Is the geodesic exponential map for a Lie group with the (-)-connection a diffeomorphism? This connection is one of two flat connections introduced by Cartan and Schouten on a Lie group and has torsion $T(X,Y)=-[X,Y]$.
[Math] The (-)-Connection on a Lie Group
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Related Solutions
Yes.
Let $\nabla$ be an arbitrary connection on the tangent bundle of a Riemannian manifold $(M,g)$. The standard trick for expressing the Levi-Civita connection in terms of $g$ gives you, for any 3 vector fields $X$, $Y$, $Z$: $$Xg(Y,Z)+ Yg(Z,X)- Zg(X,Y)= N(X,Y,Z) $$ $$+ g(T(X,Z),Y)+ g(T(Y,Z),X)- g(T(X,Y),Z) $$ $$ +2 g(\nabla_X Y,Z)- g([X,Y],Z) + g([X,Z],Y) + g([Y,Z],X),$$
where $$ T(X,Y)=\nabla_X Y- \nabla_Y X -[X,Y]$$ is the torsion of $\nabla$ and $$ N(X,Y,Z)= \nabla_Xg(Y,Z)+ \nabla_Yg(Z,X)-\nabla_Zg(X,Y). $$ This is the "non-metricity": $N=0\Leftrightarrow \nabla g=0$.
Now, turning to the case at hand: we define the $\pm$ and $0$ connections by $$ (\nabla_X Y)_e=\epsilon [X,Y],$$ $ \epsilon = 1, 0, \frac{1}{2}$ respectively, so the torsion is $$T(X,Y) = (2\epsilon -1)[X,Y]= \pm[X,Y]\textrm{ or } 0, $$ hence the names of the connections. But then you get $$ 0 = N(X,Y,Z) -2\epsilon\left[ g([Z,Y],X) + g(Y,[Z,X]) \right],$$ and the second summand is zero due to bi-invariance, so $N=0$.
(this is also related to The (-)-Connection on a Lie Group, Metric Connections on a Lie Group)
If we want to speak about bi-invariant metric connections,then we cannot drop the assumption of compactness. Actually, a compact connected Lie group $G$ with a bi-invariant Riemannian metric $\rho$ can be viewed as a Riemannian symmetric space of the form $((G\times G)/{\rm diag}(G), \rho)$. If this Lie group is in addition simple, then it can be considered as a compact, isotropy irreducible, Riemmanian symmetric space, the so-called of Type II in Helgason's book. The classical Cartan-Schouten theorem is about a compact simple Lie group $G$, and it states that the unique flat bi-invariant metric connections on $G$ are the so-called +1 and -1 connections, say $\nabla^{\pm 1}$. They have non-zero (skew)torsion $T^{\pm 1}(X, Y)=\pm [X, Y]$. Moreover, $\nabla^{\pm 1}T^{\pm 1}=0$. In general, one can construct a 1-dimensional family $\{\nabla^{t} : t\in R\}$ of bi-invariant metric canonical connections on $G$, which joins the Levi-Civita connection and the $\pm 1$-connections. This family occurs by a reductive decomposition $\frak{g}\oplus\frak{g}=\Delta_{\frak{g}}\oplus{\frak{m}}_{t}$, which generalizes (and includes) the classical Cartan decomposition of $G$ (the latter induces the L-C connection on $G$). For example, see Section 4/page 8 of the following paper
http://arxiv.org/pdf/1111.5044.pdf
Notice that by the term canonical, we usually mean these (bi-invariant) connections on $G\cong (G\times G)/{\rm diag}(G)$, say $\nabla$, for which the $\nabla$-parallel tensor fields are exactly the $(G\times G)$-invariant tensor fields.
This 1-parameter family of bi-invariant metric connections on $G$, has non-trivial (parallel) skew-torsion (except the trivial case of the Levi-Civita connection) and only the values $\pm 1$ give rise to flat metric connections. For all the other values of the parameter $t$, the associated curvature is non-zero.
