As an ambitious young researcher, you can miss the wood for the trees. When you’re presented with a theorem, your inclination is to dive straight into the proof, and start grappling with the toughest ideas in there. But when you’re addressing a general audience, you have to step back, pause, and ask “What is the point of this theorem? Where did it come from? Why should we care?” You’re forced to adopt a different perspective on the subject, and that’s refreshing.
“The parametric engines are jammed! Orthogonal vector’s gone! I’m almost out of ideas!”
I have a guest post at The Aperiodical reporting on the London Mathematical Society’s birthday party last week, where Doctor Who was a recurring theme.
There are many songs in the world about love and loss, heartbreak and heart-ache. There are altogether fewer about algebraic geometry in the style of Alexander Grothendieck. Here is my attempt to fill that gap:
Disclaimer: I hope none of this needs saying, but just in case of misinterpretation:
1. No views expressed therein are attributable to any organisation to which I am affiliated.
2. Most of the views expressed therein are not attributable to me either, but are a deliberate exaggeration, for comic effect, of a common initial reaction to one’s first meeting with Grothendieck-style algebraic geometry.
3. It is not intended as a serious critique of any mathematician or school of mathematics!
Lyrics
When I was a young boy doing maths in class
I thought I knew it all.
Every test that I took, I was sure to pass.
I felt pride, and there never came a fall.
Up at university, I found what life is for:
A world of mathematics, and all mine to explore.
Learning geometry and logic, I was having a ball.
Until I hit a wall…
For I adore Euler and Erdős,
Élie Cartan and Ramanujan
Newton and Noether. But not to sound churlish
There’s one man I cannot understand.
No, I can’t get to grips with Grothendieck,
My palms feel sweaty and my knees go weak.
I’m terrified that never will I master the technique
Of Les Éments de Géométrie Algébrique.
He’s a thoughtful and a thorough theory-builder, sans pareil.
But can anybody help me find the secret, s’il vous plaît
Of this awe-inspiring generality and abstraction?
I have to say it’s driving me to total distraction.
For instance… A Euclidean point is a location in space, and that we can all comprehend.
René Descartes added coordinates for the power and the rigour they lend.
Later came Zariski topology, where a point’s a type of algebraic set
Of dimension nought. Well, that’s not what I thought. But it’s ok. There’s hope for me yet!
But now and contra all prior belief
We hear a point’s a prime ideal
In a locally ringed space, overlaid with a sheaf.
Professor G, is truly this for real?
No, I can’t make head nor tail of Grothendieck
Or Deligne, or Serre, or any of that clique.
I’ll have to learn not to care whenever people speak
Of Les Fondements de la Géométrie Algébrique.
But don’t take me for a geometrical fool.
I can do much more than merely prove the cosine rule.
I’ll calculate exotic spheres in dimension 29
And a variety of varieties, projective and affine.
I’m comfortable with categories (though not if they’re derived)
I’ll tile hyperbolic space in dimension 25
I can compute curvature with the Gauss-Bonnet law
And just love the Leech Lattice in dimension 24.
But algebro-geometric scheming
Leaves me spluttering and screaming.
And in logic too, you may call me absurd
But I wouldn’t know a topos, if trampled by a herd.
I’ve tried Pursuing Stacks but they vanished out of sight,
I’ve fought with étale cohomology with all my might.
And Les Dérivateurs. It’s 2000 pages long.
I reach halfway through line 3, before it all goes badly wrong.
No, I’ll never get to grips with Grothendieck
And I’m frightened that I’m failing as a mathematics geek.
All the same, I can’t deny the lure and the mystique
Of Le Séminaire de Géométrie Algébrique.
– Richard Elwes, 2015
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Last week, we Leeds mathematicians were treated to a talk by Professor GreenProfessor Green with the delightfully elementary title of Points and Lines. It reported on joint work of Ben Green and Terence Tao. Here I’m going to talk about the background to their work. As always, any errors are mine.
The idea is that we will be presented with some points, specifically a finite collection of points in the plane. We want to predict the answer to questions of the form: how many lines are there through exactly (n) many of these points? We want to do this while knowing as little as possible about the details of what we’ll be given. Actually, on a first pass, these questions are pretty trivial:
There are guaranteed to be infinitely many lines through zero points since when you’re positioning a line, a finite set of points is very easy to miss.
