The Hermitian Sandwich on a Krein Space

Introduction

On a Krein space $K$ with fundamental symmetry $J$ and indefinite adjoint $\dagger$, the Hermitian sandwich of a bounded operator $T$ by an operator $Q$ is

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

the two-sided operator built from $Q$ and its indefinite adjoint. It is the indefinite counterpart of the sandwich $\Theta_x$ of a Hilbert algebra, and it is the operation that transports operators along the symmetry group of the form: for a $J$-unitary $Q$ the sandwich is the inner automorphism $\mathrm{Ad}_Q$, it preserves the indefinite form and the $J$-positivity, and it acts on self-adjointness and on the kernel and the image in a way that can be computed exactly.

The contrast with the Hilbert case is the interesting part. In Hilbert space the sandwich by a unitary $U$ is a unitary equivalence, so it preserves the Hilbert norm, the spectrum, the positive cone and every spectral subspace. In a Krein space the sandwich by a $J$-unitary $Q$ preserves the indefinite form but not the Hilbert inner product, because a $J$-unitary operator need not be Hilbert-unitary: it is a form-isometry of the indefinite geometry, and the Hilbert norm of the sandwich generally changes. What is preserved is the indefinite geometry — the form, the $J$-positivity, the $J$-self-adjointness, the $J$-unitarity — and what is not preserved is the Hilbert metric; the sandwich therefore fails to transport the spectrum, and this failure is the reason the indefinite theory needs the modular and spectral tools of Spectral Theory on Krein Spaces rather than the Hilbert spectral theorem.

This article fixes the dagger sandwich, its adjoint and composition laws, the preservation of the form and of $J$-positivity, the kernel and the image, the automorphism statement on the unitary slice, and the contrast with the Hilbert case.

The indefinite adjoint is J-Self-Adjoint and J-Unitary Operators; the fundamental symmetry and the form are The Fundamental Symmetry and Krein Spaces; the $J$-positive cone and the order are The J-Positive Cone and the J-Order and Krein Algebras; the Hilbert-algebra sandwich is The Adjoint of the Sandwich on a Hilbert Algebra and The Hermitian Sandwich on a Hilbert Algebra with Hermitian Adjoint; the spectral consequences are Spectral Theory on Krein Spaces. Those are cited. The space is $K$ with form $[\cdot,\cdot]$, fundamental symmetry $J$, Hilbert adjoint $*$ and indefinite adjoint $\dagger$.

The Dagger Sandwich

Definition. For bounded operators $Q, T$ the Hermitian sandwich (or dagger sandwich) is

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

Proposition (adjoint and composition laws). For all bounded $Q, S, T$,

$$ H_{Q}(T)^{\dagger} = H_{Q}(T^{\dagger}) , \qquad H_{Q}(H_{S}(T)) = H_{QS}(T) , \qquad H_{Q}(ST) = H_{Q}(S)H_{Q}(T) \text{ if } Q^{\dagger}Q = 1 . $$

Proof. $(QTQ^{\dagger})^{\dagger} = Q^{\dagger\dagger}T^{\dagger}Q^{\dagger} = QT^{\dagger}Q^{\dagger}$; the composition is associativity; the multiplicativity uses $Q^{\dagger}Q = 1$ to cancel the middle pair.

Proposition (the parameter rule). For a scalar $\alpha$ one has $H_{\alpha Q}(T) = |\alpha|^{2}H_{Q}(T)$; the sandwich is a semilinear, homogeneous quadratic operation in the parameter and linear in the argument.

Proof. $(\alpha Q)^{\dagger} = \bar\alpha Q^{\dagger}$ and $(\alpha Q)T(\bar\alpha Q^{\dagger}) = |\alpha|^{2}QTQ^{\dagger}$.

Remark (the Hilbert-algebra sandwich is the definite case). When $J = \mathrm{id}$ the indefinite adjoint is the Hilbert adjoint, the dagger sandwich is the sandwich $\Theta_Q(T) = QTQ^{*}$ of a Hilbert algebra, and every statement below reduces to the Hilbert one; the indefinite theory is the same family of operations read with a form of nonzero rank.

