Biquaternion Jordan Algebra
Introduction
Every product of two biquaternions splits into a symmetric and an antisymmetric half, and each half carries a structure of its own. The antisymmetric half is the commutator, read in Biquaternion Lie Algebra; the symmetric half is the symmetrized product, and with it the biquaternion algebra is a commutative Jordan algebra. This article reads that second structure: the Jordan product, the Jordan identity, the trace form, the idempotents and the Peirce decomposition, the Hermitian subspace as a Jordan subalgebra, and the way the two halves of the product divide the distinguished subspaces between them.
The product and its two halves are from Biquaternion Multiplication; the general theory is Jordan Algebras, and the special case used throughout is its §The Symmetrisation of an Associative Algebra. The algebra, its basis and its two idempotents are from Biquaternion Algebra, Different Ways to Consider Biquaternions and Biquaternion Idempotents and Projections; the subspace-by-subspace behaviour is tabulated in Biquaternion Relations Between Subspaces.
Conventions. The biquaternion algebra is $\mathbb{B}=\mathbb{C}\otimes_{\mathbb{R}}\mathbb{H}$, with basis $e_0=1,e_1,e_2,e_3$ and central scalar imaginary $i$; a general element is $\tilde{Q}=\sum_{\mu=0}^{3} Q_\mu e_\mu$ with $Q_\mu\in\mathbb{C}$. Throughout, a biquaternion is written $\tilde{Q}=Q_0e_0+\mathbf{Q}$ with $\mathbf{Q}=\sum_{k=1}^3 Q_k e_k$, and $(\mathbf{P},\mathbf{Q})=\sum_k P_kQ_k$ is the complex bilinear dot product of the vector parts.
The Jordan Product
Definition
The symmetric part of the product, also called the symmetrized product or the Jordan product, is
$$ \tilde{P}\bullet\tilde{Q} := \tfrac{1}{2}\bigl(\tilde{P}\tilde{Q}+\tilde{Q}\tilde{P}\bigr). $$
It is $\mathbb{C}$-bilinear and commutative, and it agrees with the ordinary square on the diagonal:
$$ \tilde{P}\bullet\tilde{Q} = \tilde{Q}\bullet\tilde{P} , \qquad \tilde{P}\bullet\tilde{P} = \tilde{P}^2 . $$
In scalar–vector notation the antisymmetric terms of the product cancel and the mixed terms double:
$$ \tilde{P}\bullet\tilde{Q} = \bigl(P_0Q_0-(\mathbf{P},\mathbf{Q})\bigr) + P_0\mathbf{Q} + Q_0\mathbf{P} . $$
The full product is the sum of its two halves,
$$ \tilde{P}\tilde{Q} = \tilde{P}\bullet\tilde{Q} + \tilde{P}\wedge\tilde{Q} , \qquad \tilde{P}\wedge\tilde{Q} = \tfrac{1}{2}\bigl(\tilde{P}\tilde{Q}-\tilde{Q}\tilde{P}\bigr), $$
the second summand being the outer product of Biquaternion Multiplication, which is the commutator up to the factor $2$ and is read in Biquaternion Lie Algebra.
The Polarisation of the Square
Proposition. The Jordan product is the polarisation of the square:
$$ \tilde{P}\bullet\tilde{Q} = \tfrac{1}{2}\Bigl((\tilde{P}+\tilde{Q})^2 - \tilde{P}^2 - \tilde{Q}^2\Bigr). $$
Proof. Expand $(\tilde{P}+\tilde{Q})^2=\tilde{P}^2+\tilde{P}\tilde{Q}+\tilde{Q}\tilde{P}+\tilde{Q}^2$ by bilinearity and halve.
The square of a biquaternion is the scalar–vector expression
$$ \tilde{P}^2 = \bigl(P_0^2-(\mathbf{P},\mathbf{P})\bigr) + 2P_0\mathbf{P} , $$
a scalar plus a scalar multiple of $\mathbf{P}$, so the square carries no cross-product term; every square-root problem is therefore a problem in the Jordan product alone, as Biquaternion Square Roots of a General Element reads it.
The Jordan Algebra
Theorem. With the product $\bullet$ the biquaternion algebra is a commutative Jordan algebra: $\bullet$ is commutative and $\mathbb{C}$-bilinear, and the Jordan identity
$$ (\tilde{P}\bullet\tilde{Q})\bullet\tilde{P}^2 = \tilde{P}\bullet\bigl(\tilde{Q}\bullet\tilde{P}^2\bigr) $$
holds for all $\tilde{P},\tilde{Q}\in\mathbb{B}$.
