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

[cianfa72]

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.

 

[fresh_42]

Here is a bunch of isomorphisms related to $SU(2).$

https://www.physicsforums.com/insights/journey-manifold-su2mathbbc-part/

The answer to your question is $\varphi \circ \psi.$

 

[cianfa72]

Sorry, a basis question: take a topological space A and another set B. Define two different bijections between A and B. Then consider the two topologies induced on B by the two different bijections.

I believe the two induced topologies on B are the same since there is an homeomorphism between them (a composition of homeomorphisms is an homeomorphism).

Is the above correct ?

 

[fresh_42]

Sorry, a basis question: take a topological space A and another set B. Define two different bijections between A and B. Then consider the two topologies induced on B by the two different bijections.

I believe the two induced topologies on B are the same since there is an homeomorphism between them (a composition of homeomorphisms is an homeomorphism).

Is the above correct ?

I'm not sure I understand what you mean.

There can be many bijections between two topological spaces and none of them needs to be a homeomorphism. IIRC then there are even bijections that are continuous in one but not in the other direction, but I'm not sure. You have to be careful with the term since the continuity of the inverse is a strong requirement.

However, there are no problems with the mappings I quoted since they are basically just a rearrangement.

 

[jbergman]

Sorry, a basis question: take a topological space A and another set B. Define two different bijections between A and B. Then consider the two topologies induced on B by the two different bijections.

I believe the two induced topologies on B are the same since there is an homeomorphism between them (a composition of homeomorphisms is an homeomorphism).

Is the above correct ?

$SU(2)$'s topology is a subspace topology of $\mathbb{R}^8$ since it is a matrix subgroup of $GL(\mathbb{C}, 2)$. More later...

 

[cianfa72]

There can be many bijections between two topological spaces and none of them needs to be a homeomorphism.

My point is that, regardless of the specific bijection between a topological space $A$ and a "naked" set $B$, the two induced topologies on $B$ from $A$ using the two bijections are the same (i.e. $(B, \tau_1)$ and $(B, \tau_2)$ are homeomorphic).

 

[fresh_42]

My point is that, regardless of the specific bijection between a topological space $A$ and a "naked" set $B$, the two induced topologies on $B$ from $A$ using the two bijections are the same (i.e. $(B, \tau_1)$ and $(B, \tau_2)$ are homeomorphic).

If we have a continuous bijection $f\, : \,(A,\mathcal{T}_A)\longrightarrow B$ then the following properties are equivalent:

1. $f$ is a homeomorphism.
2. $f$ is open.
3. $f$ is closed.

The problem is therefore to make $f$ continuous and open. If we define
$$
\mathcal{T}_B:=\{O\subseteq B\,|\,O=f(U)\text{ for some }U\in \mathcal{T}_A\}
$$
then $f$ becomes open. But is it continuous? Say we have $O=f(U)\in \mathcal{T}_B.$ Then $f^{-1}(O)=f^{-1}(f(U)) = U$ since $f$ is injective. Hence, $f$ is continuous.

The answer is yes.

 

[cianfa72]

Ok, so as composition of homeomorphisms $(B,\mathcal{T}_1)$ and $(B,\mathcal{T}_2)$ are homeomorphic.

 

[fresh_42]

Ok, so as composition of homeomorphisms $(B,\mathcal{T}_1)$ and $(B,\mathcal{T}_2)$ are homeomorphic.

You said "naked $B$" - at least before you edited your question. I therefore defined the topology on $B$ so that the bijective function becomes open and continuous. I made it suited.

If you have any topology on $B$ to start with, then $f$ doesn't need to be a homeomorphism. E.g., you can cut the sphere in half. That would be still a bijection, but not continuous anymore.

In the case of $\mathbb{S}^3$ and $SU(2)$ we can find natural topologies on both (see post #5) such that the mappings I linked to become a homeomorphism. Open neighborhoods in the chart of the four real coordinates of the 3-sphere become an open neighborhood in the product topology which provides a chart for $SU(2)$. The mappings go:
$$
\mathbb{S}^3 \supseteq_\text{open} U \rightarrow \text{ chart in } \mathbb{R}^4 \stackrel{\varphi \, \psi}{\longrightarrow }\text{ chart in } \mathbb{R}^4 \rightarrow O\subseteq_\text{open} SU(2)
$$

 

[cianfa72]

E.g., you can cut the sphere in half. That would be still a bijection, but not continuous anymore.

