A finite simplicial set is a simplicial set having only a finite number of non degenerate simplicies. My question is: if $A$ and $B$ are finite simplicial sets, does this imply that the simplicial set $A^B$ is also finite?

  • 4
    $\begingroup$ Regarding the tag 'homotopy-theory', this question does not appear to be actually about homotopy theory. $\endgroup$ – Ricardo Andrade Nov 17 '13 at 22:27
  • 1
    $\begingroup$ Maybe combinatorics. $\endgroup$ – Fernando Muro Mar 4 '15 at 23:42

I think the statement is false for $ X = A = \Delta_4/\partial \Delta_4$ and $B = \Delta_1$:

As Charles Rezk already mentioned, it is enough to consider the growth of $f_{X^{\Delta_1}}(n) = |Hom(\Delta_1\times \Delta_n, X)|$. Writing the prism $\Delta_1 \times \Delta_n$ as a coequalizer over its $(n+1)$- simplices, I think we can identify $Hom(\Delta_1\times \Delta_n, X)$ with tuples $(x_0, \dotsc, x_n)$ where $x_i \in X_{n+1}$ and $ d_{i+1} x_i = d_{i+1} x_{i+1}$ whenever this makes sense. Now, we can restrict our attention to tuples where even $d_{i+1} x_i = d_{i} x_i = [\partial \Delta_4]$ holds (and for simplicity $x_n = [\partial \Delta_4]$ where $[\partial \Delta_4]$ denotes the unique degeneracy of the basepoint).

These definitely satisfy the gluing conditions, and the number of them grows exponentially: We can choose the $x_i$ independently, and for $n$ large enough there are at least 2 possibly choices for any $i< n$: If we denote the unique non-degenerate simplex in $X$ by $\iota$, then we can consider morphisms in the $\Delta$-category $\sigma : [n] \rightarrow [4]$ which are surjective and have the property that $\sigma^{-1}(\sigma(i))$ and $\sigma^{-1}(\sigma(i+1))$ are singleton sets. Then $\sigma^* \iota$ is a possible choice for $x_i$.

This gives exponential growth as a lower bound, so $X^{\Delta_1}$ can't be finite.

| cite | improve this answer | |
  • $\begingroup$ Very nice! I guess in the argument I had in mind on my comment on Charles's answer I was implicitly imagining $X$ was actually a simplicial complex. $\endgroup$ – Eric Wofsey Mar 4 '15 at 23:24
  • $\begingroup$ I think you can also get exponential growth for $X=\Delta_3/\partial \Delta_3$ by a slightly different construction. Don't require $d_{i+1}x_i=d_{i+1}x_{i+1}$ to always be $[\partial \Delta_3]$, and instead consider the set $S$ of all values of $i$ such that it is $[\partial\Delta_3]$. It is not hard to see that $S$ can be almost any subset of $\{0,\dots,n\}$, so $Hom(\Delta_1\times \Delta_n, X)$ must have at least $\approx 2^n$ elements. $\endgroup$ – Eric Wofsey Mar 4 '15 at 23:45
  • $\begingroup$ Thanks! I think I understand the argument, I just don't see why you had to assume $x_n = [\partial \Delta_4]$. I also believe it should be $\sigma : [n+1] \rightarrow [4]$ (or perhaps I misunderstand something?). $\endgroup$ – Ilan Barnea Jul 14 '15 at 6:16

I failed to figure this out. But here are some thoughts.

Say a simplicial set $X$ is finite type if it has finitely many simplices in each degree. It's not hard to prove that $X^K$ is finite type if $X$ is finite type and $K$ is finite.

For finite type $X$, let $f=f_X$ be the function $f_X(n)=|X_n|$, the number of $n$-simplices, and let $g=g_X(n)$ be the number of non-degenerate $n$-simplices. Note that $X$ is finite iff $g(n)=0$ for $n$ large.

There is a simple relationship between $f$ and $g$: $$ f(n) = \sum_{k=0}^n \binom{n}{k}g(k).$$ Since the $g(k)\geq0$, we should thus be able to say that $X$ if finite if and only if $f$ has polynomial growth, since $n\mapsto \binom{n}{k}=\frac{1}{k!}n^k+\cdots$ is polynomial of degree $k$ (and is a monotone function).

If $K$ is finite, then $K$ is a union of a finite number of non-degenerate simplices. Thus $X^K\subseteq \prod_{i=1}^d X^{\Delta_{k_i}}$. Since a finite product of finite simplicial sets is finite, it suffices to show that $X$ finite implies $X^{\Delta_k}$ finite.

It's an easy exercise to show that $\Delta_k$ is a retract of the cube $(\Delta_1)^k$. Thus, by the exponential property of function complexes, it suffices merely to show that $X$ finite implies $X^{\Delta_1}$ finite.

A naive estimate is as follows. $f_{X^{\Delta_1}}(n) = |Hom(\Delta_1\times \Delta_n, X)|$. The complex $\Delta_1\times \Delta_n$ is a union of $n+1$ copies of $\Delta_{n+1}$ along various faces. Thus there is an injection $$ Hom(\Delta_1\times \Delta_n ,X) \subseteq (X_{n+1})^{n+1}.$$ This provides an estimate $$ f_{X^{\Delta_1}}(n) \leq f_X(n+1)^{n+1}. $$ Unfortunately, the right hand side grows exponentially as a function of $n$, so this is of no use.

Note that in some cases it is easy to show that $X^K$ is finite. For instance, if $X$ is the nerve of a finite poset, then $X^{\Delta_n}$ is also the nerve of a finite poset, from which it follows that $X^K$ is finite for all finite $K$. This applies for instance to $X=\Delta_k$.

| cite | improve this answer | |
  • $\begingroup$ I don't have the energy now to write out the messy details, but I believe that your "naive estimate" can be improved to $O(n^{2dim(X)})$. For large $n$, the $(n+1)$-simplices that make up a prism $\Delta_1\times \Delta_n\to X$ must all be very degenerate and very similar to each other; once you pick the first one (about $n^{dim(X)}$ choices), the other ones must more or less look the same except that the locations of the degeneracies get to move around (which I think gives another $n^{dim(X)}$ or so choices). In particular, $X^{\Delta_1}$ should have dimension at most twice that of $X$. $\endgroup$ – Eric Wofsey Nov 17 '13 at 17:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.