I ##SU(2)## homeomorphic with ##\mathbb S^3##

[cianfa72]

It is always necessary to mention the scalar field...

You mean the scalar field that enters in the definition of vector space, I believe.

 

[fresh_42]

You mean the scalar field that enters in the definition of vector space, I believe.

I mean the one that is associated with the terms I listed. $e$ and $\pi$ are $\mathbb{Q}$-linearly independent, but $\mathbb{R}$-linearly dependent. That makes a big difference. And $GL(2,\mathbb{C})$ is four-dimensional (over $\mathbb{C}$) in the first place. Claiming it to be eight-dimensional without mentioning that we changed the perspective to a real vector space manifold is sloppy.

Sorry, I was sloppy, too.

 

[cianfa72]

Claiming it to be eight-dimensional without mentioning that we changed the perspective to a real vector space is sloppy.

Ok, got it

 

[WWGD]

I mean the one that is associated with the terms I listed. $e$ and $\pi$ are $\mathbb{Q}$-linearly independent, but $\mathbb{R}$-linearly dependent. That makes a big difference. And $GL(2,\mathbb{C})$ is four-dimensional (over $\mathbb{C}$) in the first place. Claiming it to be eight-dimensional without mentioning that we changed the perspective to a real vector space manifold is sloppy.

Sorry, I was sloppy, too.

But wait, there's more. Vector spaces are often referred to as linear manifolds.

 

[fresh_42]

But wait, there's more. Vector spaces are often referred to as linear manifolds.

Only if they carry a topological, preferably differentiable structure. This would be a tough requirement over finite fields. ;-)

 

[jbergman]

$GL(\mathbb{C},2)$ has real dimension 8 ...

It is always necessary to mention the scalar field as soon as we say "linearly (in)dependent", "basis", or "dimension" since their meaning depends on it. It is even more necessary in cases where the matrix entries are complex but we consider them as real vector spaces!

As a real smooth or topological manifold it has dimension 8. I thought that was clearly implied context by mentioning the subspace topology and previous questions about that.

 

[jbergman]

TL;DR Summary: How to define the homeomorphism between $SU(2)$ and $\mathbb S^3$

Hi,

$SU(2)$ group as topological space is homeomorphic to the 3-sphere $\mathbb S^3$.

Since $SU(2)$ matrices are unitary there is a natural bijection between them and points on $\mathbb S^3$. In order to define an homeomorphism a topology is needed on both spaces involved. $\mathbb S^3$ has the subspace topology from $\mathbb R^4$, which is the topology on $SU(2)$ ?

We can endow $SU(2)$ with a topology "induced" by the bijection (i.e. declaring open those sets having the preimage open). With this induced topology the bijection is homeomorphism by definition.

Is actually this the case? Thanks.

As I mentioned before SU(2) has the subspace topology of GL(C,2). However, the easiest way to see that is homeomorphic to the unit 3-sphere is to exhibita Lie Group isomorphism between it and $S^3$. This can easily be done by exhibiting a Lie Group homomorphism between SU(2) and the unit quaternions. A Lie Group isomorphism implies a diffeomorphism and a homeomorphism of the manifolds in question.

 

[jbergman]

As I mentioned before SU(2) has the subspace topology of GL(C,2). However, the easiest way to see that is homeomorphic to the unit 3-sphere is to exhibita Lie Group isomorphism between it and $S^3$. This can easily be done by exhibiting a Lie Group homomorphism between SU(2) and the unit quaternions. A Lie Group isomorphism implies a diffeomorphism and a homeomorphism of the manifolds in question.

One other thing to say is that a Lie Group structure is very strong. It implies in some sense that the manifold is globally self-similar and I believe homogeneous. The only distinguishing features, I believe are the number of connected components and the local structure.

 

[fresh_42]

As a real smooth or topological manifold it has dimension 8. I thought that was clearly implied context by mentioning the subspace topology and previous questions about that.

