I am trying to understand why

1) all finite-dimensional complex representations $V$ of $G$ are self dual, and

2) How the derived subgroup $[G,G]$ is a union of particular conjugacy classes.

My thoughts: on

1) by definition the dual representation of $V$ is the dual representation $\rho ^*$ on the dual vector space $V^*:Hom(V,\mathbb{C})$ of linear maps defined by $\rho(g)^*L = L \circ \rho(g)^{-1}$. Now according to the source: Def: 5.7 this preserves the duality pairing of $V$ and $V^*$ by $L(\mathbf{v}) = \rho^*(g)(L)(\rho(g)(\mathbf{v}))$.

I don't see how this works though as $\rho(g)^*L = L \circ \rho(g)^{-1}$ means that the dual representation of a group element $g$ is really the inverse representation of an element of $g$ which is then mapped to a complex number. I am a bit lost on what how this machinery works and what $\rho(g)^{-1}$ really is.

2) If we take some $x \in [G,G]$ we have that $xg = gx$ for some element $g \in G$ then of course $g^{-1}xg = x$ and {$g^{-1}xg \forall g \in G$} is the conjugacy class. There are several things rolling around in my mind, the first being that perhaps I can take the union of the representations, but I am not sure how to go about writing it out.

Any thoughts on either question is appreciated.

  • 5
    $\begingroup$ It isn't true (for a general finite group $G$) that all finite dimensional representations are self-dual. What is true is that the dual of a representation is another (not necessarily equivalent) representation. $\endgroup$ Sep 15 '14 at 19:14
  • $\begingroup$ For an example, consider any non-trivial one-dimensional complex representations of a cyclic group of order three. $\endgroup$ Sep 15 '14 at 21:37

Any normal subgroup, not only the commutator subgroup, Is the union of (disjoint) conjugacy classes.

  • $\begingroup$ (More general: Any set/subset closed under the action of a group is a union of orbits.) $\endgroup$
    – whacka
    Sep 15 '14 at 21:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.