As we all know, the space of invariant vector fields on a Lie group can be identified with the tangent space at the identity (or any other point for that matter). My question is: How does this generalize to homogeneous spaces? My guess would be that one can equate the space with the tangent space at any point point. However, it is not clear to me why everything should carry over smoothly to this more general setting.

  • $\begingroup$ This could be dumb, but how are you defining an invariant vector field on a homogeneous space? I'm trying to visualize it in the case of S^2, but I may need some coffee or something. $\endgroup$
    – j.c.
    Mar 21 '10 at 19:28
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    $\begingroup$ If you think of $S^2$ as an $SO_3$-space, the only $SO_3$-invariant vector fields are zero. Is that what you're interested in, Aston? $\endgroup$ Mar 21 '10 at 19:31
  • $\begingroup$ However, say you were looking for Jacobi fields along a closed geodesic, then they are often induced by left invariant vector fields on the the Lie group. $\endgroup$ Mar 21 '10 at 22:10

The short answer is that many (most? all?) homogeneous spaces do NOT have such a nice description. In particular, at any point $p\in M$, the set of the $G$ or $H$ invariant vectors is a strict subset of $T_pM$. (Here, and below, I'm assuming $G$ and $H$ are compact - it's all I'm really familiar with)

Here's the quick counterexample when looking for $G$ invariance: Take any $H\subseteq G$ with rank$(G)$ = rank$(H)$. Then the Euler characteristic of $G/H$ is greater than $0$. It follows that EVERY vector field has a 0, and hence, the only $G$-invariant vector field is trivial (since $G$ acts transitively).

For (much more) detail, including how it often fails even with rank$(H)<$rank$(G)$...

Suppose $M = G/H$ (and hence, $M$ is also compact), that is, I'm thinking of $M$ as the RIGHT cosets of $G$. For notation, set $e\in G$ as the identity and let $\mathfrak{g}$ and $\mathfrak{h}$ denote the Lie algebras to $G$ and $H$.

Since $G$ is compact, it has a biinvariant metric. The biinvariant metric on $G$ is equivalent to an $Ad(G)$ invariant inner product on $\mathfrak{g}$. This gives an orthogonal splitting (as vector spaces) of $\mathfrak{g}$ as $\mathfrak{h}\oplus \mathfrak{m}$. (The vector subspace $\mathfrak{m}$ is not typically an algebra)

Now, the tangent space $T_{eH}M$ is naturally be identified with $\mathfrak{m}$. In fact, this identification is $H$ invariant (where $H$ acts on $\mathfrak{m}$ via $Ad(H)$ and $H$ acts on $M$ by left multiplication: $h * (gH) = (hg)H$).

Now, when you say "invariant vector fields", you must first specify $G$ invariant or $H$ invariant. Lets focus of $G$ first, then $H$.

Typically, there are not enough $G$-invariant vector fields to span the tangent space at any point. This is because not only does $G$ act transitively on $M$, but it acts very multiply transitive (an $H$s worth of elements move any element to any other given element). Thus, a $G$-invariant vector field on $M$ is actually equivalent to an $H$ invariant vector in $T_{eH}M$.

So, for example, viewing $S^{n} = SO(n+1)/SO(n)$, You'd find that $SO(n)$ acts transitively on the unit vectors in $\mathfrak{m}$, and hence, there are no $SO(n+1)$ invariant vector fields on $S^{n}$.

Now, on to $H$ invariant vector fields. I haven't thought much about this, but even in this case, there are (often? always?) not enough. For example, viewing $S^2 = SO(3)/SO(2)$, we see that $H$ is a circle. In this case, viewing the north pole of $S^2$ as $eSO(2)$, the only $H$ invariant vector fields are the velocity vectors of rotation through the north-south axis - i.e., the flows are lines of lattitude.

Then we see that at generic points, the set of all vectors tangent to an $H$ invariant vector is a 1 dimensional vector space (and hence, doesn't span the whole tangent space), and the situation at the north and south pole is even worse: the only $H$ invariant vectors are trivial.

Even if you come up with a case where at some point, there were enough $H$ invariant vector fields, you can not translate them around all of $M$, since $H$ does NOT act transitively on $M$.

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    $\begingroup$ In other words: the tangent vector space to $G/H$ at $eH$ is the quotient of the Lie algebra of $G$ by the Lie algebra of $H$. $G$ acts on its Lie algebra by the adjoint action and the restriction of this to $H$ descends to act on the quotient vector space. The vectors fixed under this action are exactly the ones that extend to invariant vector fields. $\endgroup$ Mar 22 '10 at 1:20
  • $\begingroup$ Can one construct a basis of all vector fields using the invariant ones, like in the Lie group case? $\endgroup$ Mar 22 '10 at 12:27
  • $\begingroup$ @Aston: No, the point is that for G-invariance, the only invariant vector field is typically the 0 field. For H-invariance, it's usually (always?) the case, that they don't span at even a single point. $\endgroup$ Mar 22 '10 at 13:20
  • $\begingroup$ @Ben: Yes, that's a great summary! I guess that as a grad student, I still write in the "Grad student/homework" mode with significantly more detail than is usually presented in papers (or on MO). See, for example, most of my other posts ;-). $\endgroup$ Mar 22 '10 at 13:22
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    $\begingroup$ Well, as a grad student I really like "Grad student/homework" answers. $\endgroup$ Mar 22 '10 at 20:44

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