Krein Algebras

Introduction

A Hilbert algebra carries an involution and a positive definite form tied to the product by the adjoint axiom. A Krein algebra is the same data with the positivity removed: a $\ast$-algebra with an indefinite sesquilinear form that the involution still turns into the adjoint of multiplication. The form is allowed to take both signs, so the algebra has neutral and negative elements, and the positivity returns only through the choice of a fundamental symmetry $J$.

The construction is then exactly the construction of The Fundamental Symmetry, carried inside the algebra: if $J$ is an involutive operator of the algebra, self-adjoint for the indefinite form and positive, and if moreover $J$ is compatible with the involution, then $\langle x,y\rangle = [Jx,y]$ is a positive definite form and the algebra becomes a Hilbert algebra; conversely every Krein algebra acquires its Hilbert algebras by choosing such a $J$. This is the sense in which the Krein algebra is not a new kind of object but a Hilbert algebra together with a chosen symmetry: the Hilbert algebra is the case $J = \mathrm{id}$.

What the choice of $J$ adds is an order. An element is $J$-positive when it is Hermitian and $[xv,v]\geq0$ for every $v$, that is when its left multiplication is $J$-self-adjoint and $JL_x$ is positive for the definite form; the resulting cone is convex, proper, and stable under the inner conjugation, but it is not generated by the elements $y^{\dagger}y$, because a scalar square may be negative. This article fixes the axioms, the $J$-positivity condition, the passage to Hilbert algebras through $J$, the $J$-positive cone and the $J$-order.

The Hilbert algebras, the involution, the adjoint axiom, the trace form and the positive cone are Hilbert Algebras; the indefinite form and its fundamental decomposition are Indefinite Inner Product Spaces and The Fundamental Symmetry; the representational use of the construction – the indefinite GNS construction and the operator algebras – is The Indefinite GNS Construction and Krein–von Neumann Algebras. Those are cited. The base is $\mathbb{R}$ or $\mathbb{C}$ with its conjugation, the involution is $\dagger$ and the indefinite form is $[\cdot,\cdot]$.

The Axioms

Definition. A Krein algebra is a unital associative algebra $A$ over $F$ with

an involution $\dagger$: an additive map with $(xy)^{\dagger} = y^{\dagger}x^{\dagger}$, $(x^{\dagger})^{\dagger} = x$, $1^{\dagger} = 1$;
an indefinite Hermitian form $[\cdot,\cdot]$: sesquilinear, $[x,y] = \sigma([y,x])$, non-degenerate, and indefinite;
the adjoint axiom: $[xy,z] = [y,x^{\dagger}z]$ for all $x,y,z$.

So a Krein algebra is a $\ast$-algebra with an indefinite form tied to the product exactly as the Hilbert algebra ties a positive definite form to it.

Proposition (the left regular representation). The adjoint axiom says that for each $x$ the operator $L_x$ of left multiplication has the operator $L_{x^{\dagger}}$ as its adjoint for the indefinite form:

$$ [L_xy,z] = [y,L_{x^{\dagger}}z] = [y,x^{\dagger}z] . $$

So the left regular representation is a $\ast$-representation for the indefinite form, exactly as it is for the positive definite one.

Proof. The first identity is the adjoint axiom read as an operator identity; the second is the definition of $L_{x^{\dagger}}$.

Proposition (the trace form). Let $\tau$ be a linear form with $\tau(xy) = \tau(yx)$, Hermitian in the sense $\tau(x^{\dagger}) = \sigma(\tau(x))$, and with $\tau(x^{\dagger}x)\in F$ non-zero for $x \neq 0$. Then $[x,y] = \tau(x^{\dagger}y)$ is a Hermitian form satisfying the adjoint axiom, and it is positive definite exactly when $\tau(x^{\dagger}x) > 0$ for $x\neq0$, that is when $\tau$ is positive in the sense of Hilbert Algebras. So the indefinite forms of the category are the trace forms of the non-positive traces.

Proof. The adjoint axiom is immediate, $[xy,z] = \tau(z^{\dagger}xy) = \tau((x^{\dagger}z)^{\dagger}y) = [y,x^{\dagger}z]$, and it needs only the linearity of $\tau$. Hermiticity does need the Hermitian property of $\tau$: $[y,x] = \tau(x^{\dagger}y)$ is not in general the conjugate of $[x,y] = \tau(y^{\dagger}x)$, and it is when $\tau(a^{\dagger}) = \sigma(\tau(a))$, since then $\sigma([x,y]) = \sigma(\tau(y^{\dagger}x)) = \tau((y^{\dagger}x)^{\dagger}) = \tau(x^{\dagger}y) = [y,x]$. The definiteness is the positivity of $\tau$.

