S^{1}$ because $A\mapsto Ae_0$ is unitary of norm $A$. The kernel is $\{N=1\}\cap U(\mathbb{B})$, which is the unit quaternions.

The Structure of the Unitary Group

Theorem (the product decomposition). $U(\mathbb{B})=S^{1}\cdot S^{3}$ with $S^{1}\cap S^{3}=\{\pm e_0\}$, where $S^{1}=U(1)e_0$ is the centre circle and $S^{3}$ the unit quaternions; the multiplication map $S^{1}\times S^{3}\to U(\mathbb{B})$ is a surjective homomorphism with kernel $\{(e_0,e_0),(-e_0,-e_0)\}$, so

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

with the diagonal action, and this quotient is homeomorphic to $S^{1}\times S^{3}$.

Proof. The normal form writes an element as $A\tilde q$ with $A\in S^{1}$ and $\tilde{q}\in S^{3}$, and the intersection of the two subgroups is $\{A\in S^{1}: A\in\mathbb{H}\}=\{\pm e_0\}$; the product map is a homomorphism because $S^{1}$ is central. The quotient description is the first isomorphism theorem for Lie groups, and the homeomorphism $U(\mathbb{B})\cong S^{1}\times S^{3}$ is the section $\tilde{U}\mapsto(N(\tilde{U}),N(\tilde{U})^{-1/2}\tilde{U})$.

Remark (not an isomorphism of groups). The homeomorphism $U(\mathbb{B})\cong S^{1}\times S^{3}$ is not a group isomorphism: its centre is connected, that of $S^{1}\times S^{3}$ is not, and the product map above is two-to-one.

Theorem (the maximal torus and the flag variety). The central scalars of modulus one form a maximal torus $T^{2}\cong S^{1}\times S^{1}$; every element of $U(\mathbb{B})$ lies in some maximal torus; and

$$ U(\mathbb{B})/T^{2}\cong P^{1}\cong S^{2}, $$

so $U(\mathbb{B})$ is a fibre bundle over $S^{2}$ with fibre $T^{2}$. It does not deformation retract onto $T^{2}$.

Proof. The diagonal matrices are the maximal tori of $U(2)$, and $U(2)/T^{2}$ is the complete flag variety of $\mathbb{C}^{2}$, which is $P^{1}$; a retraction onto a maximal torus would force $\pi_1(U(\mathbb{B}))\cong\pi_1(T^{2})$, but these are $\mathbb{Z}$ and $\mathbb{Z}^{2}$.

The Retraction of the Unit Group onto the Unitary Group

Theorem (polar decomposition). Every $\tilde{A}\in\mathbb{B}^{\times}$ has a unique decomposition

$$ \tilde{A}=\tilde{U}\tilde{P},\qquad \tilde{U}\in U(\mathbb{B}),\quad \tilde{P}\ \text{Hermitian and positive definite}, $$

with $\tilde{P}=(\tilde{A}^{*}\tilde{A})^{1/2}$; the map $\tilde{A}\mapsto(\tilde{U},\tilde{P})$ is a homeomorphism of $\mathbb{B}^{\times}$ onto $U(\mathbb{B})\times\{\text{Hermitian positive definite}\}$.

Proof. $\tilde{A}^{*}\tilde{A}$ is Hermitian and positive definite for $\tilde{A}$ a unit, because $\mathrm{Sc}(\tilde{A}^{*}\tilde{A})=\|\tilde{A}\|_E^{2}>0$ and the Hermitian elements that are positive for the dagger form the cone of the units (Positivity and the Hermitian Cone of the Biquaternion Algebra with Hermitian Adjoint, §The Hermitian Cone); a positive definite Hermitian element has a unique positive definite square root $\tilde{P}$, and $\tilde{U}=\tilde{A}\tilde{P}^{-1}$ satisfies $\tilde{U}^{*}\tilde{U}=\tilde{P}^{-1}\tilde{A}^{*}\tilde{A}\tilde{P}^{-1}=e_0$. Uniqueness is the standard argument: $\tilde{P}^{2}=\tilde{A}^{*}\tilde{A}$ has one positive definite root.

