User leon lampret - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T16:38:55Z http://mathoverflow.net/feeds/user/11317 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/127293/stanley-reisner-ring-of-a-simplicial-complex-is-a-functor Stanley-Reisner ring of a simplicial complex is a functor? Leon Lampret 2013-04-11T23:17:52Z 2013-04-14T13:42:35Z <p>Let $K$ bea field and $[n]={1,\ldots,n}$ and $K[x]=K[x_1,\ldots,x_n]$. For $\sigma={i_1,\ldots,i_k}\subseteq [n]$, denote $x_\sigma=x_{i_1}\cdots x_{i_k}=\prod_{i\in\sigma}x_i\in K[x]$. Let $\Delta$ and $\Delta'$ be (abstract finite) simplicial complexes on $[n]$ and $[n']$ respectively. Let $f: \Delta\rightarrow \Delta'$ be a simplicial map, i.e. a map $f:[n]\rightarrow[n']$ for which $\forall \sigma\in\Delta: f(\sigma)\in\Delta'$ holds. </p> <p>The <em>Stanley-Reisner ideal</em> of $\Delta$ is $I_\Delta:=\langle\langle x_\sigma;\; [n]\supseteq\sigma\notin\Delta\rangle\rangle$, and <em>Stanley-Reisner ring</em> of $\Delta$ is $$K[\Delta]:=K[x]/I_\Delta= K[x_1,\ldots,x_n \:|\: x_\sigma;\; [n]\supseteq\sigma\notin\Delta].$$</p> <p>The <em>coordinate ring</em> of an affine variety $A\subseteq\mathbb{A}^n_K$ is $K[A]=K[x]/I_A$, where $I_A$ is the vanishing ideal ${f\in K[x]; f|_A=0}$. Thus the notation is the same as that of the Stanley-Reisner ring. Since the coordinate ring $K[-]$ is a functor from the category of affine varieties over an algebraically closed field to the category of finitely generated commutative unital reduced $K$-algebras, defined for $f: A\rightarrow A'$ by $K[f]: K[A']\rightarrow K[A]$ that sends $\alpha:A'\rightarrow K$ to $\alpha\circ f: A\rightarrow K$. This is in fact an antiequivalence of categories.</p> <p>The immediate impulse is to try to make the Stanley-Reisner ring an antiequivalence of categories. But I'm having trouble making it even a functor. I didn't find anything in the literature (Bruns &amp; Herzog, Stanley, Herzog &amp; Hibi, Miller &amp; Sturmfels) regarding functors.</p> <p><strong>1st try:</strong> We let $K[f]: K[\Delta]\rightarrow K[\Delta']$ send $x_i\mapsto x_{f(i)}$. But then for $\sigma\notin\Delta$, this map sends $x_\sigma\mapsto x_{f(\sigma)}$ (here $f(\sigma)$ is considered as a multiset, i.e. if $f(i)=f(j)$, then $x_{f(\sigma)}$ contains both $x_{f(i)}$ and $x_{f(j)}$), and we do not necessarily have $f(\sigma)\notin\Delta'$, meaning that $K[f]$ is not well defined on the quotient of $K[x_1,\ldots,x_n]$.</p> <p><strong>2nd try:</strong> We let $K[f]: K[\Delta']\rightarrow K[\Delta]$ send $x_{i'}\mapsto x_{f^{-1}(i')}$, where $f^{-1}$ denotes the preimage. Then for $\sigma'\notin\Delta'$, this map sends $x_{\sigma'}\mapsto x_{f^{-1}(\sigma')}$. We wish to have $f^{-1}(\sigma')\notin\Delta$. If this does not hold, $f^{-1}(\sigma')\in\Delta$, then $\sigma'\supseteq f(f^{-1}(\sigma'))\in\Delta'$, so I don't get any contradiction, and we don't have well-definedness. </p> <p><strong>Question:</strong> How can the Stanle-Reisner ring be made into a functor (preferrably in a way that it becomes an antiequivalence between the category of simplicial complexes and category of finitely generated commutative unital monomially related reduced $K$-algebras)?</p> <p><strong>Question:</strong> Can the Stanley-Reisner ideal be seen as an ideal of polynomial functions $\alpha: \Delta\rightarrow K$?