Split-Complex Automorphisms and Derivations

Introduction

The split-complex algebra $\mathbb{D} = \mathbb{R}[x]/(x^2-1)$ is the two-dimensional real algebra with basis $1$, $j$, $j^2 = +1$. This article describes its two standard invariants: the group of $\mathbb{R}$-algebra automorphisms and the Lie algebra of $\mathbb{R}$-linear derivations. It is the two-dimensional counterpart of Biquaternion Automorphisms and Derivations, and the degeneration is sharp. The automorphism group collapses to the two-element group of the identity and the conjugation, and the derivation space collapses to zero: the split-complex algebra, like the complex field, is separable (étale) over $\mathbb{R}$, and a separable algebra of finite rank has no derivations at all. So the count that is two-dimensional here is the Lie algebra of the unit group, not the derivation space, and the article states that distinction explicitly rather than blurring it.

The article owns the automorphism group $\operatorname{Aut}_{\mathbb{R}}(\mathbb{D})$, the derivation space $\operatorname{Der}_{\mathbb{R}}(\mathbb{D})$, the reason the derivation space vanishes, the two-dimensional space of twisted (conjugation-)derivations that replaces it, and the comparison with $\mathbb{C}$ and with $\mathbb{B}$. It assumes the algebra, the idempotents and the norm of Split-Complex Algebra, Split-Complex Idempotents and Projections and Split-Complex Norm and Invertibility, and it cites the general theory of automorphisms and derivations of an algebra as the companion article Automorphisms and Derivations of Algebras. No physics is invoked and no new results are claimed beyond the explicit computations for $\mathbb{D}$.

Conventions. The algebra is $\mathbb{D} = \mathbb{R}[x]/(x^2-1)$, commutative, of real dimension $2$; basis $1$, $j$, $j^2 = +1$; general element $A = a+j a'$; conjugate $\bar A = a-j a'$; idempotents $\Pi_\pm = \tfrac12(1\pm j)$; idempotent coordinates $A_\pm = a\pm a'$; norm $N(A) = a^2-a'^2$. The centre is written $Z(\mathbb{D})$.

Standing Facts: Commutativity, the Centre and Separability

Three structural facts govern the two invariants below. They are recorded here and used throughout.

Commutativity and the centre. The algebra $\mathbb{D}$ is commutative, so its centre is the whole algebra:

$$ Z(\mathbb{D}) = \mathbb{D}, \qquad \dim_{\mathbb{R}} Z(\mathbb{D}) = 2. $$

Consequently every inner derivation $\operatorname{ad}_A(B) = [A,B]$ vanishes identically, and $\mathbb{D}$ is neither simple nor central simple: it has the nontrivial two-sided ideals $\mathbb{R}\Pi_1$ and $\mathbb{R}\Pi_2$, as developed in Split-Complex Ideals and Peirce Decomposition. The centre being two-dimensional and not $\mathbb{R}$ is the statement, over $\mathbb{R}$, that the algebra is generated by an element of square $+1$ rather than being a field; over any base it is the statement that the algebra splits as $\mathbb{R}\oplus\mathbb{R}$.

Separability. The polynomial $x^2-1$ is separable over $\mathbb{R}$, its roots $\pm 1$ being distinct, equivalently its derivative $2x$ is invertible at each root. The algebra extension $\mathbb{R}\to\mathbb{D}$ is therefore étale (separable of finite rank), and this is the reason the derivation space vanishes in the next section but one.

The idempotent form. In the idempotent basis $\{\Pi_1, \Pi_2\}$ the algebra is the direct product

$$ \mathbb{D} \;\cong\; \mathbb{R}\Pi_1 \oplus \mathbb{R}\Pi_2 \;\cong\; \mathbb{R}\times\mathbb{R}, \qquad A \longmapsto (A_+, A_-), $$

an isomorphism of $\mathbb{R}$-algebras, because multiplication is componentwise. This form exhibits $\mathbb{D}$ as a product of two copies of the field $\mathbb{R}$, and it is the form in which both the automorphisms and the (non-existent) derivations are most easily read.

The Automorphism Group

Definition. An $\mathbb{R}$-algebra automorphism of $\mathbb{D}$ is a bijective $\mathbb{R}$-linear map $\sigma : \mathbb{D} \to \mathbb{D}$ with $\sigma(AB) = \sigma(A)\sigma(B)$ and $\sigma(1) = 1$. These maps form a group under composition, written $\operatorname{Aut}_{\mathbb{R}}(\mathbb{D})$.

