The Four Adjoints of the Biquaternion Algebra in Examples

Introduction

The biquaternion algebra carries four pairings, and every pairing carries its own adjoint of a linear map. Three of the pairings carry a named operator family of the corpus, and those families are named by their adjoints: Association, Signed Inner Conjugation and Hermitian Adjoint, the last with a signed version. The fourth pairing, the general quaternionic sesquilinear form, carries the indefinite Hermitian sandwich instead, named by its Krein adjoint; it is the form of the Krein articles, and its family is not one of the four names. The names look unrelated and the suffixes look arbitrary, so this article sets the four adjoints side by side and works them out on explicit maps and explicit matrices, in the coefficient basis $e_t,\vec e_x,\vec e_y,\vec e_z$.

The result is a one-line rule, and the examples below are the evidence for it: each of the four operator names records an adjoint, and the adjective names the insertion of the sign. Association is the transpose of the general plain bilinear form, the reversal of a product with no conjugation anywhere, and it is the name of that adjoint because the adjoint carries no conjugation to name. Signed inner conjugation is the name of the adjoint of the general quaternionic bilinear form: there the adjoint of $L_{\tilde Q_6}$ is the left multiplication by $\tilde Q_6^{\natural}$, the signed left multiplication, and the sandwich built from it is the inner conjugation up to the norm. Hermitian adjoint is the name of the adjoint of the general plain sesquilinear form, which is the conjugate linear $\ast$; and signed Hermitian adjoint is the same adjoint read on the operators that carry the sign, which is why the Hermitian layer carries two families where the plain layer carries one. The word signed always has the same meaning in the corpus: the grade involution $\natural$, the conjugation that negates the vector part and fixes the centre, has been inserted.

The forms are The Four Forms and the Four Algebraic Norms over the Biquaternion Algebra; the four conjugations are The Group of Involutions; the plain transpose is the general transpose of a bilinear form of Bilinear Forms, §Isometries, and its operator family is Two-Sided Operators on the Biquaternion Algebra; the family of the second form is Two-Sided Operators on the General Quaternionic Algebra. The table of §Which Switch Does What reads the two involutions on the four adjoints: the sign character $\alpha$ exchanges the side, and the coefficient involution $\sigma=\bar{\cdot}$ acts on the parameter. The switches themselves, with the twisted right factor and the four forms read at the unit, belong to the operator layer, in Introduction to the Operators of the Biquaternion Algebra, §The Two Switches and the Twisted Right Factor. The article is pure algebra; no argument uses a distance, a limit or a derivative.

Conventions and the Four Forms

Theorem (the four forms and their Gram matrices). The four pairings of the algebra are

$$ \langle\tilde Q_1,\tilde Q_2\rangle=\mathrm{Sc}(\tilde Q_1\tilde Q_2), \qquad \langle\tilde Q_1,\tilde Q_2\rangle_{\natural}=\mathrm{Sc}(\tilde Q_1^{\natural}\tilde Q_2), $$ $$ \langle\tilde Q_1,\tilde Q_2\rangle_{\ast}=\mathrm{Sc}(\tilde Q_1\tilde Q_2^{\ast}), \qquad \langle\tilde Q_1,\tilde Q_2\rangle_{\natural\ast}=\mathrm{Sc}(\tilde Q_1^{\natural}\tilde Q_2^{\ast}), $$

and in the basis $e_t,\vec e_x,\vec e_y,\vec e_z$ their Gram matrices are $\mathrm{E}=\operatorname{diag}(1,-1,-1,-1)$, $\mathrm{I}_4$, $\mathrm{I}_4$ and $\mathrm{E}$. The first two are bilinear and the last two are sesquilinear, conjugate linear in the second argument. All four are non-degenerate and no two of them agree.

Proof. Each pairing is the scalar part of one of the four general products of The Four General Products and Their Relations over the Biquaternion Algebra, and $\mathrm{Sc}(e_\mu e_\nu)=\varepsilon_\mu\delta_{\mu\nu}$ with $\varepsilon=(1,-1,-1,-1)$; the natural and the Hermitian conjugations are diagonal in the basis, of diagonal entries $\varepsilon$ and $-\varepsilon$ respectively, which gives the four Gram matrices. The diagonal forms $\sum_\mu\varepsilon_\mu Q_\mu^{2}$, $\sum_\mu Q_\mu^{2}$, $\sum_\mu|Q_\mu|^{2}$ and $\sum_\mu\varepsilon_\mu|Q_\mu|^{2}$ separate the four pairings on the units.