</p> 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/105664/morse-theory-and-homology-of-an-algebraic-surface-example Morse theory and homology of an algebraic surface (example) Leon Lampret 2012-08-28T00:21:10Z 2012-08-30T14:29:20Z <p>Let $T_n$ denote the $n$-th <a href="http://en.wikipedia.org/wiki/Chebyshev_polynomial" rel="nofollow"><em>Chebyshev polynomial</em></a> and define $$f_n(x,y,z):=T_n(x)+T_n(y)+T_n(z)\;\;\;\text{ and}$$ $$Z_n:=\mathcal{Z}(f_n) \subseteq \mathbb{R}^3,$$ the <a href="http://mathworld.wolfram.com/ChmutovSurface.html" rel="nofollow"><em>Banchoff-Chmutov surface</em></a>, where in general, $\mathcal{Z}(f_1,\ldots,f_k)$ denotes the zero set of polynomials $f_1,\ldots,f_k$, i.e. ${(x,y,z) \in \mathbb{R}^3; f_1(x,y,z)=\ldots=f_k(x,y,z)=0}$.</p> <p>Let us prove, that this is a surface. By the implicit function theorem, it suffices to prove that the points, where $[D_x{f_n},D_y{f_n},D_z{f_n}]$ is zero, do not lie in $Z_n$ (here $D_x$ is just the partial derivative). This is quivalent to showing that the set $$\mathcal{Z}(f_n,D_xf_n,D_yf_n,D_zf_n)=\mathcal{Z}(T_n(x) + T_n(y) + T_n(z),D_xT_n(x),D_yT_n(y),D_zT_n(z))$$ is empty. This can be done by using (from wiki page) $D_xT_n(x) = nU_{n-1}(x)$ and Pell's equation $T_n(x)^2 - (x^2 - 1)U_{n-1}(x)^2 = 1$, to obtain $\mathcal{Z}(1 + 1 + 1) = \emptyset$.</p> <p>Let us observe the height function $Z_n \rightarrow \mathbb{R}$, $(x,y,z) \mapsto ax + by + cz = [a,b,c][x,y,z]^t$. It is linear, so its derivative is $[a,b,c] :T_pZ_n \rightarrow T_p\mathbb{R} = \mathbb{R}$. Its critical points are therefore those, where the tangent plane $T_pZ_n$ has normal $[a,b,c]$. But the tangent plane of $\mathcal{Z}(f)$ always has normal $[D_xf,D_yf,D_zf]$. Thus the critical points of our height function are those $x,y,z$ where $[D_xf_n,D_yf_n,D_zf_n]=[a,b,c]$, i.e. the critical points are $$\mathcal{Z}(f_n,T_n(x) - a,T_n(y) - b,T_n(z) - c).$$ Now I don't know how to check if these critical points are nondegenerate. I don't even have local parametrizations to work with. </p> <p><strong>Question:</strong> Can one calculate the homology $H_\ast(Z_n)$ by using the elementary methods from Morse theory (i.e. structural theorem, handle decomposition, Morse inequalities, Morse complex)?</p> http://mathoverflow.net/questions/101204/order-density-of-smooth-functions-among-continuous-functions Order density of smooth functions among continuous functions? Leon Lampret 2012-07-03T04:07:20Z 2012-07-03T04:49:32Z <p>Let $\mathcal{C}^0([a,b],\mathbb{R})$ be the space of all continuous functions $f:[a,b]\rightarrow\mathbb{R}$ and $\mathcal{C}^\infty([a,b],\mathbb{R})$ the subspace of all smooth functions. Define $f\leq g:\Leftrightarrow(\forall x: f(x)\leq g(x))$ and $f\ll g:\Leftrightarrow(\forall x: f(x)&lt; g(x))$. Is the following true:</p> <p>$$\forall f,h\in\mathcal{C}^0([a,b],\mathbb{R}) \;\exists g\in\mathcal{C}^\infty([a,b],\mathbb{R}): f\ll h\Rightarrow f\leq g\leq h,$$ i.e. between any two continuous functions $f\ll h:[a,b]\rightarrow\mathbb{R}$, there exists a smooth function $g$?</p> <p>Does this yield that $(\mathcal{C}^r([a,b],\mathbb{R}),\leq)$ is not a lattice, for any $r\in\mathbb{N}$ (I'd like to construct a function $g$ between $|x|$ and supposed $x\vee-x$).