The Multiplication Operators of the Algebra of the Reduced Biquaternions

Introduction

This article reads the multiplication of the general plain algebra through its multiplication operators: for an element $\breve Q$ the left multiplication is the map $L_{\breve Q}:\breve Q_{2}\mapsto \breve Q \breve Q_{2}$, the right multiplication is $R_{\breve Q}:\breve Q_{2}\mapsto \breve Q_{2}\breve Q$, and the multiplication is studied by studying the operators it produces. Because the algebra is commutative the two operators coincide, $L_{\breve Q}=R_{\breve Q}$, and this is the first of the article's specificities: the block has one operator per element and not two. The map $\breve Q\mapsto L_{\breve Q}$ is the regular representation of the algebra, and it is injective, so the algebra is realised as a commutative algebra of operators on itself; the image is the two-dimensional space of the $2\times2$ matrices of the form $\left(\begin{smallmatrix}Q_{\blacktriangle}&Q_{\blacktriangledown}\\ Q_{\blacktriangledown}&Q_{\blacktriangle}\end{smallmatrix}\right)$, not the whole endomorphism algebra.

The results. In the basis $1,j$ the operator $L_{\breve Q}$ has the matrix $$ \mathsf L(\breve Q)=\begin{pmatrix}Q_{\blacktriangle}&Q_{\blacktriangledown}\\ Q_{\blacktriangledown}&Q_{\blacktriangle}\end{pmatrix}, $$ symmetric with the two equal diagonal entries $Q_{\blacktriangle}$ and the two equal off-diagonal entries $Q_{\blacktriangledown}$; in the idempotent basis $\Pi_+,\Pi_-$ it is the diagonal matrix $$ \operatorname{diag}\bigl(Q_{\blacktriangle}+Q_{\blacktriangledown},\ Q_{\blacktriangle}-Q_{\blacktriangledown}\bigr), $$ with the two eigenvalues $Q_{+}=Q_{\blacktriangle}+Q_{\blacktriangledown}$ and $Q_{-}=Q_{\blacktriangle}-Q_{\blacktriangledown}$ and the two idempotent lines as eigenspaces. Its trace is $2Q_{\blacktriangle}$, its determinant is $\langle \breve Q\rangle_{\mathrm{ts}}=Q_{\blacktriangle}^{2}-Q_{\blacktriangledown}^{2}$, and its characteristic polynomial is $\lambda^{2}-2Q_{\blacktriangle}\lambda+\langle \breve Q\rangle_{\mathrm{ts}}$; the determinant is the norm, so $L_{\breve Q}$ is invertible exactly when $\breve Q$ is a unit, and $L_{\breve Q}$ is never nilpotent for $\breve Q\neq0$. The map $L$ is an injective algebra homomorphism, $L_{\breve Q_{1}}L_{\breve Q_{2}}=L_{\breve Q_{1}\breve Q_{2}}$, and its image is the commutative two-dimensional subalgebra of $M_{2}(\mathbb{C})$ of the matrices of the displayed form. The enveloping operator algebra acts through the left multiplications alone, $L_{\breve Q_{1}}R_{\breve Q_{2}}=L_{\breve Q_{1}\breve Q_{2}}$, so its image is that same two-dimensional space; the derivations of the algebra into itself are all zero, and its $\mathbb{C}$-algebra automorphisms are the two elements $\mathrm{id}$ and ${}_{\mathrm{ts}}$.

The Multiplication Operators

The Definition and the Coincidence of the Two Operators

The left and the right multiplication of $\breve Q$ coincide, because the algebra is commutative.

Definition. For $\breve Q\in\mathbb{B}_{\mathrm{r}}$ the left multiplication by $\breve Q$ is the endomorphism $$ L_{\breve Q}:\mathbb{B}_{\mathrm{r}}\to\mathbb{B}_{\mathrm{r}},\qquad L_{\breve Q}(\breve Q_{2})=\breve Q \breve Q_{2}, $$ and the right multiplication is $R_{\breve Q}(\breve Q_{2})=\breve Q_{2}\breve Q$. Both are $\mathbb{C}$-linear, because the product is $\mathbb{C}$-bilinear.

