Involutive Categories and the Dagger Functor

Introduction

A dagger category, also called an involutive category, is a category together with a rule that assigns to each morphism $f : A \to B$ a morphism $f^{\dagger} : B \to A$, the adjoint of $f$, in a way that is an involution on the morphisms: $f^{\dagger\dagger} = f$, the identities are fixed, and the adjoint of a composite reverses the factors, $(g \circ f)^{\dagger} = f^{\dagger} \circ g^{\dagger}$. The rule is a contravariant functor ${}^{\dagger} : \mathcal{C}^{\mathrm{op}} \to \mathcal{C}$, the dagger functor, which is the identity on objects and is its own inverse. The morphisms fixed by the involution are the self-adjoint morphisms, and the morphisms invertible with inverse their adjoint are the unitary ones. This article develops that involution on the morphisms, its fixed part and the structure of the unitary and self-adjoint morphisms.

The article presupposes Universal Properties and Categories — categories, functors, natural transformations, the opposite category and monomorphisms — and the involutive structures of this category: the converse of a relation, treated in Relation Algebras and the Converse, which is the dagger of the category of relations, and inversion in a groupoid, treated in Groupoids with an Involution, which follows in this group.

Three boundaries are observed. The adjoint in the operator sense, on a Hilbert space with an inner product, is Part II and the * Operator Theory group of this category, and is named only; here "adjoint" means the morphism assigned by the dagger and needs no form. The star of an algebra with involution is Involutive Algebras and the - Algebras group, a different layer, and is named only. The 2-categorical and higher structure of dagger categories is not developed; only the involution and its fixed part are treated.

Dagger Categories

The Dagger Functor

Definition. A dagger category is a category $\mathcal{C}$ together with a functor ${}^{\dagger}: \mathcal{C}^{\mathrm{op}} \to \mathcal{C}$ that is the identity on objects and satisfies $({}^{\dagger}) \circ ({}^{\dagger})^{\mathrm{op}} = \mathrm{id}_{\mathcal{C}}$; equivalently, a rule $f \mapsto f^{\dagger}$ with

$$ \mathrm{id}_A^{\dagger} = \mathrm{id}_A, \qquad (g \circ f)^{\dagger} = f^{\dagger} \circ g^{\dagger}, \qquad f^{\dagger\dagger} = f $$

for all objects $A$ and all composable morphisms $f, g$. The morphism $f^{\dagger}$ is the adjoint of $f$, and ${}^{\dagger}$ is the dagger functor.

A contravariant functor out of $\mathcal{C}^{\mathrm{op}}$ into $\mathcal{C}$ is a covariant functor out of $\mathcal{C}$, so a dagger functor is equivalently a covariant functor ${}^{\dagger} : \mathcal{C} \to \mathcal{C}$ that is the identity on objects, is the identity on the assignment $A \mapsto A$, and satisfies $g \circ f \mapsto f^{\dagger} \circ g^{\dagger}$; the two descriptions differ only in which way the composition is read, and the article uses the contravariant one.

Proposition. The dagger functor is an isomorphism of categories $\mathcal{C}^{\mathrm{op}} \to \mathcal{C}$ whose inverse is the same functor read as $\mathcal{C} \to \mathcal{C}^{\mathrm{op}}$; it is an involution of the category, and it carries every morphism $f : A \to B$ to a morphism $B \to A$.

Proof. The two identities $f^{\dagger\dagger} = f$ and $(g \circ f)^{\dagger} = f^{\dagger} \circ g^{\dagger}$ say that the two readings are mutually inverse on morphisms; being the identity on objects, the functor is bijective on morphisms of each hom-set, hence an isomorphism of categories, and applying it twice is the identity.

Example (relations). The category $\mathbf{Rel}$ has sets as objects and relations as morphisms, composition being relational composition. Its dagger is the converse $R^{\dagger} = R^{-1}$ of Relation Algebras and the Converse: the diagonal is fixed, the converse of a composite reverses the factors, and the converse is an involution. The category $\mathbf{Rel}$ is the standard dagger category.

Example (the discrete dagger). Any category becomes a dagger category with $f^{\dagger} = f$ for every morphism, that is, with the identity rule; this dagger is the discrete one, and it is the involution that fixes every morphism. A dagger category in which every morphism is fixed is exactly a category with the discrete dagger, so the discrete dagger carries no information.

Unitary and Self-Adjoint Morphisms

Definition. Let $\mathcal{C}$ be a dagger category.

$f : A \to B$ is an isometry if $f^{\dagger} \circ f = \mathrm{id}_A$;
$f$ is a coisometry if $f \circ f^{\dagger} = \mathrm{id}_B$;
$f$ is unitary if it is both, that is, if $f^{\dagger}$ is a two-sided inverse of $f$;
$f$ is self-adjoint if $f^{\dagger} = f$, which requires $f$ to be an endomorphism.

Proposition. In a dagger category the unitary morphisms are closed under composition and under adjoint, the identities are unitary, and every isomorphism that is unitary has unitary inverse; so the unitary morphisms form a groupoid with the same objects. A morphism is unitary if and only if it is an isomorphism and its inverse is its adjoint.

