5
$\begingroup$

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?

$\endgroup$
1
  • $\begingroup$ 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

8
$\begingroup$

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

$\endgroup$
2
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.