The General Plain Sesqualgebra in the $4\times4$ Matrix Element Representation $M_4(\mathbb{C})_L$

Introduction

The left regular representation $\mathsf{M}_4$ of Introduction to the 4×4 Matrix Element Representation $M_4(\mathbb{C})_L$ of Biquaternions is built from the multiplication and carries the Hermitian conjugation as the conjugate transpose, $\mathsf{M}_4(\tilde{Q}^{*})=\mathsf{M}_4(\tilde{Q})^{\dagger}$, so it is a $*$-representation of the sesqualgebra. This article is the companion of The General Plain Sesqualgebra in the $2\times2$ Matrix Element Representation $M_2(\mathbb{C})$, and it reads the group Topology on the Introduction to the General Plain Sesqualgebra of Biquaternions on the regular matrices: the sesquilinear product becomes the product with the conjugate-transposed second factor, and the general plain sesquilinear form becomes the Hilbert–Schmidt pairing of the regular matrices, positive definite, with the Frobenius norm as its carrier and a double cover of the Lorentz group as its algebraic content.

The regular representation adds one thing to the $2\times2$ reading: the positive definite form makes the Hermitian subspace a definite space on which the norm-one group acts, and that action is the two-to-one cover $SL_2(\mathbb{C})\to SO^{+}(1,3)$ of the identity component of the Lorentz group. The Frobenius norm of the regular matrix is twice the definite norm of the element.

The Representation

Definition. The left regular matrix is the isomorphism written $\mathsf{M}_4$. It converts a biquaternion into a $4 \times 4$ complex matrix,

$$ \mathsf{M}_4:\mathbb{B}\longrightarrow M_4(\mathbb{C}), $$

and it is fixed by its values on the basis, the rest following by $\mathbb{C}$-linearity: in the basis $e_t,\vec e_x,\vec e_y,\vec e_z$ its $m$-th column is the coordinate column of the product $\tilde{Q}e_m$. It satisfies

$$ \mathsf{M}_4(\tilde{Q})\mathsf{M}_4(\tilde{Q}_2)=\mathsf{M}_4(\tilde{Q}\tilde{Q}_2),\qquad \mathsf{M}_4(\tilde{Q}^{*})=\mathsf{M}_4(\tilde{Q})^{\dagger},\qquad \operatorname{Tr}\mathsf{M}_4(\tilde{Q})=4Q_t,\qquad \det\mathsf{M}_4(\tilde{Q})=\langle\tilde{Q}\rangle_{\natural}^2 , $$

so $\mathsf{M}_4$ is a $*$-representation for the Hermitian conjugation and the conjugate transpose.

The Sesquilinear Product in the Regular Representation

Theorem (the product is the conjugate-transposed product of the regular matrices). For all biquaternions,

$$ \boxed{\; \mathsf{M}_4(\tilde{Q}\star\tilde{Q}_2)=\mathsf{M}_4(\tilde{Q})\,\mathsf{M}_4(\tilde{Q}_2)^{\dagger} \;}, $$

so the sesquilinear product is read on the regular matrices by conjugating and transposing the second factor and multiplying.

Proof. $\tilde{Q}\star\tilde{Q}_2=\tilde{Q}\tilde{Q}_2^{*}$, the representation is multiplicative, and $\mathsf{M}_4(\tilde{Q}^{*})=\mathsf{M}_4(\tilde{Q})^{\dagger}$ by the $*$-property.

The square and the unit. At $\tilde{Q}_2=\tilde{Q}$ the theorem gives the square $\mathsf{M}_4(\tilde{Q}\star\tilde{Q})=\mathsf{M}_4(\tilde{Q})\mathsf{M}_4(\tilde{Q})^{\dagger}$, a positive semidefinite matrix whatever $\tilde{Q}$ is, which is the matrix form of the fact that $\tilde{Q}\tilde{Q}^{*}$ is a sum of Hermitian squares; the identity matrix $I_4=\mathsf{M}_4(e_t)$ is a right unit alone of the product, its left action being the conjugate transpose.

