Using the orbit-stabilizer theorem to identify groups

[aalma]

TL;DR

Using the orbit-stabilizer theorem to identify groups.

I want to identify:

1. $S^n$ with the quotient of $O(n + 1,R)$ by $O(n,R)$.
2. $S^{2n+1}$ with the quotient of $U(n + 1)$ by $U(n)$.

The orbit-stabilizer theorem would give us the result, but my problem is to apply it. My problem is how to find the stabilizer.
In 1 how to define the action of $O(n+1,R)$ on $S^n$ and then conclude that $stab(x)=O(n,R)$ for $x \in S^n$. And similar in 2?

It would be helpful and appreciated if you simplify/explain the method for showing this.

 

[fresh_42]

Here are some explicit homomorphisms if that helps to figure out the action:
https://www.physicsforums.com/insights/journey-manifold-su2mathbbc-part/

 

[aalma]

Thanks. I think I got 1: the stabilizer is the block matrix H=
[A 0]
[0 1]
Where $A$ is in $O(n)$ so $H$ is the collection of $n+1\times n+1$ orthogonal matrices.
In 2, I say that $U(n+1)$ acts on $S^{2n+1}$ as follows:
For a matrix $A$ in $U(n+1)$ and a vector $e_{2n+1}$ in $S^{2n+1}$ we consider $Ae_{2n+1}$ then the map
$U(n+1)\to S^{2n+1}$ given by $A\to Ae_{2n+1}$.
But then is the multiplication valid here?

 

[fresh_42]

Thanks. I think I got 1: the stabilizer is the block matrix H=
[A 0]
[0 1]
Where $A$ is in $O(n)$ so $H$ is the collection of $n+1\times n+1$ orthogonal matrices.
In 2, I say that $U(n+1)$ acts on $S^{2n+1}$ as follows:
For a matrix $A$ in $U(n+1)$ and a vector $e_{2n+1}$ in $S^{2n+1}$ we consider $Ae_{2n+1}$ then the map
$U(n+1)\to S^{2n+1}$ given by $A\to Ae_{2n+1}$.
But then is the multiplication valid here? Since $A$ is $n+1 ×n+1$ but $v$ is $2n+1 × 1$ ?!

That looks a bit confusing.

An action of a group $G$ on a set $X$ is a group homomorphism $G\rightarrow \operatorname{GL}(X).$ What is your group homomorphism in this case? The natural operation would be $\operatorname{O}(n+1,\mathbb{R}) \rightarrow \operatorname{GL}(n+1,\mathbb{R}).$
$$
H = \{A\in SO(n+1)\, : \,A.\mathfrak{e}_{n+1}=\mathfrak{e}_{n+1}\} \leq SO(n+1)
$$
is the stabilizer of $\mathfrak{e}_{n+1}\in \mathbb{R}^{n+1}.$

Now show that $H\cong \operatorname{SO}(n,\mathbb{R})$ and $\operatorname{SO}(n+1,\mathbb{R})/H \cong S^n.$

 

[fresh_42]

The German Wikipedia page has a nice explanation of how it should be done:
https://de.wikipedia.org/wiki/Gruppenoperation#Stabilisator
so maybe you can work with the translate function in Chrome. The English page is less constructive
https://en.wikipedia.org/wiki/Group_action#Orbits_and_stabilizers
at least as I could tell.

 

[aalma]

That looks a bit confusing.

An action of a group $G$ on a set $X$ is a group homomorphism $G\rightarrow \operatorname{GL}(X).$ What is your group homomorphism in this case? The natural operation would be $\operatorname{O}(n+1,\mathbb{R}) \rightarrow \operatorname{GL}(n+1,\mathbb{R}).$
$$
H = \{A\in SO(n+1)\, : \,A.\mathfrak{e}_{n+1}=\mathfrak{e}_{n+1}\} \leq SO(n+1)
$$
is the stabilizer of $\mathfrak{e}_{n+1}\in \mathbb{R}^{n+1}.$

Now show that $H\cong \operatorname{O}(n,\mathbb{R})$ and $\operatorname{O}(n+1,\mathbb{R})/H \cong S^n.$

I think you're right. But I can look at the action $\phi:O(n+1)\times S^n\to S^n$ given by $\phi(A,v)=Av$. So I need to see that it's transitive that is for every $v, w\in S^n$ there exists $A\in O(n+1)$ such that $Av=w$. How can I find such $A$?

