If a $d \log$ form is exact, is it zero?

Let $T = \mathrm{Spec}\ \mathbb{C}[x_1^{\pm 1}, x_2^{\pm 1}, \ldots, x_n^{\pm 1}]$ be an algebraic torus and $X$ a closed subvariety. Let $\eta$ be a differential form on $T$ of the form $$\sum_I a_I\cdot \bigwedge_{i \in I} d \log x_i$$ where $I$ runs over subsets of $\{ 1,2, \ldots, n\}$ and $a_I$ are various constants. If the cohomology class of $\eta$ restricts to $0$ on $X$, is $\eta$ identically zero?

Cases where this is true: $X$ a subtorus (easy), $X$ a curve (easy), $X$ cut out by linear equations in the $x_i$ (Orlik-Solomon).

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This is a sort of Hodge theorem for subvarieties of tori, I guess? That each cohomology class should have but one representative of this form? –  Allen Knutson May 9 at 1:42
If we add extra variables, $X$ can always be seen as the intersection of a subtorus and a subvariety cut out by linear equations. If it is a complete intersection when so described, I think we can wedge with a form that is perpendicular to the subtorus to reduce to the linear equations case, but that might be nonsense. –  Will Sawin May 9 at 4:53
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2 Answers

To expand on my comment: Suppose $X$ is a complete intersection of a subtorus and a linear subspace. View the form as a form on the linear subspace. If its restriction to a subvariety like the $X$ is cohomologous to $0$, then its cup product with the class of $X$ is cohomologous to $0$. But the class of $X$, which is the pullback of the class of the subtorus, can be represented by a form of this type. One simply pulls back the "volume form" ($I=\{1,\dots,n\}$) from the quotient torus.

So the original form wedge this new form is cohomologous to $0$ on the linear subspace, so by Orlik-Solomon it is identically zero. Now looking locally at a point of the intersection, the new form is nonzero and completely transverse to $X$, so if wedging with it produces the zero form, then restricting to $X$ will as well.

If $X$ is any complete intersection, then by adding variables we can make it a complete intersection of a torus and a linear subspace. Simply replace each monomial that occurs in the polynomial equations of $X$ with a new variable, making them linear equations, and then set the new variables equal to the old monomials, creating new toric equations. Clearly this preserves the property of being a complete intersection.

EDIT: We use the duality between regular cohomology and compactly supported cohomology. Let $a$ be the original form, let $b$ be the new form, and let $c$ be any compactly supported form. By non-compact Poincare duality, showing that $a \wedge b \wedge c$ integrates to $0$ for all $c$ of the appropriate degree is sufficient to show $a\wedge b$ is $0$. This is the same as showing $(c \wedge a) \wedge b$ integrates to $0$. But this is another example of integrating a compact form against a non-compact form, so it's another Ponicare duality. But we can also express Poincare duality as integrating non-compact forms over compact submanifolds, or non-compact forms over compact submanifolds. Clearly $b$ corresponds to the class of $X$, by pullback from the torus , so we need $c \wedge a$ to be cohomologous to $0$ when restricted to $X$, which happens as long as $c$ is cohomologous to $0$ when restricted to $X$.

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Wait, I'm confused. When you say that $\omega|_X$ is cohomologous to $0$ if and only if $\omega \wedge \eta$ is cohomologous to $0$ for a properly chosen $\eta$, you seem to be using some sort of Poincare duality. But this is not a compact setting... –  David Speyer May 9 at 17:02
I think I can do it using duality between compact and regular cohomology. –  Will Sawin May 9 at 18:44
Please add the details when you figure them out! –  David Speyer May 10 at 11:56
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I think this works. We prove a slightly stronger statement by induction on $\dim X$: Let $X$ be a quasi-projective variety over $\mathbb{C}$. Let $x_1$, $x_2$, ..., $x_n$ be units in $H^0(X, \mathcal{O})$. By "$d \log x_i$" I mean the class in $H^1(X)$ pulled back from the generator of $H^1(\mathbb{G}_m)$ by the map $x_i : X \to \mathbb{G}_m$. (I say this to address the possibility that $X$ isn't smooth, since I'm not sure whether Kahler differentials give cohomology classes in general.) Then, as before, I claim that:

If a polynomial $\eta$ in the $d \log x_i$ is exact, then it is $0$.

