The Krein Gram Matrix and the Restrictions of the Form

Introduction

The Krein form $[\tilde{Q},\tilde{Q}']=\sum_{\mu}\varepsilon_{\mu}\bar Q_{\mu}Q'_{\mu}$ of The Biquaternion Krein Form and Its Signature is a Hermitian form of signature $(1,3)$ over $\mathbb{C}$ (respectively $(2,6)$ over $\mathbb{R}$), and its coefficients are the four signs $\varepsilon=(1,-1,-1,-1)$. This article collects the matrix side of that form: its Gram matrix in the coefficient basis, which is the sign matrix $E$ itself; the changes of basis that make the form canonical; the Gram matrices and signatures of its restrictions to the six distinguished subspaces of Biquaternion Relations Between Subspaces; the orthogonality pattern of those restrictions; and the discriminant. The article is the matrix companion of the two articles of the same layer, Krein Orthogonality and the Fundamental Decomposition and The Isotropic Structure of the Krein Form, and it fixes the conventions for all three.

Conventions. $e_0=1$, $e_k^{2}=-e_0$, central scalar imaginary $i$, $\mathrm{Sc}$ the scalar part, $E=\mathrm{diag}(1,-1,-1,-1)$, and $\Phi$ the $2\times2$ matrix representation of Biquaternion 2×2 Matrix Element Representation, so that $\mathrm{Sc}(\tilde R\tilde S)=\tfrac12\operatorname{Tr}(\Phi(\tilde R)\Phi(\tilde S))$ and $\Phi(\tilde{Q}^{\natural})=\operatorname{adj}\Phi(\tilde{Q})$.

The Gram Matrix in the Coefficient Basis

Definition. The Gram matrix of the Krein form in the basis $e_0,e_1,e_2,e_3$ is $G_{\mu\nu}=[e_{\mu},e_{\nu}]$.

Theorem (the Gram matrix is the sign matrix). In the coefficient basis

$$ G=\mathrm{diag}(1,-1,-1,-1)=E , $$

so the diagonal entries are $+1,-1,-1,-1$, the off-diagonal entries vanish, and $\det G=-1$. In the real basis $e_0,e_1,e_2,e_3,ie_0,ie_1,ie_2,ie_3$ of $\mathbb{B}_{\mathbb{R}}$ the Gram matrix is

$$ G_{\mathbb{R}}=\mathrm{diag}(1,-1,-1,-1,\,1,-1,-1,-1), $$

of signature $(2,6)$.

Proof. $[e_{\mu},e_{\nu}]=\sum_{\lambda}\varepsilon_{\lambda}\overline{(e_{\mu})_{\lambda}}(e_{\nu})_{\lambda}=\varepsilon_{\mu}\delta_{\mu\nu}$: the units are orthogonal and each has its own sign. The determinant is $\varepsilon_0\varepsilon_1\varepsilon_2\varepsilon_3=-1$. In the real basis the coefficient $i$ contributes $|i|^{2}=1$ to every entry, so the Gram matrix repeats the block $E$ on the two coefficient blocks, whence the signature $(2+0,\,6)$ by counting the signs.

Remark (the three Gram matrices). The three pairings of The Three Pairings of the Biquaternion Algebra have Gram matrices $\mathrm{I}_4$ (bilinear), $\mathrm{I}_4$ (Hermitian) and $E$ (Krein): the Krein form is the only one of the three whose coefficient basis is already orthogonal with signs. The matrix $E$ is also the Gram matrix of the product form $\mathrm{Sc}(\tilde{P}\tilde{Q})$, and it is the fundamental symmetry of the next article written in the coefficient basis.

The Krein Form in the Matrix Representation

Proposition (the form as a matrix trace). For all biquaternions,

$$ [\tilde{Q},\tilde{Q}']=\tfrac12\operatorname{Tr}\bigl(\operatorname{adj}\Phi(\tilde{Q})^{\dagger}\,\Phi(\tilde{Q}')\bigr) =\tfrac12\operatorname{Tr}\bigl(\Phi(\tilde{Q}^{\natural})^{\dagger}\Phi(\tilde{Q}')\bigr), $$

where $\dagger$ is the conjugate transpose in the matrix algebra.

Proof. $[\tilde{Q},\tilde{Q}']=\mathrm{Sc}(\bar{\tilde{Q}}\tilde{Q}')=\mathrm{Sc}\bigl((\tilde{Q}^{\natural})^{*}\tilde{Q}'\bigr)$, because $(\tilde{Q}^{\natural})^{*}=\overline{\tilde{Q}}$; the trace identity $\mathrm{Sc}(\tilde R\tilde S)=\tfrac12\operatorname{Tr}(\Phi(\tilde R)\Phi(\tilde S))$ of The Forms in the Matrix Representation of the Biquaternion Algebra then gives the result, and $\Phi(\tilde{Q}^{\natural})=\operatorname{adj}\Phi(\tilde{Q})$ with $\Phi$ a $*$-homomorphism gives the second form.

