Split-Quaternion Relations Between Subspaces

Introduction

The four distinguished subspaces of $\mathbb{H}_{\mathrm{s}}$ — the scalar line $S$, the vector subspace $V$ and the two split-complex planes $\mathbb{D}_2, \mathbb{D}_3$ — are defined by different conditions, but they are not independent: they contain one another, they intersect, they sum to subspaces of $\mathbb{H}_{\mathrm{s}}$, and they are cut out of the four real coordinates by the involutions. This article collects those relations in one place, after the individual articles Split-Quaternion Scalar and Vector Subspaces and Split-Quaternion Split-Complex Subspaces, and before the structural article Split-Quaternion Involution Lattice, which treats the involutions themselves.

Every number below — dimensions of intersections, of sums, and the entries of the tables — is derived from the definitions by comparison of coefficients, so the article doubles as an index of the four subspaces and of the coordinate blocks.

Conventions. The algebra is $\mathbb{H}_{\mathrm{s}}$, with basis $1, e_1, e_2, e_3$, the relations $e_1^2 = -1$, $e_2^2 = e_3^2 = +1$, $e_3 = e_1 e_2$, $e_1 e_2 = -e_2 e_1$, and a general element $\tilde q = q_0 e_0 + q_1 e_1 + q_2 e_2 + q_3 e_3$. The conjugation ${}^{\natural}$, the principal involution $\alpha$ and the reversal $\rho$ are as in Split-Quaternion Subspaces and the Involutions, and $N$, $B$ are the split-quaternion norm and its polarisation.

The Three Decompositions

Two of the three involutions split $\mathbb{H}_{\mathrm{s}}$ into distinguished subspaces, and the third splits it into two further subspaces:

involution	fixed space	anti-fixed space	decomposition
${}^{\natural}$	$S = \mathbb{R}\cdot 1$	$V = \operatorname{span}\{e_1,e_2,e_3\}$	$\mathbb{H}_{\mathrm{s}} = S \oplus V$
$\alpha$	$\mathbb{D}_3 = \operatorname{span}\{1,e_3\}$	$\operatorname{span}\{e_1,e_2\}$	$\mathbb{H}_{\mathrm{s}} = \mathbb{D}_3 \oplus \operatorname{span}\{e_1,e_2\}$
$\rho$	$\operatorname{span}\{1,e_1,e_2\}$	$\mathbb{R} e_3$	$\mathbb{H}_{\mathrm{s}} = \operatorname{span}\{1,e_1,e_2\} \oplus \mathbb{R} e_3$

The three decompositions are the scalar–vector decomposition, and the two decompositions associated with the principal involution and the reversal. Every element has three readings, one per row. The finest common refinement of the three is the coordinate-block decomposition of the next section.

The Four Subspaces at a Glance

subspace	defining condition	real basis	$\dim_{\mathbb{R}}$	subalgebra	norm
$S$	$\tilde{q}^{\natural} = \tilde q$	$1$	$1$	yes, $\cong \mathbb{R}$	$q_0^2$, positive definite
$V$	$\tilde{q}^{\natural} = -\tilde q$	$e_1,e_2,e_3$	$3$	no (Lie)	$q_1^2-q_2^2-q_3^2$, signature $(2,1)$
$\mathbb{D}_2$	closed under products of $1, e_2$	$1,e_2$	$2$	yes, $\cong \mathbb{D}$	$q_0^2-q_2^2$, signature $(1,1)$
$\mathbb{D}_3$	$\alpha(\tilde q) = \tilde q$	$1,e_3$	$2$	yes, $\cong \mathbb{D}$	$q_0^2-q_3^2$, signature $(1,1)$

Only three of the four are closed under multiplication: the scalar line and the two split-complex planes. The vector subspace is not a subalgebra but is closed under the commutator, and it is the only one of the four that carries a nonzero bracket — the other three are commutative. The norms are the restrictions of $N$; all four are non-degenerate, the first definite and the other three indefinite.

The Four Coordinate Blocks

The four real coordinates of a split-quaternion are $(q_0,q_1,q_2,q_3)$ in the basis $1, e_1, e_2, e_3$. They group into four one-dimensional coordinate blocks,

$$ \mathbb{H}_{\mathrm{s}} = \langle 1 \rangle \oplus \langle e_1 \rangle \oplus \langle e_2 \rangle \oplus \langle e_3 \rangle, $$

where $\langle \cdot\rangle$ denotes the real span of the displayed elements. Each of the blocks $\langle 1\rangle$, $\langle e_2\rangle$ and $\langle e_3\rangle$ is the intersection of the distinguished subspaces containing it, and $\langle e_1\rangle$ is the complement of $\mathbb{D}_2 + \mathbb{D}_3$ in $\mathbb{H}_{\mathrm{s}}$; each of the four distinguished subspaces is a sum of blocks:

$$ S = \langle 1 \rangle, \qquad V = \langle e_1 \rangle \oplus \langle e_2 \rangle \oplus \langle e_3 \rangle, $$

$$ \mathbb{D}_2 = \langle 1 \rangle \oplus \langle e_2 \rangle, \qquad \mathbb{D}_3 = \langle 1 \rangle \oplus \langle e_3 \rangle . $$

The blocks are the building units of the article: every intersection below is a sum of blocks, which is why the dimension arithmetic is so simple.

