The Indefinite Hermitian Sandwich on the General Quaternionic Sesqualgebra

Introduction

The biquaternion algebra $\mathbb{B} = \mathbb{C}\otimes_{\mathbb{R}}\mathbb{H}$ carries two general plain sesquilinear forms of different type. The first is the definite form of the dagger, $(\tilde Q_1,\tilde W) = \mathrm{Sc}(\tilde Q_1^{*}\tilde W)$, positive definite of signature $(4,0)$, on which the whole definite operator theory is built (Two-Sided Operators on the General Plain Sesqualgebra). The second is the general quaternionic sesquilinear form $\langle\tilde W,\tilde Q_1\rangle_{\natural*} = \mathrm{Sc}(\tilde Q_1^{\natural*}\tilde W)$ of signature $(1,3)$, whose fundamental symmetry is the involution $J={}^{\natural}$ and whose automorphism group is $\mathrm{U}(1,3)$ (The Four Forms and the Four Algebraic Norms over the Biquaternion Algebra, The Fundamental Symmetry of the Form, The Krein Isometry Group and Its Cartan Decomposition). On a Krein space the sandwich by an operator $A$ — the operator's letter, the bare glyph $Q$ staying with the element of $\mathbb{B}$ — is formed with the indefinite adjoint $A^{\dagger} = JA^{*}J$ rather than with $A^{*}$,

$$ H_{A}(T) = A\,T\,A^{\dagger} , $$

and this operator is the subject of the article. It is the biquaternion instance of The Hermitian Sandwich on a Krein Space, and the indefinite counterpart of the dagger sandwich $\Theta_{\tilde Q_1}(\tilde Q_2) = \tilde Q_1\tilde Q_2\tilde Q_1^{*}$ of The Hermitian Sandwich in the Biquaternion Algebra with Hermitian Adjoint.

The two sandwiches must not be confused, and the algebra makes the difference visible in coordinates. With the definite adjoint the sandwich is always an operator of the algebra's own family $\{\Theta_{\tilde Q_1}\}$ and the operator is quadratic in the parameter; with the indefinite adjoint the sandwich of a $J$-unitary parameter is an inner automorphism of the operator algebra, and the parameter families that are $J$-unitary are exactly the ones computed in The Krein Isometry Group and Its Cartan Decomposition. The article determines the sandwich laws, the preservation of the indefinite form, of $J$-self-adjointness and of $J$-positivity, the transport of the kernel and the image, the parameters coming from the algebra, and the contrast with the definite case.

The forms and the algebra are The Biquaternion Algebra and Its Vector Spaces; the definite form and the dagger sandwich are The Four Forms and the Four Algebraic Norms over the Biquaternion Algebra and Two-Sided Operators on the General Plain Sesqualgebra; the indefinite form and its symmetry are The Four Forms and the Four Algebraic Norms over the Biquaternion Algebra and The Fundamental Symmetry of the Form; its adjoint and the positive cone of the indefinite order are J-Self-Adjoint and J-Unitary Operators on the General Quaternionic Sesqualgebra and The Krein Isometry Group and Its Cartan Decomposition; the group of $J$-unitary parameters is The Krein Isometry Group and Its Cartan Decomposition; and the general operator is The Hermitian Sandwich on a Krein Space.

$$ \langle\tilde W,\tilde Q_1\rangle_{\natural*} = (J\tilde Q_1,\tilde W) , \qquad J^{2} = \mathrm{id}, \qquad E^{2} = \mathrm{I}_4 . $$

The Krein adjoint of a $\mathbb{C}$-linear operator is $A^{\dagger} = JA^{*}J$, where $A^{*}$ is its adjoint for the definite form. Since all operators here are finite-dimensional, existence and uniqueness of adjoints raise no question.

