Another mile on the treadmill this morning while I listened to the NPR show on the cost of limb amputation, cleaned some dishes, and interspersed the day with short intervals of another 1/4 acre of very difficult mowing (lots of sticks and rocks) with a lawn mower that cuts out under the slightest load (about every 20 feet), getting the spanish grammar computer program to run, and finally after multiple efforts trying to boot my computer with the IPAD worth of screen on it and a working sound card, sitting down to listen to the first Harvard lecture on group theory. I thought that the idea of being able to look for conserved structures across groups was a pretty fascinating one – one that I guess that I’m dealing with in the context of languages. For example, two-to-one mapping of a concept from a language that has 2 genders to one that has none. I thought the professor was pretty good. He introduced a lot of concepts, and talked about the founders of the field Abel and Gallois. Both of them died in their 20’s, one a Norwegian who died from poverty and lack of a job, and the other one wrote most of his treatise on finite field theory the night before he died in a duel. It struck me that this may not be so unlike trying to study physiology with a chronic degenerative disease. If your Abel to solve the problem, you die of poverty and lack of a job. Otherwise, your big breakthroughs involve duels where you might die. Staying motivated, and not fleeing are big challenges, either way.

So, the basics of group under an operator are that:

1) it must return an element of the group under that operator.

2) it must associate.

3) it must have an identity.

4) every element must have an inverse within the group.

If the elements of the group are commutative under the operator, then the group is said to be Abelian.

I’m not sure if point 4 is technically the same as the transformation being reversible, or if I’m confusing transformation with set in my thinking. I wish I could get the book.

I thought it was pretty funny that the German mathematician Zagier said that he learned to think of all groups as integers, with the exception that the rule A+B=B+A doesn’t work in high math. He never understood why, he just accepted that that’s the way it works with high math. Germans like rules, but some of them have been pretty deep thinkers too. He was only 12 or 13 when he took the course, so his understanding kind of reduces everything to a pretty simple point of view.

Update 5/17/2013 Finished mowing one side of the lawn. Someone came and put in the NID so that I might eventually get a phone line and internet access. There are 2 wires, I guess a power wire that runs 48V and boosts to 72V at a remote central box if the phone rings, and then the signal itself which comes from 2 wires that come off of a bundle of 12 (I guess one is signal in and one signal out, but I am not sure about this). The other wires I guess are for high definition video. I still have to put in a line from the phone jack to the NID, drilling a hole through the trailer, etc. Checking the back of the phone jack, it is wired, although the phone connection on the trailer outside is just a label (no outlet) above the vent. I guess you have to remove the vent, drill the hole for the connection, and reach under to pull the wire. Otherwise, crawling under the trailer lacks appeal to me. It requires a flashlight, and also a lot of resistant clothing to scorpions. One sting would probably kill me.

I listened to the lecture again and took 3 pages of notes.

I need to work on the very last problem. I’m not sure that I understand what is supposed to be proved, i.e. is it that the permutation operation is not commutative for n>=3? That doesn’t really make sense to me.

From a theoretical point of view, the idea of transformations and maps could expand verbally to:

A –translate to French –> B –translate to German—> C—–translate to English—–>D

is not necessarily the same as:

A –translate to German –> B –translate to French—> C—–translate to English—–>D

If the above is true, then I guess the group of concepts would be Abelian with respect to the first 2 transformations, but every word in D would have to be in A for it to be a group to begin with. I need to do some homework to make things more concrete.

I thought of another couple of examples.

ex. 1. This would be anatomy, or possibly a model for thinking about infections across species, if A were an infection that could cross species and starts in a human A.

A –translate to cat –> B –translate to dog—> C—–translate to human—–>D

ex. 2. This could be a religion, where A begins with a Jewish idea.

A–translate to Christian –> B –translate to Muslim—> C—translate to Jewish–>D

Some more thinking about groups. I guess a little more concretely, and accurately:

Groups may work well with physical objects.

1) Take a cube of clay, and mold it into a sphere. Every atom in the cube maps to an atom in the sphere, although I’m not sure that any of the points are reversible. Yes they are. Every atom can be put back to its original atom using a translocation – although not necessarily the same one for every point. So, the coordinates of each point are transformed, although not necessarily following the same rule of transformation.