Theorem (strong deformation retraction). For $t\in[0,1]$ put $\tilde{P}_t=(1-t)\tilde{P}+t e_0$ and $\tilde{H}(t,\tilde{A})=\tilde{U}\tilde{P}_t$. Then $\tilde{H}$ is a strong deformation retraction of $\mathbb{B}^{\times}$ onto $U(\mathbb{B})$:

$$ \tilde{H}(0,\tilde{A})=\tilde{A},\qquad \tilde{H}(1,\tilde{A})=\tilde{U},\qquad \tilde{H}(t,\tilde{U})=\tilde{U}\ \text{for}\ \tilde{U}\in U(\mathbb{B}). $$

Hence $\mathbb{B}^{\times}\simeq U(\mathbb{B})$, and $\pi_n(\mathbb{B}^{\times})\cong\pi_n(U(\mathbb{B}))$ for all $n$.

Proof. The eigenvalues of $\tilde{P}_t$ are $(1-t)\lambda+t$ with $\lambda>0$, hence positive, so $\tilde{P}_t$ is positive definite and $\tilde{U}\tilde{P}_t\in\mathbb{B}^{\times}$; the positive definite square root depends continuously on $\tilde{A}$ (The Two-Sided Operators on the Biquaternion Algebra with Hermitian Adjoint, §The Operator of an Element), so $\tilde{H}$ is continuous; the three identities are immediate from $\tilde{P}_0=\tilde{P}$, $\tilde{P}_1=e_0$ and the fact that a unitary element has $\tilde{P}=e_0$.

Corollary (connectedness). $\mathbb{B}^{\times}$ is connected, and $U(\mathbb{B})$ is connected.

Proof. $\tilde{H}$ joins every unit to a unitary element, and $U(\mathbb{B})\cong(S^{1}\times S^{3})/\{\pm\}$ is a continuous image of the connected group $S^{1}\times S^{3}$.

The Retraction of the Norm-One Group onto $S^{3}$

Theorem. The unit quaternions $S^{3}$ are a strong deformation retract of the norm-one group $\mathbb{B}^{\times}_1=\{N=1\}$, so $\mathbb{B}^{\times}_1\simeq S^{3}$: it is connected and simply connected, with $\pi_3\cong\mathbb{Z}$ and vanishing $\pi_1,\pi_2$.

Proof. Let $\tilde{A}\in\mathbb{B}^{\times}_1$ with polar decomposition $\tilde{A}=\tilde{U}\tilde{P}$. Then $1=N(\tilde{A})=N(\tilde{U})N(\tilde{P})$, with $|N(\tilde{U})|=1$ and $N(\tilde{P})$ a positive real, so $N(\tilde{U})=N(\tilde{P})=1$ and $\tilde{U}\in S^{3}$ by the normal form of the previous section. For $t\in[0,1]$ the element

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

is Hermitian positive definite of norm one (the denominator is a positive real), the map $\tilde{H}(t,\tilde{A})=\tilde{U}\tilde{P}_t$ is continuous and lands in $\mathbb{B}^{\times}_1$, and $\tilde{H}(0,\tilde{A})=\tilde{A}$, $\tilde{H}(1,\tilde{A})=\tilde{U}\in S^{3}$, $\tilde{H}(t,\tilde{U})=\tilde{U}$.

Remark. $\mathbb{B}^{\times}_1$ is not homeomorphic to $S^{3}$: it is a non-compact real $6$-manifold. It is a closed subgroup of $\mathbb{B}^{\times}$, of real dimension $6$ (Biquaternion Lie Group and Exponential Structure, §The Subgroups and the Real Forms).

