I'm hoping to determine the geodesic equation for a right invariant Randers metric $F(x) = \sqrt{a(x,x)} + b(x)$ on $SU(N)$. In my special case the navigation data $(h,W)$ for the Randers metric are such that $h$ is the biinvariant metric and $W$ is right invariant.

The geodesic spray coefficients induced by a Randers metric are known and can be found in "Finsler Geometry, An Approach via Randers Spaces" as formula (2.30) and in many other places. However, formulating the problem on a Lie group seems to encounter the problem that no coordinates are holonomic coordinates for any basis for the tangent space in which either the metric $h$ or $W$ take a practical form.

Is there a coordinate free way to solve this problem which results in a first order equation for the tangent vector to a geodesic involving only objects in the Lie algebra $\mathfrak{su}(N)$?


You are asking about a particular case of the general right invariant Lagrangian for curves on a Lie group. This is a well-known story, but I can summarize it here:

Let $G$ be a Lie group with Lie algebra ${\frak{g}}=T_eG$ and dual ${\frak g}^\ast$, with the canonical pairing $\langle,\rangle:{\frak g}^\ast\times{\frak g} \to \mathbb{R}$. Let $\mathrm{ad}$ and $\mathrm{ad}^\ast$ be the adjoint and co-adjoint representations, respectively, so that, for example $$ \langle \mathrm{ad}^\ast(x)\xi,y\rangle = -\langle\xi,\mathrm{ad}(x),y\rangle = -\langle\xi,[x,y]\rangle. $$ (Some people often forget about this minus sign, which is why I am reminding you of it now.)

Now, let $F:{\frak g}\to\mathbb{R}$ be a function that is smooth away from $0\in{\frak g}$ and has the property that $L = F^2$ is strictly convex on $\frak g$. Then we want to know the geodesics of the right-invariant functional $$ \lambda(\gamma) = \int_a^b F\bigl(\rho(\dot\gamma(t))\bigr)\ dt $$ where $\gamma:[a,b]\to G$ is a differentiable curve and $\rho:TG\to{\frak g}$ is the canonical right-invariant $1$-form on $G$. Usually, to get a convex functional (and fix the parametrization), we instead consider the energy functional $$ E(\gamma) = \int_a^b \bigl(F\bigl(\rho(\dot\gamma(t))\bigr)\bigr)^2\ dt =\int_a^b L\bigl(\rho(\dot\gamma(t))\bigr)\ dt. $$

Here is the standard formula: Let $L':{\frak g}\to {\frak g}^\ast$ be the Legendre transform of $L$, i.e., $d L = \langle L'(p), d p\rangle$.
Then a curve $\gamma:[a,b]\to G$ satisfies the Euler-Lagrange equations if and only if $p(t)=\rho\bigl(\dot\gamma(t)\bigr)$ satisfies the Euler equation $$ \frac{d\ }{dt}\bigl(L'(p(t))\bigr) = -\mathrm{ad}^\ast\bigl(p(t)\bigr)\bigl(L'(p(t))\bigr). $$

You should have no difficulty specializing this to your case.

| cite | improve this answer | |
  • $\begingroup$ If one doesn't square the function $F$, in the special case of $F$ a randers norm, does the final equation still hold as stated? $\endgroup$ – Benjamin Dec 31 '13 at 21:24
  • 1
    $\begingroup$ Well, if you don't square the norm, the problem is that the Legendre transform for $F$ (as opposed to $F^2$) isn't locally invertible, so there are some issues there, which I'd have to think about, but I think it might be OK. I'm not sure why you wouldn't want to square the norm, though, since it makes the functional nondegenerate and has the same (unparametrized) geodesics; working with the energy (aka action) instead just fixes the parametrization. $\endgroup$ – Robert Bryant Dec 31 '13 at 23:55
  • $\begingroup$ Revisiting this after some time, what exactly do you mean by the "canonical right-invariant 1-form". What is the canonical form you are referring to? $\endgroup$ – Benjamin Jan 23 '18 at 1:30
  • $\begingroup$ @Benjamin: This is standardly treated in many differential geometry textbooks. If $G\subset\mathrm{GL}(n,\mathbb{R})$ is a matrix group and $g:G\to {GL}(n,\mathbb{R})$ is the inclusion, regarded as a matrix-valued function on $G$, then the canonical right-invariant form is $\rho = \mathrm{d}g\,g^{-1}$. More abstractly, if $R_a:G\to G$ is right multiplication by $a\in G$, then the value of $\rho$ at $a\in G$ is the linear isomorphism of vector spaces defined by $$ \rho_a(v) = (R_{a^{-1}})'(a):T_aG\to T_eG \simeq\frak{g,}$$ where we are identifying $T_eG$ with the Lie algebra of $G$. $\endgroup$ – Robert Bryant Jan 23 '18 at 12:22

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.