Accessibility

G is a finitely presented group.

X is its presentation complex (a simplicial 2-complex). Since G is finitely presented, the number of vertices of X is finite. Suppose $$u_1 , \cdots , u_q$$ be the vertices of X.

$$\tilde {X}$$ be its universal cover.

Fix lifts of the vertices of X. Suppose they are $$v_1 , \cdots , v_q$$.

Choose $$\mathcal {A}$$ : a preferred set of subgroups of G, stable under conjugation and taking subgroups. We will be working with trees, whose edge stabilizers come from$$\mathcal {A}$$

Let $$T_1 \leftarrow \cdots \leftarrow T_{k+1} \leftarrow T_{k} \leftarrow \cdots$$ be refinements of $$\mathcal {A}$$ -trees.

This means there is a collapse map from $$T_{k+1} \to T_k$$ for each k.

Collapse Map

A collapse map (is a G-equivariant map) that maps some edges to vertices. On the remaining tree, it is a graph isomorphism.

Pre-image of midpoint of every edge $$e$$ of $$T_k$$ is the midpoint of an edge e’ in $$T_{k+1}$$ that maps isomorphically onto e.

We will construct equivariant maps $$f_k$$ from $$\tilde {X}$$ to $$T_k$$ for all k inductively.

Here is the recipe:

• Map the (favorite) vertices $$v_i$$ of $$\tilde {X}$$ to vertices of $$T_1$$.
• Extend equivariantly to all vertices of $$\tilde {X}$$. The map is now defined on 0- skeleton.
That is if $$v_i \to f_1 (v_i)$$ then $$gv_i \to f_1 (gv_i) = g f(v_i)$$
• In order to extend the map to 1-skeleton, first note that we know where the endpoints of each edge are going.
• $$\tilde {e}$$ be an edge of $$\tilde {X}$$. Let v, w be its endpoints. Since $$T_1$$ is a tree, there is an unique edge path connecting $$f_1 (v)$$ to $$f_1 (w)$$
• Map $$\tilde {e}$$ to that path.
• For $$T_2$$, first send v to any vertex $$t_v$$  in the set $$p_1^{-1} ( f_1 (v) )$$. Similarly w is mapped to any vertex $$t_w$$  in $$p_1^{-1} (f_1 (w) )$$
• Restrict $$p_1$$ to the segment $$t_v t_w$$. It is a collapse map.
• In particular, the pre-image of the midpoint of an edge $$e_k$$ of $$T_k$$ is the midpoint of the edge of $$T_{k+1}$$ mapping onto $$e_k$$.
• Since $$f_1$$ is constant or injective on $$\tilde {e}$$ this allows us to define $$f_{2}$$ on $$\tilde {e}$$ as a map which is either constant or injective.
• Extend $$f_{2}$$ equivariantly on the 1-skeleton.
• Finally, we will extend $$f_2$$ to the two simplices. Whenever it is not constant, the image is tripod. Pre-images of midpoints of edges of  $$T_2$$ are straight arcs joining two sides of the triangle.

Pattern of Dunwoody

The pre-images of midpoints of $$T_k$$ is $$\tilde {\tau_k } \subset \tilde {X}$$

Also note that $$\tilde {\tau_k} \subset \tilde {\tau_{k+1}}$$

We will project the Dunwoody pattern to X (the presentation complex we started off with). This is a finite graph.

$$S_k$$ be the tree dual to the pattern $$\tilde {\tau_k} \in \tilde {X}$$. There is a natural induced map that sends $$S_k \to T_k$$, sending edge to edge. Hence edge stabilizers of $$S_k$$ are in $$\mathcal{A}$$.

Cut points in Boundary – 1

Proceedings from Top Secret Topology Seminar.

Disclaimer: Handle with care. Several unproved (possibly untrue) statements lie ahead.

Le G be a word-hyperbolic group and $$\partial G$$ its boundary.

Theorem (Bestvina, Messa, 1991)

If G has one end, then $$\partial G$$ is connected and locally connected.

The second part (locally connected), was proved under the assumption that $$\partial G$$ contains no cut points. The theory of $$\mathbb{R}$$treeswas used to establish:

Theorem (Bowditch, Swarup)

If G has one end, then $$\partial G$$ contains no cut points.

