Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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.

share|improve this question
    
1) I don't know what the +/- connections are, and I suspect I'm not the only one. Perhaps you could provide definitions? 2) Have you tried to work it out from scratch? It should all reduce to a calculation involving the group and Lie algebra at the origin. –  Deane Yang Jan 27 '12 at 7:42
2  
The +/-/0 terminology is used in section 11 of Nomizu's paper "Invariant affine connections on homogeneous spaces" where he attributes the distinction, and possibly also the notation, to Cartan (I've not followed the references). A left invariant affine connection is determined by specifying a bilinear mapping from the Lie algebra to itself. As one takes as this bilinear mapping respectively minus the Lie bracket, the Lie bracket, or half the Lie bracket one gets the -/+/0 connection. –  Dan Fox Jan 27 '12 at 8:11
    
Kobayashi and Numizu also mention these connections briefly in Chapter X (vol.2). They reference Cartan and Schouten. –  Peter Dalakov Jan 27 '12 at 12:57
2  
You should be careful to specify which Lie groups you mean. Not all Lie groups carry bi-invariant metrics (of any signature). Thus, for example, the connected, nonabelian Lie group of dimension 2 does not have such a metric (or even a bi-invariant volume form, for that matter). In this example, the $(0)$-connection does not preserve any metric. As Élie Cartan remarked when he defined these connections, the $(+)$ and $(-)$ connections on any Lie group are flat, so they each necessarily preserve (many) metrics on the group. If the $(0)$-connection preserves a metric, it is bi-invariant. –  Robert Bryant Jan 27 '12 at 21:15
    
Good point Robert. If I understand you correctly, you're saying that, in general, Cartan connections need not be metric connections. –  Oliver Jones Jan 27 '12 at 21:35

2 Answers 2

up vote 6 down vote accepted

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 $N=0$.

share|improve this answer
    
Thanks Peter, very helpful. –  Oliver Jones Jan 27 '12 at 21:42

If I understand correctly, the answer is Yes.

the +/-/0 connections can be defined by, if $X,Y$ is the left invariant vector $$\nabla_{X}Y=a[X,Y]$$ where $a=1,-1,0$.

The connection is metric for left invariant metric iff $$0=\langle\nabla_{X}Y,Z\rangle+\langle Y,\nabla_{X}Z\rangle.$$

This is trival for the bi-invariant metric.

share|improve this answer
    
Thanks Shu. But as Robert pointed out, I had implicitly assumed the existence of a bi-invariant metric. –  Oliver Jones Jan 27 '12 at 21:39

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.