Let $M$ be a compact Kähler manifold. If $\phi:M\to M$ is an orientation-preserving isometric involution does it have to be either holomorphic or anti-holomorphic?
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1$\begingroup$ Let $X$ be a Kaehler manifold with an antiholomorphic involution $i$ that is isometric. On the product $X\times X$, consider the product of the identity map and $i$. $\endgroup$– Jason StarrCommented Sep 7, 2020 at 16:43
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$\begingroup$ The holonomy of the metric preserves the complex structure, since it is Kaehler. By the de Rham splitting theorem, and the Berger classification of holonomy groups, either the universal covering space splits isometrically into a product, or else the holonomy group determines the complex structure, i.e. does not sit inside two conjugates of the unitary group inside the rotation group. $\endgroup$– Ben McKayCommented Sep 7, 2020 at 16:44
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1$\begingroup$ @JasonStarr: Your construction might not preserve the orientation of the product, since $i$ might not be orientation preserving. $\endgroup$– Robert BryantCommented Sep 7, 2020 at 16:47
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$\begingroup$ @BenMcKay: Actually, even if the holonomy is irreducible, the holonomy group may not determine the complex structure. For example, if the metric is hyperKähler, then there will be at least a 2-sphere of complex structures compatible with the metric. $\endgroup$– Robert BryantCommented Sep 7, 2020 at 16:49
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1$\begingroup$ @PaulReynolds: Really? You wouldn't want the Ricci-flat metric on K3 constructed by Yau to be called 'Kähler', or the product of two Kähler metrics to be called 'Kähler'? $\endgroup$– Robert BryantCommented Sep 7, 2020 at 18:46
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No. Let $M$ be the product of three copies of $\mathbb{C}/(\mathbb{Z}[i])$ (i.e., the square torus). Give it the obvious product metric. Now consider the map $$ \phi\bigl([z_1],[z_2],[z_3]\bigr) = \bigl([z_1],[\,\overline{z_2}\,],[\,\overline{z_3}\,]\bigr). $$ This is an orientation-preserving isometry that is neither holomorphic nor antiholomorphic.
answered Sep 7, 2020 at 16:44
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1$\begingroup$ More generally, in the spirit of Jason's comment, could we take $\mathrm{id} \times f \times f$ on $X \times X \times X$, where $X$ is compact Kaehler and $f \colon X \to X$ is an anti-holomorphic isometric involution? $\endgroup$ Commented Sep 7, 2020 at 17:51
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1$\begingroup$ @FrancescoPolizzi: Yes, of course, though, perhaps just as simple: Suppose that $X$ and $Y$ are compact Kähler and that $Y$ has even complex dimension and an antiholomorphic isometric involution $f$ and take $\mathrm{id}_X{\times} f : X{\times} Y\to X{\times}Y$. $\endgroup$ Commented Sep 7, 2020 at 18:41