Proof. The algebra $\mathbb{B}$ is associative, and the symmetrisation $a\bullet b=\tfrac12(ab+ba)$ of any associative algebra satisfies the Jordan identity, by the associativity of the underlying product (Jordan Algebras, §The Symmetrisation of an Associative Algebra). Commutativity and bilinearity are built into the definition. Verified on random pairs.
A Jordan algebra obtained by symmetrising an associative algebra is called special, and the associative algebra is its envelope. The special Jordan algebra here is $\mathbb{B}$ with $\bullet$, and its envelope is the associative algebra $\mathbb{B}$ itself, so the envelope is finite-dimensional and four-dimensional over $\mathbb{C}$. No exceptional Jordan algebra arises: the exceptional ones have no such envelope, whereas every biquaternion computation is an ordinary associative computation in the envelope followed by symmetrisation.
Powers and the Subalgebra Generated by One Element
A Jordan algebra is power-associative, so every power $\tilde{Q}^n$ is well defined without brackets, $\tilde{Q}^{n+1}=\tilde{Q}\bullet\tilde{Q}^n$, and the powers commute. The subalgebra generated by a single element $\tilde{Q}$ is the span of its powers,
$$ \mathbb{C}[\tilde{Q}] = \mathrm{span}_{\mathbb{C}}\{\tilde{Q},\tilde{Q}^2,\tilde{Q}^3,\dots\}, $$
which is commutative and closed under $\bullet$.
The Trace Form
Proposition. The trace of the Jordan product is the trace of the product, and it is the symmetric $\mathbb{C}$-bilinear function
$$ \mathrm{Tr}\bigl(\tilde{P}\bullet\tilde{Q}\bigr) = \mathrm{Tr}\bigl(\tilde{P}\tilde{Q}\bigr) = 2\bigl(P_0Q_0-(\mathbf{P},\mathbf{Q})\bigr). $$
Proof. The trace is linear and vanishes on the outer product, $\mathrm{Tr}(\tilde{P}\tilde{Q}-\tilde{Q}\tilde{P})=0$, so the trace of the two halves is the same; the trace of a product is twice its scalar part, and the scalar part of $\tilde{P}\tilde{Q}$ is $P_0Q_0-(\mathbf{P},\mathbf{Q})$.
Proposition (the trace form is associative). For all $\tilde{P},\tilde{Q},\tilde{R}\in\mathbb{B}$,
$$ \mathrm{Tr}\Bigl(\bigl(\tilde{P}\bullet\tilde{Q}\bigr)\bullet\tilde{R}\Bigr) = \mathrm{Tr}\Bigl(\tilde{P}\bullet\bigl(\tilde{Q}\bullet\tilde{R}\bigr)\Bigr). $$
Proof. Expanding both products by the definition gives $\tfrac14$ times the sum of four terms each; the two sums are $\mathrm{Tr}(\tilde{P}\tilde{Q}\tilde{R})+\mathrm{Tr}(\tilde{Q}\tilde{P}\tilde{R})+\mathrm{Tr}(\tilde{R}\tilde{P}\tilde{Q})+\mathrm{Tr}(\tilde{R}\tilde{Q}\tilde{P})$ and $\mathrm{Tr}(\tilde{P}\tilde{Q}\tilde{R})+\mathrm{Tr}(\tilde{P}\tilde{R}\tilde{Q})+\mathrm{Tr}(\tilde{Q}\tilde{R}\tilde{P})+\mathrm{Tr}(\tilde{R}\tilde{Q}\tilde{P})$, and the two agree term by term under the cyclic invariance $\mathrm{Tr}(XYZ)=\mathrm{Tr}(ZXY)$ of the trace.
The trace form is the symmetric bilinear function attached to the Jordan algebra. Read as a form rather than as a trace it is the object of Biquaternion Norm and Invertibility; the norm, its polar form and the topology they carry are not used here, and only the algebraic identity $\mathrm{Tr}(\tilde{P}\bullet\tilde{Q})=\mathrm{Tr}(\tilde{P}\tilde{Q})$ is needed below.
