Does this attempt to get a nontrivial theory of orientation for complex manifolds actually work? Consider the following construction of the orientations of a smooth manifold:

Let $M$ be an $n$-dimensional smooth manifold, $E = \Omega^n(M)$ its space of top-level forms, and $E_0 \subset E$ the whole bundle minus its zero section. Let $\mathbb R^+$ act fiberwise multiplicatively on $E_0$ in the obvious way. Then $M$'s orientations are precisely the global sections of the quotient bundle $O = E_0 / \mathbb R^+$.

Every complex manifold has a canonical orientation in this sense, of course. I want to tweak the construction so as to obtain a nontrivial theory of orientability of complex manifolds. Suppose I tried the following:

Let $M$ be an $n$-dimensional complex manifold, $E = \Omega^n(M)$ be its space of complex top-level forms, $E_0 \subset E$ the whole bundle minus its zero section. Let $\mathbb R^+$ act fiberwise multiplicatively on $E_0$ in the obvious way. Then $M$'s orientations are precisely the global sections of the quotient bundle $O = E_0 / \mathbb R^+$.

Note that, this time, $E_0$'s fibers are punctured planes $\mathbb C^\star$, hence $O$'s fibers are unit circles $S^1 = \mathbb C^\star / \mathbb R^+$. Thus, a “complex orientation” of $M$ is only defined up to rotation. Now my questions are

*

*Is this theory still trivial, i.e., does every complex manifold admit a canonical orientation in this sense? I think the answer is no (a compact Riemann surface that is not a torus should be a counterexample, because then zero is not a canonical divisor), but I would still like confirmation by a more knowledgeable third party.


*Is this theory still useful for investigating the topology of complex manifolds? For example, I would like to be able to classify the topological obstructions to a complex manifold being “complex orientable”.
 A: I will give a (very) rough sketch of an argument showing that your notion of orientation for complex manifolds is equivalent to the notion I gave in my comment. I admittedly don't have sources for some unverified claims, but I worked out the details to a few of them (using things like the Borel construction aka homotopy quotient) and I think they all go through. You might be able to find much of this in Ebert's course notes on Index Theory.
As I mentioned in my comment you can consider the following form of orientation: given a complex $n$-manifold $M$ and a classifying map $c\colon M \to BU(n)$ of $TM$, a complex orientation of $M$ is a (homotopy class of) lift of $c$ to $BSU(n)$. (This interpretation parallels the interpretation of orientations for real vector bundles as lifts of classifying maps along $BSO(n) \to BO(n)$.)  If $\kappa \colon U(n) \to U(n)$ denotes the conjugate-transpose map (which is a homeomorphism and anti-homomorphism) then the dual of the tangent bundle $T^*M$ is classified by $B\kappa \circ c$, and $c$ admits a lift to $BSU(n)$ iff $B\kappa \circ c$ does.
To relate this notion of orientation to yours, for a complex manifold $M$ of complex dimension $n$ let $$det(M) = \Lambda^n(T^*M)$$
denote the determinant line bundle of its cotangent bundle, so that $\Omega^n(M) =\Gamma det(M)$. If we choose a hermitian metric then the unit sphere bundle $S = S(det(M))$ is isomorphic to your bundle $O$, and if $T^*M$ is classified by $c\colon M \to BU(n)$ then $det(M)$ and $S$ are both classified by $Bdet\circ c \colon M \to BU(1)$. More specifically $S\cong (Bdet \circ c)^* EU(1)$ and $det(M)$ is the associated complex line bundle.
Since we have the fibration sequence $BSU(n) \to BU(n) \stackrel{Bdet}{\to} BU(1)$ then as a circle bundle over $BU(n)$ we have $BSU(n) \simeq (Bdet)^* EU(1)$, so in fact $S \cong c^* BSU(n)$. Finally, using the definition of the pullback you can then see that sections of $S$ are canonically identified with lifts of $c$ to $BSU(n)$.

I should add, just to actually answer your question and justify why this interpretation is useful, in this context you can show a complex vector bundle $E$ admits an orientation in the "lifting of classifying map" sense iff $c_1(E)=0$ (this essentially boils down to the fact that $c_1(E) = c_1(det(E)))$, again paralleling the situation for real vector bundles where orientability is equivalent to $w_1 = 0$. This verifies your suspicion about complex surfaces, since $c_1(T)=0$ but $c_1(\Sigma) \neq 0$ for all other closed complex surfaces, since in this case the first Chern number is equal to the Euler characteristic.