The Coefficient Model of the Form

Proposition (the coefficient model). Under the $\mathbb{C}$-linear identification $\tilde Q_1\mapsto(Q_{1t},Q_{1x},Q_{1y},Q_{1z})$ of $\mathbb{B}$ with $\mathbb{C}^{4}$, the general quaternionic sesquilinear form is the standard form of signature $(1,3)$,

$$ \langle\tilde W,\tilde Q_1\rangle_{\natural*} = \overline{Q_{1t}}W_t - \overline{Q_{1x}}W_x - \overline{Q_{1y}}W_y - \overline{Q_{1z}}W_z , $$

and the fundamental symmetry is the diagonal operator $J = \mathrm{diag}(1,-1,-1,-1)$, self-adjoint for both forms, with $J^{2}=\mathrm{id}$ and with the twisted form $\langle\tilde W,\tilde Q_1\rangle_{\natural*} = (J\tilde Q_1,\tilde W)$.

Proof. Immediate from the definitions in the coefficient basis; the diagonal of the definite form is $\mathrm{I}_4$ and the diagonal of the general quaternionic sesquilinear form is $E$, so $\langle\tilde W,\tilde Q_1\rangle_{\natural*} = \tilde Q_1^{\mathsf T*}E\tilde W = (E\tilde Q_1,\tilde W) = (J\tilde Q_1,\tilde W)$.

Remark (four forms, not one). The quaternion norm $\langle\tilde Q_1,\tilde Q_1\rangle_{\natural} = \sum_\mu Q_{1,\mu}^{2}$ is a complex quadratic form, isotropic on the null cone of Quaternionic Norm Regularity and Degeneracy in the Biquaternion Algebra; the definite form $(\cdot,\cdot)$ is positive definite; the general quaternionic sesquilinear form $\langle\cdot,\cdot\rangle_{\natural*}$ is an indefinite Hermitian structure of signature $(1,3)$, split as $\mathbb{B} = \mathbb{C}_{\mathbb{B}}\perp_{K}\mathbb{V}_{\mathbb{B}}$ into the centre line and the vector subspace; and the general quaternionic bilinear form of The Four Forms and the Four Algebraic Norms over the Biquaternion Algebra is another object again. The four must not be interchanged, and it is the general quaternionic sesquilinear form that defines the adjoint of this article.

The Dagger Sandwich

Definition. For operators $A,T\in\mathrm{End}_{\mathbb{C}}(\mathbb{B})$ the indefinite Hermitian sandwich, or dagger sandwich, is

$$ H_{A}(T) = A\,T\,A^{\dagger} , \qquad A^{\dagger} = JA^{*}J . $$

In the coefficient basis, where $A$ and $T$ are $4\times4$ complex matrices, this reads $H_{A}(T) = ATEA^{*}E$ with $E = \mathrm{diag}(1,-1,-1,-1)$.

Proposition (the adjoint law). For all operators $A,T$,

$$ H_{A}(T)^{\dagger} = H_{A}\bigl(T^{\dagger}\bigr) . $$

Proof. $H_A(T)^{\dagger} = (ATA^{\dagger})^{\dagger} = A^{\dagger\dagger}T^{\dagger}A^{\dagger} = AT^{\dagger}A^{\dagger}$, because the operation $T\mapsto T^{\dagger}$ is an involution and an anti-involution: $J^{2}=\mathrm{id}$ gives $A^{\dagger\dagger}=A$. The right-hand side is $H_A(T^{\dagger})$ by definition.

Proposition (composition and multiplicativity). For all operators $A,S,T$,

$$ \boxed{\; H_{A}\bigl(H_{S}(T)\bigr) = H_{AS}(T) , \qquad H_{A}(ST) = H_{A}(S)\,H_{A}(T) \iff A^{\dagger}A = \mathrm{id} \;}. $$

Proof. The composition is associativity: $A(S\,T\,S^{\dagger})A^{\dagger} = (AS)T(AS)^{\dagger}$. For the second, $H_A(S)H_A(T) = ASA^{\dagger}ATA^{\dagger} = AS(A^{\dagger}A)TA^{\dagger}$, which equals $ASTA^{\dagger}=H_A(ST)$ identically exactly when $A^{\dagger}A=\mathrm{id}$.

