Complex Automorphisms and Derivations

Introduction

The complex algebra $\mathbb{C}$ carries two standard invariants of its algebra structure: the group of algebra automorphisms and the Lie space of derivations. Both depend on the ground field, so the two views are kept separate and the field is named at each step. The complex case is the degenerate base of the family: the automorphism group over $\mathbb{R}$ is the two-element group generated by complex conjugation, the automorphism group over $\mathbb{C}$ is trivial, and the derivation space vanishes over both fields. These are the smallest possible values, and their smallness is a mathematical statement about a commutative field rather than an omission.

The conventions are those of Complex Algebra and Complex Subspaces: basis $1$, $i$ with $i^2 = -1$; a general element $A = a + i a'$; the unique nontrivial involution written $\bar{\cdot}$ and called complex conjugation; norm $N(A) = A\bar A = a^2+a'^2$. The comparison throughout is with the biquaternion algebra $\mathbb{B} = \mathbb{C} \otimes_{\mathbb{R}} \mathbb{H}$, for which $\operatorname{Aut}_{\mathbb{C}}(\mathbb{B}) = PGL(2,\mathbb{C})$ and $\operatorname{Der}_{\mathbb{C}}(\mathbb{B}) \cong \mathrm{SL}(2,\mathbb{C})$; the present article is the one-dimensional account of the same two invariants.

No physics is invoked and no new results are claimed. Everything below is standard structure theory of the field $\mathbb{C}$ and of the quadratic extension $\mathbb{C}/\mathbb{R}$.

Standing Facts: The Ideal Lattice, the Centre and the Real Subspace

The automorphism group and the derivation space of $\mathbb{C}$ are governed by three structural facts, recorded here and used throughout.

No ideals. Every nonzero element of $\mathbb{C}$ is invertible, so the only two-sided ideals of $\mathbb{C}$ are $0$ and $\mathbb{C}$: the algebra is a field, in particular a division algebra with no proper nonzero ideal. The ideal lattice is the two-point lattice $\{0, \mathbb{C}\}$, the smallest possible.

The centre. The centre of $\mathbb{C}$ is $\mathbb{C}$ itself, since the algebra is commutative:

$$ Z(\mathbb{C}) = \{A \in \mathbb{C} : AB = BA \ \text{for all}\ B\} = \mathbb{C}. $$

Hence, over the ground field $\mathbb{C}$, the algebra is central simple — simple, finite-dimensional, with centre exactly the ground field — while over $\mathbb{R}$ it is simple but not central, its centre $\mathbb{C}$ being strictly larger than $\mathbb{R} 1$. This is exactly the distinction that holds for $\mathbb{B}$: $\mathbb{B}$ is central simple over $\mathbb{C}$ and merely simple over $\mathbb{R}$. The two algebras differ not in that distinction but in what the invariant groups turn out to be.

The real subspace. The image of $\mathbb{R}$ in $\mathbb{C}$ under $a \mapsto a$ is the fixed field of the nontrivial automorphism, a real-closed subfield of $\mathbb{C}$ over which $\mathbb{C}$ is a degree-two Galois extension. It is called the real subspace and written $\mathbb{R}_{\mathbb{C}}$; its structure is the subject of Complex Subspaces. It is the fixed field of complex conjugation and the subfield of reference, but it is not the only subfield of $\mathbb{C}$ of index two: a wild automorphism carries $\mathbb{R}$ to a different real-closed subfield $\sigma(\mathbb{R})$ over which $\mathbb{C}$ is again quadratic, as discussed in Galois Theory of ℂ/ℝ.

Automorphisms over $\mathbb{R}$

Throughout this section the ground field is $\mathbb{R}$.

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

Determination by the image of $i$. An $\mathbb{R}$-algebra automorphism $\sigma$ is $\mathbb{R}$-linear, hence determined by its value on the single extra basis element $i$, since $\sigma(a+i a') = a + a'\,\sigma(i)$ for $a, a' \in \mathbb{R}$. The value $\sigma(i)$ obeys

$$ \sigma(i)^2 = \sigma(i^2) = \sigma(-1) = -1, $$

so $\sigma(i)$ is a square root of $-1$ in $\mathbb{C}$. The two square roots are $\pm i$, and both choices extend to an automorphism.

Theorem. $\operatorname{Aut}_{\mathbb{R}}(\mathbb{C}) = \{\operatorname{id}, \bar{\cdot}\} \cong \mathbb{Z}/2$.

