[last updated 2000216]
nstage trees, nfibonacci numbers, and the cohomology of k(a,n), part 2
in part 1 i suggested that the cohomology of the eilenbergmaclane
space k(a,n) could be understood in a way involving nstage trees and
nfibonacci numbers. here i demonstrate this phenomenon in the
special case of the cohomology groups of k(z/2,2) with coefficients in
the field z/2; i describe a certain cochain complex whose cohomology
groups are the desired cohomology groups.
the dimensions of the vector spaces in the cochain complex are the
fibonacci numbers 1, 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, and so on. thus
the coboundary matrixes are approximate golden rectangles, which can
be constructed according to a certain recursive scheme not described
here, or by a method involving 2stage trees as follows:
the matrixes are indexed by the nondegenerate 2stage trees with
given numbers of edges, and the (j,k) entry of the coboundary matrix
is zero unless k is a 2stage tree with x and y adjacent to each other
and j is the 2stage tree with one less edge than k obtained by
joining together the x branch with the y branch, in which case the
matrix entry is the binomial coefficient ((x+y)!/(x!*y!)) modulo 2.
i hope to eventually give a more thorough explanation of how and why
this method of calculating the cohomology groups of k(a,n) works but
for now i just list for the particular example considered the first
few coboundary matrixes and their ranks, and the dimensions of the
cohomology groups calculated from them.
(note: unlike in part 1, it seems convenient here for some reason to
count the empty list as a 2stage tree with no edges.)
number of
2stage trees
n with n edges
0 1 {empty list}
1 0 {}
2 1 {1}
3 1 {2}
4 2 {11,3}
5 3 {21,12,4}
6 5 {111,31,22,13,5}
7 8 {211,121,41,112,32,23,14,6}
8 13 {1111,311,221,131,51,212,122,42,113,33,24,15,7}
empty
list

rank of coboundary matrix = 0
dimension of 0th cohomology = 1  0  0 = 1
1
_____
rank of coboundary matrix = 0
dimension of 1th cohomology = 0  0  0 = 0
2
_____

1  0

rank of coboundary matrix = 0
dimension of 2th cohomology = 1  0  0 = 1
11 3
__________

2  0 0

rank of coboundary matrix = 0
dimension of 3th cohomology = 1  0  0 = 1
21 12 4
_______________

11  0 0 0

3  1 1 0

rank of coboundary matrix = 1
dimension of 4th cohomology = 2  0  1 = 1
111 31 22 13 5
_________________________

21  0 0 0 0 0

12  0 0 0 0 0

4  0 0 0 0 0

rank of coboundary matrix = 0
dimension of 5th cohomology = 3  1  0 = 2
211 121 41 112 32 23 14 6
________________________________________

111  0 0 0 0 0 0 0 0

31  1 1 0 0 0 0 0 0

22  0 0 0 0 0 0 0 0

13  0 1 0 0 0 0 0 0

5  0 0 1 0 0 0 1 0

rank of coboundary matrix = 3
dimension of 6th cohomology = 5  0  3 = 2
1111 311 221 131 51 212 122 42 113 33 24 15 7
_________________________________________________________________

211  0 0 0 0 0 0 0 0 0 0 0 0 0

121  0 0 0 0 0 0 0 0 0 0 0 0 0

41  0 0 0 0 0 0 0 0 0 0 0 0 0

112  0 0 0 0 0 0 0 0 0 0 0 0 0

32  0 0 0 0 0 1 1 0 0 0 0 0 0

23  0 0 1 0 0 1 0 0 0 0 0 0 0

14  0 0 0 0 0 0 0 0 0 0 0 0 0

6  0 0 0 0 0 0 0 1 0 0 1 0 0

rank of coboundary matrix = 3
dimension of 7th cohomology = 8  3  3 = 2