3
$\begingroup$

Let $X$ be a projective scheme and $X \subset \mathbb{P}^n$ for some positive integer $n$. Let $j:Z \hookrightarrow X$ be a closed subscheme. Is it true that $H^0(j^*\mathcal{N}_{X|\mathbb{P}^n}) \cong H^0(j^{-1}\mathcal{N}_{X|\mathbb{P}^n})$?

EDIT Assume $Z$ is an irreducible component of $X$.

$\endgroup$
3
  • $\begingroup$ Your editing doesn't change the problem, Anton's example still applies. $\endgroup$
    – abx
    Commented Jan 14, 2014 at 8:47
  • $\begingroup$ @abx: Sorry. Meant to write irreducible component. $\endgroup$
    – user45397
    Commented Jan 14, 2014 at 8:53
  • 1
    $\begingroup$ OK, in this new version at least it makes sense. $\endgroup$
    – abx
    Commented Jan 14, 2014 at 8:55

1 Answer 1

6
$\begingroup$

Take $X=\mathbb{P}^1\subset \mathbb{P}^2$ and let $Y$ be a single point $y\in X$. Then $j^*$ gives you a trivial bundle corresponding to $\mathrm{k}$, while $j^{-1}$ is isomorphic to the full stalk $\mathcal{N}_y\simeq \mathcal{O}_{X,y}\simeq \mathrm{k}[x]_{(x)}$ (here $\mathrm{k}$ denotes the base field and $\mathcal{N}$ is the normal bundle). This is exactly why one considers the pullback functor in algebraic geometry and not the inverse image functor.

$\endgroup$
1
  • $\begingroup$ @Fonarev: Thank you for the answer. I have edited the question. $\endgroup$
    – user45397
    Commented Jan 14, 2014 at 8:40

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .