Biquaternion Ideals and Peirce Decomposition

Introduction

The biquaternion algebra $\mathbb{B} = \mathbb{C} \otimes_{\mathbb{R}} \mathbb{H}$ is four-dimensional over $\mathbb{C}$ and eight-dimensional over $\mathbb{R}$, as established in the article on biquaternion algebra. This article studies the ideal structure of $\mathbb{B}$ and the decomposition of $\mathbb{B}$ relative to a family of idempotents.

Two facts organize the discussion. The algebra $\mathbb{B}$ is simple: its only two-sided ideals are $0$ and $\mathbb{B}$. Its ideal theory is therefore a theory of one-sided ideals: the main results are the classification of the left ideals — $0$, the minimal ones, and $\mathbb{B}$ — and the Peirce decomposition into four one-dimensional corners relative to a pair of orthogonal idempotents.

Because the biquaternions carry both a complex and a real structure, every statement is tagged with the field over which it is made. The sections before the last work over $\mathbb{C}$; §The Real Structure discusses the $\mathbb{R}$-view, where the ideal lattice itself is unchanged but the real structure leaves further traces. The notation follows the article on biquaternion algebra: $e_0$ is the unit, $e_1, e_2, e_3$ are the quaternion units with $e_1 e_2 = e_3$, $i$ is the central scalar imaginary, and a general element is $\tilde{Q} = \sum_{\mu=0}^{3} Q_\mu e_\mu$ with $Q_\mu \in \mathbb{C}$. The definitions of the article on rings apply to the noncommutative algebra $\mathbb{B}$, where "left" and "right" must be distinguished.

Ideals in an Algebra

Let $A$ be an associative unital algebra over a field $k$. An additive subgroup $I \subseteq A$ is a left ideal if $A I \subseteq I$, a right ideal if $I A \subseteq I$, and a two-sided ideal if it is both. The distinction matters only when $A$ is noncommutative; for $\mathbb{B}$ the three notions genuinely differ. The ideals $0$ and $A$ are called trivial.

A base-field remark will be used repeatedly. If $A I \subseteq I$, then $I$ is automatically a $k$-subspace, since $\lambda \tilde R = (\lambda 1) \tilde R \in I$ for $\lambda \in k$, $\tilde R \in I$, as $\lambda 1 \in A$. So the left ideals of a unital algebra do not depend on which field of scalars inside the center is used to view it; in particular the ideal lattice of $\mathbb{B}$ is the same in the $\mathbb{C}$-view and the $\mathbb{R}$-view (§The Real Structure).

For a two-sided ideal $I$, the quotient algebra $A/I$ is the set of cosets with the induced operations, well defined precisely because $I$ absorbs multiplication on both sides. Kernels of algebra homomorphisms are two-sided ideals, and the first isomorphism theorem gives $A/\ker\varphi \cong \operatorname{im}\varphi$. For a left ideal only, $A/I$ is still a left $A$-module but not in general an algebra. Thus the left ideals govern module theory and the two-sided ideals govern quotient algebras.

The Two-Sided Ideals: Simplicity of $\mathbb{B}$

Theorem. Over $\mathbb{C}$, the only two-sided ideals of $\mathbb{B}$ are $0$ and $\mathbb{B}$. Equivalently, $\mathbb{B}$ is a simple $\mathbb{C}$-algebra.

Proof. Let $I \neq 0$ be a two-sided ideal. It is a $\mathbb{C}$-subspace, because multiplication by the central element $i$ is left multiplication by an element of $\mathbb{B}$. Fix $0 \neq \tilde{Q} = \sum_\mu Q_\mu e_\mu \in I$. For every unit $\tilde B \in \mathbb{B}$ the conjugate $\tilde B\tilde{Q}\tilde B^{-1}$ lies in $I$, since $I$ is two-sided.