Proposition. $L_{\breve Q}=R_{\breve Q}$ for every $\breve Q$.

Proof. The multiplication is commutative, $\breve Q \breve Q_{2}=\breve Q_{2}\breve Q$ for all $\breve Q_{2}$, so the two maps agree at every element.

Corollary (the operator and the element). An ideal generated by $\breve Q$ is the image of $L_{\breve Q}$, and the annihilator of $\breve Q$ is its kernel, so the left and the right readings give one ideal and one annihilator; the ideal theory of Ideals and Peirce Decomposition of the Algebra of the Reduced Biquaternions is the image–kernel theory of a single operator.

The Matrix of the Operator

In the basis $1,j$ the operator is the symmetric matrix $\left(\begin{smallmatrix}Q_{\blacktriangle}&Q_{\blacktriangledown}\\ Q_{\blacktriangledown}&Q_{\blacktriangle}\end{smallmatrix}\right)$; in the idempotent basis it is diagonal.

Theorem (the matrix). In the complex basis $\{1,j\}$ the matrix of $L_{\breve Q}$ is $$ \mathsf L(\breve Q)=\begin{pmatrix}Q_{\blacktriangle}&Q_{\blacktriangledown}\\ Q_{\blacktriangledown}&Q_{\blacktriangle}\end{pmatrix}, $$ and in the idempotent basis $\{\Pi_+,\Pi_-\}$ it is $$ \boxed{\; \mathsf L_{\Pi}(\breve Q)=\begin{pmatrix}Q_{\blacktriangle}+Q_{\blacktriangledown}&0\\ 0&Q_{\blacktriangle}-Q_{\blacktriangledown}\end{pmatrix}=\operatorname{diag}(Q_{+},Q_{-}) \;}. $$

Proof. For the first, $L_{\breve Q}(1)=\breve Q=Q_{\blacktriangle}+jQ_{\blacktriangledown}$ has coordinates $(Q_{\blacktriangle},Q_{\blacktriangledown})$ and $L_{\breve Q}(j)=\breve Q j=Q_{\blacktriangle}j+Q_{\blacktriangledown}j^{2}=Q_{\blacktriangledown}+Q_{\blacktriangle}j$ has coordinates $(Q_{\blacktriangledown},Q_{\blacktriangle})$, so the two columns are $(Q_{\blacktriangle},Q_{\blacktriangledown})$ and $(Q_{\blacktriangledown},Q_{\blacktriangle})$. For the second, $L_{\breve Q}(\Pi_+)=\breve Q\Pi_+=Q_{+}\Pi_+$ and $L_{\breve Q}(\Pi_-)=\breve Q\Pi_-=Q_{-}\Pi_-$, so the operator is diagonal with entries $Q_{+}$ and $Q_{-}$ in the idempotent basis.

Corollary (the eigenspaces). The idempotent lines $\mathbb{C}\Pi_+$ and $\mathbb{C}\Pi_-$ are the eigenspaces of $L_{\breve Q}$, with eigenvalues $Q_{+}=Q_{\blacktriangle}+Q_{\blacktriangledown}$ and $Q_{-}=Q_{\blacktriangle}-Q_{\blacktriangledown}$; in particular the operator is diagonalisable for every $\breve Q$, and its two eigenvalues are the idempotent coordinates.

The Trace, the Determinant and the Spectrum

The trace of the operator is $2Q_{\blacktriangle}$, its determinant is the norm $\langle \breve Q\rangle_{\mathrm{ts}}$, and its characteristic polynomial is $\lambda^{2}-2Q_{\blacktriangle}\lambda+\langle \breve Q\rangle_{\mathrm{ts}}$.

