If $M$ is a compact Riemann surface, then $H^1(M,\mathbb R)\cong H^1(M,\mathcal O)$

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?

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

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.
• 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 $\dim_{\mathbb{R}}H^1(X, \mathcal{O}) = 2$. 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.