This thread is in response to questions in the Speculations forum about exactly what a metric is and this level of detailand discussion is neither speculation nor really on topic for the older thread.

I am posting it in Physics as most instances of metric at SF are connected with coordinates.

Also Maths is rather dry with its format of "Let us state an axiom or axioms and see what we can do with them."

Physics can be more chatty.

First a little backgorund.

"A set with the notion of distance between its elements." from Wiki.

Yes indeed a set with a notion of distance between its elements.

But we really need more than a notion, so let us examine the sort of things we want 'distance' to do for us or not do for us.

As so often in maths we gather togther all the objects we want to find a distnce between and put them in a set.

Voila we have a basic manifold or space.

We will call the set X and the elements x1, x2 x3 etc

Note these elements may be coordinate points in some coordinate system in which case our distance is coordinate distance.

Or they may be binary strings or actual words in characters in which case our distance is known as the Hamming distance.

So what do we want from our notion ?

Well here are some suggestions.

Our first notion of distance is that it is between two elements. So our set must contain at least two elements.

We do not want any elements to be 'left out.'

That is our distance determining function (D) must apply between each and every pair of elements (xn , xm)

That is we we do not want the distances to be undefined or undefinable between any pairs of elements.

We do not want any distance to be infinite (or do we?)

Ideally we would like the distance to appear as a number that we can append a physical unit to.

The function, D, must allow repeat values of distance from all different to all the same and everything in between.

We want the distance from xn to xm to be the same as the distance from xm to xn.

We want the distance to be zero if xn and xm are the same point or element. That is D(xn, xm) = 0

But we really do not want to cope with negative distances so we specify that D(xn , xm) is greater than or equal to zero.

These can all be written very compactly as a couple of mathematical axioms, but I am going to add one further dersire that is very useful, but not essential.

We want our D(xn , xm) to be the shortest or least value and that any distance via an intermediary point is greater than this.

These conditions form the basis of Riemannian Geometry (and thus Euclidian Geometry)

For Relativity we need to relax the restriction on negative values.

I think this has rapidly swept through the questions in the anti-time thread so next time will be for examples and further answers along with clearer maths.

Thanks @studiot .
looking forward to it.

On 9/10/2025 at 7:26 PM, studiot said:

These conditions form the basis of Riemannian Geometry (and thus Euclidian Geometry)

For Relativity we need to relax the restriction on negative values.

This is an interesting and important topic, so if I may be allowed to pick it up......

First recall that for every vector space \(V\) there exists a dual space \(V^*\), this being the vector space of all linear functions \(V \to \mathbb{R}\), so that for any \(v \in V\) and any \(\varphi \in V^*\) then \(\varphi(v) = \alpha \in \mathbb{R}\).

Suppose now an arbitrary manifold \(M\). Here our vector space may be taken to be, in some vague sense, "tangent" to it at some defined point, say \(p\). Accordingly, it is called the "tangent space" and is written as \(T_p(M\) and its dual as \(T^*(M)\). But still we will have that \(\phi(v) = \alpha \in \mathbb{R}\).

Now every vector space must have a basis that allows us to write, for any vector, \(v=\sum\nolimits _j\beta^j e_j\) where as beforthe set \({\beta^j}\) are scalars and the set \({e_J}\) are the basis vectors. in the specific case of our tangent dual space, the basis is the set \({dx^k}\), which I can prove but it would take too long (even if any one were interested

Let's now introduce a new operation \(T^*_p(M) \otimes T^*_p(M): T_p(M) \times T_p(M) \to \mathbb{R}\) so that in our specific case, \(\varphi \otimes \phi = \sum\nolimits _{jk} g_j g_k dx^j\otimes dx^k\)

By convention, the producct \(g_jg_k\) is usually written as \(g_{jk}\) and since the product \(dx^dx^k\) ican be taken to be the infinitessemal coordinate "distance", the construction \(g_{jk}(dx^j dx^k\) = ds^2\) is called the line element, and the product \(g_{jk}\) is referred to as the metric tensor.

A Riemann manifold is simply one that boasts a field of such tensors.

When we recall that, in any field, not every element need to be the same (though it might be), we can start to see how this beast is so important in the relativistic theory of gravitation.

