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Ian Wright

Next year's plans

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I'm going all applied and physical.

In parallel I'm going to do M820 (on my own, not officially with the OU) and study physics, particularly quantum mechanics, a bit of electromagnetism, and relativity.  I've done all the maths to enable me to do this now.  I've previously covered vector calculus which is required for electromagnetism, and this year I've covered differential geometry in a fairly abstract way, such that the way it's presented for general relativity seems pretty easy.  M820 will cover Lagrangian formalism, and once I've done that I'll go back to my M827 text and re-read the bit about Lagrangians and Hamiltonians in the setting of a differentiable manifold.

All this stems from buying Roger Penrose's 'Road to Reality' tome, which I foolishly thought I could just read and then know everything.  It's taken me 5 years of reasonably intense mathematical studying to get to the point where most of it makes sense, although I've still got a long way to go.

Year after that?  Probably start working on my proof of the Riemann hypothesis big grin

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Ian Wright

Quit the OU?

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Yet another change of plan - I think I'll bring my OU journey to an end, but certainly not my maths studies.  I don't think I get value for money any more.

Each module costs me about £1200.  To complete the MSc, given inflation and the increased fee for the dissertation, I'll therefore be looking at between £5000 and £6000 over the next 2 or 3 years.  What do I actually get, for that outlay?

I've made a list, comprising:

  • A qualification - but I don't actually need one, and will not do anything with it in any case
  • Tutorials - all online, and rarely that helpful
  • A tutor - all he does is mark the TMAs.  The one I've had this year has been slow to respond to queries, not that I send him that many, and I can manage generally on my own.
  • An exam - it's quite nice to have official affirmation that you've learnt something properly, but is it really necessary?
  • Chat forums - the course specific ones are very quiet, and I'm generally a giver not a taker when it comes to it, so again I can do without
  • Motivation - having spent the money you're less likely to quit, but I've proved to myself this year that I can study independently
  • Course materials - are helpful, but with the right textbooks and online research, are they essential?

So It's likely I'll not be signing up for any modules next year, and will claim my post-graduate certificate and be done with it.  I've got a good few years of study materials and textbooks to play with, and thanks to the OU I've developed the mathematical maturity to be able to study effectively.

 

So, au revoir OU - if I get withdrawal symptoms there's always next year!

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Ian Wright

Manifolds, prime number theorem

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So chapter 10 of Pirani and Crampin is where it all comes together.  I haven't studied it properly yet but did read through it quickly - it basically goes over most of the work done in the first 9 chapters of the book and puts it into the context of an abstract differentiable manifold, rather than either an affine space or an embedded surface in 3D space.  And the results are all the same!  There's a bit of abstraction and technical detail using bump functions to justify this, but in reality and from a calculation perspective I don't think there's anything new.  It's a good revision chapter.

Before taking up the baton of differential geometry again this week I all but finished the analytic number theory course, completing the proof of the prime number theorem.  Feels like an achievement.  I've understood evey bit and followed all the proofs, but if I was asked to write it out from memory - I'd have no chance!

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Ian Wright

Differential Geometry - it's still hard!

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Have now worked through 9 chapters of the course book and am about halfway through TMA03.  Finally getting to the good stuff - surfaces, curvature, Riemann tensor, parallel transport; all the machinery for general relativity.

I'm still finding the textbook hard.  It makes massive jumps, and hides many of the important results in the exercises.  I've never worked through any maths as slowly as this.  The other thing that is lacking is worked examples.  There are solutions in the course notes to many of the exercises, but again these are pretty brief.  Calculations in differential geometry always seem to involve masses of tedious algebra with multiple subscripts, superscripts and exponents, which confuses things horribly.

I'm definitely getting there though.  Will be back to number theory again soon, but will be looking forward to chapter 10 - manifolds!

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Ian Wright

Prime Number Theorem

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Edited by Ian Wright, Wednesday 24 April 2013 at 23:00

So now I know how it's all proved.  Not yet halfway through M829 and we've reached the climax.  Well perhaps not, but I undersand it all in principle - I think.

In M823 we covered a few equivalent statements to the actual theorem.  The prime number theorem states that pi of x tilde operator x divided by log of x for large x .  This can be shown to be equivalent to psi of x tilde operator x or psi sub one of x tilde operator one divided by two times x squared where psi of x equals n ary summation over n less than or equals x over normal cap lamda of n and psi sub one is its integral.  Oh and normal cap lamda of n is Mangoldt's function where normal cap lamda of n equals log of p whenever n equals p super m and zero otherwise.

Next, an integral expression is constructed that is equal to psi sub one of x .  This expression contains the zeta function and its derivative in the form zeta super prime of x solidus zeta of x .  Using the residue theorem, and knowledge of the function's poles, this can be integrated and shown, as x approaches infinity, to equal one divided by two times x squared .

And the prime number theorem is proved.  That only took about 4 and a half years of studying maths to understand.

