I'm a physicist trying to make sense of some topics of group theory and their applications to quantum mechanics. Let me state the things I think I understand
Special Orthogonal Group $SO(3)$
This is the group of orthogonal rotations with unit determinant in 3-dimensions. We can parametrize them with a unit vector $\hat{n}$ and an angle $\theta=[-\pi,\pi]$ where $\hat{n}$ tells us the rotation axis and $\theta$ the rotation angle as long as we identify rotations with $\theta=\pi$ and $\theta=-\pi$. In other words, the group manifold is a 3-dimensional ball where antipodal points on the surface are identified.
Because we can parametrize the unit vector $\hat{n}$ with two other angles $\phi$ and $\psi$ that are coordinates on the 2-sphere $S_2$, the group manifold of $SO(3)$ is also parametrized by the three angles $(\phi,\psi,\theta)$ parametrizing a 3-sphere, as long as we only consider one hemisphere (since antipodal points are still identified).
Special unitary group $SU(2)$
This is the group of unitary matrices with unit determinant in 2-dimensions. We can parametrize its elements as
\begin{bmatrix}\alpha&\beta\\-\beta^*&\alpha^*\end{bmatrix}
such that $|\alpha|^2+|\beta|^2=1$. parametrizing this with $\alpha=a+ib$ and $\beta=c+id$ we get the manifold with $a^2 +b^2 +c^2 +d^2=1$ which is just a 3-sphere.
In physics, it's common to parametrize $SU(2)$ rotations in exponential form using Pauli matrices as
\begin{equation} U(\vec{\theta})=e^{i\frac{1}{2}\vec{\sigma}\cdot \vec{\theta}} \end{equation}
where $\vec{\sigma}=(\sigma_1,\sigma_2,\sigma_3)$ are the usual Pauli matrices and $\vec{\theta}=(\theta_1,\theta_2,\theta_3)$ are three angles parametrizing how much we are rotating in each direction.
Here's where I'm confused. First, because this is supposed to be the effect of physical rotations on spinors, I'd think that the angles $\theta$ range in the usual way from $-\pi$ to $\pi$. However, it's easy to see that plugging $\theta=2\pi$ for any of the angles in $\vec{\theta}$ we get $U=-\mathbb{I}$, which can't be reached if we only let $\theta$ range from $-\pi$ to $\pi$. This makes me think that the manifold can be parametrized with angles ranging from $-2\pi$ to $2\pi$. However, this brings me to my second problem: I still have in the back of my mind that I will want to take the quotient group $SU(2)/Z_2$ and show that the manifold we get is the one for $SO(3)$. However, $SO(3)$ has radius $\pi$, not $2\pi$!
In short, my question is: Given the exponential representation of elements of $SU(2)$ that I wrote above, what is the range of the coordinates $\vec{\theta}$?
My try at a solution
I think that the elements in $SU(2)$ can be parametrized as
\begin{equation} U=e^{i\frac{1}{2}\vec{\sigma}\cdot\hat{n}\theta} \end{equation}
where $\hat{n}$ is a unit vector and $\theta=[-2\pi,2\pi]$ measures the rotation. Here we identify elements $\theta+4\pi=\theta$. I'm already not convinced with this, since we know that the manifold of $SU(2)$ is a 3-sphere, which can be parametrized by one angle ranging from $0$ to $2\pi$ and another two angles ranging from $0$ to $\pi$. two of those can parametrize $\hat{n}$ and the last one can only go from $0$ to $\pi$, not to $2\pi$. Forgetting about this for a moment, when we do the quotient group $SU(2)/Z_2$ we are identifying any elements on $SU(2)$ that differ by multiplication by an element of $Z_2=\{\mathbb{I},-\mathbb{I}\}$. That means that any two elements parametrized by the same $\hat{n}$ and angles $\theta_1$ and $\theta_2$ that differ in $2\pi$ should be identified as the same. We therefore get a manifold parametrized by a unit vector $\hat{n}$ and an angle $\theta=[-\pi,\pi]$ with elements differing in $2\pi$ idenfitied. Does this make sense?
My problem with this solution is also that if $SU(2)$ is parametrized in this way (that is, with a 3-dimensional ball of radius $2\pi$ and opposite points on its surface identified), then it wouldn't be simply connected!
I think I figured it out!
Indeed, any element $U\in SU(2)$ can be represented as
\begin{equation} U=e^{i\frac{1}{2}\vec{\sigma}\cdot \hat{n}\theta} \end{equation}
where $\hat{n}$ is a 3-dimensional unit vector and $\theta$ is an angle where we identify $\theta+4\pi=\theta$. Therefore the manifold can be thought of as a 3-dimensional ball of radius $2\pi$ where the surface is identified with the element $-\mathbb{I}$. This should be isomorphic to $S_3$.
Concerns about $SU(2)$ not being simply connected
My concern about this not being simply connected was funded by the usual construction one does for $SO(3)$ where one builds a curve that starts at the identity, goes all the way to the surface of the manifold, comes on the antipodal point and ties it back to the identity. However, this is not a problem for $SU(2)$ because the manifold of $SU(2)$ doesn't have just an antipodal map, instead, every element at the surface $\theta=2\pi$ is the element $-\mathbb{I}$: We identify the whole surface as one element! So if we build a curve that starts at the identity, goes to the surface, comes on the antipodal point and goes back to the identity, we can easily deform it to a single point since every point in the surface is the same element.
My concern about the quotient group
The quotient group $SU(2)/ Z_2$ takes the manifold parametrized by a 3-dimensional unit vector $\hat{n}$ and an angle $\theta=[0,2\pi]$ and identifies any elements with the same $\hat{n}$ that differ by $2\pi$. That creates a manifold that is essentially two times the manifold of $SO(3)$! Which is the reason it's called a double cover! It starts at the identity, then builds a ball of radius $\pi$ with the antipodal points identified, and then it keeps going to radius $2\pi$ when it's the identity again! Since the manifold of $SO(3)$ can be thought of as a ball of radius $\pi$ with antipodal points on its surface identified, it's clear that the manifold of $SU(2)/Z_2$ is just twice this.