Let $G$ be a Lie group with Lie algebra $g$. As it is well known the MaurerCartan form $ω:TG\rightarrow g$ transports any vector $X\in T_{x}G$ to the start $l_{x^{1}*}(X)\in g$, $l_{x^{1}}$ denoting the left translation. Let $σ:[0,1]\rightarrow G$ a smooth path on $G$. It there a way to define path integration on $G$ such that $\int_{σ}{ω}=σ(1)σ(0)^{1}$?

1$\begingroup$ This is only a heuristic comment: if such a formula existed it would show that the integral of $\omega$ depended only on the endpoints of the path so $\omega$ would be closed. This would violate the MaurerCartan equation $d\omega+\frac{1}{2}[\omega,\omega]=0$. $\endgroup$– Liviu NicolaescuNov 21, 2018 at 6:27

$\begingroup$ @LiviuNicolaescu, yes this is the case for Rvalued forms and usual integration. But here we have a (possibly non abelian Lie group) and we are asking for an integration independent of path, although $ω$ may be not exact. However, $ω$ is integrable. A motivation for my question was the book: Sharpe, Differential Geometry (Springer) pages 118126. $\endgroup$– AllotriosNov 21, 2018 at 15:35
1 Answer
The tangent bundle of the Lie group is canonically trivialized, by left or right translations (depending on your conventions). The MaurerCartan $1$form defines a connection on the tangent bundle and the MaurerCartan equation state that this connection is flat. Thus the parallel transport of this connection along a loop depends only on the homotopy type of that loop. In particular, if the group is simply connected, the parallel transport along any loop is trivial. Thus, the parallel transport along a path depends only on its endpoints. The integral of $\omega$ along a path should be defined to be this parallel transport. (I think this parallel transport is ${\rm Ad}_{\sigma(1)\sigma(0)^{1}}$, depending on how you choose your conventions: Lie algebra  space of left/right invariant vector fields.)
Remark. Let me point out a nice fact. Pick an ${\rm Ad}$invariant homogeneous polynomial of degree $k$ on the Lie algebra of $G$. On $TG$ there are two natural connections: the trivial connection $\nabla^0$ and the connection $\nabla^\omega$ determined by the MaurerCartan form $\omega$. Both connections are flat!.
The ChernWeil construction associates to the polynomial $P$ and each $G$connection $\nabla$ on $TG$ an closed form $P(\nabla)$ of degree $2k$. This form vanishes if $\nabla is flat$. Moreover, for any connections $\nabla',\nabla$ on $TG$ there exists a canonical transgression form $TP(\nabla',\nabla)$ of degree $2k1$ such that $$ P(\nabla')P(\nabla)= dTP(\nabla',\nabla). $$ Applying this to our connections $\nabla^0,\nabla^\omega$ we deduce that $TP(\nabla^\omega, \nabla^0)$ is a closed form of degree $2k1$. According to C. Chevalley the close forms $TP(\nabla^\omega, \nabla^0)$ generate the cohomology ring with real coefficients of the Lie group when $P$ runs through the Adinvariant homogeneous polynomials on the Lie algebra. For more details I refer to Remark 8.1.15 of this book for more details.

$\begingroup$ Thank you for your answer and your book. It clarified lot of things. Though, may i ask: So, if , instead of the above tangent bundle on $G$, we consider the trivial primary fiber bundle $p: G\times G \rightarrow G$ with connection form the pullback $p^*ω$ of $ω$, then this new connection is again flat, so what you just said above should be applied also here. But now the parallel transport would be $σ(1)σ(0)^{1}$. Do you think this is right? $\endgroup$ Nov 21, 2018 at 18:38

$\begingroup$ Yes, but this time instead of adjoint action you have left multiplication action. I'm almost sure about this but you would have to check this yourself. $\endgroup$ Nov 21, 2018 at 18:44