# [SOLVED] show SO(3) a manifold inside R^6 and prove SO(3) embeds in R^5

• January 24th 2007, 12:18 PM
metanosis
[SOLVED] show SO(3) a manifold inside R^6 and prove SO(3) embeds in R^5
have done some research, and here is what I know;

to show SO(3) a manifold "inside" R^6(am guessing inside means a submanifold), the hint was that only two vectors are needed to describe the tangent space of the 2-sphere, ie if given a radial vector, and a tangent vector, the other orthonormal vector is then completely specified. more specifically;
the Jacobian takes tangent spaces to matrices smoothly,

so am trying to write the diffeo from the tangent space to the sphere S^2(T_xS^2)
into SO(3). This was from the hint. It seems the jacobian would fill this bill.

Also found that RP^3 can be constructed from the unit Ball, and that can then construct a diffeo from RP^3 to SO(3)(take unit Ball in above to have radius Pi), but don't see how this would help to show a submanifold,
when I look at the definition below;

Starting from the definition of a submanifold;
A C^infty (infinitely differentiable) manifold is said to be a submanifold of a C^infty manifold M^' if M is a subset of M^' and the identity map of M into M^' is an embedding. Well clearly SO(3) is a subset of R^6, so need show the identity map from SO(3) into R^6 is an embedding. Here it seems that one could just add zeros to the vectors in SO(3), but this must be too simple.

An embedding is a proper immersion, and proper means only that the inclusion
map is closed(I think). Well, not too sure how to tie all this together.

To show SO(3) embeds in R^5, must exhibit an embedding, and this wouold be harder than above, not sure how to proceed.

thanks in advance for any help, suggestions you may be able to render.
metanosis
• June 14th 2007, 05:25 AM
Rebesques
Quote:

and that can then construct a diffeo from RP^3 to SO(3)
Wouldn't help here, but it's a useful fact anyway. :rolleyes:

Quote:

Well clearly SO(3) is a subset of R^6, so need show the identity map from SO(3) into R^6 is an embedding
As u say! But the following
Quote:

Here it seems that one could just add zeros to the vectors in SO(3), but this must be too simple.
can't be done, as SO(3) as vectors, is contained in R^9. We must show three dimensions vanish.

We prove the embedding in the following steps:

(i) $SO(3)\subset \mathbb{R}^6$

Note that all rotations leave a line fixed (Euler's theorem). This is a nice fact, but irrelevant here :(

(ii) All elements of $SO(3)$ can be written as $(x_1,x_2,x_3,x_4,x_5,x_6)$, $x_i \in \mathbb{R}$.

Actually, the x's in the previous relation are related, so call this relation $f(x_1,\ldots,x_6)=0$.

Apply the standard matrix operations to $S\in SO(3)$ to get a matrix of the form $[\underline{u} \ \underline{v} \ \underline{0} ]=:\phi_S(S)$. Here $\underline{u},\underline{v}$ are 3x1 vectors, with entries $x_1$ to $x_3$ and $x_4$ to $x_6$ respectively.

The operation is a diffeomorphism between SO(3) and $\Theta=\{\phi_S(S): \ S\in SO(3)\}=\{(x_1,x_2,\ldots,x_6): \ x_i\in \mathbb{R}, \ f(x_1,\ldots,x_6)=0\}$. (Ok, this is long, but at least it's not difficult)

(iii) $\Theta$ can be embedded in $\mathbb{R}^6$.

For this, we show that the differential ${\rm d}\jmath$ of the inclusion map $\jmath:\Theta\rightarrow\mathbb{R}^6$ has maximum rank (=6 here) everywhere.

Consider $P=(x_1,x_2,\ldots,x_6)\in\Theta.$ Then for $v\in \mathbb{R}^6$,

${\rm d}\jmath_P(v)=\langle\nabla\jmath(P), v\rangle=\langle I_6, v\rangle$

where $I_6$ is the 6x6 unitary matrix. So ${\rm d}\jmath(P)$ has matrix $I_6$ for all $P\in \Theta$, and thus has rank 6.