</p> <p>I haven't found this result anywhere in my calculus books, so I'd appreciate any references.</p> http://mathoverflow.net/questions/50382/how-to-triangulate-real-projective-spaces-as-simplicial-complexes-in-mathematica/54729#54729 Answer by Leon Lampret for How to triangulate real projective spaces (as simplicial complexes in Mathematica)? Leon Lampret 2011-02-08T05:08:40Z 2011-02-08T05:08:40Z <p>I've managed to find my mistake and here is the final (if I'm not mistaken) correct solution.</p> <pre><code>ClearAll[faces, skeleton, subseteq, subset, sBarycentricSubdivision, \ scBarycentricSubdivision, sphere, projectiveSpace] faces[s_, k_] := Reverse@Subsets[s, {k + 1}] skeleton[sc_, k_] := Union[Sort /@ Flatten[faces[#, k] &amp; /@ sc, 1]] subseteq[{},A_List] := True subseteq[{x_,y___}, A_List] := MemberQ[A, x] &amp;&amp; subset[{y}, A] subseteq[A_List,B_List, C__List] := subset[A,B] &amp;&amp; subset[B,C] subset[A_List, B_List] := subseteq[A, B] &amp;&amp; Complement[B, A]!={} subset[A_List, B_List, C__List] := subset[A, B] &amp;&amp; subset[B, C] sBarycentricSubdivision[{x_Integer}] := {{{x}}} sBarycentricSubdivision[s_List] := Subsets[s, {1, Length@s}] //Subsets[#,{Length@s}]&amp; // Select[#,subset@@#&amp;]&amp; scBarycentricSubdivision[sc_List] := Flatten[sBarycentricSubdivision /@ sc, 1] sphere[n_Integer] := skeleton[{Range[n+2]},n] projectiveSpace[n_Integer] := Module[ {sc1, vertices1, k, identification, sc2, vertices2, sc3}, sc1 = scBarycentricSubdivision[sphere[n]]; vertices1 = Sort@DeleteDuplicates@ Flatten[sc1, 1]; (*the vertices (lists) are ordered antipodally with respect \ to complement, ie. vertices1[i]={1,...,n+2}-vertices1[-i]*) k = Length@vertices1; identification = Table[vertices1[[-i]] -&gt; vertices1[[i]], {i, 1, k/2}]; sc2 = DeleteDuplicates[Sort /@ (sc1/.identification)]; vertices2 = Flatten[sc2, 1] //Sort //DeleteDuplicates; sc3 = Map[Position[vertices2, #][[1, 1]] &amp;, sc2, {2}] //Sort // DeleteDuplicates; sc3 ] </code></pre> <p>And now, the command projectiveSpace[2] indeed returns a triangulation of the projective plane with 7 vertices and 12 faces. Also, the homology groups of $RP^0$,..., $RP^3$ all seem to be correct.</p> <p>Thank you, mr. Palmieri, greetings from Slovenia.</p> http://mathoverflow.net/questions/50382/how-to-triangulate-real-projective-spaces-as-simplicial-complexes-in-mathematica How to triangulate real projective spaces (as simplicial complexes in Mathematica)? Leon Lampret 2010-12-25T23:24:18Z 2011-02-08T05:08:40Z <p>Hello!</p> <p>I have written a program in Mathematica 7, which calculates for a (finite abstract) simplicial complex all its homology groups. I would really like to test it on the projective spaces, but cannot find a way to triangulate them. There seem to be very few articles on this matter and none of them states directly how the actual triangulations might be achieved, so I turn to MO for help.</p> <p><strong>How can the real projective spaces $$RP^n \approx B^n/_{x\:\sim-x;\;\; x\in\partial B^n}$$ be triangulated as simplicial complexes?</strong> A simplicial complex is presented as a list of simplices that aren't a face of any bigger simplex (the "main simplices"). Each k-simplex is just an ordered list of some k+1 integers. </p> <p>For example, {0,1,2} is a 2-simplex, as is {1,3,15}, etc. Examples of simplicial complexes: {{0,...,n}} is a n-dimensional ball, {{1,2},{1,3},{2,3}} is an empty triangle, skeleton[{Range[n+2]},n] is a n-sphere, etc.