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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.

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Please leave a comment for the downvote so that I can improve my question. –  Makoto Kato Dec 9 '12 at 18:31
    
I think that you are perhaps looking for K-theory: 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? –  Alex Youcis Dec 9 '12 at 18:32
    
@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. –  Makoto Kato Dec 9 '12 at 18:52
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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? –  Aaron Mazel-Gee Dec 10 '12 at 8:59
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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. –  Aaron Mazel-Gee Dec 10 '12 at 13:11

1 Answer 1

Ring theory is always lurking more or less visible in algebraic topology. As your question is quite broad, I'll try and give a quick overview with a few references. Since homological algebra takes place often over R-modules or it can be reduced to this case somehow, it might be a bit difficult to define what you mean by "homological algebra on rings". Nonetheless, here are a few examples that I think fit your description:

  • The Hochschild homology of an $R-R$-bimodule reflects some ring-theoretic stuff. For instance, $H_1(R,R)$ of a $k$-algebra $R$ is the module of differentials $\Omega_{R/k}$. If $\mathbb{Q}\subseteq R$ then there is an algebraic decomposition of this homology that is analogous to the Hodge decomposition in complex manifold theory.

  • Taking the above example further if $k$ is a ring and $X$ a simplicial set, the cyclic homology (Hochschild homology taking into account a cyclic action on the corresponding simplicial set) of the simplicial module $k[X]$ is the same as the $S^1$-equivariant homology of the geometric realisation of $X$ with coefficients in $k$.

(Hochschild and cyclic homology are related to $K$-theory and the Lie algebra of matrix algebras as well. For the above examples, see Chapter 9 of Weibel's book "An Introduction to Homological Algebra" and all of Loday's book "Cyclic Homology")

  • As mentioned in the comments, $K$-theory is like homology on rings. Moreover, algebraic topology is clearly interested in vector bundles; on a nice space $X$ the category of rank $n$ real vector bundles on $X$ is equivalent to the category of rank $n$ finitely generated projective modules over the continuous functions $C(X,\mathbb{R})$. The group $K_0(C(X))$ is the Grothendieck group of the isomorphism classes of fg projectives.

  • $K$-theory of course has its roots and many applications to algebraic topology and algebraic geometry. Plentiful examples can be found in "The Handbook of $K$-Theory" (http://www.math.uiuc.edu/K-theory/handbook/)

(In fact, the higher $K$-groups of a ring $R$ can be defined as the homotopy groups of a certain simplicial resolution associated to $R$; simplicial resolutions are like chain complexes in an abelian category, and in fact for abelian categories nonnegative chain complexes and simplicial sets coincide. So it really is like a homology of rings, as opposed to doing abelian homological algebra in the category of modules.)

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