User - MathOverflow

What is the intuition behind the Freudenthal suspension theorem

2011-02-23T19:25:42Z

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 $n\geq k+2$.

My questions is: What is the intuition behind the proof of the Freudenthal suspension theorem?

Which properties of finite simplicial sets can be computed?

2010-04-22T18:22:00Z

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.

The first problem is how to input the simplicial set (maybe that's not really a problem).

Does one know terminating algorithms for the following problems then? Are there other problems for which anything is known in this direction?

~~Is a given finite simplicial set a Kan complex?~~
Given two finite simplicial sets $X$ and $Y$, are $|X|$ and $|Y|$ homeomorphic?
Given two finite simplicial sets $X$ and $Y$, are $|X|$ and $|Y|$ homotopy equivalent?
~~What are the homotopy/homology groups of a given Kan complex~~?
Given a finite simplicial set $X$, what are the homotopy/homology groups of $|X|$?

etc.

Simplicial complexes vs. geometric realization of abstract simplicial complexes

2010-05-06T14:11:37Z

A finite abstract simplicial complex 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. $(\{a,b,c\},\{\emptyset,\{a\},\{b\},\{c\},\{a,b\}\})$.

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 standard $n$-simplex. A topological space homeomorphic to the standard $n$-simplex is called an $n$-simplex. For $n\geq 1$ the $n+1$ faces of any $n$-simplex are $n-1$ simplices.

A finite topological simplicial complex 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

$X=\cup_k F_k(\Delta^{i_k})$
$F_k\neq F_l$ if $k\neq l$
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 $F_l(\Delta^{i_l})=A$
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.

I hope, the definitions are correct.

There is the notation of a geometric realization 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 $D$.

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.

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?

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:

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 same 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 opposite directions. How can I see that this condition gives the right concept of orientability? Why "opposite"?

Why do Delta-sets not allow quotients?

2010-04-28T14:31:16Z

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).

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 $\Delta'op\to Sets$ has colimits?

(This question was already asked in a comment on Allen Hatcher's answer to this 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.)

Convergence of spectral sequences of cohomological type

2010-03-16T21:38:53Z

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 homological situation.

An exact couple consists of bigraded $R$-modules $A$ and $E$ and bigraded $R$-module homomorphisms $i$, $j$, and $k$, such that

$$ \begin{array}{rcl}A&\xrightarrow{i}&A\newline {\scriptsize k}\nwarrow&&\swarrow{\scriptsize j}\newline&E&\end{array} $$

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

$$ \begin{array}{rcl}A^{(r)}&\xrightarrow{(1,-1)}&A^{(r)}\newline {\scriptsize (-1,0)}\nwarrow&&\swarrow{\scriptsize (-(a-1+r),(a-1+r))}\newline&E^{(r)}&\end{array} $$

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.

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: $$ \begin{array}{lcccccr} \to H_{p+q}(C_p,C_{p-1})&\xrightarrow{k}&H_{p+q-1}(C_{p-1})&\xrightarrow{i}&H_{p+q-1}(C_p)&\xrightarrow{j}&H_{p+q-1}(C_p,C_{p-1})\to\newline \to H_{p+q}(C_{p-1},C_{p-2})&\xrightarrow{k}&H_{p+q-1}(C_{p-2})&\xrightarrow{i}&H_{p+q-1}(C_{p-1})&\xrightarrow{j}&H_{p+q-1}(C_{p-1},C_{p-2})\to\newline \to H_{p+q}(C_{p-2},C_{p-3})&\xrightarrow{k}&H_{p+q-1}(C_{p-3})&\xrightarrow{i}&H_{p+q-1}(C_{p-2})&\xrightarrow{j}&H_{p+q-1}(C_{p-2},C_{p-3})\to \end{array} $$

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.

Here is Hatcher's illuminating argument for convergence. I have never seen this so clearly presented.

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 $H_{p+q}(C)$.)

