Let $\pi:\mathcal X\to B$ be a deformation of a compact complex manifold $X=\pi^{-1}(0)$, then for any $t\in B$, the first Chern class $c_1(X)=c_1(X_t)$?

I know the Chern class of a manifold depends on the complex structure, for example, the same diffeomorphism type of $\mathbb C^1$ with a different complex structure may have a different Chern class, but also the same differential manifold with different complex sturctures may have the same Chern class, for example, the deformations of a Calabi-Yau manifold have the same Chern class $c_1=0$ while the complex structures are different. So my question is: is the Chern class a deformation invariant?

  • $\begingroup$ I always thought Chern classes $c_i(X) \in H^{2i}(X, \mathbb Z)$ are actually topological invariants? $\endgroup$ Commented Oct 13, 2021 at 12:04

1 Answer 1


This is actually true for all Chern classes, but you must first say how you identify $H^*(X,\mathbb{Z})$ and $H^*(X_t,\mathbb{Z})$. There is no problem for small deformations, that is if $B$ is a ball (say). Then the restriction maps $H^*(\mathscr{X},\mathbb{Z})\rightarrow H^*(X,\mathbb{Z})$ and $H^*(\mathscr{X},\mathbb{Z})\rightarrow H^*(X_t,\mathbb{Z})$ are isomorphisms, so you can consider everything in $H^*(\mathscr{X},\mathbb{Z})$. But now from the exact sequence $0\rightarrow \pi ^*T_B\rightarrow T_{\mathscr{X}}\rightarrow T_{\mathscr{X}/B}\rightarrow 0$ and the triviality of $T_B$ you get that $c_i(X)$ and $c_i(X_t)$ are the restrictions of the same class, namely $c_i(T_{\mathscr{X}})$.

  • $\begingroup$ You said $c_i(X_t)$ is the restriction of $c_i(T_{\mathcal X})$, do you mean $c_i(X_t)=c_i(T_{\mathcal X}|_{X_t})$? $\endgroup$
    – Tom
    Commented Oct 14, 2021 at 8:57
  • $\begingroup$ Yes. Just restrict the exact sequence above to $X_t$. $\endgroup$
    – abx
    Commented Oct 14, 2021 at 12:41

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