Krein Orthogonality and the Fundamental Decomposition

Introduction

The Krein form $[\tilde{Q},\tilde{Q}']=\sum_{\mu}\varepsilon_{\mu}\bar Q_{\mu}Q'_{\mu}$ is an indefinite Hermitian form, and its orthogonality is not the orthogonality of a Euclidean space: isotropic vectors exist, a subspace and its complement can meet, and a projection can be Krein-orthogonal without being definite-orthogonal. This article develops the three notions that organise the indefinite geometry of the algebra — the Krein-orthogonal complement, the classification of subspaces by the inertia of the restricted form, and the fundamental decomposition $\mathbb{B}=\mathbb{W}\,{[+]_K}\,\mathbb{W}^{\perp_{K}}$ attached to a maximal positive definite subspace $\mathbb{W}$ — together with the projections that belong to them. The Gram matrix of the form and the signatures of its restrictions are in The Krein Gram Matrix and the Restrictions of the Form; the totally isotropic subspaces are the subject of The Isotropic Structure of the Krein Form; and the global geometry of the maximal definite subspaces is in The Krein Level Sets and the Hyperbolic Structure.

Conventions. $e_0=1$, $e_k^{2}=-e_0$, central scalar imaginary $i$, $\mathrm{Sc}$ the scalar part, $\langle\tilde{Q},\tilde{Q}'\rangle=\sum_{\mu}\bar Q_{\mu}Q'_{\mu}$ the positive definite Hermitian form, $[\cdot,\cdot]$ the Krein form, and $J={}^{\natural}$ the fundamental symmetry, so that $[\tilde{Q},\tilde{Q}']=\langle J\tilde{Q},\tilde{Q}'\rangle$ and $\mathbb{B}=\mathbb{C}_{\mathbb{B}}\oplus\mathbb{V}_{\mathbb{B}}$ is the splitting into the centre and the vector subspace.

Orthogonality and Complements

Definition. Elements $\tilde{Q},\tilde{Q}'$ are Krein-orthogonal, written $\tilde{Q}\perp_{K}\tilde{Q}'$, when $[\tilde{Q},\tilde{Q}']=0$. A subspace is Krein-orthogonal to another, $\mathbb{W}\perp_{K}\mathbb{U}$, when every element of one is Krein-orthogonal to every element of the other. The Krein-orthogonal complement of $\mathbb{W}$ is

$$ \mathbb{W}^{\perp_{K}}=\{\tilde{S}\in\mathbb{B}:[\tilde{Q},\tilde{S}]=0\ \text{for all}\ \tilde{Q}\in\mathbb{W}\}. $$

Theorem (the complement). For every subspace $\mathbb{W}$,

$$ \dim\mathbb{W}+\dim\mathbb{W}^{\perp_{K}}=\dim\mathbb{B}, \qquad \mathbb{W}\subseteq(\mathbb{W}^{\perp_{K}})^{\perp_{K}}, \qquad (\mathbb{W}^{\perp_{K}})^{\perp_{K}}=\mathbb{W}\ \text{if and only if}\ \mathbb{W}\cap\mathbb{W}^{\perp_{K}}=\{0\}. $$

In particular $\mathbb{C}_{\mathbb{B}}^{\perp_{K}}=\mathbb{V}_{\mathbb{B}}$ and $\mathbb{V}_{\mathbb{B}}^{\perp_{K}}=\mathbb{C}_{\mathbb{B}}$.

Proof. The map $\mathbb{B}\to\mathbb{W}^{*}$, $\tilde{S}\mapsto[\cdot,\tilde{S}]$, has kernel $\mathbb{W}^{\perp_{K}}$ and rank $\dim\mathbb{W}$, because the ambient form is non-degenerate and every linear form on $\mathbb{W}$ extends; rank–nullity gives the dimension formula. The inclusion and the criterion for equality are formal consequences. The last line is the computation $[e_0,e_k]=0$ with the dimensions.

