A Lie group $G$ can be considered as a reductive homogeneous space in at least two different ways; $G/\{e\}$ and $G\times G/G^*$. In the first case, the canonical connection associated with the reductive decomposition has zero curvature and non-zero torsion (if $G$ isn't abelian). This connection coincides with the Cartan (-)-connection. In the second case, the canonical connection has zero torsion and generally nonzero curvature.

Can we express a Lie group as a reductive homogeneous space for which the canonical connection has both nonzero curvature and torsion? We could take as our connection a combination of the two connections described above. However, how do we know that this will be a canonical connection with respect to some reductive decomposition?

  • $\begingroup$ Apparently there's one more case: mathoverflow.net/questions/126673/the-connection-on-a-lie-group $\endgroup$ Feb 5, 2014 at 1:12
  • $\begingroup$ @QiaochuYuan Could you elaborate? $\endgroup$ Feb 5, 2014 at 2:39
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    $\begingroup$ There are some missing assumptions on $G$ for last sentence in the opening paragraph to be correct. The Levi-Civita connection is by definition a metric, torsion-free connection and the statement the OP writes is true for Lie groups with a bi-invariant metric. (Most Lie groups, by any measure, do not have such metrics.) $\endgroup$ Feb 5, 2014 at 11:20
  • $\begingroup$ @José You can ignore the reference to the Levi-Civita connection because it's not relevant to my question. Simply put, I'm asking what connections on a Lie group are canonical w.r.t. some reductive decomposition. $\endgroup$ Feb 5, 2014 at 21:56

1 Answer 1


(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


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 $\frak{g}\oplus\frak{g}=\Delta_{\frak{g}}\oplus{\frak{m}}_{t}$.

  • $\begingroup$ By the canonical connection I mean the canonical connection of the second kind in the sense of Nomizu. $\endgroup$ Jul 8, 2014 at 3:01

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