The $J$-Positivity Condition

Definition. A fundamental symmetry of the algebra is an $F$-linear operator $J$ on $A$ with

$$ J^{2} = \mathrm{id}, \qquad [Jx,y] = [x,Jy], \qquad [Jx,x] > 0 \text{ for } x\neq0, \qquad J(x^{\dagger}) = (Jx)^{\dagger}, $$

the last condition being the compatibility condition: $J$ is compatible with the involution. (The name $J$-positivity is kept for the cone of the later section, where the third and the fourth conditions are both used.)

Proposition (the induced positive definite form). With $J$ as above, $\langle x,y\rangle = [Jx,y]$ is a positive definite Hermitian form and it satisfies the adjoint axiom; so $(A,\dagger,\langle\cdot,\cdot\rangle)$ is a Hilbert algebra, and the indefinite form is recovered by $[x,y] = \langle Jx,y\rangle$.

Proof. Positivity and Hermitian symmetry are The Fundamental Symmetry; the adjoint axiom is $\langle xy,z\rangle = [Jxy,z] = [y,x^{\dagger}Jz] = [y,Jx^{\dagger}z] = \langle y,x^{\dagger}z\rangle$, using $J$ compatible with $\dagger$ to move $J$ past $x^{\dagger}$.

Remark (why compatibility with $\dagger$ is the condition). Without $J(x^{\dagger}) = (Jx)^{\dagger}$ the induced form would be positive definite but would not satisfy the adjoint axiom, and the algebra would lose its $\ast$-structure; the condition is exactly what makes the transport of The Fundamental Symmetry take $\ast$-algebras to $\ast$-algebras. It also makes the $J$-positive cone of the next section a $\ast$-invariant set.

The Relation to Hilbert Algebras through $J$

Theorem (the passage). The following data are equivalent:

a Krein algebra $(A,\dagger,[\cdot,\cdot])$ together with a fundamental symmetry $J$ compatible with $\dagger$;
a Hilbert algebra $(A,\dagger,\langle\cdot,\cdot\rangle)$ together with an involutive $\langle\cdot,\cdot\rangle$-self-adjoint operator $J$ commuting with $\dagger$, such that $[x,y] := \langle Jx,y\rangle$ is indefinite.

Proof. From 1 to 2 is the previous proposition and the definition of $J$. From 2 to 1, the form $[x,y] = \langle Jx,y\rangle$ is Hermitian, non-degenerate and indefinite, and the adjoint axiom follows from the self-adjointness and the commutation with $\dagger$: $[xy,z] = \langle Jxy,z\rangle = \langle y,x^{\dagger}Jz\rangle = \langle y,Jx^{\dagger}z\rangle = [y,x^{\dagger}z]$.

Corollary (the definite case). A Krein algebra is a Hilbert algebra exactly when it admits the fundamental symmetry $J = \mathrm{id}$; in that case the indefinite form is positive definite and the two notions coincide.

Proof. $J = \mathrm{id}$ satisfies the conditions exactly when $[x,x] = \langle x,x\rangle > 0$, that is when the form is definite.

Remark (one concept once). The theorem is the reason this article does not restate the Hilbert algebra: the involution, the adjoint axiom, the trace form and the positive cone are owned by Hilbert Algebras, and this article adds only the indefinite form and the operator $J$ that reconciles the two.

The $J$-Positive Cone and the $J$-Order

Definition. An element $x \in A$ is $J$-positive when $x = x^{\dagger}$ and

$$ [xv, v] \geq 0 \qquad \text{for every } v \in A . $$

By the adjoint axiom this says that the left multiplication $L_x$ is self-adjoint for the indefinite form and that the operator $J L_x$ is positive for the definite form $\langle u,v\rangle = [Ju,v]$; this is the sense in which the positivity is a $J$-positivity. The $J$-positive cone $A_J^{+}$ is the set of $J$-positive elements, and the $J$-order is $a\leq_J b$ when $b - a\in A_J^{+}$.

Proposition (the cone is convex, invariant and proper). $A_J^{+}$ is a convex cone contained in the Hermitian elements, stable under the inner conjugation $x\mapsto y^{\dagger}xy$ for every $y$, and proper: $x\in A_J^{+}$ and $-x\in A_J^{+}$ only for $x = 0$. Hence $\leq_J$ is a partial order on the Hermitian elements of $A$.

Proof. Convexity is $[(\alpha x+\beta y)v,v] = \alpha[xv,v]+\beta[yv,v]$ for $\alpha,\beta\geq0$. For the invariance, $[(y^{\dagger}xy)v,v] = [x(yv),yv]\geq0$ by the adjoint axiom with $x = y^{\dagger}$, so $y^{\dagger}xy\in A_J^{+}$ when $x$ is. For properness, $[xv,v]\geq0$ and $[(-x)v,v] = -[xv,v]\geq0$ together give $[xv,v] = 0$ for every $v$, whence $L_x = 0$ and $x = 0$ in a unital algebra. Reflexivity and transitivity are then the axioms of the order of a cone.

