From: kathleen-s@home.com (Kathleen) Subject: lowly little undergrad taking her 1st algebraic topology course, needless to say, I need help! Date: 27 Jan 2001 23:09:45 -0500 Newsgroups: sci.math Summary: Computing the homology of surfaces (connected sums) Is the homology of T^2#P^2 (that's the connect sum between the torus and the projective plane) homeomorphic to the n-fold connected sum of tori, the n-fold connected sum of projective planes or S^2? ... my text made reference to some thm that stated that every compact surface is homeomorphic to one of the above spaces. I just wondered b/c right now I'm staring at a rather scary looking triangulation of T^2#P^2 and am thinking I am not doing a direct calculation of its homology groups - that would be far too tedious. I would like to know what I'm aiming for ahead of time though. We were shown a sort of "short cut" (looking at cycles & chains being carried by subcomplexes etc) for calculating these sorts of things ... I think I shall try that. I'll be sure to post again when I next get stuck. Thanks ============================================================================== From: Robin Chapman Subject: Re: lowly little undergrad taking her 1st algebraic topology course, needless to say, I need help! Date: Sun, 28 Jan 2001 09:58:56 GMT Newsgroups: sci.math In article , kathleen-s@home.com (Kathleen) wrote: > > > Is the homology of T^2#P^2 (that's the connect sum between the torus > and the projective plane) homeomorphic to the n-fold connected sum of > tori, the n-fold connected sum of projective planes or S^2? One of the things that homology of compact surfaces detects is orientability. The H_2 of a surafce is nonzero iff the surface is orientable. Is T^2#P^2 orientable? -- Robin Chapman http://www.maths.ex.ac.uk/~rjc/rjc.html "His mind has been corrupted by colours, sounds and shapes." The League of Gentlemen Sent via Deja.com http://www.deja.com/ ============================================================================== From: Alan D Burns Subject: Re: lowly little undergrad taking her 1st algebraic topology course, needless to say, I need help! Date: Mon, 29 Jan 2001 11:48:07 +0000 Newsgroups: sci.math In article <950qgv$e6q$1@nnrp1.deja.com>, Robin Chapman writes >In article , > kathleen-s@home.com (Kathleen) wrote: >> >> >> Is the homology of T^2#P^2 (that's the connect sum between the torus >> and the projective plane) homeomorphic to the n-fold connected sum of >> tori, the n-fold connected sum of projective planes or S^2? > >One of the things that homology of compact surfaces detects is >orientability. The H_2 of a surafce is nonzero iff the surface >is orientable. Is T^2#P^2 orientable? > You should be able to use Van Kampfen's theorem to compute the fundamental group of T^2#P^2. This then determines the complex surface up to homeomorphism, from the classification theorem. H_1 is just the abelianisation of the fundamental group. H_2 is Z or O, as above. Alternively, if you want to avoid the fundamental group, use the Mayor- Vietoris sequence to compute the homology of T^2#P^2 from that of T^2 and P^2 respectiviely. -- Alan D Burns ============================================================================== From: hook@nas.nasa.gov (Ed Hook) Subject: Re: lowly little undergrad taking her 1st algebraic topology course,needless to say,I need help! Date: 29 Jan 2001 16:06:19 -0500 Newsgroups: sci.math In , Kathleen writes: |> |> Is the homology of T^2#P^2 (that's the connect sum between the torus |> and the projective plane) homeomorphic to the n-fold connected sum of |> tori, the n-fold connected sum of projective planes or S^2? Well, T^2#P^2 is homeomorphic to a sphere with 3 crosscaps, so the correct choice up above is the 3-fold connected sum of P^2's. |> |> ... my text made reference to some thm that stated that every compact |> surface is homeomorphic to one of the above spaces. I just wondered |> b/c right now I'm staring at a rather scary looking triangulation of |> T^2#P^2 and am thinking I am not doing a direct calculation of its |> homology groups - that would be far too tedious. I