1
$\begingroup$

I want to construct free $S^1$ action on $\mathbb{R}P^n$ and $\mathbb{C}P^n$.

For $n=2m-1$, consider $S^n ⊂ C^m$. Then $S^1$ freely act on $S^n$ by $(ξ, (z_1 , z _2 , · · · , z _m )) → (ξz_1 , ξz_2 , · · · , ξz_m )$ where $ξ ∈ S^1 $. Then $S^1/\mathbb{Z}_2\cong S^1$ will act freely on $\mathbb{R}P^n = S^n /\mathbb{Z}_2$.

I want to know what are all other free $S^1$ action on $\mathbb{R}P^n$?

Also How to construct a free $S^1$ action on $\mathbb{C}P^n$? Thank you in advance...

$\endgroup$
8
  • 10
    $\begingroup$ The Euler characteristic of $RP^{2n}$ and $CP^n$ are nonzero, so every vector field vanishes somewhere. So they have no free $S^1$-action. It remains to classify free $S^1$-action on $S^{2m-1}$. $\endgroup$ – YCor Nov 10 '17 at 9:02
  • $\begingroup$ @YCor Thank you.I did not know this fact. It will be very helpful if you kindly elaborate your answer. $\endgroup$ – Shivani Sengupta Nov 10 '17 at 9:39
  • 2
    $\begingroup$ See en.wikipedia.org/wiki/Poincare-Hopf_theorem. It works in the smooth case. I guess it's still true in the topological context that a connected topological manifold (or even a finite CW-complex?) with a free continuous action of the circle group has zero Euler characteristic, but this requires another argument. $\endgroup$ – YCor Nov 10 '17 at 12:48
  • $\begingroup$ @YCor Sorry I didn’t understand how vanishing of every vector field is related to non existence of free S^1 action? $\endgroup$ – Shivani Sengupta Nov 10 '17 at 14:01
  • $\begingroup$ Let $f:(t,x)\mapsto tx=f(t,x)$ be the free action, $t\in S^1$, $x\in X$, and consider the vector field $x\mapsto\frac{\partial}{\partial t}f(t,x)$. $\endgroup$ – YCor Nov 10 '17 at 14:28

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.