This is my last week in Sydney. The year-long drought in Australia has finally been broken by a series of rainstorms, but the sky was clear as I walked to my office tonight, and I saw the Milky Way really well! It's so much more prominent in the southern sky, since you can see the center of the Galaxy better.
Some issues of This Week's Finds are mainly for explaining things to other people, while others are mainly for myself. I'm afraid this Week is one of the latter. But I'll try to start by explaining what I'm up to.
Conversations with Tony Smith and Thomas Larsson have been making me think more about the biggest exceptional Lie group, that magnificent 248-dimensional monstrosity called E8. This plays a significant role in string theory and some other attempts to wrap everything we know about physics into a big, glorious Theory of Everything. None of these attempts have succeeded in predicting anything new that's actually been observed (ahem), but I still think it's worth pondering the group E8.
Why? First of all, it's a beautiful thing in itself. Second, it has strong ties to many "exotic" things in mathematics, including:
... in short, a whole zoo of strange creatures! Third, if the laws of physics are indeed structures of "exceptional beauty" rather than "classical beauty" - see "week106" for an explanation of what I mean by that - then it's natural to hope that E8 plays an important role.
How do we get our hands on E8? It's a bit tricky. To understand a group, it's always best to see it as the symmetries of something. Often we try to see it as the symmetries of some vector space equipped with extra structure. But for E8, the smallest vector space that will do the job is 248-dimensional - it's the Lie algebra of E8 itself! In mathspeak, the smallest nontrivial irrep of E8 is the adjoint rep.
But in normal Engish, the problem is this: it's hard to construct E8 as the symmetries of anything simpler than itself. It reminds me of Baron von Munchausen pulling himself out of the swamp by his own bootstraps.
One possible way around this is to construct E8 as the symmetries of something other than a vector space - for example, some manifold equipped with extra structure. Here there is some hope: the compact real form of E8 is the isometry group of a 128-dimensional Riemannian manifold called the "octooctonionic projective plane". The reason for this name is that around 1956, Boris Rosenfeld claimed that you can construct this manifold as a projective plane over the "octooctonions": the octonions tensored with themselves. Unfortunately, while there's definitely something to this idea, I don't think anyone knows how to make it precise without first constructing E8. Maybe someday....
Recently, some mathematical physicists have been studying a construction of E8 as the symmetries of a 57-dimensional manifold equipped with extra structure:
1) Murat Gunaydin, Koepsell and Hermann Nicolai, Conformal and quasiconformal realizations of exceptional Lie groups, Commun. Math. Phys. 221 (2001), 57-76, also available as hep-th/0008063
2) Thomas A. Larsson, Structures preserved by exceptional Lie algebras, available as math-ph/0301006.
When I heard this, the number 57 instantly intrigued me - and not just because Heinz advertises "57 varieties", either! No, the reason is that the smallest nontrivial of irrep of E8's little brother E7 is 56-dimensional: it's a vector space equipped with extra structure making it into the so-called "Freudenthal algebra". When you study this subject long enough, you realize that strange numbers can serve as clues to hidden relationships... and guess what: there's one here! I'll say a bit more about it later.
(By the way, the story behind Heinz's "57 varieties" is that Henry John Heinz saw an ad for 21 styles of shoe, and liked the gimmick - but the numbers 5 and 7 held a special significance for him and his wife. If you don't believe me, send a letter to Heinz Consumer Affairs, P.O. Box 57, Pittsburgh, PA 15230 and ask them!)
Another way to get ahold of the group E8 is starting with its "root lattice", the so-called E8 lattice. There are different ways to describe this. Perhaps the most efficient is to say that it's the densest lattice packing of spheres in 8 dimensions! If I were about to drown and needed to define the E8 lattice before I went under, this is how I'd do it. Unfortunately this leaves the recipient of the message with a lot of work: they have to find the lattice meeting this description.
