Subject: Re: two more on BSU(n) From: Brayton Gray Date: Thu, 1 Dec 2005 17:03:02 -0600 This question is similar to the real case where O(n) is isomorphic to SO(n) x Z2 iff n is odd, since -Identity is central and has determinant -1 precisely when n is odd. In the complex case, the same argument works localized at p if n is a unit mod p; since the matrix with u on the diagonal is central, all we need is that u--->u^n is a homotopy equivalence. On Thursday, Dec 1, 2005, at 14:45 US/Central, Don Davis wrote: > Two responses to yesterday's remark on BSU(n)...........DMD > ___________________________________________________ > > 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 > >