# one-parameter subgroup and geodesics on Lie group

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

• Is this the same as extension problem for local Lie groups? What happens in dims 1 & 2? Nov 22, 2011 at 10:40

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.

• How does the above construction change if $G$ is non-compact, for instance $\mathrm{Sp}(2N,\mathbb{R})$?
– LFH
May 24, 2017 at 14:41
• @LFH If $G$ is semisimple, the Killing form is nondegenerate so we can still define $f$ but it is not necessarily definite. I think Koszul formula yields the same expression for $\nabla_XY$ as above. May 26, 2017 at 22:03
• I see, but Sp(2N,R) is not semi-simple. I'm currently Reading Arnold's book on classical mechanics (App 2) where this corresponds to stable rotations of a rigid body. He does not make any reference to the Killing form, but rather states that subgroup geodesics correspond to critical points of the dual inner product on the co-adjoint orbits on the dual Lie algebra $g^*$. Do you know if this is equivalent to your description?
– LFH
May 27, 2017 at 1:45
• Ok, I tried to write down my thoughts in my answer below - I'm curious if you agree with my statements and if you think it's equivalent to your statement - I'm a little confused...
– LFH
May 27, 2017 at 2:44
• $Sp(2N,\mathbb{R})$ is semisimple. May 27, 2017 at 4:46

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 $X$.

I may have a similar answer with an alternative derivation. Given a group $G$ with Lie algebra $\mathfrak{g}$ and right-invariant inner product $\langle A,B\rangle_g=\langle Ag^{-1},Bg^{-1}\rangle_e$ where $A,B\in T_gG$, $Ag^{-1}=(R_{g^{-1}})_*A$, $Bg^{-1}=(R_{g^{-1}})_*B$ and $e$ being the neutral element. Right-translations of a group are generated by left-invariant vector fields which are therefore Killing vector fields. Let us assume that $M_I\in\mathfrak{g}$ for $I\in\{1,\cdots,\dim{G}\}$ is a basis of the Lie algebra. This means we have the left-invariant vector fields \begin{align} X_I=gM_I=L_gM_I\in T_gG\,. \end{align} It is a well-known fact that the inner product of (affinely parametrized) geodesics $u: [0,1]\to G: t\mapsto u(t)$ have conserved quantities \begin{align} C_I&=\langle X_I,\dot{u}(t)\rangle_{u(t)}\\ &=\langle uM_I,\dot{u}\rangle_u\\ &=\langle uM_Iu^{-1},\dot{u}u^{-1}\rangle_e\,. \end{align} For a Lie group with right-invariant metric, knowing the $\dim{G}$ conserved quantities $C_I$ and initial point of a geodesics $u(0)$ determine the geodesics uniquely.

Let us check what the condition for a 1-parameter subgroup $u(t)=e^{tA}$ with $A\in\mathfrak{g}$ is to be a geodesic, or equivalently to have the $C_I$ being conserved. Plugging $u(t)$ into above formula gives \begin{align} C_I=\langle e^{tA}M_Ie^{-tA},A\rangle_e\,. \end{align} We can take the derivative $d/dt$ on both sides and find \begin{align} 0=\langle e^{tA}[A,M_I]e^{-tA},A\rangle_e\,. \end{align} This formula is equivalent to $\frac{d}{dt}\lVert e^{tM_I}Ae^{-tM_I}\rVert=\langle[M_I,A],A\rangle=0$ that Claudio wrote above and that Denis guessed. It's just an alternative derivation without making reference to the Killing form or its properties...

• This seems essentially right (+1). I did not yet check everything though, but I know that this property of one-parameter groups hold on every Lie group, be it semi-simple or not which seems to be your claim. Feb 25, 2021 at 15:15