# How would you define a geodesic in $\operatorname{SO}(2n)$ and $\operatorname{SO}(2n+1)$?

Consider me a beginner. I am trying to find the intrinsic diameter of $\operatorname{SO}(n)$ and $\operatorname{SO}(2n+1)$, but I am unsure how to define a metric between two different points in either of these spaces. When I asked my professor about the metric he gave me the answer "Distance along shortest path in $\operatorname{SO}(n)$, which sits in $\mathbb{R}^{n^2}$".

From what I understand, $\operatorname{SO}(n)$ is the group of $n\times n$ orthogonal matrices with determinant $1$. They can be parametrized as rotation matrices, so you can look at them as points on an $S^{n-1}$ sphere, I think. Would I define my path to be the path between points on these spheres?

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$\text{SO}(n)$ is not homeomorphic to a sphere for $n \ge 3$. I think your professor wants you to put the standard Euclidean metric on $\mathbb{R}^{n^2}$ (but you really should ask because this question is unclear). More intrinsically you can consider the Hilbert-Schmidt inner product $\text{tr}(A^{\dagger} B)$ which induces the standard Euclidean metric on $\mathbb{R}^{n^2}$ (and this induces a different intrinsic metric on $\text{SO}(n)$). –  Qiaochu Yuan Dec 3 '12 at 20:16
@QiaochuYuan could you describe the induced intrinsic metric? –  draks ... Dec 3 '12 at 20:58

You are correct in your interpretation, but incorrect with regards to the spheres. Think about $\mathrm{SO}(n)$ as a subset of $\mathbb{R}^{n^2}$. The distance between two matrices $A, B \in \mathrm{SO}(n)$ will be the shortest length of a path $\gamma : [0,1] \rightarrow \mathbb{R}^{n^2}$ that connects $A$ and $B$, so $\gamma(0) = A$ and $\gamma(1) = B$, which passes only through elements of $\mathrm{SO}(n)$. That is, $\gamma(t) \in \mathrm{SO}(n)$ for all $t \in [0,1]$, the path is through orthogonal matrices.

Like Qiaochu said above, $\mathrm{SO}(n)$ is not a topological sphere if $n > 1$, but even if it was, if you work with your definition of the distance, it wouldn't help you much. For example, even though $\mathrm{SO}(2)$ is homeomorphic to a circle through $$\theta \mapsto \left( \begin{array}{cc} \cos(\theta) & \sin(\theta) \\ -\sin(\theta) & \cos(\theta) \end{array} \right) \mapsto (\cos(\theta), \sin(\theta), -\sin(\theta), \cos(\theta)),$$ you need to think about it as sitting in $\mathbb{R}^4$ and minimize the distance in $\mathbb{R}^4$ with respect to all paths that pass through $\mathrm{SO}(2)$. Alternatively, if you are familiar with Riemannian metrics, you can work infinitesimally and pull everything back to your preferred description using any coordinates you want.

To expand on my last point, I'll demonstrate it for $\mathrm{SO}(2)$. Let us denote the coordinates in $\mathbb{R}^4$ by $(x^1, x^2, x^3, x^4)$ and we will identify the circle $S^1$ with $[0,2\pi]$ with ends point identified. Consider the map $\varphi : [0, 2\pi] \to \mathbb{R}^4$ given by $\varphi(\theta) = (\cos(\theta), \sin(\theta), -\sin(\theta), \cos(\theta))$. This map gives a diffeomorphism between $S^1$ and $\mathrm{SO}(2)$ inside $\mathbb{R}^4$. But from the existence of this map, you cannot conclude that the image, geometrically, looks like a circle. A priori, the image might look like a very complicated loop which doesn't resemble geometrically circle at all, something like:

If that is the case, the fact that you have a map from the nice circle to $\mathrm{SO}(2)$ wouldn't a priori help you at all to answer questions such as what is the diameter of $\mathrm{SO}(2)$.

How can you understand the geometry of the image? You can pull back the standard Riemannian metric to $S^1$ and think about it as a one-dimensional Riemannian manifold $(S^1, g)$. All the metric properties of $\mathrm{SO}(2) \subset \mathbb{R}^4$, as a metric space with the metric induced from the metric on $\mathbb{R}^4$, and much more, can be then deduced from the one-dimensional Riemmanian manifold $(S^1, g)$. You can forget about the embedding!

The standard Riemannian metric on $\mathbb{R}^4$ is given by $\sum_{i=1}^4 dx^i \otimes dx^i$. We have $$x_1 = \cos(\theta), x_2 = \sin(\theta), x_3 = -\sin(\theta), x_4 = \cos(\theta).$$ Taking the differentials, we have $$dx_1 = -\sin(\theta) d\theta, dx_2 = \cos(\theta) d\theta, dx_3 = -\cos(\theta) d\theta, dx_4 = -\sin(\theta).$$ So the pullback of the metric is just $$g = (-\sin(\theta) d\theta) \otimes (-\sin(\theta) d\theta) + (\cos(\theta) d\theta) \otimes (\cos(\theta) d\theta) + (-\cos(\theta) d\theta) \otimes (-\cos(\theta) d\theta) + (-\sin(\theta) d\theta) \otimes (-\sin(\theta) d\theta) = 2 d\theta \otimes d\theta.$$ This means that as a Riemannian manifold, $(S^1,g)$ looks exactly like the circle of radius $r = \sqrt{2}$. In particular, the length of the loop $\mathrm{SO}(2) \subset \mathbb{R}^4$ is $2\sqrt{2}\pi$ and the diameter of this metric space is $d = 2\sqrt{2}$.

So in fact, the image doesn't look as a complicated loop but "similar" to the good old circle of radius $\sqrt{2}$. If it would look like a complicated loop, the pullback metric wouldn't be as nice but would look like $g = f(\theta) d\theta \otimes d\theta$ for some non-constant function $f(\theta)$, and still, you could deduce from the model $(S^1,g)$ all the information you want about $\mathrm{SO}(2) \subset \mathbb{R}^4$.

In this simple example, you can of course see it directly, as the image is $$\{ \cos(\theta)(1, 0, 0, 1) + \sin(\theta)(0, 1, -1, 0) \}.$$ The vectors $(1,0,0,1)$ and $(0,1,-1,0)$ are orthogonal, and of length $\sqrt{2}$, so you get the result immediately, but the idea of the Riemannian approach is to forget about the embedding into $\mathbb{R}^{n^2}$ and encode everything in the (pullback) metric, working with $(\mathrm{SO}(n), g)$, a Riemannian manifold of dimension $\frac{n(n-1)}{2}$. For example, $\mathrm{SO}(3)$ is diffeomorphic to $\mathbb{RP}^3$. If you compute the pullback metric, you might discover that it looks like a familiar metric on $\mathbb{RP}^3$ and then deduce the metric properties from those of $(\mathbb{RP}^3, g)$.

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+1 nice. Could you explain your last sentence a little further? –  draks ... Dec 3 '12 at 20:55
Thank you for the answer. In $R^{n^2}$, do we compute the Euclidean distance as if the matrix was a long vector, adding up the squares of the components and taking a square root? –  Mike Flynn Dec 4 '12 at 2:58
draks, I've added a further explanation, but one should be familiar with Riemannian metrics to fully appreciate it. Mike, yes. –  levap Dec 4 '12 at 12:10