Unique connection on a connected Lie group $G$ with certain properties So let $G$ be a connected lie group with lie algebra $\mathfrak{g}$. I would like to show that there exists a unique connection $\nabla $ on $TG$ that is invariant under left and right translations and under inversion. Now I was able to prove that such a connection exists by setting $\nabla_X Y=\frac{1}{2}[X,Y]$ . Now I am not sure how to see the uniqueness statement, might be because the lie group is connected and so it's generated by a neighborhood of the identity but I am not sure if this is helpful.
Then it's easy to check that for this connection the exponential map for a lie group and the exponential map for a Riemannian manifold are the same , and that the geodesics are left-translations by the one-parameter subgroups of $G$.
I would also like to see that parallel transport along the curve $exp(tX)$ is $\tau_t(v)=dL_{\exp(tX/2)}(dR_{\exp(tX/2})v)$. I am not being able to check this either, I have tried using local coordinates but everything just becomes a big mess and I am not sure that is helpfull , I just can't seem to able to see that $\nabla_{\frac{d}{dt}}s=0$, where $s$ is the vector field along the curve $s(t)=dL_{\exp(tX/2)}(dR_{\exp(tX/2})v)$. Does anyone have any advice for this ?
Thanks in advance.
 A: Are you sure that there is a unique connection invariant under left- and right- translation and inversion? It seems to me that any scalar multiple of the connection you wrote has those properties.
The connection on $G$ defined by $\nabla_X Y = \frac{1}{2} [X,Y]$ for left-invariant vector fields $X,Y$ is the unique torsion-free connection with $1$-parameter subgroups as geodesics [see Nomizu "Invariant affine connections on homogeneous spaces"]. As you showed, this connection is also invariant under right-translation.
It would be satisfying to compute the parallel transport from the connection, but since you already guessed it, you can verify it with an easy computation. For a left-invariant vector field $v$ transported via $\tau$ along the curve $e^{tX}$ we have
$$ \tau_t^0 (v_{e^{tX}})= \mathrm{d} l_{e^{-tX/2}} \mathrm{d} r_{e^{-tX/2}} (v_{e^{tX}}) =  \mathrm{d} l_{e^{tX/2}} \mathrm{d} r_{e^{-tX/2}} (v_1) = \mathrm{Ad}(e^{tX/2}) (v_1) $$
so
$$ \frac{\mathrm{d}}{\mathrm{d}t} \tau_t^0 (v_{e^{tX}}) \bigg|_{t=0} = \frac{\mathrm{d}}{\mathrm{d}t} \mathrm{Ad}(e^{tX/2}) (v_1) \bigg|_{t=0} = \frac{1}{2}[X,v] $$
as desired. In other words, $\tau$ satisfies defining properties of the parallel transport with respect to this connection.
Also, a word about one more point of confusion. For a general Lie group, there may not exist any bi-invariant metric, so this connection would not be the Levi-Civita connection of a metric in that case. Of course you can still talk about the exponential map of this connection but it's not really a "Riemannian" exponential map. It is true that the exponential map of this connection restricts to the Lie theoretic exponential map at the identity.
