The Clifford Structure of the Biquaternion Algebra

Introduction

The biquaternion algebra is a Clifford algebra in several compatible ways. It is the even subalgebra of the Minkowski Clifford algebra, $\mathbb{B}\cong\mathrm{Cl}^{+}_{3,1}$; it is a three-dimensional positive definite Clifford algebra, $\mathbb{B}\cong\mathrm{Cl}_{3,0}$ as real algebras; and it is the two-dimensional complex Clifford algebra, $\mathbb{B}\cong\mathbb{C}\mathrm{l}_2$, as complex algebras. This article reads those identifications, the volume element that makes the algebra complex, and the dictionary between the four Clifford grades and the four biquaternion components. The identifications are the form-level content of the biquaternion norm: the vector space of $\mathrm{Cl}_{3,0}$, with its positive definite form, is the imaginary quaternion space of the algebra, so the Clifford structure is the norm read as a pairing between the algebra and itself. A final section reads the four conjugations of the algebra as the intrinsic order-two maps of this Clifford algebra.

The article is the Clifford entry of the Topology slot, after Biquaternion Norm and Invertibility, whose polar form it uses throughout. The classification of the low-dimensional Clifford algebras, the centrality of the volume element of an odd-dimensional algebra, the grade decomposition and the Hodge star are Part II results; they are cited and not re-derived. The matrix model, the idempotents and the Peirce decomposition are Biquaternion 2×2 Matrix Element Representation and Biquaternion Ideals and Peirce Decomposition; the spinor module attached to $\mathbb{B}\cong\mathrm{Cl}^{+}_{3,1}$ is Biquaternion Spin Geometry.

Conventions. The biquaternion 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 $Q_\mu\in\mathbb{C}$. The quaternion products are $e_1e_2=e_3$, $e_2e_3=e_1$, $e_3e_1=e_2$ and $e_k^2=-e_0$, so that $e_1e_2e_3=-e_0$. The biquaternion norm and its polar form are $N(\tilde{Q})=\sum_\mu Q_\mu^2$ and $B(\tilde{P},\tilde{Q})=\sum_\mu P_\mu Q_\mu$ (Biquaternion Norm and Invertibility). A Clifford algebra is written $\mathrm{Cl}_{p,q}$ when $p$ generators square to $+1$ and $q$ to $-1$, in the series convention $v^2=q(v)\cdot1$.

The Even Part of the Minkowski Clifford Algebra

Theorem (the Clifford identification). The biquaternion algebra is the even subalgebra of the real Clifford algebra of signature $(3,1)$: $$ \mathbb{B}\cong\mathrm{Cl}^{+}_{3,1}. $$

Proof. Let $\Gamma_1,\Gamma_2,\Gamma_3,\Gamma_4$ generate $\mathrm{Cl}_{3,1}$, so that $\Gamma_k^2=+1$ for $k=1,2,3$, $\Gamma_4^2=-1$, and $\Gamma_j\Gamma_k=-\Gamma_k\Gamma_j$ for $j\neq k$. In the even part the three products $\Gamma_1\Gamma_2$, $\Gamma_2\Gamma_3$, $\Gamma_3\Gamma_1$ pairwise anticommute and each squares to $-1$, since $(\Gamma_i\Gamma_j)^2=-\Gamma_i^2\Gamma_j^2=-1$ when $i,j\leq3$; they therefore generate a copy of the quaternion algebra. The products $\Gamma_k\Gamma_4$ square to $+1$, since $(\Gamma_k\Gamma_4)^2=-\Gamma_k^2\Gamma_4^2=+1$. The even part is spanned by $1$ and these six bivectors, so it is eight-dimensional over $\mathbb{R}$ and is generated by a quaternion subalgebra together with the central element $\Gamma_1\Gamma_2\Gamma_3\Gamma_4$; that is $\mathbb{H}\otimes_{\mathbb{R}}\mathbb{C}=\mathbb{B}$ after the identification of the last element with the scalar imaginary.