Idempotents and the Peirce Decomposition
Jordan Orthogonality
An idempotent of the Jordan algebra is an element $\tilde{E}$ with $\tilde{E}\bullet\tilde{E}=\tilde{E}$, and two idempotents are Jordan orthogonal when $\tilde{E}\bullet\tilde{F}=0$. The two idempotents of Biquaternion Idempotents and Projections,
$$ \tilde{\Pi}_1=\tfrac12(e_0+ie_3), \qquad \tilde{\Pi}_2=\tfrac12(e_0-ie_3), $$
are idempotents of the Jordan algebra, they are Jordan orthogonal, and they sum to the unit:
$$ \tilde{\Pi}_1\bullet\tilde{\Pi}_2=0 , \qquad \tilde{\Pi}_1+\tilde{\Pi}_2=e_0 . $$
A family of pairwise Jordan-orthogonal idempotents summing to the unit is complete, and the number of its members is the degree of the Jordan algebra. The two idempotents above are such a family, and a third can never be added: in the envelope $\mathbb{B}\cong M_2(\mathbb{C})$ a complete family of orthogonal idempotents has at most two members, their images being complementary subspaces of the defining two-dimensional module $S$ of Modules over the Biquaternion Algebra. The biquaternion Jordan algebra therefore has degree two.
The Peirce Spaces
For an idempotent $\tilde{E}$ let $\mathrm{J}_\lambda(\tilde{E})$ be the $\lambda$-eigenspace of the multiplication operator $\tilde{X}\mapsto\tilde{E}\bullet\tilde{X}$. In a Jordan algebra this operator has the eigenvalues $1,\tfrac12,0$ and the algebra splits as
$$ \mathbb{B} = \mathrm{J}_1(\tilde{E}) \oplus \mathrm{J}_{1/2}(\tilde{E}) \oplus \mathrm{J}_0(\tilde{E}), $$
the Peirce decomposition (Jordan Algebras, §The Peirce Spaces). For $\tilde{E}=\tilde{\Pi}_1$ the three spaces are computed from the products alone.
Proposition. For $\tilde{E}=\tilde{\Pi}_1$ the Peirce spaces are
| Peirce space | elements | dimension over $\mathbb{C}$ |
|---|---|---|
| $\mathrm{J}_1(\tilde{\Pi}_1)$ | $\mathbb{C}\tilde{\Pi}_1$ | $1$ |
| $\mathrm{J}_{1/2}(\tilde{\Pi}_1)$ | $\mathrm{span}_{\mathbb{C}}\{e_1,e_2\}$ | $2$ |
| $\mathrm{J}_0(\tilde{\Pi}_1)$ | $\mathbb{C}\tilde{\Pi}_2$ | $1$ |
Proof. The relations $\tilde{\Pi}_1\bullet\tilde{\Pi}_1=\tilde{\Pi}_1$ and $\tilde{\Pi}_1\bullet\tilde{\Pi}_2=0$ put $\tilde{\Pi}_1$ in the first space and $\tilde{\Pi}_2$ in the third. For the vector units, $\tilde{\Pi}_1\bullet e_1=\tfrac12 e_1$ and $\tilde{\Pi}_1\bullet e_2=\tfrac12 e_2$, so $e_1$ and $e_2$ span the middle space. The three spaces are independent and their dimensions sum to $1+2+1=4=\dim_{\mathbb{C}}\mathbb{B}$, so they exhaust the algebra. Verified on the basis.
The dimensions $1+2+1$ are the Jordan counterpart of the two minimal left ideals of Biquaternion Ideals and Peirce Decomposition: the middle space is two-dimensional there as well, and the two one-dimensional ends are the two idempotents.
The Hermitian Subspace as a Jordan Subalgebra
The Hermitian subspace $\mathbb{M}_+$ is closed under the Jordan product. If $\tilde{P},\tilde{Q}\in\mathbb{M}_+$ then $\tilde{P}\tilde{Q}+\tilde{Q}\tilde{P}$ is again Hermitian, so $\tilde{P}\bullet\tilde{Q}\in\mathbb{M}_+$, and with the induced product $\mathbb{M}_+$ is itself a commutative Jordan algebra, this time over $\mathbb{R}$.
It is of real dimension four and of degree two, and it is isomorphic to the Jordan algebra $H_2(\mathbb{C})$ of The Six Subspaces and the Jordan Algebra; that article reads the same structure from the side of the Hermitian elements, and Jordan Algebras treats the degree-two algebra in general under the heading of the spin factor. The Hermitian subspace is therefore a Jordan subalgebra of $\mathbb{B}$ of the special kind, the one real form on which the symmetrized product closes.
The Two Halves of the Product
The product of two biquaternions carries two structures at once: the outer product makes $\mathbb{B}$ a Lie algebra, read in Biquaternion Lie Algebra, and the Jordan product makes $\mathbb{B}$ a commutative Jordan algebra, read here. The two halves are the two parts of a single bilinear product, the general case being The Lie–Jordan Decomposition of a Bilinear Product and the Jordan Triple System.
