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Let $X$ be a compact complex manifold, $L$ be a holomorphic line bundle on $X$, then the exponential exact sequence $0\to \mathbb Z\hookrightarrow \mathcal O\to \mathcal O^*\to 0$ induces the map $c:H^1(X,\mathcal O^*)\to H^2(X,\mathbb Z)$, it is well-known that the line bundle $L$ can be seen as an element in $H^1(X,\mathcal O^*)$, the image $c(L)\in H^2(X,\mathbb Z)$ is called the Chern class of $L$.

According to Atiyah's paper in 1957, p.196, the exact sequence $0\to \mathbb C\hookrightarrow \mathcal O\stackrel{d}\to \Omega^1\to 0$ together with the exponential exact sequence above induce the map $\mathcal O^*\to \Omega^1:f\mapsto \frac{1}{2\pi i}d\text{log}f$, which induces the map $\sigma:H^1(X,\mathcal O^*)\to H^1(X,\Omega)$, we call the image $\sigma(L)\in H^1(X,\Omega^1)$ the Atiyah class of the line bundle $L$.

In Atiyah's paper, the author assumed $X$ to be a compact Kähler manifold, then he concludes that the Atiyah class equals to Chern class, here my question is: can this condition be weakened a bit? For example, $X$ is a $\partial\bar\partial$-manifold or the Frölicher spectral sequence degenerates at $E_1$? Or equivalently, what's the sufficient and necessary condition of the Atiyah class coincides with the Chern class?

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  • $\begingroup$ Why is the sequence involving $\Omega^1$ exact? $\endgroup$
    – S.D.
    Oct 29, 2022 at 9:06
  • $\begingroup$ @S.D. It is taken from Atiyah's paper, p.196. Actually, I have the same suspect, I think $\Omega^1$ should be replaced by $Z^{1,0}:=A^{1,0}\cap\ker d$, maybe in the Kähler case (as in Atiyah's paper), $\Omega^1=Z^{1,0}$? so the author takes $\Omega^1$ for $Z^{1,0}$? $\endgroup$
    – Tom
    Oct 29, 2022 at 9:16
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    $\begingroup$ I wanted to just point out that on page 196, Atiyah just said that it is a commutative diagram of sheaves. Does not say anything about exactness! But anyway, your main question is different. $\endgroup$
    – S.D.
    Oct 29, 2022 at 9:25
  • $\begingroup$ @S.D. but he put $\to 0$ after $\Omega^1$, and wrote $0\to \mathbb C\hookrightarrow \mathcal O\stackrel{d}\to \Omega^1\to 0$, doesn't it mean the exactness? $\endgroup$
    – Tom
    Oct 29, 2022 at 9:32
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    $\begingroup$ I agree it's confusing, but, if I put $0 \to \mathbb C \hookrightarrow \mathcal O \xrightarrow d \Omega^1$ or $0 \to \mathbb C \hookrightarrow \mathcal O \xrightarrow d \Omega^1 \to 0$ in a diagram, and assert only its commutativity, then the latter is redundant, but surely makes no claim about exactness? $\endgroup$
    – LSpice
    Oct 31, 2022 at 16:21

2 Answers 2

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$\def\ZZ{\mathbb{Z}}\def\CC{\mathbb{C}}\def\cO{\mathcal{O}}$To spell out my comment a little more, let $Z^1$ be the sheaf of $\partial$-closed holomorphic $(1,0)$-forms. Since "holomorphic" means $\overline{\partial}$-closed, this can also be described as the space of closed $(1,0)$forms. Then we have a commutative diagram, with exact rows and columns:

