User - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T10:18:32Z http://mathoverflow.net/feeds/user/4676 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/56435/what-is-the-intuition-behind-the-freudenthal-suspension-theorem What is the intuition behind the Freudenthal suspension theorem ferret 2011-02-23T19:25:42Z 2012-01-07T22:24:49Z <p>The Freudenthal suspension theorem states in particular that the map $$\pi_{n+k}(S^n)\to\pi_{n+k+1}(S^{n+1})$$ is an isomorphism for $n\geq k+2$.</p> <p>My questions is: What is the intuition behind the proof of the Freudenthal suspension theorem?</p> http://mathoverflow.net/questions/22232/which-properties-of-finite-simplicial-sets-can-be-computed Which properties of finite simplicial sets can be computed? ferret 2010-04-22T18:22:00Z 2010-11-10T10:47:34Z <p>A simplicial set $X$ is a a combinatorial model for a topological space $|X|$, its realization, and conversely every topological space is weakly equivalent to such a realization of a simplicial set. I am wondering, which properties of finite simplicial sets can be effectively (or even theoretically) computed using a computer program. Maybe there are some implementations and I am not aware of their existence, so I would be very pleased about any information.</p> <p>The first problem is how to input the simplicial set (maybe that's not really a problem).</p> <p>Does one know terminating algorithms for the following problems then? Are there other problems for which anything is known in this direction?</p> <ol> <li><s>Is a given finite simplicial set a Kan complex?</s></li> <li>Given two finite simplicial sets $X$ and $Y$, are $|X|$ and $|Y|$ homeomorphic? </li> <li>Given two finite simplicial sets $X$ and $Y$, are $|X|$ and $|Y|$ homotopy equivalent?</li> <li><s>What are the homotopy/homology groups of a given Kan complex</s>? </li> <li><strong>Given a finite simplicial set $X$, what are the homotopy/homology groups of $|X|$</strong>?</li> </ol> <p>etc.</p> http://mathoverflow.net/questions/23713/simplicial-complexes-vs-geometric-realization-of-abstract-simplicial-complexes Simplicial complexes vs. geometric realization of abstract simplicial complexes ferret 2010-05-06T14:11:37Z 2010-05-07T03:06:14Z <p>A <em>finite abstract simplicial complex</em> is a pair $D=(S,D)$ where $S$ is a finite set and $D$ is a non-empty subset of the power set of $S$ closed under the subset operation, e.g. $(\{a,b,c\},\{\emptyset,\{a\},\{b\},\{c\},\{a,b\}\})$.</p> <p>For $n\geq 0$ the topological space $\Delta^n=\{(x_0,...,x_n)\in\mathbb{R}^{n+1}\mid x_i\geq 0, \sum x_i =1\}$ is called the <em>standard $n$-simplex</em>. A topological space homeomorphic to the standard $n$-simplex is called an <em>$n$-simplex</em>. For $n\geq 1$ the $n+1$ faces of any $n$-simplex are $n-1$ simplices.</p> <p>A <em>finite topological simplicial complex</em> is a pair $(X,F)$ where $X$ is a topological space and $F=(F_1,...,F_m)$ is a finite sequence of embeddings $F_k:\Delta^{i_k}\to X$ such that</p> <ul> <li>$X=\cup_k F_k(\Delta^{i_k})$</li> <li>$F_k\neq F_l$ if $k\neq l$ </li> <li>for every $1\leq k\leq m$ with $i_k\geq 1$ and for every face $A$ of the $i_k$-simplex $F_k(\Delta^{i_k})$ there is a $1\leq l\leq m$ with $F_l(\Delta^{i_l})=A$</li> <li>for every $1\leq k\neq k'\leq m$ the simplex $F_k(\Delta^{i_k})\cap F_{k'}(\Delta^{i_{k'}})$ is a face of each of them.</li> </ul> <p>I hope, the definitions are correct.</p> <p>There is the notation of a <em>geometric realization</em> of a finite abstract simplicial complex: Let $D=(S,D)$ be a finite abstract simplicial complex. Then choose a total order on $S$, w.l.o.g. $S=\{1,...,M\}$. The colimit of the functor sending an element $\{0,...,n\}$ of the poset $D$ (considered as a category) to $\Delta^n$ is the geometric realization $|D|$ of $D$.</p> <p><s>If I am not mistaken there are finite topological simplicial complexes which are not the geometric realization of a finite abstract simplicial complex. This is because the choice of the total order determines an orientation of the realization. I think the projective plane for example is not in the image of the realization functor.