Remark (the four forms are one family). Any two of the four are recovered from the others by the two sign characters of the argument, $\langle\tilde Q_1,\tilde Q_2\rangle_{\natural}=\langle\tilde Q_1^{\natural},\tilde Q_2\rangle$, $\langle\tilde Q_1,\tilde Q_2\rangle_{\ast}=\langle\tilde Q_1,\bar{\tilde Q}_2\rangle_{\natural}$, so the four forms differ only in which of the identity, $\natural$, $\bar{\cdot}$ and $\ast$ is inserted into which slot. The four adjoints below differ in exactly the same way, and that is the whole point of the article.

The Four Adjoints

Definition. For each of the four pairings $\phi=\langle\cdot,\cdot\rangle,\langle\cdot,\cdot\rangle_{\natural},\langle\cdot,\cdot\rangle_{\ast},\langle\cdot,\cdot\rangle_{\natural\ast}$ and each $\mathbb{C}$-linear map $F$ of $\mathbb{B}$, the adjoint of $F$ for $\phi$ is the unique $\mathbb{C}$-linear map $F^{\phi}$ with

$$ \phi\bigl(F\tilde Q_1,\tilde Q_2\bigr)=\phi\bigl(\tilde Q_1,F^{\phi}\tilde Q_2\bigr)\qquad\text{for all }\tilde Q_1,\tilde Q_2\in\mathbb{B}. $$

The form is non-degenerate, so the adjoint exists and is unique, and $F\mapsto F^{\phi}$ is conjugate linear in the sesquilinear cases and linear in the bilinear ones.

The corpus writes the first adjoint $F^{\approx}$ and calls it the associate; it writes the second $F^{\natural}$ and calls it the adjoint of the quaternionic form; the third is the Hermitian adjoint $F^{\ast}$. This article writes $F^{\phi}$ with the form for the subscript, and uses $\approx$ for the adjoint of the plain form and $\natural$ for the adjoint of the quaternionic form.

Theorem (the adjoints of the one-sided and two-sided operators). For the left multiplication $L_{\tilde Q_6}(\tilde Q)=\tilde Q_6\tilde Q$, the right multiplication $R_{\tilde Q_3}(\tilde Q)=\tilde Q\tilde Q_3$, and their product $L_{\tilde Q_6}R_{\tilde Q_3}$,

$$ (L_{\tilde Q_6})^{\approx}=R_{\tilde Q_6},\qquad (R_{\tilde Q_3})^{\approx}=L_{\tilde Q_3},\qquad (L_{\tilde Q_6}R_{\tilde Q_3})^{\approx}=L_{\tilde Q_3}R_{\tilde Q_6}; $$ $$ (L_{\tilde Q_6})^{N}=L_{\tilde Q_6^{\natural}},\qquad (R_{\tilde Q_3})^{N}=R_{\tilde Q_3^{\natural}},\qquad (L_{\tilde Q_6}R_{\tilde Q_3})^{\natural}=L_{\tilde Q_6^{\natural}}R_{\tilde Q_3^{\natural}}; $$ $$ (L_{\tilde Q_6})^{\ast}=L_{\tilde Q_6^{\ast}},\qquad (R_{\tilde Q_3})^{\ast}=R_{\tilde Q_3^{\ast}},\qquad (L_{\tilde Q_6}R_{\tilde Q_3})^{\ast}=L_{\tilde Q_6^{\ast}}R_{\tilde Q_3^{\ast}}; $$ $$ (L_{\tilde Q_6})^{\natural\ast}=R_{\bar{\tilde Q_6}},\qquad (R_{\tilde Q_3})^{\natural\ast}=L_{\bar{\tilde Q_3}},\qquad (L_{\tilde Q_6}R_{\tilde Q_3})^{\natural\ast}=L_{\bar{\tilde Q_3}}R_{\bar{\tilde Q_6}}. $$

Proof. For the first pair, $\mathrm{Sc}(\tilde Q_6\tilde Q_1\tilde Q_2)=\mathrm{Sc}(\tilde Q_1\tilde Q_2\tilde Q_6)$ by the cyclicity of $\mathrm{Sc}$, which is $\langle L_{\tilde Q_6}\tilde Q_1,\tilde Q_2\rangle=\langle\tilde Q_1,R_{\tilde Q_6}\tilde Q_2\rangle$, and the two-sided formula follows from the reversal of the adjoint of a composite, $(FG)^{\approx}=G^{\approx}F^{\approx}$. For the second, $\mathrm{Sc}((\tilde Q_6\tilde Q_1)^{\natural}\tilde Q_2)=\mathrm{Sc}(\tilde Q_1^{\natural}\tilde Q_6^{\natural}\tilde Q_2)=\langle\tilde Q_1,L_{\tilde Q_6^{\natural}}\tilde Q_2\rangle_{\natural}$, and the factors are not reversed because the adjoint of a left multiplication is again a left multiplication; the two-sided formula is $R_{\tilde Q_3^{\natural}}L_{\tilde Q_6^{\natural}}=L_{\tilde Q_6^{\natural}}R_{\tilde Q_3^{\natural}}$. The third is the same computation with the conjugate linear $\ast$ in place of $\natural$. For the fourth, $\mathrm{Sc}((\tilde Q_6\tilde Q_1)^{\natural}\tilde Q_2^{\ast})=\mathrm{Sc}(\tilde Q_1^{\natural}\tilde Q_6^{\natural}\tilde Q_2^{\ast})$ and $\mathrm{Sc}(\tilde Q_1^{\natural}\tilde Q_6^{\natural}\tilde Q_2^{\ast})=\mathrm{Sc}(\tilde Q_1^{\natural}(\tilde Q_2\bar{\tilde Q_6})^{\ast})$ by the cyclic property together with $(\tilde Q_2\bar{\tilde Q_6})^{\ast}=\tilde Q_6^{\natural}\tilde Q_2^{\ast}$, so the adjoint of $L_{\tilde Q_6}$ is $R_{\bar{\tilde Q_6}}$; the two-sided formula again reverses. Every identity was verified on the sixteen pairs of basis elements.

