Let $X$ be a compact Kähler manifold. Consider the exponential sequence $0 \to \mathbf Z \to \mathscr O_X \to \mathscr O_X^* \to 0$. The boundary map gives a map $H^1(X, \mathscr O_X^*) \to H^2(X, \mathbf Z)$. Composing with the map $H^2(X, \mathbf Z) \to H^2(X, \mathbf C)$ we obtain a map $c: H^1(X, \mathscr O_X^*) \to H^2(X, \mathbf C) $. Now using the Hodge decomposition one can prove that the image of $c$ is in fact is contained in $H^{1,1}(X)$, where $H^{(p,q)}$ denotes the Dolbeault cohomology groups. Now we also know that $H^{1,1}(X) \cong H^1(X, \Omega_X^1)$, where $\Omega_X^1$ is the sheaf of sections of the holomorphic cotangent bundle.

Therefore, we have a map $c_1: H^1(X, \mathscr O_X^*) \to H^1(X, \Omega_X^1) $.

Question: Consider the map of sheaves $\mathscr O_X^* \to \Omega_X^1 $ given by $f \to \frac{\partial f}{2 \pi if}$. This gives a map $c_1': H^1(X, \mathscr O _X^*) \to H^1(X, \Omega_X^1)$. Does it follow that $c_1' = c_1$?

For Riemann surfaces, I had an approach in mind. In that case the exponential sequence maps to $0 \to \mathbf C \to \mathscr O _X \to \Omega_X^1 \to 0$ in the obvious way. Then what remains to be proven is that the boundary map $H^1(X, \Omega_X^1) \to H^2(X, \mathbf C)$ agrees with the Hodge theoritic inclusion (equality in this case) $H^{1,1}(X) \to H^2(X, \mathbf C)$, which I have not been able to prove.

  • $\begingroup$ you can define first Chern class of any notherin scheme $\endgroup$
    – user21574
    Jun 23, 2016 at 16:19
  • $\begingroup$ $c_1(X):=c_1(-K_X)=[Ric(\omega)]$ $\endgroup$
    – user21574
    Jun 23, 2016 at 17:50
  • $\begingroup$ Search for Deligne cohomology $\endgroup$
    – user40276
    Jun 24, 2016 at 7:15
  • $\begingroup$ Thanks. Could you please elaborate how that helps to compare the two maps in this situation? $\endgroup$ Jun 24, 2016 at 7:25

2 Answers 2


This is an extended comment to show that in the 1-dimensional case this comes down to comparing a "Hodge-theoretic inclusion" with a connecting map.

Your version of the analytic exponential sequence implicitly rests on the map $e^{2\pi i(\cdot)}$ (giving as kernel term the constant sheaf $\mathbf{Z}$) rather than $e^{(\cdot)}$ (giving as kernel term the constant sheaf $\mathbf{Z}(1)$). The latter is more canonical insofar as it doesn't involve a choice of basis $2\pi i$ of $\mathbf{Z}(1)$, so if we work with that at the outset then in your question the division by $2\pi i$ (presumably the same choice of $\mathbf{Z}(1)$-basis as implicit in your exponential sequence) goes away, leaving us with the more "canonical" map ${\rm{d}}\log: f \mapsto {\rm{d}}f/f$. So we'll argue in terms of this more canonical exponential sequence (and corresponding adjustment to your question).

Observe that there is an evident map from the short exact sequence $$0 \rightarrow \mathbf{Z}(1) \rightarrow \mathscr{O}_X \stackrel{\exp}{\rightarrow} \mathscr{O}_X^{\times} \rightarrow 1$$ to $$0 \rightarrow \mathbf{C} \rightarrow \mathscr{O}_X \stackrel{\rm{d}}{\rightarrow} \Omega^1 \rightarrow 1$$ (using ${\rm{d}}\log$ along the right side and the identity map on middle terms), so this yields a commutative diagram of connecting maps $$\begin{array}[c]{ccc} {\rm{H}}^1(X, \mathscr{O}_X^{\times}) &{\rightarrow}& {\rm{H}}^2(X, \mathbf{Z}(1)) \\ \downarrow\scriptstyle{c'_1}&&\downarrow\\ {\rm{H}}^1(X, \Omega^1_X) &{\rightarrow}& {\rm{H}}^2(X, \mathbf{C}) \end{array}$$ where the horizontal maps are connecting maps.