Proposition (the parameter rule). For a complex scalar $\alpha$ and all $T$,

$$ H_{\alpha A}(T) = \lvert\alpha\rvert^{2}\,H_{A}(T) . $$

Proof. $(\alpha A)^{\dagger} = J(\alpha A)^{*}J = \bar\alpha\,JA^{*}J = \bar\alpha\,A^{\dagger}$, so $(\alpha A)T(\bar\alpha A^{\dagger}) = \lvert\alpha\rvert^{2}ATA^{\dagger}$.

Remark (linear in the argument, quadratic in the parameter). The assignment $T\mapsto H_A(T)$ is linear in the argument and homogeneous of degree two in the parameter, and it is blind to the phase of the parameter. This is the operator-level form of the rule that the two-sided operator $\Theta_{\tilde Q_1}$ is quadratic in $\tilde Q_1$ (Two-Sided Operators on the General Plain Sesqualgebra), and it is the reason the sandwich is a map of the projective parameter, not a representation of the algebra.

Preservation of the Form and of J-Positivity

Theorem (form preservation by $J$-unitary parameters). Let $A$ be $J$-unitary, $A^{\dagger}A = AA^{\dagger} = \mathrm{id}$. Then for every operator $T$ and all $\tilde Q_1,\tilde W\in\mathbb{B}$,

$$ \boxed{\; \langle H_{A}(T)\tilde W,H_{A}(T)\tilde Q_1\rangle_{\natural*} = \langle TA^{\dagger}\tilde W,TA^{\dagger}\tilde Q_1\rangle_{\natural*} , \qquad \langle \tilde W,H_{A}(T)\tilde Q_1\rangle_{\natural*} = \langle A^{\dagger}\tilde W,TA^{\dagger}\tilde Q_1\rangle_{\natural*} \;}, $$

the first identity being the second applied twice; and $H_{A}(T)$ is a $J$-unitary operator whenever $T$ is.

Proof. The one-argument identity is the reading of the elementary rule $\langle\tilde S,A\tilde Q_3\rangle_{\natural*} = \langle A^{\dagger}\tilde S,\tilde Q_3\rangle_{\natural*}$, which holds for every operator $A$ and follows from $\langle\tilde S,\tilde Q_3\rangle_{\natural*}=(J\tilde Q_3,\tilde S)$ and $(A\tilde Q_3,\tilde S)=(\tilde Q_3,A^{*}\tilde S)$. With $\tilde Q_3 = TA^{\dagger}\tilde Q_1$ it gives $\langle\tilde W,ATA^{\dagger}\tilde Q_1\rangle_{\natural*} = \langle A^{\dagger}\tilde W,TA^{\dagger}\tilde Q_1\rangle_{\natural*}$; replacing $\tilde W$ by $H_A(T)\tilde W = ATA^{\dagger}\tilde W$ and applying the rule once more gives the first identity. If $T$ is a $J$-unitary operator then $\langle TA^{\dagger}\tilde W,TA^{\dagger}\tilde Q_1\rangle_{\natural*}=\langle A^{\dagger}\tilde W,A^{\dagger}\tilde Q_1\rangle_{\natural*}$, and $A^{\dagger}$ is $J$-unitary with $A$, so this is $\langle\tilde W,\tilde Q_1\rangle_{\natural*}$.

Remark (the correction of the general display). The general article states the second identity with $T\tilde Q_1$ in place of $TA^{\dagger}\tilde Q_1$; that reading is not correct, as the elementary rule shows, and the version above is the one the general proof uses. In the biquaternion algebra the distinction is not cosmetic, because $(L_{\tilde Q_1})^{\dagger}=R_{\bar{\tilde Q_1}}$ for an algebra element and the parameter cannot be moved across $T$ freely.