Proof:

Step 1: Assume that $$\partial G$$ has a cut point. We will show that G is more than one-ended leading to proof by contradiction.

Basic Notions 1

• Word Hyperbolic Group
• $$\partial G$$
• End of a Group
• Cut Point
• Locally Connected

$$\partial G$$ is compact metric space

• G is word hyperbolic implies $$\tilde {G} = G \cup \partial G$$ is a compact metric space and $$\partial G$$ is a closed (hence compact) subset of $$\tilde {G}$$. (Bridson, Chapter III.H , Proposition 3.7, Pg 429)

Sequence of ideas leading to the definition of Dendrite (from BowB)

Uniquely Arc Connected Space

A uniquely arc-connected space, T, is a hausdorff topological space in which every pair of distinct points are joined by a unique arc (subset homeomorphic to a closed real interval).

Equivalently, it can be defined as a path connected Hausdorff space which contains no topologically embedded circle.

Real Tree

A real tree is a uniquely arc connected space T such that for every $$x \in T$$ and every neighborhood V of x, there is a neighborhood U of x such that if $$y \in U$$ then $$[x, y] \subseteq V$$

Topologically Dense

Suppose $$R \subseteq T$$. We use the term topologically dense to mean that every open subset of T contains a point of R.

Separable Real Tree

A real tree T is separable if it contains a countable dense subset.

Dendron

A dendron is a compact real tree

Dendrite

A dendrite is a separable dendron.

Hence a dendrite is a uniquely arc connected, hausdorff, topological space which is compact and has a countable dense subset (hence is separable).

A dendrite is metrizable, locally connected (theorem).

Yet another definition of dendrite: a compact separable $$\mathbb{R}$$ Tree.

We want to make $$\partial G$$ into a dendrite.

Claim: G acts naturally and minimally on $$\partial G$$ (recall G is word – hyperbolic).

Minimal Group Action

The action of G on X is minimal if Gx is dense in X for each $$x \in X$$. This is equivalent to saying that X has no proper G – invariant closed subset of X. (why is it equivalent?)

Hence if there is one cut point in $$\partial G$$ then there exists many cut points in $$\partial G$$.

Define an equivalence relation $$\sim$$ on $$\partial G$$. Two points $$x, y \in M$$ are not equivalent if there is a collection C of cut points in $$\partial G$$ that each separate x from y and which is order isomorphic to the rationals.

Next assertion follows from a Theorem by Bowditch (Bowb, Theorem 5.2, Pg. 52)

Assertion: $$\partial G / \sim$$ is a dendrite D.

Picture to keep in mind

Claim: $$\partial G / \sim$$ is not a point

This follows from: (Bowb, Theorem 6.1, Pg. 52)

We will be considering the action of G on D \ {endpoints}, which is an $$\mathbb{R}$$ tree.

Basic Notions 2

1. Order Isomorphic
2. Quotient Topology
3. End points of a Dendrite

Summary:

• G is an one ended hyperbolic group, $$\partial G$$ is it’s Gromov Boundary.
• Assume $$\partial G$$ has a cut point x .
• Since natural action of G on its boundary is minimal, hence Gx is dense in $$\partial G$$; that is there are many cut points.
• Quotient out the points in $$\partial G$$ which are not separated by an order isomorphic set of cut points.
• The quotient space is a dendrite D (compact, uniquely arc connected, separable, metric space)
• D is not trivial.
• Remove the endpoints of D to get a real tree T.

Coming up next

We will be analyzing the action of G on this real tree to find that G splits over a finite group.

[Bowb] B.H. Bowditch, Treelike structures arising from continua and convergence groups, Memoirs Amer. Math. Soc. , Vol 139, Number 662, May 1999.

Free Groups, Gromov Hyperbolicity – (translated excerpt from Harpe)

Soit T un arbre simplicial muni d’une distance pour laquelle chaque arête est isométrique au segment [0,1] de la droite réelle, et pour laquelle la distance entre deus points est la borne inférieure des longueurs des chemins joignant ces points. Tout triangle de T est dégénéré au sens oú chacun de ses côtés est contenu dabs la réunion des deux autres. Un tel arbre est donc hyperbolique puisqu’il satisfait à la définition 27 avec $$\delta = 0$$

Let T be a simplicial tree (endowed) with a distance (function) for which each edge is isometric to the segment [0,1] of the real line, and for which the distance between two points is the lower bound of the lengths of the paths joining these points. Every triangles of T is degenerate in the sense that each of its sides is contained in the union of two others. Such a tree is hyperbolic since it satisfies definition 27 with $$\delta = 0$$

Pour entrer dans notre conception générale, il est important que la notion d’hyperbolicité soit invariante par quasi-isomeétrie. Bien que ce soit le cas, il se trouve que la démonstration de cette assertion n’est pas banale; nous las donnons au CHAPITRE 5.

