Let $G$ be a finite group. What is meant by two finite dimensional vector spaces (over $\mathbb{C}$) $V$ and $W$ being "isomorphic as representations of $G$"? To show that we have such an isomorphism, wouldn't it suffice to just show that $\dim V = \dim W$?


No, it doesn't suffice that the representation spaces have equal dimension. What is usually meant by isomorphism of representations, is that there exists an equivariant isomorphism between the representation spaces, i.e. one that also preserves the action of $G$. In other words, two representations $\pi: G \to GL(V)$ and $\pi': G \to GL(W)$ are isomorphic if there exists an isomorphism $A: V \to W$ such that $A(\pi(g)v) = \pi'(g) Av$.

  • $\begingroup$ Martin's answer is perfect, but let me just add that, in universal algebra, we would simply say the two representations are equivalent iff the resulting algebras $\langle V, \{\pi(g) : g\in G\}\rangle$ and $\langle W, \{\pi'(g) : g\in G\}\rangle$ are isomorphic (which is exactly what Martin's condition says -- i.e. the map $A$ is an algebra isomorphism). $\endgroup$ Feb 12 '12 at 1:53
  • $\begingroup$ @Martin Taking what you said further, suppose $G$ has two generators $x,y$ (say). If we know the two objects (modules or vector spaces) are isomorphic when we only consider the $x$-action (say). Then to show they are isomorphic when considering $x$ and $y$, is it sufficient to just find an isomorphism $A$ such that $A(\pi(y)v)=\pi^\prime(y)Av$? Or does this isomorphism need to work for both $x,y$? $\endgroup$ Sep 25 '16 at 19:45

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.