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Let $c_k \in H^{2k}(M, \mathbb{Z})$ be the $k$-th Chern class of the tangent bundle of a Hermitian manifold $M$. Is $c_k$ necessarily vertical, i.e. $$ c_k = \sum_{i_1,\dots, i_{k}} \alpha_{i_1 \dots i_{k}} \, \omega^{i_1} \wedge \dots \wedge \omega^{i_k} $$ for $\omega^i$ a basis of $H^{2}(M, \mathbb{Z})$ and some suitable coefficients $\alpha_{i_1 \dots i_{k}}$?

If this is not true in general, does it hold for Kähler (or Calabi-Yau) manifolds?

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    $\begingroup$ Aren't there loads of manifolds of pretty much whatever type you mention, for which degree two cohomology is zero, but for which there are non-zero Chern classes for the tangent bundle? $\endgroup$ Apr 12 at 23:07
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    $\begingroup$ @dave-benson: I thought that for $M$ a Kähler manifold $H^2$ is always non-zero? $\endgroup$
    – Severin
    Apr 13 at 7:55
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    $\begingroup$ For a compact Kähler manifold $H^2$ is nonzero, because its symplectic structure gives a nontrivial class, but for noncompact Kähler manifolds it can be zero: for example, $\mathbb{C}^n$. $\endgroup$
    – John Baez
    Apr 13 at 13:12

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For a counter-example (with real coefficients), take for $M$ the Grassmannian $\mathbb{G}(p,p+q)$ with $p\neq q$, and $p,q\geq 2$. If I computed correctly: $$c_2(M) = \frac{1}{2}\left[(p-q)^2-(p-q)+2\right] c_1^2-(p-q)c_2\, , $$where $c_1,c_2$ are the Chern classes of the tautological quotient bundle. Now $H^2(M,\mathbb{R})$ is one-dimensional, generated by $c_1$. Since $c_2$ and $c_1^2$ are linearly independent in $H^4(M,\mathbb{R})$, $ c_2(M)$ does not belong to the image of $H^2(M,\mathbb{R})^{\otimes 2}$.

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  • $\begingroup$ Great, thank you! I believe this answers my question. I'm still wondering though if anything can be said for compact Calabi-Yau manifolds. $\endgroup$
    – Severin
    Apr 13 at 13:27
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    $\begingroup$ Just take for $M$ the intersection of $\mathbb{G}(p,n)$ with a general hypersurface of degree $n$ in the Plücker embedding. By Lefschetz the restriction maps on $H^2$ and $H^4$ are isomorphisms; $c_2(M)$ is the restriction of $c_2(\mathbb{G})$ plus some multiple of $c_1^2$, hence the same argument applies. $\endgroup$
    – abx
    Apr 14 at 5:27
  • $\begingroup$ Not sure I fully understand. If $c_1(M) = 0$, $c_1^2$ and $c_2$ won't be linearly independent in $H^4(M, \mathbb{R})$ and $c_1$ also won't generate $H^2(M, \mathbb{R})$. Or is it enough to have this in the ambient space? $\endgroup$
    – Severin
    Apr 14 at 10:29
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    $\begingroup$ I was keeping the assumptions and notation of my answer: $c_1,c_2$ are the Chern classes of the tautological quotient bundle, and not the Chern classes of $M$. Also, I assume $p,q \geq 2$ and $p\neq q$, hence $\dim(M)=pq\geq 6$. $\endgroup$
    – abx
    Apr 14 at 11:49
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    $\begingroup$ Yes, I think you can take a complete intersection, e.g. of degree $(2,3)$ in $\mathbb{G}(2,5)$. The restriction map on $H^4$ is still injective, and that's what you need. $\endgroup$
    – abx
    Apr 14 at 14:16
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abx's counterexample is correct. It might be worth remembering the splitting principle, though: Let $E$ be any rank $n$ vector bundle on $M$, and let $F(E)$ be the bundle of complete flags in $E$, so $\pi : E \to M$ is a fiber bundle whose fibers are flag manifolds. Then $\pi^{\ast}(E)$ has a filtration whose subquotients are line bundles $L_1$, $L_2$, ..., $L_n$, so $\pi^{\ast}(c_k(E))$ is a polynomial in the classes $c_1(L_j) \in H^2(F(E), \mathbb{Z})$. Moreover, $\pi^{\ast} : H^{\ast}(M) \to H^{\ast}(F(E))$ is injective.

You can often use this to move any computation that you want to do over to $H^{\ast}(F(E))$, and then write the Chern class in the way you want to.

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