4
$\begingroup$

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.

$\endgroup$
8
  • $\begingroup$ 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? $\endgroup$
    – Sidharth Ghoshal
    Apr 25, 2016 at 14:03
  • $\begingroup$ hmm... the original comment appears to have disappeared. $\endgroup$
    – Sidharth Ghoshal
    Apr 25, 2016 at 14:04
  • $\begingroup$ (I did not see the word "abelian", that was my mistake.) $\endgroup$
    – Captain Lama
    Apr 25, 2016 at 14:04
  • $\begingroup$ 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 ? $\endgroup$
    – reuns
    Apr 25, 2016 at 14:05
  • 1
    $\begingroup$ @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 $\endgroup$
    – reuns
    Apr 25, 2016 at 14:09

1 Answer 1

Reset to default
2
$\begingroup$

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.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .