Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let $G$ be a compact lie group and $H$ a closed subgroup and hence think of $G/H$ as a homogeneous space.

Then how are the Killing fields on $G/H$ the projection of the right-invariant vector fields on $G$?

In the same vein I would like to know why the following construction works:

If one looks at the tangent vectors at identity on $G$ which are "transverse" to $H$ and then exponentiate it down and flow along it and project it down to $G/H$ then on $G/H$ you would be flowing along the integral curves of the vielbeins on $G/H$.

This gives a computation approach to writing down the vielbeins on $G/H$.

I am thinking of $G/H$ to have the metric induced on it by the bi-invariant metric on $G$.

share|improve this question
    
What is a ``vielbein'? –  Richard Montgomery Jan 4 '10 at 23:59
    
I am not aware of a very precise definition but as used in Physics contexts, a "vielbien" is a basis of never-zero continuous (most often smooth) vector fields on a riemannian manifold such that it diagonalizes the metric at every point. Like any riemannian Lie Group would have such global vielbiens since its tangent bundle is trivial. Anyone has a more precise definition? –  Anirbit Jan 5 '10 at 8:06
1  
I think a vielbien is what is called an orthonormal frame, as in the method of moving frames. –  Willie Wong Mar 21 '10 at 21:06
    
Willie Wong is right. The etymology is the following: "vielbein" is the generalisation to arbitrary dimensions of "vierbien" (= tetrad). (In German, "vier"=4, "viel"=many.) –  José Figueroa-O'Farrill Jul 11 '10 at 17:18

4 Answers 4

up vote 6 down vote accepted

I think that if you generalize that statement a little it becomes clearer (also the proof).

Let $G$ be any Lie group (not necessarily compact) with a closed subgroup $H$ and a metric (not necessarily positive definite) on $G$ which is $G$-left-invariant and $H$-right-invariant (not necessarily bi-invariant).

These conditions are equivalent to picking a metric (quadratic form) at $Lie(G)$ (the lie algebra of $G$, thought of as the tangent space at the identity) which is invariant under the Adjoint representation of $G$ restricted to $H$. You extend this metric from the identity to all of $G$ by left translations.

Example: $G=SL(2,R)$, $H=SO(2)$, with the Killing metric on $G$ (bi-invariant but not positive definite). In this case $G/H$ is the hyperbolic plane. Also any semi-simple $G$ with the Cartan-Killing metric and a maximal compact $H$ (then $G/H$ is called a symmetric space).

Another example is $G=SO(3)$, $H=SO(2)$ (standard embedding) with left-invariant metric which is not necessarily right-invariant, but $H$-right-invariant. This is a model for a rigid body motion whose ellipsoid of inertia is axially symmetric.

From these conditions you get that the metric descends to $G/H$ ($G$ modulo right traslations by $H$), and that left translations by $G$, which by definition act by isometries on $G$, descend to isometries on $G/H$ (since left and right translations commute, by associativity).

If you want the metric on $G/H$ to be riemannian (ie positive definite) then you need to ask that $Lie(G)/Lie(H)$ is positive definite. This holds in the examples above.

Next pick any vector $v\in Lie(G)$ and extend it to a right invariant vector field $X$ on $G$.

Exercise: the flow of $X$ is given by the action of the 1-parameter subgroup of $G$ generated by $v$, $g_t=exp(tv)$, acting by left translations on $G$.

Since left translations are isometries of $G$ it follows that $X$ is Killing. Since $X$ is right invariant it descends to a vector field $\tilde X$ on $G/H$ and the left translations by $g_t$ descend to the flow of $\tilde X$, which is by isometries, so $\tilde X$ is Killing.

Note that $v\in Lie(G)$ doesn't have to be transverse to $Lie(H)$. Picking $v\in Lie (H)$ generates Killing fields $\tilde X$ with fixed point $[H]\in G/H$.