Proposition (preservation of $J$-self-adjointness, $J$-positivity and $J$-unitarity). For every operator $A$ and every $\tilde Q_1$,

$$ \langle\tilde Q_1,H_{A}(T)\tilde Q_1\rangle_{\natural*} = \langle A^{\dagger}\tilde Q_1,T\,A^{\dagger}\tilde Q_1\rangle_{\natural*} . $$

Consequently, if $A$ is $J$-unitary then $H_{A}$ maps $J$-self-adjoint operators to $J$-self-adjoint operators, the $J$-positive cone to itself, and $J$-unitary operators to $J$-unitary operators.

Proof. The displayed identity is the elementary rule with $\tilde Q_3=TA^{\dagger}\tilde Q_1$ and $\tilde S=\tilde Q_1$, and it holds for every $A$. If $T$ is $J$-self-adjoint then $H_A(T)^{\dagger}=H_A(T^{\dagger})=H_A(T)$ by the adjoint law; if $T$ is $J$-positive the display gives $\langle\tilde Q_1,H_A(T)\tilde Q_1\rangle_{\natural*}=\langle A^{\dagger}\tilde Q_1,TA^{\dagger}\tilde Q_1\rangle_{\natural*}\geq0$; and if $T$ is $J$-unitary then the multiplicativity of the second proposition applies to $T^{*}$ too.

Theorem (kernel and image). Let $A$ be invertible. Then for every $T$,

$$ \boxed{\; \ker H_{A}(T) = \bigl(A^{\dagger}\bigr)^{-1}\bigl(\ker T\bigr) , \qquad \mathrm{im}\,H_{A}(T) = A\bigl(\mathrm{im}\,T\bigr) \;}, $$

and for $J$-unitary $A$ these read $\ker H_{A}(T) = A\ker T$ and $\mathrm{im}\,H_{A}(T) = A\,\mathrm{im}\,T$.

Proof. $ATA^{\dagger}\tilde Q_1 = 0$ iff $TA^{\dagger}\tilde Q_1=0$ because $A$ is invertible, so $\tilde Q_1\in\ker H_A(T)$ iff $A^{\dagger}\tilde Q_1\in\ker T$. The image statement is the same computation with the roles reversed, using the surjectivity of $A^{\dagger}$. For $J$-unitary $A$ one has $A^{\dagger}=A^{-1}$, hence $A\ker T$ and $A\,\mathrm{im}\,T$.

Corollary (what the sandwich preserves). For $J$-unitary $A$ the sandwich preserves the dimension of the kernel, the codimension and the dimension of the image, hence the rank of $T$; it preserves the $J$-orthogonal complement of the image and not the definite one.

Proof. $A$ is a bijective $J$-unitary operator, so it preserves dimensions and the indefinite orthogonality; it does not preserve $(\cdot,\cdot)$-orthogonality because a $J$-unitary operator need not be unitary, as the next sections show.

The Parameters Coming from the Algebra

The parameters of the theory are the operators, and the algebra supplies three natural families of them: the left multiplications $L_{\tilde Q_1}$, the right multiplications $R_{\tilde Q_1}$ and the two-sided operators $\Theta_{\tilde Q_1}(\tilde Q_2)=\tilde Q_1\tilde Q_2\tilde Q_1^{*}$. Their Krein adjoints and their $J$-unitarity are computed in One-Sided Operators on the General Plain Sesqualgebra and The Krein Isometry Group and Its Cartan Decomposition; the results are collected here in the notation of the sandwich.

Proposition (the Krein adjoints of the algebra families). For every $\tilde Q_1\in\mathbb{B}$,

$$ \bigl(L_{\tilde Q_1}\bigr)^{\dagger} = R_{\bar{\tilde Q_1}} , \qquad \bigl(R_{\tilde Q_1}\bigr)^{\dagger} = L_{\bar{\tilde Q_1}} , $$

with $\bar{\tilde Q_1} = \sum_\mu\overline{Q_{1,\mu}}e_\mu$ the coefficient conjugation.

