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Let a piecewise linear $d$-dimensional complex $M \subset \mathbb{R}^m, m \geq d$ be obtained by attaching $K$ convex $d$-dimensional polytopes $\{ C_i \}_{i=1}^K$ (cells), along their facets without deforming or banding those polytopes. That is: $$ M = C_1 \cup C_2 \cup C_3 \cup \cdots \cup C_K. $$ Since each cell $C_i$ is convex, they are topological disks and intersection of any two cells, $C_i\cap C_j$, is also a topological disk.

I'm wondering if there is an easy way to show the following bound on the sum of the Betti numbers of $M$:

$$ \sum_{N=0}^d\beta_N(M) \stackrel{?}{\leq} K \tag{*} $$

Intuitively, it seems to me that the largest possible sum is obtain in case where all cells $\{C_i\}_{i=1}^K$ are not attached to one another, but each is just "floating" in the space, and therefore each $C_i$ contributes a single Homology $\beta_0(C_i)=1,~~ \beta_N(C_i) = 0, N >0 $ to the sum in (*). It seems wasteful to try to attache the cells to introduce higher Homologies, to get loops, holes etc...

My goal is to see whether this intuition is correct? Any help or any suggestions, comments will be appreciated.


I have only limited familiarity with topology, and was wondering if this is a known fact or easily provable one. I did some literature review and was thinking about two approaches listed below.


First approach.

After reading the excellent notes by Forman on Morse theory on Cell complexes I was trying to use Corollary 3.7 (The Weak Morse Inequalities) on page 108:

Corollary 3.7 For every $N$ $$ \beta_N(M) < m_N(f). $$Where $\beta_N(M)$ is the $N$th Betti number of cell complex $M$ and $m_N(f)$ is the number of critical points of index $N$ of (discrete) Morse function $f$ on $M$.

Since each facet of $M$ can contribute at most one critical point to $f$. This corollary seems to imply the bound:

$$ \sum_{i=0}^d\beta_N(M) \leq \# \{\text{all facets of } M\} $$ However, this clearly involves a lot of over counting. So was wondering if one can find Morse function $f$ (in a sense of Furman) on $M$ such that on each cell $C_i$ the function has few critical points? To get $\sum_{N=1}^d m_N(f) \leq m K$, for some small $m$, perhaps $m=1$ and show (*), this is somewhat related to this math stackexchange question.

Second approach.

Alternatively, I was thinking that maybe just application of Mayer-Vietoris can yield $(*)$? But was not able to progress much. Not sure how to use convexity of each cell $C_i$ to get simple MV sequence.

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In fact, all what you need is a finite CW complex $X$ with the total number of cells equal to $K$. (Your polyhedral complex would be a very special case.) Let $c_i$ denote the number of $i$-dimensional cells. Then $H_i(X)$ is isomorphic to the cellular $i$-th homology group $$ H^{cw}_i(X)= Z^{cw}_i(X)/B^{cw}_i(X).$$ We have: $Z^{cw}_i(X)\cong {\mathbb Z}^{c_i}$, hence, $\beta_i(X)\le c_i$ and, thus, $$ \sum_{i} \beta_i(X)\le \sum_{i} c_i = K. $$
As a reference for CW complexes and cellular homology, see for instance, Hatcher's "Algebraic Topology"

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  • $\begingroup$ I'm not sure I understand the equality $\sum_i c_i =K$. I think we have by construction $c_d =K$, but for $i < d$ what is the bound on $c_i$? how can we bound $c_i$, i.e., the number of lower-dimensional cells? For example, if cell $C_i$ is a $d$-simplex, doesn't it also yield a lot of lower-dimensional simplices formed by faces of $C_i$? $\endgroup$ – them Dec 5 '19 at 2:52
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    $\begingroup$ @them: Oh, I see, I misunderstood your definition of $K$, I thought that it the total number of cells used to define $X$. Then my argument does not give the correct inequality. $\endgroup$ – Moishe Kohan Dec 5 '19 at 3:13

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