Algebraic Morse theory - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-20T11:22:27Z http://mathoverflow.net/feeds/question/118946 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/118946/algebraic-morse-theory Algebraic Morse theory Leon Lampret 2013-01-15T06:35:22Z 2013-02-07T06:34:06Z <p>In 2005, prof. Emil Skoldberg developed a theory, similar to Forman's Discrete Morse Theory, but suited for arbitrary based chain complexes, in his <a href="http://www.maths.ed.ac.uk/~aar/papers/skoldberg.pdf" rel="nofollow"><em>Morse Theory from an algebraic viewpoint</em></a>. I'm going through the paper and am having some difficulties. I'd be most grateful for an answer to my question 2 below.</p> <p><strong>Question 1:</strong> On p. 116, in the definition of a <em>Morse matching</em>, there is written:</p> <blockquote> <p>We call a partial matching $M$ on the digraph $G_K$ a <em>Morse matching</em> 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 $\alpha\succ\gamma$ <strong>whenever</strong> there is a path $\alpha^{(n)}\to\beta\to\gamma^{(n)}$ in $G^M_K$.</p> </blockquote> <p>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 $\Leftrightarrow$?</p> <p><strong>Edit:</strong> Which definition is the right one (are all of them ok?): for $\alpha,\beta\in I_n$, we let:</p> <ol> <li>$\alpha\succeq\beta$ iff there exists a directed path in $G_K^M$ from $\alpha$ to $\beta$;</li> <li>$\alpha\succeq\beta$ iff there exists a directed path in $G_K^M$ from $\alpha$ to $\beta$ with vertices in $I_{n+1}\cup I_n$;</li> <li>$\alpha\succeq\beta$ iff there exists a directed path in $G_K^M$ from $\alpha$ to $\beta$ with vertices in $I_n\cup I_{n-1}$;</li> </ol> <p><strong>Question 2:</strong> 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?</p> <p><strong>Question 3:</strong> In Corollary 3, in the first sum, $\sigma$ ranges through $M^0_{n-1}$, right?</p> <p><strong>Question 4:</strong> 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, $\tilde{d}\rho = \rho(d-d\phi d)\rho = \rho d\rho-\rho d\phi d\rho \overset{???}{=} \rho d$.</p> <p><strong>Question 5:</strong> In general, there does not hold $\tilde{d}|<em>{\pi(K)}=d|</em>{\pi(K)}$, right?</p> <p><strong>Question 6:</strong> 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> <p>P.S. I might later add additional questions regarding p.116-122.</p> http://mathoverflow.net/questions/118946/algebraic-morse-theory/119014#119014 Answer by Vidit Nanda for Algebraic Morse theory Vidit Nanda 2013-01-15T18:53:59Z 2013-02-07T06:34:06Z <p>It's always nice to see people working on discrete Morse theory.</p> <p><strong>Answer 1</strong></p> <p>It is an "if and only if". Meaning: the partial order $\prec$ is <em>defined</em> by $\alpha \prec \gamma$ if and only if $\gamma$ precedes $\alpha$ in a path of the matching. The idea goes back to Forman's "<a href="https://drona.csa.iisc.ernet.in/~vijayn/courses/TopoForVis/papers/FormanDiscreteMorseTheory.pdf" rel="nofollow">Morse theory for cell complexes</a>" 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 "<a href="http://www.math.ucdavis.edu/~deloera/MISC/BIBLIOTECA/trunk/ChariManoj/Chari3.pdf" rel="nofollow">On discrete Morse functions and combinatorial decompositions</a>". </p> <p>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.</p> <p><strong>Answer 2</strong></p> <p>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. </p> <p><strong>Answer 3</strong></p> <p>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.</p> <p><strong>Answer 4</strong></p> <p>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 <em>$\rho$ and $\pi$ are inverses</em>, 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.</p> <p><strong>Answer 5</strong></p> <p>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 $\pi(K)$.</p> <p><strong>Answer 6</strong></p> <p>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...</p> http://mathoverflow.net/questions/118946/algebraic-morse-theory/119461#119461 Answer by Ronnie Brown for Algebraic Morse theory Ronnie Brown 2013-01-21T12:02:55Z 2013-01-21T12:02:55Z <p>It may be useful to note in relation to the work of Forman that his notion of discrete vector field on a complex is equivalent to the notion of marked cone complex in David W. Jones' 1984 PhD thesis on Poly T-complexes available from <a href="http://pages.bangor.ac.uk/~mas010/doctorates.html" rel="nofollow">here</a>, also published as "A general theory of polyhedral sets and the corresponding $T$-complexes" <em>Dissertationes Math. (Rozprawy Mat.)</em> 266 (1988) 110.</p> <p>So it would be interesting if this is related also to Skoldberg's work. </p>