How can I prove that the following 2 prehomogeneous vector spaces are not isomorphic? 1)$(GL_n,\Lambda_1\oplus \Lambda_1,\mathbb{C}^n \oplus \mathbb{C}^n)$ 2)$(GL_n,\Lambda_1\oplus \Lambda_1^*,\mathbb{C}^n \oplus \mathbb{C}^n)$ where $\Lambda_1$ is the standard representation of $GL_n$ on $\mathbb{C}^n$. in the case of prehomogeneous vector spaces the notion of isomorphism is given by:

Two triplets $(G, \rho, V)$ and $(G', \rho', V')$ are isomorphic if there exist a rational isomorphism $\sigma : \rho(G) \to \rho'(G')$ and an isomorphism $\tau : V \to V'$, both defined over $\mathbb{C}$, such that $$\tau(\rho(g)x)=\sigma\rho(g)(\tau(x))$$ for all $g\in G$ and $x\in V$. That is the following diagram is commutative for all $g\in G$:

$$\require{AMScd} \begin{CD} V @>{\tau}>> V'\\ @V{\rho(g)}VV @VV{\sigma\rho(g)}V \\ V @>{\tau}>> V' \end{CD} $$

  • $\begingroup$ What is $\rho(\tau(x))$? The domain of $\rho$ is something like $GL(V)$, while $\tau(x)$ is some element of $V'$. Even if you replace $\rho$ with $\rho'$ I don't understand. Is the RHS of your equation supposed to be $\sigma(\rho'(g)\tau(x))$? $\endgroup$ – David Hill Apr 12 '11 at 17:37
  • $\begingroup$ sorry I forgot a $g$ in the LHS. $\endgroup$ – Michele Torielli Apr 12 '11 at 20:58
  • $\begingroup$ it was RHS. sorry again $\endgroup$ – Michele Torielli Apr 13 '11 at 10:09

Your two representations of $GL_n$ are not isomorphic, because one of them contains $\Lambda_1$ with multiplicity 2 and the other with multiplicity 1, $\Lambda_1^*$ and $\Lambda_1$ being non-isomorphic. More generally, if $V$ and $W$ are finite-dimensional rational representations of $G=GL_n$ with $V$ irreducible then the multiplicity of $V$ in $W,$ $$\dim\operatorname{Hom}_{G}(V,W),$$ is an invariant of $W$ modulo isomorphism.

By the way, the second space is not prehomogeneous, because the bilinear pairing between $\Lambda_1$ and $\Lambda_1^*$ is a non-constant polynomial $GL_n$-invariant.

  • $\begingroup$ Thank you. just one question, when you say that $\Lambda_1$ and $\Lambda^*_1$ are not isomorphic, do you mean as representation?because as triplets $(GL_n,\Lambda_1,\mathbb{C}^n)$ and $(GL_n,\Lambda^*_1,\mathbb{C}^n)$ are isomorphic. $\endgroup$ – Michele Torielli Apr 12 '11 at 21:01
  • 1
    $\begingroup$ Yes, as representations of $GL_n.$ Even if you allow outer automorphisms of $G=GL_n,$ the two spaces you asked about in the main question would not be isomorphic: the first one is isotypic, but the second is not. (Additionally, for $n\geq 2$ the first one is prehomogeneous and the second is not.) $\endgroup$ – Victor Protsak Apr 12 '11 at 21:11
  • $\begingroup$ thanks. can I ask you what is the definition of isotypic? $\endgroup$ – Michele Torielli Apr 13 '11 at 8:05

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.