Proof. In the coefficient basis $L_{\tilde Q_1}$ is left multiplication by $\tilde Q_1$ and the Krein adjoint is $E\,L_{\tilde Q_1}^{*}\,E$; the definite adjoint of $L_{\tilde Q_1}$ is $L_{\tilde Q_1^{*}}$ (The Adjoint of the Left Multiplication on a Hermitian Algebra), and conjugating it by $E$ replaces the Hermitian star by the coefficient conjugation, which is the natural involution composed with the star. The computation is the operator form of the identity $(L_{\tilde Q_1})^{\dagger} = R_{\bar{\tilde Q_1}}$ used throughout the indefinite theory of the algebra.

Theorem ($J$-unitarity of the algebra parameters). For $\tilde Q_1\in\mathbb{B}$,

$$ L_{\tilde Q_1}\ \text{is }J\text{-unitary} \iff \tilde Q_1\in S^{1}e_t , \qquad R_{\tilde Q_1}\ \text{is }J\text{-unitary} \iff \tilde Q_1\in S^{1}e_t , \qquad \Theta_{\tilde Q_1}\ \text{is }J\text{-unitary} \iff \lvert \langle\tilde Q_1,\tilde Q_1\rangle_{\natural}\rvert = 1 . $$

Proof. This is the theorem of The Krein Isometry Group and Its Cartan Decomposition, repeated because the sandwich needs it: $L_{\tilde Q_1}^{\dagger}L_{\tilde Q_1}$ is the operator $\tilde Q_2\mapsto\tilde Q_1\tilde Q_2\bar{\tilde Q_1}$, which is the identity only for central phases; and $\Theta_{\tilde Q_1}^{\dagger}\Theta_{\tilde Q_1} = \Theta_{\langle\tilde Q_1,\tilde Q_1\rangle_{\natural}e_t} = \lvert \langle\tilde Q_1,\tilde Q_1\rangle_{\natural}\rvert^{2}\mathrm{id}$.

Corollary (the two regimes of the sandwich). For a central phase $\tilde Q_1 = ce_t$, $\lvert c\rvert=1$, the parameter $L_{\tilde Q_1}=c\,\mathrm{id}$ gives the trivial sandwich $H_{L_{\tilde Q_1}}(T)=\lvert c\rvert^{2}T=T$, so the only sandwich coming from a left or right multiplication by a $J$-unitary element is the identity. The interesting sandwiches come from the norm-one slice: for $\lvert \langle\tilde Q_1,\tilde Q_1\rangle_{\natural}\rvert=1$ the operator $\Theta_{\tilde Q_1}$ is $J$-unitary, the sandwich

$$ H_{\Theta_{\tilde Q_1}}(T) = \Theta_{\tilde Q_1}\circ T\circ\Theta_{\tilde Q_1}^{\dagger} $$

is an automorphism of $\mathrm{End}_{\mathbb{C}}(\mathbb{B})$ preserving the indefinite form, and

$$ \bigl(\Theta_{\tilde Q_1}\bigr)^{\dagger} = \Theta_{\tilde Q_1^{-1}} . $$

Proof. The triviality of the central-phase sandwich is the parameter rule with $\lvert c\rvert=1$. For the second, the theorem makes $\Theta_{\tilde Q_1}$ $J$-unitary on the norm-one slice, so $\Theta_{\tilde Q_1}^{\dagger}=\Theta_{\tilde Q_1}^{-1}=\Theta_{\tilde Q_1^{-1}}$, the last equality by the composition law $\Theta_{\tilde Q_1}\Theta_{\tilde Q_1^{-1}}=\Theta_{e_t}=\mathrm{id}$ of Two-Sided Operators on the General Plain Sesqualgebra.

Remark (the kernel of the sandwich map). The sandwich map $A\mapsto H_A$ is a homomorphism from the $J$-unitary group into the automorphism group of $\mathrm{End}_{\mathbb{C}}(\mathbb{B})$ whose kernel is the circle of scalars of modulus one: $H_A(T)=T$ for all $T$ forces $A$ central, and the $J$-unitary central operators are the phases. The algebra's own parameters generate inside the $J$-unitary group the subgroup $\mathrm{U}(1)\times PU(2)\cong \mathrm{U}(1)\times SO(3)$ of The Krein Isometry Group and Its Cartan Decomposition, and the sandwich restricts to the automorphisms $\Theta_{\tilde Q_1}$ themselves, which are inner and depend only on the class of $\tilde Q_1$ in $PU(2)$.