Corollary (the three trace expressions). The bilinear form, the Hermitian form and the Krein form are the three pairings

$$ \tfrac12\operatorname{Tr}\bigl(\Phi(\tilde P)\operatorname{adj}\Phi(\tilde Q)\bigr), \qquad \tfrac12\operatorname{Tr}\bigl(\Phi(\tilde P)^{\dagger}\Phi(\tilde Q)\bigr), \qquad \tfrac12\operatorname{Tr}\bigl(\operatorname{adj}\Phi(\tilde P)^{\dagger}\Phi(\tilde Q)\bigr), $$

so that each pairing is a Hilbert–Schmidt pairing of $\Phi(\tilde P)$ with $\Phi(\tilde Q)$, preceded by adjugation for the first and by adjugation of the conjugate transpose for the third.

The Canonical Basis and Sylvester's Law

Definition. A basis is orthogonal for the Krein form when $[u_j,u_k]=0$ for $j\neq k$, and orthonormal when in addition each $[u_j,u_j]=\pm1$. A subspace $\mathbb{W}$ is positive definite when $[\tilde{Q},\tilde{Q}]>0$ for its nonzero elements, negative definite when $[\tilde{Q},\tilde{Q}]<0$, and definite when it is one of the two.

Theorem (Sylvester). Every basis can be replaced by an orthonormal one, and the numbers $p$ of $+1$ and $q$ of $-1$ are invariants of the form: over $\mathbb{C}$, $(p,q)=(1,3)$; over $\mathbb{R}$, $(p,q)=(2,6)$. The coefficient basis is already orthonormal with $p$ entries $+1$ and $q$ entries $-1$, and the maximal positive definite subspaces are the subspaces $\mathbb{C}e_0$ over $\mathbb{C}$ and the real plane $\mathrm{span}_{\mathbb{R}}\{e_0,ie_0\}$ over $\mathbb{R}$, of dimensions $p$.

Proof. Diagonalising by congruence and then scaling each basis vector to have square $\pm1$ gives an orthonormal basis, and inertia is invariant by Sylvester's law (Quadratic Forms and Polarisation, §Sylvester's Law of Inertia). For the maximality, a positive definite subspace meets the maximal negative definite subspace $\mathbb{V}_{\mathbb{B}}$ only at $0$, so its complex dimension is at most the codimension $1$ of $\mathbb{V}_{\mathbb{B}}$; the centre $\mathbb{C}e_0$ achieves the bound, and every line $\mathbb{C}(e_0+\tilde V)$ with $\|\tilde V\|_E<1$ is another maximal positive definite subspace (The Fundamental Symmetry of the Biquaternion Algebra). Over $\mathbb{R}$ the maximal positive definite subspaces are the real planes such as $\mathrm{span}_{\mathbb{R}}\{e_0,ie_0\}$.

The Restrictions to the Six Subspaces

The six subspaces are defined in Biquaternion Relations Between Subspaces: the centre $\mathbb{C}_{\mathbb{B}}=\mathbb{C}e_0$, the vector subspace $\mathbb{V}_{\mathbb{B}}=\mathrm{span}_{\mathbb{C}}\{e_1,e_2,e_3\}$, the quaternion subspace $\mathbb{H}_{\mathbb{B}}=\mathrm{span}_{\mathbb{R}}\{e_0,e_1,e_2,e_3\}$, the anti-quaternion subspace $i\mathbb{H}_{\mathbb{B}}$, the Hermitian subspace $\mathbb{M}_{+}=\{\tilde{Q}:\tilde{Q}^{*}=\tilde{Q}\}$, and the anti-Hermitian subspace $\mathbb{M}_{-}=\{\tilde{Q}:\tilde{Q}^{*}=-\tilde{Q}\}$.

Theorem (the restrictions). With $Q_{\mu}=q_{\mu}+iq'_{\mu}$ the Krein form restricts as follows.

subspace	real dimension	$[\tilde{Q},\tilde{Q}]$	signature
$\mathbb{C}_{\mathbb{B}}$	$2$	$q_0^2+(q'_0)^2$	$(2,0)$
$\mathbb{V}_{\mathbb{B}}$	$6$	$-\sum_{k}\bigl(q_k^2+(q'_k)^2\bigr)$	$(0,6)$
$\mathbb{H}_{\mathbb{B}}$	$4$	$q_0^2-q_1^2-q_2^2-q_3^2$	$(1,3)$
$i\mathbb{H}_{\mathbb{B}}$	$4$	$(q'_0)^2-(q'_1)^2-(q'_2)^2-(q'_3)^2$	$(1,3)$
$\mathbb{M}_{+}$	$4$	$q_0^2-(q'_1)^2-(q'_2)^2-(q'_3)^2$	$(1,3)$
$\mathbb{M}_{-}$	$4$	$(q'_0)^2-q_1^2-q_2^2-q_3^2$	$(1,3)$

