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I am interested in the Haar measure on $U(n)$ resp. $O(n)$. I had the idea that you cannot "restrict" the Haar measure on $U(n)$ to Borel sets on $O(n)$, otherwise you would get a zero measure (i. e. not a Haar measure). How does it work? Note that $O(n)$ is not a connected subgroup (since matrices of det 1 and det -1 are allowed), therefore, we cannot simply use this.

In addition, these groups are no vector spaces, which would allow us to apply a simple dimension argument, but are subgroups of $GL(n, \mathbb{C})$ resp. $GL(n, \mathbb{R})$. Therefore, they are subsets, submanifolds of different manifolds, but I see no way to apply a dimension argument ("Dimension of O(n) is lower than that of $U(n)$, therefore the theorem follows.")

In addition, is a Haar measure (like in "On LCH topological groups, there always exists a Haar measure up to a positive scalar factor.") always a positive measure?

I sadly do not know Lie Theory nor Harmonic Analysis.

Is there even a constructive description of both of the Haar measures (on $O(n)$ as well as on $U(n)$)?

user7427029
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  • Related: https://mathoverflow.net/questions/76295/intuition-for-haar-measure-of-random-matrix – Jose Avilez Dec 25 '21 at 16:42
  • It is not clear what you are looking for. For example, take $U(1) \subset U(1) \times U(1)$. What do you want to say about the Haar measures of $U(1)$ and $U(1) \times U(1)$? – Arctic Char Dec 25 '21 at 18:02
  • I think one may proceed as follows, but I hope someone will correct me if it is incorrect. Choose a non-zero element $\Omega \in \Lambda^d(\mathfrak{g})^$ where $d$ is the dimension of a compact Lie group $G$, $\mathfrak{g}$ is the associated Lie algebra, which you can think of as the tangent space $T_1(G)$ to $G$ at the identity element $1 \in G$. Then define a left-invariant volume form $\widetilde{\Omega}$ on $G$ as follows. Just define it to be, at $g \in G$, equal to $L_{g^{-1}}^(\Omega)$. I think that the associated measure will also be right-invariant. Can someone confirm? – Malkoun Dec 26 '21 at 00:11
  • You are correct that there is not simple way to "restrict" the Haar measure of a Lie group to obtain the Haar measure of a subgroup of lower dimension, but beyond that, what are you asking about? When you ask "How does it work?" what is "it"? Haar measures on Lie groups can be locally described by differential forms, but the resulting forms will have different rank. – Kajelad Dec 26 '21 at 01:53

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