A more user-friendly description is this. In any dimension we can make a "checkerboard" with alternating red and black hypercubes, and we get a lattice by taking the centers of all the red ones. In n dimensions this is called the Dn lattice. We can pack spheres by centering one at each point of this lattice and making them just big enough so they touch. There will of course be some space left over. But when we get up to dimension 8, there's enough room left over so we can slip another identical array of spheres in the gaps between the ones we've got! This gives the E8 lattice.
We can translate this into formulas without too much work. The Dn lattice consists of all n-tuples of integers that sum to an even integer: requiring that they sum to an even integer picks out the center of every other hypercube in our checkerboard. Then, to get E8, we take the union of two copies of the D8 lattice: the original one and another one shifted by (1/2, ..., 1/2).
(Actually this "doubled Dn" is interesting in any dimension, and it's called Dn+. In 3 dimensions this is how carbon atoms are arranged in a diamond! But Dn+ is only a lattice in even dimensions. In dimensions that are multiples of 4, it's an "integral" lattice, meaning that the dot product of any two vectors in the lattice is an integer. It's also "unimodular", meaning that the volume of the unit cell is 1. And in dimensions that are multiples of 8, it's also "even", meaning that the dot product of any vector with itself is even. D8+ = E8 is the only even unimodular lattice in 8 dimensions; in 16 dimensions there are just two: E8 x E8 and D16+. As explained in "week95", these give two versions of heterotic string theory.)
Summarizing, we can say E8 consists of all 8-tuples of real numbers (x1, ..., x8) that sum to an even integer and that are either all integers or all integers plus 1/2.
Using this description it's easy to see that when you pack spheres in an E8 lattice, each sphere touches 240 others. The reason is that the shortest nonzero vectors in this lattice, the so-called "roots", have length-squared equal to 2, and there are 240 of them:
(1,1,0,0,0,0,0,0) and all permutations thereof: there are 8 choose 2 = 28 of these (-1,-1,0,0,0,0,0,0) and all permutations thereof: there are 8 choose 2 = 28 of these (1,-1,0,0,0,0,0,0) and all permutations thereof: there are twice 8 choose 2 = 56 of these (1/2,1/2,1/2,1/2,1/2,1/2,1/2,1/2): there is 1 of these (-1/2,-1/2, 1/2,1/2,1/2,1/2,1/2,1/2): there are 8 choose 2 = 28 of these (-1/2,-1/2,-1/2,-1/2, 1/2,1/2,1/2,1/2): there are 8 choose 4 = 70 of these (-1/2,-1/2,-1/2,-1/2,-1/2,-1/2, 1/2,1/2): there are 8 choose 2 = 28 of these (-1/2,-1/2,-1/2,-1/2,-1/2,-1/2,-1/2,-1/2): there is 1 of thesefor a total of
28 x 6 + 70 + 2 = 168 + 72 = 240roots.
There's also another description of the E8 lattice, which I've been meaning to understand for ages, but which always scared me. You can think of 8-dimensional space as the octonions. The unit octonions are closed under multiplication and taking inverses. If you take the E8 lattice, rescale it so the roots have length one, and rotate it correctly, you get a collection of 240 unit octonions that are closed under multiplication! It then follows that the octonions in the E8 lattice are closed under addition and multiplication; these are called the "Cayley integral octonions".
This sounds like just the sort of thing I'd like; the problem is the phrase "rotate it correctly". First, you have to rotate the rescaled E8 lattice so that it contains the octonion 1. That already means that the coordinate system used above is not the one we usually use for octonions, where
(x0,...,x7) = x0 + x1 e1 + ... + x7 e7
with e1,...,e7 being the unit imaginary octonions, which we multiply using the standard octonion multiplication table. And just rotating the lattice any old way so that it contains 1 = (1,0,0,0,0,0,0,0) is not good enough; you have to do it the right way to get a lattice closed under multiplication.