Remark (the competing labelling). The notation $\mathrm{Cl}_{3,1}$ records three generators of square $+1$ and one of square $-1$; the opposite convention is written $\mathrm{Cl}_{1,3}$, with one generator of square $+1$ and three of square $-1$. The two even subalgebras are the same algebra, $$ \mathrm{Cl}^{+}_{3,1}\cong\mathrm{Cl}^{+}_{1,3}\cong M_2(\mathbb{C})\cong\mathbb{B}, $$ because the change of labelling reverses the sign of the form on one generator, and the even part of an even-dimensional Clifford algebra is insensitive to that sign. The full algebras, by contrast, differ: $\mathrm{Cl}_{3,1}\cong M_4(\mathbb{R})$ and $\mathrm{Cl}_{1,3}\cong M_2(\mathbb{H})$. The two labellings therefore agree on $\mathbb{B}$ and disagree on its ambient algebra, which is why this article fixes $\mathrm{Cl}_{3,1}$ and records $\mathrm{Cl}_{1,3}$ as the competing form. The sign consequence is explicit: in $\mathrm{Cl}_{3,1}$ it is the bivectors $\Gamma_i\Gamma_j$ with $i,j\leq3$ that square to $-1$ and generate the quaternion copy, while the bivectors $\Gamma_k\Gamma_4$ square to $+1$ and generate a copy of $\mathrm{Cl}_{3,0}$; reversing the labelling exchanges the two roles but leaves the even part, and so the algebra, unchanged. The general classification of the full algebras is The Clifford, Pin and Spin Groups with Signed Inner Conjugation and Real Spinors and Reality Conditions with Inner Conjugation.

Remark (the signature is a choice of generators). The two labellings above fix a signature; a Clifford algebra does not. In a Euclidean $\mathrm{Cl}_{4,0}$, with generators $e_0,e_1,e_2,e_3$ all of square $+1$, put $$ \gamma_0 = e_0, \qquad \gamma_k = e_k e_0 \quad (k=1,2,3). $$ Then $\gamma_k^2 = -e_k^2e_0^2 = -1$ and $\gamma_0^2 = +1$, and the $\gamma_\mu$ anticommute, so $\gamma_\mu\gamma_\nu+\gamma_\nu\gamma_\mu = 2\eta_{\mu\nu}$ with $\eta = \mathrm{diag}(+,-,-,-)$: the four elements generate the same sixteen-dimensional algebra and carry the Minkowski relation. A Minkowski vector space is thus a choice of generators inside a Euclidean Clifford algebra, and the choice is invisible to the full algebra while it changes the even part. The volume element of the Euclidean generators, $\omega = e_0e_1e_2e_3$, satisfies $\omega^2 = (-1)^{6}(+1)^4 = +1$, so $\mathrm{Cl}^{+}_{4,0}$ contains a central element of square $+1$ and splits; the volume element of the Minkowski generators is $\gamma_0\gamma_1\gamma_2\gamma_3 = e_1e_2e_3$, of square $-1$, and the even part of the Minkowski grading is spanned by $1$ and the products of pairs, that is by $1, e_1, e_2, e_3, e_1e_2, e_1e_3, e_2e_3, e_1e_2e_3$ — the algebra generated by $e_1, e_2, e_3$, which is $\mathrm{Cl}_{3,0}\cong M_2(\mathbb{C})\cong\mathbb{B}$ by the companion identification above. Both statements were checked by direct multiplication of the sixteen basis blades of $\mathrm{Cl}_{4,0}$. The identification of this article is therefore not that $\mathbb{B}$ is the even part of the four-dimensional Clifford algebra, but that it is the even part for the choice of generators whose volume element squares to $-1$; the corpus's fixed signature is a selection of that kind, and a programme that lets the signature vary — as extended relativity in Clifford spaces does, where the signature is relative to a chosen slice through the polyvector arena — is choosing the generators per slice rather than once. The arena alternative is recorded in Curved Spacetime and the Biquaternion Framework.

The Real and the Complex Identifications

Theorem (the companion identifications). There are algebra isomorphisms $$ \mathbb{B}\cong\mathrm{Cl}_{3,0}\quad(\text{as real algebras}),\qquad \mathbb{B}\cong\mathbb{C}\mathrm{l}_2\quad(\text{as complex algebras}), $$ and the second is the complexification of the quaternion algebra, $\mathbb{B}\cong\mathbb{H}\otimes_{\mathbb{R}}\mathbb{C}$.

Proof. For the first, let $\gamma_1,\gamma_2,\gamma_3$ generate $\mathrm{Cl}_{3,0}$, with $\gamma_k^2=+1$ and $\gamma_j\gamma_k=-\gamma_k\gamma_j$ for $j\neq k$, and let $e_k$ be the quaternion units of $\mathbb{B}$, with $e_k^2=-e_0$. The assignment $\gamma_k\mapsto ie_k$ sends the generators to elements of square $(ie_k)^2=i^2e_k^2=(-1)(-1)=+1$ that anticommute, since $i$ is central and the $e_k$ anticommute; it extends to an algebra isomorphism because both algebras are eight-dimensional and the images generate. For the second, $\mathrm{Cl}_{3,0}$ has a central volume element of square $-1$, which supplies a complex structure on the real algebra; the complex algebra so obtained has complex dimension two, a two-dimensional space of generators, and is therefore $\mathbb{C}\mathrm{l}_2$, which is $M_2(\mathbb{C})$. The identification with $\mathbb{H}\otimes_{\mathbb{R}}\mathbb{C}$ is the definition of $\mathbb{B}$ combined with the real description $\mathrm{Cl}_{3,0}\cong\mathbb{H}\oplus i\mathbb{H}$ read as a complex algebra.