The Intersections

The ten pairwise intersections of the four subspaces, with the dimensions (the diagonal carries the dimensions of the subspaces themselves):

	$S$	$V$	$\mathbb{D}_2$	$\mathbb{D}_3$
$S$	$1$	$0$	$1$	$1$
$V$	$0$	$3$	$1$	$1$
$\mathbb{D}_2$	$1$	$1$	$2$	$1$
$\mathbb{D}_3$	$1$	$1$	$1$	$2$

Explicitly,

$$ S \cap V = \{0\}, \quad S \cap \mathbb{D}_2 = S, \quad S \cap \mathbb{D}_3 = S, \quad V \cap \mathbb{D}_2 = \mathbb{R} e_2, $$

$$ V \cap \mathbb{D}_3 = \mathbb{R} e_3, \quad \mathbb{D}_2 \cap \mathbb{D}_3 = S . $$

Three features are worth isolating:

The scalar line is contained in both split-complex planes, so it meets them in dimension $1$; it meets the vector subspace only in the origin, being the complement of $V$ in the scalar–vector decomposition.
The vector subspace meets each split-complex plane in a single line, spanned by the common basis element $e_2$ or $e_3$, and the two planes meet each other in the scalar line.
No two of the four subspaces are equal, and no pair except the obvious containment pairs is nested: $S \subset \mathbb{D}_2$, $S \subset \mathbb{D}_3$, and $S \subset S \oplus V$ but $S \not\subseteq V$, $\mathbb{D}_2 \not\subseteq \mathbb{D}_3$.

The Sums

The sums of the pairs, with the same ordering:

	$S$	$V$	$\mathbb{D}_2$	$\mathbb{D}_3$
$S$	$1$	$4$	$2$	$2$
$V$	$4$	$3$	$4$	$4$
$\mathbb{D}_2$	$2$	$4$	$2$	$3$
$\mathbb{D}_3$	$2$	$4$	$3$	$2$

The dimension of a sum is the sum of the dimensions minus the dimension of the intersection. The pairs that span the whole algebra are $S + V = \mathbb{H}_{\mathrm{s}}$, $V + \mathbb{D}_2 = \mathbb{H}_{\mathrm{s}}$, and $V + \mathbb{D}_3 = \mathbb{H}_{\mathrm{s}}$; the pair $\mathbb{D}_2 + \mathbb{D}_3 = \operatorname{span}\{1,e_2,e_3\}$ has dimension $3$ and misses the line $\mathbb{R} e_1$; and $S + \mathbb{D}_k = \mathbb{D}_k$ because $S$ lies inside each split-complex plane. The sum $V + \mathbb{D}_k$ spans the algebra because $V$ contributes $e_1$ and $\mathbb{D}_k$ contributes $1$, and together with $e_2$ or $e_3$ they exhaust the four coordinates.

The Involutions as Sign Patterns

Each involution preserves each of the four subspaces, and its restriction to a subspace is diagonal in the displayed bases with signs $\pm 1$. The following table records the multiplicity of the sign $-1$ on each subspace, in the basis of the table above:

involution	$S$	$V$	$\mathbb{D}_2$	$\mathbb{D}_3$
${}^{\natural}$	$0$ of $1$	$3$ of $3$	$1$ of $2$	$1$ of $2$
$\alpha$	$0$ of $1$	$2$ of $3$	$1$ of $2$	$0$ of $2$
$\rho$	$0$ of $1$	$1$ of $3$	$0$ of $2$	$1$ of $2$

The vanishing rows are the definitions: ${}^{\natural}$ acts as the identity exactly on $S$, $\alpha$ exactly on $\mathbb{D}_3$, and $\rho$ exactly on $\operatorname{span}\{1,e_1,e_2\} \supset \mathbb{D}_2$. The full multiplicities give the anti-fixed spaces: ${}^{\natural}$ negates all of $V$, $\alpha$ negates the plane $\operatorname{span}\{e_1,e_2\}$, and $\rho$ negates the line $\mathbb{R} e_3$. The remaining entries are mixed restrictions. The composition rule ${}^{\natural} = \alpha\rho$ is visible row by row: the coordinates negated by ${}^{\natural}$ are exactly those negated by one of $\alpha, \rho$ but not both, so the multiplicities compose by symmetric difference.

