The Isotropic Structure of the Krein Form

Introduction

An indefinite Hermitian form has a null set beyond the origin, and the shape of that set and of the totally isotropic subspaces it carries is the geometry of the form. For the Krein form $[\tilde{Q},\tilde{Q}']=\sum_{\mu}\varepsilon_{\mu}\bar Q_{\mu}Q'_{\mu}$ the null set is the real quadric $[\tilde{Q},\tilde{Q}]=0$ of dimension $7$, it is smooth away from the origin, its link is the product $S^{1}\times S^{5}$, and its maximal totally isotropic subspaces have dimension the Witt index $\min(p,q)$, that is $1$ over $\mathbb{C}$ and $2$ over $\mathbb{R}$. This article treats that structure: the isotropic elements, the totally isotropic subspaces, the isotropic lines and their boundary sphere, and the comparison with the other null set of the algebra, the zero-divisor cone of the norm $N$. The orthogonality and the subspaces are in Krein Orthogonality and the Fundamental Decomposition; the level sets of positive and negative sign are in The Krein Level Sets and the Hyperbolic Structure; and the projective geometry of the norm's cone is in Biquaternion Topology.

Conventions. $e_0=1$, $e_k^{2}=-e_0$, central scalar imaginary $i$, $\mathrm{Sc}$ the scalar part; the Krein form is $[\tilde{Q},\tilde{Q}']=\sum_{\mu}\varepsilon_{\mu}\bar Q_{\mu}Q'_{\mu}$ with $\varepsilon=(1,-1,-1,-1)$, the Hermitian form is $\langle\tilde{Q},\tilde{Q}'\rangle=\sum_{\mu}\bar Q_{\mu}Q'_{\mu}$ with $\|\tilde{Q}\|_E^{2}=\langle\tilde{Q},\tilde{Q}\rangle$, and the norm is $N(\tilde{Q})=\sum_{\mu}Q_{\mu}^{2}$.

Isotropic Elements

Definition. An element $\tilde{Q}$ is Krein-isotropic, or Krein-null, when $[\tilde{Q},\tilde{Q}]=0$; the Krein null set is

$$ \mathcal{K}=\{\tilde{Q}:[\tilde{Q},\tilde{Q}]=0\}=\Bigl\{\tilde{Q}:\sum_{\mu}\varepsilon_{\mu}|Q_{\mu}|^{2}=0\Bigr\}. $$

Theorem (the shape of the null set). In the splitting $\tilde{Q}=c+v$ of the algebra into its centre and vector parts, with $c\in\mathbb{C}_{\mathbb{B}}$ the scalar part and $v\in\mathbb{V}_{\mathbb{B}}$ the vector part,

$$ \mathcal{K}=\{\tilde{Q}:\|c\|_E=\|v\|_E\}; $$

it is a closed real algebraic cone of apex the origin, homogeneous of degree two, of real dimension $7$, irregular only at the apex, and it is connected.

Proof. $[\tilde{Q},\tilde{Q}]=\|c\|_E^{2}-\|v\|_E^{2}$ by the sign vector of the coefficient basis, so the equation is $\|c\|_E=\|v\|_E$. The set is the zero set of a single non-constant real polynomial, hence of dimension $7$ at its smooth points; its gradient with respect to the four real coordinates of $c$ and the six of $v$ is $2(c,-v)\neq0$ off the origin, so the apex is the only singular point and the cone is a real $7$-manifold away from it. Connectivity follows because the map $\tilde{Q}\mapsto(\|c\|_E,\|v\|_E)$ makes the cone for each $t>0$ the product of the spheres of radius $t$ in the two factors, glued along $t$, that is $S^{1}\times S^{5}$.

Theorem (the link). The intersection of the null set with the Euclidean unit sphere $\|\tilde{Q}\|_E=1$ is diffeomorphic to $S^{1}\times S^{5}$; the punctured cone $\mathcal{K}\setminus\{0\}$ retracts onto it and is homotopy equivalent to $S^{1}\times S^{5}$.

Proof. On the unit sphere $\|c\|_E^{2}+\|v\|_E^{2}=1$ together with $\|c\|_E=\|v\|_E$ forces $\|c\|_E=\|v\|_E=1/\sqrt2$, so the link is $S^{1}\times S^{5}$ with the radii $1/\sqrt2$; the retraction is $\tilde{Q}\mapsto\tilde{Q}/\|\tilde{Q}\|_E$.

Example. $e_0+e_1$ has $[e_0+e_1,e_0+e_1]=1-1=0$ and $N=2$, so it is isotropic for the Krein form and not for the norm. The element $e_1+ie_2$ has $N=1+i^{2}=0$ and $[e_1+ie_2,e_1+ie_2]=-2$, the reverse situation.

The Comparison with the Norm Cone

Definition. The norm cone of the algebra is the zero set $\mathcal{N}=\{\tilde{Q}:N(\tilde{Q})=0\}$ of the norm, the union of the origin and the zero-divisor set of Biquaternion Zero Divisors; it is a complex cone of real dimension $6$.

