I About the definition of a Manifold

cianfa72

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Locally euclidean request in manifold definition
Hi,

I'm a bit confused about the locally euclidean request involved in the definition of manifold (e.g. manifold ): every point in $X$ has an open neighbourhood homeomorphic to the Euclidean space $E^n$.

As far as I know the definition of homeomorphism requires to specify a topology for both spaces (here the open neighbourhood as a space itself and the Euclidean space $E^n$). Are the two spaces "silently" supposed to be endowed each with the subspace topology ?

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Orodruin

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Euclidean space includes the definition of a metric and therefore a corresponding metric topology. It is unclear what you mean by subspace in this regard. Neither the manifold nor Euclidean space need to be thought of as subspaces of a larger space. Although you can consider the case of submanifolds, there is nothing in the definition of a manifold that requires you to do so.

It should also be mentioned that, while Euclidean space is equipped with a metric, there is no requirement that a general manifold is or, in the case when it is, that such a metric needs to be mapped to the Euclidean metric by the homeomorphism.

• Klystron

cianfa72

Euclidean space includes the definition of a metric and therefore a corresponding metric topology. It is unclear what you mean by subspace in this regard. Neither the manifold nor Euclidean space need to be thought of as subspaces of a larger space. Although you can consider the case of submanifolds, there is nothing in the definition of a manifold that requires you to do so.
Sure that's clear but is not my point.

My doubt is actually more basic (I'm not a mathematician): manifold definition requires homeomorphisms between manifold open sets and euclidean open sets. Each homeomorphism definition requires a domain with its own topology and the same for the codomain.
My question is: which are the underlying topologies involved in each homeomorphism definition ?

Orodruin

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For the coordinate function (i.e., the map from the manifold to $U \subset \mathbb R^n$), the topology in codomain $U$ is given by the standard topology on $\mathbb R^n$. This directly tells you what the (local) topology on the manifold needs to be.

cianfa72

For the coordinate function (i.e., the map from the manifold to $U \subset \mathbb R^n$), the topology in codomain $U$ is given by the standard topology on $\mathbb R^n$. This directly tells you what the (local) topology on the manifold needs to be.
quoting the definition: "A topological space X is called locally Euclidean if there is a non-negative integer n such that every point in X has a neighbourhood which is homeomorphic to the Euclidean space $E^n$"

Thus starting from the very fact topological space $X$ has its own topology take an open neighbourhood $O$ of a manifold point and consider the homeomorphism to the Euclidean space. The open neighbourhood $O$ as a set is the domain of the homeomorphism, but which is the topology to "attach" to it ? I believe it is the subset (or subspace) topology when the open neighbourhood $O$ is viewed as subset of the manifold $X$

Orodruin

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The manifold can either come with a topology or have a topology induced by the mapping from $U$. In the latter case it is quite clear that the resulting mapping is a homeomorphism, otherwise you have to check that it is. It is not sufficient that the set $O$ itself is open in the manifold topology. There is a topology on $O$ induced by the topology on the manifold. So
I believe it is the subset (or subspace) topology when the open neighbourhood O is viewed as subset of the manifold X
is correct in that case.

fresh_42

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I believe it is the subset (or subspace) topology when the open neighbourhood is viewed as subset of the manifold X
Correct. The essential point, however, is that the topology on $X$ cannot be arbitrary, since it is required to look like a Euclidean topology. This is a strong restriction.

The basic idea of manifolds is not so much the topologies on them, but rather to get rid of the view that a manifold has to be a surface in a surrounding space. It is as if there is only the manifold and nothing else. Now to do analysis on it, we need the charts, the local homeomorphisms. That is: we go down to the chart, perform our calculations, and return up to the manifold.

• Orodruin

cianfa72

Correct. The essential point, however, is that the topology on $X$ cannot be arbitrary, since it is required to look like a Euclidean topology. This is a strong restriction.
I do not understand why this point about the subspace topology is not highlighted in the definitions I read...