Corollary (the plain adjoint swaps, the signed adjoint conjugates). The first two lines above read

$$ \boxed{\; (L_{\tilde Q_6}R_{\tilde Q_3})^{\approx}=L_{\tilde Q_3}R_{\tilde Q_6}, \qquad (L_{\tilde Q_6}R_{\tilde Q_3})^{\natural}=L_{\tilde Q_6^{\natural}}R_{\tilde Q_3^{\natural}} \;}. $$

The associate reverses the two parameters and leaves them alone; the $\natural$-adjoint keeps the parameters in place and replaces each by its image under the sign $\natural$. So the word association names a transpose, and it is the suffix of a family exactly when the form of that family is the plain one; the word signed inner conjugation names the adjoint of the $\natural$-form, and no factor is reversed there. The third adjoint conjugates in place with the conjugate linear $\ast$, and the fourth swaps and conjugates.

Remark (why this answers the naming question). Each of the first three names is the name of an adjoint of one form: $\approx$ is printed as "Association", $\langle\cdot,\cdot\rangle_{\natural}$ as "Signed Inner Conjugation", $\ast$ as "Hermitian Adjoint". The form decides the adjoint, and the adjoint supplies the name. The last adjoint above, for the general quaternionic sesquilinear form, is the Krein adjoint of the indefinite theory; its family is the indefinite Hermitian sandwich, and it is named after the Krein structure rather than after one of the four operator names. The suffix "Signed Hermitian Adjoint" belongs to the Hermitian-algebra layer, where it names the operators with the sign inserted on the same form as the Hermitian family.

Association as a Transpose

The first of the four adjoints is the one the corpus has named, and it carries a calculus of its own. Write it $\approx$ for a general $\mathbb{C}$-linear map $F$, not only for a one-sided or a two-sided operator.

Proposition (the associate is an involution that reverses composition). For $\mathbb{C}$-linear maps $F,G$ of $\mathbb{B}$,

$$ (F^{\approx})^{\approx}=F,\qquad (F\circ G)^{\approx}=G^{\approx}\circ F^{\approx}, $$

and on a pair of elements the defining property reads

$$ \mathrm{Sc}\bigl(F(\tilde Q_1)\,\tilde Q_2\bigr)=\mathrm{Sc}\bigl(\tilde Q_1\,F^{\approx}(\tilde Q_2)\bigr)\qquad\text{for all }\tilde Q_1,\tilde Q_2\in\mathbb{B}. $$

Proof. The involution and the reversal are the uniqueness of the adjoint applied twice, as in the two-sided line of the theorem above; the second display is the definition with $\langle\tilde Q_1,\tilde Q_2\rangle=\mathrm{Sc}(\tilde Q_1\tilde Q_2)$ substituted.

Proposition (association reverses the factors of the Conway basis). On the Conway operator basis $e_n[\,]e_m$, the elementary operator being $(e_n[\,]e_m)(\tilde Q)=e_n\tilde Q e_m$,

$$ \bigl(e_n[\,]e_m\bigr)^{\approx}=e_m[\,]e_n, \qquad\text{so}\qquad \Bigl(\sum_{n,m}A_{nm}\,e_n[\,]e_m\Bigr)^{\approx}=\sum_{n,m}A_{nm}\,e_m[\,]e_n . $$

The edge rows $L_a=\sum_n a_n\,e_n[\,]e_t$ and $R_b=\sum_m b_m\,e_t[\,]e_m$ give back $L_a^{\approx}=R_a$ and $R_b^{\approx}=L_b$.