Proof. By the display above, $\sigma(i) = i$ or $\sigma(i) = -i$. The first choice gives $\sigma(a+i a') = a+i a' = \operatorname{id}$, the second gives $\sigma(a+i a') = a-i a' = \bar{\cdot}$. Both are $\mathbb{R}$-algebra automorphisms: the identity trivially, conjugation because $\overline{AB} = \bar A\bar B$, and the group is $\mathbb{Z}/2$ since conjugation is an involution and there are no further elements.

The two automorphisms. The identity fixes every element; complex conjugation $\bar A = a - i a'$ fixes the real subspace $\mathbb{R}_{\mathbb{C}}$ pointwise and negates the imaginary subspace $i\mathbb{R}_{\mathbb{C}}$.

Automorphisms over $\mathbb{C}$

Now the ground field is $\mathbb{C}$: an automorphism is required to be $\mathbb{C}$-linear, not merely $\mathbb{R}$-linear, which by the usual convention means $\sigma(\lambda A) = \lambda\,\sigma(A)$ for $\lambda \in \mathbb{C}$.

Theorem. $\operatorname{Aut}_{\mathbb{C}}(\mathbb{C}) = \{\operatorname{id}\}$, the trivial group.

Proof. A $\mathbb{C}$-algebra automorphism fixes $1$ by definition and is $\mathbb{C}$-linear, so it fixes every element of the form $\lambda \cdot 1 = \lambda$ with $\lambda \in \mathbb{C}$; since every element of $\mathbb{C}$ has that form, $\sigma = \operatorname{id}$.

The result is the exact analogue of Skolem–Noether for a one-dimensional central simple algebra. In the biquaternion case, by contrast, the algebra is four-dimensional and central simple over $\mathbb{C}$, every automorphism is inner, and the group is $PGL(2,\mathbb{C})$ of dimension $3$. Here the algebra is the ground field itself, the group of units modulo scalars is $GL(1,\mathbb{C})/\mathbb{C}^\times = 1$, and the automorphism group is trivial. Complex conjugation does not appear in $\operatorname{Aut}_{\mathbb{C}}(\mathbb{C})$: it is not $\mathbb{C}$-linear, since $\overline{i \cdot 1} = -i$ while $i\cdot \overline{1} = i$. This mirrors the biquaternion remark that the quaternion conjugation is an anti-automorphism and the complex conjugation is an automorphism but not $\mathbb{C}$-linear; here the only candidate nontrivial automorphism is exactly the non-$\mathbb{C}$-linear one.

Derivations

Definition. An $\mathbb{R}$-linear derivation of $\mathbb{C}$ is an $\mathbb{R}$-linear map $D : \mathbb{C} \to \mathbb{C}$ with

$$ D(AB) = D(A)\,B + A\,D(B) \qquad (A, B \in \mathbb{C}). $$

The set of such maps is a real vector space, written $\operatorname{Der}_{\mathbb{R}}(\mathbb{C})$. A $\mathbb{C}$-linear derivation requires in addition $D(\lambda A) = \lambda D(A)$ for $\lambda \in \mathbb{C}$, and the set of these is written $\operatorname{Der}_{\mathbb{C}}(\mathbb{C})$. Both spaces carry the commutator bracket and are Lie algebras.

Every derivation vanishes on the base field. Every derivation satisfies $D(1) = 0$, since $D(1) = D(1\cdot 1) = D(1) + D(1) = 2D(1)$ in characteristic not $2$, and hence $D(a) = 0$ for all $a \in \mathbb{R}$.

Theorem. $\operatorname{Der}_{\mathbb{R}}(\mathbb{C}) = 0$; a fortiori $\operatorname{Der}_{\mathbb{C}}(\mathbb{C}) = 0$.

Proof. A derivation $D$ is $\mathbb{R}$-linear and $\mathbb{C} = \mathbb{R} 1 \oplus \mathbb{R} i$, so $D$ is determined by the single value $D(i)$. From $i^2 = -1$,

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

so $D(i) = 0$ and hence $D = 0$. A $\mathbb{C}$-linear derivation is in particular an $\mathbb{R}$-linear one, so it too vanishes.