Yes. But as this is a public thread with many readers who do not actively participate, I thought I should clarify "implied by context" to avoid potentially false claims like "I have read on PF that $GL(2,\mathbb{C})$ is eight-dimensional".

 

[cianfa72]

Claiming it to be eight-dimensional without mentioning that we changed the perspective to a real vector space manifold is sloppy.

Ok, since $GL(2,\mathbb{C})$ is not a vector space on field $\mathbb{C}$ itself.

 

[fresh_42]

Ok, since $GL(2,\mathbb{C})$ is not a vector space on field $\mathbb{C}$ itself.

Yes. But its dimension is defined by the dimension of its tangent space, so it wasn't completely wrong, just sloppy. Since I have started nitpicking I thought I should be more accurate.

 

[cianfa72]

Sorry to be boring, I would try to recap my understanding.

There is a bijection between the set of 2x2 complex matrices and the vector space $\mathbb{C}^4$. The determinant of the 2x2 complex matrices is a continuous map from $\mathbb{C}^4$ (with its standard topology) to $\mathbb C$. Now the set $\{\mathbb{C} - 0\}$ is open in $\mathbb C$, hence its preimage under the determinant map is open in $\mathbb{C}^4$. Therefore using the bijection we can claim that $GL(2, \mathbb C)$ has complex dimension 4 (or real dimension 8).

$SU(2)$ as the set of unitary 2x2 complex matrices, has a bijective map with $\mathbb{S^3}$. Using this bijection we can transport the topology of $\mathbb{S^3}$ on $SU(2)$ turning the bijection in an homeomorphism. This way we define on $SU(2)$ the unique topology (up to homeomorphisms) such that it is homeomorphic with $\mathbb{S^3}$.

Does it sound right ? Thank you.

 

[jbergman]

Sorry to be boring, I would try to recap my understanding.

There is a bijection between the set of 2x2 complex matrices and the vector space $\mathbb{C}^4$. The determinant of the 2x2 complex matrices is a continuous map from $\mathbb{C}^4$ (with its standard topology) to $\mathbb C$. Now the set $\{\mathbb{C} - 0\}$ is open in $\mathbb C$, hence its preimage under the determinant map is open in $\mathbb{C}^4$. Therefore using the bijection we can claim that $GL(2, \mathbb C)$ has complex dimension 4 (or real dimension 8).

$SU(2)$ as the set of unitary 2x2 complex matrices, has a bijective map with $\mathbb{S^3}$. Using this bijection we can transport the topology of $\mathbb{S^3}$ on $SU(2)$ turning the bijection in an homeomorphism. This way we define on $SU(2)$ the unique topology (up to homeomorphisms) such that it is homeomorphic with $\mathbb{S^3}$.

Does it sound right ? Thank you.

This is not how I would show there is a homeomorphism between the 2. I have to head to work but will try and write something up later today.

 

[martinbn]

Sorry to be boring, I would try to recap my understanding.

There is a bijection between the set of 2x2 complex matrices and the vector space $\mathbb{C}^4$. The determinant of the 2x2 complex matrices is a continuous map from $\mathbb{C}^4$ (with its standard topology) to $\mathbb C$. Now the set $\{\mathbb{C} - 0\}$ is open in $\mathbb C$, hence its preimage under the determinant map is open in $\mathbb{C}^4$. Therefore using the bijection we can claim that $GL(2, \mathbb C)$ has complex dimension 4 (or real dimension 8).

This is right.

$SU(2)$ as the set of unitary 2x2 complex matrices, has a bijective map with $\mathbb{S^3}$. Using this bijection we can transport the topology of $\mathbb{S^3}$ on $SU(2)$ turning the bijection in an homeomorphism. This way we define on $SU(2)$ the unique topology (up to homeomorphisms) such that it is homeomorphic with $\mathbb{S^3}$.

