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In the book "Lectures on Differential Geometry" by Sternberg page 233

"Given a representation,p, of a Lie group G (in particular the adjoint representation) on a vector space F, if p(x) is compact then there exist an invariant scalar product <.,.> on F satisfying = , for all x e G, v, w e V " ie G has bi-invariant metric in the case of p(x) is adjoint represantation. Why p(x) must be compact??? for adjoint representation, what is intuitive reason for non-existance of bi-invariant metric for SE(n) n>1 Lie group? suppose a rigid body in 3d space. attitude of this rigid body can be measured with respect to the inertial reference frame (right invariant) and body frame (left invariant). if this rigid body rotate by rotation matrix R1 and translate by t1 what happen to left invariant metric and right invariant metric and why these quantities are not equal!?

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  • $\begingroup$ The theorem does not say "If $G$ is noncompact then no bi-invariant metric exists". It only says "If $G$ is compact then a bi-invariant metric exists". $\endgroup$ – Lee Mosher Apr 15 '16 at 14:38
  • $\begingroup$ Dear Prof Mohser, Thanks very much for your response, yes you siad right and this condition is sufficient. my purpose is to find pesudo-bi invariant metric on non compact lie group. In many references bi-invariant metric is found for compact lie group! :( and prove that there isn't any bi- invariant metric for example SE(n) lie group. $\endgroup$ – mary Apr 16 '16 at 6:29
  • $\begingroup$ What is "pesudo-bi invariant metric"? Do you mean "biinvariant pseudoriemannian metric"? Please, rewrite your question as currently it is totally unclear. Use TeX. $\endgroup$ – Moishe Kohan Apr 16 '16 at 16:18

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