The Skew Field of a Ring with Involution

Introduction

For a ring $A$ with an involution $\sigma$ the skew elements are those with $\sigma(x) = -x$, and they stand against the symmetric elements, the fixed ones. When $2$ is invertible the two sets are the eigenspaces of $\sigma$ for $-1$ and $+1$, and they decompose the ring, $A = \mathrm{Sym}(A,\sigma)\oplus\mathrm{Skew}(A,\sigma)$; what makes the pair interesting is that the two sets do not multiply among themselves in the naive way and that, with the commutator and the anti-commutator, they form exactly the two halves of a graded Lie structure. The word skew field in the title names this set of skew elements; it is not a division ring, and the collision with the established meaning of the phrase is flagged and avoided below, where the set is written $\mathrm{Skew}(A,\sigma)$.

This article reads the skew elements against the symmetric ones, computes the products of two skew elements (their anti-commutator is symmetric and their commutator skew), records the resulting graded Lie and Jordan structure, and derives the vanishing of a trace on the skew elements for a trace-preserving involution. It assumes Involutive Rings for the involution, the fixed set and the eigenspace decomposition, and Lie Algebras and Jordan Algebras only as the names of the structures that the products suggest, since those categories come later in this Part. Throughout, $A$ is a ring with $1 \neq 0$, $\sigma$ is an involution, $2$ is invertible in $A$ when the eigenspace decomposition is used, and

$$ \mathrm{Sym}(A,\sigma) = \{x : \sigma(x) = x\}, \qquad \mathrm{Skew}(A,\sigma) = \{x : \sigma(x) = -x\}. $$

The Two Eigenspaces

Proposition. $\mathrm{Sym}(A,\sigma)$ and $\mathrm{Skew}(A,\sigma)$ are additive subgroups of $A$, they intersect in $\{0\}$, and for $2$ invertible

$$ A = \mathrm{Sym}(A,\sigma)\oplus\mathrm{Skew}(A,\sigma), \qquad x = \tfrac12(x+\sigma(x)) + \tfrac12(x-\sigma(x)). $$

An element is symmetric and skew at once exactly when $2x = 0$; in particular the intersection is trivial in characteristic not two and equals $\{x : 2x = 0\}$ in general.

Proof. This is the eigenspace decomposition of the order-two additive map $\sigma$ of Involutive Rings: the sum of the two projections is the identity, each projection lands in the named set, and an element in both satisfies $x = -x$.

Remark (the name). The set $\mathrm{Skew}(A,\sigma)$ is not a field or a division ring, and the phrase "skew field" is used here in the sense of "the skew elements" only. To avoid the collision with the standard meaning, the article writes $\mathrm{Skew}(A,\sigma)$ throughout and never calls it a field.

The Products

Proposition. For homogeneous $x, y$ the symmetry of a product is governed by the parities:

$$ \mathrm{Sym}\cdot\mathrm{Sym}\subseteq\mathrm{Sym} \iff \text{the factors commute}, \qquad \mathrm{Sym}\cdot\mathrm{Skew}\subseteq\mathrm{Skew} \iff \text{the factors commute}, $$

and the two products of a symmetric and a skew element have a fixed behaviour through the commutator and the anti-commutator, not through the product itself.

Proof. For $x, y$ symmetric, $\sigma(xy) = \sigma(y)\sigma(x) = yx$, which equals $xy$ exactly when they commute. For $x$ symmetric and $y$ skew, $\sigma(xy) = \sigma(y)\sigma(x) = -yx$, which equals $-xy$, the skew condition, exactly when $yx = xy$.

Theorem (the commutator and the anti-commutator). For all homogeneous pairs the following hold with no commutativity hypothesis:

$$ [x,y] = xy-yx \in \mathrm{Skew} \ \text{if } x, y \in \mathrm{Skew}, \qquad [x,y] \in \mathrm{Sym} \ \text{if } x \in \mathrm{Sym},\ y \in \mathrm{Skew}, $$

$$ \{x,y\} = xy+yx \in \mathrm{Sym} \ \text{if } x, y \in \mathrm{Skew}, \qquad \{x,y\} \in \mathrm{Skew} \ \text{if } x \in \mathrm{Sym},\ y \in \mathrm{Skew} . $$

In particular $\mathrm{Skew}(A,\sigma)$ is closed under the commutator and is a Lie subalgebra of $A$ under $[x,y] = xy-yx$, while $\mathrm{Sym}(A,\sigma)$ is closed under the anti-commutator and is a Jordan subalgebra under $x\bullet y = xy+yx$.

Proof. In each case apply $\sigma$ to the combination. For skew $x, y$: $\sigma(xy-yx) = \sigma(y)\sigma(x)-\sigma(x)\sigma(y) = (-y)(-x)-(-x)(-y) = yx-xy = -(xy-yx)$, so the commutator is skew; and $\sigma(xy+yx) = yx+xy$, so the anti-commutator is symmetric. For symmetric $x$ and skew $y$: $\sigma(xy-yx) = (-y)x-x(-y) = -yx+xy = xy-yx$, symmetric; and $\sigma(xy+yx) = -yx-xy = -(xy+yx)$, skew. The closure statements are the two rows.

Corollary (the graded structure). With $\mathrm{Sym}$ in even parity and $\mathrm{Skew}$ in odd parity, the two brackets $[\,,\,]$ and $\{\,,\,\}$ make $A$ a graded Lie-and-Jordan structure: the bracket of two even or two odd elements is odd, the bracket of an even and an odd element is even, the anti-commutator of two odd or of an even and an odd element is odd, and the anti-commutator of two even elements is even. This is the sign structure recorded in Superalgebras and Graded Structures, read on the symmetric and the skew parts rather than on a $\mathbb{Z}/2$-grading of the ring.

