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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 $g\in G$: $\begin{equation} \xymatrix{V \ar[d]_{\rho(g)} \ar[r]^\tau &V' \ar[d]^{\sigma \rho(g)} \\ V \ar[r]^\tau &V'} \end{equation}$

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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 $\sigma(\rho'(g)\tau(x))$? –  David Hill Apr 12 '11 at 17:37
    
sorry I forgot a $g$ in the LHS. –  Michele Torielli Apr 12 '11 at 20:58
    
it was RHS. sorry again –  Michele Torielli Apr 13 '11 at 10:09
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1 Answer

up vote 6 down vote accepted

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.

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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. –  Michele Torielli Apr 12 '11 at 21:01
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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.) –  Victor Protsak Apr 12 '11 at 21:11
    
thanks. can I ask you what is the definition of isotypic? –  Michele Torielli Apr 13 '11 at 8:05
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