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Reading about Sasakian manifolds one come across two slogans:

A) "A Sasakian manifold is an odd-dimensional analogue of a Kahler manifold."

B) "A Sasakian manifold sits between two Kahler manifolds - one above and one below."

I would like to understand the second slogan for the motivating example of the three sphere $S^3$. What are the two Kahler manifolds that it sits between? I would guess that below is the projective line $\mathbb{CP}^1$. But I cannot guess what lies above.

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I will answer your question for $S^{2n+1}$, since there is no difference between the case $n=1$ and the case of general $n$.

Let $(M^{2n+1},g,\theta)$ be a Sasakian manifold. One definition of a Sasakian manifold is that its metric cone is Kähler; this is the "one above". Here the metric cone is the manifold $M\times(0,\infty)$ with metric $dt^2+t^2g$. Thus in the case of $S^{2n+1}$, the metric cone is $\mathbb{C}^{n+1}\setminus\{0\}$ with the flat metric (written in spherical coordinates).

Taking the quotient by the $S^1$-action on $(M^{2n+1},\theta)$ determined by the Reeb vector field gives the "Kähler manifold below". For $S^{2n+1}$, the $S^1$-action is scalar multiplication by $e^{i\phi}$ (regarding $S^{2n+1}$ as the unit sphere in $\mathbb{C}^{n+1}$), so the quotient is $\mathbb{C}P^n$.

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  • $\begingroup$ Will the metric cone be non-compact in general? $\endgroup$
    – Fofi Konstantopoulou
    Commented Apr 29, 2020 at 20:32
  • $\begingroup$ Does this not work for n =0? S1 x R does not seem to be homeomorphic to C. $\endgroup$
    – Asvin
    Commented Apr 29, 2020 at 22:24
  • $\begingroup$ @Asvin I think it should be $[0,\infty)$, with $M \times 0$ quotiented out. This part goes to the origin. $\endgroup$
    – Ryan
    Commented Apr 29, 2020 at 23:26
  • $\begingroup$ I see. That makes more sense, thanks! $\endgroup$
    – Asvin
    Commented Apr 29, 2020 at 23:39
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    $\begingroup$ The usual definition does not include $0$ in the second factor, the reason being that $(M\times[0,\infty)/\sim,dt^2+t^2g)$ is not smooth at the origin unless $M$ is a round sphere (here $\sim$ identifies $M\times\{0\}$ to a point). I have edited the original answer to reflect that the origin is missing. Also, the metric cone is always noncompact, because of the second factor. $\endgroup$
    – Jeffrey Case
    Commented Apr 29, 2020 at 23:45

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