Comparison of the Six Subspaces
Introduction
The six distinguished subspaces of $\mathbb{B}$ — the centre, the vector subspace, the quaternion and anti-quaternion subspaces, and the Hermitian and anti-Hermitian subspaces — are defined and tabulated one by one in Introduction to the Six Subspaces. They are not independent objects: they intersect, they sum to subspaces of $\mathbb{B}$ larger than themselves, they are built from four common pieces, and the four conjugations permute and negate them.
This article collects those relations in one place. Every number below is derived from the definitions by the comparison of coefficients, so the article doubles as the index of the six. Its tool is the four coordinate blocks, which are the pieces into which the eight real coordinates of $\mathbb{B}$ group and of which every one of the six is a sum.
The frame is the three decompositions of the algebra, established in Biquaternion Algebra:
| involution | fixed space | anti-fixed space | decomposition |
|---|---|---|---|
| quaternion conjugation ${}^{\natural}$ | $\mathbb{C}_{\mathbb{B}}$ | $\mathrm{Vect}(\mathbb{B})$ | $\mathbb{B} = \mathbb{C}_{\mathbb{B}} \oplus \mathrm{Vect}(\mathbb{B})$ |
| complex conjugation $\bar{\cdot}$ | $\mathbb{H}_{\mathbb{B}}$ | $i\mathbb{H}_{\mathbb{B}}$ | $\mathbb{B} = \mathbb{H}_{\mathbb{B}} \oplus i\mathbb{H}_{\mathbb{B}}$ |
| Hermitian conjugation ${}^{*}$ | $\mathbb{M}_+$ | $\mathbb{M}_-$ | $\mathbb{B} = \mathbb{M}_+ \oplus \mathbb{M}_-$ |
The reversal $\flat = -{}^{*}$ gives no fourth decomposition: its two spaces are those of ${}^{*}$ in the other order. The three decompositions are the scalar–vector, the quaternion and the Hermitian decomposition of $\mathbb{B}$, and their three pairs are the only pairs of distinct subspaces that meet in the origin.
The Six Subspaces and the Coordinate Blocks
The relations between the six subspaces are read off from four subspaces of $\mathbb{B}$, the coordinate blocks, into which the eight real coordinates group.
Definition. The coordinate blocks are the four real subspaces
$$ A_1 = \mathbb{R}e_0 , \qquad A_2 = \mathbb{R}(ie_0) , \qquad B_1 = \operatorname{span}_{\mathbb{R}}\{e_1, e_2, e_3\} , \qquad B_2 = \operatorname{span}_{\mathbb{R}}\{ie_1, ie_2, ie_3\} , $$
the two scalar blocks $A_1, A_2$ of dimension $1$ and the two vector blocks $B_1, B_2$ of dimension $3$. They are the pieces into which the eight real coordinates of $\mathbb{B}$ group: the two scalar blocks carry the two real coordinates of the scalar coefficient $Q_0$, and the two vector blocks the six real coordinates of the vector part.
Each block is the intersection of the three subspaces that contain it:
| block | basis | $\dim_{\mathbb{R}}$ | it is the intersection |
|---|---|---|---|
| $A_1$ | $e_0$ | $1$ | $\mathbb{C}_{\mathbb{B}} \cap \mathbb{H}_{\mathbb{B}} \cap \mathbb{M}_+$ |
| $A_2$ | $ie_0$ | $1$ | $\mathbb{C}_{\mathbb{B}} \cap i\mathbb{H}_{\mathbb{B}} \cap \mathbb{M}_-$ |
| $B_1$ | $e_1, e_2, e_3$ | $3$ | $\mathrm{Vect}(\mathbb{B}) \cap \mathbb{H}_{\mathbb{B}} \cap \mathbb{M}_-$ |
| $B_2$ | $ie_1, ie_2, ie_3$ | $3$ | $\mathrm{Vect}(\mathbb{B}) \cap i\mathbb{H}_{\mathbb{B}} \cap \mathbb{M}_+$ |
Theorem. Each of the six subspaces is the sum of two blocks:
| subspace | blocks | $\dim_{\mathbb{R}}$ |
|---|---|---|
| $\mathbb{C}_{\mathbb{B}}$ | $A_1 \oplus A_2$ | $2$ |
| $\mathrm{Vect}(\mathbb{B})$ | $B_1 \oplus B_2$ | $6$ |
| $\mathbb{H}_{\mathbb{B}}$ | $A_1 \oplus B_1$ | $4$ |
| $i\mathbb{H}_{\mathbb{B}}$ | $A_2 \oplus B_2$ | $4$ |
| $\mathbb{M}_+$ | $A_1 \oplus B_2$ | $4$ |
| $\mathbb{M}_-$ | $A_2 \oplus B_1$ | $4$ |
Proof. Each subspace is defined by conditions on the real coordinates: the centre by the vanishing of the six vector coordinates, the vector subspace by the vanishing of the two scalar ones, the quaternion subspace by the vanishing of the three imaginary parts of the vector coordinates and of the imaginary part of the scalar one, and the other three by the corresponding sign choices. Reading the conditions block by block gives the table.
