The Three Pairings of the Biquaternion Algebra

Introduction

The biquaternion algebra carries not one but three distinguished pairings of its elements, and the whole Krein layer of this category is the study of the third. The bilinear form is written $N(\cdot,\cdot)$ in The Bilinear Form on the Biquaternion Algebra and $B$ here. The three are indexed by the involution group of the algebra, the Klein four-group generated by the natural conjugation ${}^{\natural}$ and the complex conjugation $\bar{\cdot}$: each non-identity involution $\sigma$ gives the pairing

$$ \Phi_{\sigma}(\tilde{Q},\tilde{Q}')=\mathrm{Sc}\bigl(\sigma(\tilde{Q})\,\tilde{Q}'\bigr), $$

and the three resulting forms are the bilinear form $B$ of The Bilinear Form on the Biquaternion Algebra, the Hermitian form $\langle\cdot,\cdot\rangle$ of The Hermitian Form on the Biquaternion Algebra, and the Krein form $[\cdot,\cdot]$ of The Biquaternion Krein Form and Its Signature. This article sets them side by side: their conjugation, their linearity, their symmetry, their Gram matrix in the coefficient basis, their signature, and the isometry group each one defines. The comparison is the natural entry into the Krein category, because it shows exactly what the Krein form shares with its two siblings and where it differs.

The fourth conjugation, the reversal $\flat=-\!{}^{*}$, gives a fourth pairing, the negative of the Hermitian one; together with the identity conjugation, which gives the unsymmetric product form $\mathrm{Sc}(\tilde{Q}\tilde{Q}')$, the five pairings exhaust the maps $\sigma\mapsto\Phi_{\sigma}$ built from the four conjugations of Biquaternion Involution Lattice.

Conventions. $\mathbb{B}=\mathbb{C}\otimes_{\mathbb{R}}\mathbb{H}$, $\tilde{Q}=\sum_{\mu}Q_{\mu}e_{\mu}$ with $Q_{\mu}\in\mathbb{C}$, units $e_0=1$ and $e_k^{2}=-e_0$, central scalar imaginary $i$, and $\mathrm{Sc}$ the scalar part. The sign vector is $\varepsilon=(1,-1,-1,-1)$ and $\mathbb{B}$ is identified with $\mathbb{C}^{4}$ by $\tilde{Q}\mapsto(Q_0,Q_1,Q_2,Q_3)$.

The Involution Group and the Pairing Map

Definition. An involution of $\mathbb{B}$ is a $\mathbb{R}$-linear map $\sigma$ with $\sigma^{2}=\mathrm{id}$. It is linear if $\sigma(\lambda\tilde{Q})=\lambda\sigma(\tilde{Q})$ for $\lambda\in\mathbb{C}$ and antilinear if $\sigma(\lambda\tilde{Q})=\bar\lambda\sigma(\tilde{Q})$; it is an anti-automorphism if $\sigma(\tilde{Q}\tilde{R})=\sigma(\tilde{R})\sigma(\tilde{Q})$.

Theorem (the four conjugations). $\mathbb{B}$ has four distinguished involutions:

name	notation	formula	linearity	product
natural conjugation	${}^{\natural}$	$\tilde{Q}^{\natural}=Q_0e_0-Q_1e_1-Q_2e_2-Q_3e_3$	linear	anti-automorphism
complex conjugation	$\bar{\cdot}$	$\bar{\tilde{Q}}=\bar Q_0e_0+\bar Q_1e_1+\bar Q_2e_2+\bar Q_3e_3$	antilinear	automorphism
Hermitian conjugation	${}^{*}$	$\tilde{Q}^{*}=\overline{\tilde{Q}^{\natural}}$	antilinear	anti-automorphism
reversal	$\flat$	$\tilde{Q}^{\flat}=-\tilde{Q}^{*}$	antilinear	anti-automorphism up to sign

They satisfy ${}^{*}=\bar{\cdot}\circ{}^{\natural}$ and $\flat=-{}^{*}$, and $\{\mathrm{id},{}^{\natural},\bar{\cdot},{}^{*}\}$ is the Klein four-group of pairwise commuting involutions.