Vanishing of the inner derivations. For a commutative algebra every inner derivation vanishes:

$$ \operatorname{ad}_A(B) = AB - BA = 0 \qquad (A, B \in \mathbb{C}), $$

so the kernel of $\operatorname{ad} : \mathbb{C} \to \operatorname{Der}_{\mathbb{R}}(\mathbb{C})$ is all of $\mathbb{C}$, and the map itself is the zero map. In the biquaternion case $\operatorname{ad} : \mathbb{B} \to \operatorname{Der}_{\mathbb{C}}(\mathbb{B})$ has kernel the centre and induces $\operatorname{Der}_{\mathbb{C}}(\mathbb{B}) \cong \mathbb{B}/\mathbb{C}_{\mathbb{B}} \cong \mathrm{SL}(2,\mathbb{C})$; here the source and the target both collapse to the commutativity of the field.

Dimension. $\dim_{\mathbb{R}} \operatorname{Der}_{\mathbb{R}}(\mathbb{C}) = 0$ and $\dim_{\mathbb{C}} \operatorname{Der}_{\mathbb{C}}(\mathbb{C}) = 0$.

The Lie Algebra of the Trivial Group

The derivations of an algebra are the Lie algebra of its automorphism group. For $\mathbb{C}$ this consistency is visible on both sides.

Over $\mathbb{C}$, $\operatorname{Aut}_{\mathbb{C}}(\mathbb{C}) = 1$ and $\operatorname{Der}_{\mathbb{C}}(\mathbb{C}) = 0$: the Lie algebra of the trivial group is the zero Lie algebra, of dimension $0$. Over $\mathbb{R}$, the group $\operatorname{Aut}_{\mathbb{R}}(\mathbb{C}) = \mathbb{Z}/2$ is discrete; its identity component is the trivial group, and the Lie algebra of a discrete group is $0$, matching $\operatorname{Der}_{\mathbb{R}}(\mathbb{C}) = 0$. Thus the identity

$$ \operatorname{Lie} \operatorname{Aut}(\mathbb{C}) = \operatorname{Der}(\mathbb{C}) = 0 $$

holds over both fields, in the degenerate sense that both sides are the zero space and both groups have trivial identity component. In the biquaternion case the corresponding identity is $\operatorname{Lie}\operatorname{Aut}_{\mathbb{C}}(\mathbb{B}) = \operatorname{Der}_{\mathbb{C}}(\mathbb{B}) = \mathrm{SL}(2,\mathbb{C})$, of dimension $3$; the complex case is the same identity with the dimension reduced to $0$.

The Relation to the Galois Theory of $\mathbb{C}/\mathbb{R}$

The automorphism group computed above is exactly the Galois group of the field extension $\mathbb{C}/\mathbb{R}$:

$$ \operatorname{Aut}_{\mathbb{R}}(\mathbb{C}) = \operatorname{Gal}(\mathbb{C}/\mathbb{R}) \cong \mathbb{Z}/2 . $$

The extension is separable (the base field has characteristic $0$), normal (it is the splitting field of $x^2+1$), of degree $2$, and its Galois group is generated by complex conjugation. In the fundamental theorem of Galois theory the subgroups $\{1\}$ and $\mathbb{Z}/2$ correspond respectively to the two intermediate fields

$$ \mathbb{C} \longleftrightarrow \{1\}, \qquad \mathbb{R}_{\mathbb{C}} \longleftrightarrow \mathbb{Z}/2, $$

so the fixed field of the full group is the real subspace and the fixed field of the trivial subgroup is $\mathbb{C}$ itself. The fixed-field correspondence is the subspace structure of Complex Subspaces read as a theorem of Galois theory: $\mathbb{R}_{\mathbb{C}}$ is the fixed subspace of the nontrivial automorphism, and $i\mathbb{R}_{\mathbb{C}}$ is the anti-fixed one. The field-theoretic development of the extension is the subject of the companion article Galois Theory of ℂ/ℝ; the present article records the automorphism group as an algebra invariant and stops at the boundary of that article.

Worked Examples

Conjugation is an automorphism. For $A = 3+4i$ and $B = 1-2i$,

$$ \bar A\bar B = (3-4i)(1+2i) = 3+6i-4i-8i^2 = 11+2i, \qquad \overline{AB} = \overline{11-2i} = 11+2i, $$

so conjugation preserves the product of the worked elements; it fixes the real subspace pointwise and negates the imaginary subspace, and it is an involution.

