Taylor Polynomials in Global Diplomacy

Theorem: Taylor polynomials have the power to end all war.

Proof: Replace the world with its first Taylor polynomial. Since this is linear, its second derivative is identically zero. Since the second derivative measures acceleration, to a first approximation the world can not accelerate.

By Newton's Second Law, since the acceleration of the world is zero, so is the net force exerted on it. In other words, the world experiences no force.

It is well-known in the literature that war is an application of force.

Since the world experiences no force, it follows that there can be no war. QED.

Balinese Candle Trick

This is a practical demonstration that the Lie group SO(n,R) has a fundamental group of order 2 for large n.

"You hold the coffee cup with your right hand underneath it, straight out in front of you. Now bring it left, under your underarm, awkwardly around front with your elbow straight up in the air. That's 360 degrees, and you're a pretzel. Keep going around counterclockwise, this time swinging your arm around over your head. At 720 degrees the coffee cup is back where it started, unspilled, and your arm is straight once more. Keep going round and round until you believe it."

Try it. It's blowing my mind, man.

Mathspeak

Right now I'm working on something called string topology. The first question everyone asks about string topology is whether it has anything to do with string theory. The answer, as far as I can tell, is no. The whole 'string' thing is a rather confusing piece of math terminology, one of several. Lie super-algebras, for example, do not fight crime. And quantum groups have nothing to do with dirty hippies beating off to The Tao of Physics. Mathematicians seem to use these terms in ways no other human beings do. As far as I can tell, their mathematical meanings are as follows...

1. super-: Z/2Z-graded
2. string: give me more money
3. quantum: I'm hip; have sexual relations with me

I hope this helps.

Integration and Inaccessible Cardinals

Here's a zippy little article on the evolution of integration that I found while preparing for tomorrow's multivariable calculus lecture. (I'll be showing the young 'uns the triple integral, and was wondering if I could make the idea of Jordan measure comprehensible at this level.) The bit at the end is the interesting part--I hadn't realised that the existence or nonexistence of certain measures could have set-theoretical consequences. Like inaccessible cardinals. Who knew?

Crochet the Lorenz Manifold

Teresa Nielsen Hayden likes, among other things, to knit. I like, among other things, to do math. Sometimes, these overlap. Like now! There's a link on her sidebar today to another intersection of mathematics with the fibrous arts. You, too, can now crochet your own model of the two-dimensional stable manifold of the origin of the Lorenz system, that chaotic dynamical weather model system thingy that spits out all those pretty butterfly pictures of strange attractors. I'm still not really sure what a strange attractor is; not my field, y'see. I keep meaning to look it up...

Actually, why don't I do that right now? Yes, now!

This is about as useful as a completely pointless thing to a person.

Wikipedia to the rescue!

Now I know. Now you do, too. We learned together. Isn't that sweet? Give me a hug. Give me a hug or I'll beat a hug out of you, so help me God. Don't make me take my belt off. I'll do it. Don't push me. That's it. You're asking for it. I'm getting Mediaeval in a moment, I swear. Just as soon as I'm done being a total sissy. Should be any moment now. Really. Any moment...now...

Really soon now...

Don't go! I'm about to get all primitive and savage on your unprotected buttocks.

Come back!

I made scones...

Hello?

Proof by Picture

The best part of algebraic topology is when you can solve a problem by doodling. In particular, if you stare at this picture with an open mind, you should soon become convinced that the Thom space of the tangent bundle to the n-sphere is, in fact, homeomorphic to the quotient space SⁿxSⁿ/A, where A is the antidiagonal.

The picture illustrates how to map the total space of this tangent bundle (considered as a subspace of SⁿxRⁿ⁺¹) homeomorphically onto the space SⁿxSⁿ-A, by stereographically projecting, from the point -x, each tangent space T_xSⁿ onto Sⁿ-{-x}. The Thom space of the bundle can be identified with the one-point compactification of the total space, and a little point-set topology will convince you that, if X is compact and B is a closed subset, the one-point compactification of X-B is X/B.

QED

Charming Silliness

Behold another miracle of the Information Age: the MASSIVE database, with information on over 1700 songs about mathematics and the sciences. And links to the lyrics of many. It seems as if there ought to be more Tom Lehrer numbers, though...But perhaps I'm on crack. As the saying goes.

The new apartment still has not burned down. This is going very well indeed.

Return of the Revenge of the Night of the Living Blog Strikes Back

Back to blogging!

I finished up all my summer obligations on Monday. I'd been teaching integral calculus, the second quarter of the U of Washington's three-quarter calculus sequence. It was a bit depressing: while some people did well, lots of people did, as it were, not so well, and they did not so well in ways that really bugger belief. I think that a fair number of people just didn't have the kind of math background necessary to do a course in calculus: most especially, they were terrible at algebra. This fits in with a general pattern I've noticed, through my time as a TA. People get into the UW with good grades and high test scores and try to jump right into calculus and still don't really know how to do simple algebra. We're not talking finitely-generated abelian groups here; we're talking simple simple, high school algebra. Nobody has ever taught these kids how to do math. (Their grammar's really weak, too.) The impression I'm left with is that high schools now are, by and large, even bigger jokes than they were when I was a teenager. This is quite sad.

More worryingly, lots of people seemed to have a lot of difficulty with slicing arguments. Every problem in integral calculus essentially works in the same way: take your problem, slice it up into little chunks or rectangles or cross-sections, solve the problem for each chunk or whatever, add up all these infinitesimal solutions to give a global solution. We used this sort of argument in every single application of integration we considered. It's a general template that should make integration problems really really easy to set up, and yet a lot of people never seemed to 'get' it. They'd write down the formulae that popped out in the end, but they'd have endless difficulty using them because they didn't seem to think about where these formulae came from. I probably didn't do a very good job explaining this because it just seems so natural to me, but even so...

They weren't picking up on the underlying logic.

Let's be honest: most people will never use any kind of (remotely) sophisticated mathematics in their lives. They won't have to find roots of polynomials. They won't need to integrate trigonometric functions. They won't need to optimise. Basic arithmetic will suit them just fine. But the habits of thought they could pick up from basic and well-designed primary- and secondary-school math courses would be invaluable. It's not the content that's most important for the average hominid, but the technique. The most important thing someone can pick up from even a cursory study of mathematics is the ability to differentiate between sound and faulty logic, or reasonable and bullshit arguments. It's so easy to learn these habits of rigorous thought by doing mathematics; everything in mathematics is so clean, and it's so easy to separate the wheat from the Chalabi. Mathematics is a wonderful, ideal playground for the exercise of the rational faculties, and once the habits of logical thought are ingrained, it becomes almost trivial to import them into real-world problem-solving and decision-making.

Mathematics trains people to be rational entities.

Or it should. America seems to be failing miserably at this.

Americans just don't know how to think. It's really the only conclusion one can reach, especially if one has been paying any attention to politics. This whole Swift Boat Veterans for Utter Bullshit thing, for example. Or Ann 'the Man' Coulter, who still apparently has some kind of career. Or this ghastly Michelle Malkin creature, peddling on the basis of shoddy dishonest research a book defending the internment of Japanese-Americans during World War II, who tried to imply on cable television that John Kerry's wounds in Vietnam were self-inflicted. (At least the host smacked her down for that one.) Or listen to any single thing President Bush has ever said in his life. It's all bullshit. The right wing (I'm generalising, of course) in America isn't offering up any kind of rational, defensible argument in its own favour, but simply pouring thick liquid bullshit like foul, foul syrup all over everything in sight. You can't debate people who work like this. You can't ever win. What else can you do but try to marginalise these people and exclude them from the discourse, and what kind of way is that to run a democracy?

This country is so fucked.

If only people learned more math...

Vector Bundles

'O Mister Hog,' cry the frenzied masses, surging like tidal jam, 'where have you been? What have you done? What great work of hoggishness has consumed thee?'

'O dear sweet readers,' I reply, 'o both of you, such wonders have I seen! I have crossed the Timeless Void that yawns sleepily beyond the outermost Sphere of the Fixed Stars, to reach the Realms of Light; and I took tea with Jesus Christ, Whose blood ransomed us from the cruel God of these Aeons. In the Realms of Light they drink only organic lapsang souchong, and they drink it very quickly, for the background radiation vapourises unshielded tea in half a microsecond flat. The crumpets are armoured with neutronium, and the butter is a quark-gluon plasma.

'And I asked the Saviour to share with me the wisdom of the Alien God, His Father.

'And Jesus said unto me, "I can suck my own dick."

'And I saw that this was Truth.'

The trip back from the Realms of Light took quite a long time, what with security and so forth. I was held up for two hours at the Sphere of Ialdabaoth, Archon of Saturn, when some lady started panicking over a mariachi band she claimed was using the lavatory 'swarthily' and menacing her with their moustaches. I got in a good bit of reading...

'O Mister Hog,' I hear you cry like tiny baby trumpets made of crab, 'what have you been reading?'

I hear this question; and I hear that this question is good.

'Several things!'

I've been reading John Milnor's book on characteristic classes, which is to do with the cohomology of vector bundles. Most of you probably aren't very interested in that. Fortunately for all of us, it's fun to try and picture certain vector bundles (and related spaces) in our heads, so let's all do that instead! You too, Jesus.

'Holy pigfuck,' exclaims the Christ. 'Hglaghlaghlaglug.'

Imagine you have some kind of topological space sitting around. Now imagine that, at every point in that space, you have a line, or a plane, or some higher-dimensional hyperplane, and that you can weave all these lines or planes together without overlapping to form a new space sort of sitting over your old one. That is essentially a vector bundle over your original space.

Imagine a circle. At every point on the circle, there is a tangent line. We'd like to assemble all of these tangent lines into a new object, the tangent bundle to the circle. We don't want any of the lines to overlap, though. It's easy to piece them together in four-dimensional space, but that's pretty impossible to really visualise. I think of it as living in three-dimensional space by giving all the lines a little twist. Imagine you tweak each line ever so slightly out of the plane of the circle, so that they never overlap; start from any one point on the circle, and twist the fibres more and more as you go until you go a quarter of the way around, when the tangent line is twisted vertically. Then you can keep twisting as you go until, halfway around, the tangent line is horizontal again. Now take this helical mess, and imagine yourself pulling on the first, horizontal tangent line until it becomes vertical. Since all the tangent lines fit together, all the other lines have to move, too; the effect sort of ripples along the circle, leaving all the tangent lines vertical in its wake. When the dust settles, you're left with a cylinder, living happily in three-dimensional space. You have just discovered that the circle is parallelisable.

