J-Normal Operators and the Indefinite Spectral Theorem

Introduction

In a positive definite space the normal operators are the ones a spectral theorem can reach, and normality forces the spectrum into a symmetry: normal operators have no reason to be self-adjoint, but they are diagonalised by unitaries, and their spectra are conjugate-symmetric. This article asks what survives of that circle of ideas for the Krein form. The answer is a partial one, and its partiality is the point: the $J$-normal operators are still those commuting with their $J$-adjoint, the algebra's three operator families are all $J$-normal, the $J$-self-adjoint operators with positive definite partner $JT$ still have real spectra and diagonalisations, and the spectral symmetries still hold — but $J$-normality no longer forces a real spectrum, a $J$-self-adjoint operator need not be diagonalisable, and the $J$-unitary group is not compact. The $J$-adjoint itself is J-Self-Adjoint and J-Unitary Operators on the Biquaternion Algebra; the spectra of the algebra's operators are computed in The Indefinite Spectra of the Operators on the Biquaternion Algebra; and the group of $J$-unitaries is The Krein Isometry Group and Its $J$-Contractions.

Conventions. $\mathbb{B}$ is the biquaternion algebra with its Hermitian form $\langle\tilde{Q},\tilde{Q}'\rangle=\sum_{\mu}\bar Q_{\mu}Q'_{\mu}$ (positive definite, Gram matrix $\mathrm{I}_4$ in the coefficient basis), its Krein form $[\tilde{Q},\tilde{Q}']=\sum_{\mu}\varepsilon_{\mu}\bar Q_{\mu}Q'_{\mu}$ (Gram matrix $E=\mathrm{diag}(1,-1,-1,-1)$), and its fundamental symmetry $J={}^{\natural}$. The Krein adjoint of a $\mathbb{C}$-linear operator $T$ is $T^{\dagger}=JT^{*}J$, characterised by $[T\tilde{Q},\tilde{Q}']=[\tilde{Q},T^{\dagger}\tilde{Q}']$, where $T^{*}$ is the adjoint for $\langle\cdot,\cdot\rangle$. The algebra's operators are $L_{\tilde{Q}}$, $R_{\tilde{Q}}$ and $\Theta_{\tilde{Q}}$, with $(\Theta_{\tilde{Q}})(\tilde{P})=\tilde{Q}\tilde{P}\tilde{Q}^{\dagger}$.

J-Normality

Definition. A $\mathbb{C}$-linear operator $T$ is $J$-normal when

$$ T^{\dagger}T=TT^{\dagger}, $$

that is, when $T$ commutes with its Krein adjoint. It is $J$-self-adjoint when $T^{\dagger}=T$ and $J$-unitary when $T^{\dagger}T=TT^{\dagger}=\mathrm{id}$.

Theorem (closure properties). Every $J$-self-adjoint and every $J$-unitary operator is $J$-normal. The $J$-normal operators are closed under the Krein adjoint, under powers and inverses, and under $\mathbb{C}$-linear combinations of mutually commuting ones; the $J$-normal operators that are also $J$-self-adjoint form a real vector space, and those that are also $J$-unitary form a group.

Proof. For $J$-self-adjointness, $T^{\dagger}T=T^{2}=TT^{\dagger}$; for $J$-unitarity both products are the identity. The adjoint of $T^{\dagger}$ is $T$, and $T^{\dagger}$ trivially commutes with $T$ whenever $T$ does; powers of a $J$-normal operator commute with each other and with their adjoints by induction; inverses follow from $T^{\dagger}T=TT^{\dagger}$ by inversion; and commuting normal operators add, because every term in the expansion of $(S+T)^{\dagger}(S+T)$ and $(S+T)(S+T)^{\dagger}$ can be reordered to the same expression.

Theorem (the definite model of $J$-normality). In the coefficient basis, with $T^{*}$ the conjugate transpose and $E=\mathrm{diag}(1,-1,-1,-1)$, the operator $T$ is $J$-normal if and only if

$$ T^{*}\,E\,T\,E=E\,T\,E\,T^{*}, $$

equivalently if and only if $T^{*}$ commutes with $J TJ$.

