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here is my question:

Suppose that the torus $T^n = (S^1)^n$ acts on $\mathbb{C}^n$ by linear transformations $$(e^{i \theta_1},...,e^{i \theta_n}).(z_1,...,z_n) = (e^{i \theta_1}.z_1,...,e^{i \theta_n}.z_n),$$ and consider its diagonal action on $\mathbb{C}^{2nN} = (\mathbb{C}^n)^{2N}$, which we endow with its standard hermitian structure.

Given a $T^n$-invariant hermitian form $\mathcal{Q} : \mathbb{C}^{2nN} \to \mathbb{R}$ (or more precisely the quadratic form associated to a hermitian matrix), is it possible to find a decomposition $$\mathbb{C}^{2nN} = \overset{2nN}{\underset{k=1}{\bigoplus}}\mathbb{C}.v_k$$ of $\mathbb{C}^{2nN}$ into complex lines such that the two following requirements hold:

  1. each line $\mathbb{C}.v_k$ is $T^n$-invariant and $T^n$ acts on it by one of its standard characters $\chi(e^{i \theta_1},...,e^{i \theta_n}) = e^{i \theta_j}$, for some $j=1,...,n$;
  2. the hermitian form associated with $\mathcal{Q}$ is diagonal in the basis $(v_1,...,v_{2nN})$.

Thanks for your help !

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  • $\begingroup$ Is your hermitian form $\mathcal Q$ positive definite or at least non-degenerate? $\endgroup$ Commented May 30, 2018 at 14:09
  • $\begingroup$ Hi Mikhail. Yes, we could suppose that it is non-degenerate. Would it help ? $\endgroup$
    – BrianT
    Commented May 30, 2018 at 16:41
  • $\begingroup$ It turns out that I can do it for any invariant hermitian form, not necessarily non-degenerate. $\endgroup$ Commented May 30, 2018 at 17:26

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We need some notation. I write $V$ for your $\Bbb C^{2nN}$, $T$ for your $T^n$, and $\rho\colon T\to GL(V)$ for the representation of $T$ in $V$. I define the character $\chi_j$ of $T$ by $$\chi_j(e^{i \theta_1},...,e^{i \theta_n}) = e^{i \theta_j}.$$ Further, for each $j$ I define a subspace $V_j$ of $V$ by $$V_j=\{x\in V\ |\ \rho(t) x=\chi_j(t)x\quad\text{for all }t\in T\}.$$ Let $j'\neq j$, $x\in V_j$, $x'\in V_{j'}$. Since the hermitian form $\mathcal Q$ is $T$-invariant, we have $$\mathcal Q(\rho(t) x,\rho(t) x')=\mathcal Q( x, x').$$ On the other hand, it follows from the definition of $V_j$ and $V_{j'}$ that $${\mathcal Q}(\rho(t) x,\rho(t) x')={\mathcal Q}(\,\chi_j(t) x,\,\chi_{j'}(t) x'\,)=\chi_j(t)\chi_{j'}(t)^{-1}{\mathcal Q}(x,x').$$ Since $\chi_j\neq \chi_{j'}$, we see that ${\mathcal Q}(x,x')=0$. This the subspaces $V_j$ and $V_{j'}$ are orthogonal with respect to ${\mathcal Q}$. Let ${\mathcal Q}_j$ denote the restriction of ${\mathcal Q}$ to $V_j$. We diagonalize each ${\mathcal Q}_j$ and thus obtain a desired diagonalization of ${\mathcal Q}$.

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  • $\begingroup$ Thank you very much for your answer. How do we ensure that the diagonal matrix is still invariant by the torus action ? $\endgroup$
    – BrianT
    Commented May 30, 2018 at 18:33
  • $\begingroup$ Each basis vector $v$ is contained in some $V_j$, hence, $\rho(t)v=\chi_j(t)v$. $\endgroup$ Commented May 30, 2018 at 18:51

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