Proof. Multiplicativity and the kernel statement are the propositions of the first section; the subgroup statement is the corollary of the cited article, and the inner reading is the composition law, since $H_{\Theta_{\tilde Q_1}}(\Theta_{\tilde Q_2}) = \Theta_{\tilde Q_1}\Theta_{\tilde Q_2}\Theta_{\tilde Q_1}^{\dagger}$ is conjugation in the group of $J$-unitary operators.

Contrast with the Definite Sandwich

Remark (two adjoints, two sandwiches). The definite sandwich $\Theta_{\tilde Q_1}(\tilde Q_2)=\tilde Q_1\tilde Q_2\tilde Q_1^{*}$ of The Hermitian Sandwich in the Biquaternion Algebra with Hermitian Adjoint uses the adjoint $Q^{*}$ of the definite form $(\cdot,\cdot)$ and is multiplicative in its argument exactly when $\tilde Q_1^{*}\tilde Q_1=e_t$, that is on the unitary slice $\mathrm{U}(2)$; the indefinite sandwich $H_A(T)=ATA^{\dagger}$ uses the adjoint of the general quaternionic sesquilinear form and is multiplicative exactly when $A^{\dagger}A=\mathrm{id}$, that is on the Krein-unitary group $\mathrm{U}(1,3)$. The identity that separates them is that the adjoint of a left multiplication for the general quaternionic sesquilinear form changes sides, $(L_{\tilde Q_1})^{\dagger}=R_{\bar{\tilde Q_1}}$, whereas for the definite form it stays, $(L_{\tilde Q_1})^{*}=L_{\tilde Q_1^{*}}$.

Theorem (the failures of the indefinite sandwich). For a $J$-unitary $A$ that is not unitary, the sandwich $H_A$ does not preserve the norm of the definite form, its orthogonal complements, its positive cone or its spectrum. It preserves the general quaternionic sesquilinear form, $J$-self-adjointness, $J$-positivity, $J$-unitarity, and the kernel and the image of every operator.

Proof. A $J$-unitary $A$ satisfies $A^{*}JA=J$ and is unitary only if additionally $A^{*}A=\mathrm{id}$. When it is not, $H_A$ changes the norm of operators, and their spectrum is not preserved, while all the indefinite statements are the theorems of the previous sections.

Remark (the algebra reason). In the biquaternion algebra the failure is visible inside the algebra. The definite adjoint of a two-sided operator is $\Theta_{\tilde Q_1}^{*}=\Theta_{\tilde Q_1^{*}}$, so $\Theta_{\tilde Q_1}^{*}\Theta_{\tilde Q_1}=\Theta_{\tilde Q_1^{*}\tilde Q_1}$ is the identity exactly when $\tilde Q_1^{*}\tilde Q_1$ is a central phase, that is exactly when $\tilde Q_1$ is a positive scalar multiple of a unitary element; while $\Theta_{\tilde Q_1}$ is $J$-unitary exactly on the norm-one slice $\lvert \langle\tilde Q_1,\tilde Q_1\rangle_{\natural}\rvert=1$. The two conditions differ: an explicit element on the norm-one slice with $\tilde Q_1^{*}\tilde Q_1$ not central was checked to give $\Theta_{\tilde Q_1}$ $J$-unitary and not unitary. The Lorentz group $SL(2,\mathbb{C})$ acts on the algebra by $J$-unitary operators and changes the definite structure: no positive definite form is invariant under it.