It is not true that you can always twist the fibres of a vector bundle until they all become parallel. Imagine now the sphere. At every point on the surface of the sphere, we have a two-dimensional tangent plane; we can assemble all of these tangent planes to form the tangent bundle to the sphere. None of these tangent planes are supposed to overlap; I imagine them sort of fitting together like the petals of a rose, and again I use a sort of 'twistiness' to imagine their necessary extension into higher dimensions. Just as there was a circular hole in the centre of the tangent bundle to the circle, there is a spherical hole in the centre of this bundle. I try to picture myself both inside this cavity looking out at the bundle, and outside the bundle looking in. From the inside, it looks to me like a curvilinear zero-gravity Gothic cathedral made of glass, tinted, for some reason, with patches and streaks of blue. The planes seen at strange angles and edge-on seem to form an infinite web of buttresses, and everything tries to curve away. From the outside, it is more like a green glass carnation mating with a spiral galaxy, but with perfect spherical symmetry, and infinite in every direction. I can't twist this structure to arrange all the tangent planes in a nice parallel way without it breaking somewhere; wherever I start, the parallelising ripple always tears a hole at the antipodal point. This intuitive picture of mine actually is fairly accurate: the tangent bundle to the sphere is not parallelisable in the technical sense, but if you delete any one point from the sphere and its corresponding tangent plane, what's left is.

I'm not sure where the colours come from...

Projective space is deeply involved in the study of vector bundles. You can construct the projective plane as follows: take a sphere. Imagine taking a point on the sphere, and its antipodal point, and pulling them together to meet somewhere inside the sphere. Now do it with another pair of points, but make sure they meet somewhere else. Do this with every single point on the sphere, each point and its antipodal point meeting each other but meeting no other points. It's a weird, collapsed sphere that can't properly live in three dimensions, but I imagine it as looking a bit like a seashell, all curled up on itself. And pink. I've no idea why, but this projective plane is tinted a distinct pink. Like a salmon. Now imagine that, back on your original sphere, you attached to every point a line connecting the point to its antipodal point, and stretching off to infinity in both directions beyond the sphere. Picture what happens to all these lines as the sphere collapses into the projective plane; what pops out is called the canonical line bundle over the projective plane, and I still can't imagine what it might look like, but I'm working on it. I think it's rose red.

A Brief Thought on Category Theory

I need to invent a construction called the 'associated category'. Then, when writing on the blackboard, I can abbreviate it 'ass cat'.

A Dorkish Observation

So I've been playing a lot of Sleuth lately; to help myself keep track of everything, I tend to draw a directed graph for every game, with one vertex for each suspect, and an arrow from one vertex to another if the first suspect tells me the motive of the second. In this setup, a very simple proposition follows immediately...

Prop'n: In any game of Sleuth, the corresponding (connected) directed graph contains at least one cycle, and therefore the graph has nontrivial homology.

pf: Someone must supply your client's motive. QED.

Topology Seminar

The topology seminar has come and gone...How did it go, I hear you ask? I've really no idea...

Ursula told us how to go about knitting Klein bottles and Mobius strips, and directed us to this website for further details, and to see one person's valiant attempt to knit the projective plane. Her Klein bottle was quite splendid: I am tempted to learn how to knit so that I can make one of my own. (You know I'll try wearing it as a hat.) I'd always pictured the projective plane as looking a little bit swirlier, though...I think of it as being some kind of funky seashell. There's really no mathematical reason why I should; I just do.

I'm not sure how well my talk went over; it may have been nonsense. I took a lot longer than I'd expected, even without giving the details. And my subject matter was, to say the least, eclectic. I spoke on the de Rham cohomology of compact Lie groups; my goal was to relate them to objects defined only in terms of the Lie algebras. To everyone's shock and disgust, I started off doing Riemannian geometry, with metrics, Hodge stars, harmonic forms...Used a lot of Lie derivatives...I finally related that to topology and de Rham cohomology with the Hodge Decomposition Theorem, which (as a corollary) tells one that the k^th de Rham cohomology group of a compact Riemannian manifold is very isomorphic to the space of harmonic k-forms; in the bi-invariant metric I put on my group G, harmonic forms turned out to be precisely bi-invariant forms, as one would hope. I used that to translate de Rham cohomology into the cohomology of the Lie algebra, and, upon translation, found that the Hodge decomposition gave a completely natural and completely representation-theoretic description of both Lie derivatives and the cohomology spaces: the cohomology spaces turned out to be the submodules of invariants in the cochain spaces.

So de Rham cohomology on a compact Lie group is really a special case of the representation theory of Lie algebras. I find this splendid.

Geometry, topology, representation theory...I even fielded a question on measures from someone in the audience. I got some weird, weird looks from the topologists, let me tell you...Analysis is rubbish.

I hope they realise that I really am one of them. They should not be afraid. I'm just...interdisciplinary, that's all.

Another seminar tonight...This time on fluid dynamics, in particular the dynamics of really cheap gin and tonics as they pass through my esophagus. The Rosebud has a happy hour running from 4 'til midnight daily: $2 well drinks. The price is right. This is especially splendid.

Fun with Maths

Why do cicadas have prime-number (13- or 17-year) life-cycles?

Glenn Webb, a mathematician at Vanderbilt University in Nashville, Tennessee, has demonstrated mathematically that prime-numbered lifecycles could help cicadas avoid damaging “resonances” with the two- and three-year population fluctuations of their predators. These would result in lots of predators being around in years when there were lots of prey. Dr Webb's model shows that, over a 200-year period, average predator populations during hypothetical outbreaks of 14- and 15-year cicadas would be up to 2% higher than during outbreaks of 13- and 17-year cicadas. That may not sound like much, but it is enough to drive natural selection towards a prime-numbered life-cycle.

Only one predator—or, rather, parasite—is known to have overcome this anti-resonance strategy, by developing its own 17-year clock. Massospora cicadina, a fungus, lives in cicada larvae and passes between adults when they mate. But according to Gene Kritsky, an entomologist at the College of Mount St Joseph, in Cincinnati, Ohio, natural selection is working against this resonance too. In 2000, he recorded thousands of Brood X members emerging four years early—in other words, shifting to a 13-year cycle that Massospora is not equipped to match. Lo and behold, when Dr Kritsky examined several dozen members of the “accelerated” Brood X that emerged in 2000, he found only one infected female among them, and she had but one fungal spore. By contrast, he found from 50 to 300 spores in each cicada female from another brood that emerged on time that year in North Carolina. Brood X, it seems, is splitting up, and a new 13-year cicada population is evolving.

The article's also got a really rocking photo of a cicada. I do miss hearing them over the summers...

(I found the article thanks to the fine folks at Crooked Timber.)

Chalabi

According to this Salon article, Iraqi con-man, blackguard, scoundrel and swindler Ahmad Chalabi has a Ph.D. in mathematics from the University of Chicago, with a thesis entitled 'On the Jacobson Radical of a Group Algebra'.

Here he is in the Mathematics Genealogy Project. He can trace his mathematical lineage back to Schur, Frobenius, Weierstrass, and eventually Gauss...Though who can't, really?

This doesn't change the fact that Ahmad Chalabi is a sack of crap who bears a lot of the responsibility for the disasterous way things have gone in Iraq.

On the bright side, Art Garfunkel got an MA in mathematics from Columbia, and seems to have pursued but not completed his Ph.D. Art Garfunkel has never conned a world power into launching a poorly-thought-out war of aggression followed by an even more half-assed occupation full of repression, torture, mercenaries, and the utter obliteration of any slight chance for an acceptable outcome. Perhaps Art Garfunkel and Ahmad Chalabi were born to be one another's arch-nemeses, Art the defender of all that is good and true, Ahmad a creature of purest evil...

Octopus Jesus does not like Ahmad Chalabi. Octopus Jesus does, however, like 'Bridge Over Troubled Waters'.

Invariant de Rham Cohomology

Finished!

On a compact, connected Lie group, every closed differential form is cohomologous to a left-invariant form, and you can compute the cohomology groups using only the left-invariant subcomplex, which is precisely the exterior algebra of the (dual of the) Lie algebra, and thus amenable to completely algebraic treatment.

Topology=Algebra.

And I did it in a cute little way, too! With respect to the unique bi-invariant Riemannian metric on the group, the harmonic forms all turn out to be left-invariant. QE freaking D, baby.

...Unless I'm wrong, of course.

Oddly enough, I got a lot of my inspiration from Samuel Goldberg's Curvature and Homology, despite the fact that I couldn't even read most of his coordinate-heavy proofs. (Though perhaps 'couldn't be bothered to try and read' would be more accurate. Coordinates are the Devil's assgoblins.)

A Joke

This is Dave's joke; I can't take credit for it myself, much as I'd like to...

Q: What was Chairman Mao's favourite induced map?
A: The Great Push Forward.

The Statistics of Divinity

Via the notorious Stet, I see that a certain gentleman calculates that there is a 67% chance God exists.

His methodology puts me very much in mind of the following.

Suppose we're given two variables, A and B.

• Let A=B.
• A(A-B)=A²-AB=A²-B²=(A+B)(A-B).
• Cancelling, A=A+B.
• Let B=1.
• Then subtracting A from each side, 0=1. Call this equation (*).
• Winston Churchill has one head. Therefore, by (*), Winston Churchill has no head.
• Winston Churchill has no green leafy top. Therefore Winston Churchill has one green leafy top.
• Multiply each side of the equation (*) by two. Then two equals zero. Since Winston Churchill has two arms and two legs, Winston Churchill has no arms and no legs.
• Multiply each side of (*) by the waist size of Winston Churchill. Then the waist size of Winston Churchill is zero: Winston Churchill tapers to a point.
• Multiply each side of (*) by l, where l is the wavelength in nanometers of any photon reflected off of Winston Churchill. Then l=0 nm. Now multiply each side of (*) by 620 nanometers: 620 nm=0 nm. Therefore, we obtain l=620 nm. Winston Churchill is orange.
• Winston Churchill has no head, no arms, no legs, and a green leafy top; he tapers to a point; and he's orange. Therefore Winston Churchill is a carrot.

Which just goes to show that if you plug batshit into an equation, you shouldn't be surprised when you get pureed hogsbollocks out.

This message has been brought to you by the National God Almighty I Need A Drink Foundation.

Casimir

Ha! I have done it! I think...

A nice, clean, totally non-computational proof that the homology groups of a semisimple Lie algebra with coefficients in a nontrivial irreducible module vanish!