Preservation of the Form and of J-Positivity

Theorem (form preservation by $J$-unitary parameters). Let $Q$ be $J$-unitary, $Q^{\dagger}Q = QQ^{\dagger} = 1$. Then for every $T$ and all $x, y$

$$ [H_{Q}(T)x, H_{Q}(T)y] = [TQ^{\dagger}x, TQ^{\dagger}y] , \qquad [H_{Q}(T)x,y] = [TQ^{\dagger}x, Q^{\dagger}y] , $$

and $H_{Q}(T)$ is $J$-isometric whenever $T$ is; in particular the $J$-unitary group acts on the operators by sandwiches preserving the indefinite geometry.

Proof. The elementary rule is $[Qx,y] = [x,Q^{\dagger}y]$, which follows from $[x,y] = \langle Jx,y\rangle$ and $(Qx,y) = (x,Q^{*}y)$. With $x$ replaced by $TQ^{\dagger}x$ it gives $[QTQ^{\dagger}x,y] = [TQ^{\dagger}x,Q^{\dagger}y]$, the second identity; the first follows by applying the rule once more with $y$ replaced by $H_{Q}(T)y$. If $T$ is $J$-isometric then $[TQ^{\dagger}x,TQ^{\dagger}y] = [Q^{\dagger}x,Q^{\dagger}y]$, and $Q^{\dagger}$ is $J$-unitary with $Q$, so this is $[x,y]$.

Proposition (preservation of $J$-self-adjointness and $J$-positivity). If $T$ is $J$-self-adjoint then $H_{Q}(T)$ is $J$-self-adjoint; if $T$ is $J$-positive and $Q$ is $J$-unitary then $H_{Q}(T)$ is $J$-positive; and if $T$ is $J$-unitary then $H_{Q}(T)$ is $J$-unitary.

Proof. The first statement is $H_Q(T)^{\dagger} = H_Q(T^\dagger) = H_Q(T)$ for $J$-self-adjoint $T$; for the second, $[H_Q(T)x,x] = [TQ^{\dagger}x,Q^{\dagger}x]\geq0$ by the second identity of the theorem and the $J$-positivity of $T$; the third is the isometry statement applied twice.

Proposition (the $J$-adjoint sandwich). With $Q = J$ one has $J^{\dagger} = J$ and $H_{J}(T) = JTJ$; this sandwich is the dictionary between the Hilbert data and the indefinite ones: $H_{J}(T)$ is Hilbert-self-adjoint exactly when $T$ is, and $H_{J}(T)\geq0$ in the Hilbert sense exactly when $T\geq0$. The link with the indefinite notions is the dictionary $T$ is $J$-self-adjoint $\iff$ $JT$ is Hilbert-self-adjoint, and in that case $H_{J}(T) = JTJ = T^{*}$.

Proof. $H_J(T) = JTJ$ and $(JTJ)^{*} = JT^{*}J$, so $H_J(T)$ is Hilbert-self-adjoint iff $T^{*}=T$; further $\langle JTJx,x\rangle = \langle T(Jx),Jx\rangle$ and $J$ is surjective, so $H_J(T)\geq0$ iff $T\geq0$. Finally $T$ is $J$-self-adjoint iff $JT^{*}J=T$, i.e. $T^{*}=JTJ=H_J(T)$, iff $JT^{*} = TJ$, which says $JT$ is Hilbert-self-adjoint.

The Kernel and the Image

Theorem (kernel and image for invertible parameters). Let $Q$ be invertible. Then for every $T$

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

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

Proof. $QTQ^{\dagger}x = 0 \iff TQ^{\dagger}x = 0$, so $x\in\ker H_Q(T) \iff Q^{\dagger}x\in\ker T$; the image statement is the same computation with the roles reversed, and $Q$ maps the image of $T$ onto the image of the sandwich.

Corollary (rank and index). For $J$-unitary $Q$ the sandwich preserves the dimension of the kernel, the codimension of the image and hence the Fredholm index of $T$; it does not preserve the Hilbert-orthogonal complement of the image, only the $J$-orthogonal one.

Proof. $Q$ is an isomorphism of the vector space $K$ and an isometry of the indefinite form; the kernel and image statements are the theorem, and the orthogonality statement is that $Q$ preserves $[\cdot,\cdot]$ but not $\langle\cdot,\cdot\rangle$.