The pattern of the table is the reason the six subspaces are neither more nor fewer than six. Each of the four four-dimensional subspaces takes one scalar block and one vector block, and there are exactly four ways of choosing one block from each of the two pairs; the centre takes both scalar blocks and the vector subspace both vector blocks. The three decompositions are exactly the three ways of splitting the four blocks into two complementary pairs: $\{A_1, B_1\}$ against $\{A_2, B_2\}$ gives the quaternion decomposition, $\{A_1, B_2\}$ against $\{A_2, B_1\}$ gives the Hermitian one, and $\{A_1, A_2\}$ against $\{B_1, B_2\}$ gives the scalar–vector one. A four-element set has exactly three pairings into two pairs, which is why there are exactly three decompositions and no fourth.
The Intersections
Theorem. Two distinct subspaces of the six meet in the blocks they have in common; consequently the dimension of their intersection is $0$, $1$ or $3$, and never $2$, $4$, $5$ or $6$.
Proof. Each subspace is a sum of two blocks, and two of them share either no block, one scalar block, or one vector block. Sharing no block gives intersection $\{0\}$ and dimension $0$; sharing one scalar block gives dimension $1$; sharing one vector block gives dimension $3$. Two distinct subspaces cannot share two blocks, since two blocks determine a subspace of the list.
The fifteen dimensions, the diagonal carrying the dimensions of the subspaces themselves:
| $\mathbb{C}_{\mathbb{B}}$ | $\mathrm{Vect}(\mathbb{B})$ | $\mathbb{H}_{\mathbb{B}}$ | $i\mathbb{H}_{\mathbb{B}}$ | $\mathbb{M}_+$ | $\mathbb{M}_-$ | |
|---|---|---|---|---|---|---|
| $\mathbb{C}_{\mathbb{B}}$ | $2$ | $0$ | $1$ | $1$ | $1$ | $1$ |
| $\mathrm{Vect}(\mathbb{B})$ | $0$ | $6$ | $3$ | $3$ | $3$ | $3$ |
| $\mathbb{H}_{\mathbb{B}}$ | $1$ | $3$ | $4$ | $0$ | $1$ | $3$ |
| $i\mathbb{H}_{\mathbb{B}}$ | $1$ | $3$ | $0$ | $4$ | $3$ | $1$ |
| $\mathbb{M}_+$ | $1$ | $3$ | $1$ | $3$ | $4$ | $0$ |
| $\mathbb{M}_-$ | $1$ | $3$ | $3$ | $1$ | $0$ | $4$ |
Three features of the table are used again and again in the thematic articles.
- The two members of a decomposition meet in the origin. The entries $0$ are exactly the three pairs $(\mathbb{C}_{\mathbb{B}}, \mathrm{Vect}(\mathbb{B}))$, $(\mathbb{H}_{\mathbb{B}}, i\mathbb{H}_{\mathbb{B}})$ and $(\mathbb{M}_+, \mathbb{M}_-)$, and no other pair meets in the origin.
- The centre meets each of the other four subspaces in a scalar line: $A_1 = \mathbb{R}e_0$ with the quaternion and Hermitian subspaces, $A_2 = \mathbb{R}(ie_0)$ with the anti-quaternion and anti-Hermitian ones.
- The vector subspace meets each of the other four in a vector triple: $B_1$ with the quaternion and anti-Hermitian subspaces, $B_2$ with the anti-quaternion and Hermitian ones.
In particular, no two of the six subspaces are equal, since equal subspaces would meet in their common dimension, and no off-diagonal entry of the table is $2$, $4$ or $6$.