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Ian Wright

Frobenius' theorem

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I'm posting this not really for information, but as I'm trying to get my own head around it.  It's a theorem in differential geometry, giving conditions for the existence of integral submanifolds of a given space, given a 'distribution' of that space.

So what the hell is all that about?  Start with a 2-dimensional space (eg a sheet of paper) which has a vector field defined on it, such that each point (on the paper) has a little arrow defined at it.  A bit like a weather map with wind speed drawn on it.  It's pretty intuitive that you can join up those arrows with smooth curves, such that at each point there is a single curve that goes through it, and the curves fill the whole space (page).  This is demonstrably true as long as the vector is never zero.

Frobenius' theorem takes that up to an arbitrary dimension.  Given any space, a 'distribution' defines a 'plane' at each point - a plane here can be of arbitrary dimension, not just 2.  Frobenius' theorem gives the conditions that mean that the distribution define a 'foliation' of the space, such that the defined planes join to form a surface (or hyper-surface) through each point.  In 3D this is understandable - at each point in a 3D space a 2D plane can be defined, and these planes may be tangential to a 2D curved surface in the space, such that the whole space is filled by these curved 'sheets' - this is what is meant by a foliation.

How do you define the plane at each point?  There are 3-ways.  You can define a set of basis vectors at each point (in a 3D space defining a 2D plane would mean giving 2 vectors).  You can define a 'characterising (n-m) form' or you can define a set of basis constraint 1-forms.  What the hell are 'forms'?  basically they are formulae that map vectors to the real numbers.  A 1-form takes a single vector as an argument, a 2-form takes 2 vectors etc.  'Forms' form a vector space themselves, and in the normal 3D Euclidean space we're all familiar with a 1-form is just a vector, and it defines a plane via the set of vectors it has a zero dot product with.

So, given a space of arbitrary dimension, and a rule for defining a characterising form at each point, Frobenius' theorem tells us whether or not the distribution gives a foliation.  If the form at each point is theta then Frobenius theorem states that d times theta logical and theta equals zero .  This gives necessary and sufficient conditions for the foliation to exist.

Why is this important?  Don't know yet, but it is important I think with regard to the solution of sets of differential equations.  I think I'm getting it though!

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Ian Wright

How plans change

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Just found an old notebook from before I started my OU studies, and in it were my plans for my open degree.

2008/2009 - MST121 and MS221

2010 - M208

2011 - MST209

2012 - DD202

2013 - M337 and SM358

2014 - M338 and S357

So what happened to all that?  If I'd stuck to that plan I'd have done a 60 point economics course last year and would now be starting complex analysis and quantum mechanics.  Next year would have to have changed as M338 is no more, like S357.

It started off right, but I found the workload in the first year so light that I did 208 and 209 together, advancing things by a year.  I then decided that I'd just do maths, as they dropped the old relativity course and I wasn't that interested in economics anyway.  I have the DD202 course books (£10 from eBay) and may read them one day.  I still plan to learn quantum mechanics and relativity one day, probably once I've got the MSc.

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Ian Wright

Galois finished

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Pretty much finished Stewart's book.  I understand the basics of Galois theory now.  I've followed most of the proofs, I know enough about rings and fields for my liking at this stage, I understand why we can't trisect an angle or duplicate a cube, and I understand why a general quintic equation is not soluble.

Abstract algebra is back on the shelf, for now at least.

Still got a bit more set theory to finish off, but I've strayed into knot theory for bedtime reading, and have started my quest to understand differential geometry - the old OU course on elementary differential geometry which I plan to cover in the next few weeks before starting on the old OU MSc differential geometry.  I think I understand 1-forms now, which is a start.

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Ian Wright

Banach-Tarski paradox

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First take a sphere S2 in 3D space.  Next consider the free group G generated by 2 elements, i.e. <a,b> which consists of all possible combinations of a, b, a-1, b-1.  Each element of G can be written in a unique way.  G can be split into 5 disjoint sets: G = {e} U S(a) U S(a-1) U S(b) U S(b-1) where S(x) represents the set of elements in G whose 'string' starts with x.

It is clear that aS(a-1) is equal to {e} U S(a-1) U S(b) U S(b-1) (the a 'cancels' the a-1 at the start of each element in S(a-1) leaving all elements starting with either a-1, b or b-1.  Similarly bS(b-1) is equal to {e} U S(b-1) U S(a) U S(a-1).

So aS(a-1) U S(a) = G.  Same for bS(b-1) U S(b).

Now consider a and b to be rotations of the sphere about orthogonal axes by some irrational number of degrees - say 1 radian.  Every element of G is a combination of rotations about these 2 axes, and is thus a rotation.  Each element of G (except e) will therefore have 2 fixed points on the sphere.  Let the set of all these fixed points be Q, which is countable (as the elements of G are countable, being...equinumerous with N?...Wait!  This is wrong!