Thus implying that the cohomology groups vanish too, by Poincare duality.

This seems like such an easy proof...

UPDATE (7 March): No, I was wrong. Sorry.

Lie Algebra Cohomology

Here are some documents online dealing with cohomology of Lie algebras. I'll bet you aren't going to look at them. That makes me sad.

Here's one brief summary stressing connections to de Rham cohomology.

This one describes itself as 'quick and dirty', just like me.

Here's someone's thesis, which is mainly on cohomology of restricted Lie algebras, but has a section on the more standard case as well.

I've got Jacobson's Lie Algebras...We'll see how that goes. I should check out Knapp.

More Manifolds

Remember that exercise I mentioned, about the path components of a certain subset of the orthogonal group O(n)?

Consider the subspace X_n of orthogonal matrices with square the identity: in other words, symmetric orthogonal matrices. Let O(n) act on these by conjugation: certainly any orthogonal conjugate of a matrix in X_b is still in X_n. Consider the conjugacy classes. Since symmetric matrices are orthogonally diagonalisable, each conjugacy class contains a diagonal matrix; since it's orthogonal, its diagonal entries are +1 and -1 with certain multiplicities. Two such matrices are similar iff they have the same number of +1 entries; thus we have distinct classes C_k, k=0...n, where C_k is the conjugacy class of the diagonal matrix D_k with k +1 entries followed by n-k -1 entries. The elements of this conjugacy class are uniquely determined by their 1-eigenspaces, which are k-planes in Rⁿ: in fact, C_k is the Grassmannian G_kRⁿ.

Note that each conjugacy class is path-connected as a subspace of X_n, since the special orthogonal group SO(n) is path-connected and acts transitively on each class. Note also that the trace function takes on the constant value 2k-n on each C_k. Therefore, the C_k are exactly the path components of X_n.

Since O(n) acts transitively on each C_k with the isotropy group of D_k a closed Lie subgroup--the same as the isotropy group of the standard k-plane in the Grassmannian; you can find it explicitly without much fuss--we can give C_k a unique smooth manifold structure so that the O(n)-action is smooth; with this structure, C_k is diffeomorphic to O(n) mod the isotropy group, which is also diffeomorphic to the Grassmannian; so C_k is diffeomorphic to the Grassmannian of k-planes (equivariantly, even). It's not too hard to go on and show that with this smooth structure, C_k is an embedded submanifold of O(n).

So there.

I Rock

I am the Manifolds King! I make all manifolds my bitches! (Yea, even space-time itself.)

The problem was this: let X_n={A in O(n) : A²=I}. That is, it's the set of all symmetric orthogonal nxn matrices. Your mission, if you choose to accept it, is to show that all the path components of X_n (as a subspace of O(n)) are embedded submanifolds of O(n), and O(n)-homogeneous spaces, and in fact figure out which familiar homogeneous spaces they are.

I exhibited the path components in what I thought was an awfully nice way...And as soon as I saw them, I had a sudden flash of intuition revealing exactly what these manifolds had to be. It just seemed right, even though there didn't seem to be any geometric or topological reason for it...But I could see how I might build a mapping between them. When I worked through the details, it turned out I was right. Not only did I have intuition, I had correct intuition! That is just about the coolest thing ever.

I'm not going to tell you what they turned out to be. The problem's not due until Monday, and I don't want to give anything away if, by some bizarre twist of Fate or Chance someone else in my Manifolds class should read this before then. I do not want to spoil anyone's fun. For it was fun! Yes, it was fun when I finished writing it down...Heady, even.

There were a lot of details to wade through, it's true, the Equivariant Rank Theorem spraying left and right, remembering that all symmetric matrices are orthogonally diagonalisable, exhibiting the path components as different level sets for the trace map, a few commutative diagrams...

But the payoff was pretty.

Also, I think the set of all Borel subgroups of a complex semisimple Lie group, since it carries a transitive action of the group by conjugation, and the isotropy group of any Borel is the Borel itself (since it's its own normaliser), should have the structure of a smooth manifold, via identifying it with the coset space of any handy Borel.

Local Degree

It's been a good long while since we've had a maths post here, and I think we're long overdue. You've probably been tossing and turning for hours upon hours in the dead of night, fisted by an overpowering urge to feast upon fatty scraps of technical detail rent from the flesh of my budding mathematical career. And I, dear reader, shall indulge you.

I've been working through Allen Hatcher's Algebraic Topology with some other grads here and one of our faculty.

(Incidentally, my roommate may possibly have just related the complex K-theory of a space to its first cohomology with coefficients in the infinite unitary group U. Which would be crazy. I'm excited.)

Flip to Chapter 2, on homology. Now read it. Come back when you're finished.

All done? Swell.

By now you surely understand why it's important that we can compute degrees of maps between spheres: we need it to do cellular homology. And by now you surely understand what the degree of a map f from the n-sphere to itself is: the n^th homology group of the n-sphere is just the group Z of integers, and f induces a homomorphism from Z to itself, which must be multiplication by some integer, which we dub the degree of f. Unfortunately, computing the degree of a map directly is usually horribly difficult, at least for me. Hatcher offers a sort of short-cut, defining local degrees and proving that we can obtain the degree of a map, under suitable hypotheses, as the sum of some local degrees. Doubly unfortunately, local degrees as he defines them aren't much easier to compute than full-fledged degrees. Which is sad. But, mirabile dictu, there is another way one could define local degree, which is quite tractable!

Suppose f is a smooth map of the n-sphere to itself. Suppose y is a regular value of f: at every point in the (finite) inverse image, the pushforward (or Jacobian, or derivative, or [insert notation here]) of f has rank n, making it a linear isomorphism. Then at every point x in the inverse image, f is locally a diffeomorphism: we can choose some neighbourhood V of y, with f^-1V a disjoint collection of neighbourhoods of the x's. Given such a neighbourhood U of an x, f induces a map from H_n(U,U-x) to H_n(V,V-y), both groups being Z; so this map is multiplication by some integer, the local degree of f at x. If f is a diffeomorphism on U, then this map is invertible, so the local degree must be +1 or -1. We can also consider the pushforward of f at x; it's a linear isomorphism of the tangent space at x to the tangent space at y, so we can easily check whether it preserves or reverses the orientations on these tangent spaces defined by the standard orientation of the sphere. I claim the local degree at x will be +1 precisely when the pushforward preserves orientations, and -1 when it reverses orientations. Note that this should be given simply by the sign of the determinant of the Jacobian in appropriate coordinate charts, which is immensely amenable to computation.

Why should this be true?

Let's sweep a lot of crap under the rug right now. Let's choose V and our charts so that the image of y is 0, and the image of V is a teeny tiny disk about 0. And let's also choose charts so that the image of x in its chart is likewise 0. Just for shits. So let's look at everything in these coordinates; if F is the coordinate representation of f, then F(0)=0, and for any z in the image of U, we can write F(z)=F(0)+DF(0)z+o(z)=DF(0)z+o(z) where DF(0) is the Jacobian of F at 0, and the 'little o' function dies off very rapidly near the origin.

Here's the kicker: the map DF(0)z+to(z) should be a homotopy between F and DF(0), at least on a small enough neighbourhood. So DF(0) and F must have the same local degrees at 0. Since DF(0) is just an invertible linear map, and GL(n,R) has exactly two path components, DF(0) should be homotopic either to the identity or to the diagonal matrix with all diagonal entries 1 except for the first, which will be -1. Depending on the sign of its determinant. The local degree of DF(0) is then the sign of its determinant, and so we obtain the local degree of F; assuming we've chosen our charts so as not to fuck anything up, we now have the local degree of f at x, too.

Wave your hands in the air! Wave them like you don't care!

So there. That is enough math-posting for today. Fill in the details at your leisure; correct any egregious mistakes in the above. Perhaps I will try to do this rigourously at some point in the near future. It depends on how distracted I get.

We now return you to your regularly-scheduled programming. (You were looking at porn, weren't you?)

What TAs Want

Sometimes, when a student asks me a question on something they really, really, really ought to know if they're taking and expecting to pass the class that they are, what I really want to say to them is 'I will burn your soul with my fury.'

But I don't.

Another Free Book

In the ongoing quest to sodomise the Internet with free, advanced mathematics textbooks, here is a preprint, in PostScript format, of a book on complex geometry. I've only just started reading the first bit, which is quite interesting; it connected a function's being holomorphic with a certain diagonalisation of its Jacobian matrix. Which I thought was fun.

My Gift to You

Allen Hatcher's book Algebraic Topology is free for the downloading, in PDF format, online. Gosh bless the Internet.

I've ordered a paper copy, so I can read it in the bathroom.

That's a compliment.

Universal Cover

It is after 5pm Pacific Time, so at last my lips are unsealed...I can reveal to you at last the shocking, explosively revelatory news that the universal covering space of RP²vRP² looks like an infinitely long caterpillar, or snowman, or string of anal beads. To see why this should be, imagine someone hands you a wedge of two projective planes, and asks you to cover them. A projective plane is just a crushed sphere, so the first thing you'd probably think to do is to blow up one of your planes into a sphere again. You want to build a two-sheeted covering of your original space; your blown-up sphere covers one of the planes twice, but the other plane is covered only once so far, by itself, so to balance things out we need to attach another copy of the projective plane to the sphere antipodally from the first. Then you'd say to yourself: Well, Self, the obvious thing to do now is to blow up one of these projective planes again...You'll get a sphere there, connected at one end to another sphere, which is connected at its other end to a projective plane; and now, to keep the covering even, you have to attach both another sphere and then a projective plane to your new sphere, antipodally, so you always get a symmetrical structure...If you keep blowing your planes up indefinitely, you'll wind up with the Snowman Space, which turns out to be simply connected and a countable cover of the original space. Keen, eh?

This take-home Manifolds final was my only exam. I love graduate school. I'm now free of all responsibilities 'til January. And you know what that means: video games that involve smashing dirty zombies with a pipe.

LaTeX

If you ever find yourself forced to type up an algebraic geometry assignment in LaTeX, and you realise you need to find out how to write matrices, please, please, do not do a Google search on 'latex matrix'. I'm serious.

The Octonions

If you don't read John Baez's survey article on the octonions, then the terrorists have already won.

Can you think of a better way to spend a chill winter's night than meditating upon the octonions with a snifter of brandy and a spaniel, or perhaps some kind of cat? Possibly a rabbit? Or a Sigmund Freud action figure? (They also have Pope Innocent III.)