Proof. $\langle(a[\,]b)\tilde Q_1,\tilde Q_2\rangle=\mathrm{Sc}(a\tilde Q_1b\tilde Q_2)$ and the scalar part is invariant under the cyclic permutation of its three factors, $\mathrm{Sc}(a\tilde Q_1b\tilde Q_2)=\mathrm{Sc}(\tilde Q_1b\tilde Q_2a)=\langle\tilde Q_1,(b[\,]a)\tilde Q_2\rangle$; the identity is satisfied by $b[\,]a$ and it is unique by non-degeneracy. The edge rows are the special cases $b=e_t$ and $a=e_t$. The Conway basis is The Enveloping Operator Algebra of the Biquaternion Algebra and the Bi-module Structure, §The Conway Operator Basis.

Proposition (the matrix of the associate). Let $F$ have matrix $M$ in the coefficient basis $e_t,\vec e_x,\vec e_y,\vec e_z$. Then the matrix of the associate is

$$ \boxed{\; M^{\approx}=\mathrm{E}\,M^{\mathsf T}\mathrm{E} \;}, $$

the sign matrix $\mathrm{E}$ conjugating the plain transpose.

Proof. Write the form as $\langle\tilde Q_1,\tilde Q_2\rangle=Q_1^{\mathsf T}\mathrm{E}Q_2$ on the coordinate columns. Then $\langle F\tilde Q_1,\tilde Q_2\rangle=(MQ_1)^{\mathsf T}\mathrm{E}Q_2=Q_1^{\mathsf T}M^{\mathsf T}\mathrm{E}Q_2$ and $\langle\tilde Q_1,F^{\approx}\tilde Q_2\rangle=Q_1^{\mathsf T}\mathrm{E}M^{\approx}Q_2$ for all $Q_1,Q_2$, so $M^{\mathsf T}\mathrm{E}=\mathrm{E}M^{\approx}$, that is $M^{\approx}=\mathrm{E}M^{\mathsf T}\mathrm{E}$ because $\mathrm{E}^{-1}=\mathrm{E}$.

Remark (when the associate is the plain transpose). The two agree exactly when $M^{\mathsf T}$ commutes with $\mathrm{E}$. An operator that fixes the scalar unit $e_t$ and preserves the vector subspace has the block form $\operatorname{diag}(1,M)$ with $M$ a $3\times3$ matrix, and the conjugation by $\mathrm{E}$ acts as the sign $-1$ on the two off-diagonal blocks and as the identity on the two diagonal ones, so it is invisible there and $M^{\approx}=M^{\mathsf T}$. For a general map the two differ. The formula is the coordinate form of the general statement that the transpose of a bilinear form is taken with the Gram matrix conjugating both sides, $M\mapsto G^{-1}M^{\mathsf T}G$ (Bilinear Forms, §Isometries, and Quadratic Forms and Polarisation, §Isometry of Forms).

Proposition (association against the Hermitian adjoint). Let $\bar F$ be the map obtained from $F$ by conjugating every coefficient. Then

$$ \bar F^{\approx}=\mathrm{E}\,F^{\ast}\,\mathrm{E}, $$

$F^{\ast}$ being the adjoint for the positive definite Hermitian form $\langle\cdot,\cdot\rangle_{\ast}$ of the theorem above.

Proof. Conjugating the coefficients conjugates the entries of the matrix, taking $M$ to $\bar M$, so $\bar F^{\approx}$ has matrix $\mathrm{E}\bar M^{\mathsf T}\mathrm{E}$. The Hermitian adjoint has matrix $M^{\dagger}=\overline{M^{\mathsf T}}$, the conjugate transpose, so $\mathrm{E}F^{\ast}\mathrm{E}=\mathrm{E}\overline{M^{\mathsf T}}\mathrm{E}$, the same matrix.

Corollary (the collapse on the vector-preserving unitaries). On the operators of the block form $\operatorname{diag}(1,N)$ the conjugation by $\mathrm{E}$ is invisible, so the proposition reads $\bar F^{\approx}=F^{\ast}$, and for a unitary $F$ it reads $F^{\ast}=F^{-1}$. On those operators the plain transpose, the associate and the Hermitian adjoint are one involution, which is why the source literature writes the three interchangeably; for a general linear map they are three, and the corpus keeps the association, the transpose for the indefinite form of Gram matrix $\mathrm{E}$, apart from the dagger, the adjoint for the positive definite form of Gram matrix $\mathrm{I}_4$.

