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I've seen that there always exists a bi-invariant (i.e., left and right invariant metric ) on a compact Lie group.

However, most of the metrics are not left or right invariant ( I assume),

Can this (lack of) abundance of left or right invariant metrics be quantified precisely in anyway? For example, in terms of the infinite dimensional Frechet manifold of metrics on an arbitrary compact Lie group?

Specific examples are welcome.

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  • $\begingroup$ The first sentence is wrong: true for compact groups though. I do not understand the question itself. $\endgroup$ – Moishe Kohan Sep 21 '16 at 20:30
  • $\begingroup$ @MoisheCohen Thank you, fixed it. $\endgroup$ – Ashley Sep 21 '16 at 21:00
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    $\begingroup$ There is one left invariant metric for each inner product on the Lie algebra of the group, and likewise for the right invariant metrics. The left invariant metrics form a tiny subset of the set of all metrics on the group, since they are essentially a finite dimensional subspace. It is pretty obvious that its interior es empty, for example: just perturb a left invariant metric very little around a point. $\endgroup$ – Mariano Suárez-Álvarez Sep 21 '16 at 21:08
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If you have a left invariant metric on a Lie group, then you can simply project it (pull it back) onto the Lie algebra by left translations. The projected metric is a positive definite dot product on the Lie algebra. Conversely, if you get any positive definite dot product on the Lie algebra you can spread it (push it forward) around the Lie group with left translations, obtaining a left invariant metric on the whole group. So as you can see, the space of left invariant metrics on a Lie group is isomorphic to the finite dimensional space of positive definite symmetric bilinear forms on the Lie algebra.

Analogous arguments lead to the conclusion that a bi-invariant metric on the group can be projected (pulled back) to an $\text{Ad}$-invariant dot product on the Lie algebra. Conversely, if you have an $\text{Ad}$-invariant dot product on the Lie algebra, you can spread it around the group by say left translations (push it forward to every point on the group by the left translation) you get a bi-invariant metric on the group. So the space of bi-invariant metrics on a compact Lie group is isomorphic to the space of $\text{Ad}$-invariant dot products on the Lie algebra.

Left invariant metrics on a Lie group play an important role in the theory of Lagrangian mechanics on Lie groups. They are usually the kinetic energy term of the Lagrangian of the system. For instance, look at the Euler top (and maybe the Lagrnage top). The Euler top models the motion of a rigid body in the absence of external forces (i.e. freely rotating body around a fixed point). It's dynamics is described as a motion of a point on $\text{SO}(3)$ (or one can also do it on $\text{SU}(2)$). It only has kinetic energy, so its Lagrangian is a left invariant metric on the rotation group (left invariant, because we look at the motion from a coordinate frame firmly attached to the body and moving with it). Consequently, the left invariance of the Lagrnagian (the metric) allows you to project the dynamics onto the Lie algebra and obtain Euler's equations.

The existence of bi-invariant metrics is guaranteed for compact Lie groups. By simply taking a positive definite symmetric dot product on the lie algebra, acting on it by adjoint representation and then integrating over the elements of the group with respect to the Haar measure of the group, i.e. $\big(v, w\big)_G = \int_{g \in G} \big(\text{Ad}(g)v, \, \text{Ad}(g)w\big) \, dg$ one gets an $\text{Ad}$-invariant dot product on the Lie algebra. If you spread the latter around the group by say left translations (push it forward to every point on the group) you get a bi-invariant metric. This construction works because compact Lie groups (as well as semi-simple Lie groups) have a bi-invariant Haar measure (unique up to positive scaling if I am not wrong).

If the Lie group is semi-simple, then the Killing form on the Lie algebra gives rise to a bi-invariant metric on the group because then the Killing form is positive definite and adjoint invariant so you can push it forward all over the group.

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