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Jul 14, 2015 at 17:26 comment added Eric Wofsey 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).
Jul 14, 2015 at 17:06 comment added Eric Wofsey 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 $x_{i+1}$).
Jul 14, 2015 at 16:38 comment added Ilan Barnea Very nice! Thanks! What about $\Delta_2/\partial\Delta_2$? Your argument certainly doesn't apply then.
Jul 14, 2015 at 16:32 vote accept Ilan Barnea
Jul 14, 2015 at 16:09 history answered Eric Wofsey CC BY-SA 3.0