One Operator, Four Adjoints

The example works one operator $F=L_{\tilde Q_6}R_{\tilde Q_3}$ with $\tilde Q_6=e_t+2\vec e_x$ and $\tilde Q_3=e_t+i\vec e_z$ through all four adjoints. The matrices are written in the basis $e_t,\vec e_x,\vec e_y,\vec e_z$, the columns being the images of the four units.

$$ F=\mathsf{M}_4(\tilde Q_6)\mathsf{M}_4^{R}(\tilde Q_3)= \begin{pmatrix} 1&-2&-2i&-i\\ 2&1&i&-2i\\ -2i&-i&1&-2\\ i&-2i&2&1 \end{pmatrix}, \qquad F^{\approx}=\mathsf{M}_4(\tilde Q_3)\mathsf{M}_4^{R}(\tilde Q_6)= \begin{pmatrix} 1&-2&2i&-i\\ 2&1&-i&-2i\\ 2i&i&1&2\\ i&-2i&-2&1 \end{pmatrix}, $$ $$ F^{\natural}=\mathsf{M}_4(\tilde Q_6^{\natural})\mathsf{M}_4^{R}(\tilde Q_3^{\natural})= \begin{pmatrix} 1&2&-2i&i\\ -2&1&-i&-2i\\ -2i&i&1&2\\ -i&-2i&-2&1 \end{pmatrix}, \qquad F^{\ast}=\mathsf{M}_4(\tilde Q_6^{\ast})\mathsf{M}_4^{R}(\tilde Q_3^{\ast})= \begin{pmatrix} 1&2&2i&-i\\ -2&1&i&2i\\ 2i&-i&1&2\\ i&2i&-2&1 \end{pmatrix}, $$ $$ F^{\natural\ast}=\mathsf{M}_4(\bar{\tilde Q_3})\mathsf{M}_4^{R}(\bar{\tilde Q_6})= \begin{pmatrix} 1&-2&-2i&i\\ 2&1&i&2i\\ -2i&-i&1&2\\ -i&2i&-2&1 \end{pmatrix}. $$

The five matrices are distinct. Reading the four adjoint matrices against $F$: the associate $F^{\approx}$ differs from $F$ by the reversal of the two factors, $2\vec e_x$ and $i\vec e_z$ changing places; the $\natural$-adjoint differs by the replacement $\tilde Q_6\mapsto\tilde Q_6^{\natural}=e_t-2\vec e_x$ and $\tilde Q_3\mapsto\tilde Q_3^{\natural}=e_t-i\vec e_z$; the Hermitian adjoint by $\tilde Q_6\mapsto\tilde Q_6^{\ast}=e_t-2\vec e_x$ and $\tilde Q_3\mapsto\tilde Q_3^{\ast}=e_t-i\vec e_z$ as well, the two differing only in the bar on the coefficients of $F$ itself; and the fourth adjoint by both operations, the swap of $L_{\bar{\tilde Q_3}}R_{\bar{\tilde Q_6}}$ and the two conjugations. None of the four adjoints equals $F$: the example is self-adjoint for no pairing, which is the generic case.

Remark (the sign is a bar and a swap in the entries). The four matrices above are the same matrix up to the conjugations and the transposition along the anti-diagonal of the parameter pair, which is what the corollary predicts; the reader who wants the transposition alone can compare $F$ with $F^{\approx}$, and the reader who wants the conjugation alone can compare $F$ with $F^{\natural}$.

Signed, Unsigned and Associated

The second meaning of the word signed lives in the operators rather than in the adjoints. Two maps carry the same name in the corpus and differ by the sign $\natural$:

$$ \mathrm{Ad}_{\tilde Q_6}(\tilde Q)=\tilde Q_6\tilde Q\tilde Q_6^{-1}, \qquad \mathrm{Ad}^{\alpha}_{\tilde Q_6}(\tilde Q)=\tilde Q_6^{\natural}\tilde Q\tilde Q_6^{-1}, $$

the unsigned inner conjugation and the signed inner conjugation; the second is the first with the grade involution $\alpha=\natural$ inserted on the left factor, and it is the operator of the general theory of The Clifford, Pin and Spin Groups with Signed Inner Conjugation, where $\mathrm{Ad}^{\alpha}_{U}(v)=\alpha(U)vU^{-1}$.

Proposition (the difference is the vector part). For an invertible $\tilde Q_6$ with scalar part $Q_{6t}$ and vector part $\vec Q_6$,

$$ \boxed{\; \mathrm{Ad}^{\alpha}_{\tilde Q_6}(\tilde Q)-\mathrm{Ad}_{\tilde Q_6}(\tilde Q)=\bigl(\tilde Q_6^{\natural}-\tilde Q_6\bigr)\tilde Q\tilde Q_6^{-1}=-2\,\vec Q_6\,\tilde Q\tilde Q_6^{-1} \;}, $$

moreover $\mathrm{Ad}^{\alpha}_{\tilde Q_6}=\mathrm{Ad}_{\tilde Q_6}$ when $\tilde Q_6$ is central, and $\mathrm{Ad}^{\alpha}_{\tilde Q_6}=-\mathrm{Ad}_{\tilde Q_6}$ when $\tilde Q_6$ lies in the vector subspace $V$. The second case is the insertion of the sign at its sharpest: for $\tilde Q_6\in V$ the whole operator is negated.

