Quadratic form of the square of the distance function

Question

I'm looking for a special kind of metric (say Riemannian manifold $(M, g)$), where square of distance function $d^2:M\times M\to R$ between an arbitrary points $x$ and basepoint of $x_0$ is quadratic form. Are there certain criteria for quadratic form of the square of the distance function?!

Answer 1

OK, here are a few comments that might help you think about this.

1. The term "quadratic form" only makes sense if the underlying space $M$ is a vector space (or at least a module over a commutative ring). Since you also want it to be a smooth manifold, it should be a finite-dimensional real vector space. Up to choice of basis any such space is just $\mathbb R^n$, so we might as well stipulate that $M=\mathbb R^n$ from the start.
2. As I mentioned in my comment, a quadratic form on $\mathbb R^n$ is a function of one vector, while a metric on $\mathbb R^n$ would be a function of two vectors, so it doesn't make sense to ask for "a metric whose squared distance function is a quadratic form."
3. From your comment, it sounds as if one thing you're looking for is a metric on $\mathbb R^n$ with the property that the function $f(x) = d(0,x)^2$ is a quadratic form. Because this condition only depends on the distance function from the origin and is insensitive to distances between other points, there are lots of such metrics. One way to construct such metrics is to switch to spherical coordinates $(\rho,\theta^1,\dots,\theta^{n-1})$, and write $$ g = d\rho^2 + \sum_{\alpha,\beta=1}^{n-1} g_{\alpha\beta}(\rho,\theta)d\theta^\alpha\,d\theta^\beta, $$ for a smooth positive definite matrix-valued function $(g_{\alpha\beta})$. There are some subtle conditions that $g_{\alpha\beta}$ has to satisfy near the origin in order for the resulting metric to be smooth, but for example, any Riemannian metric in polar normal coordinates will satisfy these conditions.

If you want the distance functions from other points besides $0$ to be given by quadratic forms, then there will certainly be more stringent conditions. The appropriate condition in that case would be that for each $y\in\mathbb R^n$, the function $f_y(x) = d(y,y+x)$ is a quadratic form. (You can't ask that $d(y,x)$ be a quadratic form in $x$, because it doesn't vanish when $x=0$.) My guess is that in that case, the only such metrics are ones with constant coefficients in Cartesian coordinates, but I don't have a handy proof to offer.