</p> <p>I have already found a concrete triangulation for the real projective plane, but nothing more general.</p> <p>It would also be <strong>appreciated, that the actual triangulation is reasonably small</strong> (not necessarily minimal), so that the program calculates homology groups fast enough. Also, <strong>answers in code / pseudocode are much desirable</strong> (projectiveSpace[n_]:=???).</p> <p>P.S. I have already written some functions, which can be used (sc...simplicial complex):</p> <ol> <li>skeleton[sc,k] ...list of all k-faces of all simplices</li> <li>sum[sc1,sc2]...topological sum (disjoint union)</li> <li>connected sum[sc1,sc2] ...removes two k-simplicices and adds a tunnel between them</li> <li>product[sc1,sc2]...staircase triangulation of a product (reorders the vertices)</li> <li>cone[sc]</li> <li>suspension[sc]</li> </ol> http://mathoverflow.net/questions/54653/algorithm-that-decreases-the-size-of-the-simplicial-triangulation Algorithm that decreases the size of the simplicial triangulation Leon Lampret 2011-02-07T15:58:25Z 2011-02-07T19:50:34Z <p>Hello!</p> <p>Let X be a topological space. We are considering only abstract simplicial complexes, i.e. a finite list of finite lists of integers. <strong>Is there any algorithm (more or less efficient?), that would for a given triangulation of X produce another SMALLER simplicial complex, that is also homeomorphic to X?</strong> </p> <p>My main need for such an algorithm is coming from the desire to implement a more efficient algorithm for computing <strong>homology</strong> of a space.</p> <p>The algorithm doesn't need to produce a minimal triangulation (not at all), but it has to be fast enough that it can be used in my homology program (faster than the integer smith normal form algorithm).</p> <p>Currently, the f vector of my triangulation of e.g. the 4-torus is {81, 1215, 4050, 4860, 1944}, and of the 5-torus is {243, 7533, 43740, 94770, 87480, 29160}, which is ridiculously large (I used the staircase triangulation of the product to construct torii). So there is a great need to decrease size of the triangulation without losing topological information.</p> <p>I have searched the literature and ended up empty handed...</p> <p>P.S. code in Mathematica would be ideal, but I'd be happy with (pseudo)code of any sort.</p> http://mathoverflow.net/questions/50382/how-to-triangulate-real-projective-spaces-as-simplicial-complexes-in-mathematica/50653#50653 Answer by Leon Lampret for How to triangulate real projective spaces (as simplicial complexes in Mathematica)? Leon Lampret 2010-12-29T13:13:20Z 2010-12-29T13:13:20Z <p>OK, I've managed to write the whole thing, but it doesn't work quite right. Here is the code (s...simplex, sc...simplicial complex):</p> <pre><code>ClearAll[faces, skeleton, subseteq, subset, barycentricSubdivisionOfSimplex, barycentricSubdivision, sphere, projectiveSpace] faces[s_, k_] := Reverse@Subsets[s, {k + 1}] skeleton[sc_, k_] := Union[Sort /@ Flatten[faces[#, k] &amp; /@ sc, 1]] subseteq[{}, A_List] := True