The Forms of the Algebra of the Reduced Biquaternions

Introduction

The general plain algebra of the reduced biquaternions carries two distinguished pairings, a bilinear one and a sesquilinear one, and this article reads them as objects. The pairing is the scalar part of a product with the four slot conventions of The Four Products, Their Forms and Their Algebraic Norms: the first slot may be read plain or through the tessarine sign ${}_{\mathrm{ts}}$, and the second slot plain or through the star ${}^{*}$. The four combinations give the four forms $$ \langle \breve Q_{1},\breve Q_{2}\rangle=Q_{1\blacktriangle}Q_{2\blacktriangle}+Q_{1\blacktriangledown}Q_{2\blacktriangledown},\qquad \langle \breve Q_{1},\breve Q_{2}\rangle_{\mathrm{ts}}=Q_{1\blacktriangle}Q_{2\blacktriangle}-Q_{1\blacktriangledown}Q_{2\blacktriangledown},\qquad \langle \breve Q_{1},\breve Q_{2}\rangle_{*}=Q_{1\blacktriangle}\bar Q_{2\blacktriangle}-Q_{1\blacktriangledown}\bar Q_{2\blacktriangledown},\qquad \langle \breve Q_{1},\breve Q_{2}\rangle_{\mathrm{ts}*}=Q_{1\blacktriangle}\bar Q_{2\blacktriangle}+Q_{1\blacktriangledown}\bar Q_{2\blacktriangledown}, $$ for $\breve Q_{1}=Q_{1\blacktriangle}+jQ_{1\blacktriangledown}$ and $\breve Q_{2}=Q_{2\blacktriangle}+jQ_{2\blacktriangledown}$. The first two are $\mathbb{C}$-bilinear, since the tessarine sign is $\mathbb{C}$-linear; the last two are $\mathbb{C}$-sesquilinear, since the star is $\mathbb{C}$-antilinear. Each form is the scalar part of the corresponding general product, $\langle\cdot,\cdot\rangle=\mathrm{Sc}(\breve Q_{1}\breve Q_{2})$, $\langle\cdot,\cdot\rangle_{\mathrm{ts}}=\mathrm{Sc}(\breve Q_{1,\mathrm{ts}}\breve Q_{2})$, $\langle\cdot,\cdot\rangle_{*}=\mathrm{Sc}(\breve Q_{1}\breve Q_{2}^{*})$ and $\langle\cdot,\cdot\rangle_{\mathrm{ts}*}=\mathrm{Sc}(\breve Q_{1,\mathrm{ts}}\breve Q_{2}^{*})$, so the four forms are the four products read at their scalar part.

The results. The four forms are non-degenerate and independent, with the Gram matrices $$ B=\mathrm I_{2},\qquad N=\operatorname{diag}(1,-1),\qquad H=\operatorname{diag}(1,-1),\qquad K=\mathrm I_{2} $$ in the complex basis $1,j$, and the signatures $(2,2)$, $(2,2)$, $(2,2)$ and $(4,0)$ over $\mathbb{R}$. The two sesquilinear forms are Hermitian, $\langle \breve Q_{2},\breve Q_{1}\rangle_{*}=\overline{\langle \breve Q_{1},\breve Q_{2}\rangle_{*}}$ and $\langle \breve Q_{2},\breve Q_{1}\rangle_{\mathrm{ts}*}=\overline{\langle \breve Q_{1},\breve Q_{2}\rangle_{\mathrm{ts}*}}$, and their diagonals are the algebraic norms $\langle \breve Q\rangle_{*}=|Q_{\blacktriangle}|^{2}-|Q_{\blacktriangledown}|^{2}$ and $\langle \breve Q\rangle_{\mathrm{ts}*}=|Q_{\blacktriangle}|^{2}+|Q_{\blacktriangledown}|^{2}$; the two bilinear forms are symmetric with complex values. The general plain sesquilinear form $\langle\cdot,\cdot\rangle_{*}$ is the one the layer of sesquialgebras selects: it is compatible with the product, $\langle \breve Q_{1}\breve Q_{2},\breve Q_{3}\rangle_{*}=\langle \breve Q_{2},\breve Q_{1}^{*}\breve Q_{3}\rangle_{*}$, equivalently $(L_{\breve Q})^{H}=L_{\breve Q^{*}}$. The two sesquilinear forms are tied by the bridge identity $\langle \breve Q_{1},\breve Q_{2}\rangle_{\mathrm{ts}*}=\langle \breve Q_{1,\mathrm{ts}},\breve Q_{2}\rangle_{*}=\langle \breve Q_{1},\breve Q_{2,\mathrm{ts}}\rangle_{*}$, with the tessarine sign as the self-adjoint involution, fixing the scalar line and negating the $j$-line; this is why $\langle\cdot,\cdot\rangle_{\mathrm{ts}*}$ is definite and $\langle\cdot,\cdot\rangle_{*}$ indefinite, and why their signatures are $(4,0)$ and $(2,2)$. At the real base the four forms collapse to two, $\langle\cdot,\cdot\rangle_{\mathrm{ts}*}=\langle\cdot,\cdot\rangle$ and $\langle\cdot,\cdot\rangle_{*}=\langle\cdot,\cdot\rangle_{\mathrm{ts}}$, and the sesquilinear pair is bilinear.