Average the conjugates over the finite group $\{\pm e_0, \pm e_1, \pm e_2, \pm e_3\}$. Conjugation by $e_\mu$ fixes $e_0$ and $e_\mu$ and reverses $e_\nu$ for $\nu \neq \mu$, so the group elements of quaternion part $\pm e_\mu$ all give the same conjugate, and the average is twice

$$ Q_0 e_0 + \tfrac12\big(Q_0 e_0 + Q_1 e_1 - Q_2 e_2 - Q_3 e_3\big) + \tfrac12\big(Q_0 e_0 - Q_1 e_1 + Q_2 e_2 - Q_3 e_3\big) + \tfrac12\big(Q_0 e_0 - Q_1 e_1 - Q_2 e_2 + Q_3 e_3\big) = 4 Q_0 e_0, $$

divided by $8$, giving $Q_0 e_0 \in I$. If $Q_0 \neq 0$ then $e_0 = Q_0^{-1}(Q_0 e_0) \in I$ and $I = \mathbb{B}$. If $Q_0 = 0$ then $\tilde{Q}$ is a nonzero element of the vector subspace, and conjugating it by the real unit quaternions rotates it: the conjugates run over a sphere in $\mathrm{Vect}(\mathbb{B})$, whose real span is all of $\mathrm{Vect}(\mathbb{B})$, so $\mathrm{Vect}(\mathbb{B}) \subseteq I$. In particular $e_1 \in I$, hence $e_1^2 = -e_0 \in I$ and again $I = \mathbb{B}$.

Consequences over $\mathbb{C}$:

The only quotient algebras of $\mathbb{B}$ are $\mathbb{B}$ and $0$.
Every nonzero element generates $\mathbb{B}$ as a two-sided ideal.
The center of $\mathbb{B}$ is the scalar copy of $\mathbb{C}$, so $\mathbb{B}$ is a central simple $\mathbb{C}$-algebra; in particular it is not a product $A_1 \times A_2$ of nonzero algebras, since each factor would give a nontrivial two-sided ideal.
The many one-sided ideals of §Minimal Left and Right Ideals and §The Lattice of Left Ideals are all non-two-sided, so they do not contradict simplicity.

Artinian, Semisimple, and Length Two

A nonzero module is simple if it has no submodules other than $0$ and itself. A composition series is a finite strictly increasing chain $0 = M_0 \subset M_1 \subset \cdots \subset M_r = M$ with simple successive quotients $M_i/M_{i-1}$; the number $r$ is the length of $M$. A ring is left artinian if every descending chain of left ideals stabilizes, and semisimple if its left regular module is a direct sum of simple modules.

Over $\mathbb{C}$, every descending chain of left ideals of $\mathbb{B}$ is a descending chain of $\mathbb{C}$-subspaces, so it stabilizes; thus $\mathbb{B}$ is artinian. By Wedderburn–Artin, a unital ring is semisimple if and only if it is artinian with zero Jacobson radical, and every simple artinian ring is semisimple. Hence $\mathbb{B}$ is semisimple, and every left ideal is a direct sum of minimal left ideals.

All simple left $\mathbb{B}$-modules are isomorphic, since a simple artinian ring has a unique simple module up to isomorphism. With the idempotents $\tilde\Pi_1, \tilde\Pi_2$ of §Idempotents and Orthogonal Idempotents, the algebra as a left module over itself reads $\mathbb{B} = \mathbb{B}\tilde\Pi_1 \oplus \mathbb{B}\tilde\Pi_2$, with $\mathbb{B}\tilde\Pi_1$ and $\mathbb{B}\tilde\Pi_2$ both minimal left ideals. Hence $0 \subset \mathbb{B}\tilde\Pi_1 \subset \mathbb{B}$ is a composition series, and the length of $\mathbb{B}$ as a left module over itself is $2$, with both factors isomorphic. The right regular module has length $2$ as well. All of this is over $\mathbb{C}$.

Idempotents and Orthogonal Idempotents

Let $A$ be an associative unital algebra. An element $e \in A$ is an idempotent if $e^2 = e$, and two idempotents $e, f$ are orthogonal if $ef = fe = 0$. A nonzero idempotent $e$ is primitive if it is not a sum of two nonzero orthogonal idempotents, and for a semisimple algebra

$$ e \text{ primitive} \iff Ae \text{ is a minimal left ideal} \iff eAe \text{ is a division ring}. $$

Over $\mathbb{C}$, the idempotents

$$ \tilde\Pi_1 = \frac{e_0 + i e_3}{2}, \qquad \tilde\Pi_2 = \frac{e_0 - i e_3}{2} $$

satisfy $\tilde\Pi_1^2 = \tilde\Pi_1$, $\tilde\Pi_2^2 = \tilde\Pi_2$, $\tilde\Pi_1\tilde\Pi_2 = \tilde\Pi_2\tilde\Pi_1 = 0$ and $\tilde\Pi_1 + \tilde\Pi_2 = e_0$. They are primitive, and they generate the off-diagonal elements of the next section.