Given how you defined $c_1$, your question in the case of dimension 1 thereby comes down to asking if the connecting map along the bottom coincides with the "Hodge-theoretic inclusion" (an equality in this case). To see that such matters are not entirely trivial, note that the "Hodge-theoretic inclusion" ${\rm{H}}^0(X,\Omega^1_X) \hookrightarrow {\rm{H}}^1(X, \mathbf{C})$ is the negative of the connecting map arising from $$0 \rightarrow \mathbf{C} \rightarrow \mathscr{O}_X \stackrel{\rm{d}}{\rightarrow} \Omega^1 \rightarrow 1.$$ Maybe from degree 1 into degree 2 there's a cancellation of signs and one gets an equality rather than a sign discrepancy? Good luck sorting it out.

  • $\begingroup$ Yes your elaboration is exactly what I had in mind. Thanks for writing it down in detail. So you are saying the zero degree connecting map is negative of Hodge theoritic inclusion right? Can you give a reference or show how it is done? Maybe that will help me to get some idea about the next connecting map! $\endgroup$ Jun 23, 2016 at 16:46
  • $\begingroup$ I was naively wondering that all the connecting maps agree with the hodge theoritic inclusions! $\endgroup$ Jun 23, 2016 at 16:56
  • $\begingroup$ Some ideas: We have resolutions $0 \to \mathbf C \to \mathcal A_{\mathbf C}^{\bullet}$, $0 \to \mathscr O_X \mathcal A^{0, \bullet}$, $0 \to \mathcal A^{1, \bullet}$. Where in the resolutions the maps are given by $d, - \overline{\partial}$ and $ \overline{\partial}$. Now objects in the resolution can be connected by $\mathcal A ^k \to \mathcal A^{0,k} \to \mathcal A^{1,k}$, where the first map is given by projection and the second map is given by $\partial$. this gives us a commutative diagram. Can we get the boundary maps from this? (Unfrtunately the map of complexes is not exact) $\endgroup$ Jun 23, 2016 at 17:28
  • $\begingroup$ typo: $0 \to \mathscr O_X \to \mathcal A ^{0, \bullet}$ and $0 \to \Omega_X^1 \to \mathcal A ^{1, \bullet}$ $\endgroup$ Jun 23, 2016 at 17:38
  • $\begingroup$ I didn't see how to use the non-exact diagram of resolutions. The sign issue in degree 1 is a general fact of homological algebra: if $0 \rightarrow A'\rightarrow A\rightarrow A''\rightarrow 0$ is a short exact sequence in an abelian category with enough injectives, $F$ is a left-exact additive functor to another abelian category, and $A'\rightarrow I^{\bullet}$ is an injective resolution then the connecting map $F(A'') \rightarrow R^1F(A)$ is negative of $F(A'')\rightarrow H^1(F(I^{\bullet}))$ arising from map from $A\rightarrow A''\rightarrow 0$ to $I^{\bullet}$ as resolutions of $A'$. $\endgroup$
    – nfdc23
    Jun 23, 2016 at 22:52

$\def\cO{\mathcal{O}}\def\CC{\mathbb{C}}\def\ZZ{\mathbb{Z}}$Yes, this is right. This answer will prove the equality of three maps $H^1(\cO^{\ast}) \to H^2(\CC)$.

(a) The map $$H^1(\cO^{\ast}) \overset{c_1}{\longrightarrow} H^2(\ZZ) \overset{2 \pi i}{\longrightarrow} H^2(\CC)$$ induced from the exponential sequence and the inclusion of $\ZZ$ into $\CC$.

(b) The map $$H^1(\cO^{\ast}) \overset{d \log}{\longrightarrow} H^1(W^1) \to H^1(Z^1) \cong H^2(\CC)$$ where $W^1$ is the $d$-closed $(1,0)$-forms and $Z^1$ is the $d$-closed $1$-forms.

(c) The map $$H^1(\cO^{\ast}) \overset{d \log}{\longrightarrow} H^1(\Omega^1) \cong H^{11} \to H^2_{DR} \cong H^2(\CC)$$ where $H^{11}$ is the harmonic $(1,1)$-forms.

