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.