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Hi,

Given a Matrix Lie Group, I would like to know if the one-parameter subgroups (which can be written as $\exp^{tX}$) are the same as the geodesics (locally distance minimizing curves). Geodesics depends on the metric used so perhaps a more precise formulation of this question is:

Given a Matrix Lie Group, under what metric the one-parameter subgroups are the same as the geodesics ?

Thanks, Frank

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Is this the same as extension problem for local Lie groups? What happens in dims 1 & 2? –  rauindia Nov 22 '11 at 10:40

2 Answers 2

up vote 7 down vote accepted

Let $G$ be a compact connected semisimple Lie group and fix a left-invariant Riemannian metric $B$ on $G$. Of course, $B$ is completely determined by its value at the identity. Since $G$ is compact and semisimple, the negative of its Cartan-Killing form, which we denote by $\beta$, is a positive definite inner product; the extension of $\beta$ to a left-invariant Riemannian metric is in fact bi-invariant. Next, we diagonalize $B$ with respect to $\beta$, namely, let $f$ be the positive definite symmetric endomorphism of the Lie algebra $\mathfrak g$ of $G$ such that $B(X,Y)=\beta(f(X),Y)$ for all $X$, $Y\in\mathfrak g$. One computes easily from the Koszul formula for the Levi-Civita connection associated to $B$ that $\nabla_XY=\frac12(\mathrm{ad}_XY+f^{-1}\mathrm{ad}_Xf(Y)+f^{-1}\mathrm{ad}_Yf(X))$ for left-invariant vector fields $X$, $Y\in\mathfrak g$. In particular $\nabla_XX=f^{-1}\mathrm{ad}_Xf(X)$ and we see that the one-parameter associated to $X$ is a geodesic if and only if $\nabla_XX=0$ if and only if $[X,f(X)]=0$. In particular, this condition is satisfied if $X$ is an eigenvector of $f$.

Note that the Koszul formula above also gives $B(\nabla_XX,Z)=B([Z,X],X])=\frac12\frac{d}{dt}||\mathrm{Ad}_{\exp tZ}X||^2$ at $t=0$, which checks Denis guess that $t\mapsto\exp(tX)$ is a geodesic if and only if $X$ is a critical point of the norm-square in its adjoint orbit. In particular, there are infinitely many one-parameter groups which are geodesics if the rank of $G$ is bigger than one.

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A natural metric is the left-invariant one. Then the equation for geodesics is the so-called 'Euler equation'. Examples are

  • the equation of the free motion of a solid body (group: $SO_3(\mathbb R)$),
  • the equation of a the motion of an incompressible inviscid fluid in a bounded domain (group: measure-preserving diffeormorphism).

In general, a geodesic passing through identity is not a one-parameter subgroup. Because such a geodesic is defined by its tangent at the identity, this is equivalent to saying that a one-parameter subgroup is not always a geodesic. A one-parameter subgroup that is also a geodesic corresponds to a permanent regime in the examples above. For the fluid case, have a look to my paper Sur le principe variationnel des équations de la mécanique des fluides parfaits. RAIRO Modél. Math. Anal. Numér. 27 (1993), no. 6, 739–758.

The following is a guess, not a claim: a one-parameter subgroup $e^{tX}$ is a geodesic if and only if $\|X\|$ is critical within the adjoint orbit of $X$.

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