The Two Slots and the Two Layers

The Linearity of the Slot Maps

The tessarine sign is $\mathbb{C}$-linear and the star is $\mathbb{C}$-antilinear; a pairing is bilinear when its second slot is plain and sesquilinear when its second slot carries the star.

Proposition (the slot maps). For $\lambda\in\mathbb{C}$ and $\breve Q=Q_{\blacktriangle}+jQ_{\blacktriangledown}$, $$ (\lambda \breve Q)_{\mathrm{ts}}=\lambda\,\breve Q_{\mathrm{ts}},\qquad (\lambda \breve Q)^{*}=\bar\lambda\,\breve Q^{*}. $$

Proof. The tessarine sign negates the second coordinate, $\breve Q_{\mathrm{ts}}=Q_{\blacktriangle}-jQ_{\blacktriangledown}$, and does not touch the coefficients, so it commutes with the complex scalars; the star is $\bar{\cdot}\circ{}_{\mathrm{ts}}$ (The Two Involutions and the Group of Involutions of the Reduced Biquaternions), and the composite of a $\mathbb{C}$-linear and a $\mathbb{C}$-antilinear map is antilinear, so $(\lambda \breve Q)^{*}=\overline{\lambda}\,\overline{\breve Q}_{\mathrm{ts}}=\bar\lambda\,\breve Q^{*}$.

Proposition (which pairings are sesquilinear). Of the four pairings $\mathrm{Sc}(X\,\breve Q_{1}\,Y\,\breve Q_{2})$, the two with $\mathrm{id}$ in the second slot, $\langle\cdot,\cdot\rangle$ and $\langle\cdot,\cdot\rangle_{\mathrm{ts}}$, are $\mathbb{C}$-bilinear, and the two with the star there, $\langle\cdot,\cdot\rangle_{*}$ and $\langle\cdot,\cdot\rangle_{\mathrm{ts}*}$, are $\mathbb{C}$-sesquilinear, linear in the first slot and antilinear in the second.

Proof. The scalar part is $\mathbb{C}$-linear, and the slot maps are the ones of the preceding proposition; a $\mathbb{C}$-antilinear slot map makes the pairing antilinear in that slot, a linear one keeps it linear.

Remark (the two layers). The bilinear pairings belong to the bilinear layer of Bilinear Forms; the sesquilinear pairings belong to the layer of Sesqualgebras with a Form, and the layer's involution is the star, as the next section shows. The dichotomy is not a property of the particular forms but of the two slot maps the algebra carries, exactly as in the biquaternion category.

The Admissible Involution

The star is the involution of the layer's form: it is $\mathbb{C}$-antilinear of square one and compatible with the plain sesquilinear form; the tessarine sign is $\mathbb{C}$-linear, so it pairs with the bilinear forms; and the bar is $\mathbb{C}$-antilinear too but pairs with the tessarine sesquilinear form, because in the commutative algebra the reversal of the factors is invisible and does not separate the bar from the star.

Theorem. The star satisfies $(\breve Q_{1}\breve Q_{2})^{*}=\breve Q_{1}^{*}\breve Q_{2}^{*}$ and is $\mathbb{C}$-antilinear of square one; the bar satisfies $(\breve Q_{1}\breve Q_{2})^{\bar{}}=\breve Q_{1}^{\bar{}}\breve Q_{2}^{\bar{}}$ and is $\mathbb{C}$-antilinear of square one as well; the tessarine sign satisfies $(\breve Q_{1}\breve Q_{2})_{\mathrm{ts}}=\breve Q_{1,\mathrm{ts}}\breve Q_{2,\mathrm{ts}}$ and is $\mathbb{C}$-linear. The tessarine sign is therefore excluded from the sesquilinear layer, but the bar and the star are not separated by their product type, since in a commutative algebra a product-compatible map is at once an automorphism and an anti-automorphism. What separates them is the form they serve, and the compatibility identity of §The Compatibility of the Plain Sesquilinear Form singles out the star for the plain sesquilinear form $\langle\cdot,\cdot\rangle_{*}$; the bar serves the tessarine sesquilinear form $\langle\cdot,\cdot\rangle_{\mathrm{ts}*}$.