Another comment is that this construction doesn't generate in general all the Killing fields on $G/H$. Take for example $G$ compact with bi-invariant metric and $H$ trivial. The construction misses all the left-invariant vector fields on $G$ (generating right translations).

share|improve this answer
    
Very nice answer! –  José Figueroa-O'Farrill Jan 4 '10 at 16:40

(1) Projections of right-invariant vector fields are Killing fields. The converse hold say if $G$ is the group of isometries on the space (in general it is not true, take $G=S^3$ and $H=\{e\}$.)

(2) Yes, it is true --- it is a projection of $G$-action on $G/H$...

share|improve this answer
    
It would be of great help if you could explain the constructions. The one reference I found in a paper was very convoluted. –  Anirbit Jan 3 '10 at 19:37
    
There is no "construction", it is just translation from one language to an other --- you only keep in mind that Killing field is an "infinitesimal isometry" and Lie algebra is "infinitesimal action". –  Anton Petrunin Jan 3 '10 at 21:45

First I would like to recommend the appendix of article by : Roberto Camporesi for a clear exposition of this material.

The explanation of both questions comes from the following two facts:

The metric on G/H is induced from the Cartan-Killing form on the tangent space of G is G invariant.

The generators of the Lie algebra of H are orthogonal to coset generators of G/H (at the origin) under the Cartan-Killing metric.

H is compact therefore its action on the tangent space of G/H (at the origin) will be unitary. In fact this action will be by means of some orthogonal rotation

As a consequence, the action of G on G/H will translate the tangent space to the tangent space of the translated point accompanied by some H-rotation which doesn't change inner products.

share|improve this answer
    
There is no bi-invariant metric on $G/H$, only a $G$-invariant metric. –  José Figueroa-O'Farrill Jan 4 '10 at 5:04
    
Thanks, corrected –  David Bar Moshe Jan 4 '10 at 6:58
    
Precisely thats the paper by Camporesi that I was reading and this point in his writing wasn't clear to me! –  Anirbit Jan 4 '10 at 9:43
    
On G, the right invariant vector fields represent the left G action on itself (and vice versa). On the left coset space G/H, G acts on G/H from the left only. Another way to see that is by picking a coset representative, i.e., a section xi: G/H-->G. Let's take xi in the fundamental reprepresenation of G. It is not difficult to see that it is possible to define the inverse of the induced metric by trace(xi^-1*dxixi^-1*dxi). the G-action on G/H can be written: g.xi = gxi*h(g,xi) where h(g,xi) is an H-valued cocycle. Now the invariance of the induced metric can be checked explicitely. –  David Bar Moshe Jan 4 '10 at 10:05

I do not have a copy of Camporesi's paper, so I cannot comment on that. However, when I was learning this stuff, I found it much easier to understand if I assumed that G is a matrix group (i.e, a subgroup of GL(n)). Key examples include:

H = O(n), G = group of rigid motions (embedded in GL(n+1)), G/H = Euclidean space H = O(n), G = O(n+1), G/H = sphere H = O(n), G = O(n,1), G/H = hyperbolic space

There are corresponding examples for complex projective and hyperbolic spaces.

Also, I liked the way this stuff was presented in the book by Cheeger and Ebin.

share|improve this answer
    
Did you mean the book "Comparison Theorems in Riemannian Geometry" by Cheeger and Eben ? (AMS Chelsea Publication) I recently located that book and has one chapter on Homogeneous Spaces. –  Anirbit Feb 25 '10 at 9:53
    
Yes, that is the book I meant. I like the co-ordinate free vector field approach taken, and I like its brevity. By the way, a "vielbein" is just a "moving frame", i.e., a set of tangent vector fields on an open subset of the manifold that form an orthonormal basis of the tangent space at each point on which it is defined. Using such an object, you can also do the calculations using the dual 1-forms and the Maurer-Cartan equations (the equations that tell you what the exterior derivative of the left-invariant 1-forms of G are). –  Deane Yang Feb 25 '10 at 13:56

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.