I've run across a simplicial complex which, according to Sage, seems to have a very easily-described homology. However, proving this fact has been rather difficult.
Fix $s\ge 2$ (though I would be very interested if anyone has an answer that is only for $s=2$; feel free to just assume that if you like). The vertices of this simplicial complex are certain subsets of $[n] = \{1,2,\ldots,n\}$, where $n>s$ and faces are formed from sets of subsets that have nonempty intersection. Let $p$ be the largest prime not more than $n$. The complex $\Delta_{s,n}$ has the following properties:
- For all subsets $X\subseteq \{2,\ldots,n\}$ of size $s-1$, we have $\{1\}\cup X\in \Delta_{s,n}$. Denote by $S_1$ the set of all such $X$; $S_1$ is a simplex, since all of these sets contain $1$.
- The complementary set $S_2 = \Delta_{s,n}\setminus S_1$ is such that all of its elements contain $p$ but do not contain $1$. Furthermore, for any set of $s-1$ elements of $S_1$, there is some set in $S_2$ that intersects each of these $s-1$ elements (pairwise).
Computationally, it seems that all the reduced homology groups are zero except for the $(s-1)$st, which equals the free abelian group of rank ${n-2\choose s-1}$.
A guess at a basis in the case $s=2$. I'll write, for example, $12$ instead of $\{1,2\}$. We have $$\Delta_{s,n} = \{12, 13, 14, \ldots, 1n\} \cup S_2,$$ where every element in $S_2$ contains $p$ but not $1$. Of course, any $1$-cycle that is completely contained in either $S_1$ or $S_2$ is the boundary of a $2$-simplex. For every pair of integers $i$ and $j$ both not $1$, there is some element $X$ of $S_2$ that contains both $i$ and $j$, so that $[1i, 1j, X]$ is a $1$-cycle in the simplicial complex without an interior (since $1i \cap 1j = 1$ and $1\notin X$). Lastly, if we take two sets $X$ and $Y$ in $S_2$ and there exists some $i$ such that $1i\cap X$ and $1i\cap Y$ are both nonempty, then it follows that $i\in X$ and $i\in Y$, so this $1$-cycle is the boundary of a $2$-simplex. What we have just shown is that the only $1$-cycles that are not the boundary of a $2$-simplex have two endpoints in $S_1$ and one endpoint in $S_2$.
For any pair of integers $(i,j)$, let $f(i,j)$ be some element in $S_2$ that contains both $i$ and $j$ (if there are multiple choices, just pick one). From here, my guess at a basis of the first homology group is the following: $$\eqalign{ [13, f(2,3)] - [12, f(2,3)] + [13, f(2,3)],\quad [14, f(2,4)] - [12, f(2,4)] + [14, f(2,4)],\cr [15, f(2,5)] - [12, f(2,5)] + [15, f(2,5)], \ldots [1n, f(2,n)] - [12, f(2,n)] + [1n, f(2,n)].\cr }$$ If this were correct, then there would be $n-2 = {n-2\choose 2-1}$ elements in the basis. Then we need to prove that the other reduced homology groups are zero, which shouldn't be as difficult. But I do not see any obvious way to show that this basis is spanning and minimally so, and I have heard that anytime you're doing explicit homology computations with elements in this way, you're probably barking up the wrong tree, so I would be grateful if anyone had any pointers or alternative approaches!
Edit. I believe I have something resembling a proof in the case $s=2$ using Mayer-Vietoris sequences, but I would really appreciate if someone could check it, as it is my first time using this theorem. So the version I saw looks like this: $$\cdots \to \tilde H_1(A\cap B) \to \tilde H_1(A)\oplus \tilde H_1(B) \to \tilde H_1(X)\to \tilde H_0(A\cap B) \to \tilde H_0(A)\oplus \tilde H_0(B) \to \tilde H_0(X)\to 0$$ Where $A$ and $B$ are such that $A\cup B =X$. What I thought to do was to make $A$ the simplex consisting of everything that contains $1$, and then $B$ everything else (another simplex with lots of little $1$-cells and $2$-cells sticking out). Importantly, $B$ does not have any $1$-cycles that are not the boundary of a $2$-simplex, since any such $1$-cycle must have two vertices in $A$. So $A$ and $B$ are both contractible. Now $A\cap B$ consists of the disjoint union of $n-1$ points, so $\tilde H_0(A\cap B) = {\bf Z}^{n-2}$. And putting this all together, we find that $$\cdots \to 0\to 0\to \tilde H_1(\Delta_{2,n}) \to {\bf Z}^{n-2} \to 0\to 0\to 0,$$ from which we conclude that $\tilde H_1(\Delta_{2,n}) = {\bf Z}^{n-2}$. Does this proof check out or have I made some error somewhere? I know this post has been a bit messy, so thanks for your time if you have made it to the end here!
