User leon lampret - MathOverflow

Stanley-Reisner ring of a simplicial complex is a functor?

2013-04-11T23:17:52Z

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.

The Stanley-Reisner ideal of $\Delta$ is $I_\Delta:=\langle\langle x_\sigma;\; [n]\supseteq\sigma\notin\Delta\rangle\rangle$, and Stanley-Reisner ring of $\Delta$ is $$K[\Delta]:=K[x]/I_\Delta= K[x_1,\ldots,x_n \:|\: x_\sigma;\; [n]\supseteq\sigma\notin\Delta].$$

The coordinate ring 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.

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 & Herzog, Stanley, Herzog & Hibi, Miller & Sturmfels) regarding functors.

1st try: 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 $K[x_1,\ldots,x_n]$.

2nd try: 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.

Question: 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)?

Question: Can the Stanley-Reisner ideal be seen as an ideal of polynomial functions $\alpha: \Delta\rightarrow K$?

Algebraic Morse theory

2013-01-15T06:35:22Z

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.

Morse theory and homology of an algebraic surface (example)

2012-08-28T00:21:10Z

Let $T_n$ denote the $n$-th Chebyshev polynomial 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 Banchoff-Chmutov surface, where in general, $\mathcal{Z}(f_1,\ldots,f_k)$ denotes the zero set of polynomials $f_1,\ldots,f_k$, i.e. ${(x,y,z) \in \mathbb{R}^3; f_1(x,y,z)=\ldots=f_k(x,y,z)=0}$.

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 $\mathcal{Z}(1 + 1 + 1) = \emptyset$.

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.

Question: 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)?

Order density of smooth functions among continuous functions?

2012-07-03T04:07:20Z

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 $f\ll g:\Leftrightarrow(\forall x: f(x)< g(x))$. Is the following true:

$$\forall f,h\in\mathcal{C}^0([a,b],\mathbb{R}) \;\exists g\in\mathcal{C}^\infty([a,b],\mathbb{R}): f\ll h\Rightarrow f\leq g\leq h,$$ i.e. between any two continuous functions $f\ll h:[a,b]\rightarrow\mathbb{R}$, there exists a smooth function $g$?

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 $x\vee-x$).

I haven't found this result anywhere in my calculus books, so I'd appreciate any references.

Answer by Leon Lampret for How to triangulate real projective spaces (as simplicial complexes in Mathematica)?

2011-02-08T05:08:40Z

I've managed to find my mistake and here is the final (if I'm not mistaken) correct solution.

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] & /@ sc, 1]]

subseteq[{},A_List] := True
subseteq[{x_,y___}, A_List] := MemberQ[A, x] && subset[{y}, A]
subseteq[A_List,B_List, C__List] := subset[A,B] && subset[B,C]
subset[A_List, B_List] := subseteq[A, B] && Complement[B, A]!={}
subset[A_List, B_List, C__List] := subset[A, B] && subset[B, C]

sBarycentricSubdivision[{x_Integer}] := {{{x}}}
sBarycentricSubdivision[s_List] := 
 Subsets[s, {1, Length@s}] //Subsets[#,{Length@s}]& //
  Select[#,subset@@#&]& 
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]] -> vertices1[[i]], {i, 1, k/2}];
  sc2 = DeleteDuplicates[Sort /@ (sc1/.identification)];
  vertices2 = Flatten[sc2, 1] //Sort //DeleteDuplicates;
  sc3 = Map[Position[vertices2, #][[1, 1]] &, sc2, {2}] //Sort //
    DeleteDuplicates;
  sc3
  ]

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.

Thank you, mr. Palmieri, greetings from Slovenia.

How to triangulate real projective spaces (as simplicial complexes in Mathematica)?

2010-12-25T23:24:18Z

Hello!

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.

How can the real projective spaces $$RP^n \approx B^n/_{x\:\sim-x;\;\; x\in\partial B^n}$$ be triangulated as simplicial complexes? 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.

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.

I have already found a concrete triangulation for the real projective plane, but nothing more general.

It would also be appreciated, that the actual triangulation is reasonably small (not necessarily minimal), so that the program calculates homology groups fast enough. Also, answers in code / pseudocode are much desirable (projectiveSpace[n_]:=???).