(Incidentally, Innocent's scroll reads 'Sons of *Something*, kiss my ass.' Although I'm not sure the word 'asinus' carried anal connotations...I would try to find out via Google, but I'd probably just wind up with porn.)

Fundamental Groups of Topological Groups are Abelian

It's a fact: the fundamental group of any topological group is abelian.

Let X be a topological group, with identity element e. Let f and g be paths at e(continuous maps from the unit interval I into X, whose value at both 0 and 1 is e). Denote the ordinary multiplication of paths with a stop: f.g is just the path obtained by following f, and then following g, and reparametrising appropriately to obtain another path. Now, there is another natural multiplication on paths, induced by the group structure of X: denote this by *, where (f*g)(t)=f(t)g(t), where juxtaposition represents the group multiplication. Since multiplication is a continuous operation, this is indeed a path at e. Now, I put it to you that this operation is well-defined not only on paths, but on homotopy classes of paths: suppose f is homotopic to f^~ via a path homotopy F, and g is homotopic to g^~ via a path homotopy G. Then let H(s,t)=F(s,t)G(s,t); this is again a path homotopy, between f*g and f^~*g^~. Which is all well and good.

Now, let c_e denote the constant path at the identity. It should not surprise you to learn that for any paths f and g at e (not path classes, mind you), f.g=(f.c_e)*(c_e.g). Now let's pass to path homotopy classes: [f].[g]=[f.c_e]*[c_e.g]=[f]*[g]. Eureka! The new operation * is absolutely identical to the old operation . on the level of path classes. Which is good. We might as well forget about . altogether, and treat our fundamental group now as having its operation defined by *. Which makes everything very easy indeed. Inversion is a continuous group operation, so for any path f at e, the inverse path f^-1 defined by f^-1(t)=(f(t))^-1 is again a path at e. To show that the fundamental group of X is abelian, one must find a homotopy between f*g and g*f for any paths f and g at e.

But this is the very essence of simplicity! We want to get our homotopy H(s,t) by multiplying f(t)g(t) on the left by something that is the identity when s=0 and is f^-1(t) when s=1, to kill off the left f; and we want to multiply on the right by something that is the identity when s=0, and is f(t) when s=1. Furthermore, we need to fix H(s,0)=H(s,1)=e for all s. There's an obvious way to accomplish this. Just let H(s,t)=f^-1(st)f(t)g(t)f(st). When s=0, this is just f(t)g(t). When s=1, this is g(t)f(t).

And we're done!

It's a snap.

Noetherian Pun

Artesian wells are all Noetherian.

They satisfy the ascending chain condition.

Homeomorphisms

Today I solved a topology problem--I figured out how to create a homeomorphism of any connected manifold to itself moving any point to any other, arbitrary, point--by thinking about General Tommy Franks farting into a jar. Or, alternatively, lava lamps. I will draw you a picture tomorrow.

It worked much better than my original inspiration, which involved farting into goldfish bowls. And not surprisingly; farting into goldfish bowls is just silly.

Baseball

So, baseball. Yeah. How about that? Did you see the way those guys hit that creamy orb with a stick?

I don't actually know anything about baseball, but everyone seems to be talking about it right now, and I didn't want to feel left out.

I do know how to play baseball mod 2.

There are only two bases, home and first. Only one strike and you're out, but that's not so bad, because after zero balls you get a walk. But that doesn't make things as easy as you might think, because there is only one person on each team. Or there would be one person on each team, if teams were involved. But a game of baseball mod 2 is played between zero teams. All the professional baseball mod 2 players stay at home making obscene phone calls to each other all day long.

Another Bad Math Pun

Sensible mathematicians are all pro-Choice.

Seifert-van Kampen

The Seifert-van Kampen Theorem is something to do with galaxies full of pork and beans exploding. (Well, that's what you get.)

Post-Prelim

You can be sure as heck that the integral of that analysis prelim converges, because I dominated it. Ha! I'm a riot. I'm 50,000 people running amok through Seattle in 1999 to protest the WTO.

I made that exam my bitch.

I went into it feeling pretty low, too. I jerked into wakefulness this morning to find I'd overslept by an hour, having somehow deactivated my alarm in my sleep, and barely made it to the exam on time, after feeling yesterday that the algebra exam had broken something in my head because nothing (mathematical) I tried to read made any sense...The contour integral problem seemed impenetrable, and I found out why: when the exam was 7/8 over, we were told that in fact there was a typo which made the problem impossible. There was screaming, yea, like unto all the damn'd souls of Hell. But on the bright side, we got an extra half hour. I needed four of the eight problems to pass; I blew my own mind and did six. Which means that even if I accidentally answered two of the questions with Sumerian pornography, I'll still pass. I am a happy man.

If--and I say if--I passed Monday's algebra exam, I am done with prelims for all time. That's a very nice thought. All warm and soft and cuddly, like a kitten. An adorable little kitten.

Three cheers for kittens!

Hell

Wow. A four-hour exam really gallops by when (you're made to feel as if) your entire future career depends on it.

I think I passed, barely.

It was a horrible exam. Even by prelim standards. Whoever wrote it had better feel awfully, awfully sorry for what they've done, and I'd hope they've learned their lesson and will never, ever do it again. One of the problems was so obscure it was just silly.

On the bright side, it's over now. Although I still have to take another one Wednesday. Also on the bright side, it's after noon, so it's acceptable for me to start drinking now and beat the rush.

PS...Pain.

PPS...Kill me^*.

*But not until after I've found out if I actually passed or not. Unless you really just can't wait.

Prelims

I have an algebra prelim in eleven hours. Sweet Jesus' cock, I am going to fucking die^*.

*Note I am not actually going to fucking die. I've been obsessively studying for this thing. I've done the last eleven years' worth of prelims. I can fart the Sylow Theorems in my sleep. I can compute Galois groups with one foot up my ass. By any objective standard, I'm pretty well-prepared. That doesn't change the fact that a horrific amount of pressure goes with these exams, or that what they test is essentially meaningless. At this level, the only sensible exams one could give are take-home things, to be done at one's leisure, with a certain amount of research and some nice cups of tea and raspberry-filled butter cookies. Something involving highly nontrivial results, not just tricksy applications of low-level theorems in ways that reward rote memorisation of old exams.

On the bright side, I totally know more than enough to teach something equivalent to my undergraduate abstract algebra class now. The structure theorem for modules over a PID is a natural for that setting.

I'm going to sleep now, on the grounds that the sooner I do, the sooner I can wake up, take this bitch, and get it all over with.

The world is a vampire^**.

**Note^*** the world is not actually a vampire.

***Do footnotes make you want to play Johnny Cash's cover of 'Hurt' over and over and over?^****

****No, they don't. It's prelims that do that.

PS...Prelims.

More Dirty Math

The finite simple group ²F₄(2)' is also known as the Tits Group. If you run a Google search on Tits Group, your first hit will be an entry at Mathworld. Then will come a great deal of porno. I believe this group to be named after Jacques Tits, a Belgian mathematician who should never, ever have his own web page. Ever.

If one were to write a book about Jacques Tits' influence on module theory, one could call it Tits and Ass.

Terrible Math Riddles My Roommate Just Told Me

Q: What's yellow, and equivalent to the Axiom of Choice?
A: Zorn's Lemon.

Q: What's yellow, and complete?
A: A Bananach space.

Q: What's yellow, and has a convergent power series expansion?
A: A bananalytic function.

Q: What's purple, and commutes?
A: An abelian grape.

Q: What's purple, commutes, and is occasionally worshipped?
A: A finitely-venerated abelian grape.

Q: What's non-orientable, and lives in the ocean?
A: Möbius Dick.

Academic Life

The worst thing about being a mathematics graduate student--apart, of course, for the immense pressure of qualifying exams, which consume years of useful mathematical time that could be spent, as it is in physics, joining a research group and working towards publication--is the grading. TAdom is how we earn our keep; it is an inevitable fact of life, like death and taxes, and talk radio. In addition to draining away our sorely-needed time and energy, grading also saps our hope. Students very rarely do well on exams. At least that has been my experience here. After a while, one begins to take a sadistic delight in splattering red, red ink across their helpless, virginal papers...

Every now and again, though, one gets a pleasant surprise. Mostly, these pleasant surprises are pictures.

Perhaps this is less common outside of maths...

Most students in the sort of courses we get to TA are not math people. Most likely they don't enjoy math in the least, and are only taking the class because it's required. Math does not come intuitively to them. Sometimes, they just give up. And sometimes when they do, they flaunt their other talents, as if to convince me that they're not total failures. I love it when students, in lieu of a solution, draw me pictures. I had a very good weeping puppy once...Many, many tearful cartoon faces...Just today, I had a rather fine one, a sketch of a squatty-headed and thuggish young man saying 'Sorry, I'm not good at math!'

I wanted to give him points for that.

But I could not.

Life is cruel.

Another highlight can be the student evaluation forms, after each term ends. Occasionally, their written comments can delight. Spring quarter, in response to a question about what detracted most from their learning in the class, one person wrote 'SARS,' and another 'Ugly girls.' There was also 'I don't like the kid with the curly hair or the one with the big teeth.' And, dear to my heart, someone even used the phrase 'Unless I'm smoking crack' in their comments.

The Herglotz Formula

Let R be a connected open region in the plane, and let f_n be a sequence of holomorphic functions on R such that {Re(f_n)} converges uniformly on compact subsets to a harmonic function u. Suppose there exists z₀ in R such that v(z₀)=lim Im(f_n(z₀)) exists. I wish to prove that {f_n} converges uniformly on compact subsets.

Why do I wish to prove such a thing? Because I'm preparing to take my Ph.D. qualifying exams in September, which make strong women weep and strong men befoul themselves, and the more of these problems I can do, the better my chances are. Also, I had trouble finding references to the Herglotz integral formula online...Maybe someone out there will find this useful. Also I think the name 'Herglotz' is just really keen. Apparently people liked him. I'm a big fan of his hair. Can't find a biography of him at the University of St Andrews site, which is a shame...I like them.

{f_n} is normal on R iff it is normal on all disks D whose closures are contained in R. It is normal in such a disk, by the Arzela-Ascoli Theorem, if it is equicontinuous on compact subsets, and pointwise bounded. Let K be a compact subset of D, and let z and w be any points of K. We may map D linearly to the open unit disk U, and apply the Herglotz integral formula, which is just like the Poisson formula, only for analytic rather than harmonic functions:

Thus on each compact subset K of a disk D in R, there is a constant C with |f_n(z)-f_n(w)| less than C|z-w| for all n and for all z, w in K.