Proof. The centre and the vector subspace are spanned by $e_0,ie_0$ and by $e_k,ie_k$; there $[e_{\mu},e_{\mu}]=\varepsilon_{\mu}$ and $[ie_{\mu},ie_{\mu}]=\varepsilon_{\mu}$, which gives $\pm1$ on each basis vector and the first two rows. A real quaternion has $Q_{\mu}=q_{\mu}$, a purely imaginary one has $Q_{\mu}=iq'_{\mu}$; on the Hermitian subspace $Q_0=q_0$ and $Q_k=iq'_k$, on the anti-Hermitian subspace $Q_0=iq'_0$ and $Q_k=q_k$; substituting $\varepsilon$ in $[\tilde{Q},\tilde{Q}]=\sum_{\mu}\varepsilon_{\mu}|Q_{\mu}|^{2}$ gives the last four rows.

Corollary (the common signature of the four (1,3) subspaces). The quaternion subspace, the anti-quaternion subspace and the two sectors carry the same signature $(1,3)$: each is a copy of Minkowski space, with the Krein form as its interval form. The two remaining subspaces are definite, the centre positive and the vector subspace negative.

Remark (the Krein form versus the norm on the same subspace). The comparison with the restriction of the norm $N=\sum Q_{\mu}^{2}$, tabulated in The Bilinear Form on the Biquaternion Algebra, is instructive: on $\mathbb{H}_{\mathbb{B}}$ the norm is $N=\sum q_{\mu}^{2}$ of signature $(4,0)$ while the Krein form is $\sum_{\mu}\varepsilon_{\mu}q_{\mu}^{2}$ of signature $(1,3)$; the two differ by the sign vector $\varepsilon$, which is exactly the difference between the bilinear and the Krein pairing. On the two sectors the norm is already of signature $(1,3)$ and $(3,1)$, and the Krein form again differs by the sign of the vector part.

Orthogonality of the Distinctions

Definition. Two subspaces are Krein-orthogonal, written $\mathbb{W}\perp_{K}\mathbb{U}$, when $[\tilde{Q},\tilde{T}]=0$ for all $\tilde{Q}\in\mathbb{W}$ and $\tilde{T}\in\mathbb{U}$. The Krein-orthogonal complement is $\mathbb{W}^{\perp_{K}}=\{\tilde{T}:[\tilde{Q},\tilde{T}]=0\ \forall\tilde{Q}\in\mathbb{W}\}$.

Theorem (the orthogonal splitting). The Krein form splits the algebra as an orthogonal sum

$$ \mathbb{B}=\mathbb{C}_{\mathbb{B}}\perp_{K}\mathbb{V}_{\mathbb{B}}, \qquad \mathbb{C}_{\mathbb{B}}^{\perp_{K}}=\mathbb{V}_{\mathbb{B}},\qquad \mathbb{V}_{\mathbb{B}}^{\perp_{K}}=\mathbb{C}_{\mathbb{B}}, $$

and the four subspaces of signature $(1,3)$ are non-degenerate, $\mathbb{W}^{\perp_{K}}=\{0\}$, and are pairwise not Krein-orthogonal.

Proof. $[e_0,e_k]=0$ and $[ie_0,e_k]=0$ give the splitting and the two complements. In coefficients $[e_{\mu},\tilde{T}]=\varepsilon_{\mu}T_{\mu}$ and $[ie_{\mu},\tilde{T}]=-i\varepsilon_{\mu}T_{\mu}$, so an element orthogonal to a basis of a subspace is orthogonal to all of $\mathbb{B}$: the four subspaces of signature $(1,3)$ are therefore non-degenerate. For the failure of mutual orthogonality, $[e_0,ie_0]=i\neq0$ shows that $\mathbb{H}_{\mathbb{B}}$ and $i\mathbb{H}_{\mathbb{B}}$ are not orthogonal, and $[ie_0,e_0]=-i\neq0$ does the same for $\mathbb{M}_{+}$ and $\mathbb{M}_{-}$.

The Discriminant

Definition. The discriminant of a non-degenerate symmetric or Hermitian form of rank $n$ and Gram matrix $G$ is the class of $(-1)^{n(n-1)/2}\det G$ modulo squares of the field.

Theorem (the discriminant of the form and of its restrictions). Over $\mathbb{R}$ the discriminant of the Krein form, and of each of the six restrictions, is the class of $-1$; over $\mathbb{C}$ every one of them has trivial discriminant, since $-1=(i)^{2}$ is a square.

