T minus 86 days and counting

I received confirmation the day before yesterday that the SAAS are paying my course fees for M381 and M336 this October and so I am all set for my next year's work. I am very grateful to the Scottish system of support for the likes of me.

I am progressing ok. I finished unit 3 of the number theory half of M381 on congruence. This was fairly straight forward and didn't offer any particular difficulties. In the mean time I have also finished unit IB3 of M336 on frieze groups and again this wasn't particularly challenging. I can now see where this course is going with its classification of friezes and wallpaper patterns. I did always wonder what Alan, our tutor for M208, was always going on about when he said that there were 17 different types of wallpaper pattern and now I have an inkling of what he may mean. I have also now got through a fair chunk of unit 1 of the mathematical logic half of M381 on computability. This is ok as it is basic linear programming.

Going back to friezes, I thought I would show you an example of a frieze that can be found in our flat.

My Dad (who knows these things because he is an architect) pointed out to me that this type of cornice is called "egg and dart" which I think is an apt description. From a mathematical point of view, if you take the decoration bit in the middle as the frieze (ignoring the fact that it obviously doesn't extend indefinitely to the left and right), then it has a number of different symmetries. For example, if you take a basic element of an egg bounded by half a dart on either side, then you can translate this element horizontally left and right by multiples of the distance between two adjacent darts. This is what defines a frieze. There are also, obviously, reflections in a vertical axis (either through a dart or through the centre of an egg). There are, however, no rotational symmetries or reflections or glide reflections in a horizontal axis.

Each frieze can be analysed according to its group of symmetries and an algorithm is given in the unit for identifying the type of frieze according to its group. This is a type 2 frieze or in international notation a pm11. There are seven types of frieze in all, as defined by their symmetry group.

Generating subgroups

I have just finished going through book 2 of M336 Groups and Geometry and I am making reasonable progress. As usual I seem to be working at roughly half the speed that you are supposed to but I expect I will get through enough material to make life easier when the course actually starts in October. So now I have read four books in total from M336 and M381 and I am alternating between the two courses; i.e. when I finish a book from one course I start on a book from the other course. I will be looking at book 3 from M381 next which is on congruence.

Before I leave book 2 of M336, there are a few things I want to say about it. Firstly, I have noticed that the exercises in these level 3 courses require much more from the student. At level 2 it was pretty much spoon feeding - i.e. the books would give examples of how to do something and then there would be exercises where you repeat what you just learnt. Here at level 3 the questions are more demanding as the way to do an exercise is often left up to the student. I welcome this as I like thinking things through for myself, but at times it can be very time consuming. What has surprised me is that a certain amount of 'hand-waving' has appeared both in the text and in the answers to exercises; i.e. rather than thrashing out an answer exhaustively, a bit of 'well, the answer is like this because of such and such.' It can be a bit frustrating if the details are skipped over and a bit of shock after the intense rigour of M208.

I want to put down here some bits of this book which I thought ventured slightly into the hand-waving arena. This part of the book was the most difficult and the most time consuming and so it is worth going through so that later I have something to look back on for revision.

How can you generate subgroups? In M208 we have already met cyclic subgroups which are generated from integer powers of an element in a group. For example, if we consider the group of symmetries of the regular hexagon and label a rotation about its centre through π/3 as r then we can generate a cyclic subgroup from r as

$$<r>=\left\{r^{0}, r, r^{2}, r^{3}, r^{4}, r^{5}\right\}$$

Now the new idea is to expand this notion of generating subgroups to include two or more distinct elements of a group. In the text it says that "Informally, we define the subgroup generated by x,y,... to be the set obtained by forming all possible combinations of copies of x,y,.. and their inverses." In the margins it does say that a formal definition is given later in unit IB4 but I found it difficult to understand this without some formal definition. Here is my version of a formal definition for two distinct elements of a group x and y:-

$$<x,y>=\left\{uv:u,v \in \left\{ (x)^{i},(y)^{i}:i,j \in Z \right\} \right\}$$

Now it is clear that integer powers of x and integer powers of y are just the cyclic subgroups <x> and <y> and so we can rewrite this as

$$<x,y>=\left\{uv:u,v \in <x>\cup<y> \right\}$$

In other words we are taking combinations of elements taken from a union of the subgroups generated by x and y. This does not mean that the resulting subset is a subgroup. This still has to be verified by seeing if the subset satisfies the subgroup properties. One thing that becomes clear straight away is that <x,y> contains the elements of <x> and the elements of <y>. This is because if u and v are drawn from <x>, say, then because <x> is a cyclic subgroup, then the combinations of uv are also in <x>. What is new are products of elements which are not common to both sets.

