# Is every 3-Sasakian a Sasakian-Einstein manifold?

Hi, a short question: Is every 3-Sasakian manifold a Sasaki-Einstein manifold? If not, do you have an example? If yes, how can I prove this?

Thanks and best regards

-

Yes, because a manifold is Sasaki-Einstein if and only if its metric cone is Ricci-flat Kähler, whereas the cone of a 3-sasakian manifold is hyperkähler.

See, for instance, Bär's "Real Killing spinors and holonomy" published in CMP.

-
Okay, I thought this... If someone is hyperkähler, is this Ricci-flat Kähler, too? What I mean: What is the connection between hyperkähler and Ricci-flat Kähler? Thanks! –  user7028 Nov 22 '10 at 18:27
A hyperkähler manifold is Ricci-flat Kähler in a variety of ways -- that variety being $\mathbb{CP}^1$. –  José Figueroa-O'Farrill Nov 22 '10 at 21:41