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

]]>@Tim_Porter: I totally agree…

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

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 $sk^n(S) \simeq S$.

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

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

$\mathbb{Z}(\Delta[1])_2$ 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 $(z_0,z_1,z_2,z_3) \in \mathbb{Z}(\Delta[1])_2$ with $z_1 \neq 0$ and $z_2 \neq 0$. 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 $\geq 2$ 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?

]]>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.

]]>every*

]]>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.)

]]>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 ;-)

]]>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?

]]>Todd, the isomorphism $hom([m],[n]) \leftrightarrow \mathbb{Z}^{\binom{n+m+1}{n}}$ 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 $\hom([m],[n])$ in such a way that the resulting face and degeneracy maps between the $\mathbb{Z}^{\binom{n+m+1}{n}}$’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.

]]>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.

]]>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

$\mathbb{Z}(\hom([m], [n]) \to \bigoplus_{i=1}^{\binom{m+n+1}{n}} \mathbb{Z}$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.

]]>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 $\mathbb{Z}(\Delta[0])$ and $\mathbb{Z}(\Delta[1])$. 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 $\mathbb{Z}(\Delta[2])$ like the one for $\mathbb{Z}(\Delta[1])$, but obviously that doesn’t mean that there is none. Since the elements of the sequences $m \rightarrow \Delta[n]_m$

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

But that’s not what I’m actually looking for. Now that I know what $\mathbb{Z}(\Delta[n])$ 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.

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

]]>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.)

]]>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 $\hom([m], [n])$ (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* $\mathbb{Z}(S)$ to be the set of functions $f: S \to \mathbb{Z}$ such that $f(s) = 0$ for all but finitely many $s \in S$. Or just the set of functions $f: S \to \mathbb{Z}$, if $S$ happens to be finite (which is the case here). Such $f$ are added in the obvious pointwise way, so $\mathbb{Z}(S)$ is an abelian group.

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

$(\mathbb{Z}(h)(f))(t) = \sum_{h(s) = t} f(s)$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.

]]>Can someone please verify if my understanding of the simplicial abelian group $\mathbb{Z}(\Delta[n])$ 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 $\mathbb{Z}(S) \simeq \oplus_{j=1}^s\mathbb{Z}$ and if $f: S \rightarrow T$ is a set map into another finite set of cardinality $\tau$, then it lifts to a unique abelian group mophism $\mathbb{Z}(f): \mathbb{Z}(S) \rightarrow \mathbb{Z}(T)$. On the element level this morphism goes like $\mathbb{Z}(f)(z_{s_1},\ldots,z_{s_\sigma}) = (\sum_{f(s_j) = t_1} z_{s_j}, \ldots,(\sum_{f(s_j) = t_\tau} z_{s_j})$, where $\sum_{f(s_j) =t_k}z_{s_j} =0$ if there are no $s \in S$ with $f(s)=t_k$.

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

$d_j : \mathbb{Z}^{\binom{n+m+1}{n}} \rightarrow \mathbb{Z}^{\binom{n+m}{n}} \;; \; (z_{f_1},\ldots,z_{f_{\binom{n+m+1}{n}}}) \rightarrow (\sum_{f\delta^j = f'_1} z_{f}, \ldots,\sum_{f\delta^j = f'_{\binom{n+m}{n}}} z_{f})$

$s_j : \mathbb{Z}^{\binom{n+m+1}{n}} \rightarrow \mathbb{Z}^{\binom{n+m+2}{n}} \; ; \; (z_{f_1},\ldots,z_{f_{\binom{n+m+1}{n}}}) \rightarrow (\sum_{f\sigma^j = f''_1} z_{f}, \ldots,\sum_{f\sigma^j = f''_{\binom{n+m+2}{n}}} z_{f})$

with $f_i : [m] \rightarrow [n]$, $f'_i : [m-1] \rightarrow [n]$ and $f''_i : [m+1] \rightarrow [n]$ as well as $\delta^j : [m-1] \rightarrow [m]$ and $\sigma^j : [m+1] \rightarrow [m]$ 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 $\mathbb{Z}(\Delta[1])$ is in dimension $m$ given by $\mathbb{Z}^{m+2}$ with faces and degeneracies:

$d_j: \mathbb{Z}^{m+2} \rightarrow \mathbb{Z}^{m+1} \; ; \; (z_0, \ldots, z_{m+1}) \rightarrow (z_0, \ldots, z_j+z_{j+1}, \ldots, z_{m+1})$

$s_j: \mathbb{Z}^{m+2} \rightarrow \mathbb{Z}^{m+3} \; ; \; (z_0, \ldots, z_{m+1}) \rightarrow (z_0, \ldots, z_j,0,z_{j+1}, \ldots, z_{m+1})$

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

]]>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.

