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Let $R$ be a commutative ring. For awhile I have been trying to motivate to myself more fully the definition of and various structures on the category $\text{Ch}(R)$ of chain complexes of $R$-modules (and various subcategories thereof). One significant piece of motivation is the Dold-Kan correspondence, at least when $R = \mathbb{Z}$, which tells us that studying connective chain complexes is like studying linearized homotopy theory (or linearized higher category theory). This is a great idea, but I don't have much intuition for what's going on in the proof of Dold-Kan, and I don't see how one could have predicted in advance that something like Dold-Kan might be true just by looking at all the definitions in the right way. I like the idea of linearized homotopy but I don't know what the conceptual path is from linearized homotopy to, for example, the braiding $a \otimes b \mapsto (-1)^{|a| |b|} b \otimes a$.

Consider also the differential. I can think of various ways to motivate $d^2 = 0$, and I don't quite know how they fit together. For example, one can talk about boundaries of manifolds with boundary, the exterior derivative, and Stokes' theorem. If one starts from the simplicial / higher-categorical perspective, the differential encodes something like the generalized source / target of a higher morphism, and somehow the fact that this generalized source / target ought to satisfy a natural "gluing law" (for example if $a \to b$ is a $1$-morphism then $d(a \to b + b \to c)$ ought to equal $d(a \to c)$) is equivalent to it squaring to zero. I can sort of picture how this works in low dimensions but I don't completely grasp what the exact relationship between these two ideas is.

Keeping in mind the symmetric monoidal structure, the differential behaves like an element of a super Lie algebra, concentrated in degree $-1$, acting on a representation (see for example Theo Johnson-Freyd's MO answer here). The action of a super Lie algebra should be related to infinitesimal symmetry coming from a super Lie group, but I don't have a clear idea of what this super Lie group is or what it has to do with homotopy theory. This seems to have something to do with the supergeometric definition of differential forms, but I don't really know anything about this.

Algebraically, the relation $d^2 = 0$ seems to come from at least two different ideas: first-order approximation, and odd things anti-commuting with themselves. Both of these ideas seem relevant to what I'm confused about, but I can't put them together into a cohesive story.

So what is that cohesive story?

Edit: If the bulk of the question seems sort of silly to you, feel free to focus on that last bit about super Lie algebras. I remember hearing that this has something to do with the action of the automorphism group of the odd real line; I would appreciate if someone could clarify that for me.

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In all likelyhood, no once could have predicted D-K to be true (in fact, no one did...) I cannot see what would constitute an answer to your question. History provides what I consider the motivation for complexes, but you want an a priori motivation ---something I find quite unexistent, because it simply clashes with the way mathematics is done. One can produce various a posteriori a priori (this is not a typo) motivations for the rule of signs (Verdier(-Deligne?) has a cute one, for example, involving cocycles) but that is only synthetic, and if you want to... –  Mariano Suárez-Alvarez Jun 21 '11 at 13:16
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...explain to someone why the signs are there, it is pretty absurd to start with "well, there is this cocycle..." –  Mariano Suárez-Alvarez Jun 21 '11 at 13:17
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What do you mean by "first principles"? Chain complexes and maps of them were around before "category" had its modern meaning. So isn't this like trying to motivate the existence of Stonehenge from Keynesian economics? Anyone could try of course but perhaps it's not all that informative. Perhaps take a look at the first 20 pages of Dieudonne's history of algebraic and differential topology ? –  Ryan Budney Jun 21 '11 at 19:05
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@Ryan: maybe "first principles" was a poor choice of words. I would say it is more like trying to motivate representation theory from quantum mechanics. Of course the former predates the latter; nevertheless I think nowadays the latter can serve as a very good motivation for thinking that the former is a natural thing to study. –  Qiaochu Yuan Jun 21 '11 at 21:23
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I think this is a great question, in that a good answer would be really valuable in terms of understanding the concepts that we work with, rather than merely accepting them. –  Daniel Moskovich Jun 22 '11 at 13:27
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7 Answers 7

$ d^2 $ and homology are almost the most basic and natural quests in mathematics. In general, the image of a map can be thought of as "things that can be constructed". In general, the kernel of a map is "things we can test". So $ \ker d = \operatorname{im }d$ is one of the basic quests in mathematics, linearized: "let us find a construction for all the things that satisfy a certain criterion".

