Homology with Coefficients - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T07:16:03Z http://mathoverflow.net/feeds/question/18667 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/18667/homology-with-coefficients Homology with Coefficients Tony Huynh 2010-03-18T21:59:40Z 2010-03-20T20:57:27Z <p>We can define the (first) homology of a surface $S$ by working with graphs embedded in $S$. That is, we take any (oriented) graph which is 2-cell embedded in $S$, and take cycles modulo boundaries in the usual way. Here, I am talking about homology with coefficients in $\mathbb{Z}$. The group that we get is independent of the graph, so is indeed a topological invariant of the surface.</p> <p>I work with <em>group-labelled graphs</em>, which are oriented graphs with their edges labelled from a finite abelian group $\Gamma$. Proceeding as above, group-labelled graphs allow us to define <em>group-labelled surfaces</em>. That is, let $G$ be a $\Gamma$-labelled graph 2-cell embedded in a surface $S$. If each face of the embedding has group-value zero (the labels of edges on the boundary of the face sum to zero), then this gives a well-defined map on homology. In fact, the embedding of $G$ in S induces a homomorphism from $H_1(S)$ to $\Gamma$. So, we can forget about the $\Gamma$-labelled graph and just study this homomorphism. </p> <p><strong>My question is:</strong> how does this construction relate to taking homology with coefficients from $\Gamma$? </p> <p>Someone once told me that what I am really doing is working with <em>cohomology</em> with coefficients in $\Gamma$, but I didn't really get this. Can someone please clarify? </p> http://mathoverflow.net/questions/18667/homology-with-coefficients/18675#18675 Answer by Mariano Suárez-Alvarez for Homology with Coefficients Mariano Suárez-Alvarez 2010-03-18T22:40:25Z 2010-03-18T22:40:25Z <p>Browsing a good introduction to algebraic topology up to the <a href="http://en.wikipedia.org/wiki/Universal_coefficient_theorem#Universal_coefficient_theorem_for_cohomology" rel="nofollow">Universal coefficient theorem for cohomology</a> would be a good plan. Hatcher's book, which you can get online, for example.</p> http://mathoverflow.net/questions/18667/homology-with-coefficients/18684#18684 Answer by Johannes Hahn for Homology with Coefficients Johannes Hahn 2010-03-19T00:20:49Z 2010-03-19T00:20:49Z <p>Hi Tony.</p> <p>This is not really a homology-question, the core of it is the fundamental group. The homomorphism you are using is used in the study of <a href="http://en.wikipedia.org/wiki/Van_Kampen_diagram" rel="nofollow">Van Kampen diagrams</a>. Consider a presentation $G=\langle A|R\rangle$. A Van Kampen diagram on $S$ is a labeled graph like you have defined it. The only difference is that in a Van Kampen diagram all labels are generators (or their inverses) $a^{\pm 1}$ and not arbitrary words (although you could define it in this general way without problems because of the Van Kampen lemma).</p> <p>Then every path in this graph has a group word written on it and "reading the word along a path" is a homomorphism {Paths}$\to G$ with respect to composition of paths. It turns out, that this is compatible with homotopy of paths so this induces a homomorphism $\pi_1(S,x_0)\to G$.</p> <p>This is the general version of your homomorphism: If your $G$ happens to be abelian, then this homomorphism factorizes through $\pi_1(S,x_0)^{ab}$ which is $H_1(S)$ by the Hurewicz theorem.</p> <p>This point of view clarifies some connections between the geometry of Van Kampen diagrams and group theoretic questions.</p> <p>For example the Van Kampen lemma tells you that a group word is trivial if and only if there is a Van Kampen diagram on this disk with this word written on the boundary.</p> <p>Another fact is this one: If there are no nontrivial "reduced" Van Kampen diagrams on the torus, then every two commuting elements of $G$ generate a cyclic subgroup (i.e. $xyx^{-1}y^{-1}=1$ has only the trivial solutions $x=a^k, y=a^m$ for some $a\in G$.). In a similar spirit one can prove: If there are no nontrivial reduced Van Kampen diagrams on the real projective plane, then there are no involutions in $G$ (i.e. $x^2=1$ has only the trivial solution $x=1$), and if there are no nontrivial reduced Van Kampen diagrams on Klein's bottle, then the only element that is conjugated to its own inverse is the identity (i.e. $yxy^{-1}=x^{-1}$ has only the trivial solution $x=1$).</p> <p>This connection between geometry and group properties becomes less obscure, if one knows the fundamental groups of the disk (1), the torus ($\langle x,y | xyx^{-1}y^{-1}=1\rangle$), the real projective plane ($\langle x | x^2=1\rangle$) and Klein's bottle ($\langle x,y | yxy^{-1}=x^{-1}\rangle$).</p> http://mathoverflow.net/questions/18667/homology-with-coefficients/18878#18878 Answer by Theo Johnson-Freyd for Homology with Coefficients Theo Johnson-Freyd 2010-03-20T20:57:27Z 2010-03-20T20:57:27Z <p>The reason that this is <em>cohomology</em> and not <em>homology</em> is that you are looking at functions from the cells to the group.</p>