Volume of $SO(n)\subset\mathbb R^{n^2}$, again

Question

I posted this on MSE, but no answer is received, so I post this here.

The problem of finding the $n(n-1)/2$-dimensional volume of the set $SO(n)\subset\mathbb R^{n^2}$ is asked before in this MO post. In the comment section, it is suggested that the answer is $\prod^{n-1}_{k=1}2^{(k-1)/2}\cdot \textrm{volume}(S^k)$ (where $S^k$ denotes the unit $k$-sphere). It was pointed out that, as I quote:

"... the factor of $2^{(k-1)/2}$ that you have to put in at each level because the natural map $\pi:SO(k)→S^{k−1}$ is not a Riemannian submersion, but is a Riemannian submersion scaled by a factor of $1/{\sqrt 2}$, i.e., each horizontal vector for this bundle is shrunk by a factor of $\sqrt 2$ by the differential of $\pi$."

My question is:

Explicitly, what is the map $\pi$? How to use $\pi$ to calculate the volume of $SO(n)$? How to compute $d\pi$? Thanks.

Answer 1

Maybe this will help: Regard $\mathrm{SO}(n)\subset M_{n,n}(\mathbb{R})$ as the set of $n$-by-$n$ matrices $a$ that satisfy ${}^ta\,a=\mathrm{I}_n$ and $\det(a)=1$. Then $\mathrm{SO}(n)$ is a smooth, connected submanifold of $M_{n,n}(\mathbb{R})$ of dimension $\frac12n(n{-}1)$.

Give $M_{n,n}(\mathbb{R})$ the positive definite inner product $\langle x,y\rangle = \mathrm{tr}(\,{}^txy\,)$ for $x,y\in M_{n,n}(\mathbb{R})$. (See the note at the end about the effects of this choice of inner product.)

Let $E:\mathrm{SO}(n)\to M_{n,n}(\mathbb{R})$ be the inclusion map, thought of as a $M_{n,n}(\mathbb{R})$-valued function on $\mathrm{SO}(n)$. Then the induced Riemannian metric on $\mathrm{SO}(n)$ is given by $$ g = \langle \mathrm{d}E,\mathrm{d}E\rangle = \mathrm{tr}\bigl({}^t(\mathrm{d}E)\,\mathrm{d}E\,\bigr) = \mathrm{tr}\bigl({}^t(E^{-1}\mathrm{d}E)\,E^{-1}\mathrm{d}E\,\bigr) = \mathrm{tr}\bigl({}^t\omega\,\omega\,\bigr) = - \mathrm{tr}\bigl(\omega\,\omega\,\bigr), $$ where $\omega = E^{-1}\mathrm{d}E=- {}^t\omega$ is a $1$-form with values in skew-symmetric $n$-by-$n$ matrices, i.e., $\omega = (\omega_{ij})$ where $\omega_{ij}=-\omega_{ji}$. In particular, $$ g = -\sum_{i,j}\omega_{ij}\,\omega_{ji} = 2\sum_{i<j} (\omega_{ij})^2 = \sum_{i<j} \bigl(\sqrt2\,\omega_{ij}\bigr)^2. $$

Meanwhile, writing $E = (e_1\ e_2\ \ldots\ e_n)$ where $e_i$ is the $i$-th column and hence is a smooth map $e_i:\mathrm{SO}(n)\to S^{n-1}\subset\mathbb{R}^n$, one has $\mathrm{d}e_i = e_j\,\omega_{ji}$ (sum on $j$) for each $i$. In particular, the pullback of the metric on $S^{n-1}$ by the submersion $e_n:\mathrm{SO}(n)\to S^{n-1}$ is the quadratic form $$ g_n = \mathrm{d}e_n\cdot \mathrm{d}e_n = (e_j\,\omega_{jn})\cdot (e_i\,\omega_{in}) = \sum_{i<n} (\omega_{in})^2. $$ Thus, $e_n:\mathrm{SO}(n)\to S^{n-1}$ is a Riemannian submersion (with fibers isometric to $\mathrm{SO}(n{-}1)$) only after scaling the metric $g$ by a factor of $1/2$.

Consequently, we have the formula $$ \mathrm{vol}\bigl(\mathrm{SO}(n)\bigr) = (\sqrt 2)^{n-1}\mathrm{vol}(S^{n-1}) \mathrm{vol}\bigl(\mathrm{SO}(n{-}1)\bigr), $$ which, by induction, yields $$ \mathrm{vol}\bigl(\mathrm{SO}(n)\bigr) = (\sqrt 2)^{n(n-1)/2}\,\mathrm{vol}(S^{n-1})\mathrm{vol}(S^{n-2})\cdots \mathrm{vol}(S^{1}). $$

N.B. If one doesn't like the factors of $\sqrt 2$ that appear in this formula, one can fix this problem by instead taking the metric on $M_{n,n}(\mathbb{R})$ to be the positive definite inner product $\langle x,y\rangle = \tfrac12\,\mathrm{tr}(\,{}^txy\,)$ for $x,y\in M_{n,n}(\mathbb{R})$. In that case, the $\sqrt2$ factors go away. Some sources use this convention, others do not. If this is important in one's calculations, one should always check the convention of the source.