P.S. I have already written some functions, which can be used (sc...simplicial complex):

skeleton[sc,k] ...list of all k-faces of all simplices
sum[sc1,sc2]...topological sum (disjoint union)
connected sum[sc1,sc2] ...removes two k-simplicices and adds a tunnel between them
product[sc1,sc2]...staircase triangulation of a product (reorders the vertices)
cone[sc]
suspension[sc]

Algorithm that decreases the size of the simplicial triangulation

2011-02-07T15:58:25Z

Hello!

Let X be a topological space. We are considering only abstract simplicial complexes, i.e. a finite list of finite lists of integers. 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?

My main need for such an algorithm is coming from the desire to implement a more efficient algorithm for computing homology of a space.

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).

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.

I have searched the literature and ended up empty handed...

P.S. code in Mathematica would be ideal, but I'd be happy with (pseudo)code of any sort.

Answer by Leon Lampret for How to triangulate real projective spaces (as simplicial complexes in Mathematica)?

2010-12-29T13:13:20Z

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):

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] & /@ sc, 1]]

subseteq[{}, A_List] := True
subseteq[{x_, y___}, A_List] := MemberQ[A, x] && subset[{y}, A]
subseteq[A_List, B_List, C__List] := subset[A, B] && subset[B, C]
subset[A_List, B_List] := subseteq[A, B] && Complement[B, A] != {}
subset[A_List, B_List, C__List] := subset[A, B] && subset[B, C]

barycentricSubdivisionOfSimplex[{x_Integer}] := {{{x}}}
barycentricSubdivisionOfSimplex[s_List] := 
 Subsets[s, {1, Length@s}] // Subsets[#, {Length@s}] & // 
  Select[#, subset @@ # &] & 
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[#] <= Quotient[n + 2, 2], #, True, 
      Complement[Range[n + 2], #]] &, sc1, {2}];
  vertices = Flatten[sc2, 1] // Sort // DeleteDuplicates;
  sc3 = Sort /@ Map[Position[vertices, #][[1, 1]] &, sc2, {2}] // 
     Sort // DeleteDuplicates;
  Which[n == 0, {{1}}, True, sc3]
  ]

I basically followed the instructions of mr. Palmieri. The problem is, that the homology of my projectiveSpace[n] for n=0,...,4 returns

{{1}, {{}}}
{{1, 1}, {{}, {}}}
{{1, 0, 3}, {{}, {2}, {}}}
{{1, 0, 0, 1}, {{}, {2}, {}, {}}}
{{1, 0, 0, 0, 10}, {{}, {2}, {}, {2}, {}}}

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.

Any help would be much appreciated.

Answer by Leon Lampret for Most memorable titles

2010-12-04T23:05:03Z

Mathematical Fallacies, Flaws and Flimflam was definitely by far the most memorable title I have ever read. Also A Taste of Topology seemed tasty.

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.

etc.

Comment by Leon Lampret

2013-04-14T02:22:48Z

What is $X_\Delta$? What is $N$ and $F^N$?

Comment by Leon Lampret

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!

Comment by Leon Lampret

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&Welker, Minimal Resolutions via Algebraic Discrete Morse Theory, assume that all modules are free of rank $1$? Why does Kozlov, Combinatorial Algebraic Topology §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.

Comment by Leon Lampret

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.

Comment by Leon Lampret

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.

Comment by Leon Lampret

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 $Z_n\!\cap\!\{z\!=\!\zeta\}=\{T_n(x)\!+\!T_n(y)\!=\!-T_n(\zeta)\}$?

Comment by Leon Lampret

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.

Comment by Leon Lampret

2012-08-30T01:22:35Z

Yes, now it agrees with my intuition. Thank you for your answer!

Comment by Leon Lampret

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.

Comment by Leon Lampret

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?

Comment by Leon Lampret

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.

Comment by Leon Lampret

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?

Comment by Leon Lampret

2012-07-03T04:36:02Z

Could you please provide any reference for this theorem? I haven't found this in Duistermaat & Kolk's Multidimensional Real Analysis 1 & 2, nor in Callahan's Advanced Calculus, 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 $\mathcal{C}^r([−1,1],\mathbb{R})$?

Comment by Leon Lampret

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 arxiv.org/PS_cache/math/pdf/9911/9911256v1.pdf would be the place to start. Thanks.

Comment by Leon Lampret

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...