Equation for geodesic in manifold of orthogonal matrices

Question

From the following post

Geodesic of Stiefel manifold

it reads that a geodesic (under the canonical metric) in the manifold of orthogonal matrices can be expressed as

$Y(t) = Q e^{Xt} I_{np}$

for some matrix $Q$ and $X$, and for $I_{np} = I$, since $n = p$.

1) Does it follow that the geodesic between two orthogonal matrices $Y_1$ and $Y_2$ is given by $Y(t) = Y_1(Y^\top_2 Y_1)^{-t}$, for $t \in [0,1]$?

2) Does this curve have constant speed?

Answer 1

First, you have to restrict your attention to $\operatorname{SO}(n)$ (that is, the group of orthogonal matrices with determinant $+1$), because otherwise not all pairs of matrices in the group can be connected by a path.

With that understood, there are two interrelated problems with your proposed formula.

First, in general, "the geodesic between two orthogonal matrices $Y_1$ and $Y_2$" does not make sense, because there will in general be many different geodesics between the same two matrices. (Think of the case of $\operatorname{SO}(2)$, the circle group, in which any two points are connected by a sequence of geodesics with increasing lengths.)

Even if you restrict attention to minimizing geodesics (ones that are the shortest paths between their endpoints), there is not always a unique such geodesic. Again the circle group provides an example: between any two points that are diametrically opposite each other, there are two distinct minimizing geodesics.

The second problem is how to interpret the expression $(Y^\top_2 Y_1)^{-t}$ when $t$ is an arbitrary real number. In general, such an expression is defined to mean $\exp (-t \log(Y^\top_2 Y_1))$. But the matrix $(Y^\top_2 Y_1)$ will not in general have a unique logarithm, so we have to figure out how to interpret that expression.

What is true is that if $Y_1$ and $Y_2$ are sufficiently close to each other, then there will be a unique minimizing geodesic joining them, and that geodesic will be given by the expression $Y(t) = Y_1 \exp (-tX)$, where $X$ is the smallest skew-symmetric matrix such that $\exp(X) = Y^\top_2 Y_1$. If you interpret $\log(M)$ when $M$ is close to the identity to mean the smallest matrix $X$ such that $\exp(X)= M$, and $M^{-t}$ to mean $\exp(-t\log(M))$, then your formula is valid.

EDIT:

Let me address the questions you raised in your comments. You can find justifications for most of these claims in my book Introduction to Riemannian Manifolds (2nd ed.).

(1) is it true that any geodesic in the manifold of orthogonal matrices can be written in the form $Y(t) = Q \exp(-tX)$ for some skew-symmetric matrix and some orthogonal matrix $Q$?

Yes. Because the canonical metric is bi-invariant, the geodesics starting at the identity are exactly the one-parameter subgroups, which are the curves of the form $Y(t) = \exp(tX)$ for $X$ in the Lie algebra of $O(n)$. Then, because the metric is left-invariant, the geodesics starting at $Q$ are exactly the left translates of those starting at $I$.

(2) is there some simple formula for ONE possible minimum distance geodesic between two orthogonal matrices $Y_1$ and $Y_2$ that have determinant $+1$?

Probably not in the form you're hoping for, because such a formula cannot be continuous everywhere -- for example, in $\operatorname{SO}(2)$, if you fix $Y_1$ and let $Y_2$ move across the antipodal point, the minimizing geodesic suddenly switches from clockwise to counterclockwise (or vice versa).

However, there are sets $\operatorname{C}(I)\subset \operatorname{SO}(n)$ called the cut locus of $I$ and $\operatorname{ID}(I)\subset T_I \operatorname{SO}(n)$ called the injectivity domain of $I$, with the following properties: (1) $\operatorname{ID}(I)$ is open and star-shaped with respect to $0$; (2) $\operatorname{SO}(n) \smallsetminus \operatorname{C}(I)$ is an open dense subset containing $I$; (3) $\exp: \operatorname{ID}(I)\to \operatorname{SO}(n) \smallsetminus \operatorname{C}(I)$ is a diffeomorphism; and (4) for each $M\in \operatorname{SO}(n) \smallsetminus \operatorname{C}(I)$ there is a unique $X\in \operatorname{ID}(I)$ such that $\exp X = M$ (which we can write as $X = \log M$); and (5) for each such $M$, the curve $Y(t) = \exp (t \log M)$ for $t\in [0,1]$ is the unique minimizing geodesic from $I$ to $M$. It then follows that as long as $Y_1^\top Y_2\in \operatorname{SO}(n) \smallsetminus \operatorname{C}(I)$, the curve $Y(t) = Y_1 \exp (t \log Y_1^\top Y_2)$ is the unique minimizing geodesic from $Y_1$ to $Y_2$.

(3) are the expressions above (and the one I suggested) for constant speed geodesics?

Yes. When I say "geodesic," I mean a curve that satisfies the geodesic equation, and these are all automatically constant-speed curves.

(4) what is the reason behind the "smallest skew-symmetric matrix"? Why does X need to be small for the geodesic to minimize distance? Is it because we want the integral of the velocity over time (which is proportional to X) to be small?

The speed of a curve $t \mapsto \exp (tX)$ is exactly $|X|$ (where the norm is defined by $|X|^2 = \operatorname{trace}(X^\top X)$), and thus the length of the curve with parameter interval $[0,1]$ is exactly $|X|$.