Likewise, given a construction for things that satisfy a certain property (name, given $A\stackrel{d_1}{\longrightarrow}B\stackrel{d_2}{\longrightarrow}C$, where $d_1$ is the construction and $d_2=0$ is the property), the homology $\ker d_2/\operatorname{im }d_1$ measures how successful you had been - to what extent you were able to construct all the things you wanted to construct. Again a very basic mathematical quest.

So two-step complexes, $d^2=0$, and homology are as natural as anything. What puzzles me is that way too often two-step complexes have a natural extension to become many-step complexes. I have no good philosophical explanation for why that should be the case.

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That's a very nice way to put it! What examples of often do you have in mind in your last paragraph? –  Mariano Suárez-Alvarez Jun 22 '11 at 17:36
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This seems like a very nice point of view for thinking about the relation between (co)homology and deformation / obstruction. But can you expand a little on what you mean by "things that can be constructed" and "things we can test"? –  Qiaochu Yuan Jun 22 '11 at 19:09
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Things that can be constructed: Express an element of B as the image of an element in A. Things we can test: Ask the boolean question "is it in ker d_2?". –  Daniel Moskovich Jun 22 '11 at 19:22
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If by many-step complexes you mean chain complexes (with $d^2=0$ and not something with $d^N=0$, which I gather is what Mariano thought you meant), then I think Voevodsky would say that the reason they arise so often is that mathematics is inherently homotopical (he currently is working on formulating explicitly homotopical foundations for mathematics). –  Omar Antolín-Camarena Jul 8 '11 at 19:01
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If the most basic question in math is "how can we solve this problem we have characterized?" then the next one is "how can we simplify the solutions?" and this is where multi-step complexes come from. You may be able to construct the solutions but your construction will have redundancies, which may or may not be explained by some stuff mapping into $A$, and so on. Likewise, if $C$ is the "terms in which you describe the problem", then you can ask for a characterization of "the terms that are really used" (i.e. coming from $B$) in terms of a map from $D$ elsewhere. –  Ryan Reich Feb 17 '13 at 1:43
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"...utinam intelligere possim rationacinationes pulcherrimas quae e propositione concisa DE QUADRATUM NIHILO EXAEQUARI fluunt."

Henri CARTAN

[...if I could only understand the beautiful consequences following from the concise proposition $d^2=0$ ]

Extracted from Cartan's laudatio on receiving his Doctor Honoris Causa degree from Oxford in 1980. It is the epigraph of Gelfand-Manin's Methods of Homological Algebra.

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1890, maybe?$ $ –  Ryan Reich Jun 22 '11 at 18:45
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Dear Ryan: no, Henri Cartan [received][1] his honorary degree on June 25,1980. What gave you and your upvoter the idea that it happened in 1890 (before Cartan's birth in 1904)? [1]blms.oxfordjournals.org/content/13/3/263.abstract –  Georges Elencwajg Jun 22 '11 at 20:15
    
@Georges: it was somewhat possible that you were talking about the senior Cartan... –  Qiaochu Yuan Jun 23 '11 at 20:25
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Dear Qiaochu, it was not possible because I explicitly wrote Henri Cartan in order to prevent any confusion with his father Elie Cartan. And even Elie was born only in 1869 and couldn't plausibly have gotten an honorary degree when he was 21 years old (he only got his regular doctorate in 1894, aged 25). Anyway, no harm was done by Ryan's comment. Quite the contrary: the chronology of this remarkable filiation is now displayed! –  Georges Elencwajg Jun 23 '11 at 21:32
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It is not clear what constitutes an answer. Historically chain complex technology came out of topological homology from simplicial complexes, then via homological algebra. The objects initially are positive or negatively graded as that is from the topology. The $d^2=0$ is direct from the combinatorics. (and this is even clearer if one takes total orders on the vertices and converts from simplicial complexes to simplicial sets). Morphisms then are given directly by the combinatorics, and models of mapping spaces and constructions such as suspension lead to considering chain complexes with both positive and negative terms. The conventions on orientation lead to the tensor product differential etc.

All that interacted via de Rham etc. with the intuitions from differential geometry which is where some of your other ideas fit. The idea of simplicial approximation is then central as that leads to the need for homotopies etc. and also as before to $d^2 =0$.

