[GAP Forum] Chain Complexes: The Projective plane

[Marc Bogaerts]

# In the posting about the Moebius strip we saw that it had a unique
# border and that it was a 1-cycle:
border:=SingularFaces(c2);
[ [ 0, 1, -1, 0 ], [ -1, 0, 0, 1 ], [ 1, 0, -1, 0 ], [ 0, 1, 0, -1 ] ]

# this corresponds top the cycle 2,4,5,6
# we get the set of faces already defined:
faces:=GeneratorsOfBoundaries(1,c2);
[ [ 1, 1, 1, 1, 0, 0 ], [ 1, 0, -1, 0, -1, 1 ] ]
# Let's find the coefficients of boundary in the basis defined by the 
edges M:
M:=GeneratorsOfBoundaries(0,c1);
[ [ 1, -1, 0, 0 ], [ 0, 1, -1, 0 ], [ 0, 0, 1, -1 ], [ -1, 0, 0, 1 ],
   [ 1, 0, -1, 0 ], [ 0, 1, 0, -1 ] ]

u:=[1..6]*0;
[ 0, 0, 0, 0, 0, 0 ]
u{[2,4,5,6]}:=[1,-1,-1,-1];
[ 1, -1, -1, -1 ]
u{[2,4,5,6]}:=[1,b,c,d];
[ 1, b, c, d ]
u*M;
[ -b+c, d+1, -c-1, b-d ]

u{[2,4,5,6]}:=[1,-1,-1,-1];
[ 1, -1, -1, -1 ]
# control if it is the right combination
u*M;
[ 0, 0, 0, 0 ]

Now we glue a 2-disk to the 2-skeleton of the Moebius strip:
Add(faces,u);

# and construct the projective space:
c2:=ChainComplex(2,c1,faces);
dim 0:[ 1 .. 4 ]
dim 1:[ [ 1, -1, 0, 0 ], [ 0, 1, -1, 0 ], [ 0, 0, 1, -1 ], [ -1, 0, 0, 1 ],
   [ 1, 0, -1, 0 ], [ 0, 1, 0, -1 ] ]
dim 2:[ [ 1, 1, 1, 1, 0, 0 ], [ 1, 0, -1, 0, -1, 1 ], [ 0, 1, 0, -1, -1, 
-1 ]
  ]
Basis of 3-cells: [  ]

Remark that there are no 2-cycles, the projective plane is non orientable.
HomologyGroups(c2);
[ [ 0 ], [ 2 ], [  ] ]

The first homology group is Z/2Z as expected
SingularFaces(c2);
[  ]
# The projective space has no border.

Marc Bogaerts.