I am wondering why we have the isomorphism stated in the title. Concretely, we have the following exact sequences of sheaves: $$0\rightarrow\mathbb R\rightarrow\mathcal O\rightarrow\mathcal O/\mathbb R\rightarrow 0$$ Where $\mathbb R$ denotes the constant sheaf over $M$ and the first map is the canonical inclusion. If we pass to the long exact sequence in cohomology, we have that $$\dots\rightarrow H^0(M,\mathcal O/\mathbb R)\rightarrow H^1(M,\mathbb R)\rightarrow H^1(M,\mathcal O)\rightarrow H^1(M,\mathcal O/\mathbb R)\rightarrow\dots$$ is exact, but somehow, the groups at the extremes should be trivial whenever $M$ is a compact Riemann surface.

I know this has to do with Hodge theory over compact Kähler manifolds, but although I have been searching for a reference of this fact, I haven't found it.

How can we deduce this result from Hodge theory?

Thanks in advance for your answers.

  • $\begingroup$ Is there a mistake in the title? Who’s $X$? $\endgroup$
    – Plop
    Commented Feb 6, 2022 at 0:46

2 Answers 2


This is not a very natural isomorphism, I guess. I'm not sure where you found this. One is a real vector space and the other is a complex vector space, so in what sense are these isomorphic? I will give the argument to deduce that the two have the same dimension as real vector spaces. I will use the Dolbeault isomorphism $H^{p,q}(M) \cong H^q(M,\Omega^p)$.

If the genus of $M$ is $g$, then $\dim_{\Bbb R} H^1(M,\Bbb R) = 2g$ and $g=\dim_{\Bbb C} H^0(M,\Omega^1) = \dim_{\Bbb C} H^{1,0}(M)$. On the other hand, it follows from harmonic theory (taking complex conjugates of harmonic representatives) that $H^1(M,\mathscr O) \cong H^{0,1}(M) \cong \overline{H^{1,0}(M)}$, and so $\dim_{\Bbb C} H^1(M,\mathscr O) = g$, as well.

You can deduce the statement slightly more indirectly from the Hodge decomposition: $H^1(M,\Bbb C) \cong H^{1,0}(M)\oplus M^{0,1}(M)$. Then $$H^1(M,\Bbb R)\otimes\Bbb C \cong H^1(M,\mathscr O)\oplus \overline{H^1(M,\mathscr O)}.$$ The claim on dimensions follows immediately from this.

  • $\begingroup$ If you are curious, this appears in Nitsure's online PDF slides about the Narasimhan-Seshadri theorem. He uses that the map I presented is an isomorphism in order to prove the theorem in the simplest case, that is, when we are dealing with line bundles. $\endgroup$
    – Akerbeltz
    Commented Feb 1, 2022 at 18:24
  • $\begingroup$ Interesting. In my 45 years doing complex geometry, I don't think I ever bumped into this one. I still wonder in what category the isomorphism is defined. $\endgroup$ Commented Feb 1, 2022 at 18:35
  • 1
    $\begingroup$ It's worth noting that this result (the two real vector spaces have the same dimension) is true on compact Kähler manifolds, but it is no longer true in the non-Kähler case. For example, on a Hopf surface $X$, we have $\dim_{\mathbb{R}}H^1(X; \mathbb{R}) = 1$ and $\dim_{\mathbb{R}}H^1(X, \mathcal{O}) = 2$. $\endgroup$ Commented Feb 1, 2022 at 20:59

The first three terms of the long exact sequence already form the short exact sequence $0 \rightarrow \mathbb{R} \rightarrow \mathbb{C} \rightarrow \mathbb{C}/\mathbb{R} \rightarrow 0$, since the global holomorphic functions on a compact complex manifold are constant. Therefore, the map $H^0(M,\mathcal{O}/\mathbb{R}) \rightarrow H^1(M,\mathbb{R})$ factors by $0$ and thus the map $H^1(M,\mathbb{R}) \rightarrow H^1(M,\mathcal{O})$ is injective. Hodge decomposition and Serre duality imply that both are vector spaces of the same dimension, therefore the map is an isomorphism.


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