Subject: Re: two postings From: wilker@math.purdue.edu Date: Thu, 1 Dec 2005 09:54:48 -0500 (EST) >> >> Subject: Re: four postings >> From: Clarence W Wilkerson Jr >> Date: Mon, 28 Nov 2005 12:26:18 -0500 >> >> BU(n) and BSU(n) x BS1 are not homotopy equivalent at the prime $p=2$, >> but are at >> other primes. The mod 2 Steenrod algebra should detect the >> non-equivalence. >> In particular, I suspect that Sq2 behaves differently on the two spaces. >> >> Clarence Actually, Andy Baker has reminded me that I needed to say that BU(2n) and BSU(2n) x BS1 are not h.e. at p=2. The point is that as groups U(n) = (S1 x SU(n))/ diagonal Z/nZ . Therefore one needs 2 | n to get the counterexample. It's similar at other primes. ___________________________________________________________ Subject: BSU(n) From: Tom Goodwillie Date: Thu, 1 Dec 2005 10:34:16 -0500 > > BU(n) and BSU(n) x BS1 are not homotopy equivalent at the prime $p=2$, but are at > other primes. The mod 2 Steenrod algebra should detect the non-equivalence. > In particular, I suspect that Sq2 behaves differently on the two spaces. > > Clarence Sq2 suffices for this when n is even but not when n is odd. Note: Wlog a homotopy equivalence BU(n)-->BSU(n) x BU(1) will make the determinant map BU(n)-->BU(1) correspond to the projection, so composing with the other projection it will yield a left inverse to the usual map BSU(n)-->BU(n). So we want to show that there is no such right inverse. Maybe this can be done by showing that the canonical surjection of graded rings H^*BU(n)-->H^*BSU(n) has no left inverse which (after reducing mod 2) respects the action of the Steenrod operations. If n is even then it does not have a left inverse respecting Sq2, but if n is odd it does. Maybe you can finish the job with higher Sq's or with K-theory. Note: Of course, when n goes to infinity we get BU ~ BSU x BU(1), using the H-space structure on BU. Tom