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Francois Ziegler
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Let your manifold be $X=G/H$. First of all, since it is simply connected, we can write it as $K/U$ where $K$ and $U=K\cap H$ are compact in $G$ (Montgomery’s theorem, 1950). Next, since $K/U$ is homogeneous symplectic, one knows that $U$ is the centralizer of a torus $S\subset K$.1) In particular $U$ contains any maximal torus containing $S$, i.e. $U$ is an equal rank subgroup of $K$. And finally, one knows that equal rank subgroups satisfy $χ(K/U)\ne0$: e.g. Samelson (1958), or Mostow (2005).


1) That is clear, with $S$ the closure of $\exp(\mathbf Rx)$, if we already know that $X\simeq$ the (co)adjoint orbit of some $x\in\mathfrak k^*\simeq\mathfrak k$. But it can also be proved a priori : Borel–Weil (1954, Thm 1), or in more detail Matsushima (1957, Thm 1).

Let your manifold be $X=G/H$. First of all, since it is simply connected, we can write it as $K/U$ where $K$ and $U=K\cap H$ are compact in $G$ (Montgomery’s theorem, 1950). Next, since $K/U$ is homogeneous symplectic, one knows that $U$ is the centralizer of a torus $S\subset K$.1) In particular $U$ contains any maximal torus containing $S$, i.e. $U$ is an equal rank subgroup of $K$. And finally, one knows that equal rank subgroups satisfy $χ(K/U)\ne0$: e.g. Samelson (1958), or Mostow (2005).


1) That is clear, with $S$ the closure of $\exp(\mathbf Rx)$, if we already know that $X\simeq$ the (co)adjoint orbit of some $x\in\mathfrak k^*\simeq\mathfrak k$. But it can also be proved a priori : Borel–Weil (1954, Thm 1).

Let your manifold be $X=G/H$. First of all, since it is simply connected, we can write it as $K/U$ where $K$ and $U=K\cap H$ are compact in $G$ (Montgomery’s theorem, 1950). Next, since $K/U$ is homogeneous symplectic, one knows that $U$ is the centralizer of a torus $S\subset K$.1) In particular $U$ contains any maximal torus containing $S$, i.e. $U$ is an equal rank subgroup of $K$. And finally, one knows that equal rank subgroups satisfy $χ(K/U)\ne0$: e.g. Samelson (1958), or Mostow (2005).


1) That is clear, with $S$ the closure of $\exp(\mathbf Rx)$, if we already know that $X\simeq$ the (co)adjoint orbit of some $x\in\mathfrak k^*\simeq\mathfrak k$. But it can also be proved a priori : Borel–Weil (1954, Thm 1), or in more detail Matsushima (1957, Thm 1).

Source Link
Francois Ziegler
  • 31.5k
  • 6
  • 121
  • 176

Let your manifold be $X=G/H$. First of all, since it is simply connected, we can write it as $K/U$ where $K$ and $U=K\cap H$ are compact in $G$ (Montgomery’s theorem, 1950). Next, since $K/U$ is homogeneous symplectic, one knows that $U$ is the centralizer of a torus $S\subset K$.1) In particular $U$ contains any maximal torus containing $S$, i.e. $U$ is an equal rank subgroup of $K$. And finally, one knows that equal rank subgroups satisfy $χ(K/U)\ne0$: e.g. Samelson (1958), or Mostow (2005).


1) That is clear, with $S$ the closure of $\exp(\mathbf Rx)$, if we already know that $X\simeq$ the (co)adjoint orbit of some $x\in\mathfrak k^*\simeq\mathfrak k$. But it can also be proved a priori : Borel–Weil (1954, Thm 1).