Why isn't this a Lie group?

[yamata1]

TL;DR Summary

I'm trying to grasp Lie algebra with a non-example. Why isn't it a Lie group? The matrices seem invertible but is it a smooth manifold ?

"The group given by $H = \left\{ \left( \begin{array} { c c } { e ^ { 2 \pi i \theta } } & { 0 } \\ { 0 } & { e ^ { 2 \pi i a \theta } } \end{array} \right) | \theta \in \mathbb { R } \right\} \subset \mathbb { T } ^ { 2 } = \left\{ \left( \begin{array} { c c } { e ^ { 2 \pi i \theta } } & { 0 } \\ { 0 } & { e ^ { 2 \pi i \phi } } \end{array} \right) | \theta , \phi \in \mathbb { R } \right\}$ with $a \in \mathbb { P } = \mathbb { R } \backslash \mathbb { Q }$ a fixed irrational number,is a subgroup of the torus $\mathbb { T}^2$ that is not a Lie group when given the subspace topology.

The group

can, however, be given a different topology, in which the distance between two points

is defined as the length of the shortest path in the group

joining

to

. In this topology,

is identified homeomorphically with the real line by identifying each element with the number

in the definition of

. With this topology,

is just the group of real numbers under addition and is therefore a Lie group."

Why isn't it " a Lie group when given the subspace topology" ?

 

[fresh_42]

Consider an open set in either topology. One is connected, the other one is not.

 

[yamata1]

Consider an open set in either topology. One is connected, the other one is not.

How do we change the topology of the subgroup from the original group? Could you show a simple proof of why it becomes connected ?

 

[fresh_42]

Look at the second component. This is a discrete set, namely $[0,2\pi]-(\mathbb{Q}\cap [0,2\pi])$. All these points are open sets in the subspace topology, or combined with the first component, the union of open circular stripes. However, there is no way to guarantee that functions on two overlapping charts behave equally, Between two charts is always another chart such that the chart maps do not need to be continuous on their intersection.

 

[yamata1]

Look at the second component. This is a discrete set, namely [0,2π]−(Q∩[0,2π]). All these points are open sets in the subspace topology, or combined with the first component, the union of open circular stripes. However, there is no way to guarantee that functions on two overlapping charts behave equally, Between two charts is always another chart such that the chart maps do not need to be continuous on their intersection.

I'm a bit lost, is $[0,2\pi]-(\mathbb{Q}\cap [0,2\pi])$ equal to H or $\mathbb{P}$ ?I've never worked with topological groups,how does a disconnected topological group or set become connected?

 

[fresh_42]

A basis of the subspace topology on $H$ is given by
$$
\mathcal{T}_H=\left\{\left.\begin{pmatrix}e^{2\pi i \theta}&0\\0&e^{2\pi i a }\end{pmatrix}\, \right| \, \theta \in (\alpha,\beta) \subseteq \mathbb{R}\, , \,a\in \mathbb{P}\right\}
$$
Unions of these sets are the subspace topology on $H$. Let us use two overlapping such sets
\begin{align*}
U_1&:= \left\{\left.\begin{pmatrix}e^{2\pi i \theta}&0\\0&e^{2 \pi i a}\end{pmatrix}\, \right| \, 0< \theta < 1\, , \,0<a<2/3\, , \, a\notin \mathbb{Q}\right\}\\
U_2&:= \left\{\left.\begin{pmatrix}e^{2\pi i \theta}&0\\0&e^{2\pi i a}\end{pmatrix}\, \right| \, 0< \theta < 1\, , \,1/3<a<1\, , \,a\notin \mathbb{Q} \right\}
\end{align*}
The second coordinate is expected to make problems. We now need the definition of a Lie group to see why this isn't one.

Which properties do you have that define a Lie group?

 

[yamata1]

Which properties do you have that define a Lie group

1-a finite-dimensional real smooth manifold
2- the group operations of multiplication and inversion are smooth maps.

 

[fresh_42]

Let's consider the manifold property. Can you define a chart at $(\theta , a )=\left(\dfrac{1}{2},\dfrac{1}{\sqrt{2}}\right)\in U_2\,$ such that multiplication with $\begin{bmatrix}e^{2\pi i / n}&0\\0&e^{\sqrt{2}\pi i /n}\end{bmatrix}$ becomes smooth?

 

[yamata1]

I can't find a proper homeomorphism in $U_2$.

 

[fresh_42]

Yes, it is not locally Euclidean because of the gaps at the rationals. No matter how tiny the neighborhood is chosen, there are still gaps.

 

[yamata1]

How do we change the topology to make H a Lie group ?

 

[fresh_42]

I don't think this is possible, except dropping the constraint $a\notin \mathbb{Q}$. With $a=\theta \in \mathbb{R}$ we get a one parameter subgroup.

 

[yamata1]

"The group

can, however, be given a different topology, in which the distance between two points

is defined as the length of the shortest path in the group

joining

to

"
What does this give us in formal math ?

 

[Infrared]

@fresh_42 I think you're misreading the definition. The constant $a$ is fixed. The space $H$ "should be" one-dimensional. You wouldn't vary both $\theta$ and $a$ in a coordinate chart.

The subset $H\subset T^2$ is not a manifold because small open neighborhoods of every point are disconnected. I think you should try to draw a picture of this. Geometrically, $H$ is a continuous path in $T^2$ that gets arbitrarily close to every point in $T^2$, infinitely often.

The map $\theta\mapsto\begin{pmatrix}e^{2\pi i \theta} & 0\\0 & e^{2\pi i a\theta}\end{pmatrix}$ gives a bijection from $\mathbb{R}$ to $H.$ You can give $H$ the topology that makes this map a homeomorphism- this is the metric topology you describe in your last post.

 

[yamata1]

The map $\theta\mapsto\begin{pmatrix}e^{2\pi i \theta} & 0\\0 & e^{2\pi i a\theta}\end{pmatrix}$ gives a bijection from $\mathbb{R}$ to $H.$ You can give $H$ the topology that makes this map a homeomorphism- this is the metric topology you describe in your last post.

How can we write the topology change for H as a topological group ?

 

[Infrared]

I'm not sure what you mean by "topology change".

Call the elements of $H$ by $A_\theta=\begin{pmatrix}e^{2\pi i \theta} & 0\\0 & e^{2\pi i a\theta}\end{pmatrix}.$ The metric you describe in your post 13 is $d(A_\theta,A_\phi)=|\theta-\phi|.$ This is a valid metric on $H$ because the map $\theta\mapsto A_\theta$ is a bijection (from $\mathbb{R}$ to $H$), so it defines a topology on $H.$ It is different from the subspace topology from the torus.

 

[yamata1]

I'm not sure what you mean by "topology change".

Call the elements of $H$ by $A_\theta=\begin{pmatrix}e^{2\pi i \theta} & 0\\0 & e^{2\pi i a\theta}\end{pmatrix}.$ The metric you describe in your post 13 is $d(A_\theta,A_\phi)=|\theta-\phi|.$ This is a valid metric on $H$ because the map $\theta\mapsto A_\theta$ is a bijection (from $\mathbb{R}$ to $H$), so it defines a topology on $H.$ It is different from the subspace topology from the torus.

So giving a different metric gives us the right topology to make H a Lie group.
Thank you.

 

[WWGD]

One result I remember is that given $G$ a Lie Group; here $\mathbb T^2$, a subset is a Lie Group if it's both a subgroup and topologically closed in $G$.