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• CommentRowNumber1.
• CommentAuthorMirco Richter
• CommentTimeFeb 28th 2012

Is the set of natural transformations between two simplicial (abelian) groups a simplicial group or a chain complex?

• CommentRowNumber2.
• CommentAuthorUrs
• CommentTimeFeb 29th 2012
• (edited Feb 29th 2012)

It comes to you manifestly as a simplicial abelian group. But since, by the Dold-Kan correspondence, simplicial abelian groups are equivalent to chain complexes, you can equivalently regard it as a chain complex.

• CommentRowNumber3.
• CommentAuthorMike Shulman
• CommentTimeFeb 29th 2012

But for simplicial nonabelian groups, the simplical set of maps is not in general even a simplicial group, for the same reason that the set of group homomorphisms between nonabelian groups is not a group — nonabelian groups not being a commutative theory. Apologies if this is irrelevant; I wasn’t sure how to interpret your parenthetical “(abelian)”.

• CommentRowNumber4.
• CommentAuthorUrs
• CommentTimeFeb 29th 2012
• (edited Feb 29th 2012)

Hi Mike, your link was broken. I have now made commutative theory redirect to commutative algebraic theory.

• CommentRowNumber5.
• CommentAuthorMirco Richter
• CommentTimeFeb 29th 2012

@Mike: You interpreted the parenthetical (abelian) just right. I thought that it should work in the abelian but not in the non abelian situation, but wasn’t sure.

@Urs: Can you tell explicit how to get the faces and degeneracies in this complex from those of the domain and codomain?

• CommentRowNumber6.
• CommentAuthorUrs
• CommentTimeFeb 29th 2012
• (edited Feb 29th 2012)

Mirco, it’s the usual internal hom of simplicial sets (see closed monoidal structure on presheaves):

for $X,Y\in$ sSet (possibly $X,Y\in \mathrm{sAb}\to \mathrm{sSet}$), their internal hom $\left[X,Y\right]$ or ${Y}^{X}$ (“function complex”) is the simplicial set

$\left[X,Y\right]:\left[n\right]↦{\mathrm{Hom}}_{\mathrm{sSet}}\left(\Delta \left[n\right]×X,Y\right)\phantom{\rule{thinmathspace}{0ex}}.$

The face and degeneray maps on this are the evident ones induced from the cosimplicial simplicial set of simplicial simplices: $\Delta \left[-\right]:\Delta \to \mathrm{sSet}$.

If $Y$ has abelian group structure, then this inherits a simplicial abelian group structure by pointwise group operations (just as functions with values in an abelian group form an abelian group).

• CommentRowNumber7.
• CommentAuthorMirco Richter
• CommentTimeFeb 29th 2012
• (edited Feb 29th 2012)

And this is the same as ${\mathrm{hom}}_{\mathrm{sAb}}\left(X,Y\right)$? For simplicial abelian groups $X$ and $Y$? (I mean the same as the set of simplicial morphisms from $X$ to $Y$)

• CommentRowNumber8.
• CommentAuthorUrs
• CommentTimeFeb 29th 2012
• (edited Feb 29th 2012)

What I wrote down is the very definition of the simplicial set of homomorphisms of simplicial sets (if that’s what you mean by “simplicial morphisms”). Which is canonically a simplicial abelian group if $Y$ is. I thought that’s what you are asking about.

• CommentRowNumber9.
• CommentAuthorUrs
• CommentTimeFeb 29th 2012
• (edited Feb 29th 2012)

Oh, sorry, you mean if that’s also group homomorphisms. So above I wrote ${\mathrm{Hom}}_{\mathrm{sSet}}$, but you can just as well replace that with ${\mathrm{Hom}}_{\mathrm{sAb}}$

• CommentRowNumber10.
• CommentAuthorMirco Richter
• CommentTimeFeb 29th 2012
• (edited Feb 29th 2012)

I mean the set of natural transformations (a.k.a. simplicial morphisms) from one abelian simplicial group into another. Does it have a simplicial structure? If not does it at least have the structure of a chain complex?

(’Simplicial morphisms’ because they are morphisms between simplicial sets (simplicial ab. groups here). Like linear morphisms are the morphisms between linear sets ect. :-) )

• CommentRowNumber11.
• CommentAuthorUrs
• CommentTimeFeb 29th 2012

Okay, looking back I see that I didn’t give a good reply, since I kept talking about just homs of the underlying simplicial sets.

$\left(ℤ\left[-\right]⊣\mathrm{Forget}\right):\mathrm{Set}\stackrel{ℤ\left[-\right]}{\to }\mathrm{Ab}$

immediately lifts this to what, I think, you are after:

Proposition The category $\mathrm{sAb}$ of simplicial abelian groups is canonically a closed monoidal category where for $A,B\in \mathrm{sAb}$ their intenal hom $\left[A,B\right]$ is the simplicial abelian group given by

$\left[A,B\right]:\left[n\right]↦{\mathrm{Hom}}_{\mathrm{sAb}}\left(A\otimes ℤ\left(\Delta \left[n\right]\right),B\right)\phantom{\rule{thinmathspace}{0ex}}.$

I’ll add this to the nLab entry simplicial abelian group now.

• CommentRowNumber12.
• CommentAuthorUrs
• CommentTimeFeb 29th 2012

Okay, I made a short note here.

This deserves to be expanded and better organized. But I really need to be looking into something else now.

If you need a textbook account, try for instance chapter 4 here.

• CommentRowNumber13.
• CommentAuthorMirco Richter
• CommentTimeFeb 29th 2012
• (edited Feb 29th 2012)

Thank you. Maybe we can talk later on.

