J-Self-Adjoint and J-Unitary Operators

Introduction

On a Krein space $K$ with fundamental symmetry $J$, form $[x,y] = \langle Jx,y\rangle$ and Hilbert adjoint $*$, the adjoint for the indefinite form — written $\dagger$ and called the $J$-adjoint — is $T^{\dagger} = JT^{*}J$. It is the unique operator satisfying $[Tx,y] = [x,T^{\dagger}y]$ for all vectors, it is involutive, and it reverses products, $ (ST)^{\dagger} = T^{\dagger}S^{\dagger}$. An operator is $J$-self-adjoint when $T^{\dagger} = T$, $J$-unitary when $T^{\dagger}T = TT^{\dagger} = 1$, and $J$-normal when $T^{\dagger}T = TT^{\dagger}$; these are the indefinite counterparts of self-adjointness, unitarity and normality, and the whole theory of the group rests on the translation between them and their Hilbert analogues.

The translation is simple and useful: $T$ is $J$-self-adjoint exactly when $JT$ is Hilbert-self-adjoint; $T$ is $J$-isometric exactly when it preserves the indefinite form, $[Tx,Ty] = [x,y]$; and $T$ is a $J$-projection exactly when it is a Hilbert projection commuting with $J$. What is not simple is the spectral consequence: a $J$-self-adjoint operator need not have real spectrum, and the failure is the entrance to the spectral theory of Spectral Theory on Krein Spaces. The $J$-positive cone, $\{T = T^{\dagger} : [Tx,x]\geq0\}$, is strictly larger than the Hilbert positive cone and contains $J$ itself, so positivity in the indefinite sense is a genuinely weaker notion.

This article fixes the $J$-adjoint, $J$-self-adjointness, $J$-isometry and $J$-unitarity, $J$-projections and $J$-normality, the $J$-positive cone, and the failure of the real spectrum in general.

The Krein space, the fundamental symmetry and the form are Krein Spaces and The Fundamental Symmetry; the $J$-positive cone and the order are Krein Algebras and The J-Positive Cone and the J-Order; the spectral theory is Spectral Theory on Krein Spaces; the boundedness of $J$ and the topology of $K$ are Indefinite Inner Product Spaces. Those are cited. The space is $K$ with form $[\cdot,\cdot]$, fundamental symmetry $J$, Hilbert adjoint $*$ and indefinite adjoint $\dagger$.

The Indefinite Adjoint

Definition. The $J$-adjoint of a bounded operator $T$ is $T^{\dagger} = JT^{*}J$.

Proposition (characterisation and calculus). $T^{\dagger}$ is the unique bounded operator with

$$ [Tx,y] = [x,T^{\dagger}y] \qquad \text{for all } x, y\in K ; $$

it satisfies $(T^{\dagger})^{\dagger} = T$, $(ST)^{\dagger} = T^{\dagger}S^{\dagger}$, $(\alpha T + \beta S)^{\dagger} = \bar\alpha T^{\dagger} + \bar\beta S^{\dagger}$, and $\|T^{\dagger}\| = \|T\|$ for the Hilbert norm.

Proof. $[Tx,y] = \langle JTx,y\rangle = \langle x,T^{*}Jy\rangle = \langle x,J(JT^{*}J)y\rangle = [x,T^{\dagger}y]$; uniqueness follows because the form is nondegenerate; the calculus is the corresponding calculus of the Hilbert adjoint conjugated by $J$.

Proposition (the $J$-adjoint is a Hilbert adjoint in disguise). $T$ is $J$-self-adjoint if and only if $JT$ is Hilbert-self-adjoint; $T$ is $J$-isometric if and only if $T$ preserves the form, $T^{*}JT = J$, and for a bounded $T$ on a Krein space this is already equivalent to $T$ being $J$-unitary.

Proof. $T^{\dagger} = JT^{*}J$, so $T$ is $J$-self-adjoint, $T = JT^{*}J$, exactly when $JT = J^{2}T^{*}J = T^{*}J$; and $(JT)^{*} = T^{*}J^{*} = T^{*}J$, so the condition is $JT = (JT)^{*}$, that is, $JT$ Hilbert-self-adjoint. For the second statement, the form is preserved, $[Tx,Ty] = [x,y]$ for all $x,y$, exactly when $T^{\dagger}T = 1$, by nondegeneracy; and $T^{\dagger}T = JT^{*}JT = 1$ is equivalent, after multiplication on the left by $J$, to $T^{*}JT = J$. A bounded operator on a Banach space that has a one-sided inverse is invertible, so $T^{\dagger}T = 1$ also gives $TT^{\dagger} = 1$, and a $J$-isometry is $J$-unitary.

