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Let $\rho : \mathbb{R}^n\to \mathfrak{so}(2m)$ be a faithful representation of the commutative Lie algebra $\mathbb{R}^n$ into the Lie algebra of skew-symmetric matrices. There is an orthonormal basis $d_1,\ldots,d_{2m}$ of $\mathbb{R}^{2m}$ such that for all $i=1,\ldots,n$ and $j=1,\ldots,m$ $$\rho(e_i)(d_{2j-1})=\lambda_{ij}d_{2j}\ \text{and}\ \rho(e_i)(d_{2j})=-\lambda_{ij}d_{2j-1}$$ for some real numbers $\lambda_{ij}$. The degeneracy property of this representation is as follows \begin{equation}\label{deg}\tag{1} \lambda_{ki}^2=\lambda_{kj}^2\ \text{for all}\ k=1,\ldots,n\ \text{and for all basis of}\ \mathbb{R}^n. \end{equation} If $v_k=\sum_{p=1}^na_{pk}e_p$ $k=1,\ldots,n$ is another basis of $\mathbb{R}^n$, then \begin{align*} \text{$\eqref{deg}$}& \Longleftrightarrow \left(\sum_{p=1}^na_{pk}\lambda_{pi}\right)^2-\left(\sum_{p=1}^na_{pk}\lambda_{pj}\right)^2=0\ \text{for all}\ k=1,\ldots,n\\ & \Longleftrightarrow \sum_{p=1}^na_{pk}^2(\lambda_{ki}^2-\lambda_{kj}^2)+2\sum_{1\leq p<q\leq n}a_{pk}a_{qk}\left(\lambda_{pi}\lambda_{qi}-\lambda_{pj}\lambda_{qj}\right)=0\ \text{for all}\ k=1,\ldots,n\\ & \Longleftrightarrow \sum_{1\leq p<q\leq n}a_{pk}a_{qk}\left(\lambda_{pi}\lambda_{qi}-\lambda_{pj}\lambda_{qj}\right)=0\ \text{for all}\ k=1,\ldots,n \end{align*}

So I am studying the following question: $$\sum_{1\leq i<j\leq n}a_{ik}a_{jk}x_{ij}=0\ \text{for all invertible matrix}\ A=(a_{ij})$$

  1. Does this imply that $x_{ij}=0$ for all $1\leq i<j\leq n$?
  2. Is it possible to write the equality above using some quadratic form?
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Suppose $V = \mathbb{R}^n$ has a basis $(e_1,\dots,e_n)$. Your assumption is that you have a family of linear maps $\lambda_1,\dots,\lambda_m \in V^*$ which are defined such that $\lambda_r(e_i) = \lambda_{ir}$ for any $1 \leqslant r \leqslant m$. Working inside the algebra of all functions $\mathrm{Fun}(V,\mathbb{R})$, with multiplication given by pointwise evaluation, we have a function

$$q_{rs} = \lambda_r\lambda_r - \lambda_s\lambda_s$$

for any $1 \leqslant r,s \leqslant m$. This is, indeed, a quadratic form on $V$.

If $(e_1^*,\dots,e_n^*)$ is the dual basis of $V^*$ then $\lambda_r = \sum_{i=1}^n \lambda_{ir}e_i^*$ and

$$\lambda_r\lambda_r = \sum_{i=1}^n \lambda_{ir}^2e_i^* + 2\sum_{1 \leqslant i < j \leqslant n} \lambda_{ir}\lambda_{jr}e_i^*e_j^*$$

Your assumption that $\lambda_{ir}^2 = \lambda_{is}^2$ for any $1 \leqslant r,s \leqslant m$ means that

$$q_{rs} = 2\sum_{1 \leqslant i < j \leqslant n} (\lambda_{ir}\lambda_{jr}-\lambda_{is}\lambda_{js})e_i^*e_j^*$$

As you do, we could divide by $2$ to get a quadratic form $q_{rs}'$ so that $q_{rs} = 2q_{rs}'$.