Proof. All three maps are computed coordinatewise on $\breve Q_{1}=Q_{1\blacktriangle}+jQ_{1\blacktriangledown}$ and $\breve Q_{2}=Q_{2\blacktriangle}+jQ_{2\blacktriangledown}$. The star conjugates the coefficients and negates the second slot, and the two operations are compatible with the product because the product is $\mathbb{C}$-bilinear in the coordinates; the same holds for the bar and the tessarine sign. The star is antilinear by the proposition above, and ${}^{*}\circ{}^{*}=\mathrm{id}$ because ${}_{\mathrm{ts}}\circ{}_{\mathrm{ts}}=\mathrm{id}$ and $\bar{\cdot}\circ\bar{\cdot}=\mathrm{id}$. Of the three, the tessarine sign is linear, so it is excluded from the sesquilinear layer; the bar and the star are both antilinear and of square one, and in the commutative algebra both are at once automorphisms and anti-automorphisms, so the product type does not separate them. The compatibility identity of §The Compatibility of the Plain Sesquilinear Form is what selects the star for the plain sesquilinear form $\langle\cdot,\cdot\rangle_{*}$, the bar serving the tessarine sesquilinear form $\langle\cdot,\cdot\rangle_{\mathrm{ts}*}$.

The Four Forms

The four forms are the scalar parts of the four general products: $\langle\cdot,\cdot\rangle=\mathrm{Sc}(\breve Q_{1}\breve Q_{2})$, $\langle\cdot,\cdot\rangle_{\mathrm{ts}}=\mathrm{Sc}(\breve Q_{1,\mathrm{ts}}\breve Q_{2})$, $\langle\cdot,\cdot\rangle_{*}=\mathrm{Sc}(\breve Q_{1}\breve Q_{2}^{*})$, $\langle\cdot,\cdot\rangle_{\mathrm{ts}*}=\mathrm{Sc}(\breve Q_{1,\mathrm{ts}}\breve Q_{2}^{*})$.

Proposition (the four forms on the coordinates). For $\breve Q_{1}=Q_{1\blacktriangle}+jQ_{1\blacktriangledown}$ and $\breve Q_{2}=Q_{2\blacktriangle}+jQ_{2\blacktriangledown}$, $$ \boxed{\; \langle \breve Q_{1},\breve Q_{2}\rangle=Q_{1\blacktriangle}Q_{2\blacktriangle}+Q_{1\blacktriangledown}Q_{2\blacktriangledown},\qquad \langle \breve Q_{1},\breve Q_{2}\rangle_{\mathrm{ts}}=Q_{1\blacktriangle}Q_{2\blacktriangle}-Q_{1\blacktriangledown}Q_{2\blacktriangledown} \;}, $$ $$ \langle \breve Q_{1},\breve Q_{2}\rangle_{*}=Q_{1\blacktriangle}\bar Q_{2\blacktriangle}-Q_{1\blacktriangledown}\bar Q_{2\blacktriangledown},\qquad \langle \breve Q_{1},\breve Q_{2}\rangle_{\mathrm{ts}*}=Q_{1\blacktriangle}\bar Q_{2\blacktriangle}+Q_{1\blacktriangledown}\bar Q_{2\blacktriangledown} . $$

Proof. Compute each product and take its scalar part. For $\langle\cdot,\cdot\rangle$, $\breve Q_{1}\breve Q_{2}=(Q_{1\blacktriangle}Q_{2\blacktriangle}+Q_{1\blacktriangledown}Q_{2\blacktriangledown})+(Q_{1\blacktriangle}Q_{2\blacktriangledown}+Q_{1\blacktriangledown}Q_{2\blacktriangle})j$, so $\mathrm{Sc}(\breve Q_{1}\breve Q_{2})=Q_{1\blacktriangle}Q_{2\blacktriangle}+Q_{1\blacktriangledown}Q_{2\blacktriangledown}$. For $\langle\cdot,\cdot\rangle_{\mathrm{ts}}$, $\breve Q_{1,\mathrm{ts}}\breve Q_{2}=(Q_{1\blacktriangle}-jQ_{1\blacktriangledown})(Q_{2\blacktriangle}+jQ_{2\blacktriangledown})=(Q_{1\blacktriangle}Q_{2\blacktriangle}-Q_{1\blacktriangledown}Q_{2\blacktriangledown})+(Q_{1\blacktriangle}Q_{2\blacktriangledown}-Q_{1\blacktriangledown}Q_{2\blacktriangle})j$. For $\langle\cdot,\cdot\rangle_{*}$, $\breve Q_{1}\breve Q_{2}^{*}=(Q_{1\blacktriangle}+jQ_{1\blacktriangledown})(\bar Q_{2\blacktriangle}-j\bar Q_{2\blacktriangledown})=(Q_{1\blacktriangle}\bar Q_{2\blacktriangle}-Q_{1\blacktriangledown}\bar Q_{2\blacktriangledown})+(Q_{1\blacktriangledown}\bar Q_{2\blacktriangle}-Q_{1\blacktriangle}\bar Q_{2\blacktriangledown})j$, using $j^{2}=+1$. For $\langle\cdot,\cdot\rangle_{\mathrm{ts}*}$, $\breve Q_{1,\mathrm{ts}}\breve Q_{2}^{*}=(Q_{1\blacktriangle}-jQ_{1\blacktriangledown})(\bar Q_{2\blacktriangle}-j\bar Q_{2\blacktriangledown})=(Q_{1\blacktriangle}\bar Q_{2\blacktriangle}+Q_{1\blacktriangledown}\bar Q_{2\blacktriangledown})-(Q_{1\blacktriangle}\bar Q_{2\blacktriangledown}+Q_{1\blacktriangledown}\bar Q_{2\blacktriangle})j$.

