Are Chern classes always vertical?

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?

• 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? Apr 12 at 23:07
• @dave-benson: I thought that for $M$ a Kähler manifold $H^2$ is always non-zero? Apr 13 at 7:55
• 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, $\mathbb{C}^n$. Apr 13 at 13:12

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 $$H^2(M,\mathbb{R})^{\otimes 2}$$.

• Great, thank you! I believe this answers my question. I'm still wondering though if anything can be said for compact Calabi-Yau manifolds. Apr 13 at 13:27
• 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.
– abx
Apr 14 at 5:27
• 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? Apr 14 at 10:29
• 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 $\dim(M)=pq\geq 6$.
– abx
Apr 14 at 11:49
• 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.
– abx
Apr 14 at 14:16

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.