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Let $\rho_\epsilon(a)=\begin{bmatrix}\epsilon & 0\\0 & \epsilon^{-1}\end{bmatrix}$ and $\rho_\epsilon(b)=\begin{bmatrix}0 & 1\\1 & 0\end{bmatrix}$

I can prove that $\rho_\epsilon$ is complex irreducible (as $\epsilon\neq-1,1$) representation of the dihedral group. But I need some help with the following questions.

1) Let $\rho$ be a complex irrep of the dihedral group. Prove that $\exists \epsilon$ such as $\rho$ is isomorphic to $\rho_\epsilon$

2)Let $\phi$ be a natural 2-dimensional real representation of the dihedral group as transformations that make regular n-gon. Find an $\epsilon$ that $\phi$ is isomorphic to $\rho_\epsilon$

I have no idea how to start it. I am sorry for my English

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The cyclic group generated by $a$ is of order $n,$ so its image under a representation has to be cyclic of order dividing $n.$ Can you show that any cyclic subgroup of the $GL(2, \mathbb{C})$ can be put in the form $\rho_\epsilon(a)$ for some root of unity $\epsilon.$ Similarly, the image of $b$ has to be cyclic of order $2$ (to have irreducibility -- you need to prove it), so it is conjugate to either $-Id$ or your $\rho_\epsilon(b).$ The first would violate irreducibility.

For the second question, $\rho_\epsilon(a)$ is cyclic of order $n.$ You can figure out the rest.

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  • $\begingroup$ A cyclic subgroup of the complex numbers is a set of scalars, which $\rho_\epsilon(a)$ is not. Perhaps you mean that any matrix $A$ such that $A^n=I$ is conjugate to $\rho_\epsilon(a)$. At this point we have $\rho(a)\sim\rho_\epsilon(a)$ and $\rho(b)\sim\rho_\epsilon(b)$, so in some coordinates we have $\rho(a)=\rho_\epsilon(a)$ and in some coordinates we have $\rho(b)=\rho_\epsilon(b)$. What we need is coordinates where both equalities hold at the same time. $\endgroup$
    – anon
    Commented Jan 14, 2014 at 1:13
  • $\begingroup$ @anon I am half-asleep, will revise. $\endgroup$
    – Igor Rivin
    Commented Jan 14, 2014 at 1:29

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