To enter into our general conception, it is important that the notion of hyperbolicity is invariant by quasi-isometry. Although this is the case, it turns out that the demonstration of this is assertion is not trivial; we give them in CHAPTER 5.

29. – Théoréme. Si deux espaces métriques géodésiques sont quasi – isométriques et si l’un d’eux est hyperbolique, il en est de même de l’autre.

Le théorème justifie la définition suivante

30.- Définition. Un groupe de type fini $$\Gamma$$ est hyperbolique si le graphe de Cayley défini par $$\Gamma$$ and a système fini de générateurs de $$\Gamma$$ est hyperbolique.

Le théorème 29 montre entre auters que le graph de Cayley $$\mathcal{G} (\Gamma, S)$$ est hyperbolique pour tout système fini S de générateurs de $$\Gamma$$ dès qu’il l’est pour un système particular $$S_0$$. Il montre aussi que deux groupes quasi-isométrique sont en même temps hyperboliques sont immédiats au vu numéros 4 (graphe de Cayley d’un groupe libre) et 28.

29. – Theorem. If two geodesic metric spaces are quasi-isometric and if one of them is hyperbolic, so is the other.

The theorem justifies the following definition.

30.- Definition. A finite type group $$\Gamma$$ is hyperbolic if the Cayley Graph defined by $$\Gamma\ ) and a finite system of generators of \(\Gamma$$ is hyperbolic.

The theorem 29 shows among others that the Cayley Graph $$\mathcal {G} (\Gamma, S )$$ is hyperbolic for all finite system S of generators of $$\Gamma$$ if it is for a particular system $$S_0$$. It also shows that two quasi – isometric groups are at same time hyperbolic are immediate as seen in number 4 (Cayley graph of a free group) and 28.

31. – Proposition. Tout groupe libre est hyperbolique.

Le numéro 27 est une définition possible de l’hyperbolicité d’un espace métrique, mais ce n’est pas la définition las mieux adaptée à tous les besoins. Pour cette raison, nous donnons dès le CHAPITRE 2 plusieurs définitions équivalentes, en espérant aider ainsi le lecteur à se familiariser avec cette notion. Puis, comme les groupes libres ne suffisent pas longtemps comme illustration, nous consacrons l’essentiel du CHAPITRE 3 à l’une des principales sources d’examples, celle qui d’ailleurs a suggéré l’appellation “hyperbolique”.

31. – Proposition. Every free group is hyperbolic.

The number 27 is one possible definition of the hyperbolicity of a metic space, but this is not the definition best suited for every needs. For this reason, we give in the Chapter 2 several equivalent definition, hoping thus to help the reader to become familiar with this notion. Then, as the free groups are large enough as illustration, we dedicate the most of Chapter 3 to one of the main sources of examples, the one that also suggested the name ‘hyperbolic’.

Mazur Manifold – Class Lecture

Let G = < A | R > be a presentation of a group.

Tietze Transformations (finite versions) are ways that a group presentation can be altered without changing the group.

$$T_I$$ – add a relator that is a consequence of the other relators.

example: $$aba^{-1}b^{-1} , aba$$ implies $$a^2b$$

$$T_{II}$$ – add a new generator while simultaneously adding a relator that expresses that new generator as a word in the existing generators.

example: $$<a, b | aba^{-1}b^{-1}, aba>$$ transformed into $$<a, b, c| c= a^2b^3, aba^{-1}b^{-1} , aba >$$

$$T_{I}^{-1}, T_{II}^{-1}$$ – reverses of the above moves.

Fact: Finite presentation < A | R > and  < B | S > represent isomorphic groups if one may be obtained from the other via a finite sequence of Tietze Transformations.

Note: Also true for non-finitely presented groups if we generalize the Tietze Transformation.

