How can one find generators of basic differential forms on homogeneous spaces?

Dear all,

In short, my problem is that I would like to have a better control of the 1-forms on a homogeneous space. Contrary to the group case, the module of differential form is not trivialisable.

To be more precise about my specific problem, let $G$ be compact simply connected semisimple (with algebra $\mathcal{G}$), and $T$ a maximal torus (with algebra $\mathcal{T}$). Then by standard associated bundle theory, one has $\Omega^1 (G/T) \simeq \{ T-equivariant\ maps\ G\to \mathcal{T}^0 \}$ , where $\mathcal{T}^0$ are the elements of $\mathcal{G}^\ast$ vanishing on $\mathcal{T}$. Here, $T$ acts on the right, by multiplication on $G$, and by $Ad^\ast$ on $\mathcal{T}^0$.

Now I'm looking for generators (as $C^\infty(G/T)-$module) of the RHS of this version of 1-forms (ie in terms of equivariant maps).

For instance, for the different and simple problem of $G$, the requirement of equivariance is trivial, and generators are given by constant maps to $\mathcal{G}^*$.

Any reference/advice is very welcome.

Thanks, Amin

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Pardon if I don't really get the language here... what do you mean by "the generators"? Vector fields satisfying the Lie algebra? – H. Arponen Jun 22 '12 at 19:28
I was hoping for a finite, more or less canonical, collection of forms such that any form can be written as as $C^\infty(G/T)$ linear combination of them (or at least identify a canonical finite dimensional vector space of such forms). I have given a bit of loose second thought about it, and I think that one way to do this would be to use coadjoint orbits identification, but I'm really not sure this would be very manageable... – Amin Jun 22 '12 at 21:38

1 Answer

This actually can be done in much greater generality. Let $G$ be a compact group and $K \subseteq G$ a closed subgroup. Then for any finite-dimensional representation $(V,\pi)$ of $K$ you can form the associated bundle $G \times_K V$ over $G/K$. Sections of this bundle are given by functions $f : G \to V$ satisfying the equivariance condition $$f(xs) = \pi(s)^{-1} f(x)$$ for all $x \in G$ and $s \in K$. You are now asking for a collection of sections that generates the module of all sections of this bundle. One way to do this is via Frobenius reciprocity.

Frobenius reciprocity implies that there is a representation $(W,\sigma)$ of $G$ such that $V \subseteq W$ and $\sigma(s)v = \pi(s)v$ for $v \in V$. Since $G$ is compact, there is a $G$-invariant inner product on $W$. Let $P$ be the orthogonal projection of $W$ onto $V$ with respect to this inner product.

Then choose an orthonormal basis $(w_i)$ for $W$. For each $i$, define a function $\eta_i : G \to V$ by $$\eta_i(x) = P \sigma(x)^{-1}w_i.$$ It is not too hard to show that each $\eta_i$ is actually a section of the associated bundle.

To show that these sections generate the module, define a $C^\infty(G/K)$-valued pairing on the module $\Gamma = \Gamma^\infty(G/K, G \times_K V)$ by $$\langle \zeta, \xi \rangle(x) = \langle \zeta(x), \xi(x) \rangle,$$ for sections $\zeta, \xi$, where $\langle , \rangle$ was the invariant inner product on $W$ that we chose above. Then it turns out that for any $\xi \in \Gamma$, we have $$\xi = \sum_i \langle \xi, \eta_i \rangle \eta_i.$$ When verifying this you need to use the fact that since the inner product on $W$ is $G$-invariant, any element $x$ of $G$ takes the orthonormal basis $(w_i)$ to another orthonormal basis.

Anyway, this is only as canonical as choosing an orthonormal basis of $W$, so it may not be what you want. But it is at least a way of getting a nice generating set.

Edit: I should explain what this is really doing. Note that a homogeneous vector bundle bundle $G \times_K V$ over $G/K$ will be a trivial bundle if and only if the representation $(V,\pi)$ of $K$ is actually the restriction of a representation of $G$ on $V$. To see this, note that if this is the case, then in the construction above we can take $W = V$, the projection $P$ is just the identity operator, and the global sections $\eta_i$ vanish nowhere and form a global frame for the bundle.

In the case when we have to take $W$ to be strictly larger than $V$, what we are doing is embedding the homogeneous bundle $G \times_K V$ into the trivial bundle $G \times_K W$. The projection $P$ tells us how to cut down fiberwise from the trivial bundle to the nontrivial one.

I should add that in your case, you can take the larger vector space $W$ to be $\mathcal{G}^\ast$ itself with the coadjoint action of $G$, and you can take the invariant inner product to be the one induced by the Killing form on $\mathcal{G}$.

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Great answer, thanks a lot, that's exactly the kind of thing I was looking for ! – Amin Jun 23 '12 at 7:02
By the way, this is written up in the beginning of Marc Rieffel's paper "A Global View of Equivariant Vector Bundles and Dirac Operators on Some Compact Homogeneous Spaces" if you want to see more details. – MTS Jun 23 '12 at 7:07
That's good to know, will check that immediately ! – Amin Jun 23 '12 at 7:57