Proof. These are the definitions and composition rules of Biquaternion Involution Lattice, §The Four Conjugations and §Two Composition Rules; linearity is read from the real coordinates there.

Definition (the pairing map). To an involution $\sigma$ associate the pairing

$$ \Phi_{\sigma}(\tilde{Q},\tilde{Q}')=\mathrm{Sc}\bigl(\sigma(\tilde{Q})\,\tilde{Q}'\bigr). $$

Proposition (standard properties of the pairing map). For every involution $\sigma$ the pairing $\Phi_{\sigma}$ is $\mathbb{R}$-bilinear, is linear in the second argument, and is conjugate-linear in the first exactly when $\sigma$ is antilinear. Its Gram matrix in the basis $e_{\mu}$ is $G_{\mu\nu}=\varepsilon^{(\sigma)}_{\mu}\varepsilon_{\mu}\,\delta_{\mu\nu}$, where $\sigma(e_{\mu})=\varepsilon^{(\sigma)}_{\mu}e_{\mu}$.

Proof. $\mathbb{R}$-bilinearity and the second-argument linearity are immediate from the definition. For the Gram matrix, $\sigma(e_{\mu})=\varepsilon^{(\sigma)}_{\mu}e_{\mu}$ with $\varepsilon^{(\natural)}=\varepsilon$, $\varepsilon^{(\bar{\cdot})}=(1,1,1,1)$, $\varepsilon^{(*)}=\varepsilon$ and $\varepsilon^{(\flat)}=-\varepsilon$; then $\Phi_{\sigma}(e_{\mu},e_{\nu})=\varepsilon^{(\sigma)}_{\mu}\,\mathrm{Sc}(e_{\mu}e_{\nu})=\varepsilon^{(\sigma)}_{\mu}\varepsilon_{\mu}\delta_{\mu\nu}$, because $e_{\mu}e_{\nu}=0$ off the diagonal in the scalar part. The conjugate-linearity in the first argument is the antilinearity of $\sigma$, since $\mathrm{Sc}$ is $\mathbb{C}$-linear.

The Three Pairings

Theorem (the three forms). The three non-identity involutions of the group $\{\mathrm{id},{}^{\natural},\bar{\cdot},{}^{*}\}$ give the three pairings

$$ B(\tilde{Q},\tilde{Q}')=\mathrm{Sc}\bigl(\tilde{Q}^{\natural}\tilde{Q}'\bigr)=\sum_{\mu}Q_{\mu}Q'_{\mu}, \qquad \langle\tilde{Q},\tilde{Q}'\rangle=\mathrm{Sc}\bigl(\tilde{Q}^{*}\tilde{Q}'\bigr)=\sum_{\mu}\bar Q_{\mu}Q'_{\mu}, $$

$$ [\tilde{Q},\tilde{Q}']=\mathrm{Sc}\bigl(\bar{\tilde{Q}}\,\tilde{Q}'\bigr)=\sum_{\mu}\varepsilon_{\mu}\bar Q_{\mu}Q'_{\mu}, $$

with the diagonal values

$$ B(\tilde{Q},\tilde{Q})=N(\tilde{Q})=\sum_{\mu}Q_{\mu}^{2}, \qquad \langle\tilde{Q},\tilde{Q}\rangle=\|\tilde{Q}\|_{E}^{2}=\sum_{\mu}|Q_{\mu}|^{2}, \qquad [\tilde{Q},\tilde{Q}]=\sum_{\mu}\varepsilon_{\mu}|Q_{\mu}|^{2}. $$

Proof. Substituting the definitions of ${}^{\natural}$, ${}^{*}$ and $\bar{\cdot}$ and using $\mathrm{Sc}(e_{\mu}e_{\nu})=\varepsilon_{\mu}\delta_{\mu\nu}$ gives each identity; the diagonal values follow, and the first is the biquaternion norm of Biquaternion Norm and Invertibility.

