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.

  • $\begingroup$ 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! $\endgroup$ – user7028 Nov 22 '10 at 18:27
  • 1
    $\begingroup$ A hyperkähler manifold is Ricci-flat Kähler in a variety of ways -- that variety being $\mathbb{CP}^1$. $\endgroup$ – José Figueroa-O'Farrill Nov 22 '10 at 21:41

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