# $S^3$ as a Sasakian Manifold

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.

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 $$\mathbb{C}P^n$$.
• @Asvin I think it should be $[0,\infty)$, with $M \times 0$ quotiented out. This part goes to the origin. – Ryan Apr 29 '20 at 23:26
• 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. – Jeffrey Case Apr 29 '20 at 23:45