Is there a map of spectra implementing the inverse of the Thom isomorphism? - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T21:46:18Z http://mathoverflow.net/feeds/question/73744 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/73744/is-there-a-map-of-spectra-implementing-the-inverse-of-the-thom-isomorphism Is there a map of spectra implementing the inverse of the Thom isomorphism? roger123 2011-08-26T08:28:15Z 2011-08-26T09:01:19Z <p>In the top answer to the question <a href="http://mathoverflow.net/questions/46787/is-there-a-map-of-spectra-implementing-the-thom-isomorphism" rel="nofollow">"Is there a map of spectra implementing the Thom isomorphism?"</a> it is explained (with a reference to Rudyaks book) that from a rank $r$ vector bundle $\mu:V\to X$, a spectrum $E$ with multiplication $m:E\wedge E\to E$ and a Thom class $t:X^\mu\to \Sigma^r E$ one gets the Thom isomorphism:</p> <p>The (cohomological) Thom isomorphism sends a map $x:X_+\to E^n$ of spectra to the composition $$X^\mu\to X_+\wedge X^\mu\xrightarrow{x\wedge t}E^n\wedge E^r\xrightarrow{m}E^{n+r}.$$ The induced isomorphism on homotopy groups is then $E^{n}(X_+)\cong E^{n+r}(X^\mu)$.</p> <blockquote> <p>Is it possible to give an inverse map, i.e. a map sending a $X^\mu\to E^{n+r}$ to an $X_+\to E^n$ on the level of spectra, too?</p> </blockquote> <p>I am not asking for a proof of the Thom isomorphism. Even with the existence of such a map one has to prove that the correspondence induces an isomorphism on homotopy groups.</p> http://mathoverflow.net/questions/73744/is-there-a-map-of-spectra-implementing-the-inverse-of-the-thom-isomorphism/73747#73747 Answer by Johannes Ebert for Is there a map of spectra implementing the inverse of the Thom isomorphism? Johannes Ebert 2011-08-26T09:01:19Z 2011-08-26T09:01:19Z <p>You find the inverse Thom isomorphism if you generalize the construction from my answer to the <a href="http://mathoverflow.net/questions/46787/is-there-a-map-of-spectra-implementing-the-thom-isomorphism/46809#46809" rel="nofollow">old question</a> to stable vector bundles. The point is that if you have two stable vector bundles $\eta$ and $\mu$ over $X$, then you get a diagonal map</p> <p>$$X^{\eta\oplus\mu} \to X^{\eta} \wedge X^{\mu}.$$</p> <p>Now if there is a Thom class $X^{\mu} \to \Sigma^n E$, then there is a Thom class $X^{-\mu} \to \Sigma^{-n} E$; and the inverse to the Thom isomorphism maps an element $X^{\mu}\to \Sigma^k E$ to the composition</p> <p>$$X \to X^{\mu} \wedge X^{-\mu} \to \Sigma^k E \wedge \Sigma^{-n} E \to \Sigma^{k-n} E.$$</p>