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Let $X$ be a complex compact Kähler manifold, $\Omega_X$ its cotangent bundle and $D$ the Chern connection.

It is, I believe, a standard fact that parallel $k$-forms $\alpha \in \mathcal{A}^k(X)$, i.e. $D(\alpha)=0$, are closed, see for example

https://math.stackexchange.com/questions/1672761/riemannian-manifold-alpha-in-omegapm-parallel-implies-alpha-is-close?rq=1

It is even true that parallel forms are harmonic, see for example

https://math.stackexchange.com/questions/479991/geometric-interpretation-parallel-forms-are-harmonic

I have found in a book the statement that "any form $\alpha \in H^0(X, \Omega_X^k)$ is parallel". Could someone explain why this is the case? I know that these forms are harmonic with respect to any Kähler metric. How do I show that they are also parallel?

If needed, we can assume that $X$ is simply-connected, but I would not see how that helps apart from the fact that one only has to consider $k>1$.

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    $\begingroup$ 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. $\endgroup$
    – Michael Albanese
    Commented May 24, 2022 at 14:39
  • $\begingroup$ 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? $\endgroup$
    – Nico Berger
    Commented May 24, 2022 at 14:53
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    $\begingroup$ 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. $\endgroup$
    – Ben McKay
    Commented May 25, 2022 at 13:54

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