In 2005, prof. Emil Skoldberg developed a theory, similar to Forman's Discrete Morse Theory, but suited for arbitrary based chain complexes, in his Morse Theory from an algebraic viewpoint. I'm going through the paper and am having some difficulties. I'd be most grateful for an answer to my question 2 below.
Question 1: On p. 116, in the definition of a Morse matching, there is written:
We call a partial matching $M$ on the digraph $G_K$ a Morse matching if
for each edge $\alpha\to\beta\in M$ the corresponding component $d_{\beta,\alpha}$ is an
isomorphism, and furthermore, there is a well-founded partial order $\preceq$
on each $I_n$ such that $\alpha\succ\gamma$ whenever there is a path $\alpha^{(n)}\to\beta\to\gamma^{(n)}$ in $G^M_K$.
Is $\preceq$ defined by "exists a path $\alpha^{(n)}\to\beta\to\gamma^{(n)}$ in $G^M_K$", or is that just a necessary condition on $\preceq$? More precisely, the word "whenever" in the above quote, is that meant as $\Leftarrow$ or $\Leftrightarrow$?
Edit: Which definition is the right one (are all of them ok?): for $\alpha,\beta\in I_n$, we let:
- $\alpha\succeq\beta$ iff there exists a directed path in $G_K^M$ from $\alpha$ to $\beta$;
- $\alpha\succeq\beta$ iff there exists a directed path in $G_K^M$ from $\alpha$ to $\beta$
with vertices in $I_{n+1}\cup I_n$; - $\alpha\succeq\beta$ iff there exists a directed path in $G_K^M$ from $\alpha$ to $\beta$
with vertices in $I_n\cup I_{n-1}$;
Question 2: In the proof of Theorem 2 on p. 121. How do Lemmas 3 and 4 imply that for $x\in K_\alpha$ with $\alpha \in M_n^0$ there holds the equality $$\rho\pi(x)=x?$$ We have $\rho\pi(x)=\rho(x)-\rho\phi d(x)-\rho d\phi(x)$. Since $x \in C_n$ and $\rho$ is a projection, we have $\rho(x)=x$. By Lemma 3, we have $d\phi(x)= 0$. By Lemma 4, we have $\phi d(x) = \sum_{\beta\preceq\alpha}y_\beta=:(\ast)$ for some $y_\beta \in K_\beta$, but why is $(\ast)=0$ when $\alpha$ is critical?
Question 3: In Corollary 3, in the first sum, $\sigma$ ranges through $M^0_{n-1}$, right?
Question 4: If I understand correctly, the proof of Theorem 2 shows that if $\pi(K)$ has the induced boundary operator $d|_{\pi(K)}$ and $C$ has the operator $\tilde{d} := \rho(d-d\phi d) = \rho d \pi$, then the maps $\pi: C\longrightarrow \pi(C)=\pi(K)$ and $p: \pi(K)=\pi(C)\longrightarrow C$ are inverse to each other. Furthermore, $\pi\tilde{d} = \pi\rho(d-d\phi d) = d-d\phi d = d(\mathrm{id}-\phi d-d\phi) = d\phi$, so $\pi$ is a chain map. However, $\tilde{d}\rho = \rho(d-d\phi d)\rho = \rho d\rho-\rho d\phi d\rho \overset{???}{=} \rho d$.
Question 5: In general, there does not hold $\tilde{d}|_{\pi(K)}=d|_{\pi(K)}$, right?
Question 6: In the proof of Corollary 3, by Lemma 5 we have $\tilde{d}(x)$ $=$ $\rho(d-d\phi d)(x)$ $=$ $\rho(\sum_{\alpha\to\beta}d_{\beta\alpha}(x)-d\phi\sum_{\alpha\to\beta}d_{\beta\alpha}(x))$ $=$ $\rho\sum_{\alpha\to\beta}(d_{\beta\alpha}(x)-d\phi d_{\beta\alpha}(x))$ $=$ $\rho\sum_{\alpha\to\beta}(d_{\beta\alpha}(x)-d\sum_{\alpha'\in I_n,\gamma\in\Gamma_{\alpha',\beta}} m(\gamma)d_{\beta\alpha}(x))$. How do I continue to get $\rho\sum_{\sigma\in I_{n-1},\gamma\in\Gamma_{\sigma,\alpha}} m(\gamma)(x)$?
P.S. I might later add additional questions regarding p.116-122.
Best Answer
It's always nice to see people working on discrete Morse theory.
Answer 1
It is an "if and only if". Meaning: the partial order $\prec$ is defined by $\alpha \prec \gamma$ if and only if $\gamma$ precedes $\alpha$ in a path of the matching. The idea goes back to Forman's "Morse theory for cell complexes" where it is not explicitly stated as a partial order. I think the origins of the partial order are in Chari's reformulation of discrete Morse functions as acyclic matchings, see page 7 of "On discrete Morse functions and combinatorial decompositions".
Regarding your edited version of this question: the degrees of the vertices in the path don't really matter. If $\alpha, \beta$ are both in $I_n$ then $\alpha \succeq \beta$ if there is an alternating path from $\alpha$ to $\beta$ of either the type $(n,n-1,\ldots,n)$ or the type $(n,n+1,\ldots,n)$: in either case, you have $\alpha \succeq \beta$. In the first case, you will have $\alpha, \beta \in M^- \cup M^0$ and in the second case you will have $\alpha, \beta \in M^+ \cup M^0$. In general, this partial order might relate $\alpha$ and $\beta$ lying across different $I_n$'s too.
Answer 2
As you noted, $\phi(dx_\alpha) = \sum_{\beta \prec \alpha}y_\beta$ and your question asks why this quantity should be trivial. In fact, it is not trivial in general, but its image under $\rho$ is trivial (which is all you need): by definition the image of $\phi$ is never critical, so $\rho\circ\phi$ is trivial.
Answer 3
Yes, $\sigma$ ranges over all critical cells of dimension $(n-1)$. This is a common problem with notation: I suspect that the author intended to specify the multiplicity $m(\gamma:\sigma\to\alpha)$ when writing $m(\gamma)$ where $\gamma$ is a path of the matching. In this case, one should define $m(\gamma) = 0$ whenever the source cell $\sigma$ is non-critical, which allows you to sum over arbitrary cells.
Answer 4
There is a typo at the end of your calculation: one has $\pi\tilde{d} = d\pi$ instead of $d\phi$. If you already accept that $\rho$ and $\pi$ are inverses, then there is really no trouble: $$\tilde{d}\rho = (\rho d \pi) \rho = \rho d(\pi \rho) = \rho d $$ Maybe you mean to ask some other question here: even the proof that $\pi$ is a chain map (which you worked out in your question) follows easily by a similar argument if you accept that $\rho$ is its inverse.
Answer 5
You are right, this is not true in general: the image of $\tilde{d}$ must be critical (that is, in the span of $M^0$) whereas there is no such requirement on the image of $d$ restricted to $\pi(K)$.
Answer 6
The result follows from recursive application of Lemma 5. Instead the calculation that you have performed so far, look at $(d - d\phi d)x$ as $(1 - d \phi) (dx)$ and write $dx$ out as a linear combination of $(n-1)$-dimensional cells. By Lemma 3, only the cells from $M^-$ will contribute. Now apply Lemma 5 to each piece of this combination...