The General Plain Sesqualgebra in the $2\times2$ Matrix Element Representation $M_2(\mathbb{C})$

Introduction

The reading group Topology on the Introduction to the General Plain Sesqualgebra of Biquaternions reads the algebra through the general plain sesquilinear form $\langle\tilde{Q}_1,\tilde{Q}_2\rangle_{*}=\mathrm{Sc}(\tilde{Q}_1\tilde{Q}_2^{*})=\sum_\mu Q_{1\mu}\overline{Q_{2\mu}}$ and through the sesquilinear product $\tilde{Q}_1\star\tilde{Q}_2=\tilde{Q}_1\tilde{Q}_2^{*}$ that it polarises, whose algebra is Introduction to the General Plain Sesqualgebra. This article reads that group through the $2\times2$ matrix realization $\mathsf{M}_2$ of Introduction to the 2×2 Matrix Element Representation $M_2(\mathbb{C})$ of Biquaternions, and it is the first of the two representation articles of the group; the companion The General Plain Sesqualgebra in the $4\times4$ Matrix Element Representation $M_4(\mathbb{C})_L$ repeats the reading on the left regular representation.

The Hermitian conjugation is the characteristic operation of the group, and in the model it is the conjugate transpose: $\mathsf{M}_2(\tilde{Q}^{*})=\mathsf{M}_2(\tilde{Q})^{\dagger}$. The sesquilinear product is therefore the matrix product with the conjugate transpose of the second factor, $\mathsf{M}_2(\tilde{Q}_1\star\tilde{Q}_2)=\mathsf{M}_2(\tilde{Q}_1)\mathsf{M}_2(\tilde{Q}_2)^{\dagger}$, and the general plain sesquilinear form is the Hilbert–Schmidt pairing of the matrices. The form is positive definite, so its matrix reading is the one of the five that is definite rather than indefinite, and the Frobenius norm is its carrier.

The Representation

Definition. The matrix realization is the isomorphism written $\mathsf{M}_2$. It converts a biquaternion into a $2 \times 2$ complex matrix,

$$ \mathsf{M}_2:\mathbb{B}\longrightarrow M_2(\mathbb{C}), $$

and it is fixed by its values on the basis, the rest following by $\mathbb{C}$-linearity: $\mathsf{M}_2(e_t)=I$ on the identity and $\mathsf{M}_2(\vec e_k)=-i\sigma_k$ on the three vector units.

The invariants and the two conjugations are

$$ \operatorname{Tr}\mathsf{M}_2(\tilde{Q})=2Q_t,\qquad \det \mathsf{M}_2(\tilde{Q})=\sum_\mu Q_\mu^2,\qquad \mathsf{M}_2(\tilde{Q}^{\natural})=\operatorname{adj}\mathsf{M}_2(\tilde{Q}),\qquad \mathsf{M}_2(\tilde{Q}^{*})=\mathsf{M}_2(\tilde{Q})^{\dagger} . $$

The realization is a $*$-isomorphism: it is multiplicative and it carries the Hermitian conjugation of the algebra to the conjugate transpose of the matrices.

The Sesquilinear Product in the Model

Theorem (the product is the conjugate-transposed matrix product). For all biquaternions,

$$ \boxed{\; \mathsf{M}_2(\tilde{Q}_1\star\tilde{Q}_2)=\mathsf{M}_2(\tilde{Q}_1)\,\mathsf{M}_2(\tilde{Q}_2)^{\dagger} \;}, $$

the ordinary matrix product with the conjugate transpose in the second slot.

Proof. $\tilde{Q}_1\star\tilde{Q}_2=\tilde{Q}_1\tilde{Q}_2^{*}$, $\mathsf{M}_2$ is multiplicative and $\mathsf{M}_2(\tilde{Q}^{*})=\mathsf{M}_2(\tilde{Q})^{\dagger}$.

