Lie group and corresponding smooth manifold, and also why $SO(3)$ have a 3-dimensional manifold embedded in 4-dimensional Real space? I think I have some loop holes on a connecting smooth manifold to a lie group.
I state what my concepts are,
Lie groups are expressed as manifold because the parameters in corresponding metric form a parameter space which can be visualised as manifold.
For example $SU(2)$ this can rotate vectors in complex 2 dimensional space. Each vector in 2 dimensional complex space has 4 parameters $(x,y,z,w)$. To rotate them the metric also contain these 4 parameter's.
But for $SU(2)$ determinant must one. This gives a condition on parameters 
$$x^2 + y^2 + z^2 + w^2  =1. $$
This $S^3$ is a smooth manifold for $SU(2)$ and also each point in this manifold corresponds to a group element in $SU(2)$.
Similarly thinking $SO(3)$ rotate vectors in 3-dimensional space.
So metrix must contain 3 parameters let them   ' $ (x,y,z) $  '  a relation connecting this parameters should form some 2 dimensional manifold embedded in $R^3$.
But some resources show me that manifold of $SO(3)$ is 3 dimensional and embedded in $R^4$.I don't understand how the manifold of $SO(3)$ is 3-dimensional one. 
 A: The three parameters of the rotation group can be taken to be the  Euler angles $\theta$, $\phi$, $\psi$ and if we write the SU(2) element as 
$$
U=x_0{\mathbb I}+ i\sigma_1 x_1+i\sigma_2 x_2+i\sigma_3 x_3
$$
with the $x_i$ real and obeying $x_0^2+x_1^2+x_2^2+x_3^2=1$ so they define a point on the three sphere $S^3$, the relation is 
$$
x_0= \cos\theta/2\cos(\psi+\phi)/2\\
x_1= \sin\theta/2\sin(\phi-\psi)/2\\
x_2=- \sin\theta/2 \cos(\phi-\psi)/2\\
x_3= - \cos\theta/2 \sin(\psi+\phi)/2
$$
The complete $S^3$ is covered if $0<\phi<2\pi$, $0<\theta<\pi$, $0<\psi<4\pi$ and we can think of the Euler angles as being an anlogue the spherical polar coordinate angles. Antipodal points on $S^3$ correspond to the same rotation in SO(3). 
A: *

*On one hand
$$SU(2)~=~\left\{\left. \begin{bmatrix} a & b \\ -b^{*} & a^{*}  \end{bmatrix}\right|  a,b\in \mathbb{C}, |a|^2+|b|^2=1\right\}~\cong~S^3~\subseteq~\mathbb{R}^4.$$

*On the other hand, identify the Lie algebra $$su(2)~:=~\{\sigma\in{\rm Mat}_{2\times 2}(\mathbb{C})\mid \sigma^{\dagger}=\sigma, {\rm tr}(\sigma)=0\}~={\rm span}_{\mathbb{R}}\{\sigma_1,\sigma_2,\sigma_3\}~\cong~\mathbb{R}^3$$ with 3D space equipped with the Euclidean norm $||\sigma||^2=-\det(\sigma)$. The Lie group $SU(2)$ acts on the Lie algebra $su(2)$ via the adjoint representation ${\rm Ad}(g)(\sigma)=g\sigma g^{\dagger}$. It is length-preserving map, i.e. ${\rm Ad}(g)$ is an orthogonal transformation. One may show that ${\rm Ad}:SU(2)\to SO(3)$ is a 2:1 Lie group homomorphism. 
A: Althought the group manifolds may be isomorphoic, it takes a $2$ to $1$ homomorphoric mapping of the $SU(2)$ group onto $SO(3)$ group in order to use the $SO(3)$ representation. There are respresentations of spinors which have no $SO(3)$ representation. 
In essence, if $A\in SU(2)$ maps onto $R(A) \in SO(3)$, then $R(A)=R(-A)$ and choose the Pauli spin matrices.
And regarding your comment on the $4$ dimensional space, you might be confusing it with the Lorentz group $SO(3,1)$ and $SU(2)\times SU(2)$ - which has $6$ generators acting on $4$ vectors.
