Let $\Delta$ be a simplicial complex on $n$ vertices, and $\phi$ a simplicial map that identifies two vertices $x$ and $y$ of $\Delta$. I want to show that the Betti numbers of $\phi(\Delta)$ cannot increase much from those of $\Delta$.

For instance, trivially $b_0(\phi(\Delta))\leq b_0(\Delta)$, and I could prove $b_1(\phi(\Delta))\leq b_1(\Delta)+1$. It is simply because $\mathrm{Ker}_{\phi(\Delta)}\partial_1\setminus\mathrm{Ker}_{\Delta}\partial_1=\phi\circ \partial_1^{-1}(\{c[x]-c[y]\mid c\in\mathbb{F}\})$. The same argument seems to only yield $b_k(\phi(\Delta))\leq b_k(\Delta)+O(n^k)$ for larger $k$.

So my questions are, are results with better dependence on $n$ known for higher-order Betti numbers? Can we bound $b_2(\phi(\Delta))$ in terms of both $b_2(\Delta)$ and $b_1(\Delta)$ (and $n$)? Is it true that the sum of Betti numbers cannot increase much under identifying two vertices?


Let $\Delta$ be the complex with vertices $$ x, y, a, b_1, b_2, \dots, b_k, c_1, c_2, \dots, c_k $$ generated by the following $2$-simplices: $$ (a,b_i,c_i), (a,b_i, y), (b_i, c_i, y), (a, c_i, x) $$ For any individual $i$, these 2-simplices generate a contractible subcomplex $\Delta_i$ whose realization is $D^2$ because all of the simplices contain $a$. The intersection of the subcomplexes $\Delta_i$ and $\Delta_j$ is the pair of edges $(a,x)$ and $(a,y)$. The geometric realization of $\Delta$ is a union of discs, glued along a common edge. It is contractible and $\Delta$ has Betti numbers $1, 0, 0, \dots$

The quotient complex $\phi \Delta$ has vertices $x, a, b_i, c_i$ and is generated by $2$-simplices $$ (a,b_i,c_i), (a,b_i, x), (b_i, c_i, x), (a, c_i, x). $$ Each value of $i$ gives the boundary of a tetrahedron $T_i$. The intersection of $T_i$ with $T_j$ is the edge $(a,x)$. The geometric realization of $\phi \Delta$ is a union of $k$ 2-spheres, glued along a common edge, and its Betti numbers are $1, 0, k, 0, \dots$.

(This is really $k$ copies of the example you described in a now-deleted comment.)

As a result, the Betti numbers of $\phi \Delta$ aren't bounded by those of $\Delta$.

Here is a more general result.

Consider the standard simplex with vertices $(v_0, \dots, v_k)$, and let $K$ and $L$ be subcomplexes of it.

Let $X$ be the subcomplex of $(x, v_0, \dots, v_k)$ generated by the following simplices:

  • $(x)$

  • $(v_0, \dots, v_k)$

  • $(x, c_1, \dots, c_l)$ whenever $(c_1, \dots, c_l)$ is a simplex of $K$

Then $X$ is the mapping cone of the map $K \to \Delta^k$ and $$ \widetilde H_*(X) \cong \widetilde H_{*-1}(K). $$

Similarly, let $Y$ be the subcomplex of $(y,v_0,\dots,v_k)$ whose reduced homology groups are a shift of the homology groups of $L$.

Let $\Delta$ be the union of $X$ and $Y$, as a subcomplex of $(x,y,v_0,\dots,v_k)$. The intersection of $X$ and $Y$ is the simplex $(v_0,\dots,v_k)$ which is contractible, and so $$ \widetilde H_*(\Delta) \cong \widetilde H_{*-1}(K) \oplus \widetilde H_{*-1}(L). $$ The complex $\phi \Delta$ is the subcomplex of $(x, v_0, \dots, v_k)$ generated by $(x)$, $(v_0,\dots,v_k)$, simplices associated to $K$, and simplices associated to $L$. As a result, it is also a mapping cone: the mapping cone of the map $K \cup L \to \Delta^k$, and so $$ \widetilde H_*(\phi \Delta) \cong \widetilde H_{*-1}(K \cup L). $$ Therefore, it suffices to find complexes $K$ and $L$ with not much homology but whose union has large homology.

Now let's do a specific example to show that we can get large growth.

For any $d \geq 0$, let $K$ be the subcomplex generated by the $d$-simplices containing $v_0$, and let $L$ be the subcomplex generated by the $d$-simplices containing $v_1$. Then $K$ linearly contracts onto $v_0$ and $L$ linearly contracts onto $v_1$, so $\widetilde H_*(K) = \widetilde H_*(L) = 0$, and so $$\widetilde H_*(\Delta) = 0.$$ Also, by the Mayer-Vietoris sequence, we have $$\widetilde H_*(\phi \Delta) = \widetilde H_{*-1}(K \cup L) = \widetilde H_{*-2}(K \cap L).$$

The complex $K \cap L$ is the union of all of the $(d-1)$-simplices of $(v_2,\dots,v_k)$--the $(d-1)$-skeleton of $\Delta^{k-2}$. Because of this there is a long exact sequence $$ 0 \to \widetilde H_{d-1}(K \cap L) \to C_{d-1}(\Delta^{k-2}) \to \dots \to C_1(\Delta^{k-2}) \to C_0(\Delta^{k-2}) \to \Bbb Z \to 0, $$ the rank of $H_{d+1}(\phi \Delta)$ is the alternating sum of the ranks, which is the alternating sum of the number of simplices in $\Delta^{k-2}$:

$$ \begin{align*} b_{d+1} &= \text{rank of }H_{d+1}(\phi \Delta)\\ &= \text{rank of }\widetilde H_{d-1}(K \cap L)\\ &= num\binom{k-1}{d} - \binom{k-1}{d-1} + \binom{k-1}{d-2} - \dots + (-1)^{d-1} \binom{k-1}{1} + (-1)^d \\ &= \binom{k-2}{d}. \end{align*} $$ In particular, this is polynomial of degree $d$. If $n = k+3$ is the number of vertices and $i = d+1$, then the growth of the $i$'th Betti number can be $O(n^{i-1})$.

  • $\begingroup$ Thank you for the answer. The example here shows a increment of $O(n)$ in $b_2$. I didn't make it clear in the question, but I actually want to know how small the dependence on $n$ could be. Is $O(n)$ always true? $\endgroup$ – Willard Zhan Nov 7 '18 at 10:22

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