The basic idea of manifolds is not so much the topologies on them, but rather to get rid of the view that a manifold has to be a surface in a surrounding space. It is as if there is only the manifold and nothing else. Now to do analysis on it, we need the charts, the local homeomorphisms. That is: we go down to the chart, perform our calculations, and return up to the manifold.
Sure I got it !

fresh_42

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I do not understand why this point about the subspace topology is not highlighted in the definitions I read...
If we have a topological space and refer to subsets, we don't change the topology. The aspect to consider a subset as the entire space for your now no more local homeomorphism is artificial. The natural view is to consider the homeomorphism as a function which is only locally defined, will say on an open set $\mathcal{O}\subseteq X$. There is simply no need to transfer from $X$ to $\mathcal{O}$. That's why.

Your consideration comes into play if we talk about manifolds with boundaries. For a boundary we need something outside. Now a manifold has no outside, just as the universe doesn't expand into something. There is no something. To solve the issue, we lend the concept from the Euclidean space: if the charts have a boundary, then the upload of it to the manifold is called boundary there. In this case we have to take care about subset topologies. In the general case we do not.

• Klystron

cianfa72

If we have a topological space and refer to subsets, we don't change the topology. The aspect to consider a subset as the entire space for your now no more local homeomorphism is artificial. The natural view is to consider the homeomorphism as a function which is only locally defined, will say on an open set $\mathcal{O}\subseteq X$. There is simply no need to transfer from $X$ to $\mathcal{O}$. That's why.
Not sure to fully get your point. Could you elaborate a bit please ?
Thanks.

fresh_42

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If we have a chart $\xi\, : \, X \longrightarrow \mathbb{R}^n$ then it is locally defined, that is its domain is an open set $\mathcal{O}\subseteq X$. This is a local homeomorphism. We do not need to switch to a view $\xi\, : \,\mathcal{O} \longrightarrow \mathbb{R}^n$ where the locally defined map $\xi$ becomes a globally defined map on a topological space $\mathcal{O}$, what the transfer to the subset topology would be. Quite the opposite is true! What we really consider is a family of such charts, an atlas. The main requirement for an atlas is, that overlapping charts yield the same results for our calculations, i.e. it doesn't matter if we perform a calculation in one chart or the other. Therefore we need to consider the entire family of charts. If we had transferred to the subspace topology for each, we would have trouble to handle the overlappings as they were now in two different spaces! But as open sets in still the same topological space $X$, we don't have to bother any subspace topologies at all.

Or for short: We only need open neighborhoods in $\mathcal{O}$ and they are open in the subspace topology as well as in the original topology. So why bother a change?

cianfa72

If we have a chart $\xi\, : \, X \longrightarrow \mathbb{R}^n$ then it is locally defined, that is its domain is an open set $\mathcal{O}\subseteq X$. This is a local homeomorphism. We do not need to switch to a view $\xi\, : \,\mathcal{O} \longrightarrow \mathbb{R}^n$ where the locally defined map $\xi$ becomes a globally defined map on a topological space $\mathcal{O}$, what the transfer to the subset topology would be.
Thus we have a chart $\xi$ from $X$ to $\mathbb{R}^n$ defined just only on the set $\mathcal{O}$ open in $X$. In that case on which $\mathbb{R}^n$ values are mapped by $\xi$ points not in $\mathcal{O}$ ?

fresh_42

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In that case on which $\mathbb{R}^n$ values are mapped by $\xi$ points not in $\mathcal{O}$?
What do you mean? $\xi(X-\mathcal{O})$ is not defined, so we do not care. Same as $x\longmapsto \frac{1}{x}$ isn't defined for $x=0$. The image of $\xi$ is an open set in $\mathbb{R}^n$.

