# A cohomology group- What does the $F^2$ denote?

I ran across another denotion I am not familiar with while studying Bloch's conjecture, with the 2nd cohomology group being given as $w'\in F^2H^2(X,\mathbb{C})$. I understand what the part on and after the second cohomology denotes,but not the $F^2$. The article is avaliable here for further reference: http://arxiv.org/pdf/alg-geom/9503008v1.pdf

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It looks like a notation for some sort of filtration on cohomology. –  Andrew Oct 17 '12 at 19:36

Let $X$ be a smooth, proper complex variety (possibly weaker conditions can be imposed to get this to work, but certainly the situation in that paper). We could spend a long time unraveling all the details that go into the Hodge decomposition, but roughly here is what it means.

Given $H^n(X, \mathbb{C})$ you can decompose this complex vector space into a weight $n$ Hodge structure $\displaystyle H^n(X, \mathbb{C})\simeq \bigoplus_{p+q=n}H^q(X, \Omega^p)$.

A Hodge decomposition is decomposing a vector space $\displaystyle V\simeq \bigoplus_{p+q=n} H^{p,q}$ that satisfy a bunch conditions such as $\overline{H^{q,p}}=H^{p,q}$. You can look these up by searching for "Hodge structure."

It turns out that it is exactly equivalent to giving a decreasing filtration on $V$, $F^0V=V\supset F^{1}V\supset \cdots \supset F^nV=0$.

In your case this has a particularly nice description. $F^0H^2(X,\mathbb{C})=H^2(X, \mathbb{C})$, $F^1H^2(X, \mathbb{C})=H^2(X, \mathcal{O})\oplus H^1(X, \Omega^1)$, and $F^2H^2(X, \mathbb{C})=H^2(X, \mathcal{O}_X)$ which is identified with the so-called $(0,2)$-forms (locally expressible using only the antiholomorphic coordinates).

Anyway, I just did this from memory, and sometimes conventions differ, so you should look up Hodge decomposition for cohomology of complex manifolds and figure out exactly what the filtration they are talking about is. I have no doubt that if it is written without explanation, then it is the Hodge filtration.

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