subseteq[{x_, y___}, A_List] := MemberQ[A, x] &amp;&amp; subset[{y}, A] subseteq[A_List, B_List, C__List] := subset[A, B] &amp;&amp; subset[B, C] subset[A_List, B_List] := subseteq[A, B] &amp;&amp; Complement[B, A] != {} subset[A_List, B_List, C__List] := subset[A, B] &amp;&amp; subset[B, C] barycentricSubdivisionOfSimplex[{x_Integer}] := {{{x}}} barycentricSubdivisionOfSimplex[s_List] := Subsets[s, {1, Length@s}] // Subsets[#, {Length@s}] &amp; // Select[#, subset @@ # &amp;] &amp; barycentricSubdivision[sc_List] := Flatten[barycentricSubdivisionOfSimplex /@ sc, 1] sphere[n_Integer] := skeleton[{Range[n + 2]}, n] projectiveSpace[n_Integer] := Module[{sc1, sc2, vertices, sc3}, sc1 = barycentricSubdivision[sphere[n]]; sc2 = Map[ Which[Length[#] &lt;= Quotient[n + 2, 2], #, True, Complement[Range[n + 2], #]] &amp;, sc1, {2}]; vertices = Flatten[sc2, 1] // Sort // DeleteDuplicates; sc3 = Sort /@ Map[Position[vertices, #][[1, 1]] &amp;, sc2, {2}] // Sort // DeleteDuplicates; Which[n == 0, {{1}}, True, sc3] ] </code></pre> <p>I basically followed the instructions of mr. Palmieri. The problem is, that the homology of my projectiveSpace[n] for n=0,...,4 returns </p> <pre><code>{{1}, {{}}} {{1, 1}, {{}, {}}} {{1, 0, 3}, {{}, {2}, {}}} {{1, 0, 0, 1}, {{}, {2}, {}, {}}} {{1, 0, 0, 0, 10}, {{}, {2}, {}, {2}, {}}} </code></pre> <p>which is obviously incorrect for n=2,4 (the first list represents the betti numbers, the second the torsion coefficients). However, quite a few betti numbers and torsion coefficients seem to be correct, so I'm guessing the error is minor. </p> <p>Any help would be much appreciated.</p> http://mathoverflow.net/questions/44326/most-memorable-titles/48323#48323 Answer by Leon Lampret for Most memorable titles Leon Lampret 2010-12-04T23:05:03Z 2010-12-04T23:05:03Z <p><a href="http://www.amazon.com/Mathematical-Fallacies-Flaws-Flimflam-Spectrum/dp/0883855291" rel="nofollow"><strong>Mathematical Fallacies, Flaws and Flimflam</strong></a> was definitely by far the most memorable title I have ever read. Also <strong>A Taste of Topology</strong> seemed tasty. </p> <p>But I would also like to stress, that to me, the books that have the most 'classical' and 'general' titles, seem the most appealing. Eg. </p> <ul> <li><a href="http://www.amazon.com/Graph-Theory-Graduate-Texts-Mathematics/dp/1849966907/ref=sr_1_2?s=books&amp;ie=UTF8&amp;qid=1291503199&amp;sr=1-2" rel="nofollow"><strong>Graph Theory</strong></a>, </li> <li><a href="http://www.amazon.com/Algebraic-Topology-Allen-Hatcher/dp/0521795400/ref=sr_1_1?s=books&amp;ie=UTF8&amp;qid=1291503414&amp;sr=1-1" rel="nofollow"><strong>Algebraic Topology</strong></a>, </li> <li><a href="http://www.amazon.com/Algebraic-Geometry-Graduate-Texts-Mathematics/dp/1441928073/ref=sr_1_1?s=books&amp;ie=UTF8&amp;qid=1291503472&amp;sr=1-1" rel="nofollow"><strong>Algebraic Geometry</strong></a>,</li> <li><a href="http://www.amazon.com/Abstract-Algebra-Graduate-Texts-Mathematics/dp/1441924507/ref=sr_1_1?ie=UTF8&amp;s=books&amp;qid=1291503684&amp;sr=1-1" rel="nofollow"><strong>Abstract Algebra</strong></a>,</li> </ul> <p>etc.