Proof. The first two displays are the definition, and $\tilde Q_6^{\natural}-\tilde Q_6=-2\vec Q_6$ with $\vec Q_6=\tilde Q_6-Q_{6t}e_t$. If $\vec Q_6=0$ the difference vanishes; if $Q_{6t}=0$ then $\tilde Q_6^{\natural}=-\tilde Q_6$.

Worked example ($\tilde Q_6=e_t+\vec e_x$, mixed, neither central nor in $V$). The two operators and their difference are

$$ \mathrm{Ad}_{\tilde Q_6}= \begin{pmatrix} 1&0&0&0\\ 0&1&0&0\\ 0&0&0&-1\\ 0&0&1&0 \end{pmatrix}, \qquad \mathrm{Ad}^{\alpha}_{\tilde Q_6}= \begin{pmatrix} 0&1&0&0\\ -1&0&0&0\\ 0&0&1&0\\ 0&0&0&1 \end{pmatrix}, $$ $$ \mathrm{Ad}^{\alpha}_{\tilde Q_6}-\mathrm{Ad}_{\tilde Q_6}= \begin{pmatrix} -1&1&0&0\\ -1&-1&0&0\\ 0&0&1&1\\ 0&0&-1&1 \end{pmatrix}. $$

The unsigned operator is the identity on the plane $\mathbb{C}\{e_t,\vec e_x\}$ and the quarter turn $\vec e_y\mapsto \vec e_z$, $\vec e_z\mapsto-\vec e_y$ on $\mathbb{C}\{\vec e_y,\vec e_z\}$; the signed operator interchanges the two roles, it is a quarter turn on $\mathbb{C}\{e_t,\vec e_x\}$ and the identity on $\mathbb{C}\{\vec e_y,\vec e_z\}$. The difference is twice the vector part, $2\vec Q_6=2\vec e_x$, and it reads as the matrix above.

Worked example (the two extreme cases). For $\tilde Q_6=ie_t$, central, both operators are the identity: $\mathrm{Ad}^{\alpha}_{ie_t}=\mathrm{Ad}_{ie_t}=\mathrm{id}$. For $\tilde Q_6=\vec e_x\in V$, of norm one, with $\vec e_x^{-1}=-\vec e_x$,

$$ \mathrm{Ad}_{\vec e_x}=\operatorname{diag}(1,1,-1,-1), \qquad \mathrm{Ad}^{\alpha}_{\vec e_x}=\operatorname{diag}(-1,-1,1,1)=-\mathrm{Ad}_{\vec e_x}. $$

The unsigned operator fixes $e_t$ and $\vec e_x$ and negates the plane $\mathbb{C}\{\vec e_y,\vec e_z\}$; the signed operator does the opposite. Both are isometries of the plain form, both of determinant one on the four-dimensional space, because the space has even dimension.

Corollary (the sign costs multiplicativity). The unsigned inner conjugation is an algebra automorphism for every unit, $\mathrm{Ad}_{\tilde Q_6}(\tilde Q_1\tilde Q_2)=\mathrm{Ad}_{\tilde Q_6}(\tilde Q_1)\mathrm{Ad}_{\tilde Q_6}(\tilde Q_2)$. The signed inner conjugation with a parameter in $V$ is anti-multiplicative, $\mathrm{Ad}^{\alpha}_{\tilde Q_6}(\tilde Q_1\tilde Q_2)=-\mathrm{Ad}^{\alpha}_{\tilde Q_6}(\tilde Q_1)\mathrm{Ad}^{\alpha}_{\tilde Q_6}(\tilde Q_2)$; both statements were verified on one hundred pairs. This is the concrete cost of the insertion, and it is the reason the signed operators of the corpus are read for their own laws and not as automorphisms.

Remark (two places for the sign, and the norm reads it). The corpus inserts the sign on the left factor in $\mathrm{Ad}^{\alpha}$, and it writes the twisted two-sided operator $\Theta_{\tilde Q_6}=L_{\tilde Q_6}R_{\tilde Q_6^{\natural}}$, with the sign on the right factor. The two are exchanged by the inverse and by the norm:

$$ \boxed{\; \Theta_{\tilde Q_6}=\langle\tilde Q_6\rangle_{\natural}\,\mathrm{Ad}_{\tilde Q_6}, \qquad \mathrm{Ad}^{\alpha}_{\tilde Q_6}=\langle\tilde Q_6\rangle_{\natural}\,L_{\tilde Q_6^{-1}}R_{\tilde Q_6^{-1}}, \qquad \langle\tilde Q_6\rangle_{\natural}=\tilde Q_6\tilde Q_6^{\natural} \;}, $$

verified on one hundred invertible elements. On the shell $\langle\tilde Q_6\rangle_{\natural}=\pm1$ the first line reads $\Theta_{\tilde Q_6}=\pm\mathrm{Ad}_{\tilde Q_6}$: the same operator is the inner automorphism or its negative according to the sign of the norm, which is the second reading of the word signed: the operator $\Theta_{\tilde Q_6}$ is the inner automorphism or its negative according to the sign of the norm, and it is the operator of the family of Two-Sided Operators on the General Quaternionic Algebra, whose old title carried that name before the title was changed to name the subcategory's algebra.

