Involutions of the Endomorphism Ring

Introduction

An endomorphism ring carries an intrinsic involution as soon as the object it acts on carries a nondegenerate pairing: the adjoint of an endomorphism is the unique map that transfers the pairing from one side to the other, and the assignment $T \mapsto T^{*}$ is additive, anti-multiplicative and of order two, that is, an involution of the ring. The unitary elements, those with $T^{*}T = TT^{*} = \mathrm{id}$, are exactly the endomorphisms that preserve the pairing, so the involution of the endomorphism ring is the algebraic form of the preservation of a form; this article fixes the involution and the unitary group at the ring level, and the linear-space version of the same construction is Involutions of the Endomorphism Algebra.

This article defines the nondegenerate reflexive pairing on an additive group, constructs the adjoint involution of the endomorphism ring, proves the four laws (additivity, anti-multiplicativity, order two, compatibility with the scalars), identifies the unitary elements and records the two kinds that the symmetry of the pairing produces. It assumes Groups for the additive group, Rings for the endomorphism ring and Involutive Rings for the involution and the unitary elements; the vector-space and module refinements are forward references. Throughout, $(V,+)$ is an additive group, $R$ is a commutative ring with $1 \neq 0$, $B : V\times V \to R$ is a biadditive perfect pairing, $E = \operatorname{End}(V)$ is the endomorphism ring, and $\sigma$ is an involution of $R$ when the coefficients are twisted.

The Pairing and the Adjoint

Definition. A biadditive pairing $B : V\times V \to R$ is

(a) nondegenerate if $B(x,y) = 0$ for all $y$ implies $x = 0$ and $B(x,y) = 0$ for all $x$ implies $y = 0$;

(b) reflexive if $B(x,y) = 0$ exactly when $B(y,x) = 0$;

(c) perfect if the map $V \to \operatorname{Hom}(V,R)$, $x \mapsto B(x,-)$, is a bijection.

A perfect pairing is nondegenerate and reflexive, and for each $y$ the functional $x \mapsto B(Tx,y)$ is additive.

Theorem (the adjoint exists). For every $T \in E$ and every $y \in V$ there is a unique $T^{*}y \in V$ with

$$ B(Tx,y) = B(x,T^{*}y) \quad \text{for all } x \in V . $$

The assignment $T \mapsto T^{*}$ is well defined, and it is an involution of the ring $E$: it is additive, $(ST)^{*} = T^{*}S^{*}$, $\mathrm{id}^{*} = \mathrm{id}$, and $(T^{*})^{*} = T$.

Proof. For fixed $y$ the map $x \mapsto B(Tx,y)$ is additive $V \to R$, hence an element of $\operatorname{Hom}(V,R)$; since $B$ is perfect there is a unique $T^{*}y$ with $B(x,T^{*}y) = B(Tx,y)$ for all $x$, which proves existence and uniqueness. Uniqueness in $y$ makes $T^{*}$ a well-defined map $V \to V$; it is additive because $B(x,T^{*}(y+y')) = B(Tx,y+y') = B(Tx,y)+B(Tx,y') = B(x,T^{*}y+T^{*}y')$ for all $x$, and the pairing is nondegenerate, so $T^{*}(y+y') = T^{*}y+T^{*}y'$. For the anti-multiplicativity, $B(x,(ST)^{*}y) = B(STx,y) = B(S(Tx),y) = B(Tx,S^{*}y) = B(x,T^{*}S^{*}y)$ for all $x$, so $(ST)^{*}y = T^{*}S^{*}y$. The unit is fixed because $B(x,y) = B(x,\mathrm{id}^{*}y)$ for all $x$ gives $\mathrm{id}^{*} = \mathrm{id}$. Finally $B(x,(T^{*})^{*}y) = B(T^{*}x,y)$ for all $x$; by reflexivity this vanishes exactly when $B(y,T^{*}x)$ vanishes, which by the defining identity equals $B(Ty,x)$, and by reflexivity again this vanishes exactly when $B(x,Ty)$ vanishes; since the pairing is nondegenerate, $(T^{*})^{*}y = Ty$ for all $y$, that is $(T^{*})^{*} = T$.

Remark (compatibility with the coefficients). If $R$ carries an involution $\sigma$ and the pairing satisfies $B(\lambda x,y) = \lambda B(x,y)$, $B(x,\lambda y) = \sigma(\lambda)B(x,y)$, then the adjoint is $\sigma$-semilinear in the coefficient of a scalar endomorphism: for the endomorphism $\lambda\,\mathrm{id}$ one has $(\lambda\,\mathrm{id})^{*} = \sigma(\lambda)\,\mathrm{id}$. The coefficient involution enters only through the sesquilinearity of the pairing, and the whole construction is that of Rings with a Semilinear Involution when $\sigma \neq \mathrm{id}$.

The Unitary Elements

Definition. The unitary elements of $(E,B)$ are

$$ U(V,B) = \{T \in E^\times : T^{*}T = TT^{*} = \mathrm{id}\}. $$

Proposition. $U(V,B)$ is a subgroup of the unit group $E^\times$, and $T \in E^\times$ is unitary exactly when $T$ preserves $B$,

$$ T \in U(V,B) \iff B(Tx,Ty) = B(x,y) \ \text{for all } x, y \in V . $$

Proof. If $T$ is unitary then $B(Tx,Ty) = B(T^{*}Tx,y) = B(x,y)$, using the defining identity for $T^{*}$ with the pair $(Tx,y)$ replaced by $(x,Ty)$; conversely, if $T$ preserves $B$ then $B(x,T^{*}Ty) = B(Tx,Ty) = B(x,y)$ for all $x$, so $T^{*}T = \mathrm{id}$ by nondegeneracy, and the same argument on the other side gives $TT^{*} = \mathrm{id}$. The product of two unitary elements is unitary because the adjoint is anti-multiplicative, $(ST)^{*}(ST) = T^{*}S^{*}ST = T^{*}T = \mathrm{id}$, and the inverse of a unitary element is unitary because $T^{*} = T^{-1}$.