Remark. The three descriptions are not interchangeable. As a real algebra $\mathbb{B}$ carries a real structure — a signature, a conjugation, reality conditions — and is the Clifford algebra of a three-dimensional positive definite form; as a complex algebra it carries a complex-linear structure and is the Clifford algebra of a two-dimensional complex form. The complexification $\mathbb{C}\mathrm{l}_3=\mathrm{Cl}_{3,0}\otimes_{\mathbb{R}}\mathbb{C}$ is $M_2(\mathbb{C})\times M_2(\mathbb{C})$, the two-factor algebra of odd complex dimension, and is not $\mathbb{B}$; the identification is with $\mathbb{C}\mathrm{l}_2$, one dimension lower.

The Volume Element and the Central Scalar

Theorem. Let $\gamma_1,\gamma_2,\gamma_3$ generate $\mathrm{Cl}_{3,0}$ and let $\omega=\gamma_1\gamma_2\gamma_3$ be the volume element. Then $\omega$ is central, $\omega^2=-1$, and under the isomorphism $\gamma_k\mapsto ie_k$ it is the central scalar imaginary of the algebra: $$ \omega\longmapsto(ie_1)(ie_2)(ie_3)=i^3e_1e_2e_3=(-i)(-e_0)=i\,e_0 . $$

Proof. The volume element of an odd-dimensional Clifford algebra is central: in a product of $n$ generators the monomial $\omega$ can be moved past each generator at the cost of the sign $(-1)^{n-1}$, so for $n=3$ one has $\omega\gamma_k=(-1)^{n-1}\gamma_k\omega=\gamma_k\omega$ for every $k$, and $\omega$ commutes with all of them. Its square is $\omega^2=(-1)^{n(n-1)/2}\prod_k\gamma_k^2=(-1)^3(+1)^3=-1$. The image is computed directly from $e_1e_2e_3=-e_0$.

Corollary. The centre of $\mathbb{B}$ is $\mathbb{C}e_0$, and the central scalar imaginary is the Clifford volume element. The algebra is therefore central simple over $\mathbb{C}$: it is $M_2(\mathbb{C})$, with one isomorphism class of simple module, of complex dimension two.

Remark (the complex structure). Because $\omega$ is central and of square $-1$, it defines a complex structure on $\mathrm{Cl}_{3,0}$ as a real algebra, and this is the complex structure that makes $\mathbb{B}$ a complex algebra. The volume element supplies the scalar imaginary inside the algebra, which is why the real and complex descriptions agree in this dimension and no exterior complexification is needed.

Remark (the volume element in $\mathrm{Cl}_{3,1}$). In the ambient algebra the product of all four generators, $\Omega=\Gamma_1\Gamma_2\Gamma_3\Gamma_4$, satisfies $\Omega^2=(-1)^{6}(+1)^3(-1)=-1$ and is central in the even part; it is the element that the central scalar imaginary of $\mathbb{B}$ becomes under the identification of the theorem above. In an even-dimensional Clifford algebra the volume element is not central in the full algebra, and it is only on the even part — the biquaternion algebra — that it commutes with everything.

Remark (duality and the Hodge star). Multiplication by the volume element is the duality of the Clifford algebra: it maps a vector to a bivector and a scalar to a pseudoscalar, and conversely, so that it exchanges the grades in pairs. In the physics articles the same operation appears, with a sign convention, as the Hodge dual of a bivector, $\tilde F_\star=-i\tilde F$; the general theory is The Volume Element, Duality and the Hodge Star.

The Grades and the Biquaternion Components

Under the identification $\mathbb{B}\cong\mathrm{Cl}_{3,0}$ with $\gamma_k\mapsto ie_k$, the four Clifford grades are the four biquaternion components. The generators are $\gamma_k=ie_k$, so grade one is the imaginary pure quaternions $\mathrm{span}_{\mathbb{R}}\{ie_1,ie_2,ie_3\}$; grade two is $\mathrm{span}_{\mathbb{R}}\{e_1,e_2,e_3\}$, since $(ie_j)(ie_k)=-e_je_k$ is a real pure quaternion; grade zero is the real scalar line $\mathbb{R}e_0$; and grade three is the imaginary scalar line $\mathbb{R}ie_0$, spanned by the volume element. The correspondence is summarised as $$ \tilde{P}\longleftrightarrow S(\tilde{P})e_0+\underbrace{V(\tilde{P})}_{\text{grade 2}}+\underbrace{iV(\tilde{P})}_{\text{grade 1}}+iS(\tilde{P})e_0\ \ (\text{grade 3}), $$ so that the quaternion vector part is the grade-two part and the imaginary quaternion part is the grade-one part. The four grades are exactly the four components isolated by the character projections of the involution analysis of Biquaternion Algebra. The general theory of the grading and of the geometric product is The Geometric Product and the Grade Decomposition.

