User somnath basu - MathOverflow

How transitive are the actions of symplectomorphism groups ?

2011-04-17T04:02:44Z

This question is motivated by the classical fact from differential geometry :

Let $M$ be a smooth manifold of dimension at least $2$. Then for any $n$ the diffeomorphism group $\textrm{Diff}(M)$ acts transtively on the configuration space of $n$-points in $M$ or equivalently it acts $n$-transitively on $M$.

As I recall, it is known that the symplectomorphism group $(M,\omega)$ acts transitively on $M$, which is assumed to be symplectic. My question is then the following :

Let $(M,\omega)$ be a symplectic manifold.
(i) When does $\textrm{Symp}(M,\omega)$ act $n$-transtively for $n\geq 2$ ?
(ii) If the answer above is NOT ALWAYS then what is known ?

As some background, the usual way one proves (rather the only way I know how to prove this) the first fact is by showing the following :
(i) for two sets of distinct $n$ points in $M$ given by $\{p_1,\ldots,p_n\}$ and $\{q_1,\ldots,q_n\}$ which are close, we find disjoint disks $D_i$'s containing $p_i,q_i$. This requires dimension at least $2$. Use some diffeomorphism of $D_i$ that is smoothly identity at the boundary and looks like a rotation inside $D_i$ that swaps $p_i$ and $q_i$.
(ii) Define the natural equivalence relation on $n$-tuples and observe that the configuration space of $n$-points in $M$. By (ii) each equivalence class is open. It is alsoclosed being the complement of open sets. Since the configuration space is connected (this requires dimension at least $2$) this means there is only one equivalence class.

Does this idea work in the symplectic setting - perhaps by taking paths $\gamma_i$ from $p_i$ to $q_i$ and getting Hamiltonian vector fields via $\omega(\gamma_i',\cdot)$ ?

Answer by Somnath Basu for Homology of classifying space of spin group BSpin(n)

2012-11-30T03:02:47Z

Consider the classifying space $EG$ of a given group $G$. In your case, $G={Spin}(n)$. We have a fibration $G\to EG\to BG$ where $EG$ is contractible and $G$ acts freely on it. Therefore, the long exact sequence in homotopy groups tells you that $\pi_j(BG)\cong \pi_{j-1}(G)$. But $G={Spin}(n)$ is a Lie group which is simply connected. It is also classically known that $\pi_2(G)=0$ for finite dimensional Lie groups $G$. And, we also know that $\pi_3({Spin}(n))=\mathbb{Z}$. This implies that $\pi_j(B {Spin}(n))=0$ for $j=0,1,2,3$. Moreover, $\pi_4(B {Spin}(n))\cong H_4(B{Spin}(n);\mathbb{Z})$ by Hurewicz theorem. This is also isomorphic to $\pi_3({Spin}(n))$. The universal coefficient theorem now tells you that $H^4(B{Spin}(n);\mathbb{Z})=\mathbb{Z}$.

Whitehead product with identity on homotopy groups of spheres

2012-03-18T06:12:43Z

For $n\geq 2$ let $(S^n,p)$ be the $n$-sphere with a base point $p$. Let $1:S^n\to S^n$ denote the identity map. Let us define the map

$Wh_1: \pi_i(S^n,p)\to \pi_{i+n-1}(S^n,p), \alpha\mapsto [\alpha,1],$

where $[\cdot,\cdot]$ is the Whitehead bracket.

Question 1 : What is known about $Wh_1$ ?

Question 2 : If we let $Wh_f$ denote the corresponding map on homotopy groups for $f:(S^m,p)\to (S^n,p)$ then what is known about $Wh_f$ ?

Combinatorics of the Stasheff polytopes

2009-11-21T15:34:47Z

First a little background for those unaware. The Stasheff polytopes (or associahedra) are certain convex polytopes that arise in the theory of $A_\infty$-algebras. There is one polytope for each $n\geq 2$ and is denoted by $K_n$. The $K_n$'s essentially encode the homotopies, higher homotopies and so on of the associativity relation. One way to describe $K_n$, which is has dimension $(n-2)$, is to take all rooted binary trees with $n$ leaves and take a suitable convex hull. For example, $K_2={\ast}$, $K_3$ is an interval while $K_4$ is a pentagon.

It is known that the number of vertices $v$ of $K_n$ is the $(n-1)^{th}$ Catalan number, i.e., $v=\frac{1}{n}{2n-2 \choose n-1}$. What can one say about the number of edges of $K_n$ and generally about counting faces of all codimension?

Answer by Somnath Basu for Applications of string topology structure

2012-03-05T03:00:14Z

As a shameless plug, I may say that in my thesis we do show that string topology, interpreted in a broader context, is NOT a homotopy invariant. What we do is the following : instead of looking at loops in $M$ we think of them as arcs in $M\times M$ with its boundary in the diagonal $M$ that sits inside $M\times M$. Now we look at the space of such arcs $\mathcal{S}(M)$, which, when they intersect the diagonal at intermediate stages, do so transversely. One can then define a suitable coalgebra structure which is NOT a homotopy invariant. In particular, this structure distinguishes the Lens spaces $L(7,1)$ from $L(7,2)$, which are homotopy equivalent but NOT homeomorphic.

