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May 25, 2022 at 13:54 comment added Ben McKay The proof surely uses the Bochner technique: comparing two different Laplace-like operators, differing by Ricci, and uses the vanishing of Ricci on the hyperkaehler.
May 25, 2022 at 13:17 history edited Nico Berger
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May 24, 2022 at 14:53 comment added Nico Berger Thank you! In "Calabi-Yau Manifolds and Related Geometries" by Gross, Huybrechts, Joyce, Prop. 23.3, one wants to show that $H^0(X, \Omega_X^k)$ is one-dimensional, if $k$ even, and vanishes otherwise for $X$ hyperkähler. The proof says "Any form in $H^0(X, \Omega_X^2)$ is parallel" and later that for $k>2$ one proceeds analogously. From the holonomy principle one can deduce the assertion about the dimension for parallel sections. I thought one uses the claim I mentioned in my question to deduce the desired properties of $H^0(X, \Omega_X^k)$. Is the claim maybe true for Calabi-Yau manifolds?
May 24, 2022 at 14:39 comment added Michael Albanese As stated, your claim is false. For example, the existence of a non-zero parallel form $\alpha \in H^0(X, \Omega^n_X)$ implies that the holonomy is contained in $SU(n)$, but this is only true for Calabi-Yau manifolds.
May 24, 2022 at 13:57 history edited Nico Berger CC BY-SA 4.0
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May 24, 2022 at 13:38 history asked Nico Berger CC BY-SA 4.0