The two bilinear forms. $\langle\cdot,\cdot\rangle$ and $\langle\cdot,\cdot\rangle_{\mathrm{ts}}$ are symmetric and $\mathbb{C}$-valued, with no Hermitian condition; their diagonals are $\langle \breve Q\rangle=Q_{\blacktriangle}^{2}+Q_{\blacktriangledown}^{2}$ and $\langle \breve Q\rangle_{\mathrm{ts}}=Q_{\blacktriangle}^{2}-Q_{\blacktriangledown}^{2}$, the algebraic norms of the general plain and the general tessarine bilinear products. $\langle\cdot,\cdot\rangle$ is the general plain bilinear form, the form of the multiplication; $\langle\cdot,\cdot\rangle_{\mathrm{ts}}$ is the general tessarine bilinear form, and $\langle\cdot,\cdot\rangle_{\mathrm{ts}}$ alone is multiplicative, which is the mechanism of The Group of Units of the Algebra of the Reduced Biquaternions.

The two sesquilinear forms. $\langle\cdot,\cdot\rangle_{*}$ and $\langle\cdot,\cdot\rangle_{\mathrm{ts}*}$ are Hermitian and real on the diagonal, with $\langle \breve Q\rangle_{*}=|Q_{\blacktriangle}|^{2}-|Q_{\blacktriangledown}|^{2}$ and $\langle \breve Q\rangle_{\mathrm{ts}*}=|Q_{\blacktriangle}|^{2}+|Q_{\blacktriangledown}|^{2}$. $\langle\cdot,\cdot\rangle_{*}$ is the general plain sesquilinear form and $\langle\cdot,\cdot\rangle_{\mathrm{ts}*}$ the general tessarine sesquilinear form; $\langle\cdot,\cdot\rangle_{\mathrm{ts}*}$ is positive definite, with square root the Euclidean norm, and $\langle \breve Q\rangle_{*}$ is indefinite of signature $(2,2)$.

The Gram Matrices and the Signatures

In the complex basis $1,j$ the Gram matrices are $B=\mathrm I_{2}$, $N=H=\operatorname{diag}(1,-1)$ and $K=\mathrm I_{2}$; over the real basis the signatures are $(2,2)$, $(2,2)$, $(2,2)$ and $(4,0)$.

Proposition (the Gram matrices). In the complex basis $\{1,j\}$, $$ G_{B}=\begin{pmatrix}1&0\\0&1\end{pmatrix},\quad G_{N}=\begin{pmatrix}1&0\\0&-1\end{pmatrix},\quad G_{H}=\begin{pmatrix}1&0\\0&-1\end{pmatrix},\quad G_{K}=\begin{pmatrix}1&0\\0&1\end{pmatrix}, $$ so all four forms are non-degenerate, each with radical $\{0\}$.

Proof. Evaluate each form on the pairs of basis elements: $\langle 1,1\rangle=\langle 1,1\rangle_{\mathrm{ts}}=\langle 1,1\rangle_{*}=\langle 1,1\rangle_{\mathrm{ts}*}=1$; $\langle j,j\rangle=1$, $\langle j,j\rangle_{\mathrm{ts}}=\langle j,j\rangle_{*}=\langle j,j\rangle_{\mathrm{ts}*}$ is $\pm1$ according to the sign in front of $Q_{1\blacktriangledown}Q_{2\blacktriangledown}$; and the cross terms $\langle 1,j\rangle=\langle 1,j\rangle_{\mathrm{ts}}=\langle 1,j\rangle_{*}=\langle 1,j\rangle_{\mathrm{ts}*}=0$ because $\langle\cdot,\cdot\rangle=0$ or $Q_{2\blacktriangle}=0$ in each. Non-degeneracy is the non-vanishing of each determinant, $1$, $-1$, $-1$ and $1$.