Theorem. Every $\mathbb{R}$-algebra automorphism of $\mathbb{D}$ is either the identity or the conjugation, and these two maps are automorphisms:

$$ \operatorname{Aut}_{\mathbb{R}}(\mathbb{D}) = \{\mathrm{id}, \bar{\cdot}\} \cong \mathbb{Z}/2, \qquad \bar{\cdot} : a+j a' \longmapsto a-j a'. $$

Proof. Let $\sigma$ be an automorphism. Since $\sigma(1) = 1$ and $\sigma$ respects products,

$$ \sigma(j)^2 = \sigma(j^2) = \sigma(1) = 1, $$

so $u = \sigma(j)$ satisfies $u^2 = 1$ in $\mathbb{D}$. Writing $u = a+j a'$, the equation $u^2 = (a^2+a'^2) + 2a j a' = 1$ forces $a a' = 0$ and $a^2+a'^2 = 1$, whose solutions are $u \in \{1, -1, j, -j\}$. The cases $u = \pm 1$ are excluded because $\sigma$ is injective: $u = 1$ would give $\sigma(j-1) = 0$, and $u = -1$ would give $\sigma(j+1) = 0$, with $j \mp 1 \neq 0$. Hence $u = \pm j$, and $\sigma(a+j a') = a \pm j a'$. Both signs occur, as the identity and the conjugation, and each is bijective and multiplicative.

In the idempotent basis the statement is a permutation of the two components: an automorphism fixes $1 = \Pi_1 + \Pi_2$ and $\Pi_1\Pi_2 = 0$, so it either fixes both idempotents or swaps them, and the swap is exactly the conjugation,

$$ \bar \Pi_1 = \Pi_2, \qquad \bar \Pi_2 = \Pi_1 . $$

So $\operatorname{Aut}_{\mathbb{R}}(\mathbb{D}) \cong \mathbb{Z}/2$ is the group generated by the swap of the two idempotents, and the two element group $\mathbb{Z}/2$ is the whole of it. The map is discrete, so its Lie algebra — the derivations that exponentiate to automorphisms — is zero, in agreement with the vanishing theorem below.

Remark (anti-automorphisms). The conjugation is simultaneously an automorphism and an anti-automorphism, because $\mathbb{D}$ is commutative; there is no distinction here between the two notions, unlike in the biquaternion algebra, where quaternion conjugation is an anti-automorphism that is not an automorphism. The unique nontrivial automorphism is its own inverse and has fixed algebra $\mathbb{R}_{\mathbb{D}}$ and anti-fixed line $j\mathbb{R}_{\mathbb{D}}$, as in Split-Complex Subspaces.

The Derivation Space Vanishes

Definition. An $\mathbb{R}$-linear derivation of $\mathbb{D}$ is an $\mathbb{R}$-linear map $D : \mathbb{D} \to \mathbb{D}$ satisfying the Leibniz rule

$$ D(AB) = D(A)\,B + A\,D(B) $$

for all $A, B \in \mathbb{D}$. The set of all such maps is a real vector space, written $\operatorname{Der}_{\mathbb{R}}(\mathbb{D})$, and it is a Lie algebra under the commutator bracket $[D_1, D_2] = D_1\circ D_2 - D_2\circ D_1$.

Theorem. The derivation space of the split-complex algebra is zero:

$$ \operatorname{Der}_{\mathbb{R}}(\mathbb{D}) = 0 . $$

Proof (direct). A derivation satisfies $D(1) = D(1\cdot 1) = D(1) + D(1)$, hence $D(1) = 0$, and by $\mathbb{R}$-linearity $D$ vanishes on every real number. Apply $D$ to the defining relation $j^2 = 1$:

$$ 0 = D(j^2 - 1) = D(j)j + jD(j) = 2j\,D(j), $$

using commutativity. Since $2j$ is a unit of $\mathbb{D}$ (its inverse is $\tfrac12 j$, because $N(2j) = -4 \neq 0$), it follows that $D(j) = 0$. Hence $D$ vanishes on the basis $\{1, j\}$, so $D = 0$.

Proof (separability). The algebra is the quotient $\mathbb{D} = \mathbb{R}[x]/(x^2-1)$ by a squarefree polynomial. The module of Kähler differentials is

$$ \Omega_{\mathbb{D}/\mathbb{R}} = \mathbb{D}\,dx \big/ (2x\,dx) = \mathbb{D}/(2x)\,\mathbb{D}\, dx = 0, $$

because $2x = 2j$ generates the unit ideal of $\mathbb{D}$. The universal derivation $\mathbb{D}\to\Omega_{\mathbb{D}/\mathbb{R}}$ is therefore zero, and derivations correspond to $\mathbb{D}$-linear maps $\Omega_{\mathbb{D}/\mathbb{R}} \to \mathbb{D}$, so $\operatorname{Der}_{\mathbb{R}}(\mathbb{D}) = \operatorname{Hom}_{\mathbb{D}}(\Omega_{\mathbb{D}/\mathbb{R}}, \mathbb{D}) = 0$.

