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?