Hence, we represent ${\EuScript{J}}(v)$ by a non-degenerate symmetric bilinear form $<{\EuScript{J}}_x v,v>_x$. Now we use Lagrange’s method to diagonalize this bilinear form, obtaining a base $\{u_1,\dots,u_n\}$ of $E_x$ such that $$\begin{aligned}
  {\EuScript{J}}_x(\sum_{i}\alpha_iu_i)=\sum_{i=1}^q -\lambda_i\alpha_i^2 +
  \sum_{j=q+1}^n \lambda_j\alpha_j^2, \quad
  (\alpha_1,\dots,\alpha_n)\in{{\mathbb R}}^n.\end{aligned}$$ Replacing each element of this base according to $v_i=|\lambda_i|^{1/2}u_i$ we deduce that $$\begin{aligned}
{\EuScript{J}}_x(\sum_{i}\alpha_iv_i)=\sum_{i=1}^q -\alpha_i^2 +
  \sum_{j=q+1}^n \alpha_j^2, \quad
  (\alpha_1,\dots,\alpha_n)\in{{\mathbb R}}^n.\end{aligned}$$ Finally, we can redefine $<,>$ so that the base $\{v_1,\dots, v_n\}$ is orthonormal. This can be done smoothly in a neighborhood of $x$ in $M$ since we are assuming that the quadratic forms are non-degenerate; the reader can check the method of Lagrange in a standard Linear Algebra textbook and observe that the steps can be performed with small perturbations, for instance in [@Maltsev63].

In this adapted inner product we have that $J_x$ has entries from $\{-1,0,1\}$ only, $J_x^*=J_x$ and also that $J_x^2=J_x$.

Having fixed the orthonormal frame as above, the *standard negative subspace* at $x$ is the one spanned by $v_{1},\dots, v_{q}$ and the *standard positive subspace* at $x$ is the one spanned $v_{q+1},\dots,v_n$.

### Positive and negative cones {#sec:positive-negative-co}

Let ${\EuScript{C}}_\pm=\{C_\pm(x)\}_{x\in U}$ be the family of positive and negative cones $$\begin{aligned}
  C_\pm(x):=\{0\}\cup\{v\in E_x: \pm{\EuScript{J}}_x(v)>0\}  \quad x\in U\end{aligned}$$ and 