Of course, this new structure is not related to the loop product or the BV operator as per the question asked. Moreover, this structure is defined on a much smaller space then $LM$. However, if you take the point of view that string topology is broadly the study of loops in a manifold then this is a new and interesting algebraic structure.

How are the classifying space of $E_8$ and $K(\mathbb{Z},4)$ related?

2011-01-17T03:02:03Z

I recently heard the following fact :

Up to the $15$th skeleton, the classifying space $BE_8$ and $K(\mathbb{Z},4)$ are homotopy equivalent?

I have two questions on this :

(1) Is there any easy way to see this? Of course, knowing the first fourteen homotopy groups of $E_8$ is enough but then the question is how does one compute them?

(2) Is there any feasible explanation that suggests that $4$th cohomology classes (possibly related to gerbes), i.e., elements in $H^4(X;\mathbb{Z})$, arise from physical considerations and if $X$ is of dimension $14$ or less then we're classifying $E_8$-bundles on $X$, thereby suggesting that $E_8$ arises out of physical considerations?

The last question is a little vague but any pointers would be great!

Homotopy type of the self-homotopy equivalences of a bouquet of spheres

2011-11-23T03:37:31Z

Before I state the questions I have in mind, let me give some background. If one considers $S^2$ then it is known due to Kneser that $\textrm{Homeo}^{+}(S^2)$ has the homotopy type of $SO(3)$. By Smale's work we also know that $\textrm{Diff}^{+}(S^2)$ is homotopic to $SO(3)$. However, when we work in the homotopy category this changes. Later Hansen considered $\textrm{Aut}_0(S^2)$, the connected component of identity in the space of all self homotopy equivalences. He showed that its homotopy type is that of $SO(3)\times \mathbf{\Omega}$, where $\mathbf{\Omega}$ is the universal cover of the connected component of the constant loop in the double loop space $\Omega^2 S^2$.

Although this is a very nice and interesting fact, it's turns out to be hard to generalize his method of proof for higher spheres. For one, the homotopy type of $\textrm{Diff}(S^n)$ is not really known for $n\geq 7$ if I'm not mistaken. For another, Hansen crucially uses the fact that $S^2$ is the base of the usual fibration $SO(2)\to SO(3)\to S^2$ and the fact $SO(2)=S^1$ has no higher homotopy. This indubitably fails for higher spheres!

Question 1 What is known about the homotopy type of $\textrm{Aut}_0(S^n)$ for $n\geq 3$?

I should say that the rational homotopy type is fairly easily calculable via Sullivan's minimal models. So, I'm looking for a bit more here.

Question 2 What is the homotopy type of the identity component of self homotopy equivalences of $\vee_k S^2$, a bouquet of $2$-spheres?

Of course, one can ask this question for higher spheres but bearing in mind question 1, I decided I would be happy with an answer for $S^2$. If it helps, the homotopy groups of $\vee_k S^2$ can be calculated by Hilton-Milnor theorem. This fits in a long exact sequence of groups associated to $\textrm{Aut}_0^\ast(\vee_k S^2)\to \textrm{Aut}_0(\vee_k S^2)\to \vee_k S^2$ where the last map is evaluation of an automorphism at the common point of the bouquet and $\textrm{Aut}^\ast_0(\vee_k S^2)$ consists of based maps. But this doesn't seem to lead anywhere!

Answer by Somnath Basu for Homology of loop space

2011-11-17T06:05:34Z

This may not be exactly what you're looking for but rational homotopy theory provides an answer.

Step $1$ : Let $X$ be a nilpotent space, i.e., $\pi_1(X)$ is nilpotent and $\pi_i(X)$ is a nilpotent module over $\pi_1(X)$ for $i\geq 2$. Note that this is satisfied in your case. We also assume that $X$ has finite (rational) cohomology in each dimension.

Step $2$ : Methods from Sullivan's rational homotopy theory tells us that one can find a model for the rational cohomology of $X$, i.e., find a differential graded algebra $(\mathcal{A},d)$ and a map of algebras $\varphi:(\mathcal{A},d)\to \big(H^\ast(X;\mathbb{Q}),0\big)$ which is a quasi-isomorphism, i.e., an isomorphism when we pass to cohomology. This algebra $\mathcal{A}$ is usually written as $\Lambda V$ for a graded vector space $V$. Elements in $V$ are called indecomposables and one can take $V$ as the direct sum of the duals of $V_i:=\pi_i(X)\otimes\mathbb{Q}$ and hence the name rational homotopy theory.

Remark : A space $X$ is called formal if there is a model (these are often called Sullivan models) for $H^\ast(X;\mathbb{Q})$ with vanishing differential. H-spaces ans spheres are examples of such spaces.

Step $3$ : It can be shown (refer Rational Homotopy Theory by Felix, Halperin and Thomas) that a model $\Lambda V$ for $X$ defines a model for $\Omega X$ by shifting the underlying vector space $V$ down by $1$. It turns out that the differential for $\Lambda(s^{-1} V)$ can be taken to be zero as $\Omega X$ is an $H$-space.

