I sometimes create small problems (mostly at pre-college level), to have fun. Here are three recent ones. Readers may give it a try or point out issues with problem-statement or indicate that it is trivial.
Please note that some version of these problems may exist somewhere as there is a tendency of familiar ideas to echo through the brain. May be I have read something like this somewhere in the past. At any rate, this is just some musing for fun.
Problem 1
A connected, directed graph is called a ‘Calcutta Graph’ if degree of every vertex is 3, each vertex has exactly two incoming edges and one outgoing edge.
Suppose G is a Calcutta Graph with n vertices. Then at most how many directed circuits can G have?
Problem 2
A $n \times n$ grid is made up of $n^2$ squares. In each square you are allowed to draw one diagonal. If $k$ diagonals line-up, end-to-end, then they make a path of length $k$. The figure below shows a $4 \times 4$ grid with a path of length $6$.
Find the maximum number of paths of length $2n$ in a $n \times n$ grid.
Problem 3
Suppose $n$ circles and $n$ (infinite) lines are drawn in the plane such that
No three lines pass through a single point
No three circles pass through a single point
Every pair of circles intersect each other at two distinct points
Every pair of lines intersect each other at one point
Every line cuts every circle at two distinct points.
The graduate-school days are zooming away quickly from my life. It seems that the space of human memory is hyperbolic in nature. Things get thin and small at an exponential rate. I defended my thesis in July 2020 and reached India in August of the same year. The pandemic was in full swing. It was a conscious choice to return to my aging family who needed support.
Almost all of 2021 was spent on the paper that Chris and I were working on. It is an extension of the results in my doctoral thesis. We showed that connected boundary of a relatively hyperbolic group is locally connected. This removed some of the tameness restrictions that Bowditch’s theorem has on the peripheral subgroups. At the end of 2021, my advisor suggested that I should work on some projects alone.
For a few days, I felt like a radarless ship in the ocean of mathematics. Since I was not associated with any university at the time, research had to be a solo adventure. I decided to build a small research group at Cheenta. It is the organisation that I developed from scratch since 2010.
Cheenta was conceived as a training school for math olympiads for school students. Subsequently we have also accepted college students for university level programs. We already have a strong alumni and student base spread all around the world. I could easily get a few people who became curious about geometric group theory.
We started meeting weekly. In order to keep a psychological leverage, I put the meeting time on Tuesdays at 10:30 PM IST or 11 AM CST. In graduate school, that was the time when I met my advisor weekly. My brain-clock responded to this procedure and a group of 7 students was assembled for weekly adventures in geometric group theory.
2022 was also productive. I managed to prove a small theorem related to Dehn fillings and connectedness of Bowditch boundary. The entire team participated in a translation project of the famous green book by Ghys and Harpe from French to English. I also started collaborating with Arka Banerjee on another problem related to embedding of hyperbolic plane in relatively hyperbolic groups.
I hope 2023 will be productive. I want to understand how small cancellation theory, and splittings of relatively hyperbolic groups interact. It could be a powerful source of examples in group theory. Another area that interests me is the theory of hierarchically hyperbolic groups and spaces.
There are a few obstacles for my research activities. Books and journals are not easily available outside the university system. Access to conferences is hard. I was invited to speak at a conference in Ohio (to be held in April). However due to VISA and funding issues I was forced to decline the offer. There are few positive ends as well. My work at Cheenta allows me to have flexible work-hours and financial security. It also helps me to stay with my family at home.
Lets hope 2023 will be productive with what I have.
This is an ongoing survey of olympiad problems. Source material is INMO, USAMO and IMO. The goal is to indicate key ideas involved in the proof.
USAMO 2010 Problem 1
Let $A X Y Z B$ be a convex pentagon inscribed in a semicircle of diameter $A B$. Denote by $P, Q, R, S$ the feet of the perpendiculars from $Y$ onto lines $A X, B X$, $A Z, B Z$, respectively. Prove that the acute angle formed by lines $P Q$ and $R S$ is half the size of $\angle X O Z$, where $O$ is the midpoint of segment $A B$.
Key Concepts
Angle at centre is twice the angle at circumference
External angle in a cyclic quadrilateral is equal to the interior opposite angle
Simson Line
Cyclic Pentagon
Idea of Proof
If we use Simson line (feet of perpendiculars from a point in the circumference, in this case $Y$, to three sides of a triangle (in this case $\Delta A X B$ and $\Delta A Z B$) are collinear), then we immediately conclude that $PQ$ and $RS$ meet at a point $M$ on the diameter $AB$. Here is part of the picture.
