6
$\begingroup$

The following is a very naive construction, and I am almost embarrassed to ask questions about it.

Suppose I have a connected simple graph $\Gamma = (\Gamma_0,\Gamma_1)$ (i.e., vertex set, edge set) with a total ordering of its vertices and a Lie group $G$. I can then form the short cochain complex $C^i(\Gamma;G) =$ the functions $\Gamma_i\to G$, with smooth coboundary operator $$ \delta:C^0(\Gamma;G) \to C^1(\Gamma;G) $$ defined by $\delta(g)(\alpha) = g(d_0(\alpha))g(d_1(\alpha))^{-1}$, where $d_0,d_1$ are given by taking faces of edges ordered by the total ordering. In fact, $\delta$ is an orbit map associated with an action of $C^0(\Gamma;G)$ on $C^1(\Gamma;G)$ defined by $$ (g\cdot f)(\alpha) := g(d_0(\alpha))g(d_1(\alpha))^{-1}f(\alpha) $$ This leads to the following definition of $H^1(\Gamma;G)$ as the orbit space of this action.

When $G$ is discrete, this is a studied notion going back to Olum in the 1950s. But I am interested in the non-discrete case.

For example, if $G = S^1$, one has $H^1(\Gamma;S^1) \cong T^n$ where $T^n$ is the $n$-torus with $n$ is the first Betti number of $\Gamma$.

Questions: Is the above notion to be found in the literature? If so, where? Has it been generalized to other degrees for arbitrary simplicial complexes, or spaces?

I did a google search and a mathscinet search but I could not find any references.

Improve this question
$\endgroup$
6
  • $\begingroup$ If you want higher dimensional cohomology groups, you need $G$ abelian, see my answer mathoverflow.net/questions/36466/… $\endgroup$
    – David Roberts ♦
    Commented Jan 4, 2012 at 2:18
  • $\begingroup$ David: If $X$ is a simplicial set, then one can form the cosimplicial topological group $n \mapsto G^{X_n}$, where the target is the functions $X_n\mapsto G$ (here $G$ can be any topological group). Couldn't one take something like a normalized Moore complex of this to obtain a cochain complex of topological groups? (I.e., the intersection of the kernels of all the coface operators except the last in each degree, where the coboundary is given by the last coface operator.) $\endgroup$
    – John Klein
    Commented Jan 4, 2012 at 12:21
  • $\begingroup$ In your definition of $(g\cdot f)(\alpha)$, did you want to put the $f$ factor between the $g$ and $g^{-1}$ factors? $\endgroup$
    – Kevin Walker
    Commented Jan 4, 2012 at 15:58
  • $\begingroup$ Kevin: you are probably right. $\endgroup$
    – John Klein
    Commented Jan 4, 2012 at 17:12
  • $\begingroup$ If I try to form the analogue of the normalized cochain complex using a cosimplicial group I don't see why the iterated boundary map should be trivial. $\endgroup$
    – Justin Noel
    Commented Jan 4, 2012 at 19:56

0

Reset to default

You must log in to answer this question.

Browse other questions tagged .