Compute the degree of map Let $S:SU(2)\rightarrow SU(2)$ be defined as $S(X)=X^{4}$. Compute
the degree of $S$.
Now, $SU(2)$ is homeomorphic to $S^{3}$, so the degree can be taken
as:$$
\int_{S^{3}}S^{*}\omega=(\deg S)\int_{S^{3}}\omega$$
where $\omega$ is a nontrivial 3-form on $H^{3}(S^{3})$. Explicitly,
I know that this is: $$
\omega=\sum_{i=0}^{k}(-1)^{i}x^{i}dx^{0}\wedge\dots\wedge dx^{i-1}\wedge dx^{i+1}\wedge\dots\wedge dx^{k}$$
where $k=3$. Now, $\int_{S^{3}}\omega$ is equal to $4$ times the
volume of $B^{4}$. I am having some trouble computing $\int_{S^{3}}S^{*}\omega$
which would enable me to find $\deg S$.
 A: The left-invariant metric on $SU(2)$ is defined as $ds^2 = -2 \operatorname{Tr}( g^{-1} \mathrm{d} g \cdot g^{-1} \mathrm{d} g )$. 
Let us choose the following group element parameterization:
$$
   g(\theta, \psi, \phi) = \exp( \psi \tau_3 ) \exp( \theta \tau_2 ) \exp( \phi \tau_3 )
$$
where
$$
  \tau_1 = \left(
\begin{array}{cc}
 0 & -\frac{i}{2} \\
 -\frac{i}{2} & 0 \\
\end{array}
\right) \quad\quad 
   \tau_2 = \left(
\begin{array}{cc}
 0 & -\frac{1}{2} \\
 \frac{1}{2} & 0 \\
\end{array}
\right) \quad\quad
  \tau_3 = \left(
\begin{array}{cc}
 -\frac{i}{2} & 0 \\
 0 & \frac{i}{2} \\
\end{array}
\right)
$$
Matrices $\tau_i$ are realizations of $\mathfrak{su}(2)$ Lie algebra, i.e. $\left[ \tau_1, \tau_2 \right] = \tau_3$ and cyclic variants of that. With this representation
$$
   g(\theta, \psi, \phi) = \left(
\begin{array}{cc}
 \cos \left(\frac{\theta }{2}\right) e^{-\frac{1}{2} i (\psi +\phi )} & \sin
   \left(\frac{\theta }{2}\right) \left(-e^{\frac{1}{2} i (\phi -\psi )}\right) \\
 \sin \left(\frac{\theta }{2}\right) e^{-\frac{1}{2} i (\phi -\psi )} & \cos
   \left(\frac{\theta }{2}\right) e^{\frac{1}{2} i (\psi +\phi )} \\
\end{array}
\right)
$$
Here $0 \le \theta 2\pi$, $0 \le \phi, \psi < 4 \pi$. Some algebra leads to 
$$
   \mathrm{d}s^2 = \mathrm{d} \theta^2 + \mathrm{d} \phi^2 + \mathrm{d} \psi^2 + 2 \cos \theta \mathrm{d} \psi \mathrm{d} \psi = h_{i,j} \mathrm{d}x^i \otimes \mathrm{d}x^j
$$
Then the volume element is $\mathrm{d}V = \sqrt{\det h} \cdot \mathrm{d} \theta \, \mathrm{d}\psi \, \mathrm{d}\psi = \vert \sin\theta \vert \cdot \mathrm{d} \theta \, \mathrm{d}\psi \, \mathrm{d}\psi$. The volume of $SU(2)$ then is 
$$
   \int_0^{2 \pi} \mathrm{d} \theta \int_0^{4 \pi} \mathrm{d} \phi \int_0^{4 \pi} \mathrm{d} \psi \cdot \vert \sin \theta \vert = (4 \pi)^2
$$
We can now repeat this same process using 
$$
g^{-4} \cdot \mathrm{d} g^4 = \sum_{k=0}^3 g^{-k} ( g \cdot \mathrm{d} g) g^k
$$ 
The algebra gets quite more involved, with 
$$
   \sqrt{ \det \tilde{h} } = 8 \vert \sin \left(\frac{\theta }{2}\right) \cos ^3\left(\frac{\theta }{2}\right) \cos
   ^2 \left(\frac{\psi +\phi }{2}\right) k(\theta, \phi, \psi) \vert
$$
where $k(\theta, \phi, \psi) = (\cos (\theta -\psi -\phi )+\cos (\theta
   +\psi +\phi )+2 \cos (\theta )+2 \cos (\psi +\phi )-2)^2$.
Carrying out the integration
$$
\int_0^{2 \pi} \mathrm{d} \theta \int_0^{4 \pi} \mathrm{d} \phi \int_0^{4 \pi} \mathrm{d} \psi \cdot \sqrt{ \det \tilde{h}} = 64 \pi^2 = ( 8 \pi )^2
$$
As the consequence, according to your formula, 
$$
\deg S = \frac{(8 \pi)^2}{(4 \pi)^2} = 4
$$
So it turned out an heavy lifting exercise which confirm intuition behind Ryan's answer in comments.
