Biquaternion 2×2 Matrix Element Representation

Introduction

The biquaternion algebra $\mathbb{B} = \mathbb{C} \otimes_{\mathbb{R}} \mathbb{H}$ is the four-dimensional complex algebra with basis $e_0 = 1, e_1, e_2, e_3$, where $e_k^2 = -e_0$ and $e_j e_k = -e_k e_j$ for $j \neq k$, and with a central scalar imaginary $i$ satisfying $i^2 = -1$. A general element is $\tilde{Q} = \sum_{\mu=0}^{3} Q_\mu e_\mu$ with $Q_\mu \in \mathbb{C}$. The algebra, its conjugations, its six distinguished subspaces and its determinant $N$ are those of Biquaternion Algebra.

This article presents the matrix realization of $\mathbb{B}$ together with the module on which the matrices act. It is deeper than the coefficient realization of Biquaternion Four-Vector Element Representation by exactly one level of structure: the four-vector article lists the four coefficients, while the matrix realization exhibits an action. The word representation is therefore used here in both of its senses at once, the concrete realization and the technical representation of an algebra on a vector space, and the module $V = \mathbb{C}^2$ is introduced as the object on which the action is defined. The two realizations agree on the four parameters: they are related by an explicit $\mathbb{C}$-linear isomorphism, stated below.

The article owns the isomorphism $\mathbb{B} \cong M_2(\mathbb{C})$, the trace and the determinant, the matrix form of the four conjugations, the six distinguished subspaces in matrix form, the characteristic polynomial and the spectrum, the Cayley–Hamilton identity, the structural consequences of the isomorphism, and the simple module $V$ with its left action. It deliberately does not treat the spinor reading of $V$, which belongs to Biquaternion Spin Geometry; it does not treat the module-theoretic treatment with matrix units, which belongs to Modules over the Biquaternion Algebra; it does not re-derive the Clifford identification of The Clifford Structure of the Biquaternion Algebra; and it does not reprove the classification of the simple modules or Schur's lemma, which belong to Modules over the Biquaternion Algebra. It does not treat the Jordan structure of the Hermitian subspace, which belongs to The Six Subspaces and the Jordan Algebra and Biquaternion Jordan Algebra, nor the exponential, which belongs to Biquaternion Elementary Functions. No physical vocabulary is used: in particular the matrices below are not gamma matrices and the module is not a spinor of a physical field.

The Isomorphism

Definition. The matrix realization of $\mathbb{B}$ is the $\mathbb{C}$-linear map

$$ \Phi : \mathbb{B} \longrightarrow M_2(\mathbb{C}) $$

determined on the basis by

$$ \Phi(e_0) = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}, \qquad \Phi(e_1) = \begin{pmatrix} 0 & -i \\ -i & 0 \end{pmatrix}, \qquad \Phi(e_2) = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}, \qquad \Phi(e_3) = \begin{pmatrix} -i & 0 \\ 0 & i \end{pmatrix}, $$

and on the central scalar by $\Phi(i e_\mu) = i \Phi(e_\mu)$. The four images are the basis images; the map is fixed by them. On a general element it reads

$$ \Phi(\tilde{Q}) = Q_0 \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} + Q_1 \begin{pmatrix} 0 & -i \\ -i & 0 \end{pmatrix} + Q_2 \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix} + Q_3 \begin{pmatrix} -i & 0 \\ 0 & i \end{pmatrix} = \begin{pmatrix} Q_0 - i Q_3 & -i Q_1 - Q_2 \\ -i Q_1 + Q_2 & Q_0 + i Q_3 \end{pmatrix}. $$

Remark (the choice, stated). The assignment above is the corpus's choice, and it is held fixed so that the matrix realization agrees with Worked Examples in the Biquaternion Algebra and with The Clifford Structure of the Biquaternion Algebra. It is convenient for computation because the three images of $e_1, e_2, e_3$ are $-i$ times the Pauli matrices, and this is mentioned once as a shorthand and nothing more: the algebra treated here is $\mathbb{B}$, not the algebra generated by the Pauli matrices, and no structural statement below is made about the Pauli matrices. Every proposition is a statement about $\mathbb{B}$ and about $M_2(\mathbb{C})$. The choice is not canonical; a different choice of isomorphism differs from this one by conjugation by an invertible matrix, and the invariants below are unchanged.

Remark (linear isomorphism, not a permutation). The four-vector $Q^\mu$ of Biquaternion Four-Vector Element Representation and the matrix $\Phi(\tilde{Q})$ carry the same four complex parameters, but the passage from one to the other is not a rearrangement of them. Each entry of $\Phi(\tilde{Q})$ is a fixed linear combination of the four components with coefficients among $1, -1, i, -i$; for instance the entry $\Phi(\tilde{Q})_{12} = -i Q_1 - Q_2$ involves two components at once. The map between the four-vector and the matrix is a $\mathbb{C}$-linear isomorphism, not a permutation of coordinates.

Remark (other realizations from the light-cone basis). A second family of $2\times2$ models comes from the isotropic (light-cone) basis used in The Chiral Algebra of Biquaternions and the Cyclic Representation of the Dirac Equation. Writing a biquaternion in the null coordinates $(\alpha, \beta, \xi, \eta)$ there, two of that article's four products become ordinary matrix products under two different maps: the outer product under $M = \left(\begin{smallmatrix}\xi & \alpha\\ \beta & \eta\end{smallmatrix}\right)$, and the diagonal product under $F \leftrightarrow \left(\begin{smallmatrix}f_\alpha & -f_\beta\\ -f_\xi & f_\eta\end{smallmatrix}\right)$. The source claims a third, the inner product under $\left(\begin{smallmatrix}\alpha & -\eta\\ \xi & -\beta\end{smallmatrix}\right)$ with subtractions in place of the additions; that claim is false — an exhaustive search over the possible maps and pairing rules finds no realization of the inner product — and only the outer and the diagonal maps are isomorphisms $\mathbb{B} \cong M_2(\mathbb{C})$. They differ from $\Phi$ by the change of basis between the isotropic basis and the corpus's basis, and they are named here only to place the corpus's chosen $\Phi$ among the others; the two articles' products are not the same objects.