Worked Examples

The Fundamental Symmetry

For $A = J$ one has $J^{\dagger}=J$ and

$$ H_{J}(T) = JTJ , \qquad \bigl(H_{J}(T)\bigr)^{*} = J\,T^{*}J = H_{J}(T^{*}) , $$

so $H_{J}(T)$ is self-adjoint for the definite form exactly when $T$ is, and $\langle H_{J}(T)\tilde Q_1,\tilde Q_1\rangle = \langle T(J\tilde Q_1),J\tilde Q_1\rangle$ shows that $H_{J}(T)\geq0$ for the definite form exactly when $T\geq0$. The link with the indefinite data is the dictionary of J-Self-Adjoint and J-Unitary Operators on the General Quaternionic Sesqualgebra: $T$ is $J$-self-adjoint exactly when $JT$ is self-adjoint for the definite form, and in that case $H_{J}(T)=T^{*}$.

The Boost

On the two-dimensional sub-block spanned by $e_t,\vec e_x$ let

$$ \mathsf{B}_{t} = \begin{pmatrix}\cosh t & \sinh t\\ \sinh t & \cosh t\end{pmatrix} , \qquad t\in\mathbb{R}, $$

extended by the identity on $\vec e_y,\vec e_z$. Then $\mathsf{B}_{t}$ is $J$-unitary, $\mathsf{B}_{t}^{\dagger}=\mathsf{B}_{t}^{-1}=\mathsf{B}_{-t}$, and $\mathsf{B}_t$ is not unitary, since $\mathsf{B}_t^{*}\mathsf{B}_t = \mathsf{B}_t^{2} = \mathsf{B}_{2t}$. For $\tilde Q_1 = e_t$ one has $\mathsf{B}_te_t = \cosh t\,e_t + \sinh t\,\vec e_x$ and

$$ \langle \mathsf{B}_te_t,\mathsf{B}_te_t\rangle_{\natural*} = 1 = \langle e_t,e_t\rangle_{\natural*} , \qquad (\mathsf{B}_te_t,\mathsf{B}_te_t) = \cosh^{2}t + \sinh^{2}t = \cosh 2t . $$

So the sandwich preserves the general quaternionic sesquilinear form and changes the norm of the definite form, by the factor $\cosh 2t$ per unit of the parameter; the boost is the model of the failure theorem, and it is the reason the indefinite sandwich preserves the indefinite form only.

The Phase Parameters

For $\tilde Q_1 = ce_t$ with $\lvert c\rvert=1$ the left multiplication is $L_{\tilde Q_1}=c\,\mathrm{id}$, $J$-unitary and central, and $H_{L_{\tilde Q_1}}(T)=\lvert c\rvert^{2}T=T$ by the parameter rule. The sandwich therefore sees the projective class of the parameter, and the circle $S^{1}e_t$ is exactly the kernel of $A\mapsto H_A$, as the kernel computation of the previous section asserts.

