I Why is the subgroup H not a Lie Group under the Subspace Topology?

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I was reading a Wikipedia page where it's given an example of a group that's not a Lie Group. Here's the page https://en.wikipedia.org/wiki/Lie_group ; refer to "Counterexample".

If we work with the topology of ##\mathbb{T}^2## it seems obvious that a map from some ##\mathbb{R}^m## would not be continuous, as a point of ##H## would have as its neighbour a point of ##\mathbb{T}^2## which is not a point of ##H##. Ok... But Wikipedia says that ##H## is not a Lie group (though it's a group) given the Subspace Topology.

Now imagine that ##\mathbb{T}^2## is given the trivial topology. Then ##H## would have the Subspace Topology ##\{H, \emptyset \}##. It seems obvious that a homeomorphism from ##\mathbb{R}## to ##H## can be carried out. So why ##H## is not a Lie Group given the Subspace Topology?
 
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The example assumes the natural from the embedding in ##\mathbb{R}^3## induced subspace topology, not any subspace topology. Also the differentiable structure on either ##\mathbb{T}^2## or ##H## might become a bit of a problem with the trivial topology. I haven't checked whether there is a trivial solution to the problem, but anyway, it's not part of the example meant.
 
fresh_42 said:
The example assumes the natural from the embedding in ##\mathbb{R}^3## induced subspace topology
Did you mean ##\mathbb{R}^2##? And how do you notice that from the text?
fresh_42 said:
might become a bit of a problem with the trivial topology
Could you provide me with a link or give me an example of how a choice of topology can dramatically affect the differentiable structure?
 
kent davidge said:
Did you mean ##\mathbb{R}^2##?
No, I meant ##\mathbb{R}^3## but the plane should be fine, too.
And how do you notice that from the text?
  • It is the natural choice, and all others would have been mentioned.
  • From the attached image.
  • From the description in the text, as it works with it.
  • From the condition that ##a## is irrational, which guarantees distances.
Could you provide me with a link or give me an example of how a choice of topology can dramatically affect the differentiable structure?
I only know of definitions which require Hausdorff, and in one text even a countable basis, but this is for convenience. So just choose a non Hausdorff space. Or more extremely: a differentiable (analytic) structure to be a differentiable (analytic) manifold requires local homeomorphisms to Euclidean spaces. So with a discrete topology I can't imagine how this will get something differentiable. And without being a manifold, how should differentiability even be defined?
 
Got it. Thanks.
 
A sphere as topological manifold can be defined by gluing together the boundary of two disk. Basically one starts assigning each disk the subspace topology from ##\mathbb R^2## and then taking the quotient topology obtained by gluing their boundaries. Starting from the above definition of 2-sphere as topological manifold, shows that it is homeomorphic to the "embedded" sphere understood as subset of ##\mathbb R^3## in the subspace topology.

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