The Isomorphism Is an Algebra Isomorphism

Theorem. The map $\Phi$ is an algebra isomorphism $\mathbb{B} \cong M_2(\mathbb{C})$.

Proof. The three images of $e_1, e_2, e_3$ satisfy the quaternion relations. Each squares to $-I$:

$$ \Phi(e_1)^2 = \begin{pmatrix} 0 & -i \\ -i & 0 \end{pmatrix}^2 = \begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix} = -I, $$

and likewise for $e_2$ and $e_3$, since each of these matrices has square $-I$ by direct computation. The cyclic products match:

$$ \Phi(e_1)\Phi(e_2) = \begin{pmatrix} 0 & -i \\ -i & 0 \end{pmatrix}\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix} = \begin{pmatrix} -i & 0 \\ 0 & i \end{pmatrix} = \Phi(e_3), $$

and the reversed product is $-\Phi(e_3)$, so the relations $e_j e_k = -e_k e_j$ for $j \neq k$ are respected. Since $\mathbb{B}$ is generated by $e_0, e_1, e_2, e_3$ subject only to these relations, the $\mathbb{C}$-linear map extending the assignment is multiplicative, that is, an algebra homomorphism. It is bijective because the four images are linearly independent in the four-dimensional space $M_2(\mathbb{C})$: they are a basis, so $\Phi$ is a linear isomorphism as well as a homomorphism.

Corollary (the centre). The centre of $M_2(\mathbb{C})$ is the set of scalar matrices $\lambda I$, $\lambda \in \mathbb{C}$, so the centre of $\mathbb{B}$ is the scalar subspace $\mathbb{C}_{\mathbb{B}} = \mathbb{C} e_0$.

Proof. A matrix commuting with every matrix is a scalar matrix; this is the standard form of Schur's lemma and is recorded in Matrix Algebras. Transport along $\Phi$.

Corollary (simplicity). The algebra $\mathbb{B}$ is simple: its only two-sided ideals are $0$ and $\mathbb{B}$.

Proof. Let $I$ be a nonzero two-sided ideal of $M_2(\mathbb{C})$ and let $M \in I$ have a nonzero entry $M_{kl}$. With the matrix units $E_{ij}$ one has $E_{ik} M E_{lj} = M_{kl} E_{ij}$, so $E_{ij} \in I$ for every pair $(i,j)$; the matrix units span $M_2(\mathbb{C})$, so $I = M_2(\mathbb{C})$. Transport along $\Phi$.

The lattice of one-sided ideals of $\mathbb{B}$ is larger than this corollary suggests, because simplicity is a statement about two-sided ideals only. The full lattice is the subject of Biquaternion Ideals and Peirce Decomposition.

The Trace and the Determinant

Proposition (trace and determinant). For every biquaternion $\tilde{Q}$,

$$ \operatorname{Tr}(\Phi(\tilde{Q})) = 2Q_0, \qquad \det(\Phi(\tilde{Q})) = N(\tilde{Q}) = Q_0^2 + Q_1^2 + Q_2^2 + Q_3^2 . $$

Proof. The trace is the sum of the diagonal entries,

$$ \operatorname{Tr}(\Phi(\tilde{Q})) = (Q_0 - i Q_3) + (Q_0 + i Q_3) = 2Q_0 . $$

The determinant is

$$ \det(\Phi(\tilde{Q})) = (Q_0 - i Q_3)(Q_0 + i Q_3) - (-i Q_1 - Q_2)(-i Q_1 + Q_2) = (Q_0^2 + Q_3^2) - \bigl( (-iQ_1)^2 - Q_2^2 \bigr) = Q_0^2 + Q_1^2 + Q_2^2 + Q_3^2, $$

since $(-i Q_1)^2 = -Q_1^2$ and the cross terms of each product cancel within that product. The result is the determinant of Biquaternion Algebra and of Biquaternion Four-Vector Element Representation.

Corollary (invertibility). A biquaternion $\tilde{Q}$ is invertible if and only if $\det\Phi(\tilde{Q}) = N(\tilde{Q}) \neq 0$, and then $\Phi(\tilde{Q}^{-1}) = \Phi(\tilde{Q})^{-1}$. The zero divisors of $\mathbb{B}$ are exactly the nonzero elements whose matrix is singular.

Proof. A square matrix over a field is invertible if and only if its determinant is nonzero, and $\Phi$ is an isomorphism, so invertibility transports. The matrix is singular precisely when $\det\Phi(\tilde{Q}) = N(\tilde{Q}) = 0$.

The classification of the zero divisors is the subject of Biquaternion Zero Divisors; what the matrix realization adds is the reading of the criterion in terms of the determinant and the rank, developed in §The Rank-One Elements and the Outer Product.

Remark (the determinant is multiplicative, the trace is not). The determinant is multiplicative, $\det\Phi(\tilde{P}\tilde{Q}) = \det\Phi(\tilde{P})\det\Phi(\tilde{Q})$, so $N$ is multiplicative, $N(\tilde{P}\tilde{Q}) = N(\tilde{P})N(\tilde{Q})$. The trace is additive but not multiplicative, and its value $2Q_0$ depends only on the scalar part.

Remark (the two dimensions). The dimension of the simple module introduced below is $2$ over $\mathbb{C}$, and the size of the matrix algebra is $2 \times 2$. These are the same two for the same reason: the algebra of endomorphisms of a $2$-dimensional space is the algebra of $2 \times 2$ matrices, and the isomorphism $\Phi$ is exactly the identification of $\mathbb{B}$ with that endomorphism algebra.

Worked example. For the element $\tilde{Q} = (2+i)e_0 + (1-i)e_1 + 3e_2 + ie_3$ the image is

$$ \Phi(\tilde{Q}) = (2+i)I_2 + (1-i)(-i\sigma_1) + 3(-i\sigma_2) + i(-i\sigma_3) = \begin{pmatrix} 3+i & -4-i \\ 2-i & 1+i \end{pmatrix}, $$

whose trace is $4+2i = 2Q_0$ and whose determinant is $(3+i)(1+i) - (-4-i)(2-i) = 11+2i = N(\tilde{Q})$, so the element is a unit with inverse $\tilde{Q}^{\natural}/N$.