Proof. $T^{\dagger}=ET^{*}E$, so $T^{\dagger}T=ET^{*}ET$ and $TT^{\dagger}=TET^{*}E$. Multiplying the equality $ET^{*}ET=TET^{*}E$ by $E$ on the left and on the right gives $T^{*}ETE=ETET^{*}$, which is $T^{*}(ETE)=(ETE)T^{*}$.

Remark (why the definite case is not recovered). For the Hermitian adjoint the condition is $T^{*}T=TT^{*}$; here the conjugation by $E$ intervenes on both sides, so $J$-normality is not the normality of $T$ for $\langle\cdot,\cdot\rangle$ and not the normality of $JT$. The two notions coincide only on the operators that commute with $E$ in the appropriate sense.

The Spectral Theorem for J-Self-Adjoint Operators

Theorem (the dictionary). The map $T\mapsto JT$ is a bijection from the $J$-self-adjoint operators onto the self-adjoint operators for $\langle\cdot,\cdot\rangle$, with inverse $S\mapsto JS$; consequently every $J$-self-adjoint operator is $J$ times a Hermitian operator, and its spectrum is the spectrum of that Hermitian operator read through $J$.

Proof. $T^{\dagger}=T\iff ET^{*}E=T\iff (ET)^{*}=ET$, and $S=ET$ is Hermitian exactly when $S^{*}=S$; the map is its own inverse up to the involutions $E$.

Theorem (the definite-spectrum case). Let $T$ be $J$-self-adjoint and suppose that $JT$ is positive definite for $\langle\cdot,\cdot\rangle$. Then $T$ has real spectrum and is diagonalisable over $\mathbb{C}$.

Proof. Write $S=JT$, positive definite and self-adjoint. Then $T=ES$ and

$$ S^{1/2}\,T\,S^{-1/2}=S^{1/2}ES^{1/2}, $$

an operator that is self-adjoint for $\langle\cdot,\cdot\rangle$, since $S^{1/2}$ and $E$ are; it is therefore diagonalisable with real eigenvalues (Self-Adjoint Elements and the Positive Cone), and $T$, being similar to it, is diagonalisable with the same real spectrum.

Remark (the semidefinite case fails the conclusion). The hypothesis of positive definiteness cannot be weakened to positive semidefiniteness. With $E=\mathrm{diag}(1,-1,-1,-1)$ and $S=|\tilde{Q}\rangle\langle\tilde{Q}|$ the rank-one operator of the isotropic element $\tilde{Q}=e_0+e_1$, the operator $T=ES$ is $J$-self-adjoint and lies in the $J$-positive cone, while

$$ T^{2}=0,\qquad T\neq0 , $$

so $T$ is nilpotent, has spectrum $\{0\}$, and is not diagonalisable. Thus a $J$-self-adjoint, $J$-nonnegative operator of the algebra need not have a $J$-orthonormal eigenbasis.

The Spectral Symmetries

Theorem (the symmetry of the $J$-self-adjoint spectrum). If $T$ is $J$-self-adjoint then $T$ and $T^{*}$ have the same characteristic polynomial, so the spectrum of $T$, with multiplicities, is invariant under conjugation:

$$ \lambda\in\mathrm{spec}(T)\iff\bar\lambda\in\mathrm{spec}(T). $$

Proof. $T=ES$ with $S$ self-adjoint, so $T^{*}=(ES)^{*}=SE$; the matrices $ES$ and $SE$ have the same characteristic polynomial, because $AB$ and $BA$ do for square matrices. Since $T^{*}$ is the conjugate transpose of $T$, its spectrum is the conjugate of the spectrum of $T$.

Theorem (the symmetry of the $J$-unitary spectrum). If $T$ is $J$-unitary then its spectrum is invariant under inversion in the unit circle,

$$ \lambda\in\mathrm{spec}(T)\iff\frac{1}{\bar\lambda}\in\mathrm{spec}(T), $$

with the same multiplicity; in particular an eigenvalue on the unit circle has its conjugate present as well, but the spectrum is in general not contained in the unit circle.