The right way is described in Conway and Sloane's book (see "week20"). These days you can even look it up on the web:
3) Neil J. A. Sloane, Index of Lattices, the E8 lattice: coding version, http://www.research.att.com/~njas/lattices/E8_code.html
However, it always scared me, because the description involved the "Hamming code H(8,4,4)". You see, lattices are closely connected to coding theory - not coding in the sense of cryptography, but coding in the sense of efficient data transmission. In a code like this you want to pack information as efficiently as possible while keeping some error-correction ability, and mathematically this is related to the problem of densely packing spheres in higher-dimensional space! This is all very cool, but I don't understand it very well... and more importantly, whenever I looked at the description of the Hamming code H(8,4,4), I could "understand it" in the sense of nodding in mute assent, but not in the sense of seeing how it was related to anything.
Luckily, I now see how to get around this. Instead of describing the Cayley integral octonions using the theory of codes, I now see how to describe them using the octonion multiplication table! I'm sure everyone else already knew this - but they never told me.
Here's how it goes. First you have to remember your multiplication table - the octonion multiplication table, that is. Draw an equilateral triangle, draw a line from each corner to the midpoint of the opposite side, and inscribe a circle in the triangle. Then label the corners, the midpoints of the edges and the center of the triangle with the unit imaginary octonions, any way you like:
There are 6 straight lines and a circle here: we call these all "lines", and call this gadget the "Fano plane". There are 7 points and 7 lines: each point lies on 3 lines, and each line goes through 3 points... very nice.
I won't describe how to use this picture to multiply octonions, since I already did that in "week104", and we won't need that here.
Now let me describe the Cayley integral octonions. I'll actually describe all 240 of them that have length 1. Integer linear combinations of these give the Cayley integral octonions - or in other words, a rescaled version of the E8 lattice.
First, we include ±ei for i=0,...,7. Second, we include
(± 1 ± ei ± ej ± ek)/2whenever ei, ej and ek are imaginary octonions that all lie on the same line in the above chart. Third, we include
(± ei ± ej ± ek ± el)/2whenever ei, ej, ek and el are imaginary octonions that all lie off the same line in the above chart.
It's easy to see that all these octonions have length 1. It's also easy to count them! There are 2 x 8 = 16 of the first form, 24 x 7 = 112 of the second form, and 24 x 7 = 112 of the third form, for a total of 240.
It's harder to check that these 240 guys are closed under multiplication. You can save some work by noticing that each line in the Fano plane gives a copy of the quaternions sitting inside the octonions. Moreover, the 24 quaternions of the form
±1, ±i, ±j, ±k, (± 1 ± i ± j ± k)/2are closed under multiplication - these are just the unit vectors among the "Hurwitz integral quaternions", which form a D4 lattice in the quaternions (see "week91"). So, each line in the Fano plane gives a copy of the integral quaternions sitting inside the integral octonions. Even better - I'm sorry, this is getting a bit technical, but I need to write it down or I'll forget! - if we do the Cayley-Dickson construction (see "week59") to any of these copies of the integral quaternions, we get a bigger set of integral octonions that's also closed under addition and multiplication. Unfortunately, this bunch is just a copy of D4 × D4 sitting inside E8, not the whole E8. E8 is the union of all these D4 × D4's, one for line in the Fano plane. So, I have to calculate more to finish convincing myself that the Cayley integral octonions are closed under multiplication - or equivalently, that the 240 guys listed above are closed under multiplication.
[Note: I later realized that they are not closed under multiplication! We have a perfectly fine E8 lattice, so everything that follows is okay... but it's not the Cayley integral octonions! I'll explain this next week.]
Anyway: this probably makes no sense to you, but I'm happy as a clam! So what can I do with them, for example?
Well, I can see some ways to make E8 into a graded Lie algebra!
I guess I should start by saying some general stuff about graded Lie algebras, which explains why this is interesting.