Remark (the adjoint depends on $J$). The indefinite adjoint is not intrinsic to the topology of the Krein space: changing the fundamental symmetry changes the form and with it the adjoint. This is the first place where the choice of $J$ is visible at the level of operators, and it is the reason the whole theory of the group is a theory of the pair (Krein space, fundamental symmetry).

J-Self-Adjointness

Definition. $T$ is $J$-self-adjoint when $T^{\dagger} = T$; it is $J$-skew-adjoint when $T^{\dagger} = -T$; and every $T$ decomposes as $T = \frac{1}{2}(T + T^{\dagger}) + \frac{1}{2}(T - T^{\dagger})$ into a $J$-self-adjoint and a $J$-skew-adjoint part.

Proposition (the real space of $J$-self-adjoint operators). The $J$-self-adjoint operators form a real vector space, closed under the $J$-adjoint; they are exactly the operators $T$ with $JT$ Hilbert-self-adjoint; and $T^{\dagger}T$ is $J$-self-adjoint for every $T$.

Proof. $T^{\dagger} = T$ is a real-linear condition; the identification with $JT = (JT)^{*}$ is the previous proposition; and $(T^{\dagger}T)^{\dagger} = T^{\dagger}(T^{\dagger})^{\dagger} = T^{\dagger}T$.

Proposition (the quadratic form is real but not sign definite). If $T$ is $J$-self-adjoint then $[Tx,x]$ is real for every $x$, and the numerical range of $T$ is symmetric with respect to the real axis; but $[Tx,x]$ need not have a constant sign, and the range can contain both signs.

Proof. $[Tx,x] = \overline{[x,Tx]} = \overline{[T^{\dagger}x,x]} = \overline{[Tx,x]}$ by $J$-self-adjointness; the indefiniteness is that of the form itself, since $T = J$ is $J$-self-adjoint with $[Jx,x] = \langle x,x\rangle>0$ for $x\neq0$ while $T = -J$ is also $J$-self-adjoint with $[-Jx,x]<0$.

J-Isometry, J-Unitarity, J-Projection, J-Normality

Definition. $T$ is $J$-isometric when $T^{\dagger}T = 1$, $J$-unitary when $T^{\dagger}T = TT^{\dagger} = 1$, $J$-normal when $T^{\dagger}T = TT^{\dagger}$, and a $J$-projection when $P^{2} = P = P^{\dagger}$.

Theorem (isometry is form preservation). $T$ is $J$-isometric if and only if $[Tx,Ty] = [x,y]$ for all $x, y$; $T$ is $J$-unitary if and only if it is $J$-isometric and surjective. The $J$-unitary operators form a group, the $J$-unitary group $\mathcal{U}_{J}(K)$, and the $J$-isometric operators are the isometries of the form.

Proof. $[Tx,Ty] = [x,T^{\dagger}Ty]$, so preservation of the form is $T^{\dagger}T = 1$; surjectivity of a $J$-isometry gives an inverse which is $J$-isometric by the same identity applied to the inverse, and the group axioms follow from the composition rules.

Theorem ($J$-projections are the $J$-orthogonal projections onto non-degenerate subspaces). An operator $P$ is a $J$-projection if and only if it is an idempotent whose kernel is the $J$-orthogonal complement of its range, $\ker P = (\mathrm{ran}\,P)^{\perp_{J}}$; equivalently, if and only if $\mathrm{ran}\,P$ is a non-degenerate subspace and $P$ is the projection onto it along $(\mathrm{ran}\,P)^{\perp_{J}}$. The $J$-projections that are in addition orthogonal for the Hilbert structure, equivalently those commuting with $J$, are the special case in which the range is $J$-invariant.

Proof. If $P = P^{2} = P^{\dagger}$, take $x = Pu \in \mathrm{ran}\,P$ and $y = (1-P)v \in \ker P$: then $[x,y] = [Pu,(1-P)v] = [P^{\dagger}Pu,(1-P)v] = [Pu,(P-P^{2})v] = 0$, so $\mathrm{ran}\,P \perp_{J} \ker P$. If $x \perp_{J} \mathrm{ran}\,P$, then $[Px,u] = [x,P^{\dagger}u] = [x,Pu] = 0$ for every $u$, so $Px = 0$ by the non-degeneracy of the form; hence $\ker P = (\mathrm{ran}\,P)^{\perp_{J}}$, and $\mathrm{ran}\,P \cap \ker P = \{0\}$ makes the range non-degenerate. Conversely, if $\mathrm{ran}\,P$ is non-degenerate and $P$ is the projection onto it along $(\mathrm{ran}\,P)^{\perp_{J}}$, so that $K = \mathrm{ran}\,P \oplus \ker P$ is $J$-orthogonal, then $P^{2} = P$ and $[Px,y] = [Px,Py] = [x,Py]$ for all $x, y$, the middle step because the two factors are totally $J$-orthogonal to each other; hence $P^{\dagger} = P$. Finally, $P$ is a Hilbert-orthogonal projection exactly when $\ker P = (\mathrm{ran}\,P)^{\perp}$, and since $(\mathrm{ran}\,P)^{\perp_{J}} = J(\mathrm{ran}\,P)^{\perp}$, this holds exactly when $(\mathrm{ran}\,P)^{\perp}$ is $J$-invariant, equivalently when the range is; then $P$ commutes with $J$.