I suggest looking back at Cartan and Eilenberg, or MacLane's "Homology" and how they motivate things. That will give a historical perspective on the origins of the ideas. (There is a good history of homological ideas see http://www.math.uiuc.edu/K-theory/0245/.) I am sure you want more than this, since you seem to be searching for `why' not just the history, but I think the history is a good place to start. Many of the formulae and conventions come from a clear historical context and the origins of a formula reveal a lot about the motivation and intuitions of the originator!

(Edit added several days later:) You suggest the idea of group as being now 'symmetry' of 'thing' then one modern way forward would be to go from group to category, (way of comparing things), to 2-category (ways of comparing ways of comparing things) etc, until you get some form of infinity category theory. If I remember rightly, the Abelian group objects in various forms of infinity category form categories equivalent to chain complexes (in positive degrees) or Ch-enriched categories if you use a different interpretation. That gives one chain complexes and then the various ideas on boundary of boundary relate to the various models of infinity category, e.g. quasi-complex should give simplicial Abelian groups, etc. The `extra structure' is then also there as dg-algebras are the vertex 'groups' in one of these contexts, and other laxer models lead to A-infinity categories etc. (This brings in ideas from several answers, but hopefully it provides an extra link in the motivational route.)

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I'm aware of how to motivate everything from simplicial complexes, and I've read Weibel's History of homological algebra. But I am not really looking for a historical explanation. As an analogy, the historical explanation for groups is that Galois considered the symmetries of the roots of a polynomial (or something like that), but nowadays we can just say "synmmetry is everywhere" and point to lots of examples Galois didn't have in mind. I'm lacking a similarly broad range of examples of "concrete chain complexes" from which a theory of "abstract chain complexes" may be abstracted. The... –  Qiaochu Yuan Jun 21 '11 at 12:56
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...problem is that all the "concrete chain complexes" I can think of have too much structure: they are all dg-algebras or dg-coalgebras, I think. It's a little like motivating algebras using group algebras; you can do it, but you'll get confused if you take group algebras as your "prototype" because they have extra structure that the typical algebra doesn't: they're Hopf algebras. –  Qiaochu Yuan Jun 21 '11 at 12:58
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I thought the historical perspective would be known to you, however don't think that for "concrete chain complexes" the historical perspective does not give you insight. One of the historical threads is via the Cech nerve construction and that leads directly into many of the resolution ideas that are used everywhere today. These nerve based examples do not have any really extra structure except that they are free. I would almost say "nerves are everywhere" in some shape or form as the hypercoverings used in Alg. Geom. are descended from them. You mentioned simplicial things and that is ... –  Tim Porter Jun 21 '11 at 15:28
    
area that gives loads of concrete examples.. without extra structure. –  Tim Porter Jun 21 '11 at 15:29
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You've raised a number of good questions, one of which is to better understand the graded-geometric aspects of having a boundary operation. I will try to address this part of your question, but I will begin with an apology: yours is largely a "why" question, whereas I will give only a "what" answer, motivated only after you have decided that chain complexes, etc., are useful. Also, my answer will be a bit long and rambly, and will probably raise more questions than it answers.

Notation: graded always means $\mathbb Z$-graded, with the Koszul signs (which I do not have a good way to motivate right now). Eventually I will stop even saying "graded": if I say "vector space" or "commutative algebra" or whatever, I probably mean "in the category of graded ...". Also, "vector space" almost certainly always means "abelian group" or "$R$-module" for your favorite commutative ring $R$, or ...

First, I don't know what your favorite conventions are: do you prefer that a chain complex have $\partial$ increasing or decreasing degree? Well, no matter. Let $\mathfrak Q$ denote the line ($\otimes$-invertible object) in degree $\pm 1$, so that a chain complex structure on a graded vector space $V$ includes a map of graded vector spaces $\mathfrak Q \to \operatorname{End}(V)$. (I use Q because it is the first letter of the word "homology".) Then $\mathfrak Q$ has an abelian group structure, given by $+$, and its Lie algebra is the vector space $\mathfrak Q$ with the trivial bracket. We've seen already that a chain complex is the same as a representation of the Lie algebra $\mathfrak Q$. Because this Lie algebra acts nilpotently on all representations, any such action integrates to an action of the group $\mathfrak Q$. So you can equivalently describe chain complexes as the representations of a group.