I thought I understood this, but it doesn't work.  The elements of G aren't countable, are they?  Perhaps they are.  Back to the drawing board.  Rest of the proof involves the action of G on the set of points in the sphere, writing the set of points as a disjoint set of the orbits of G acting on S-Q, invoking the axiom of choice to select one point from each orbit which are put into a set M.  This set is then composed with the disjoint bits of G to decompose the point-set of the sphere into 4 bits, two of which are translated across (say MS(a) and MS(a-1)) and then MS(a-1) is rotated by rotation a, and hey presto you've got your original sphere back.  Do the same with the non-translated bits and you've got 2 spheres.

Few technical details with regard to the set Q - you have to rotate it by something not a member of G or something like that, and to extend the transformation to a solid ball rather than a hollow sphere you have to include the centre as a separate point and use the projections of each point to the centre, but it's largely the same.

It's all nonsense anyway - axiom of choice is clearly false, certainly when it comes to maths that could be applied to the real world.  At least that's my opinion.

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Ian Wright

Galois Theory

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I get it - I think.  Nearly finished Stewrart's book - not properly worked through some of the later chapters but have read them fairly carefully, and followed (most of) the proofs.

The gist of it is a correspondence between a group (the Galois group formed from automorphisms of a field) and extensions of the field.  If the field of rational numbers (Q) is extended by adjoining the element i, the automorphisms of Q are the identity and complex conjugation (a+bi maps to a-bi).  These form a group isomorphic to Z2.  With more complex field extensions you get more complex groups.  The correspondence is one-one if the field extension is both normal and separable.

The roots of a polynomial with coefficients in Q probably won't be in Q because they will involve nth roots.  Probably the most significant result from Galois theory is that a polynomial is solvable by radicals if and only if the Galois group of the field extension that includes the roots is soluble.  For a group to be soluble if has to have a chain of normal subgroups such that adjacent quotient groups are Abelian.  The Galois group of the general polynomial of degree n is the permutation group Sn. Quadratics, cubics and quartics are soluble because their Galois groups are soluble.  The general quintic equation is not solvable as S5 is not soluble.  It has one normal subgroup, the alternating group A5, which itself is simple (has no normal non-trivial proper subgroups).  So that's why you can't solve a general quintic.

I've still got a bit of work to do yet with a few exercises, and a bit more study of the more abstract Galois theory in finite fields, but overall, I think I get it.

Galois theory really is a great piece of classical mathematics, giving the answers to the celebrated problems of antiquity (squaring the circle - not possible because pi is transcendental and you can only construct fields extensions of order 2n using ruler and compasses), polynomial solutions and the fundamental theorem of algebra.  It's also a great way of consolidating group theory knowledge and learning about other algebraic structures - rings, fields, integral domains and vector spaces are all there.  I highly recommend anyone who knows maths to read up on it.

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Ian Wright

Axiom of Choice

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Edited by Ian Wright, Saturday 3 November 2012 at 23:27

I've been doing well with my private study - am making a proper go of learning Galois theory and boning up on set theory, which leads me to the title of this post.

The Axiom of Choice (AC).  Introduced initially by Zermelo when formulating the axioms of set theory, so that they would cover all of Cantor's work on infinite sets and avoid paradoxes such as Russell's.  It soon became apparent that mathematicians had been using AC for years without really knowing it.

I don't accept it.  If accepting it means you can prove that a sphere can be cut into no more than 9 pieces which are then reassembled into 2 spheres the same size as the first one (Banach-Tarski paradox/theorem whichever way you look at it), then it either has to be wrong or its level of abstraction is so far removed from reality that it's irrelevant.

You need the axiom of choice to fully appreciate Cantor's infinite cardinals.  Again, there's got to be something wrong here, and it must be with the priniciple that a bijection between infinite sets means they are equinumerous.  How can the set of natural numbers be equinumerous with the set of even natural numbers?  I know there's a bijection n -> 2n, but by any reasonable reckoning there are clearly half as many even natural numbers as there are natural numbers.

I've come to the conclusion that infinity is just infinity, which just means without bound.  This works for epsilon-delta treatments in real analysis (but for sequential continuity you need AC).  Complex analysts happily define the set double-struck cap c union open normal infinity close and then divide by zero.  I'm happy with that.  I'm not happy with saying there are more irrational numbers than rationals on the one hand, and then saying that between any two rational numbers there is an irrational, and between any two irrationals there is a rational.  Makes no sense.  I follow the proof, just don't accept the philosophy.

I'll soon finish the set theory book.  I'm then going to study the Banach-Tarski proof (which is on about page 2 of my Axiom of Choice book) just so I understand what I don't believe.  Mohammed put it best when he said (probably in the Koran somewhere) that one should study witchcraft, but not practice it.

More on Galois theory anon - algebra gets more interesting, and so far seems devoid of some of the nonsense that seems to plague foundational mathematics such as set theory and logic.

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