Answer by Somnath Basu for Realisability cohomological class as product or as immersed sphere

2011-10-23T01:13:43Z

Consider $M=SU(4)$. Rationally, the cohomology ring $H^\ast(M;\mathbb{Q})$ is the product $H^\ast(S^3;\mathbb{Q})\otimes H^\ast(S^5;\mathbb{Q})\otimes H^\ast(S^7;\mathbb{Q})$. Consider $a$ to the product of the generators of $S^3$ and $S^5$. The dual is the generator of $S^7$ which doesn't split. However, $\pi_8(SU(4))=\mathbb{Z}_{4!}$ which is torsion and hence a sphere cannot possibly generate $a$.

Tutte polynomials of appropriate Cayley graphs

2011-10-12T04:41:00Z

I was quite intrigued by Tutte polynomials in a recent talk I had been to. It was introduced as a polynomial associated to a undirected finite graph. For a graph $G=(V,E)$ we form the polynomial

$T_G(x,y)=\sum_{A\subseteq E} (x-1)^{k(A)-k(E)}(y-1)^{k(A)+|A|-|V|}$

where $k(A)$ is the number of components of the graph $(G,A)$.

I have heard and read about the significance of this polynomial. However, I'm completely new to combinatorics and graph theory and have no prior knowledge of what is or isn't known about these things. I have two questions which should be easy enough for the experts.

(1) Let us start with a finite group $G$, chose a presentation and consider its Cayley graph $\Gamma$. Now consider the Tutte polynomial $T_\Gamma$ of $\Gamma$. We suppose that $G$ is the fundamental group of some reasonable space $M$ (it always is $\pi_1$ of some space but what I mean here is the space is a manifold or very broadly an object where distances and volumes make sense).

Question Does the Tutte polynomial $T_\Gamma$ encode any information about the topological invariants of $M$?

(2) In my own naive way I was trying to see if anything similar (like the formula at the outset) works for infinite graphs. The existence of such a thing, at least, looks very interesting as Tutte polynomials generally have a lot of significance. If such a thing doesn't exist, one may begin to study properties of the Tutte polynomials associated to finite subgraphs of an infinite graph and analyze the asymptotics of the these properties on the poset of finite subgraphs.

Question Is there a known way to define Tutte polynomials for infinite graphs? If so, how is defined and what is known?

Question If the above answer is yes then what does $T_\Gamma$ encode about the growth of $G$ when $G$ is the fundamental group of a manifold? Note that by the $\check{S}varc$-$Milnor$ lemma this growth is asymptotic to the growth (in the sense of geometry) of the universal cover of $M$.

REMARK : I edited the first question as the growth of a finite group is not terribly interesting. The original first question should make more sense now, as presented, as the last question.

Answer by Somnath Basu for bundle over a chain

2011-04-23T02:44:50Z

If you have a $G$-bundle over $M$ (without boundary) then this corresponds to the homotopy class of a map $\gamma:M\to BG$. It is known that for $G$ simply connected, any $G$-bundle over $M$ is trivializable. One way to see this is by obstruction theory. The other is to notice that $\pi_i(BG)=\pi_{i-1}(G)$ and for $i=1,2,3$ this is zero. Therefore, one can get a cellular model for $BG$ which has no $3$-cells. Therefore, $\gamma$ is homotopic to a cellular map $\gamma':M\to BG$ which is necessarily constant. Since $\gamma_\ast[M]=0$ in $H_3(BG)$ by Hurewicz and the previous observations, there is a singular $4$-chain $B$ with boundary $\gamma_\ast[M]$. Then using the necessary pullbacks you get what you want.

Answer by Somnath Basu for Which topics/problems could you show to a bright first year mathematics student?

2011-03-13T04:01:21Z

We have been teaching an annual 8 lecture series mathematics course on topology of surfaces, Euler characteristic and loops on surfaces. It is more of discussion session than a lecture. Although it is primarily intended for a certain group of female undergraduates who are part of some program, I believe the topics could be introduced to any group of undergraduates who are interested enough! Here's the link for those interested :

http://www.math.sunysb.edu/~basu/courses/WSE187-spring11/WSE187.html

We assume no basic background (of calculus or linear algebra) on their part. It is very hands on and they try to develop a concept of what a surface is and what numerical invariants help distinguish between them. Among other things, we encourage them to use play-doh to model filled-in surfaces and help them visualize deformations. We also have fun cutting Mobius strips along various curves and slice bagels into links! From my experience (and I have only taught it twice) some of these hands-on events always excite even the ones who aren't showing much interest. After all, it's fun for them to learn that the soccer ball has only so many pentagons and hexagons essentially due to Euler! They have to give a presentation at the end figuring out and explaining (from basic principles) how some problem (assigned to them) can be solved via topology - the solution could be in terms of pictures, play-doh or any other material if need be. For the interested, have a look at the presentations from last year :

http://www.math.sunysb.edu/~basu/courses/WSE187-spring10/Lectures_WSE187.html

Lastly, although we haven't done it the course referred to above, including something leading up to the four colour theorem should be plenty of fun too, both for the teachers and the students!

Answer by Somnath Basu for Euler characteristic, Gauss-Bonnet, and a product formula

2011-03-03T22:33:00Z

I understand you're trying to prove using the Pfaffian but what's wrong with the triangulation proof of what you want? Fibre bundles $F\to E\to B$ are locally trivial and triangulate your base and the fibres then it should be clear that $\chi(E)=\chi(F)\chi(B)$.