As far as I understand those things, the internal hom (That is what Urs points to here) is in general not the same thing as the ’external’ hom. And in my situation the set of simplicial morphism from one abelian group into another (Just the plain set of natural transformations) is just the external hom. Hence the answer is only applicable to my situation when there is a theorem stating that we have an isomorphism between the internal and the external hom. Only then I can transfer the simplicial structure from the internal hom to what I’m thinking about.

Or am I on the wrong road?

• CommentRowNumber14.
• CommentAuthorUrs
• CommentTimeFeb 29th 2012

What is the external hom for you here? Taken at face value, the external hom would be the set of morphisms ${\mathrm{Hom}}_{\mathrm{sAb}}\left(A,B\right)$. This can’t be what you mean, since you are looking for simplicial structure on the hom. The internal hom is given by that external hom in degree 0 and adds the simplicial structure that you are looking for.

• CommentRowNumber15.
• CommentAuthorMirco Richter
• CommentTimeFeb 29th 2012
• (edited Feb 29th 2012)

That ${\mathrm{Hom}}_{\mathrm{sAb}}\left(A,B\right)$ is what I mean. (I wasn’t aware that the capital H indicates this. I thought the convention was the other way around. Sry)

Actually I was asking if there is a simplicial structure on ${\mathrm{Hom}}_{\mathrm{sAb}}\left(A,B\right)$ and if not, if there is at least the structure of a chain complex on it. (For the internal hom the situation is clear.)

So guessing from you replay there is not a simplicial structure on it.

But at least ${\mathrm{Hom}}_{\mathrm{sAB}}\left(A,B\right)$ is a graded vector space (isn’t it?), so the question remains whether or not there is a differential on it…

• CommentRowNumber16.
• CommentAuthorUrs
• CommentTimeFeb 29th 2012
• (edited Feb 29th 2012)

If you want a simplicial structure on the collection of morphisms, and want it to be a simplicial abelian group or equivalently a chain complex, then what you want is precisely the internal hom $\left[-,-\right]$ that I gave you in #11.

• CommentRowNumber17.
• CommentAuthorMirco Richter
• CommentTimeFeb 29th 2012

Ok, but I have the external Hom in my particular situation and I want to figure out whether or not there is a natural differential on it.

• CommentRowNumber18.
• CommentAuthorUrs
• CommentTimeFeb 29th 2012
• (edited Feb 29th 2012)

No. Somehow this discussion is going in circles. If you want the differential, you need that internal hom. Why not just take that?

What is the situation that you are looking at?

• CommentRowNumber19.
• CommentAuthorMirco Richter
• CommentTimeFeb 29th 2012
• (edited Feb 29th 2012)

Ok so there is no natural differential on the external Hom. Then indeed I have to change to the internal hom. The point why I hesitate is just because I’m not an Homotopy expert and the definition of the internal hom looks plain scary to me at first look.

Working just with the natural transformations between simplicial abelian groups is very easy, because you only have “linear” maps between the dimension groups, satisfying the commutation with faces and degeneracies. But when changing to the internal hom, I can’t see what the elements explicit are.

Maybe there is an Exercise or something where I can work this out in a test case?

• CommentRowNumber20.
• CommentAuthorUrs
• CommentTimeFeb 29th 2012
• (edited Feb 29th 2012)

But when changing to the internal hom, I can’t see what the elements explicit are.

I gave you the explicit formula in #11. That expresses the internal hom in terms of the external hom.

If you can’t parse that formula, let me know which bits are the problem.

• CommentRowNumber21.
• CommentAuthorTodd_Trimble
• CommentTimeFeb 29th 2012
• (edited Feb 29th 2012)

It should be pointed out that the construction of internal abelian group homs is not particularly special to simplicial sets. If you have any cartesian closed category with equalizers $C$, you can form the internal hom of abelian group objects, which will be another abelian group object.

You can figure out what it looks like just by generalizing from the $\mathrm{Set}$-case. If $A$, $B$ are abelian groups, then the internal hom $\mathrm{Hom}\left(A,B\right)$ will be a subset of the full set of functions $f:UA\to UB$ ($UA$ being the underlying set), satisfying $f\left(a+a\prime \right)=f\left(a\right)+f\left(a\prime \right)$. So we just need to write down an equalizer that expresses this equation.

Let $T=UF$ be the monad for $F⊣U$. The equalizer we want internalizes the commutativity of the diagram

$\begin{array}{ccc}TUA& \stackrel{Tf}{\to }& TUB\\ {\xi }_{A}↓& & ↓{\xi }_{B}\\ A& \underset{f}{\to }& B\end{array}$

where the $\xi$’s are the $T$-algebra structure maps. And so we write down

$\begin{array}{ccc}U{B}^{UA}& \to & TU{B}^{TUA}\\ & U{B}^{{\xi }_{A}}↘& ↓{\xi }_{B}^{TUA}\\ & & U{B}^{TUA}\end{array}$

where the horizontal arrow comes about via the strength on $T$ (or the fact that $T$ is $C$-enriched). The equalizer of this diagram is the internal hom.

The same construction works perfectly well for any commutative algebraic theory $T$, or any monoidal monad $T$ on $C$.

• CommentRowNumber22.
• CommentAuthorMirco Richter
• CommentTimeFeb 29th 2012

ok. Thanks. I’ll think about it and let you know if there are further problems in the construction of my int hom….

• CommentRowNumber23.
• CommentAuthorMike Shulman
• CommentTimeFeb 29th 2012

Mirco, it sounds to me that we may have a confusion in language. I think the phrase “a set X has a simplicial structure” has questionable meaning. One can talk about a simplicial set whose 0-simplices are X, and that’s what we have here, since the natural “underlying set” functor from simplicial sets (which produces the “underlying ordinary category” from any simplicially enriched category, such as sAb) just takes the set of 0-simplices. But a simplicial set is a sequence of sets with operations between them, not a single set with operations on it, so I think it’s better to ask whether a set “can be extended to a simplicial set” rather than whether it “has a simplicial structure”.

