1
$\begingroup$

I am reading Tom Dieck's page 537 and I am not sure what the vertical map that I put in the title is in the diagram in the bottom of the page. This map is labeled Thom Isomorphism. Here $MSO(k)$ is the thom space of the tautological bundle over $BSO(n)$. There is a thom isomorphism from $H_{j-1}(MSO(k))\to H_{j-k-1}(BSO(k))$ giving by capping with the $k$ dimensional cohomology class of $H^k(MSO(K))$.

The suspension isomorphism on reduced homology is from $H_j(MSO(k)) \to H_{j-k}(\Sigma BSO(k))$. I am skeptical that $H_{j-k}(\Sigma BSO(k))=H_{j-k}(BSO(k))$. For example $\Sigma BSO(2)=\Sigma BS^1=\Sigma K(\mathbb{Z},2)=\Sigma \Omega K(\mathbb{Z},3)$ which admits an evaluation map that admits isomorphisms on low dimensional homotopy groups , to $K(\mathbb{Z},3)=$ which doesn't have the same low dimensional homoptopy groups as $BSO(2)=K(\mathbb{Z},2)$

Another random idea is I know that the thom isomorphism commutes with the boundary map and hence suspension. Do you have any ideas on finding this map?


edit: I put the wrong map in the title and consequentially my post made no sense. It should be readable now.

$\endgroup$
0
$\begingroup$

It is just the map from $H_j(\Sigma MSO(k))=H_j(M(SO(k) \oplus SO(1))) \to H_{j-(k+1)}(B(SO(k) \oplus SO(1)))=H_{j-(k+1)}(B(SO(k)))$

But I don't know why the dimensions are not matching, for the degree of the thom isomorphism.

I personally think that tom tieck got his indices mixed up and that his book has a typo.

$\endgroup$

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.