Theorem (the two cones are different). The Krein null set and the norm cone are distinct: neither is contained in the other. Their intersection $\mathcal{K}\cap\mathcal{N}$ is a real algebraic cone of dimension $5$, and it contains the two complex lines $\mathbb{C}(e_0+ie_1)$ and $\mathbb{C}(e_0-ie_1)$.

Proof. $e_0+e_1$ is in the Krein null set and not in the norm cone, $e_1+ie_2$ in the norm cone and not in the Krein null set, so neither inclusion holds. For the dimension, in the affine chart $Q_0=1$ the equations $N=0$ and $[\tilde{Q},\tilde{Q}]=0$ are one complex and one real equation on the six real coordinates $Q_1,Q_2,Q_3$, that is three real equations in six unknowns, so the projective intersection has real dimension $3$ and the cone over it has real dimension $5$. The two displayed lines are isotropic for both forms: $N(e_0\pm ie_1)=1+(i)^{2}=0$ and $[e_0\pm ie_1,e_0\pm ie_1]=1-1=0$.

Remark (the two cones agree on a real slice). On the real quaternion subspace $\mathbb{H}_{\mathbb{B}}$ the Krein form is the interval form and the norm is the Euclidean form, $[\tilde{Q},\tilde{Q}]=q_0^{2}-\sum_kq_k^{2}$ and $N(\tilde{Q})=\sum_{\mu}q_{\mu}^{2}$; the Krein null set is the light cone of the slice, the norm cone meets the real slice only at the origin, and the two agree nowhere except at $0$.

Totally Isotropic Subspaces

Definition. A subspace $\mathbb{W}$ is totally isotropic when $[\tilde{Q},\tilde{Q}']=0$ for all $\tilde{Q},\tilde{Q}'\in\mathbb{W}$, equivalently when $\mathbb{W}\subseteq\mathbb{W}^{\perp_{K}}$; it is maximal when it is contained in no larger one. The Witt index of the Krein form is the common dimension of the maximal totally isotropic subspaces.

Theorem (the Witt index and the maximal isotropic subspaces). The Witt index is $\min(p,q)$: it is $1$ over $\mathbb{C}$ and $2$ over $\mathbb{R}$. A maximal totally isotropic complex subspace is $\mathbb{C}(e_0+e_1)$, and a maximal totally isotropic real subspace is the plane

$$ \mathbb{W}_{\mathrm{iso}}=\mathrm{span}_{\mathbb{R}}\{e_0+e_1,\ i(e_0+e_1)\}, \qquad e_0+e_1\ \text{and}\ i(e_0+e_1)\ \text{isotropic, mutually orthogonal}. $$

Every totally isotropic subspace is contained in a maximal one, and the isometry group $U(1,3)$ acts transitively on the maximal ones.

Proof. If $\mathbb{W}$ is totally isotropic then it is orthogonal to itself, so $\mathbb{W}\subseteq\mathbb{W}^{\perp_{K}}$, and the restriction of the form to $\mathbb{W}+\mathbb{W}^{\perp_{K}}/\mathbb{W}^{\perp_{K}}$ makes the dimension of $\mathbb{W}$ at most $\min(p,q)$: an isotropic subspace meets the maximal positive definite subspaces and the maximal negative definite subspaces only at $0$. The two displayed spaces attain $\min(p,q)=1$ over $\mathbb{C}$ and $\min(p,q)=2$ over $\mathbb{R}$, since $[e_0+e_1,e_0+e_1]=0$, $[e_0+e_1,i(e_0+e_1)]=i-i=0$ and the plane is $2$-dimensional over $\mathbb{R}$. The extension and transitivity are Witt's extension theorem for Hermitian forms (Witt's Theorems).

Remark (the isotropic and the null elements). An element $\tilde{Q}$ is isotropic exactly when the line $\mathbb{C}\tilde{Q}$ is totally isotropic (over $\mathbb{R}$, the real plane $\mathrm{span}_{\mathbb{R}}\{\tilde{Q},i\tilde{Q}\}$ is); the null set of the first section is the union of the totally isotropic complex lines, and the totally isotropic subspaces are the subspaces all of whose elements are isotropic.

The Isotropic Lines and the Boundary Sphere

Definition. The isotropic lines of $\mathbb{C}^{4}$ for the Krein form are the complex lines $\mathbb{C}\tilde{Q}$ spanned by nonzero isotropic elements.

Theorem (the isotropic lines form $S^{5}$). The isotropic complex lines are exactly the lines $\mathbb{C}(e_0+\tilde{V})$ with $\|\tilde{V}\|_E=1$, so the assignment $\tilde{V}\mapsto\mathbb{C}(e_0+\tilde{V})$ is a bijection from the unit sphere of $\mathbb{V}_{\mathbb{B}}\cong\mathbb{C}^{3}$ — a copy of $S^{5}$ — onto the set of isotropic lines.