• CommentRowNumber24.
• CommentAuthorMirco Richter
• CommentTimeFeb 29th 2012

Ok. When I say A simplicial set, I mean that sequence of sets, so in my terms a simplicial set is a set of sets together with the structure maps, its elements are the dimension parts. But ok, to be absolutely precise a simplicial set is not a set but the image of a functor hence a category. So it was sloppy to call it that way.

But what I explicitly NOT mean by a set with simplicial structure is just the degree zero part! So please note that whenever I said ’set with a simplicial structure’ I mean ’simplicial set’ i.e. a sequence of sets with structure maps.

• CommentRowNumber25.
• CommentAuthorMirco Richter
• CommentTimeFeb 29th 2012
• (edited Feb 29th 2012)

So when I ask whether or not the external Hom has a simplicial structure I mean the following:

${\mathrm{Hom}}_{\mathrm{sAB}}\left(A,B\right)$ is the set of abelian simplicial morphism. That means it is a sequence of sets\

${\mathrm{Hom}}_{\mathrm{sAB}}\left(A,B{\right)}_{k}\subset {\mathrm{Hom}}_{\mathrm{AB}}\left({A}_{k},{B}_{k}\right)$.

(Because every simplicial morphism $f:A\to B$ is determined by its dimension $k$ parts ${f}_{k}:{A}_{k}\to {B}_{k}$.)

And now the question is if we can put a simplicial structure on that sequence. This just means whether or not there are appropriate face and degeneracy maps between the ${\mathrm{Hom}}_{\mathrm{sAB}}\left(A,B{\right)}_{k}$ ?? I.e are there maps

${d}_{j}:{\mathrm{Hom}}_{\mathrm{sAB}}\left(A,B{\right)}_{k}\to {\mathrm{Hom}}_{\mathrm{sAB}}\left(A,B{\right)}_{k-1}$

${s}_{j}:{\mathrm{Hom}}_{\mathrm{sAB}}\left(A,B{\right)}_{k}\to {\mathrm{Hom}}_{\mathrm{sAB}}\left(A,B{\right)}_{k+1}$

satisfying the coherency equations. Hopefully now it is clear!

And if this is in general not true,then is there a differential $D:{\mathrm{Hom}}_{\mathrm{sAB}}\left(A,B\right)\to {\mathrm{Hom}}_{\mathrm{sAB}}\left(A,B\right)$? That means is there a sequence of maps ${D}_{k}:{\mathrm{Hom}}_{\mathrm{sAB}}\left(A,B{\right)}_{k+1}\to {\mathrm{Hom}}_{\mathrm{sAB}}\left(A,B{\right)}_{k}$ linear with ${D}_{k-1}{D}_{k}=0$ ?

• CommentRowNumber26.
• CommentAuthorTodd_Trimble
• CommentTimeFeb 29th 2012

Ah, I didn’t see what the confusion was before. A transformation of simplicial abelian groups can be interpreted as a tuple $⟨{f}_{k}{⟩}_{k}$ of abelian group homomorphisms ${A}_{k}\to {B}_{k}$, it is true, but the only way I see how to get a set of morphisms ${A}_{k}\to {B}_{k}$ from the set ${\mathrm{Hom}}_{\mathrm{sAB}}\left(A,B\right)$ is to take those ${A}_{k}\to {B}_{k}$ which are the ${k}^{\mathrm{th}}$ component of some simplicial homomorphism $f:A\to B$, in other words by taking the image of the ${k}^{\mathrm{th}}$ projection map. And this gives you nothing interesting. For example, there aren’t meaningful well-defined face maps

${\pi }_{k}\left({\mathrm{Hom}}_{\mathrm{sAB}}\left(A,B\right)\right)\to {\pi }_{k-1}\left({\mathrm{Hom}}_{\mathrm{sAB}}\left(A,B\right)\right).$

(Or at least I invite you to try to think of some, and see that they won’t work.)

I think you should probably work with the internal hom, as Urs and Mike suggested. If it helps, the exponential ${Y}^{X}$ between two simplicial sets (that has ordinary simplicial maps $X\to Y$ as its points or 0-cells) should be thought of as a function space, and paths = 1-cells in that function space, $I\to {Y}^{X}$, are equivalent to homotopies $I×X\to Y$, are equivalent to functions to the path space $X\to {Y}^{I}$. This is the meaning of the formula Urs gave you, that $n$-cells in ${Y}^{X}$ are tantamount to maps ${\Delta }_{n}×X\to Y$ of simplicial sets.

• CommentRowNumber27.
• CommentAuthorUrs
• CommentTimeFeb 29th 2012
• (edited Feb 29th 2012)

That means it is a sequence of sets\

No. It means that it is a subset of the product of all the sets ${\mathrm{Hom}}_{\mathrm{Ab}}\left({A}_{k},{B}_{k}\right)$.

Notice: product, not sequence = coproduct.

[edit: I see I overlapped with Todd’s latest comment.]

• CommentRowNumber28.
• CommentAuthorMirco Richter
• CommentTimeFeb 29th 2012
• (edited Feb 29th 2012)

Ah yes… That was a real mistake in the way I thought about simplicial morphism. So I move on to the construction of the internal hom.

….

Later …

• CommentRowNumber29.
• CommentAuthorMirco Richter
• CommentTimeMar 5th 2012
• (edited Mar 5th 2012)

So on the set level $\left(ℤ\left(\Delta \left[n\right]\right){\right)}_{m}\simeq {ℤ}^{\left(\genfrac{}{}{0}{}{n+m+1}{n}\right)}$. Now are there fancy explicit terms for the faces ans degeneracies known?