Remark (the naming). The dictionary is where the literature is easily misled. The scalar/vector terminology standard for quaternions labels the three quaternion units $e_k$ as vectors, whereas in the Clifford reading they are directed areas, of grade two; the geometric grade-one vectors are the imaginary quaternions $ie_k$. Sangwine, Ell and Le Bihan record that the axial and polar terminology of physics has caused further confusion, because the two terms suggest different types of vector for what is one algebraic object, and Suter's summary is quoted with approval: "a quaternion is a scalar plus a bivector". This article uses the geometric names throughout: scalar, vector, bivector, pseudoscalar for grades zero, one, two, three.

Remark. The antisymmetric part of the biquaternion product is the outer product of the Clifford algebra, and it reads back on the vector parts as the cross product: $\tilde{P}\wedge\tilde{Q}=\tfrac12(\tilde{P}\tilde{Q}-\tilde{Q}\tilde{P})=V(\tilde{P})\times V(\tilde{Q})$ for every pair of biquaternions, not only for vector-like ones. The identity and the naming consequences are in Biquaternion Multiplication.

Remark (an open point). The geometric reading of the full product is incomplete. The paper on which the identification is recorded writes the general product as $\tilde{P}\tilde{Q}=S(\tilde{P})S(\tilde{Q})+S(\tilde{P})V(\tilde{Q})+S(\tilde{Q})V(\tilde{P})+V(\tilde{P})V(\tilde{Q})$, splits the last term into its inner and outer parts, and then states explicitly that "a deeper analysis of the biquaternions as a geometric algebra requires further work". The corpus records the grades, the outer product and the duality, and claims no interpretation of a general product beyond them.

The Involutive Structure and the Conjugations

The Four Conjugations

The biquaternion algebra carries four distinguished order-two maps, the conjugations

$$ {}^{\natural}, \qquad \bar{\cdot}, \qquad {}^{*} = {}^{\natural}\circ\bar{\cdot}, \qquad {}^{\flat} = -{}^{*}, $$

and $\{\mathrm{id}, {}^{\natural}, \bar{\cdot}, {}^{*}\}$ is a group isomorphic to the Klein four-group. The formulas, the fixed spaces $\mathbb{C}_{\mathbb{B}}$, $\mathbb{H}_{\mathbb{B}}$, $\mathbb{M}_{+}$, $\mathbb{M}_{-}$ and $\mathrm{Vect}(\mathbb{B})$, and the composition table are Biquaternion Involution Lattice, and only what the Clifford reading uses is recalled here. Under the product the four are not of one kind:

conjugation	linearity over $\mathbb{R}$	product rule	kind
quaternion conjugation ${}^{\natural}$	$\mathbb{C}$-linear	$(\tilde{Q}\tilde{R})^{\natural} = \tilde{R}^{\natural}\,\tilde{Q}^{\natural}$	anti-automorphism of order two
complex conjugation $\bar{\cdot}$	conjugate-linear	$(\tilde{Q}\tilde{R})^{*} = \bar{\tilde{Q}}\bar{\tilde{R}}$	automorphism of order two
Hermitian conjugation ${}^{*}$	conjugate-linear	$(\tilde{Q}\tilde{R})^{\dagger} = \tilde{R}^{}\tilde{Q}^{}$	anti-automorphism of order two
reversal ${}^{\flat}$	conjugate-linear	$(\tilde{Q}\tilde{R})^{\flat} = -\tilde{R}^{\flat}\tilde{Q}^{\flat}$	order two, anti-multiplicative up to sign

Two of them are involutions of the algebra, that is anti-automorphisms of order two; one is a conjugate-linear involutive automorphism; and the fourth is an order-two map that is anti-multiplicative only up to the central sign, hence not an involution. The group has a second reading: with $\mathbb{B} = \mathbb{H}\otimes_{\mathbb{R}}\mathbb{C}$ the four maps are $\mathrm{id}\otimes\mathrm{id}$, ${}^{\natural}_{\mathbb{H}}\otimes\mathrm{id}$, $\mathrm{id}\otimes\mathrm{conj}$ and ${}^{\natural}_{\mathbb{H}}\otimes\mathrm{conj}$, so the involution group of the algebra is the product of the quaternion conjugation of $\mathbb{H}$ and the complex conjugation of $\mathbb{C}$.