</s></p> <p><s>My question is: Is there a reasonable notation of a geometric realization for abstract simplicial complexes which has exactly the topological simplicial complexes as its image or do I have a wrong understanding somehow?</s></p> <p>I have realized that the original question does not make sense. Please let me ask if this is the right way to understand the situation:</p> <p>A finite triangulation of a space is the same as a "finite topological simplicial complex". Every finite triangulation is the realization of a finite abstract simplicial complex. The realization of a finite abstract simplicial complex comes with a "direction" of each 1-simplex such that the neighbouring edges are pointing in the <strong>same</strong> directions (they are glued together in this way). The triangulation is orientable if and only if one can permute these "directions" of the 1-simplices such that all the neighbouring edges are pointing in <strong>opposite</strong> directions. How can I see that this condition gives the right concept of orientability? Why "opposite"?</p> http://mathoverflow.net/questions/22855/why-do-delta-sets-not-allow-quotients Why do Delta-sets not allow quotients? ferret 2010-04-28T14:31:16Z 2010-04-28T15:06:12Z <p>A $\Delta$-set is a contravariant functor from the category $\Delta'$ of order-preserving injections to the category of sets (this is essentially what Allen Hatcher calls a $\Delta$-complex).</p> <p>A main reason for working with simplicial sets instead of $\Delta$-sets should be that they allow quotients (see e.g. Allen Hatcher's nice appendix "CW complexes with simplicial structure" to his Algebraic Topology book: "A major disadvantage of $\Delta$-complexes is that they do not allow quotient constructions"), How does this go well with the fact that the category of functors $\Delta'op\to Sets$ <strong>has</strong> colimits?</p> <p>(This question was already asked in a comment on Allen Hatcher's answer to <a href="http://mathoverflow.net/questions/6281/definition-of-simplicial-complex/6302#6302" rel="nofollow">this</a> question on the definition of simplicial complexes. I apologize for asking it twice but there has been no answer given and I am afraid that the reason is - if it's not the silliness of my question - that the comment appears only after pressing the "more comments" button. However, I apologize.)</p> http://mathoverflow.net/questions/18429/convergence-of-spectral-sequences-of-cohomological-type Convergence of spectral sequences of cohomological type ferret 2010-03-16T21:38:53Z 2010-03-18T13:44:01Z <p>Following the first chapter of Hatcher's great book "Spectral Sequences in Algebraic Topology", I got into problems with spectral sequences of cohomological type. Fix a ring $R$ once and for all. Please let me first recapitulate the <strong>homological situation</strong>.</p> <p>An <em>exact couple</em> consists of bigraded $R$-modules $A$ and $E$ and bigraded $R$-module homomorphisms $i$, $j$, and $k$, such that</p> <p><code>$$\begin{array}{rcl}A&amp;\xrightarrow{i}&amp;A\newline {\scriptsize k}\nwarrow&amp;&amp;\swarrow{\scriptsize j}\newline&amp;E&amp;\end{array}$$</code></p> <p>is exact at every corner. Its derived pair with $A'=i(A)$ and $E'=H(E)$ with respect to $d=j\circ k$ is again exact. Before you wonder about the indices in what comes next, please continue reading up to the canonical example. This will explain the degrees, if I have not made a mistake. Let $a\in \mathbb{N}$ (in the example below we have $a=1$). Set $r=0$ for the moment. An exact couple with bidegrees</p> <p><code>$$\begin{array}{rcl}A^{(r)}&amp;\xrightarrow{(1,-1)}&amp;A^{(r)}\newline {\scriptsize (-1,0)}\nwarrow&amp;&amp;\swarrow{\scriptsize (-(a-1+r),(a-1+r))}\newline&amp;E^{(r)}&amp;\end{array}$$</code></p> <p>induces a spectral sequence $E_{p,q}^{r+a}=E_{p,q}^{(r)}$ of homological type with differentials $d^{r+a}=j^{(r)}\circ k^{(r)}$. Here ${\scriptsize something}^{(r)}$ denotes the corresponding term in the $r$-th derived couple. The $r$-th derived couple has bidegrees as in the diagram above.