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

[kent davidge]

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?

 

[fresh_42]

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.

 

[kent davidge]

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?

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?

 

[fresh_42]

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?

 

[kent davidge]

Got it. Thanks.