We have said that non-finite dimensional Hopf algebras may not have a linear dual, but that if they are graded we can fake it with a graded dual. For the non-graded case, how much information is lost if we define the dual to be the “hopfication” of the group algebra of the group of characters? Obviously the dual of this dual would not be the original algebra…

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Can you explain in a nutshell what a group algebra is? I found this: http://en.wikipedia.org/wiki/Group_algebra.

Also, why is it so obvious that the dual of the dual would not be the original algebra? (Does this have to do with the dimension of the dualized algebra?)

By Montgomery in Hopf Algebras and their Actions on Rings, any group can be turned into a hopf algebra () by using the group operation as the product, making the each group element grouplike for the coproduct, defining the counit of a group element to be 1, and taking the antipode of a group element to be its inverse in the group.

This “dualization” taken twice would not return the original algebra since typically we use linear maps for the dual, but we are restricting to algebra maps for the group of characters. Hence any non-multiplicative element of the would-be dual is omitted from the “psuedo-dual”, and so we lost some information. I’m curious how much was lost…

see an earlier exchange with Darij where he schooled me to the subject…

https://hopfcombinatorics.wordpress.com/2012/04/09/dual-of-a-hopf-algebra/

Hmmm . . . I still don’t see why it is so obvious that you don’t get the original algebra. When you construct a Hopf algebra out of your group of characters (in the way you suggest above), although your dual will consist of linear maps, it seems conceivable that elements in your group of characters could show up there as well –they’re linear maps too, after all. Unless I am missing something very small, it seems that no new elements are thrown in as a result of dualizing . . .

(Or do the arrows go back, so that we can be sure that elements in the group of characters do not show up in the dualized algebra?)

We know that for the linear dual of the linear dual, we get the original algebra. Since the group of characters consists of algebra maps, some of the linear maps are missing, and so when we take again the algebra maps of the algebra maps, we might be missing some, unless there is one of those inclusion-reversing things is happening…

you always get a linear dual space but in the case of a countably infinite basis the dual space is not isomorphic to the original ( too big uncountable) .

And I don’t think a dual space can ever be Smaller than the original, so again won’t be isomorphic.

But by ”dual” in quotes you mean just consider algebra maps, and your guessing that all the extra maps in the dual were not algebra maps?

I don’t think we always get a linear dual hopf algebra, since a non-finite-dimensional algebra doesn’t give us a coalgebra structure for our dual…(see lecture 6)

by “psuedo-dual” I meant the group Hopf algebra formed by the group of characters. I haven’t yet checked if bialgebra maps form a group as well…

The linear dual definitely contains maps not in the set of characters. ( see darij’s note https://hopfcombinatorics.wordpress.com/2012/04/09/dual-of-a-hopf-algebra/) Any linear but non multiplicative map is not a character.

Sorry when I saw linear dual I did not automatically add ” hopf algebra ” as I was obviously supposed to do.

I was not implying the dual space only includes characters.

Sorry when I saw linear dual I did not automatically add ” hopf algebra ” as I was obviously supposed to do.

I was not implying the dual space only includes characters

Need to read previous posts

I think my question is answered.in your conversation with darij