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A (Riemnnian) symmetric space is locally characterized by the first covariant derivative of its curvature tensor, namely, a manifold $M$ is locally a symmetric space if and only if $\nabla R=0$ (where $R$ stands for the Riemannian curvature tensor).

Is there a similar characterization for homogeneous manifolds?

More specifically, I wonder if the condition $(\nabla_XR)(Y\,X)X=0$ (for all $X,Y\in TM$) is sufficient.

I apologize if it is well-known.

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    $\begingroup$ See page 413, Theorem 2.1 here : seminariomatematico.unito.it/rendiconti/cartaceo/50-4/411.pdf $\endgroup$
    – Holonomia
    Commented Feb 2, 2016 at 12:02
  • $\begingroup$ @Holonomia : Thank you for the comment. This is certainly an answer to the question, but I still wonder if the equality above makes any sense. $\endgroup$
    – Llohann
    Commented Feb 2, 2016 at 12:11
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    $\begingroup$ It seems to me that the equality above is closer to properties (C) and (P) in Theorem 1, page 59 here ac.els-cdn.com/092622459290009C/… $\endgroup$
    – Holonomia
    Commented Feb 2, 2016 at 12:32
  • $\begingroup$ @Holonomia Since I found the answeres in through your comments, wouldn't you like to write it down as an answer so I could accept it? (The identiy I was seeking makes the space symmetric as in Lemma 5.1 of link.springer.com/article/10.1007%2FBF01457081) $\endgroup$
    – Llohann
    Commented May 31, 2016 at 12:18
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    $\begingroup$ I am happy my comments were helpful for you. It is not necessary to write my answer down. Your comment it is enough to me. $\endgroup$
    – Holonomia
    Commented Jun 1, 2016 at 13:37

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