2
$\begingroup$

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 $\mathfrak{su}(N)$?

Improve this question
$\endgroup$

1 Answer 1

Reset to default
2
$\begingroup$

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.

Improve this answer
$\endgroup$
4
  • $\begingroup$ 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
    Commented Dec 31, 2013 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
    Commented Dec 31, 2013 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
    Commented Jan 23, 2018 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
    Commented Jan 23, 2018 at 12:22

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .