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.


1 Answer 1


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 $H^k(\mathbf A^n, \mathbf Q) = 0$.

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

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