Are there any general results on when conjugate representations of a real Lie algebra are equivalent? I'm inclined to say that they are often not, but this is merely going on my case by case experience.
In particular if we know that the fundamental and antifundamental representations of the Lie algebra are inequivalent, can we deduce that all conjugate representations are? I feel that this should be possible, but can't get started with a proof.
Has anyone got either (a) some hints to get me started or (b) a good reference which might be able guide me through such a problem?
I genuinely don't know at present whether this is trivial or difficult, so any advice would be much appreciated.
Any hermitian inner product on a complex vector space $V$ gives an isomorphism between $\bar V$ and $V^*$. Furthermore, if $V$ has a hermitian inner product that is preserved by the representation of the Lie group/algebra, then this gives an isomorphism of representations between $\bar V$ and $V^*$. Thus in such cases you are asking about representations being self-dual (which is equivalent to the representation having an invariant bilinear form).
For semi-simple Lie groups, every representation has an invariant hermitian inner product (the representation comes from the simply connected compact real form and you can average over this group to get an invariant hermitian inner product-- see Weyl's unitarian trick).
You can determine if a representation of a semi-simple Lie algebra is self-dual by using weights. If $\mu$ is the highest weight of $V$ then the highest weight of $V^*$ is $-\omega_0 \mu$, where $\omega_0$ is the unique element of the Weyl group mapping the positive Weyl chamber to its negative. The element $\omega_0$ is $-1$ for the groups $SU(2)$, $SO(2n+1)$, $Sp(n)$, and $SO(4n)$. So for all of these groups $V \simeq V^* (\simeq \bar V)$. A reference for this is section 6.4 (and it's exercises) of Sepanski's Compact Lie groups.