2
$\begingroup$

I am reading Claire Voisin's presentation of the Hodge structure of a blowup. It is somewhat faithfully copied on page 76-78 of these notes (except compared to the notation below, which follows Voisin's, $Z$ becomes $Y$ -- most of the time, at least). My question is about the second-to-last displayed line on page 77 of those notes.

Let $X$ be a Kahler manifold, $Z$ a submanifold of (complex) codimension $r$, and $U=X-Z$. Then she claims that by excision and Thom isomorphism, $$H^{k-1}(X,U)\cong H^{k-2r}(Z),$$ coefficients in $\mathbf Z$ throughout.

I could understand this isomorphism if the index $k-1$ were replaced by $k$ in the following way: excision to a tubular neighborhood of $Z\subset X$ gives an isomorphism $H^k(X,U)\cong H^k(N_{Z/X},N_{Z/X}-\{0\})$, where $N_{Z/X}$ is the normal bundle of $Z$ in $X$ and $N_{Z/X}-\{0\}$ is that same bundle minus the zero section. Then as $Z$ has complex codimension $r$, $N_{Z/X}$ is an oriented vector bundle of real rank $2r$. The Thom isomorphism then gives $$H^k(N_{Z/X},N_{Z/X}-\{0\})\cong H^{k-2r}(Z).$$

The issue is that the claimed isomorphism is $H^{k-1}(X,U)\cong H^{k-2r}(Z)$, not $H^k(X,U)\cong H^{k-2r}(Z)$.

$\endgroup$
1
  • $\begingroup$ Indeed, it should be $H^k(X,U)=H^{k-2r}(Z)$ $\endgroup$
    – Doug
    Sep 27, 2020 at 16:31

2 Answers 2

1
$\begingroup$

Ok, this is my very first time giving an answer in this website, so I ask for some understanding on my technical limitations.

I just finished reading that proof, and is full of mistakes. The key point is that the part of the argument where she identifies j_* with a morphism between the cohomology of E and the blow-up is not right, by two points. One was pointed out by you, and the other, as someone said earlier, being the fact that the l.e.s. of relative cohomology is also wrong.

I was actually interested to check if any of this could possibly be salved to mixed Hodge structures, so I will keep trying to come up or find a corrected proof, if I do I will try to remember to post it here.

$\endgroup$
1
  • $\begingroup$ Thanks @Jaime for your response! Yes, any corrections are very helpful to have online. $\endgroup$
    – Tomo
    Jul 7, 2017 at 0:46
0
$\begingroup$

The indices of her LES of cohomology for the pair are messed up also in a way which appears to cancel this typo.

$\endgroup$

You must log in to answer this question.

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