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Let $M \to B$ be a surjective morphism of smooth algebraic varieties. Is it true that the normal bundle to any fiber $F=f^{-1}(b)$, $b \in B$ is trivial of rank equal to the $\dim B$?

The idea is to think about this normal bundle as pullback of the normal bundle to $b$ which is just just the tangent space $T_b$ at the point $b$.

In more details, let's consider the following commutative diagram

$\require{AMScd} \begin{CD} F @>{i}>> M\\ @V{p}VV @V{\pi}VV \\ b @>{j}>> B \end{CD}$

The canonical map $T_M \to \pi^* T_B$ is surjective, let coherent sheaf $V$ be the kernel of this map, so $$ 0 \to V \to T_M \to \pi^* T_B \to 0. $$

Sheaf $V$ is the sheaf of all vertical vector fields and $T_F \cong i^*V$. Then using the definition of normal bundle as the cokernel of the canonical map $T_F \to i^* T_M$ i.e. $$ 0 \to T_F \to i^* T_M \to N_{F/M} \to 0, $$ we can conclude that $N_{F/M} \cong i^* \pi^* T_B \cong p^* j^* T_B \cong p^* T_b$.

Am I missing some subtle points in this argument?

Alex
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  • I think that checks out. You're really just restricting the usual exact sequence $$ 0 \to T_{M/B} \to T_M \to \pi^* T_B \to 0 $$ to a fiber $F$, which is the same as pulling it back by your $i$. But your diagram is commutative, so $\pi \circ i = j \circ p$, so $i^* \pi^* T_B = p^* j^* T_B$, and $j^* T_B = T_{B,b}$, so $p^j^T_B$ is trivial because it's the pullback by $p$ of a vector space. – Gunnar Þór Magnússon Oct 09 '16 at 17:25

1 Answers1

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This fact is OK for the general fiber (see for example this post). But for special fibers it is not true. For example blow up a smooth variety B of dimension n at a point and let $\pi:X \to B$ be the blow-up map with exceptional divisor $E$. It is well known that $E \cong {\mathbb P}^{n-1}$ and $N_{E/X} \cong {\mathcal O}_{{\mathbb P}^{n-1}}(-1)$. And in fact, in this case, the exact sequence of tangent bundles is $$0 \to T_X \to \pi^*T_B \to T_E \to 0.$$

Angelo
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  • What is correct general exact sequence that works for general fiber and blow-up example (and other such examples)? – Alex Dec 17 '22 at 16:50
  • @Alex: I do not understand what you are asking – Angelo Dec 18 '22 at 19:12
  • I think my argument works for smooth morphisms and you gave an example of non-smooth map. In that case ses for tangent bundles is replaced by long exact sequence with Andre-Quillen cohomology and that is the general exact sequence I was asking... – Alex Dec 20 '22 at 08:29
  • @Alex: yes, your argument works for smooth morphisms. As for the case of a non smooth morphism, as I said, the assertion is false for special fibers. So it is not clear to me what you are asking. Also could you provide a reference? – Angelo Dec 21 '22 at 11:35
  • In general (non-smooth map and spaces) we need cotangent complex and ses above replaced by triangle in the derived category https://stacks.math.columbia.edu/tag/08P5 – Alex Dec 21 '22 at 11:47