# Coefficient of an 1-form in position-representation of momentum operator where configuration space is NOT $\mathbb{R}^m$

I found this in the book Geometric Phase in Quantum Systems by A. Bohm et al.

Where the position space representation of the momentum operator carries a (Where exactly my doubt is) coefficient of 1-form with the condition

$$\partial_i \omega_j - \partial_j \omega_i =0 \implies d\omega=0$$

The author(s) argued about $$Poincare \ lemma$$ and how, for $$\mathbb{R}^m$$ configuration space the term can be $$gauged \ away$$.

I understand usual momentum operator representation without this 1-form, and this is very non-trivial for me.

Can someone please explain me how this comes and what it means in details?

1. For $$M=\mathbb{R}^m$$, starting from the canonical commutation relations (CCRs), we have the Stone-von Neumann theorem, which proves the existence of the standard Schrödinger position representation (i.e. without the 1-form $$\omega$$) up to unitary equivalence, cf. e.g. this Phys.SE post.
2. However conjugating with unitary multiplication operators naturally induces an exact $$\omega$$ one-form, cf. e.g. my Phys.SE answer here.
3. Therefore it is quite natural to consider a 1-form $$\omega$$ from the onset, even for a topologically non-trivial $$m$$-dimensional configuration space $$M$$.
4. The CCRs then imply that the 1-from $$\omega$$ must be a closed one-form. The cohomology class $$[\omega]$$ is conserved under conjugating with unitary multiplication operators.
5. Since $$M=\mathbb{R}^m$$ is a contractible space, we may use Poincare Lemma to show that the cohomology class $$[\omega]=[0]$$ is trivial.