What is the relationship between SO(n) and S^n?

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pivoxa15
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Homework Statement


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?

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.
 
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pivoxa15 said:

Homework Statement


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?

No, the last two are not correct.

[tex]S^2 \cong SO \left( 3 \right) / SO \left( 2 \right)[/tex]

[tex]S^3 \cong SO \left( 4 \right) / SO \left( 3 \right) \cong SU \left( 2 \right)[/tex]

In general,

[tex]S^n \cong SO \left( n+1 \right) / SO \left( n \right)[/tex]

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

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

[tex]x^T x = 1.[/tex]

Set [itex]y = Ax[/itex], with [itex]A[/itex] in [itex]SO \left( n+1 \right)[/itex], and calculate the length of [itex]y[/itex].
 
George Jones said:
No, the last two are not correct.

[tex]S^2 \cong SO \left( 3 \right) / SO \left( 2 \right)[/tex]

[tex]S^3 \cong SO \left( 4 \right) / SO \left( 3 \right) \cong SU \left( 2 \right)[/tex]

In general,

[tex]S^n \cong SO \left( n+1 \right) / SO \left( n \right)[/tex]

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

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


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

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