Similarly, there will be infinitely many lines which hit exactly one point, since once you’ve decided which to hit, you can easily tweak that line to miss the rest.
Must there be any lines which hit exactly two points? Not necessarily. If all the points lie on a single straight line (and if there are more than two of them), then any line either goes through 0, 1, or all of them.
Arranging the points around a circle illustrates that there also need not be any lines which hit exactly 3 points. The same argument works for n= 4,5,6…
So far, so straightforward: we can predict the answer exactly for (n=0,1) and not at all otherwise. But let’s think again: in the case of (n=3), there’s obviously nothing special about a circle. There are numerous other curves which would do just as well: a parabola, hyperbola, squircle, catenary, … all of these and many others share the property than any straight line hits them at most twice.
However, for the case (n=2), there is something undeniably special about demanding that all our points lie along a single straight line. So the question arises: is this configuration the only obstacle? Must it be the case that either all the points lie along a line, or we will be able to find a line hitting exactly two of them? This problem was first posed, so far as we know, by J.J. Sylvester in the back pages of the Educational Times in 1893, under the unassuming title of mathematical question 11851. (The Royal Society’s Sylvester Medal is depicted above, of which Ben Green is the most recent winner, for another famous result of his jointly with Tao.)
The question was later re-asked by Paul Erdős and then solved in 1940 by Tibor Gallai, who established that the answer was yes: if a set of points do not all lie on a line, then there must be some line through exactly two of them.
A later proof by Leroy Kelly is, as Ben put it, a `classic of mathematics’ and actually induced an audible gasp from the audience. So let’s see it:
Kelly’s proof (sketch)
We assume that our points don’t all lie along a single line. That means we can definitely find two of them which lie on a line L, along with some other point X lying off L, as shown. In fact, there are likely to be numerous such configurations; we choose the one where the perpendicular distance from X to L is the least possible.
The claim is then that L contains only those two points. Suppose not. Then there must be two points A & B both lying on L and on the same side of the perpendicular line XL. Now, let M = XA. This line also contains two points (obviously), but the distance from B to M is less than that from X to L, which is a contradiction. QED
Clever, right? Anyway, for reasons unexplained, lines which pass through exactly two of our specified collection of points are known as ordinary lines. The Sylvester-Gallai theorem above therefore says that (so long as our points are not colinear) an ordinary line will always exist. But the next question is: how many must there be? In most cases there will be lots: if you try this for a random scattering of (m) points, then in all likelihood, whenever you draw a line through two points it will miss all the others. Thus there are around (frac{1}{2}m^2) ordinary lines.
But it is not hard to come up with sets with fewer. For instance, take (m-1) points all lying along a line, and add in one extra point (X) off it. Then the only ordinary lines are those which connect (X) to another point, giving (m-1) many.
A cunning construction by Károly Böröczky brings this down even further: a set of (m) points with only (frac{1}{2}m) many ordinary lines.[1]
Now we come to the theorem of Green and Tao. In their paper, they have shown that Böröczky’s examples are as good as it gets. For large enough values of (m), they prove that every non-colinear set of (m) points must have at least (frac{1}{2}m) many ordinary lines.
How large is `large enough’? The bound they establish is in the vicinity (10^{10^{10}}), but the true answer may be as low as 14. For smaller sets, it is possible to do better. The left-hand picture here shows a set of 7 points with only 3 ordinary lines.
And how is it proved? Well, I’m not going to go into details, but the key step is the unexpected (to me) connection of the property `has few ordinary lines’ with the property `can be covered with a single cubic curve’.
[1] For the cognoscenti: it achieves this by moving the action to the real projective plane, arranging (m) points symmetrically around the circle, and placing another (m) on the line at infinity at positions corresponding to the directions determined pairs of points from the first set. Then the only ordinary lines are the (m) many tangents to the circle at points. Finally, since no-one mentioned anything about lines at infinity being permitted, the whole set-up needs to be transported back to the plain old Euclidean plane, which can be achieved via a projective transformation. This however will distort the circle into an ellipse. See page 11 of their paper for a picture. )
If you shuffle a deck of cards perfectly eight times, you get back exactly to where you started. Here’s a video of the magician Adam West doing it:
In this post, we’ll see why this works. First let’s tighten up our terminology: a “perfect shuffle” here is really what magicians call a Faro Out-Shuffle: you split the deck in half, and then interleave the two halves. To make things easier, let’s work with a deck of ten cards, initially in numerical order.