The two actions of the unit. The identity matrix acts by

$$ I\star Y=IY^{\dagger}=Y^{\dagger},\qquad Y\star I=YI^{\dagger}=Y, $$

so $e_t$ is a right unit and not a left unit of the product: the right action of the unit is the identity while the left action is the conjugate transpose. In the notation of the two multiplication operators, $\mathsf{M}_2(L_{\tilde{Q}_4}(\tilde{Q}))=\mathsf{M}_2(\tilde{Q}_4)\mathsf{M}_2(\tilde{Q})^{\dagger}$ and $\mathsf{M}_2(R_{\tilde{Q}_4}(\tilde{Q}))=\mathsf{M}_2(\tilde{Q})\mathsf{M}_2(\tilde{Q}_4)^{\dagger}$. The absence of a two-sided unit is the failure of the conjugate transpose to be the identity, and every asymmetry of the group has this origin. The involution ${}^{*}$ is the conjugate transpose, so an element is self-adjoint exactly when its matrix is Hermitian and the Hermitian subspace $\mathbb{M}_+$ is the set of Hermitian matrices of $M_2(\mathbb{C})$.

The units and the zero divisors. An element is a unit of the algebra exactly when its matrix is invertible, so the units are $GL_2(\mathbb{C})$ and the zero divisors are the singular matrices; the norm-one group $\{\tilde{Q}_4:\langle\tilde{Q}_4,\tilde{Q}_4\rangle_{\natural}=1\}$ is $SL_2(\mathbb{C})$.

The Sesquilinear Objects in the Model

The ternary product and the quadratic representation. The ternary product is the matrix product

$$ \mathsf{M}_2\bigl(\{\tilde{Q}_1,\tilde{Q}_2,\tilde{Q}_3\}\bigr)=\mathsf{M}_2(\tilde{Q}_1)\,\mathsf{M}_2(\tilde{Q}_2)^{\dagger}\,\mathsf{M}_2(\tilde{Q}_3),\qquad\text{so}\qquad \{X,Y,Z\}=XY^{\dagger}Z , $$

the middle model of the matrix algebra, and its quadratic representation is $Z\mapsto ZY^{\dagger}Z$.

The associator. The associator is

$$ [X,Y,Z]=X\bigl(Y^{\dagger}Z^{\dagger}-ZY^{\dagger}\bigr), $$

the two groupings of the product being $(X\star Y)\star Z=XY^{\dagger}Z^{\dagger}$ and $X\star(Y\star Z)=XZY^{\dagger}$.

The sandwich, the commutator and the symmetrised product. The sandwich operator $S_{\tilde{Q}_1,\tilde{Q}_2}$ is $X\mapsto \mathsf{M}_2(\tilde Q_1)X^{\dagger}\mathsf{M}_2(\tilde Q_2)^{\dagger}$, the commutator is the antisymmetric part $X\mapsto XY^{\dagger}-YX^{\dagger}$, and the symmetrised product is $\tfrac12(XY^{\dagger}+YX^{\dagger})$, whose diagonal is the Hermitian part of $XY^{\dagger}$. The rank of an element is the rank of its matrix, and the rank of a sandwich is at most the smaller of the ranks of its two parameters.

The idempotents. The idempotents of the multiplication are the Hermitian idempotents: $\tilde\Pi\star\tilde\Pi=\tilde\Pi$ is $\mathsf{M}_2(\tilde\Pi)\mathsf{M}_2(\tilde\Pi)^{\dagger}=\mathsf{M}_2(\tilde\Pi)$, that is $\mathsf{M}_2(\tilde\Pi)$ is Hermitian of square itself. Their matrix images are $0$, $I$ and the orthogonal projections onto the lines of $\mathbb{C}^{2}$; the rank-one ones form the complex projective line $\mathbb{CP}^{1}\cong S^{2}$.

Simplicity. The multiplication has exactly the two two-sided ideals of $M_2(\mathbb{C})$, because a nonzero two-sided ideal of a full matrix algebra contains a matrix unit and hence the identity; the centre is $\mathbb{C}I=\mathbb{C}e_t$, and the only central idempotents are $0$ and $I$.