Let me describe this group a bit more. First, it is a group where? I'd like to be able to talk about "$+$ is a group structure on the set of points in $\mathfrak Q$", but that's nonsense. Instead, let $R$ be any (graded) commutative ring. Then there is a presheaf on affine schemes that assigns to $R$ the set of elements of $R$ which are homogeneous for the $\mathbb Z$-grading with grading $\mp 1$; i.e. this is the set of maps of graded vector spaces $\mathfrak Q^{\otimes -1} \to R$, where $\mathfrak Q^{\otimes -1}$ is the vector space which is $\otimes$-inverse to $\mathfrak Q$. This presheaf is in fact an affine scheme, because we have a "free" functor $\operatorname{Sym}$ from vector spaces to commutative rings: its represented by the commutative ring consisting of polynomials in one variable $\mathfrak q$, where $\mathfrak q$ has the grading of a basis element of $\mathfrak Q^{-1}$. Just to confuse notation, I will write $\mathfrak Q$ for both the invertible vector space above, and also for $\mathfrak Q = \operatorname{Spec}(\mathbb Z[\mathfrak q])$. Note that I do mean to impose that $\mathfrak q^2 = 0$, which follows from the grading of $\mathfrak q$ if $2$ is invertible in $\mathbb Z$. (So really I would be happiest picking a field of characteristic $0$, and replacing $\mathbb Z$ with that field.) Anyway, the affine scheme $\mathfrak Q$ is in fact an affine algebraic group. As a group scheme, it is the scheme that assigns to a commutative ring $R$ not just the set of grading-$(\mp 1)$ elements, but the abelian group of those (with $+$); as a Hopf algebra, the group structure is encoded by the comultiplication $\mathfrak q \mapsto 1\otimes \mathfrak q + \mathfrak q \otimes 1$.

Can we recognize this group $\mathfrak Q$ as arising from topology? No. Indeed, for any topological space $X$, its cohomology is supported entirely on the side of grading-$0$ that the vector space $\mathfrak Q$ is on, whereas the coordinate $\mathfrak q$ then has the opposite grading. In particular, the scheme $\mathfrak Q$ is inditinguishable from the trivial scheme when tested against the cohomology rings of spaces.

Nevertheless, it is tempting to try to describe $\mathfrak Q$ in topological language. The temptation comes, in part, from a famous theory of Quillen and Sullivan, which says that (not too large) toplogical spaces, if you can only see them "rationally", are the same as affine schemes over $\mathbb Q$. So henceforth I will work over the rational numbers, and wish that I hadn't picked the letter Q above. Oh, well, that's what fonts are for. (There are many conditions necessary to make the equivalence precise, and most of my examples will not satisfy those conditions, but c'est la vie.)

Let $G$ be a group. Then it has a classifying space $\mathrm B G$, with the property that $\pi_n(\mathrm B G) = \pi_{n-1} G$, and with a distinguished point $\ast \to \mathrm B G$. In general, $\mathrm B G$ has no group structure left over (the group structure on $G$ has been "used up" to create the classifying space). If instead of saying "group" I had said "homotopy-associative group", which is the correct notion of "group" in Spaces, then I would go on to describe "homotopy-$E_n$ groups", and observe that if $G$ has a homotopy-$E_n$ group structure, then $\mathrm B G$ has a homotopy-$E_{n-1}$ structure. (In particular, a (homotopy) $E_0$-structure on a space $X$ is the same as a distinguished map $\ast\to X$.) Rather than describing this theory, let me simply observe that if $G$ is a commutative group, then $\mathrm B G$ also has a commutative group structure.

Let's work rationally, and let $\mathbb G_a$ denote the additive group of the ground field $\mathfrak k$. Then there's a sequence of abelian groups $\mathrm B^n \mathbb G_a$, with $\pi_m \mathrm B^n \mathbb G_a =\mathbb G_a$ when $m=n$, and $=0$ otherwise. You already know some of these groups. In particular, the circle $S^1$ is (integrally!) a $\mathrm K(\mathbb Z,1)$, and so rationally $S^1 = \mathrm K(\mathbb G_a,1)$, since $\mathbb G_a = \mathbb G_a \otimes \mathbb Z$; so $S^1 = \mathrm B \mathbb G_a$. In general, if memory serves, $\mathrm B^n\mathbb G_a$ is the $n$-sphere, since I'm working rationally. (Yes, the $n$-sphere is a group.)

