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*](http://www.maths.ed.ac.uk/%7Eaar/papers/skoldberg.pdf). 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:


1. $\alpha\succeq\beta$ iff there exists a directed path in $G\_K^M$ from $\alpha$ to $\beta$;
2. $\alpha\succeq\beta$ iff there exists a directed path in $G\_K^M$ from $\alpha$ to $\beta$
with vertices in $I\_{n+1}\cup I\_n$;
3. $\alpha\succeq\beta$ 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.



Algebraic Morse theory
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](https://drona.csa.iisc.ernet.in/~vijayn/courses/TopoForVis/papers/FormanDiscreteMorseTheory.pdf)" 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](http://www.math.ucdavis.edu/~deloera/MISC/BIBLIOTECA/trunk/ChariManoj/Chari3.pdf)". 


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...