The square and its invariants. At a single element $\tilde{Q}$ the value is $\tilde{Q}\star\tilde{Q}=\tilde{Q}\tilde{Q}^{*}$, of matrix $\mathsf{M}_2(\tilde{Q})\mathsf{M}_2(\tilde{Q})^{\dagger}$, and

$$ \operatorname{Tr}\mathsf{M}_2(\tilde{Q}\star\tilde{Q})=2\sum_\mu|Q_\mu|^{2},\qquad \det \mathsf{M}_2(\tilde{Q}\star\tilde{Q})=\bigl|\langle\tilde{Q},\tilde{Q}\rangle_{\natural}\bigr|^{2}, $$

both real and nonnegative; the trace vanishes exactly at $\tilde{Q}=0$ and the determinant exactly at the zero divisors. This is the matrix form of the positivity of the square: the trace of the square is twice the squared definite norm of the eight real coordinates, and the determinant of the square is the squared modulus of the norm.

The Batch in the Model

The model collects the objects of the group in one dictionary.

The ordinary matrix algebra, with no transpose, is the bilinear reading of the algebra; the table is the sesquilinear reading, and the transpose is the whole difference between the two columns.

The General Plain Sesquilinear Form in the Model

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

$$ \boxed{\; \langle\tilde{Q}_2,\tilde{Q}_1\rangle_{*} = \tfrac12\operatorname{Tr}\bigl(\mathsf{M}_2(\tilde{Q}_1)^{\dagger}\mathsf{M}_2(\tilde{Q}_2)\bigr) \;}, $$

the Hilbert–Schmidt pairing of the matrices; since $\mathsf{M}_2(\tilde{Q})^{\dagger}=\mathsf{M}_2(\tilde{Q}^{*})$, the right-hand side is $\tfrac12\operatorname{Tr}(\mathsf{M}_2(\tilde{Q}_1^{*}\tilde{Q}_2))=\mathrm{Sc}(\tilde{Q}_1^{*}\tilde{Q}_2)$ by the trace identity $\mathrm{Sc}(\tilde{Q}_2\tilde{S})=\tfrac12\operatorname{Tr}(\mathsf{M}_2(\tilde{Q}_2)\mathsf{M}_2(\tilde{S}))$.

Theorem (the Frobenius norm). On the diagonal,

$$ \langle\tilde{Q},\tilde{Q}\rangle_{*} = \tfrac12\operatorname{Tr}\bigl(\mathsf{M}_2(\tilde{Q})^{\dagger}\mathsf{M}_2(\tilde{Q})\bigr) = \tfrac12\bigl\lVert \mathsf{M}_2(\tilde{Q})\bigr\rVert_F^2 = \sum_\mu|Q_\mu|^2 = \lVert\tilde{Q}\rVert_E^2 , $$

so that $\lVert \mathsf{M}_2(\tilde{Q})\rVert_F=\sqrt2\,\lVert\tilde{Q}\rVert_E$, and the Hilbert–Schmidt inner product of the matrix algebra is one half of the squared Frobenius norm.

Theorem (the trace and the determinant of a value). For all biquaternions,

$$ \operatorname{Tr}\mathsf{M}_2(\tilde{Q}_1\star\tilde{Q}_2)=2\langle\tilde{Q}_1,\tilde{Q}_2\rangle_{*},\qquad \det \mathsf{M}_2(\tilde{Q}_1\star\tilde{Q}_2)=\langle\tilde{Q}_1,\tilde{Q}_1\rangle_{\natural}\overline{\langle\tilde{Q}_2,\tilde{Q}_2\rangle_{\natural}}, $$

the trace of a value twice its Hermitian form and the determinant of a value the norm of the first factor times the conjugate of the norm of the second. The determinant is multiplicative and $\det \mathsf{M}_2(\tilde{Q})^{\dagger}=\overline{\det \mathsf{M}_2(\tilde{Q})}$, so the value of two norm-one elements again has determinant one and the product of two units is a unit.

The signature and the null set. The form is positive definite of signature $(8,0)$ on the eight real coordinates, so its null set is $\{0\}$ alone: no non-zero biquaternion is isotropic. The definite form is the companion of the four indefinite ones of the algebra.

The automorphism group and the unit groups. The form lives on the four complex coordinates, so its automorphism group is the unitary group $\mathrm{U}(4)$, of real dimension $16$; among these, the maps $X\mapsto UXV^{\dagger}$ form the subgroup $\mathrm{U}(2)\times \mathrm{U}(2)$ modulo the common centre, which is the part that also preserves the algebra structure of $M_2(\mathbb{C})$. Under $\mathsf{M}_2$ the group of units of the algebra is $GL_2(\mathbb{C})$, the norm-one group is $SL_2(\mathbb{C})$, and the unitary slice is $\mathrm{U}(2)$, whose special part is $SU(2)\cong\mathrm{Spin}(3)$.