The Rank-One Elements and the Outer Product

The corollary above says that the zero divisors are the nonzero singular matrices; the matrix form also says what such a matrix is.

Proposition. For $\tilde{Q}\neq0$ the following are equivalent: $N(\tilde{Q})=0$; $\Phi(\tilde{Q})$ is singular; $\Phi(\tilde{Q})$ is an outer product $$ \Phi(\tilde{Q})=uv^{T},\qquad u=\binom{\alpha}{\beta}\neq0,\quad v=\binom{\gamma}{\delta}\neq0, $$ the pair $(u,v)$ being determined up to $(u,v)\mapsto(\lambda u,\lambda^{-1}v)$.

Proof. The first two are equivalent by $\det\Phi(\tilde{Q})=N(\tilde{Q})$, and $\tilde{Q}\neq0$ makes the matrix nonzero; a nonzero singular two-by-two matrix has rank one. Let $M=\Phi(\tilde{Q})$ have rank one and let $M_{pq}\neq0$ be an entry. Every two-by-two minor of $M$ vanishes, so $M_{iq}M_{pj}=M_{ij}M_{pq}$ for all $i,j$; hence, with $u$ the $q$-th column and $v^{T}$ the $p$-th row divided by $M_{pq}$, $$ u=\binom{M_{1q}}{M_{2q}},\qquad v^{T}=\frac{1}{M_{pq}}\begin{pmatrix}M_{p1}&M_{p2}\end{pmatrix}, $$ one has $(uv^{T})_{ij}=M_{iq}M_{pj}/M_{pq}=M_{ij}$, so $M=uv^{T}$, and $u\neq0$ because its $p$-th entry is $M_{pq}\neq0$. A different pivot gives a pair differing by precisely the rescaling: the $j$-th columns of $uv^{T}=u'(v')^{T}$ give $u'=\lambda u$ for $\lambda=v_{j}/v'_{j}\neq0$, and the remaining entries give $v'=\lambda^{-1}v$.

Worked example. For $\tilde{Q}=e_2+ie_3$ the image is $\Phi(\tilde{Q})=\begin{pmatrix}1&-1\\1&-1\end{pmatrix}$, of rank one; the pivot $M_{11}=1$ gives $u=(1,1)^{T}$, $v=(1,-1)^{T}$, and the pivot $M_{12}=-1$ gives $u=(-1,-1)^{T}$, $v=(-1,1)^{T}$, the same pair rescaled by $\lambda=-1$.

The Characteristic Polynomial and the Spectrum

Definition. The spectrum of a biquaternion $\tilde{Q}$ is the spectrum of the matrix $\Phi(\tilde{Q})$, that is, its two eigenvalues in $\mathbb{C}$. This is the unqualified meaning of the word in the series; the other inequivalent spectra are the subject of Biquaternion Spectral Theory.

Proposition (characteristic polynomial). The characteristic polynomial of $\Phi(\tilde{Q})$ is

$$ \det\bigl(\lambda I - \Phi(\tilde{Q})\bigr) = \lambda^2 - 2Q_0\,\lambda + N(\tilde{Q}), $$

so the spectrum is $\{Q_0 + iB,\; Q_0 - iB\}$ with $B = \sqrt{Q_1^2 + Q_2^2 + Q_3^2}$, the trace is $2Q_0$, and the determinant is $N(\tilde{Q})$.

Proof. For a $2 \times 2$ matrix the characteristic polynomial is $\lambda^2 - \operatorname{Tr}(M)\lambda + \det(M)$, and $\operatorname{Tr}\Phi(\tilde{Q}) = 2Q_0$, $\det\Phi(\tilde{Q}) = N(\tilde{Q})$ by the trace-and-determinant proposition. The eigenvalues are the two roots $Q_0 \pm \sqrt{Q_0^2 - N(\tilde{Q})}$, and $Q_0^2 - N(\tilde{Q}) = -(Q_1^2 + Q_2^2 + Q_3^2) = -B^2$.

Remark (similarity). Two biquaternions are similar, $\tilde{Q}' = S\tilde{Q}S^{-1}$ with $S$ invertible, exactly when their matrices are similar, so the spectrum is a similarity invariant. The similarity classification, the eigenvalue dichotomy, and the resolvent are the subject of Biquaternion Spectral Theory, which also records the several inequivalent meanings of "spectrum".

The Cayley–Hamilton Identity

Proposition. Every biquaternion satisfies the Cayley–Hamilton identity

$$ \tilde{Q}^2 - 2Q_0\,\tilde{Q} + N(\tilde{Q})\,e_0 = 0 . $$

Proof. Cayley–Hamilton for the $2 \times 2$ matrix $\Phi(\tilde{Q})$ reads $\Phi(\tilde{Q})^2 - \operatorname{Tr}\Phi(\tilde{Q})\,\Phi(\tilde{Q}) + \det\Phi(\tilde{Q})\,I = 0$. Substituting $\operatorname{Tr}\Phi(\tilde{Q}) = 2Q_0$ and $\det\Phi(\tilde{Q}) = N(\tilde{Q})$ and transporting back along the injective map $\Phi$ gives the identity.

Remark. This is the identity that reduces every power of $\tilde{Q}$ to a combination of $e_0$ and $\tilde{Q}$, and it is the starting point for the closed forms of the exponential and the trigonometric functions in Biquaternion Elementary Functions.

The Conjugations in Matrix Form

The four conjugations of $\mathbb{B}$ act on the matrices, and three of the four are read off at once: quaternion conjugation is the adjugate, Hermitian conjugation is the conjugate transpose, and anti-Hermitian conjugation is its negative. Complex conjugation is the one that requires care.

