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.

| cite | improve this answer | |
  • $\begingroup$ 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. $\endgroup$ – Mathgymnast Feb 3 '19 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 '19 at 15:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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