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Let $G$ be a finite group and $H_1$ and $H_2$ are two proper subgroups of $G$. Also, let $\rho:G \rightarrow \mathbb{C}^m \times \mathbb{C}^m$ be an irreducible non-trivial representation of $G$. Let $V_1$ and $V_2$ be subspaces of $\mathbb{C}^m$ such that $V_1 \cap V_2=0$. Further, $V_1$ is the maximal subspace such that for every element $h_1 \in H_1$ and $v \in V_1$, $\rho(h_1) v = v$. Similarly, $V_2$ is the maximal subspace such that for every element $h_2 \in H_2$ and $v \in V_2$, $\rho(h_2) v = v$.

Is it true that $V_1$ and $V_2$ are orthogonal to each other (w.r.t the standard inner product on $\mathbb{C}^m$) ? If not, can you provide a counterexample?

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orthogonal with respect to what? –  Jonas Hartwig Nov 21 '10 at 20:05
    
@Jonas : orthogonal w.r.t. the standard inner product function. –  Anindya De Nov 21 '10 at 20:52
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What the hell is $\rho:G\to\mathbb C^m\times \mathbb C^m$ supposed to mean? Maybe $\rho:G\to \mathrm{End}\mathbb C^m$ ? –  darij grinberg Nov 21 '10 at 20:57
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Anyway, by the definition of $V_1$, an irrep of $G$ occurs in $V_1$ iff it occurs in $\mathbb C^m$ and trivializes $H_1$ (i. e., the group $H_1$ acts trivially in this irrep). Similarly, an irrep of $G$ occurs in $V_2$ iff it occurs in $\mathbb C^m$ and trivializes $H_2$ (i. e., the group $H_2$ acts trivially in this irrep). Thus, no irrep can occur in both $V_1$ and $V_2$ (because then it would trivialize both $H_1$ and $H_2$, and thus occur in $V_1\cap V_2$, contradicting $V_1\cap V_2=0$). Therefore, $V_1$ and $V_2$ are orthogonal with respect to any $G$-invariant (!) scalar product. –  darij grinberg Nov 21 '10 at 21:03
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Of course, they need not be orthogonal with respect to some random (e. g., standard) scalar product (in fact, there is no reason why the standard scalar product here should be better than any randomly chosen one - think of representations of $G$ as some vector spaces, not necessarily $\mathbb C^m$). –  darij grinberg Nov 21 '10 at 21:04
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closed as off-topic by Ricardo Andrade, Andrey Rekalo, Stefan Kohl, Chris Godsil, David White Nov 28 '13 at 15:57

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1 Answer

No. Let $G$ be the dihedral group of order $6$ acting on its reflection representation $V_{\mathbb{R}}=\mathbb{R}^2$ (as symmetries of an equilateral triangle) and let $V$ be the complexification of this representation. Let $H_1$ and $H_2$ be the subgroups of order $2$ generated by distinct reflections. Then the fix spaces of $H_1$ and $H_2$ are not orthogonal wrt the $G$-inv. inner product, but they intersect trivially.

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Yes, this disproves it for bilinear inner products. –  darij grinberg Nov 22 '10 at 12:18
    
...and for sesquilinear (linear in first variable, conjugate linear in the 2nd) ones. –  Sheikraisinrollbank Nov 22 '10 at 13:32
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