I have a doubt which I hope the MO community can quickly resolve.

The associative grassmannian is an eight-dimensional homogeneous space $G_2/SO(4)$. It can be identified with the space of quaternion subalgebras of the octonions. It has a $G_2$-invariant riemannian metric making it a rank-2 riemannian symmetric space. The Poincaré polynomial of its de Rham cohomology is $1 + t^4 + t^8$. This is also the Poincaré polynomial for the quaternionic projective plane $\mathbb{HP}^2$, also an eight-dimensional homogeneous space $Sp(3)/Sp(2)\times Sp(1)$. It is the space of quaternionic lines in $\mathbb{H}^3$. It has an $Sp(3)$-invariant riemannian metric making it into a rank-1 riemannian symmetric space.

The difference in the ranks shows that as riemannian symmetric spaces, the associative grassmannian and the quaternionic projective plane are not isometric, but perhaps one is obtained from the other by squashing (also known as the canonical variation). Anyway, my real question is whether they are diffeomorphic.

Question: Are the associative grassmannian and the quaternionic projective plane diffeomorphic?

Thanks in advance.


According to

Characteristic Classes and Homogeneous Spaces, I

A. Borel and F. Hirzebruch,

Section 17, they are not even homotopy equivalent: $G_2 / SO(4)$ has mod $2$ homology in degree 2, whereas $\mathbb{HP}^2$ does not.

  • $\begingroup$ Many thanks! This is the answer I was hoping for. Cheers, José $\endgroup$ Apr 13 '11 at 8:55

To expand on Oscar's answer: the principal $SO(4)$ bundle over $G_2/SO(4)$ gives us $$ \qquad \qquad \qquad \ldots \to \pi_2(G_2) \to \pi_2(G_2/SO(4)) $$ $$ \to \pi_1(SO(4)) \to \pi_1(G_2) \to \pi_1(G_2/SO(4)) $$ $$ \to \pi_0(SO(4)) \to \ldots \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad$$ and $\pi_2(G_2) = \pi_1(G_2) = \pi_0(SO(4)) = 1$, so $\pi_1(G_2/SO(4)) = 1$, $\pi_2(G_2/SO(4)) = Z_2$.


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.