Proof. An isotropic line $\mathbb{C}\tilde{Q}$ has $[\tilde{Q},\tilde{Q}]=0$, hence $\|c\|_E=\|v\|_E$, so it is not contained in $\mathbb{V}_{\mathbb{B}}$ and admits a representative with nonzero scalar part; scaling it gives exactly one representative of the form $e_0+\tilde{V}$. For that representative $[e_0+\tilde{V},e_0+\tilde{V}]=1-\|\tilde{V}\|_E^{2}$, so the line is isotropic exactly for $\|\tilde{V}\|_E=1$; and distinct points $\tilde{V}$ of the unit sphere give distinct lines, since $e_0+\tilde{V}$ and $e_0+\tilde{V}'$ span the same line only if $\tilde{V}=\tilde{V}'$. This is the boundary case of the parametrisation of the maximal positive definite subspaces in Krein Orthogonality and the Fundamental Decomposition, which uses the open ball $\|\tilde{V}\|_E<1$.

Corollary (the two links). The punctured real cone $\mathcal{K}\setminus\{0\}$ is homotopy equivalent to $S^{1}\times S^{5}$, and the space of isotropic complex lines is $S^{5}$: passing from one to the other is dividing by the action of the circle of complex phases, which acts freely on the isotropic elements.

Worked Examples

An isotropic element of the centre–vector type. $\tilde{Q}=e_0+e_1$: isotropic, norm $2$, in the light cone of the real quaternion slice at the boundary.

A zero divisor that is isotropic. $\tilde{Q}=e_0+ie_1$: both $N=0$ and $[\tilde{Q},\tilde{Q}]=0$; the line it spans is a doubly null line.

A zero divisor that is not isotropic. $\tilde{Q}=e_1+ie_2$: $N=0$, $[\tilde{Q},\tilde{Q}]=-2$, a strictly negative element.

A timelike element. $\tilde{Q}=e_0+0.5e_1$: $[\,\tilde{Q},\tilde{Q}]=0.75>0$, so it lies inside the positive region and spans a maximal positive definite line.

A maximal isotropic plane. $\mathbb{W}_{\mathrm{iso}}=\mathrm{span}_{\mathbb{R}}\{e_0+e_1,\ i(e_0+e_1)\}$: real dimension $2$, over $\mathbb{C}$ it is the single line $\mathbb{C}(e_0+e_1)$, and it is maximal because the Witt index over $\mathbb{R}$ is $2$.

The link at a point. For $\tilde{Q}=(e_0+e_1)/\sqrt2$ one has $\|\tilde{Q}\|_E=1$ and $[\,\tilde{Q},\tilde{Q}]=0$: the point $S^{1}\times S^{5}$ of the link is reached by normalising, and the circle is the phase of the complex coefficient.

Summary

The Krein null set is $\mathcal{K}=\{\|c\|_E=\|v\|_E\}$, a real algebraic cone of dimension $7$, smooth away from its apex, connected, with link and punctured homotopy type $S^{1}\times S^{5}$. It is distinct from the complex norm cone of the zero divisors: neither contains the other, the two meet in a cone of real dimension $5$, and the lines $\mathbb{C}(e_0\pm ie_1)$ are null for both forms. A totally isotropic subspace is one contained in its own Krein-orthogonal complement; the Witt index is $\min(p,q)$, that is $1$ over $\mathbb{C}$ and $2$ over $\mathbb{R}$; a maximal isotropic complex subspace is the line $\mathbb{C}(e_0+e_1)$, a maximal isotropic real subspace is the plane it spans with $i(e_0+e_1)$; every isotropic subspace extends to a maximal one and the isometry group acts transitively on the maximal ones. The isotropic complex lines are the points of $S^{5}$, the boundary sphere of the positive half of the ball; the centre $S^{1}$ is the difference between the punctured cone and the space of isotropic lines.

Summary of Notation

Symbol	Meaning
$\mathcal{K}=\{\tilde{Q}:[\tilde{Q},\tilde{Q}]=0\}=\{\lVert c\rVert_E=\lVert v\rVert_E\}$	The Krein null set
$S^{1}\times S^{5}$	The link of the null cone and the homotopy type of $\mathcal{K}\setminus\{0\}$
$\mathcal{N}=\{N(\tilde{Q})=0\}$	The norm cone of the zero divisors
$\mathbb{C}(e_0\pm ie_1)$	Doubly null isotropic lines
$\min(p,q)=1$ over $\mathbb{C}$, $2$ over $\mathbb{R}$	The Witt index
$\mathbb{W}_{\mathrm{iso}}=\mathrm{span}_{\mathbb{R}}\{e_0+e_1,i(e_0+e_1)\}$	The maximal isotropic real plane
$\mathbb{W}\subseteq\mathbb{W}^{\perp_{K}}$	Totally isotropic subspace
$S^{5}$	The space of isotropic complex lines