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A braided rigid category has a Drinfeld morphism

Statement

In a braided rigid monoidal category there is a natural isomorphism

$u_{X} : X \to X^{\lor \lor},$

called the Drinfeld morphism (or Drinfeld isomorphism), defined by

$X \overset{1_{X} \otimes {coev}_{X^{\lor}}}{\to} X \otimes X^{\lor} \otimes X^{\lor \lor} \overset{c_{X, X^{\lor}} \otimes 1_{X^{\lor \lor}}}{\to} X^{\lor} \otimes X \otimes X^{\lor \lor} \overset{{ev}_{X} \otimes 1_{X^{\lor \lor}}}{\to} X^{\lor \lor} .$

Write

$d_{X, Y} : X^{\lor \lor} \otimes Y^{\lor \lor} \tilde{\to} (X \otimes Y)^{\lor \lor}$

for the monoidal comparison of the double-dual functor. Then

$d_{X, Y} \circ (u_{X} \otimes u_{Y}) = u_{X \otimes Y} \circ c_{Y, X} \circ c_{X, Y} .$

Thus $u$ need not be monoidal: the double braiding is precisely its monoidality obstruction.

Facts & Assumptions

Given: A braided rigid monoidal category with braiding $c$ and chosen left duals.

[F1]

Bruguières--Virelizier, Lemma 8.1, proves under the bare braided-autonomous hypotheses that the displayed natural transformation is an isomorphism, gives its explicit inverse, and proves its tensor relation; Remark 8.2 identifies symmetry as exactly the case in which it is monoidal. Shibata--Shimizu, Section 6.4, independently uses the same map as the Drinfeld isomorphism of an arbitrary braided rigid monoidal category and uses $u^{- 1}$ in the pivotal/twist correspondence. EGNO formula (8.30) and Proposition 8.9.3 give the same map and tensor relation in their strict chosen-dual convention.

[L1]

A braiding is natural in both variables and satisfies the hexagon identities (Braiding).

[L2]

With chosen left duals, the double-dual assignment is a monoidal endofunctor, with comparison maps $d_{X, Y}$ of the displayed type (The double dual is a monoidal functor).

Proof

technique · direct

1.1

The displayed composite is well typed because rigidity provides ${coev}_{X^{\lor}} : 1 \to X^{\lor} \otimes X^{\lor \lor}$ and ${ev}_{X} : X^{\lor} \otimes X \to 1$ , while the braiding supplies the middle swap. Naturality of the braiding in [L1] makes the assignment $X \mapsto u_{X}$ natural in $X$ .

givenF1L1

1.2

After inserting the monoidal comparison $d_{X, Y}$ from [L2], both sides of the displayed tensor relation are morphisms $X \otimes Y \to (X \otimes Y)^{\lor \lor}$ . The relation in [F1] is exactly this compatibility in a convention that suppresses the comparison. In particular, it records the precise obstruction to $u$ being monoidal: the double braiding appears between $u_{X \otimes Y}$ and the transported map $u_{X} \otimes u_{Y}$ .

F1L2

2.1

The explicit inverse in [F1] is defined under the same braided-rigid hypotheses as the displayed map, so every $u_{X}$ is an isomorphism. Together with step 1.1, this makes $u$ a natural isomorphism in every braided rigid monoidal category; step 1.2 records why it need not be monoidal.

step 1.1step 1.2F1∎

Depends on

Used by

Dependency tree · two levels

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Sources

P. Etingof, S. Gelaki, D. Nikshych, and V. Ostrik, Tensor Categories, formula (8.30), Proposition 8.9.3, and Proposition 8.10.6 (standard reference, not scraped)
A. Bruguières and A. Virelizier, Hopf monads, Lemma 8.1 and Remark 8.2 (standard reference, not scraped)
T. Shibata and K. Shimizu, Modified traces and the Nakayama functor, Section 6.4 (standard reference, not scraped)