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

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

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

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.

• Fair answer. I upvoted, but I prefer not to use matrices. Aug 29, 2019 at 0:38