The classification of the idempotents of $\mathbb{B}$ — the trivial idempotents, the bijection with the roots of $-1$, the Hermitian projections and the dimension of the idempotent set — is the subject of Biquaternion Idempotents and Projections.

The Off-Diagonal Elements

The two off-diagonal corners are spanned by the elements

$$ \tilde R = \frac{i e_1 - e_2}{2}, \qquad \tilde T = \frac{i e_1 + e_2}{2}. $$

Then $\{\tilde\Pi_1, \tilde R, \tilde T, \tilde\Pi_2\}$ is a $\mathbb{C}$-basis of $\mathbb{B}$, and the multiplication is

$$ \tilde\Pi_1\tilde R = \tilde R = \tilde R\tilde\Pi_2, \qquad \tilde\Pi_2\tilde T = \tilde T = \tilde T\tilde\Pi_1, \qquad \tilde R\tilde T = \tilde\Pi_1, \qquad \tilde T\tilde R = \tilde\Pi_2, $$

together with $\tilde R\tilde\Pi_1 = \tilde\Pi_2\tilde R = \tilde\Pi_1\tilde T = 0$, $\tilde T\tilde\Pi_2 = 0$, and $\tilde R^2 = \tilde T^2 = 0$.

The Peirce Decomposition

The Peirce decomposition is the decomposition of an algebra relative to a family of orthogonal idempotents; it is the algebraic form of a block decomposition of a matrix.

One idempotent. Let $e \in A$ be idempotent and $f = 1-e$. Every $a \in A$ expands as $a = eae + eaf + fae + faf$, giving the direct sum

$$ A = eAe \oplus eAf \oplus fAe \oplus fAf, $$

whose summands are the Peirce spaces. The corner $eAe$ is a subalgebra with identity $e$; the other corners are only one-sided pieces. The expansion, the directness of the four terms and the product rule below are proved in Unital Algebras, §Idempotents and the Peirce Decomposition, where the case of a central idempotent is also separated: for a central $e$ the off-diagonal corners vanish and the sum is a product of algebras. That special case does not arise here, since the idempotents of this article lie in a simple algebra and no nontrivial idempotent of $\mathbb{B}$ is central.

A complete orthogonal family. If $\{e_1, \dots, e_n\}$ is complete and pairwise orthogonal, the same expansion gives

$$ A = \bigoplus_{i,j=1}^{n} e_i A e_j, \qquad (e_i A e_j)(e_k A e_l) \subseteq \delta_{jk}\, e_i A e_l. $$

Each diagonal corner $e_i A e_i$ is an algebra with identity $e_i$, and each off-diagonal piece is a bimodule over the corresponding corners.

The biquaternion case. Take $e_1 = \tilde\Pi_1$, $e_2 = \tilde\Pi_2$ from §Idempotents and Orthogonal Idempotents. The Peirce decomposition of $\mathbb{B}$ is

$$ \mathbb{B} = \tilde\Pi_1\mathbb{B}\tilde\Pi_1 \oplus \tilde\Pi_1\mathbb{B}\tilde\Pi_2 \oplus \tilde\Pi_2\mathbb{B}\tilde\Pi_1 \oplus \tilde\Pi_2\mathbb{B}\tilde\Pi_2, $$

and by the multiplication table of §The Off-Diagonal Elements each summand is one-dimensional over $\mathbb{C}$:

$$ \tilde\Pi_1\mathbb{B}\tilde\Pi_1 = \mathbb{C}\tilde\Pi_1, \qquad \tilde\Pi_1\mathbb{B}\tilde\Pi_2 = \mathbb{C}\tilde R, \qquad \tilde\Pi_2\mathbb{B}\tilde\Pi_1 = \mathbb{C}\tilde T, \qquad \tilde\Pi_2\mathbb{B}\tilde\Pi_2 = \mathbb{C}\tilde\Pi_2. $$