Next notice that we have a cyclic pentagon $A P Y R M$. Now angle chasing shows $\angle Z M Y = \angle Z A Y$ and so on. This proof is suggested in Evan Chen’s notes.
Alternatively we may do angle chasing. Here the key idea is to show $M C Y D$ is a cyclic quadrilateral.
INMO 2014 Problem 5
In an acute-angled triangle $A B C$, a point $D$ lies on the segment $B C$. Let $O_1, O_2$ denote the circumcentres of triangles $A B D$ and $A C D$, respectively. Prove that the line joining the circumcentre of triangle $A B C$ and the orthocentre of triangle $O_1 O_2 D$ is parallel to $B C$.
Key Concepts
Similarity of triangles
Cyclic pentagon
Angles at orthocenter
Angle at the centre of a circle is twice the angle at circumference
Idea of Proof
Notice the $H$ and $O$ are orthocenter and circumcenter of two different triangles $\Delta O_1 O_2 D$ and $\Delta ABC$. One recurring theme in geometry is to bind special points in different triangles via cyclic pentagons. A major step in this problem is to identify $A O_1 H O O_2$ as a cyclic pentagon. In fact notice that $\Delta A O_1 O_2$ is similar to $\Delta A B C$ is similar to $O_1 O_2 D$.
INMO 2016 Problem 1
Let \( ABC \) be triangle in which \( AB = AC \). Suppose the orthocentre of the triangle lies on the in-circle. Find the ratio \( \frac{AB}{BC} \).
Idea of proof
Suppose \( H \) is the orthocenter and \( I \) is the incenter. Suppose \(AD\) is the perpendicular cum angle bisector passing through \( H \) and \( I\) meeting \( BC \) at \( D \).
Then \( \tan \frac{B}{2} = \frac{r}{BD} \), the inradius.
We can also show \( \tan \frac{A}{2} = \frac{2r}{BD} \).
Since \( A = \pi – 2B \), doing some trigonometry we obtain the value of \( \cos B \). The final answer is equivalent to \( \frac{1}{2 \cos B } \) which turns out to be \( \frac {3}{4} \).
INMO 2014 Problem 1
In a triangle \( ABC \), let \(D\) be a point on the segment \(BC\) such that \(AB + BD = AC + CD\). Suppose that the points \(B, C\) and the centroids of triangles \(ABD\) and \(ACD\) lie on a circle. Prove that \(AB = AC\).
Key Concepts
Parallel lines divide sides in equal ratios
Apollonious Theorem
Triangular Inequality
Centroid divides median in \( \frac{2}{1} \) ratio.
Proof Idea
Suppose \( T \) is the midpoint of \(AD\). If \(G_1\) and \( G_2 \) are the two medians, we show that \(G_1 G_2\) is parallel to \( BC \).
Then \(B G_1 G_2 C\) is a cyclic trapezium. This implies \(BT = CT\)
Now apply Apollonious theorem to deduce that \(AB^2 + BD^2 = AC^2 + CD^2\)
Finally notice that \(AB + BD = AC + CD\) from given hypothesis. Transposing we get \(AB^2 – AC^2 = CD^2 – BD^2\). Hence either \(CD – BD\) cancels out with \(AB – AC \) provided both the non-zero, or we have \(AB + AC = BD + CD\). The second case is impossible as it violates triangular inequality. Hence the first case holds.
Mathematics is all about the beauty of patterns and their reasonable connections. How about connecting patterns from seemingly different domains of the subject? This is a note borne out of a Geometry workshop at Cheenta where we tried exactly that. The audience comprised of 9 to 11 years old students.
The purpose of this note is to share some teaching methods in mathematics. A detailed discussion on this method is available in another note.
We begin the discussion with triangular numbers.
One dot gives the number 1. We may have think of this as the 1-dotted triangle. The shape of this triangle is still not very triangular.
Next we add a 2-dotted row to make a 3-dotted triangle. Thus the second triangle has 3 dots. Now it looks like a triangle!
Next step makes the evolution of the triangular shape apparent. We add 3-dot row beneath in the 3-dotted triangle to get the 6-dotted triangle. Thus the third triangle has 6 dots.