Edited September 16Sep 16 by Xerxes

1 hour ago, Xerxes said:

This is an interesting and important topic, so if I may be allowed to pick it up......

Yes I agree and I was beginning to fear that no one else except the redoubtable MigL and myself were interested.

So thank you +1

Interestingly since the time you posted this is the first time I have seen your post in full and I was going to say that I cannot correct your Tex.

But I see that some fairy godmoderator has done it.

There is lots to say about a metric, not the least that we should not only concentrate on distance functions but also ask about area/volume/shape/ direction/angle.

Further we should be careful when defining our base manifold.

Are you aware of the story of the sea-mile ?

Oh, good Lord, Xerxes.
I started having flash-backs of being in 3rd year Uni math class.

18 hours ago, Xerxes said:

This is an interesting and important topic, so if I may be allowed to pick it up......

First recall that for every vector space V there exists a dual space V∗, this being the vector space of all linear functions V→R, so that for any v∈V and any φ∈V∗ then φ(v)=α∈R.

Suppose now an arbitrary manifold M. Here our vector space may be taken to be, in some vague sense, "tangent" to it at some defined point, say p. Accordingly, it is called the "tangent space" and is written as Tp(M and its dual as T∗(M). But still we will have that ϕ(v)=α∈R.

Now every vector space must have a basis that allows us to write, for any vector, v=∑jβjej where as beforthe set βj are scalars and the set eJ are the basis vectors. in the specific case of our tangent dual space, the basis is the set dxk, which I can prove but it would take too long (even if any one were interested

Let's now introduce a new operation T∗p(M)⊗T∗p(M):Tp(M)×Tp(M)→R so that in our specific case, φ⊗ϕ=∑jkgjgkdxj⊗dxk

By convention, the producct gjgk is usually written as gjk and since the product dxdxk ican be taken to be the infinitessemal coordinate "distance", the construction gjk(dxjdxk = ds^2\) is called the line element, and the product gjk is referred to as the metric tensor.

A Riemann manifold is simply one that boasts a field of such tensors.

When we recall that, in any field, not every element need to be the same (though it might be), we can start to see how this beast is so important in the relativistic theory of gravitation.

Physicists be like, "dual space? That's nothing but a fancy name for covariant vectors".

Mathematicians be like, "No, no, no, physicists! You must take the dual space seriously. It has a life of its own!"

Bilinears meet you half-way; wink, wink.

Very interesting topic BTW, @studiot .

Edited September 17Sep 17 by joigus
correction

1 hour ago, joigus said:

Physicists be like, "dual space? That's nothing but a fancy name for covariant vectors".

Mathematicians be like, "No, no, no, physicists! You must take the dual space seriously. It has a life of its own!"

Bilinears meet you half-way; wink, wink.

Very interesting topic BTW, @studiot .

Thank you, agreed.

As to the cautionary tale of the sea mile that no one seems interested in,

A few hundred years ago we could determine latitude pretty well but not longitude.
At that time we had a properly spherical model of our globe.

Maritime nations, in particular England, introduced the 'sea mile' or 'nautical mile' as a measure of distance ... at sea where the surface of the model globe was considered uniformly spherical.

The sea mile was defined as 'One minute of arc, measured along a meridion of longitude'.
Although they did no know which meridion, they assumed all segments of all meridions were equivalent.

They chose meridions of longitude because they were all supposed to be great arcs, whereas the equator is the only parallel of latitude that is a great arc.

Mixing these up still leads to some unfortunate descriptions (particularly when AI is asked) about the arc being one of latitude to this day.

The point is that they did not know the flattening of the Earth distorts these great arcs of longitude, especially near the poles.

This was sufficient for the 'metric' to vary from place to place, a most undesirable requirement.

Subsequently an average was agreed internationally, but it still means that the sea mile still does not conform to its original definition anywhere.

Thanks Studiot.

I gotta start posting this stuff in the 'Today I Learned' thread.

First let me apologize for my last post- it came out way more symbolic than I had intended. More like a student tutorial as MigL said. Not really what you want in a discussion forum. See if I can do better......

We have our metric tensor field (that is, if you think I was making any sort of sense).

Let's assume, as Einstein did, that gravitation and acceleration are in some sense equivalent. Let's further assume, again as Einstein did, that within a small enough spacetime interval, an accelerating body is in inertial (unaccelerated) motion at each point in its trajectory. So that at each point the Special Theory applies, and the different coordinates there are related by a local orthogonal tranformation (a Lorentz transformatin).