So the Peirce decomposition of $\mathbb{B}$ is

$$ \mathbb{B} = \mathbb{C}\tilde\Pi_1 \oplus \mathbb{C}\tilde R \oplus \mathbb{C}\tilde T \oplus \mathbb{C}\tilde\Pi_2. $$

The diagonal part $\tilde\Pi_1\mathbb{B}\tilde\Pi_1 \oplus \tilde\Pi_2\mathbb{B}\tilde\Pi_2 = \mathbb{C}\tilde\Pi_1 \oplus \mathbb{C}\tilde\Pi_2$ is a two-dimensional commutative subalgebra isomorphic to $\mathbb{C} \times \mathbb{C}$. Each diagonal corner is a division ring, namely $\mathbb{C}$, which is the primitivity criterion of §Idempotents and Orthogonal Idempotents.

A null (light-cone) variant of the same split is used in The Chiral Algebra of Biquaternions and the Cyclic Representation of the Dirac Equation. The idempotents there are the uniform nullquaternions $N = \tfrac12(1, \mathbf n)$ and $\bar N = \tfrac12(1, -\mathbf n)$, and the two Peirce components of a biquaternion are its signed parts, whose sign the source reads as the chirality of the Weyl spinor. The pair is orthogonal and complete in the article's own product — the outer product, for which $N \odot N = N$, $\bar N \odot \bar N = \bar N$, $N \odot \bar N = 0$ and $N + \bar N = e_0$ — and the article's matrix isomorphism carries the outer product to the ordinary matrix product and $N, \bar N$ to the two standard diagonal idempotents. It is the same orthogonal pair as above, transported to the light-cone coordinates with the product changed; these same elements are not idempotent under the Hamilton product of this article, so the chiral split is naturally stated in that paper's algebra rather than in this one.

The Decomposition into Minimal Left and Right Ideals

The two idempotents group the basis into one-sided ideals in a second way. The two left ideals $\mathbb{B}\tilde\Pi_1, \mathbb{B}\tilde\Pi_2$ and the two right ideals $\tilde\Pi_1\mathbb{B}, \tilde\Pi_2\mathbb{B}$ are one-sided, and

$$ \mathbb{B} = \mathbb{B}\tilde\Pi_1 \oplus \mathbb{B}\tilde\Pi_2 = (\mathbb{C}\tilde\Pi_1 \oplus \mathbb{C}\tilde T) \oplus (\mathbb{C}\tilde R \oplus \mathbb{C}\tilde\Pi_2), $$

$$ \mathbb{B} = \tilde\Pi_1\mathbb{B} \oplus \tilde\Pi_2\mathbb{B} = (\mathbb{C}\tilde\Pi_1 \oplus \mathbb{C}\tilde R) \oplus (\mathbb{C}\tilde T \oplus \mathbb{C}\tilde\Pi_2). $$

The first exhibits $\mathbb{B}$ as a direct sum of the two minimal left ideals; the second exhibits it as a direct sum of the two minimal right ideals. The two groupings of the same four basis elements differ: the Peirce decomposition groups $\tilde\Pi_1$ with $\tilde R$ and $\tilde\Pi_2$ with $\tilde T$, whereas the left-ideal decomposition groups $\tilde\Pi_1$ with $\tilde T$ and $\tilde\Pi_2$ with $\tilde R$. Each left ideal is two-dimensional over $\mathbb{C}$; each right ideal is its dual. Since $\mathbb{B}$ is simple, a one-sided ideal is never two-sided; for instance $\mathbb{B}\tilde\Pi_1$ is not stable under right multiplication by $\tilde R$.