Proposition (quaternion conjugation is the adjugate). For every biquaternion $\tilde{Q}$,

$$ \Phi(\tilde{Q}^{\natural}) = \operatorname{adj}\bigl(\Phi(\tilde{Q})\bigr), \qquad \det\Phi(\tilde{Q}^{\natural}) = N(\tilde{Q}^{\natural}) = N(\tilde{Q}), $$

where $\operatorname{adj}(M)$ is the adjugate of $M$, the transpose of the cofactor matrix. Equivalently, $\Phi(\tilde{Q}^{\natural}) = \epsilon\, \Phi(\tilde{Q})^{\mathsf{T}} \epsilon^{-1}$ with $\epsilon = i\sigma_2$.

Proof. The adjugate of a $2 \times 2$ matrix satisfies $M \operatorname{adj}(M) = \det(M) I$ and has entries the cofactors. Replacing $Q_1, Q_2, Q_3$ by their negatives in the explicit formula for $\Phi(\tilde{Q})$ and comparing with the cofactor matrix gives the identity directly. For the second form, note that $\epsilon \Phi(\tilde{Q})^{\mathsf{T}}\epsilon^{-1}$ equals the adjugate for every $2 \times 2$ matrix $M$, since $\operatorname{adj}(M) = \det(M) M^{-1}$ when $M$ is invertible and the identity is polynomial in the entries. The determinant identity is $N(\tilde{Q}^{\natural}) = N(\tilde{Q})$, the invariance of the determinant $N$ under quaternion conjugation.

Remark (the delicate point: complex conjugation is not entrywise). Complex conjugation does not act entrywise on $\Phi(\tilde{Q})$. The realization is built with the same $i$ that conjugates the coefficients, and the two operations compete: conjugating the entries of $\Phi(\tilde{Q})$ sends $-i\sigma_1 \mapsto +i\sigma_1$, which is the negative of the image of $e_1$, while the image of $e_2$ is a real matrix and is preserved. Entrywise conjugation is therefore not $\Phi(\bar{\tilde{Q}})$, and it is not the image of any of the four involutions of the algebra. The correct correspondence dresses the conjugation with the antisymmetric form

$$ \epsilon = \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix} = i\sigma_2 = \Phi(-e_2), \qquad \epsilon^{\mathsf{T}} = -\epsilon, \qquad \epsilon^2 = -I : $$

$$ \Phi(\bar{\tilde{Q}}) = \epsilon \, (\Phi(\tilde{Q}))^{\natural} \, \epsilon^{-1}, \qquad (\Phi(\tilde{Q}))^{\natural} \neq \Phi(\bar{\tilde{Q}}) \text{ in general}. $$

For $\tilde{Q} = e_1$ the two sides differ by a sign: $\overline{\Phi(e_1)} = +i\sigma_1 = -\Phi(e_1)$, while $\epsilon \overline{\Phi(e_1)}\epsilon^{-1} = \Phi(e_1)$ because $e_1^* = e_1$. For $\tilde{Q} = e_2$ the matrices agree entrywise, since $\Phi(e_2)$ is real. This is the single place in the matrix realization where the plausible guess is wrong, and it is the reason the correspondence is stated with the antisymmetric form and not with transposition alone.

Proposition (Hermitian and anti-Hermitian conjugation). For every biquaternion $\tilde{Q}$,

$$ \Phi(\tilde{Q}^{*}) = \Phi(\tilde{Q})^\dagger, \qquad \Phi(\tilde{Q}^\flat) = -\Phi(\tilde{Q})^\dagger, $$

where ${}^\dagger$ on the right is the conjugate transpose.

Proof. Hermitian conjugation is the composite ${}^{\natural} \circ \bar{\cdot}$, so $\Phi(\tilde{Q}^{*}) = \Phi\bigl(\bar{\bar{\tilde{Q}}}\bigr) = \epsilon\,\Phi(\bar{\tilde{Q}})^{\mathsf{T}}\,\epsilon^{-1}$, and substituting the correspondence for $\bar{\cdot}$ gives

$$ \Phi(\tilde{Q}^{*}) = \epsilon \bigl( \epsilon (\Phi(\tilde{Q}))^{\natural} \epsilon^{-1} \bigr)^{\mathsf{T}} \epsilon^{-1} = \epsilon^{2}\,((\Phi(\tilde{Q}))^{\natural})^{\mathsf{T}}\,\epsilon^{-2}, $$

where the inner transposition passes through the two factors because $\epsilon^{-1} = \epsilon^{\mathsf{T}}$ and hence $(\epsilon M \epsilon^{-1})^{\mathsf{T}} = \epsilon M^{\mathsf{T}} \epsilon^{-1}$ for every $M$. Since $\epsilon^{2} = -I$ and $(-I)^{-1} = -I$, the two outer factors multiply to $I$, so $\Phi(\tilde{Q}^{*}) = ((\Phi(\tilde{Q}))^{\natural})^{\mathsf{T}} = \Phi(\tilde{Q})^\dagger$. The statement for $\flat$ is the definition $\flat = -{}^{*}$.

Proposition (the congruence of a unit). For every unit $\tilde{Q}$ and every $\tilde U$, $$ \Phi\bigl(\tilde{Q}\tilde U\tilde{Q}^{*}\bigr)=\Phi(\tilde{Q})\,\Phi(\tilde U)\,\Phi(\tilde{Q})^{\dagger}, $$ read off from the multiplicativity of $\Phi$ and from $\Phi(\tilde{Q}^{*})=\Phi(\tilde{Q})^{\dagger}$. The identity is the action of the group of units on the matrix algebra by $*$-congruence: it preserves the Hermitian matrices as a set, multiplies the determinant by $\lvert N(\tilde{Q})\rvert^{2}$ and fixes the rank of $\Phi(\tilde U)$.

The Six Subspaces in Matrix Form

The six distinguished subspaces of Biquaternion Four-Vector Element Representation have the following matrix characterizations, each developed in its own article in the Focus on Subspaces group of the series. Each is defined before it is used, and the table records the image under $\Phi$ of the subspace, not a new object.

