Let $G$ be a Lie group with bi-invariant metric $\langle , \rangle$ and $X,Y,Z$ left invariant vector fields in $G$, how to conclude that $X\langle Y,Z\rangle=0$?
A question about left invariant vector fields
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lie-groups
riemannian-geometry
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0What have you tried until now? Where are you stuck? Why do not you take a look at Manfredo Perdigao do Carmo 'Riemannian Geometry'? There is a traitment of this point in §2.6 on Lie Groups on pages 40-41. Cf. for example here: http://books.google.it/books?id=ct91XCWkWEUC&pg=PA40&lpg=PA40&dq=bi-invariant+metric+and+left+invariant+vector+fields&source=bl&ots=ewy957RfTH&sig=a05P2qVBTvcfze9n-c37Rne4YHA&hl=it&sa=X&ei=a2rKT42yM4jf4QTKvrQQ&ved=0CIQBEOgBMAk#v=onepage&q=bi-invariant%20metric%20and%20left%20invariant%20vector%20fields&f=false – 2012-06-02
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0I don't know where to use the condition "left invariant". I read that chapter, but I still don't know the answer... – 2012-06-02
1 Answers
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The function $$ G \ni g \mapsto \langle Y_g, Z_g \rangle $$ is constant since $$ \langle Y_g, Z_g\rangle = \langle dL_g Y_e, dL_g Z_e\rangle = \langle Y_e, Z_e\rangle. $$ So differentiating it by $X$ gives 0. Note we have used the facts that $Y$ and $Z$ are left-invariant but $X$ need not be.
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0Why is $\langle dL_g Y_e, dL_g Z_e\rangle = \langle Y_e, Z_e\rangle$? – 2012-10-09
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1That's what it means for the metric to be biinvariant. – 2012-10-09