Suppose in some disk D there is a point z with {f_n(z)} bounded. For all z in D, there is a compact curve g contained in D (a straight line segment, even) connecting z and z. By the above, |f_n(z)-f_n(z)| is less than C|z-z| for all n, so |f_n(z)| is less than sup|f_n(z)|+C|z-z| for all n: {f_n} is pointwise bounded on D. If E is the set of points z in R for which {f_n(z)} is bounded, this shows E is open (and nonempty, since z₀ is in it). Now suppose w is in R-E; since R is open, there is a disk D containing w whose closure lies in R, which contains an even smaller disk whose closure is a compact subset of D containing w. For all z in this smaller disk, |f_n(z)-f_n(w)| is less than C|z-w| for all n; since {f_n(w)} is not bounded, neither is {f_n(z)}. Thus R-E is open, so E is closed. Since R is connected and E is nonempty, open and closed, E=R: {f_n} is always pointwise bounded. Hence the family is normal on all disks D, and thus on all of R. It contains a subsequence converging normally to some analytic function f on R. Of necessity, Re(f)=u. Suppose g is any other normal subsequential limit of {f_n}. Then Re(g)=u also, so f-g is an analytic, pure imaginary function: the Cauchy-Riemann equations imply f-g is an imaginary constant A. But since lim Im(f_n(z₀)) exists, the imaginary parts of f and g are both equal at z₀, so A=0. Thus any normally convergent subsequence converges to f. Suppose {f_n} as a whole did not converge to f uniformly on some compact subset K: then on this subset, there would be e greater than zero such that for all N, there exists an n greater than N with sup|f_n-f| greater than e. Therefore we can construct a subsequence {f_nk} with sup|f_nk-f| greater than e for all k. But this subsequence in turn contains a subsequence converging normally, necessarily to f, since {f_n} is a normal family, which is a contradiction. Therefore {f_n} is uniformly convergent on compacta.

Artinian Modules

This one bothered me to no end until I looked it up...It seemed so simple, yet I couldn't see how to do it, no matter how I banged my head on the chalkboard. So, without further ado:

Say a module M is Artinian if it satisfies the descending chain condition on submodules: every chain of submodules
MÊM₁ÊM₂Ê...
stabilises: for some n, M_n=M_n+m for all m.

Let R be a ring, M a left R-module, K a submodule of M such that K and M/K are both Artinian. Then M is also Artinian.

Proof:

We have a short exact sequence 0®K®M®M/K®0, where the first map, i, is inclusion, and the second, p, is the natural projection onto the quotient. Let
MÊM₁ÊM₂Ê...
be any descending chain of submodules in M. Then this chain induces two other chains,
KÊKÇM₁ÊKÇM₂Ê...
and
M/KÊp(M₁)Êp(M₂)Ê....

By assumption, each of these induced chains terminates: for some n, and all k,
KÇM_n=KÇM_n+k and p(M_n)=p(M_n+k).
If xÎM_n, then p(x)Îp(M_n)=p(M_n+k)
so $yÎM_n+k with p(y)=p(x):
x-yÎkerpÇM_n=KÇM_n=KÇM_n+k.
Therefore xÎM_n+k, or M_n=M_n+k. QED!

Bad Math Pun Thing

This isn't mine...Don't shoot the messenger.

'There are only 10 types of people in this world, those who understand binary, and those who don't.'

Iterated Complex Mappings

Make sure the Symbol font is enabled on your browser...I'll use ¶ for the complex differentiation operator.

Let G be a bounded connected (open) region in the complex plane, and let f:G®G be an analytic mapping. Suppose there exists z₀ÎG with f(z₀)=z₀, and that |¶f(z₀)| is less than 1. Define f₁(z)=f(z), and inductively, f_n+1(z)=f(f_n(z)). Note that this makes sense, since the range of f_n is contained in the domain of f, and that all the f_n are analytic, and, by induction, f_n(z₀)=z₀ for all n. The goal is to show that f_n converges to the constant function z₀ uniformly on compact subsets of G.

Differentiate:

¶f_n+1(z)=¶[f(f_n(z))]=¶f(f_n(z))¶f_n(z)

By induction, we have

¶f_n(z)=Õ¶f(f_k(z)), k=0..n-1

Now, since G is bounded, and f_n maps G into G, the functions G are all uniformly bounded by some constant. Therefore, since the f_n are all analytic, the family {f_n} is normal: applying the Arzela-Ascoli Theorem, we learn that {f_n} is equicontinuous on each compact subset of G. (Incidentally, I've heard the term 'compacta' used for 'compact subsets', and I like it.) Now, the fun begins. Since |¶f(z₀)| is less than 1, and |¶f| is continuous, $e,h, both positive, with h less than 1, with |¶f(z)| less than h for all z with |z-z₀| less than e. Since {f_n} is equicontinuous in a neighbourhood of z₀, $d positive with |f_n(z)-f_n(z₀)|=|f_n(z)-z₀| less than e for all z with |z-z₀| less than d. Therefore, in this d-disk about z₀, |¶f(f_n(z)| is less than h, for all n. Therefore |¶f_n(z)| is less than hⁿ, which tends to 0 as n approaches infinity. Therefore ¶f_n tends to 0 uniformly on this disk.

Must be off now, but I'll finish later.

UPDATE: Okay, I solved it in the shower just now. Since {f_n} is a normal family, it contains a subsequence {f_nk} converging uniformly on compacta to some function g. By Weierstrass' Theorem, g is analytic, and ¶f_nk converges uniformly to ¶g on compacta. But these derivatives are converging to 0 on an open neighbourhood of z₀, so, since nonzero analytic functions have only isolated zeroes, these derivatives are converging to 0, and hence ¶g=0: since G is connected, and g(z₀)=z₀, g is the constant function z₀. Now, fix a compact subset K of G. {f_n} is equicontinuous on K, so "e $d so that if |z-z₀| is less than d, then for all n, |f_n(z)-f_n(z₀)|=|f_n(z)-z₀| is less than e. Furthermore, there exists n_K such that |f_nK(z)-z₀| is less than d for all z in K. Thus, for all z in K, for all m, |f_m(f_nK(z))-z₀|=|f_nK+m(z)-z₀| is less than e. Therefore, for all large n, |f_n(z)-z₀| is less than e for all z in K: f_n converges uniformly to z₀ on all compact subsets of G. QED

Linear Algebra

Let V be an n-dimensional vector space over a field F, and let T be a linear transformation of V to itself. Let us say a vector v in V is cyclic for T if {v,Tv,T²v,...,T^n-1v} is a basis of V. I put it to you that if T has a cyclic vector, then the only other linear transformations with which T commutes are polynomials in T.

Let S be a linear transformation with ST=TS. Since {v,Tv,T²v,...,T^n-1v} is a basis for V, whatever Sv might be, we can expand it as Sv=a₀v+a₁Tv+...+a_n-1T^n-1v=(a₀+a₁T+...+a_n-1T^n-1)v=p(T)v, where p(x) is the polynomial a₀+a₁x+...+a_n-1x^n-1. Now let us compute the matrix for S in our cyclic basis: the i^th column of this matrix S' is the image of the i^th basis vector under S.

S'=[Sv|S(Tv)|S(T²v)|...|S(T^n-1v)]
=[Sv|T(Sv)|T²(Sv)|...|T^n-1(Sv)]
=[p(T)v|Tp(T)v|T²p(T)v|...|T^n-1p(T)v]
=[p(T)v|p(T)(Tv)|p(T)(T²v)|...|p(T)(T^n-1v)]

S', the matrix for S in our basis, is thus, we see, precisely the matrix for the linear transformation p(T) in our basis. Therefore, S=p(T): S is a polynomial in T.

This cyclic basis puts T into rational canonical form, with a single block: the matrix for T in this basis is [Tv|T(Tv)|...|T(T^n-1v)] or

0	0	...	0	?
1	0	...	0	!
0	1	...	0	*
.	.	...	.	.
0	0	...	1	#

Conversely, suppose M is any linear transformation whose matrix is in rational canonical form with a single block. Then there is a basis {e₁,e₂,...,e_n} of V such that the matrix of M in this basis, which is [Me₁|Me₂|...|Me_n] is precisely of the above form: thus Me₁=e₂, Me₂=e₃, and so forth down the line to Me_n-1=e_n. Thus we have a basis of V of the form {e₁,Me₁,...,M^n-1e₁}, so e₁ is cyclic for M. Thus a linear transformation has a cyclic vector if and only if its matrix in rational canonical form consists of a single block.

Now suppose does not have a cyclic vector: then, in rational canonical form, T consists of at least two blocks. Let S be a transformation with a block diagonal matrix in this basis with blocks the same sizes as those of T. Then the product of the matrices is the block diagonal matrix whose blocks are the products of the corresponding blocks of S and T, so S and T commute if and only if each block in S is a polynomial in the corresponding block of T (since each rational canonical block is cyclic). If T contains any zero blocks, let the corresponding blocks of S be diagonal with all entries nonzero and distinct, and let the other blocks be zero. ST=TS=0 so S and T commute, but S is not a polynomial in T, since any such would have all its diagonal entries in T's zero blocks equal. If not...

This may be wrong; I just came up with it. Let A and B be the final two blocks of T; then the minimal polynomial of A divides the minimal polynomial of B, which is the minimal polynomial of T. Let S be block diagonal, with all its blocks zero but the last two, which are A and B+tI for some scalar t. ST=TS, but I put it to you that S is not a polynomial in T. Suppose that it were: q(T)=S. Then q(A)=A, and q(B)=B+tI. Then A satisfies the polynomial q(x)-x, and B satisfies the polynomial q(x)-x-t. Therefore, the minimal polynomial of A must divide q(x)-x, and must also divide the minimal polynomial for B, which divides in turn q(x)-x-t. Choose t so that q(x)-x and q(x)-x-t are relatively prime; we reach a contradiction. Therefore T has a cyclic vector iff the only operators it commutes with are polynomials in T.