 

[fresh_42]

How can I find such $A$?

Visually, it is easy. You have two points ($v,w$) on a ball ($S^n$) and rotations ($\operatorname{SO}(n+1)$) at hand to do so. A "soft" argumentation is, that there exists an invertible matrix $A\in \operatorname{GL}(n+1)$ such that $A.v=w.$ Since $\|v\|=\|w\|$ we can do this with a matrix that leaves lengths, and angles invariant, i.e. an orthogonal matrix.

Polar coordinates seem to be suitable for a formal proof, possibly an induction over $n.$

You can chose $v$ to be the pole and simply skip the rotation axis to $w.$

 

[martinbn]

I think you're right. But I can look at the action $\phi:O(n+1)\times S^n\to S^n$ given by $\phi(A,v)=Av$. So I need to see that it's transitive that is for every $v, w\in S^n$ there exists $A\in O(n+1)$ such that $Av=w$. How can I find such $A$?

Pick coordinates so that your two points are in the $x_1x_2$ plane. In the plane the problem is clear. There is an element of $O(2)$ that takes one to the other. For your $A$, take blockdiagonal with the 2x2 in the upper left and 1's on the rest of the diagonal.

 

[mathwonk]

you could use induction for transitivity, essentially what fresh is suggesting. i.e. if O(n) is transitive on S(n-1), then O(n+1) is transitive on the points of the hemisphere in S(n) perpendicular to e1. But given any two points of S(n), n≥2, we can pass a hyperplane through them, and choose a new orthonormal basis so that hyperplane is spanned by e2,...,en+1, and use the same argument.

Here we think of S(n) as living in an abstract coordinate free space equipped with an inner product, and O(n+1) as length preserving linear transformations of that space. They are represented by matrices only when we choose a basis, which we may change as desired from time to time.

for this however, you must start the induction by proving it for n = 0 and 1. I.e. it is not true for S(1) that any 2 points lie in a hyperplane, so the case of O(1) being transitive on S(0) does not help prove O(2) is transitive on S(1). It should not be too hard to show rotations act transitively on the circle, using sines and cosines, or just geometry. I.e. the first column of the matrix should be the desired destination of e1, and the second should be perpendicular to that, and also length one.

martinbn's succinct remarks should also be helpful.

 

[martinbn]

Extend $v$ to an orthonormal basis so that it is the first vector, extend $w$ to one as well. There is a matrix that takes one basis to the other. It will be orthogonal.

 

[aalma]

Extend $v$ to an orthonormal basis so that it is the first vector, extend $w$ to one as well. There is a matrix that takes one basis to the other. It will be orthogonal.

If I extend $v$ to an orthonormal basis ${v_1,...,v_{n+1}}$ and write it as axmatrix whose column are these vectors so is it just enough to say that any orthogonal matrix-orthogonal transformation $A$ acting on this matrix would give us an orthogonal matrix whose column is the orthonormal basis ${Av_1,...,Av_{n+1}}={w_1,...,w_{n+1}}$?
I just do not see how our matrix must look!.

 

[martinbn]

If I extend $v$ to an orthonormal basis ${v_1,...,v_{n+1}}$ and write it as axmatrix whose column are these vectors so is it just enough to say that any orthogonal matrix-orthogonal transformation $A$ acting on this matrix would give us an orthogonal matrix whose column is the orthonormal basis ${Av_1,...,Av_{n+1}}={w_1,...,w_{n+1}}$?
I just do not see how our matrix must look!.

I am not sure what you are saying. Take the standard basis $\{e_i\}$, then the matrix $A_v$ that takes it to the basis $\{v_i\}$ has columns the vectors $v_i$, so it is orthogonal. Same fore the matrix $A_w$ that takes $\{e_i\}$ to $\{w_i\}$. The matrix that takes the $\{v_i\}$ to $\{w_i\}$ is $A_wA_v^{-1}$.