3
$\begingroup$

Consider complex affine space $\mathbb{C}^n$ and let $U$ be a Zariski open subset of $\mathbb{C}^n$. By a celebrated result of Deligne, the cohomology $H^i(U)$ has a canonical Hodge structure. In particular, $H^i(U)$ has a weight filtration and a subalgebra of pure classes (since a cohomology class can't have lower weight than expected, only higher). I believe it's true that

The pure subalgebra of $H^i(U)$ is exactly the identity.

This is as far from being pure as possible.

What I hope to get from the collective intelligence of the internet is somewhere where this fact is written. I want to emphasize that what I am really hoping to get is a reference, since (as you can see below) I basically know how the proof should go.

In hopes of getting either confirmation or a mistake pointed out, let me write a proof:

By Alexander duality $\tilde H^i(U)\cong H_{n-i-1}^{BM}(X)$ where $X=\mathbb{C}^n\setminus U$. This is an isomorphism of Hodge structures after Tate twist by $n$. The weights of $H_{n-i-1}^{BM}(X)$ lie in $[-n+i+1,0]$, so those of $\tilde H^i(U)$ lie in $[i+1,n]$.

As a second-best request, does anyone know of a reference for the version of Alexander written above? It's dual to way things are usually written.

$\endgroup$

1 Answer 1

13
$\begingroup$

Maybe I'm missing something, but I think this should be a simpler proof. Let $j \colon U \hookrightarrow \mathbf P^n$ be the natural compactification, and let $k > 0$. Then $W_k H^k(U,\mathbf Q)$ is the image of $j^\ast \colon H^k(\mathbf P^n,\mathbf Q) \to H^k(U,\mathbf Q)$ (Hodge II, Corollaire 3.2.17). But $j^\ast$ factors through $H^k(\mathbf A^n, \mathbf Q) = 0$.

$\endgroup$
1
  • $\begingroup$ Sweet. That's easy enough that I don't need a reference. $\endgroup$
    – Ben Webster
    May 28, 2012 at 22:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.