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Is there a specific formula/method to find geodesics for a Homogeneous space? (excluding general methods applicable to arbitrary riemannian manifold)

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For sol space, there is a simple ODE, but there is no known explicit formula for the geodesics. –  Ian Agol Feb 15 '11 at 2:22
An arbitrary homogeneous space or do you have some specific ones in mind? –  Deane Yang Feb 15 '11 at 2:34
I also recommend consulting Cheeger-Ebin, "Comparison Theorems in Riemannian Geometry", which has been reprinted by the AMS. –  Deane Yang Feb 15 '11 at 2:34
@ Deane for an arbitrary homogeneous manifold. –  sam Feb 15 '11 at 10:28

2 Answers 2

up vote 10 down vote accepted

I assume you mean riemannian homogeneous space and you are talking about geodesics relative to the Levi-Civita connection.

If so, you can always try to find geodesics which are homogeneous; that is, geodesics which correspond to the orbits of one-parameter subgroups.

Let $M = G/H$ and let $\mathfrak{g} = \mathfrak{h} \oplus \mathfrak{m}$ be a reductive split (always possible for riemannian homogeneous spaces, since $H$ is compact) of the Lie algebra of $G$. By homogeneity, you can look for geodesics passing through the identity coset $o \in M$ and then transport them using the $G$-action to any other point. To specify a geodesic through $o$ it is necessary and sufficient to specify the tangent vector at $o$, which we can think of as element $X \in \mathfrak{g}$. Then the orbit $\exp(t X) \cdot o$ is a geodesic if and only if

$$ \left< [X,Z]_{\mathfrak{m}},X_{\mathfrak{m}} \right> = 0 $$

for all $Z \in \mathfrak{m}$, and where the subscript means the component along $\mathfrak{m}$.

Although this is not guaranteed to find you all geodesics, there are some homogeneous spaces (so-called geodesic orbit or g.o. spaces) for which all geodesics are of this type.

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I don't think you can find a reductive split for any Riemannian homogeneous space since $H$ does not need to be compact. It only needs to be closed in $G$. –  Bart Apr 3 '11 at 9:25
@Bart: by Riemannian, I assumed the OP meant positive-definite. In that case $H$ is a subgroup of an orthogonal group and hence compact. If by Riemannian, the OP meant also pseudo-riemannian, then you are certainly right and there need not be a reductive split. –  José Figueroa-O'Farrill Apr 3 '11 at 12:32
@José, maybe I am missing something here, but I don't see what the positivity of the OP (orthogonal product?) has to do with $H$ being a subgroup of the orthogonal group? Any $G/H$ with $H$ a closed Lie subgroup of $G$ is a homogeneous space. For sure, one can put a positive OP on $G/H$. –  Bart Apr 3 '11 at 16:35
@Bart: (OP = original post). Let $G/H$ be a homogeneous space with $H$ the stabiliser of a point $o$ (the identity coset). If $G/H$ admits a $G$-invariant positive-definite metric, then its restriction to the tangent space at $o$ is H-invariant. This means that the linear isotropy representation of $H$ is contained in the orthogonal group. I don't think that the OP would be asking about riemannian metrics on a homogeneous space which are not G-invariant, for then the question would not have an interesting answer. – –  José Figueroa-O'Farrill Apr 3 '11 at 18:30
@José: Thanks for the explanation! (Maybe you can specify in the answer that you consider the $G$-invariant metric.) –  Bart Apr 4 '11 at 8:34

One method is to use the generalization of the principle of conservation of momemntum and angular momentum. If $\phi_t$ is any one parameter group of isometries of any Riemannian manifold, with time derivative expressed as a vector field $X$, then for any geodesic $g$, the inner product $\left \langle X, \dot g \right \rangle $ is constant along the geodesic, where $\dot g$ denotes the unit tangent vector to $g$ (assuming $g$ is parametrized by arclength). In Euclidean space, applied to 1-parameter groups of translations, this gives the principle of conservation of momemntum for a geodesic; applied to 1-parameter groups of rotations, it gives the principle of conservation of angular momentum.

In a homogeneous space, there are at least enough such vector fields to form a basis at each point. In an $n$-dimensional homogeneous space, if there are more than $n$ linearly independent infinitesimal isometries, then the equations, for a typical initial vector, are consistent only in a proper subset, and on that subset, they typically define a vector field.

This may give you all the information you need, without even integrating the vector field. But in any case, it cuts the differential equation down to many fewer degrees of freedom.

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Sorry, I don't understand what $\langle X,Dg\rangle$ means. What is $Dg$? –  Dmitri Feb 15 '11 at 19:06
Thanks a lot for changing $Dg$ for $\dot g$ ! Now I understand, this is really cool :) –  Dmitri Feb 15 '11 at 23:12

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