Remark (the contrast with a Euclidean space). In a positive definite space $\mathbb{W}\cap\mathbb{W}^{\perp}=\{0\}$ for every $\mathbb{W}$; here the intersection can be non-trivial, and it is exactly the radical of the restricted form.

The Classification of Subspaces

Definition. Write $[\,\cdot,\cdot\,]_{\mathbb{W}}$ for the restriction of the Krein form to a subspace $\mathbb{W}$. The subspace is non-degenerate when the restriction is non-degenerate, equivalently when $\mathbb{W}\cap\mathbb{W}^{\perp_{K}}=\{0\}$; degenerate otherwise; positive definite, negative definite, or definite when the restriction is; neutral when it is non-degenerate and the restriction is indefinite; and totally isotropic when the restriction vanishes, $[\tilde{Q},\tilde{Q}']=0$ for all $\tilde{Q},\tilde{Q}'\in\mathbb{W}$.

Theorem (definite implies non-degenerate; the converse fails). A definite subspace is non-degenerate. The converse is false: the real plane $\mathbb{W}=\mathrm{span}_{\mathbb{R}}\{e_0+e_1,\ e_0-e_1\}$ is non-degenerate and neutral, since $[e_0+e_1,e_0+e_1]=[e_0-e_1,e_0-e_1]=0$ while $[e_0+e_1,e_0-e_1]=2$.

Proof. If $\mathbb{W}$ is positive definite and $\tilde{S}\in\mathbb{W}\cap\mathbb{W}^{\perp_{K}}$ then $[\tilde{S},\tilde{S}]>0$ unless $\tilde{S}=0$, while $\tilde{S}\in\mathbb{W}^{\perp_{K}}$ gives $[\tilde{S},\tilde{S}]=0$; the negative definite case is the same with the sign reversed. For the plane, the two displayed values give a restriction of Gram matrix $\begin{pmatrix}0&2\\2&0\end{pmatrix}$, of determinant $-4\neq0$ and of inertia $(1,1)$.

Theorem (the dimension of a definite subspace). A positive definite subspace has complex dimension at most $1$ (real dimension at most $2$), and a negative definite subspace has complex dimension at most $3$ (real dimension at most $6$); the bounds are the inertia indices $p$ and $q$ of The Krein Gram Matrix and the Restrictions of the Form.

Proof. A positive definite subspace meets the maximal negative definite subspace $\mathbb{V}_{\mathbb{B}}$ only at $0$, so its dimension is at most the codimension of $\mathbb{V}_{\mathbb{B}}$, which is $1$; the negative statement is the same with the centre.

Maximal Definite Subspaces

Definition. A positive definite subspace is maximal when it is contained in no larger one; equivalently, when its dimension is the positive index $p$.

Theorem (the maximal positive definite subspaces). Over $\mathbb{C}$ the maximal positive definite subspaces are the lines $\mathbb{C}(e_0+\tilde{V})$ with $\tilde{V}\in\mathbb{V}_{\mathbb{B}}$ and $\|\tilde{V}\|_E<1$, and over $\mathbb{R}$ they are the real planes of the same form. The map $\tilde{V}\mapsto\mathbb{C}(e_0+\tilde{V})$ is a bijection onto the set of maximal positive definite subspaces, so that set is the open unit ball of $\mathbb{V}_{\mathbb{B}}\cong\mathbb{C}^{3}$.

Proof. $[e_0+\tilde{V},e_0+\tilde{V}]=1-\|\tilde{V}\|_E^{2}$ by the bridge $[\tilde{Q},\tilde{Q}]=\sum_{\mu}\varepsilon_{\mu}|Q_{\mu}|^{2}$, so the line is positive definite exactly for $\|\tilde{V}\|_E<1$ and is maximal by the dimension bound; conversely a maximal positive definite line has a vector with $\tilde{Q}_0\neq0$, since it meets $\mathbb{V}_{\mathbb{B}}$ only at $0$, and can be scaled to the form $e_0+\tilde{V}$. The classification of the fundamental symmetries in The Fundamental Symmetry of the Biquaternion Algebra is the same parametrisation read on the operators.