Like ${d}_{j}:{ℤ}^{p}\to {ℤ}^{q};a↦{d}_{j}\left({a}_{1},\dots ,{a}_{p}\right)=\mathrm{something}$

for $p=\left(\genfrac{}{}{0}{}{n+m+1}{n}\right)$ and $q=\left(\genfrac{}{}{0}{}{n+m}{n}\right)$. Doing this as an exercise for a particular (and arbitrary by myself) chosen order on the elements of $\Delta \left[n{\right]}_{m}$ and hence on the coordinates of ${ℤ}^{p}$ gave me rather involved results when $n$ gets bigger. So I guess it is hard if possible to give closed forms for the faces and degeneracies for arbitrary $n$.

But nevertheless, maybe there is, so I better ask ! (At least I don’t need them necessarily but thinking about this helps me getting familiar with these things)

• CommentRowNumber30.
• CommentAuthorTodd_Trimble
• CommentTimeMar 6th 2012

Mirco, I sense that your questions can be answered, but before going any further, can I ask you where you are headed with all of this? What is it that you wish to accomplish? :-)

Please feel free to start a new thread if you don’t think this one would be appropriate.

• CommentRowNumber31.
• CommentAuthorMirco Richter
• CommentTimeMar 6th 2012
• (edited Mar 6th 2012)

@Todd: This is just my afford to understand the inner hom for abelian simplicial groups in detail. Since it is based on the products with the $ℤ\left(\Delta \left[n\right]\right)$ my thought is, that understanding the latter explicit is a good way to start. Whether or not this is reasonable, I don’t know. (If the explicit faces and degeneracies become too involved then likely it is not.)

From another point of view I think it would be desirable to have those basic constructions explicit (if possible). In my believe this will help to develop a better access to these kind of mathematics from other branches of science where it maybe helps making progress. Personally I found that it is really hard to get into the higher category and simplicial stuff, mostly because it lacks explicit constructions and introductory examples by now.

• CommentRowNumber32.
• CommentAuthorTodd_Trimble
• CommentTimeMar 6th 2012
• (edited Mar 6th 2012)

In my view, the constructions discussed in this and related threads are explicit. As I understand it, you want to write down formulas for the set of $n$-cells of the internal hom, and face and degeneracy maps. I’m claiming it’s already been done, but you have to know how to interpret the results.

I might get to saying more on this later, but not now – I have other things I need to do.

But I hope that at some point you become convinced that category theory is in large degree a way of making mathematics simpler, easier to digest, etc., instead of making life harder, more inaccessible, more obfuscatory, etc. (I hope you won’t mind my saying that knowing the cardinality of hom sets, while nice, is not particularly useful for the present problem; what is more useful is being able to name the elements in an effective way and say what certain operations do to them in terms of how they are named.)

You should know that category theorists, more often than not, delight in being explicit, and to a large degree react against a lot of mathematics as it was once practiced, with large doses of axiom of choice (where unnecessary) and mere existence statements. Another way of saying it is that category theorists delight in making their mathematics constructive, so that it is easier to transport it to wide contexts.

• CommentRowNumber33.
• CommentAuthorUrs
• CommentTimeMar 6th 2012
• (edited Mar 6th 2012)

Re #29:

So I guess it is hard if possible to give closed forms for the faces and degeneracies for arbitrary n.

As Todd says, it’s indeed very simple.

Let’s try to figure out where you are, so that we can pick you up there. You just need these two steps:

1. give an explicit description of the morphisms of simplicial sets $\Delta \left[k\right]\to \Delta \left[l\right]$;

2. understand for a given function of sets $f:S\to T$ what the corresponding morphism of free abelian groups $ℤ\left(S\right)\to ℤ\left(T\right)$ is.

You should let us know which of these two steps is clear to you, and which not. Is the first step clear, for instance?

if not, try, as a warmup, to describe all the possible morphisms $\Delta \left[1\right]\to \Delta \left[2\right]$. First do it pictorially, if that helps, then try to find a formal language for it that suits you.

• CommentRowNumber34.
• CommentAuthorMirco Richter
• CommentTimeMar 7th 2012

As I understand it, you want to write down formulas for the set of n-cells of the internal hom, and face and degeneracy maps

No! Obviously we can’t write the faces and degeneracies of $\mathrm{hom}\left(A,B\right)$ explicit because they depend on $A$ and $B$. But when $\mathrm{hom}\left(A,B{\right)}_{n}={\mathrm{Hom}}_{\mathrm{sAb}}\left(A\otimes ℤ\left(\Delta \left[n\right]\right),B\right)$, then there is one more thing we can do to make this as explicit as possible:

We can try to unravel $ℤ\left(\Delta \left[n\right]\right)$ because (1) it is necessary in the definition of $\mathrm{hom}\left(A,B\right)$ and (2) it remains the same once and for all. Hence (in future applications, when $A$ and $B$ are not abstract but concrete given from a context) everyone who wants to compute ’his’ $\mathrm{hom}\left(A,B\right)$ first has to compute $ℤ\left(\Delta \left[n\right]\right)$. From this point of view it is just a matter of efficiency to do this just once and add this as additional information to the definition of the inner hom in case of simpl. abelian groups. Then this additional information reflects the fact, that we look not just on the general simplicial inner hom, but on a simplicial inner hom with additional structure. (Abelian group structure in this case.)

• CommentRowNumber35.
• CommentAuthorMirco Richter
• CommentTimeMar 7th 2012
• (edited Mar 7th 2012)

@Urs:

1.) For the morphisms $f:\Delta \left[n\right]\to \Delta \left[m\right]$ I just know the usual definition that is determined from the general construction of the Yoneda embedding i.e. $f\left(g\right)=f\circ g$. In addition as far as I know the functor $\left[n\right]\to \Delta \left[n\right]$ is a cosimplicial set, so the morphisms should be compositions of codegeneracies and cofaces. What they look like I don’t know yet.