Proposition (the contrast with the Hilbert case). In a Hilbert algebra the positive cone is generated by the elements $y^{\dagger}y$; in the indefinite case it is not, because $y^{\dagger}y$ need not be $J$-positive: $[y^{\dagger}y\,v,v] = [yv,yv]$ by the adjoint axiom, and the scalar square $[yv,yv]$ may be negative.

Proof. The identity is the adjoint axiom with $x = y^{\dagger}$; the negativity is the indefiniteness of the form. So the involution alone no longer produces the cone, which is why it depends on the chosen symmetry $J$.

Proposition (the bridge of orders). The $J$-order and the $J'$-order for two fundamental symmetries agree on the Hermitian elements commuting with both $J$ and $J'$, and disagree in general; the orders are parametrised by the fundamental symmetries, and the Hilbert order is the member $J = \mathrm{id}$.

Proof. On the common commutant both orders are the Hilbert order of the induced definite form; an element of the angular graph of $J'$ with respect to $J$ is positive for one and not for the other.

Remark (what the cone is for). The cone $A_J^{+}$ is the object on which the representational theory of the indefinite case is built: the $J$-positive linear functionals, the indefinite GNS construction and the Krein–von Neumann algebras all start from it, and the order it defines is the indefinite replacement of the Hilbert order of a $\ast$-algebra.

Worked Cases

The Indefinite Algebra

Let $A = \mathbb{C}^{2}$ with the pointwise product, the conjugation involution $(a,b)^{\dagger} = (\bar a,\bar b)$ and the form $[(a,b),(c,d)] = a\bar c - b\bar d$. The adjoint axiom holds, so $A$ is a Krein algebra, and the operator $J(a,b) = (a,-b)$ is a fundamental symmetry compatible with the involution.

The Failure of the Squares

In the same algebra let $y = (1,i)$, so $y^{\dagger}y = (1,1)$. The element $(1,1)$ is Hermitian but not $J$-positive: with $v = (0,1)$ one has $[(1,1)v,v] = [(0,1),(0,1)] = -1 < 0$. So $y^{\dagger}y\notin A_J^{+}$, and the involution does not generate the cone, the sharpest difference from Hilbert Algebras.

The Definite Case

If the form of the Krein algebra is positive definite, the fundamental symmetry $J = \mathrm{id}$ is available and the $J$-positive cone is the ordinary positive cone of Hilbert Algebras, generated by the squares; this is the boundary case of the theory.

Summary

A Krein algebra is a $\ast$-algebra with an indefinite Hermitian form tied to the product by the adjoint axiom $[xy,z] = [y,x^{\dagger}z]$, so that the left regular representation is a $\ast$-representation for the indefinite form; its indefinite forms are the trace forms of non-positive traces. A fundamental symmetry of the algebra is an involutive operator $J$, self-adjoint for $[\cdot,\cdot]$, positive, and compatible with the involution, $J(x^{\dagger}) = (Jx)^{\dagger}$; the last is the $J$-positivity condition, and it is exactly what makes $\langle x,y\rangle = [Jx,y]$ a Hilbert algebra form satisfying the adjoint axiom. So a Krein algebra with a compatible $J$ is the same thing as a Hilbert algebra with an involutive self-adjoint $\dagger$-commuting operator $J$, and the Hilbert algebra is the case $J = \mathrm{id}$: the indefinite theory is the definite theory with a symmetry chosen. An element is $J$-positive when it is Hermitian and $[xv,v]\geq0$ for every $v$; the $J$-positive cone $A_J^{+}$ is convex, proper, and stable under the inner conjugation $x\mapsto y^{\dagger}xy$, and the $J$-order $a\leq_J b$ is the order it defines on the Hermitian elements. Unlike the Hilbert case the cone is not generated by the squares $y^{\dagger}y$, since $[y^{\dagger}y\,v,v] = [yv,yv]$ may be negative; the orders for different $J$ agree on the common commutant and are parametrised by the symmetries. The involution, the adjoint axiom and the positive cone are Hilbert Algebras; the indefinite form and $J$ are Indefinite Inner Product Spaces and The Fundamental Symmetry; the representations are The Indefinite GNS Construction and Krein–von Neumann Algebras.

Summary of Notation

Symbol	Meaning
$\dagger$, $[\cdot,\cdot]$	Involution and indefinite Hermitian form
$[xy,z] = [y,x^{\dagger}z]$	Adjoint axiom
$L_x$, $L_{x^{\dagger}}$	Left multiplication and its adjoint for the form
$[x,y] = \tau(x^{\dagger}y)$	Indefinite trace form
$J$, $J(x^{\dagger}) = (Jx)^{\dagger}$	Fundamental symmetry; $J$-positivity condition
$\langle x,y\rangle = [Jx,y]$	Induced Hilbert algebra form
$[xv,v]\geq0$ for all $v$	$J$-positivity of $x = x^{\dagger}$
$A_J^{+}$	$J$-positive cone
$y^{\dagger}xy\in A_J^{+}$	Invariance of the cone under inner conjugation
$a\leq_J b \iff b-a\in A_J^{+}$	$J$-order