# Parallel transport on the Group manifold

We think of the group as a manifold G (called a Group manifold), whose points are the elements of our Lie group. More generally, we could think of any manifold H on which the elements act as smooth transformation

The infinitesimal group elements are to be pictured as particular vector fields on G (or, indeed, H). That is, we think of ‘moving G’ infinitesimally along the relevant vector field $$\zeta$$ on G, in order to express the transformation that corresponds to pre-multiplying each element of the group by the infinitesimal element represented by $$\zeta$$.

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The relevant notion of ‘parallelism’ comes from the group action, supplying the needed notion of ‘parallel transport’, which actually gives a connection with torsion but no curvature.

Page-312, 313 Roger Penrose's Road to Reality

Could someone explain what Penrose is meaning by the bolded part? I understand the group action is to move the points under flow of vector field but what does mean for the parallelism to come from it?

I have read these answers but found the discussion to be too algebraic. Could a more visual understanding of what's going on be provided?

• How about Alexander Thumm's answer in the linked question? I think this answer follows the definition of parallel transport in a more friendly way Aug 17 at 13:34
• Helpful, but I felt that it doesn't really a give a geometric picture for how the parallel transport or curvature is considered here @onriv Aug 17 at 15:13

The connection that Penrose is referencing is called the Maurer-Cartan connection, defined by the Maurer-Cartan form $$\theta\in\Omega^1(G,\mathfrak{g})$$. The assertion that the connection has no curvature is known as the Maurer-Cartan equation, which states that $$d\theta+\frac{1}{2}[\theta,\theta]=0$$ Edit: The Maurer-Cartan form is defined in the most natural way you could imagine, since we have a group structure on $$G$$. Denote left multiplication with $$g$$ by $$L_g$$, which is a map $$G\to G$$. We define the $$1$$-form as $$\theta_g(v):=dL_{g^{-1}}v\in T_eG=\mathfrak{g}$$, which gives rise to a Lie algebra valued $$1$$-form. This allows us to identify the fibres of the tangent bundle with one another, which is equivalent to the data of a connection which we denote $$\nabla$$. A vector field $$X$$ is called "parallel" or "horizontal" along a curve $$\gamma:I\to G$$ if it satisfies $$\nabla_{\dot{\gamma}}X=0$$ The way in which one might think about this geometrically, is as follows. On Euclidean space, it is quite clear when we should call a function flat in a certain direction: when its directional derivative vanishes. However, on vector bundles over manifolds, there is generally no canonical way to choose a directional derivative, i.e. a way to differentiate sections along vector fields. This is what a connection does. The way we can think about parallel sections, then, is by thinking about them as being constant with respect to the chosen connection.

On a Lie group, we have a canonical way of choosing such a connection, because the (left or right) invariant vector fields give us a trivialisation of $$TG$$, and this connection is the Maurer-Cartan connection. Perhaps even more concretely, the connection is defined by declaring that the invariant vector fields are its flat/parallel sections. The connection is determined by this, using the Leibniz rule and the trivialisation of $$TG$$ by invariant vector fields. Note: the Maurer-Cartan connection typically does not coincide with the trivial connection on a trivial bundle $$G\times\mathbb{R}^k$$, although both connections are flat. For example, work out the cases $$G=(\mathbb{R},+)$$ and $$G=(\mathbb{R}_{>0},\times)$$, which are isomorphic as Lie groups. Draw the flat sections, as functions $$f:\mathbb{R}\to\mathbb{R}$$ in each case.

• Thank you for sharing your knowledge. Could you please add more about how to think about the connetion and the underlying geometric intuition of it, if any? Aug 17 at 18:56
• I have updated the answer to reflect more geometric intuition, I hope that helps. Aug 17 at 21:17
• Is the Maurer-Cartan connection a metric connection? Aug 18 at 4:24
• @onriv Yes: Since left-invariant vector fields are parallel, so are all left-invariant tensors, including all left-invariant metrics. Aug 18 at 4:50

It's also helpful to contrast with the situation on an arbitrary smooth manifold $$M=M^n$$. In the category of smooth manifolds, every smooth manifold $$M$$ has an associated tangent bundle $$TM$$, where each tangent space $$T_xM \cong \mathbb R^n$$, but there is not a canonical isomorphism of tangent spaces $$T_xM\cong T_yM$$.

In Riemannian geometry, we add a Riemannian metric to $$M$$ (i.e. a certain two-form with some conditions imposed on it) to allow us to measure the angles and distances between vectors in the tangent space to $$M$$. We can go even further by equipping $$M$$ with an affine connection $$\nabla$$, and an associated parallel-transport map $$P_{x\to y}:T_xM\cong T_yM$$, which we remember is determined by the extra data of the connection $$\nabla$$. The parallel-transport map is a canonical isomorphism of tangent spaces, but only up to the choice of a connection $$\nabla$$. It would be especially nice if there were a natural canonical choice for a canonical isomorphism of tangent spaces.

Any Lie group $$G$$, which is a smooth manifold with a smoothly compatible group structure, does have a natural notion of parallelism (i.e. natural identification of tangent spaces) which is described nicely in the other answer.