4
$\begingroup$

Problem. I'm working on an exercise from a text in Riemannian geometry, that tells me to do this:

Let $\mathbf{O}(n)=\{A\in\mathbb{R}^{n\times n}:A^t A=I\}$ be equipped with its usual (left-invariant!) Riemannian metric $g_p(X_p,Y_p)=\operatorname{tr}(X_p^t Y_p)$ for $p\in \mathbf{O}(n)$ and $X_p,Y_p\in T_p\mathbf{O}(n)$.

Show that for a $C^2$-curve $\gamma\colon I\to \mathbf{O}(n)$, it holds that $\gamma$ is geodesic if and only if $\gamma^t \ddot{\gamma}=\ddot{\gamma}^t \gamma$.

I think I solved it while trying to explain where I was stuck (it's always surprising how helpful it can be to just try to formulate a question), so I will post my suggested solution as an answer to this post, in case somebody else needs it or somebody else has something to add.

$\endgroup$
3
$\begingroup$

Solution. By definition, $$\text{$\gamma$ is a geodesic $\iff$ $\tilde{\nabla}_\dot{\gamma}\dot{\gamma}=0$}\,,$$ where $\tilde{\nabla}$ is the Levi-Civita connection of $\mathbf{O}(n)$.

Since $\mathbf{O}(n)$ is a submanifold of $\mathbb{R}^{n\times n}$, and $g$ is induced by the usual metric on $\mathbb{R}^{n\times n}$, it holds that $$\tilde{\nabla}_\dot{\gamma}\dot{\gamma}=(\nabla_\dot{\gamma}\dot{\gamma})^\top=(\partial_\dot{\gamma}\dot{\gamma})^\top=\ddot{\gamma}^\top,$$ where $\nabla=\partial$ is the Levi-Civita connection of $\mathbb{R}^{n\times n}$ and $\small \top$ denotes the tangential part. This means that $$\tilde{\nabla}_\dot{\gamma}\dot{\gamma}=0\iff \ddot{\gamma}^\top=0\,.$$ Next, note that $\ddot{\gamma}(s)^\top=0$ is equivalent to $\ddot\gamma(s)\perp T_{\gamma(s)}\mathbf{O}(n)$, which just the same as saying $$\ddot\gamma(s)\in N_{\gamma(s)}\mathbf{O}(n)=\{\gamma(s)Y:Y\in\mathrm{Sym}(\mathbb{R}^n)\}$$ [cf. the end of Ch. 5 here]. Using that $\gamma(s)\in\mathbf{O}(n)$, this can be rewritten as $\gamma(s)^t\ddot\gamma(s)\in \mathrm{Sym}(\mathbb{R}^n)$, from which we finally conclude that $$\ddot\gamma^\top =0 \iff \gamma^t \ddot{\gamma}=(\gamma^t \ddot{\gamma})^t=\ddot{\gamma}^t \gamma.\quad\square$$


Remark. As was pointed out in the comments, we can use this result to show that the geodesics of $\mathbf{O}(n)$ are exactly its one-parameter subgroups composed with left translations, i.e. curves of the form $s\mapsto p\exp(sX)$ for $p\in \mathbf{O}(n)$ and $X\in T_e\mathbf{O}(n)$.

We first show that all such curves are geodesics. Let $p\in \mathbf{O}(n)$ and $X\in T_e\mathbf{O}(n)$ and form $\gamma\colon \mathbb{R}\to \mathbf{O}(n)$ defined by $\gamma(s)=p\exp(sX)$. By using the fact that $p^tp=e$ and $(X^2)^t=(X^t)^2=(-X)^2=X^2$, it's easy to show that $$\gamma(s)^t\ddot\gamma(s)=\exp(sX)^t X^2\exp(sX)= \ddot\gamma(s)^t\ddot\gamma(s)\,,$$ and hence that $\gamma$ is a geodesic.

We now show that all geodesics of $\mathbf{O}(n)$ are (at least locally) of this form. For this end, let $\alpha\colon I\to \mathbf{O}(n)$ be an arbitrary geodesic with $0\in I$ and $\alpha(0)=p$. Since $\alpha$ is a curve in $\mathbf{O}(n)$, we must have $\dot\alpha(0)\in T_p\mathbf{O}(n)$, i.e. $\dot\alpha(0)=pX$ for some $X\in T_e\mathbf{O}(n)$. But by the uniqueness of geodesics, this must mean that $\alpha$ is locally equal to the geodesic $\gamma\colon s\mapsto p\exp(sX)$, since $\gamma(0)=\alpha(0)$ and $\dot\gamma(0)=\dot\alpha(0)$. $\quad\square$

