From: hook@nas.nasa.gov (Ed Hook) Subject: Re: Connected sums of surfaces question Date: 30 Jan 2001 12:19:39 -0500 Newsgroups: sci.math Summary: Elimination of non-orientable summands in connected sums of surfaces In , Joe writes: |> |> Hi all, |> |> I've completely forgotten where I saw this result, and what exactly it |> said - any help with the correct statement and/or a reference would be |> greatly appreciated! |> |> |> First, some notation (if there is a standard notation for this sort of |> thing, I'd be greatful to know that too): |> |> Let S(n,k) be the connected sum of the orientable closed surface of |> genus n (i.e. a doughnut with n 'holes') with k copies of the |> projective plane (n is assumed to be non-zero, k could be zero). The standard notation for this would be something like (n) (k) T^2#T^2#...#T^2#P^2#...#P^2 ; it's not clear that your version isn't better :-) |> |> I remember a result which says something like |> |> S(n,3) = S(n+1,2), |> |> or |> |> S(n,3) = S(n+1,1) |> |> or something along those lines - that is, any connected sum of an |> orientable closed surface and some projective planes can be obtained |> as a connected sum using no more than two copies of the projective |> plane. That's true. To fancy it up a bit, you can look at the set of [homeo/diffeo]morphism classes of closed 2-manifolds. This is a commutative monoid wrt. the operation of connected sum, with identity = [S^2]. It is generated as a monoid by [T^2] and [P^2], subject to the single relation [T^2] # [P^2] = [P^2] # [P^2] # [P^2] . This amounts to your "S(n,3) = S(n+1,1)" up above - that's because S(n,k) = n[T^2] # k[P^2]. It's interesting to attempt to understand/prove that relation "intuitively" - what it says is that attaching a crosscap to a torus is indistinguishable from attaching one to a Klein bottle. The "visual" proof of this: start with the 2-sphere, attach a handle to get a torus and then replace a disk with a Moebius strip to end up with T^2#P^2. Now slide one end of the handle along the surface until you reach the embedded Moebius strip, at which point you slide the handle's end once around the Moebius strip. At the end of that motion, push the end of the tube back into the complement of the strip. When you've finished, you discover that the handle is now attached "the other way" -- there are really only two ways to attach a handle to an S^2, one of them giving a T^2 and the other a Klein bottle, and the presence of the imbedded Moebius strip in this surface gives you a means of swapping between the two means of attachment. And, once you managed this feat, you've shown [T^2#P^2] = [K^2#P^2]. (If anyone knows a proof that doesn't come off sounding like mysticism, I'd *love* to see it :-) As for references, I'm a little out-of-date, but I think that you can probably find most of what you need to know in Hirsch's "Differential Topology" and/or Andrew Wallace's "Differential Topology: First Steps" (I'm not at all sure that's the correct title -- it was a very nice little paperback published by Benjamin many years ago, when I was young and dinosaurs ruled the Earth ...) -- Ed Hook | Copula eam, se non posit Computer Sciences Corporation | acceptera jocularum. NAS, NASA Ames Research Center | All opinions here are Internet: hook@nas.nasa.gov | mine alone