How the Product and the Commutator Act

The Scalar Line

$S$ is the centre, so it acts on every subspace by scalar multiplication: $S \cdot V \subseteq V$, $S \cdot \mathbb{D}_k \subseteq \mathbb{D}_k$, and $[S, \tilde q] = 0$ for all $\tilde q$. As a product, $S \times S \to S$ is closed.

The Vector Subspace

The product of two vectors is not a vector: it splits as

$$ \mathbf{u}\mathbf{v} = -B(\mathbf{u},\mathbf{v})\cdot 1 + \tfrac{1}{2}[\mathbf{u},\mathbf{v}], $$

with scalar part $-B(\mathbf{u},\mathbf{v}) \in S$ and the remainder $\tfrac{1}{2}[\mathbf{u},\mathbf{v}] \in V$; so $V \cdot V \subseteq S \oplus V = \mathbb{H}_{\mathrm{s}}$, with equality. The commutator stays in $V$, and in the basis $e_1, e_2, e_3$,

$$ [e_1,e_2] = 2 e_3, \qquad [e_2,e_3] = -2 e_1, \qquad [e_3,e_1] = 2 e_2, $$

so $[V,V] \subseteq V$ and $V \cong \mathrm{SL}_2(\mathbb{R})$. The bracket is alternating and satisfies the Jacobi identity. It is the antisymmetrisation of the product on the vector subspace.

The Split-Complex Planes

Each split-complex plane is a commutative subalgebra, so $\mathbb{D}_k \cdot \mathbb{D}_k \subseteq \mathbb{D}_k$ and $[\mathbb{D}_k, \mathbb{D}_k] = 0$. Across the planes the product leaves both: since $e_2 e_3 = -e_1$ and $e_3 e_2 = e_1$, one has $\mathbb{D}_2 \cdot \mathbb{D}_3 = \mathbb{H}_{\mathrm{s}}$ and $\mathbb{D}_2 \cdot \mathbb{D}_3 \not\subseteq \mathbb{D}_2 \cup \mathbb{D}_3$. The product of a scalar with a plane stays in the plane, and the two planes share only the scalar line.

The Operation Table

product	$S$	$V$	$\mathbb{D}_2$	$\mathbb{D}_3$
$S \cdot$	$S$	$V$	$\mathbb{D}_2$	$\mathbb{D}_3$
$V \cdot$	$V$	$\mathbb{H}_{\mathrm{s}}$	$\mathbb{H}_{\mathrm{s}}$	$\mathbb{H}_{\mathrm{s}}$
$\mathbb{D}_2 \cdot$	$\mathbb{D}_2$	$\mathbb{H}_{\mathrm{s}}$	$\mathbb{D}_2$	$\mathbb{H}_{\mathrm{s}}$
$\mathbb{D}_3 \cdot$	$\mathbb{D}_3$	$\mathbb{H}_{\mathrm{s}}$	$\mathbb{H}_{\mathrm{s}}$	$\mathbb{D}_3$

The commutator table is the antisymmetrisation of this one: it vanishes on the diagonal pairs among $S, \mathbb{D}_2, \mathbb{D}_3$ (all commutative), equals the bracket of $V$ restricted to the vector–vector entry, and is nonzero only where the product table has an off-diagonal entry that is not symmetric.

Worked Verifications

Example (an intersection of dimension one). Take $V \cap \mathbb{D}_2$. An element is $q_1 e_1 + q_2 e_2 + q_3 e_3 = q_0' + q_2' e_2$; comparing coefficients gives $q_1 = 0$, $q_3 = 0$, $q_0' = 0$ and $q_2 = q_2'$, so the intersection is $\mathbb{R} e_2$, of dimension $1$, as tabulated.

Example (a pair that spans the algebra). The four elements $e_1, e_2, e_3$ span $V$ and $1, e_2$ span $\mathbb{D}_2$; together they include $1, e_1, e_2, e_3$, a basis, so $V + \mathbb{D}_2 = \mathbb{H}_{\mathrm{s}}$, in agreement with $\dim V + \dim \mathbb{D}_2 - \dim(V \cap \mathbb{D}_2) = 3 + 2 - 1 = 4$.

Example (a product that leaves its subspace). $e_2 \in \mathbb{D}_2$ and $e_3 \in V$; their product is $e_2 e_3 = -e_1 \notin \mathbb{D}_2$ and $\notin \mathbb{D}_3$ (indeed $e_1 \in V$), consistent with the entry $\mathbb{H}_{\mathrm{s}}$ in the product table.