For starters, I'm not talking about Z/2-graded Lie algebras, also known as "Lie superalgebras"; I'm talking about taking a plain old Lie algebra L and writing it as a direct sum of subspaces L(i), one for each integer i, such that
[L(i), L(j)] is contained in L(i+j).If only the middle 3 of these subspace are nonzero, like so:
L = L(-1) ⊕ L(0) ⊕ L(1)we say L is "3-graded". If only the middle 5 are nonzero, like so:
L = L(-2) ⊕ L(-1) ⊕ L(0) ⊕ L(1) ⊕ L(2)we say L is "5-graded". And so on. In these situations, some nice things happen.
First of all, L(0) is always a Lie subalgebra of L. Second of all, it acts on each other space L(i) by means of the bracket. Third of all, if L is 3-graded, we can give L(1) a product by picking any element k of L(-1) and defining
x o y = [[x,k],y]This product automatically satisfies two of identities defining a Jordan algebra:
x o y = y o x x o ((x o x) o y) = (x o x) o (x o y)so 3-graded Lie algebras are a great source of Jordan algebras. Fourth of all, in this situation L(0) acts on L(1) by means of the bracket operation, so we get a Lie algebra of "infinitesimal symmetries" of our Jordan algebra, too. Fifth of all, if L is 5-graded, we get a more fancy algebraic structure called a "Kantor triple system", but I'm not ready to talk about these, and you're probably not ready to listen, either!
There's a lot more to say about this stuff, but let's just see a bit about how it works for E8. We've got two nice pictures of the 240 roots of the E8 lattice; you should imagine these as the dazzling vertices of a beautiful diamond in 8 dimensions. To get a grading on E8, all we need to do is slice this diamond with evenly spaced parallel hyperplanes in such a way that each vertex of the diamond, as well as its center, lies on one of these hyperplanes. There are different ways to do this, so you should imagine yourself as a gem cutter, turning around this diamond, looking for nice ways to slice it.
For example, if we use our picture of the E8 lattice as 8-tuples that sum to an even integer are either all integers or all half-integers, one obvious way to slice the diamond is to let each slice go through those roots where the first coordinate takes on some fixed value. The first coordinate can be 1, 1/2, 0, -1/2, or -1, so we get a 5-grading. Let's work out how many roots there are of each kind:
The number of roots with a "1" as the first component is 7 + 7 = 14. The number of roots with a "1/2" as the first component is 1 + (7 choose 5) + (7 choose 3) + (7 choose 1) = 1 + 21 + 35 + 7 = 64. The number of roots with a "0" as the first component is 84. The number of roots with a "-1/2" as the first component is 1 + (7 choose 5) + (7 choose 3) + (7 choose 1) = 1 + 21 + 35 + 7 = 64. The number of roots with a "-1" as the first component is 7 + 7 = 14.Since I'm lazy, I figured out the number of roots with a "0" as the first component by totalling up all the rest and subtracting that from 240. That's how I got the number 84.
Now, whenever you have a simple Lie algebra it's a direct sum of "root spaces", one for each root, together with an n-dimensional subspace called the Cartan algebra, where n is the called the "rank" of the Lie algebra. The rank of E8 is 8, so its dimension is 240 + 8 = 248. When we taking our way of slicing the diamond and convert it into a grading of E8, the roots in the ith slice form a basis of L(i), except we also have to count the Cartan as part of L(0). Thus in this example the dimension of L(0) is not just 84 but 84 + 8 = 92. Some basic stuff about simple Lie algebra guarantees that this trick always works: we get
[L(i), L(j)] is contained in L(i+j)as desired.
So, in this example we get a 5-grading where
E8 = L(-2) ⊕ L(-1) ⊕ L(0) ⊕ L(1) ⊕ L(2) 248 = 114 + 64 + 92 + 64 + 14where I'm writing the dimension of each vector space direct below it.
Now, L(0) is a Lie algebra, but which one? To figure this out we need to think about how this diamond-cutting trick worked. At least in this case - and in fact it often works like this - the roots in the 0th slice are just the roots of a simple Lie algebra of rank one less than the one we started with. Since the Cartan of this smaller Lie algebra is one dimension smaller, it turns out that L(0) equals this smaller Lie algebra plus a one-dimensional abelian subalgebra - namely u(1).