Proof. For the form itself $n=4$ and $\det G=-1$, so $(-1)^{6}(-1)=-1$. The centre has $n=2$ and $\det=1$, giving $(-1)^{1}\cdot1=-1$; the vector subspace has $n=6$ and $\det(-I_6)=+1$, giving $(-1)^{15}\cdot1=-1$; each $(1,3)$ subspace has Gram matrix $\operatorname{diag}(1,-1,-1,-1)$ of determinant $-1$, giving $-1$. Over $\mathbb{C}$ the class of $-1$ is the class of $1$ because $i^{2}=-1$.

Remark. The invariance of the sign is not accidental: the six restrictions are the forms of a Minkowski plane, a six-dimensional definite space and four copies of Minkowski space, and each of these has discriminant $-1$ over $\mathbb{R}$. The unimodularity of the coefficient lattice, $\det G=\pm1$, is the same statement read on the units.

Worked Examples

A positive vector. $\tilde{Q}=e_0+2ie_0$: $[\tilde{Q},\tilde{Q}]=|1|^{2}+|2|^{2}=5>0$, in the centre, the maximal positive definite subspace over $\mathbb{C}$.

A negative vector. $\tilde{Q}=e_1+e_2$: $[\tilde{Q},\tilde{Q}]=-1-1=-2<0$, in the vector subspace.

A Minkowski causal vector. $\tilde{Q}=e_0+0.5e_1$, a real quaternion: $[\tilde{Q},\tilde{Q}]=1-0.25=0.75>0$, timelike in the $(1,3)$ slice.

A null element of the norm only. $\tilde{Q}=e_1+ie_2$: $[\,\tilde{Q},\tilde{Q}]=-2$ while $N(\tilde{Q})=0$; the element is a zero divisor and definite for the Krein form.

A null element of the Krein form only. $\tilde{Q}=e_0+e_1$: $[\tilde{Q},\tilde{Q}]=0$ while $N(\tilde{Q})=2$.

A real quaternion of each sign. $[\tilde{Q},\tilde{Q}]$ on $\mathbb{H}_{\mathbb{B}}$ is the interval form: positive on $e_0$, negative on $e_1$, null on $e_0+e_1$ — the three causal types of the Minkowski slice in one basis.

Summary

In the coefficient basis the Krein form has the Gram matrix $G=\mathrm{diag}(1,-1,-1,-1)=E$, of determinant $-1$; in the real basis it is $\mathrm{diag}(1,-1,-1,-1,1,-1,-1,-1)$ of signature $(2,6)$. In the $2\times2$ matrix representation the form is the Hilbert–Schmidt pairing $[\tilde{Q},\tilde{Q}']=\tfrac12\operatorname{Tr}(\operatorname{adj}\Phi(\tilde{Q})^{\dagger}\Phi(\tilde{Q}'))$, the adjugate and the conjugate transpose marking the Krein case among the three pairings. Sylvester's law gives $(p,q)=(1,3)$ over $\mathbb{C}$ and $(2,6)$ over $\mathbb{R}$, the coefficient basis already being orthonormal. The restrictions to the six subspaces are the positive definite centre $(2,0)$, the negative definite vector subspace $(0,6)$, and four copies of Minkowski space $(1,3)$ on the quaternion subspace, the anti-quaternion subspace and the two sectors; the centre and the vector subspace are mutual Krein-orthogonal complements, while the four $(1,3)$ subspaces are non-degenerate and not mutually orthogonal. The discriminant is the class of $-1$ over $\mathbb{R}$ for the form and for each restriction, and is trivial over $\mathbb{C}$.

Summary of Notation

Symbol	Meaning
$G=E=\mathrm{diag}(1,-1,-1,-1)$	The Gram matrix of the Krein form in the coefficient basis
$G_{\mathbb{R}}=\mathrm{diag}(1,-1,-1,-1,1,-1,-1,-1)$	The Gram matrix over $\mathbb{R}$; signature $(2,6)$
$\tfrac12\operatorname{Tr}(\operatorname{adj}\Phi(\tilde{Q})^{\dagger}\Phi(\tilde{Q}'))$	The Krein form in the $2\times2$ matrix representation
$(p,q)=(1,3)$ over $\mathbb{C}$, $(2,6)$ over $\mathbb{R}$	The inertia (Sylvester)
$\mathbb{C}_{\mathbb{B}}$, $\mathbb{V}_{\mathbb{B}}$, $\mathbb{H}_{\mathbb{B}}$, $i\mathbb{H}_{\mathbb{B}}$, $\mathbb{M}_{+}$, $\mathbb{M}_{-}$	The six subspaces and their restrictions
$\mathbb{W}^{\perp_{K}}$	The Krein-orthogonal complement
$(-1)^{n(n-1)/2}\det G$	The discriminant; the class of $-1$ over $\mathbb{R}$