The Frobenius Norm of the Realization

Proposition (the scale of the realization). For every biquaternion,

$$ \bigl\|\mathsf{M}_2(\tilde{Q})\bigr\|_F=\sqrt{2}\,\bigl\|\tilde{Q}\bigr\|_E , $$

so $\mathsf{M}_2$, regarded as a real-linear map $\mathbb{B}_{\mathbb{R}}\to M_2(\mathbb{C})_{\mathbb{R}}\cong\mathbb{R}^{8}$, is bijective and multiplies all lengths by the fixed constant $\sqrt2$. The scale $\sqrt2$ is the trace normalisation $\operatorname{Tr}\mathsf{M}_2(\tilde{Q})=2Q_t$, and with the normalised map $\mathsf{M}_2/\sqrt2$ the two norms agree, $\lVert \mathsf{M}_2(\tilde{Q})/\sqrt2\rVert_F=\lVert\tilde{Q}\rVert_E$.

The Unit Group in Matrix Form

Proposition (the unit groups in matrix form). Under $\mathsf{M}_2$,

$$ \boxed{\; \mathbb{B}^{\times}\cong GL_2(\mathbb{C}),\qquad \{\tilde{Q}:\langle\tilde{Q},\tilde{Q}\rangle_{\natural}=1\}\cong SL_2(\mathbb{C}) \;}, $$

and their unitary slices are $\mathrm{U}(2)$ and $SU(2)$, with $SU(2)\cong\mathrm{Spin}(3)$ and $SU(2)/\{\pm I\}\cong SO(3)$. The unit-norm group is the group that acts on the Hermitian subspace by $*$-congruence, $\tilde{Q}\mapsto\tilde{Q}_4\tilde{Q}\tilde{Q}_4^{*}$, which in the model is $X\mapsto AXA^{\dagger}$ on the Hermitian matrices. Three representatives in the model are

$$ A=\begin{pmatrix}1&1\\0&1\end{pmatrix},\quad \det A=1\ (\text{the unit-norm group}),\qquad U=i\sigma_1=\begin{pmatrix}0&i\\i&0\end{pmatrix},\quad \det U=1,\ UU^{\dagger}=I_2\ (\text{the unitary slice}), $$

$$ \begin{pmatrix}\exp(i\theta)&0\\0&\exp(-i\theta)\end{pmatrix}\in SU(2),\qquad\text{and}\qquad A\,\sigma_3\,A^{\dagger}=\begin{pmatrix}1&1\\0&1\end{pmatrix}\begin{pmatrix}1&0\\0&-1\end{pmatrix}\begin{pmatrix}1&0\\1&1\end{pmatrix}=\begin{pmatrix}0&-1\\-1&-1\end{pmatrix}, $$

the last Hermitian as the congruence action requires.

Worked Examples

A definite element. Let $\tilde{Q}=e_t+i\vec e_x$. Then $\langle\tilde{Q},\tilde{Q}\rangle_{*}=|1|^2+|i|^2=2$ and $\lVert \mathsf{M}_2(\tilde{Q})\rVert_F=2=\sqrt2\cdot\sqrt2$; the element is a zero divisor of the plain product, yet it is not isotropic for the general plain sesquilinear form, and its matrix is singular.

A Hermitian element. Let $\tilde{Q}=e_t+i\vec e_z$, so $Q_t=1$, $Q_z=i$. Then the coefficients of $\tilde{Q}^{*}$ are $\varepsilon_\mu\overline{Q_\mu}=(1,0,0,(-1)(-i))=Q_\mu$, so $\tilde{Q}^{*}=\tilde{Q}$ and the element is Hermitian; in the model $\mathsf{M}_2(\tilde{Q})=I+\sigma_3=\operatorname{diag}(2,0)$, a Hermitian matrix, and $\langle\tilde{Q},\tilde{Q}\rangle_{*}=2=\tfrac12\operatorname{Tr}(\mathsf{M}_2(\tilde{Q})^{\dagger}\mathsf{M}_2(\tilde{Q}))$.