Subspace	Image under $\Phi$	Description
$\mathbb{C}_{\mathbb{B}}$	$\{ \lambda I : \lambda \in \mathbb{C} \}$	the scalar matrices
$\mathrm{Vect}(\mathbb{B})$	$\{ M : \operatorname{Tr} M = 0 \}$	the traceless matrices, $\mathrm{SL}_2(\mathbb{C})$
$\mathbb{H}_{\mathbb{B}}$	$\left\{ \begin{pmatrix} z & w \\ -\bar{w} & \bar{z} \end{pmatrix} : z, w \in \mathbb{C} \right\}$	the fixed points of $M \mapsto \epsilon\overline{M}\epsilon^{-1}$
$i\mathbb{H}_{\mathbb{B}}$	$\left\{ \begin{pmatrix} u & v \\ \bar{v} & -\bar{u} \end{pmatrix} : u, v \in \mathbb{C} \right\}$	$i$ times the preceding subspace
$\mathbb{M}_+$	$\{ M : M = M^\dagger \}$	the Hermitian matrices
$\mathbb{M}_-$	$\{ M : M = -M^\dagger \}$	the anti-Hermitian matrices

Proof. The centre and the traceless part are read from the trace and the definition of the centre. For the quaternion subspace, $\tilde{Q}$ has real coefficients if and only if $\Phi(\tilde{Q})$ is fixed by $\epsilon \overline{(\cdot)}\epsilon^{-1}$, which is the condition $M_{21} = -\overline{M_{12}}$ and $M_{22} = \overline{M_{11}}$; write $z = M_{11}$ and $w = M_{12}$. The anti-quaternion subspace is the multiple of the quaternion subspace by $i$, giving $M_{21} = \overline{M_{12}}$ and $M_{22} = -\overline{M_{11}}$, and writing $u = M_{11}$, $v = M_{12}$ gives the displayed form. For $\mathbb{M}_+$ and $\mathbb{M}_-$, the fixed and anti-fixed spaces of ${}^{*}$ are exactly the matrices equal to, respectively the negative of, their conjugate transpose, by the proposition above.

The two middle rows are the matrix form of the real and imaginary coefficient conditions: the quaternion subspace consists of the matrices of the displayed pattern, in which the off-diagonal entries are related by $M_{21} = -\overline{M_{12}}$ and the diagonal entries by $M_{22} = \overline{M_{11}}$, and the anti-quaternion subspace has the same pattern with the two diagonal signs exchanged, $M_{21} = \overline{M_{12}}$ and $M_{22} = -\overline{M_{11}}$. The Hermitian and anti-Hermitian subspaces are the real and imaginary parts of the coefficient space in the matrix picture, as recorded in Biquaternion Four-Vector Element Representation.

Remark (the two named images). Two entries of the table carry standard names. The traceless matrices are exactly the Lie algebra $\mathrm{SL}(2,\mathbb{C}) = \mathrm{span}_{\mathbb{C}}\{e_1, e_2, e_3\}$, so $\Phi(\mathrm{Vect}(\mathbb{B})) = \mathrm{SL}(2,\mathbb{C})$, whose compact real form $\mathrm{SU}(2)$ is the image of the real vector triple; and the quaternion subspace $\mathbb{H}_{\mathbb{B}}$ maps onto a four-dimensional real subspace of $M_2(\mathbb{C})$ closed under multiplication, whose nonzero elements are invertible, so that it is a realization of the quaternion division algebra inside $M_2(\mathbb{C})$, with $\det\Phi(\tilde{Q}) = N(\tilde{Q}) > 0$ on it. The Lie-algebra structure is developed in Biquaternion Lie Algebra, and the quaternion subspace in Introduction to the Six Subspaces.

Explicit Patterns and Invariants

The table above gives each image by its defining condition; solving that condition gives the explicit pattern and the trace and determinant. With $\tilde{Q} = Q_0e_0 + \mathbf{Q}$ and the coefficient notation $Q_\mu = q_\mu + iq'_\mu$,

subspace	explicit pattern of $\Phi(\tilde{Q})$	$\operatorname{Tr}\Phi(\tilde{Q})$	$\det\Phi(\tilde{Q})$
$\mathbb{C}_{\mathbb{B}}$	$Q_0 I$	$2Q_0$	$Q_0^2 = N$
$\mathrm{Vect}(\mathbb{B})$	$\begin{pmatrix} -iQ_3 & -iQ_1 - Q_2 \\ -iQ_1 + Q_2 & iQ_3 \end{pmatrix}$, traceless	$0$	$N$
$\mathbb{H}_{\mathbb{B}}$	$\begin{pmatrix} z & w \\ -\bar{w} & \bar{z} \end{pmatrix}$, $z = q_0 - iq_3$, $w = -q_2 - iq_1$	$2q_0$	$\lvert z\rvert^2 + \lvert w\rvert^2 = N > 0$
$i\mathbb{H}_{\mathbb{B}}$	$\begin{pmatrix} u & v \\ \bar{v} & -\bar{u} \end{pmatrix}$, $u = q'_3 + iq'_0$, $v = q'_1 - iq'_2$	$2iq'_0$	$-\bigl(\lvert u\rvert^2 + \lvert v\rvert^2\bigr) = N < 0$
$\mathbb{M}_+$	$M = M^\dagger$, Hermitian	$2q_0 \in \mathbb{R}$	$N \in \mathbb{R}$
$\mathbb{M}_-$	$M = -M^\dagger$, anti-Hermitian	$2iq'_0$	$N \in \mathbb{R}$

The quaternion subspace is the fixed space of the antilinear conjugation $M \mapsto \epsilon\overline{M}\epsilon^{-1}$ with $\epsilon = i\sigma_2 = \begin{pmatrix} 0 & 1 \\ -1 & 0\end{pmatrix}$; the anti-quaternion subspace has the transpose-conjugate pattern, with $M_{21} = \overline{M_{12}}$ and the two diagonal signs exchanged. Two elements illustrate the degenerate cases: $\Phi(e_0 + ie_1) = \begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix}$, a Hermitian matrix of rank one, determinant zero and square equal to twice itself, so that half of it is an idempotent; and $\Phi(ie_0 + e_1) = \begin{pmatrix} i & -i \\ -i & i \end{pmatrix}$, a singular anti-Hermitian matrix of trace $2i$, corresponding to the null element $ie_0 + e_1$. On the quaternion subspace the image is a realization of the quaternion division algebra, with nonzero determinant; on the anti-quaternion subspace the determinant is strictly negative, so that subspace misses the singular matrices altogether, which is the matrix statement of the absence of zero divisors there. The determinant of the traceless image identifies the null elements with the singular traceless matrices and the nilpotent ones with the rank-one matrices of zero square.