More precisely, assume that you have triangulated finely enough so that the simplices in your base are already in local charts. For simplicity, let $B$ be a $k$-dimensional manifold. If $\Delta_k\subset B$ then the number of vertices in $\pi^{-1}(\Delta_k)$ is the product of the number of vertices of $\Delta_k$ and the number of vertices $v(F)$ of $F$. The number of edges of $\pi^{-1}(\Delta_k)$ equals $e(\Delta_k)v(F)+v(\Delta_k)e(F)$. You continue in this way and an alternate sum tells you that $\chi(\pi^{-1}(\Delta_k))=\chi(\Delta_k)\chi(F)=\chi(F)$. When you glue fibres over charts (in this case faces of $\Delta_k$) then the Euler characteristic formula $\chi(X_1\cup X_2)=\chi(X_1)+\chi(_2)-\chi(X_1\cap X_2)$ tells you that $\chi(\pi^{-1}(\Delta_k\cup\Delta_k'))=\chi(\Delta_k)\chi(F)+ \chi(\Delta_k')\chi(F)-\chi(\Delta_{k-1})\chi(F)$. Then another alternating sum tells you that you should ultimately sum $1$'s (with signs) over the simplices of your base and finally multiply $\chi(F)$ which gives you the formula you want.

EDIT : In case this answer isn't not helpful, feel free to comment. I initially wanted to put it up as a comment but ended up clicking on the answer button!

Answer by Somnath Basu for Spaces with same homotopy and homology groups that are not homotopy equivalent?

2011-01-26T21:52:34Z

Following up on John's comment, one can consider $S^2$-fibrations over $S^2$. There are two of them since such fibrations are classified by $\pi_1(\textrm{Diff}^{+}(S^2))=\mathbb{Z}_2$. One of them is $S^2\times S^2$ while the other can be shown to be the connected sum of $\mathbb{CP}^2$ and $\overline{\mathbb{CP}}^2$. These two spaces have the same homology. They have the same homotopy groups since they both form the base of a $S^1$-fibration with total space $S^2 \times S^3$. However, the intersection forms are not equivalent and hence they are not homotopy equivalent.

Answer by Somnath Basu for Why must a reducible flat SU(2)-connection over a homology sphere be trivial?

2011-01-26T07:02:39Z

The following is probably well known. For example, one can find it in Fukaya's notes on Floer homology, $A_\infty$-categories and topological field theory.

Let $P$ be the trivial principal $SU(2)$-bundle on $M$ and let $\mathcal{A}(M)$ (resp. $\mathcal{A}^{\textrm{flat}}(M)$) denote the space of connections (resp. flat connections) on $P$. For $a\in\mathcal{A}(M)$ let $\mathcal{G}^a$ denote the stabilizer of $a$ under the action of the gauge group $\mathcal{G}(M)$. Then $\mathcal{G}^a=\pm 1, U(1)$ or $SU(2)$.

One can elaborate on a proof of this later. May be from this one can deduce that there are no flat connections with stabilizers $U(1)$ and the ones that have full stabilizers are the ones that you're looking for. On the other hand, following what Dan had said before, one has the following :

$\mathcal{A}^{\textrm{flat}}(M)/\mathcal{G}(M)\cong \textrm{Hom}(\pi_1(M),SU(2))/SU(2)$.

Here the right hand quotient is to be interpreted as all homomorphisms up to conjugation. Again, this may lead to some proof but it's been a while since I have seen these things. I'll put up something once I think it through a bit.

Answer by Somnath Basu for Euler characteristic of orbifolds

2011-01-13T21:20:55Z

Just to add to Johannes answer, when you have a map $f:M\to M$ and you want to compute $\sum_i (-1)^i \textrm{Tr}_{H_i(M)}(f_\ast)$, one may proceed as follows : imagine $i$-cycles in $M$, i.e., elements in $H_i(M)$, arising as either $i$-cycles in $M^f$ which are fixed by $f_\ast$ or they are not fixed by $f_\ast$. One can do this rigorously by choosing an appropriate triangulation of $M$, homotoping $f$ to be a simplicial map and choosing a basis for $H_i(M)$ for all $i$ simultaneously. It then follows that in this basis, the trace picks up $1$ for each element in the basis of $H^i(M^f)$, which when summed up with alternating signs produces $\chi(M^f)$. The classical Lefschetz fixed point formula can be proven in exactly the same way.

Answer by Somnath Basu for Fibrewise homotopy-equivalence of unit sphere bundles vs isomorphism of tangent bundles

2011-01-09T23:39:40Z

The rational homotopy groups of $HomEq(S^{m-1})$ can be calculated via (Sullivan) minimal models (refer page 314 of D. Sullivan's Infinitesimal computations in topology). In short, if I'm not mistaken one can show that $\pi_i(HomEq(S^{2n})\otimes\mathbb{Q})=\mathbb{Q}$ if $i=0,4n-1$ and $\pi_i(HomEq(S^{2n+1})\otimes\mathbb{Q})=\mathbb{Q}$ if $i=0,2n+1$. One the other hand, $Spin(2n+2)$ is rational homotopy equivalent to $S^3\times S^7\times\cdots\times S^{4n-1}\times S^{2n+1}$ while $Spin(2n+1)$ is rationally $S^3\times S^7\times\cdots\times S^{4n-1}$. This should imply at least that, even rationally, the map $O_m\to HomEq(S^{m-1})$ is not a homotopy equivalence in general. May be more is known about the specifics of this map.