You mean take for instance a 2-sphere $A$ with standard topology and an homeomorphism $f$ to another 2-sphere $B$ endowed with the same standard topology. Then cut $B$ in half to get $B'$. $f$ continues to be a bijection with $B'$ but it is no longer an homemorphism when $B'$ is endowed with the subspace topology from $\mathbb R^3$. Nevertheless if we endow the set $B'$ with the induced topology by $f$ from topological space $A$, then $A$ and $B'$ are homeomorphic by definition of induced topology through a bijection.

In the case of $\mathbb{S}^3$ and $SU(2)$ we can find natural topologies on both (see post #5) such that the mappings I linked to become a homeomorphism.

Ok, so $SU(2)$ has a natural topology as subspace topology from $\mathbb R^8$.

Open neighborhoods in the chart of the four real coordinates of the 3-sphere become an open neighborhood in the product topology which provides a chart for $SU(2)$.

Sorry, given the fact that a 3-sphere is a three-dimensional object, any chart has to map an open subset of it into an open subset of $\mathbb R^3$ and not of $\mathbb R^4$.

 

[fresh_42]

You mean take for instance a 2-sphere $A$ with standard topology and an homeomorphism $f$ to another 2-sphere $B$ endowed with the same standard topology. Then cut $B$ in half to get $B'$. $f$ continues to be a bijection with $B'$ but it is no longer an homemorphism when $B'$ is endowed with the subspace topology from $\mathbb R^3$. Nevertheless if we endow the set $B'$ with the induced topology by $f$ from topological space $A$, then $A$ and $B'$ are homeomorphic by definition of induced topology through a bijection.

I do not know what exactly you mean by "induced". That's why I explicitly defined it. Given a topology on A, and a bijection f, we can define a specific topology on B which makes f open, and thus a homeomorphism.

Bring your own topology on B and nothing general can be said.

Ok, so $SU(2)$ has a natural topology as subspace topology from $\mathbb R^8$.
Sorry, given the fact that a 3-sphere is a three-dimensional object, any chart has to map an open subset of it into an open subset of $\mathbb R^3$ and not of $\mathbb R^4$.

Right. I forget to cut out the condition $\|z\|^2=1.$

The topology on $M=\mathbb{S}^3$ and also on $M=SU(2)$ is given by the subspace topology:
$$
\{U\,|\, \exists\, O\subseteq \mathbb{R}^4 \text{ open }\, : \,U=O\cap M \wedge \|z\|^2=1\,\forall \,z\in O\}
$$
The charts are the local tangent spaces at points $p\in M$ that are three-dimensional, but embedded in $\mathbb{R}^4$ or even in $\mathbb{R}^8$ if we let it play in $GL(2,\mathbb{R}^2).$

 

[cianfa72]

I do not know what exactly you mean by "induced". That's why I explicitly defined it. Given a topology on A, and a bijection f, we can define a specific topology on B which makes f open, and thus a homeomorphism.

Bring your own topology on B and nothing general can be said.

Yes, induced topology on set $B$ as in your definition of topology in your post#7. Then bringing the $(A,\mathcal{T}_A)$ topology on $B$ (making the bijection $f$ homeomorphism by definition) there are not different topologies $\mathcal{T}_B'$ on $B$ (up to homeomorphism) such that $(A,\mathcal{T}_A)$ and $(B,\mathcal{T}_B')$ are homeomorphic.

Right. I forget to cut out the condition $\|z\|^2=1.$

The topology on $M=\mathbb{S}^3$ and also on $M=SU(2)$ is given by the subspace topology:
$$
\{U\,|\, \exists\, O\subseteq \mathbb{R}^4 \text{ open }\, : \,U=O\cap M \wedge \|z\|^2=1\,\forall \,z\in O\}
$$
The charts are the local tangent spaces at points $p\in M$ that are three-dimensional, but embedded in $\mathbb{R}^4$ or even in $\mathbb{R}^8$ if we let it play in $GL(2,\mathbb{R}^2).$

Sorry, what is $z$ in $\|z\|^2=1$ ?