The Simple Module

The matrix realization is the realization in which the module of the algebra is exhibited concretely.

Definition. The simple module of $\mathbb{B}$ is the complex vector space

$$ V = \mathbb{C}^2 = \left\{ v = \begin{pmatrix} v_1 \\ v_2 \end{pmatrix} : v_1, v_2 \in \mathbb{C} \right\} $$

with the left action

$$ \tilde{Q} \cdot v = \Phi(\tilde{Q})\, v . $$

Proposition. The action above makes $V$ a left $\mathbb{B}$-module, and $V$ is simple: it has no nonzero proper submodule.

Proof. The action is $\mathbb{C}$-linear in $v$ because $\Phi(\tilde{Q})$ is a matrix, and it is compatible with multiplication because $\Phi$ is a homomorphism: $\tilde{P}\cdot(\tilde{Q}\cdot v) = \Phi(\tilde{P})\Phi(\tilde{Q})v = \Phi(\tilde{P}\tilde{Q})v = (\tilde{P}\tilde{Q})\cdot v$, and $e_0 \cdot v = v$. For simplicity, a submodule is a subspace of $\mathbb{C}^2$ stable under every matrix in $M_2(\mathbb{C})$, because $\Phi$ is surjective; a nonzero stable subspace contains a nonzero vector, and applying the matrix units $E_{11}, E_{12}, E_{21}, E_{22}$ to it produces both standard basis vectors, so the subspace is all of $\mathbb{C}^2$.

Theorem (uniqueness of the simple module). Up to isomorphism, $V$ is the only simple left $\mathbb{B}$-module. Consequently every finitely generated $\mathbb{B}$-module is a direct sum of copies of $V$.

Proof. Over the field $\mathbb{C}$, the algebra $\mathbb{B} \cong M_2(\mathbb{C})$ is semisimple, every simple module is a minimal left ideal, and all minimal left ideals of a full matrix algebra are isomorphic to the column space; the classification and Schur's lemma are the subject of Modules over the Biquaternion Algebra, and the statement is cited from there.

Remark (the minimal left ideals are the column spaces). The standard minimal left ideal of $M_2(\mathbb{C})$ is the column space

$$ M_2(\mathbb{C}) E_{11} = \left\{ \begin{pmatrix} \alpha & 0 \\ \beta & 0 \end{pmatrix} : \alpha, \beta \in \mathbb{C} \right\}, $$

which is $V$ by the map $\begin{pmatrix} \alpha & 0 \\ \beta & 0 \end{pmatrix} \mapsto \begin{pmatrix} \alpha \\ \beta \end{pmatrix}$. Under the inverse of $\Phi$ this is the minimal left ideal $\mathbb{B} \tilde\Pi_1$ of $\mathbb{B}$, where

$$ \tilde\Pi_1 = \tfrac{1}{2}(e_0 + i e_3), \qquad \Phi(\tilde\Pi_1) = E_{11}, $$

and the complementary ideal is $\mathbb{B} \tilde\Pi_2$ with $\tilde\Pi_2 = \tfrac{1}{2}(e_0 - i e_3)$ and $\Phi(\tilde\Pi_2) = E_{22}$. The two ideals are isomorphic as left modules and satisfy $\mathbb{B} = \mathbb{B}\tilde\Pi_1 \oplus \mathbb{B}\tilde\Pi_2$. The idempotents and the resulting Peirce decomposition are the subject of Biquaternion Ideals and Peirce Decomposition and are cited here only for the identification of $V$ with $\mathbb{B}\tilde\Pi_1$.

Remark (the endomorphism algebra). Two facts complete the picture of $V$. First, the commutant is one-dimensional, $\operatorname{End}_{\mathbb{B}}(V) = \mathbb{C}\,\mathrm{id}$, so $V$ is absolutely irreducible; this is Schur's lemma, and the element of $\operatorname{End}_{\mathbb{B}}(V)$ is the scalar action $\lambda v$, the diagonal copy of $\mathbb{C}$ in $M_2(\mathbb{C})$. Second, the double centralizer theorem gives $\mathbb{B} = \operatorname{End}_{\mathbb{C}}(V)$: the algebra is not merely contained in the endomorphisms of $V$, it is all of them. Both statements are proved in Modules over the Biquaternion Algebra, where the module is exhibited with the matrix units and shown to reconstruct the algebra; that article writes the simple module $S$ and reads the regular module as $S \oplus S$, so its $S$ is the module called $V$ here. The reading of $V$ as a pair of chiral spinors with their dual and conjugate is the subject of Biquaternion Spin Geometry, which also writes the module $S$.

Remark (the underlying real space). Regarded over $\mathbb{R}$ by restriction of scalars, $V$ becomes a real vector space

$$ S = \operatorname{Res}_{\mathbb{C}/\mathbb{R}} V, \qquad \dim_{\mathbb{R}} S = 2 \dim_{\mathbb{C}} V = 4 . $$

The real dimension of $S$ is four, and the left action of the eight-dimensional real algebra $\mathbb{B}$ on it is the restriction of the complex-linear action; the real structure carried by $S$ is treated in Biquaternion Ideals and Peirce Decomposition, §The Real Structure. The letter $S$ names the realification here, not the simple module: this article writes the simple module $V$, and Modules over the Biquaternion Algebra writes it $S$.