Proposition (the three Gram matrices). In the coefficient basis the Gram matrices are

$$ G_{B}=\mathrm{I}_4,\qquad G_{\langle\cdot,\cdot\rangle}=\mathrm{I}_4,\qquad G_{[\cdot,\cdot]}=\mathrm{E}=\mathrm{diag}(1,-1,-1,-1). $$

Proof. The Gram entries are $\varepsilon^{(\sigma)}_{\mu}\varepsilon_{\mu}$ by the proposition above: for ${}^{\natural}$ and ${}^{*}$ this is $\varepsilon_{\mu}^{2}=1$, for $\bar{\cdot}$ it is $\varepsilon_{\mu}$.

Remark (the identity and the reversal). Completing the picture, the identity conjugation gives the unsymmetric product pairing $\mathrm{Sc}(\tilde{Q}\tilde{Q}')$ with Gram matrix $\mathrm{E}$, and the reversal gives $-\langle\tilde{Q},\tilde{Q}'\rangle$, negative definite. Neither is used in the category; they are listed because they show that the sign vector $\varepsilon$ belongs to the identity and to the complex conjugation, while the natural conjugation is the one that removes it.

The Forms in Comparison

The three forms are collected in one table; each entry is defined before it is used, and the signatures are those of the underlying real quadratic or Hermitian form.

form	conjugation	in the first argument	symmetry	Gram matrix	signature over $\mathbb{R}$	diagonal object
$B$ bilinear	${}^{\natural}$ (linear)	linear	symmetric, $\mathbb{C}$-bilinear	$\mathrm{I}_4$	$(4,4)$ for $\mathrm{Re}\,B$	the norm $N$
$\langle\cdot,\cdot\rangle$ Hermitian	${}^{*}$ (antilinear)	conjugate-linear	Hermitian, positive definite	$\mathrm{I}_4$	$(8,0)$	$\\|\tilde{Q}\\|_{E}^{2}$
$[\cdot,\cdot]$ Krein	$\bar{\cdot}$ (antilinear)	conjugate-linear	Hermitian, indefinite	$\mathrm{E}$	$(2,6)$	the Krein square

Proof of the signatures. $B$ is a complex bilinear form; writing $Q_{\mu}=q_{\mu}+iq'_{\mu}$ gives $\mathrm{Re}\,B=\sum_{\mu}(q_{\mu}^{2}-{q'_{\mu}}^{2})$, of signature $(4,4)$ over the eight real coordinates. The Hermitian form is $\sum_{\mu}|Q_{\mu}|^{2}$, positive definite of rank $8$. The Krein form is $\sum_{\mu}\varepsilon_{\mu}|Q_{\mu}|^{2}$ with the signs $\varepsilon$, hence of signature $(2,6)$: the positive part is the centre and the negative part the vector subspace (The Biquaternion Krein Form and Its Signature, §The Signature).

Theorem (the three forms are mutually non-degenerate and no two agree). All three pairings are non-degenerate; $B$ is not sesquilinear, $\langle\cdot,\cdot\rangle$ is definite and $[\cdot,\cdot]$ is not, and $B$ is $\mathbb{C}$-valued while the other two are real on the diagonal. Nevertheless they determine one another by the identities

$$ [\tilde{Q},\tilde{Q}']=\bigl\langle\tilde{Q}^{\natural},\tilde{Q}'\bigr\rangle, \qquad B(\bar{\tilde{Q}},\tilde{Q}')=\bigl\langle\tilde{Q},\tilde{Q}'\bigr\rangle, \qquad [\tilde{Q},\tilde{Q}']=B\bigl((\bar{\tilde{Q}})^{\natural},\tilde{Q}'\bigr). $$

