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I just started learning about discrete Morse Theory and I got stuck on a corollary that in the book I'm reading is described as simply following from a lemma.

Denote by $P$ an almost linear metric space with distinguished point p. We say that a nonempty fininte metric space $P$ is almost linear if there is a point $p \in P$ (distinguished point) such that $P\setminus \{p\}$ is isometric to a finite subset of $\mathbb{R}$. We consider the discrete gradient vector field $X$, defined as follows on the Vietoris-Rips complex $\mathcal{V}(P)_r$. We use the convention that $p$ is the smallest point and for a simplex $[a_1,\ldots,a_j]$: $a_1<\ldots<a_j$.

Definition of $X$: If a simplex $\sigma = [a_1,a_2,\ldots,a_j]$ in $\mathcal{V}(P)_r$ has a coface $a_0 \cup \sigma:= [a_0,a_1,a_2,\ldots,a_j]$ with $a_0 < a_1$, then $X$ matches $\sigma$ to $a_0 \cup \sigma$ with $a_0$ as small as possible. $X$ matches no other simplices.

One can show that for $j \geq 2$, a simplex $[a_1,\ldots,a_j]$ is critical in $X$ if and only if the following conditions are satisfied:

  • $a_1 \neq p$,
  • $[q,a_1,\ldots,a_j] \notin \mathcal{V}(P)_r$ for any $q < a_1$,
  • $[p,a_2,a_3,\ldots,a_j] \in \mathcal{V}(P)_r$.

Now I want to show, using the above, that if $[a_1,a_2,a_3,\ldots,a_j]$ is a critical simplex in $X$, and under the assumption that there does not exist a vertex $b$ such that $[p,b] \in \mathcal{V}(P)_r$ and $a_1 < b < a_2$, then the simplex $[a_1,a_3,\ldots,a_j]$ is critical in $X$.

The first item follows directly by definition and so does the third since $[p,a_3,\ldots,a_j]$ is a subsimplex of $[p,a_2,a_3,\ldots,a_j]$. However, I struggle with showing that the second item is satisfied, as well. Moreover, I do not see where the "b-condition" comes into play. I tried to perform a proof by contradiction but I did not succeed.

I really appreciate any hint on how I could prove that the second item is satisfied by $[a_1,a_3,\ldots,a_j]$!

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  • $\begingroup$ Appreciated the link to the reference. I think you should probably include the definition $p < a$ for $a\in P-\{p\}$ in the question. $\endgroup$
    – user176372
    Commented Jun 25 at 22:40
  • $\begingroup$ I adapted it, thanks. $\endgroup$
    – moschops
    Commented Jun 26 at 6:49

1 Answer 1

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You're looking to prove that if $q$ violates condition (2) for $[a_1,a_3,\dots,a_j]$ then it also violates (2) for $[a_1,a_2,\dots,a_j]$, which reduces to showing that $d(q,a_2)\leq 2r$.

If $q$ is not the distinguished point $p$, then $2r\geq d(q,a_3) = d(q,a_2)+d(a_2,a_3)$, with the last equality following because the metric space is almost linear.

So the case you need to check more carefully is $q=p$, the distinguished point.

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  • $\begingroup$ Thank you @user176372! The "b-condition"-part makes sense. By writing showing that $q$ violates condition (2) for $[a_1,a_2,\ldots,a_j]$ reduces to showing $d(q,a_1) \leq r$, you probably mean that it reduces to showing $d(q,a_2) \leq r$, right (since $d(a_1,q)$ is given by assumption)? After having showed that $d(q,a_2) = d_\mathbb{R}(q,a_2)$, using your hint, in order to derive a contradiction, I now want to show that $d(q,a_2) \leq r$. My idea was juggling around with triangle inequalities, but this did not work and I do not see how I could deduce it based on the assumptions made. $\endgroup$
    – moschops
    Commented Jun 25 at 9:54
  • $\begingroup$ @moschops looks right, I'll edit when I'm at a computer. In that case, $r\geq d(q,a_3) = d(q,a_2) + d(a_2,a_3)$ probably works. The points on the line don't just satisfy a triangle inequality, they satisfy a triangle equality. $\endgroup$
    – user176372
    Commented Jun 25 at 15:09
  • $\begingroup$ This is what I first thought as well, however, I guess we might have $d(q,a_3)=d(q,p)+d(p,a_3)<d(q,a_2)+d(a_2,a_3)$... Or am I missing something? $\endgroup$
    – moschops
    Commented Jun 25 at 15:16
  • $\begingroup$ @moschops let me come back to this. If the definition of a nearly linear space is that the distances between the line points are precisely given by real line distances, then I am over thinking it. In that case, if $q$ is not the distinguished point, then the above equality and assumptions gets you the conclusion without your "b-condition". What book is this from, if you don't mind? $\endgroup$
    – user176372
    Commented Jun 25 at 15:27
  • $\begingroup$ I am talking about this paper arxiv.org/pdf/1804.01398, p. 24, Def. The Discrete Gradient Vector Field Y. I see, all points besides one can be embedded into the real line but the space can not necessarily be embedded into $\mathbb{R}^2$. $\endgroup$
    – moschops
    Commented Jun 25 at 15:42

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