By Theo Johnson-Freyd


2020-03-24 18:49:02 8 Comments

If I had a vector space with a linear endomorphism $D$ satisfying $D^2 = 0$, I might call it a differential and study its (co)homology $\operatorname{ker}(D) / \operatorname{im}(D)$. I might say that $D$ is exact if this (co)homology vanishes. I would especially do this if $D$ increased by 1 some grading on my vector space.

But I don't have this structure. Instead, I have a vector space with an endomorphism $D$, which increases by 1 some grading, satisfying $D^3 = 0$ but $D^2 \neq 0$. Then there are two possible "(co)homologies": $\operatorname{ker}(D) / \operatorname{im}(D^2)$ and $ \operatorname{ker}(D^2)/\operatorname{im}(D) $. It is an amusing exercise that if one of these groups vanishes, then so does the other, so that it makes sense to talk about $D$ being "exact".

Surely this type of structure has appeared before. Does it have a standard name? Where can I read some elementary discussion?

6 comments

@AlexArvanitakis 2020-04-18 17:18:54

What about converting this trifferential to an ordinary differential?

If $D$ acts on a space $X$, let $X_1$ and $X_2$ be two copies of $X$. Then define a differential $Q$ on the direct sum $X_1\oplus X_2$ by $$ Q X_1 \subseteq X_2\,,\quad QX_2 \subseteq X_1\,,\qquad Qx_1=Dx_1\in X_2 \forall x_1\in X_1,\quad Qx_2=D^2 x_2 \in X_1\forall x_2\in X_2\,.$$

Clearly $D^3=0\iff Q^2=0$, no?

EDIT: Assuming $D$ is degree $1$ on $X$, then if you shift the degree on $X_2$ by $+1$ relative to $X$ and assuming $X_1$ has the same degree assignment as $X$ does then I think $Q$ is also of degree 1. Nevermind that.

@Mike Miller 2020-04-18 18:14:10

$\text{ker}(Q) = \text{ker}(D) \oplus \text{ker}(D^2)$. $\text{Im}(Q) = \text{Im}(D^2) \oplus \text{Im}(D)$. So your homology groups are (the direct sum of) those of Kapranov, which are in the case of $D^p = 0$, given as $_k H(X) = \text{ker}(D^k)/\text{Im}(D^{p-k})$.

@AlexArvanitakis 2020-04-18 18:23:17

@MikeMiller: thanks, that's good to know. The obvious generalisation to $D^p=0$ seems to have the same homology groups so your formula for $_kH(X)$ is a nice check

@Severin Bunk 2020-04-17 14:22:31

You could also call this a "curved differential", or a "curved complex", motivated by the differential induced by a connection with non-trivial curvature: consider a vector bundle $E \to M$ with connection on a manifold $M$. This induces a differential $d_E \colon \Omega^p(M;E) \to \Omega^{p+1}(M;E)$ on forms on $M$ with coefficients in $E$. The differential satisfies $d_E^2 = F_E \wedge (-)$, where $F_E$ is the curvature 2-form of the connection. By the Bianchi identity it always follows that $d_E^3 = 0$. I think this would also be in agreement with the notion of curved $L_\infty$-algebras.

@Jeremy Rickard 2020-03-26 15:31:48

Such objects, for more general values of $3$, or at least the graded version (i.e., $\mathbb{Z}$-graded objects where $D$ is a degree one map with $D^N=0$) have attracted some interest in the representation theory of finite dimensional algebras in recent years, under the name of "$N$-complexes".

The fairly recent paper

Iyama, Osamu; Kato, Kiriko; Miyachi, Jun-Ichi, Derived categories of $N$-complexes, J. Lond. Math. Soc., II. Ser. 96, No. 3, 687-716 (2017). ZBL1409.18013.

may be of interest to you for the fairly lengthy list of relevant references in the introduction.

