Relation between maximum value of 1st Betti number and number of faces

Question

Let $K$ be an abstract simplicial complex, where $\{0\}, \{1\}, \ldots, \{n-1\} \in K$, and $\{i,j\} \in K$ for $0 \leq i < j < n$ (indicating that the 1-skeleton forms a complete graph). It's important to note that $K$ has no 2-simplices. However, we should observe that the number of 2-faces in $K$ is given by $\binom{n}{2+1}$.

The number of 1-dimensional holes, represented by the Betti number $\beta_1$, must be upper-bounded by the number of 2-faces, as each 'hole' corresponds to a missing higher-order simplex that covers such a hole. Surprisingly, a post on this link highlights a scenario where the number of 1-holes is strictly less than the number of 2-faces, as demonstrated in the case of the tetrahedron.

A quick simulation, considering the simplicial complex $K$ with $n = 2$ through $20$, reveals that $\beta_1$ is significantly smaller than $\binom{n}{2+1}$.

Vertices of the simplicial complex	Number of 2-faces	Max value of the Betti number $\beta_1$
2	0	0
3	1	1
4	4	3
5	10	6
6	20	10
7	35	15
8	56	21
9	84	28
10	120	36
11	165	45
12	220	55
13	286	66
14	364	78
15	455	91
16	560	05
17	680	20
18	816	36
19	969	53
20	1140	71

What is the maximum value of $\beta_1$ for a simplicial complex with $n$ vertices? Can this formula be generalized to $\beta_k$ for any arbitrary $k$?

Answer 1

The maximum value of $\beta_1$ for a simplicial complex with $n$ vertices is $\binom{n}{2} - n + 1$. This is achieved with the complete graph on $n$ vertices, so we have the maximum $\binom{n}{2}$ edges and no $2$-faces. To compute the homology of this, collapse a spanning tree to a point. Since this tree is a nice contractible subspace, this does not change the homotopy type and hence doesn't change the homology. In a complete graph on $n$ vertices, a spanning tree will have $n-1$ edges: pick a vertex $v$ and add an edge from $v$ to each of the other $n-1$ vertices. Once this has been collapsed to a point, the resulting space is a bouquet of circles, one for each remaining edge. We started with $\binom{n}{2}$ edges and collapsed $n-1$ to a point, so we are left with $\binom{n}{2} - n + 1$ edges, and hence this many circles. So this is the first Betti number.

I didn't know the formula for $\max \beta_k$ for large values of $k$, but it should be achieved with the $k$-skeleton of an $(n-1)$-simplex. Some computations in SageMath:

sage: def max_betti(n, k): # n=vertices, k=dimension
....:     K = simplicial_complexes.Simplex(n-1)
....:     return K.n_skeleton(k).betti(k)
....: 
sage: [max_betti(n,2) for n in (1..15)]
[0, 0, 0, 1, 4, 10, 20, 35, 56, 84, 120, 165, 220, 286, 364]
sage: [max_betti(n,3) for n in (1..15)]
[0, 0, 0, 0, 1, 5, 15, 35, 70, 126, 210, 330, 495, 715, 1001]
sage: [max_betti(n,4) for n in (1..15)]
[0, 0, 0, 0, 0, 1, 6, 21, 56, 126, 252, 462, 792, 1287, 2002]

The first of these seems to be https://oeis.org/A000292. The second is https://oeis.org/A000332. The third is https://oeis.org/A000389.