Proof (idempotent basis). Under the isomorphism $\mathbb{D}\cong\mathbb{R}\times\mathbb{R}$ a derivation corresponds to a pair of derivations of $\mathbb{R}$, since

$$ D(A_+, A_-) = \bigl(D_1(A_+), D_2(A_-)\bigr) $$

for the two components, and $\operatorname{Der}_{\mathbb{R}}(\mathbb{R}) = 0$ because a derivation of $\mathbb{R}$ vanishes on $\mathbb{Q}$ and hence, being $\mathbb{R}$-linear, on $\mathbb{R}$. So each $D_i = 0$ and $D = 0$.

Corollary. Every $\mathbb{R}$-linear derivation of $\mathbb{D}$ is the zero map, so $\operatorname{Der}_{\mathbb{R}}(\mathbb{D}) = 0$; equivalently, the algebra is rigid, and the exponential of any derivation is the identity automorphism.

Remark (inner derivations). Because $\mathbb{D}$ is commutative, $\operatorname{ad}_A = 0$ for every $A$, so the inner derivations are only the zero derivation, $\operatorname{Inn}(\mathbb{D}) = 0$; and since $\operatorname{Der}_{\mathbb{R}}(\mathbb{D}) = 0$, there are no outer derivations either. The whole derivation Lie algebra is the zero Lie algebra of dimension $0$, in contrast with the biquaternion case, where $\operatorname{Der}_{\mathbb{R}}(\mathbb{B}) \cong \mathrm{SO}(1,3)$ has real dimension $6$.

The Two-Dimensional Twisted Derivation Space

The ordinary derivation space of $\mathbb{D}$ is zero, and no genuine two-dimensional derivation space exists. What is two-dimensional, and is the closest structural analogue, is the space of twisted derivations with respect to the conjugation $\sigma = \bar{\cdot}$.

Definition. A $\sigma$-derivation (or twisted derivation) of $\mathbb{D}$, for the automorphism $\sigma = \bar{\cdot}$, is an $\mathbb{R}$-linear map $D : \mathbb{D}\to\mathbb{D}$ satisfying the twisted Leibniz rule

$$ D(AB) = D(A)\,B + \sigma(A)\,D(B) $$

for all $A, B \in \mathbb{D}$. The set of all such maps is written $\operatorname{Der}_{\mathbb{R}}(\mathbb{D}, \sigma)$.

Theorem. The space of $\sigma$-derivations is two-dimensional and is identified with $\mathbb{D}$ itself:

$$ \operatorname{Der}_{\mathbb{R}}(\mathbb{D}, \sigma) \;\cong\; \mathbb{D} \;\cong\; \mathbb{R}^2, \qquad D \longmapsto D(j). $$

Proof. Let $D$ be a $\sigma$-derivation. From the twisted rule applied to $1\cdot 1$,

$$ D(1) = D(1\cdot 1) = D(1) + \sigma(1)D(1) = 2D(1), $$

so $D(1) = 0$, and $D$ is $\mathbb{R}$-linear over $\mathbb{R}$ because $\sigma$ fixes $\mathbb{R}$: $D(rx) = D(r)x + \sigma(r)D(x) = rD(x)$ for $r \in \mathbb{R}$. Apply $D$ to the defining relation, using $\sigma(j) = -j$ and commutativity:

$$ D(j^2 - 1) = D(j)\,j + \sigma(j)\,D(j) - D(1) = D(j)\,j - j\,D(j) = 0, $$

so the relation imposes no constraint on $u = D(j)$. Conversely, for any $u \in \mathbb{D}$, the map

$$ D_u(a+j a') = a'\,u $$

is well defined, $\mathbb{R}$-linear, satisfies $D_u(1) = 0$, and satisfies the twisted rule because the only relation of $\mathbb{D}$ is $j^2 = 1$, on which the twisted rule gives $D_u(j^2) = u\,j - j\,u = 0 = D_u(1)$. Hence $D\mapsto D(j)$ is a linear isomorphism from $\operatorname{Der}_{\mathbb{R}}(\mathbb{D},\sigma)$ onto $\mathbb{D}$.