$$\begin{matrix} && && 0 && 1 && \\ && && \downarrow && \downarrow && \\ 0 &\longrightarrow& \mathbb{Z} &\overset{2 \pi i}{\longrightarrow}& \mathbb{C} &\overset{\exp}{\longrightarrow}& \mathbb{C}^{\times} &\longrightarrow& 1 \\ & &=& &\downarrow& &\downarrow& \\ 0 &\longrightarrow& \mathbb{Z} &\overset{2 \pi i}{\longrightarrow}& \cO &\overset{\exp}{\longrightarrow}& \cO^{\times} &\longrightarrow& 1 \\ & && &\phantom{\partial} \downarrow \partial& &\phantom{\partial \log}\downarrow \partial \log& \\ && && Z^1 &=& Z^1 && \\ && && \downarrow && \downarrow && \\ && && 0 && 0 && \\ \end{matrix}$$ Using each of the $4$ short exact sequences, we have a diagram $$ \begin{matrix} H^1(\cO^{\times}) &\longrightarrow& H^2(\ZZ) \\ \downarrow && \downarrow \\ H^1(Z^1) &\longrightarrow& H^2(\CC) \\ \end{matrix}$$ and, with care, one can check that it commutes.

We also have a short exact sequence $0 \to Z^1 \to \Omega^1 \overset{\partial}{\longrightarrow} Z^2 \to 0$, where $Z^2$ is the closed $(2,0)$-forms. So we can extend this diagram to $$ \begin{matrix} H^1(\cO^{\times}) &\longrightarrow& H^2(\ZZ) \\ \downarrow && \downarrow \\ H^1(Z^1) &\longrightarrow& H^2(\CC) \\ \downarrow && \\ H^1(\Omega^1) && \\ \end{matrix}$$ The Atiyah class is the image of a class from $H^1(\cO^{\times})$ in $H^1(\Omega^1)$; the Chern class is the image in $H^2(\CC)$. This much is true without assuming anything about your complex manifold.

If you want to consider the Atiyah class and the Chern class to be "the same", then it seem like you want to ask is "when does it make sense to think of $H^1(\Omega^1)$ as a subspace of $H^2(\CC)$?" If your spectral sequence degenerates at $E^1$, then the map $H^1(Z^1) \to H^2(\CC)$ is an injection and $H^1(Z^1) \to H^1(\Omega^1)$ is a surjection, so $H^1(\Omega^1)$ is a subquotient of $H^2(\CC)$. But I don't know of anything which lets you identify $H^1(\Omega^1)$ with a subspace of $H^2(\CC)$ without Hodge theory. (With Hodge theory, on a compact Kahler manifold, $H^1(Z^1)$ is the piece $H^{20} \oplus H^{11}$ in the Hodge filtration, and the maps $H^1(Z^1) \to H^2(\CC)=H^{20} \oplus H^{11} \oplus H^{02}$ and $H^1(Z^1) \to H^1(\Omega^1)=H^{11}$ are the obvious ones.)

