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

1. [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

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

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
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.