</p> http://mathoverflow.net/questions/127293/stanley-reisner-ring-of-a-simplicial-complex-is-a-functor/127302#127302 Comment by Leon Lampret Leon Lampret 2013-04-14T02:22:48Z 2013-04-14T02:22:48Z What is $X_\Delta$? What is $N$ and $F^N$? http://mathoverflow.net/questions/118946/algebraic-morse-theory/119014#119014 Comment by Leon Lampret Leon Lampret 2013-03-19T02:37:10Z 2013-03-19T02:37:10Z @Vel Nias: It wasn't until now that I managed to go through the article and your answer. Thank you, much obliged! http://mathoverflow.net/questions/118946/algebraic-morse-theory/119014#119014 Comment by Leon Lampret Leon Lampret 2013-01-16T22:30:04Z 2013-01-16T22:30:04Z Thank you very much, this helped me a lot. I'm aware that there is an analogous theory by Jollenbeck and Welker, though the fact that the results are stated in lesser generality has put me off from reading it (though I will, in the near future). Why does Jollenbeck&amp;Welker, <i>Minimal Resolutions via Algebraic Discrete Morse Theory</i>, assume that all modules are free of rank $1$? Why does Kozlov, <i>Combinatorial Algebraic Topology</i> &#167;11.3, assume that $R$ is commutative and modules are free of finite rank? Judging by the results from Skoldberg's article, those are all unnecessary assumptions. http://mathoverflow.net/questions/118946/algebraic-morse-theory Comment by Leon Lampret Leon Lampret 2013-01-16T22:18:00Z 2013-01-16T22:18:00Z @arsmath: I first asked on math.stackexchange, with no intention of asking here, but then there were no responses, and furthermore, there were votes to close the question down, so I became pessimistic about getting an answer there, and I asked here. Besides, each question has a link to the other, so I don't see where the problem is. http://mathoverflow.net/questions/105664/morse-theory-and-homology-of-an-algebraic-surface-example/105706#105706 Comment by Leon Lampret Leon Lampret 2012-11-16T00:00:53Z 2012-11-16T00:00:53Z Yes, now I understand it. Thank you for your illustrative and elegant solution. It's somewhat surprising how nicely all calculations go through. I imagine that, given a different family of polynomials instead of $T_n$, we would have much more problems. http://mathoverflow.net/questions/105664/morse-theory-and-homology-of-an-algebraic-surface-example/105706#105706 Comment by Leon Lampret Leon Lampret 2012-09-10T00:21:55Z 2012-09-10T00:21:55Z Furthermore, you claim that $Z_n\!\cap\!\{z\!=\!\zeta\}=\{T_n(x)\!+\!T_n(y)\!=\!0\}$, but don't we have $Z_n\!\cap\!\{z\!=\!\zeta\}=\{T_n(x)\!+\!T_n(y)\!=\!-T_n(\zeta)\}$? http://mathoverflow.net/questions/105664/morse-theory-and-homology-of-an-algebraic-surface-example/105706#105706 Comment by Leon Lampret Leon Lampret 2012-09-10T00:20:28Z 2012-09-10T00:20:28Z @Liviu Nicolaescu: Regarding the connectivity of $Z_n$, why does it suffice to prove that the level set $Z_n\!\cap\!\{z\!=\!\zeta_i\}$ is connected for every $\zeta_i\!=\!h(\text{saddle point})$? For example, if we have a disjoint union of an upright torus and a sphere, and if the sphere is small enough that both saddle points of the torus are above the sphere and the bottom of the torus is below the bottom of the sphere, then both level surfaces of corresponding to saddle points are connected (they are the wedge of two circles), yet the surface is not connected. http://mathoverflow.net/questions/105664/morse-theory-and-homology-of-an-algebraic-surface-example/105706#105706 Comment by Leon Lampret Leon Lampret 2012-08-30T01:22:35Z 2012-08-30T01:22:35Z Yes, now it agrees with my intuition. Thank you for your