We have a symmetric bilinear form on $\mathbb R^n$ defined by
$$e_1\cdot e_1=e_2\cdot e_2=e_3\cdot e_3=\cdots=0$$
$$e_1\cdot e_2=e_1\cdot e_3=e_2\cdot e_3=\cdots=1.$$
The signature is the numbers of vectors in an orthogonal basis squaring to $+1,-1,0$.
For $n=1$, this is just $e_1\cdot e_1=0$, so the signature is $(0,0,1)$.
For $n=2$, we can find an orthogonal basis $(e_1\pm e_2)/\sqrt2$ which shows that the signature is $(1,1,0)$.
For $n=3$, my answer here shows that the signature is $(1,2,0)$.
Would the general case for $n>1$ have $(1,n-1,0)$ ?
Best Answer
Yes.
If we start with an orthogonal basis $\{\sigma_1,\tau_2,\tau_3,\cdots,\tau_n\}$ and try to construct an isomorph of the original basis, the idea is to place $\{e_i\}$ on the null cone as the vertices of a regular $(n-1)$-simplex.
For the reverse process, define
$$\sigma_1=\frac{e_1+e_2+\cdots+e_n}{\sqrt{n(n-1)}}$$
and then just Gram-Schmidt to get $\{\tau_i\}$ from $\{e_i\}$. Anything orthogonal to $\sigma_1$ has negative square:
$$e_i\cdot\sigma_1=\frac{n-1}{\sqrt{n(n-1)}}$$
$$x=x_1e_1+x_2e_2+\cdots+x_ne_n$$
$$x\cdot\sigma_1=\frac{n-1}{\sqrt{n(n-1)}}\big(x_1+x_2+\cdots+x_n\big)=0$$
$$x\cdot x=2x_1x_2+2x_1x_3+2x_2x_3+\cdots+2x_{n-1}x_n$$
$$=\big(x_1+x_2+\cdots+x_n\big)^2-\big(x_1\!^2+x_2\!^2+\cdots+x_n\!^2\big)$$
$$=0^2-\big(x_1\!^2+x_2\!^2+\cdots+x_n\!^2\big)\quad<0$$
so the signature is indeed $(1,n-1,0)$.