1
$\begingroup$

Let $(M, J, \omega)$ be a compact Kähler manifold. Let $\phi:M\to M$ is an orientation-preserving isometric involution.

Given a point $p\in M$ must there exist a decomposition $T_pM=\oplus_i W_i$ with each $W_i$ being preserved by both $J$ and $\phi^*J$ such that $J|_{W_i}=\pm \phi^*J|_{W_i}$?

$\endgroup$

1 Answer 1

2
$\begingroup$

No. Here is a construction.

It is not hard to see that there is an orientation-preserving isometry $L:\mathbb{H}\to\mathbb{H}$ (where $\mathbb{H}\simeq\mathbb{R}^4$ is the ring of quaternions) such that $L^2=1$, namely, $$ L(x) = \tfrac12 (j+k)\,x\,(j+k). $$ This isometry satisfies $L(jx)=kL(x)$.

Now let $\Lambda\subset\mathbb{H}$ be a lattice that is preserved by $L$ (for example, it could be generated by lattices in the two $2$-dimensional eigenspaces of $L$), and let $M = \mathbb{H}/\Lambda$. Then $L$ induces an orientation-preserving isometric involution $\phi$ of $M$.

Moreover, left multliplication by $j$ (respectively, $k$) defines a translation-invariant orthogonal complex structure $J$ (respectively, $K$) on $\mathbb{H}$ that descends to $M$, and, by the equation $L(jx)=kL(x)$, we have $\phi^*(J) = K$.

Note that, with the quotient flat metric $g$ on $M$, we have that $(M,J,g)$ and $(M,K,g)$ are Kähler manifolds and $\phi$ is an orientation-preserving isometric involution of $g$.

For any tangent vector $v$, the vectors $v$, $Jv$, and $Kv$ are linearly independent, hence there are no $2$-dimensional subspaces of the tangent space at any point that are both $J$ and $K$ invariant. Thus, the only possibility would be to take $W_i = T_pM$, but $J\not=\pm K$ on $T_pM$ for any $p\in M$.

$\endgroup$
2
  • $\begingroup$ is a counterexample necessarily hyperkähler? $\endgroup$
    – user164740
    Commented Sep 7, 2020 at 20:19
  • 1
    $\begingroup$ @JoeT: Well, no, because you can take the product of a hyperKähler counterexample with non-hyperKähler example for which there is such a decomposition in such a way that the product is neither hyperKähler nor does it have a decomposition of the kind you are asking for. If $M^n$ is holonomy-irreducible, then the only possibility for a counter-example to your decomposition is that it be hyperKähler, since, if it is holonomy-irreducible and Kähler but not hyperKähler, then the holonomy is either $\mathrm{U}(n)$ or $\mathrm{SU}(n)$ and, in that case, the only possibility is $\phi^*(J)=\pm J$. $\endgroup$ Commented Sep 7, 2020 at 20:32

You must log in to answer this question.