Merry Christmas

Hi, everyone, Hope you will enjoy a happy Christmas!

This is the second day after I return back my home university in China. I still have to work hard for my final exmanation in my home university, because I have two final test after two weeks. 

Guess what, the winding number for my  itinerary( from China to U.S and from U.S to China)about any point in northern Pacific is exactly 1!

Office hour availability this week

I will be in my office on Monday and Wednesday afternoons from 1:30, you are welcome to come talk with me about the course work or your project (or anything else).  I will be there until 3:30, if nobody has shown up by then I may go home, but you can also reserve time with me later in the afternoon if you need to schedule a meeting.   Just send me email to set it up.

John

Room 101

“You asked me once, what was in Room 101. I told you that you knew the answer already. Everyone knows it. The thing that is in Room 101 is the worst thing in the world.” George Orwell, Nineteen Eighty-Four.

“No, it’s just a math question.” John Roe, 2013

Here’s an image of the exam schedule in case anybody needs it.

2013-12-04 12.07.10

Proof of Lover and hater problem based on Brouwer fixed theorem

Here I want to share an interesting proof of the Lover and Hater Problem with you  without using the theory of winding number.

In our course, we met with Lover and hater problem first, and after a long time we study the Brouwer Fixed Theorem. In fact, we can prove the Brouwer fixed theorem with a purely analytical method, if you want to see that, download this PDF file.

Click to access brouwer.pdf

Suppose we already know the Brouwer fixed theorem, as a special case, we have

Theorem. (Brouwer fixed point theorem) Let \(f: [-1,1]^2 \to [-1,1]^2\) be continuous, then there exists \(x \in [-1,1]^2\) such that \(f(x) = x\).

Using this theorem, we can prove the Lover and hater problem in a simple and analytical way.

Problem. Let \( c: [-1,1] \to [-1,1]^2\) and \(d: [-1,1] \to [-1,1]^2\) be continuous functions such that \(c(-1) \in \{-1\} \times \mathbf{R}, c(1) \in \{1\} \times \mathbf{R}, d(-1) \in \mathbf{R} \times \{1\}, d(1) \in \mathbf{R} \times \{-1\} \), then there exists \( s,t \in [-1,1]\) such that \( c(s) =d(t) \).

This is just another statement of the Lover and Hater problem.

Proof. (Sketch)  Let

\(c(t)=(x_c(t),y_c(t)), d(t)=(x_d(t), y_d(t)).\)

Define a map

\( r: \mathbf{R}^2 – \{(0,0)\} \to [-1,1]^2, r(x,y) = \frac{(x,y)}{\textrm{max}\{|x|,|y|\}}.\)

Then it’s easy to check that \(r\) is continuous and the image of \(r\) is the boundary of the square \([-1,1]^2\).

Consider the map

\(g: [-1,1]^2 \to [-1,1]^2, g(s,t)=r(d(t)-c(s)).\)

Then \(g\) is continuous, now by the Brouwer Fixed point theorem, we know there is a point \((s,t) \in [-1,1]^2\) such that \( g(s,t)=(s,t)\).

But the image of \(r\) is the boundary of the square \([-1,1]^2\), so we have either \(|s|=1\) or \(|t|=1\). Without loss of generality, suppose \(|s|=1\), if \(s=1\), then we have

\(|x_d(t)-x_c(1)|= x_d(t)-x_c(1)=x_d(t)-1 \le 0\),

which is a contradiction!

Similarly if \(s=-1\), then we have

\(|x_d(t)-x_c(-1)|= -(x_d(t)-x_c(-1))=-1-x_d(t) \le 0\),

which is also a contradiction!

Hence we complete our proof.

Welcome back, and plans for the final class week.

Welcome back!  I hope that you have all enjoyed the break and are looking forward to the final week of our class.   Here is some information about that.