Summary

On the biquaternion algebra the general quaternionic sesquilinear form $\langle\tilde W,\tilde Q_1\rangle_{\natural*}=\mathrm{Sc}(\tilde Q_1^{\natural*}\tilde W)$ has signature $(1,3)$, with fundamental symmetry $J={}^{\natural}$ and indefinite adjoint $A^{\dagger}=JA^{*}J$. The indefinite Hermitian sandwich $H_{A}(T)=ATA^{\dagger}$ is linear in the argument and quadratic in the parameter, satisfies $H_A(T^{\dagger})=H_A(T)^{\dagger}$ and $H_A(H_S(T))=H_{AS}(T)$, and is multiplicative in the argument exactly when $A^{\dagger}A=\mathrm{id}$. For a $J$-unitary parameter it preserves the general quaternionic sesquilinear form to the twisted arguments, $\langle H_A(T)\tilde W,H_A(T)\tilde Q_1\rangle_{\natural*}=\langle TA^{\dagger}\tilde W,TA^{\dagger}\tilde Q_1\rangle_{\natural*}$, it preserves $J$-self-adjointness, $J$-positivity and $J$-unitarity, it transports the kernel and the image as $\ker H_A(T)=(A^{\dagger})^{-1}\ker T$ and $\mathrm{im}\,H_A(T)=A\,\mathrm{im}\,T$, and it is an automorphism of the operator algebra. The parameters coming from the algebra are governed by the Krein adjoints $(L_{\tilde Q_1})^{\dagger}=R_{\bar{\tilde Q_1}}$ and $(R_{\tilde Q_1})^{\dagger}=L_{\bar{\tilde Q_1}}$, by the $J$-unitarity criteria of The Krein Isometry Group and Its Cartan Decomposition — central phases for the one-sided families and the norm-one slice for the two-sided family — and by the identity $(\Theta_{\tilde Q_1})^{\dagger}=\Theta_{\tilde Q_1^{-1}}$ on that slice. The contrast with the definite sandwich $\Theta_{\tilde Q_1}(\tilde Q_2)=\tilde Q_1\tilde Q_2\tilde Q_1^{*}$ is that a $J$-unitary operator need not be unitary: the indefinite sandwich preserves the form, the indefinite positivity, self-adjointness and the kernel and image, but not the norm of the definite form, its decompositions, its positive cone or its spectrum, so it is a symmetry of the indefinite form and not of the definite one. The general theory is The Hermitian Sandwich on a Krein Space; the definite instance is The Hermitian Sandwich in the Biquaternion Algebra with Hermitian Adjoint and Two-Sided Operators on the General Plain Sesqualgebra; and the indefinite positivity and the $J$-unitary group are The Krein Isometry Group and Its Cartan Decomposition, and the indefinite spectral theory is J-Self-Adjoint and J-Unitary Operators on the General Quaternionic Sesqualgebra, §The Indefinite Spectral Data.

Summary of Notation

Symbol	Meaning
$(\tilde Q_1,\tilde W)=\mathrm{Sc}(\tilde Q_1^{*}\tilde W)$	Definite form, Gram $\mathrm{I}_4$, signature $(4,0)$
$\langle\tilde W,\tilde Q_1\rangle_{\natural}=\mathrm{Sc}(\tilde Q_1^{\natural}\tilde W)$	general quaternionic sesquilinear form, Gram $E$, signature $(1,3)$
$J={}^{\natural}$, $E=\mathrm{diag}(1,-1,-1,-1)$	Fundamental symmetry, $\langle\tilde W,\tilde Q_1\rangle_{\natural*}=(J\tilde Q_1,\tilde W)$
$A^{\dagger}=JA^{*}J$	Indefinite adjoint
$H_{A}(T)=ATA^{\dagger}$	Indefinite Hermitian sandwich
$H_{A}(T^{\dagger})=H_{A}(T)^{\dagger}$	Sandwiches commute with adjunction
$H_{A}(H_{S}(T))=H_{AS}(T)$	Composition
$H_{A}(ST)=H_{A}(S)H_{A}(T)\iff A^{\dagger}A=\mathrm{id}$	Multiplicativity
$\langle\tilde W,H_A(T)\tilde Q_1\rangle_{\natural}=\langle A^{\dagger}\tilde W,TA^{\dagger}\tilde Q_1\rangle_{\natural}$	Form preservation, one argument
$\langle\tilde Q_1,H_A(T)\tilde Q_1\rangle_{\natural}=\langle A^{\dagger}\tilde Q_1,TA^{\dagger}\tilde Q_1\rangle_{\natural}$	Preservation of $J$-positivity
$\ker H_A(T)=(A^{\dagger})^{-1}\ker T$, $\ \mathrm{im}\,H_A(T)=A\,\mathrm{im}\,T$	Kernel and image
$(L_{\tilde Q_1})^{\dagger}=R_{\bar{\tilde Q_1}}$, $(R_{\tilde Q_1})^{\dagger}=L_{\bar{\tilde Q_1}}$	The algebra families
$\Theta_{\tilde Q_1}$ $J$-unitary $\iff\lvert \langle\tilde Q_1,\tilde Q_1\rangle_{\natural}\rvert=1$	The norm-one slice
$(\Theta_{\tilde Q_1})^{\dagger}=\Theta_{\tilde Q_1^{-1}}$ on the slice	Krein adjoint of the two-sided family