The question is about the family of tensors that are naturally associated to any nice Lie group. Take the MauerCartan form, $\omega=g^{1} dg$ and I would like to make the covariant index of this oneform explicit, $\omega_a =g^{1}\partial_a g$. Then the Riemannian, left/right invariant, metric is just $g_{ab}=tr(\omega_a\omega_b)$. More generally one can construct a family of leftright invariant tensors $T_{a_1...a_s}=tr(\omega_{a_1}...\omega_{a_s})$. Does this family have any meaning/application/nice properties?

$\begingroup$ Hmmm. You have some confusion here, and I'm not sure exactly what you mean by 'make the covariant index...explicit'. However, you should know that, if you just take any basis of leftinvariant forms, then the formula you wrote down for a metric $g$ will not be biinvariant. (In fact, many Lie groups do not even admit a biinvariant volume form, let alone a metric, so you will need to say what you mean by 'nice', too.) The biinvariant tensors on a given Lie group are controlled by its adjoint representation, and that gives the whole story there. Yes, these have lots of applications. $\endgroup$– Robert BryantAug 6, 2013 at 9:36

$\begingroup$ Indeed, the question is vague and a lot was put on 'nice Lie group term'. My interest is in higher rank tensors one can have and I will accept any Lie group that is needed for them to exists. $\endgroup$– Eugene StarlingAug 6, 2013 at 14:13
2 Answers
Sounds like you are in physics; particle physicists often assume that all Lie groups are compact. Compact Lie groups admit biinvariant metrics. The tensors you are considering are essentially the characteristic polynomials in the adjoint representation, so if you insert wedge product signs (work with the associated alternating tensors) then they represent certain of the ChernWeil invariant differential forms that give rise to characteristic classes, if I understand your notation correctly. There may be some other ChernWeil forms (like the Pfaffian, if $G=SO(2n)$).

$\begingroup$ Thank you for the comment. What if I symmetrize instead of taking alternating tensors? $\endgroup$ Aug 6, 2013 at 14:14
Expanding on Ben's answer: If you symmetrize, you get the left (or bi) invariant extension of the characteristic polynomials.
Example: Let $G=SO(n,\mathbb R)$ be a simple matrix group. Then you get (symmetrized version) the Newton polynomials in the eigenvalues $\sum \lambda_i^p$ of $g^{1} X$ if $(g,X)$ is a tangent vector with foot point $X$ of $SO(n)$ in $Mat(n)$.