2.) if $#\left(S\right)=\sigma$ and $#\left(T\right)=\theta$ are of finite cardinality and $f:S\to T$ is a set map, then $ℤ\left(S\right)\simeq {\oplus }_{s=1}^{\sigma }ℤ$ and $ℤf$ is a function ’on the coordinates’ that is $ℤf\left({x}_{{s}_{1}},\dots ,{x}_{{s}_{\sigma }}\right)=\left({\sum }_{f\left({s}_{i}\right)={t}_{1}}{x}_{f\left({s}_{i}\right)},\dots ,{\sum }_{f\left({s}_{i}\right)={t}_{\theta }}{x}_{f\left({s}_{\sigma }\right)}\right)$.

Now using the latter and $p:=#\left(\Delta \left[n{\right]}_{k}\right)=\left(\genfrac{}{}{0}{}{k+n+1}{n}\right)$ as well as $q:=#\left(\Delta \left[n{\right]}_{k-1}\right)=\left(\genfrac{}{}{0}{}{k+n}{n}\right)$ with $S=\left\{f:\left[k\right]\to \left[n\right]\right\}$ we have an explicit definition of the faces ${d}_{j}:ℤ\left(\Delta \left[n{\right]}_{k}\right)\to ℤ\left(\Delta \left[n{\right]}_{k-1}\right)$ given by

($A$) ${d}_{j}:{ℤ}^{p}\to {ℤ}^{q};\left({x}_{{f}_{1}},\dots ,{x}_{{f}_{p}}\right)↦\left({\sum }_{{d}_{\left(}{j}_{i}\right)=f{\prime }_{1}}{x}_{{d}_{j}\left({f}_{i}\right)},\dots ,{\sum }_{{d}_{\left(}{j}_{i}\right)=f{\prime }_{q}}{x}_{{d}_{j}\left({f}_{i}\right)}\right)$ and analog for the degeneracies inside a particular $\Delta \left[n\right]$. This is true for any order we have defined on the elements of $\Delta \left[n{\right]}_{k}$ i.e. whatever $f\in \Delta \left[n{\right]}_{k}$ we call ${f}_{1}$ and so on.

That is exactly where I’m :-) … So the question in #29 was IF we can go further:

Is there an order on the $f$’s such that we could write the faces and degeneracies even without explicit reference to the degeneracies and faces and the elements $f\in \Delta \left[n{\right]}_{k}$ ?

IF this is possible then for example in $ℤ\left(\Delta \left[n\right]{\right)}_{k}$ the face d_j would look like: ${d}_{j}:ℤ\left(\Delta \left[n{\right]}_{k}\right)\to ℤ\left(\Delta \left[n{\right]}_{k-1}\right);\left({x}_{1},\dots ,{x}_{p}\right)=\left({\sum }_{?=1}{x}_{i},\dots ,{\sum }_{?=q}{x}_{i}\right)$ where the questionmark indicates that I have no idea how to do it. (In addition this depends on both $k$ and $n$). In $\Delta \left[1\right]$ and of course $\Delta \left[0\right]$ this is quite easy. But in $\Delta \left[2\right]$ things getting more complicated…

But as I said, this is just an additional thought and if nothing like this is known the ($A$) is good enough ..

• CommentRowNumber36.
• CommentAuthorTodd_Trimble
• CommentTimeMar 7th 2012

No! Obviously we can’t write the faces and degeneracies of $\mathrm{hom}\left(A,B\right)$ explicit because they depend on $A$ and $B$.

I meant write faces and degeneracies of $\mathrm{hom}\left(A,B\right)$ in terms of the faces and degeneracies on $A$ and $B$. Obviously.

• CommentRowNumber37.
• CommentAuthorMirco Richter
• CommentTimeMar 7th 2012

Ok … I answered the question late at night.

• CommentRowNumber38.
• CommentAuthorTodd_Trimble
• CommentTimeMar 7th 2012

Ok … I answered the question late at night.

In the interest of having a productive scientific discussion, I would just like to remind everyone to observe some basic precepts, as laid out long ago here – this is a note to self as much as to anyone else.

To continue: to write down explicit formulas for the internal hom, a logical starting place is the cartesian closed category structure on the presheaf category ${\mathrm{Set}}^{{\Delta }^{\mathrm{op}}}$. Suppose $X$ and $Y$ are simplicial sets. We define

$\left({Y}^{X}{\right)}_{n}=\mathrm{Nat}\left(\Delta \left(-,n\right)×X,Y\right)$

If $f:\left[n\right]\to \left[m\right]$ is any morphism in $\Delta$, the function $\left({Y}^{X}{\right)}_{f}:\left({Y}^{X}{\right)}_{m}\to \left({Y}^{X}{\right)}_{n}$ takes a natural transformation $\varphi :\Delta \left(-,m\right)×X\to Y$ to the composite

$\Delta \left(-,n\right)×X\stackrel{\Delta \left(-,f\right)×1}{\to }\Delta \left(-,m\right)×X\stackrel{\varphi }{\to }Y.$

This is impeccably explicit, provided that you know what $\Delta \left(-,f\right)$ is. I think you do know, but it doesn’t hurt to repeat that it’s the morphism of simplicial sets $\Delta \left(-,\left[n\right]\right)\to \Delta \left(-,\left[m\right]\right)$ which, at the set of $j$-cells, takes an ordinal map $g:\left[j\right]\to \left[n\right]$ to $f\circ g:\left[j\right]\to \left[m\right]$.