Can you guess how to create the next triangle?
Obviously we add a 4-dot row beneath the 6-dotted triangle. This gives us the fourth triangle in the sequence which is a 10-dotted triangle.
Students quickly catch on and they create fifth triangle which is 15-dotted, sixth triangle which is 21-dotted and seventh triangle which is 28-dotted. By this time, the process of designing the next triangle is understood by most students. It takes only one more indulgence to expose the series form of the number of dots.
1st triangle has 1 dot
2nd triangle has 1+2 dots
3rd triangle has 1+2+3 dots
4th triangle has 1+2+3+4 dots
Can you find the number of dots in the 20th triangle? Well it must be 1+2+3+…+20 dots. How do we sum these numbers quickly and efficiently?
At this juncture we let the cat out of the bag and introduce the students to the genius of Carl Freidrich Gauss. Write the sum backward beneath the original sum. Each column adds up to 21. There are 20 columns. Hence the sum of twenty 21’s is 420. But we added each number twice hence the sum we are looking for is 210!
As the kids get marvelled by this little trick, we quickly switch gears and look at a more geometric problem.
If you put 1 line in the plane how many regions do we have? Clearly two.
Next put another line in the plane. This second line must cut through the first line. How many regions do we have now? Four.
Let us put another line in the plane. This line must cut the other two lines and must not pass through the previous intersection point. How many regions do we have now? Students take a little time to label the regions and come up with the right answer: seven.
We continue the process of drawing by adding the fourth, fifth and the sixth line. Each time we ensure that the new line cuts all previous line. Moreover the new line must not pass through any of the old intersection points. How many regions are produced in each step?
1st line produces 2 regions
2nd line produces 2 more regions. In total we have 4 regions now.
3rd line produces 3 more regions. In total we have 7 regions now.
4th line produces 4 more regions. In total we have 11 regions now.
Students quickly notice that 5th line produces 5 new regions, 6th line produces 6 new regions and so on.
The punch line is this: total number of regions produced by n lines is exactly 1 more than the nth triangular number.
The splitting of the plane by lines (which is of more universal appeal) has this striking connection with a sequence of integers related to dotted triangles.
The spirit of the discussion should be experimental in nature. We constantly ask the students questions like:
Can you draw the next dotted triangle?
How many dots are there in the 10th dotted triangle?
Can you observe a pattern?
Can you find the number of regions created by four lines?
Draw draw draw… observe observe observe… analyze and conclude.
Yesterday we were discussing the Heisenberg group in our weekly Geometric group theory workshop. The game of Mario came up! If you have not played Mario, then here is a the only thing you need to know. Mario can jump and hit floating bricks. These bricks may pop open to produce gold coins.
The method of simplifying products of elements in Heisenberg group is similar. First a few definitions.
Let \( \Gamma \) be a group generated by \( \{a, b, c\} \). Moreover set of relators contain \( ac = ca, bc = cb, ba = cab \). We say \(\Gamma\) is the Heisenberg group.
One way to think about this is: \( c\) commutes with everything, \(a\) and \( b\) do not commute but they are \( c\) away from actually commuting. Hence we do not have \( ba = ab \) but we have \( ba = c \cdot ab \).
We claim that every element of Heisenberg group looks like \( a^m b^n c^p \) where \( (m, n, p) \in \mathbb{Z}^3 \). We play a game of Mario with the elements of Heisenberg group to illustrate this.
Choose two elements \( a^m b^n c^p \) and \( a^r b^s c^t \) in \( \Gamma \). Lets multiply them and see what happens.
\[ a^m b^n c^p \cdot a^r b^s c^t \]
Since \( c \) commutes with everything, hence we can pull the \( c \)’s to the right and get the first simplification.
Now we focus on the \( b^n \cdot a^r \) in the middle of the expression. Think of each \( b\) as Mario and each of the \(a\)’s in \(a^r \) as a gold coin bearing brick. We will push the \( b\) to the right replacing \( b \cdot a \) by \(c \cdot a \cdot b\). The gold coin is \(c\). That is if we push the \(b\) to the right through one \(a\) then one \(c \) is produced and \(ba\) becomes \(ab\).
Now we can push that \( b \) through all the \( r \) copies of \(a\) so that it moves to the right and sits with \( b^s \). In this process \( r \) gold coins (\(c\)’s) will be produced. Hence we have \( c^r \) produced.