Proof. If $f$ and $g$ are unitary and composable then $(g \circ f)^{\dagger} = f^{\dagger} \circ g^{\dagger}$ is a two-sided inverse of $g \circ f$, because $f^{\dagger}, g^{\dagger}$ are the inverses of $f, g$; the adjoint of a unitary morphism is unitary with the same witness. The last statement is the definition of the inverse in a category.

Proposition. The self-adjoint endomorphisms of an object are closed under addition when an additive structure is present and are closed under the adjoint; they are not closed under composition in general. The self-adjoint morphisms of $\mathbf{Rel}$ are the symmetric relations, and the composition of two symmetric relations need not be symmetric, as Relation Algebras and the Converse records.

Proof. If $f^{\dagger} = f$ and $g^{\dagger} = g$ and $f, g$ are composable endomorphisms, then $(g \circ f)^{\dagger} = f^{\dagger} \circ g^{\dagger} = f \circ g$, which need not equal $g \circ f$; the example of the symmetric relations on a three-element set, whose composite is not symmetric, is carried over.

Definition. A dagger functor between dagger categories is a functor $F$ with $F(f^{\dagger}) = F(f)^{\dagger}$ for every morphism $f$. A dagger functor carries isometries to isometries, coisometries to coisometries, unitaries to unitaries and self-adjoint morphisms to self-adjoint morphisms.

Proof. Apply $F$ to the defining identities and use $F(\mathrm{id}) = \mathrm{id}$ and the compatibility $F(f^{\dagger}) = F(f)^{\dagger}$.

The Fixed Part and the Unitary Groupoid

The fixed part of the dagger is the class of self-adjoint morphisms, and on each object it is the set of self-adjoint endomorphisms, a subset of the endomorphism monoid closed under the adjoint and, in the examples with an additive structure, under sums; the unitary groupoid is the groupoid of unitary morphisms. The two are related by the involution: a unitary endomorphism $u$ is self-adjoint exactly when $u = u^{\dagger} = u^{-1}$, that is, when $u$ is an involution of the object.

Definition. An involution on an object $A$ of a dagger category is a unitary endomorphism $u : A \to A$ with $u \circ u = \mathrm{id}_A$ that is self-adjoint; equivalently $u = u^{\dagger} = u^{-1}$.

Proposition. In a dagger category the involutions on an object are exactly the unitary self-adjoint endomorphisms, and they are the elements of order two of the unitary groupoid at the object.

Proof. A unitary self-adjoint $u$ satisfies $u^{-1} = u^{\dagger} = u$, so $u^2 = \mathrm{id}$; conversely $u^2 = \mathrm{id}$ and $u^{\dagger} = u^{-1} = u$.

Example (a group as a dagger category). A group $G$ becomes a dagger category with one object and all morphisms invertible, the dagger being inversion $g^{\dagger} = g^{-1}$: inversion is an involution, $\mathrm{id}^{\dagger} = \mathrm{id}$, and $(gh)^{-1} = h^{-1}g^{-1}$. The unitary morphisms are all the elements, the self-adjoint elements are those of order at most two, and the involutions are exactly the elements of order two. The dagger of a one-object groupoid is treated again in Groupoids with an Involution.

Example (a groupoid as a dagger category). More generally every groupoid, in the sense of a category in which every morphism is invertible, is a dagger category with inversion as dagger; the fixed morphisms are the identities and the elements of order two, and the unitary morphisms are all morphisms.

Summary

A dagger category is a category with a contravariant involution on its morphisms: $f^{\dagger\dagger} = f$, identities fixed, $(g \circ f)^{\dagger} = f^{\dagger} \circ g^{\dagger}$. The rule is the dagger functor, an isomorphism $\mathcal{C}^{\mathrm{op}} \to \mathcal{C}$ that is its own inverse; it is the formal counterpart on the morphisms of the converse of a relation and of the inversion in a groupoid.

The unitary morphisms, those with $f^{\dagger} = f^{-1}$, form a groupoid with the same objects; the self-adjoint morphisms, those with $f^{\dagger} = f$, form the fixed part of the involution and are closed under the adjoint but not under composition; and the self-adjoint unitaries are the involutions of the objects, the elements of order two of the unitary groupoid. The category of relations with the converse is the standard example, and the group with inversion is the one-object example.

Summary of Notation

Symbol	Meaning
$\mathcal{C}$, ${}^{\dagger}$	A dagger category and its dagger functor
$f^{\dagger}$	The adjoint of $f$, the image of the involution
self-adjoint	$f^{\dagger} = f$; the fixed part of the dagger
isometry	$f^{\dagger} \circ f = \mathrm{id}$
coisometry	$f \circ f^{\dagger} = \mathrm{id}$
unitary	$f^{\dagger} = f^{-1}$; the unitary morphisms form a groupoid
involution on $A$	A unitary self-adjoint $u : A \to A$; $u = u^{\dagger} = u^{-1}$, $u^2 = \mathrm{id}$