Fiber bundle understanding of the wavefunction Usually people say that given a wavefunction $\Psi$ although $|\Psi(\cdot, t)|^2$ is the probability density for the position random variable at time $t$, the wavefunction $\Psi$ itself has no physical meaning. It occurs, however, that if $M$ is spacetime, then we know that $\Psi : M\to \mathbb{C}$. In that setting, if one considers $E = M\times \mathbb{C}$ and $\pi : E\to M$ given by $\pi(a, z)=a$ then $(E,M,\pi)$ is the trivial bundle with typical fibre $\mathbb{C}$
In that case if we let $s\in \Gamma(E)$ be a section of the bundle, we have $s(a)=(a,\xi(a))$. In that case the wavefunction can be thought of as a section of the bundle $\Psi(a)=(a,\xi(a))$ where $\xi(a)$ is what we called wavefunction earlier.
In that case, it seems we are, at each point of spacetime, plugging one copy of the plane and laying down a vector on the plane. This vector is the wavefunction at the point.
It is tempting, therefore, to generalize this a little further and consider a general vector bundle $(E,M,\pi)$ where $M$ is spacetime and $\mathbb{C}$ is the typical fibre, only this time we allow twists. Then locally, given an event $a\in M$ there is one open set $U\subset M$ with a local trivialization $\varphi : \pi^{-1}(U)\to U\times \mathbb{C}$ where the arguments work as before.
So my question is: this way to view the wavefunction has any advantage? The generalization to allow non-trivial bundles give something useful? This point of view allows one to understand the meaning of the wavefunction itself?
 A: Your introductory claims are not correct. The usual wavefunction is not a function of spacetime, for two reasons: It is not Lorentz invariant (time is singled out, so it is not a function of a point on a spacetime manifold, and it is not even time dependent at all in the Heisenberg picture) and it can depend on more than one "set" of positions - for a system of $n$ particles in $d$ spatial dimensions, the wavefunction will depend on $n\cdot d$ position variables. Thus, viewing it as a section of something over spacetime is inappropriate.
If you want to understand a quantum state as the section of a complex line bundle, that is possible through the procedure of geometric quantization.
Let $X$ be the classical phase space with symplectic form $\omega$.
The actual line bundle you have to consider is the prequantum line bundle, which is a line bundle over the phase space equipped with a $\mathrm{U}(1)$-connection $\nabla$ such that the curvature of the connection is the symplectic form $\omega$. If you have now a choice of polarization on the phase space, i.e. a split of the coordinates into positions $q$ and momenta $p$, the actual wavefunctions of quantum mechanics are the sections of the prequantum line bundle which only depend on position, i.e. are constant on surfaces of constant $q$.1
If you're wondering where time evolution is here: We have only said what the space of states is. The time evolution is of course given by the unitary operator generated by the Hamiltonian (which is encoded in $\omega$) acting on this space of states. Generically, the action of the quantum version of a phase space observable $f$ is given by the covariant derivative (defined by the connection $\nabla$) of the section along the vector field $X_f$ associated to $f$ by $\mathrm{d}f(\dot{}) = \omega(X_f,\dot{})$. More precisely, $f$ acts on a section $\psi$ as
$$ \hat{f}(\psi) = \mathrm{i}\nabla_{X_f}\psi + f\cdot\psi$$
where $\hat{f}$ is now the (pre-)quantum operator associated to $f$.
Now, lastly, for the (non-)triviality of the bundle:
Nothing forbids the prequantum line bundle from being non-trivial, but the classical phase space needs to have non-trivial cohomology for that - complex line bundles are completely characterized by their first Chern class, which is an element in $H^2(X,\mathbb{Z})$. Thus, in most situations you will encounter in typical quantum mechanics applications (where the phase space is just a symplectic vector space, and hence in particular contractible, so most cohomologies vanish), the prequantum line bundle will indeed be trivial.

1Also, they should be square integrable in an appropriate sense. This requires discussing how we equip the space of sections with a Hilbert space structure and what measure we integrate against - these choices are done by a metaplectic correction.