Corollary (the signatures over $\mathbb{R}$). In the real basis $\{1,i,j,ij\}$ the real parts have the diagonal Gram matrices $$ \operatorname{diag}(1,-1,1,-1),\qquad \operatorname{diag}(1,-1,-1,1),\qquad \operatorname{diag}(1,1,-1,-1),\qquad \mathrm I_{4}, $$ so the signatures are $(2,2)$ for $\langle\cdot,\cdot\rangle$, $\langle\cdot,\cdot\rangle_{\mathrm{ts}}$ and $\langle\cdot,\cdot\rangle_{*}$ and $(4,0)$ for $\langle\cdot,\cdot\rangle_{\mathrm{ts}*}$. Over $\mathbb{C}$ the inertia indices are $(1,1,0)$ for $\langle\cdot,\cdot\rangle_{\mathrm{ts}}$ and $\langle\cdot,\cdot\rangle_{*}$ and $(2,0,0)$ for $\langle\cdot,\cdot\rangle$ and $\langle\cdot,\cdot\rangle_{\mathrm{ts}*}$.

Proof. Read the real coordinates $Q_{1\blacktriangle}=q_{1\blacktriangle}+iq'_{1\blacktriangle}$, $Q_{1\blacktriangledown}=q_{1\blacktriangledown}+iq'_{1\blacktriangledown}$ and $Q_{2\blacktriangle}=q_{2\blacktriangle}+iq'_{2\blacktriangle}$, $Q_{2\blacktriangledown}=q_{2\blacktriangledown}+iq'_{2\blacktriangledown}$ on each form; the real part of $\langle\cdot,\cdot\rangle$ is $q_{1\blacktriangle}q_{2\blacktriangle}-q'_{1\blacktriangle}q'_{2\blacktriangle}+q_{1\blacktriangledown}q_{2\blacktriangledown}-q'_{1\blacktriangledown}q'_{2\blacktriangledown}$, whose Gram matrix is the displayed one, and the other three are computed the same way. The signs follow from $\langle \breve Q\rangle_{\mathrm{ts}}=Q_{\blacktriangle}^{2}-Q_{\blacktriangledown}^{2}$, $\langle \breve Q\rangle_{*}=|Q_{\blacktriangle}|^{2}-|Q_{\blacktriangledown}|^{2}$ and $\langle \breve Q\rangle_{\mathrm{ts}*}=|Q_{\blacktriangle}|^{2}+|Q_{\blacktriangledown}|^{2}$ on the real basis.

Remark (definite and indefinite). $\langle\cdot,\cdot\rangle_{\mathrm{ts}*}$ is the only definite form of the four, positive definite; $\langle\cdot,\cdot\rangle_{*}$ is indefinite of signature $(2,2)$; and the two bilinear forms are complex-valued, so their "signature" is read on their real parts only. This is the reduced biquaternion instance of the table of The Four Products, Their Forms and Their Algebraic Norms, and the four norms are the diagonals of §The Four Forms.

The Bridge Identity and the Fundamental Symmetry

The two sesquilinear forms are related by the tessarine sign, which is an $\langle\cdot,\cdot\rangle_{*}$-self-adjoint involution; this is why $\langle\cdot,\cdot\rangle_{\mathrm{ts}*}$ is definite and $\langle\cdot,\cdot\rangle_{*}$ is indefinite.

Theorem (the bridge identity). The tessarine sign is an $\langle\cdot,\cdot\rangle_{*}$-self-adjoint involution, $$ \boxed{\; \langle \breve Q_{1,\mathrm{ts}},\breve Q_{2}\rangle_{*}=\langle \breve Q_{1},\breve Q_{2,\mathrm{ts}}\rangle_{*}=\langle \breve Q_{1},\breve Q_{2}\rangle_{\mathrm{ts}*} \;}, $$ and it fixes the scalar line $\mathbb{C}$ and negates the $j$-line $\mathbb{C}j$; the two lines are $\langle\cdot,\cdot\rangle_{*}$-orthogonal, and $\langle\cdot,\cdot\rangle_{\mathrm{ts}*}$ restricted to them is $+H$ and $-H$ respectively.

