1
$\begingroup$

I want to find example of two spaces with same cohomology ring but different homotopy group.

I know spaces with same homotopy groups but different homology groups and vice versa. But currently I have not found example of spaces with same cohomology ring but different homotopy group. Thanks!

$\endgroup$
  • 2
    $\begingroup$ Any homology sphere and the actual sphere. $\endgroup$ – user98602 Aug 25 '16 at 0:51
  • $\begingroup$ @MikeMiller I am sorry, could you explain more about that? Thanks. $\endgroup$ – user330928 Aug 25 '16 at 1:07
  • $\begingroup$ @MikeMiller I get your point, but I still can not find a homology sphere...... Could you please give an example? $\endgroup$ – user330928 Aug 25 '16 at 1:23
0
$\begingroup$

Mike's example in the comments is fine, but here is a simply connected example.

Let $X$ denote the homogeneous space $U(3)/U(1)^2$ where $U(1)^2$ sits in $U(3)$ as matrices of the form $\operatorname{diag}(z,w,1)$ and let $Y$ denote $S^2\times S^5$. I claim that the cohomology rings of $X$ and $Y$ are isomorphic, but that $\pi_4$ distinguishes them.

To compute the cohomology ring of $X$, note that there is an intermediate subgroup $U(1)^2\subseteq U(2)\subseteq U(3)$ giving rise to a homogeneous fibration $$U(2)/U(1)^2\rightarrow U(3)/U(1)^2\rightarrow U(3)/U(2).$$

Using the fact that $U(2)/U(1)^2 \cong S^2$ and $U(3)/U(2) = S^5$, we see that $U(3)/U(1)^2$ has the structure of an $S^2$ bundle over $S^5$. From the Gysin sequence, this is easily enough to show that that cohomology ring of $U(3)/U(1)^2$ is isomorphic to that of $S^2\times S^5$.

Now, from the fibration $U(1)^2\rightarrow U(3)\rightarrow X$, we get an LES in homotopy groups. Since $U(1) = S^1$ has vanishing higher homotopy groups, this implies $\pi_k(X)\cong \pi_k(U(3))$ for any $k\geq 3$. Further, since $U(3)$ is diffeomorphic (but not Lie isomorphic) to $SU(3)\times S^1$, $\pi_k(X)\cong \pi_k(SU(3))$. Then, according to this paper by Mimura and Toda, $\pi_4(SU(3)) = 0$, so $\pi_4(X) = 0$.

On the other hand, $\pi_4(Y)\cong \pi_4(S^2)\times \pi_4(S^5)\cong \mathbb{Z}_2\times \{0\}$.

$\endgroup$
  • $\begingroup$ An analogous argument shows $Sp(3)/Sp(1)^2$ has cohomology ring isomorphic to that of $S^4\times S^{11}$. In his thesis, Kamerich shows that $\pi_{10}$ distinguishes these spaces. Also, I'm curious if there are lower dimensional examples of closed simply connected manifolds with isomorphic cohomology rings and different higher homotopy groups. $\endgroup$ – Jason DeVito Aug 25 '16 at 2:46

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.