2) Another way of thinking about groups operators may be, that one defines a group with certain rules, and then considers all of the ways of transforming the individuals of the group, without breaking the rules of the group, or allowing any of them to leave.

Some more thoughts on groups in physical space:

One could consider the human body as a group with a symmetry operator at the plane bissecting the body along the spine. Every point on the left side technically has an equivalent point on the right side through it’s mirror image. This is low resolution, and technically not true for the gut where symmetry is broken with the liver, stomach, pancreas, appendix, etc. If you translate a point, and then mirror invert, it is Abelian in the sense that one could also mirror invert and translate and end up in the same place. In terms of leaving the group, if one considers the coordinate system as self-referencial and not cartesian, one technically never leaves the group.

Update 5/20: I spent a little time tackling some of the class homework. Maybe I can find the book at a library. The in-class homework isn’t yet complete, but it explores some of the confusion I have about some of the ideas.

The first proof yields a polynomial in g or h for the entire subset of the negative of (AB not equal to BA), with 2 potentially unique solutions for e. There are 4 more constraints if one wants to make all of the matrices invertible, i.e. ad-bc and eh-fg both not equal to zero, and (ae+bg)(ef+dh)-(ce+dg)(af+bh) and (ea+fc)(gb+hd)-(ga+hc)(eb+fd) not equal to zero.

Ok. There was a mistake in the above problem. Notably, I misread a c as an e in the last line of the 4 nonlinear equations. e is not equal to c. One does not substitue and obtain a quadratic equation in G=sqrt(g).

Instead, everything reduces to 2 equations:

bg=fc and f/b=(h-e)/(d-a). Testing this with random values. It works.

I feel challenged to try to find 2 invertible matrices that will produce a noninvertible matrix upon multiplication. That is, to explore leaving the group by a legal operation within, and to understand or explore if there is any structure to the set of matrices that would allow escape, if it is indeed possible.

Update 5/28: OK. It was pretty easy to determine that the above cannot be done. In fact, if my poor memory serves me (I no longer have any way of listening to the lecture), he did this in class. For 2×2 one can work it out. More generally, because of the multiplication of determinants, one can know that this is generally true regardless of the dimension of the problem. The more profound question (always 🙂 ) is WHY?????? Is matrix (lack of inversion competency) synonymous with dimension collapse, and dimension collapse in one of 2 multiplying matrices always associated with dimension collapse in the product of 2 matrices?

Maybe this weekend, I will have enough gas to go to the local college library and try to find the homework problems in the book. Otherwise, I have found another book at the local public library on advanced algebra. I will work through and read this book to clarify concepts.

The second exploration or requested proof at the end of class, I am really not sure about. The conclusion should be modified to state that those elements that change within the outermost range of the operation must obey the same symmetry law (of inversion in 1-d).

So, here is the conflict. Am I exploring inversion, and not the commutativity of the operators? I think that I am. It may be that only those operators that share the same point of symmetry are commutative (and this subset is Abelian). Another set of of commutative operators are those that do not involve any of the same members, so to make things concrete:

T={1, 2, 3, 4, 5, 6, 7, …}

f(T)={4, 2, 1, 3, 5, 6, 7, …}

g(f(T))={4, 2, 1, 3, 7, 6, 5,…}

yields the same as:

g(T)={1, 2, 3, 4, 7, 6, 5, …}

f(g(T))={4, 2, 1, 3, 7, 6, 5, …}

and so g and f would be transforms that are in the Abelian set. Also,

h(T)={1, 2, 4, 3, 5, 6, 7, …}

i(h(T))={6, 5, 3, 4, 2, 1, 7, …}

yields the same as:

i(T)={6, 5, 4, 3, 2, 1, 7, …}

h(i(T))={6, 5, 3, 4, 2, 1, 7,…}

so i and h would be in the Abelian set as a pair as would f and g, but not necessarily combined as i and f, or i and g. Since the set as a whole has to be Abelian, I’m not sure what this means.

Trying to think in 3-d space. If one takes every point on the left side of the body, and mirror images around a central plane bisecting the body, one stays in the group and every point is reversible. If one translates vertically by jumping, one leaves the group, but comes back when gravity finishes working – maybe to the same position.

I don’t think that the symmetry point is required to be an element. It can be in between elements.

A couple of more interesting math classes: algorithms in French, a German math philosophy class, and a modelling class. One at a time is enough thinking for me.