Conormal bundle of the strict transform

Question

I'm trying to understand the proof of Theorem II.1.7 in Kollar's Rational curves on algebraic varieties. In particular, there is a claim in there that I can't make sense of.

The setting is the following (we slightly simplify the statement). Let $C/S$ be a flat projective curve without embedded points and $Y/S$ a smooth scheme over $S$. Suppose that $B/S\subset C/S$ is the image of a flat section $p$, and that $C/S$ is smooth along $B$. We are given a morphism $g:B/S\to Y/S$. Let $s$ be a point in $S$, and $f_s:C_s\to Y_s$ a morphism respecting $G_s$.

The thesis is that the Zariski tangent space to $\mbox{Hom}(C_s,Y_s,G_s)$ at $[f_s]$ is isomorphic to $$ H^0(C_s,f_s^*T_{Y_s}\otimes I_{B_s}) $$ where $I_{B_s}=\mathcal{O}_{C_s}(-p(s))$ is the ideal sheaf of $B$ at the fiber above $s$.

By a previous result of the book (Thm. I.2.15 and following), we can prove this by showing that $\mbox{Hom}(C,Y,G)$ is isomorphic to an open set of a certain Hilbert scheme, constructed as follows.

Let $X=C\times_S Y$ and $\gamma:C_s\simeq\Gamma\subset X$, where $\Gamma$ is the graph of $f_s$. Let $I$ be the ideal sheaf of $\Gamma$ in $\mathcal{O}_X$. By construction (we have a splitting of the conormal bundle sequence since we are in a product) we get

$$ \gamma^*(I/I^2)\simeq f_s^*\Omega_{Y_s}. $$ Now let $X'$ be the blowup of $X$ along $p(S)$ (how does this live in $X$?), and $\Gamma'$ be the strict transform of $\Gamma$, still isomorphic to $C_s$ via $\gamma'$. Let $I'$ be its ideal sheaf in $\mathcal{O}_{X'}$. He claims that

$$ \gamma'^*(I'/I^{'2})\simeq \gamma(I/I^2)\otimes \mathcal{O}_{C_s}(-p(s))\simeq f_s^*\Omega_{Y_s}\otimes I_{B_s} $$

and this first isomorphism is the one I can't motivate.

The correct Hilbert scheme is then $\mbox{Hilb}(X'/S)$. Another thing I don't understand is that to get the $H^0$ in the thesis I would have to dualize $f_s^*\Omega_{Y_s}\otimes I_{B_s}$, so shouldn't I expect to have a $I_{B_s}^*$ there?

Answer 1

There are definitely some typos in that proof, but for your question you may assume that $S=\{s\}$ which will take care most of them.

If you write down the restriction of differential forms sequence (the dual of the normal bundle sequence) for both $\Gamma\subset X$ and $\Gamma'\subset X'$ and compare them, then you get the following short exact sequence: $$ 0\to \gamma^*(I/I^2) \to \gamma'^*(I'/I'^2) \to \gamma'^*\left(\Omega_{X'/X}\right)\to 0 $$

$\Omega_{X'/X}$ is explicitly computable (HW!), but you only need that it is a rank $r=\dim X-1=\mathrm{codim}(\Gamma, X)=\mathrm{rank}(\gamma^*(I/I^2))$ vector bundle on the exceptional divisor of $X'\to X$. So, this pull-back is a (sort of) skyscraper sheaf $k^{\oplus r}$ sitting at the intersection of $\Gamma'$ with that exceptional divisor. Comparing this to the short exact sequence above shows that the first sheaf is just the ideal sheaf of this intersection times the second sheaf.

At this point one can also answer the question in your last sentence. There are some more typos in the proof. As you can see what we get is not exactly what is claimed, but rather that $$ \gamma'^*(I'/I^{'2})\simeq \gamma^*(I/I^2)\otimes \mathcal{O}_{C_s}(p(s))\simeq f_s^*\Omega_{Y_s}\otimes I_{B_s}^*. $$ Hence when at the end you take the dual you do get what is claimed in the statement of the theorem, so the world is back to order.