An easy way to compute the curvature and the torsion (or its covariant derivative) is by using the correspondence between bi-invariant affine connections on $G$ and bilinear maps $\lambda : \frak{g}\times\frak{g}\to\frak{g}$ which are ${\rm Ad}(G)$-equivariant, i.e. $\lambda({\rm Ad}(g)X, {\rm Ad}(g)Y)={\rm Ad}(g)\lambda(X, Y)$ for any $X, Y\in\frak{g}$ and $g\in G$. Then
$$R(X, Y)=[\Lambda(X), \Lambda(Y)]-\Lambda([X, Y])$$ $$T(X, Y)=\Lambda(X)Y-\Lambda(Y)X-[X, Y],$$ where $\Lambda :\frak{g}\to{\rm End}(\frak{g})$ is the equivariant endomorphism associated to $\lambda$, i.e. $\Lambda(X)Y=\lambda(X, Y)$. It is easy to see that $\lambda$ induces a bi-invariant metric connection on $G$, if and only if $\Lambda(X)\in\frak{so}(\frak{g})$ for any $X\in\frak{g}$, i.e. $$\langle \Lambda(X)Y, Z\rangle+\langle Y, \Lambda(X)Z\rangle =0 \quad \forall \ X, Y, Z\in\frak{g}.$$
For example, the 1-parameter family of bi-invariant canonical metric connections on $G$ is induced by the bilinear map $\lambda(X, Y)=((1-t)/2)[X, Y]$ (up to scalar and sign), but it depends how we consider the reductive decomposition $\frak{g}\oplus\frak{g}=\Delta_{\frak{g}}\oplus{\frak{m}}_{t}$.
Best Answer
The $(+)$/0/$(-)$-connections on a Lie group $G$ are the left-invariant connections (I will identify connections with covariant derivatives) defined on left-invariant vector fields by $\nabla_XY=c[X,Y]$ with respectively $c=1,\frac{1}{2},0$. Their name is related to the fact that their torsion is $T(X,Y)=d[X,Y]$ with respectively $d=1,0,-1$.
The name is also, and perhaps more importantly, related to the fact that in the general theory of invariant connections on reductive homogeneous spaces (see Kobayashi-Nomizu Vol. I Chap. X.2, applied here to $(G\times G)/\operatorname{diag}(G)$) they come from a choice of $\mathfrak k$-invariant subspace of $\mathfrak g\times\mathfrak g$ supplementary to $\mathfrak k=\{(X,X)\mid X\in \mathfrak g\}$ given by $\mathfrak m_+=\{(0,X)\mid X\in\mathfrak g\}$, $\mathfrak m_0=\{(X,-X)\mid X\in\mathfrak g\}$ or $\mathfrak m_-=\{(X,0)\mid X\in\mathfrak g\}$, where $\mathfrak g=\operatorname{Lie}(G)$.
The geodesics of the $(+)$/0/$(-)$-connections on a Lie group are the one-parameter subgroups. To see that, just check the geodesic equation with $\gamma(t)=\exp(tX)$: since $\dot\gamma(t)={L_{\exp(tX)}}_*X$, we get $\nabla_{\dot\gamma(t)}\dot\gamma(t)={L_{\exp(tX)}}_*\nabla_XX={L_{\exp(tX)}}_*c[X,X]=0$.
To have one more justification, notice that if the difference $S(X,Y)=\nabla_XY-\nabla^\prime_XY$ between two connections is an antisymmetric tensor, then these connections have the same (parametrized) geodesics (to see it, write the geodesic equation $\nabla'_{\dot\gamma}\dot\gamma=0$ for $\nabla'$ in terms of that for $\nabla$ and of $S$). Since the three connections at hand differ by a multiple of the Lie bracket, they have the same geodesics.
So the exponential map of the $(-)$-connection is a diffeomorphism whenever the exponential map of the Lie group is a diffeomorphism. And this happens exactly when both
The Lie groups with these properties are called strongly exponential and are all solvable (the condition on the eigenvalues rules out the existence in $\operatorname{Lie}(G)$ of any compact semisimple subalgebra, which in turn rules out the existence of any semisimple subalgebra).