answer! http://mathoverflow.net/questions/105664/morse-theory-and-homology-of-an-algebraic-surface-example/105706#105706 Comment by Leon Lampret Leon Lampret 2012-08-29T16:17:46Z 2012-08-29T16:17:46Z Umm, $Z_4$ has just one hole drilled in each of the three directions, and hopefully, we agree that it has genus $5$, hence $\chi=-8$. However, $n^2-\frac{n^3}{4}\neq-8$ for $n=4$. Something must be wrong. http://mathoverflow.net/questions/105664/morse-theory-and-homology-of-an-algebraic-surface-example/105706#105706 Comment by Leon Lampret Leon Lampret 2012-08-29T13:58:48Z 2012-08-29T13:58:48Z Hmm, judging from the pictures, if I'm not mistaken, for $n=2(k+1)$, we have a cube (sphere) with $k^2$ holes drilled in each direction $x,y,z$, so I would say that $Z_n$ has genus $k^2+k^2(k+1)+k^2(k+1)=k^2(2k+3)$, for$k\in\mathbb{N}$. For $n=2,4,6,8$, this gives genus $0,5,28,81$. Thus the Euler characteristic is $\chi=2-2g=2-2k^2(2k+3)$, i.e. for $n=2,4,6,8$, we get $2,-8,-54,-160$. But $n^2-\frac{n^3}{4}$ gives $2,0,-18,-64,-150$. Where did I go wrong? http://mathoverflow.net/questions/101204/order-density-of-smooth-functions-among-continuous-functions/101205#101205 Comment by Leon Lampret Leon Lampret 2012-07-03T05:03:09Z 2012-07-03T05:03:09Z Thank you very much for everything, much appreciated. I never had much dealings with bump functions, but now they seem quite useful. http://mathoverflow.net/questions/101204/order-density-of-smooth-functions-among-continuous-functions/101205#101205 Comment by Leon Lampret Leon Lampret 2012-07-03T04:55:07Z 2012-07-03T04:55:07Z Yes, you are right, bump functions are what I really needed. Thank you! May I humbly ask for just one vote up, because I currently can't ask questions (as comments) in other threads? http://mathoverflow.net/questions/101204/order-density-of-smooth-functions-among-continuous-functions/101205#101205 Comment by Leon Lampret Leon Lampret 2012-07-03T04:36:02Z 2012-07-03T04:36:02Z Could you please provide any reference for this theorem? I haven't found this in Duistermaat &amp; Kolk's <i>Multidimensional Real Analysis 1 &amp; 2</i>, nor in Callahan's <i>Advanced Calculus</i>, which are the more advanced calculus books that I have. Should I look among Functional Analysis books? Also, what of the nonexistence of $x\vee -x$ in $\mathcal{C}^r([−1,1],\mathbb{R})$? http://mathoverflow.net/questions/54653/algorithm-that-decreases-the-size-of-the-simplicial-triangulation Comment by Leon Lampret Leon Lampret 2011-02-08T06:29:34Z 2011-02-08T06:29:34Z Hmm, yes, finding different non-homeomorphic but homologically equivalent triangulations indeed sounds the appropriate way to do it. I'll check out your suggestions in the future. I'm guessing this emis.de/journals/SLC/wpapers/s48forman.pdf or this <a href="http://arxiv.org/PS_cache/math/pdf/9911/9911256v1.pdf" rel="nofollow">arxiv.org/PS_cache/math/pdf/9911/9911256v1.pdf</a> would be the place to start. Thanks. http://mathoverflow.net/questions/50382/how-to-triangulate-real-projective-spaces-as-simplicial-complexes-in-mathematica/50653#50653 Comment by Leon Lampret Leon Lampret 2010-12-30T16:29:32Z 2010-12-30T16:29:32Z I intended to accept it right after the solution was fully developed, but at this stage, the code still doesn't work properly. I accepted as you suggested, but I still seek a working code in Mathematica...