Proof. $T$ is $J$-unitary exactly when $T^{*}ET=E$, that is $T^{*}=ET^{-1}E$; so $T^{*}$ is similar to $T^{-1}$, whence the characteristic polynomials of $T^{*}$ and $T^{-1}$ agree, and $\mathrm{spec}(T^{-1})=\{\lambda^{-1}:\lambda\in\mathrm{spec}(T)\}$ while $\mathrm{spec}(T^{*})=\overline{\mathrm{spec}(T)}$. The last clause is the hyperbolic boost of the next sections, with eigenvalues $e^{\pm t}$.

Remark (no unit-circle confinement). The symmetry is inversion, not confinement: the boost $T$ with $e_0\mapsto\cosh t\,e_0+\sinh t\,e_1$, $e_1\mapsto\sinh t\,e_0+\cosh t\,e_1$ and $e_2,e_3$ fixed is $J$-unitary with eigenvalues $e^{t},e^{-t},1,1$, so a $J$-unitary operator leaves the unit circle as soon as it moves the Minkowski slice (The Krein Isometry Group and Its $J$-Contractions).

The Normality of the Algebra's Operators

Theorem (the three families are $J$-normal). For every biquaternion $\tilde{Q}$ the left multiplication $L_{\tilde{Q}}$, the right multiplication $R_{\tilde{Q}}$ and the sandwich $\Theta_{\tilde{Q}}$ are $J$-normal. Precisely,

$$ (L_{\tilde{Q}})^{\dagger}=R_{\bar{\tilde{Q}}},\qquad (R_{\tilde{Q}})^{\dagger}=L_{\bar{\tilde{Q}}},\qquad (\Theta_{\tilde{Q}})^{\dagger}=\Theta_{{}^{\natural}\tilde{Q}}, $$

and in each case the two factors of $T^{\dagger}T$ commute.

Proof. $[L_{\tilde{Q}}\tilde{P},\tilde{U}]=[\tilde{P},R_{\bar{\tilde{Q}}}\tilde{U}]$ was computed in J-Self-Adjoint and J-Unitary Operators on the Biquaternion Algebra; then $L_{\tilde{Q}}^{\dagger}L_{\tilde{Q}}=R_{\bar{\tilde{Q}}}L_{\tilde{Q}}$ and $L_{\tilde{Q}}L_{\tilde{Q}}^{\dagger}=L_{\tilde{Q}}R_{\bar{\tilde{Q}}}$, equal because left and right multiplications commute. The right case is the same. For the sandwich, $\Theta_{\tilde{Q}}^{\dagger}\Theta_{\tilde{Q}}=\Theta_{{}^{\natural}\tilde{Q}}\Theta_{\tilde{Q}}=\Theta_{{}^{\natural}\tilde{Q}\,\tilde{Q}}=\Theta_{N(\tilde{Q})e_0}$, and $\Theta_{N(\tilde{Q})e_0}=|N(\tilde{Q})|^{2}\mathrm{id}$; the same value is obtained in the other order, because $\tilde{Q}$ and ${}^{\natural}\tilde{Q}$ have the same norm.

Remark (a product of a left and a right multiplication need not be normal). The composite $L_{\tilde{Q}}R_{\tilde{R}}$ has Krein adjoint $L_{\bar{\tilde{R}}}R_{\bar{\tilde{Q}}}$, and the normality condition is the commutation of the two parameters; it fails for general $\tilde{Q},\tilde{R}$, so the $J$-normal operators are not the whole of the algebra's operator span.

The Limits of the Analogy

Theorem (normality does not force a real spectrum). A $J$-normal operator of the algebra can have a non-real spectrum. For $\tilde{Q}=e_1$ the operator $L_{e_1}$ is $J$-normal with

$$ \mathrm{spec}(L_{e_1})=\{i,-i\}, $$

each eigenvalue of multiplicity two.

Proof. $L_{e_1}$ is $J$-normal by the theorem above. As a $\mathbb{C}$-linear operator on $\mathbb{B}\cong\mathbb{C}^{4}$ it is the left multiplication of the matrix $\Phi(e_1)$, whose eigenvalues are $i$ and $-i$; and the left regular representation of the algebra is the direct sum of two copies of the defining representation, so each eigenvalue occurs twice (The Indefinite Spectra of the Operators on the Biquaternion Algebra).