The base case, $\dim X=0$, is trivial.

Reduction We may (and do) assume that $X$ is smooth.

Proof Let $\tilde{X} \to X$ be a resolution of singularities. Since the class of $\eta$ is $0$ in $H^{\ast}(X)$, it pulls back to $0$ in $H^{\ast}(\tilde{X})$. So $\eta$ pulls back to $0$ on $\tilde{X}$, and is thus $0$. $\square$

Now, choose a simple normal crossing compactification $\bar{X}$ of $X$. Let $D_1$, $D_2$, ..., $D_m$ be the components of $D$. Our next goal is to prove

Claim For any component $D_1$ of $D$, the form $\eta$ has no pole on $D_1$.

Reduction We may assume that $x_2$, $x_3$, ...., $x_n$ don't have poles or zeroes along $D_1$, and $x_1$ vanishes on $D_1$.

Proof Let $d_i$ be the order to which $x_i$ vanishes on $D_1$. Let $d = GCD(d_1, d_2, \dots, d_m)$. Then by applying a monomial transformation to the $x$'s, we may assume that $x_1$ vanishes to order $d$ on $D_1$. $\square$

Let $Y = D_1 \setminus \bigcup_{i \geq 2} D_1 \cap D_i$ and let $Z$ be the union of $X$ and $Y$ inside $\bar{X}$. So $Y$ is a smooth hypersurface in $Z$. We claim that the $x_i$ extend to homolomorphic function on $Z$, and that $x_2$, ..., $x_n$ extend to nonvanishing functions. Proof: Suppose $x_i$ does not extend to $Z$. Then $x_i$ has a pole along some hypersurface in $Z$. By the previous Reduction, that hypersurface is not $Y$, so it must meet $X$. But $x_i$ is well defined on $X$. For $i \geq 2$, this argument also shows that $x_i^{-1}$ extends.

Proof of Claim Let $\omega = \mathrm{Res}_{Y} \eta$. The form $\omega$ is a $d \log$ form on $Y$, in the ring generated by $d \log x_2$, $d \log x_3$, ..., $d \log x_n$. The residue map is a well defined map $H^k(X) \to H^{k-1}(Y)$. More specifically, let $U$ be a tubular neighborhood of $Y$, so $\partial U \subset X$ is a circle bundle over $Y$. Then $\mathrm{Res}$ is the composition of restriction $H^k(X) \to H^k(\partial U)$ and the Gysin map $H^k(\partial U) \to H^{k-1}(Y)$.

So $\omega$ is exact on $Y$. By induction, $\omega=0$. We have now established that the residue of $\eta$ along $Y$ is $0$. But $\eta$ clearly has at most a first order pole on $Y$, so it has no pole at all. $\square$.

Proof of theorem We have shown that $\eta$, as a form on $\bar{X}$, does not blow up on any of $D_i$. Since $\bar{X}$ is smooth, this shows that $\eta$ extends to $\bar{X}$. But, by Hodge theory, on a smooth projective variety, any exact holomorphic form is $0$.

Alternate proof As above, reduce to $X$ smooth; let $\overline{X}$ be a normal crossing compactification; let $D = \overline{X} \setminus X$. Let $\eta$ be a $k$-form. The condition that $\eta$ is generated by $d \log$ forms implies that $\eta$ is an element of $H^0(\overline{X}, \Omega^k(\log D))$. The condition that $\eta$ is exact says that the image of $\eta$ in $H^k(X, \mathbb{C})$ is $0$.

Conveniently, there is a spectral sequence $H^q(\overline{X}, \Omega^p(\log D)) \Rightarrow H^{p+q}(X, \mathbb{C})$ which degenerates at $E_1$. (I'm looking at Voisin's book, volume 1, Theorem 8.35, she cites Deligne Theorie de Hodge II.) So $H^0(\overline{X}, \Omega^k(\log D))$ injects into $H^k(X, \mathbb{C})$, which was the desired claim.

Thomas Lam and I have finally started writing the paper where we needed this lemma; this is probably the proof we will use.

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