11
$\begingroup$

Let $M^{2n}$ be $2n$ dimensional toric manifold over a simple polytope $P^n$. Let $\pi : M^{2n} \longrightarrow P^n$ be the orbit map of the torus action. Let $F^k$ be a $k$ dimensional face of $P^n$. Let Int$(F^k)$ be the relative interior of $F^k$.

How is $\pi |_V : V \longrightarrow$ Int$(F^k)$ a fibre bundle with fibre $(S^1)^k$? Where $V = \pi^{-1}($Int$(F^k))$

I can see why the fibre above each point of this map should be $(S^1)^k$ but I also need local triviality and to find transition functions which I am unsure how to do.

I think the anwer to this question will give me the answer to another question that I had asked earlier.

Thanks!

EDIT : as per QiaochuYuan's suggestion I have tried Ehresmann's lemma according to which I only need to show that $\pi |_V$ is a proper submersion. I have been unable to show it is a submersion. Also, To show it is proper, since Int$(F^k)$ is locally compact and Hausdorff it is enough to say inverse of singletons is compact and the map is closed. Here $\pi^{-1}$ of a point in Int$(F^k)$ is $(S^1)^k$ which is compact. But I've gone no where in proving it is closed :(

$\endgroup$
2
  • 1
    $\begingroup$ Have you tried applying Ehresmann's theorem? $\endgroup$ Jun 27, 2015 at 7:32
  • $\begingroup$ @QiaochuYuan as per your suggestion I have tried this theorem (Thank you!). According to it I only need to show that $\pi |_V$ is a proper submersion. But I am unable to do so. $\endgroup$
    – R_D
    Jun 27, 2015 at 15:53

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.