MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $G$ be a connected, simply-connected complex semisimple Lie group, and let $P\subseteq G$ be a parabolic subgroup. Suppose that $V$ is a $1$-dimensional complex $P$-representation and consider the associated complex line bundle $G\times_P V\rightarrow G/P$. Let $X\subseteq G\times_P V$ denote the complement of the zero-section. I would like to compute $H^*(X;\mathbb{C})$ in terms of the given data.

My first attempt has involved identifying $X$ up to homotopy with a circle bundle over $G/P$. The Gysin sequence is then applicable. However, it becomes necessary to understand the maps $$H^i(G/P)\xrightarrow{\cup e} H^{i+2}(G/P),$$ where $e\in H^2(G/P)$ is the Euler class of the circle bundle. If one fixes a maximal torus and Borel $T\subseteq B\subseteq P$, then I think the Euler class equals the weight of the $T$-rep $V$ (where we are identifying $H^2(G/P)$ with $H^2(G/B)^{W_P}=Hom(T,\mathbb{C}^*)^{W_P}$).

I would appreciate any advice concerning my approach. If you like, you may take $G$ to be simple, and you may take the $T$-weight to be the highest root.

I would also appreciate any alternatives to my approach.

Fundamentally, though, my question is: How should one use the Hard Lefschetz Theorem to understand the maps $H^i(G/P)\xrightarrow{\cup e} H^{i+2}(G/P)$? I would really appreciate a reference on this.

share|cite|improve this question
Definitely your Euler class definition (which I would rather call first Chern class, for no strongly arguable reason) gives a $W$-equivariant morphism from $(T^*)^{W_P}$ to $H^2(G/P)$, indeed to $H^2_G(G/P)$, so your guess must be right up to a multiple. For the inverse take your $G$-equivariant line bundle $L$ to $wt(L|_{P/P})$, so the multiple must be $\pm 1$, which is good enough I hope. – Allen Knutson Sep 7 '13 at 15:05
Thank you, that is very helpful! For my purposes, it suffices to know this up to sign. I'll now try to use the Gysin sequence to find $H^*$ of the circle bundle. It would be nice if taking the cup product with the first Chern class were injective (on sufficiently low-degree cohomology), but I think this will depend heavily on the weight. I'll need to think about this carefully. – Peter Crooks Sep 7 '13 at 15:28
If it comes from a regular dominant weight then it's a hyperplane class, and Hard Lefschetz will tell you it's injective up to middle dimension. The highest root usually isn't $P$-regular. Probably you'd want to think about fibering $G/P$ over $G/P'$ where the highest root is $P'$-regular, i.e. in the interior of the wall $(T^*_+)^{W_P}$ of the Weyl chamber. – Allen Knutson Sep 7 '13 at 16:01
Juteau studies this situation in detail (when P is the parabolic corresponding to the stabilizer of the highest root) in His main concert is to do it integrally, when the hard Lefschetz theorem only tells you part of the story. He finds a very satisfactory description of the middle cohomology in terms of the root system. – Geordie Williamson Sep 7 '13 at 17:48
It occurs to me that you don't have a question. Maybe you could ask one, such that Geordie's comment could become an answer. – Allen Knutson Sep 8 '13 at 6:39
up vote 3 down vote accepted

Here is my comment as an answer.

An excellent reference / example for these ideas is Juteau's paper "Cohomology of the minimal nilpotent orbit" ( The minimal nilpotent orbit can be described as the the complement of the zero section in a line bundle on $G/P$, where $P$ is the standard parabolic subgroup corresponding to the simple roots orthogonal to the highest root. Hence calculating its cohomology is an example of your question.

Juteau is interested in doing this calculation integrally. This is quite subtle, as this paper shows. Rationally it is much simpler (by the hard Lefschetz theorem as you remark). Over $\mathbb{Q}$ the first half of the cohomology is canonically isomorphic to the primitive part of the cohomology. See Remark 2.4 in Juteau's paper.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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