# Cauchy-Schwarz for metrics with arbitrary signatures

When the norm of a vector is always greater than or equal to zero, the Cauchy-Schwarz inequality holds, but what if we look at a metric with an arbitrary signature? Then the inner product of a vector with itself could be negative. Is there any Cauchy-Schwarz inequality for an arbitrary metric? I suspect it would look something like this:

$$g_{ij}^2\leq |g_{ii}g_{jj}|,$$

but I am not sure if that is necessarily the case (let alone how to go about proving it). Here, $g$ is the matrix representation of the metric tensor.

## 2 Answers

For the $(1,3)$ metric $x\cdot y = x_0y_0-\sum_{i=1}^3 a_ib_i$ the following reverse Cauchy-Schwarz inequality holds for timelike and lightlike vectors: $$|x\cdot y| \ge \|x\| \|y\| \quad \text{ if } \|x\|,\|y\|\ge 0 \tag{1}$$ Proof: If $x$ and $y$ are collinear, the inequality is trivial. Otherwise, their linear span is a plane, and any plane contain some spacelike vectors. Therefore, the polynomial $p(t)=\|x+ty\|^2$ changes sign. Looking at its discriminant, we arrive at (1).

Source: page 185 of this book.

The same argument work if only one of two vectors has nonnegative norm, but then (1) is trivial. If both vectors are spacelike, the sign in (1) could go either way, depending on whether the linear span crosses the light cone.

An obvious generalization: if the set $\{x:\|x\|\ge 0\}$ does not contain any 2-dimensional subspaces, (1) holds.

Take, for example, the following hyperbolic scalar product on $\mathbb{C}^2$:

$$[x, y] = y^* J x, \quad J = \mathop{\rm diag}(1,-1),$$

and let

$$x = \begin{bmatrix} 1 \\ 1 \end{bmatrix}, \quad y = \begin{bmatrix} 1 \\ -1 \end{bmatrix}.$$

Then

$$[x,x] = [y,y] = 0, \quad \text{but} \quad [x,y] = [y,x] = 2,$$

so I don't think you can make a Cauchy–Schwarz-like inequality.

More generally, if you have a product induced by a nonsingular indefinite $J$, you can always find $x,y$ such that $[x,x] = [y,y]$ (meaning that $x$ and $y$ are degenerate), but $[x,y] = [y,x] \ne 0$. This should be quite simple to prove.