# Is a metric on a metric space a bilinear form?

I've just finished a course on bilinear forms and am now starting a cause on topological spaces and was just wondering; for a metric space which is made up of a set $M$ and a metric function $d$ such that $$d:M \times M \to \mathbb{R}$$ defines the distance between points in $M$ etc etc..

Is this 'distance function' a simply a bilinear form? And if not why not? What goes wrong?

Thanks!

-
Bilinear requires that you have an addition, which you do not here. – Tobias Kildetoft Apr 30 '12 at 12:02

Unless $M$ has a vector space structure, you don't even have a usual notion of linearity. Even if $M$ is a vector space, the metric (almost always) can't be bilinear because it has to always be positive, and if $d$ were bilinear, then $d(-x,y)=-d(x,y)$. Thus if you have two distinct points in $M$, you can't have a bilinear metric.

-

This might be slightly more involved than what you are encountering right now, but there is a link between these two notions : first, a positive-definite bilinear form yields a norm on vector spaces, so this is a function on a vector space, but a distance function on a vector space is not necessarily of this form. Then, on a smooth manifold (a particularly well-behaved kind of topological space), there is the notion of a tangent space, hence tangent vector to a curve, and then, integrating the "norm" (i.e. a smoothly-varying bilinear form on the tangent bundle) of the tangent vector to a curve along the whole curve gives its length, and then, tanking the minimum of all lengths between two points give the distance between these two points.

-

A metric is just a function, it is nothing to do with an "operation" on the space. Linearity is usually to do with functions that respect operations.

-

While not being bilinear, one instance of metrics playing nicely with operations is the notion of a left-invariant metric on a topological group. Here left-invariance means that if you have a topological group $G$ with a metric $d$, then for all $g,h,k \in G$ we have $d(h,k)=d(gh,gk)$. And you can have your metric be right-invariant or bi-invariant as well.

-