Remark (what is preserved and what is not). The sandwich by a $J$-unitary $Q$ preserves the kernel dimension, the range and its codimension, and the indefinite geometry; it does not preserve the Hilbert norm, the Hilbert-orthogonal complements, the spectrum or the Hilbert positive cone. This is the exact sense in which the indefinite sandwich is a symmetry of the form and not of the metric.

The Unitary Slice and Automorphisms

Theorem (the sandwich is the inner automorphism on the $J$-unitary group). For $J$-unitary $Q$ the map $T\mapsto H_{Q}(T) = QTQ^{\dagger}$ is an algebra automorphism of $B(K)$ preserving the $J$-adjoint operation, with inverse $H_{Q^{\dagger}}$; the assignment $Q\mapsto H_{Q}$ is a group homomorphism from the $J$-unitary group $\mathcal{U}_{J}(K)$ into $\mathrm{Aut}(B(K))$ with kernel the scalars of modulus one.

Proof. Multiplicativity and preservation of the $J$-adjoint are the laws of the first section with $Q^{\dagger}Q = 1$; the inverse statement is $H_{Q}H_{Q^{\dagger}} = H_{QQ^{\dagger}} = \mathrm{id}$; the kernel statement is that $QTQ^{\dagger} = T$ for all $T$ forces $Q$ to be a scalar.

Proposition (what the automorphism does to the $J$-self-adjoint part). The automorphism $H_{Q}$ maps the real space of $J$-self-adjoint operators onto itself and the $J$-positive cone onto itself; on the $J$-unitary group it is the conjugation $U\mapsto QUQ^{\dagger}$, which is the defining action of $\mathcal{U}_{J}(K)$ on itself.

Proof. Preservation of $J$-self-adjointness and of $J$-positivity is the theorem of the previous section; the action on the unitary group is multiplicativity.

Remark (the sandwich generates the inner symmetries). Every $J$-unitary equivalence of operators is a sandwich, and every sandwich by a $J$-unitary element is a symmetry of the indefinite geometry. The inner symmetries of the theory are therefore exactly the sandwiches, and the outer ones — the modular and spectral operations that do not come from an element — are what the theory cannot express as a sandwich; this is the same dichotomy as in the Hilbert-algebra case, where the sandwiches give the inner automorphisms and the modular flow is generally outer.

Contrast with the Hilbert Case

Proposition (Hilbert space). If $J = \mathrm{id}$ then $K$ is a Hilbert space, every $Q$ with $Q^{*}Q = QQ^{*} = 1$ is unitary and $H_{Q}(T) = QTQ^{*}$; the sandwich is a unitary equivalence, so it preserves the Hilbert norm, the spectrum, the Hilbert positive cone, the Hilbert-orthogonal complements and the whole Hilbert spectral theory.

Proof. With $J = \mathrm{id}$ the indefinite adjoint is the Hilbert adjoint and the statements are the definitions of unitary equivalence.

Theorem (Krein space: the failures). For a $J$-unitary $Q$ that is not Hilbert-unitary, the sandwich $H_{Q}$ does not preserve the Hilbert norm, the Hilbert-orthogonal direct sum decompositions, the Hilbert positive cone or the spectrum; it preserves the indefinite form, the $J$-self-adjointness, the $J$-positivity, the $J$-unitarity and the kernel and image of every operator.

Proof. A $J$-unitary $Q$ satisfies $Q^{*}JQ = J$ and is Hilbert-unitary only if moreover $Q^{*}Q = 1$; when it is not, the sandwich changes the Hilbert norm, and the spectrum is not preserved because a $J$-self-adjoint operator has conjugate-symmetric spectrum while its sandwich by a merely form-preserving $Q$ need not; the positive statements are the theorems above.

Remark (why the contrast matters). The Hilbert sandwich is the natural notion of equivalence in Hilbert space, and it preserves everything. The indefinite sandwich is the natural notion of equivalence in a Krein space, and it preserves only the indefinite structure; the resulting equivalence relation on $J$-self-adjoint operators is coarser than similarity, and the classification of $J$-self-adjoint operators up to it is the classification of their invariants under form-preserving change of basis. This is why the spectral theory of Spectral Theory on Krein Spaces cannot be imported from the Hilbert theory by transport along a sandwich.

