Professor Sullivan discussed the two main problems we are faced with in mathematics
1. Recognition - Given an object, determine what we are looking at.
2. Classification - Given certain specifications, list all the possible objects.
When discussing these problems it is necessary to define what we mean by two maps being equivalent. This definition depends on what situation we are in and what we are studying.
I. Maps between two different spaces.
[Definition] Two maps f: X -> Y and g: X' -> Y' are said to be equivalent if there are structure preserving bijections Ixx': X->X' and Iyy':Y->Y' such that Iyy'*f = g*Ixx'.
What a structure preserving bijection is depends on the spaces we are dealing with. For example, bijections for sets, isomorphisms for vector spaces, groups, homeomorphisms for topological spaces, diffeomorphisms for smooth manifolds, etc.
Since from this point of view X,X' are equivalent and Y,Y' are equivalent it suffices to just consider g: X->Y
[Examples]
1) If X and Y are finite sets all the important information of a map are contained in the following data:
-Cardinality of domain
-Cardinality of range
-Cardinality of preimage of each point in range
2) If X and Y are finite dimensional vector spaces the problem is even simpler. All we need is:
-Dimension of domain
-Dimension of range
-Dimension of image
II. Maps between a space and itself.
[Definition] Similarly, we define maps f: X -> X and g: X' -> X' to be equivalent if there is a single structure preserving bijection Ixx': X->X' such that Ixx'*f = g*Ixx'.
Thus, in this case we are looking for solutions to g = I^(-1)*f*I. This is in contrast to case I where we are looking for solution to g = Ixx'^(-1)*f*Iyy'. In case II there are less unknowns, so the problem is such more difficult and its study is much more fruitful.
[Examples]
1) Finite Dimensional Vector Space
The problem of finding equivalent maps translates to determine conjugacy invariants of matrices. Some familiar conjugacy invariants are a matrix's determinant and trace (and in fact each coefficient of a matrix's characteristic polynomial).
Thus, just by considering this simple example, we can see the strong difference in cases I and II. In case I we had a finite number of criteria by which every map could be classified. In case II, given a map between an n-dimensional vector space and itself, we have an n-dimensional family of conjugacy invariants.
2) We can consider power of maps now and study how successive powers of a map behave.
3) X is a finite set and our map is a bijection.
(In terms of group theory, out problem would translate to determining the conjugacy classes of Sn - the group of permutation)
We know that one particular bijection between a set in itself is a a "rotation" of itself elements. Namely, if we label our elements 1 through n, the bijection would send 1->2, 2->3, ..., n-1->n, n->1. (There is only one such bijection up to isomorphism).
We can then study every bijection by breaking it up into what are called cycles, which is a group of elements that act as a rotation. To determine the cycles of a bijection we start with an element and see where this element goes under powers of our bijection. We continue looking at powers of the bijection until we come back to the element we started with. The elements that we cycled through, in order, are the member of the first cycle.
We continue this process with another element not in the first cycle to determine the second cycle. This process ends after a finite amount of time.
Looking at the cycles of a bijection gives us a "picture" of the bijection. For example, suppose our set has 3 elements. The possible bijections are as follows:
-3 cycles of length 1. This would be the identity map as each element remains fixed under the map.
-1 cycle of length 3. This would be what was described as a rotation above.
-1 cycle of length 1, 1 cycle of length 2. The map would fix one element and interchange the other two.
To get an idea of how big the class are that we are looking at the number of bijections is n!, which is approximately (n/e)^n. The number of different pictures (or in different language the number of conjugacy classes of Sn) is approximately e^(n/2).
Now the discussion turned to how the Greeks compared the length of two objects.
How do we compute the ratio of two objects such as the ones below?
|-----------------|
|---|
We put as many of the smaller pieces as we can in the bigger piece. For example, we can put 4little pieces inside the smaller piece
|-----------------|
|---|---|---|---|
We then take the remaining part of the bigger piece and see how many times it can fit inside the smaller piece. For example, we can put 1.
|----|
|--|
You then continue this process definitely or indefinitely to conclude that the ratio of the original lengths is (3 + (1/(1+(1/(x+ ...)))) where x would be the next step of the process.
This process could equivalently be discussed by studying rotations of an arc around a circle.
[Theorem] A continued fraction of a number X is eventually periodic iff X is the root of a quadratic equation with integer coefficients.
[Problem for homework] Given two curves on a surface, is there always a homeomorphism of a surface to itself that sends one curve to the other. You can assume that the two curves have minimal self intersection.
The undergraduate then began the present the progress on their projects.
Keren discussed her problem of studying the connections between hyperbolic length of curves and the combinatorial length of curves. She explained how the problem involves computing the eigenvalues and eigenvector of matrices representing the transformations of the surfaces. She successfully determined which matrices properly describe the transformation and calculated the eigenvalues and the slope of he eigenvectors.
Ren explained her project of examining surfaces transformation of the plane with 3 holes removed. She is studying two particular transformations: (Tab) interchanging points a and b, with 1 going over the top and (Tcb) interchanging points b and c, with 3 going over the top. She is then studying the composition Tab*Tcb*Tab*Tcb*... and how a curve changes under these compositions.
If we let the generators of the plane with 3 holes removed be {a,b,c} , then she explained how Tab changes b at a and changes a to b along with conjugation by a^(-1). Similarly, Tcb changes b to c and changes c to b along with conjugation by c.
Ren also said that under these compositions the combinatorial length of a curve increases like the Fibonacci numbers and there is never cancellation in the words.
No comments:
Post a Comment