Does it sound right ? Thank you.

No, $SU(2)$ has a topology and with it it is homeomorphic to the three sphere. Of course this the same topology as the one you get from the bijection, but you don't topologize it through the bijection. All sets involved in this thread have the cardinality of the continuum. Thus are bijective. To give them topology using some of these bijections is pointless.

 

[cianfa72]

No, $SU(2)$ has a topology and with it it is homeomorphic to the three sphere. Of course this the same topology as the one you get from the bijection, but you don't topologize it through the bijection.

That’s the point I'm still confused. You mean $SU(2)$ has got its own topology (with this topology $SU(2)$ happen to be homeomorphic to $\mathbb S^3$).

So, how is defined the $SU(2)$ topology ?

 

[martinbn]

That’s the point I'm still confused. You mean $SU(2)$ has got its own topology (with this topology $SU(2)$ happen to be homeomorphic to $\mathbb S^3$).

So, how is defined the $SU(2)$ topology ?

This was already said, but it should have been your starting point. What is the definition of this group?

 

[cianfa72]

This was already said, but it should have been your starting point. What is the definition of this group?

It is the group of (complex) unitary matrices with determinant 1. As set it is a subset of $GL(2,\mathbb C)$, hence we can define its topology as the subspace topology from $GL(2,\mathbb C)$. The topology of $GL(2,\mathbb C)$ is defined through the "natural" map with $\mathbb C^4$.

 

[martinbn]

It is the group of (complex) unitary matrices with determinant 1. As set it is a subset of $GL(2,\mathbb C)$, hence we can define its topology as the subspace topology from $GL(2,\mathbb C)$. The topology of $GL(2,\mathbb C)$ is defined through the "natural" map with $\mathbb C^4$.

Then why are you confused? This is a topology which a priori has nothing to do with the sphere's topology.

 

[cianfa72]

This is a topology which a priori has nothing to do with the sphere's topology.

ok, so the above "procedure" to assign a topology on $SU(2)$ is valid.

Then using the identification $\mathbb C^4 \cong \mathbb R^8$ we get that $GL(2, \mathbb C)$ is homeomorphic to an open set of $\mathbb R^8$. Now how can we write down the homeomorphism between $SU(2)$ and $\mathbb S^3$ ?

 

[martinbn]

The unitary group consists if 4 complex numbers. The unitarity condition implies that two of them are just the conjugate of the other two. So it is 2 complex or 4 real. The determinant condition gives that the sum of their squares is 1. So it is the three sphere. And this gives you the map and its inverse explicitly in terms of continuous functions.

 

[cianfa72]

I was thinking about another way to look at it. Your conditions of complex entries of unitary matrices (each entry seen as a pair of real numbers) amounts to 4 linear conditions on $\mathbb R^8$ (each defines an hyperplane in $\mathbb R^8$). The intersection of such loci and the locus defined by the condition on determinant 1 is a $\mathbb S^3$ sphere. Using the homeomorphism between $GL(2,\mathbb C)$ and $\mathbb R^8$ we can restrict it to an homeomorphism between the subset $SU(2)$ endowed with the subspace topology from $GL(2,\mathbb C)$ and $\mathbb S^3$ endowed with the subspace topology from $\mathbb R^8$ (the restriction of an homeomorphism to subsets endowed with the corresponding subspace topologies is an homeomorphism as well).

 

[martinbn]

They are not linear. They are quadratic.

 

[cianfa72]

They are not linear. They are quadratic.

Sorry, the conditions to be complex coniugate (given using pairs of real numbers in $\mathbb R^8$) are linear.

 

[martinbn]

Sorry, the conditions to be complex coniugate (given using pairs of real numbers in $\mathbb R^8$) are linear.

The conditions for the matrix to be unitary have the entries multiplied with each other. Not linear.

 

[cianfa72]

The conditions for the matrix to be unitary have the entries multiplied with each other. Not linear.