@Vladimir Dotsenko 2020-03-26 10:58:37

The situation similar to what you are describing happens when people talk about the so-called N-Koszul algebras, originally defined by R. Berger in his paper "Koszulity for nonquadratic algebras" (J. Algebra 239 (2001), 705-734). Basically, for an algebra with homogeneous relations of degree N, the way one constructs a complex that determines whether the algebra is "homologically well behaved" (an analogue of the Koszul complex of a quadratic algebra) is obtained from a $D^N=0$ situation by considering the chain complex where the differential is equal to $D$ in odd places and to $D^{N-1}$ in even places. Apart from the paper of Kapranov mentioned in comments, this is the most common source of this kind of phenomena I am aware of.

@Robert Bryant 2020-03-24 19:50:48

Many years ago, when I was a graduate student, I remember seeing a couple of papers on the homologies of operators satisfying $\partial^p=0$, generalizing the case $p=2$. I seem to remember that they were by somebody like Steenrod, and it might evan have been in the Annals, sometime in the 40s or 50s.

Unfortunately, I'm at home now and not able to access MathSciNet to look it up. However, I do remember that there was something like a set of axioms, generalizing the Steenrod axioms, for the various '$p$-homologies' that could be associated to topological spaces using such operators.

I forget now why I was interested in them. When I'm back in my office (maybe tomorrow), I'll try to find it on MathSciNet.

@Phil Tosteson 2020-03-24 19:59:33

Here is one by Kapranov arxiv.org/abs/q-alg/9611005

@Dan Petersen 2020-03-24 20:23:05

The paper you're thinking of is probably by Mayer, "A new homology theory", Annals 1942. There was a resurgence of interest in these objects after the paper of Kapranov that Phil mentioned and if you follow the citation trail from Kapranov's paper on Google scholar you'll find plenty of references.

@Robert Bryant 2020-03-24 20:39:44

@DanPetersen: Thanks (and thanks to Phil, too)! Yes, Mayer, that's who it was, and now that you mention the title, that's the paper I had in mind. I hadn't thought about that stuff in more than 40 years. You saved me the trouble of rooting around in MathSciNet.

@Mozibur Ullah 2020-03-24 21:12:51

I recall looking through Kapronovs paper before I actually understood anything about homology or cohomology.

@Achim Krause 2020-03-24 19:10:38

I would just call it a module over the truncated polynomial algebra $k[D]/D^3$. Your two flavors of homology appear as positive odd-degree and even-degree groups in $\operatorname{Ext}^*_{k[D]/D^3}(k,M)$. (This is $2$-periodic in positive degrees.) The "exactness" should hold precisely if $M$ is free as such a module.

The same interpretation works also for usual chain complexes, by the way. The algebra appearing there is the exterior algebra $k[D]/D^2$, and the $\operatorname{Ext}$-groups are actually $1$-periodic in positive degrees and recover the usual notion of homology.

If you want to be fancy, you can localize the derived category of $k[D]/D^3$ (or $k[D]/D^2$) by killing free modules. If you do this correctly, you obtain the so-called "stable module category" (a stable $\infty$ or dg-category), in which the mapping complex from $k$ to $M$ is actually a fully periodic version of the above $\operatorname{Ext}$, so in some sense is precisely described by your two different "homologies" of $M$.

@Peter LeFanu Lumsdaine 2020-03-25 11:34:04

That interpretation of homology as $\mathrm{Ext}_{k[D]/D^2}$ is very neat! Do you know any treatment that develops this interpretation?

@Pedro Tamaroff 2020-03-25 14:45:36

@PeterLeFanuLumsdaine The Ext interpretation is not terribly surprising. Note that the minimal resolution of $\mathbb k$ over $\Lambda=\mathbb k[t]/(t^2)$ is periodic given by multiplication by $t$ on $\Lambda$ everywhere. Hence, if $M$ is a chain complex, then $\mathsf{Ext}_\Lambda(\mathbb k, M)$ identifies naturally with the complex where we have $M$ everywhere and the map is $d :M\to M$, so that its homology is just $H_*(M)$ in each degree. It is also true that this is given by $\mathsf{Tor}^\Lambda(\mathbb k,M) $.

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