The two automorphisms on a non-real element. $\operatorname{id}(i) = i$ and $\bar{\cdot}(i) = -i$; there is no third possibility, since an automorphism must send $i$ to a square root of $-1$ and there are exactly two. On the generic element $a+i a'$ the two automorphisms act as $a+i a' \mapsto a+i a'$ and $a+i a' \mapsto a-i a'$.

No nonzero derivation. Suppose $D$ is a derivation with $D(i) = c \in \mathbb{C}$. Then $0 = D(-1) = 2i c$ forces $c = 0$; hence $D = 0$ on $i$ and on $\mathbb{R}$, and therefore on all of $\mathbb{C}$. There is no derivation of $\mathbb{C}$, over $\mathbb{R}$ or over $\mathbb{C}$.

The inner derivations. For every $A \in \mathbb{C}$, $\operatorname{ad}_A = 0$; for instance with $A = i$ and $B = 3+4i$,

$$ [i, 3+4i] = i(3+4i) - (3+4i)i = (3i-4) - (3i-4) = 0, $$

the two terms being equal by commutativity.

Summary

The two ground fields give the following table; the field is stated explicitly in every entry.

Structure	Over $\mathbb{C}$	Over $\mathbb{R}$
Algebra	$\mathbb{C}$, complex dimension $1$	$\mathbb{C}$, real dimension $2$
Ideals	$\{0\}$ and $\mathbb{C}$ only	$\{0\}$ and $\mathbb{C}$ only
Centre	$\mathbb{C}$, dimension $1$ over $\mathbb{C}$	$\mathbb{C}$, dimension $2$ over $\mathbb{R}$
Central simple?	yes, central simple over $\mathbb{C}$	no: simple, but centre $\mathbb{C} \neq \mathbb{R}$
Automorphism group	$\{\operatorname{id}\}$, trivial	$\{\operatorname{id}, \bar{\cdot}\} \cong \mathbb{Z}/2$, discrete
Derivation space (the Lie algebra of the automorphism group)	$\operatorname{Der}_{\mathbb{C}}(\mathbb{C}) = 0$	$\operatorname{Der}_{\mathbb{R}}(\mathbb{C}) = 0$

In summary: over $\mathbb{R}$ the automorphism group is the two-element Galois group $\operatorname{Gal}(\mathbb{C}/\mathbb{R})$ generated by complex conjugation, with the real subspace as its fixed field; over $\mathbb{C}$ the automorphism group is trivial, because a $\mathbb{C}$-linear automorphism of the field fixes the only basis element $1$. The derivation space vanishes over both fields, since the value $D(i)$ is forced to be zero by the relation $i^2 = -1$, and every inner derivation vanishes because the algebra is commutative. In the biquaternion algebra the two invariants are $PGL(2,\mathbb{C})$ and $\mathrm{SL}(2,\mathbb{C})$, of dimension $3$; the complex field is the case in which the derivation space and the $\mathbb{C}$-linear automorphism group both collapse to the zero object, and the surviving structure is the Galois group of the quadratic extension.

Summary of Notation

symbol	meaning
$\mathbb{C}$	the complex algebra, basis $1$, $i$, $i^2 = -1$
$A = a+i a'$	a complex number
$\bar A = a - i a'$	complex conjugation, the nontrivial $\mathbb{R}$-automorphism
$\operatorname{id}, \bar{\cdot}$	the two elements of $\operatorname{Aut}_{\mathbb{R}}(\mathbb{C}) \cong \mathbb{Z}/2$
$\operatorname{Aut}_{\mathbb{R}}(\mathbb{C}) \cong \mathbb{Z}/2$	automorphisms over $\mathbb{R}$, the Galois group $\operatorname{Gal}(\mathbb{C}/\mathbb{R})$
$\operatorname{Aut}_{\mathbb{C}}(\mathbb{C}) = \{\operatorname{id}\}$	automorphisms over $\mathbb{C}$
$\operatorname{Der}_{\mathbb{R}}(\mathbb{C}) = 0$	derivations over $\mathbb{R}$
$\operatorname{Der}_{\mathbb{C}}(\mathbb{C}) = 0$	derivations over $\mathbb{C}$
$\operatorname{ad}_A(B) = [A,B] = AB - BA$	inner derivation, identically zero
$\mathbb{R}_{\mathbb{C}}, i\mathbb{R}_{\mathbb{C}}$	real subspace (fixed field) and imaginary subspace