We will need to do this \(n \) times (for each of the Mario’s or \(b\)’s in \(b^n \)). Hence \( n \times r \) gold coins or \(c\)’s are produced in the process. After this is done we have the following expression.
We use \(NFNN\) as an abbreviation for ‘not far not near’.
Annulus \(A(w, r, R)\) in a metric space \((X, d)\), for \(w \in X, 0 < r < R < \infty \) is defined as the set \[A =\{ x \in X | r \leq d(w, x) \leq R \} \].
Definition
Suppose \(P\) is a finitely generated, one-ended group. Let \(\Gamma(P, S)\) be its Cayley graph with respect to some finite, symmetric generating set \(S\). We say \(P\) is \(M-NFNN\) if there exists an integer \(M\) such that given any two points \(x,y \in \Gamma(P)\) with \(2M \leq r_x \leq r_y\), where \(r_x = d(e,x)\) and \(r_y = d(e,y)\), there exists an arc in \(A(e, \frac{r_x}{3}, 2 r_y) \subset \Gamma(P)\) that connects \(x\) and \(y\).
Question
Characterise NFNN groups.
Remark
Finitely presented groups are \(M-NFNN\) for \(M\) equal to the largest length of a relator.
I was speaking to Mr. Paul at a petrol pump in Kolkata. He lives in Birati and has an eight year old daughter. Mr. Paul noticed that the little girl has a keen interest in mathematics. People in Kolkata are chatty. Perhaps my attire and demeaner gives away my profession. He enquired briefly about the nature of my work (‘ki koren’) and then earnestly requested my counsel: ‘how should I help my daughter with mathematics?
I tried to explain a few of my ideas. ‘She should do interesting problems, parents should informally chat about such ideas at dinner table, she should not enter into formal training’ were my usual suggestions. Mr. Paul told me about the kid’s involvement with abacus. ‘She is in 4th year. I am worried if it is creating excessive pressure on her brain.’
Many kids do abacus at an early age. This ancient Chinese calculation tool has really made some inroads in Indian households. Personally I do not believe that calculation based exercises should have such a center-stage in children’s lives. It creates a wrong impression about the more important aspects of mathematics; deduction, imagination, construction and so on.
‘Why don’t you use’… I paused a little for the appropriate phrase…’creative problems?’
I had the math circle experience of eastern Europe in my mind. Additionally I was thinking about the hands-on, constructivist approach suggested by Cedric Villani, the holistic approach to education suggested by Tagore and so on. I needed something catchy, possibly in at most two words. as an alternative to ‘abacus’ and routine mathematics.
Apparently Mr. Paul found the phrase actionable. ‘Where can I find this thing; the creative problems.’
The Thousand Flowers program and Math Olympiad program at Cheenta certainly uses what I have in mind. It has been implemented systematically over the last few years. We certainly noticed that the effect of this program on children is quite profound. It transforms the way they approach mathematics.
This program at Cheenta draws from many resources. However there is no one-stop shop for all of these ideas and tools. Over the last few years I have contemplated several times about creating a book that will incorporate the ideas and tools that we use in this program.
The ‘Creative Math’ project is an attempt to write such a book. Ideally there should be 12 books corresponding to 12 years of schooling for a child. It should contain theory, problems and projects that students, teachers and parents can use. The thrust of these books is to introduce non-routine problem solving and an attitude for research in children. Moreover, it should encourage activities that promotes a culture that celebrates intellectual and spiritual happiness over material well-being. After all, as Gabor Szego pointed out, a deep involvement in mathematical enquiry over a sustained period of time is possible only when there is a cultural preference of such pursuits.
The goals of this GeoGebra model are the following:
1) Input a hyperbolic isometry of the hyperbolic plane using the matrix representation. In other words input a real matrix with determinant 1 and trace more than 2.
2) Programmatically draw the axis.
3) Mark any point P on the hyperbolic plane (upper half plane)
4) Show the movement of the point P when powers of the isometry are applied sequentially
This is a personal musing. Possible errors, uncredited excerpts lie ahead.
We constructed sequence of equivariant maps \(f_k\) from the universal cover \( \tilde {X} \) to the sequence of refinements \(T_k\).
The construction was complete up to the 1-skeleton. We want to extend the maps to 2-skeleton in a certain way. To motivate the extension, let us consider the following example.