A product read in matrices. Let $\tilde{Q}_1=\vec e_x$ and $\tilde{Q}_2=\vec e_y$. Then $\tilde{Q}_1\star\tilde{Q}_2=\vec e_x \vec e_y^{*}=-\vec e_z$, and in the model $\mathsf{M}_2(\vec e_x)\mathsf{M}_2(\vec e_y)^{\dagger}=(-i\sigma_1)(i\sigma_2)=\sigma_1\sigma_2=i\sigma_3=-\mathsf{M}_2(\vec e_z)$, as the theorem requires.

The Matrices

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

$$ \mathsf{M}_2(e_t)=\begin{pmatrix}1&0\\0&1\end{pmatrix},\qquad \mathsf{M}_2(\vec e_x)=\begin{pmatrix}0&-i\\-i&0\end{pmatrix},\qquad \mathsf{M}_2(\vec e_y)=\begin{pmatrix}0&-1\\1&0\end{pmatrix},\qquad \mathsf{M}_2(\vec e_z)=\begin{pmatrix}-i&0\\0&i\end{pmatrix}. $$

A general element has the matrix $\mathsf{M}_2(\tilde Q)$ of trace $2Q_t$ and determinant $\sum_\mu Q_\mu^2$; the Hermitian conjugation is the conjugate transpose, $\mathsf{M}_2(\tilde Q^{*})=\mathsf{M}_2(\tilde Q)^{\dagger}$.

The product, entry by entry. The product is the matrix product with the conjugate transpose in the second slot, and at the pair $\vec e_x,\vec e_y$ it is

$$ \mathsf{M}_2(\vec e_x)\mathsf{M}_2(\vec e_y)^{\dagger} =\begin{pmatrix}0&-i\\-i&0\end{pmatrix}\begin{pmatrix}0&1\\-1&0\end{pmatrix} =\begin{pmatrix}i&0\\0&-i\end{pmatrix}=-\mathsf{M}_2(\vec e_z), $$

which is $\vec e_x\star \vec e_y=\vec e_x \vec e_y^{*}=-\vec e_x \vec e_y=-\vec e_z$, since a vector unit is anti-Hermitian, $\vec e_y^{*}=-\vec e_y$ and $\mathsf{M}_2(\vec e_y)^{\dagger}=-\mathsf{M}_2(\vec e_y)$.

The right unit, entry by entry. The identity matrix acts on the left by the conjugate transpose and on the right by the identity, $$ I\star Y=IY^{\dagger}=Y^{\dagger},\qquad Y\star I=YI^{\dagger}=Y, $$ and at $Y=\mathsf{M}_2(\vec e_x)$ the two differ, $I\star\mathsf{M}_2(\vec e_x)=\mathsf{M}_2(\vec e_x)^{\dagger}=\begin{pmatrix}0&i\\i&0\end{pmatrix}=-\mathsf{M}_2(\vec e_x)$ against $\mathsf{M}_2(\vec e_x)\star I=\mathsf{M}_2(\vec e_x)$: the two actions of the unit are the two opposite entries of the basis.

The idempotents, entry by entry. The matrix image of the idempotent $\tfrac12(e_t+i\vec e_z)$ is the rank-one projector onto the first coordinate line, $$ \mathsf{M}_2\bigl(\tfrac12(e_t+i\vec e_z)\bigr)=\tfrac12\bigl(I+i\mathsf{M}_2(\vec e_z)\bigr)=\tfrac12(I+\sigma_3) =\begin{pmatrix}1&0\\0&0\end{pmatrix}, $$ Hermitian of square itself; the other rank-one idempotents of the multiplication have the other orthogonal projections of $\mathbb{C}^2$ as their matrix images, and together these images form the projective line $\mathbb{CP}^{1}$.

