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Math Help - Expressing a character in terms of abstract settings

  1. #1
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    Smile Expressing a character in terms of abstract settings

    Hi all

    Recently I have been studying the character

    \chi(G) = \# \{(x,y) \in G \times G | g= x^{-1} y^{-1} xy \}

    There is a lemma that states that this is equal to

    \sum_{i=1}^n \frac{\chi_i(g)}{\chi_i(1)} |G|

    Ranging over all the n characters of G.

    I have been trying to ascertain when this character is equal to the sum of permutation characters for a given group G. Computationally this is very easy, using GAP I am able to solve the SLEs and find the above character in terms of perm characters. So I have proved that this character is a sum of perm characters for some specific sizes of some classes groups, through pretty much brute force.

    However I am looking to look at this character in terms of more abstract settings, which may be able to help me prove more. I am sure this character is somehow related to the centre of the group (as for abelian groups it is trivial) but am unsure how.

    I realise this is a very vague question, but i'm not really looking for any specific answers, just any initial ideas that people have on the character or any musings on the subject, as very little info is available in texts.

    Many thanks for any help,

    Matt
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  2. #2
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    Thought an example might help: S_3

    \begin{bmatrix} <br />
  6&  0&  0 \\<br />
    3&  1&  0 \\<br />
    2&  0&  2 \\<br />
    1&  1&  1 <br />
\end{bmatrix}<br />

    Is a table of the permutation characters for S_3

    Then using the above formula we get the character \chi_c(g)=

    [18,0,9]

    Solving this we get \chi_c(g)=3P_2+6P_3-3P_3

    Where P_i are the permutation characters for S_3.
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  3. #3
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    Quote Originally Posted by Brothergomez View Post
    Thought an example might help: S_3

    \begin{bmatrix} <br />
  6&  0&  0 \\<br />
    3&  1&  0 \\<br />
    2&  0&  2 \\<br />
    1&  1&  1 <br />
\end{bmatrix}<br />

    Is a table of the permutation characters for S_3

    Then using the above formula we get the character \chi_c(g)=

    [18,0,9]

    Solving this we get \chi_c(g)=3P_2+6P_3-3P_3

    Where P_i are the permutation characters for S_3.
    I have some questions on your problem,

    In your matrix,

    \begin{bmatrix} <br />
  6&  0&  0 \\<br />
    3&  1&  0 \\<br />
    2&  0&  2 \\<br />
    1&  1&  1 <br />
\end{bmatrix}
    It seems like [6 0 0] corresponds to the regular representation, where each column denotes a conjugacy class (1^3), (2, 1), (3).
    Similarly, [3 1 0] corresponds to the defining representation, and [1 1 1] corresponds to the trivial representation for S_3.

    Anyhow, where does [2 0 2] come from?

    \chi(G) = \# \{(x,y) \in G \times G | g= x^{-1} y^{-1} xy \}

    There is a lemma that states that this is equal to

    \sum_{i=1}^n \frac{\chi_i(g)}{\chi_i(1)} |G|
    Can you also give some description of your lemma or provide reference for that?
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