Geometric models for classifying spaces of a group

2010-10-29T00:12:56Z

For any given topological group $G$ we have Segal's construction/definition of $BG$. I'm recalling it in case the details turn out to be relevant.

Form the disjoint union of $G^n\times\Delta_n$ for $n\geq 0$ and identify points via $(d_i\cdot,\cdot)\sim (\cdot,\partial_i \cdot)$ where $\partial_i:\Delta_{n-1}\to\Delta_n$ are the face maps and $d_i :G^n\to G^{n-1}$ maps $(g_1,\ldots,g_n)\to (g_1,\ldots , g_i g_{i+1},\ldots,g_n)$ for $i\neq 0,n$. When $i=0$ (resp. $n$) the map $d_i$ is the projection on to the first (resp. last) $n-1$ factors.

On the other hand, one can consider $G$ as a topological category with one object and $G$ as its morphisms. One then considers the topological nerve $\mathcal{N}(G)$ which has one $0$-simplex, one $1$-simplex for each element of $G$, $2$-simplices are given by triangles labelled by $g_1,g_2,g_3$ such that the interior of the triangle corresponds to a path $h$ joining $g_1 g_2$ and $g_3$. Higher simplices are defined similarly. If one uses the natural gluing on this collection, one arrives at the geometric realization of $\mathcal{N}(G)$. Let's call it $\mathcal{B}G$. So here's my question, which may very well be known.

How are the two constructions $BG$ and $\mathcal{B}G$ related?

In fact, this prompts a more general question :

Given a topological category (in the sense of Segal) $\mathcal{C}$, i.e., $\mathit{Ob}$ and $\mathit{Mor}$ are topological spaces, one can form the topological nerve $\mathcal{N}(\mathcal{C})$, as explained above, and then take its realization. On the other hand, the usual nerve $N(\mathcal{C})$ of $\mathcal{C}$ is a simplicial object in $\mathbf{Top}$ and we can take its realization. How are these two realizations related?

I would love to know the answer to the above question for the usual notion of topological category too, viz., where only morphisms $\mathit{Mor}(x,y)$ is required to be a (compacty generated Hausdorff) space for any $x,y\in\mathit{Ob}$.

EDIT : May be I hadn't explained the topological nerve definition which Harry succintly does in his comment below. It is also the same definition given in Higher Topos Theory by J. Lurie.

Answer by Somnath Basu for Examples of categorification

2010-10-25T23:24:17Z

In my limited experience of categories, I liked Quillen's notion of homotopy fibre (his paper Higher Algebraic K-Theory I) for a functor between categories modelling the homotopy fibre of any map.

Answer by Somnath Basu for How and how much do the notations and diagrams influence our understanding of mathematical concepts?

2010-10-23T20:09:04Z

This example illustrates a point of view that I learnt from Dennis Sullivan. In general, when speaking of free (or projective) resolutions of $R$-modules one writes

$\cdots\to F_n\to\cdots\to F_1\to F_0\stackrel{\varepsilon}{\rightarrow} M\to 0$

where $F_n$'s are free (or projective) $R$-modules. For calculating $Ext$ or $Tor$ we use one such resolution of $M$ and then show that the resulting answer doesn't depend on the resolution chosen. The resolution above can be rewritten as a map $\varphi$ between two chain complexes of $R$-modules :

$\cdots\to F_n\to\cdots\to F_1\to F_0\to 0$

$\cdots\to 0\hspace{0.2cm} \to \hspace{0.2cm} \cdots\hspace{0.2cm}\to 0\to M\to 0$

where the only non-trivial vertical map is $\varepsilon:F_0\to M$. Notice that $\varphi$ is a quasi-isomorphism and saying that the derived functors are independent of the resolution chosen is akin to saying that any two chain complex of $R$-modules representing $M$ are quasi-isomorphic. To an algebraic topologist (and possibly for others too) this is so much more natural.

This point of view emphasizes that the non-triviality of the module $M$ gets coded into the differentials between $F_n$'s. The $F_n$'s themselves contain almost no information since the rank is the only possible invariant for $F_n$ and even that be made to change by adding a copy of $R$ to $F_{n+1}$ and $F_n$.

Hopf algebras as cohomology of $\mathbb{CP}^\infty$, $\Omega S^3$ and related $H$-spaces

2010-09-20T21:33:10Z

Let me begin by a couple of questions :

Consider a graded abelian group $V=\oplus_{i\geq 0} V_i$ such that $V_{2i}=\mathbb{Z}$ and $V_{\textrm{odd}}=0$. What are the possible Hopf algebra structures on it?

One can ask a slightly stronger question :

When does a given Hopf algebra structure on $V$ (as above) arise as the integral cohomology of a $H$-space?

