# Connections on a Lie Group

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?

• Apparently there's one more case: mathoverflow.net/questions/126673/the-connection-on-a-lie-group – Qiaochu Yuan Feb 5 '14 at 1:12
• @QiaochuYuan Could you elaborate? – Oliver Jones Feb 5 '14 at 2:39
• 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.) – José Figueroa-O'Farrill Feb 5 '14 at 11:20
• @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. – Oliver Jones Feb 5 '14 at 21:56

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

• This post was flagged for its being edited quite a few times. (Each edit bumps the post to the front page; many such bumps are often annoying to the community.) – Todd Trimble Jul 7 '14 at 23:00
• By the canonical connection I mean the canonical connection of the second kind in the sense of Nomizu. – Oliver Jones Jul 8 '14 at 3:01
• @Todd Trimble, sorry, I have just tried to write it in a more detailed way. Giving answers, certified in a professional way, I don't think that it is so bad, in fact I thought that this was one of the goals of Mathoverflow, isnt it? If someone of the members in the community does not like that the post is repeating (bumping), then can avoid it, or Matheoverflow has to find a way to fix this problem (for you and probably other members). – 314159. Jul 8 '14 at 11:18
• Believe me, I sympathize, and this phenomenon has been discussed many times. I just wanted to report it neutrally; it might help either to work in your favorite text editor, or to let a little more time elapse between edits. – Todd Trimble Jul 8 '14 at 11:45