Goal: To construct a compact contractible 4 – manifold, that is not homeomorphic to a $$B^4$$.

We will construct a Mazur 4-manifold. We begin with a 4 -dimensional solid torus $$S^2 \times B^3$$.

We will attach a single 2-handle which kills the fundamental group but does not yield a 4-ball.

Clearly $$\partial (S^1 \times B^3 ) = S^1 \times S^2$$

Begin by creating a picture of $$S^2 \times S^2$$ by doing a single 0 – Dehn surgery on the punctured unknot.

Next, we attach a 2 – handle whose attaching sphere (i.e. circle) lies in the complement of that unknot and hence in $$S^1 \times S^2$$.

That circle is pictured.

For meridional curve in the corresponding integral Dehn surgery, choose a simple closed curve that stays parallel to $$\Gamma$$ in the given picture.

$$\pi_1 (S^3 – \Gamma \cup \Theta ) = < x_1 , … , x_9 | r_1 , … , r_9 >$$

$$\pi_1(S^1 \times S^2 – \Gamma ) = < x_1 , …, x_9 | r_1, … r_9 , x_5 x_2^{-1}x_1^{-1} >$$

$$\pi_1 ( \partial M^4 ) = <x_1 , … , x_9 | r_1, …, r_9, x_5x_2^{-1}x_1^{-1}, x_7^{-1} x_5^{-1} x_7 x_3^{-1} x_2^{-1} x_7^{-1} >$$

Perform Tietze Transformations

Let $$\beta = x_7 , \lambda = x_2 , \alpha = \beta \lambda$$

Perform Tietze Transformations

$$\pi_1 ( \partial M^4 ) \equiv < \alpha , \beta | \beta^5 = \alpha^7 , \beta^4 = \alpha^2 \beta \alpha^2 >$$

Goal: To show that this group is non-trivial.

First, we will introduce an additional relator $$\beta^5 = 1$$

The resulting group is the quotient group $$\pi_1 (\partial M^4 ) / << \beta^5 >>$$. If it is non-trivial then so is $$\pi_1 ( \partial M^4 )$$.

We now have $$< \alpha, \beta | \alpha^7 , \beta^5, \beta^4 = \alpha^2 \beta \alpha^2 >$$

Genus Two Symmetry

Thurston 1.2.6 (a)

Julian Rosen created a photo and answer:

Imagine three lines of longitude on a unit sphere (embedded in ℝ3), at angles 0, 2π/3, and 4π/3. Let G be the union of those three lines of longitude. Then G has three-fold rotational symmetry and the set of points in ℝ3 that are distance exactly 14 from G form a genus 2 surface.

1.2.6 (b)

একটি সূত্রের জন্ম

(ফরাসি গণিতজ্ঞ, সেড্রিক ভিলানি, ‘থিয়োরেম ভিভান্ত’ গ্রন্থ রচনা করেন ২০০৮ সালে। বইটি খানিকটা ডায়রি লেখার ঢঙে রচিত। মূল ফরাসি থেকে বাংলায় অংশ বিশেষ অনুবাদ করছি স্রেফ খেলার ছলে। কোনো বাণিজ্যিক উদ্দেশ্য নেই।)।

লিয়ঁ, ২৩শে মার্চ, ২০০৮

রবিবার দুপুর ১টা। প্রায় জনহীন গবেষণাগারে দু’জন ব্যস্ত গণিতজ্ঞ ছাড়া বোধকরি আর কেউ নেই। লিয়ঁর এই গবেষণাকেন্দ্রে আমি গত আট বছর ধরে আছি। ইকোল নরম্যাল সুপিরিয়রের তিন তলার এই ঘর আমাদের যাবতীয় নিবিড় চিন্তনের সূতিকাগার।

এই ঘরে একটা বেশ মস্ত আরামকেদারা আছে। সেখানে বসে আমি মাকড়ষার মত আঙুল ছড়িয়ে তাল ঠুকছিলাম টেবিলে। ঠিক যেমন আমার পিয়ানো শিক্ষক আমায় শিখিয়েছিলেন।

