# Group of rotations of S^n

1. Jan 26, 2008

### pivoxa15

1. The problem statement, all variables and given/known data
Group of rotations of S^1 = SO(2)=S^1 conincidently
Group of rotations of S^2 = SO(3)
Group of rotations of S^3 = SO(4)

Correct?

3. The attempt at a solution
SO(3) is the group of all rotations in R^3 so it can rotate all elements of S^2 which is part of R^3. Although I can't show it rigorously.

Would it be enough to say since SO(3) consists of all rotations in R^3, it can offcouse rotate all elements having unit distance from the origin.

Similar argument made to S^1 and S^3.

Last edited: Jan 26, 2008
2. Jan 27, 2008

### George Jones

Staff Emeritus
No, the last two are not correct.

$$S^2 \cong SO \left( 3 \right) / SO \left( 2 \right)$$

$$S^3 \cong SO \left( 4 \right) / SO \left( 3 \right) \cong SU \left( 2 \right)$$

In general,

$$S^n \cong SO \left( n+1 \right) / SO \left( n \right)$$

$$S^{2n + 1} \cong SU \left( n+1 \right) / SU \left( n \right).$$

 Sorry, I think I misunderstood what you wrote. Note that "=" should not be substituted for the word "is."

I think what you meant is something like the following.

Consider $S^n$ as a subset of $\mathbb{R}^{n+1}$. Show that each element of $SO \left( n+1 \right)$ maps $S^n$ to $S^n$. A general mapping from $\mathbb{R}^{n+1}$ to $\mathbb{R}^{n+1}$ might map an element of $S^n$ to an element of $\mathbb{R}^{n+1}$ that is not in $S^n$, so there is something "special" going on here.[/edit]

Last edited: Jan 27, 2008
3. Jan 27, 2008

### George Jones

Staff Emeritus
Continuing, let $x$ be an element of $S^n$, so $x$ is an element of $\mathbb{R}^{n+1}$ such that

$$x^T x = 1.$$

Set $y = Ax$, with $A$ in $SO \left( n+1 \right)$, and calculate the length of $y$.

4. Jan 30, 2008

### pivoxa15

With this one, $$S^{2n + 1} \cong SU \left( n+1 \right) / SU \left( n \right).$$

does it mean each coset representation is a matrix in SU(2)?