## Sunday, 17 June 2012

### M338 TMA03

Finished this today had to ask for a small extension due to work, but don't know if it's been granted as my tutor hasn't replied to my request. Anyway I'll submit it tomorrow and see what happens.

This block was a bit more straightforward than Block A as it concerns what most people consider topology to be namely the invariance of certain features of a solid under a transformation. Thus the usual example given is that that a cup with a handle is topologically equivalent to a ring doughnut but not a solid doughnut as they both have a hole and can be deformed into each other. There is also a discussion of the Mobius band.

I have to admit that my heart was sinking when I first started this block as it seemed to be all about visualisation and introducing boundaries in order to classify various combinations of holes and twists in a surface. As there seemed to be no systematic way of doing this it was all a bit vague.

However by the second half of block 2 an algebraic method of charactersing surfaces was introduced and I was much happier. There are a number of ways of systematically obtaining the characteristic properties of a solid and these are essentially algebraic if a bit fiddly. The TMA was mostly concerned with the algebraic aspects of the block. There are three main characteristics of a surface

The Euler Characteristic = V - E +F where V is the number of vertices E = the number of edges and F = is the number of faces. This is invariant for topologically equivalent surfaces.
The Boundary number which is the number of separarate pieces forming the boundary of a surface
The orientability number of a surface essentially the number of twists.

Associated with each surface is an N sided  polygon with arrows along each side denoting the orientation with a number of edges of the polygon having the same name.  An edge expression is obtained by  labelling the sides in order for a given direction of the arrow and labelling the side by its inverse if the arrow is pointing in an opposite direction. Various identifications give rise to different solids. Given such an edge expression one can then go onto calculate the characteristic numbers of a surface.  This process is quite straightforward but can be quite fiddly.  Also the edge equations can be reduced to canonical form which enables the classification of the surface to be made easily both directly from the edge expression and also there is a method of classifying the solids in terms of the  connected sums of constituent surfaces eg Torus's Closed disk's etc.

The final block concerned a discussion of the coloring theorems. One of the key quantities is the chromaticiy number which is related to the number of handles of a given surface and gives the minimum number of colours required to colour the parts of the surface so that no regions next to each other have the same colour. As some readers will probably know the minimum number of colours to colour a map is 4. But this wasn't actually proven until the 1970's and only by a computer so it's doubtful to old fashioned mathematicians whether it counts as a proof at all.

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

Question 1 concerned identifying the edge expression  of a hexagon, obtaining the edge expressions and the
characteristic numbers. This question seemed quite straightforward

Question 2 was a question on the subdivisions of a surface for a given Euler characteristic this is quite straightforward

Question 3 The bulk of the TMA with a whopping 40% of the marks concerned obtaining a single edge expression from a number of constituent ones, then deducing the characteristic numbers. Then peforming a canonical transformation and obtained the connected sum form for the surface and then using that to deduce the characteristic numbers. Fortunately the numbers seemed to tie up so I'm consisitent if not correct.

Question 4 The final question involved obtaining some inequalities for the number of handles h in terms of the chromaticity number. This seemed quite straigthforward or at least the first part. However the last part which asked us to find the range of h for a large chromaticity numbers. On the face of it this seemed quite straightforward but the sting in the tail was that dreaded phrase "justifying your answers fully" as there were 10 marks for this part. I'm sure we were supposed to do more than solve the inequality for the various values of h. But I couldn't see what.

So on the whole I think I've done reasonably well assuming my tutor accepts my late submission but that last part has me worried that I've missed something quite fundamental.

1. Hi Chris.

It seems like we had similar experiences of this TMA as well. I was puzzling over that "justifying fully" part of Q4 for quite some time, trying to come up with something, anything, more to say about it that would be worth a few marks.

In fact, a lot of this TMA had me worrying that the number of points allocated was more than was warranted by the work required; I was, therefore, frequently left thinking I'd surely missed something.

All that said, we've been told that it was, indeed, the "easy" part of the course. Time will tell I guess, when we get our marks back.

I hope your tutor had no problem with the last minute extension. I expect they were okay with it.

Neil H

2. Alan will be OK with it.
And, yes, what the hell was wanted there?
arb
nellie

3. the New Scientist has a reader contribution where he discovered there is a field of maths called "pointless topology"

Which is rather nice...