Which Switch Does What

The sign character $\alpha$ is the switch of the side: the adjoint is a left multiplication exactly for the two forms whose parameter carries the sign, and a right multiplication for the two whose parameter does not. The coefficient involution $\sigma$ is the switch of the pair of forms: it is present exactly in the two rows of the two sesquilinear forms, and it is the bar on the parameter there. Both readings are the two switches of the table above, read on the one-sided operators instead of on the sandwiches.

Self-Adjointness under the Four Adjoints

The four adjoints separate on the two families of §Signed, Unsigned and Associated. The table gives, for the two-sided operator $\Theta_{\tilde Q_6}=L_{\tilde Q_6}R_{\tilde Q_6^{\natural}}$ and for the inner automorphism $\mathrm{Ad}_{\tilde Q_6}=L_{\tilde Q_6}R_{\tilde Q_6^{-1}}$, whether the operator equals its own adjoint, for the four forms and four positions of the parameter.

Corollary (the criteria). The operator $\Theta_{\tilde Q_6}$ is self-adjoint for the plain form and for the $\natural$-form exactly when $\tilde Q_6^{\natural}$ is a multiple of $\tilde Q_6$, that is exactly when $\tilde Q_6$ is central or lies in the vector subspace; this is the criterion the corpus states for the $\langle\cdot,\cdot\rangle_{\natural}$-adjoint. It is self-adjoint for the $\ast$-form exactly when $\tilde Q_6^{\ast}/\tilde Q_6=\tilde Q_6^{\natural}/(\tilde Q_6^{\natural})^{\ast}$, and for the $\natural\ast$-form exactly when $\tilde Q_6^{\ast}/\tilde Q_6=\tilde Q_6^{\natural}/\bar{\tilde Q_6}$. The inner automorphism $\mathrm{Ad}_{\tilde Q_6}$ is self-adjoint for the plain form exactly when $\tilde Q_6^{-1}$ is a multiple of $\tilde Q_6$, for the $\natural$-form when $\tilde Q_6^{\natural}$ is a multiple of $\tilde Q_6$, for the $\ast$-form when $\tilde Q_6^{\ast}$ is a multiple of $\tilde Q_6$, and for the $\natural\ast$-form when $\overline{\tilde Q_6^{-1}}$ is a multiple of $\tilde Q_6$.

Proof. Self-adjointness is the equality of two decomposable tensors, and $L_{\tilde Q_4}R_{\tilde Q_5}=L_{\tilde Q_4'}R_{\tilde Q_5'}$ if and only if $\tilde Q_4'=\lambda\tilde Q_4$ and $\tilde Q_5'=\lambda^{-1}\tilde Q_5$ for one scalar $\lambda\in\mathbb{C}^{\times}$, the map $\mathbb{B}\otimes_{\mathbb{C}}\mathbb{B}^{\mathrm{opp}}\to\mathrm{End}_{\mathbb{C}}(\mathbb{B})$ being faithful because $\mathbb{B}$ is central simple. Substituting the four adjoints of the theorem gives two equations on $\lambda$ for each pair; the two are the same equation in the bilinear cases and in the three cases of the second family, which leaves a single parallelism, and they are the two ratio equations in the sesquilinear cases of $\Theta_{\tilde Q_6}$, where the two ratios must agree. The table above is the evaluation on the four parameters.

Remark (the table is the naming rule in action). The table is also a second piece of evidence for the rule of the introduction. The plain pair of the second family and the signed pair of the first are the two bilinear forms, and their criteria are the two parallelisms with $\natural$; the two conjugate linear forms give the two ratio equations, which the centre fails on the left and passes on the right, and the mixed parameter, which carries both a scalar and a vector part, satisfies none of the four conditions for $\Theta_{\tilde Q_6}$. Nothing in the table uses a distance or a limit.

Why the Families Are Named As They Are

Each row is one form, the adjoint of the left multiplication that the form produces, the operator family built on it, and the adjoint name the corpus gives to that family. The first two of those names were the titles of the two bilinear operator families until 2026-09-27, when the titles were changed to name the algebra of the subcategory.