]]>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 $Set^{\Delta^{op}}$. Suppose $X$ and $Y$ are simplicial sets. We define

$(Y^X)_n = Nat(\Delta(-, n) \times X, Y)$If $f: [n] \to [m]$ is any morphism in $\Delta$, the function $(Y^X)_f: (Y^X)_m \to (Y^X)_n$ takes a natural transformation $\phi: \Delta(-, m) \times X \to Y$ to the composite

$\Delta(-, n) \times X \stackrel{\Delta(-, f) \times 1}{\to} \Delta(-, m) \times X \stackrel{\phi}{\to} Y.$This is impeccably explicit, provided that you know what $\Delta(-, f)$ is. I think you do know, but it doesn’t hurt to repeat that it’s the morphism of simplicial sets $\Delta(-, [n]) \to \Delta(-, [m])$ which, at the set of $j$-cells, takes an ordinal map $g: [j] \to [n]$ to $f \circ g: [j] \to [m]$.

If $X$ and $Y$ carry simplicial abelian group structures, then the internal hom $[X, Y]$, 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 $[X, Y]$ is simply gotten by restricting the face and degeneracy operations on $Y^X$, since $[X, Y]$ is by construction a subobject of $Y^X$. For example, to define a face operation

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

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

Is there anything about these definitions that is inexplicit?

]]>Ok … I answered the question late at night.

]]>No! Obviously we can’t write the faces and degeneracies of $hom(A,B)$ explicit because they depend on $A$ and $B$.

I meant write faces and degeneracies of $\hom(A, B)$ in terms of the faces and degeneracies on $A$ and $B$. *Obviously*.

@Urs:

1.) For the morphisms $f: \Delta[n] \rightarrow \Delta[m]$ I just know the usual definition that is determined from the general construction of the Yoneda embedding i.e. $f(g) = f \circ g$. In addition as far as I know the functor $[n] \rightarrow \Delta[n]$ 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 $#(S)=\sigma$ and $#(T)=\theta$ are of finite cardinality and $f : S \rightarrow T$ is a set map, then $\mathbb{Z}(S) \simeq \oplus_{s=1}^{\sigma}\mathbb{Z}$ and $\mathbb{Z}f$ is a function ’on the coordinates’ that is $\mathbb{Z}f(x_{s_1},\ldots,x_{s_{\sigma}}) = (\sum_{f(s_i)=t_1}x_{f(s_i)},\ldots,\sum_{f(s_i)=t_\theta}x_{f(s_\sigma)})$.

Now using the latter and $p:= #(\Delta[n]_k) = \binom{k+n+1}{n}$ as well as $q:= #(\Delta[n]_{k-1}) = \binom{k+n}{n}$ with $S = \{f: [k] \rightarrow [n] \}$ we have an explicit definition of the faces $d_j : \mathbb{Z}(\Delta[n]_k) \rightarrow \mathbb{Z}(\Delta[n]_{k-1})$ given by

($A$) $d_j: \mathbb{Z}^p \rightarrow \mathbb{Z}^q; (x_{f_1},\ldots,x_{f_p}) \mapsto (\sum_{d_(j_i)=f'_1}x_{d_j(f_i)},\ldots,\sum_{d_(j_i)=f'_q}x_{d_j(f_i)})$ and analog for the degeneracies inside a particular $\Delta[n]$. This is true for any order we have defined on the elements of $\Delta[n]_k$ i.e. whatever $f \in \Delta[n]_k$ 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 $f \in \Delta[n]_k$ ?

IF this is possible then for example in $\mathbb{Z}(\Delta[n])_k$ the face d_j would look like: $d_j: \mathbb{Z}(\Delta[n]_k) \rightarrow \mathbb{Z}(\Delta[n]_{k-1}); (x_{1},\ldots,x_{p}) = (\sum_{?=1}x_{i},\ldots,\sum_{?=q}x_{i})$ where the questionmark indicates that I have no idea how to do it. (In addition this depends on both $k$ and $n$). In $\Delta[1]$ and of course $\Delta[0]$ this is quite easy. But in $\Delta[2]$ 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 ..

]]>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 $hom(A,B)$ explicit because they depend on $A$ and $B$. But when $hom(A,B)_n = Hom_{sAb}(A \otimes \mathbb{Z}(\Delta[n]),B)$, then there is one more thing we can do to make this as explicit as possible:

We can try to unravel $\mathbb{Z}(\Delta[n])$ because (1) it is necessary in the definition of $hom(A,B)$ 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’ $hom(A,B)$ first has to compute $\mathbb{Z}(\Delta[n])$. 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.)

]]>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:

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

understand for a given function of sets $f : S \to T$ what the corresponding morphism of free abelian groups $\mathbb{Z}(S) \to \mathbb{Z}(T)$ 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 $\Delta[1] \to \Delta[2]$. First do it pictorially, if that helps, then try to find a formal language for it that suits you.

]]>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.

@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 $\mathbb{Z}(\Delta[n])$ 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.

So your answer implies that there really are explicit formulars?

]]>