So the two-dimensional object attached to the derivation picture of $\mathbb{D}$ is the twisted derivation space, identified with the algebra itself by evaluation at $j$. The identification is natural: the basis $\{1, j\}$ of $\mathbb{D}$ corresponds under $D\mapsto D(j)$ to the two twisted derivations

$$ D_1 = D_{1} : a+j a' \longmapsto a', \qquad D_2 = D_{j} : a+j a' \longmapsto j a', $$

with $D_1(j) = 1$ and $D_2(j) = j$. Neither is an ordinary derivation, precisely because $D_1(j) \neq 0$ and the ordinary theorem forbids it.

Remark (no contrast with $\mathbb{C}$ in this respect). The complex field has ordinary derivation space $\operatorname{Der}_{\mathbb{R}}(\mathbb{C}) = 0$ and, with respect to its conjugation, a twisted derivation space $\operatorname{Der}_{\mathbb{R}}(\mathbb{C},\sigma) \cong \mathbb{C}$ by the same computation. So at the level of derivations the split-complex algebra and the complex field behave identically; what separates them is the algebra, a product $\mathbb{R}\times\mathbb{R}$ against a field, and not the derivation calculus. The genuine contrast in this circle of ideas is with the noncommutative biquaternion algebra, whose derivation space is nonzero and six-dimensional.

Comparison with the Biquaternion Case

The biquaternion algebra $\mathbb{B} = \mathbb{C}\otimes_{\mathbb{R}}\mathbb{H}$ is simple, central simple over $\mathbb{C}$, of centre $\mathbb{C}_{\mathbb{B}}\cong\mathbb{C}$; its automorphism group over $\mathbb{R}$ is $\mathrm{PGL}(2,\mathbb{C})\rtimes\mathbb{Z}/2$ and its derivation space over $\mathbb{R}$ is $\mathrm{SO}(1,3)$ of real dimension $6$, spanned by the inner derivations $\operatorname{ad}_a = [a,\cdot]$ and identified with the bivector subspace. The split-complex algebra is the commutative, separable degeneration, and the whole passage is summarised as follows.

feature	$\mathbb{B}$	$\mathbb{D}$
dimension, commutativity	$4$ over $\mathbb{C}$, noncommutative	$2$ over $\mathbb{R}$, commutative
centre	$\mathbb{C}_{\mathbb{B}}\cong\mathbb{C}$, $1$-dimensional	$\mathbb{D}$ itself, $2$-dimensional
simple?	yes (simple; central over $\mathbb{C}$)	no ($\mathbb{R}\Pi_\pm$ are ideals)
$\operatorname{Aut}$ over $\mathbb{R}$	$\mathrm{PGL}(2,\mathbb{C})\rtimes\mathbb{Z}/2$, real dim $6$, two components	$\mathbb{Z}/2$, dim $0$, discrete
generators of $\operatorname{Aut}$	inner automorphisms, plus the conjugate-linear coset	the identity and the swap of $\Pi_1,\Pi_2$
$\operatorname{Der}$ over $\mathbb{R}$	$\mathrm{SO}(1,3)$, real dim $6$, all inner	$0$
reason for the derivation space	noncommutativity $\Rightarrow$ nonzero $\operatorname{ad}_A$; not separable	commutativity and separability $\Rightarrow$ rigidity
twisted derivations	$\sigma$-derivations for the conjugations	$\operatorname{Der}_{\mathbb{R}}(\mathbb{D},\bar{\cdot}) \cong \mathbb{D}$, dim $2$

The pattern is that over $\mathbb{R}$ the biquaternion algebra is noncommutative and has a large inner derivation space, while the split-complex algebra is commutative and separable and has none; the automorphism groups descend from the projective linear group to the two-element group generated by the idempotent swap.

Worked Examples

The nontrivial automorphism as a swap. On the element $A = 4+3j = 7\Pi_1 + 1\Pi_2$, the conjugation gives $\bar A = 4-3j = 1\Pi_1 + 7\Pi_2$, so the two idempotent coordinates $7$ and $1$ are exchanged. On the diagonal, $\bar A = A$ precisely when $a' = 0$; on the anti-diagonal, $\bar A = -A$ precisely when $a = 0$. So the automorphism fixes the real line pointwise and negates the split imaginary line.

The automorphism group is genuine $\mathbb{Z}/2$. The two automorphisms $\mathrm{id}$ and $\bar{\cdot}$ satisfy $\bar{\cdot}\circ\bar{\cdot} = \mathrm{id}$ and are distinct, since $\overline{j} = -j \neq j$; there is no automorphism sending $j$ to $1$ or to $-1$, because it would fail to be injective. Hence $\operatorname{Aut}_{\mathbb{R}}(\mathbb{D})$ has order $2$ exactly.

