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.

• Can you please include the references where you've seen that the parameter space of $SO(3)$ is 3 dimensional embedded in $R^4$ Apr 4 '20 at 16:47
• 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. Apr 4 '20 at 17:31
• Would Mathematics be a better home for this question? Apr 4 '20 at 18:03
• I don't need too much mathematical explanation using technical terms.What I need is something geometrical or physical explanation@Qmechanic Apr 4 '20 at 18:06
• 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? Apr 4 '20 at 20:02

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).

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.

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.

• So manifold of $SO (3)$ is embedded in a space of how many dimensions?@Cinaed Simson Apr 5 '20 at 4:33
• @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. Apr 5 '20 at 5:46
• 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 Apr 5 '20 at 6:22
• @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. Apr 5 '20 at 19:27