If $X$ and $Y$ carry simplicial abelian group structures, then the internal hom $\left[X,Y\right]$, as a simplicial set, is a subobject of ${Y}^{X}$. A very explicit construction of such subobjects, stated for the theory of rings but clearly generalizable to any algebraic theory such as the theory of abelian groups, was given here. The point I wish to make now is that the definition of the face and degeneracy operations on $\left[X,Y\right]$ is simply gotten by restricting the face and degeneracy operations on ${Y}^{X}$, since $\left[X,Y\right]$ is by construction a subobject of ${Y}^{X}$. For example, to define a face operation

${d}^{j}:\left[X,Y{\right]}_{m}\to \left[X,Y{\right]}_{m-1}$

(corresponding to an inclusion of ordinals ${\iota }^{j}:\left[m-1\right]\to \left[m\right]$), just treat an element $\varphi \in \left[X,Y{\right]}_{m}$ as an element of $\left({Y}^{X}{\right)}_{m}$, and define ${d}^{j}\left(\varphi \right)=\left({Y}^{X}{\right)}_{{\iota }^{j}}\left(\varphi \right)$. This is guaranteed to land in $\left[X,Y{\right]}_{m-1}$, by construction of $\left[X,Y\right]$.

This way of going about it will relieve you of having to think about applying the free abelian group $ℤ\left(-\right)$ to representables, even though that’s nothing to worry about either.

Is there anything about these definitions that is inexplicit?

• CommentRowNumber39.
• CommentAuthorMirco Richter
• CommentTimeMar 7th 2012
• (edited Mar 7th 2012)

Thanks Todd and Urs for that great amount of time and effort you put into this. I’ll take a back seat and lets these things act on me now.

• CommentRowNumber40.
• CommentAuthorMirco Richter
• CommentTimeMar 13th 2012
• (edited Mar 13th 2012)

Can someone please verify if my understanding of the simplicial abelian group $ℤ\left(\Delta \left[n\right]\right)$ is correct: (It looks like that here in Berlin/Germany there is nobody I can talk to about these things personal)

First of all if $S$ is a finite set of cardinality $\sigma$ then $ℤ\left(S\right)\simeq {\oplus }_{j=1}^{s}ℤ$ and if $f:S\to T$ is a set map into another finite set of cardinality $\tau$, then it lifts to a unique abelian group mophism $ℤ\left(f\right):ℤ\left(S\right)\to ℤ\left(T\right)$. On the element level this morphism goes like $ℤ\left(f\right)\left({z}_{{s}_{1}},\dots ,{z}_{{s}_{\sigma }}\right)=\left({\sum }_{f\left({s}_{j}\right)={t}_{1}}{z}_{{s}_{j}},\dots ,\left({\sum }_{f\left({s}_{j}\right)={t}_{\tau }}{z}_{{s}_{j}}\right)$, where ${\sum }_{f\left({s}_{j}\right)={t}_{k}}{z}_{{s}_{j}}=0$ if there are no $s\in S$ with $f\left(s\right)={t}_{k}$.

If this is true, the rest is easy: $ℤ\left(\Delta \left[n\right]\right)$ is the simplicial abelian group that in dimension $m$ is given by the abelian group ${ℤ}^{\left(\genfrac{}{}{0}{}{n+m+1}{n}\right)}$ and the faces and degeneracies are explicit ( :-) –overstressed term I know ) given by

${d}_{j}:{ℤ}^{\left(\genfrac{}{}{0}{}{n+m+1}{n}\right)}\to {ℤ}^{\left(\genfrac{}{}{0}{}{n+m}{n}\right)}\phantom{\rule{thickmathspace}{0ex}};\phantom{\rule{thickmathspace}{0ex}}\left({z}_{{f}_{1}},\dots ,{z}_{{f}_{\left(\genfrac{}{}{0}{}{n+m+1}{n}\right)}}\right)\to \left({\sum }_{f{\delta }^{j}=f{\prime }_{1}}{z}_{f},\dots ,{\sum }_{f{\delta }^{j}=f{\prime }_{\left(\genfrac{}{}{0}{}{n+m}{n}\right)}}{z}_{f}\right)$

${s}_{j}:{ℤ}^{\left(\genfrac{}{}{0}{}{n+m+1}{n}\right)}\to {ℤ}^{\left(\genfrac{}{}{0}{}{n+m+2}{n}\right)}\phantom{\rule{thickmathspace}{0ex}};\phantom{\rule{thickmathspace}{0ex}}\left({z}_{{f}_{1}},\dots ,{z}_{{f}_{\left(\genfrac{}{}{0}{}{n+m+1}{n}\right)}}\right)\to \left({\sum }_{f{\sigma }^{j}=f{″}_{1}}{z}_{f},\dots ,{\sum }_{f{\sigma }^{j}=f{″}_{\left(\genfrac{}{}{0}{}{n+m+2}{n}\right)}}{z}_{f}\right)$

with ${f}_{i}:\left[m\right]\to \left[n\right]$, $f{\prime }_{i}:\left[m-1\right]\to \left[n\right]$ and $f{″}_{i}:\left[m+1\right]\to \left[n\right]$ as well as ${\delta }^{j}:\left[m-1\right]\to \left[m\right]$ and ${\sigma }^{j}:\left[m+1\right]\to \left[m\right]$ as usual.

– No need for an argumentation: This is complicated and of doubtable value – It is just to be absolute sure what I’m doing here.

For example $ℤ\left(\Delta \left[1\right]\right)$ is in dimension $m$ given by ${ℤ}^{m+2}$ with faces and degeneracies:

${d}_{j}:{ℤ}^{m+2}\to {ℤ}^{m+1}\phantom{\rule{thickmathspace}{0ex}};\phantom{\rule{thickmathspace}{0ex}}\left({z}_{0},\dots ,{z}_{m+1}\right)\to \left({z}_{0},\dots ,{z}_{j}+{z}_{j+1},\dots ,{z}_{m+1}\right)$

${s}_{j}:{ℤ}^{m+2}\to {ℤ}^{m+3}\phantom{\rule{thickmathspace}{0ex}};\phantom{\rule{thickmathspace}{0ex}}\left({z}_{0},\dots ,{z}_{m+1}\right)\to \left({z}_{0},\dots ,{z}_{j},0,{z}_{j+1},\dots ,{z}_{m+1}\right)$

If there are no serious errors in here somewhere, this is what I mean by explicit.

