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Recently, answering a question here, Dror Bar-Natan observed that «way too often two-step complexes have a natural extension to become many-step complexes». By such a thing I mean (and I think Dror means too!) a "complex" in which $d^N=0$ and not $d^2=0$, as usual. I like to think of $N$-complexes are modules over the quiver $$\cdots\to\bullet\to\bullet\to\bullet\to\bullet\to\cdots$$ bound by the relations that say that the product of $N$ consecutive arrows is zero.

Many-step complexes have appeared in recent work of Mikhail Kapranov (arXiv:q-alg/9611005, which appears not to be published, according no MathSciNet?) and [Kassel, C.; Wambst, M. Algèbre homologique des N-complexes et homologie de Hochschild aux racines de l’unité. Publ. Res. Inst. Math. Sci. 34, (1998), 91–114], for example, but they certainly predate that: they appear already in [Mayer, W. A new homology theory. I, II. Ann. of Math. 43, (1942). 370–380, 594–605] and in [Spanier, E.H. The Mayer homology theory, Bull. Amer. Math. Soc. 55 (1949), 102- 112], and maybe earlier than that.

Is there a reasonable analogue for what an $N$-step simplicial complex might be?

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  • $\begingroup$ What precisely is an $N$-step complex? Is it simply a chain complex in which the groups are trivial outside a certain range of dimensions? $\endgroup$ Commented Jun 23, 2011 at 23:07
  • $\begingroup$ @Tom - I think it is a diagram of abelian groups (say - obvious extension to abelian or even pointed categories) $\mathbb{Z} \to Ab$ such that $d_i\circ \ldots\circ d_{i+N-1}$ is the zero map for all $i$, where $d_i:G_i \to G_{i+1}$. $\endgroup$
    – David Roberts
    Commented Jun 24, 2011 at 1:03
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    $\begingroup$ Oh! That's not how I interpreted what Dror wrote. $\endgroup$ Commented Jun 24, 2011 at 1:49
  • $\begingroup$ Yes, I thought by "step" Dror meant the number of chain groups in a chain complex... $\endgroup$ Commented Jun 24, 2011 at 14:20

2 Answers 2

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One approach would be to take Dold-Kan as a motivating principle, i.e. to say that a good notion of n-step simplicial object will be equivalent to n-step complexes via an equivalence which "normalizes" simplicial objects to get complexes, and which associates simplices to a complex by considering maps from the normalized k-simplex, i.e.

$X_.\mapsto (NX.,d)$

$(C_.,d)\mapsto KC_.$

where $KC_k:=Hom_{n-step~complexes}(N\Delta^k,C)$

In proposition 0.2 of his article, Kapranov gives an assignment of an n-step complex to any simplicial vector space. This seems like a reasonable candidate for the assignment on objects of a normalization functor N, and the test would be whether defining K as above plays well with the homological properties of the n-step complex.

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Oh, there are many papers recently devoted to $N$-complexes, in particular in connection with theoretical physics. A lot of these papers have simplicial stuff in it. I don't have access to MathSciNet now so some of my suggestions you have to look up yourself as I sometimes don't know the exact titles.

Here goes:

  • Tikaradze: "Homological cobstructions on $N$-complexes" Journal of pure and applied algebra (I'm pretty sure that there are some simplical stuff in this.)

  • Michel Dubois-Violette has a lot of papers that are very relevant. Especially a paper called $d^N=0$ or something to that effect in $K$-theory.

  • Angel and Diaz: "Differential graded algebras" Journal Pure and Applied Algebra I think.

  • Connes (Alain, that is) et al has some papers on so called Homogenous algebras and Yang-Mills algebras, you can look up.

  • Berger and Marconnet: "Koszul and Gorenstein properties of Homogenous algebras" Algebras and representation theory.

  • Same goes for Richard Kerner and Victor Abramov

  • And, if I may bang my own drum, I ever so briefly dabbled a bit in this area a few years back: Larsson and Silvestrov: "On $q$-differential graded algebras and $N$-complexes". Nothing very deep though ;)

Finally, and maybe most importantly, my friend Goro Kato tried to construct a derived category of $N$-complexes but managed to show instead that in reality one gains almost nothing with $N$-complexes instead of ordinary $2$-complexes, at least not from a homological perspective: From an $N$-complex one can construct, essentially in a unique manner, an ordinary $2$-complex. I don't remember the title but MathSciNet should solve that easily.

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  • $\begingroup$ I am familiar with several of these... Do you know if any of them actually has an $N$-step analogue of simplicial sets/complexes? $\endgroup$ Commented Jul 8, 2011 at 15:55
  • $\begingroup$ I'm pretty sure Tikaradze's paper has some simplicial complexes in it (my memory could fail me here). Also, I think Dubois-Violette's paper in K-theory has some simplicial complexes appearing. Once again, I could be wrong. $\endgroup$ Commented Jul 8, 2011 at 16:30

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