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Intuitively what is meant by a left invariant vector field on a manifold?

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    $\begingroup$ You probably need the manifold to be a Lie group for this notion to make sense. (There has to be some action in the picture so that the vector field has something to be invariant under.) $\endgroup$
    – Matt E
    Jan 8, 2011 at 14:19
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    $\begingroup$ Alternatively you could talk about a vector field on a manifold that's invariant under a group action. But I think calling it "left invariant" without any additional context would be seen as a little unusual. Sam, I'm a little confused by your question. If you do a google search on "left invariant vector field" the 2nd link is a presentation of Alexey Bolsinov's that's quite good. In particular it describes how the flow generated by a left-invariant vector field is right multiplication. $\endgroup$ Jan 8, 2011 at 20:07
  • $\begingroup$ Matt Yes, the manifold is assumed to be a Lie group. Ryan yes, i came across the slides and seems good. $\endgroup$
    – 911
    Jan 9, 2011 at 4:20
  • $\begingroup$ could you please give the link for the presentation of Alexey Bolsinov? I am confused by this definition, and cant find this doc. $\endgroup$
    – Martial
    Feb 10, 2015 at 14:06
  • $\begingroup$ I'm also interested in that property for the flow... I could not find the precise file but it could be one of those, maybe around lecture 10. (go to "files") $\endgroup$
    – Noix07
    May 19, 2017 at 21:44

1 Answer 1

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To talk about left invariance, you probably want to assume your manifold is a Lie group, so that the vector field is left invariant under the (derivative of) the group action. Intuitively, this means that the vector field is entirely determined by the vector at the unit element of the Lie group. Given any other element of the Lie group, say $g$, the vector at $g$ has to be $(l_g)_*(X_0)$, where $(l_g)_*$ is the derivative of left-multiplication by $g$ and $X_0$ is the vector at the unit element. So the Lie group action allows you to take a single vector and distribute it out over the manifold in a smooth, nonvanishing way.

The simplest example is Euclidean space $\mathbb R^n$ regarded as an abelian Lie group under addition. In this case, a left invariant vector field is simply a constant vector field in the usual calculus sense. All the vectors point in the same direction. The map $(l_g)_*$ identifies the tangent spaces at each point in the usual "parallel transport" way.

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    $\begingroup$ The manifold is a Lie group. Jim can you re-elaborate on what's meant by $l_g(X_0)$ where $l_g$ is the derivative of left-multiplication by $g$. O'Neill's Semi-Riemannian geometry book says left multiplication merely permutes the tangent vectors constituting a vector field X. So is below interpretation correct? Lie group action allows to take a single vector and distribute it out over the manifold. So under this group action the existing vector fields at each point on the manifold merely gets permuted. So the set of all these vectors remain the same. $\endgroup$
    – 911
    Jan 9, 2011 at 4:47
  • $\begingroup$ I think you've got a reasonable first understanding. $\endgroup$ Jan 9, 2011 at 13:43
  • $\begingroup$ your answers are very helpful. I have one question: could, in your answer, "left invariant" be changed to be "right invariant $r_g(X_0)$"? I don't think it makes some difference. Thank you. $\endgroup$
    – Martial
    Feb 5, 2015 at 15:03
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    $\begingroup$ You could push the vectors in $T_e G$ around by right translation as well. In this case the right invariant vector fields are anti-isomorphic to the left invariant ones as Lie algebras. $\endgroup$
    – ಠ_ಠ
    Aug 15, 2016 at 9:46
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    $\begingroup$ For some reason I was confused by what it means to say that left multiplication "permutes" the tangent vectors. For posterity: the pushforward of left multiplication is a bijection on $X$. $\endgroup$
    – A_P
    Oct 9, 2019 at 0:29

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