• CommentRowNumber41.
• CommentAuthorTodd_Trimble
• CommentTimeMar 13th 2012
• (edited Mar 13th 2012)

There seem to be some typos, since you have some $f$’s without subscripts. Otherwise it looks like there are no serious errors.

It’s perhaps not as “explicit” as you might think, since you have not specified a particular way of linearly ordering the elements in $\mathrm{hom}\left(\left[m\right],\left[n\right]\right)$ (saying which is the first ${f}_{1}$, the second ${f}_{2}$, etc.). But I would immediately add: you don’t have to, and (more strongly) you probably shouldn’t. It seems to me likely to involve unnatural choices which would add nothing to the discussion. The fact that you haven’t specified such a choice of ordering makes me think that you probably agree with me – that it’s beside the point.

I have an alternate suggestion. You could define $ℤ\left(S\right)$ to be the set of functions $f:S\to ℤ$ such that $f\left(s\right)=0$ for all but finitely many $s\in S$. Or just the set of functions $f:S\to ℤ$, if $S$ happens to be finite (which is the case here). Such $f$ are added in the obvious pointwise way, so $ℤ\left(S\right)$ is an abelian group.

If $h:S\to T$ is a function, then $ℤ\left(h\right):ℤ\left(S\right)\to ℤ\left(T\right)$ is defined in just the way you said: for any $f:S\to ℤ$, $ℤ\left(h\right)\left(f\right)$ is the function $T\to ℤ$ defined by the rule

$\left(ℤ\left(h\right)\left(f\right)\right)\left(t\right)=\sum _{h\left(s\right)=t}f\left(s\right)$

Look, Ma! No subscripts! :-)

Eilenberg used to give people a hard time about deploying subscripts when it was often totally unnecessary, and I think he had a serious point. You can read one such story here, especially pages 21 and 22.

• CommentRowNumber42.
• CommentAuthorDavidRoberts
• CommentTimeMar 13th 2012
• (edited Mar 13th 2012)

Hi Todd, that last link you gave resolves to http://nforum.mathforge.org/discussion/3608/simplicial-group-structure-on-simpl-morphism-sets/www.nap.edu/html/biomems/seilenberg.pdf (something is going wrong as this is the second such link error I’ve seen.)

• CommentRowNumber43.
• CommentAuthorTodd_Trimble
• CommentTimeMar 13th 2012

I fixed it (I think) while you were writing – try again.

• CommentRowNumber44.
• CommentAuthorMirco Richter
• CommentTimeMar 13th 2012
• (edited Mar 13th 2012)

Ok, so from the point where you already know what is going on here, the index free descriptions/equations are short and you can focus on other things. But on the other side, if you don’t know, the subscript expressions are much faster to understand.

Finally I must say that for me the only reason against a subscript description is that it seems to be only very simple for $ℤ\left(\Delta \left[0\right]\right)$ and $ℤ\left(\Delta \left[1\right]\right)$. The latter subscript expression given at the end of #40 looks very simple and pretty to me.

Now I failed giving a nice expression for $ℤ\left(\Delta \left[2\right]\right)$ like the one for $ℤ\left(\Delta \left[1\right]\right)$, but obviously that doesn’t mean that there is none. Since the elements of the sequences $m\to \Delta \left[n{\right]}_{m}$

can be seen as the ’figurate numbers’, maybe there is an order on them giving a nice expression like that of $ℤ\left(\Delta \left[1\right]\right)$.

But that’s not what I’m actually looking for. Now that I know what $ℤ\left(\Delta \left[n\right]\right)$ means, I’m fine.

Another point is maybe, that today with TeX and Computers everywhere, indexed formulars just looks much nicer, then in the days of Eilenberg.

• CommentRowNumber45.
• CommentAuthorTodd_Trimble
• CommentTimeMar 13th 2012
• (edited Mar 13th 2012)

But on the other side, if you don’t know, the subscript expressions are much faster to understand.

Well, if you are speaking only for yourself here, then I can’t and won’t argue with that. But if you are speaking on behalf of everyone who is not already accustomed to such things, then permit me to express strong doubts about that.

Again, you have not specified an isomorphism

$ℤ\left(\mathrm{hom}\left(\left[m\right],\left[n\right]\right)\to \underset{i=1}{\overset{\left(\genfrac{}{}{0}{}{m+n+1}{n}\right)}{⨁}}ℤ$

you are using. A student might legitimately wonder what you have in mind, how much of a difference it makes, and so on. For that reason alone, I don’t see that using subscripts makes anything faster to understand; it just raises questions where there ought to be none. From the point of view of a categorist (and many other modern-day mathematicians who have learned to use basis-free descriptions of linear algebra constructions when it makes sense to do so), such choices are utterly irrelevant and arguably unnatural, and this is what is really behind my suggestion. Besides that, the suggestion is in my opinion much freer from notational clutter, and gets right down to the essential point.

• CommentRowNumber46.
• CommentAuthorTodd_Trimble
• CommentTimeMar 13th 2012

Another point is maybe, that today with TeX and Computers everywhere, indexed formulars just looks much nicer, then in the days of Eilenberg.

I don’t agree with that either: same amount of notational clutter. I only agree that TeX and LaTeX are wonderful word document systems, and have made the world a much better place.