Proof. Non-degeneracy follows from the invertible Gram matrices $\mathrm{I}_4$ and $\mathrm{E}$. For the first identity, $\langle\tilde{Q}^{\natural},\tilde{Q}'\rangle=\sum_{\mu}\overline{\varepsilon_{\mu}Q_{\mu}}Q'_{\mu}=\sum_{\mu}\varepsilon_{\mu}\bar Q_{\mu}Q'_{\mu}=[\tilde{Q},\tilde{Q}']$, using $\varepsilon_{\mu}^{2}=1$ and $\varepsilon_{\mu}\in\mathbb{R}$; the second is $\sum_{\mu}\overline{\bar Q_{\mu}}Q'_{\mu}=\sum_{\mu}Q_{\mu}Q'_{\mu}$; the third composes the two, since $(\bar{\tilde{Q}})^{\natural}=\overline{(\tilde{Q}^{\natural})}$ is the element whose coefficients are $\varepsilon_{\mu}\bar Q_{\mu}$.

The Adjoints and the Three Isometry Groups

Each form has its own adjoint on the operators and its own group of isometries. The first structural fact is a dichotomy in the involution group.

Theorem (the adjoint dichotomy). Let $\sigma$ be an involution of the group and $\Phi_{\sigma}$ its pairing, and let $L_{\tilde{Q}}$ be the left multiplication. Then

$$ \Phi_{\sigma}\bigl(L_{\tilde{Q}}\tilde{P},\tilde{S}\bigr)=\Phi_{\sigma}\bigl(\tilde{P},L_{\sigma(\tilde{Q})}\tilde{S}\bigr) \ \ \text{if }\sigma\text{ is an anti-automorphism}, $$ $$ \Phi_{\sigma}\bigl(L_{\tilde{Q}}\tilde{P},\tilde{S}\bigr)=\Phi_{\sigma}\bigl(\tilde{P},R_{\sigma(\tilde{Q})}\tilde{S}\bigr) \ \ \text{if }\sigma\text{ is an automorphism}, $$

so the adjoint of a left multiplication is a left multiplication for the bilinear and Hermitian forms, and a right multiplication for the Krein form. Consequently

$$ \bigl(L_{\tilde{Q}}\bigr)^{B}=L_{\tilde{Q}^{\natural}},\qquad \bigl(L_{\tilde{Q}}\bigr)^{\langle\cdot,\cdot\rangle}=L_{\tilde{Q}^{*}},\qquad \bigl(L_{\tilde{Q}}\bigr)^{[\cdot,\cdot]}=R_{\bar{\tilde{Q}}}. $$

Proof. $\Phi_{\sigma}(L_{\tilde{Q}}\tilde{P},\tilde{S})=\mathrm{Sc}(\sigma(\tilde{Q}\tilde{P})\tilde{S})$. If $\sigma$ is an anti-automorphism this is $\mathrm{Sc}(\sigma(\tilde{P})\sigma(\tilde{Q})\tilde{S})=\mathrm{Sc}(\sigma(\tilde{P})(\sigma(\tilde{Q})\tilde{S}))$, which is $\Phi_{\sigma}(\tilde{P},L_{\sigma(\tilde{Q})}\tilde{S})$. If $\sigma=\bar{\cdot}$ is the automorphism, it is $\mathrm{Sc}(\bar{\tilde{Q}}\,\bar{\tilde{P}}\tilde{S})$; by the cyclicity $\mathrm{Sc}(PQR)=\mathrm{Sc}(QRP)$ this is $\mathrm{Sc}(\bar{\tilde{P}}\tilde{S}\bar{\tilde{Q}})=\mathrm{Sc}(\bar{\tilde{P}}(\tilde{S}\bar{\tilde{Q}}))=\Phi_{\sigma}(\tilde{P},R_{\bar{\tilde{Q}}}\tilde{S})$. The three adjoints are the substitutions $\sigma={}^{\natural},{}^{*},\bar{\cdot}$.

Remark. The third line is the structural surprise of J-Self-Adjoint and J-Unitary Operators on the Biquaternion Algebra: because the Krein form is built on the automorphism $\bar{\cdot}$, its adjoint changes the side. The reason is exactly the dichotomy above: the cyclicity of $\mathrm{Sc}$ transports the parameter to the other side.