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  • $\begingroup$ I guess that they both are realizations of some "motivic Chern class" (but I don't know anything about motivic cohomology for complex manifolds). $\endgroup$
    – Z. M
    Oct 31, 2022 at 18:34
  • $\begingroup$ I think this is the kind of answer I'm looking for, so you mean $H^1(\Omega^1)$ can be treated as a subspace of $H^2(X,\mathbb C)$ if and only if $X$ is a compact Kähler manifold? $\endgroup$
    – Tom
    Nov 1, 2022 at 10:29
  • $\begingroup$ Yes, or a little more weakly, I don't know of any other setting in which there is a natural way to treat $H^1(\Omega^1)$ as a subspace of $H^2(\mathbb{C})$. $\endgroup$ Nov 1, 2022 at 16:09
  • $\begingroup$ Have you considered of $\partial\bar\partial$-manifolds? For this kind of manifolds, Bott-Chern cohomology and Dolbeault cohomology are isomorphic, and the Frölicher spectral sequence degenerates at $E_1$ (see Angella & Tomassini's 2013paper p.73 and Remark 5.21 of DGMS75, Real homotopy theory of Kähler manifolds). So there is $H^k(X,\mathbb C)=\bigoplus_{p+q=k}H^{p,q}_{\bar\partial}(X)=\bigoplus_{p+q=k} H_{BC}^{p,q}(X)$, it seems this kind of manifolds have a chance. $\endgroup$
    – Tom
    Nov 2, 2022 at 15:27
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    $\begingroup$ For a $\partial\bar\partial$-manifold, the natural map $f:H^{p,q}_{BC}(X)\to H_{\bar\partial}^{p,q}(X)$ is an isomorphism, and there is a natural map $g:H^{p,q}_{BC}(X)\to H^{p+q}_{dR}(X,\mathbb C)$, note that $H^k_{dR}(X,\mathbb C)\cong \bigoplus_{p+q=k}H^{p,q}_{BC}(X)$, $H^{p,q}_{BC}(X)$ can be treated as a subspace of the de Rham cohomology, so $g\circ f^{-1}:H^{p,q}_{\bar\partial}(X)\to H^{p+q}_{dR}(X,\mathbb C)$ maps $H^{p,q}_{\bar\partial}(X)$ to a subspace of $H^{p+q}_{dR}(X,\mathbb C)$. $\endgroup$
    – Tom
    Nov 3, 2022 at 3:46
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I guess this is always true, if you adjust the statement appropriately.

Consider the Bott–Chern cohomology $H^*_{BC}(M):=\dfrac{\ker d\cap \ker d^c}{\operatorname{im} dd^c}$. Since the curvature of a line bundle is a closed $(1,1)$-form, its Chern class can be considered as an element of the Bott–Chern cohomology. There are natural maps from $H^{1,1}_{BC}$ to the Dolbeault cohomology and to de Rham cohomology; the de Rham image of the curvature is $c_1(L)$, and the Dolbeault image is the cohomology class of $\bar\partial \partial\log\lvert f\rvert$, which is equal to the Dolbeault representative of the Atiyah class.

This is also seen from the commutative square that Atiyah writes

$$\require{AMScd}\begin{CD} H^1(O^*_M) @>>> H^2(M,{\Bbb Z}) \\ @VVV @VVV \\ H^1(\Omega^1_M) @>>> H^2(M,{\Bbb C}) \end{CD}$$

which is valid in general situation, non-compact or non-Kähler as well.

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  • $\begingroup$ TeX note: \mathop affects only the spacing (and that possibly not as you want); it does not affect the typesetting. For example, compare $\mathop{im} dd^c$ \mathop{im} dd^c to $\operatorname{im} dd^c$ \operatorname{im} dd^c, where the latter is probably what you meant. I have edited accordingly. \mathop is for operators like \sum and \lim (the "large operator" class), not for arbitrary operator names. $\endgroup$
    – LSpice
    Oct 31, 2022 at 16:26
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    $\begingroup$ It seems to me like, in the bottom left, $\Omega^1$ should be replaced by $Z^1$, the sheaf of closed holomorphic $(1,0)$-forms. There is a boundary map $H^1(Z^1) \to H^2(\mathbb{C})$, coming from the short exact sequence $0 \to \mathbb{C} \to \mathcal{O} \to Z^1 \to 0$, and there is a map $H^1(Z^1) \to H^1(\Omega^1)$, coming from the inclusion $Z^1 \to \Omega^1$, but I don't think there is a natural map $H^1(\Omega^1) \to H^2(\mathbb{C})$ in general. $\endgroup$ Oct 31, 2022 at 16:44
  • $\begingroup$ I can't see why the Dolbeault image of the Bott-Chern class is equal to the Dolbeault representative of the Atiyah class, can you elaborate on it a bit? $\endgroup$
    – Tom
    Nov 1, 2022 at 9:13
  • $\begingroup$ they are both represented by $\partial \bar\partial f$ (up to a sign and a constant) $\endgroup$ Nov 1, 2022 at 17:37

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