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Math Help - Group Representation Theory

  1. #1
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    Group Representation Theory

    (1) Let G:= <g|g^4 = 1> and F be the field of real numbers. Show that the representation

    p:g -> (0 -1)
    (-1 0)

    is irreducible over F.

    (2) Let G be a finite group, x & y two characters of G

    (i) If x is linear, show the x.y is irreducible <=> y is irreducible

    (ii) If y(1) > 1, show that y .y is always irreducible.

    (iii)Let x be an element of Irreducible G and no other character in Irreducible G have the same degree as x. Suppose that there is a linear character y such that for some g, y(G) is not equal to 1. Show that x(g) = 0.
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Turloughmack View Post
    (1) Let G:= <g|g^4 = 1> and F be the field of real numbers. Show that the representation

    p:g -> (0 -1)
    (-1 0)

    is irreducible over F.
    I think you may be over thinking this. So, I assume that you mean \rho:\mathbb{Z}_4\to\text{GL}_2(\mathbb{R}) and it is the homomorphism generated by 1\mapsto \begin{pmatrix}0 &1\\-1 & 0\end{pmatrix} (since as you can quick check, for the matrix you put the space \text{span}(1,1) is a non-trivial \rho-invariant subspace). Assume then that \mathbb{R}^2 had a non-trivial invariant subspace S, then by a dimension argument we see that \dim_\mathbb{R}(S)=1 and thus S=\text{span}(x) for some common eigenvector x of \rho_0,\cdots,\rho_3. That said, this is impossible since both the eigenvectors for \rho(1) are non-real. Get the idea?



    (2) Let G be a finite group, x & y two characters of G

    (i) If x is linear, show the x.y is irreducible <=> y is irreducible

    What field are you working over here? If it's \mathbb{C} then everything's easy because you can check easily that \left\langle \lambda\chi,\lambda\chi\right\rangle=1\left\langle \chi,\chi\right\rangle=1 (this is because \lambda [what I'm calling the linear character] is irrelevant since |\lambda(g)|=1 and so won't affect the inner product).

    If the field is \mathbb{R} then this is simple since by definition an irreducible character \chi is a character of an irrep \rho and a linear character \lambda is the character of a degree one irrep \psi and thus \lambda\chi is the character of \psi\otimes\rho. Thus, it suffices to show that \psi\otimes\rho is irreducible if and only if \rho is irreducible. But, by assumption you can assume the representation space of \psi\otimes\rho is (up to equivalence) \mathbb{R}\otimes\mathbb{R}^m etc. You can take it from there?


    (ii) If y(1) > 1, show that y .y is always irreducible.
    Is this really what you want to say here?


    (iii)Let x be an element of Irreducible G and no other character in Irreducible G have the same degree as x. Suppose that there is a linear character y such that for some g, y(G) is not equal to 1. Show that x(g) = 0.
    [/quote]

    You know (from i) that \lambda\chi is an irreducible character of G with the same degree as \chi. Thus, by assumption you have that \lambda\chi=\chi and thus, in particular \lambda(g)\chi(g)=\chi(g) since \lambda(g)\ne 1 you may conclude.
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  3. #3
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    Thanks Drexel28, this helps alot!
    For (ii) I actually mean "If y(1) > 1, show that y .y(BAR) is always irreducible"
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