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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.

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    $\begingroup$ Can you please include the references where you've seen that the parameter space of $SO(3)$ is 3 dimensional embedded in $R^4$ $\endgroup$
    – RedGiant
    Apr 4 '20 at 16:47
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    $\begingroup$ I say because manifold of $SU (2)$ is diffeomorphic to manifold of $SO (3)$.So for such an isomorphism between $SU (2)$ and $SO (3)$ implies smooth manifold of $SO (3)$ is also a 3 manifold like $SU (2)$. I know there are some loop holes in my concept. Can you help me @RedGiant. $\endgroup$
    – ROBIN RAJ
    Apr 4 '20 at 17:31
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    $\begingroup$ Would Mathematics be a better home for this question? $\endgroup$
    – Qmechanic
    Apr 4 '20 at 18:03
  • $\begingroup$ I don't need too much mathematical explanation using technical terms.What I need is something geometrical or physical explanation@Qmechanic $\endgroup$
    – ROBIN RAJ
    Apr 4 '20 at 18:06
  • $\begingroup$ The rotation group Lie algebra is 3-dimensional, and the most popular parameterization of the logarithm of a rotation is ω $\cdot$ L, where the 3d Euler vector parameter ω $=\theta \mathbf {n}$, θ being the rotation angle (compact circle) and n the unit vector characterizing the axis of rotation, so a 2-sphere. Is this what you are after visualizing? $\endgroup$ Apr 4 '20 at 20:02
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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).

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

  2. 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.

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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.

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  • $\begingroup$ So manifold of $SO (3) $ is embedded in a space of how many dimensions?@Cinaed Simson $\endgroup$
    – ROBIN RAJ
    Apr 5 '20 at 4:33
  • $\begingroup$ @ROBINRAJ: The dimension of the represention space for $SO(3)$ is $3$. The Lie alegras are isomorphoric, but the matrices $A$ in $SU(2)$ are $2x2$ complex matrices - and the matrices $A$ in $SO(3)$ are $3x3$ real matrices. In order to find a representation in $3$ dimensional space for $SU(2)$, a guy name Cromwell figured out how to preform a homomorporic mapping - which means same structure, i.e., it doesn't change the structure of the Lie algebra - and does just that - provides a representation of $SU(2)$ in $3$ dimensional space. $\endgroup$ Apr 5 '20 at 5:46
  • $\begingroup$ What type of manifold $SO (3) $ make in $R^3$ I mean shape of manifold(for $SU (2) $ shape of manifold is $S^3$)@Cinaed Simson $\endgroup$
    – ROBIN RAJ
    Apr 5 '20 at 6:22
  • $\begingroup$ @ROBINRAJ: A manifold is topological space which is locally Euclidean. Topologically, both $SU(2)$ and $SO(3)$ are compact and $SU(2)$ is simply connected. The representation here is in $R^{3}$ - there are infinite number of representation. Geometrically they're circles which fill up a sphere in $R^3$ - where $SU(2)$ is the double covering of $SO(3)$. Your rolling question has not include a single physics question. If you have a physics question, open another question. $\endgroup$ Apr 5 '20 at 19:27

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