- #1

friend

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Can the topology of a manifold determine the metric use on it? Or is the metric completely independent of topology and coordinates systems?

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- Thread starter friend
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- #1

friend

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Can the topology of a manifold determine the metric use on it? Or is the metric completely independent of topology and coordinates systems?

- #2

WannabeNewton

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- #3

pasmith

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Every metric space is a topological space; the metric naturally determines a topology on the space by generating one through the base of open balls of the metric.

It is also possible for different metrics to generate the same topology.

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friend

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- #5

pasmith

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Metrics and topologies are both structures which can be put on arbitrary non-empty sets in order to define a notion of "closeness". Having done that one can then define notions of convergence of sequences and continuity of functions.

There is a sense in which topological spaces are more general than metric spaces, in as much as every metric space is a topological space but not every topological space can be a metric space.

- #6

friend

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And more generally, can you have a function on points without a metric between them. I mean, if there is a continuous function, then df/dx can be defined, and dx is a distance, right?

- #7

economicsnerd

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Do the separation axioms on a topology imply a metric? If there is a continuous function from one point or set to another, doesn't that continuous function imply a measure between points, namely the continuous function has a different value for one point than for another, perhaps only for two point very close to each other?

Urysohn's metrization theorem does exactly this. It says that if a topology on [itex]X[/itex] is Hausdorff (a separation condition), regular (a separation condition), and second-countable (a condition stating that the topology isn't too rich), then there exists a metric on on [itex]X[/itex] giving rise to said topology.

- #8

friend

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Urysohn's metrization theorem does exactly this. It says that if a topology on [itex]X[/itex] is Hausdorff (a separation condition), regular (a separation condition), and second-countable (a condition stating that the topology isn't too rich), then there exists a metric on on [itex]X[/itex] giving rise to said topology.

When you say, "giving rise to said topology", that seems confusing to me. Your first statements make it sound as though the topology already exists, with certain properties such as Hausdorff, etc. So would it be sufficient to say, ",then there exists a metric on X", without the rest of the sentence?

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- #9

pwsnafu

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When you say, "giving rise to said topology", that seems confusing to me. Your first statements make it sound as though the topology already exists, with certain properties such as Hausdorff, etc. So would it be sufficient to say, ",then there exists a metric on X", without the rest of the sentence?

The phrase "giving rise to a topology" means there exists a metric, such that the topology induced by said metric, is equivalent to the original topology you started with.

So to be precise: you have a set ##T## with topology ##\mathcal{T}##. Now ##\mathcal{T}## is metrisable if and only if there exists metric ##d## (which induces a metric ##\mathcal{D}## via open balls) satisfying ##\mathcal{T}=\mathcal{D}##.

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WannabeNewton

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And more generally, can you have a function on points without a metric between them. I mean, if there is a continuous function, then df/dx can be defined, and dx is a distance, right?

A function is a purely set theoretic concept so no there is no need for a notion of distance whatsoever. Also, you need the notion of differentiability to define derivatives, continuity isn't enough. For continuity you just need a topology but for differentiability you need a ##C^n## differentiable structure (##n \geq 1##). Finally, ##dx## has absolutely no relation to distance.

- #11

friend

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The phrase "giving rise to a topology" means there exists a metric, such that the topology induced by said metric, is equivalent to the original topology you started with.

So to be precise: you have a set ##T## with topology ##\mathcal{T}##. Now ##\mathcal{T}## is metrisable if and only if there exists metric ##d## (which induces a metric ##\mathcal{D}## via open balls) satisfying ##\mathcal{T}=\mathcal{D}##.

"a metric", you say. Is there a way to discern what kind of metric it is for a given topology, or what signature the metric might have? Or does this theorem only state that a metric can be constructed. Or does this mean that you can construct a metric of your chosing?

- #12

gufiguer

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So to be precise: you have a set ##T## with topology ##\mathcal{T}##. Now ##\mathcal{T}## is metrisable if and only if there exists metric ##d## (which induces atopology##\mathcal{D}## via open balls) satisfying ##\mathcal{T}=\mathcal{D}##.

For no more cofusing arise :)

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