Matrix Units, the One-Sided Ideals, and the Projective Line

The four matrix units are realized by the two standard idempotents together with two off-diagonal elements,

$$ E_{11} = \Phi(\tilde\Pi_1) = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}, \qquad E_{22} = \Phi(\tilde\Pi_2) = \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix}, $$ $$ E_{12} = \Phi(\tilde U), \qquad E_{21} = \Phi(\tilde S), \qquad \tilde U = \frac{i e_1 - e_2}{2}, \qquad \tilde S = \frac{i e_1 + e_2}{2}, $$

so that $\{\tilde\Pi_1, \tilde U, \tilde S, \tilde\Pi_2\}$ is a $\mathbb{C}$-basis of $\mathbb{B}$ and the multiplication is the table of the matrix units,

$$ \tilde\Pi_1\tilde U = \tilde U = \tilde U\tilde\Pi_2, \qquad \tilde\Pi_2\tilde S = \tilde S = \tilde S\tilde\Pi_1, \qquad \tilde U\tilde S = \tilde\Pi_1, \qquad \tilde S\tilde U = \tilde\Pi_2, $$

together with $\tilde U\tilde\Pi_1 = \tilde\Pi_2\tilde U = \tilde\Pi_1\tilde S = 0$, $\tilde S\tilde\Pi_2 = 0$, and $\tilde U^2 = \tilde S^2 = 0$.

The matrix units group the basis into one-sided ideals. The two left ideals $\mathbb{B}\tilde\Pi_1, \mathbb{B}\tilde\Pi_2$ are the column spaces and the two right ideals $\tilde\Pi_1\mathbb{B}, \tilde\Pi_2\mathbb{B}$ are the row spaces, and

$$ \mathbb{B} = \mathbb{B}\tilde\Pi_1 \oplus \mathbb{B}\tilde\Pi_2 = (\mathbb{C}\tilde\Pi_1 \oplus \mathbb{C}\tilde S) \oplus (\mathbb{C}\tilde U \oplus \mathbb{C}\tilde\Pi_2), $$ $$ \mathbb{B} = \tilde\Pi_1\mathbb{B} \oplus \tilde\Pi_2\mathbb{B} = (\mathbb{C}\tilde\Pi_1 \oplus \mathbb{C}\tilde U) \oplus (\mathbb{C}\tilde S \oplus \mathbb{C}\tilde\Pi_2). $$

The two groupings of the same four basis elements differ: the Peirce decomposition groups $\tilde\Pi_1$ with $\tilde U$ and $\tilde\Pi_2$ with $\tilde S$, whereas the left-ideal decomposition groups $\tilde\Pi_1$ with $\tilde S$ and $\tilde\Pi_2$ with $\tilde U$. Each left ideal is two-dimensional over $\mathbb{C}$ and isomorphic to the simple module $V$; each right ideal is isomorphic to the dual $V^{*}$.

The minimal left ideals are parametrized by the projective line. For a $\mathbb{C}$-subspace $W \subseteq V = \mathbb{C}^2$ put

$$ L_W = \{\, M \in M_2(\mathbb{C}) : M|_W = 0 \,\}, $$

the matrices annihilating $W$; this is a left ideal, since $\ker(AM) \supseteq \ker M$ for every $A$, and $W \mapsto L_W$ reverses inclusions. One has $L_0 = \mathbb{B}$, $L_{\mathbb{C}^2} = 0$, and $L_W$ is minimal exactly when $\dim_{\mathbb{C}} W = 1$. Conversely every left ideal arises this way, because a left ideal is a submodule of the regular module $\mathbb{B} \cong V \oplus V$, and the only submodules of $V \oplus V$ are $0$, a copy of $V$, and the whole module. So the left ideals of $\mathbb{B}$ are exactly $0$, the whole algebra, and the minimal ideals $L_W$ indexed by the projective line

$$ \mathbb{P}^1(\mathbb{C}) = \{\, W \subseteq \mathbb{C}^2 : W \text{ a one-dimensional } \mathbb{C}\text{-subspace} \,\}, $$

which form the middle layer of the lattice; they are pairwise incomparable, and because the regular module has length two each is maximal as well as minimal. Right ideals admit the same description, with lines in the dual space and the rows as coordinate members.

Structural Consequences of the Isomorphism

The matrix realization places the structure of $\mathbb{B}$ in view, and the consequences below are the ones the rest of the corpus uses.

Theorem (simple, central, with prescribed centre). The algebra $\mathbb{B}$ is simple; its centre is the scalar subspace $\mathbb{C}_{\mathbb{B}} = \mathbb{C} e_0$, isomorphic to $\mathbb{C}$; and $\mathbb{B}$ is central simple over $\mathbb{C}$.

Proof. Simplicity and the centre are the two corollaries after the isomorphism theorem. Central simple means simple with centre exactly the base field, and the centre is $\mathbb{C} e_0 \cong \mathbb{C}$.

Theorem (endomorphism algebra). The isomorphism $\Phi$ identifies $\mathbb{B}$ with the endomorphism algebra of the two-dimensional complex space $V$, so that $\mathbb{B} = \operatorname{End}_{\mathbb{C}}(V)$ under the identification.

Proof. The algebra $M_2(\mathbb{C})$ is by definition the algebra of $\mathbb{C}$-linear endomorphisms of $\mathbb{C}^2$, and $\Phi$ is an isomorphism onto $M_2(\mathbb{C})$.

Corollary (zero divisors). The zero divisors of $\mathbb{B}$ are exactly the nonzero elements $\tilde{Q}$ with $\det\Phi(\tilde{Q}) = N(\tilde{Q}) = 0$, that is, the nonzero elements whose matrix is singular. In matrix terms they are the elements of rank one, and they form the nonzero part of the determinantal variety $\det = 0$.

Proof. A nonzero matrix over a field is singular exactly when it is a zero divisor in the matrix algebra, and rank is one for the nonzero singular $2 \times 2$ matrices. The identification with the vanishing of the determinant $N$ is the determinant statement above.