The motivation behind this question is purely my own curiosity. While discussing how to distinguish $\Omega S^3$ and $\mathbb{CP}^\infty$ rationally, we saw that the rational cohomology or the rational homotopy groups are unable to detect the difference. However, $H^\ast(\Omega S^3;\mathbb{Z})=\Gamma_{\mathbb{Z}}[\alpha]$, the divided polynomial algebra generated by $\alpha$ (of degree $2$) while $H^\ast(\mathbb{CP}^\infty;\mathbb{Z})=\mathbb{Z}[u]$ is the polynomial algebra generated by $u$ (of degree $2$). Moreover, a polynomial algebra such as $\mathbb{Z}[u]$ has a comultiplication map given by $u\stackrel{\Delta}{\longrightarrow}1\otimes u+u\otimes 1$ and extended naturally. One can check that the dual (as a Hopf algebra) of $\mathbb{Z}[u]$ is isomorphic to $\Gamma_{\mathbb{Z}}[u^\ast]$, where $u^\ast$ is the dual of $u$.

From what I could conclude by playing around with coefficients is that for each prime $p$ and a positive integer $r$ one can cook up a Hopf algebra structure on $V$. I don't know if they come from a space. However, these structure constants must be compatible with the action of the Steenrod algebra (or the mod $p$ version) if $V=H^\ast(X;\mathbb{Z})$ for some $H$-space $X$. I vaguely remember that compatibility with the Steenrod algebra is not sufficient and Adams operations provide further obstructions (although I may be wrong on this point). This leads me to :

Is there a (list of) necessary and sufficient criteria (in general or at least in this case) which tells us when a given Hopf algebra structure on graded vector space arises as $H^\ast(X;\mathbb{Z})$ for some $H$-space?

This may be well known (and perhaps classical) to homotopy theorists and any reference to known results are good enough for me.

Answer by Somnath Basu for Obstruction Cocycles

2010-09-12T20:29:36Z

You may find this and this by Tony Phillips useful.

Answer by Somnath Basu for Euler Characteristic of a manifold with non-vanishing vector field,

2010-09-11T18:17:37Z

This is basically a watered down version of Poincare-Hopf theorem. Assuming that your compact manifold $M$ has a tringulation. At each of the simplex put a vector field $V$ which is zero at the centre of its subsimplices and flows out from the centre of the higher simplex to its boundary simplices. This vector field has finitely many zeroes and the index of $V$ is precisely $\chi(M)$. Now notice that the index of a vector field doesn't depend on the vector field. This is because it's topologically the intersection number of the image of $V$ in $TM$ and $M$ sitting as the zero section and $V$ is homotopic to $M$. In other words, given your non-vanishing vector field $V'$ one can linealy homotope this to $V$ whose index calculates $\chi(M)$.

Answer by Somnath Basu for Equivariant maps inducing isomorphism in integral cohomology

2010-08-20T20:07:03Z

I don't know of a reference off-hand but here's one way to think about it. First, one can think of $H^i(X;\mathbb{Z})$ as $[X,K(\mathbb{Z},n)]$, the set of homotopy classes of maps. Notice that a cellular model for $K(\mathbb{Z},n)$ can be taken to be $S^n$ union higher cells that kill off the higher homotopy groups. Second, any map $f:X\to Y$ can be replaced by an inclusion $\iota:X\to M_f$, where $M_f$ is the mapping cylinder and it has the same homotopy type as that of $Y$. This works in the equivariant setting also. The third fact is that if any equivariant map $f:X\to Y$ induces an isomorphism in cellular cohomology and $f$ acts freely on both $X$ and $Y$ then $f$ induces an isomorphism on equivariant cohomology as well. The equivariant cohomology can be thought of as maps from spaces to $K(\mathbb{Z},n)$ up to equivariant homotopy.

Now think of $f:X\to Y$ as in inclusion and there is a long exact sequence in cohomology $\cdots\to H^\ast(X;\mathbb{Z})\to H^{\ast+1}(Y,X;\mathbb{Z})\to H^{\ast+1}(Y;\mathbb{Z})\to H^{\ast+1}(X;\mathbb{Z})\to H^{\ast+2}(X;\mathbb{Z})\to\cdots$ which tells you in your case that $H^{i}(Y,X;\mathbb{Z})=0$ if $i>i_0$. The kernel of $H^{i_0}(Y;\mathbb{Z})\to H^{i_0}(X;\mathbb{Z})$ is just the image of $H^{i_0}(Y,X;\mathbb{Z})$ in $H^{i_0}(Y;\mathbb{Z})$. Thinking of $H^{i_0}(Y,X;\mathbb{Z})$ as relative homotopy classes of maps from $(Y,X)$ to $K(\mathbb{Z},i_0)$. These maps only probe the $i_0$ skeleton of $(Y,X)$ because $H^i(Y,X;\mathbb{Z})=0$ if $i>i_0$. And since any map can be made homotopic to a cellular map we need only study homotopy classes of maps from the $i_0$-skeleton of $(Y,X)$ to the $i_0$-skeleton of $K(\mathbb{Z},i_0)$ which is $S^{i_0}$. These are precisely the different ways of factoring a given equivariant map $X\to S^{i_0}$ via $X\stackrel{f}{\to} Y\to S^{i_0}$ (all upto equivariant homotopy).

Answer by Somnath Basu for What would be your suggestion of textbooks in Lie groups and Galois theory?