The Correspondence with the Clifford Anti-Involutions

The Clifford algebra of the previous sections carries three order-two maps that need no involution of a base ring: reversion $X^{r}$, the grade involution $\alpha$ and their composite, the Clifford conjugation $X^{\natural} = \alpha(X^{r})$; the three together are the sense in which a Clifford algebra is always involutive, and they are the subject of Involutive Clifford Algebras.

Theorem. Under the isomorphism $\mathbb{B}\cong\mathrm{Cl}_{3,0}$ with $\gamma_k\mapsto ie_k$, the four conjugations of $\mathbb{B}$ are the intrinsic order-two maps of the Clifford algebra:

conjugation of $\mathbb{B}$	Clifford partner	sign on a $k$-blade	signs on grades $0,1,2,3$
quaternion conjugation ${}^{\natural}$	Clifford conjugation $X^{\natural} = \alpha(X^{r})$	$(-1)^{k(k+1)/2}$	$+,-,-,+$
complex conjugation $\bar{\cdot}$	grade involution $\alpha$	$(-1)^{k}$	$+,-,+,-$
Hermitian conjugation ${}^{*}$	reversion $X^{r}$	$(-1)^{k(k-1)/2}$	$+,+,-,-$
reversal ${}^{\flat}$	$-$reversion $-X^{r}$	$-(-1)^{k(k-1)/2}$	$-,-,+,+$

Proof. Reversion fixes the generators $\gamma_k$ and negates the bivectors, the grade involution negates the generators and fixes the bivectors, and the Clifford conjugation is their composite. Under $\gamma_k\mapsto ie_k$ the generators are the grade-one elements $ie_k$, the bivectors are the grade-two elements, and the volume element $\gamma_1\gamma_2\gamma_3$ is the central scalar imaginary $i$, of grade three. The conjugation fixing the whole grade-one part is ${}^{*}$, since ${}^{*}(ie_k) = ie_k$ for every $k$; the conjugation negating it and fixing the grade-two part is $\bar{\cdot}$, since $\bar{\cdot}(ie_k) = -ie_k$ and $\bar{\cdot}(e_k) = e_k$; their composite is ${}^{\natural}$, which agrees with both on the generators and is separated from them on the bivectors, where it acts as $-1$ while $\bar{\cdot}$ acts as $+1$. The signs on the four grades, displayed in the last column, agree on both sides, and an order-two map that is multiplicative or anti-multiplicative is determined by its values on a basis of blades.

Remark (the two readings of one conjugation). In the identification with the even part of the Minkowski algebra, $\mathbb{B}\cong\mathrm{Cl}^{+}_{1,3}$, the two intrinsic anti-involutions of the even part, reversion and Clifford conjugation, coincide there, because they differ by the grade involution, which is the identity on the even part; both equal quaternion conjugation. The other two conjugations use the complex structure that the volume element supplies, and the correspondence of the theorem is therefore read in the $\mathrm{Cl}_{3,0}$ picture, where the volume element is the central scalar that makes the three maps distinct.

The Fixed Spaces as Sums of Grades

The four grades of the last section are the real spaces

$$ \text{grade }0 = \mathbb{R}e_0, \qquad \text{grade }1 = \mathrm{span}_{\mathbb{R}}\{ie_1,ie_2,ie_3\}, \qquad \text{grade }2 = \mathrm{span}_{\mathbb{R}}\{e_1,e_2,e_3\}, \qquad \text{grade }3 = \mathbb{R}ie_0 , $$

of real dimensions $1,3,3,1$.

Corollary. The fixed and anti-fixed spaces of the conjugations are the sums of the grades:

conjugation	Clifford partner	fixed space	anti-fixed space
quaternion conjugation ${}^{\natural}$	Clifford conjugation	$\mathbb{C}_{\mathbb{B}} = \text{grade }0\oplus\text{grade }3$	$\mathrm{Vect}(\mathbb{B}) = \text{grade }1\oplus\text{grade }2$
complex conjugation $\bar{\cdot}$	grade involution	$\mathbb{H}_{\mathbb{B}} = \text{grade }0\oplus\text{grade }2$	$i\mathbb{H}_{\mathbb{B}} = \text{grade }1\oplus\text{grade }3$
Hermitian conjugation ${}^{*}$	reversion	$\mathbb{M}_{+} = \text{grade }0\oplus\text{grade }1$	$\mathbb{M}_{-} = \text{grade }2\oplus\text{grade }3$
reversal ${}^{\flat}$	$-$reversion	$\mathbb{M}_{-} = \text{grade }2\oplus\text{grade }3$	$\mathbb{M}_{+} = \text{grade }0\oplus\text{grade }1$

Proof. Each row is the sign of the theorem read on the grade.

