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Let $X$ be a smooth quasiprojective scheme defined over $\mathbb{C}$, and let $\Omega^{\bullet}_X$ denote its cotangent complex, explicitly, we have:

$\Omega^{\bullet}_X:=\mathcal{O}_X\longrightarrow \Omega^1_X\longrightarrow...\longrightarrow \Omega^{\dim X}_X$.

We define the naive filtration, $\Omega^{\bullet,\ge p}_X$, as the subcomplex, given explicitly by:

$\Omega^{\bullet,\ge p}_X := 0\longrightarrow ...\longrightarrow 0\longrightarrow \Omega_X^p\longrightarrow...\longrightarrow \Omega_X^{\dim X}.$

Now, the Hodge filtration is defined by:

$\mathcal{F^p}H^i_{\text{dR}}(X) := \text{Im}(H^i(X, \Omega_X^{\bullet, \ge p})\longrightarrow H^i(X, \Omega_X^{\bullet})).$

My issue is that even though I clearly see that $\Omega_X^{\bullet, \ge p}$ is a subcomplex of $\Omega_X^{\bullet}$, I don't see why $\mathcal{F}^pH^i_{\text{dR}}(X)$ forms a vector subspace of $H^i_{\text{dR}}(X)$ and isn't the whole thing.

For example, in the case where $X$ is affine, consider $\mathcal{F}^1H^1_{\text{dR}}(X)$. In this case I believe the de Rham complex is acyclic so de Rham cohomology is simply the hypercohomology of said complex, in particular, it seems that $H^1_{\text{dR}}(X, \Omega_X^{\bullet, \ge 1}) = H^0(X, \Omega^1_X)$, however, $H^1_{dR}(X, \Omega_X^{\bullet})$ is isomorphic to a quotient of $H^0(X, \Omega^1_X)$ by the span of exact 1-forms.

It seems my reasoning implies that the first Hodge filtration of the $H^1_{\text{dR}}$ of an affine curve is trivial. Is this correct, or am I missing something?

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    $\begingroup$ Minor terminological quibble: the complex $\Omega^\bullet_X$ is usually called the de Rham complex. Generally, the term "cotangent complex" refers to something a bit different. In this case, the cotangent complex of $X/\mathbb{C}$ is $\Omega_{X} (= \Omega^1_X)$. $\endgroup$
    – Brian Shin
    Commented May 4 at 1:52

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The naive Hodge filtration of a smooth affine variety is, indeed, the whole thing. We always have the short exact sequence of complexes: $$0 \to \Omega^{\bullet, \geq p} \to \Omega^{\bullet} \to \Omega^{\bullet, <p} \to 0$$ so we have a long exact sequence of hypercohomology $$\cdots \to \mathbb{H}^p(X,\Omega^{\bullet, \geq p}) \to \mathbb{H}^p(X,\Omega^{\bullet}) \to \mathbb{H}^p(X,\Omega^{\bullet, < p}) \to \cdots. \qquad (\ast)$$

Now, if $X$ is affine and $C^{\bullet}$ is a complex of quasi-coherent sheaves in degree $\leq k$, then $\mathbb{H}^i(X, C^{\bullet})$ vanishes for $i>k$. (This is the hypercohomology version of "sheaf cohomology of quasi-coherent sheaves on an affine variety vanishes".) So $\mathbb{H}^p(X,\Omega^{\bullet, < p})=0$ in this case, and $\mathbb{H}^p(X,\Omega^{\bullet, \geq p}) \to \mathbb{H}^p(X,\Omega^{\bullet})$ is surjective.

If you identify algebraic de Rham with differential geometric de Rham (which is very deep), then this is a very non-obvious fact: It says that, when $X$ is affine, every class in $H_{DR}^p(X, \mathbb{C})$ can be represented by a closed algebraic (in particular, holomorphic) $p$-form.


For an example where the naive Hodge filtration is more interesting, consider a (smooth, irreducible) projective curve $X$ of genus $g$. The cotangent complex has two terms, $\mathcal{O} \to \Omega^1$. Taking $p=1$, the sequence $(\ast)$ above becomes $$\cdots \to \mathbb{H}^1(\Omega^1[1]) \to H^1_{DR} \to \mathbb{H}^1(\mathcal{O})) \to \cdots$$ or $$\cdots \to H^0(X, \Omega^1) \to H^1_{DR}(X) \to H^1(X, \mathcal{O}) \to \cdots. \qquad (\dagger)$$

Hodge theory tells us that the arrows from and to the "$\cdots$" in $(\dagger)$ are $0$, so we have a short exact sequence: $$0 \to H^0(X, \Omega^1) \to H^1_{DR}(X) \to H^1(X, \mathcal{O}) \to 0.$$ The middle term has dimension $2g$, the left and right terms have dimension $g$.


I like Section 1 of Kedlaya's "$p$-adic cohomology: from theory to practice" as an introduction to this material.


Oh, I should add that the mixed Hodge structure on $H_{DR}(X)$ isn't defined using the naive Hodge filtration. That's why it's called the naive filtration, because it doesn't give the mixed Hodge structure correctly (except when $X$ is projective).

To get the mixed Hodge structure, you take a normal crossing compactification $\overline{X}$ of $X$, with $\overline{X} \setminus X = D$, you consider the complex $\Omega^{\bullet}(\log D)$ on $\overline X$, and you filter that by $\Omega^{\geq p}(\log D)$. I'm not sure what the best reference is, but Wikipedia has the basic definitions.

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  • $\begingroup$ Incredibly clear. Many thanks Professor Speyer! $\endgroup$
    – kindasorta
    Commented May 2 at 18:02
  • $\begingroup$ The last link to wiki is broken. $\endgroup$
    – Z. M
    Commented May 3 at 15:02
  • $\begingroup$ @Z.M Thanks, not sure what happened there. $\endgroup$ Commented May 3 at 16:11

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