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In the literature, is there a non-commutative analogue of the fact that every Riemannian manifold whose isometry group has sharp dimension must be a constant curvature manifold?

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    $\begingroup$ What do you mean by "sharp dimension"? Do you mean that $M$ is $n$-dimensional and $dim(Isom(M,g))=n(n+1)/2$? Such manifolds need not be flat, they are manifolds of constant curvature. $\endgroup$
    – Moishe Kohan
    Commented Mar 14, 2020 at 3:48
  • $\begingroup$ I mean so. Tes you are right I revise the question $\endgroup$
    – Ali Taghavi
    Commented Mar 14, 2020 at 4:11
  • $\begingroup$ @MoisheKohan thanks for your helpful comment. $\endgroup$
    – Ali Taghavi
    Commented Mar 14, 2020 at 16:00

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