The Hermitian and anti-Hermitian subspaces are therefore the two halves of the grade decomposition: $\mathbb{M}_{+}$ the grades zero and one, $\mathbb{M}_{-}$ the grades two and three. The real quaternion form $\mathbb{H}_{\mathbb{B}}$ pairs the scalar with the bivectors, and the centre $\mathbb{C}_{\mathbb{B}}$ pairs it with the pseudoscalar.

The Two Norms

Theorem. For $\tilde{Q} = \sum_\mu Q_\mu e_\mu$,

$$ N(\tilde{Q}) = \tilde{Q}\tilde{Q}^{\natural} = \tilde{Q}^{\natural}\tilde{Q} = \Bigl(\sum_\mu Q_\mu^{2}\Bigr)e_0, \qquad \langle \tilde{Q},\tilde{Q}\rangle = \mathrm{Sc}\bigl(\tilde{Q}^{*}\tilde{Q}\bigr) = \sum_\mu \bar Q_\mu Q_\mu = \sum_\mu |Q_\mu|^{2}. $$

The first is complex, central, multiplicative, $N(\tilde{P}\tilde{Q}) = N(\tilde{P})N(\tilde{Q})$, isotropic, and it is the reduced norm of the algebra, that is the determinant under $\mathbb{B}\cong M_2(\mathbb{C})$; the second is real and positive definite, and it is the Hermitian form of the dagger. The conjugations preserve the units and the zero divisors,

$$ N(\tilde{Q}^{\natural}) = N(\tilde{Q}), \qquad N(\bar{\tilde{Q}}) = (N(\tilde{Q}))^{\natural}, \qquad N(\tilde{Q}^{*}) = (N(\tilde{Q}))^{\natural}, $$

so the group of units $N\neq0$, the norm cone $N = 0$ and the zero-divisor cone are stable under all four.

Proof. $\tilde{Q}^{\natural} = Q_0e_0 - Q_1e_1 - Q_2e_2 - Q_3e_3$, and the product with $\tilde{Q}$ reduces to $\sum_\mu Q_\mu^{2}$ because $e_k^{2} = -e_0$. The multiplicativity follows from ${}^{\natural}$ being an anti-automorphism: $N(\tilde{P}\tilde{Q}) = \tilde{P}\tilde{Q}\,(\tilde{Q})^{\natural}\,\tilde{P}^{\natural} = \tilde{P}N(\tilde{Q})\tilde{P}^{\natural} = N(\tilde{Q})N(\tilde{P})$, with $N(\tilde{Q})$ central. For the Hermitian form, $e_k^{\dagger} = -e_k$ and $\bar Q_\mu Q_\mu = |Q_\mu|^{2}$, and the sum of the squares is positive unless $\tilde{Q} = 0$.

Proposition. The element $\tilde{Q}^{*}\tilde{Q}$ is fixed by ${}^{*}$ and has scalar part $\sum_\mu|Q_\mu|^{2}$, so it is Hermitian and positive semidefinite; it is a scalar only for special $\tilde{Q}$, so the definite norm is read from its scalar part and not from the product alone. Complex conjugation produces no norm: $\tilde{Q}\bar{\tilde{Q}}$ is fixed by $\bar{\cdot}$ but need not be central, and $\tilde{Q} = e_1 + ie_2$ gives $(e_1 + ie_2)(e_1 + ie_2)^{*} = -2 - 2ie_3$.

Proof. The first statement is $(\tilde{Q}^{*}\tilde{Q})^{\dagger} = \tilde{Q}^{*}\tilde{Q}$ and the computation of the scalar part; the second is the displayed computation.

On the unitary slice $U = \{\tilde{Q} : \tilde{Q}^{*}\tilde{Q} = 1\}$ the dagger is the inverse and the Hermitian sandwich is the inner conjugation; $U$ is $U(2)$, its determinant-one part is $SU(2)\cong\mathrm{Spin}(3)$, and $N(\tilde{Q}^{*}\tilde{Q}) = |N(\tilde{Q})|^{2}$ shows that $|N| = 1$ on $U$.

The Type of the Conjugations

Proposition. The centre of $\mathbb{B}$ is $\mathbb{C}_{\mathbb{B}} = \text{grade }0\oplus\text{grade }3 = \mathbb{C}e_0$. Quaternion conjugation acts as the identity on the centre, so it is of the first kind; as an involution of the complex algebra $\mathbb{B}\cong M_2(\mathbb{C})$ it has $\dim_{\mathbb{C}}\mathrm{Sym} = 1$, which is $n(n-1)/2$ for $n = 2$, so it is of symplectic type. Complex conjugation and Hermitian conjugation act on the centre as complex conjugation, so they are of the second kind, with fixed field the real scalars, of index two. The reversal acts on the centre as $\lambda\mapsto-\bar\lambda$ and is not an anti-automorphism, so it carries no kind.