In this example this smaller Lie algebra is so(14), which has dimension 91. L(1) is a 64-dimensional chiral spinor rep of so(14), and L(2) is the 14-dimensional vector rep... and similarly for L(-1) and L(-2). So we get a very "14-dimensional" picture of E8:
E8 = [vectors] ⊕ [spinors] ⊕ [so(14) ⊕ u(1)] ⊕ [spinors] ⊕ [vectors]But we get a more exciting way of slicing the diamond if we use the picture of E8 as the Cayley integral octonions! Let's do this, and let each slice go through those roots where the "real part" x0 of our octonion
x0 + x1 e1 + ... + x7 e7takes on some fixed value. This value can be 1, 1/2, 0, -1/2, or -1, so we again get a 5-grading. Let's count the number of roots in each slice:
The number of roots with real part 1 is 1. The number of roots with real part 1/2 is 56. The number of roots with real part 0 is 126. The number of roots with real part -1/2 is 56. The number of roots with real part -1 is 1.Here I got 56 roots with real part 1/2 by multiplying the number of lines in the Fano plane by the number of sign choices in
(1 ± ei ± ej ± ek)/2Similarly for the roots with real part -1/2. I got 126 roots with real part 0 by subtracting all the other numbers on my list from 240.
So, we get a 5-grading of E8 like this:
E8 = L(-2) ⊕ L(-1) ⊕ L(0) ⊕ L(1) ⊕ L(2) 248 = 1 + 56 + 134 + 56 + 1since 126 + 8 = 134.
This shows how to get E8 to act on a 57-dimensional manifold: we form the group E8, and form the subgroup G whose Lie algebra is L(-2) ⊕ L(-1) ⊕ L(0), and the quotient E8/G will be a 57-dimensional space on which E8 acts! In fact this space is the smallest "Grassmannian" of E8, as explained in "week181" - look at the picture of the E8 Dynkin diagram near the end.
My goal in life is now to define a set of algebraic varieties, one for each root in L(1) and L(2), so I can write a paper entitled "57 Varieties" and get sued for trademark infringement by Heinz.
In the above grading of E8, the Lie algebra L(0) is the direct sum of E7 and u(1). This is no surprise if you know that the dimension of E7 is 133... but the reason it's true is that if you take the roots of E8 that are orthogonal to any one root, you get the roots of E7. So, we get a very E7-ish description of E8:
E8 = [trivial] ⊕ [Freudenthal] ⊕ [E7 ⊕ u(1)] ⊕ [Freudenthal] ⊕ [trivial]Here the "Freudenthal algebra" is the 56-dimensional irrep of E7, which has an invariant symplectic structure and ternary product satisfying some funky equations which get turned into the definition of... a Freudenthal algebra!
There are a lot of other games we can play like this, but like solitaire they're not too fun to watch, so I'll just mention one more, and then give a bunch more references.
Above we have seen the roots of E7 as the imaginary Cayley integral octonions of norm 1. These form a 7-dimensional gemstone with 126 vertices, and we can repeat the same "gem-slicing" trick on a smaller scale to get gradings of the Lie algebra E7. If we do this in a nice way, we get a 3-grading of E7:
E7 = L(-1) ⊕ L(0) ⊕ L(1) 133 = 27 + 79 + 27Since E7's baby brother E6 is 78-dimensional, it's no surprise that the Lie algebra L(0) is E6 plus u(1). Since 3-gradings tend to give us Jordan algebras, it's no suprise that L(1) is the exceptional Jordan algebra h3(O) consisting of all 3x3 hermitian octonionic matrices. E6 acts as the group of all transformations of h3(O) preserving the determinant, and in fact h3(O) is an irrep of E6. L(-1) is just the dual of this rep. So, we get a very octonionic description of E7:
E7 = h3(O)* ⊕ [E6 ⊕ u(1)] ⊕ h3(O)Now, since E6 sits in E7 which sits in E8, just like nested Russian dolls, we can take our previous description of E8:
E8 = [trivial] ⊕ [Freudenthal] ⊕ [E7 + u(1)] &oplus [Freudenthal] ⊕ [trivial]and decompose everything in sight as irreps of E6. If we do this, the only new exciting thing that happens is that the Freudenthal algebra decomposes into a copy of the exceptional Jordan algebra, a copy of its dual, and two copies of the trivial rep:
[Freudenthal] = [trivial] ⊕ h3(O)* ⊕ h3(O) ⊕ [trivial]At least I think this is right: people sometimes write elements of the Freudenthal algebra as 2×2 matrices
a x y bwhere a,b are real and x,y lie in h3(O), but I suspect they're "cheating" a bit and identifying h3(O) with its dual.
