It is well known that the stationary curves (say $\Xi(t)$) of a regular Lagrangian $\mathcal{L}$ on a compact, semi-simple Lie group $G$ have the property that $\xi(t) = \frac{d \Xi(t)}{dt} \Xi(t)^{-1}$ solves the Euler Poincaré equations:

$\frac{d}{dt}\frac{\partial \ell}{\partial \xi} = -ad^*_{\xi} \left(\frac{\partial \ell}{\partial \xi}\right)$

or in coordinates:

$\frac{d}{dt}\frac{\partial \ell}{\partial \xi^d} = -C^{b}_{ad}\frac{\partial \ell}{\partial \xi^b} \xi^a$

wherein $\ell$ is the restriction of $\mathcal{L}$ to the lie algebra of $G$, say $\mathfrak{g}$, and $C$ are the structure constants of $\mathfrak{g}$. I also further suppose that $\mathcal{L}$ is homogeneous degree, this entails that action of any curve does not depend on the parametrisation of that curve, i.e. only the image of the curve is taken into account. If I wanted to find constrained extremal curves of $\mathcal{L}$ (for example, determining the geodesics of a sub-Riemanian metric) I would typically impose a constraint using a Lagrange multiplier, $\lambda$ say.

I could solve the standard Euler Lagrange equations for $\mathcal{L}(\Xi, \frac{d \Xi}{dt}) + \lambda(f(\Xi, \frac{d \Xi}{dt}) - c)^2$ including an equation for the variation by $\lambda$. However, I want to know if it's possible to write this problem as an Euler Poincaré equation in the case that $f$ depends only on $\frac{d \Xi(t)}{dt} \Xi(t)^{-1}$. That is to say, the constraint is also right invariant.

I'm specifically interested in the geodesics of a Finsler metric on $SU(n)$ which are parallel to a linear affine distribution.


1 Answer 1


Generally no. The Euler-Poincaré equation derives all of its structure from the Lie bracket, and an arbitrary constraint does not respect this structure. The most simple counter-example is probably a subspace constraint. Let $V \subset \mathfrak{g}$ be a subspace which is not a sub-algebra. This subspace generates a constraint distribution on $TG$ in the obvious way, which we'll denote by $VG$. The constrained Euler-Lagrange equations on $TG$ are given by

$$\frac{d}{dt} \left( \frac{\partial L}{\partial \dot{g}} \right) - \frac{\partial L}{\partial g} \in (VG)^\perp \\ \dot{g} \in VG$$

Here $(VG)^\perp \subset T^*G$ is the annihilator of $VG$. These equations can be reduced to $\mathfrak{g}$. Here they take the form

$$\frac{d}{dt} \left( \frac{\partial \ell}{\partial \xi} \right) \pm {\rm ad}^*_\xi \left( \frac{\partial \ell}{\partial \xi} \right) \in V^\perp \\ \xi \in V$$

Here $V^\perp \subset\mathfrak{g}^*$ is the annihilator of $V$. In any case, the above equation is not an Euler-Poincaré equation, and It can not be transformed into one (although, one may call it a "non-holonomic Euler-Poincar\'e equation).

  • $\begingroup$ Thanks for that. Does what you've said apply to the case of sub finsler geodesic that I mentioned? I've obtained an equation for then only in the lie algebra but I wanted to check it as I have doubts! $\endgroup$
    – Benjamin
    Aug 19, 2014 at 0:40
  • $\begingroup$ I guess so. The use of a Finsler metric and linear constraints wont change things. The lagrangian in the above equation is arbitrary. The use of affine constraints does change the equations (not sure how to write it though). However, "linear" is a special case of "affine", so the argument should still work. $\endgroup$
    – hoj201
    Aug 19, 2014 at 1:01
  • $\begingroup$ It is the affine case that I realy need to work! $\endgroup$
    – Benjamin
    Aug 19, 2014 at 1:02
  • $\begingroup$ after reading into this more, I would point out this work: arxiv.org/pdf/math-ph/0107024v1.pdf $\endgroup$
    – Benjamin
    Aug 23, 2014 at 14:40
  • $\begingroup$ and "Various aspects of n-dimensi onal rigid body dynamic" has a great section on essentially exactly this issue. $\endgroup$
    – Benjamin
    Aug 23, 2014 at 14:51

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