Proof. The centre is the fixed space of quaternion conjugation, of complex dimension one; the dimension count and the classification of the two kinds and the two types are Involutive Clifford Algebras; the last statement is the product rule of the first table.

Remark (the dagger of the general theory). Over a base ring with an involution $\sigma$ the article Involutive Clifford Algebras forms the dagger $X^{\dagger} = \sigma(\alpha(X^{r}))$. On $\mathrm{Cl}_{3,0}$ the base is $\mathbb{R}$ with the identity involution, so that dagger is $\alpha(X^{r})$, the Clifford conjugation, which under $\gamma_k\mapsto ie_k$ is the quaternion conjugation and not the Hermitian conjugation ${}^{*}$ of the biquaternion literature, which is reversion. The glyph ${}^{*}$ therefore denotes two different maps in the two articles, and this article keeps the biquaternion meaning.

Examples

Example (the conjugations of a null element, verdict: ${}^{\natural}$ is $\mathbb{C}$-linear, the other three conjugate-linear). Let $\tilde{Q} = e_1 + ie_2$. Then $\tilde{Q}^{\natural} = -e_1 - ie_2$, $\bar{\tilde{Q}} = e_1 - ie_2$, $\tilde{Q}^{*} = -e_1 + ie_2$ and $\tilde{Q}^{\flat} = e_1 - ie_2$. The two norms separate: $N(\tilde{Q}) = 1 + i^{2} = 0$, while $\sum_\mu|Q_\mu|^{2} = 2$. Verdict: $\tilde{Q}$ is a zero divisor whose Hermitian norm is strictly positive, so the isotropic reduced norm and the definite Hermitian norm disagree as sharply as they can; the map ${}^{\natural}$ is $\mathbb{C}$-linear, the maps $\bar{\cdot}$, ${}^{*}$ and ${}^{\flat}$ are conjugate-linear, and the table above is the complete record of the four.

Example (the correspondence on the generators and the bivectors, verdict: four $\mathbb{R}$-linear order-two maps). Under $\gamma_k\mapsto ie_k$ the generator $\gamma_1$ is the grade-one element $ie_1$, fixed by ${}^{*}$ and negated by ${}^{\natural}$ and $\bar{\cdot}$; the bivector $\gamma_1\gamma_2$ is the grade-two element $-e_3$, fixed by $\bar{\cdot}$ and negated by ${}^{\natural}$ and ${}^{*}$; the volume element $\gamma_1\gamma_2\gamma_3$ is the grade-three element $i$, fixed by ${}^{\natural}$ and ${}^{\flat}$ and negated by $\bar{\cdot}$ and ${}^{*}$. Verdict: all four conjugations are $\mathbb{R}$-linear and of order two, ${}^{\natural}$ and ${}^{*}$ are anti-automorphisms, $\bar{\cdot}$ is an automorphism, and the example exhibits the sign of each on one element of the three nonzero grades.

Summary

The biquaternion algebra is a Clifford algebra in three compatible ways: as the even part of the Minkowski Clifford algebra, $\mathbb{B}\cong\mathrm{Cl}^{+}_{3,1}\cong\mathrm{Cl}^{+}_{1,3}$; as the real Clifford algebra of a three-dimensional positive definite form, $\mathbb{B}\cong\mathrm{Cl}_{3,0}$; and as the two-dimensional complex Clifford algebra, $\mathbb{B}\cong\mathbb{C}\mathrm{l}_2=\mathbb{H}\otimes_{\mathbb{R}}\mathbb{C}$. The identifications agree on the even part and disagree on the ambient algebra, where $\mathrm{Cl}_{3,1}\cong M_4(\mathbb{R})$ and $\mathrm{Cl}_{1,3}\cong M_2(\mathbb{H})$.

The volume element $\omega=\gamma_1\gamma_2\gamma_3$ of $\mathrm{Cl}_{3,0}$ is central, has square $-1$, and is the central scalar imaginary $i$ of the algebra, so that it supplies the complex structure and the algebra needs no exterior complexification. Its counterpart in $\mathrm{Cl}_{3,1}$ is the product $\Omega=\Gamma_1\Gamma_2\Gamma_3\Gamma_4$, of square $-1$, central in the even part. Multiplication by the volume element is the duality that exchanges the grades, and it is the Hodge star of the physics convention, $\tilde F_\star=-i\tilde F$.