Theorem. For $\breve Q=Q_{\blacktriangle}+jQ_{\blacktriangledown}$, $$ \boxed{\; \operatorname{Tr}L_{\breve Q}=2Q_{\blacktriangle},\qquad \det L_{\breve Q}=Q_{\blacktriangle}^{2}-Q_{\blacktriangledown}^{2}=\langle \breve Q\rangle_{\mathrm{ts}},\qquad \chi_{L_{\breve Q}}(\lambda)=\lambda^{2}-2Q_{\blacktriangle}\lambda+\langle \breve Q\rangle_{\mathrm{ts}} \;}. $$

Proof. Read the matrix $\operatorname{diag}(Q_{+},Q_{-})$ in the idempotent basis: the trace is $Q_{+}+Q_{-}=2Q_{\blacktriangle}$, the determinant is $Q_{+}Q_{-}=(Q_{\blacktriangle}+Q_{\blacktriangledown})(Q_{\blacktriangle}-Q_{\blacktriangledown})=Q_{\blacktriangle}^{2}-Q_{\blacktriangledown}^{2}$, and the characteristic polynomial is $(\lambda-Q_{+})(\lambda-Q_{-})=\lambda^{2}-(Q_{+}+Q_{-})\lambda+Q_{+}Q_{-}$.

Corollary (no nilpotent operator). $L_{\breve Q}$ is nilpotent if and only if $\breve Q=0$. It is invertible if and only if $\langle \breve Q\rangle_{\mathrm{ts}}\neq0$, that is if and only if $\breve Q$ is a unit; and it is a scalar operator, $L_{\breve Q}=Q_{\blacktriangle}\,\mathrm{id}$, if and only if $Q_{\blacktriangledown}=0$, that is if and only if $\breve Q$ is a scalar.

Proof. A nilpotent operator has both eigenvalues zero, $Q_{+}=Q_{-}=0$, hence $Q_{\blacktriangle}=Q_{\blacktriangledown}=0$. The determinant is the product of the eigenvalues, so the operator is invertible exactly when $Q_{+}Q_{-}\neq0$, which is the criterion of The Group of Units of the Algebra of the Reduced Biquaternions. The operator is scalar exactly when its two eigenvalues are equal, $Q_{+}=Q_{-}$, that is $Q_{\blacktriangledown}=0$.

Composition and the Regular Representation

The map $\breve Q\mapsto L_{\breve Q}$ is an injective algebra homomorphism whose image is the two-dimensional commutative algebra of the matrices $\left(\begin{smallmatrix}Q_{\blacktriangle}&Q_{\blacktriangledown}\\ Q_{\blacktriangledown}&Q_{\blacktriangle}\end{smallmatrix}\right)$.

Theorem (the regular representation). For all $\breve Q_{1},\breve Q_{2}$, $$ \boxed{\; L_{\breve Q_{1}}L_{\breve Q_{2}}=L_{\breve Q_{1}\breve Q_{2}}=L_{\breve Q_{2} \breve Q_{1}}=L_{\breve Q_{2}}L_{\breve Q_{1}},\qquad L_{1}=\mathrm{id} \;}, $$ and the map $L:\breve Q\mapsto L_{\breve Q}$ is an injective algebra homomorphism from $(\mathbb{B}_{\mathrm{r}},\cdot)$ into $\mathrm{End}_{\mathbb{C}}(\mathbb{B}_{\mathrm{r}})$.

Proof. $L_{\breve Q_{1}}L_{\breve Q_{2}}(\breve Q_{3})=\breve Q_{1}(\breve Q_{2}\breve Q_{3})=(\breve Q_{1} \breve Q_{2})\breve Q_{3}=L_{\breve Q_{1} \breve Q_{2}}(\breve Q_{3})$ by associativity, and $\breve Q_{1} \breve Q_{2}=\breve Q_{2}\breve Q_{1}$ by commutativity, which is the second equality; $L_{1}(\breve Q)=\breve Q$. The kernel is $\{\breve Q:L_{\breve Q}=0\}$, and $L_{\breve Q}=0$ evaluated at $1$ gives $\breve Q=0$, so $L$ is injective.

