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I've just started reading a book on functional analysis, and first definition given there is for a metric and metric space:

Let $\mathfrak{M}$ be an arbitrary set. A function $\rho\colon \mathfrak M\times\mathfrak M\to[0,\infty)$ is called metric if it has the following properties:

1) $\rho(x,y)=0 \iff x=y$ (axiom of identity)

2) $\rho(y,x)=\rho(x,y)\;\forall x,y\in\mathfrak M$ (axiom of symmetry)

3) $\rho(x,y)\le\rho(x,z)+\rho(z,y)\;\forall x,y,z\in\mathfrak M$ (triangle inequality)

The pair $(\mathfrak M,\rho)$ is called metric space.

First and second identities don't make any surprise, I understand them. But what about image of $\rho$ and third inequality? They don't seem to hold in Minkowski space, where if we check interval as candidate for metric, we get $$s^2=t^2-x^2-y^2-z^2,$$ which can be negative and violate triangle inequality (and if take square root, it becomes complex and inequality makes no sense in this case).

So this clearly isn't a metric. But is Minkowski space then not a metric space? What is it then?

Ruslan
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    from wikipedia (search metric): "In differential geometry, one considers metric tensors, which can be thought of as "infinitesimal" metric functions. They are defined as inner products on the tangent space with an appropriate differentiability requirement. While these are not metric functions as defined in this article, they induce metric functions by integration." – Dubious Oct 14 '13 at 18:59

2 Answers2

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It's a real vector space with an indefinite bilinear form.

Rasmus
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Minkowski space is a metric space, but the metric is not the "Minkowski metric".

Robert Israel
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    OK, but what metric would make sense for Minkowski space? Which one is actually used, e.g. in special relativity? – Ruslan Oct 15 '13 at 06:32
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    The topology is the standard one of ${\mathbb R}^4$. Which metric (in the "metric space" sense) you use for that is not particularly relevant. – Robert Israel Oct 15 '13 at 06:52
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    I suspect if you wanted a metric on a Minkowski space you'd want one that's Lorentz-invariant. I also suspect that there exist no such metric. Indeed wikipedia says: "Minkowski space is, in particular, not a metric space". Therefore I believe this answer is more confusing than useful. – Rotsor Apr 29 '17 at 23:53