How to probe the group SU(2) is simply connected?

[wdlang]

why group SO(3) is not

any good reference on the relation of SU(2) and SO(3)?

 

[Fredrik]

You can prove that SU(2) is homeomorphic to a 3-sphere, like this:

Write down the most general complex 2×2 matrix U and find out what relationships between its components you can derive from the conditions $U^\dagger U=I$ and $\det U=1$. You should end up with the condition that defines the unit 3-sphere.

The relationship between SO(3) and SU(2) can be found by first noting that $$\mathbb R^3$$ is isomorphic to the 3-dimensional real vector space of complex 2×2 traceless self-adjoint matrices, and then showing that if X is a member of that space, and U is a member of SU(2), then

$$X\mapsto UXU^\dagger$$

is a proper rotation, i.e. a member of SO(3). Since you can change the sign of the U without changing the result, there are two members of SU(2) for each member of SO(3).

The first thing you should know is that any 2x2 traceless self-adjoint matrix can be written as

$$\begin{pmatrix}x_3 & x_1-ix_2\\ x_1+ix_2 & -x_3\end{pmatrix}=x_i\sigma_i$$

so the set of Pauli spin matrices is just a basis of the (real) vector space of (complex) 2x2 traceless self-adjoint matrices.

Not sure what the best reference is if you don't want to figure out the details for yourself. I think Weinberg's QFT book covers this pretty well (vol. 1, chapter 2), but he's actually doing it to find the relationship between SO(3,1) and SL(2,C), so he's doing essentially the same thing with the "traceless" condition dropped, and U not necessarily unitary. This brings a fourth basis vector into the picture: the 2×2 identity matrix.

 

[wdlang]

You can prove that SU(2) is homeomorphic to a 3-sphere, like this:


The relationship between SO(3) and SU(2) can be found by first noting that $$\mathbb R^3$$ is isomorphic to the 3-dimensional real vector space of complex 2×2 traceless self-adjoint matrices, and then showing that if X is a member of that space, and U is a member of SU(2), then

$$X\mapsto UXU^\dagger$$

is a proper rotation, i.e. a member of SO(3). Since you can change the sign of the U without changing the result, there are two members of SU(2) for each member of SO(3).


Not sure what the best reference is if you don't want to figure out the details for yourself. I think Weinberg's QFT book covers this pretty well (vol. 1, chapter 2), but he's actually doing it to find the relationship between SO(3,1) and SL(2,C), so he's doing essentially the same thing with the "traceless" condition dropped, and U not necessarily unitary. This brings a fourth basis vector into the picture: the 2×2 identity matrix.

yes, thanks a lot

your explanation is very good

though the argument is presented in many books, your interpretation tells the inside