• CommentRowNumber47.
• CommentAuthorMirco Richter
• CommentTimeMar 13th 2012
• (edited Mar 13th 2012)

Todd, the isomorphism $\mathrm{hom}\left(\left[m\right],\left[n\right]\right)↔{ℤ}^{\left(\genfrac{}{}{0}{}{n+m+1}{n}\right)}$ is exactly the point here. Obviously there are many of them and the last question that remains is whether or not we can chose one (not necessarily the same) for any $\mathrm{hom}\left(\left[m\right],\left[n\right]\right)$ in such a way that the resulting face and degeneracy maps between the ${ℤ}^{\left(\genfrac{}{}{0}{}{n+m+1}{n}\right)}$’s become ’nice’ relative to these isomorphisms.

And because I don’t know whether or not this is possible, I leave this piece of information out.

And in the argumentation pro vs. contra indexing, I think there are good reasons for both descriptions. For example in tensor calculus when doing a real calculation like calculating space time deformation around a black hole of given mass ect., then the index description is preferable because you really have to insert numbers into equations and so on. But understanding the tensor calculus as it is, a index free description is preferable. The point I would like to make here is, that when the tensor calculus is USED in other branches of science they almost ever have to use indexes beyond the pure understanding of the theory.

When I do mathematics I prefer to know both as far as possible.

• CommentRowNumber48.
• CommentAuthorTodd_Trimble
• CommentTimeMar 13th 2012

When I do mathematics I prefer to know both as far as possible.

Sure. I know where you’re coming from.

I think the core of the ’disagreement’, if there is one, is that you seem to be rejecting the descriptions I have offered as being ’explicit’, whereas in my mind they are completely explicit and unambiguous, and clean to boot.

Of course you may well succeed in obtaining something ’nice’ according to your lights – let me not stand in the way of that! – but I’m still unclear as to what your ultimate plans are with all of this. If there were some calculation you plan to undertake where the conceptual descriptions fall short, and you actually need to work with a nice ordered basis to see your way clear to the end (this is a plural ’you’), then I would be more supportive. But until I’m convinced of such a pragmatic need, I probably wouldn’t bother myself.

So: why are you going on about homs of simplicial abelian groups? Is there some specific problem you are trying to solve? What are your research plans regarding this, if I may be so bold to ask?

• CommentRowNumber49.
• CommentAuthorMirco Richter
• CommentTimeMar 13th 2012
• (edited Mar 13th 2012)

As I said, its just an exercise I’m doing beside regular things. So you are probably right, that it is not worth to worry about.

And please note that I’m not at all rejecting your descriptions as not being explicit. At that level of abstraction I think they are fine. I just said, that when descending to particular calculations sometimes more details are needed (like bases in LA).

So category theory is great of course, but it doesn’t fit best on any level of abstraction ;-)

• CommentRowNumber50.
• CommentAuthorTodd_Trimble
• CommentTimeMar 14th 2012

but it doesn’t fit best on any level of abstraction

?

Do you mean, doesn’t fit best on every level of abstraction? (Still don’t know what is meant, but it might not be important enough to worry about.)

• CommentRowNumber51.
• CommentAuthorMirco Richter
• CommentTimeMar 14th 2012

every*

• CommentRowNumber52.
• CommentAuthorTim_Porter
• CommentTimeMar 14th 2012

Mirco, my own experience with indices and choice of bases is that they should be put off until the very last possible time. They lead to a much greater chance of slips and in any reasonably long calculation if the result looks wrong it is very very difficult to find if there is a slip or if the calculation is giving you something interesting and new (but unexpected). One the other hand in calculations using shuffles etc, you need the indices AND a conceptual framework (usually categorical, combinatorial and geometric, all at the same time) for seeing how to pair up terms, but the indices are there to help not to hinder the concepts behind the calculations are what really makes things tick.

• CommentRowNumber53.
• CommentAuthorMirco Richter
• CommentTimeMar 14th 2012
• (edited Mar 14th 2012)

@Tim_Porter: I totally agree…

But back on topic i.e. on the understanding of the free simplicial abelian group $ℤ\left(\Delta \left[n\right]\right)$:

Is the following true: If a simplicial set $S$ has only degenerate simplices in any dimension above lets say dimension $m$ (that is every such element is in the image of a degeneracy map) then ${\mathrm{sk}}^{n}\left(S\right)\simeq S$.

Suppose this is true and we call such a simplicial set degenerate from dimension m. Now $\Delta \left[n\right]$ is degenerate from dimension $n$,i.e ${\mathrm{sk}}^{m}\left(\Delta \left[n\right]\right)=\Delta \left[n\right]$ for all $m\ge n+1$. Does the same hold for $ℤ\left(\Delta \left[n\right]\right)$ ? Is it degenerate from dimension n, too?

I think it should be, but if we take the expression of $ℤ\left(\Delta \left[1\right]\right)$ given in #4 at the end we have the following:

$ℤ\left(\Delta \left[1\right]{\right)}_{2}$ should be degenerate, that is every element should be in the image of a degeneracy map. But this is not the case! To see that take $\left({z}_{0},{z}_{1},{z}_{2},{z}_{3}\right)\in ℤ\left(\Delta \left[1\right]{\right)}_{2}$ with ${z}_{1}\ne 0$ and ${z}_{2}\ne 0$. Then this is neither in the image of ${s}_{0}$ nor in the image of ${s}_{1}$, hence it is not degenerate, right? It is the sum of degenerated elements, but that is not the same, I think.

So in any dimension $\ge 2$ any element is the sum of degenerated elements. In contrast in dimension 1 not every element is the sum of degenerate elements.

So what is going on here?

• CommentRowNumber54.
• CommentAuthorMirco Richter
• CommentTimeMar 14th 2012

Is it because ${\mathrm{sk}}^{n}$ is defined to be ’freely filled’ with degenerate simplices in dimension above $n$ and this ’freely’ means in our context that ${\mathrm{sk}}^{n}$ is defined by arbitrary sums of degenenerate simplices?