A direct check of rigidity. Suppose $D$ is an ordinary derivation and, tentatively, $D(j) = \lambda \in \mathbb{D}$. The relation $j^2 = 1$ gives $2j\lambda = 0$, and multiplying by the inverse $\tfrac12 j$ of $2j$ gives $\lambda = 0$. So the single relation kills the candidate completely: there is no freedom, and $\operatorname{Der}_{\mathbb{R}}(\mathbb{D}) = 0$.

A twisted derivation. For $u = 1+2j$ the twisted derivation $D_u$ is $D_u(a+j a') = a'(1+2j)$, so $D_u(1) = 0$, $D_u(j) = 1+2j$, and the twisted rule $D_u(xy) = D_u(x)y + \bar x\,D_u(y)$ holds on the basis. Its image is the line $\mathbb{R}(1+2j)$, so $D_u$ is a rank-one twisted derivation; the map $u\mapsto D_u$ recovers $u = D_u(j)$.

Summary

The split-complex algebra has exactly two $\mathbb{R}$-algebra automorphisms, the identity and the conjugation $\bar A = a-j a'$, which is the swap of the two idempotents $\Pi_1\leftrightarrow \Pi_2$; the group is $\mathbb{Z}/2$, discrete, with zero-dimensional Lie algebra. The derivation space vanishes, $\operatorname{Der}_{\mathbb{R}}(\mathbb{D}) = 0$, and the result has three proofs: the direct one, in which $D(j^2-1) = 2jD(j) = 0$ and $2j$ is a unit force $D(j)=0$; the separable one, in which the algebra is étale over $\mathbb{R}$ and its Kähler differentials vanish; and the idempotent one, in which $\mathbb{D}\cong\mathbb{R}\times\mathbb{R}$ has a derivation space that is a product of the (zero) derivation spaces of the two copies of $\mathbb{R}$. There are no inner derivations other than zero, and no outer ones.

The two-dimensional space attached to the derivation picture is therefore not the ordinary derivation space but the space of twisted derivations with respect to the conjugation, $\operatorname{Der}_{\mathbb{R}}(\mathbb{D},\bar{\cdot}) \cong \mathbb{D}$, identified with the algebra by evaluation at $j$ and spanned by $D_1 : A \mapsto a'$ and $D_2 : A \mapsto j a'$. At the derivation level $\mathbb{D}$ behaves exactly like the complex field, which also has vanishing ordinary derivations and a two-dimensional twisted derivation space with respect to its conjugation; the two differ as algebras, a product of fields against a field. The contrast of the theory is with the noncommutative biquaternion algebra, whose automorphism group is $\mathrm{PGL}(2,\mathbb{C})\rtimes\mathbb{Z}/2$ and whose derivation space is the six-dimensional $\mathrm{SO}(1,3)$. The two-dimensional quantity that is genuinely attached to the algebra is the Lie algebra of its unit group, $\mathrm{Lie}(\mathbb{D}^\times) = \mathbb{D}$ as an abelian Lie algebra, treated in Split-Complex Exponential and Lie Group Structure, not the derivation space.

Summary of Notation

Symbol	Meaning
$\mathbb{D}$	Split complex algebra, $\mathbb{R}[x]/(x^2-1)$
$A = a + j a'$	General split complex number
$\bar{\cdot} = \sigma$	Conjugation, $a+j a' \mapsto a-j a'$; the nontrivial automorphism
$\Pi_\pm = \tfrac12(1\pm j)$	Idempotents, swapped by $\bar{\cdot}$
$Z(\mathbb{D}) = \mathbb{D}$	Centre; the whole algebra, dimension $2$
$\operatorname{Aut}_{\mathbb{R}}(\mathbb{D})$	Automorphism group $\{\mathrm{id}, \bar{\cdot}\}\cong\mathbb{Z}/2$
$\operatorname{Der}_{\mathbb{R}}(\mathbb{D})$	Derivation space; $= 0$
$\operatorname{ad}_A(B) = [A,B]$	Inner derivation; zero for commutative $\mathbb{D}$
$\operatorname{Der}_{\mathbb{R}}(\mathbb{D},\sigma)$	Twisted ($\sigma$-)derivations; $\cong\mathbb{D}$
$D_u(a+j a') = bu$	The twisted derivation with $D_u(j) = u$
$\Omega_{\mathbb{D}/\mathbb{R}}$	Module of Kähler differentials; $= 0$
$\mathrm{SO}(1,3)$	Derivation space of $\mathbb{B}$ over $\mathbb{R}$, for comparison