0
$\begingroup$

Let $C$ be the set of points $(a,b,c,d) \in \mathbb{C}^4$ which satisfy

1) $ \left|a\right|^2+\left|c\right|^2=\left|b\right|^2+\left|d\right|^2 =1 $.

2) $ a\bar{b}+c\bar{d}=0 $

There is a (component-wise) $S^1$ action on $C$ and let $S$ be the quotient ($S$ is a 3-manifold).

Is $S$ orientable or not ?

Thanks.

$\endgroup$
3
  • 2
    $\begingroup$ Isn't $S=U(2)/U(1)$? $\endgroup$ Jan 30, 2012 at 21:51
  • $\begingroup$ So then $S=S^3$, which is orientable. $\endgroup$ Jan 30, 2012 at 21:58
  • $\begingroup$ OH you are right; Indeed it is! Peculiar ! For me it was the set of mobius transformations that commute with $c(z)=\frac{-1}{\bar{z}}$ and writing $c$ as $c[z,w]=[-\bar{w},\bar{z}]$, we can see where does $U(2)$ come from. $\endgroup$ Jan 30, 2012 at 22:08

1 Answer 1

5
$\begingroup$

Identifying your points with matrices $M$ with column vectors $(a,b)^T$ and $(c,d)^T$, your equations come from the components of $M M^\dagger=I$ where $M^\dagger$ denotes the conjugate transpose. So $C$ is $U(2)$ and $S$ is $SU(2)$. Since $SU(2)$ is diffeomorphic to $S^3$, it is orientable.

$\endgroup$
3
  • $\begingroup$ so that's what Muro said above. you did not need to duplicate it. $\endgroup$ Jan 31, 2012 at 0:38
  • 2
    $\begingroup$ Yes, but it hadn't been answered yet. Also, it is unclear to me which appeared first, although I gather you received notifications regarding the comments. Now at least the question appears as answered. $\endgroup$ Jan 31, 2012 at 0:51
  • $\begingroup$ @Mohammad Muro's comment wasn't visible to me when I posted. $\endgroup$
    – Adam
    Jan 31, 2012 at 16:51

Not the answer you're looking for? Browse other questions tagged or ask your own question.