[GAP Forum] Symmetric bicharacters

Dan Lanke dan_lanke at yahoo.com
Thu Oct 9 16:31:51 BST 2008

Dear Shaun,

Thanks for your  message.
I think there is no isomorphism between 
Hom( A, Hom(A, C^*)) and Hom( A \otimes A, C^* ). 
There is an isomorphism between Hom(A, Hom(A,C^*))
and the group of all functions T: A x A --> C^*
that are multiplicative in both components, i.e.,
T(ab, c) = T(a, c)T(b, c) and T(a, bc) = T(a, b)T(a,c). 
Am I right?
My previous question is equivalent to the question:

How to create the group of all non-degenerate functions T : A x A --> C^* that are multiplicative in both components such that T(a, b) = T(b, a), for all a,b \in A.
Non-degenerate here means that: If T(a, b) = 1 for all b \in A, then a=identity.


--- On Thu, 10/9/08, Shaun V. Ault <ault at fordham.edu> wrote:
From: Shaun V. Ault <ault at fordham.edu>
Subject: Re: [GAP Forum] Symmetric bicharacters
To: dan_lanke at yahoo.com
Date: Thursday, October 9, 2008, 7:36 AM

Dear Dan,

Perhaps you meant "homomorphisms"?  Surely there are no isomorphisms
f : A
--> Hom(A, C^*), since there is an isomorphism on homomorphism sets:

adj :  Hom( A, Hom(A, C^*) )  \cong  Hom( A \otimes A, C^* ),

and the condition that (f(a))(b) = (f(b))(a) for all a, b in A implies that
adj(f) is a homomorphism f' with the property that f'(a \otimes b)
= f'(b
\otimes a) for all a in A.  Such an f' will not be injective unless A
trivial, and in that case, f' will not be surjective.

On the other hand, the homomorphisms A --> Hom(A, C^*) with the above
property can be identified with the representations A \otimes A -->
such that a \otimes b induces the same transformation on C as b \otimes
for any pair a, b in A.  Unfortunately, I don't know enough representation
theory to say much more about the latter.

Hope this helps,

Shaun V. Ault
Department of Mathematics
Fordham University
441 E.  Fordham Rd.
Bronx, NY 10458

             Dan Lanke                                                 
             <dan_lanke at yahoo.                                         
             com>                                                       To
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                                       [GAP Forum] Symmetric bicharacters
             10/08/2008 11:11                                          
             Please respond to                                         
             dan_lanke at yahoo.c                                         

Dear GAP Forum,

Let A be a finite abelian group. Let C^* denote the multiplicative group of
non-zero complex numbers. Let Hom(A, C^*) denote the group of all
homomorphisms from A to C^*.

I would like to create the group of all isomorphisms f : A --> Hom(A,  C^*)
that satisfy

(f(a))(b) = (f(b))(a), for all  a,b \in A.

Could you please point me in the right direction?

Many thanks,

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