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I would like to compute the generators of the first cohomology group of this simplicial complex:

Graph

with coefficients in the Klein group $\mathbb{Z}_2 \times \mathbb{Z}_2$. I have no idea how to do that since I can't express the matrix corresponding to the boundary maps, like I would do if the coefficients were in $\mathbb{Z}$. How can I perform this calculation ?

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Are there any 2-simplices glued into the gaps, or is this just a graph? –  Kevin Carlson Nov 26 '12 at 16:51
    
This is just a graph, only vertices and edges... –  JeanThiviers Nov 26 '12 at 16:54

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We have a simplicial cochain complex on no generators in dimensions greater than $1$, $6$ $0$-dimensional generators $e_i$, and $7$ $1$-dimensional generators $[e_i, e_j]$. Name the degree-$4$ vertex $e_4,$ the three to its left $e_1,e_2,e_3$, the two to its right $e_5,e_6$.

The next thing to observe is that every element of the Klein $4$-group $V$ is its own inverse, so that just as in computing $\mathbb{Z}_2$-coefficients we can rewrite the coboundary formulae like so: $\delta(\varphi)([e_i,e_j])=\varphi(e_i)-\varphi(e_j)=\varphi(e_i)+\varphi(e_j)$. In short, the orientation of the $1$-cells will have no impact.

Now in computing $H^1$ of our complex, we've only got to compute the image of $\delta$. From the definition of $\delta$ we get for fixed $g\in V$ the cochain that is $g$ on $[e_i,e_j]$ and $[e_j,e_k]$ and vanishing elsewhere is a coboundary when $i=1,j=2,k=4$, $i=1,j=3,k=4,$ $i=1,j=2,k=3, i=4,j=5,k=6,$ or $i=4,j=6,k=5$. Furthermore by applying $\delta$ to the cochain that is $g$ at all four left-hand vertices, we see the cochain that's $g$ just on $[e_4,e_5]$ and $[e_4,e_6]$ is a coboundary; similarly for $[e_2,e_4]$ with $[e_3,e_4]$.

Contrary to my earlier revisions, there are no coboundaries with support on a single 1-simplex, and this is actually easy to see: $\delta(e_i)$ is supported on either two or four 1-simplices, and if $e_i,e_j$ are the two vertices of an edge, $\delta(e_i+e_j)$ is also supported on two 1-simplices, so that every coboundary is supported on an even number of edges.

So we'll generate the cohomology by some subset of the 1-simplices. It's not hard to see that all the 1-simplices in the quadrilateral and in the triangle are cohomologous by applying the pairs of 1-simplices we've already seen are coboundaries. So we need at most two generators. In fact we do need both: the argument above about parity of support of coboundaries applies to each cycle separately as well as to the whole graph. I mean to say that e.g. $[e_1,e_2]$ can only be cohomologous to cochains supported on 1 or 3 of the edges of the quadrilateral, so in particular not to any cochain supported only on the triangle.

So, we have $H^1=V\oplus V$, with generators the equivalence classes of $[e_5,e_6]$ and of $[e_1,e_2]$.

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Is the first $\phi$ in your third paragraph supposed to be $0$ on all the other elements $[e_i,e_j]$? If so, what happens on $[e_3,e_4]$? –  Jason DeVito Nov 26 '12 at 18:20
    
Thanks for the check, Jason. I've fixed it. –  Kevin Carlson Nov 26 '12 at 18:26
    
I'm afraid I'm still confused. We have $\delta(\phi(e_1) + \phi(e_2) + \phi(e_3) + \phi(e_4) + \phi(e_5))$ gives $g$ when applied to both $[e_4,e_6]$ and $[e_5,e_6]$), right? (I'll try to work out exactly what the right combination is, now that I have some time.) –  Jason DeVito Nov 26 '12 at 20:26
    
Yes, I think what's really the case is that we can get all and only pairs of edges as coboundaries, which makes so much sense. Thanks again for taking more care than I did. –  Kevin Carlson Nov 26 '12 at 20:29
    
I like it now ;-). Thanks for carrying out all the details! –  Jason DeVito Nov 26 '12 at 21:11

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