Homotopy Groups, Generators and the Universal Cover

Theorem. The homotopy invariants are

$$ \mathbb{B}^{\times}\simeq U(\mathbb{B})\simeq S^{1}\times S^{3},\qquad \mathbb{B}^{\times}_1\simeq S^{3}, $$

$$ \pi_1(\mathbb{B}^{\times})\cong\mathbb{Z},\quad \pi_2(\mathbb{B}^{\times})=0,\quad \pi_3(\mathbb{B}^{\times})\cong\mathbb{Z},\qquad \pi_1(\mathbb{B}^{\times}_1)=0,\quad \pi_2(\mathbb{B}^{\times}_1)=0,\quad \pi_3(\mathbb{B}^{\times}_1)\cong\mathbb{Z}, $$

and the universal covers are $\widetilde{\mathbb{B}^{\times}}\cong\widetilde{U(\mathbb{B})}\cong\mathbb{R}\times S^{3}$, while $S^{3}$ and $\mathbb{B}^{\times}_1$ are their own universal covers.

Proof. The two retractions give the homotopy equivalences; the homotopy groups of $S^{3}$ are $\pi_1=\pi_2=0$, $\pi_3\cong\mathbb{Z}$, and those of $S^{1}\times S^{3}$ follow by the product formula. The universal cover of $S^{1}\times S^{3}$ is $\mathbb{R}\times S^{3}$ because $S^{3}$ is simply connected, and the same holds for the group of units by the homotopy equivalence.

Generators. $\pi_3(U(\mathbb{B}))\cong\mathbb{Z}$ is generated by the image of the class of $\mathrm{id}_{S^{3}}$ under $S^{3}\hookrightarrow U(\mathbb{B})$, which induces an isomorphism on $\pi_3$; $\pi_1(U(\mathbb{B}))\cong\mathbb{Z}$ is generated by the central loop $\gamma(t)=e^{2\pi it}e_0$, and $N_*:\pi_1(U(\mathbb{B}))\to\pi_1(S^{1})$ is an isomorphism, so a generator is a loop along which the biquaternion norm winds once.

Proof. The inclusions $S^{3}\subset U(\mathbb{B})\subset\mathbb{B}^{\times}$ and the retractions give the first statement; the second is the identification of $\pi_1$ with the fundamental group of the centre circle via the bundle $S^{1}\to U(\mathbb{B})\to S^{3}$... precisely, $U(\mathbb{B})\cong(S^{1}\times S^{3})/\{\pm\}$ and the projection to the first factor realises $N$.

Corollary (Hurewicz). $H_1(\mathbb{B}^{\times})\cong H_1(U(\mathbb{B}))\cong\mathbb{Z}$, $H_1(S^{3})=H_1(\mathbb{B}^{\times}_1)=0$, and $\pi_n(U(\mathbb{B}))\cong\pi_n(S^{3})$ for $n\geq2$.

Proof. Immediate from the theorem and the Hurewicz theorem in degree one.

The Unitary Group and the Lorentz Group

The compact slice is not only the compact real form of the algebra; it is the double cover of the rotation group of the Hermitian subspace.

Theorem. The inner conjugation $\tilde{U}\mapsto\Theta_{\tilde{U}}$, $\Theta_{\tilde{U}}(\tilde{R})=\tilde{U}\tilde{R}\tilde{U}^{\dagger}$, restricts to an action of $U(\mathbb{B})$ on the Hermitian subspace $\mathbb{M}_{+}$ by isometries of the interval form, and the map

$$ U(\mathbb{B})\longrightarrow SO(3),\qquad \tilde{U}\longmapsto \Theta_{\tilde{U}}|_{\mathbb{M}_{+}}, $$

is a surjective homomorphism with kernel $U(1)e_0$, so that $U(\mathbb{B})/U(1)\cong PU(2)\cong SO(3)$.

Proof. $\Theta_{\tilde{U}}$ is an algebra automorphism and a Euclidean isometry, and it preserves $\mathbb{M}_{+}$ because it preserves the dagger; it preserves the interval form because it is an algebra automorphism and the form is $N$ on $\mathbb{M}_{+}$. The kernel consists of the $\tilde{U}$ acting trivially on $\mathbb{M}_{+}$, and an automorphism fixing the Hermitian subspace is inner by a central element, so the kernel is the central circle; the quotient $PU(2)$ is the adjoint form of $U(2)$, isomorphic to $SO(3)$. On the unit quaternions the same map is the classical double cover $S^{3}\to SO(3)$.