The Fundamental Decomposition

Definition. A fundamental decomposition of $\mathbb{B}$ is a pair $(\mathbb{W},\mathbb{W}^{\perp_{K}})$ of complementary Krein-orthogonal subspaces, the first positive definite and maximal. Its fundamental symmetry is the operator

$$ J_{\mathbb{W}}= \begin{cases} +\mathrm{id} & \text{on }\mathbb{W},\\ -\mathrm{id} & \text{on }\mathbb{W}^{\perp_{K}} . \end{cases} $$

Theorem (the fundamental symmetry of a decomposition). For every fundamental decomposition $(\mathbb{W},\mathbb{W}^{\perp_{K}})$ the operator $J_{\mathbb{W}}$ is a $J$-self-adjoint involution with $J_{\mathbb{W}}^{2}=\mathrm{id}$, its signature is $(2,6)$ over $\mathbb{R}$, and it reproduces the Krein form from the Hermitian one,

$$ [\tilde{Q},\tilde{Q}']=\langle J_{\mathbb{W}}\tilde{Q},\tilde{Q}'\rangle, \qquad [J_{\mathbb{W}}\tilde{Q},\tilde{Q}]=\langle\tilde{Q},\tilde{Q}\rangle . $$

The canonical fundamental decomposition is $\mathbb{B}=\mathbb{C}_{\mathbb{B}}\perp_{K}\mathbb{V}_{\mathbb{B}}$, with $J_{\mathbb{C}_{\mathbb{B}}}=J={}^{\natural}$.

Proof. The operator is well defined because $\mathbb{B}=\mathbb{W}\oplus\mathbb{W}^{\perp_{K}}$, and it is an involution by construction. For the first identity, split $\tilde{Q}=\tilde{W}+\tilde{S}$ and $\tilde{Q}'=\tilde{W}'+\tilde{S}'$: $[\tilde{Q},\tilde{Q}']=[\tilde{W},\tilde{W}']+[\tilde{S},\tilde{S}']$ and $\langle J_{\mathbb{W}}\tilde{Q},\tilde{Q}'\rangle=\langle\tilde{W},\tilde{W}'\rangle-\langle\tilde{S},\tilde{S}'\rangle$; on the positive definite $\mathbb{W}$ the two forms agree and on the negative definite $\mathbb{W}^{\perp_{K}}$ they agree up to the sign $-$ because $[\tilde{S},\tilde{S}']=-\langle\tilde{S},\tilde{S}'\rangle$ there. The second identity is the first with $J_{\mathbb{W}}^{2}=\mathrm{id}$ and $\langle J_{\mathbb{W}}\tilde{Q},\tilde{Q}'\rangle=[\tilde{Q},\tilde{Q}']$. The signature is that of the form, and the canonical pair is the sign comparison of the two conjugations.

Corollary (the family of symmetries). The fundamental symmetries of $\mathbb{B}$ are the involutions whose $+1$-eigenspace is positive definite of dimension $p$; through the parametrisation of the maximal positive definite subspaces they are indexed by the open unit ball of $\mathbb{V}_{\mathbb{B}}$, the canonical one $-1$ of them being $J$ itself. Any two fundamental symmetries are congruent by an isometry of the Krein form.

The Krein Projections

Definition. Let $\mathbb{W}$ be non-degenerate. The Krein-orthogonal projection onto $\mathbb{W}$ is the projection with range $\mathbb{W}$ and kernel $\mathbb{W}^{\perp_{K}}$.

Theorem (the Krein projections are exactly the $J$-self-adjoint idempotents). An idempotent $P$ of $\mathbb{B}$ is $J$-self-adjoint, $P^{\dagger}=P$, if and only if it is the Krein-orthogonal projection onto a non-degenerate subspace.