Corollary (contrast with the definite theory). In a positive definite space a normal operator with real eigenvalues is self-adjoint and a normal operator with non-real eigenvalues is not; here $L_{e_1}$ is $J$-normal, is not $J$-self-adjoint, and its spectrum is imaginary. Normality, self-adjointness and reality of the spectrum are three independent conditions in the indefinite case, whereas in the definite case normality and reality together imply diagonalisability by unitaries.

Worked Examples

$J$ itself. $J={}^{\natural}$ is $J$-self-adjoint and $J$-unitary, hence $J$-normal; its spectrum is $\{+1,-1\}$, real, and the multiplicities $(2,6)$ are the inertia of the form.

A central scalar. $\tilde{Q}=te_0$, $t\in\mathbb{R}$: $L_{\tilde{Q}}=t\,\mathrm{id}$ is $J$-self-adjoint with spectrum $\{t\}$ of multiplicity four, the positive definite case of the spectral theorem with $JT=tJ$ definite exactly for $t\neq0$ up to sign.

A vector element. $L_{e_1}$: $J$-normal and not $J$-self-adjoint, with the imaginary spectrum $\{i,-i\}$.

A sandwich. $\Theta_{e_1}$: $J$-normal, $J$-self-adjoint and $J$-unitary at once, since $e_1$ lies in the vector subspace and $|N(e_1)|=1$; its spectrum is $\{+1,-1\}$, each of multiplicity two.

A $J$-positive nilpotent. $T=ES$ with $S=|\tilde{Q}\rangle\langle\tilde{Q}|$ and $\tilde{Q}=e_0+e_1$: $J$-self-adjoint, in the $J$-positive cone, with $T^{2}=0$ and spectrum $\{0\}$; not diagonalisable.

A $J$-unitary with non-real spectrum. $L_{e^{i\theta}e_0}$: $J$-unitary with spectrum $\{e^{i\theta}\}$ on the circle, showing that confinement to the circle takes place exactly for the scalar phases.

Summary

A $\mathbb{C}$-linear operator is $J$-normal when it commutes with its Krein adjoint $T^{\dagger}=JT^{*}J$; in the coefficient basis this is $T^{*}ETE=ETET^{*}$. The $J$-self-adjoint and the $J$-unitary operators are $J$-normal, and so are all three families of the algebra, $L_{\tilde{Q}}$, $R_{\tilde{Q}}$ and $\Theta_{\tilde{Q}}$, because the Krein adjoint of a left multiplication is a right one and the two commute. The map $T\mapsto JT$ puts the $J$-self-adjoint operators in bijection with the Hermitian ones, and when $JT$ is positive definite the operator is diagonalisable with real spectrum; positive semidefiniteness is not enough, since $T=J|\tilde{Q}\rangle\langle\tilde{Q}|$ with $\tilde{Q}$ isotropic is $J$-self-adjoint, $J$-nonnegative and nilpotent. The spectra carry two symmetries: conjugation for $J$-self-adjoint operators and inversion $\lambda\mapsto1/\bar\lambda$ for $J$-unitary ones, without confinement to the unit circle. The analogy with the definite theory stops at normality: $L_{e_1}$ is $J$-normal with the non-real spectrum $\{i,-i\}$, so normality, self-adjointness and reality of the spectrum are independent conditions here.

Summary of Notation

Symbol	Meaning
$T^{\dagger}=JT^{*}J$	The Krein adjoint
$T^{\dagger}T=TT^{\dagger}$	$J$-normality
$T^{}ETE=ETET^{}$, or $[T^{*},JTJ]=0$	$J$-normality in the coefficient basis
$T\mapsto JT$	The bijection with the Hermitian operators
$\mathrm{spec}(T)=\overline{\mathrm{spec}(T)}$	The symmetry for $J$-self-adjoint $T$
$\lambda\leftrightarrow1/\bar\lambda$	The symmetry for $J$-unitary $T$
$L_{\tilde{Q}}$, $R_{\tilde{Q}}$, $\Theta_{\tilde{Q}}$	The three $J$-normal families
$J\lvert\tilde{Q}\rangle\langle\tilde{Q}\rvert$ with $\tilde{Q}=e_0+e_1$	$J$-self-adjoint, $J$-nonnegative, nilpotent