2010-08-17T15:09:10Z

Patrick Morandi's Field and Galois Theory is a good book for beginners. He gives lots of examples and has interesting exercises too. For a later reading though, I would suggest the Galois theory section in Lang's Algebra.

I really liked Hsiang's Lectures in Lie Groups although it may be a bit short for a full course. And Kirillov Jr.'s book Introduction to Lie Groups and Lie Algebras (also available as a published book) is a very good introduction to the topic with plenty of nice examples in the exercises. And lastly, Serre's Complex Semisimple Lie Algebras is great once you manage to get through it, i.e., it's a gem but not for the first reading!

Answer by Somnath Basu for prime ideals in C([0,1])

2010-08-16T23:20:12Z

I think this paper may be of some help. Although this may not be the earliest source, it has a proposition there which says that given a point $p\in [0,1]$, one can find non-maximal prime ideals $\mathfrak{p}_1,\mathfrak{p}_2$ such that $\mathfrak{m}_p$, the maximal ideal at $p$, is the sum of $\mathfrak{p}_1$ and $\mathfrak{p}_2$. They take a sequence of points $x_n$ converging to $p$ and then use two different ultrafilters on $D= \{ x_n \}_{n\geq 1}$ to define the two prime ideals.

Answer by Somnath Basu for What are your favorite puzzles/toys for introducing new mathematical concepts to students?

2010-08-14T21:55:51Z

A few months back we taught a course on curves and surfaces to undergraduates and asked them to slice a bagel into two linked halves as in here. Of course, you need at least two bagels per student since inevitably most of them end up cutting the first bagel into two unlinked pieces.

The 15-tile sliding puzzle (may be a bit outdated by now) is also a good way to introduce permutation groups and even permutations in particular.

And lastly, the game of Sim (not to be confused with sim city) where two players take turns in drawing edges in red and blue on set of 6 vertices. The rule is that if an edge already exists between a and b then one cannot draw another one. The aim is to avoid a triangle in your own colour. It is known that this game always has a winner. Obvious generalizations to more colours and more vertices lead to Ramsey theory. I actually took this route while lecturing to high school kids and they get into it if you start your talk by playing a few games on the blackboard.

Answer by Somnath Basu for When is there a deRham duality relation between the fundamental class and a top form.?

2010-04-29T01:34:28Z

Consider a suitably small tubular neighbourhood $\mathcal{N}$ of $\mathbb{CP}^1$, thought of as sitting inside $\mathbb{CP}^2$. Then $\mathcal{N}$ locally looks like $\mathbb{CP}^1\times D_2$. The volume form $\omega$ of $\mathbb{CP}^1$ is not necessarily a $2$-form in $\mathbb{CP}^2$. However, one can imagine changing it so that we have new $2$-form $\widetilde{\omega}$ in $\mathbb{CP}^2$, supported in $\mathcal{N}$, such that $\widetilde{\omega}$ restrcited to ${p}\times D_2$ (for $p\in\mathbb{CP}^1$) looks like a smooth bump function which integrates to $1$. This can be taken to be $w_1$ in your case. Now assume you take your copy of $\mathbb{CP}^1$ inside $\mathbb{CP}^2$ and perturb it a bit (i.e., make it transversal to itself) to get another copy. Apply what we said before and get $w_2$ supported in a suitable tubular neighbourhood of this perturbed copy. Now integrating $w_1\wedge w_2$ over $\mathbb{CP}^2$ gives you an integration over balls around points where self-intersections occur. The normalization were so chosen that it counts the intersection number of $\mathbb{CP}^1$ with itself.

Answer by Somnath Basu for Rational homotopy theory of a punctured manifold

2010-04-09T23:35:38Z

One can certainly go the other way if you're wiling to let $N$ be $M$ minus a disk. I'll just recall what is already done in Felix-Halerpin-Thomas's "Rational Homotopy Theory".

Take a minimal model $(\Lambda V,d)$ for $N$. Since $M$ can be thought of as attaching a disk on a certain boundary $S^{n-1}$ of $N$, let $f:(S^n,p_0)\to(N,x_0)$ be the attaching map. Then $M=N\cup_{f} D^{n}$. Define a commutative cochain algebra $(\Lambda V\oplus \mathbb{Q} u,D)$ such that (i) $U$ has degree $n$, (ii) $\Lambda V$ is a subalgebra with $u^2=0=u\cdot \Lambda^{+}V$ and (iii) $Du=0$ and $Dv=dv+\left\langle v,f\right\rangle u$ if $v\in V$. Here $v$ is identified with an element of $A_{PL}(X)$, the space of polynomial differential forms, and the pairing is the usual pairing of cohomology and homology. Then $(\Lambda V\oplus \mathbb{Q} u,D)$ is a commutative model for $M$.

I hope this helps.

Answer by Somnath Basu for Cohomology classes annihilated by pullbacks

2010-03-07T22:22:17Z

Here's my two cents although it's rather sketchy.