আমার বাম দিকে, টেবিলে, একটা কম্পিউটার রাখা। ডানদিকে, একটা মস্ত ক্যাবিনেট। সেখানে কয়েকশো গণিত ও পদার্থবিদ্যার বই ঠাসা। আমার ঠিক পেছনে, টানা বই-এর তাক। সেখানে হাজার হাজার পাতার প্রবন্ধ, আদ্যিকালের গবেষণাপত্রের ফোটোকপি রাখা। এসব সেই সময়ের সংগ্রহ যখন আমার বেতন নিতান্ত অল্প ছিল। পয়সা খরচ করে বই-তেষ্টা মেটানো তখন কল্পনাতীত। এছাড়া তাকে সারদিয়ে অনেকগুলি খসড়া, হাতে লেখা ক্লাসনোট, সেমিনার নোট রাখা। কথ অগুনতি ঘণ্টা যে আমি গবেষণাপত্র পাঠ শুনে কাটিয়েছি তার ইয়ত্তা নেই। এগুলি তারই সাক্ষ্য দিচ্ছে।

সামনের ডেস্কে বসে আছে গ্যাসপার্ড। গ্যাসপার্ড আমার ল্যাপটপের নাম। প্রবাদপ্রতীম গণিতজ্ঞ গ্যাসপার্ড মঞ্জের নামে রাখা। ডেস্কের ওপর বেশ কিছু কাগজ রাখা। সেখানে দিন দুনিয়ার আট মুলুক থেকে আনা অজস্র গাণিতিক আঁকিবুকি কিলবিল করছে।

সামনের দেওয়াল জুড়ে মস্ত হোয়াইট বোর্ড। আমার সহচর ক্লেমেন্ট মোহুট একটা মার্কার পেন নিয়ে, বোর্ডের সামনে দাঁড়িয়ে। তার চোখে বিদ্যুৎ খেলছে যেন।

Motivation

Locally finite: Each vertex is attached to only finitely many simplices.

Why locally finite: To make sure it is a CW complex. Notice that the closure-finite criteria require each cell of a CW complex to meet only finitely many other cells.

Hence we do not have a situation like this:

Suppose K is any finite subcomplex of X.

We are interested in the number of infinite connected components of X – K. Call that number n(K).

Example:

Consider the Real line (with usual triangulation). If we remove one vertex (shown in the picture), then n(I) = 2 (as there are two open rays or two infinite connected components).

From $$\mathbb{R}^n, n \ge 2$$ onward, if you remove a finite subcomplex K, that can be enclosed within a (large enough) closed ball. Hence there can only one infinite connected component of $$\mathbb{R}^n – K , n \ge 2$$

This number n(K) can be different for different K.  Take for example the infinite binary tree.

Removing the top vertex, the number of infinite connected components are 2. But removing up to level 2, we will have 4 infinite connected components. And as we go down the levels, n(K) will grow.

So, we try with all possible such finite subcomplexes (possibly infinitely many of them) and check each time how many infinite connected components of X – K is there.

This is precisely what leads us to ends of X.

Ends of (X) =  E(X) = sup n(K)

Example: $$E(\mathbb{R}) = 2, E(\mathbb{R^2}) = 1, E(T) = \infty, \textrm{T = infinite binary tree}$$

Star of K: If K is a (finite) subcomplex of a simplicial complex X, then st(K) or star of K is defined to be the interior of all simplices which meet the vertices in K.

Since K is locally finite st(K) is an open finite subset of K.

If two point $$v_1, v_2$$ in X – st(K) can be connected by a path in X – st(K) then they can be connected via an edge path in X – st(K). Hence they are in a path component of X – K. Hence it is sufficient to look into the 1-skeleton of X – K to find out n(K).

Cohomology

Recall that given a simplicial complex, one can create Chain complexes and Co-Chain Complexes.

A chain complex, $$C_0 (X)$$ of vertices is simply the set of formal sums of vertices in X. In other words, if $$v_1, v_2, v_3$$ are vertices in X then $$v_1 + v_2 – v_3$$ is an element of $$C_0 (X)$$. In fact it has all such finite formal sums.

Once the chain complex is defined, one may quickly define co-chain complexes. They are formal sums of homomorphisms from $$C_0(X) \to G$$ where G is a suitable group. Usually $$G = \mathbb{Z}_2, \textrm {or} G = \mathbb{Z}$$. It depends on what we are trying to achieve.