The McLaughlin Group

Not only is the McLaughlin Group a weekly public affairs television programme, it's also a finite simple group of order 898,128,000. To prevent confusion, here are a few simple ways to tell the two apart. The television programme is a flimsy and ephemeral product of human civilisation which will not long endure in this universe. The finite simple group is a logical necessity inhabiting a timeless Platonic realm of abstraction and pure truth. The television programme predicts that Pope John Paul II will win a Nobel Peace Prize. The finite simple group takes no stand on the matter. The television programme contains only very pedestrian and boring subgroups, while the finite simple group contains the Mathieu group M22 as a maximal subgroup. The television programme is generated by John McLaughlin (as far as I can tell, no relation to the jazz artist of the same name), whereas a 22-dimensional representation of the finite group over the finite field F₇ is generated by the matrices

0100000000000000000000          3600000000000000000000
1000000000000000000000          0010000000000000000000
0001000000000000000000          0000100000000000000000
0010000000000000000000          0000010000000000000000
0000001000000000000000          0000000100000000000000
0000000010000000000000          0000000001000000000000
0000100000000000000000          0000000000100000000000
0000000000010000000000          4310500400000000000000
0000010000000000000000          0000000000001000000000
0000000000000100000000          0000000000000010000000
0000000000000001000000          0000000000000000100000
0000000100000000000000          0000000000000000010000
2561116463431403000000          0000000000000000001000
0000000001000000000000          0000000000000000000100
0000000000000000000010          0000000000000000000001
0000000000100000000000          3634503603202620362340
0043661213350404100000          6150405500433261654462
3416205006500132010040          5556523141121520054163
5225106102260525001050          3450340335252250144500
2543522352640541000130          2506015620240224344010
0000000000000010000000          2654224150141552331230
0052522353240445000031          0001000000000000000000.

The McLaughlin Group, and the McLaughlin group. Do not confuse the two.

More Banach Space Guff

Why are all finite-dimensional subspaces of a Banach space closed?

I hear you shouting this question to the uncaring heavens.

At last, your curiosity will be satisfied.

I will actually prove something stronger: let M be a closed subspace of a Banach space X (over a field K). Let x be any vector in X not lying in M¹. Let N be the direct sum of M and the subspace Kx spanned by x. I put it to you that N is also closed. If N is all of X, then there's nothing to prove. So let Y=X\N, and suppose Y is nonempty. Since M is closed, by a corollary to the Hahn-Banach Theorem, for every y in Y we can find a functional f_y in X^* with f_y(y)=1 and f_y(z)=0 for all z in M. Likewise we can find a functional g with g(x)=1 and g(z)=0 for all z in M. Now let h_y=f_y-f_y(x)g. Then h_y(z)=0 for all z in M, and h_y(x)=f_y(x)-f_y(x)g(x)=0, so h_y(z)=0 for all z in N: N is contained in the kernel ker h_y, for every y. Thus N is contained in the intersection of all the ker h_y, as y ranges over Y: this is an intersection of closed sets, hence a closed set. And I put it to you that this intersection is exactly N, which will complete the proof. For suppose there existed y' in Y which lay in this intersection. Then we'd have h_y(y')=0 for all y. But then f_y(y')-f_y(x)g(y')=0, or f_y(y'-g(y')x)=0, so y'-g(y')x lies in ker f_y for all y in Y; therefore it's in the intersection of all such kernels. But the intersection of all the ker f_y is truly contained in N: for any vector in Y=X\N, we constructed an f_y which was 1 at that vector. Hence y'-g(y')x lies in N, and since x lies in N, y' lies in N, a contradiction. QED

Now, the zero subspace {0} is closed. By the above, then, any one-dimensional subspace is closed: it's the direct sum of {0} with Kx for some x. By induction, therefore, all finite-dimensional subspaces are closed.

This seemed like the natural way, to me, to prove this when it was assigned as homework in the winter: if you need a vector subspace for some reason, look at kernels of linear transformations. But for some reason, analysts don't find it especially clear. Your mileage, as the old saw goes, may vary.

***
1. I left out the last clause originally. Silly me. I tacitly assumed all along that x did not lie in M but forgot to specify that. (22 June)

Division by Zero

I recently read the short story collection Stories of Your Life and Others, the first book of Seattle-area resident Ted Chiang, and my one-word summary of it would be 'rocking'. It contains some extremely excellent short science fiction, including 'Understand', which you can read for yourself online thanks to the nice folk at Infinity Plus. 'Understand' is a tale of intelligence augmentation, and it's up there in swellness with Thomas Disch's Camp Concentration. There is also, and this won the collection a special place in my heart, a story, 'Seventy-Two Letters', about golems. Also the more-or-less eponymous tale 'Story of Your Life', which reminded me of, and suggested sharp constrasts with, Kurt Vonnegut's Slaughterhouse-Five, or more accurately vice-versa, since I read Chiang first; Chiang's piece differs primarily in being much less depressing. About all this I may say more in the fullness of time, but for now I wanted to have some words about the one story in the collection that I found less than entirely satisfying: 'Division by Zero'.

The premise of 'Division by Zero' is that a gifted mathematician develops a new formalism which leads her to discover that arithmetic as a system is inconsistent: she finds a way to prove, rigourously, that all numbers are in fact equal, and thus that arithmetic, and by extension pretty much all of pure mathematics, is meaningless. It sends her into bleakest despair and basically ruins her life. Which is understandable; to a mathematician, being told that mathematics just doesn't work would be about the worst thing imaginable. The thing is, I can't bring myself to suspend my disbelief. It's hard to swallow, in a way that golems or Babylonians building a tower all the way to Heaven aren't. With golems, if you make one simple assumption--that Kabbalism can replace the physical sciences--the rest follows logically. (Which is one of the best things about Chiang: everything follows so wonderfully logically!) Here...I think that is not quite the case. Maybe I'm saying that because I am, or hope to be, a mathematician myself; the idea of maths just not being true...Well. It'd be brown trousers time, for sure.

I am, like many mathematicians, a Platonist at heart. I don't believe there is literally a world of idealised mathematics made solid floating around somewhere in the formless void, but I can't help but view mathematics more as a process of discovery than of invention; true theorems weren't any less true before they were formulated and proved, after all. If they are true now, they have always been true, and they're universally true, even multiversally true: true independently of the physical world and any of its parameters. (Modulo the appropriate definitions, of course.) So the idea of maths not holding water...It's like destroying an entire universe, albeit one I perceive only indirectly. A universe, by the way, without George Bush in it.

But I think I have some firm, logical basis for my unease, beyond my epistemological and ontological qualms. Renee, the unfortunate mathematician in the story, develops a new formalism which allows her to rewrite unspecified axiom systems, and thereby prove, in essence, that 1=2. I am not a logician (an algebraist, actually, probably a representation theorist), but I do know a little set theory (which no-one else in the department seems to enjoy, damn them) and I'm familiar with the axiomatics of arithmetic. An axiom system for workaday arithmetic, called the Peano postulates, was devised by, of all things, a chap called Peano in 1889. It's possible Renee was working with these, but unlikely: the Peano postulates aren't very fundamental. If you assume ZFC set theory (Zermelo-Fraenkel, with Choice), or an equivalent, which you're more or less obliged to do to do any modern mathematics whatsoever (unless you're some kind of silly Intuitionist) you can prove the Peano axioms as theorems about the set of natural numbers; if the Peano postulates are inconsistent, then ZFC is inconsistent as well. So Renee has discovered that ZFC is self-contradictory, that 1=2 when it is axiomatic that 1 and 2 are, by construction, very different entities. This would be bad. But not that bad. This doesn't mean that mathematics in general or arithmetic in particular are themselves, in their Platonic forms, inconsistent: it means the axiom systems Renee and I have been using are, which is not at all the same thing. You can always replace your axioms. Maths has been bitch-slapped like this before, and come out all the stronger for it. Cantor's original naive set theory was inconsistent, but that didn't destroy set theory; it just meant it needed different foundations, like ZFC. If ZFC were itself inconsistent, while I'd be quite shocked, I'd also be confident that a replacement scheme could be devised. I can't imagine it'd drive anyone to suicide...

I just don't buy it. But do buy the book. The rest is, as I said, very fine.

(And what's so implausible about transfinite induction?)

Anyone interested in learning about set theory from a naive, or non-axiomatic, perspective, which is the best way to see it first, would do well to seek out Paul Halmos's Naive Set Theory, which is accessible to the tenacious layman. For a much shorter and less comprehensive introduction with more sex in it, read on...

(Edited slightly on 17 June for clarity and extra bitch-slapping.)

The most basic area of mathematics, or the most basic interesting area, if you ask me (and I acknowledge in advance that you didn't) is set theory, which is, as you might have guessed, the study of sets. Sets are exactly what you'd think they are, things made up of other things. A set is like a box, filled with these sub-things, which mathematicians like to call elements. There are many sorts of boxes. Boxes may be wooden. They may be cardboard. They can in fact be made of metal, and if you really wanted to you could probably make some sort of a box out of crack. Sets aren't quite like that. To a mathematician, a set is completely defined by its elements, the things it contains. If you have two sets and they both have precisely the same elements, then they're the same set. It's like numbers; you can have one apple or one orange or one orgasm, , and in the real world these are (usually) very different things, but abstractly, it's still just the number one. Sets are usually written as a list of their elements between curly brackets, like this: {1,2,3,4} or as a description of their elements, also between curly brackets, like this: {all positive integers less than five}. While I've described these two sets in completely different ways, you'll notice that they are, in fact, the same set. They both have exactly the same elements: 1, 2, 3, and 4. The fancy math term for this idea, that two sets are the same if they have the same elements, is extensionality. To save time and space, sets will usually be given names, like A={1,2,3,4}, and then the symbol A can be used in any expression in which I want to talk about that set. Having said nothing more, I can already display one interesting set factoid for you: let x be anything you want, a number, a letter, a symbol, a thing. It doesn't matter. Sets don't care what you put in them. Then {x,x} and {x} are equal: they're the same set. Why? Because every element of the first set, which has to be x, is an element of the second set, and vice-versa. Repeating an element doesn't change the set, so I'll just delete any references to repeated elements in my sets. And it's entirely possible to have a set that doesn't have any elements at all: {}. This is called the empty set, and I'll name it 0: 0={}. You'll see why.

There are only a few basic things you can do with sets, according to the very strict rules laid out in the axiomatic version of the theory which I'm not going into here, but as it turns out you don't really need anything complicated. First off, if you have two sets, you can take their union. If A and B are both sets, then their union is written A U B, and this is the set whose elements are all the elements of A, together with all the elements of B, and nothing else. If A={1,2,3,4} and B={1,2,5}, then A U B = {1,2,3,4,5}, since we can ignore repeated elements. We can also take intersections, though there isn't a good way to type them out here, and I won't need intersections for my purposes. The intersection of two sets is the set of all things that are elements of both of these sets at once. You can also construct subsets; a subset of some given set A is a set whose elements are also all elements of A. With my A above, {1,2} is a subset of A, and I'll write this {1,2} < A although normally it's curvier. And we can build new sets from old ones! A set doesn't care what you stick in it, after all, so why not build a set whose elements are other sets? So {A,B} is a set...But be careful. {A,B} is not at all the same set as A U B. The elements of A U B are 1, 2, 3, 4, and 5; the elements of {A,B} are A and B.