Let's take an example. If we consider the regular hexagon again and call the symmetry that is a reflection in the horizontal axis s, then s is self-inverse and

$$<s>=\left\{e,s\right\}$$

If we also consider the half-turn about the centre r³ then this is also self-inverse and

$$<r^{3}>=\left\{e,r^{3}\right\}$$

But what about <r³,s>? We see from the above discussion that this is going to contain e, r³, and s but it also must contain combinations of elements which are not common to both sets, namely sr³ and r³s. These turn out to be the same element since sr³ is a reflection of some sort which is self-inverse so

$$sr^{3}=(sr^{3})^{-1}=r^{-3}s^{-1}=r^{3}s$$

Hence we have the subset

$$<r^{3},s>=\left\{e, r^{3}, s, r^{3}s\right\}$$

which does indeed turn out to be a subgroup of the symmetries of the regular hexagon and is isomorphic to the Klein group.

Tilings

I am still ploughing through the first unit of M336 'Groups and Geometry' and there have been a lot of new ideas to digest concerning the topic of tilings. Sometimes I find the definitions hard to fix in my mind and it is a bit like learning a new language. For example, I found adjacency and incidence bothersome and I keep having to go back to the definitions in order to understand what they mean.

A tiling can be divided into three different parts; the tiles themselves, the vertices and the edges. Each of these parts has a particular meaning. The vertices are points on the boundary where three or more tiles meet and the edges are points on the boundary which join vertices and are where two tiles meet. Now the edges don't have to be straight lines, they can be curved. Also, if the tiles are polygons, the tiles can have corners which are not necessarily vertices and sides that can be made up of more than one edge (and vice versa). All very confusing. Some of this confusion is removed by demanding that a tiling is edge-to-edge, i.e. that each side of the polygon corresponds to exactly one edge and vice versa.

Now adjacency is a relationship between the same parts of a tiling whereas incidence is a relationship between different parts of a tiling. Now here is the definition of these two concepts.

Two distinct tiles are adjacent if they share a common edge. Two distinct vertices are adjacent if they are joined by an edge. Two distinct edges are adjacent if they share a common vertex and bound a common tile.

The edges and vertices on the boundary of a tile T in a tiling are incident with T. Similarly, T is incident with the edges and vertices on its boundary. Also, if an edge E joins vertices V and W, then E is incident with V and W and these vertices are incident with E.

Another important definition is the degree of a tile or vertex. The degree of a tile in a tiling is the number of other tiles to which it is adjacent. The degree of a vertex in a tiling is the number of tiles (or alternatively the number of edges) with which it is incident.

Some examples of Archimedean tilings can be found here. An Archimedean tiling is an edge-to-edge tiling with regular polygons that is vertex-uniform. Only 11 such tilings can be constructed (allowing for one to be a reflection of the other). The first three are the regular tilings and the other 8 are semi-regular. Under each tile is a number. For example under the snub hexagonal tiling is the number 3⁴.6. In OU parlance this is the vertex type (3,3,3,3,6). If V is any vertex in a tiling, then the vertex type of V is given by listing the degrees of the tiles incident with V, starting with any one of them and proceeding in either direction in clockwise or anticlockwise order. So you can see how you have to understand all these definitions!

If you look at the snub hexagonal tiling you notice that all the vertices look the same. Choosing any vertex, then incident with it are four equilateral shaped tiles and one hexagonal tile. If we look at the equilateral triangle tile, then there are three tiles to which it is adjacent, that is its degree is 3 (since this is an edge-to-edge tiling there are as many adjacent tiles as there are sides of the equilateral triangle). If we look at the hexagonal tile then its degree is obviously six. Thus going round the vertex that we chose in either direction we end up with a vertex type of (3,3,3,3,6) (allowing for cyclic permutations of these numbers). A tiling is vertex-uniform if all the vertex types in the tiling are the same.

Not only can you define a vertex type but you can also define a tile type. It is similar in definition to a vertex type, but instead you list the degrees of the vertices incident with a tile in a tiling. The tile type of our snub hexagonal tiling is not uniform. The tile type for the hexagon is [5,5,5,5,5,5] (notice that square brackets are used) but the tile type for the equilateral triangle is [5,5,5].

My method for remembering how to find a vertex type or a tile type is that for vertex types it is tile-tile (i.e. you are counting tiles around tiles incident to a vertex) and for tile types it is vertex-vertex (i.e. you are counting vertices around vertices incident to a tile).

First books

Well, I have made a start on the course books for 'Groups and Geometry' (M336) and 'Number theory and mathematical logic' (M381). I began with M381 and the first book, which I have now gone through, 'Foundations' deals with four topics; Numbers from Patterns, Mathematical Induction, Divisibility and the Linear Diophantine Equation, much of which I have already encountered. The material is well written and the proofs are straight-forward but the exercises are a jump up in difficulty. I found myself struggling with a few and one I just couldn't see how to do at all. Still, it is to be expected in going up from level 2 to level 3.

