Are metric tensors (or even pseudo-metric tensors) really metrics?

301 Views Asked by At

In some discussions of metric tensors $g : TM_p \times TM_p \rightarrow \mathbb R$ (and even of pseudo-metric tensors, for instance in this answer), and of the corresponding first fundamental forms ${\rm d}s^2$, these (mathematical) objects are referred to as "metric".

Is this serious and correct terminology?

Is it really true that metric tensors, or (even) pseudo-metric tensors, and/or the corresponding first fundamental forms are metrics (in the specific well-known sense), by which accordingly $g : X \times X \rightarrow \mathbb R$ for some suitable set $X$, such that $\forall x, y, z \in X$:

?

3

There are 3 best solutions below

2
On BEST ANSWER

Yes, it is true, but note that the metric tensor $g$ defines the distance function on $TM$ and hence a norm, it is not itself the distance function.

In particular, for $TM$ we can define

$$d(\vec{x}, \vec{y}) \equiv \sqrt{g(\vec{x} - \vec{y}, \vec{x} - \vec{y})} \quad\forall \vec{x}, \vec{y} \in TM$$

It's relatively easy to show that if $g$ is a metric (as opposed to a pseudometric) this definition of $d$ works as you would expect.

Note that I am answering about the relationship between $g$ and $d$ on $TM$: $g$ also makes $M$ a metric space, but that's a slightly different argumentt.

2
On

It works if the metric has only positive eigenvalues. It is not the case, for example, for Minkowski space where $g(V,V)=0$ for some vectors (light-like ones).

3
On

From the metric we get a curve length, if the manifold is connected we can define a distance function as the infimum of curve lengths between two points (this exists if the metric is positive definite). It is easily checked that this defines a metric in the usual sense.

A Lorentzian metric does not define a metric in the usual sense, as points may be connected by a zero length curve.