5
$\begingroup$

I was wondering whether $(\mathbb{CP}^n, g_{FS})$, where $g_{FS}$ denotes the Fubini-Study metric, is holonomy irreducible, that is, whether or not the tangent space can be split up into subspaces invariant under the holonomy action. I computed that the holonomy group is $U(n)$. My line of thinking was the following:

If $(\mathbb{CP}^n, g_{FS})$ would not be irreducible, then the de Rham Theorem would imply (because $(\mathbb{CP}^n, g_{FS})$ is simply connected and complete) that $(\mathbb{CP}^n, g_{FS})$ is globally a product, which in turn implies that its holonomy group is a product.

So there are two things that might go wrong: Either $(\mathbb{CP}^n, g_{FS})$ can not be a product manifold, that is, either $\mathbb{CP}^n$ can not even topologically be or the metric $g_{FS}$ can not split; or $U(n)$ can not be split into a product.

But $U(n)$ is reducible, so the latter might be possible. What I am left with is the first option and I have currently no idea how to show that that's not possible. So my question is:

Why can the Fubini-Study metric $g_{FS}$ not be realized as a product metric, $h_1 + h_2 = g$? Or: Why can $\mathbb{C}\mathbb{P}^n$ not be written as a product?

$\endgroup$
4
  • 2
    $\begingroup$ Your two questions are not equivalent: for $(\mathbb{C}P^n,g_{FS})$ to be a Riemannian product is stronger than for $\mathbb{C}P^n$ to be a product manifold. Both statements are false. $\endgroup$
    – Didier
    Sep 6, 2021 at 8:18
  • 1
    $\begingroup$ @user1729 thanks for feedback, I edited my question and filled it with more information. $\endgroup$
    – Mathy
    Sep 6, 2021 at 16:58
  • 1
    $\begingroup$ @Mathy Thanks, it's much better now! $\endgroup$
    – user1729
    Sep 6, 2021 at 17:25
  • $\begingroup$ @Didier Could you explain why? Thanks! $\endgroup$
    – Mathy
    Sep 8, 2021 at 5:57

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.