Corollary (the image). The image of $L$ is the set of the $2\times2$ matrices of the form $\left(\begin{smallmatrix}Q_{\blacktriangle}&Q_{\blacktriangledown}\\ Q_{\blacktriangledown}&Q_{\blacktriangle}\end{smallmatrix}\right)$ with $Q_{\blacktriangle},Q_{\blacktriangledown}\in\mathbb{C}$, a commutative two-dimensional subalgebra of $M_{2}(\mathbb{C})$ with the unit matrix, in which the two matrices $\left(\begin{smallmatrix}1&0\\ 0&1\end{smallmatrix}\right)$ and $\left(\begin{smallmatrix}0&1\\ 1&0\end{smallmatrix}\right)$ play the roles of $1$ and $j$. It is not the whole of $M_{2}(\mathbb{C})$, which has complex dimension four.

The Kernel and the Invertibility

The kernel of $L_{\breve Q}$ is the annihilator of $\breve Q$; it is nonzero exactly when $\breve Q$ is a zero divisor, and then it is the other idempotent line.

Theorem (the kernel). For $\breve Q=Q_{\blacktriangle}+jQ_{\blacktriangledown}$:

• if $\breve Q$ is a unit, $L_{\breve Q}$ is an automorphism and $\ker L_{\breve Q}=0$;
• if $\breve Q$ lies on the idempotent line $\mathbb{C}\Pi_+$ and $\breve Q\neq0$, then $\ker L_{\breve Q}=\mathbb{C}\Pi_-$;
• if $\breve Q$ lies on $\mathbb{C}\Pi_-$ and $\breve Q\neq0$, then $\ker L_{\breve Q}=\mathbb{C}\Pi_+$;
• if $\breve Q=0$, $\ker L_{\breve Q}=\mathbb{B}_{\mathrm{r}}$.

Proof. The kernel is the set of $\breve Q_{2}$ with $\breve Q \breve Q_{2}=0$, the annihilator of $\breve Q$, computed in Annihilators of the Algebra of the Reduced Biquaternions. For $\breve Q=Q_{+}\Pi_+$ with $Q_{+}\neq0$ the annihilator is $\mathbb{C}\Pi_-$, and symmetrically; for a unit the annihilator is $0$; for $\breve Q=0$ it is everything.

The Bi-module Structure and the Enveloping Algebra

The algebra is a bi-module over itself through the two multiplications, but they coincide, so every sandwich operator is a left multiplication and the enveloping algebra is two-dimensional.

Definition. The enveloping operator algebra is generated by the left and the right multiplications, $\mathbb{B}_{\mathrm{r}}^{\mathrm e}=\mathbb{B}_{\mathrm{r}}\otimes_{\mathbb{C}}\mathbb{B}_{\mathrm{r}}^{\mathrm{opp}}$, acting on $\mathbb{B}_{\mathrm{r}}$ by the sandwich $(\breve Q_{1}\otimes \breve Q_{2}^{\mathrm{opp}})\cdot \breve Q_{3}=\breve Q_{1} \breve Q_{3} \breve Q_{2}$.

Theorem (the sandwich). Because the algebra is commutative, $$ (\breve Q_{1}\otimes \breve Q_{2}^{\mathrm{opp}})\cdot \breve Q_{3}=\breve Q_{1} \breve Q_{2}\,\breve Q_{3}=L_{\breve Q_{1} \breve Q_{2}}(\breve Q_{3}), $$ so the sandwich operator of $\breve Q_{1}\otimes \breve Q_{2}^{\mathrm{opp}}$ is the left multiplication by $\breve Q_{1} \breve Q_{2}$, and the image of $\mathbb{B}_{\mathrm{r}}^{\mathrm e}$ in $\mathrm{End}_{\mathbb{C}}(\mathbb{B}_{\mathrm{r}})$ is the two-dimensional space $\{L_{\breve Q}:\breve Q\in\mathbb{B}_{\mathrm{r}}\}$ of §Composition and the Regular Representation.

