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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

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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.

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  • $\begingroup$ 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
    Commented Nov 22, 2010 at 18:27
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    $\begingroup$ A hyperkähler manifold is Ricci-flat Kähler in a variety of ways -- that variety being $\mathbb{CP}^1$. $\endgroup$
    – José Figueroa-O'Farrill
    Commented Nov 22, 2010 at 21:41

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