Why the sum is direct, and the dimensions. Every $\tilde{Q} \in \mathbb{B}$ satisfies $\tilde{Q} = \tilde{Q}(\tilde\Pi_1 + \tilde\Pi_2) = \tilde{Q}\tilde\Pi_1 + \tilde{Q}\tilde\Pi_2$, so the two left ideals span. Their intersection is zero: if $\tilde{Q}\tilde\Pi_1 = \tilde{P}\tilde\Pi_2$, then multiplying on the right by $\tilde\Pi_1$ and using $\tilde\Pi_1^2 = \tilde\Pi_1$ and $\tilde\Pi_2\tilde\Pi_1 = 0$ gives $\tilde{Q}\tilde\Pi_1 = 0$. The basis above reads $\mathbb{B}\tilde\Pi_1 = \mathbb{C}\tilde\Pi_1 \oplus \mathbb{C}\tilde T$, of dimension $2$ over $\mathbb{C}$ and $4$ over $\mathbb{R}$, with $\mathbb{B}\tilde\Pi_2 = \mathbb{C}\tilde R \oplus \mathbb{C}\tilde\Pi_2$ for the second; together they account for $4 + 4 = 8 = \dim_{\mathbb{R}} \mathbb{B}$. Each is a minimal left ideal, by the primitivity of $\tilde\Pi_1$ and $\tilde\Pi_2$ (§Idempotents and Orthogonal Idempotents, §Minimal Left and Right Ideals).

The module structure. With the basis $\{\tilde\Pi_1, \tilde T\}$ the left $\mathbb{B}$-module $\mathbb{B}\tilde\Pi_1$ is $\mathbb{C}^2$, and the central element $i$ acts on it as the scalar $i$: $$ i\,(\alpha \tilde\Pi_1 + \beta \tilde T) = (i\alpha)\tilde\Pi_1 + (i\beta)\tilde T . $$ So the simple module underlying each minimal left ideal is the standard two-dimensional one, with $i$ acting by the identity matrix.

Minimal Left and Right Ideals

A minimal left ideal is a nonzero left ideal containing no nonzero proper left ideal; equivalently, a simple submodule of the left regular module. A minimal right ideal is defined the same way on the right.

Over $\mathbb{C}$, since $\mathbb{B}$ is semisimple, every left ideal of $\mathbb{B}$ is a direct sum of minimal left ideals, and every minimal left ideal is of the form $\mathbb{B}e$ for a primitive idempotent $e$. The examples are the left ideals $\mathbb{B}\tilde\Pi_1$ and $\mathbb{B}\tilde\Pi_2$ of §The Decomposition into Minimal Left and Right Ideals, and all minimal left ideals are isomorphic as left $\mathbb{B}$-modules: a general one is obtained from $\mathbb{B}\tilde\Pi_1$ by an algebra automorphism, so it has the same shape in a suitable basis. Dually, every minimal right ideal is isomorphic to $\tilde\Pi_1\mathbb{B}$. Thus there is one isomorphism class of simple left modules and one of simple right modules. Being nonzero proper one-sided ideals, none of them is two-sided — which is exactly why their abundance is compatible with §The Two-Sided Ideals: Simplicity of $\mathbb{B}$.

The Lattice of Left Ideals

Over $\mathbb{C}$, every left ideal of the simple artinian algebra $\mathbb{B}$ is a direct sum of minimal left ideals, so the left ideals are exactly $0$, the minimal ones, and $\mathbb{B}$. The minimal left ideals form the middle layer of the lattice,

$$ 0 \;\subset\; \{\, L : L \text{ a minimal left ideal} \,\} \;\subset\; \mathbb{B}, $$

the elements of the middle layer being pairwise incomparable, each covering $0$ and covered by $\mathbb{B}$. Because the length of $\mathbb{B}$ as a left module over itself is $2$, every minimal left ideal is also a maximal left ideal, nothing lying strictly between it and $\mathbb{B}$. The minimal left ideals are parametrized by the projective line $\mathbb{P}^1(\mathbb{C})$. Right ideals admit the same description, with the dual parametrization.

The parameter space deserves a caution. In the Wedderburn–Artin decomposition $M_n(D)$ of a simple artinian algebra, the minimal one-sided ideals are parametrized by the projective space $\mathbb{P}^{n-1}(D)$ over the division ring $D$, not over an arbitrary base field. Here $D = \mathbb{C}$ and $n = 2$, so the parameter space is $\mathbb{P}^1(\mathbb{C})$. In particular, regarding $\mathbb{B}$ as an $\mathbb{R}$-algebra does not replace this by $\mathbb{P}^1(\mathbb{R})$: the one-sided ideals are still parametrized by $\mathbb{P}^1(\mathbb{C})$. The real projective line appears only as the subfamily that the real structure preserves, not as a set of fixed minimal left ideals: §The Real Structure shows that no minimal left ideal is stable under coefficient conjugation.

