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 $\zeta$. enter image description here


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?

  • $\begingroup$ 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 $\endgroup$
    – onRiv
    Aug 17, 2022 at 13:34
  • $\begingroup$ 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 $\endgroup$ Aug 17, 2022 at 15:13

2 Answers 2


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.

  • $\begingroup$ 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? $\endgroup$ Aug 17, 2022 at 18:56
  • $\begingroup$ I have updated the answer to reflect more geometric intuition, I hope that helps. $\endgroup$ Aug 17, 2022 at 21:17
  • $\begingroup$ Is the Maurer-Cartan connection a metric connection? $\endgroup$
    – onRiv
    Aug 18, 2022 at 4:24
  • 1
    $\begingroup$ @onriv Yes: Since left-invariant vector fields are parallel, so are all left-invariant tensors, including all left-invariant metrics. $\endgroup$
    – Kajelad
    Aug 18, 2022 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.


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