# Trivial canonical bundle

Let $X$ be a complex compact Kähler minimal surface with zero algebraic dimension and $H^2(X,\mathcal{O}_X) \ne 0$. We know that according to Enriques–Kodaira classification, $X$ is either a torus or a K3 surface. In particular, its canonical bundle is trivial.

My question is, are there are any direct ways to show that the canonical bundle $K_X$ is trivial without using the classification of complex compact surface? Thank you!

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Do you mean zero Kodaira dimension? I don't think "algebraic dimension" is standard because it sounds like it means "dimension as an algebraic variety". Doesn't the term Kahler mean there is an everywhere non-degenerate global 2-form? This should trivialize the canonical bundle. – Matt Aug 21 '12 at 22:55
This is weird. Cross posted, but by different people?! mathoverflow.net/questions/105198/trivial-canonical-bundle – Matt Aug 21 '12 at 23:00
Algebraic dimension is the transcendence degree of the function field of $X$ over $\mathbb{C}$. Kähler forms are of type $(1,1)$, whereas a section of the canonical bundle is a holomorphic form of type $(2,0)$. I didn't see anyway of trivializing $K_X$ from the existence of Kähler forms – Jacob Ikabruob Aug 22 '12 at 5:39
I only know this from algebraic geometry, so assume your complex surface is algebraic. Then: if the transendence degree of the function field over $\mathbb{C}$ is zero, then your manifold is zero dimensional, hence a finite collection of points...... The dimension of the manifold agrees with the transendence degree, so the function field of your surface has transendence degree 2 over $\mathbb{C}$... I'm also pretty sure you mean Kodaira dimension.. – Joachim Aug 22 '12 at 9:01
We are in the complex geometry context, there could exist manifolds which dimension doesn't agree with the algebraic one. By the way, manifolds with equal dimension and algebraic dimension are called Moishezon manifolds. – Jacob Ikabruob Aug 22 '12 at 11:54