But what do I mean by 1, anyhow? How do you know 1 isn't the same thing as this set A? And how do you know that 1 isn't the same thing as 2? It sounds at first like a silly question, but it's important. What do we mean by numbers? I can now tell you exactly what numbers are, in very concrete terms. They're sets. I already defined 0 to be the empty set, {}. It's a set with zero elements, which seems like a sensible thing for zero to be. Now, let's define 1={0}: 'one' is going to be the set whose one element is zero. And we can bootstrap ourselves onwards: let 2={0,1}, and 3={0,1,2}, and so on. If you wanted to write these out more explicitly, we'd have 1={0}={{}}; 2={0,1}={ { },{{ }} }, but that's awfully awkward. Just saying 'and so on' isn't very precise, though, so I'll put it like this: if we've defined a natural number n, define its successor, the next natural number, to be n+1 = n U {n}; the elements of n+1 are going to be all the elements of n, together with n itself. This is just what I did for 1 and 2 and 3, you'll note. And having defined all the natural numbers 0, 1, 2, ... in this way, we can define arithmetic on them, too, using more set theory, and more induction, which I've skipped over entirely and waved my hands at in the above. As I said, this isn't meant to be rigourous. Just fun. For we're right on the cusp of infinity already! Let w={0,1,2,3,...}={all natural numbers}. (This ought to be a lower-case Greek omega, in case you aren't seeing it properly.) w is a set that has infinitely many elements, by which I mean that if you took an element away, you'd still have just as many left as you started with, but w behaves almost as if it were a natural number itself. In particular, we can find its successor, just by using the definition above: w+1 = {0,1,2,3,...,w}. w+1 is the set containing all natural numbers, and also w. This is a perfectly well-defined set. It looks bigger than w, since it contains everything w does and more besides. In fact, they're precisely the same size. And what do I mean by 'size' when talking about infinite sets?

Two sets are said to have the same size, or contain the same number of elements, or, in mathspeak, have equal cardinality (the cardinality of a set is the number of elements it contains, which may be bigger than finite...), if you can cook up a function or rule for associating with each and every element of the one set precisely one element of the other set, and vice-versa: you can think up a way to pair up the elements of the two sets so that you run out of elements of both sets at the same time. For example, {1,2,3} and {5,990,5324635} have the same cardinality since I can pair up 1 and 5, 2 and 990, and 3 and 5324635, and I've used up all the elements of both sets. {1,2,3} and {1,2} do not have the same cardinality. A set is finite if, as you'd expect, it has a finite number of elements, or, to be more precise, it has the same cardinality as one of the natural numbers; that natural number is then said to be the cardinality of the set. {1,2,3} has cardinality 3, since 3={0,1,2}, and we can pair up 1 with 1, 2 with 2, and 3 with 0. If a set is too big to be paired up with any natural number, it is infinite. w and w+1 are both infinite. The cardinality (or size) of w is called À₀, alef-null (or aleph-nought, because aleph or alef is the first letter of the Hebrew alphabet, and has lots of mystic connotations, and the man who invented set theory and the idea of cardinality, Georg Cantor, was an odd sort and wound up going mad, poor chap), and À₀ is the smallest of the transfinite cardinal numbers, that is, cardinalities of infinite sets. For natural, finite numbers, n and n+1 clearly never have the same cardinality; yet infinite sets are much, much odder, and it turns out that the cardinality of w+1 is also À₀, and here's why: look at w+1={0,1,2,3,...,w} and w={0,1,2,3,...}. Pair up w and 0. Now, for each natural number n in w+1, pair it up with n+1 in w. We pair up 1 and 2, 2 and 3, and so on, and so forth. We use each and every element of both sets exactly once. Thus, they both have cardinality À₀. There are lots more...Look at the set of all even numbers, {0,2,4,6,...}. This looks like it should have only half as many elements as w does, but they're actually exactly the same size, À₀. To see this, pair up 0 and 0, 1 with 2, 2 with 4, and so on...Associate to every natural number n in w the even number 2n. Or the set of all perfect squares, {0,1,4,9,16...}. Just match up each number in w to its square in {0,1,4,9,16,...}. Or the set {1,10,100,1000,...}; match up n with 10ⁿ. In fact, any infinite subset of w has À₀ elements, just like w itself. And there are much bigger sets, too, that still have only À₀ elements. Consider the integers: {....,-2,-1,0,1,2,...}, the set of all positive and negative numbers. Pair up all the nonnegative integers with the even numbers: 0 with 0, 1 with 2, 2 with 4, and so on. Then pair up all the negative numbers with the odd numbers: -1 with 1, -2 with 3, -3 with 5, and so on...There's never any danger of running out, since there are À₀ odd number and À₀ even numbers. So there are À₀ integers, too.

One of the standard examples in set theory, a bit like Schroedinger's Cat is in physics, is called Hilbert's Hotel, since it was cooked up by a German named David Hilbert who was one of the greatest mathematicians of his time. This Hotel has À₀ rooms, you see, numbered 0, 1, 2, and so on. This hotel, let's say, is poised on a particularly green and friendly hillside in the Bavarian Alps, and one day a weary hiker comes upon it and decides to stay for their famous chocolate cake. (I'm elaborating a bit.) So the hiker goes up to the desk clerk, and asks for a room. 'I'm sorry, sir,' the clerk tells him, 'but all our rooms are full.' 'Just ask everyone to move one room up,' the hiker suggests. So the clerk sends the man in room 0 off to room 1, the couple in room 1 off to room 2, and so on, leaving room 0 vacant, which the hiker happily takes. Some time later, the entire nation of China stops in. 'I'm sorry,' the clerk says, 'but all our rooms are full.' Chairman Mao suggests that the clerk simply ask everyone to move up 1.2 billion rooms, and then there's plenty of room. And so it was. Some time later, À₀ people turn up for the universe's biggest Star Trek convention. 'Jesus Fuck!' cries the clerk, throwing up his hands. 'Can't you people read? No vacancy!' The infinite number of Trekkies confer for a moment, and then send up Leonard Nimoy. 'It would be most logical if you asked all your guests to move to the room with twice the number of their current room.' So the clerk left the televangelist and the Vaseline-smeared goat where they were in room 0, and sent the blow-up doll in room 1 to room 2, and the party of nuns in room 2 to room 4, and so on, and sure enough there were precisely À₀ rooms open, just enough to fit all the Trekkies in.

At this point you might come to suspect that all infinite sets have À₀ elements, and so where's the point in giving this cardinality a special name? This is not, in fact, true, and this is where it gets fun. Hilbert's Hotel may have infinitely many rooms, but there is a quantity of guests even it can't fit, no matter how it shuffles people around.

Consider the set of all the real numbers between 0 and 1, that is, all the decimals 0.abcde... where a, b, c, and so on are all digits, between 0 and 9. (Where I forbid you from doing anything silly like ending a decimal with an infinite string of 9s.) Suppose there were À₀ of them: then we could pair them all up with the natural numbers, without missing any decimals. Pair them up; let's write the decimal we paired up with 0 as 0.a₁₁a₁₂a₁₃... where all the a_1is are digits. And let's write the decimal we paired up with 1 as 0.a₂₁a₂₂a₂₃a₂₄..., and the decimal we paired up with n-1 as 0.a_n1a_n2a_n3..., for any n. Let's write them out in an array, one above the other, in this order:

0.a₁₁a₁₂a₁₃a₁₄...
0.a₂₁a₂₂a₂₃a₂₄...
0.a₃₁a₃₂a₃₃a₃₄...
...

Now I'm going to write out a new decimal, 0.b₁b₂b₃b₄... where all the b_js are going to be digits. I'm going to do it like this. If a₁₁ is not 6, then let's let b₁ be 6. If a₁₁ is 6, let b₁ be 0. And similarly, if a₂₂ is not 6, then let b₂ be 6, and if a₂₂ is 6, let b₂ be 0. If a_nn is not 6, let b_n be 6, and if a_nn is 6, let b_n be 0. What we get is a perfectly decent decimal number, which is certainly a real number between 0 and 1, and so it must be in our list somewhere since we listed out absolutely all the decimals possible. But it isn't our first decimal, since they have different first digits. And it isn't our second decimal, since they have different second digits. And, continuing onwards, it can't be our n^th decimal for any natural number n, since they have different n^th digits. Therefore, 0.b₁b₂b₃b₄... can't have been in our list after all, and we didn't get them all. See? No matter how we pair up decimals with natural numbers, I can always construct a decimal we missed. Therefore, the set w of natural numbers and the set (0,1) of real numbers between 0 and 1 can't have the same cardinality! There must be more real numbers than natural numbers. (This is called the Diagonal Argument, and was thought up by Cantor.) But there are infinitely many real numbers, and infinitely many natural numbers! There must be, then, different levels of infinity. There's this one infinity, called À₀, to describe the natural numbers, and there must be another, in some sense bigger infinity, called c (for 'continuum', which is what the real numbers are, continuous or without holes), to describe the real numbers...

There are in fact infinitely many different levels of infinity, or transfinite numbers. (How many? It's impossible to say.) And these transfinite numbers are ordered: given any two of them, they're either equal, or the first is bigger than the second, or the second is bigger than the first, and only one of these three possibilities can be true. And as it happens, given any transfinite cardinal number, there's always a next one, too, just like there is for natural numbers. The next larger transfinite cardinal after À₀ is called À₁, and then the next bigger level of infinity after that is called À₂, and so on, and there's an À_w beyond all of those, and so on, ad infinitum as it were. Infinite levels of infinity, all of them very, very different from the ones before or after them. And how does c fit into this hierarchy, you may ask? We know c is bigger than À₀, but how much bigger? Is c equal to À₁? À₂? Or what? That is a question called the Continuum Problem, and Georg Cantor's guess that c might be equal to À₁ is called the Continuum Hypothesis. Not only, in more than a hundred years since then, has Cantor's hypothesis never been proven to be true or false, but in fact in the '30s and '60s two mathematicians, Kurt Godel (another insane genius; he believed aliens were bombarding him with a mind-control ray, so he hid under his desk at Princeton a lot, and he wound up starving himself to death because he believed he was being poisoned; he also wouldn't open his door for his students, and made them shove their papers under the crack) and Paul Cohen, working separately, in different decades even, proved that from what we know about set theory, it is in fact impossible to prove that the Continuum Hypothesis is true or false. This is an example of the phenomenon Godel is most famous for, his Incompleteness Theorem: any logical system (like set theory) sufficiently complicated to include arithmetic, that doesn't contradict itself, must contain questions that the system isn't powerful enough to answer.