If $g \in \mathrm{End}(V)$ then $ge_k = \sum_{i=1} a_{ki}e_i$ for some $a_{ki} \in \mathbb{R}$ and we have $e_i^*(ge_k)e_j^*(ge_k) = a_{ki}a_{kj}$ so you have

$$q_{rs}'(ge_k) = \sum_{1 \leqslant i < j \leqslant n} a_{ki}a_{kj}(\lambda_{ir}\lambda_{jr}-\lambda_{is}\lambda_{js})$$

So your condition reads that $q_{rs}'(ge_k) = 0$ for all $g \in \mathrm{GL}(V)$.

Now let $q : V \to \mathbb{R}$ be a quadratic form. Your question amounts to asking whether $q = 0$ given that $q(ge_i) = 0$ for all $g \in \mathrm{GL}(V)$ and $1 \leqslant i \leqslant n$. If we were working over $\mathbb{C}$ then we could show this by viewing $V$ as an affine space and using that $q$ is a polynomial function. So as $\mathrm{GL}(V)$ is dense in $\mathrm{End}(V)$ we must have $q = 0$. Probably there is an equivalent argument over $\mathbb{R}$ but I'll leave this to someone else.

One can instead argue as follows. For any $i \neq j$ we see that the map $\tau_{ij} \in \mathrm{End}(V)$ given by $\tau_{ij}e_k = e_k + \delta_{ki}e_j$ is invertible and $\tau_{ij}e_i = e_i+e_j$. Hence $q(e_i+e_j) = 0$ for any $1 \leqslant i,j \leqslant n$ (if $i=j$ then this is $q(2e_i) = 4q(e_i) = 0$).

Now let $\beta(u,v) = q(u+v) - q(u) - q(v)$ be the symmetric bilinear form determined by $q$. Then for any $1 \leqslant i,j \leqslant n$ we have

$$\beta(e_i,e_j) = q(e_i+e_j)-q(e_i) - q(e_j) = 0$$

Using the bilinearity we see that $\beta(u,v) = 0$ for any $u,v \in V$ which shows that $q(v) = \frac{1}{2}\beta(v,v) = 0$ for any $v \in V$. Thus $q = 0$ is identically zero.

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  • $\begingroup$ Many thanks Jay, but $q_{rs}$ is already identically zero: $q_{rs}(e_i)=0$ for all $i=1,\ldots,n$ also $e_i^*e_j^*=0$ $\endgroup$
    – user56980
    Jul 18 at 16:38
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    $\begingroup$ That doesn't imply $q_{rs} = 0$ because $q_{rs}$ is not linear. Take $n=2$ and let $\beta$ be the symmetric bilinear form with $\beta(e_1,e_2) = \beta(e_2,e_1) = 1$ and $\beta(e_1,e_1) = \beta(e_2,e_2) = 0$. We have a quadratic form given by $q(v) = \frac{1}{2}\beta(v,v)$ for all $v \in V$. Certainly $q(e_1) = q(e_2) = 0$ but $q(e_1+e_2) = \frac{1}{2}\beta(e_1+e_2,e_1+e_2) = 1 \neq 0$. So it is not enough to check this on a basis. $\endgroup$
    – Jay Taylor
    Jul 18 at 16:51
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    $\begingroup$ Relatedly $e_i^*e_j^*$ is also not zero. Let $v = e_i + e_j$ then $e_i^*(v) = e_j^*(v) = 1$ and so $(e_i^*e_j^*)(v) = e_i^*(v)e_j^*(v) = 1\cdot 1 = 1$. Recall we evaluate the product pointwise. In fact, in the example in the previous comment the quadratic form is $q = e_1^*e_2^*$. $\endgroup$
    – Jay Taylor
    Jul 18 at 16:55
  • $\begingroup$ Thank you. I reacted quickly, of course $q_{rs}$ is not linear! $\endgroup$
    – user56980
    Jul 18 at 17:32
  • $\begingroup$ I can't edit the nice answer $\lambda_r^2=...e_i^{*2}$ instead of $e_i^*$ $\endgroup$
    – user56980
    Jul 18 at 18:04

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