 

[fresh_42]

Yes, induced topology on set $B$ as in your definition of topology in your post#7. Then bringing the $(A,\mathcal{T}_A)$ topology on $B$ (making the bijection $f$ homeomorphism by definition) there are not different topologies $\mathcal{T}_B'$ on $B$ (up to homeomorphism) such that $(A,\mathcal{T}_A)$ and $(B,\mathcal{T}_B')$ are homeomorphic.Sorry, what is $z$ in $\|z\|^2=1$ ?

The four coordinates of a point $z\in M\subseteq \mathbb{R}^4.$

 

[cianfa72]

The topology on $M=\mathbb{S}^3$ and also on $M=SU(2)$ is given by the subspace topology:
$$
\{U\,|\, \exists\, O\subseteq \mathbb{R}^4 \text{ open }\, : \,U=O\cap M \wedge \|z\|^2=1\,\forall \,z\in O\}
$$

Sorry, which is the meaning of the operator $\wedge$ ? I'm confused about the definition of subspace topology on $M=\mathbb{S}^3$ or $M=SU(2)$ (as subsets of $\mathbb{R}^4$).

 

[fresh_42]

Sorry, which is the meaning of the operator $\wedge$ ? I'm confused about the definition of subspace topology for $M=\mathbb{S}^3$ or $M=SU(2)$ (as subsets of $\mathbb{R}^4$).

It's a logical AND, but the entire statement is redundant anyway since it is already included in $U=O\cap M.$

Let again be $M=\mathbb{S}^3$ or $M=SU(2).$ They are three-dimensional, but what does that mean? If we define them, we write points on the sphere $z\in \mathbb{S}^3$ as
$$
x_1\mathfrak{e}_1+x_2\mathfrak{e}_2+x_3\mathfrak{e}_3+x_4\mathfrak{e}_4=x_1\mathbf{1}+x_2\mathbf{i}+x_3\mathbf{j}+x_4\mathbf{k}=(x_1,x_2,x_3,x_4) \in \mathbb{R}^4
$$
and demand $x_1^2+x_2^2+x_3^2+x_4^2=1,$ i.e. as a geometric object in the four-dimensional Euclidean space. So why is it three-dimensional? There are two possible ways to see that. We could eliminate $x_4$ by $x_4=\pm\sqrt{1-(x_1^2+x_2^2+x_3^2)}$ in the definition but that would be unnecessarily difficult to handle, and it is not the way it is done in differential geometry. We have a natural embedding of the $3$-sphere in $\mathbb{R}^4$ as unit sphere. That makes calculations a lot easier. So instead of eliminating $x_4$ we remain in $\mathbb{R}^4$ and compute the tangent space at $z\in \mathbb{S}^3.$ This gives us a local diffeomorphism $\mathbb{S}^3\rightarrow T_z\mathbb{S}^3\cong \mathbb{R}^3$ which we can use as charts in an atlas of $\mathbb{S}^3.$ The charts are now a three-dimensional subspace of $\mathbb{R}^4$ which is why we call $\mathbb{S}^3$ three-dimensional. So if we keep the embedding we do not have to bother with the square root expression of $x_4.$

Similar can be said about $M=SU(2).$ Here we have
$$x_1\mathbf{1}+x_2\mathbf{i}+x_3\mathbf{j}+x_4\mathbf{k} = \begin{bmatrix}x_1\mathbf{1}+x_2\mathbf{i} & -x_3\mathbf{1}+x_4\mathbf{i} \\x_3\mathbf{1}+x_4\mathbf{i} & x_1\mathbf{1}-x_2\mathbf{i} \end{bmatrix}=(x_1,x_2,x_3,x_4) \in \mathbb{R}^4
$$
So instead of working with only three coordinates, we work with the embedding and work with all four coordinates and the subspace topology. However, the subspace topology is not really an aspect we care about. We usually work with the atlas of the manifolds anyway, and those are the three-dimensional tangent spaces.
$$
\mathbb{S}^3\cong_{diffeo}\underbrace{ T_z(\mathbb{S}^3)\cong \mathbb{R^3}\subseteq \mathbb{R}^4 \supseteq \mathbb{R}^3\cong T_z(SU(2))}_{\text{calculations}} \cong_{diffeo} SU(2)
$$

All our open sets are the direct product of open intervals around the four coordinates $x_i$ plus $x_1^2+x_2^2+x_3^2+x_4^2=1$ in all cases where we consider $M$ or $T_zM=T_{(x_1,x_2,x_3,x_4)}M.$

 

[cianfa72]

It's a logical AND, but the entire statement is redundant anyway since it is already included in $U=O\cap M.$

Yes.

