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First of all, I am so sorry if this question is not appropriate to be here. I tried to ask something similar on Math Stack Exchange but it didn't have much attention. Any comment and I delete the question.

I was reading the classical paper from Milnor entitled Curvature of Left Invariant Metrics on Lie Groups. The approach is to consider an orthonormal frame on the Lie algebra, since all geometric information is gained considering an inner product on it vector space, once we have the correspondence between left invariant metrics and inner products on the Lie algebra.

From this is easy to take information about Levi-Civita connection, Curvatures and etc. But my question is, what about the geodesics? Is there any formula that relates the frame on the Lie algebra and the geodesic equation? Is possible to describe all the geodesics just looking for the orthonormal frame on the Lie algebra?

If it is not the case, what is the general approach to obtain geodesics on Lie groups? The only way is trying to describe the general Levi-Civita connection?

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    $\begingroup$ I had thought that the geodesics were all left-right translates of one-parameter subgroups. But even if that's incorrect, searching for "geodesics on Lie groups" led me to this: massey.ac.nz/~rmclachl/RLIMS10.pdf $\endgroup$ – Allen Knutson Apr 17 '17 at 2:38
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    $\begingroup$ @AllenKnutson: If the metric is bi-invariant, then the geodesics are the left and right translates of 1-parameter subgroups, but, in general, if all you have left-invariance, this is far from the case. Even for left-invariant metrics on $\mathrm{SL}(2,\mathbb{R})$ or $\mathrm{SO}(3)$, the geodesics can be quite complicated, e.g., the equations for free motion of a rigid body in 3-space. For example, see the discussion at mathoverflow.net/questions/108280/… $\endgroup$ – Robert Bryant Apr 17 '17 at 6:24
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    $\begingroup$ Just stumbled on this question, and I think it's worth mentioning that Noether's theorem on the Euler-Lagrange equations implies that the geodesic equation has many conserved quantities in this case (which is not the case for Riemannian geodesics in general). Also, just so that Google can index it, it's probably good to mention Poinsot's name somewhere. $\endgroup$ – Gro-Tsen Aug 24 '18 at 13:16
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Let $\nabla$ be the connection induced by the Levi-Civita connection. $\nabla$ is left invariant. It thus defines a bilinear product $b$ on ${\cal G}$ the Lie algebra of $G$. Let $c(t)$ be a geodesic. We can write $\dot c(t)=dL_{c(t)}(x(t))$ where $x(t)\in {\cal G}$. It you write the equation $\nabla_{\dot c(t)}\dot c(t)=0$, you obtain

$\dot x(t)+b(x(t),x(t))=0$.

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  • $\begingroup$ What is the expression for $b$? I mean, how it is obtained? $\endgroup$ – L.F. Cavenaghi Apr 17 '17 at 14:57
  • $\begingroup$ $b$ is defined on the Lie algebra of $G$. If $u,v\in {\cal G}, \nabla_uv$ is left invariant where $\nabla$ is the Levi-Civita connection. So the restriction of $\nabla$ on ${\cal G}$ defines a bilinear product. $\endgroup$ – Tsemo Aristide Apr 17 '17 at 15:34
  • $\begingroup$ Therefore $b$ is not tensorial, right? $\endgroup$ – Llohann Apr 17 '17 at 16:54
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    $\begingroup$ One can get this equation (known as the Euler equation) explicitly in terms of the specified inner product $Q:{\frak{g}}\times{\frak{g}}\to\mathbb{R}$ on the Lie algebra as follows: For each $x\in\frak{g}$, define $\mathrm{ad}^*_Q(x):{\frak{g}}\to {\frak{g}}$ by the rule $Q\bigl(\mathrm{ad}^*_Q(x)y,z\bigr) = Q\bigl(y,[x,z]\bigr)$. Then the Euler equation is $$ \dot x(t) = \mathrm{ad}^*_Q\bigl(x(t)\bigr)x(t).$$ Note, in particular, that, when $Q$ is ad-invariant, $\mathrm{ad}^*_Q(x) = -\mathrm{ad}(x)$, so the equation becomes $$\dot x(t) = -\bigl[x(t),x(t)\bigr] = 0,$$ as expected. $\endgroup$ – Robert Bryant Apr 22 '17 at 18:20

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