Proof. Let $r$ be large and consider (I don't know if this renders correctly) the $r$-th derived couple

$$ \begin{array}{c} \to E^{r}_{p+r,q-r+1}\xrightarrow{k^{r}} A^{r}_{p+r-1,q-r+1}\xrightarrow{i^{r}}A^{r}_{p+r,q-r}\newline \xrightarrow{j^{r}}E^{r}_{p,q}\xrightarrow{k^{r}}\newline A^{r}_{p-1,q}\xrightarrow{i^{r}}A^{r}_{p,q-1}\xrightarrow{j^{r}}E^{r}_{p-r,q-1+r}\to. \end{array} $$

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 $n<0$, the last three terms are $0$. Exactness implies the result.

Now the cohomological situation.

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.

Again we have many long exact sequences: $$ \begin{array}{lcccccr} \to H^{p+q}(C_p,C_{p-1})&\xrightarrow{k}&H^{p+q}(C_p)&\xrightarrow{i}&H^{p+q}(C_{p-1})&\xrightarrow{j}&H^{p+q+1}(C_p,C_{p-1})\to\newline \to H^{p+q}(C_{p+1},C_p)&\xrightarrow{k}&H^{p+q}(C_{p+1})&\xrightarrow{i}&H^{p+q}(C_{p})&\xrightarrow{j}&H^{p+q+1}(C_{p+1},C_{p})\to\newline \to H^{p+q}(C_{p+2},C_{p+1})&\xrightarrow{k}&H^{p+q}(C_{p+2})&\xrightarrow{i}&H^{p+q}(C_{p+1})&\xrightarrow{j}&H^{p+q+1}(C_{p+2},C_{p+1})\to \end{array} $$

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 $a=1$.

An exact couple with bidegrees

$$ \begin{array}{rcl}A_{(r)}&\xrightarrow{(-1,1)}&A_{(r)}\newline {\scriptsize (0,0)}\nwarrow&&\swarrow{\scriptsize (a+r,-(a-1+r))}\newline&E_{(r)}&\end{array} $$

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.

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:

The spectral sequence $E_1(C_\bullet)$ of cohomological type converges to $H^{p+q}(C)$ if it is bounded. One has $$ E_\infty^{p,q}=ker(H^{p+q}(C)\to H^{p+q}(C_{p-1}))/ker(H^{p+q}(C)\to H^{p+q}(C_{p})). $$

Proof? Let $r$ be large and consider (I don't know if this renders correctly) the $r$-th derived couple

$$ \begin{array}{c} \to E_{r}^{p-r,q+r-1}\xrightarrow{k_{r}} A_{r}^{p-r,q+r-1}\xrightarrow{i_{r}}A_{r}^{p-r-1,q+r}\newline \xrightarrow{j_{r}}E_{r}^{p,q}\xrightarrow{k_{r}}\newline A_{r}^{p,q}\xrightarrow{i_{r}}A_{r}^{p-1,q+1}\xrightarrow{j_{r}}E_{r}^{p+r,q-r+1}\to. \end{array} $$

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 $n<0$, the second and the third term are $0$. Exactness implies that $$ E_r^{p,q}=ker(i(H^{p+q}(C_{p+r}))\to i(H^{p+q}(C_{p+r-1}))). $$ 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:

How does Lemma 1.2. of Hatcher's text works in the cohomological case?

Comment by

2010-05-06T15:55:17Z

I am totally confused now. A total order $a<b<c<d<e<f$ on $\{a,b,c,d,e,f\}$ gives in particular an orientation of all the $2$-simplices. Now (see Wikipedia on orientability) "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,(...)".

Comment by

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) "do not allow quotient constructions". How does this go well with the fact that the category of functors $\Delta'op\to Sets$ has colimits?

Comment by

2010-04-23T07:07:31Z

This was stupid, there are no non-discrete finite Kan complexes. Thanks, Tilman.

Comment by

2010-03-17T21:15:14Z

Sure, sorry, fixed.

Comment by

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 "im" gets "ker" and how this goes well with the limit). This is exactly my question. Why is the $E_r^{p,q}$ from below for $r>>0$ the same as the $E_\infty^{p,q}$ in the "convergence theorem"?