Proof. Commutativity gives $\breve Q_{1} \breve Q_{3}\breve Q_{2}=\breve Q_{3}(\breve Q_{1} \breve Q_{2})=(\breve Q_{1} \breve Q_{2})\breve Q_{3}$, which is the displayed identity; the image is the set of all $L_{\breve Q}$ because $\breve Q_{1}$ and $\breve Q_{2}$ range over the algebra.

Remark (the contrast with the biquaternion block). In the biquaternion category the enveloping algebra acts faithfully and exhausts the endomorphisms, $\mathbb{B}^{\mathrm e}\cong\mathrm{End}_{\mathbb{C}}(\mathbb{B})\cong M_{4}(\mathbb{C})$; here the commutative product collapses the sandwich to a left multiplication, and the image has complex dimension two against the four of $\mathrm{End}_{\mathbb{C}}$. The collapsing is the operator reading of the central idempotents: a sandwich can move an element on both sides, but a central element acts the same way on both sides, so the two-sided action carries no more information than the one-sided one.

The Derivations and the Automorphisms

The algebra has no nonzero derivation, and its $\mathbb{C}$-algebra automorphisms are the two maps $\mathrm{id}$ and ${}_{\mathrm{ts}}$.

Definition. A derivation of the algebra is a $\mathbb{C}$-linear map $D$ with the Leibniz rule $D(\breve Q_{1} \breve Q_{2})=D(\breve Q_{1})\breve Q_{2}+\breve Q_{1}D(\breve Q_{2})$; it is inner when $D=\mathrm{ad}_{\breve Q_{2}}:\breve Q_{1}\mapsto \breve Q_{2}\breve Q_{1}-\breve Q_{1} \breve Q_{2}$ for some $\breve Q_{2}$.

Theorem (no derivations). Every derivation of $\mathbb{B}_{\mathrm{r}}$ into itself is zero.

Proof. A derivation satisfies $D(1)=D(1\cdot1)=2D(1)$, so $D(1)=0$. Evaluating on the idempotents, $D(\Pi_\pm)=D(\Pi_\pm^{2})=2\Pi_\pm D(\Pi_\pm)$, so $(2\Pi_\pm-1)D(\Pi_\pm)=0$; but $2\Pi_\pm-1=\pm j$, a unit, so $D(\Pi_\pm)=0$. A derivation vanishing on the spanning set $1,\Pi_+,\Pi_-$ is zero.

Corollary (no inner derivations). Because the algebra is commutative every inner derivation vanishes, $\mathrm{ad}_{\breve Q_{2}}(\breve Q_{1})=\breve Q_{2}\breve Q_{1}-\breve Q_{1} \breve Q_{2}=0$; the identification of $\mathrm{ad}$ with the zero map is the whole of the derivation algebra, and the theorem shows there is nothing else either.

Theorem (the automorphisms). The $\mathbb{C}$-algebra automorphisms of $\mathbb{B}_{\mathrm{r}}$ are the two maps $$ \mathrm{id},\qquad {}_{\mathrm{ts}}:Q_{+}\Pi_++Q_{-}\Pi_-\mapsto Q_{-}\Pi_++Q_{+}\Pi_-, $$ so $\operatorname{Aut}(\mathbb{B}_{\mathrm{r}})\cong\mathbb{Z}_{2}$.

Proof. An automorphism permutes the primitive idempotents and acts on each factor $\mathbb{C}$ by a field automorphism; a $\mathbb{C}$-linear automorphism acts on each factor by the identity, so it is determined by the permutation of $\{\Pi_+,\Pi_-\}$. The identity permutation gives $\mathrm{id}$ and the transposition gives the map displayed, which is the tessarine sign, $Q_{+}\Pi_++Q_{-}\Pi_-\mapsto(Q_{+}\Pi_++Q_{-}\Pi_-)_{\mathrm{ts}}=Q_{+}\Pi_-+Q_{-}\Pi_+$. The group of permutations is $\mathbb{Z}_{2}$.

The Structure Group

The conformal group of the general plain bilinear form $\langle\cdot,\cdot\rangle$ is the union of the two lines of the tessarine sign, isomorphic to $\mathbb{C}^{\times}\times\mathbb{Z}_{2}$.

