# 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 $\deg S$.

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I fixed your displayed environments. You should use either double dollar signs $$...$$ or double the backslashes \$...\$ to achieve what you wanted. – t.b. Aug 26 '11 at 20:44
Is your goal to know the degree, or is your goal to compute the integral? I think there's more efficient ways to compute the degree that avoid computing the integral. – Ryan Budney Aug 26 '11 at 21:08
For example, consider $S^{-1}(A)$ for a generic $A \in SU(2)$. Except for some sparse exceptional cases, this is always a four element set. So the degree is even but I believe you can check the signs are all the same, so the degree should be $4$. – Ryan Budney Aug 26 '11 at 21:11
The degree of the $n$th power map in a torus $T$ is clearly $n$, so most elements in $T$ have $n$ preimages. Now a compact Lie group $G$ is a union of tori, so most elements in $G$ have $n$ $n$th roots: so Ryan's argument "works" for all compact Lie groups. – Mariano Suárez-Alvarez Aug 28 '11 at 5:12

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.

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@kobe24 Please see my rewritten answer. – Sasha Aug 27 '11 at 1:48