I then started tackling the first book of M336 on Tilings and I must say that I think I like what I have read so far. It is refreshing to enter a new area of maths that I know nothing much about. This course also seems very well written and the ideas are being built up in nice easy stages.

In the mean time I have started on the second book of M381 which is on prime numbers. As this is something that interests me greatly, I will be relishing this book and am looking forward to its contents.

I have now signed up for M381 and M336 and my next aim will be to get together the funding I need to pay for the courses.

A change of heart

Well, contrary to everything I have said before about not continuing on with my OU studies, I have had a change of heart and have decided to see if I can finish my BSc honours maths degree that I signed up for in 2009. Why this change? Well, I realise that I do have a hankering to do something in maths and I may regret it in the future if I don't try and see how far I can get. I also realise that getting ends to meet from my current work in photography is getting harder and harder and so eventually I am going to have to do something else anyway. Coming back from a week in Berlin, I realised that I just have to give it a go, regardless of the potential difficulties of approaching the time of life when most people are thinking of retirement! Still, the government wants us all to work after retirement age, so I will be trying this out. I just hope the brain cells don't wear out before I get to the point where I can contribute something to this subject.

So having made the decision, it's all kind of exciting and daunting at the same time. The first thing is to concentrate on this degree which, due to the reorganisation of these qualifications, I now have to finish by the 31/12/17. This gives me four years and, at 60 points a year, I hope to get the additional 240 points I need to pass (I already have 120 points from MST121, MS221 and M208). There are now two routes a student can follow in this degree. Route A means following what was originally prescribed and, for me, at level 2 that would mean doing the 60 point 'Mathematical methods and models' (MST209) and the 10 point summer school 'Mathematical modelling' (MSXR209). For a number of reasons, I am going to plump for route B, which means doing the new 60 point 'Mathematical methods and models' (MST210) which does not require the summer school any more. However, this does mean that I will have to wait until October 2014 to start this course and it will also mean that I will have to do the new 30 point level 1 course 'Introducing statistics' (M140).

In the mean time I need to get another 60 points under my belt starting this October and I think the best option for me having only done M208 so far, is to do the level 3 30 point courses 'Groups and Geometry' (M336) and 'Number theory and mathematical logic' (M381). This is the final presentation of these two courses before they are replaced by 'Further pure mathematics' (M303). This new course will be available from 2014 and will be an amalgamated version of M336, M381 and the now extinct 'Topology' course (M338). I think I will be pursuing a more pure maths direction in my further studies and number theory is already something I think would like to specialise in, so it all makes good sense.

As for the remaining 90 points, I have only to do 60 more points at level 3 and 30 points from a free choice of OU modules. However, I agree with Chris that I want to squeeze in the level 3 30 point 'Complex Analysis' (M337) and at the moment this is presented on alternate years (I believe), though this could change. I expect I want to get a good grounding in both pure and applied maths in case this comes in useful, so a couple of level 3 30 point applied courses may be an option.

In the mean time it is eyes down again and time to start getting ahead to alleviate future stresses of workload. So I have already got hold of some of the books for the 'Number theory and mathematical logic' course and have started on the first book. It looks like some of this will already be covered by my reading of John Stillwell's book, which I am still making some progress on, so that's all well and good.

Using group theory to prove Fermat's little theorem

Here is a demonstration of the use of group theory that I thought was quite neat. Fermat's little theorem states that if p is a prime number and a is a natural number that is not divisible by p (i.e. a is not congruent to 0 (mod p)) then
$$a^{p-1}\equiv 1 \;\mbox{(mod p)   ...(1) }$$
Now recall from M208 that if p is prime then
$$(Z^{*}_{p},\times_{p})$$
is a group of order p-1 (see the Handbook for M208 on page 30). Now we can use Lagrange's Theorem (Handbook page 35) to prove Fermat's little theorem as follows. Let a be a natural number that is not divisible by a prime number p, then a will be congruent to an element of the above group, i.e. to one of the set
$$\left\{1,2,3,..,(p-1)\right\}$$
Let this element be g so that
$$a\equiv g \;\mbox{(mod p)}$$
Now g will generate a cyclic subgroup of order n (Handbook page 28) and by Lagrange's Theorem (page 35) n will divide p-1, the order of the group. Hence
$$p-1=mn$$
for some natural number m and thus
$$a^{p-1}\equiv g^{p-1}\equiv g^{mn}\equiv(g^{n})^{m}\;\mbox{(mod p)}$$
Now by definition
$$g^{n}\equiv 1 \;\mbox{(mod p)}$$
and so
$$a^{p-1}\equiv 1^{m}\equiv 1\;\mbox{(mod p)}$$
Hence (1) is proven.

