2
$\begingroup$

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 $\deg S$.

  • 0
    I fixed your displayed environments. You should use either double dollar signs `$$...$$` or double the backslashes `\\[...\\]` to achieve what you wanted.2011-08-26
  • 1
    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.2011-08-26
  • 2
    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 $4$.2011-08-26
  • 1
    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.2011-08-28

1 Answers 1