While reading the book Introduction to the $h$-Principle by Y. Eliashberg and N. Mishachev, I noticed that the authors state, at the end of section 9.1.A, that the space of all symplectic structures on Euclidean space $\mathbb R^{2n}$, which they identify as $\text{GL}(2n,\mathbb R)/\text{Sp}(2n)$, is not contractible, whereas the space of all Euclidean structures of the aforementioned space, which can be identified with $\text{GL}(2n,\mathbb R)/\text{O}(2n)$, is contractible.

It is easy to prove both assertions, but, what exactly is the homotopy type of $\text{GL}(2n,\mathbb R)/\text{Sp}(2n)$ ? I have no idea of homotopy theory, so I am completely clueless.

Thanks in advance for your answers.

  • 6
    $\begingroup$ It's homotopy equivalent to O(2n)/U(n). That's the most concrete general answer you'll get; notice this is always disconnected with two components. In the case n = 1 what I wrote is precisely the two element set. You can use the long exact sequence of homotopy groups of the fibration U(n) -> O(2n) -> O(2n)/U(n); for instance for n > 1 you always have pi_2 O(2n)/U(n) ~ Z. $\endgroup$
    – mme
    Jan 13 at 17:15
  • 2
    $\begingroup$ Maybe it helps if I point out that as n -> infinity these groups stabilize, and are 8-periodic by Bott periodicity: for n large (probably n > k/2 suffices but I didn't check) we have that pi_k(O(2n)/U(n)) is Z/2, 0, Z, 0, 0, 0, Z, Z/2, listing off the values depending on the value of k mod 8, the first being k = 0 mod 8. $\endgroup$
    – mme
    Jan 13 at 17:21
  • $\begingroup$ Also in case it helps, for some further low $n$ these spaces can be identified with more familiar ones: For $n=2,3,4$ they are the disjoint union of two copies of $CP^1, CP^3,$ a smooth quadric in $CP^7$ respectively. $\endgroup$ Jan 14 at 5:23


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.