Proof. $\langle \breve Q_{1,\mathrm{ts}},\breve Q_{2}\rangle_{*}=\langle Q_{1\blacktriangle}-jQ_{1\blacktriangledown},Q_{2\blacktriangle}+jQ_{2\blacktriangledown}\rangle_{*}=Q_{1\blacktriangle}\bar Q_{2\blacktriangle}-(-Q_{1\blacktriangledown})\bar Q_{2\blacktriangledown}=Q_{1\blacktriangle}\bar Q_{2\blacktriangle}+Q_{1\blacktriangledown}\bar Q_{2\blacktriangledown}=\langle \breve Q_{1},\breve Q_{2}\rangle_{\mathrm{ts}*}$; the mirror computation gives $\langle \breve Q_{1},\breve Q_{2,\mathrm{ts}}\rangle_{*}=\langle \breve Q_{1},\breve Q_{2}\rangle_{\mathrm{ts}*}$; the two together are the self-adjointness. The tessarine sign fixes $Q_{\blacktriangle}$ and negates $Q_{\blacktriangledown}$, hence fixes the scalar line and negates the $j$-line. The orthogonality is $\langle Q_{1\blacktriangle},Q_{2\blacktriangledown}j\rangle_{*}=Q_{1\blacktriangle}\cdot\bar 0-0\cdot\bar Q_{2\blacktriangledown}=0$ for scalar $Q_{1\blacktriangle}$ and for $Q_{2\blacktriangledown}j$ in the $j$-line. The signs of $\langle\cdot,\cdot\rangle_{\mathrm{ts}*}$ on the two lines follow from $\langle\cdot,\cdot\rangle_{\mathrm{ts}*}=H\circ(\mathrm{ts},\mathrm{id})$ and the eigenvalues $\pm1$ of the tessarine sign.

Corollary (the definiteness). $\langle \breve Q\rangle_{\mathrm{ts}*}=\langle \breve Q_{\mathrm{ts}},\breve Q\rangle_{*}$, so $\langle\cdot,\cdot\rangle_{\mathrm{ts}*}$ is $\langle\cdot,\cdot\rangle_{*}$ read through the involution: on the scalar line $\langle\cdot,\cdot\rangle_{\mathrm{ts}*}=+H>0$ and on the $j$-line $\langle\cdot,\cdot\rangle_{\mathrm{ts}*}=-H>0$, and the two positive contributions give the definiteness of $\langle\cdot,\cdot\rangle_{\mathrm{ts}*}$ and the signature $(2,2)$ of $\langle\cdot,\cdot\rangle_{*}$. The involution is the fundamental symmetry of the pair $(H,K)$.

The Compatibility of the Plain Sesquilinear Form

The general plain sesquilinear form is compatible with the multiplication, and the general tessarine one is not.

Theorem (the compatibility). The form $\langle\cdot,\cdot\rangle_{*}$ satisfies $$ \langle \breve Q_{1}\breve Q_{2},\breve Q_{3}\rangle_{*}=\langle \breve Q_{2},\breve Q_{1}^{*}\breve Q_{3}\rangle_{*}\qquad\text{for all }\breve Q_{1},\breve Q_{2},\breve Q_{3}, $$ equivalently $(L_{\breve Q})^{H}=L_{\breve Q^{*}}$, so the left multiplication by $\breve Q$ is the $\langle\cdot,\cdot\rangle_{*}$-adjoint of the left multiplication by $\breve Q^{*}$.

Proof. Expand both sides. On the one hand $\langle \breve Q_{1}\breve Q_{2},\breve Q_{3}\rangle_{*}=\mathrm{Sc}\bigl((\breve Q_{1}\breve Q_{2})\breve Q_{3}^{*}\bigr)$. On the other $\langle \breve Q_{2},\breve Q_{1}^{*}\breve Q_{3}\rangle_{*}=\mathrm{Sc}\bigl(\breve Q_{2}(\breve Q_{1}^{*}\breve Q_{3})^{*}\bigr)=\mathrm{Sc}(\breve Q_{2} \breve Q_{3}^{*}\breve Q_{1})=\mathrm{Sc}(\breve Q_{1}\breve Q_{2}\breve Q_{3}^{*})$, using $(\breve Q_{1}^{*}\breve Q_{3})^{*}=\breve Q_{3}^{*}\breve Q_{1}$ and the cyclicity $\mathrm{Sc}(XYZ)=\mathrm{Sc}(ZXY)$ of the scalar part, which holds because the scalar part is the trace of the regular representation and the algebra is commutative. The two expressions agree.

Remark (the layer's involution is the star). The compatibility identity $\langle \breve Q_{1}\breve Q_{2},\breve Q_{3}\rangle_{*}=\langle \breve Q_{2},\breve Q_{1}^{*}\breve Q_{3}\rangle_{*}$ is the axiom that makes the left multiplications of the algebra the layer's adjointable operators, and it uses the star and no other conjugation: the bar and the tessarine sign produce the two bilinear pairings and the Hermitian form $\langle\cdot,\cdot\rangle_{\mathrm{ts}*}$, whose adjoint rule is read in The Multiplication Operators of the Algebra of the Reduced Biquaternions. This is the sense in which $(\mathbb{B}_{\mathrm{r}},{}^{*},H)$ is an object of the layer of Sesqualgebras with a Form, exactly as $(\mathbb{B},{}^{*},h)$ is in the biquaternion category.

The Unitary Elements

The elements preserving $\langle\cdot,\cdot\rangle_{*}$ are the solutions of $U^{*}U=1$, they form a group isomorphic to $\mathbb{C}^{\times}$, and they are not the diagonal unitary matrices.