Worked Cases

The Fundamental Symmetry

For $Q = J$ the sandwich is $H_{J}(T) = JTJ$; it is Hilbert-self-adjoint exactly when $T$ is and Hilbert-positive exactly when $T$ is, $H_{J}$ is an involution on the space of operators, and for $T$ $J$-self-adjoint the sandwich is the Hilbert adjoint, $H_{J}(T)=T^{*}$. This is the translation dictionary between the indefinite and the Hilbert descriptions, read as a sandwich.

Matrices

For $K = \mathbb{C}^{1,1}$ with $J = \mathrm{diag}(1,-1)$ and $Q = \left(\begin{smallmatrix}\cosh t & \sinh t\\ \sinh t & \cosh t\end{smallmatrix}\right)$, the operator $Q$ is $J$-unitary but not unitary; the sandwich $H_{Q}(J)$ is a $J$-self-adjoint operator with Hilbert norm different from that of $J$, and its Hilbert spectrum differs from $\{-1,+1\}$ while its indefinite spectrum is the same.

The Definite Case

For $J = \mathrm{id}$ the sandwich by a unitary $Q$ is conjugation by a unitary, and kernel, image, spectrum, norm and positive cone are all preserved; the contrast of the previous section collapses.

Summary

The Hermitian sandwich on a Krein space is $H_{Q}(T) = QTQ^{\dagger}$, with $Q^{\dagger} = JQ^{*}J$; it satisfies $H_Q(T^{\dagger}) = H_Q(T)^{\dagger}$, $H_Q(H_S(T)) = H_{QS}(T)$, and it is multiplicative in the argument exactly when $Q$ is $J$-isometric. For $J$-unitary $Q$ the sandwich preserves the indefinite form in the twisted sense $[H_Q(T)x,H_Q(T)y] = [TQ^{\dagger}x,TQ^{\dagger}y]$, it preserves $J$-self-adjointness, $J$-positivity and $J$-unitarity, and it is an algebra automorphism of $B(K)$; the kernel and the image are transported exactly, $\ker H_Q(T) = Q\ker T$ and $\mathrm{im}\,H_Q(T) = Q\,\mathrm{im}\,T$, so the kernel dimension, the range and the index are preserved. The contrast with the Hilbert case is that a $J$-unitary operator need not be Hilbert-unitary: the sandwich preserves the indefinite geometry — form, positivity, self-adjointness, kernel and image — but not the Hilbert norm, the Hilbert-orthogonal decompositions, the Hilbert positive cone or the spectrum, so the indefinite sandwich is a symmetry of the form rather than of the metric and cannot transport the Hilbert spectral theory. The indefinite adjoint is J-Self-Adjoint and J-Unitary Operators, the form and the symmetry are Krein Spaces and The Fundamental Symmetry, the positivity is Krein Algebras and The J-Positive Cone and the J-Order, the Hilbert-algebra sandwich is The Adjoint of the Sandwich on a Hilbert Algebra, and the spectral consequences are Spectral Theory on Krein Spaces.

Summary of Notation

Symbol	Meaning
$H_{Q}(T) = QTQ^{\dagger}$	The Hermitian sandwich
$Q^{\dagger} = JQ^{*}J$	The indefinite adjoint of the parameter
$H_{Q}(T^{\dagger}) = H_{Q}(T)^{\dagger}$	The sandwich commutes with adjunction
$[H_{Q}(T)x,y] = [TQ^{\dagger}x,Q^{\dagger}y]$	Form preservation for $J$-unitary $Q$
$J$-positivity preserved	$[H_Q(T)x,x] = [TQ^{\dagger}x,Q^{\dagger}x]$
$\ker H_{Q}(T) = Q\ker T$	Kernel for $J$-unitary $Q$
$\mathrm{im}\,H_{Q}(T) = Q\,\mathrm{im}\,T$	Image for $J$-unitary $Q$
$J = \mathrm{id}$	Hilbert case, where the sandwich is unitary equivalence