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It is well known that for a matrix Lie group the Lie exponential map is $e ^z$. This maps a tangent vector $z$ at the identity to a group element.

On the other hand the general Riemannian exponential map centered at point $x$ is given by $\exp _x \triangle$ which maps a tangent vector $\triangle$ at point $x$ (not necessarily identity element) to a group element.

Is there a relationship between these two exponential maps?

For example is below formula correct? If so, are there any conditions involved?

$\exp _x \triangle = xe ^{x^{-1}\triangle}$

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1 Answer 1

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Notice that you need to pick a metric on a Lie group for the "general Riemannian exponential map" to be defined.

If you happen to pick an invariant metric on a Lie group, then every geodesic is (locally) a translate of a 1-parameter subgroup (so essentially both exponentials are the same thing)

I don't know what happens if there are no invariant metrics.

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  • $\begingroup$ How can there be no invariant metrics? $\endgroup$
    – t.b.
    Jan 11, 2011 at 1:15
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    $\begingroup$ @Theo: A Lie group has an invariant metric iff its adjoint group has compact closure in $\operatorname{GL}(\mathfrak g)$. $\endgroup$ Jan 11, 2011 at 1:18
  • $\begingroup$ Here did you mean either left(/right)-invariant or bi-invariant? Also is there any online/book material which explains how the relation is achived? $\endgroup$
    – 911
    Jan 11, 2011 at 1:21
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    $\begingroup$ @sam: I mean bi-invariant. You can find the proof in Lie groups and compact groups by John Frederick Price, on googlebooks: I am not familiar with it, but it's what google brings me to. $\endgroup$ Jan 11, 2011 at 1:23
  • $\begingroup$ Thank you very much for clearing up this misconception! $\endgroup$
    – t.b.
    Jan 11, 2011 at 2:01

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