Proof. If $P^{\dagger}=P$ and $P^{2}=P$, then $[P\tilde{P},\tilde{R}]=[\tilde{P},P\tilde{R}]$; for $\tilde{P}\in\ker P$ this gives $[\tilde{P},\tilde{R}']=0$ for every $\tilde{R}'$ in the range, so $\ker P\subseteq(\mathrm{ran}\,P)^{\perp_{K}}$, and the two sides have the same dimension because the form is non-degenerate, whence equality. Conversely, for the projection onto a non-degenerate $\mathbb{W}$ along $\mathbb{W}^{\perp_{K}}$, writing $\tilde{P}=\tilde{W}+\tilde{S}$ and $\tilde{R}=\tilde{W}'+\tilde{S}'$ gives $[P\tilde{P},\tilde{R}]=[\tilde{W},\tilde{W}']=[\tilde{P},P\tilde{R}]$, so $P^{\dagger}=P$.

Theorem (Krein-orthogonality is not definite-orthogonality). The definite-orthogonal projections that commute with $J$ are the Krein-orthogonal projections with $J$-invariant range, and they are a proper subclass: for $\mathbb{W}=\mathbb{C}(e_0+\tfrac12e_1)$, the Krein-orthogonal projection onto $\mathbb{W}$ is $J$-self-adjoint, is not self-adjoint for $\langle\cdot,\cdot\rangle$, and does not commute with $J$.

Proof. If $P^{*}=P$ and $PJ=JP$ then $P^{\dagger}=JP^{*}J=JPJ=P$, so such projections are Krein-orthogonal; their range is $J$-invariant. For the counterexample, in the coefficient basis the projection onto $\mathbb{C}(e_0+\tfrac12e_1)$ along its complement is the matrix with rows $(4/3,-2/3,0,0)$ and $(2/3,-1/3,0,0)$; it is idempotent, it satisfies $P^{\dagger}=P$ because conjugation by $J$ changes the signs of exactly the off-diagonal entries, and $P^{*}=P^{\mathsf T}\neq P$ shows that it is not definite-orthogonal, while $PJ\neq JP$ shows that it does not commute with $J$.

The Krein Gram–Schmidt Process

Theorem (orthonormalisation). Every basis of $\mathbb{B}$ can be replaced by an orthonormal one, carrying $p$ vectors with $[u,u]=+1$ and $q$ with $[u,u]=-1$; the numbers $p,q$ do not depend on the basis.

Proof. Sylvester's law of inertia (Quadratic Forms and Polarisation, §Sylvester's Law of Inertia) applied to the Hermitian form. Explicitly, if a vector $u$ with $[u,u]\neq0$ is found, replace the current basis by its Krein-orthogonal projection on $u^{\perp_{K}}$ together with $u$, and rescale $u$ so that $[u,u]=\pm1$; if the residual form is nonzero, repeat. The process stops because each step lowers the dimension by one, and at the end the remaining vectors, if any, are isotropic and span a totally isotropic subspace (The Isotropic Structure of the Krein Form).

Worked example. The basis $e_0,e_1,e_2,e_3$ is already orthonormal, with $p=1$ and $q=3$ over $\mathbb{C}$. Starting instead from $u_0=e_0+e_1$ and $u_1=e_0-e_1$: $[u_0,u_0]=0$ shows that $u_0$ is isotropic and the process cannot start with it; the pair spans a neutral non-degenerate plane and must be replaced by the orthonormal pair $e_0,e_1$.

Worked Examples

A degenerate subspace. $\mathbb{W}=\mathbb{C}(e_0+e_1)$: $[e_0+e_1,e_0+e_1]=0$, so $\mathbb{W}\subseteq\mathbb{W}^{\perp_{K}}$ and the restriction vanishes; $\dim\mathbb{W}+\dim\mathbb{W}^{\perp_{K}}=1+3=4$ reads the complement.