Remark (the old form of the theorem was false). The statement "$J$-projections are the orthogonal projections with $PJ = JP$" fails: in $\mathbb{C}^{1,1}$ with $J = \operatorname{diag}(1,-1)$ the $J$-orthogonal projection onto the line $\mathbb{C}(1,\tfrac12)$ is $P = \tfrac{4}{3}\left(\begin{smallmatrix}1&-\tfrac12\\\tfrac12&-\tfrac14\end{smallmatrix}\right)$, which satisfies $P^{2} = P$ and $P^{\dagger} = P$ but neither $PJ = JP$ nor $P^{*} = P$. The biquaternion instance is recomputed in Indefinite Positivity and the Krein Cone of the Biquaternion Algebra and in Krein Orthogonality and the Fundamental Decomposition.

Proposition (normality). $T$ is $J$-normal when $T^{\dagger}T = TT^{\dagger}$, that is when $JT^{*}JT = TJT^{*}J$. The $J$-self-adjoint and the $J$-unitary operators are $J$-normal; the class is closed under the $J$-adjoint; and an invertible $J$-normal operator has $T^{\dagger}T^{-1}$ $J$-unitary.

Proof. The inclusions are the definitions; for the invertible case, $(T^{\dagger}T^{-1})^{\dagger}(T^{\dagger}T^{-1}) = (T^{\dagger})^{-1}T^{\dagger\dagger}T^{\dagger}T^{-1} = (T^{\dagger})^{-1}TT^{\dagger}T^{-1} = (T^{\dagger})^{-1}T^{\dagger}TT^{-1} = 1$ using $TT^{\dagger} = T^{\dagger}T$, and the reverse product is analogous, so the operator is $J$-unitary.

Remark (what normality is not). $J$-normality is not the same as the Hilbert normality of $JT$: the two conditions are $JT^{*}JT = TJT^{*}J$ and $T^{*}T = JTT^{*}J$, and neither implies the other. This is a first sign that the indefinite theory keeps the formal shape of the Hilbert theory without transporting all of its consequences.

The J-Positive Cone

Definition. The $J$-positive cone is

$$ \{T : T = T^{\dagger},\ [Tx,x]\geq0 \text{ for every } x\in K\} . $$

Proposition (translation to Hilbert positivity). $T$ is $J$-positive if and only if $T = T^{\dagger}$ and $JT\geq0$ (Hilbert-positive). So the $J$-positive cone is the image under the map $T\mapsto JT$ of the Hilbert positive cone intersected with the $J$-self-adjoint condition.

Proof. $[Tx,x] = \langle JTx,x\rangle$, and $JT$ is self-adjoint exactly when $T$ is $J$-self-adjoint, so the condition is the Hilbert-positivity of $JT$.

Proposition (the cone is larger and contains $J$). The $J$-positive cone contains the Hilbert positive cone; it strictly contains it when $\kappa>0$; and $J$ itself is $J$-positive, since $[Jx,x] = \langle x,x\rangle\geq0$, while $J$ is not Hilbert-positive when $\kappa>0$.

Proof. If $T\geq0$ in the Hilbert sense and $T = T^{\dagger}$ then $JT$ need not be positive; but for $T$ proportional to a $J$-projection commuting with $J$ the two notions agree; the element $J$ has $J^{\dagger} = J$ and $J\cdot J = 1\geq0$, so $J$ is $J$-positive, and its Hilbert spectrum contains $-1$.

Remark (positivity is J-relative). The $J$-positive cone depends on the fundamental symmetry and is not comparable with the Hilbert positive cone beyond the containment just computed; in particular the $J$-positive elements do not form a cone that doubles as an operator-theoretic positivity in the Hilbert sense. This is the reason the spectral theory of $J$-self-adjoint operators has no analogue of the spectral theorem's ordering, and it is taken up in Spectral Theory on Krein Spaces and Definitizable Operators and the Krein–Naĭmark Theorem.

The Failure of the Real Spectrum

Proposition (a $J$-self-adjoint operator need not have real spectrum). In the Krein space $\mathbb{C}^{1,1}$ with $J = \mathrm{diag}(1,-1)$ the operator

$$ T = \begin{pmatrix} 0 & 1 \\ -1 & 0\end{pmatrix} $$

is $J$-self-adjoint, since $JT = \begin{pmatrix}0 & 1\\ 1 & 0\end{pmatrix}$ is symmetric, and its spectrum is $\{i, -i\}$: non-real.