Taken together, the three statements say that $\mathbb{B}$ is the full endomorphism algebra of a two-dimensional complex space, and that all of its one-sided ideal structure and all of its degeneracies are the corresponding structure of $M_2(\mathbb{C})$ transported along a fixed linear isomorphism. The classification of the conjugations, the trace and the determinant, and the module $V$ are the parts of that transport that the companion articles use.

Summary

The biquaternion algebra is isomorphic to $M_2(\mathbb{C})$ through the map $\Phi$ fixed by $e_0 \mapsto I$ and $e_k \mapsto -i\sigma_k$, whose explicit form is $\Phi(\tilde{Q}) = \begin{pmatrix} Q_0 - iQ_3 & -iQ_1 - Q_2 \\ -iQ_1 + Q_2 & Q_0 + iQ_3 \end{pmatrix}$. The choice is stated once: the Pauli matrices are a shorthand for the images and not the organising device, and every statement is about $\mathbb{B}$ and $M_2(\mathbb{C})$. The trace is $2Q_0$ and the determinant is $N(\tilde{Q}) = Q_0^2 + Q_1^2 + Q_2^2 + Q_3^2$, so invertibility is non-vanishing determinant and the zero divisors are the singular matrices, a nonzero zero divisor being a rank-one matrix $\Phi(\tilde{Q}) = uv^{T}$. The characteristic polynomial is $\lambda^2 - 2Q_0\lambda + N(\tilde{Q})$ and the Cayley–Hamilton identity is $\tilde{Q}^2 - 2Q_0\tilde{Q} + N(\tilde{Q})e_0 = 0$, so every power reduces to a combination of $e_0$ and $\tilde{Q}$. And $\Phi$ carries the Hermitian subspace onto the Hermitian matrices.

Quaternion conjugation is the adjugate, Hermitian conjugation is the conjugate transpose, and complex conjugation is not entrywise: it is dressed with the antisymmetric matrix $\epsilon = i\sigma_2 = \Phi(-e_2)$, as $\Phi(\bar{\tilde{Q}}) = \epsilon(\Phi(\tilde{Q}))^{\natural}\epsilon^{-1}$. The six distinguished subspaces are the scalar, traceless, quaternionic, anti-quaternionic, Hermitian and anti-Hermitian matrices. The algebra is simple with centre the scalar matrices and is the full endomorphism algebra of the simple module $V = \mathbb{C}^2$ of complex dimension $2$, on which it acts by matrix multiplication; the minimal left ideals are the column spaces, and $V$ is the only simple module. The two dimensions $2$ of the module and $2$ of the matrix are the same two, because the algebra is the endomorphism algebra of the module.

Summary of Notation

Symbol	Meaning
$\mathbb{B} = \mathbb{C} \otimes_{\mathbb{R}} \mathbb{H}$	Biquaternion algebra
$e_0, e_1, e_2, e_3$	Algebra basis, $e_0 = 1$, $e_k^2 = -e_0$
$i$	Central scalar imaginary, $i^2 = -1$
$\tilde{Q} = \sum_{\mu=0}^{3} Q_\mu e_\mu$	Developed form, $Q_\mu \in \mathbb{C}$
$Q^\mu = (Q^0, Q^1, Q^2, Q^3)$	Four-vector of Biquaternion Four-Vector Element Representation
$\Phi : \mathbb{B} \to M_2(\mathbb{C})$	Matrix realization, $\Phi(e_0) = I$, $\Phi(e_k) = -i\sigma_k$
$\sigma_1, \sigma_2, \sigma_3$	Pauli matrices, a shorthand for the images of $e_1, e_2, e_3$
$N(\tilde{Q}) = \det\Phi(\tilde{Q}) = \sum_\mu Q_\mu^2$	The determinant
$\Phi(\tilde{Q}) = uv^{T}$	Rank-one form of a nonzero null element, $N(\tilde{Q}) = 0$; the columns $u = (\alpha,\beta)^{T}$ and $v = (\gamma,\delta)^{T}$ are nonzero and determined up to $(\lambda u,\lambda^{-1}v)$
$\epsilon = i\sigma_2 = \Phi(-e_2)$	Antisymmetric matrix, $\Phi(\bar{\tilde{Q}}) = \epsilon(\Phi(\tilde{Q}))^{\natural}\epsilon^{-1}$
$\operatorname{Tr}\Phi(\tilde{Q}) = 2Q_0$	Trace of the matrix realization
$\operatorname{spec}\tilde{Q} = \{Q_0 \pm iB\}$	Spectrum, the eigenvalues of $\Phi(\tilde{Q})$, $B = \sqrt{Q_1^2 + Q_2^2 + Q_3^2}$
$H_2(\mathbb{C})$	Hermitian $2 \times 2$ matrices, the image of $\mathbb{M}_+$ under $\Phi$
$\mathrm{SL}(2,\mathbb{C})$	Traceless matrices, the image $\Phi(\mathrm{Vect}(\mathbb{B}))$
${}^{\natural}, \bar{\cdot}, {}^{*}, {}^{\flat}$	Quaternion, complex, Hermitian and anti-Hermitian conjugations
$\mathbb{C}_{\mathbb{B}}, \mathrm{Vect}(\mathbb{B}), \mathbb{H}_{\mathbb{B}}, i\mathbb{H}_{\mathbb{B}}, \mathbb{M}_+, \mathbb{M}_-$	The six distinguished subspaces
$V = \mathbb{C}^2$	Simple left $\mathbb{B}$-module, complex dimension $2$
$S = \operatorname{Res}_{\mathbb{C}/\mathbb{R}} V$	Restriction of scalars, real dimension $4$
$\tilde\Pi_1 = \tfrac{1}{2}(e_0 + ie_3)$, $\tilde\Pi_2 = \tfrac{1}{2}(e_0 - ie_3)$	Orthogonal idempotents, $\Phi(\tilde\Pi_1) = E_{11}$, $\Phi(\tilde\Pi_2) = E_{22}$
$E_{ij}$	Matrix units, $E_{ij}E_{kl} = \delta_{jk}E_{il}$
$\operatorname{End}_{\mathbb{C}}(V) \cong M_2(\mathbb{C})$	Endomorphism algebra identified with $\mathbb{B}$