Results 1 to 3 of 3

Thread: Group Representation Theory

  1. #1
    Junior Member
    Joined
    Oct 2010
    Posts
    49

    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.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor Drexel28's Avatar
    Joined
    Nov 2009
    From
    Berkeley, California
    Posts
    4,563
    Thanks
    22
    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 $\displaystyle \rho:\mathbb{Z}_4\to\text{GL}_2(\mathbb{R})$ and it is the homomorphism generated by $\displaystyle 1\mapsto \begin{pmatrix}0 &1\\-1 & 0\end{pmatrix}$ (since as you can quick check, for the matrix you put the space $\displaystyle \text{span}(1,1)$ is a non-trivial $\displaystyle \rho$-invariant subspace). Assume then that $\displaystyle \mathbb{R}^2$ had a non-trivial invariant subspace $\displaystyle S$, then by a dimension argument we see that $\displaystyle \dim_\mathbb{R}(S)=1$ and thus $\displaystyle S=\text{span}(x)$ for some common eigenvector $\displaystyle x$ of $\displaystyle \rho_0,\cdots,\rho_3$. That said, this is impossible since both the eigenvectors for $\displaystyle \rho(1)$ 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 $\displaystyle \mathbb{C}$ then everything's easy because you can check easily that $\displaystyle \left\langle \lambda\chi,\lambda\chi\right\rangle=1\left\langle \chi,\chi\right\rangle=1$ (this is because $\displaystyle \lambda$ [what I'm calling the linear character] is irrelevant since $\displaystyle |\lambda(g)|=1$ and so won't affect the inner product).

    If the field is $\displaystyle \mathbb{R}$ then this is simple since by definition an irreducible character $\displaystyle \chi$ is a character of an irrep $\displaystyle \rho$ and a linear character $\displaystyle \lambda$ is the character of a degree one irrep $\displaystyle \psi$ and thus $\displaystyle \lambda\chi$ is the character of $\displaystyle \psi\otimes\rho$. Thus, it suffices to show that $\displaystyle \psi\otimes\rho$ is irreducible if and only if $\displaystyle \rho$ is irreducible. But, by assumption you can assume the representation space of $\displaystyle \psi\otimes\rho$ is (up to equivalence) $\displaystyle \mathbb{R}\otimes\mathbb{R}^m$ 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 $\displaystyle \lambda\chi$ is an irreducible character of $\displaystyle G$ with the same degree as $\displaystyle \chi$. Thus, by assumption you have that $\displaystyle \lambda\chi=\chi$ and thus, in particular $\displaystyle \lambda(g)\chi(g)=\chi(g)$ since $\displaystyle \lambda(g)\ne 1$ you may conclude.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Oct 2010
    Posts
    49
    Thanks Drexel28, this helps alot!
    For (ii) I actually mean "If y(1) > 1, show that y .y(BAR) is always irreducible"
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 0
    Last Post: May 17th 2011, 12:33 PM
  2. the adjoint representation of a Lie group
    Posted in the Advanced Algebra Forum
    Replies: 5
    Last Post: May 5th 2011, 10:13 PM
  3. Representation Theory
    Posted in the Advanced Algebra Forum
    Replies: 3
    Last Post: Apr 7th 2011, 12:55 PM
  4. representation of symmetric group
    Posted in the Advanced Algebra Forum
    Replies: 2
    Last Post: Jan 21st 2011, 03:32 PM
  5. Practical Group Representation Theory
    Posted in the Advanced Algebra Forum
    Replies: 0
    Last Post: Jun 4th 2008, 10:32 AM

Search Tags


/mathhelpforum @mathhelpforum