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

[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?".

 

[cianfa72]

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?".

See this link (2nd answer). The set of matrices $M(z,w)$ without constraint $|z|^2 + |w|^2 =1$ is homeomorphic with $\mathbb C^2 \cong \mathbb R^4$. The condition $|z|^2 + |w|^2 =1$ defines a 3-sphere $\mathbb S^3$ in $\mathbb R ^4$. Then restricting the above homeomorphism on $\mathbb S^3$ endowed with the subspace topology from $\mathbb R ^4$ we get the homeomorphism with $SU(2)$.

 

[fresh_42]

I will not respond to an again imprecisely phrased statement since $M(z,w)$ is undefined, there is no homeomorphism other than the identity, and a point on $\mathbb{S}^3$ has nothing to do with a point on $SU(2)$ without explaining the correspondence. I can understand it, nevertheless, it is sloppy. IMO, you have to learn to be way more precise than you usually are. Half of your questions resolve themselves once they are stated with the necessary precision. That does NOT mean to write more, au contraire, it is very often less!

This thread is closed.