What is the signature of this bilinear form? $e_i\cdot e_j=1-\delta_{ij}$

Question

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 $(1,n-1,0)$ ?

Answer 1

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 $(1,n-1,0)$.

Answer 2

Let us consider a $n\times n$ matrix $M$ with $0$ on the diagonal and $1$ anywhere else. This can be written as the difference between a rank-$1$ matrix made by $1$s only and the identity matrix, $M=U-I$. The spectrum of $U$ is $(n,0,0,0,\ldots)$ and the spectrum of $I$ is $(1,1,1,1,\ldots)$. They both are symmetric matrices and $I$ commutes with $U$, so the spectrum of $U-I$ is $(n-1,-1,-1,-1,\ldots)$ and the signature is straightforward to compute.