Proof. Direct computation of $JT$ and of the characteristic polynomial $\lambda^{2} + 1$.

Proposition (conjugate symmetry). The spectrum of a $J$-self-adjoint operator is symmetric with respect to the real axis: $\lambda\in\sigma(T)$ implies $\bar\lambda\in\sigma(T)$, with the same multiplicity; and the resolvent is $J$-self-adjoint at points where it is defined.

Proof. The conjugate of $T$ is similar to $T$: $\bar T = JTJ^{-1}$ for real $J$, so $\sigma(\bar T) = \sigma(T)$; the resolvent statement is $(\lambda - T)^{\dagger} = \bar\lambda - T$.

Remark (what replaces the spectral theorem). For a Hilbert-self-adjoint operator the spectrum is real and the spectral theorem produces a functional calculus. For a $J$-self-adjoint operator the spectrum is only conjugate-symmetric, the real points need not be semi-bounded, and complex points occur in conjugate pairs; the spectral theory is therefore a theory of invariant subspaces and root subspaces rather than of projections, and it is the subject of the next articles, Spectral Theory on Krein Spaces and Definitizable Operators and the Krein–Naĭmark Theorem.

Worked Cases

The Fundamental Symmetry

$J$ is $J$-self-adjoint and $J$-unitary, since $J^{\dagger} = JJ^{*}J = J$ and $J^{2} = 1$; its Hilbert spectrum contains $-\lambda$ for each $\lambda$ in the negative part of the fundamental decomposition, which shows how far the indefinite spectral data is from the Hilbert one.

The Rotated Multiplications

For $A = M_n(\mathbb{C})$ with the indefinite form defined by $J$, the $J$-self-adjoint elements are the solutions of $JT^{*}J = T$, the $J$-unitary ones solve $T^{*}JT = J$, and the trace of a $J$-positive element is real but need not be positive for the Hilbert form; these are the $J$-analogues of the Hermitian matrices and the unitary group, and they generate the $J$-version of the matrix algebra.

The One-Dimensional Case

For $K$ of dimension one with the definite form the $J$-adjoint is the Hilbert adjoint and the theory reduces to the Hilbert one; the failure of the real spectrum and the enlargement of the cone both require $\kappa>0$, so the indefinite theory is invisible in a definite space.

Summary

The $J$-adjoint on a Krein space with fundamental symmetry $J$ is $T^{\dagger} = JT^{*}J$, characterised by $[Tx,y] = [x,T^{\dagger}y]$ and reversing products; an operator is $J$-self-adjoint when $T^{\dagger} = T$, equivalently when $JT$ is Hilbert-self-adjoint, $J$-isometric when $T^{\dagger}T = 1$, equivalently when it preserves the form, $J$-unitary when $T^{\dagger}T = TT^{\dagger} = 1$, equivalently when it is a surjective form-isometry, $J$-normal when $T^{\dagger}T = TT^{\dagger}$, and a $J$-projection when $P^{2} = P = P^{\dagger}$, equivalently when $P$ is an orthogonal projection commuting with $J$. The $J$-positive cone is $\{T = T^{\dagger} : JT\geq0\}$; it contains the Hilbert positive cone, it strictly contains it when $\kappa>0$, and it contains $J$ itself, so indefinite positivity is genuinely weaker than Hilbert positivity. Unlike a self-adjoint operator, a $J$-self-adjoint operator need not have real spectrum — the operator $\left(\begin{smallmatrix}0&1\\-1&0\end{smallmatrix}\right)$ in $\mathbb{C}^{1,1}$ has spectrum $\{i,-i\}$ — the spectrum is only conjugate-symmetric, and the resulting theory of invariant and root subspaces is Spectral Theory on Krein Spaces and Definitizable Operators and the Krein–Naĭmark Theorem. The space and the symmetry are Krein Spaces and The Fundamental Symmetry, the positivity is Krein Algebras and The J-Positive Cone and the J-Order, and the adjoint for the form is Indefinite Inner Product Spaces.

Summary of Notation

Symbol	Meaning
$T^{\dagger} = JT^{*}J$	The $J$-adjoint
$[Tx,y] = [x,T^{\dagger}y]$	Characterisation of the $J$-adjoint
$T^{\dagger} = T \iff JT = (JT)^{*}$	$J$-self-adjointness
$T^{\dagger}T = 1 \iff [Tx,Ty] = [x,y]$	$J$-isometry
$\mathcal{U}_{J}(K)$	The $J$-unitary group
$P^{2} = P = P^{\dagger} \iff PJ = JP$	$J$-projection
$T$ $J$-positive $\iff JT\geq0$	The $J$-positive cone
$\left(\begin{smallmatrix}0&1\\-1&0\end{smallmatrix}\right)$	$J$-self-adjoint with spectrum $\{i,-i\}$