The units. The units of the algebra are the elements of non-zero determinant, so under $\mathsf{M}_4$ their images form the group $\mathsf{M}_4(\mathbb{B}^{\times})\cong GL_2(\mathbb{C})$ inside $GL_4(\mathbb{C})$; the norm-one group $\{\tilde{A}:\langle\tilde{A},\tilde{A}\rangle_{\natural}=1\}$ is $SL_2(\mathbb{C})$, whose regular matrices have determinant one.

The General Plain Sesquilinear Form in the Regular Representation

Theorem (the Hilbert–Schmidt pairing of the regular matrices). For all biquaternions,

$$ \boxed{\; \tfrac12\operatorname{Tr}\bigl(\mathsf{M}_4(\tilde{Q})^{\dagger}\mathsf{M}_4(\tilde{Q}_2)\bigr) = 2\,\langle\tilde{Q}_2,\tilde{Q}\rangle_{*} \;}, $$

so the Hilbert–Schmidt pairing of the regular matrices is twice the general plain sesquilinear form of the algebra.

Proof. $\mathsf{M}_4(\tilde{Q})^{\dagger}=\mathsf{M}_4(\tilde{Q}^{*})$, so the left-hand side is $\tfrac12\operatorname{Tr}(\mathsf{M}_4(\tilde{Q}^{*}\tilde{Q}_2))=2\,\mathrm{Sc}(\tilde{Q}^{*}\tilde{Q}_2)=2\langle\tilde{Q}_2,\tilde{Q}\rangle_{*}$, using $\operatorname{Tr}\mathsf{M}_4(\tilde{Q})=4\,\mathrm{Sc}(\tilde{Q})$.

Theorem (the Frobenius norm). On the diagonal,

$$ \boxed{\; \lVert\mathsf{M}_4(\tilde{Q})\rVert_F^2=\operatorname{Tr}\bigl(\mathsf{M}_4(\tilde{Q})^{\dagger}\mathsf{M}_4(\tilde{Q})\bigr)=4\sum_\mu|Q_\mu|^2=4\,\lVert\tilde{Q}\rVert_E^2 \;}, $$

so the Frobenius norm of the regular matrix is twice the definite norm of the element, and the Hilbert–Schmidt pairing reads the definite form of the algebra on the matrices.

The signature and the null set. The pairing is positive definite: on the sixteen complex coordinates of $M_4(\mathbb{C})$ it has signature $(32,0)$, and its null set is $\{0\}$ alone.

The automorphism group. The Hilbert–Schmidt pairing on $M_4(\mathbb{C})$ has automorphism group the unitary group $\mathrm{U}(16)$, of real dimension $256$; among these, the structure-preserving maps $X\mapsto UXV^{\dagger}$ form the subgroup $\mathrm{U}(4)\times \mathrm{U}(4)$ modulo the common centre, which is the part that also preserves the algebra structure.

The two-sided action and the double cover. The positive definite form makes the Hermitian subspace $\mathbb{M}_+$ a definite space of real dimension $4$, and the norm-one group $G=\{\tilde{A}:\langle\tilde{A}\rangle_{\natural}=1\}=SL_2(\mathbb{C})$ acts on it by $\tilde{Q}\mapsto\tilde{A}\tilde{Q}\tilde{A}^{*}$, preserving the determinant:

$$ \langle\tilde{A}\tilde{Q}\tilde{A}^{*},\tilde{A}\tilde{Q}\tilde{A}^{*}\rangle_{\natural}=\langle\tilde{Q},\tilde{Q}\rangle_{\natural},\qquad \tilde{A}\in G. $$

The action defines the surjective homomorphism $SL_2(\mathbb{C})\to SO^{+}(1,3)$ with kernel $\{\pm e_t\}$, the two-to-one cover of the identity component of the orthogonal group of the restricted form; it is read on the module in Modules over the Biquaternion Algebra, and the group is Biquaternion Rotations and Lorentz Transformations.

Proposition (the image is the identity component). The image of the two-sided action is the identity component $SO^{+}(1,3)$.