Motivating example
Suppose S be a four holed torus. We will cut it along curves that are homotopically non-trivial, and non parallel.
Each such cut corresponds to an action on a tree T (by Van Kampen theorem).
Adding more curves is like refining the tree.
This process must terminate and eventually we will end up drawing curves that are parallel to the previously drawn curves. This is because, each time we draw a curve, euler characteristic of the parts would add up to give the euler euler characteristic of sum (circle’s euler characteristic is 0). Hence we have a reduction in the complexity of euler characteristic.
(This was presented by Professor Jonah Gastor in the colloquium in fall )
Back to \( \tilde {X} \)
We wish to reverse this process.
Given a tree we wish to ‘cut’ the simplicial complex using 1-dimensional structure called tracks.
Tracks:
Tracks: Let L be a connected 2-dimensional complex. A track is a subset S of |L| with the following properties.
S is connected
For each two simplex \(\sigma \) of |L|, \( S \cap | \sigma | \) is a union of finitely many disjoint straight lines
Fig 6: Finitely many (2) straight lines in MOP
If \( \gamma \) is a 1-complex of L and \( \gamma \) is not a face of a 2-complex, then either \( S \cap | \gamma | = \phi \) or S consists of a single point in the interior of \( | \gamma | \) . This implies S is 1 point set.
If |L| is a 2-manifold then S is a connected 1 dimensional submanifold.
Refinement of trees will add more tracks.
We will show that eventually we will end up having parallel tracks (by a complexity reduction argument of the ran of first cohomology group).
Extension to 2-simplices
Consider the map \(f_1 \) from \( \tilde {X} \) to \( T_1 \).
We have already defined the map on the 1-skeleton. Suppose abc is a 2-simplex in \( \tilde {X} \) such that \( f_1(a), f_1(b), f_1(c) \) are not pairwise equal.
As \( T_1 \) is a tree, \( f_1(a), f_1(b), f_1(c) \) is a tripod. Suppose M is the median of the tripod. It will have unique pre-image on each edge of abc. Also midpoint of each edge of the tripod will have exactly one pre-image on two of the three edges of the 2-simplex [abc].
We join the pair by a straight arc and map that arc to the corresponding midpoint.
straight arcs in 2-simplex map to edge midpoints
We may extend the \(f_1 \) to the remainder of 2-simplex linearly. In the next step we will throw away whatever is happening in the remaining space and recreate the map.
extension to \(T_2 \)
Pattern of Dunwoody, its dual tree
Define \( \tilde {\tau}_k \subset \tilde{X} \) as the preimage (under \( f_k \) of the midpoints of all edges of \( T_k \). This is a pattern in the sense of Dunwoody
\( \tilde{\tau}_k \subset \tilde {\tau}_{k+1} \)
\( \tau_k = \pi ( \tilde{\tau}_k ) \) be the projection of in X. It is a finite graph.
\( S_k \) be the tree dual to the pattern \( \tilde{\tau}_k \)
Claim 1: \(S_k \) is an \( (\mathcal {A} ) \) tree.
Proof: Each edge in \( S_k \) is ‘cut’ by a unique track which corresponds to a midpoint of an edge in \( T_k \). This defines a map from \( S_k \to T_k \). Hence edge stabilizers of \( S_k \) are in edge stabilizers of \( T_k \) implying \(S_k \) is \( \mathcal {A} \) tree.
Tree dual to the pattern
Claim 2: \( S_k \) has finitely generated edge stabilizers.
Proof: (Forwarded by Professor Hruska, explained by Arka Banerjee)
Let A be a subcomplex of B. Consider the universal cover \( \tilde{B} \). Each component of the preimage of A in the universal cover is a cover of A corresponding to the subgroup \( N = ker (\pi_1(A) \to \pi_1(B) ) \)
Why? Some loops in A may shrink to the basepoint in B but not in A. These loops are precisely the members of N (apart from the trivial loop that shrinks both in A and in B). Pre-images of the homotopy classes of these loops (under the covering projection in the cover) are distinct homotopy classes of loops in a connected component of the pre-image of A. Hence N is the fundamental group of each connected component.
Such a component is called an “elevation” of A to the cover. (It’s not the same thing as a “lift” of A.)
