# Decomposing the Fubini-Study metric - Irreducibility of $\mathbb{CP}^n$

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?

• 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. Sep 6, 2021 at 8:18
• @user1729 thanks for feedback, I edited my question and filled it with more information. Sep 6, 2021 at 16:58
• @Mathy Thanks, it's much better now! Sep 6, 2021 at 17:25
• @Didier Could you explain why? Thanks! Sep 8, 2021 at 5:57