Summary

The $2\times2$ realization turns the Hermitian conjugation into the conjugate transpose, so the sesquilinear product is the matrix product with the conjugate transpose in the second slot, $\mathsf{M}_2(\tilde{Q}_1\star\tilde{Q}_2)=\mathsf{M}_2(\tilde{Q}_1)\mathsf{M}_2(\tilde{Q}_2)^{\dagger}$, with the identity matrix as a right unit alone (the left action of the unit is the conjugate transpose) and the Hermitian subspace the Hermitian matrices. The ternary product is the middle model $XY^{\dagger}Z$, the associator is $X(Y^{\dagger}Z^{\dagger}-ZY^{\dagger})$, the sandwich is $X\mapsto \mathsf{M}_2(\tilde Q_1)X^{\dagger}\mathsf{M}_2(\tilde Q_2)^{\dagger}$, and the idempotents are the Hermitian idempotents, whose matrix images are $0$, $I$ and the orthogonal projections onto the lines of $\mathbb{C}^{2}$, the rank-one ones forming $\mathbb{CP}^{1}$; the multiplication is simple with centre $\mathbb{C}I$, and the rank of an element is the rank of its matrix. The general plain sesquilinear form of the group is the Hilbert–Schmidt pairing $\tfrac12\operatorname{Tr}(\mathsf{M}_2(\tilde{Q}_1)^{\dagger}\mathsf{M}_2(\tilde{Q}_2))=\langle\tilde{Q}_2,\tilde{Q}_1\rangle_{*}$, whose diagonal is the definite norm, $\tfrac12\lVert \mathsf{M}_2(\tilde{Q})\rVert_F^2=\sum_\mu|Q_\mu|^2$, with $\lVert \mathsf{M}_2(\tilde{Q})\rVert_F=\sqrt2\lVert\tilde{Q}\rVert_E$; it is positive definite of signature $(8,0)$, of null set $\{0\}$. Its automorphism group is $\mathrm{U}(4)$, whose algebra-preserving part is $\mathrm{U}(2)\times \mathrm{U}(2)$ modulo the centre; the units are $GL_2(\mathbb{C})$, the norm-one group is $SL_2(\mathbb{C})$, and the unitary slice is $\mathrm{U}(2)$ with special part $SU(2)\cong\mathrm{Spin}(3)$. The determinant of a value is $\langle\tilde{Q},\tilde{Q}\rangle_{\natural}\overline{\langle\tilde{Q}_2,\tilde{Q}_2\rangle_{\natural}}$ and its trace is $2\langle\tilde{Q}_1,\tilde{Q}_2\rangle_{*}$; the realization has scale $\sqrt2$, $\lVert \mathsf{M}_2(\tilde{Q})\rVert_F=\sqrt2\,\lVert\tilde{Q}\rVert_E$.

Summary of Notation

Further Reading

• Richard S. Pierce, Associative Algebras (Graduate Texts in Mathematics 88, Springer, 1982), for the algebra $M_2(\mathbb{C})$, its idempotents, its one-sided ideals and the rank identity.
• Max-Albert Knus, Alexander Merkurjev, Markus Rost and Jean-Pierre Tignol, The Book of Involutions (American Mathematical Society Colloquium Publications 44, 1998), for the matrix algebras with an involution and the Hermitian transpose as the standard example.
• Nathan Jacobson, Structure of Rings (American Mathematical Society Colloquium Publications 37, 1956), for the simplicity of a full matrix ring, the centre and the trace.
• Roger A. Horn and Charles R. Johnson, Matrix Analysis (Cambridge University Press, second edition, 2013), for the trace, the determinant, the rank identity and the Hermitian forms of a matrix algebra.
• Irving Kaplansky, Rings of Operators (Benjamin, 1968), for the middle model $XY^{\dagger}Z$, its quadratic representation and the associated triple systems.
• Introduction to the 2×2 Matrix Element Representation $M_2(\mathbb{C})$ of Biquaternions, for the realization and its first properties
• Introduction to the 2×2 Matrix Element Representation $M_2(\mathbb{C})$ of Biquaternions, for the further reading of the isomorphism
• Introduction to the General Plain Sesqualgebra, for the sesquilinear product on the algebra
• The Four Forms and the Four Algebraic Norms over the Biquaternion Algebra, for the form on the algebra
• The Euclidean Topology of the Biquaternion Algebra, for the topology the form induces
• The General Plain Sesqualgebra in the $4\times4$ Matrix Element Representation $M_4(\mathbb{C})_L$, for the companion reading of the group
• Remarkable Subspaces under the General Plain Sesqualgebra, for the restriction theory of the form