In short, E8 contains a lot of other "exceptional" structures, all arranged in a very nice way.
Now for some references and apologies.
I didn't do justice to the stuff about Jordan algebras and 3-graded Lie algebras, because I'm still confused about certain aspects. For example, where does the unit in the Jordan algebra come from? I also didn't explain precisely what sort of "infinitesimal symmetries" we get from the action of L(0) on L(1). If we exponentiate these infinitesimal symmetries, we don't usually get automorphisms of L(1), since there's no reason for the element "k" to preserved - remember that
x o y = [[x,k],y]Instead, we get transformations that tend to preserve a "determinant" on L(1). People call L(0) the "structure algebra" of L(1) and call the corresponding group the "structure group". There's a pretty readable explanation here:
4) Kevin McCrimmon, Jordan Algebras and their applications, Bull. AMS 84 (1978) 612-627.
and hopefully even more here:
5) Kevin McCrimmon, A Taste of Jordan Algebras, Springer, Berlin, perhaps to appear in March 2003. Available for free online at http://math1.uibk.ac.at/mathematik/jordan/archive/atoja/ - but watch out, it's 545 pages long!
In fact, all this is part of a bigger relationship between 3-graded Lie algebras and so-called "Jordan triple systems" known as the Tits-Kantor-Koecher construction. Jordan triple systems are a generalization of Jordan algebras - and I'm sort of confused about why this generalization also turns up here. I guess I should read these too:
6) J. Tits, Une class d'algebres de Lie en relations avec les algebres de Jordan, Ned. Akad. Wet., Proc. Ser. A 65 (1962), 530.
7) M. Koecher, Imbedding of Jordan algebras into Lie algebra I, Am. J. Math. 89 (1967), 787.
8) Soji Kaneyuki, Graded Lie algebras, related geometric structures, and pseudo-hermitian symmetric spaces, in Analysis and Geometry on Complex Homogeneous Domains, by Faraut, Kaneyuki, Koranyi, Lu, and Roos, Birkhauser, New York, 2000.
Kaneyuki has made some nice tables of 3-gradings on simple Lie algebras, and you can see some of these here:
9) Tony Smith, Graded Lie algebras, http://www.innerx.net/personal/tsmith/GLA.html
Thomas Larsson has made a nice table of all the formally real simple Jordan algebras you get from 3-graded simple Lie algebras, and here it is, slightly modified:
Lie algebra L L'(0) dim(L(1)) Jordan algebra L(1) sl(n+1) sl(n) n Rn-1 ⊕ R so(n+2) so(n) n Rn-1 ⊕ R sp(2n) sl(n) (n2+n)/2 hn(R) so(2n) sl(n) (n2-n)/2 hn-1(R) sl(2n) sl(n)+sl(n) n2 hn(C) so(4n) sl(2n) 2n2-n hn(H) E7 E6 27 h3(O) E6 so(10) 16 h4(C)Since L(0) always contains a u(1) summand in these cases, we write
L(0) = L'(0) + u(1)so that L'(0) is the interesting part of L(0). The formally real simple Jordan algebras appearing here are all those listed in "week162" - we get all of them! In particular, Rn-1 + R is the so-called "spin factor" Jordan algebra, which appears in special relativity.
