# Deformation invariance of Chern classes

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?

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

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 $$c_i(T_{\mathscr{X}})$$.

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