Theorem (the three isometry groups). The $\mathbb{C}$-linear operators preserving each form are, in the coefficient basis,

$$ \mathrm{Isom}(B)=\{T:T^{\mathsf T}T=\mathrm{I}_4\}=O_4(\mathbb{C}),\qquad \mathrm{Isom}\bigl(\langle\cdot,\cdot\rangle\bigr)=U(4), $$ $$ \mathrm{Isom}\bigl([\cdot,\cdot]\bigr)=\{T:T^{*}\mathrm{E}T=\mathrm{E}\}=U(1,3), $$

with real dimensions $12$, $16$ and $16$. The first is non-compact and complex, the second is compact, and the third is the non-compact real form of $GL_4(\mathbb{C})$ attached to the indefinite form.

Proof. A pairing with Gram matrix $G$ is preserved by $T$ exactly when $T^{\mathsf T}G T=G$ for the $\mathbb{C}$-bilinear case and $T^{*}GT=G$ for the Hermitian case, where $T^{\mathsf T}$ is the transpose and $T^{*}$ the conjugate transpose. With $G=\mathrm{I}_4$ the first gives $O_4(\mathbb{C})$ and the second $U(4)$; with $G=\mathrm{E}$ the Hermitian case gives the isometry group of a form of signature $(1,3)$ on $\mathbb{C}^{4}$, which is $U(1,3)$. The dimensions are the classical ones: $\dim_{\mathbb{R}}O_4(\mathbb{C})=2\cdot6=12$, $\dim_{\mathbb{R}}U(4)=16$, $\dim_{\mathbb{R}}U(1,3)=16$.

Remark (the natural conjugation is common to all three). The map $J={}^{\natural}$ preserves all three forms, $B(J\tilde{Q},J\tilde{Q}')=B(\tilde{Q},\tilde{Q}')$, $\langle J\tilde{Q},J\tilde{Q}'\rangle=\langle\tilde{Q},\tilde{Q}'\rangle$ and $[J\tilde{Q},J\tilde{Q}']=[\tilde{Q},\tilde{Q}']$; it is an element of each isometry group, and in the Krein case it is the fundamental symmetry of The Fundamental Symmetry of the Biquaternion Algebra.

Proof. Each identity is $\sum_{\mu}\varepsilon_{\mu}^{2}(\cdots)=\sum_{\mu}(\cdots)$ in the appropriate coefficient expression; in the Krein case the signs occur twice.

Worked Examples

The units. $B(e_0,e_0)=1$, $\langle e_0,e_0\rangle=1$, $[e_0,e_0]=1$: the identity is positive for all three forms.

The vector units. $B(e_1,e_1)=1$ while $[e_1,e_1]=-1$ and $\langle e_1,e_1\rangle=1$: the vector direction is positive for the bilinear and Hermitian forms and negative for the Krein form. This single contrast is the origin of the $E$-matrix.

A null element. $\tilde{Q}=e_1+ie_2$: $N(\tilde{Q})=1+i^{2}=0$, so it is $B$-null and a zero divisor, while $[\tilde{Q},\tilde{Q}]=-2$ and $\langle\tilde{Q},\tilde{Q}\rangle=2$: null for one pairing and not for the others.

An isotropic element for the Krein form only. $\tilde{Q}=e_0+e_1$: $[\tilde{Q},\tilde{Q}]=1-1=0$, while $N(\tilde{Q})=2$ and $\langle\tilde{Q},\tilde{Q}\rangle=2$.

A pair on both sides. $\tilde{Q}=e_0+ie_1$: $N(\tilde{Q})=0$ and $[\tilde{Q},\tilde{Q}]=0$ at once — a zero divisor that is also Krein-isotropic.

An isometry of one form only. A hyperbolic boost, $e_0\mapsto\cosh t\,e_0+\sinh t\,e_1$ and $e_1\mapsto\sinh t\,e_0+\cosh t\,e_1$ with $e_2,e_3$ fixed, preserves the Krein form and not the other two: it is in $U(1,3)$ and in no unitary group of the definite form (The Krein Isometry Group and Its $J$-Contractions).