The Trace

Definition. A trace on $A$ is an additive map $\tau : A \to k$ into a commutative ring $k$ with $\tau(ab) = \tau(ba)$ for all $a, b$. It is $\sigma$-invariant when $\tau(\sigma(a)) = \tau(a)$.

Proposition. Let $\tau$ be a $\sigma$-invariant trace and let $2$ be invertible in $A$ and in $k$. Then

$$ \tau(\mathrm{Skew}(A,\sigma)) = 0, \qquad \tau\bigl(\mathrm{Sym}(A,\sigma)\cdot\mathrm{Skew}(A,\sigma)\bigr) = 0 . $$

Proof. For skew $x$ one has $\tau(x) = \tau(\sigma(x)) = \tau(-x) = -\tau(x)$, so $2\tau(x) = 0$ and $\tau(x) = 0$. For symmetric $x$ and skew $y$, $\tau(xy) = \tau(\sigma(xy)) = \tau(\sigma(y)\sigma(x)) = \tau((-y)x) = -\tau(yx) = -\tau(xy)$, so again $2\tau(xy) = 0$.

Corollary. For the matrix ring $M_n(k)$ with the transpose and $\tau = $ the matrix trace, which is invariant under the transpose, the skew-symmetric matrices have trace zero and the product of a symmetric and a skew matrix has trace zero; the latter is the orthogonality of the symmetric and the skew parts under the trace pairing $(X,Y)\mapsto \operatorname{tr}(XY)$. The same computation gives the vanishing of the trace on the pure quaternions for the conjugation of the quaternion algebra and on the odd part of a graded matrix algebra with the graded trace.

Examples

(a) The transpose. In $M_n(k)$ with the transpose the skew elements are the skew-symmetric matrices and the symmetric elements the symmetric matrices; the commutator of two skew-symmetric matrices is skew-symmetric, the anti-commutator is symmetric, and the trace vanishes on the skew part. For $n = 3$ the skew-symmetric matrices have dimension $3$ and form, under the commutator and the identification with $k^3$, the cross-product Lie algebra.

(b) The quaternion conjugation. In the quaternion algebra $\mathbb{H}$ over $\mathbb{R}$ with the conjugation the symmetric part is the real line and the skew part is the space of pure quaternions; the commutator of two pure quaternions is pure, $[i,j] = 2k$, and the anti-commutator is real, $\{i,j\} = 0$, so the pure quaternions form a three-dimensional Lie algebra.

(c) The group ring. In $K[G]$ with the standard involution the skew elements are those with $a_g = -\sigma(a_{g^{-1}})$, and the commutator of two of them is skew; for a finite group the symmetric part contains the inversion-stable class sums, as in Involutions of a Group Ring.

(d) A noncommutative failure. In $M_2(F)$ with the transpose the symmetric matrices $\begin{pmatrix}1&1\\1&0\end{pmatrix}$ and $\begin{pmatrix}0&1\\1&1\end{pmatrix}$ have a non-symmetric product, so $\mathrm{Sym}$ is not closed under multiplication; the closure is restored by the anti-commutator of the theorem.

Summary

For an involution $\sigma$ of a ring $A$ with $2$ invertible, the symmetric and the skew elements are the eigenspaces and $A = \mathrm{Sym}(A,\sigma)\oplus\mathrm{Skew}(A,\sigma)$; the two sets meet in $\{x : 2x = 0\}$ and the word skew field names the set of skew elements and not a division ring. The product of two elements is governed by the commutator and the anti-commutator rather than by the product itself: for two skew elements the commutator is skew and the anti-commutator is symmetric, for a symmetric and a skew element the commutator is symmetric and the anti-commutator is skew, so that $\mathrm{Skew}(A,\sigma)$ is a Lie subalgebra under $[x,y] = xy-yx$ and $\mathrm{Sym}(A,\sigma)$ is a Jordan subalgebra under $x\bullet y = xy+yx$, the two forming the halves of a graded structure. A $\sigma$-invariant trace vanishes on the skew elements and on every product of a symmetric and a skew element, so for the transpose the symmetric and the skew matrices are orthogonal under the trace pairing.

Summary of Notation

Symbol	Meaning
$\sigma$	Involution of the ring $A$
$\mathrm{Sym}(A,\sigma)$	Fixed elements; the $+1$ eigenspace of $\sigma$
$\mathrm{Skew}(A,\sigma)$	Skew elements $\{x : \sigma(x) = -x\}$; the $-1$ eigenspace
$A = \mathrm{Sym}\oplus\mathrm{Skew}$	Eigenspace decomposition, $2$ invertible
$[x,y] = xy-yx$	Commutator; makes $\mathrm{Skew}$ a Lie subalgebra
$\{x,y\} = xy+yx$	Anti-commutator; makes $\mathrm{Sym}$ a Jordan subalgebra
$[\,\mathrm{Sym},\mathrm{Skew}\,]\subseteq\mathrm{Sym}$	Graded Lie structure
$\{\mathrm{Sym},\mathrm{Skew}\}\subseteq\mathrm{Skew}$	Graded Jordan structure
$\tau$, $\sigma$-invariant trace	$\tau(ab) = \tau(ba)$, $\tau(\sigma(a)) = \tau(a)$
$\tau(\mathrm{Skew}) = 0$, $\tau(\mathrm{Sym}\cdot\mathrm{Skew}) = 0$	Vanishing of the trace, $2$ invertible