The Radical

The Jacobson radical $J(A)$ is the intersection of all maximal left ideals of $A$, equivalently of all maximal right ideals; it is a two-sided ideal. For an artinian ring, $J(A)$ is the largest nilpotent ideal and $A/J(A)$ is semisimple; an artinian ring is semisimple if and only if its radical is zero.

Over $\mathbb{C}$, the radical of $\mathbb{B}$ vanishes:

$$ J(\mathbb{B}) = 0. $$

Indeed $J(\mathbb{B})$ is a two-sided ideal, hence by simplicity is $0$ or $\mathbb{B}$; it cannot be $\mathbb{B}$, since in a unital algebra the radical is proper. This restates the semisimplicity of §Artinian, Semisimple, and Length Two in terms of the radical. Consequences over $\mathbb{C}$:

$\mathbb{B}$ has no nonzero nilpotent two-sided ideals; the nilradical is zero.
It has no nonzero nilpotent left or right ideals either, since such an ideal generates a nonzero nilpotent two-sided ideal.
The absence of a radical is not the absence of nilpotent elements: $\tilde R^2 = \tilde T^2 = 0$ (§The Off-Diagonal Elements), yet the left ideal generated by $\tilde R$ is not nilpotent.
Every left $\mathbb{B}$-module is semisimple.

The Real Structure

Everything so far was stated over $\mathbb{C}$. We now regard the same set $\mathbb{B}$ as an eight-dimensional algebra over $\mathbb{R}$ and record what changes.

(a) The ideal lattice does not change. By the base-field remark of §Ideals in an Algebra, an additive subgroup closed under left multiplication by $\mathbb{B}$ is automatically a complex subspace, because multiplication by $i$ is left multiplication by the central element $i e_0$. So an $\mathbb{R}$-left ideal is the same thing as a $\mathbb{C}$-left ideal, and the same holds on the right and for two-sided ideals. In particular, over $\mathbb{R}$: $\mathbb{B}$ is still simple, with two-sided ideals only $0$ and $\mathbb{B}$; the minimal left ideals are the same subsets, parametrized by $\mathbb{P}^1(\mathbb{C})$; the radical is still zero; and the length as a module over itself is still $2$.

(b) The algebra is not central over $\mathbb{R}$. The center of $\mathbb{B}$ is the copy of $\mathbb{C}$ spanned by $e_0$ and $i e_0$ — the complex subspace $\mathbb{C}_{\mathbb{B}}$ — a proper field extension of $\mathbb{R}$ of degree $2$. Thus $\mathbb{B}$ is a simple $\mathbb{R}$-algebra but not a central simple one. In the Wedderburn–Artin description $\mathbb{B} \cong M_n(D)$ over $\mathbb{R}$, one has $n = 2$ and $D = \mathbb{C}$.

(c) The enrichment appears after extension of scalars. The complexification of the real algebra $\mathbb{B}$ is

$$ \mathbb{B} \otimes_{\mathbb{R}} \mathbb{C} \cong (\mathbb{C} \otimes_{\mathbb{R}} \mathbb{H}) \otimes_{\mathbb{R}} \mathbb{C} \cong (\mathbb{C} \otimes_{\mathbb{R}} \mathbb{C}) \otimes_{\mathbb{R}} \mathbb{H} \cong (\mathbb{C} \oplus \mathbb{C}) \otimes_{\mathbb{R}} \mathbb{H} \cong \mathbb{B} \oplus \mathbb{B}, $$

using $\mathbb{C} \otimes_{\mathbb{R}} \mathbb{C} \cong \mathbb{C} \oplus \mathbb{C}$ and the associativity and commutativity of $\otimes_{\mathbb{R}}$. The algebra on the right is not simple: its two-sided ideals are $0$, the two summands, and the whole ring. So the real biquaternion algebra is not absolutely simple: simple over $\mathbb{R}$, but with a complexification that splits as a product of two simple algebras. This is the precise sense in which the real structure carries a richer two-sided ideal theory — not in the lattice of $\mathbb{B}$ itself, which is $\{0, \mathbb{B}\}$ in both views, but in the lattice produced by base change. In contrast, $\mathbb{H} \otimes_{\mathbb{R}} \mathbb{C} \cong \mathbb{B}$ is simple: it is the real biquaternion algebra, not $\mathbb{H}$, whose complexification splits.

