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!
On
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.
On
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.
On
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.
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.