Summary

The four conjugations of the algebra generate a Klein four-group, and each non-identity element $\sigma$ gives the pairing $\Phi_{\sigma}(\tilde{Q},\tilde{Q}')=\mathrm{Sc}(\sigma(\tilde{Q})\tilde{Q}')$. The natural conjugation ${}^{\natural}$ gives the symmetric $\mathbb{C}$-bilinear form $B=\sum Q_{\mu}Q'_{\mu}$ with diagonal the norm $N$; the Hermitian conjugation ${}^{*}$ gives the positive definite form $\langle\cdot,\cdot\rangle=\sum\bar Q_{\mu}Q'_{\mu}$; the complex conjugation $\bar{\cdot}$ gives the indefinite Krein form $[\cdot,\cdot]=\sum\varepsilon_{\mu}\bar Q_{\mu}Q'_{\mu}$. The Gram matrices are $\mathrm{I}_4$, $\mathrm{I}_4$ and $\mathrm{E}=\mathrm{diag}(1,-1,-1,-1)$, and the signatures over $\mathbb{R}$ are $(4,4)$ for the real part of $B$, $(8,0)$ and $(2,6)$. The three forms determine one another by $[\tilde{Q},\tilde{Q}']=\langle\tilde{Q}^{\natural},\tilde{Q}'\rangle$ and $B(\bar{\tilde{Q}},\tilde{Q}')=\langle\tilde{Q},\tilde{Q}'\rangle$. The adjoint of a left multiplication is a left multiplication for the two anti-automorphic involutions and a right multiplication for the automorphic one, so $(L_{\tilde{Q}})^{B}=L_{\tilde{Q}^{\natural}}$, $(L_{\tilde{Q}})^{\langle\cdot,\cdot\rangle}=L_{\tilde{Q}^{*}}$ and $(L_{\tilde{Q}})^{[\cdot,\cdot]}=R_{\bar{\tilde{Q}}}$. The isometry groups are $O_4(\mathbb{C})$, $U(4)$ and $U(1,3)$, and the natural conjugation $J={}^{\natural}$ is an isometry of all three.

Summary of Notation

Symbol	Meaning
$\Phi_{\sigma}(\tilde{Q},\tilde{Q}')=\mathrm{Sc}(\sigma(\tilde{Q})\tilde{Q}')$	The pairing of the involution $\sigma$
$\{{}^{\natural},\bar{\cdot},{}^{*},\flat\}$	The four conjugations; ${}^{}=\bar{\cdot}\circ{}^{\natural}$, $\flat=-{}^{}$
$B(\tilde{Q},\tilde{Q}')=\sum Q_{\mu}Q'_{\mu}$	The bilinear form; $B(\tilde{Q},\tilde{Q})=N(\tilde{Q})$
$\langle\tilde{Q},\tilde{Q}'\rangle=\sum\bar Q_{\mu}Q'_{\mu}$	The Hermitian form; $\langle\tilde{Q},\tilde{Q}\rangle=\lVert\tilde{Q}\rVert_E^{2}$
$[\tilde{Q},\tilde{Q}']=\sum\varepsilon_{\mu}\bar Q_{\mu}Q'_{\mu}$	The Krein form
$\mathrm{I}_4,\ \mathrm{I}_4,\ \mathrm{E}=\mathrm{diag}(1,-1,-1,-1)$	The three Gram matrices
$(4,4)$, $(8,0)$, $(2,6)$	The signatures of $B$, $\langle\cdot,\cdot\rangle$, $[\cdot,\cdot]$
$(L_{\tilde{Q}})^{B}=L_{\tilde{Q}^{\natural}}$, $(L_{\tilde{Q}})^{\langle\cdot,\cdot\rangle}=L_{\tilde{Q}^{*}}$, $(L_{\tilde{Q}})^{[\cdot,\cdot]}=R_{\bar{\tilde{Q}}}$	The three adjoints
$O_4(\mathbb{C})$, $U(4)$, $U(1,3)$	The three isometry groups