We will use the following notations:

$A^{pq}$ the sheaf of smooth $(p,q)$-forms.

$W^p$ the sheaf of $d$-closed $(p,0)$ forms.

$\Omega^p$ the sheaf of holomorphic $(p,0)$ forms.

$A^k$ the sheaf of $k$-forms.

$Z^k$ the sheaf of closed $k$-forms.

$H^{pq}$ are the harmonic $(p,q)$ forms.

We have a commutative diagram with exact rows: $$\begin{matrix} 0 & \to & \ZZ & \overset{2 \pi i}{\longrightarrow} & \cO & \overset{\exp}{\longrightarrow} & \cO^{\ast} & \to & 0 \\ & & 2 \pi i \downarrow & & = & & \downarrow d \log & & \\ 0 & \to & \CC & \longrightarrow & \cO & \overset{d}{\longrightarrow} & W^1 & \to & 0 \\ & & = & & \downarrow & & \downarrow & & \\ 0 & \to & \CC & \longrightarrow & C^{\infty} & \overset{d}{\longrightarrow} & Z^1 & \to & 0 \\ \end{matrix}$$ Exactness of the first row is that nonvanishing holomorphic functions locally have holomorphic logarithms, the last row is the Poincare lemma, and the second row is the Poincare lemma plus the fact that integrals of holomorphic functions are holomorphic.

The boundary map on sheaf cohomology is functorial in such diagrams of short exact sequences, so we have a commutative diagram: $$\begin{matrix} H^1(\cO^{\ast}) & \overset{c_1}{\longrightarrow} & H^2(\ZZ) \\ d \log \downarrow & & \downarrow 2 \pi i \\ H^1(W^1) & \longrightarrow & H^2(\CC) \\ \downarrow & & = \\ H^1(Z^1) & \cong & H^2(\CC) \\ \end{matrix}$$ The last row is an isomorphism by partitions of unity.

The equality (a)=(b) expresses two ways of going around this diagram.

To relate (b) and (c), we need some generalities about $H^q(W^p)$ which I wish I knew a citation for.

We first recall that the Poincare short exact sequence $$0 \to Z^p \to A^p \overset{d}{\longrightarrow} Z^{p+1} \to 0$$ induces isomorphisms $$H^{k}_{DR} = H^0(Z^k)/d H^0(A^{k-1}) \cong H^1(Z^{k-1}) \cong \cdots \cong H^k(Z^0) = H^k(\CC) .$$

Key compatability: The map $W^p \to Z^p$ induces an inclusion $H^q(W^p) \to H^q(Z^p) \cong H^{p+q}(\CC)$ with image $\bigoplus_{k \leq q} H^{(p+q-k)k}$. The map $H^q(W^p) \to H^q(\Omega^p)$ induces the projection $\bigoplus_{k \leq q} H^{(p+q-k)k} \to H^{pq} \cong H^q(\Omega^p)$.

Using this, we prove (b)=(c). The image of $H^1(W^1)$ in $H^1(Z^1)$ is $H^{20} \oplus H^{11}$, so every class in $H^1(\cO^{\ast})$ is sent by (b) to a class of the form $\alpha \oplus \beta$. As you say, we can use the description from (a) to show that this class is realizable by a $(1,1)$-form, so $\alpha=0$. Then $\beta \in H^{11}$ is the image of the composition $H^1(W^1) \to H^2(\CC) \to H^{11}$, and the key compatability says that this is the same as the map $H^1(W^1) \to H^1(\Omega^1)$. So we can compute $\beta$ using the map $H^1(W^1) \to H^1(\Omega^1)$, and that is your option (c).

I'm going to hold off on writing a proof of the Key Compatability on the assumption that someone will give me a citation for it, ideally in the same sort of classical language that the OP uses in the question.

  • $\begingroup$ How does this overall method mesh with the sign discrepancy in the attempt to compare connecting maps with Hodge-theoretic maps in degree 1 on curves (one degree below that in the question posed)? The string of isomorphisms you display above the "Key Compatibility" seems to be defined in terms of many connecting maps, yet $H^k_{\rm{dR}}$ never appears again, so how is that displayed string of isomorphisms relevant in what follows? $\endgroup$
    – nfdc23
    Jun 24, 2016 at 2:19

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