Iterated diagonal maps and normal bundle

Question

Let $X$ be a smooth projective variety. Let $X \times X \xrightarrow{id \times \Delta} X \times X \times X$ be the identity map on the first factor and the diagonal morphism on the second. Is it true that the pull-back of the normal bundle along $\Delta:X \rightarrow X \times X$, $\Delta^*N_{X \times X / X \times X \times X},$ is just isomorphic to the tangent bundle on $X$, $TX$? Can you see this from the (co)normal exact sequence?

Answer 1

Here is a sketch full of missing details:

Claim: If $U=\overline U\subseteq V$ and $W$ are smooth varieties and $\pi$ is the projection $U\times W\to U$, then $$N_{W\times U/W\times V} \cong \pi^*N_{U/V}$$

"Proof": The statement is local, so we can prove it for affine varieties. If $I:=I(U)$, then translating the above statement to one about modules (and replacing normal bundles by conormal bundles) becomes $$ \frac{k[W]\otimes_k I}{(k[W]\otimes_k I)^2} \cong k[W]\otimes_k \frac{I}{I^2}. $$ We only need to check that the map induced by $f\otimes g\mapsto f\otimes g$ is an isomorphism.

$$\tag*{$\blacksquare$}$$

In the claim above, let $W = X$, and let $U\to V$ be $\Delta\colon X\to X\times X$. Then we are saying that $N_{X\times X/X\times X\times X} \cong \pi_{2}^*N_{X/X\times X}$. To get the pullback $\Delta^*$ you want you only need to use that $\Delta\circ \pi_2 = \mathrm{id}_X$.