0
$\begingroup$

As the following product is a bit unfamiliar to me:

How do we compute the rational cohomology ring $H^*(M;\mathbb{Q})$ of the product of Lie groups: $M=SO(n_1)\times U(n_2)\times SU(n_3)\times (Sp(n_4)\cdot Sp(1))\times Sp(n_5)\times G_2\times Spin(7)\times \left(\frac{SU(n_6+1)}{S(U(1)\times U(n_6))}\right)\times \left(\frac{SO(n_7+1)}{SO(1)\times SO(n_7)}\right)\times \left(\frac{Sp(n_8+1)}{Sp(1)\times Sp(n_8)}\right)\times \left(\frac{F_4}{Spin(9)}\right)?$

Thoughts: Of course, $H^*(M;\mathbb{Q})$ is: $$H^*(M;\mathbb{Q})\cong H^*(SO(n_1);\mathbb{Q})\otimes H^*(U(n_2);\mathbb{Q})\otimes H^*(SU(n_3);\mathbb{Q})\otimes...\otimes H^*(F^4/Spin(9);\mathbb{Q})$$ by the Kunneth formula. This could be possibly simplified by first computing the rationalization $M_\mathbb{Q}$, which would help us in finding the equivariant cohomology ring $H^*_{\mathbb{Q}}(M;\mathbb{Q})=H^*(E\mathbb{Q}\times_{\mathbb{Q}}M;\mathbb{Q})$, noting that the rationalization $M_\mathbb{Q}$ is a product of Eilenberg–MacLane spaces (whose rational cohomology rings are better known-the problem is in finding such a product decomposition).

Improve this question
$\endgroup$
3
  • 1
    $\begingroup$ Forgive me for being pedantic, but not all of the spaces in your product are Lie groups. For example, $\mathrm{Sp}(n_8+1)/(\mathrm{Sp}(1)\times\mathrm{Sp}(n_8))$ is the quaternionic projective space $\mathbb{HP}^{n_8}$. I think you'll have that problem with all of the quotients in that formula. $\endgroup$
    – Arun Debray
    Commented Jul 28, 2018 at 4:01
  • $\begingroup$ @ArunDebray No, you are right! I should have said and quotients of Lie groups. $\endgroup$
    – Sergio Charles
    Commented Jul 28, 2018 at 4:09
  • 2
    $\begingroup$ If $G/K$ is a compact symmetric space, the real cohomology is given by the K invariant in the exterior algebra of the cotangent space at identity: $(\wedge ^* {\mathfrak p})^K$ . From this the rational cohomology is easy to compute. $\endgroup$
    – Venkataramana
    Commented Jul 28, 2018 at 5:25

1 Answer 1

Reset to default
4
$\begingroup$

See Section 3 in Fok's nice survey paper. (Cohomology and K-Theory of compact Lie groups), and all will be revealed.

NOTE: I just noticed that (as pointed out by Arun Debray) that yours is not a product of Lie groups, so maybe not all will be revealed.

Improve this answer
$\endgroup$
1
  • 2
    $\begingroup$ +1 for "not all will be revealed" $\endgroup$
    – Venkataramana
    Commented Jul 28, 2018 at 5:15

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .