# Change of Betti numbers under simplicial maps

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 $$O(n^{i-1})$$.

• 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? – Willard Zhan Nov 7 '18 at 10:22