would like to |> know what I'm aiming for ahead of time though. OK -- I'll tell you what you would find, if you could look at the answers in the back of the book. First, though, a caveat: the final result depends on just what you're allowing for the coefficients in your chains. I'll assume that the coefficients are taken from the commutative ring R. Then you get: R if k = 0 H_k(T^2#P^2) = R x R x (R/2R) if k = 1 R_2 if k = 2 0 if k > 2 Above, R_2 = { r in R | 2r = 0 }. (So, as Robin Chapman pointed out, H_2 is 0 if R = Z [which implies that the surface you're looking at here is nonorientable], while H_2 is Z/2Z if R is ...) |> |> We were shown a sort of "short cut" (looking at cycles & chains being |> carried by subcomplexes etc) for calculating these sorts of things ... |> I think I shall try that. I'll be sure to post again when I next get |> stuck. Good luck with that ... take comfort in the knowledge that there _are_ other ways to do these sorts of calculations (which are a good deal less tedious, if you have to compute things "manually" -- of course, if you've managed to lay your hands on the right sort of software, then grinding out the answer starting from a triangulation - no matter how bletcherous - is no big deal ... :-) ... ) -- 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 ============================================================================== From: hook@nas.nasa.gov (Ed Hook) Subject: Re: lowly little undergrad taking ... Date: 31 Jan 2001 12:31:29 -0500 Newsgroups: sci.math In , Kathleen writes: |> |> > OK -- I'll tell you what you would find, if you could look at |> > the answers in the back of the book. First, though, a caveat: |> > the final result depends on just what you're allowing for |> > the coefficients in your chains. |> |> |> yes, I neglected to meniton what I was looking at. I was originally |> thinking coefficients in Z b/c all of the examples I had seen up to |> this point only dealt with this, but we were shown homology with |> arbitrary coefficients today so all is good. We calculated the |> homolgy groups of the Klein bottle with coefficients in Z/2Z ... |> |> |> >I'll assume that the coefficients |> > are taken from the commutative ring R. Then you get: |> > |> > R if k = 0 |> > |> > H_k(T^2#P^2) = R x R x (R/2R) if k = 1 |> > |> > R_2 if k = 2 |> > |> > 0 if k > 2 |> > |> |> |> When I did this with Z I got |> |> Z if k=0 |> H_k(T^2#P^2) = Z x Z x (Z/2Z) if k=1 |> 0 if k>1 |> |> is this correct? Yes. So you "did this" by calculating the homology groups ? Or by replacing R above by Z ?? :-) |> |> |> > Above, R_2 = { r in R | 2r = 0 }. (So, as Robin Chapman pointed out, |> > H_2 is 0 if R = Z [which implies that the surface you're looking |> > at here is nonorientable], while H_2 is Z/2Z if R is ...) |> |> > Good luck with that ... take comfort in the knowledge that |> > there _are_ other ways to do these sorts of calculations (which |> > are a good deal less tedious, if you have to compute things |> > "manually" -- of course, if you've managed to lay your hands |> > on the right sort of software, then grinding out the answer |> > starting from a triangulation - no matter how bletcherous - |> > is no big deal ... :-) ... ) |> |> |> Where could one find such software? Good question -- back when I still did this sort of thing, people more ambitious than myself were just starting to develop such software. Since that's 25+ years ago, I was just guessing when I wrote what's up above that (by now) it should be easy to find such stuff. Now that you've flung down the gauntlet by challenging my breezy assertion [ :-) ], I actually conducted a Google search and turned up some promising stuff. For instance, if you have access to Maple, you might want to investigate the Maple package that you'll find described at www.maplesoft.com/apps/categories/mathematics/topology/html/moise1.html Or you might want to download 'homology-3.0.tar.gz' from the URL www.mi.uni-erlangen.de/~heckenb/index_en.html I thought that I'd give that one a try myself -- it looks quite serviceable ... -- 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