Let $G$ be a Lie group with Lie algebra $\mathfrak{g}$. One can put a (left) Haar measure $\mu$ on $G$ and a Lebesgue measure $\lambda$ on $\mathfrak{g}$ which are both unique up to constants.

My question is whether fixing $\mu$ on $G$ picks out $\lambda$ uniquely on $\mathfrak{g}$ and conversely.

I think a natural first step in answering this question is to consider the exponential map $\exp:\mathfrak{g}\to G$ and try to use the change of variables formula. However there are two issues that one faces:

  1. The exponential map is in general a local diffeomorphism only.
  2. The Lebesgue measure transferred from $\mathfrak{g}$ to $G$ via the change of variables formula need not be a Haar measure: $$\int_G f(g)\, d\mu(g):=\int_{\mathfrak{g}}f(\exp X) |\mathrm{Jac}\exp|\, d\lambda(X) $$

Can someone please shed some light on the relation between the measures on the Lie group and the Lie algebra and resolving the above two issues? (I think the first issue is not severe, as $\exp$ is a local diffeomorphism and that should be enough for transferring the measure from one side to the other, if at all possible for a group $G$.)

Thanks in advance for any comments/answers.

  • 1
    $\begingroup$ An immediate problem I see is that the Lebesgue measure is both left and right invariant (as a measure of the additive group of $\mathfrak{g}$, but in general the left and right invariant Haar measures on $G$ are not the same. A more serious problem is that $\exp$ is not a (local) homomorphism. $\endgroup$ Aug 3, 2015 at 15:53
  • $\begingroup$ For a $n$-dimensional Lie group $G$ a left Haar measure is (well, up to orientation) the same as a left-invariant $n$-form on $G$, which is the same as an element of $\bigwedge^n\mathfrak g^*$, which is the same (again modulo orientation) as a Lebesgue measure on $\mathfrak g$ $\endgroup$
    – user8268
    Aug 3, 2015 at 16:16
  • $\begingroup$ @user8268 I get from your comment that a choice of Haar measure on $G$ determines a Lebesgue measure on $\mathfrak{g}$ and conversely. Please correct me if I'm mistaken. Of course, that answers the question in the title of this post, but does not touch the question of how the two measures are related via the exponential map and the change of variables formula. $\endgroup$
    – Philip M
    Aug 3, 2015 at 17:30
  • 1
    $\begingroup$ @user8268 no the question is not about the formula for the Jacobian. There are in fact two related questions: 1) Does a choice of Lebesgue measure on $\mathfrak{g}$ determine a measure on $G$, and vice versa? I think the answer to this question is yes, as you have explained in your above comment. The second question pushes this a bit further: 2) Does the change of variable formula (displayed in my post) define a Haar measure on $G$? If not, under what conditions on $G$ this happens? $\endgroup$
    – Philip M
    Aug 4, 2015 at 12:59
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    $\begingroup$ @user8268 maybe one can extend the result to the case when the exponential maps the Lie algebra onto the connected component of the identity. Then one should be able to extend the measure into the other components by choosing an element of each component to translate the identity. Does the result depend on the elements one chooses? $\endgroup$ May 27, 2021 at 23:40


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