Let $G$ be a compact , connected and simply connected Lie group and $a\in\mathfrak{g}^*$ (dual of Lie algebra of Lie group $G$). Then let $O_a$ be a generic coadjoint orbit then can we say $H^2(O_a,\mathbb{Z})\cong \mathbb{Z}^n$ for some $n$. Is there any counter example?
2 Answers
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No, there are no counter-examples. Note that a generic coadjoint orbit is $G$-equivariantly diffeomorphic to $G/T$, for a maximal torus $T\subseteq G$. However, $G/T$ (also known as the full flag variety of $G_{\mathbb{C}}$) has a cell decomposition into even-dimensional cells (the Bruhat decomposition). Therefore, $G/T$ has vanishing integral cohomology in odd degrees, and each even-dimensional integral cohomology group is a free $\mathbb{Z}$-module. In particular, $H^2(G/T;\mathbb{Z})$ is a free $\mathbb{Z}$-module, so your statement is true in general.
answered Feb 5, 2014 at 20:59
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$\begingroup$ Can you explain more this part :"$G/T$ (also known as the full flag variety of $G^C$) has a cell decomposition into even-dimensional cells (the Bruhat decomposition). Therefore, $G/T$ has vanishing integral cohomology in odd degrees, and each even-dimensional integral cohomology group is a free $Z$-module" $\endgroup$– user21574Commented Feb 5, 2014 at 21:03
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2$\begingroup$ Sure. If we choose a Borel subgroup $B$ of $G_{\mathbb{C}}$ containing $T$, then we obtain a $G$-equivariant diffeomorphism $G/T\cong G_{\mathbb{C}}/B$. The latter space can be given the structure of a cell complex, in which each cell is even-dimensional. (In fact, these cells are indexed by the Weyl group.) We can therefore compute its integral cohomology using cellular cohomology. In the cellular cochain complex, every group in odd degree is 0. So, all differentials in the complex are 0, and $H^i$ is the same as the cochain complex group in position $i$. Everything in the complex is free, $\endgroup$ Commented Feb 5, 2014 at 21:10
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1$\begingroup$ so that every $H^i$ is free. $\endgroup$ Commented Feb 5, 2014 at 21:10
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The cohomology of regular coadjoint orbits of a compact Lie group can be computed using Morse theory. This approach is due to Bott and Samelson (1955). Let $A \in \mathfrak{g}^{*}$ be a regular element, $\langle\cdot,\cdot\rangle$ the Cartan-Killing form, and consider the coadjoint orbit $\mathcal{O}_{B}$ of a regular element $B \in \mathfrak{g}^{*}$, then $\langle A,\cdot\rangle: \mathcal{O}_{B}\to\mathbb{R}$ is a Morse function whose critical set is the Weyl orbit of $B$, and the index of a critical point $w(B)$ turns out to be twice the number of root hyperplanes crossed by traversing the line from $A$ to $w(B)$. This gives a cell decomposition concentrated in even degrees, so the cohomology vanishes in odd degrees and in particular $H^{2}(\mathcal{O}_{B},\mathbb{Z})$ has the form you suggest.