Yes, but they amounts basically to conditions on complex coniugate entries and on determinant 1. The conditions on complex coniugate entries define a linear subspace $A \cong\mathbb R^4$, the condition on determinant 1 defines an $\mathbb S^3$ sphere in it.

 

[martinbn]

Yes, but they amounts basically to conditions on complex coniugate entries and on determinant 1. The conditions on complex coniugate entries define a linear subspace $A \cong\mathbb R^4$, the condition on determinant 1 defines an $\mathbb S^3$ sphere in it.

Yes, but then it is the same as in post #50.

 

[cianfa72]

Yes, but then it is the same as in post #50.

Ok, this $\mathbb S^3$, as subset of $\mathbb R^8$, has the subspace topology from it and, according what we said in post #51, with this topology it is homeomorphic with $SU(2)$.

 

[jbergman]

I was thinking about another way to look at it. Your conditions of complex entries of unitary matrices (each entry seen as a pair of real numbers) amounts to 4 linear conditions on $\mathbb R^8$ (each defines an hyperplane in $\mathbb R^8$). The intersection of such loci and the locus defined by the condition on determinant 1 is a $\mathbb S^3$ sphere. Using the homeomorphism between $GL(2,\mathbb C)$ and $\mathbb R^8$ we can restrict it to an homeomorphism between the subset $SU(2)$ endowed with the subspace topology from $GL(2,\mathbb C)$ and $\mathbb S^3$ endowed with the subspace topology from $\mathbb R^8$ (the restriction of an homeomorphism to subsets endowed with the corresponding subspace topologies is an homeomorphism as well).

I get what you are thinking and it's nice and geometric but it feels clunky to me. For one, U(2) isn't really a linear subspace in $\mathbb{R}^8$ since it doesn't contain the origin.

One rigorous way to show this is to exhibit chart maps for SU(2) and $S^3$, i.e. homeomorphisms to $\mathbb{R}^3$. Then you can compose you charts with your bijection and show that it is continuous or smooth and you are done. This is basically what Martin suggested.

 

[cianfa72]

I get what you are thinking and it's nice and geometric but it feels clunky to me. For one, U(2) isn't really a linear subspace in $\mathbb{R}^8$ since it doesn't contain the origin.

What is the problem with $U(2)$ ? Yes it is not a linear subspace in $\mathbb R ^8$. The point is that the conditions to be an element in $SU(2)$ define an $\mathbb S^3$ sphere in the linear subspace say $A$ in $\mathbb R^8$.

One rigorous way to show this is to exhibit chart maps for SU(2) and $S^3$, i.e. homeomorphisms to $\mathbb{R}^3$. Then you can compose you charts with your bijection and show that it is continuous or smooth and you are done. This is basically what Martin suggested.

But we already have the homeomorphism between an open set in $\mathbb R^8$ (determinant $\neq 0$) and $GL(2,\mathbb C)$ (this is open in the space of 2×2 complex matrices). So the restriction of the homeomorphism from a subset of the open set of $\mathbb R^8$ (i.e the $\mathbb S^3$ sphere) to the corresponding subset of $GL(2,\mathbb C)$ (both endowed with the subspace topologies) is an homeomorphism as well.

 

[fresh_42]

$SU(2)$ is a simple example with natural charts, let it be embedded in $\mathbb{R}^4,\mathbb{R}^8$ or $\mathbb{C}^2$ or any of the subspace topologies $\ldots \cap \{\ldots\,|\,\|z\|^2=1\}.$ There is nothing more to do than look at the correspondences I linked to at the beginning:
https://www.physicsforums.com/insig...c-part/#3-The-3-Sphere-or-Glome-or-SU2mathbbC

All that is needed is the decision of how you want to look at it and where to start from. We run around in circles here, partly because of the words "induced" or "subspace topology" which have been used without mentioning "induced by what and how?" or "subspace of what?".