(d) The action of complex conjugation on the lattice. Complex conjugation $\tilde{Q} \mapsto \tilde{Q}^{*} = \sum_\mu \bar{Q}_\mu e_\mu$ is an $\mathbb{R}$-algebra automorphism of $\mathbb{B}$ (it is $\mathbb{C}$-antilinear), so it permutes the left ideals, $\sigma(\mathbb{B}\tilde{\chi}) = \mathbb{B}\sigma(\tilde{\chi})$. Since $\sigma(\tilde\Pi_1) = \tilde\Pi_2$ and $\sigma(\tilde\Pi_2) = \tilde\Pi_1$, it interchanges the two standard left ideals:

$$ \sigma(\mathbb{B}\tilde\Pi_1) = \mathbb{B}\tilde\Pi_2, \qquad \sigma(\mathbb{B}\tilde\Pi_2) = \mathbb{B}\tilde\Pi_1. $$

On the parametrizing projective line the induced map $t \mapsto -1/\bar{t}$ has no fixed point, so no minimal left ideal is stable under complex conjugation. The real lines still form the real projective line $\mathbb{P}^1(\mathbb{R}) \subset \mathbb{P}^1(\mathbb{C})$, which $\sigma$ preserves and acts on as the antipodal map. So although the lattice of left ideals is unchanged from $\mathbb{C}$ to $\mathbb{R}$, the real structure does not mark out conjugation-stable minimal left ideals; what it marks out is the subfamily $\mathbb{P}^1(\mathbb{R})$ that it preserves.

Summary

Statement	Over $\mathbb{C}$	Over $\mathbb{R}$
Dimension	$4$	$8$
Algebra type	simple $\mathbb{C}$-algebra, $\dim 4$	simple $\mathbb{R}$-algebra, $\dim 8$
Two-sided ideals	$0$, $\mathbb{B}$ (simple)	$0$, $\mathbb{B}$ (simple)
Central simple?	yes, center $\mathbb{C}$	no, center $\mathbb{C} \neq \mathbb{R}$
Left ideals	$0$, the $\mathbb{P}^1(\mathbb{C})$ of minimal ones, $\mathbb{B}$	same lattice
Minimal left ideals	all isomorphic	same; $\sigma$ pairs them, none stable
Minimal right ideals	all isomorphic	same
Jacobson radical	$0$	$0$
Length as module over itself	$2$	$2$
Base change $\otimes_{\mathbb{R}}\mathbb{C}$	—	not simple; splits into two simple factors

Summary of Notation

Symbol	Meaning
$\mathbb{B} = \mathbb{C} \otimes_{\mathbb{R}} \mathbb{H}$	Biquaternion algebra; dimension $4$ over $\mathbb{C}$, $8$ over $\mathbb{R}$
$e_0, e_1, e_2, e_3$	Algebra basis, $e_0 = 1$, $e_k^2 = -e_0$
$i$	Central scalar imaginary; $\mathbb{C}_{\mathbb{B}} = \mathrm{span}_{\mathbb{R}}\{e_0, ie_0\}$ is the center
$\tilde\Pi_1, \tilde\Pi_2$	Orthogonal idempotents, $\tilde\Pi_1+\tilde\Pi_2 = e_0$, $\tilde\Pi_1\tilde\Pi_2 = \tilde\Pi_2\tilde\Pi_1 = 0$
$\tilde R, \tilde T$	Nilpotent off-diagonal elements, $\tilde R = \tfrac{i e_1 - e_2}{2}$, $\tilde T = \tfrac{i e_1 + e_2}{2}$, $\tilde R^2 = \tilde T^2 = 0$
$\mathbb{B}\tilde\Pi_1 \cong \mathbb{C}^2$	The minimal left ideal, a simple left $\mathbb{B}$-module, $i$ acting as the scalar $i$
$\mathbb{P}^1(\mathbb{C})$	The projective line parameterising the minimal left ideals
$\sigma$	The real structure, pairing the two standard minimal left ideals
$\mathcal{J}$	Jacobson radical; it is $0$ for $\mathbb{B}$