Example (a mixed element and its blocks). For $\tilde q = 2 + 3e_1 - e_2 + 4e_3$ the scalar block is $2$, the vector block is $3e_1 - e_2 + 4e_3$, and the split-complex readings are $2 - e_2 + 4e_3$ in $\mathbb{D}_2$ and $2 + 3e_1 + 4e_3$ modulo the complementary block; the involution images are $\tilde{q}^{\natural} = 2 - 3e_1 + e_2 - 4e_3$, $\alpha(\tilde q) = 2 - 3e_1 + e_2 + 4e_3$, $\rho(\tilde q) = 2 + 3e_1 - e_2 - 4e_3$.

The Relation to the Corresponding Article for $\mathbb{B}$

The biquaternion article Biquaternion Relations Between Subspaces performs the same task for the six distinguished subspaces of $\mathbb{B}$, cut out by the four $\mathbb{C}$-linear and conjugate-linear involutions. The bookkeeping is analogous — decompositions, a glance table, coordinate blocks, tables of intersections, sums, sign patterns and operations — but the geometry of the tables differs, because $\mathbb{B}$ has eight real coordinates and six subspaces, whereas $\mathbb{H}_{\mathrm{s}}$ has four coordinates and four subspaces; in $\mathbb{B}$ the blocks have dimensions $1$ and $3$, while here all four blocks have dimension $1$, which is why the intersections here take only the values $0, 1, 2, 3$. The Biquaternions category is the structural model; nothing biquaternion-specific — the central unit $i$, the quaternionic half-algebra, the Hermitian and anti-Hermitian pairing — is imported here.

Summary

The four distinguished subspaces of $\mathbb{H}_{\mathrm{s}}$ are the scalar line $S = \langle 1\rangle$, the vector subspace $V = \langle e_1,e_2,e_3\rangle$, and the two split-complex planes $\mathbb{D}_2 = \langle 1,e_2\rangle$ and $\mathbb{D}_3 = \langle 1,e_3\rangle$. They are the eigenspaces of the involutions, with ${}^{\natural}$ giving $S \oplus V$, $\alpha$ giving $\mathbb{D}_3 \oplus \operatorname{span}\{e_1,e_2\}$, and $\rho$ giving $\operatorname{span}\{1,e_1,e_2\} \oplus \mathbb{R} e_3$; the finest common decomposition is the coordinate-block decomposition $\mathbb{H}_{\mathrm{s}} = \langle 1\rangle \oplus \langle e_1\rangle \oplus \langle e_2\rangle \oplus \langle e_3\rangle$.

The intersections are $S \cap V = 0$, $V \cap \mathbb{D}_k = \mathbb{R} e_k$, and $\mathbb{D}_2 \cap \mathbb{D}_3 = S$, with $S$ inside both planes. The sums show that $V$ together with either split-complex plane spans the algebra, while the two planes together span only $\operatorname{span}\{1,e_2,e_3\}$. The involutions act on each subspace by the sign patterns tabulated above, and the composition rule ${}^{\natural} = \alpha\rho$ composes the multiplicities by symmetric difference. The product is closed on $S, \mathbb{D}_2, \mathbb{D}_3$, sends $V \cdot V$ into $S \oplus V$, and sends the products across the split-complex planes into the whole algebra; the commutator vanishes on the commutative pieces and makes $V$ the Lie algebra $\mathrm{SL}_2(\mathbb{R})$.

Summary of Notation

Symbol	Meaning	Article
$\mathbb{H}_{\mathrm{s}}$	the split-quaternion algebra	Split-Quaternion Algebra
$S = \langle 1\rangle$	the scalar line	Split-Quaternion Scalar and Vector Subspaces
$V = \langle e_1,e_2,e_3\rangle$	the vector subspace	Split-Quaternion Scalar and Vector Subspaces
$\mathbb{D}_2 = \langle 1,e_2\rangle$, $\mathbb{D}_3 = \langle 1,e_3\rangle$	the split-complex planes	Split-Quaternion Split-Complex Subspaces
$\langle \cdot \rangle$	real span of the displayed elements	this article
${}^{\natural}$, $\alpha$, $\rho$	conjugation, principal involution, reversal	Split-Quaternion Subspaces and the Involutions
$B$, $N$	the polarised form and the split-quaternion norm	Split-Quaternion Norm and Invertibility
$[\tilde q,\tilde p] = \tilde q \tilde p-\tilde p\tilde q$	the commutator	Split-Quaternion Scalar and Vector Subspaces