$\endgroup$
  • 1
    $\begingroup$ Very nice! Check out en.wikipedia.org/wiki/… .. It seems you can relate geodesics to maps $t\mapsto \exp(t X)$ $\endgroup$ – Orest Bucicovschi May 1 '18 at 13:54
  • 1
    $\begingroup$ @orangeskid: Good point. I added a remark where I (try) to prove this! $\endgroup$ – Oskar Henriksson May 1 '18 at 15:34
  • 1
    $\begingroup$ Great job! I wonder if you could prove in general for any (compact) Lie group with a bi-invariant metric. Also, what would be the parallel transport along the curve $\exp(t X)$, $t\in [0,1]$ (at least for $O(n)$ ). $\endgroup$ – Orest Bucicovschi May 1 '18 at 17:52
  • $\begingroup$ Interesting question! Well, for a general matrix Lie group $G$, an analogous argument would show that $\gamma\colon I\to G$ is a geodesic iff $\gamma^{-1}\ddot\gamma\in N_eG$. So what we need to show is that, provided that $G$ is a "sufficiently nice" matrix Lie group, the curve $\gamma\colon s\mapsto p\exp(sX)$ (for any $p\in G$ and all $X\in T_eG$) satisfies $\gamma(s)^{-1}\ddot\gamma(s)\in N_eG$, or equivalently: $\exp(-sX)X^2\exp(sX)\in N_eG$. $\endgroup$ – Oskar Henriksson May 1 '18 at 19:23
  • $\begingroup$ It's easy to verify this condition for certiain special cases. That it works for $\mathbf{SO}(n)$ follows from the fact that $T_e\mathbf{SO}(n)=T_e\mathbf{O}(n)$. To see that it works for $\mathbf{U}(n)$ we repeat the argument for $\mathbf{O}(n)$ with conjugate transposes instead of usual transposes. For $\mathbf{SU}(n)$, we note that $T_e\mathbf{SU}(n)\subseteq T_e\mathbf{U}(n)$, so that $N_e\mathbf{U}(n)\subseteq N_e\mathbf{SU}(n)$. Similarly, for $\mathbf{Sp}(n)$, we note that $T_e\mathbf{Sp}(n)\subseteq T_e\mathbf{U}(2n)$. Not sure how to turn this into a general argument though... $\endgroup$ – Oskar Henriksson May 1 '18 at 19:59
0
$\begingroup$

i) Tangent space is at a identity $$ \{ v|v^T + v=0 \} $$

That is it is set of all skew symmetric matrices

Further we define an inner product $G$ : $G(v,v)={\rm tr}\ v^T v$

ii) If $c$ is a curve at $I$, then $pc(t)$ is a curve at $p$

Here $G(pc',pc')=G(c',c')$ so that $p$ is an isometry.

iii) Assume that $c$ is a geodesic of unit speed at $I$.

Hence skew symmetric part of $c''$ is $\frac{c'' - (c'')^T}{2}$ where $c'' = \frac{c'' -(c'')^T}{2} + \frac{c'' + (c'')^T}{2}$

$c$ is a geodesic iff $ c'' - (c'')^T =0$

iv) If $\gamma $ is a geodesic at $p$, then $\gamma =pc$ so that $$ p^T\gamma '' = (p^T\gamma'')^T $$

$\endgroup$
0
$\begingroup$

Here is a way to get the connection for a biinvariant metric on a Lie group.

For the Levi-Civita connection we always have $$( 2\nabla_X Y - [X,Y],Z) = X(Y,Z) + Y(X,Z)-Z(X,Y) + ([Z,X],Y)+(X,[Z,Y])$$

Assume now we are on a Lie group, the metric is left invariant, and $X$, $Y$, $Z$ are left invariant vector fields. Then we get $$( 2\nabla_X Y - [X,Y],Z) =([Z,X],Y)+(X,[Z,Y])$$

Assume moreover that $(\cdot, \cdot )$ is bi-invariant. Then we get $$\nabla_X Y = 1/2 [X,Y]$$ for any $X$, $Y$ left invariant. In particular, $$\nabla_XX=0$$ if $X$ is left invariant. The converse is true: if $\nabla_X X=0$ for any $X$ left invariant then $([Z,X],Y)+(X,[Z,Y])=0$ for any $X$, $Y$, $Z$ left invariant, that is $(\cdot, \cdot)$ is bi-invariant ( for the connected component of $G$). Note that all are equivalent to: all of the maps $t\mapsto \exp(tX)$ are geodesics.

Assume $(\cdot, \cdot)$ bi-invariant. The parallel transport of $X$ from the origin to $\exp(X)$ along the curve $t\mapsto \exp(tX)$, $0\le t \le 1$ is $L_{\exp X}\ _\star (X) = R_{\exp X}\ _\star (X)$. However, the parallel transport of $Y$ along the same curve will not be in general either of $L_{\exp X}\ _\star (Y)$ ,$ R_{\exp X}(Y)$, since the two vectors are not equal in general. I think at every $\exp(tX)$, $t\in [0,1]$ the transport of $Y$ is $$L_{\exp \frac{t}{2} X}\ _\star\circ R_{\exp \frac{t}{2} X}\ _\star(Y)$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.