7
$\begingroup$

There is a deep connection between algebraic topology and homological algebra on groups. A group $G$ can be interpreted as the fundamental group of a covering space $Y \rightarrow X$. (Co)Homology groups of $G$ can be interpreted as those of $X$. Similarly there is a deep connection between algebraic topology of Lie groups and homological algebra on Lie algebras. So a natural question is: Is there any deep connection between algebraic topology and homological algebra on rings?

EDIT I mean by "homological algebra on rings" homological algebra over the abelian categories of modules over rings.

  • 1
    Please leave a comment for the downvote so that I can improve my question.2012-12-09
  • 0
    I think that you are perhaps looking for K-theory: http://en.wikipedia.org/wiki/Algebraic_K-theory, or at least this in the vein of what you're saying. Do you perhaps mean something related to homological dimension of rings though?2012-12-09
  • 0
    @AlexYoucis I have no idea. That's why I asked. I know almost nothing about K-theory. I read the Wikipedia article, but I'm afraid I don't see a deep connection between them.2012-12-09
  • 0
    Perhaps the first appearance of rings in algebraic topology is that cohomology groups actually assemble into a graded ring. But of course there are many appearances beyond this. What do you mean by "homological algebra on rings", though, anyways? Rings don't form an abelian category.2012-12-09
  • 0
    Anyways, modern algebraic topology concerns itself considerably with *$E_\infty$-rings*, which are a vast generalization of the ordinary notion of a ring. In fact, any one of these gives rise to an "extraordinary cohomology theory" (with lots of extra structure). And the "homological algebra" that one can do with rings -- e.g., Andre-Quillen cohomology -- can be ported over to this setting. This makes many natural appearances in algebraic topology, too.2012-12-09
  • 0
    @AaronMazel-Gee Please see my **EDIT** in the question.2012-12-09
  • 2
    Okay. The only answer I can think of that $R$-modules would come up explicitly is that the homotopy category of $HR$-module *spectra* should be the derived category of $R$-modules. (In your other examples, the algebraic objects you're asking about arise geometrically somehow, and the only way I can see for a ring to arise geometrically is for it to *be* the space of interest.) Incidentally, how does the (group?) co/homology of $G$ show up in a covering space for which it's the group of deck transformations?2012-12-10
  • 0
    @AaronMazel-Gee "Incidentally, how does the (group?) co/homology of G show up in a covering space for which it's the group of deck transformations?" I'm not 100% sure but it goes like as follows. Let $Y$ be an acyclic connected $CW$ space. Suppose a discrete group $G$ acts properly on $Y$. Let $X = Y/G$. Then $H^n(X, \mathbb{Z})$ is isomorphic to $H^n(G, \mathbb{Z})$ for all $n$.2012-12-10
  • 1
    Ah. So really this is just when $Y=EG$ and $X=BG$. Then yes, this computation is actually just a recognition of the bar complex for $BG$ as also giving the appropriate (co)chain complex with which to compute the group (co)homology of $G$. You can even take this as a *definition* of group (co)homology, if you like.2012-12-10
  • 0
    I really don't understand why anyone would downvote this question......2012-12-11

1 Answers 1