# Degenerate representation

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 $$$$\label{deg}\tag{1} \lambda_{ki}^2=\lambda_{kj}^2\ \text{for all}\ k=1,\ldots,n\ \text{and for all basis of}\ \mathbb{R}^n.$$$$ If $$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

So I am studying the following question: $$\sum_{1\leq i

1. Does this imply that $$x_{ij}=0$$ for all $$1\leq i?
2. Is it possible to write the equality above using some quadratic form?

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.

• Many thanks Jay, but $q_{rs}$ is already identically zero: $q_{rs}(e_i)=0$ for all $i=1,\ldots,n$ also $e_i^*e_j^*=0$ Jul 18 at 16:38
• 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. Jul 18 at 16:51
• 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 $q = e_1^*e_2^*$. Jul 18 at 16:55
• Thank you. I reacted quickly, of course $q_{rs}$ is not linear! Jul 18 at 17:32
• I can't edit the nice answer $\lambda_r^2=...e_i^{*2}$ instead of $e_i^*$ Jul 18 at 18:04