For the more intricate relationship between 5-graded Lie algebras, Freudenthal algebras and Kantor triple systems, I should reread these:
10) I. Kantor, I. Skopets, Some results on Freudenthal triple systems, 10el. Math. Sov. 2 (1982), 293.
11) K. Meyberg, Eine Theorie Der Freudenthalschen Tripelsysteme, I, II, Ned. Akad. Wet., Proc. Ser. A 71 (1968), 162-190.
12) R. Skip Garibaldi, Structurable algebras and groups of types E6 and E7, available at math.RA/9811035.
13) R. Skip Garibaldi, Groups of type E7 over arbitrary fields, available at math.RA/9811056.
14) G. Sierra, An application of the theories of Jordan algebras and Freudenthal triple systems to particles and strings, Class. Quant. Grav. 4 (1987), 227-236.
Also, I didn't say anything yet about the connection of Lie triple systems, Jordan algebras, and Jordan triple systems to the geometry of symmetric spaces! There is in fact a dictionary relating these funny algebraic structures to very nice kinds of geometry, which motivates the Tits-Kantor-Koecher construction and its generalizations. Someday I may understand this well enough to explain it. For now, you should try to get ahold of these:
15) W. Bertram, The Geometry of Jordan and Lie structures, Lecture Notes in Mathematics 1754, Springer, Berlin, 2001.
16) Ottmar Loos, Jordan triple systems, R-spaces and bounded symmetric domains, Bull. AMS 77 (1971), 558-561.
17) Ottmar Loos, Symmetric Spaces I: General Theory, W. A. Benjamin, New York, 1969. Symmetric Spaces II: Compact Spaces and Classification, W. A. Benjamin, New York, 1969.
Unfortunately of the last two books I can get only volume I at U.C. Riverside, and only volume II here at Macquarie University! Someone should reprint both of these books: they're nice. Loos has also written a book on "Jordan pairs", but in my current state of development I find that unreadable.
Addendum: Blichfeldt proved in 1935 that E8 is a maximally dense lattice packing of spheres in 8 dimensions, and Vetcinkin proved in 1980 that it's the unique lattice packing that achieves this density in 8 dimensions. Now Cohn and Kumar have shown that the E8 packing is darn close to the densest of all sphere packings in 8 dimensions, lattice or not. No other can be more than 1 + 10-14 as dense as this one!
They also showed that in 24 dimensions no packing can be more than 1 + 10-29 times as dense as the Leech lattice, and that this is the unique best lattice packing. Of course the E8 and Leech lattices are probably the best of all sphere packings in their dimensions, but it's very hard to understand the set of all sphere packings, so even these partial results are amazing.
Here are their papers:
18) H. Cohn and A. Kumar, Optimality and uniqueness of the Leech lattice among lattices, available at math.MG/0403263.
H. Cohn and A. Kumar, The densest lattice in twenty-four dimensions, Elec. Res. Ann. 10 (2004), 58-67. Available online at http://www.ams.org/era/2004-10-07/S1079-6762-04-00130-1/
There's also a really nice overview of this topic in the American Mathematical Society Notices, which explains how people manage to prove results about all packings:
19) Florian Pfender and Günter M. Ziegler, Kissing numbers, sphere packings, and some unexpected proofs, AMS Notices 51 (September 2004), 873-883. Available online at http://www.ams.org/notices/200408/200408-toc.html
And while you're at it, read this article, which studies a question mentioned in "week20":
20) Bill Casselman, The difficulties of kissing in three dimensions, AMS Notices 51 (September 2004), 884-885. Available online at http://www.ams.org/notices/200408/200408-toc.html
namely, how to roll twelve balls in 3 dimensions around the surface of a thirteenth ball of equal size.
© 2003 John Baez