One may think of each member of 0th cochain complex as an assignment of values to each vertex in X.

In the above picture, we have assigned 0’s and 1’s to each vertex of the finite simplicial complex. This is one member of $$C^0(X)$$. Each such assignment is nothing but a function from $$C_0(X) \to \mathbb{Z}_2$$.

We collect all such functions and their formal sum. They constitute $$C^0(X)$$.

In the present context, we may think of $$\mathbb{Z}_2$$ as {off, on}. Whenever we pick a finite subcomplex K in the simplicial complex X, we are indirectly assigning 1 to each vertex in K and 0 to all the vertices in X – K. That assignment is indeed a function from the set of vertices to {0, 1} hence a member of the cochain complex.

Formally we say that the function has finite support; that is it is non-zero at only finitely many points and zero everywhere else.

Thus the member of $$C^0(X)$$ which have finite support, are our algebraic tools for ‘picking’ finite subcomplexes K from simplicial complex X. In fact, each such cochain corresponds to one such ‘pick’.

Let $$C_f^0 (X)$$ denote the subset of $$C^0 (X)$$ with finite support.

Notice that similarly we can define $$C_f^1 (X)$$. They are maps from (formal sums of) edges of X to $$\mathbb{Z}_2$$ which have 1 assigned to finitely many edges (and 0 elsewhere).

$$C_f^0 (X)$$ maps inside $$C_f^1 (X)$$ under the coboundary map.

Why? Suppose $$\phi \in C_f^0 (X)$$ . Then $$\delta \phi (v_1, v_2) = \phi (v_2) – \phi (v_1)$$ (Recall that the coboundary map assigns ‘difference’ of values at the vertices. This can be intuited as a change in height function: value at each edge is the difference of height at its vertices).

Clearly $$\phi (v_2) – \phi (v_1)$$ can be non zero if and only if one of them is 0 and the other one is 1. But as $$\phi \in C_f^0(X)$$ has finite support, it gives nonzero output only at finitely many vertices hence $$\phi (v_2) – \phi (v_1)$$ can be non zero only at finitely many edges.

‘Ends’ is the dimension of Cohomology

Recall that cohomology groups of a space X are the homology groups of Cochain Complex.

In more simpler terms, consider the following Cochain complex:

$$… \overset{\delta_2} \Leftarrow C^1(X) \overset{\delta_1} \Leftarrow C^0(X) \overset{\delta_0} \Leftarrow 0$$

Then $$H^0 (X) = \frac{ker (\delta_1)}{im (\delta_0)} = ker (\delta_1)$$

Intuition: Image of the lower level difference map contains lower level ‘difference’ data. By quotienting that out, we may focus only on higher level differences.

We previously constructed $$C_f^0 (X)$$ that singled out the finite subcomplexes. (K’s). Next, we wish to turn our attention toward X – K. Hence it makes sense to ‘quotient out’ $$C_f^0 (X)$$ from $$C^0 (X)$$. Then we will be left out with those functions which assign 1 to infinitely many vertices.

Define $$C_e^0 (X) = \frac{C^0 (X)}{C_f^0 (X)}$$

Similarly define $$C_e^1 (X) = \frac{C^1 (X)}{C_f^1 (X)}$$

Intuition: Imagine as ends. Functions in $$C_e^0 (X)$$ flow 1’s upto ends (as 1’s appear infinitely many times at the vertices).

Claim: $$dim_{\mathbb{Z}_2} H_e^0 (X) = E(X)$$

It is not hard to see why this could be true. Afterall what is $latex H_e^0 (X)$ ? It is the collection of all those $$\phi \in C_e^0 (X)$$ that maps to 0 of $latex C_e^1 (X)$.

More explicitly, these are 0-1assignments on vertices, which have 1 at infinitely many vertices, but non-zero differences at finitely many edges.

Example:

Again consider the real line. Assign 1 at all vertices leftward from U. Assign 0 W onward (to the right).

Note that this is a member of (an equivalence class of) $$C^0_e (X)$$ as it has 1 at infinitely many vertices. Call it $$\phi$$ .

But after applying the coboundary map $$\delta$$ to $$\phi$$, we get 0 at all edges except on UV.  Clearly $$\delta \phi$$ has finite support, hence is a member of $$C^1_f (X)$$. Therefore it is in the equivalence class of 0 in $$C^1_e (X)$$ (as we have quotiented out all edge-functions with finite support).