</p> <p>Here is the canonical example. Let $0=C_{-1}\subseteq C_0\subseteq\ldots C_p\subseteq\ldots =C$ be a filtration of a chain complex $C$. Take for example the singular complex of a topological space $X$ filtered by a filtration of $X$. We have many long exact sequences in homology, here are three of them: <code>$$\begin{array}{lcccccr} \to H_{p+q}(C_p,C_{p-1})&amp;\xrightarrow{k}&amp;H_{p+q-1}(C_{p-1})&amp;\xrightarrow{i}&amp;H_{p+q-1}(C_p)&amp;\xrightarrow{j}&amp;H_{p+q-1}(C_p,C_{p-1})\to\newline \to H_{p+q}(C_{p-1},C_{p-2})&amp;\xrightarrow{k}&amp;H_{p+q-1}(C_{p-2})&amp;\xrightarrow{i}&amp;H_{p+q-1}(C_{p-1})&amp;\xrightarrow{j}&amp;H_{p+q-1}(C_{p-1},C_{p-2})\to\newline \to H_{p+q}(C_{p-2},C_{p-3})&amp;\xrightarrow{k}&amp;H_{p+q-1}(C_{p-3})&amp;\xrightarrow{i}&amp;H_{p+q-1}(C_{p-2})&amp;\xrightarrow{j}&amp;H_{p+q-1}(C_{p-2},C_{p-3})\to \end{array}$$</code></p> <p>There is an exact couple as above with $a=1$ and $E_{p,q}=H_{p+q}(C_p,C_{p-1})$ and $A_{p,q}=H_{p+q}(C_p)$. Please note, that all the bigrades are correct. To follow $d^1:E_{p,q}^1\to E_{p-1,q}^1$, you start at the upper left corner, apply $k$, go one row down (the entries are equal here if you move one right) and apply $j$. One can also follow $d^2$ but this is not quite correct since you have to deal with representatives in $E^1$. Here you start at the upper left corner and get to the lower right.</p> <p>Here is Hatcher's illuminating argument for convergence. I have never seen this so clearly presented.</p> <blockquote> <p>The spectral sequence $E^1(C_\bullet)$ of homological type converges to $H_{p+q}(C)$ if it is bounded (=only finitely many non-zero entries on every fixed diagonal $p+q$). One has $$E^\infty_{p,q}=i(H_{p+q}(C_p))/i(H_{p+q}(C_{p-1})).$$ ($i$ denotes the image in the colimit $H_{p+q}(C)$.)</p> </blockquote> <p><em>Proof.</em> Let $r$ be large and consider (I don't know if this renders correctly) the $r$-th derived couple</p> <p><code>$$\begin{array}{c} \to E^{r}_{p+r,q-r+1}\xrightarrow{k^{r}} A^{r}_{p+r-1,q-r+1}\xrightarrow{i^{r}}A^{r}_{p+r,q-r}\newline \xrightarrow{j^{r}}E^{r}_{p,q}\xrightarrow{k^{r}}\newline A^{r}_{p-1,q}\xrightarrow{i^{r}}A^{r}_{p,q-1}\xrightarrow{j^{r}}E^{r}_{p-r,q-1+r}\to. \end{array}$$</code></p> <p>The first and the last term are $0$ because of the bounding assumption. We have $A_{p,q}^{r}=i(A^1_{p-r,q+r})$ and since $C_n$ is zero for $n&lt;0$, the last three terms are $0$. Exactness implies the result.</p> <p>Now the <strong>cohomological situation</strong>.</p> <p>Let $0=C_{-1}\subseteq C_0\subseteq\ldots C_p\subseteq\ldots =C$ be a filtration of a chain complex $C$. Take again for example the singular complex of a topological space $X$ filtered by a filtration of $X$. Understand the cohomology $H^n(C)$ of $C$ as $H_n(\hom(C,\mathbb{Z}))$, thehomology of the dualized complex as one does in topology.</p> <p>Again we have many long exact sequences: <code>$$\begin{array}{lcccccr} \to H^{p+q}(C_p,C_{p-1})&amp;\xrightarrow{k}&amp;H^{p+q}(C_p)&amp;\xrightarrow{i}&amp;H^{p+q}(C_{p-1})&amp;\xrightarrow{j}&amp;H^{p+q+1}(C_p,C_{p-1})\to\newline \to H^{p+q}(C_{p+1},C_p)&amp;\xrightarrow{k}&amp;H^{p+q}(C_{p+1})&amp;\xrightarrow{i}&amp;H^{p+q}(C_{p})&amp;\xrightarrow{j}&amp;H^{p+q+1}(C_{p+1},C_{p})\to\newline \to H^{p+q}(C_{p+2},C_{p+1})&amp;\xrightarrow{k}&amp;H^{p+q}(C_{p+2})&amp;\xrightarrow{i}&amp;H^{p+q}(C_{p+1})&amp;\xrightarrow{j}&amp;H^{p+q+1}(C_{p+2},C_{p+1})\to \end{array}$$</code></p> <p>One can again follow $d^1:E_{p,q}^1\to E_{p+1,q}^1$, etc. What is an exact couple of cohomological type? The only thing which fits into the picture is this: Set $A^{p,q}=H^{p+q}(C_p)$ and $E^{p,q}=H^{p+q}(C_p,C_{p-1})$ and $a=1$.