Why? After all, a lift will include A in the universal cover of its super space. These components are covers of A and is possibly larger than A
The deck transformations that stabilize the elevation of A would be \( im ( \pi_1(A) \to \pi_1(B) ) = \pi_1(A) / N \)
Why? The elements in \( im ( \pi_1(A) \to \pi_1(B) ) \) are the homotopy classes of loops in B which do not shrink to basepoint in A and in B (except the trivial one). These are precisely all those group elements that keep members of elevations of A in elevations of A.
(The fundamental group of the base space acts on the universal cover by deck transformations. The action is determined as follows. Take a base point \(x_0 \in X \) and a pre-image \( \tilde{x}_0 \) of \(x_0 \) in the universal cover \( \tilde {X} \). Then each element of \( \pi_1 (X, x_0) \) is represented by a loop f : I → X based at \( x_0 \). There is a unique lift \( \tilde{f} : I → \tilde{X} \) starting at \( \tilde{x}_0 \). Then we define the action of the homotopy class [f] on \(\tilde{x}_0 \) by \( [f] (\tilde{x}_0 = \tilde {f}(1) \). The quotient under this group action is the base space.)
If A is a finite graph, it is clear that any quotient of \( \pi_1(A) \) is finitely generated.
Why? Fundamental group of a finite graph is generated by the finite number of edges (of that graph sans the maximal tree in it). Hence it is finitely generated. Finally quotient of finitely generated group is finitely generated. The quotient group is generated by the images of the generators of G under the canonical projection.
Number of non-parallel tracks are bounded
Theorem [Dun85, Theorem 2.2]
There is a bound on the number of non-parallel tracks in X.
This implies that there exists \( k_0 \) such that for all \( k \geq k_0 \), for every connected component \( \sigma \) of \( \tau_k \) \ \( \tau_{k_0} \), there exists a connected component \( \sigma’\) of \( \tau_{k_0} \) such that \( \sigma \cup \sigma’\) bounds a product region containing no vertex of X’.
It follows that, for \( k \geq k_0 \), one can obtain \( S_k \) from \( S_{k_0} \) by subdividing edges. We then take \( S = S_{k_0} \)
Proof of Dunwoody bounded track theorem:
If |L| is a 2-manifold then S is a connected 1 dimensional submanifold.
Band: A band is a subset of B of |L| with the following properties.
B is connected
For each 2-complex \( \sigma \) 0f L, \( B \cap | \sigma | \) is a union of finitely many components each of which is bounded by two closed intervals in distinct faces of \( \sigma \) and the disjoint lines joining the endpoints of these intervals.
If \( \gamma \) is a 1-complex of L, which is not a face of 2-complex, then either \( B \cap | \gamma | = \phi \) or B consists of a subset of \( | \gamma | \) bounded by two points in the interior of \( | \gamma | \)
If B is a band then we get a track \( t(B) \) by choosing the midpoint of each component of \( | \gamma | \cap B \) for every 1-simplex \( \gamma \) of L and joining these points in the appropriate components of \( | \sigma | \cap B \) where \( \sigma \) is a 2-complex of L.
Untwisted Band: B is untwisted if B is homeomorphic to \( t(B) \times [0, 1] \) in which case \( \partial B \) has two components each homeomorphic to t(B).
Twisted Band: If B is not untwisted then it is twisted. In this case \( \partial B \) is a track which double covers t(B).
Twisted and Untwisted tracks: If S is a track then S is called twisted (untwisted) if there is a twisted (untwisted) band B such that t(B) = S.
Fig 7: Twisted tracks and bands
Parallel Tracks: Two tracks \( S_1 \) and \( S_2 \) are parallel if there is an untwisted band B such that \( \partial B = S_1 \cup S_2 \)
1-cochains with \( \mathbb{Z}_2 \) coefficients: Let S be a track in the connected 2-dimensional complex L. If \( \gamma \) is a 1-simplex of L, 1-cochain \( z(S) (\gamma\) = k \mod 2 \) where k = # \( (|\gamma| \cap S) \). That is the number of times S intersects \(\gamma \) modulo 2. (Is it hitting \(\gamma \) even or odd times ?)
If \( \sigma \) is a 2-simplex then \( \partial \sigma \cap \S \) has an even number of points.