Step one: split 1,2,3,4,5 from 6,7,8,9,10.
Step two, interleave the two: 1,6,2,7,3,8,4,9,5,10.
Notice that the outermost cards (1 and 10) remain in place. (That’s why this is an out-shuffle — interleaving the other way produces an in-shuffle.)
At the same time, notice that the card initially in 4th position is now in 7th, and vice versa. We can indicate this in cycle notation by putting the two together in brackets: (4 7).
What about the rest? The card in second position to start with is now in third place, while the card from third position is now in fifth, the card from fifth is now in ninth,… Continuing this line of thought produces the cycle (2 3 5 9 8 6).
So the shuffle is totally described by the collection of cycles (1) (10) (4 7) (2 3 5 9 8 6). (I’ve included the two trivial cycles of length one.)
Now, notice that performing any cycle of length two twice brings us back to where we started: in this case 4 goes to 7 and then back to 4. Similarly, performing a cycle of length 6 six times takes use back to the starting position. In fact then, performing this ten card shuffle six times brings everything back to where it started.
Ok. What about a full 52 card deck? Well, for cards initially in the top half of the deck (i.e. numbered 1-26), the card in (n)th position is moved to ((2n -1))th position after one shuffle, (meaning (1 to 1, 2 to 3, 3 to 5, ldots, 26 to 51)).
For cards in the second half (numbered 27-52), the card in (n)th position is moved to ((2n-52))th position (meaning (27 to 2, 28 to 4, ldots, 55 to 54, 56 to 56 )).
Repeatedly applying the first rule, we can see that 2 goes to 3 which goes to 5 which goes to 9 which goes to 17 which goes to 33. Now applying the second rule, 33 goes to 14 which goes to 27 (by the first rule) which goes back to 2 (by the second rule). Thus we have the cycle:
(2 3 5 9 17 33 14 27).
The same reasoning tells us the other cycles which make up the shuffle, and altogether they turn out to be
which is to say six cycles of length eight, one of length two, and the two trivial cycles of length 1. Each of these cycles will take us back to the starting position when applied eight times… as the video above shows!
Question: what if we do a Faro in-shuffle instead? This means interleaving the other way. On ten cards, after one shuffle the deck reads 6,1,7,2,8,3,9,4,10,5. How many applications of this rule are needed to bring us back to the starting position in this case?
I’m teaching a third year Combinatorics module this term, which has led me to revisit some old friends: permutations and combinations. I thought I knew them well, but have already learned something new!
First a quick refresher course:
Combinations: say (C(n,k)) is the the number of subsets of size (k) from a collection of size (n). So if you select 3 pieces of fruit from a total of 6 (all different), and don’t care about the order, then the number of selections you might make is (C(6,3)). The formula[1] here is (C(n,k) = frac{n!}{k!(n-k)!}) and (C(6,3)=20).
Permutations: If you do care about the order of selection (maybe you want to arrange your fruit into military columns so that “strawberry, apple, pear” is different from “apple, pear, strawberry”), then the number you need is (P(n,k)), the number of permutations of (k) objects from (n). Its formula is (P(n,k) = frac{n!}{(n-k)!}) and (P(6,3)=120).
Total numbers of combinations: A set of size (n) has (2^n) subsets in total. To say the same thing another way, (C(n,0)+C(n,1)+…+C(n,n)= 2^n). So the total number of collections of fruit you could take from the box of 6 is (2^6=64) (this includes both the empty-plate option and the take-everything option).
All of the above I have known for many years. But I recently discovered something new (to me):
Total numbers of permutations: if you pick between (0) and (n) objects, in order, from a set of size (n), how many choices can you make? To ask the same thing another way, what is (P(n,0)+P(n,1)+…+P(n,n))? And the answer turns out to be ( lfloor n! cdot e rfloor ), that is the number you get if you round (n! cdot e) down to a whole number. So the number of military columns of fruit (including the empty column) you could make from your six are ( lfloor 6! cdot e rfloor = 1957).
I think this is very cool, and the proof of this is not hard. It’s been written out very nicely here by Michael Lugo. (When I first saw this result I assumed that it would be a consequence of Stirling’s approximation – but that’s not required.)
[1] These formulae all use factorial (! ) notation where (7!=7times 6 times 5 times 4 times 3 times 2 times 1)