The Sums
The dimension of a sum is the sum of the dimensions minus the dimension of the intersection, which is the arithmetic of the two tables above:
| $\mathbb{C}_{\mathbb{B}}$ | $\mathrm{Vect}(\mathbb{B})$ | $\mathbb{H}_{\mathbb{B}}$ | $i\mathbb{H}_{\mathbb{B}}$ | $\mathbb{M}_+$ | $\mathbb{M}_-$ | |
|---|---|---|---|---|---|---|
| $\mathbb{C}_{\mathbb{B}}$ | $2$ | $8$ | $5$ | $5$ | $5$ | $5$ |
| $\mathrm{Vect}(\mathbb{B})$ | $8$ | $6$ | $7$ | $7$ | $7$ | $7$ |
| $\mathbb{H}_{\mathbb{B}}$ | $5$ | $7$ | $4$ | $8$ | $7$ | $5$ |
| $i\mathbb{H}_{\mathbb{B}}$ | $5$ | $7$ | $8$ | $4$ | $5$ | $7$ |
| $\mathbb{M}_+$ | $5$ | $7$ | $7$ | $5$ | $4$ | $8$ |
| $\mathbb{M}_-$ | $5$ | $7$ | $5$ | $7$ | $8$ | $4$ |
Exactly three pairs have sum $8$, that is, span the whole algebra, and they are the three complementary pairs of the three decompositions. Every other pair spans a proper subspace: the pairs whose intersection is a vector triple have sum $7$, and the pairs whose intersection is a scalar line have sum $5$. A reader who wants to write an element of $\mathbb{B}$ as the sum of one element of each of two prescribed subspaces therefore has exactly three pairs to choose from, and cannot do it with any other pair.
The Involutions as Sign Patterns
Each of the four conjugations preserves each of the six subspaces, since it commutes with the three commuting involutions that define them, and its restriction to a subspace is diagonal in the real basis of that subspace with eigenvalues $+1$ and $-1$ only. The following table records, for each conjugation and each subspace, the multiplicity of the eigenvalue $-1$, written as a fraction of the dimension.
| conjugation | $\mathbb{C}_{\mathbb{B}}$ | $\mathrm{Vect}(\mathbb{B})$ | $\mathbb{H}_{\mathbb{B}}$ | $i\mathbb{H}_{\mathbb{B}}$ | $\mathbb{M}_+$ | $\mathbb{M}_-$ |
|---|---|---|---|---|---|---|
| ${}^{\natural}$ | $0$ of $2$ | $6$ of $6$ | $3$ of $4$ | $3$ of $4$ | $3$ of $4$ | $3$ of $4$ |
| $\bar{\cdot}$ | $1$ of $2$ | $3$ of $6$ | $0$ of $4$ | $4$ of $4$ | $3$ of $4$ | $1$ of $4$ |
| ${}^{*}$ | $1$ of $2$ | $3$ of $6$ | $3$ of $4$ | $1$ of $4$ | $0$ of $4$ | $4$ of $4$ |
| $\flat$ | $1$ of $2$ | $3$ of $6$ | $1$ of $4$ | $3$ of $4$ | $4$ of $4$ | $0$ of $4$ |
The eight vanishing cells are the definitions of the six subspaces read back: ${}^{\natural}$ acts as the identity exactly on the centre and as its negative exactly on the vector subspace; $\bar{\cdot}$ exactly on the quaternion subspace and its negative exactly on the anti-quaternion one; ${}^{*}$ exactly on $\mathbb{M}_+$ and its negative exactly on $\mathbb{M}_-$; and $\flat$ exactly on $\mathbb{M}_-$ and its negative exactly on $\mathbb{M}_+$. The mixed cells, which are neither $0$ nor the full dimension, are the restrictions that are neither the identity nor its negative; each of them is fixed by the block membership of the subspace, since a conjugation negates a block as a whole.
The table also exhibits the composition rule ${}^{*} = \bar{\cdot}\circ{}^{\natural}$. The coordinates negated by ${}^{*}$ are exactly those negated by one of $\bar{\cdot}$ and ${}^{\natural}$ but not by both, so the multiplicities compose by symmetric difference. On the quaternion subspace, for instance, $\bar{\cdot}$ negates none of the four coordinates and ${}^{\natural}$ negates the three vector ones, so ${}^{*}$ negates the same three and the multiplicity is $3$; on $\mathbb{M}_+$, $\bar{\cdot}$ and ${}^{\natural}$ negate the same three coordinates, and those cancel, leaving multiplicity $0$.
Two consequences follow from the table alone. First, no subspace other than the centre is fixed pointwise by more than one of the four conjugations, and the six subspaces are pairwise distinct as sets. Second, a subspace is fixed pointwise by exactly those conjugations whose multiplicity of $-1$ on it vanishes.
