Volume of Group SO(3, R) endowed with a Killing metric

Question

Got this question from Modern Geometry Part 1 (Dubrovin, Fomenko, Nokinov). No idea how to start. I'm looking for the simplest, most dumbed down solution with most things explained. It may be the case that this problem is a bit above my level. The problem is placed at the end of a section "...the Theory of Integration". I assume I need to use the Killing metric and the "volume element" tensor and integrate over all parameters of SO(3)?

Answer 1

Some guidance, rather than a complete answer

Suppose you have two tangent vectors at the identity (i.e. you have a pair of $3 \times 3$ skew-symmetric matrices).

1. Can you explain why these are tangent vectors at the identity?

2. Supposing that one has entries $a_{ij}$ and the other has entries $b_{ij}$, what's the inner product of $a$ and $b$ under your metric?

You may recall that to measure the length of an arc $c$, you compute $$ \int_a^b \|c'(t)\| ~dt $$ while to compute the area of a surface (in 3-space) you compute $$ \iint \| \partial S/\partial u (u, v) \times \partial S/\partial u (u, v) \| ~du ~dv $$

In each case, you parametrize the object, and then integrate over the parameter domain. But there's an alternative. Suppose that your curve is actually a closed curve, and you have a map $$ s: S^1 \to \Bbb R^2 : t \mapsto s(t) $$ with the properties that (1) \|s'(t)\| is constant, $c$ and (2) $s$ is a 3-to-1 map. If only it was a 1-to-1 map, you'd know that the length of your curve was exactly $c$ times the length of the unit circle, i.e., $2\pi c$. But because it's a 3-to-1 map, the length of the curve must be $\frac{2}{3}\pi c$. Clever, hunh?

OK. Well, the natural covering map from $S^3$ to $SO(3)$ actually has constant volume-change factor, and it's a 2-to-1 map. So if you knew (a) the volume-change constant, and (b) the volume of $S^3$, then you'd have an answer.

Thinking of $S^3$ as the space of unit quaternions, the tangent space at $q = 1$ is spanned by the three vectors $i, j,$ and $k$. You'll need to figure out what the volume form on that tangent space, applied to those three vectors, gives. (Hint: the answer is $1$). Those vectors correspond, under the covering map, to three rotation matrices, the matrices for the maps $$ x \mapsto i^{-1} x i \\ x \mapsto j^{-1} x j \\ x \mapsto k^{-1} x k \\ $$ defined on the 3-dimensional space of pure-vector quaternions (i.e., no real part). If you apply the volume form on SO3 to those three matrices, you'll get some number $c$. And then the volume of $SO(3)$ is just $c$ times the volume of $S^3$, divided by $2$ because the covering map is 2-to-1.

Full disclosure here: I've never actually performed this computation. But this is how I'd go about it...

Post-comment addition

Thinking of $S^3$ as unit quaternions, and $\Bbb R^3$ as pure-vector quaternions, there's a map $$ p : S^3 \to L(\Bbb R^3, \Bbb R^3) : q \mapsto (x \mapsto q^{-1} x q ). $$ The map $p$ takes a unit quaternion and converts it into a linear map on the vectorspace of pure-vector quaternions. Picking a basis $b_1 = i, b_2 = j, b_3 = k$ lets us map the space of such linear maps to the set of $3 \times 3$ matrices, thus defining a map $P: S^3 \to SO(3)$. For instance, as you observe, $$ P(i) = \pmatrix{1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & -1}. $$ More importantly, $P(1)$ is the $3 \times 3$ identity matrix. Furthermore, $p$ can be extended to define a map on all of $\Bbb R^4$ (using the same formula, or if you want to avoid inverses, replacing $q^{-1}$ with $\overline{q}$); naturally, this extends $P$ as well, although the resulting matrices, for $q \notin S^3$, are generally not orthogonal, so we get a new map -- let's call it $N$, with $$ N: \Bbb R^4 \to M_{3,3} : q \mapsto C( x \mapsto \overline{q} x q), $$ where $C$ is the function that takes a linear map on 3-space and converts it to the matrix for that map wrt the basis $i,j,k$. $N$ has the property that for $q \in S^3$, we have $N(q) = P(q)$.

That means we can use the derivative of $N$ (at $1$) to compute the derivative of $P$ (at $1$) -- just ignore the data in the non-$S^3$ direction!

Let's look at $dN(1)(i)$, the derivative of $N$ at the point $1 \in S^3$, in the direction $i$. We can start with a curve $$ c(t) = 1 + ti $$ passing through $1$ at time $0$; we have $c'(0) = i$, right?

Now look at $N(c(t))$. [This is the part where you get to do some work!] It's a matrix that depends on $t$. You take the derivative of that (to find a vector in the tangent space to $M_{3,3}$), and evaluate at $t = 0$ (to find a vector in $T_{I} M_{3,3}$); that vector (i.e., $3 \times 3$ matrix!) will be tangent to $SO(3)$ at the identity, i.e., it'll be skew-symmetric. And it'll be $dN(1)(i)$. Which means that it's also $dP(1)(i)$.

Now go through that all again for $c(t) = 1 + tj$, and $c(t) = 1 + tk$, and you'll have transformed a basis of $T_1(S^3)$ -- a basis that spans a unit volume! -- into a set of three vectors in $T_I(SO(3))$. Then you use your volume-measuring skills (something to do with the Killing metric!) to determine the volume of the cube that these three vectors span. That's your "change of volume" constant.