Definition. The structure group (conformal group) of a form $g$ is the set of invertible $\breve Q_{1}$ for which there is a scalar $\lambda(\breve Q_{1})$ with $$ g(\breve Q_{1}\breve Q_{2},\breve Q_{1}\breve Q_{3})=\lambda(\breve Q_{1})\,g(\breve Q_{2},\breve Q_{3})\qquad\text{for all }\breve Q_{2},\breve Q_{3} . $$

Theorem. For the general plain bilinear form $\langle\cdot,\cdot\rangle$ the structure group is $$ \{\breve Q_{1}:B(\breve Q_{1} \breve Q_{2},\breve Q_{1} \breve Q_{3})=\lambda B(\breve Q_{2},\breve Q_{3})\}=\mathbb{C}^{\times}\!\cdot 1\ \cup\ \mathbb{C}^{\times}\!\cdot j, $$ the nonzero elements of the scalar line and of the $j$-line, with $\lambda=Q_{\blacktriangle}^{2}$ on the scalar line and $\lambda=Q_{\blacktriangledown}^{2}$ on the $j$-line; as a group it is $\mathbb{C}^{\times}\times\mathbb{Z}_{2}$.

Proof. In the basis $1,j$ the form $\langle\cdot,\cdot\rangle$ has Gram matrix $\mathrm I_{2}$ and the operator $L_{\breve Q}$ has matrix $\left(\begin{smallmatrix}Q_{\blacktriangle}&Q_{\blacktriangledown}\\ Q_{\blacktriangledown}&Q_{\blacktriangle}\end{smallmatrix}\right)$, so $B(\breve Q_{1} \breve Q_{2},\breve Q_{1} \breve Q_{3})=\breve Q_{2}^{\mathsf t}\,\mathsf L(\breve Q_{1})^{\mathsf t}\mathsf L(\breve Q_{1})\,\breve Q_{3}$ with $$ \mathsf L(\breve Q_{1})^{\mathsf t}\mathsf L(\breve Q_{1})=\begin{pmatrix}Q_{\blacktriangle}^{2}+Q_{\blacktriangledown}^{2}&2Q_{\blacktriangle}Q_{\blacktriangledown}\\ 2Q_{\blacktriangle}Q_{\blacktriangledown}&Q_{\blacktriangle}^{2}+Q_{\blacktriangledown}^{2}\end{pmatrix}. $$ This is a scalar multiple of $\mathrm I_{2}$ exactly when $Q_{\blacktriangle}Q_{\blacktriangledown}=0$, that is $Q_{\blacktriangle}=0$ or $Q_{\blacktriangledown}=0$; then $\lambda=Q_{\blacktriangledown}^{2}$ on the $j$-line and $\lambda=Q_{\blacktriangle}^{2}$ on the scalar line. The set is closed under the product and under inversion, with group law $\mathbb{C}^{\times}\times\mathbb{Z}_{2}$: the product of two elements of the scalar line stays there, the product of one of each line lies on the $j$-line, and the product of two elements of the $j$-line lies on the scalar line.

Remark (two components). The structure group has two connected components, the scalar line and the $j$-line, and they are the fixed and the anti-fixed space of the tessarine sign. It is not the whole unit group $\mathbb{C}^{\times}\times\mathbb{C}^{\times}$: conformity with $\langle\cdot,\cdot\rangle$ is stronger than invertibility, and it forces the element onto one of the two lines of the tessarine sign.

Worked Examples

The operator of a generic element. For $\breve Q=2+3j$ the matrix in the basis $1,j$ is $\left(\begin{smallmatrix}2&3\\ 3&2\end{smallmatrix}\right)$, with trace $4$, determinant $4-9=-5=\langle \breve Q\rangle_{\mathrm{ts}}$ and eigenvalues $5$ and $-1$ on the lines $\mathbb{C}\Pi_+$ and $\mathbb{C}\Pi_-$. The element is a unit because $-5\neq0$, and the operator is invertible.