Corollary (the two kinds). If the pairing is symmetric, $B(y,x) = B(x,y)$, then the involution of $E$ is the orthogonal one; if it is antisymmetric, $B(y,x) = -B(x,y)$, then it is the symplectic one. With $2$ invertible the endomorphism ring is the sum of the self-adjoint and the skew-adjoint endomorphisms, of dimensions $\tfrac12 n(n+1)$ and $\tfrac12 n(n-1)$ for a symmetric pairing on an object of finite rank $n$ in the matrix case, and exchanged for an antisymmetric one, as in Matrix Rings with an Involution.

Proof. The transpose type is computed in the matrix case in Matrix Rings with an Involution; the endomorphism ring of a finite-rank free object is the matrix ring, and the adjoint of a matrix is its transpose or its symplectic adjoint according to the symmetry, which is the statement.

Examples

(a) The standard pairing. $V = R^n$ with $B(x,y) = \sum_i x_iy_i$ and $R$ a commutative ring; the adjoint of a matrix is its transpose, $X^{*} = X^{\mathrm t}$, so the involution of the matrix ring is the transpose and $U(V,B)$ is the orthogonal group.

(b) The symplectic pairing. $V = R^{2n}$ with the pairing $B(x,y) = \sum_i(x_iy_{i+n}-x_{i+n}y_i)$; the adjoint of $X$ is $-JX^{\mathrm t}J$ with $J$ the standard symplectic matrix, the involution is the symplectic one of Matrix Rings with an Involution, and $U(V,B)$ is the symplectic group.

(c) The Hermitian pairing. $R = \mathbb{C}$ with the conjugation, $B(x,y) = \sum_i \overline{x_i}y_i$; the adjoint is the conjugate transpose and $U(V,B)$ the unitary group. The construction is the second-kind case of the remark.

(d) The regular pairing of a ring. On $V = A$ the pairing $B(x,y) = \tau(xy)$, with $\tau$ the regular trace, is symmetric and nondegenerate under the hypothesis of The Adjoint of the Left Multiplication on a Ring; the adjoint of the left multiplication $L_a$ is the right multiplication $R_a$, and the unitary elements are the units $u$ with $u^{*} = u^{-1}$.

Summary

A perfect pairing $B$ on an additive group $V$ defines an adjoint involution $T \mapsto T^{*}$ of the endomorphism ring $E = \operatorname{End}(V)$ by $B(Tx,y) = B(x,T^{*}y)$; the adjoint exists and is unique for each $T$, the assignment is additive, anti-multiplicative and of order two, and it is $\sigma$-semilinear in the coefficients when the pairing is sesquilinear with an involution $\sigma$. The unitary elements $U(V,B) = \{T : T^{*}T = TT^{*} = \mathrm{id}\}$ form a subgroup of $E^\times$ and are exactly the endomorphisms preserving the pairing; a symmetric pairing gives the orthogonal involution and an antisymmetric one the symplectic involution, with the matrix calculations and the unitary groups of Matrix Rings with an Involution and Matrix Rings and the Adjoint. The specialisation to the regular pairing $B(x,y) = \tau(xy)$ of a ring with involution gives the adjoint of the left multiplication, treated in The Adjoint of the Left Multiplication on a Ring.

Summary of Notation

Symbol Meaning
$V$, $R$ Additive group and commutative coefficient ring
$B : V\times V \to R$ Perfect (nondegenerate, reflexive) biadditive pairing
$E = \operatorname{End}(V)$ Endomorphism ring
$B(Tx,y) = B(x,T^{*}y)$ Defining identity of the adjoint
$T \mapsto T^{*}$ Adjoint involution of $E$
$U(V,B)$ Unitary group: $T^{*}T = TT^{*} = \mathrm{id}$, equivalently $B(Tx,Ty) = B(x,y)$
symmetric $B$ / antisymmetric $B$ Orthogonal / symplectic involution
$B = \tau(xy)$ on $A$ Regular pairing of a ring; $L_a^{*} = R_a$
$\sigma$ Coefficient involution; adjoint is $\sigma$-semilinear when $B$ is sesquilinear

Further Reading

  • Nathan Jacobson, Structure of Rings, American Mathematical Society Colloquium Publications 37 (1964), for the adjoint involution of an endomorphism ring and the unitary elements.
  • Max-Albert Knus, Alexander Merkurjev, Markus Rost and Jean-Pierre Tignol, The Book of Involutions, American Mathematical Society Colloquium Publications 44 (1998), for the orthogonal and symplectic involutions and their relation to the forms they preserve.
  • I. N. Herstein, Rings with Involution (University of Chicago Press, 1976), for the involution of a ring, the unitary elements and the sesquilinear case.
  • Nicolas Bourbaki, Algebra I, Chapters 1–3 (Springer, 1998), for bilinear and sesquilinear forms, nondegeneracy and the reflexive pairings.