Lie bracket is a connection?

Solution 1:

Te connection is indeed $\nabla_LM=[L,M]$ if $L$ is left-invariant; notice that if you have a vector at some point of your Lie group, it can be uniquely extended to a left invariant vector field, so the formula makes sense and defines a connection (btw $\phi L$ is not left-invariant (unless $\phi$ is constant)).

The torsion is (for left-inv. vect. fields) $T(K,L)=[K,L]-[L,K]-[K,L]=[K,L]$, so it's not $0$.

The curvature, on the other hand, is $R(K,L,M)=[K,[L,M]]+\dots=0$ by Jacobi identity. Another way to see that $R=0$ is that any right-invariant vect. field $M$ satisfies $\nabla M=0$, as left and right-inv. fields commute.

Solution 2:

The parallel transport Penrose is talking about is given by translation in the group. More precisely, the parallel transport $\Gamma(\gamma)^t_s\colon T_{\gamma(s)}G \to T_{\gamma(t)}G$ alongside a curve $\gamma$ is simply the differential of the left translation by $\gamma(t)\cdot \gamma(s)^{-1}$.

We can retrieve the connection from this as follows: Let $\gamma$ be a smooth curve, $X = \dot\gamma(0)$, and $V$ a vector field over $\gamma$ then $$\nabla_X V = \frac{\mathrm d}{\mathrm d t} \Gamma(\gamma)_t^0 V_{\gamma(t)} \bigg\vert_{t=0}.$$

It is now easy to see that $\nabla_XV = 0$ for a left-invariant vector field $V$ and $\nabla_XW = [X,W]$ for a right-invariant vector field $W$. Moreover, as this parallel transport is independent of the curve connecting two points of $G$, the curvature of its associated connection must vanish.

Of course we can define parallel transport by right translation as well and obtain a second connection on $TG$. These two connections agree if $G$ is commutative.