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Tom Goodwillie
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The mere fact that the Grassmannian $G=G_k(\mathbb C^{n+k})$ can be embedded as a variety in some projective space $\mathbb P^N$(for example by the Pluecker embedding) implies that there is a class $c\in H^2(G)$ such that the top power $c^{kn}$ is nonzero, namely the restriction of a generator of $H^2(\mathbb P^N)$. In the case of the Grassmannian, $H^2$ is one-dimensional so every nonzero class $c\in H^2(G)$ will have this property as soon as one such class does.

The mere fact that the Grassmannian $G=G_k(\mathbb C^{n+k})$ can be embedded as a variety in some projective space $\mathbb P^N$(for example by the Pluecker embedding) implies that there is a class $c\in H^2(G)$ such the top power $c^{kn}$ is nonzero, namely the restriction of a generator of $H^2(\mathbb P^N)$. In the case of the Grassmannian, $H^2$ is one-dimensional so every nonzero class $c\in H^2(G)$ will have this property as soon as one does.

The mere fact that the Grassmannian $G=G_k(\mathbb C^{n+k})$ can be embedded as a variety in some projective space $\mathbb P^N$(for example by the Pluecker embedding) implies that there is a class $c\in H^2(G)$ such that the top power $c^{kn}$ is nonzero, namely the restriction of a generator of $H^2(\mathbb P^N)$. In the case of the Grassmannian, $H^2$ is one-dimensional so every nonzero class $c\in H^2(G)$ will have this property as soon as one such class does.

Source Link
Tom Goodwillie
  • 55.9k
  • 7
  • 151
  • 240

The mere fact that the Grassmannian $G=G_k(\mathbb C^{n+k})$ can be embedded as a variety in some projective space $\mathbb P^N$(for example by the Pluecker embedding) implies that there is a class $c\in H^2(G)$ such the top power $c^{kn}$ is nonzero, namely the restriction of a generator of $H^2(\mathbb P^N)$. In the case of the Grassmannian, $H^2$ is one-dimensional so every nonzero class $c\in H^2(G)$ will have this property as soon as one does.