</p> <p>An exact couple with bidegrees</p> <p><code>$$\begin{array}{rcl}A_{(r)}&amp;\xrightarrow{(-1,1)}&amp;A_{(r)}\newline {\scriptsize (0,0)}\nwarrow&amp;&amp;\swarrow{\scriptsize (a+r,-(a-1+r))}\newline&amp;E_{(r)}&amp;\end{array}$$</code></p> <p>induces a spectral sequence $E_{r+a}=E_{(r)}$ of cohomological type with differentials $d_{r+a}=j_{(r)}\circ k_{(r)}$. Here ${\scriptsize something}_{(r)}$ denotes the corresponding term in the $r$-th derived couple. The $r$-th derived couple has bidegrees as in the diagram.</p> <p>Now I would like to establish the following result. This is done nowhere since everything is said to be "dual" to the homological case. But if you look at the indices...hmm:</p> <blockquote> <p>The spectral sequence $E_1(C_\bullet)$ of cohomological type converges to $H^{p+q}(C)$ if it is bounded. One has $$E_\infty^{p,q}=ker(H^{p+q}(C)\to H^{p+q}(C_{p-1}))/ker(H^{p+q}(C)\to H^{p+q}(C_{p})).$$</p> </blockquote> <p><em>Proof?</em> Let $r$ be large and consider (I don't know if this renders correctly) the $r$-th derived couple</p> <p><code>$$\begin{array}{c} \to E_{r}^{p-r,q+r-1}\xrightarrow{k_{r}} A_{r}^{p-r,q+r-1}\xrightarrow{i_{r}}A_{r}^{p-r-1,q+r}\newline \xrightarrow{j_{r}}E_{r}^{p,q}\xrightarrow{k_{r}}\newline A_{r}^{p,q}\xrightarrow{i_{r}}A_{r}^{p-1,q+1}\xrightarrow{j_{r}}E_{r}^{p+r,q-r+1}\to. \end{array}$$</code></p> <p>The last term is $0$ because of the bounding assumption. We have $A_{r}^{p,q}=i(A^{p+r,q-r}_1)$ and since $C_n$ is zero for $n&lt;0$, the second and the third term are $0$. Exactness implies that <code>$$E_r^{p,q}=ker(i(H^{p+q}(C_{p+r}))\to i(H^{p+q}(C_{p+r-1}))).$$</code> I cannot see how the result follows from this. Perhaps it is "only" a limit trick but I can not see it. So the question is:</p> <blockquote> <p><em>How does Lemma 1.2. of Hatcher's text works in the cohomological case?</em></p> </blockquote> http://mathoverflow.net/questions/23713/simplicial-complexes-vs-geometric-realization-of-abstract-simplicial-complexes/23718#23718 Comment by 2010-05-06T15:55:17Z 2010-05-06T15:55:17Z I am totally confused now. A total order $a&lt;b&lt;c&lt;d&lt;e&lt;f$ on $\{a,b,c,d,e,f\}$ gives in particular an orientation of all the $2$-simplices. Now (see Wikipedia on orientability) &quot;Any surface has a triangulation (...) Each triangle is oriented by choosing a direction around the perimeter of the triangle, associating a direction to each edge of the triangle. If this is done in such a way that, when glued together, neighboring edges are pointing in the same direction, then this determines an orientation of the surface. Such a choice is only possible if the surface is orientable,(...)&quot;. http://mathoverflow.net/questions/6281/definition-of-simplicial-complex/6302#6302 Comment by 2010-04-27T11:42:32Z 2010-04-27T11:42:32Z The appendix is nicely written. It seems to me that the main reason for introducing simplicial sets is: $\Delta$-complexes (= contravariant functors from the category $\Delta'$ of order-preserving injections to the category of sets) &quot;do not allow quotient constructions&quot;. How does this go well with the fact that the category of functors $\Delta'op\to Sets$ has colimits? http://mathoverflow.net/questions/22232/which-properties-of-finite-simplicial-sets-can-be-computed Comment by 2010-04-23T07:07:31Z 2010-04-23T07:07:31Z This was stupid, there are no non-discrete finite Kan complexes. Thanks, Tilman. http://mathoverflow.net/questions/18429/convergence-of-spectral-sequences-of-cohomological-type Comment by 2010-03-17T21:15:14Z 2010-03-17T21:15:14Z Sure, sorry, fixed. http://mathoverflow.net/questions/18429/convergence-of-spectral-sequences-of-cohomological-type Comment by 2010-03-17T09:39:48Z 2010-03-17T09:39:48Z Perhaps you are right but I don't see this. I don't see how the two convergence statements transform into each other (how &quot;im&quot; gets &quot;ker&quot; and how this goes well with the limit). This is exactly my question. Why is the $E_r^{p,q}$ from below for $r&gt;&gt;0$ the same as the $E_\infty^{p,q}$ in the &quot;convergence theorem&quot;?