The operator of an idempotent. For $\breve Q=\Pi_+$ the matrix is $\left(\begin{smallmatrix}1/2&1/2\\ 1/2&1/2\end{smallmatrix}\right)$, of trace $1$, determinant $0$ and eigenvalues $1$ and $0$; its image is the line $\mathbb{C}\Pi_+$ and its kernel the line $\mathbb{C}\Pi_-$, the reading of the zero divisor.

The operator of $j$. For $\breve Q=j$ the matrix is $\left(\begin{smallmatrix}0&1\\ 1&0\end{smallmatrix}\right)$, an involution of trace $0$ and determinant $-1=\langle j\rangle_{\mathrm{ts}}$; it swaps the two scalar coordinates, and its eigenvectors are $\Pi_+$ and $\Pi_-$ with eigenvalues $+1$ and $-1$. It is the conformal operator of the $j$-line of the structure group.

Summary

For an element $\breve Q=Q_{\blacktriangle}+jQ_{\blacktriangledown}$ the left and the right multiplications coincide, $L_{\breve Q}=R_{\breve Q}$, and in the basis $1,j$ the operator is the symmetric matrix $\left(\begin{smallmatrix}Q_{\blacktriangle}&Q_{\blacktriangledown}\\ Q_{\blacktriangledown}&Q_{\blacktriangle}\end{smallmatrix}\right)$, diagonal $\operatorname{diag}(Q_{\blacktriangle}+Q_{\blacktriangledown},Q_{\blacktriangle}-Q_{\blacktriangledown})$ in the idempotent basis. Its trace is $2Q_{\blacktriangle}$, its determinant is the norm $\langle \breve Q\rangle_{\mathrm{ts}}=Q_{\blacktriangle}^{2}-Q_{\blacktriangledown}^{2}$, and its eigenvalues are the idempotent coordinates on the two idempotent lines; it is invertible exactly when $\breve Q$ is a unit, nilpotent only for $\breve Q=0$, and scalar only for $Q_{\blacktriangledown}=0$. The regular representation $\breve Q\mapsto L_{\breve Q}$ is an injective algebra homomorphism onto the two-dimensional commutative subalgebra of $M_{2}(\mathbb{C})$ of those matrices, which is not the whole endomorphism algebra. The kernel of $L_{\breve Q}$ is the annihilator of $\breve Q$: $0$ for a unit, the other idempotent line for a nonzero element of a line, everything for $0$. The enveloping operator algebra acts through the left multiplications alone, $L_{\breve Q_{1}}R_{\breve Q_{2}}=L_{\breve Q_{1}\breve Q_{2}}$, so its image is the same two-dimensional space and the sandwich carries no more than one-sided multiplication; the contrast with the biquaternion $\mathbb{B}^{\mathrm e}\cong M_{4}(\mathbb{C})$ is the collapse produced by the central idempotents. The algebra has no nonzero derivation, inner or otherwise, and its $\mathbb{C}$-algebra automorphisms are $\mathrm{id}$ and the tessarine sign, a group $\mathbb{Z}_{2}$. The structure group of the general plain bilinear form is the union of the two lines of the tessarine sign, $\mathbb{C}^{\times}\times\mathbb{Z}_{2}$.

Summary of Notation

Further Reading

• Introduction to the Algebra of the Reduced Biquaternions, for the multiplication and the idempotent coordinates.
• The Specificities of the Algebra of the Reduced Biquaternions, for the projector calculus of the operators.
• The Forms of the Algebra of the Reduced Biquaternions, for the adjoint rule and the conformal group of $\langle\cdot,\cdot\rangle$.
• The Group of Units of the Algebra of the Reduced Biquaternions, for the invertibility criterion on the elements and the operators.
• Ideals and Peirce Decomposition of the Algebra of the Reduced Biquaternions, for the image and the kernel of the operator.
• Introduction to the Operators of the Biquaternion Algebra, for the non-commutative companion with its full enveloping algebra.