I am looking to find the expression of the Haar measure of the $SO(3)$ group as a function of the Lie algebra basis $$ R(x,y,z) = \text{exp}\left( x L_x + yL_y + zL_z \right) $$ where $L_x, L_y, L_z$ is the usual basis of antisymmetric matrices.

In other words, what is the function $f(x,y,z)$ such that $d\mu(R) = f(x,y,z)dxdydz$?


1 Answer 1


For a compact matrix Lie group and a local parametrization of the latter you can locally write the density of the Haar measure, in that parametrization, simply as the determinant, which in turn is simply 1 here. Then you only need to work out the contributions of your chosen parametrization.

A very similar approach was asked for in Haar measure from axis-angle representation of $SO(3)$, when $\text{SO}(3)$ was parametrized by the Euler angles.

Edit: Maybe I can add the following:

I like to reprase the invariance of the Haar measure, that is, claiming the invariance under group multiplication, into the invariance under the family of diffeomorphisms $L_g:G\to G,~h\mapsto gh$ with $g\in G$. This is trivially the same here, but there is also more stuff/more properties on/of the group "invariant under these diffeomorphisms". This is sometimes paraphrased that a Lie group, only considering it as a manifold, "locally looks everywhere the same". For example, given a basis of the Lie algebra (= tangent space at 1) it yields a basis in every tangent space, say at $g$, which moreover (in a suitable sense) depends smoothly on the base point $g$ of these tangent spaces. Thereby you also obtain, around such $g$, a coordinate system "adapted to this basis", which is nothing but the coordinate system around 1 "adapted to your original Lie algebra basis", pushforwarded by the diffeo $L_g$. In both these coordinate systems the density of the Haar measure or course then also "looks the same".

So once you have found the Haar measure and the family of diffeo's $(L_g)_g$ you can forget about the group structure and are left with a manifold and thereon a volume form which is invariant under a family of diffeo's, where this family, for each two $g,g'$, contains an element which maps $g\mapsto g'$. Hence this manifold, also as a manifold with a volume form "looks everywhere the same". If you wish (note that also above you may already need compactness or some additional properties of the group for a unique unimodular measure), you can then normalize your measure.

So the deeper connection is, in my view, that both the Lie algebra and the Haar measure are in a sense local properties of the group which are preserved under the above family of diffeo's. If you then express the measure in the coordinate system induced by your Lie algebra basis, the result is of course again invariant.

However, if you ask for a more general view particularly on the normalization of the Haar measure in dependency of the chosen coordinate system, I think you just need to compute it. Consider $\text{SO}(3)$ and $\text{O}(3)$, they have the same Lie algebra, but their Haar measures differ by a factor 2 (pulling back the Haar measure of $\text{O}(3)$ back trough the inclusion $\text{SO}(3)\to\text{O}(3)$).


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