Note that U and V are the crossover points in the above example. When we talk about the $$ker \delta_1$$, we are seeking vertex-functions in $$C^0_e (X)$$ we have finitely many crossover points (but infinitely many points with 1 assigned).

Why? Because if we cut at the crossover, we have an infinite component at least at one side of it (after all the infinitely many 1’s need to accommodated somewhere).

The recipe for finding the finite crossover but infinitely flowing vertex functions:

• Look at all vertex-functions that map to edge-functions with finite support ($$\delta^{-1} {C^1_f(X) }$$)
• Quotient out those which are 1 (on) at finitely many vertices (they do not flow to the ends)
• Final result $$\frac{\delta^{-1} {C^1_f(X) }}{C^0_f(X)}$$

The rest is easy!

Each path component of X – K must bear the same value at each vertex. Hence we can create linearly independent vertex functions corresponding to each component of X – K.

But we have taken care of all possible K as we considered all cochains.

Hence the number of linearly independent finite crossover but infinitely flowing vertex maps provide the Ends of X.

Good News: This data is cohomological. Hence it is independent of the triangulation of the space. (It is a deep theorem of algebraic topology that cohomology groups are independent of triangulation; infact can be achieved without triangulation).

Which manifold is this?

This is the first exercise from Thurston’s Three Dimensional Geometry and Topology Vol. 1.

Which manifold is this?

It is like an old trick. Try following the lines. There are actually 6 loops (circles) in this maze.

Here is a color coded picture of it.

Direction of a vector field

Let $${ f: S^n \rightarrow S^n }$$ be a map of degree zero. Show that there exists points $${ x, y \in S^n }$$ with $${ f(x) = x }$$ and $${ f(y) = – y}$$. Use this to show that if F is a continuous vector field defined on the unit ball $${ D^n }$$ in $${ R^n }$$ such that $${ F(x) \neq 0 }$$ for all x, then there exits a point in $${ \partial D^n }$$ where F points radially outward and another point in $${\partial D^n }$$ where F points radially inward.

Solution:

Suppose g is the antipodal map.

If $${ f(x) \neq x }$$ for all $${x \in S^n }$$ then $${ f \sim g }$$ . This implies $${deg (f) = deg(g) = (-1)^{n+1} \neq 0 }$$ hence giving us a contradiction.

Similarly if $${ f(y) \neq – y \forall y \in S^n }$$ then $${ g \circ f (y) \neq y }$$ for all $${ y \in S^n }$$ . This implies $${ g \circ f }$$ is homotopic to the antipodal map. But then $${ deg (g \circ f) = (-1)^{n+1} }$$ . This is impossible as $${ deg (g \circ f ) = deg (g) \cdot deg (f) = 0 }$$ . Hence we again have a contradiction.

Now define $${ \bar{F} (x) = \frac{F(x)}{||F(x)||} , x\in S^{n-1} }$$ . It is clearly a continuous vector field from $${ S^{n-1} \rightarrow S^{n-1} }$$ . (It is well defined as F(x) never vanishes. Also restriction of F to $${S^{n-1} }$$ is continuous as F is continuous. Finally as F is continuous, composing it with absolute value function is continuous. $${||F|| }$$ is continuous and F is continuous implies their ratio is continuous).

Note that $${\bar{F} = S^{n-1} \hookrightarrow D^n \rightarrow S^{n-1} }$$ where the first map is inclusion $${i }$$ and the second map is $${ \frac{F}{||F||}}$$ . Since $${ H_n (D^n ) = 0 }$$ (as n-disk is contractible), therefore $${ \left ( i \circ \frac{F}{||F||} \right)_* = 0}$$ implying $${ \bar{F}_* }$$ is of zero degree.

As $${ \bar{F} }$$ is of degree zero therefore there is a point $${ x \in S^{n-1} }$$ such that $${ \bar{F}(x) = x }$$. Hence $${ \frac{F(x)}{||F(x)||} = x \Rightarrow F(x) = c \cdot x }$$. This F points radially outward at this point (c is positive constant).

Similarly there is a point $${ y \in S^{n-1} }$$ such that $${ \bar{F}(y) = -y }$$. This is the point where F points radially inward.