# Is the simplicial set $(\Delta_3/\partial \Delta_3)^{\Delta_1}$ finite?

A simplicial set is called finite if it has only a finite number of non degenerate simplicies (or equivalently, if its number of $n$-simplicies grows polynomially with $n$). In an answer to a previous question of mine Exponentiation in finite simplicial sets it was shown that for any $n\geq 4$ the simplicial set $(\Delta_n/\partial \Delta_n)^{\Delta_1}$ is not finite. Following one of the comments there, I wanted to ask:

Is the simplicial set $(\Delta_3/\partial \Delta_3)^{\Delta_1}$ finite?

• What about (Δ^2/∂Δ^2)^{Δ^1}, is it finite or not? – Dmitri Pavlov Jul 22 '15 at 11:33
• It is finite. See comments to the answer. – Ilan Barnea Jul 22 '15 at 20:59

No, it is not; here is a slightly simpler argument than the one indicated in my comment. As in the answer to the previous question, let $X=\Delta_3/\partial\Delta_3$. It suffices to show that the cardinality of $A_n=\operatorname{Hom}(\Delta_1\times \Delta_n, X)$ grows faster than any polynomial as a function of $n$. As in that answer, an element of $A_n$ can be identified with a sequence $(x_0,\dots,x_n)$ of elements of $X_{n+1}$ such that $d_{i+1}x_i=d_{i+1}x_{i+1}$ for $0\leq i<n$.
Let $B_n\subset A_n$ be the set of such sequences for which $d_{i+1}x_i=[\partial\Delta_3]$ is the basepoint of $X_n$ for all $0\leq i<n$ and $x_0=x_n=[\partial\Delta_3]$ is the basepoint of $X_{n+1}$. If $(x_0,\dots,x_n)\in B_n$, then for each $0<i<n$, $x_i$ must be either $[\partial\Delta_3]$ or the (image in $X$ of) the unique $(n+1)$-simplex $\sigma_i$ in $\Delta_3$ such that $\sigma_i$ is not in $\partial\Delta_3$ but $d_i\sigma_i$ and $d_{i+1}\sigma_i$ are in $\partial\Delta_3$. Explicitly, this simplex is the map $\sigma_i:[n+1]\to[3]$ such that $\sigma_i(j)=0$ for $j<i$, $\sigma_i(i)=1$, $\sigma_i(i+1)=2$, and $\sigma_i(j)=3$ for $j>i+1$.
So to define an element of $B_n$, we have exactly two choices for each $x_i$ for $0<i<n$, and every possible sequence of such choices does define an element of $B_n$. Thus $B_n$ has exactly $2^{n-1}$ elements. In particular, $|A_n|\geq 2^{n-1}$ grows exponentially.
• Very nice! Thanks! What about $\Delta_2/\partial\Delta_2$? Your argument certainly doesn't apply then. – Ilan Barnea Jul 14 '15 at 16:38
• It is not hard to show $(\Delta_2/\partial\Delta_2)^{\Delta_1}$ is finite (in fact, $4$-dimensional if I'm not mistaken). Just enumerate what a sequence $(x_0,\dots,x_n)$ can look like, and you'll only find $O(n^4)$ possibilities (this is a little messy but not hard, the point being that $x_i$ very nearly determines $x_{i+1}$). – Eric Wofsey Jul 14 '15 at 17:06
• In fact, I think a similar argument can show that $X^K$ is finite whenever $\dim X\leq 2$ and $X$ and $K$ are finite, by induction on the number of nondegenerate simplices in $X$. The point is that given $(x_0,\dots, x_n)$ as above, the sequence $(x_0, d_1x_0, x_1, d_2x_1,\dots, x_n)$ can only exit and reenter the interior of any maximal face of $X$ a bounded number of times (as opposed to the situation above for $\Delta_3/\partial\Delta_3$, where all the $d_{i+1}x_i$ could be in $\partial\Delta_3$ while none of the $x_i$ are). – Eric Wofsey Jul 14 '15 at 17:26