Anyway, our group $\mathfrak Q$ has homotopy only in grading $-1$: it is a "rational $-1$-sphere". This is what I meant when I said that it cannot arrise from topology: it has homotopy groups in negative spots. (So if I were clever, I would have insisted on homological grading, whence differentials go down, and $\mathfrak Q$ is the line supported in grading $-1$, and the coordinate $\mathfrak q$ has grading $+1$; then all conventions would be consistent.) We want to write $\mathfrak Q = \mathrm B^{-1}\mathbb G_a$. Well, what functor is inverse to $\mathrm B$? It's the functor of taking based loops: $\mathfrak Q = \Omega \mathbb G_a$.

In fact, by working with graded (or dg) schemes over $\mathbb k\supseteq Q$, you can realize this equation. Let $\ast \to X$ be a pointed space. The based loop space of $X$ is the homotopy pullback:

$$ \begin{matrix} \Omega X & \to & \to & \ast \\ \downarrow & \ulcorner & & \downarrow \\ \downarrow & & & \downarrow \\ \ast & \to & \to & X \end{matrix}$$

I.e. "the intersection of the point with itself".

Working instead with affine schemes, we replace the space $X$ with its ring of functions $\mathcal O(X)$, and the point $\ast$ with $\mathbb k = \mathcal O(\ast)$, and then the map $\ast \to X$ is an augmentation of $\mathcal O(X)$. Recall that the pushout of commutative rings is the tensor product; the homotopy pushout is the derived tensor product, and so we have:

$$ \Omega X = \operatorname{Spec}( \mathbb k \otimes_{\mathcal O(X)} \mathbb k) $$

where $\otimes$ means the derived tensor product.

In our case, $X = \mathbb G_a$ and $\mathcal O(X) = \mathbb k[x]$, where $x$ is a grading-$0$ coordinate function. Then you can check that, sure enough, the derived tensor product is represented by the ring $\mathbb k[\mathfrak q]$. So in some algebraic sense we do have $\mathfrak Q = \mathrm B^{-1}\mathbb G_a$.

Conversely, what I'm saying that $\mathbb G_a = \mathrm B \mathfrak Q$. You had originally asked for a deeper understanding of chain complexes, and we agreed that (at least rationally) chain complexes were $\mathfrak Q$-representations. But these are the same as sheaves of vector bundles over $\mathrm B \mathfrak Q$, which we now recognize (if we are willing to restrict ourselves only to work over $\mathbb k \subseteq \mathbb Q$) as (the underlying space of) the group $\mathbb G_a$, which is to say the affine line.

Then again, I'm not sure what to make of this, in part because everything I've said in the last few paragraphs is technically false: Quillen and Sullivan's rational homotopy theory only really starts to work when you restrict yourself to rings and spaces that only have homotopy starting in degree $2$.

Here are a few more things to think about. From the picture of "chain complexes are representations of some particular group $\mathfrak Q$", it makes sense why you might care about "chain complexes up to homotopy": that's like deciding to care only about the quotient of any given $\mathfrak Q$-representation by the $\mathfrak Q$-action (but because we're in the 21st century, there are good ways to take this quotient).