Proof. The action $(\tilde{A},\tilde{Q})\mapsto\tilde{A}\tilde{Q}\tilde{A}^{*}$ has kernel $\{\pm e_t\}$ and its image is the subgroup of $SO^{+}(1,3)$ generated by the one-parameter families $\tilde{A}(t)=\exp(t\tilde{Q}_1)$, which are the rotations and boosts of the restricted form; the image is the whole identity component.

Worked Examples

The identity. Let $\tilde{Q}=e_t$. Then $\mathsf{M}_4(e_t)=I_4$, $\lVert\mathsf{M}_4(e_t)\rVert_F^2=4=4\lVert e_t\rVert_E^2$, and $\tfrac12\operatorname{Tr}(I_4^{\dagger}I_4)=2=2\langle e_t,e_t\rangle_{*}$.

A Hermitian element. Let $\tilde{Q}=e_t+i\vec e_z$. Then $\tilde{Q}^{*}=\tilde{Q}$ and $\mathsf{M}_4(\tilde{Q})^{\dagger}=\mathsf{M}_4(\tilde{Q})$, so the regular matrix is Hermitian; the square is $\tilde{Q}\star\tilde{Q}=\tilde{Q}^2=e_t+2i\vec e_z+(i\vec e_z)^2=2e_t+2i\vec e_z=2\tilde{Q}$, a positive semidefinite element, and the pairing at the two equal arguments is $4\lVert\tilde{Q}\rVert_E^2=8$.

A zero divisor. Let $\tilde{Q}=e_t+i\vec e_x$. Then $\det\mathsf{M}_4(\tilde{Q})=\langle\tilde{Q}\rangle_{\natural}^2=0$ and the regular matrix is singular of rank two, while $\lVert\mathsf{M}_4(\tilde{Q})\rVert_F^2=4\lVert\tilde{Q}\rVert_E^2=8$: the element is a zero divisor and yet a definite one for the general plain sesquilinear form.

The Matrices

The generators. The regular model is fixed on the basis by four matrices:

$$ \mathsf{M}_4(e_t)=\begin{pmatrix}1&0&0&0\\0&1&0&0\\0&0&1&0\\0&0&0&1\end{pmatrix},\qquad \mathsf{M}_4(\vec e_x)=\begin{pmatrix}0&-1&0&0\\1&0&0&0\\0&0&0&-1\\0&0&1&0\end{pmatrix}, $$ $$ \mathsf{M}_4(\vec e_y)=\begin{pmatrix}0&0&-1&0\\0&0&0&1\\1&0&0&0\\0&-1&0&0\end{pmatrix},\qquad \mathsf{M}_4(\vec e_z)=\begin{pmatrix}0&0&0&-1\\0&0&-1&0\\0&1&0&0\\1&0&0&0\end{pmatrix}. $$

A general element is the block matrix

$$ \mathsf{M}_4(\tilde Q)=\begin{pmatrix}A&B\\-B&A\end{pmatrix},\qquad A=\begin{pmatrix}Q_t&-Q_x\\Q_x&Q_t\end{pmatrix},\qquad B=\begin{pmatrix}-Q_y&-Q_z\\-Q_z&Q_y\end{pmatrix}, $$

The general element has trace $4Q_t$ and determinant $\langle\tilde Q\rangle_{\natural}^2$, with the Hermitian conjugation carried to the conjugate transpose, $\mathsf{M}_4(\tilde Q^{*})=\mathsf{M}_4(\tilde Q)^{\dagger}$.

The product, entry by entry. A vector unit is anti-Hermitian, so $\mathsf{M}_4(\vec e_y)^{\dagger}=-\mathsf{M}_4(\vec e_y)$, and the product at the pair $\vec e_x,\vec e_y$ is

$$ \mathsf{M}_4(\vec e_x)\mathsf{M}_4(\vec e_y)^{\dagger}=-\mathsf{M}_4(\vec e_x)\mathsf{M}_4(\vec e_y)=-\mathsf{M}_4(\vec e_z), $$

which is $\vec e_x\star \vec e_y=\vec e_x\vec e_y^{*}=-\vec e_z$, the relation $\vec e_x\vec e_y=\vec e_z$ and the anti-Hermiticity of $\vec e_y$ carrying the sign.

