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I know that we can define the killing form on a lie algebra. However, when going to the group manifold, does this give rise to a metric on the manifold? I thought that would be the case, but I cant find any useful literature how this works. Especially, I am confused since the Lie algebra seems to always be defined around the unity, but I guess on a group manifold one can use that we can always map any point to unity and thus also map the tangent space?

So Let's say I have a Lie group manifold parametrized by coordinates x0,x1,...,xn. How can I obtain the metric induced from the killing form?

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    $\begingroup$ The killing form is not always positive definite, so does not always give a Riemannian metric... $\endgroup$
    – Igor Rivin
    Jan 20, 2015 at 20:49
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    $\begingroup$ Danger: the Killing form $B$ on a nonsemisimple Lie group is degenerate, i.e. has vectors $v$ so that $B(v,w)=0$ for all $w$. So no hope of a metric. All of this is very standard in all of the textbooks. $\endgroup$
    – Ben McKay
    Jan 20, 2015 at 21:05

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Any inner product on $\mathfrak{g}$ will give rise to a metric on $G$. More specifically, you can use a left trivialization of $TG$ to get a vector bundle isomorphism $TG\cong G\times\mathfrak{g}$. The latter bundle has a metric by virtue of the inner product on $\mathfrak{g}$.

The same construction will work if one instead uses the right trivialization.

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  • $\begingroup$ So as I understand it, to phrase it differently, the left/right multiplication of a group element $a$ defines a mapping from $\mathfrak{g}$, the tangent space at the origin, to a tangent space at a point $a$, and via this we can define the inner product at the tangent space of $a$? Btw, will the left/right construction give the same metric? $\endgroup$
    – user2133437
    Jan 20, 2015 at 21:24
  • $\begingroup$ You are right, but I would prefer to say that the differential at $e$ of left (resp. right) multiplication by $a$ allows us to transfer an inner product on $\mathfrak{g}$ to the tangent space at $a$. No, in general, the metric will depend on the particular trivialization you have chosen. $\endgroup$
    – Peter Crooks
    Jan 20, 2015 at 21:37
  • $\begingroup$ Since the OP mentioned the Killing form, it would not be amiss to point out that if the inner product on $\mathfrak{g}$ is ad-invariant: $\langle[X,Y],Z\rangle = \langle X,[Y,Z]\rangle$, then the metric on $G$ will be bi-invariant, so that left- and right- trivialisations give rise to the same metric. $\endgroup$
    – José Figueroa-O'Farrill
    Jan 21, 2015 at 21:05
  • $\begingroup$ Yes, fair enough. $\endgroup$
    – Peter Crooks
    Jan 22, 2015 at 0:42
  • $\begingroup$ @JoséFigueroa-O'Farrill: Are there any statements regarding whether taking the trace in another representation will give rise to the same metric? I tried this on SL(2,R) and both the defining and adjoint representations gave the same killing form (up to an overall constant) $\endgroup$
    – user2133437
    Jan 27, 2015 at 15:07

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