Multiplication by the Central Imaginary Unit
Multiplication by the central element $ie_0$ is not a conjugation but a complex structure, and its action on the six subspaces is read directly from the block table. Since $i(q + iq') = -q' + iq$, the map exchanges the two scalar blocks and the two vector blocks:
$$ iA_1 = A_2 , \qquad iA_2 = A_1 , \qquad iB_1 = B_2 , \qquad iB_2 = B_1 . $$
A subspace made of a block together with its image is therefore stable, and a subspace made of the images of the blocks of another is exchanged with it:
$$ ie_0\,\mathbb{C}_{\mathbb{B}} = \mathbb{C}_{\mathbb{B}} , \qquad ie_0\,\mathrm{Vect}(\mathbb{B}) = \mathrm{Vect}(\mathbb{B}) , \qquad ie_0\,\mathbb{H}_{\mathbb{B}} = i\mathbb{H}_{\mathbb{B}} , \qquad ie_0\,(i\mathbb{H}_{\mathbb{B}}) = \mathbb{H}_{\mathbb{B}} , $$
$$ ie_0\,\mathbb{M}_+ = \mathbb{M}_- , \qquad ie_0\,\mathbb{M}_- = \mathbb{M}_+ . $$
The centre and the vector subspace are stable, the quaternion subspace is carried onto the anti-quaternion subspace and conversely, and the Hermitian subspace is carried onto the anti-Hermitian one and conversely. It is multiplication by the central imaginary unit, and not quaternion conjugation, that swaps the Hermitian and anti-Hermitian subspaces, and this is the reason the six subspaces fall into the four classes they do under the conjugations.
Summary
The six distinguished subspaces of $\mathbb{B}$ are organized by the four coordinate blocks $A_1 = \mathbb{R}e_0$, $A_2 = \mathbb{R}(ie_0)$, $B_1 = \operatorname{span}_{\mathbb{R}}\{e_1,e_2,e_3\}$ and $B_2 = \operatorname{span}_{\mathbb{R}}\{ie_1,ie_2,ie_3\}$, each of which is the intersection of the three subspaces containing it. Every one of the six is a sum of two blocks: the centre of the two scalar blocks, the vector subspace of the two vector blocks, and each of the four four-dimensional subspaces of one scalar and one vector block, the four ways of choosing the pair being the four such subspaces. The three decompositions of the algebra are the three pairings of the four blocks into two complementary pairs, which is why there are exactly three. Two distinct subspaces meet in the blocks they share, so the fifteen pairwise intersections have dimension $0$, $1$ or $3$: the members of a decomposition meet in the origin, the centre meets each other subspace in a scalar line, and the vector subspace in a vector triple. Exactly the three pairs of the decompositions sum to the whole algebra; the pairs sharing a vector triple sum to a subspace of dimension $7$, and those sharing a scalar line to one of dimension $5$. Each of the four conjugations preserves each subspace and acts on it diagonally with signs $+1$ and $-1$, the vanishing multiplicities of $-1$ identifying the subspaces fixed by each conjugation and the table of multiplicities exhibiting the composition ${}^{*} = \bar{\cdot}\circ{}^{\natural}$ as a symmetric difference. Multiplication by the central imaginary unit preserves the centre and the vector subspace, exchanges the quaternion with the anti-quaternion subspace, and exchanges the Hermitian with the anti-Hermitian one.
Summary of Notation
| symbol | meaning |
|---|---|
| $\mathbb{C}_{\mathbb{B}}, \mathrm{Vect}(\mathbb{B})$ | the centre and the vector subspace |
| $\mathbb{H}_{\mathbb{B}}, i\mathbb{H}_{\mathbb{B}}$ | the quaternion and anti-quaternion subspaces |
| $\mathbb{M}_+, \mathbb{M}_-$ | the Hermitian and anti-Hermitian subspaces |
| ${}^{\natural}, \bar{\cdot}, {}^{*}, \flat$ | quaternion, complex, Hermitian conjugation and reversal |
| $A_1, A_2$ | the scalar blocks $\mathbb{R}e_0$ and $\mathbb{R}(ie_0)$ |
| $B_1, B_2$ | the vector blocks $\operatorname{span}_{\mathbb{R}}\{e_1,e_2,e_3\}$ and $\operatorname{span}_{\mathbb{R}}\{ie_1,ie_2,ie_3\}$ |
| $\oplus$ | direct sum of real subspaces |
Further Reading
- Introduction to the Six Subspaces (
articles_maths/introduction-to-the-six-subspaces.md), for the six subspaces themselves, one to a section - Biquaternion Algebra (
articles_maths/biquaternion-algebra.md), for the algebra, its conjugations and the three decompositions - Biquaternion Involution Lattice (
articles_maths/biquaternion-involution-lattice.md), for the four conjugations as an abstract group and the two spaces each defines - Biquaternion Multiplication (
articles_maths/biquaternion-multiplication.md), for the product of two elements of a subspace - The Clifford Structure of the Biquaternion Algebra (
articles_maths/biquaternion-clifford-structure.md), for the grading of the algebra and the four grades