Is it a complete coincidence that the Euler characteristic of a curve is exactly twice the Euler characteristic of its structure sheaf?



Euler characteristic of a curve
To spell out potentially dense's comment, the Hodge theorem tells you that 
$$\begin{array}{rcl}
H^0(X, \mathbb{C}) &\cong& H^0(X, \mathcal{O}) \\
H^1(X, \mathbb{C}) &\cong& H^1(X, \mathcal{O}) \oplus H^0(X, \Omega^1) \\
H^2(X, \mathbb{C}) &\cong& H^1(X, \Omega^1) \\
\end{array}$$
where the left hand sides are topological cohomology and the right hand sides are sheaf cohomology.


Serre duality tells us that $H^1(X, \mathcal{O}) \cong H^0(X, \Omega^1)^{\vee}$ and $H^0(X, \mathcal{O}) \cong H^1(X, \Omega^1)^{\vee}$. So
$$\dim H^0(X, \mathbb{C}) - \dim H^1(X, \mathbb{C}) + \dim H^2(X, \mathbb{C}) = 2 \left( \dim H^0(X, \mathcal{O}) - \dim H^1(X, \mathcal{O}) \right)$$
as desired.


