A Lie group has three standard Cartan connections; the (-)-connection, the (0)-connection, and the (+)-connection. The (0)-connection is Levi-Civita with the associated metric the bi-invariant metric. The other two connections aren't Levi-Civita due to the presence of torsion. However, there's nothing to stop them a priori from being metric connections. My question is; are the minus and plus connections compatible with the bi-invariant metric? This seems reasonable but I can't find a reference.



Metric Connections on a Lie Group
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 $N=0$.


