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May 6, 2016 at 9:56 comment added Dimitri Chikhladze @FriedrichKnop that was indicated in the comment above.
May 6, 2016 at 7:53 comment added Friedrich Knop Just for dimension reasons, the tensor identity does not even hold in the category of representations of a finite group $G$. In this case, $A$ should be the ring of functions on $G$. Is it possible that you forgot an index $A$ under some of the $\otimes$-symbols?
May 5, 2016 at 22:08 comment added Marc Hoyois sent you a copy by email.
May 3, 2016 at 13:55 comment added Dimitri Chikhladze Btw, does anyone know how to get hold on A. Rosenberg "The existence of fiber functors" in the series Gelfand Mathematical Seminars?
Apr 29, 2016 at 10:46 comment added Dimitri Chikhladze @MarcHoyois That is correct. Thank you.
Apr 28, 2016 at 21:08 comment added Marc Hoyois Sorry, by "dualizable" I meant "finite free" in the last sentence.
Apr 28, 2016 at 20:57 comment added Marc Hoyois $A\otimes -$ is not strong monoidal as an endofunctor of $Ind(T)$ (your second isomorphism is incorrect!). It is strong monoidal as a functor to $A$-modules. Then $\Gamma$ is lax monoidal from $A$-modules to $\Gamma(A)$-modules, but it is strong when restricted to the full subcategory of dualizable $A$-modules, where $A\otimes -$ lands.
Apr 28, 2016 at 18:00 history asked Dimitri Chikhladze CC BY-SA 3.0