We have a natural embedding of the $3$-sphere in $\mathbb{R}^4$ as unit sphere.

Here I've a doubt. An embedding $f$ requires the $3$-sphere object to have a topology starting with independent from how it "sits" in $\mathbb{R}^4$. Or the $3$-sphere is actually defined as a geometric object in the four-dimensional Euclidean space with constraint $x_1^2+x_2^2+x_3^2+x_4^2=1$ ?

 

[fresh_42]

Here I've a doubt. An embedding $f$ requires the $3$-sphere object to have a topology starting with independent from how it "sits" in $\mathbb{R}^4$.

No. An embedding does not require a topology of the embedded object. That is exactly the point of the subspace topology, it allows a topology inherited from "above" without the need to bring its own.

Nevertheless, $\mathbb{S}^3$ and $SU(2)$ do have their own topology. The sphere has "open patches" such as open line segments of a circle in one dimension that happen to be the same as the subspace topology.

Things are a bit more tricky for $SU(2).$ Here we do not only require a topology inherited from $\mathbb{R}^4$ or any topology on $SU(2)$ alone. Here we want to have a topology that makes matrix multiplication and inversion continuous mappings in the first place! It happens to be that those open intervals around the coordinates fulfill this requirement, but it is not automatically a given. The fact that an open set around a matrix in $SU(2)$ by its topology as a topological group is also an open set in the subspace topology and vice versa is a theorem - not difficult to prove, but a theorem, not a definition.

 

[cianfa72]

Nevertheless, $\mathbb{S}^3$ and $SU(2)$ do have their own topology. The sphere has "open patches" such as open line segments of a circle in one dimension that happen to be the same as the subspace topology.

Take for instance $\mathbb{S}^1$. We can build it by means of appropriated "gluing instructions" for two open line segments in $\mathbb R^1$. How can we define which subsets are to be considered open in it (without using the natural embedding of $\mathbb{S}^1$ in $\mathbb R^2$) ?

 

[fresh_42]

Take for instance $\mathbb{S}^1$. We can build it by means of appropriated "gluing instruction" for two open line segments in $\mathbb R^1$. How can we define which subsets are to be considered open in it (without using the natural embedding in $\mathbb R^2$) ?

The difficult part is to define a circle that isn't a line in the plane! Otherwise, you could always object that I used the embedding! This is really an issue since a circle is the collection of all points that are equally far from a certain point given outside of the circle! So how do we get rid of the center?

Let us assume that we magically defined $\mathbb{S}^1$ somehow and a distance without using points outside of $\mathbb{S}^1$ like its center or the metric of the plane. Then we call a set closed if it is either $\emptyset$ or $\mathbb{S}^1$ or a set $U$ that contains all its limit points defined by the distance. Then $\{\mathbb{S}^1-U\}$ is the topology on $\mathbb{S}^1.$

If we cannot find such a definition or a distance that allows us to define limit points, then we are stuck with the embedding. The same happens with $SU(2)$ the moment we use the word matrix.

 

[cianfa72]

Then $\{\mathbb{S}^1-U\}$ is the topology on $\mathbb{S}^1.$

Ok, so in the case of $\mathbb{S}^1$ the topology is defined as the collection of all sets of type $\mathbb {S}^1 - U$ for some set $U$ closed w.r.t. the distance.

If we cannot find such a definition or a distance that allows us to define limit points, then we are stuck with the embedding. The same happens with $SU(2)$ the moment we use the word matrix.

Ah ok, so for $SU(2)$ to define its topology we use the embedding through the unitary matrices.

 

[fresh_42]

Ok, so in the case of $\mathbb{S}^1$ the topology is defined as the collection of all sets of type $\mathbb {S}^1 - U$ for some set $U$ closed w.r.t. the distance.Ah ok, so for $SU(2)$ to define its topology we use the embedding through the unitary matrices.

Yes.

The sphere is defined via its center and the constant distance of its points to the center. Even if we looked at it as a conic section, we would start from "outside". It is inherently an embedding into $\mathbb{R}^4$ right from the beginning no matter how you look at it.