Now as you know, orthogonal transformations are characterized be the fact that they preserve the metric, which is very nice locally, but doesn't apply for the different inertial systems at different points. So if we wish to do relativity, we may not assume that transformations between them are orthogonal (metric conserving).

The solution is to allow the line element that I pedanticaly showed you (or do I mean didacticaly) to vary from point to point. So we will now have a non-constant metric field, which by the assumed equivalence of acceleration and gravitation, applies in the presence of gravitating bodies.

Which is basically the General Theory

So let us look at some examples using my definition of a distance function.

[math]d:X \times X \to {\Re ^ + } \cup \{ 0\} [/math]

So my first example is as follows

1

Let X be the set of all binary words of length 3 characters.

Thus there are 8 members of this finite set { 000, 001, 010, 100, 110, 101, 011, 111 }

Then if x, y are elements of this set define d(x,y) as the number of digits in which x and y differ.

It can easily be seen that for each x there are 3 ys that differ by one character Thus each x has 2 ys at a distance of 1 from it . d = 1

similarly there are 3 words that are at a distance of 2 characters from each.

And there is 1 word that is at a distance of 3 characters

2

Let X be the set of all real numbers and d(x,y) be be the modulus or abs function.

[math]d(x,y) = \left| {x - y} \right|[/math]

3

[math]d(x,y) = \sqrt {{{(x - y)}^2}} [/math]

Again it is easy to see that all three examples satisfy the requirements

[math]d(x,y) \ge 0[/math]
[math]d(x,y) = 0if\;and\;onlyif\;x = y[/math]
[math]d(x,y) = d(y,x)[/math]
[math]d(x,z) \le d(x,y) + d(y,z)[/math]

The distance function in Example 1 is discrete

The distance function in Example 2 is is continuous but it is not differentiable over the entire domain.

The distance function in Example 3 is both continuous and differentiable, which is why it is often used to replace the absolute value function.

So in order to do calculus with our metric we must further limit allowable distance functions, which has clear implications for relativity.

Very nice studiot. An ineresting thread you started

What's even nicer is that every metric space is automaticaly a topological space with the Hausdorff property. This means, as studiot showed, that if there are no open sets \(a\in A\) and \(b\in B\) such that \(A \cap B =\emptyset\), then we must have that \(a = b \)

Furthermore, if the metric on every subset \(S\) is the Euclidean metric, then we may have a homeomorphism (i.e. a sort of isomorphism) \(h: S \to R^n\), which is essentially the definition of a manifold.

Edited September 19Sep 19 by Xerxes

Well you guys are no fun. But as this is a subject that interests me, I shall have fun on my own.

Recall that for every vector space \(V\), there exists a companion space \(V^*\), this being the space of all linear maps \(V^* \to \mathbb{R}\), so that for any \(\varphi \in V^*\) and all \(v \in V\) we will have that \(\varphi(v) = \alpha \in \mathbb{R}\).

Now it is a fact that if and only if there exists a metric in \(V\), with an inner product \((v,w) \in \mathbb{R}\) then there will be some particular \(\phi \in V^*\) such that \(\phi(w) = (v,w) \in \mathbb{R}\). I will temporarily write this as \(\phi_v\) to emphasize this fact.

This cooincides very nicely with a useful and interseting property of the metric tensor, which some people (who really are old enough to know better) call "index raising" and "index lowering".

So recall that the metric tensor is a bilinear gadget, that is it takes 2 vectors as argument and returns a scalar, say \(g(v,w) \in \mathbb{R}\). So if we feed this beast just a single vector as argument it's not going to work unless it has had already another vector to feed on. Like \(g(v, -) (w) \in \mathbb{R}\). \(g(v, -)\) is therefore a dual vector

So with an elegant sleight of hand, I make the "equivalence" \(\phi_v \equiv g(v, -)\). And for those that insist on using index notation for tensors (they really shouldn't, since in so doing they are actually referring to tensor components, which are scalars of sorts, not tensors at all), this becomes \(g_{j,k}A^k= B_j\), which is index lowering.

So there.

Edited October 7Oct 7 by Xerxes