And now you know enough about set theory to read and enjoy Rudy Rucker's White Light. See? That didn't hurt at all.

Finals

Did I mention my finals are now over? I will soon be more drunk than you can possibly imagine. Here's a question people often ask me: if X is an infinite-dimensional Banach space, why is any nonempty open set in the weak^* topology on X^* unbounded in the norm topology?

Good question! Non-math-dorks may wish to look elsewhere for a moment. It gets hairy.

The weak^* topology on X^* is generated by sets of the form U_xyr={f: |f(x)-y|‹r} for x in X, y in our field K (either the reals or complex numbers), and r›0. Thus any open set is a union of finite intersections of such sets. So if any finite intersection is unbounded, any open set will be unbounded. What do the finite intersections look like? They are sets of the form O={f: |f(x_i)-y_i|‹r_i, i=1...n}. That is, we pick out a finite number of vectors, and our set consists of all linear functionals whose value at those vectors is 'sufficiently close' to the value we specify. Since there are only a finite number of these vectors, and X is itself infinite-dimensional, we can choose a vector x in X linearly independent from x₁...x_n; normalise it, so that x has norm 1. The linear subspace N spanned by the x_i is finite-dimensional, hence closed (by a pretty argument I may or may not share with you later on). Then x does not lie in N. Hence by a corollary of the Hahn-Banach Theorem, we can find a functional f in X^* with f(y)=0 for all y in N, but f(x)=1. Now, if there is some f₀ in O, let us construct a new functional g in X^* by letting, for any M›0, g=f₀+(M-f₀(x))f. Then since g agrees with f₀ on N, g certainly lies in O, yet g(x)=f₀(x)+Mf(x)-f₀(x)f(x)=Mf(x)=M. Thus ||g|| is at least M, since ||g||=sup{|g(z)|: ||z||=1}. Since M was arbitrary, O must be unbounded. QED.

Mathematics for the Terminally Juvenile

I realised in class on Friday that one possible acronym for the Central Limit Theorem, in probability theory, is CLiT. I hate probability theory. These are the things that come to mind when I'm sitting in lecture and absolutely not paying attention in the slightest. I doodle a lot. The number of cartoon penises appearing in my notes is inversely proportional to the quantity of attention I was paying. We also heard that same day about something called the Weiner Process, which was nicely symmetrical...

The dirtiest piece of mathematical terminology I've ever heard is 'free group action'. That ought to be a porn site. The Hairy Ball Theorem is pretty decently dirty, too.

I have decided to devote my life to formulating and proving something I will call 'the Teabagging Theorem'. I think I might actually be able to get away with it; how many professional mathematicians are liable to practice the ancient art of ball-dipping, or to have seen John Waters's Pecker?

I should be studying right now.

Things I Have Learned Recently

1. 90% of the Canadian population lives within 320 kilometers of the US border. They are all ninjas.

2. Why do witches ride broomsticks? Medieval witches used hallucinogenic herbs and ointments in their rites; the hallucinogens were absorbed through the skin. They would rub these ointments on a broomstick and straddle it so they could absorb the compounds through their vulvas, which, with their armpits, were the optimal parts of the body for it.

3. Guinness is not a drink; it is a snack.

4. No number larger than 12 ought to exist. If you see one, drop your chalk and back away slowly. Whatever you're trying to do is more trouble than it is worth.

5. At the beginning of the year, the loonie, or 'Canadian dollar', was valued at 63.40 cents US. The US economy, however, is so shitcanned that the loonie was trading at over 74 cents US just a week ago.

6. There is a Finnish band called The Cybermen. Their website includes a small Doctor Who-related graphic of a Servo Robot.

7. I am a huge dork.

8. There was apparently also a punk band of the late 70s called The Cybermen.

9. Dan Savage will name a sex act after Rick Santorum.

10. Numbered lists get dull after a while.

Commutative Algebra

Commutative rings are evil. 'Commutative', after all, sounds like 'Communist'. What more proof do you need?

A Confession

I have a confession to make. Unfortunately, it probably isn't a very interesting confession for anyone who isn't keen on maths. The whole thing revolves around the differences between two subbranches of mathematics which may in fact be indistinguishable to outsiders...Spend enough time in graduate school and one begins to forget that the entire world is not in fact made up of mathematicians.

But first, a headline: 'Bosnian Police Capture The Masturbator'.

Police in Bosnia say they have caught a prolific burglar who they dubbed The Masturbator.

The man had allegedly broken into scores of offices to spend hours on telephone sex lines.

'The Masturbator' sounds to me precisely like the sort of villain we'd have seen on Doctor Who had FOX produced a new series. I mean, what sort of crack-addled glue-sniffer casts Eric Roberts, as anything other than perhaps a floor lamp?

But I digress!

I've been feeling filled to the brim with a certain mathematical angst recently, not unlike my mug is now, if by 'mathematical angst' I mean 'tea'...

...It occurs to me that not all who may be reading this necessarily know me already. I should then explain. I'm a first-year Ph.D. student in mathematics at the University of Washington, in Seattle. Ever since I decided to go into maths, I've had my heart set on being an algebraist. Modern algebra bears little resemblance to the algebra we teach to schoolchildren. Modern algebra is what you get when fiendishly clever Europeans spend 150 years trying to escape from any contact with the diurnal world.

If you were lucky, and not an American under about 21, it's plausible that in high school, or your local equivalent, you were taught at some stage various properties of arithmetic, like the Associative Law, and the Distributive Law, and so forth. Algebra is what you get when you throw away numbers and take some of these laws as axioms, then study the objects that obey them. From a very small set of axioms, one gets a very large, even pervasive, class of mathematical beasts with a very rich theory describing them. The simplest sort of algebraic object is a group. A group is a set of things--what these things are doesn't matter in particular--along with a function or rule, called a binary operation, for taking any two of these things and getting a third thing from them; if we write this operation multiplicatively, as ab=c where a, b, and c are elements of our group and we mean by this that plugging a and b into our operation, in that order, yields up c (just imagine a, b, and c are ordinary rational numbers we're multiplying together), to qualify as a group three axioms must be satisfied:

• The Associative Law: (ab)c=a(bc) for all a, b, and c in the group.
• There exists some element e in our group such that for all elements a in our group, ae=ea=a. e is called an identity.
• For every a in our group there exists an element a^-1 such that aa^-1=e=a^-1a.

A simple example is the set of integers under addition, with identity 0.

But I'm digressing! The point is, as soon as I had my first abstract algebra course as an undergraduate, I knew algebra was the field for me. So I went into maths, and got into graduate school, and have been slogging away since late September in algebra, real analysis, and complex analysis.

See, here's where it gets complicated. Although I came in knowing I wanted to be an algebraist, one needs to pass a number of qualifying, preliminary examinations ('prelims') in order to become a real, honest Ph.D. candidate. Three, to be exact. And here at the UW, we have exactly five to choose from: algebra, real analysis, complex analysis, linear analysis, and manifolds. So one is obliged to learn some analysis, no matter what. Analysis, by the way, is sort of a jazzed-up version of calculus, although it covers so much territory it's hard to sum up. So I bit the bullet and went for reals and complex, even though by longstanding tradition algebraists hate analysis. (Although it's not unusual for analysts to like algebra. Everyone likes algebra.)

And here's where the trouble begins. When we got to the Lebesgue theory of integration in real analysis...I started to enjoy it. Measure theory is keen! And functional analysis has some interesting properties...Dual spaces...Now we're on Radon measures, which have some dead nice regularity properties, the sort of measures every locally-compact Hausdorff space dreams of having...And sometimes, in sick moments of twisted brain-wrongness, I think to myself, 'Gee, maybe I'd like to be an analyst.'

I even invented a piece of terminology on the current homework set. It probably won't catch on, I know, but it couldn't hurt to try. I decided that if m is a Radon measure on a locally-compact Hausdorff space X, then any open set O in X such that m(O)=0 should be called 'melancholy'. Because it's sad. Open sets shouldn't have measure zero; open sets should be big. I think 'melancholy' is a much better term for them than the other suggestion I've heard, which is 'fr-izz-eaky'. So the next time you see a null, open set as you go about your daily business, be kind to the poor thing: it's melancholy.

I live with a burning shame now, and fear; I feel hunted, marked somehow. What if my algebra chums find out I like analysis? What will they do to me? Will I become a pariah? Neither one thing nor the other? What would my parents say? What would the Church say? Can I control this impure lust? Can I defeat it, and go on to live a healthy, algebraic life? I still like algebra, after all (well, except for filthy unclean commutative rings)...Maybe I just need to represent more Lie algebras. But what if I'm weak? And Radon measures can be so, so good...Oh, the temptation. It's like living in Far From Heaven.

Only silly.

I'm sure God will smite me soon.

Turkiesee

Several centuries later, Tennessee is still stuck in the Middle Ages.

McNelly said he shared those concerns, though he is not against evolution as a theory. Like Treadway, he said he believes students should be taught both creation and evolution theories.

``With creationism not presented as a theory, there's a large gaping hole in the books,'' McNelly said.

The large gaping hole is in the spare ass he passes off as his head. This is exactly the sort of loose, woolly, incoherent and illogical lapse of reason one finds anywhere that frowns on booze.

On the bright side, there's an entry in the index to Miles Reid's textbook Undergraduate Commutative Algebra for 'anal germs'. They appear on page 33.

Abstract Algebrass

Today, I achieved a state of perfect bliss and harmony. The cosmos and I became as one. I saw at last the goal towards which I have been struggling my entire life, what you might almost call the very purpose of my existence.

Ass is a piece of mathematical terminology.

Given a commutative ring R, and an R-module M, Ass M is the set of all assassins (or associated primes) of M in R, that is, the set of all prime ideals p in R such that there exists an x in M with p = Ann x: an assassin is a prime ideal which, happy day!, turns out to be the annihilator of some element of the module. (Which means p is the set of all elements r in R such that rx=0.)

But that isn't really important, is it?

What is important is that I can use the word 'ass' in maths.

I am complete.