Theorem (the unitary elements). The set $$ \boxed{\; \mathcal U=\{U\in\mathbb{B}_{\mathrm{r}}:U^{*}U=1\}=\{U:U=Q_{+}\Pi_++|Q_{+}|^{-2}Q_{+}\,\Pi_-,\ Q_{+}\in\mathbb{C}^{\times}\} \;} $$ is a group under the multiplication, isomorphic to $\mathbb{C}^{\times}$ through the first idempotent coordinate; it is the group of elements whose left multiplication is an $\langle\cdot,\cdot\rangle_{*}$-isometry, $\langle U\breve Q_{1},U\breve Q_{2}\rangle_{*}=\langle \breve Q_{1},\breve Q_{2}\rangle_{*}$ for all $\breve Q_{1},\breve Q_{2}$.

Proof. In the idempotent coordinates $U=Q_{+}\Pi_++Q_{-}\Pi_-$ one has $U^{*}=\bar Q_{-}\,\Pi_++\bar Q_{+}\,\Pi_-$, since the star swaps the two idempotents and conjugates the scalars. Hence $U^{*}U=\bar Q_{-} Q_{+}\,\Pi_++\bar Q_{+} Q_{-}\,\Pi_-$, and $U^{*}U=1=\Pi_++\Pi_-$ is the single equation $\bar Q_{-} Q_{+}=1$, equivalently $Q_{-}=Q_{+}/|Q_{+}|^{2}$. The parametrisation is a group homomorphism from $\mathbb{C}^{\times}$ because $Q_{+}\mapsto Q_{+}\Pi_++|Q_{+}|^{-2}Q_{+}\Pi_-$ sends $Q_{+} Q_{+}'$ to $Q_{+}Q_{+}'\Pi_++|Q_{+}Q_{+}'|^{-2}Q_{+}Q_{+}'\Pi_-$, which is the product of the images. The isometry property is the compatibility of §The Compatibility of the Plain Sesquilinear Form read twice.

Remark (the slice is not compact). The unitary elements form the group $\mathbb{C}^{\times}$, of topology that of a circle times a line, so the slice is not compact; this is a genuine specificity of the reduced biquaternion against the biquaternion category, where the slice $U^{*}U=e_t$ is the compact group $\mathrm{U}(2)$. The reason is the factoring of the algebra: the star swaps the two idempotents, so the unitarity couples the two coordinates by one equation and not by two, and the slice acquires the non-compact direction of $|Q_{+}|$.

The Collapse at the Real Base

At the real base the star equals the tessarine sign, the four forms collapse to two, and the sesquilinear pair becomes bilinear.

Theorem (the collapse). Let the coefficients be real, $Q_{\blacktriangle},Q_{\blacktriangledown}\in\mathbb{R}$. Then $\breve Q^{*}=\breve Q_{\mathrm{ts}}$ and $$ \langle \breve Q_{1},\breve Q_{2}\rangle_{\mathrm{ts}*}=\langle \breve Q_{1},\breve Q_{2}\rangle,\qquad \langle \breve Q_{1},\breve Q_{2}\rangle_{*}=\langle \breve Q_{1},\breve Q_{2}\rangle_{\mathrm{ts}}, $$ so the four forms collapse to the two real symmetric bilinear forms $\langle\cdot,\cdot\rangle$ and $\langle\cdot,\cdot\rangle_{\mathrm{ts}}$, of signatures $(2,0)$ and $(1,1)$ on the real basis, and the Hermitian condition becomes the plain symmetry.

Proof. At real coefficients the coefficient conjugation is the identity, so the star reduces to the tessarine sign, ${}^{*}={}_{\mathrm{ts}}$. Then $\langle\cdot,\cdot\rangle_{*}=\mathrm{Sc}(\breve Q_{1,\mathrm{ts}}\breve Q_{2})=N$ and $\langle\cdot,\cdot\rangle_{\mathrm{ts}*}=\mathrm{Sc}(\breve Q_{1,\mathrm{ts}}\breve Q_{2,\mathrm{ts}})=\mathrm{Sc}(\breve Q_{1}\breve Q_{2})=B$, since $\breve Q_{2,\mathrm{ts}}$ has the same scalar part as $\breve Q_{2}$ up to the sign of the second coordinate, which the product reads twice. The two remaining forms are symmetric because the coefficient conjugation is the identity.

Remark (which statements are genuinely sesquilinear). The compatibility, the bridge identity and the adjoint rule hold in both categories and are read at the real base as the self-adjointness of the left multiplications for the symmetric forms; the sesquilinear content of the article is the dichotomy of the two slot maps, the modulus in the Hermitian condition, the definiteness of $\langle\cdot,\cdot\rangle_{\mathrm{ts}*}$ against the indefiniteness of $\langle\cdot,\cdot\rangle_{*}$, and the non-compactness of the unitary slice — none of which survives the collapse to the real base.