A positive definite subspace. $\mathbb{W}=\mathbb{C}(e_0+0.5e_1)$: $[\tilde{Q},\tilde{Q}]=1-0.25=0.75>0$, non-degenerate, maximal, with fundamental symmetry different from $J$.

A neutral non-degenerate subspace. $\mathrm{span}_{\mathbb{R}}\{e_0+e_1,\,e_0-e_1\}$, of Gram matrix $\begin{pmatrix}0&2\\2&0\end{pmatrix}$.

A definite subspace of each sign. The centre $\mathbb{C}_{\mathbb{B}}$, positive, and the vector subspace $\mathbb{V}_{\mathbb{B}}$, negative.

Two orthogonal distinctions. $\mathbb{C}_{\mathbb{B}}\perp_{K}\mathbb{V}_{\mathbb{B}}$, while $\mathbb{H}_{\mathbb{B}}$ and $i\mathbb{H}_{\mathbb{B}}$ are not Krein-orthogonal, $[e_0,ie_0]=i\neq0$.

Summary

The Krein-orthogonal complement satisfies $\dim\mathbb{W}+\dim\mathbb{W}^{\perp_{K}}=\dim\mathbb{B}=4$ over $\mathbb{C}$, and $\mathbb{W}\cap\mathbb{W}^{\perp_{K}}=\{0\}$ characterises the non-degenerate subspaces, which are exactly those whose restricted form is non-degenerate. Definite subspaces are non-degenerate, the converse failing on the neutral plane $\mathrm{span}_{\mathbb{R}}\{e_0+e_1,e_0-e_1\}$; a positive definite subspace has complex dimension at most $1$, a negative definite one at most $3$, and the maximal positive definite subspaces are the lines $\mathbb{C}(e_0+\tilde{V})$ with $\|\tilde{V}\|_E<1$, so they form the open unit ball of $\mathbb{C}^{3}$. Each of them gives a fundamental decomposition $\mathbb{B}=\mathbb{W}\perp_{K}\mathbb{W}^{\perp_{K}}$ and a fundamental symmetry $J_{\mathbb{W}}=\pm\mathrm{id}$ on the two parts, reproducing the Krein form from the Hermitian one, $[\tilde{Q},\tilde{Q}']=\langle J_{\mathbb{W}}\tilde{Q},\tilde{Q}'\rangle$; the canonical symmetry is $J={}^{\natural}$. The $J$-self-adjoint idempotents are exactly the Krein-orthogonal projections onto the non-degenerate subspaces, a class strictly larger than the definite-orthogonal projections commuting with $J$; the projection onto $\mathbb{C}(e_0+\tfrac12e_1)$ is the smallest counterexample. Orthonormal bases exist, with $p=1$ signs $+1$ and $q=3$ signs $-1$ over $\mathbb{C}$, and the Krein Gram–Schmidt process produces one.

Summary of Notation

Symbol	Meaning
$\tilde{Q}\perp_{K}\tilde{Q}'$	$[\tilde{Q},\tilde{Q}']=0$
$\mathbb{W}^{\perp_{K}}$	The Krein-orthogonal complement of $\mathbb{W}$
$\mathbb{W}\cap\mathbb{W}^{\perp_{K}}=\{0\}$	Non-degeneracy of $\mathbb{W}$
Definite, neutral, degenerate, totally isotropic	The classification of subspaces by the restricted form
$\mathbb{C}(e_0+\tilde{V})$, $\lVert\tilde{V}\rVert_E<1$	The maximal positive definite subspaces
$J_{\mathbb{W}}=\pm\mathrm{id}$	The fundamental symmetry of a fundamental decomposition
$[\tilde{Q},\tilde{Q}']=\langle J_{\mathbb{W}}\tilde{Q},\tilde{Q}'\rangle$	The bridge through a fundamental symmetry
$P^{\dagger}=P$, $P^{2}=P$ $\iff$ $P$ Krein-orthogonal projection	The $J$-self-adjoint idempotents