The four Clifford grades are the four biquaternion components: grade zero the real scalar, grade one the imaginary pure quaternion, grade two the real pure quaternion, grade three the imaginary scalar. The quaternion vector part is therefore the grade-two part, and the geometric vector part the grade-one part, a reversal of names that the literature records as a trap.

The algebra is involutive in the Clifford sense. Its four conjugations ${}^{\natural}$, $\bar{\cdot}$, ${}^{*}$ and ${}^{\flat}$ are order-two maps, and under $\gamma_k\mapsto ie_k$ they are the intrinsic order-two maps of $\mathrm{Cl}_{3,0}$: quaternion conjugation is the Clifford conjugation $\alpha(X^{r})$, of signs $(-1)^{k(k+1)/2}$ on the grades; complex conjugation is the grade involution $\alpha$, of signs $(-1)^{k}$; Hermitian conjugation is reversion $X^{r}$, of signs $(-1)^{k(k-1)/2}$; and the reversal is the negative of reversion. The fixed spaces are the grade sums: $\mathbb{C}_{\mathbb{B}}$ is the grades zero and three, $\mathbb{H}_{\mathbb{B}}$ the grades zero and two, and the Hermitian and anti-Hermitian subspaces $\mathbb{M}_{+}$ the grades zero and one and $\mathbb{M}_{-}$ the grades two and three. Quaternion conjugation is of the first kind and symplectic, while complex conjugation and Hermitian conjugation are of the second kind with fixed field the real scalars; quaternion conjugation produces the isotropic reduced norm $N(\tilde{Q}) = \sum_\mu Q_\mu^{2}$ and Hermitian conjugation the positive definite Hermitian form $\sum_\mu|Q_\mu|^{2}$, whose unitary slice is $U(2)$.

Summary of Notation

Symbol	Meaning
$\mathrm{Cl}_{3,1},\ \mathrm{Cl}_{1,3}$	Real Clifford algebras of the two signature labellings; the even parts agree, $\mathbb{B}\cong\mathrm{Cl}^{+}_{3,1}\cong\mathrm{Cl}^{+}_{1,3}$
$\mathrm{Cl}_{3,1}\cong M_4(\mathbb{R}),\ \mathrm{Cl}_{1,3}\cong M_2(\mathbb{H})$	The full algebras; distinct over $\mathbb{R}$, same even part
$\mathrm{Cl}_{3,0}$	Real Clifford algebra of a three-dimensional positive definite form; $\mathbb{B}\cong\mathrm{Cl}_{3,0}$
$\mathbb{C}\mathrm{l}_2=M_2(\mathbb{C})$	Two-dimensional complex Clifford algebra; $\mathbb{B}\cong\mathbb{C}\mathrm{l}_2$
$\gamma_k\mapsto ie_k$	Isomorphism $\mathrm{Cl}_{3,0}\to\mathbb{B}$
$\omega=\gamma_1\gamma_2\gamma_3\mapsto i$	Volume element of $\mathrm{Cl}_{3,0}$; central, $\omega^2=-1$
$\Omega=\Gamma_1\Gamma_2\Gamma_3\Gamma_4\,(\Omega^2=-1)$	Volume element of $\mathrm{Cl}_{3,1}$, central in the even part
$\tilde{P}\wedge\tilde{Q}=\tfrac12(\tilde{P}\tilde{Q}-\tilde{Q}\tilde{P})=V(\tilde{P})\times V(\tilde{Q})$	Outer product; the cross product of the vector parts (Biquaternion Multiplication)
grade 0 / 1 / 2 / 3	real scalar / imaginary pure quaternion / real pure quaternion / imaginary scalar
${}^{\natural},\ \bar{\cdot},\ {}^{*},\ {}^{\flat}$	the four conjugations, identified with $\alpha(X^{r})$, $\alpha$, $X^{r}$, $-X^{r}$ on $\mathrm{Cl}_{3,0}$
$\mathbb{M}_{+},\ \mathbb{M}_{-}$	Hermitian and anti-Hermitian subspaces; grade $0\oplus$ grade $1$ and grade $2\oplus$ grade $3$
$N(\tilde{Q}) = \sum_\mu Q_\mu^{2}$	reduced norm, $= \tilde{Q}\tilde{Q}^{\natural}$, isotropic, multiplicative
$\tilde F_\star=-i\tilde F$	Hodge star in the physics convention
$\mathrm{Cl}(\mathbb{B},N)\cong\mathrm{Cl}_4(\mathbb{C})$	Clifford algebra of the biquaternion norm as a quadratic space; a different algebra (Biquaternion Norm and Invertibility)