First, I’m going to try to use this last week to give one possible answer to the “where do we go from here” question, by heading off towards some topics in homotopy theory (specifically, the Bott periodicity theorem).  The first set of lecture notes has been posted.  These lectures will move fast and many proofs will be omitted or simply sketched. (Be warned!) But I hope they will be fun and will give you a sense of larger vistas opening out. Please note that none of this material will appear on the final exam.

Second, the last homework assignment has also been posted.  This assignment doesn’t involve anything that we’re going to do this week.  In fact, all it needs from the course is some of our results about covering spaces.  But it leads you through a quite sophisticated application of topology to algebra, one that I had hoped I might get to include in the lectures but omitted in the end.  This is a long and challenging assignment (10 questions – though I have tried to spell out the steps in detail).  It will not however count for any more in the grading than a regular homework assignment, so don’t feel that you are forced to burn the midnight oil over it.  I hope you find it fun.

Third, I will be making a third set of “blog participation” grades today.

Fourth and last, as the semester draws to an end, I just want to say that you have been a great class.  I hope that you have enjoyed learning as much as I have teaching.

 

Application of winding number to Planar graph, Lover and Hater problem revisited!

         When I was preparing for the final exam of course titled ‘Discrete Mathematics’ in my home university(after the final exam of MASS Program, we still have to take the final examination in our home university on the early days in January 2014), I found a simple argument that requires the application of knowledge in winding number, including the lover and hater problem!

Definition. A graph \(G=(V,E)\) is called planar if it can be embedded in the plane, i.e., it can be drawn on the plane in such a way that its edges intersect only at their endpoints. In other words, it can be drawn in such a way that no edges cross each other

Problem. Determine whether the following graph is planar or not.

Fig1 

Figure 1.

Solution.   We can redraw the graph as in Figure 2. Then no matter what we put the three edges \(ad,be,cf\) , at least two of them must lie in the same side (exterior or interior) of the loop , hence they must have an intersection point. So the graph is non-planar.

              This is the solution in my textbook, but now I think we need to use the theory in winding number to give a more accurate argument.

Solution. Suppose it’s a planar graph, then we can draw the figure to make \(aecdbfa\) as a Jordan curve, see Figure 2.

Fig2

Figure 2.

 Now we have still three edges \(be,cf,ad\),  since there are only two path components, there must be two of them, lie in the same components.

      If two of them, say \(be,ad\) lies in the bounded component, then by the LOVER AND HATER PROBLEM  we know they must intersect. 

     If two of them say \(be,ad\) lies in the unbounded component, then since the edges(i.e. path) \(ad\) lie in the unbounded component of Jordan curve \(aecdbfa\), we can choose a point near \(a\) and lies in the path \(ad\), denote this point as \(m\). Then \(m\) lies in the bounded component of  the Jordan curve \(eafbe\), but \(d\) lies in the unbounded component of the Jordan curve \(eafbe\), hence the path \(d\)  must intersect the Jordan curve \(eafbe\), it’s clear to see that the intersection point must lie in the path \(be\), hence we proved that it’s not a planar graph.

Fig3

Figure 3.

For the brief definition and property of planar graph, you can refer to this webpage.

http://en.wikipedia.org/wiki/Planar_graph

Sphere Eversion

Happy Thanksgivukkah! Regardless of which one you’re celebrating, here’s a cool (2-part) video:

Part 1
Part 2

In discussing the problem of sphere eversion, it offers a simpler problem: turning the circle inside out. As we learned a while ago with the rotation number (they call it the turning number), there is no regular homotopy in the plane from the counterclockwise unit circle to the clockwise unit circle. This is due to the Whitney-Graustein theorem which states that two loops in the plane are regularly homotopic if and only if they have the same rotation number. ( +1 \(\neq\) -1)

The second part shows the final eversion of the sphere and conceptually generalizes the turning number in \(\mathbb{R}^3\) – it’s the Euler characteristic!

And of course, it’s really neat to see all the symmetries that occur during the eversion process.