2007
DOI: 10.1080/00927870701511996
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Categorical Morita Equivalence for Group-Theoretical Categories

Abstract: Abstract. A finite tensor category is called pointed if all its simple objects are invertible. We find necessary and sufficient conditions for two pointed semisimple categories to be dual to each other with respect to a module category. Whenever the dual of a pointed semisimple category with respect to a module category is pointed, we give explicit formulas for the Grothendieck ring and for the associator of the dual. This leads to the definition of categorical Morita equivalence on the set of all finite group… Show more

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Cited by 58 publications
(90 citation statements)
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“…It suffices to show that isomorphisms i f;g , j f;g defined by equations (30) and (31) do not depend on the choice of equivalences (28) is isomorphic to the identity as a D-bimodule functor; (3) natural isomorphisms˛f ;g;h (27) satisfying identity (29).…”
Section: An Action Determined By Amentioning
confidence: 99%
“…It suffices to show that isomorphisms i f;g , j f;g defined by equations (30) and (31) do not depend on the choice of equivalences (28) is isomorphic to the identity as a D-bimodule functor; (3) natural isomorphisms˛f ;g;h (27) satisfying identity (29).…”
Section: An Action Determined By Amentioning
confidence: 99%
“…So if we find the necessary and sufficient condition for the existence of such U ⊆ G × G and ϕ ∈ H 2 (U, C × ) we can answer the question of whether Z(G) and Z(G ) are equivalent. This question was first answered by Naidu and Nikshych [11,12] based on the classification of Lagrangian subcategories of Z(G). Here we state the necessary and sufficient conditions of Davydov which is more appropriate for us.…”
Section: When Are Z(g) and Z(g ) Equivalent?mentioning
confidence: 99%
“…Here ν is a certain 2-cocycle in Z 2 (H\G, H) coming from the G-invariance of µ and ̟ is a certain 3-cocycle on H ⋊ ν (H\G) depending on ν and on the exact sequence 1 → H → G → H\G → 1 (see [Na,Theorem 5.8] for precise definitions). The one corresponding to r 2 is equivalent to Rep(Z/2Z×Z/2Z×Z/2Z).…”
Section: Proof Let Us First Show That Altmentioning
confidence: 99%