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I was reading through Serre's Linear Representation Theory book and encountered a question to show that the set of all irreducible characters of an abelian group form a group.

The proof of closure was given, since if we have two character functions $\chi_1, \chi_2$ of irreducible representations $V,W$ then $\chi_1 * \chi_2$ is the character of the irreducible representation $V \otimes W$.

Additionally the trivial character serves as an identity element.

But I don't see how to take inverses of characters.

Attempted Ideas:

If there is a way to define an inverse tensor product, so that given $U,V$ we have that $f(U,V) = Z | Z \otimes V = U$, then I could try to compute the inverse tensor product of a representation, with the trivial representation, and hope that the character laws travel too. But I'm not sure how to define such an operator.

Sidharth Ghoshal
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  • Then what exactly does the author mean when he writes: " let $\hat{G}$ be the set of irreducible characters of $G$. If $\chi_1, \chi_2$ belong to $\hat{G}$, the same is true of their product $\chi_1 \chi_2$ "? I know that the character of the tensor product of two representations, is the product of their characters. Does that mean there is a different representation, whose character is the product of the two original characters, that is not the tensor product of the two original representations? – Sidharth Ghoshal Apr 25 '16 at 14:03
  • hmm... the original comment appears to have disappeared. – Sidharth Ghoshal Apr 25 '16 at 14:04
  • (I did not see the word "abelian", that was my mistake.) – Captain Lama Apr 25 '16 at 14:04
  • the characters are a group simply because each character $\chi$ has an inverse $\overline{\chi}$, and that $\chi_1 \chi_2$ is also a character ? – reuns Apr 25 '16 at 14:05
  • There is the dual representation. – Elle Najt Apr 25 '16 at 14:06
  • goblin, this is something I didn't know, but I think @user1952009 makes it clear since $\frac{1}{\chi} = \chi^*$ (because the traces have to be elements in the complex unit circle). I guess that answers the question. – Sidharth Ghoshal Apr 25 '16 at 14:07
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    @frogeyedpeas : sorry this is only true for characters on finite groups, the characters on $\mathbb{Z}$ are not of modulus $1$, nevertheless $1/\chi(g)$ is a character – reuns Apr 25 '16 at 14:09
  • http://math.stackexchange.com/questions/354926/complex-finite-dimensional-irreducible-representation-of-abelian-group; now use that the set of 1-dimensional complex representations is clearly a group. – Moishe Kohan Apr 25 '16 at 18:33

1 Answers1

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The inverse is given by taking the dual representation, which has character the complex conjugate of the original character. In the $1$-dimensional case this is the inverse of the original character.

Qiaochu Yuan
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