# Homogeneity of a projective vector bundle

Let $$M=G/K$$ be a $$G$$-homogeneous manifold and suppose that $$E\to G/K$$ is a homogeneous (complex) vector bundle, i.e. it is defined by a representation $$\phi : K \to \text{Aut(V)}$$ for some complex vector space $$V$$. More precisly, $$E=(G \times V)/ \sim$$, where $$(gk,v) \sim (g,\phi (k) v)$$.

As usual, the projectivisation $$\mathbb P (E)$$ of $$E$$ is defined to be $$\mathbb P (E) := (E\setminus 0)/ \mathbb C ^*$$. Clearly, the group $$G$$ acts on the manifold $$\mathbb P (E)$$ and if we endow $$\mathbb P(E)$$ with the metric induced by $$E$$ and $$M$$, this action is by isometries.

Question 1): When is $$\mathbb P (E)$$ a $$G$$-homogeneous manifold, i.e. when is the group action of $$G$$ transitive? It seems sufficient to assume that the action of $$K$$ on $$V$$ given by $$\phi$$ is transitive, but is this actually equivalent to the $$G$$-homogeneity of $$\mathbb P (E)?$$

More concretly, I am interested in the case when $$M=G/K$$ is a compact Kähler homogeneous space and $$E=T^* M$$ is the cotangent bundle of $$M$$.

Question 2): When is $$\mathbb P (T^*M)$$ $$G$$-homogeneous? For example, is this true if $$M$$ is a compact Hermitian symmetric space? Could it be possible in this case, that the irreducibility of the corresponding representation $$\phi$$ is not only necessary, but also a sufficient condition?

EDIT: One can see explicitly that this is true for $$M=\mathbb{CP}^n=U(n+1)/(U(n)\times U(1))$$: The representation $$\phi$$ corresponding to the tangent bundle is the adjoint representation of $$U(n)\times U(1)$$ on $$\mathfrak u (n+1)$$ restricted to $$V=\mathfrak u (n+1)/ \mathfrak u (n) \times \mathfrak u (1)$$. This action is transitive, which should imply that $$U(n+1)$$ acts transitively on $$\mathbb P ( TM)$$. Dualising proves the claim for $$T^*M$$.

This is why I would expect that a similar result should be true in the case of Hermitian symmetric spaces.

I would guess that the answer to these questions is well-known, but, unfortunately, I could not find any reference, where this is proven. So I would be greatful for any help.

I think the answer is no to both questions. Let $$V$$ be a symplectic vector space, of dimension $$>2$$. Take $$G=\operatorname{Sp}(V)$$ and $$M=\mathbb{P}(V)$$. The cotangent bundle $$\mathbb{P}(T^*M)$$ can be identified with the subvariety of pairs $$(p,h)\in \mathbb{P}(V)\times \mathbb{P}(V^*)$$ such that $$p\in h$$. The pairs $$(p,h)$$ such that $$h$$ is the orthogonal of $$p$$ (for the symplectic form) form a $$G$$-orbit, hence the action of $$G$$ on $$\mathbb{P}(T^*M)$$ is not transitive.
• Thank you for the quick answer! It is true that your $G$ action is not transitive. However, I do have doubts that this action is induced by the construction in my question. In fact, I think one can check by direct computation that in the case of $CP^n$, the projectivisation is $U(n+1)$ homogeneous. I will add this example to my question. – Mathgymnast Feb 3 '19 at 14:56
• I was indeed assuming that your $G$ was a complex reductive group. But you can replace $\operatorname{Sp}(V)$ by a maximal compact subgroup and the same property will hold. However I agree that this is not the standard presentation of $\Bbb{P}^n$ as a Hermitian symmetric space. – abx Feb 3 '19 at 15:43