Why is the sun hot?

This is a slight departure from my usual posts here but I wanted to counter something that appears in the course notes for S382 Astrophysics and in the book "Stellar Evolution and Nucleosynthesis" which I think is misleading. Dan has raised the subject of 'Why is the sun hot' in his blog and I want to discuss some of the issues that have been raised here rather than trashing his blog with lots of comments.

First I am going to quote some bits and pieces from "Stellar Evolution and Nucleosynthesis" by Ryan and Norton. In chapter 2 of their book on page 46 they have a section 2.6 entitled "Why are stars hot? Putting fusion in its place". They say:-

"In Chapter 1 you saw that hydrogen fusion provides the power that is radiated by the Sun and other main-sequence stars. In this Chapter, you have seen the importance of the gravitation energy released by a collapsing cloud of gas. In order to understand the roles of these two sources of energy, before we go any further, we pause to consider exactly what fusion is and is not responsible for."

They go on to discuss how much energy is released per cubic metre in the core of the Sun by nuclear fusion and this turns out to be about 300 Watts.

They go on to say:-

"Think about this number: 300 Wm^-3. Imagine putting three 100 W lightbulbs in a broom cupboard whose volume is about 1 cubic metre. Would that make the cupboard as bright and hot as the Sun?"

"No, clearly it would not. In fact, 300 W seems like a pathetically tiny power output for a volume as large as 1m³, especially in something as hot as the core of the Sun! Is the calculation grossly wrong? No, the power output per cubic metre is that small. Clearly hydrogen burning by the proton-proton chain is not much of a powerhouse! (But it is a big reservoir!)"

After some further number crunching the books says:

"You might be tempted at this stage to think that, even though the power released in fusion is very small, the energy released heats up the core of a star, and this is why stars are hot. The energy released is indeed very important, but it is not the reason that stars heat up. In fact the opposite is true; fusion prevents a star from getting hotter!"

"In fact, a star's temperature and luminosity are not determined by nuclear burning. However, they are maintained by nuclear burning. Nuclear fusion replenishes the slow leakage of energy from the star's core and eventually from its surface. As you saw in the last Section, the release of gravitational potential energy can do exactly the same thing, but the nuclear fusion energy source has the advantage that it lasts for much longer. That is, nuclear fusion greatly delays the gravitational collapse of the star."

"If someone had asked you at the beginning of this book, 'What makes stars hot?', you could have been forgiven for answering 'It is because they have thermonuclear reaction in their cores.' That answer sounds plausible, but hopefully now you can see that it is incorrect. Stars are hot because they have collapsed from large diffuse clouds and gravitational potential energy has been converted to kinetic energy. That is why stars are hot."

I think that the above discussion and conclusion are very misleading. You could come away with the idea that nuclear fusion in the Sun is not an important source of its energy output ("Clearly hydrogen burning by the proton-proton chain is not much of a powerhouse") but that release of energy from gravitational potential energy is ("the release of gravitational potential energy can do exactly the same thing"). This is of course, incorrect.

The question that needs to be asked is this. Would the Sun be hot now after having existed for approximately 4.6 billion years if nuclear reactions weren't taking place in its core? The answer to this is no. If there was no other energy source in the Sun apart from the release of gravitational potential energy, then at its current luminosity it would only shine for approximately 18 million years , a time-scale that is about 256 times too short. It follows then that the Sun would have cooled to well below its current temperature by now if heating by contraction was its only source of energy. We know, anyway, that the Sun is in equilibrium, which means that it is neither contracting nor expanding. You can't have energy released by gravitational potential if the Sun isn't contracting.

The argument about the three 100 Watt light bulbs in a cupboard is also misleading ("Would that make the cupboard as bright and hot as the Sun? No clearly, it would not!"). The cupboard could get as bright as the surface of the Sun and even as high as the core of the Sun, if the cupboard was perfectly insulated. This is how an ideal oven would work. In fact, I find the langauge used by the authors ("300 W seems like a pathetically tiny power output for a volume as large as 1m³") emotive and inappropriate for a science text.

So is nuclear fusion in the Sun much of a powerhouse? Yes, it is. Although of 300 Watts per metre cubed sounds like a small number it is sufficient to power the output of the Sun for billions of years and maintain the temperature in its core for the duration.

So why is the Sun hot? I would argue it is hot for both of the reasons discussed, neither of which can be disentagled or discounted. The collapse of the gas cloud that led to the formation of the Sun was responsible for getting the core of the Sun up to a temperature where nuclear fusion could take place and nuclear fusion is responsible for maintaining that temperature of the Sun, in both the core and on its surface, until now. To overemphasise the role of gravitational energy in the way that these authors have done in their book is, in my opinion, very misleading and it is disappointing if this is also true of the course notes for the OU Astrophysics course.