Orodruin

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Thus we have a chart $\xi$ from $X$ to $\mathbb{R}^n$ defined just only on the set $\mathcal{O}$ open in $X$. In that case on which $\mathbb{R}^n$ values are mapped by $\xi$ points not in $\mathcal{O}$ ?
On no values. A chart is only required to be defined on a subset of the manifold. The chart is not from the manifold to $\mathbb R^n$, it is from a subset of the manifold to $\mathbb R^n$.

cianfa72

What do you mean? $\xi(X-\mathcal{O})$ is not defined, so we do not care. Same as $x\longmapsto \frac{1}{x}$ isn't defined for $x=0$. The image of $\xi$ is an open set in $\mathbb{R}^n$.
but in that case I believe $x\longmapsto \frac{1}{x}$ is a map/application/function from $\mathbb{R}-\{0\}$ to $\mathbb{R}$ and not from $\mathbb{R}$ to $\mathbb{R}$. From my understanding a map/function/application has to be defined on each point of its domain

On no values. A chart is only required to be defined on a subset of the manifold. The chart is not from the manifold to $\mathbb R^n$, it is from a subset of the manifold to $\mathbb R^n$.
that's the reason why I was wondering about the topology to attach to it in defining an homeomorphism from a set open in the manifold topology to $\mathbb{R}^n$

Orodruin

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but in that case I believe $x\longmapsto \frac{1}{x}$ is a map/application/function from $\mathbb{R}-\{0\}$ to $\mathbb{R}$ and not from $\mathbb{R}$ to $\mathbb{R}$. From my understanding a map/function/application has to be defined on each point of its domain
Yes, and the domain of the coordinate chart is a subset of the manifold, not the manifold itself in exactly the same way that $1/x$ has $\mathbb R \setminus \{0\}$ as its domain. In fact, $1/x$ is a chart of the manifold $\mathbb R$ with the covered part of the manifold being everything but $x = 0$.

fresh_42

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but in that case I believe $x\longmapsto \frac{1}{x}$ is a map/application/function from $\mathbb{R}-\{0\}$ to $\mathbb{R}$ and not from $\mathbb{R}$ to $\mathbb{R}$. From my understanding a map/function/application has to be defined on each point of its domain
That's nitpicking and a notational question how you write a function. My notation doesn't require to name the domain, only the space the domain is part of. We could debate for hours whether this is right or wrong. I don't care. I will continue to distinguish the domain from the space the function originates from. If you like, you can write $\xi\, : \,\mathcal{O} \longrightarrow \mathbb{R}^n$ and give $\mathcal{O}$ the subset topology. It is just not necessary and complicates the case of overlapping charts, as you now have three potentially incompatible topologies: on either of the two open sets, and on the intersection of both. Of course they are not incompatible, but you will have to prove it. Then for every new patch you need to consider a new topology. What for? It is good for nothing, since we always deal with the topology of $X$ and nothing else. So it is far easier to say that charts are only defined on open sets of $X$ than to introduce new topologies. It is as if we argue whether the book is an atlas of a country or a collection of individual maps. And in our case, we do not allow maps to have different scales, because their intersections are required to be compatible.

• Klystron and Orodruin

cianfa72

...otherwise you have to check that it is. It is not sufficient that the set $O$ itself is open in the manifold topology. There is a topology on $O$ induced by the topology on the manifold.
Basically what you're saying is that in case the manifold is endowed with a given topology, then we have to explicitly check if the topology induced to the set $O$ by the chart (a bijection from $\mathbb R^n$ transferring to it the Euclidean topology) is the same as the topology induced by the manifold topology when $O$ is viewed as a subset of it (actually thinking of it as endowed with the subset/subspace topology)
Does it make sense ?

Orodruin

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To be honest, I do not see why you are hung up on this technical detail. When you deal with an actual manifold most of these issues are going to be essentially self evident.

• fresh_42

mathwonk

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you are right of course that the homeomorphism from an open set O in a manifold X, to E^n, should be with respect to the subspace topology on O inherited from X. We who know this are notoriously lazy about making it clear when we write. In this case of course, the subspace topology on an open set O is pretty easy, namely the open sets in O, for the subspace topology, are exactly those open sets of X which happen to be contained in O. This is because the intersection of two open sets is also open. So maybe in this case the author can be forgiven for leaving it to the reader to surmise the correct definition, as you have in fact done. So now you have learned a valuable lesson for reading math, namely when something is left out, use the most natural fill-in that you can think of, and see if it works out. To test yourself, try reading a proof that uses the topology and see whether your candidate gives the right result.

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• dextercioby

cianfa72

Thank you all for the clarification

"About the definition of a Manifold"

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