If AB, BC and CA are the sides of a triangle and S hits AB and BC, then every time it hits AB, it will also hit BC. In this particular case z(S) (AB) – z(S) (BC) + z(CA) is even or 0 mod 2. Hence it z(S) is a map which goes to 0 under \( \delta \) or difference function.
Hence z(S) is a 1-cocycle.
Claim 1: z(S) is a coboundary (an image of the differential) if and only if S separates |L|, i.e. |L| – S has two components.
Notice that z(S) is a 1-cochain. Can it be regarded as a map obtained by taking difference of the value assigned to the vertices? That is, if \( \phi : \Delta^0 \to \mathbb{Z}_2 \), then is \( \delta \phi = z(S) \) ?
This is possible if and only if S splits L into two components.
Fig 8: Track leading to co-boundary.
Claim 2: A twisted track cannot separate L.
Proof: Follow the track along from side of the track long enough to reach the other side as there is only one side per se.
Hence if S is twisted, z(S) is not a coboundary. Hence it is a non-trivial element of \( H^1 (L; \mathbb{Z}_2 ) \). Why? Notice that z(S) is a cocycle (as shown earlier). Hence it is in kernel. \( H^1 (L; \mathbb{Z}_2 ) \) is formed by quotienting out the coboundaries from the kernel. Is S is twisted, z(S) is not coboundary (but nevertheless a cocycle). Hence it is not quotiented out. Hence it represents a non trivial element of \( H^1 (L; \mathbb{Z}_2 ) \) .
In fact no non empty union of disjoint twisted tracks separates |L|. Hence the corresponding elements of \( H^1 (L; \mathbb{Z}_2 ) \) are linearly independent.
Theorem
Suppose \( \beta = \textrm{rank} H^1 (L; \mathbb{Z}_2 ) \) is finite. Let \( T = \{ t_1, \cdots , t_n \} \) be a set of disjoint tracks. Then \( |L| – \cup t_i \) has at least \( n – \beta \) components, The set T contains at most \( \beta \) twisted tracks.
Proof: Let M be the subgroup of \( H^1 (L; \mathbb{Z}_2 ) \) generated by the elements corresponding to \( z(t_1), z(t_2), \cdots , z(t_n) \). Consider the epimorphism $$ \theta : \underbrace{ \mathbb{Z}_2 \oplus \mathbb{Z}_2 \oplus \cdots \oplus \mathbb{Z}_2}_\text{n copies } \to M $$. We can think of this epimorphism as a ‘string picker’. For example (1, 0, 0 , … 0) is mapped to \(z(t_1) \) etc.
Kernel of this epimorphism are elements mapping to the identity element of \( H^1 (L; \mathbb{Z}_2 ) \); that is the coboundary maps. Basis elements of this kernel corresponds to the number of untwisted tracks. Since untwisted (disjoint) tracks splits \( |L| – \cup t_i \) into components, hence the basis elements of the kernel corresponds to the components of \( |L| – \cup t_i \)
The proposition follows. There are atmost \( \beta \) twisted, disjoint, tracks. In T, therefore there are at least n – \( \beta \) untwisted disjoint tracks. Which leads to \( n – \beta \) components of \( L – \cup t_i \)
Suppose now that L is finite. Let
\( v_L \) = number of vertices of L
\(f_L \) = number of 2 simplexes of L
and n(L) = \( 2 \beta + v_L + f_L \)
Theorem
Suppose \( t_1 , t_2 , \cdots , t_k \) are disjoint tracks in |L|. If \( k > n(L) \), then there exists \( i \neq j \) such that \( t_i \) and \( t_j \) are parallel.
Proof: If \( \sigma \) is a 2-simplex of L and D is the closure of a component of \( |\sigma| – \cup t_i \) then D is a disc. We say that D is good if \( \partial D \cap \partial |\sigma| \) consists of two components in distinct faces of \( \sigma \)
Fig 9: only one ‘not good’ disc D without a vertex
For any \( \sigma \) there are at most three D’s which contain a vertex of \( \sigma \) and at most one other component which is not good.
If \( k > n(L) \) then \( |L| – \cup t_i \) has more than \( 2 \beta + v_L + f_L – \beta \) = \( \beta + v_L + f_L \) components.
Confusion: It follows that there are at least \( \beta + 1\) components whose closures are bands.
Since |L| contains at mosts \( \beta \) disjoint twisted bands, there is at least one component whose closure is an untwisted band. The theorem follows immediately.