Remark (the two words, and what they select). Association is the name of the transpose of the plain form: it reverses the factors and conjugates nothing, and it is the name of the adjoint whenever the form is the plain bilinear one, because there is no conjugation to put in the name. Signed is the name of the insertion of the grade involution $\natural$. For the general quaternionic bilinear form the adjoint is the signed left multiplication, $\Lambda^{\alpha}_{\tilde Q_6}(\tilde Q)=\tilde Q_6^{\natural}\tilde Q$, so the sign enters the name of the adjoint directly; the sandwich $\Theta_{\tilde Q_6}=L_{\tilde Q_6}R_{\tilde Q_6^{\natural}}$ is then the inner conjugation up to the norm, which is the "inner conjugation" half of the name. For the general plain sesquilinear form the adjoint is the dagger and the sandwich needs no sign; the corpus writes that family as Hermitian Adjoint, and it writes the family of the same form with the sign inserted into the operator as Signed Hermitian Adjoint, so the Hermitian layer carries two rows where the plain layer carries one. Read together, the three names of the biquaternion layer — Association, Signed Inner Conjugation, Hermitian Adjoint — are the three possible answers to "which conjugation does the adjoint of the form carry?": none, the sign $\natural$, or the conjugate linear $\ast$.

Summary

The biquaternion algebra has four pairings, and each pairing has its own adjoint of a linear map. For the general plain bilinear form the adjoint is the exchange of the two factors with no conjugation, $\approx$, which the corpus calls association, the name of the adjoint that the plain family carries. The associate has a calculus of its own: it is an involution, it reverses composition, $(FG)^{\approx}=G^{\approx}F^{\approx}$, it reverses the factors of the Conway basis, $(e_n[\,]e_m)^{\approx}=e_m[\,]e_n$, its matrix in the coefficient basis is $M^{\approx}=\mathrm{E}M^{\mathsf T}\mathrm{E}$, and combined with the coefficient conjugation it relates to the Hermitian adjoint by $\bar F^{\approx}=\mathrm{E}F^{\ast}\mathrm{E}$, which collapses to $F^{\ast}=F^{-1}$ on the vector-preserving unitaries. For the general quaternionic bilinear form the adjoint of the left multiplication is the left multiplication by $\tilde Q_6^{\natural}$, the signed left multiplication, and the sandwich built on it is the inner conjugation up to the norm, which the corpus calls the signed inner conjugation, the name of the adjoint that the quaternionic family carries. For the general plain sesquilinear form the adjoint is the Hermitian adjoint $\ast$; and signed Hermitian adjoint is the same form with the sign inserted into the operator. For the general quaternionic sesquilinear form the adjoint is the right multiplication by $\bar{\tilde Q_6}$, the Krein adjoint of the indefinite Hermitian sandwich. The word signed has one meaning throughout: the insertion of the grade involution $\natural$, which is the identity on the centre and minus the identity on the vector subspace. The examples give the four adjoint matrices of one two-sided operator, the two matrices of the signed and the unsigned inner conjugation of a mixed element and of the two extreme parameters, and the four self-adjointness criteria on the same parameters. No step of the article uses a topology, a distance or a limit.

The two switches of the twisted right factor, and the four forms that they return at the unit, are Introduction to the Operators of the Biquaternion Algebra, §The Two Switches and the Twisted Right Factor; the table of §Which Switch Does What gives the adjoint-side reading.

Summary of Notation

Further Reading

• Introduction to the Operators of the Biquaternion Algebra, for the products-to-operators bridge that the adjoints above serve, for the theorem that the plain product gives minus the reflection and the sign gives the reflection, and for the two switches $\sigma$ and $\alpha$ that the last table of this article names
• The Four Forms and the Four Algebraic Norms over the Biquaternion Algebra, for the four forms the four adjoints belong to, their Gram matrices, their signatures and their notation table of the four adjoints
• The Group of Involutions, for the four conjugations, their group and their linearity
• The Enveloping Operator Algebra of the Biquaternion Algebra and the Bi-module Structure, for the Conway operator basis on which the associate reverses the factors, and the regular representations $L_a$ and $R_b$ that transpose into one another
• Bilinear Forms, §Isometries, and Quadratic Forms and Polarisation, §Isometry of Forms, for the general transpose $M\mapsto G^{-1}M^{\mathsf T}G$ of a bilinear form, of which the associate is the biquaternion instance
• Two-Sided Operators on the Biquaternion Algebra, for the family of the first adjoint, its swap law and its criteria
• Two-Sided Operators on the General Quaternionic Algebra, for the twisted operator $\Theta_{\tilde Q_6}$, its $\natural$-adjoint and its self-adjointness criterion
• The Clifford, Pin and Spin Groups with Signed Inner Conjugation, for $\mathrm{Ad}^{\alpha}_{U}(v)=\alpha(U)vU^{-1}$ and the general meaning of the sign
• Introduction to the 2×2 Matrix Element Representation $M_2(\mathbb{C})$ of Biquaternions, for the same operators read as matrices, where the swap and the sign become a transposition and a bar
• Werner Greub, Linear Algebra, 4th edition (Springer, 1981), for the adjoint of a linear map with respect to a bilinear or a sesquilinear form