The matrices of $SU(2)$ are complex, unitary $2\times 2$ matrices with determinant one. That is inherently an embedding into $\mathbb{C}^2\cong \mathbb{R}^4$ right from the start, too.
$$
\begin{pmatrix}x_1+ix_2&x_3+ix_4\\x_3-ix_4&x_1-ix_2\end{pmatrix}\, , \,\|x_1+ix_2\|^2+\|x_3+ix_4\|^2=1
$$
The subspace topologies are so easy and fulfill all requirements that it's not necessary to think about the topologies of the manifolds without an embedding. We are so used to defining objects by coordinates that if you asked me for an example of a manifold that is not embedded in a Euclidean space I had to think long about it and would get up with the universe. The universe is a Riemannian manifold that isn't embedded.

 

[cianfa72]

The sphere is defined via its center and the constant distance of its points to the center. Even if we looked at it as a conic section, we would start from "outside". It is inherently an embedding into $\mathbb{R}^4$ right from the beginning no matter how you look at it.

Are you talking of a generic $n$-sphere $\mathbb{S}^n$, $n \geq 1$ ? If yes that was my point: namely each of them "borns" right from the beginning as an embedding in the respective $\mathbb R^{n+1}$. Let me say the inclusion map is an embedding by very definition.

 

[cianfa72]

We are so used to defining objects by coordinates that if you asked me for an example of a manifold that is not embedded in a Euclidean space I had to think long about it and would get up with the universe. The universe is a Riemannian manifold that isn't embedded.

The universe as 4-dimensional smooth real manifold can be smoothly embedded in $\mathbb R^8$ per Whitney embedding theorem.

 

[fresh_42]

The universe as 4-dimensional smooth real manifold can be smoothly embedded in $\mathbb R^8$ per Whitney embedding theorem.

Smooth is not granted, and how the theorem deals with an expanding manifold isn't clear either.

 

[WWGD]

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.

Still, you may be going in circles. In order to determine if the preimage is open, you need to have a topology defined.

 

[cianfa72]

The difficult part is to define a circle that isn't a line in the plane! Otherwise, you could always object that I used the embedding! This is really an issue since a circle is the collection of all points that are equally far from a certain point given outside of the circle! So how do we get rid of the center?

Thinking again about it, even in the simple case of $\mathbb{S}^1$ we are in position to define it as topological space regardless of its natural embedding in $\mathbb R^2$. It goes as follows:

Take a closed interval on the real line (e.g. the closed interval $[0,1]$) with the subspace topology as subset of $\mathbb R^1$. Then define as open the sets you get identifying the point 0 with the point 1. What you get is a topology for the resulting set - namely the quotient topology from the topology defined on $[0,1]$.

 

[cianfa72]

Still, you may be going in circles. In order to determine if the preimage is open, you need to have a topology defined.

Sorry, in case of bijection $f$ (as in this case) the induced topology $\mathcal{T}_{SU(2)}$ on $SU(2)$ by $f$ from the standard topology $(\mathbb{S}^3, \mathcal{T}_{\mathbb{S}^3})$ is defined as

$$\mathcal{T}_{SU(2)}:=\{O\subseteq SU(2)\,|\,O=f(U)\text{ for some }U\in \mathcal{T}_{\mathbb{S}^3}\}$$

 

[fresh_42]

Thinking again about it, even in the simple case of $\mathbb{S}^1$ we are in position to define it as topological space regardless of its natural embedding in $\mathbb R^2$. It goes as follows:

Take a closed interval on the real line (e.g. the closed interval $[0,1]$) with the subspace topology as subset of $\mathbb R^1$. Then define as open the sets you get identifying the point 0 with the point 1. What you get is a topology for the resulting set - namely the quotient topology from the topology defined on $[0,1]$.

Yes, that's true. We can formally define $\mathbb{S}^1=[0,1]/\sim$ where $0\sim 1 .$ This would also allow to define a difference which in return allows the definition of limit points, or directly open sets.

 

[cianfa72]

$SU(2)$'s topology is a subspace topology of $\mathbb{R}^8$ since it is a matrix subgroup of $GL(\mathbb{C}, 2)$. More later...

$GL(\mathbb{C},2)$ has dimension 8, $SU(2)$ as unitary matrix subgroup is also subspace topology of $\mathbb{R}^4$.

 

[fresh_42]

$GL(\mathbb{C},2)$ has dimension 8, $SU(2)$ as unitary matrix subgroup is also subspace topology of $\mathbb{R}^4$.

$GL(\mathbb{C},2)$ 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!

 

[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. ;-)