Remark (the two covers). The unitary group therefore sits between the two classical double covers of the corpus: $S^{3}\to SO(3)$ on the compact slice, and $S^{3}\times S^{3}\to SO(4)$ for the two-sided action (Biquaternion Rotations and Lorentz Transformations). Its full conjugation action on $\mathbb{B}$ is the adjoint action of $U(2)$ on $M_2(\mathbb{C})$, whose orbits are the level sets of $\mathrm{Sc}$ and $N$.

Worked Examples

A central element. $\tilde{U}=Ae_0$ is unitary exactly when $|A|=1$, and then $N(\tilde{U})=A^{2}$, so the centre circle maps onto $S^{1}$ twice.

A unit quaternion. $\tilde{U}\in\mathbb{H}_{\mathbb{B}}$, $N(\tilde{U})=1$, is unitary, is fixed by $\Theta$ only when central, and generates the image of $\pi_3$.

A null scalar multiple. $\tilde{Q}=A(e_1+ie_2)$ with $A\neq0$ has $N(\tilde{Q})=0$, so it is not a unit and in particular not unitary; its Euclidean norm is $\sqrt2|A|$.

A negative determinant. $U(2)$ has determinant of modulus one; the slice $SU(2)=\{N=1\}\cap U(\mathbb{B})=S^{3}$ is the unit quaternions, the kernel of $N$, and the double cover of $SO(3)$.

Summary

The unitary biquaternions $U(\mathbb{B})=\{\tilde{U}^{*}\tilde{U}=e_0\}$ are the maximal compact subgroup of $\mathbb{B}^{\times}$, isomorphic to $U(2)$ through the matrix model and equal to the scalar multiples of the unit quaternions by phases. The group splits as $U(\mathbb{B})=S^{1}\cdot S^{3}$ with intersection $\{\pm e_0\}$, so $U(\mathbb{B})\cong(S^{1}\times S^{3})/\{\pm\}\cong S^{1}\times S^{3}$ as spaces, with maximal torus $T^{2}$ and flag variety $U(\mathbb{B})/T^{2}\cong S^{2}$. The polar decomposition $\tilde{A}=\tilde{U}\tilde{P}$ supplies strong deformation retractions of $\mathbb{B}^{\times}$ onto $U(\mathbb{B})$ and of $\mathbb{B}^{\times}_1$ onto $S^{3}$, whence $\mathbb{B}^{\times}\simeq S^{1}\times S^{3}$, $\mathbb{B}^{\times}_1\simeq S^{3}$, $\pi_1(\mathbb{B}^{\times})\cong\mathbb{Z}$, $\pi_2=0$, $\pi_3\cong\mathbb{Z}$, and universal cover $\mathbb{R}\times S^{3}$. The norm $N$ is the determinant and $U(\mathbb{B})/S^{3}\cong S^{1}$; the conjugation action gives $U(\mathbb{B})/U(1)\cong SO(3)$, the compact real form of the Lorentz group.

Summary of Notation

Further Reading

• The Biquaternion Unit Group as a Topological Group (articles_maths/the-biquaternion-unit-group-as-a-topological-group.md), for the bilinear-side treatment of the group of units
• The Euclidean Topology of the Biquaternion Algebra (articles_maths/the-euclidean-topology-of-the-biquaternion-algebra.md), for the ambient Euclidean structure the retractions use
• The Hermitian Form on the Biquaternion Algebra (articles_maths/the-hermitian-form-on-the-biquaternion-algebra.md), for the form that defines the unitary slice
• Positivity and the Hermitian Cone of the Biquaternion Algebra with Hermitian Adjoint (articles_maths/positivity-and-the-hermitian-cone-of-the-biquaternion-algebra-with-hermitian-adjoint.md), for the Hermitian cone and the positive square root
• Biquaternion Lie Group and Exponential Structure (articles_maths/biquaternion-lie-group-and-exponential-structure.md), for the subgroups and the real forms
• Brian C. Hall, Lie Groups, Lie Algebras, and Representations, 2nd edition (Springer, 2015), for the standard facts about $U(2)$, $SU(2)$ and their quotients
• John Stillwell, Naive Lie Theory (Springer, 2008), for the double covers of the rotation groups