The right unit, entry by entry. The identity matrix is a right unit and not a left one, $I_4\star Y=Y^{\dagger}$ against $Y\star I_4=Y$; at $Y=\mathsf{M}_4(\vec e_x)$ the two differ, the left action returning the negative, $\mathsf{M}_4(\vec e_x)^{\dagger}=-\mathsf{M}_4(\vec e_x)\neq\mathsf{M}_4(\vec e_x)$.

The Frobenius norm, entry by entry. The regular matrix of a vector unit has four entries $\pm1$ and the remaining twelve zero, so $\lVert\mathsf{M}_4(\vec e_x)\rVert_F^2=4$, in agreement with the general $\lVert\mathsf{M}_4(\tilde Q)\rVert_F^2=4\sum_\mu\lvert Q_\mu\rvert^2$; the value of the square is the positive semidefinite matrix $\mathsf{M}_4(\tilde Q)\mathsf{M}_4(\tilde Q)^{\dagger}$.

Summary

The left regular representation is a $*$-representation, $\mathsf{M}_4(\tilde{Q}^{*})=\mathsf{M}_4(\tilde{Q})^{\dagger}$, and it reads the sesquilinear product as the product with the conjugate-transposed second factor, $\mathsf{M}_4(\tilde{Q}\star\tilde{Q}_2)=\mathsf{M}_4(\tilde{Q})\mathsf{M}_4(\tilde{Q}_2)^{\dagger}$, whose square is always positive semidefinite. The general plain sesquilinear form of the group is the Hilbert–Schmidt pairing $\tfrac12\operatorname{Tr}(\mathsf{M}_4(\tilde{Q})^{\dagger}\mathsf{M}_4(\tilde{Q}_2))=2\langle\tilde{Q}_2,\tilde{Q}\rangle_{*}$, positive definite of signature $(32,0)$ on the matrices, with Frobenius norm $\lVert\mathsf{M}_4(\tilde{Q})\rVert_F=2\lVert\tilde{Q}\rVert_E$ and null set $\{0\}$; its automorphism group is $\mathrm{U}(16)$, whose algebra-preserving part is $\mathrm{U}(4)\times \mathrm{U}(4)$ modulo the centre. The two-sided action of the norm-one group $SL_2(\mathbb{C})$ on the Hermitian subspace gives the double cover $SL_2(\mathbb{C})\to SO^{+}(1,3)$, and the units of the algebra appear as the matrices of $GL_2(\mathbb{C})$.

Summary of Notation

Symbol	Meaning
$\mathsf{M}_4(\tilde{Q})$	the $4\times4$ regular matrix, $\operatorname{Tr}\mathsf{M}_4(\tilde{Q})=4Q_t$
$\mathsf{M}_4(\tilde{Q}^{*})=\mathsf{M}_4(\tilde{Q})^{\dagger}$	the Hermitian conjugation is the conjugate transpose
$\mathsf{M}_4(\tilde{Q}\star\tilde{Q}_2)=\mathsf{M}_4(\tilde{Q})\mathsf{M}_4(\tilde{Q}_2)^{\dagger}$	the sesquilinear product on the regular matrices
$\tfrac12\operatorname{Tr}(\mathsf{M}_4(\tilde{Q})^{\dagger}\mathsf{M}_4(\tilde{Q}_2))=2\langle\tilde{Q}_2,\tilde{Q}\rangle_{*}$	the general plain sesquilinear form as the Hilbert–Schmidt pairing
$\lVert\mathsf{M}_4(\tilde{Q})\rVert_F=2\lVert\tilde{Q}\rVert_E$	the Frobenius norm of the regular matrix
$SL_2(\mathbb{C})\to SO^{+}(1,3)$	the double cover of the two-sided action on the Hermitian subspace