# Do Manifolds have distance relations between points?

1. Mar 31, 2013

### me1pg

Hello everyone,

I am currently reading 'Geometrical Methods of Mathematical Physics' by Bernard Schutz and I have some questions about manifolds. I'm fairly new to Differential Geometry so bear with me!

On P33 he states that 'manifolds need have no distance relation between points, we shall need a definition of a vector which relies only on infinitesimal neighborhoods of points of M'.

My question is: how can you define the neighborhood around a point if you haven't already defined a distance relation between points? When you define a manifold are you simply defining a set of points which have a 1-1 mapping to Euclidean space or are you also defining a distance relation between the points?

Pete

2. Mar 31, 2013

### rustynail

Look up for ''Metric'' on wikipedia. Also, note that while a manifold does have a local metric, it does not necessarily have a ''global'' metric defining distances between any two points.

3. Mar 31, 2013

### micromass

A manifold does have a global metric. The problem is that it doesn't have a canonical global metric. So in general, there are many metrics that you can put on your manifold and which can give very different answers.

For example, you can look at $\mathbb{R}$ with the usual manifold structure. A possible metric is $d(x,y) = |x-y|$ and this makes $\mathbb{R}$ in a complete, unbounded metric space. But another possible metric is $e(x,y) = |atan(x) - atan(y)|$ and this makes $\mathbb{R}$ into an incomplete, bounded metric space. There is no reason why we should prefer $d$ over $e$ as both are compatible with the metric structure.

Furthermore, a metric might not be very relevant to the problem you are wanting to solve. A topology is far more flexible and far more useful in the theory of manifolds. A topology can be seen as a generalization of a metric space. A topology can be used to make sense of neighborhoods, continuity, compactness, connectedness, etc. But a topology doesn't come from a metric in general, and if it comes from a metric then the metric is usually not unique.

A far more useful tool in manifold theory is that of a "metric tensor". This yields a distance function that is actually physically relevant to the manifold and it gives much more than that, such as angles. The idea of the metric tensor is that we are able to calculate angles and distances on an infinitesimal neighborhood of a point.

4. Mar 31, 2013

### WannabeNewton

You are missing the concept of a topology. A topology is exactly what allows you to define neighborhoods of points with no need for a metric; all topological manifolds are topological spaces with the extra conditions of locally euclidean, usually Hausdorff, and second countable or separability; smooth manifolds have smooth atlases on top of this (although u can define them without using a topological manifold per say but there is a natural topology induced on them regardless that satisfies the above). There is no need for a metric here, although you can always endow one if you want.

As a side note, I would recommend not using that book. It is quite horrible for learning differential geometry.

5. Apr 1, 2013

### me1pg

OK - so we need a way of defining neighborhoods and therefore open sets without having to define a distance function first. So we can define open sets as a collection of subsets T where

1. T contains the set itself and the empty set
2. The intersection between any 2 sets in T is also in T
3. The union of any collection of subsets in T is also in T

- so T contains all possible subsets of the points on our manifold right?

We can then define the neighborhood of a point x as being any set containing an open set containing x. Am I on the right track here?

6. Apr 1, 2013

### micromass

You probably mean well, but to make sure I'll correct some statements in your post:

The elements of T are called the open subsets.

T doesn't contain all possible sets. T contains some subsets of the entire space. The elements of T are called open sets.

right.

7. Apr 1, 2013

### me1pg

Yea fair enough - T doesn't contain all possible subsets it simply has to satisfy the rules which define open sets. I suppose there may be many (possibly infinite) different ways to construct T.

By the way - thanks for your help everyone

8. Apr 1, 2013

### WannabeNewton

Not necessarily. Given the set $X$, IF your topology happened to be $\wp (X)$ then yes; this is called the discrete topology. The other extreme is the trivial topology which is just $\left \{ X,\varnothing \right \}$. However there are a plethora of topologies you can endow on a given set. The most familiar to you will be the topology generated by the basis of open balls in $\mathbb{R}^{n}$ i.e. the euclidean topology.

EDIT: I totally missed your last post above when writing this...blame micromass =D