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.


1 Answer 1


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.

  • $\begingroup$ 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. $\endgroup$ Feb 3, 2019 at 14:56
  • $\begingroup$ 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. $\endgroup$
    – abx
    Feb 3, 2019 at 15:43

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