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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.

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closed as too localized by Ryan Budney, Bruno Martelli, Andy Putman, Daniel Moskovich, Bill Johnson Jan 31 '12 at 14:27

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2  
Isn't $S=U(2)/U(1)$? –  Fernando Muro Jan 30 '12 at 21:51
    
So then $S=S^3$, which is orientable. –  Ryan Budney Jan 30 '12 at 21:58
    
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. –  Mohammad F. Tehrani Jan 30 '12 at 22:08

1 Answer 1

up vote 5 down vote accepted

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.

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so that's what Muro said above. you did not need to duplicate it. –  Mohammad F. Tehrani Jan 31 '12 at 0:38
2  
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. –  Sean Tilson Jan 31 '12 at 0:51
    
@Mohammad Muro's comment wasn't visible to me when I posted. –  Adam Jan 31 '12 at 16:51

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