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Suppose there is a Hermitian symmetric space of compact type $X$. It is realized in the following way: $X\hookrightarrow\mathbb{P}^N$ and equipped with the induced Fubini-Study metric $g$.

What's more, the isometry group $G$ of $X$ is a compact subgroup of the unitary group of $\mathbb{P}^N$.

Let $K$ denote the isotropy group. Assume $G/K\cong X$, and $G,K$ are the standard pair for Hermitian symmetric space.

My question: is $g$ is Kähler-Einstein?

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    $\begingroup$ Are you assuming that $X$ is irreducible, i.e., that $G$ is a simple group? If so, then, yes, it must be Kähler-Einstein, since there is, up to constant multiples, only one $G$-invariant metric on $X$, and, if it were not Kähler-Einstein, then $g + \epsilon\mathrm{Ric}(g)$ for $\epsilon>0$ and small, would be another $G$-invariant metric that was not a multiple of $g$. $\endgroup$
    – Robert Bryant
    Commented Oct 29, 2015 at 21:29
  • $\begingroup$ Yes! I am assuming it's irreducible! $\endgroup$
    – user42804
    Commented Oct 29, 2015 at 21:54
  • $\begingroup$ Why there is only one G-invariant metric on X? $\endgroup$
    – user42804
    Commented Oct 29, 2015 at 21:57
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    $\begingroup$ Because $K$ acts irreducibly on the tangent space to $G/K$ at the identity coset (this is a consequence of the irreducibility). Thus, there can be only one $K$-invariant inner product on the tangent space up to constant multiples, and that gives a unique $G$-invariant metric on $G/K$ (up to constant multiples). $\endgroup$
    – Robert Bryant
    Commented Oct 30, 2015 at 3:02
  • $\begingroup$ There is a fairly small list of irreducible Hermitian symmetric spaces of compact type. In algebraic geometry (e.g., positive characteristic), these are the same as the cominuscule projective homogeneous varieties. In particular, the orthogonal Grassmannian in the Pasquier-Perrin examples is not cominuscule (as it must not be, thanks to the Hwang-Mok rigidity theorem). $\endgroup$
    – Jason Starr
    Commented Nov 2, 2015 at 2:41

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