For any CW complex $X$, $H^3(X;\mathbb{Z})=[X,K(\mathbb{Z},3)]$, where $K(\mathbb{Z},3)$ comes equipped with a fibration $\mathbb{CP}^\infty\to P\to K(\mathbb{Z},3)$. The total space $P$ is contractible. Now suppose $X$ is a compact manifold of dimension $n$ which is $2$-connected and $H^3(X;\mathbb{Z})=\mathbb{Z}$. Then choosing a generator of $H^3(X;\mathbb{Z})$ corresponds to a (homotopy class of) map $f:X\to K(\mathbb{Z},3)$. The pullback bundle $f^\ast P\to X$ has the property that $H^3(f^\ast P;\mathbb{Z})=0$.

Since we need a finite dimensional manifold which $f^\ast P$ isn't, let $E$ denote the $(n+5)$-skeleta of $f^\ast P$. It is compact and locally looks like $X\times\mathbb{CP}^2$. I think(?) that $\pi:E\to X$ is a fibre bundle. Since $\pi_3$ is unchanged for $4$-skeleta or higher, it follows that $0=\pi_3(E)=\pi_3(f^\ast P)$, whence $H^3(E;\mathbb{Z})=0$.

Feel free to tweak the answer if need be.

Edit As pointed out by algori and Igor, the second paragraph doesn't give you a fibre bundle.

Comment by Somnath Basu

2013-05-14T17:59:13Z

This is easily proven by induction and/or expansion. I wouldn't call it a conjecture!

Comment by Somnath Basu

2013-04-26T17:25:40Z

@ Elena - I took the liberty to retag it as algebraic topology.

Comment by Somnath Basu

2013-04-26T17:23:47Z

This looks like a homework question to me. As an aside, why do you think $\xi$ is isomorphic to $TS^1$?

Comment by Somnath Basu

2013-04-12T12:41:35Z

Use the definition of homology! Also, please read FAQ as I believe this question is more appropriate for math.stackexchange.com/

Comment by Somnath Basu

2013-02-19T07:21:00Z

@Xiao-Gang : Write the spectral sequence (in homology) for the fibration $G\to EG\to BG$. It will follow from the $E^5$ and $E^6$ page that $H_5(BSpin(n);\mathbb{Z})=H_4(Spin(n);\mathbb{Z})$ and $H_6(BSpin(n);\mathbb{Z})=H_5(Spin(n);\mathbb{Z})$. It follows from Milnor-Moore (with your input on the homotopy groups of $Spin(n)$) that these are torsion groups. Now you just have to compute these!

Comment by Somnath Basu

2013-02-19T07:19:11Z

@Xiao-Gang : Write the spectral sequence (in homology) for the fibration $G\to EG\to BG$. It will follow from the $E^5$ and $E^$ page that $H_5(BSpin(n);\mathbb{Z})=H_4(Spin(n);\Z)$ and $H_6(BSpin(n);\mathbb{Z})=H_5(Spin(n);\Z)$. It follows from Milnor-Moore (with your input on the homotopy groups of $Spin(n)$) that these are torsion groups. Now you just have to compute these!

Comment by Somnath Basu

2012-10-28T17:37:17Z

This is standard homework question and not suitable for this forum! Try math.stackexchange.com/

Comment by Somnath Basu

2012-09-26T17:09:15Z

@ OP - If I give you the vector $(1,0,0,\cdots,0)$ can you find a unique matrix $A$ which does the job? I don't think so!

Comment by Somnath Basu

2012-09-20T04:11:04Z

This is a contentious, debatable comment that doesn't answer the question asked by OP.

Comment by Somnath Basu

2012-09-19T12:13:11Z

Thanks for your answer! That's what (that $[x,x]=0$) I was thinking naively but I dug up some literature and realized that it's not the case. It's non-zero when $n$ is odd and generates a $\mathbb{Z}_2$. When $n$ is even it generates an infinite cyclic subgroup which splits on occasion. Judging from old papers, it seems that it would be too stupid (on my part) to expect a complete answer. Among other things, it would grossly underestimate the intricacies involved in knowing the homotopy groups of spheres!

Comment by Somnath Basu

2012-08-30T03:27:33Z

@John, eigenbunny - If $M$ is not simply connected then $LM$ is not connected; therefore, $\pi_1(LM)$ is meaningless unless you specify which component. The constant section only provides a section for $L_0M \to M$, where $L_0 M$ is the component of constant loops. The components of $LM$ are indexed by the conjugacy classes in $\pi_1(M)$.

Comment by Somnath Basu

2012-08-23T21:09:06Z

It's better to say why the base point fibration (you perhaps mean the loop space fibration $\Omega M\to LM\to M$) gives a short exact sequence. In general, you get a long exact sequence; here it splits to short exact sequence as the fibration has a section. Another point to note is that $LM$ is not connected if $M$ has a fundamental group.

Comment by Somnath Basu

2012-08-23T20:09:57Z

What kind of differential forms are you thinking of when you say "de Rham cohomology" in the context Frechet manifold? In case you need this for $LM:=C^0(S^1,M)$ and for singular cohomology then a lot more can be said; the Betti numbers of $LM$ can be computed for a large class of $M$.

Comment by Somnath Basu

2012-08-22T16:00:25Z

Have a look at this : en.wikipedia.org/wiki/…

Comment by Somnath Basu

2012-05-18T14:50:30Z

@Igor - Sounds like a bloody scene from a Tarantino movie - gore 'n stains!