Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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

share|improve this question
add comment

1 Answer

up vote 2 down vote accepted

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.

share|improve this answer
    
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
1  
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
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.