It does continue to amaze me that this group is enough to do all sorts of other parts of derived geometry. For example, let $\mathfrak g$ be a Lie algebra over $\mathbb k \subseteq \mathbb Q$, and use the same notation for the formal group with Lie algebra $\mathfrak g$ (one way to define this formal group is to take the universal enveloping algebra of $\mathfrak g$, which is ind-finite if $\mathfrak g$ is finite-dimensional, and take its "filtered dual" and thereby build a profinite commutative ring; Spec of this ring is one thing I could mean by "the formal group"). Then there is a space $\mathrm B \mathfrak g$, whose sheaves are all $\mathfrak g$-modules. This space can be equivalently described by shifting $\mathfrak g$ up one degree in homological grading, and then giving it an interesting action by $\mathfrak Q$; $\mathrm B \mathfrak g$ is the "quotient" of this action. From this perspective, the functor that turns a $\mathfrak g$-representation $V$ into a sheaf over $\mathrm B\mathfrak g$, and then takes the total space of this sheaf; that functor just is taking the quotient of the representation by the $\mathfrak g$-action. But the total space is just $V \times \mathrm B \mathfrak g$, and the $\mathfrak g$-action on $V$ is encoded by a $\mathfrak Q$-action on $V \times \mathrm B \mathfrak g$ covering the $\mathfrak Q$-action on $\mathrm B \mathfrak g$. So all stacks of the form "space mod Lie algebra action" can be encoded as "(graded) space mod $\mathfrak Q$-action". (Note: you cannot encode discrete groups this way. The $\mathfrak Q$-action only sees the infinitesimal part of the $\mathfrak g$-action, but since I'm using the formal group, and not, say, the connected simply-connected Lie group, there really is only infinitesimal action.)

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Just to make sure I'm reading your post correctly, isn't there a difference between even spheres and odd in like the 8th or so paragraph. e.g. it seems to me that $K(Q,2n)$ isn't $S^{2n}$ because they have different homotopy groups... it's more like $ΩS^{2n+1}$. It's also important to assume the characteristic of your ground field is zero but I guess that's implied by working rationally –  Daniel Pomerleano Jun 23 '11 at 18:24
    
In particular $S^{2n}$ isn't a rational H-space but $S^{2n+1}$ is. –  Daniel Pomerleano Jun 23 '11 at 18:34
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Okay, you've convinced me that there should be a direct connection to supergeometry, but you haven't convinced me that you've found it. This supergroup sounds like it ought to have some intuitive definition in terms of homotopy but I'm still not seeing it from this discussion. –  Qiaochu Yuan Jun 23 '11 at 20:31
    
@Dan P: I could very well be misremembering something about spheres. Yes, I do mean to be working everywhere over a field of characteristic $0$. @Qiaochu: That's right, I could not identify a good motivation for the group that I call $\mathfrak Q$ from homotopy theory. It is not a space, because it has nontrivial $\pi_{-1}$. –  Theo Johnson-Freyd Jun 23 '11 at 23:49
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The boundary of a boundary is empty. This is true in any mathematical context. The algebraic translation of this geometric statement is a chain complex $d^2=0$. This observation has led to great achievements in all areas of mathematics. It's impossible to summarize them here. Just ask for applications in specific areas if you wish.

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I don't consider "this observation has led to great achievements" an explanation. As a very rough analogy, if I gave a primitive civilization guns, they might take over all of the other primitive civilizations in their area and declare "this gun has led to great achievements" but it wouldn't put them any closer to understanding how guns work. –  Qiaochu Yuan Jun 21 '11 at 13:02
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It is true except when dealing with N-complexes :) arXiv:q-alg/9611005 –  Mariano Suárez-Alvarez Jun 21 '11 at 13:02
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@Qiaochu: great analogy! –  Mariano Suárez-Alvarez Jun 21 '11 at 13:08
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@Qiaochu whether you consider it an explanation or not, it's true ;-) –  Fernando Muro Jun 21 '11 at 13:20
    
@Fernando Your first sentence raises several questions. First what mathematical contexts lead to a notion of 'boundary' and how should it be define in a fairly abstract way? And secondly, why does a boundary of a boundary give us nothing? –  Tim Porter Jun 23 '11 at 5:46
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The origin of $d^2 = 0$ is surely the geometry: think of a solid ball; it's boundary is a sphere, but the sphere has no boundary. More generally, "every boundary is a cycle". The first formulation of the simplicial complex theory used orientations in order to get "a boundary is a cycle" but this had problems for the singular theory. Then Eilenberg introduced the ordered theory, which led to simplicial sets, and more generally simplicial objects.

It might be helpful to see why the Dold-Kan correspondence is true by looking at generalisations and variations. See section 7 of my survey paper "Groupoids and crossed objects in algebraic topology", Homology, homotopy and applications, 1 (1999) 1-78.