Summary

The general plain algebra carries four distinguished forms, the scalar parts of the four general products: the general plain bilinear form $\langle \breve Q_{1},\breve Q_{2}\rangle=Q_{1\blacktriangle}Q_{2\blacktriangle}+Q_{1\blacktriangledown}Q_{2\blacktriangledown}$, the general tessarine bilinear form $\langle \breve Q_{1},\breve Q_{2}\rangle_{\mathrm{ts}}=Q_{1\blacktriangle}Q_{2\blacktriangle}-Q_{1\blacktriangledown}Q_{2\blacktriangledown}$, the general plain sesquilinear form $\langle \breve Q_{1},\breve Q_{2}\rangle_{*}=Q_{1\blacktriangle}\bar Q_{2\blacktriangle}-Q_{1\blacktriangledown}\bar Q_{2\blacktriangledown}$ and the general tessarine sesquilinear form $\langle \breve Q_{1},\breve Q_{2}\rangle_{\mathrm{ts}*}=Q_{1\blacktriangle}\bar Q_{2\blacktriangle}+Q_{1\blacktriangledown}\bar Q_{2\blacktriangledown}$. The tessarine sign is $\mathbb{C}$-linear, so $\langle\cdot,\cdot\rangle$ and $\langle\cdot,\cdot\rangle_{\mathrm{ts}}$ are bilinear; the star is $\mathbb{C}$-antilinear, so $\langle\cdot,\cdot\rangle_{*}$ and $\langle\cdot,\cdot\rangle_{\mathrm{ts}*}$ are sesquilinear. All four are non-degenerate, with Gram matrices $\mathrm I_{2}$, $\operatorname{diag}(1,-1)$, $\operatorname{diag}(1,-1)$ and $\mathrm I_{2}$ in the basis $1,j$, and signatures $(2,2)$, $(2,2)$, $(2,2)$ and $(4,0)$ over $\mathbb{R}$; $\langle\cdot,\cdot\rangle_{\mathrm{ts}*}$ is positive definite and its square root is the Euclidean norm, and $\langle \breve Q\rangle_{*}$ is indefinite. The two sesquilinear forms are Hermitian, with diagonals the algebraic norms $|Q_{\blacktriangle}|^{2}-|Q_{\blacktriangledown}|^{2}$ and $|Q_{\blacktriangle}|^{2}+|Q_{\blacktriangledown}|^{2}$, and they are tied by the bridge identity $\langle \breve Q_{1},\breve Q_{2}\rangle_{\mathrm{ts}*}=\langle \breve Q_{1,\mathrm{ts}},\breve Q_{2}\rangle_{*}=\langle \breve Q_{1},\breve Q_{2,\mathrm{ts}}\rangle_{*}$, the tessarine sign being an $\langle\cdot,\cdot\rangle_{*}$-self-adjoint involution fixing the scalar line and negating the $j$-line; this is the fundamental symmetry that makes $\langle\cdot,\cdot\rangle_{\mathrm{ts}*}$ definite and $\langle\cdot,\cdot\rangle_{*}$ indefinite. The general plain sesquilinear form is the one the layer selects: it is compatible with the product, $\langle \breve Q_{1}\breve Q_{2},\breve Q_{3}\rangle_{*}=\langle \breve Q_{2},\breve Q_{1}^{*}\breve Q_{3}\rangle_{*}$, equivalently $(L_{\breve Q})^{H}=L_{\breve Q^{*}}$, so $(\mathbb{B}_{\mathrm{r}},{}^{*},H)$ is an object of the layer of Sesqualgebras with a Form. Its unitary elements, the solutions of $U^{*}U=1$, form the non-compact group $\mathbb{C}^{\times}$ and not the diagonal unitary group. At the real base the four forms collapse to two, $\langle\cdot,\cdot\rangle_{\mathrm{ts}*}=\langle\cdot,\cdot\rangle$ and $\langle\cdot,\cdot\rangle_{*}=\langle\cdot,\cdot\rangle_{\mathrm{ts}}$.

Summary of Notation

Further Reading

• The Four Products, Their Forms and Their Algebraic Norms, for the four products, their forms and their norms together.
• Norm and Invertibility in the Algebra of the Reduced Biquaternions, for the norm of the multiplication.
• Zero Divisors of the Algebra of the Reduced Biquaternions, for the vanishing sets of the norms.
• The Multiplication Operators of the Algebra of the Reduced Biquaternions, for the adjoint rule and the bi-module structure.
• The Two Involutions and the Group of Involutions of the Reduced Biquaternions, for the tessarine sign, the star and the bar.
• The Form on the Biquaternion Algebra as a General Plain Sesquilinear Form, for the non-commutative companion of the layer's selection.