On the distinguished subspaces the two structures behave in opposite ways, and the contrast is sharpest on the two three-dimensional pieces of the centre decomposition of the Lie algebra. The vector subspace is closed under the outer product — it is the derived subalgebra $[\mathrm{G},\mathrm{G}]=\mathrm{Vect}(\mathbb{B})$ — and it is not closed under the Jordan product, since for two pure vectors
$$ \mathbf{P}\bullet\mathbf{Q} = -\bigl(\mathbf{P},\mathbf{Q}\bigr)e_0 , $$
a central scalar, so the Jordan product of two pure vectors leaves the vector subspace as soon as they are not orthogonal. The Hermitian subspace behaves in the mirror fashion: it is closed under the Jordan product and not under the outer product, since the outer product of two Hermitian elements is anti-Hermitian and lands in $\mathbb{M}_-$. The action of the symmetrized product on all six distinguished subspaces is the corresponding row of the tables of Biquaternion Relations Between Subspaces, cited and not repeated here.
Summary
With the Jordan product $\tilde{P}\bullet\tilde{Q}=\tfrac12(\tilde{P}\tilde{Q}+\tilde{Q}\tilde{P})$ the biquaternion algebra is a commutative Jordan algebra, special, of degree two, whose envelope is the associative algebra $\mathbb{B}$ itself. The Jordan identity holds, the trace form $\mathrm{Tr}(\tilde{P}\bullet\tilde{Q})=\mathrm{Tr}(\tilde{P}\tilde{Q})$ is symmetric and associative, and the two idempotents $\tilde{\Pi}_1,\tilde{\Pi}_2$ are Jordan orthogonal and complete, with Peirce dimensions $1+2+1$. The Hermitian subspace $\mathbb{M}_+$ is a Jordan subalgebra isomorphic to $H_2(\mathbb{C})$. The vector subspace is closed under the outer product and not under the Jordan product, the Hermitian subspace is closed under the Jordan product and not under the outer product, so the symmetric and the antisymmetric halves of the product divide the distinguished subspaces between them.
Summary of Notation
| Symbol | Meaning |
|---|---|
| $\tilde{P}\bullet\tilde{Q}=\tfrac{1}{2}(\tilde{P}\tilde{Q}+\tilde{Q}\tilde{P})$ | Jordan product, the symmetric part |
| $\tilde{P}\wedge\tilde{Q}=\tfrac{1}{2}(\tilde{P}\tilde{Q}-\tilde{Q}\tilde{P})$ | Outer product, the antisymmetric part; the commutator is twice it |
| $\mathbb{B}$ with $\bullet$ | Commutative Jordan algebra, special, degree two |
| $(\tilde{P}\bullet\tilde{Q})\bullet\tilde{P}^2=\tilde{P}\bullet(\tilde{Q}\bullet\tilde{P}^2)$ | Jordan identity |
| $\mathrm{Tr}(\tilde{P}\bullet\tilde{Q})=\mathrm{Tr}(\tilde{P}\tilde{Q})$ | Trace form, symmetric and associative |
| $\tilde{\Pi}_1,\tilde{\Pi}_2$ | Jordan idempotents, Jordan orthogonal and complete, $\tilde{\Pi}_1+\tilde{\Pi}_2=e_0$ |
| $\mathbb{C}[\tilde{Q}]$ | Subalgebra generated by one element, the span of its powers |
| $\mathrm{J}_1\oplus\mathrm{J}_{1/2}\oplus\mathrm{J}_0$ | Peirce decomposition; dimensions $1+2+1$ |
| $\mathbb{M}_+$ | Hermitian subspace, a Jordan subalgebra isomorphic to $H_2(\mathbb{C})$ |
| $\mathbf{P}\bullet\mathbf{Q}=-(\mathbf{P},\mathbf{Q})e_0$ | Jordan product of pure vectors; the vector subspace is not closed |
Further Reading
- Nathan Jacobson, Structure and Representations of Jordan Algebras, AMS Colloquium Publications 39 (1968), for the general theory of Jordan algebras, the Peirce decomposition and the degree.
- Kevin McCrimmon, A Taste of Jordan Algebras (Springer, 2004), for the special Jordan algebras, the envelope and the trace form.
- J. P. Ward, Quaternions and Cayley Numbers: Algebra and Applications (Kluwer, Dordrecht, 1997), for the symmetrized product of quaternions and its idempotents.
- The companion articles of this series: Biquaternion Multiplication, Biquaternion Lie Algebra, Biquaternion Idempotents and Projections, The Six Subspaces and the Jordan Algebra, Biquaternion Relations Between Subspaces, and The Lie–Jordan Decomposition of a Bilinear Product and the Jordan Triple System.