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But I can't help but feel like the simplicial picture would be enriched if I understood why one gets a super Lie algebra out of it. The picture I expect is roughly that there is a super Lie group acting by "odd thickening" of simplices or something like that, and taking the "derivative" of this action gives the differential (since the derivative of thickening morally ought to be the boundary). But I would love for someone to make this precise. –  Qiaochu Yuan Jun 22 '11 at 12:39
    
@ Qiaochu ... but simplices are 'really' combinatorial almost 'algebraic' things. There happens to be a model of them in spaces so it super Lie picture exists it would have to be combinatorial or algebraic as well as interpreting geometrically in some thing like your idea (?). My question above was why does the boundary of a boundary equal the empty set. This is something to do with the idea of 'boundary'. The filtered case (open-closed ETQFTs etc.) suggests that there are situations where the boundary of a boundary of filtered things may be non trivial. (Would have to thing more on that.) –  Tim Porter Jun 24 '11 at 6:08
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I'm afraid I don't have a cohesive narrative, but I do have a potentially informative fragment.

If you apply a suitable version of Tannaka-Krein reconstruction to the faithful tensor functor $\text{Ch}(R) \to \text{super-}R\text{-mod}$ that forgets integer grading and differentials, you find that the category of chain complexes of modules over a (super)commutative ring $R$ is equivalent to the category of representations of the super-group of automorphisms of the odd fat point (also known as the category of comodules of the Hopf superalgebra of functions on the automorphism super-group). As an explicit group, it can be written as $T \rtimes \mathbb{G}_m$, where $T$ is the odd group of translations. The subgroup $\mathbb{G}_m$ gives a grading on the chain complex, and adding the $T$ provides a square zero differential.

The odd fat point over $R$ is the spectrum of $R[\eta]/(\eta^2)$, where $\eta$ is odd, and maps from this object to a superscheme describe a parity-shifted tangent bundle, so it has something to do with first-order approximation. The odd fat point is equivalent to the odd line (the spectrum of $R[\eta]$) if and only if $2 \neq 0$ in $R$, so if you aren't interested in chain complexes with characteristic 2 coefficients, you may approximate the correct answer with a conceptually simpler answer. The group $T$ is isomorphic as a superscheme to the odd fat point, but its coordinate ring is endowed with a coproduct given by $\eta \mapsto 1 \otimes \eta + \eta \otimes 1$. Its tangent space is the one-dimensional odd Lie algebra. The group structure on $T$ induces a translation action on the odd fat point, making it a $T$-torsor. The group $\mathbb{G}_m$ is the spectrum of $R[x,y]/(xy-1) = R[x,x^{-1}]$ with coproduct $x \mapsto x \otimes x$, and the comodule structure on $R[\eta]$ is by the degree 1 dilation action, i.e., by $\eta \mapsto x \otimes \eta$. The full automorphism group is then the spectrum of $R[x,y,\eta]/(xy-1, \eta^2)$ with coproduct determined by $x \mapsto x \otimes x$ and $\eta \mapsto x \otimes \eta + \eta \otimes 1$.

Here's the explicit co-action on objects: for each chain complex $V^\bullet$, a pure vector $v$ of degree $n$ is taken to $x^n \otimes v + x^{n-1}\eta \otimes dv$. If you aren't used to co-actions, you can think of the $x^n$ as a way to encode the degree $n$ nature of $v$, and the $\eta$ part describes an infinitesimal shearing transformation by $d$. More precisely, given an arbitrary finite length comodule $V$ with co-action $v \mapsto \sum p_{(1)}(x,\eta) \otimes v_{(2)}$, one can isolate graded pieces as the parts paired with powers of $x$, and reconstruct the map $d$ from the parts with $\eta$. The square zero property of $\eta$ is necessary for all representations to have $d^2 = 0$, and the odd parity on $\eta$ is necessary both for tensor products of complexes to have the correct signs and for the fat point to admit a well-defined group structure when $2 \neq 0$.

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Thanks, Scott. IIRC the supergeometric construction of the complex of differential forms on a smooth manifold $M$ is to take the space of smooth functions on what I think is the internal hom $\mathbb{R}^{0|1} \Rightarrow M$. The supergroup of automorphisms of the odd real line naturally acts on this space and that gives the usual structure on the space of differential forms as in this answer. I guess one can just say "and then we relate this to the simplicial picture by Stokes' theorem" but it would be nice to have a more concrete intuition for what the supergroup is doing... –  Qiaochu Yuan Jun 24 '11 at 11:38
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