# Jets of Equivariant Vector Bundles

Let $M$ be a (compact) $G$-homogeneous space with fibre group $H$, and let ${\cal E}$ be a $G$-equivariant $k$-dimensional vector bundle over $M$ with corresponding representation $\pi:H \to$R$^k$. What I would like to know is whether all the jet bundles (see here for a definition of jet bundle) of $\cal E$ are also $G$-equivariant, and if so, can one construct their corresponding representations from $\pi$?

• The answer to the first question is yes: since $G$ acts by bundle morphisms on $\mathcal{E}$ it will induce an action on sections, which preserves tangency of sections, hence induces an action on jets. Concerning the second questions: if I'm not mistake the representation $\pi:H\to R^k$ determines at least locally up to isomorphism the action of $G$ on the bundle, hence you should also be able to reconstruct the representation of $H$ on the fibers of jets. – Michael Bächtold Oct 11 '12 at 20:09

I don't understand your terminology, but I'm gonna try to answer your question anyway. Let $M=G/H$ and let $\mathbb{E}$ be a representation of $H$. By $E$ I denote the associated homogeneous vector bundle $G/H \times_H \mathbb{E}$. The jet space of $E$ is also a homogeneous vector bundle which is induced from the $(\mathfrak{g},H)$-representation $J_{eH}(E)$, i.e. from the representation which is induced on the fiber over identity coset. There is a duality between $J_{eH}(E)$ and $\mathfrak{U(g)\otimes_{U(h)}} \mathbb{E}^*$.
• So then what is the explicit representation of $H$ on $\frak{U(g)}\otimes _{\frak{U(h)}}$ $E^*$. I'm guessing that its the tensor product of the dual rep of $H$ on $E$ with some rep of $H$ on $\frak{U(g)}$. – Jean Delinez Oct 16 '12 at 13:00
• Yes. I believe that the representation of $H$ on $\mathfrak{U(g)\otimes_{U(h)}\mathbb{E}^*$ is the tensor product of representation on $\mathfrak{U(g)}$ and the dual representation on $\mathbb{E}$. The representation of $H$ on the universal enveloping algebra is the extension of the adjoint representation of $H$ on $\mathfrak{g}$. Unfortunately, I'm quite busy at the moment so I can't check the details and to be honest I do not know of better references other than that I've already given in the answer. Hopefully somebody will fill that gap. – Vít Tuček Oct 16 '12 at 16:31