Vật lý toán hiện đại và âm dương ngũ hành.

Tôi bắt đầu đưa lên đây các vấn đề mà tôi cho rằng sẽ liên quan đến cấu trúc toán học đằng sau của ADNH. Theo quan điểm của tôi, cấu trúc đằng sau của ADNH là từ lý thuyết biểu diễn của đại số và lý thuyết biểu diễn nhóm.

Trước khi bắt đầu, tôi copy lại một số tài liệu cần thiết. Tất cả đều là tiếng anh và xin lỗi vì không dịch ra tiếng việt. Nếu bạn đủ trình độ toán học để hiểu cái này thì chắc chắn là đã đủ trình độ tiếng anh để làm việc trực tiếp.

http://math.ucr.edu/home/baez/week253.html

Đầu tiên là về E8

June 27, 2007

This Week's Finds in Mathematical Physics (Week 253)

John Baez

Yay! Classes are over! Soon I'm going to Paris for three weeks, to talk with Paul-André Melliès about logic, games and category theory. But right now I'm in a vacation mood. So, I want to take a break from the Tale of Groupoidification, and mention some thoughts prompted by the work of Garrett Lisi:

1) Garrett Lisi, Deferential Geometry, http://deferentialgeometry.org.

Garrett is a cool dude who likes to ponder physics while living a low-budget, high-fun lifestyle: hanging out in Hawaii, surfing, and stuff like that. He recently won a Foundational Questions Institute award to think about ways to unify particle physics and gravity. That's an institute devoted precisely to risky endeavors like this.

Lately he's been visiting California. So, before giving a talk at Loops '07 - a loop quantum gravity conference taking place in Mexico this week - he stopped by Riverside to explain what he's been up to.

Briefly, he's been trying to explain the 3 generations of elementary particles using some math called "triality", which is related to the octonions and the exceptional Lie groups. In fact, he's trying to use the exceptional Lie group E8 to describe all the particles in the Standard Model, together with gravity.

I'd like to know if these ideas hold water. So, I should try to explain them! But as usual, in this Week's Finds I'll wind up explaining not what Garrett actually did, but what it made me think about.

For a long time, people have been seeking connections between the messy pack of particles that populate the Standard Model and structures that seem beautiful and "inevitable".

A fascinating step in this direction was the SU(5) grand unified theory proposed in 1975 by Georgi and Glashow. So, I'll start by summarizing that... and then explain how exceptional Lie groups might get involved in this game.

What people usually call the gauge group of the Standard Model:

SU(3) × SU(2) × U(1)

actually has a bit of flab in it: there's a normal subgroup that acts trivially on all known particles. This subgroup is isomorphic to Z/6. If we mod out by this, we get the "true" gauge group of the Standard Model:

G = (SU(3) × SU(2) × U(1))/(Z/6)

And, this turns out to have a neat description. It's isomorphic to the subgroup of SU(5) consisting of matrices like this:

(g 0)

(0 h)

where g is a 3×3 block and h is a 2×2 block. For obvious reasons, I call this group

S(U(3) × U(2))

If you want some intuition for this, think of the 3×3 block as related to the strong force, and the 2×2 block as related to the electroweak force. A 3×3 matrix can mix up the 3 "colors" that quarks come in - red, green, and blue - and that's what the strong force is all about. Similarly, a 2×2 matrix can mix up the 2 "isospins" that quarks and leptons come in - up and down - and that's part of what the electroweak force is about.

If this isn't enough to make you happy, go back to "week119", where I reviewed the Standard Model and its relation to the SU(5) grand unified theory. If even that isn't enough to make you happy, try this:

2) John Baez, Elementary particles, http://math.ucr.edu/home/baez/qg-spring2003/elementary/

Okay - I'll assume that one way or another, you're happy with the idea of S(U(3) × U(2)) as the true gauge group of the Standard Model! Maybe you understand it, maybe you're just willing to nod your head and accept it.

Now, the fermions of the Standard Model form a very nice representation of this group. SU(5) has an obvious representation on C5, via matrix multiplication. So, it gets a representation on the exterior algebra Λ(C5). If we restrict this from SU(5) to S(U(3) × U(2)), we get precisely the representation of the true gauge group of the Standard Model on one generation of fermions and their antiparticles!

This really seems like a miracle when you first see it. All sorts of weird numbers need to work out exactly right for this trick to succeed. For example, it's crucial that quarks have charges 2/3 and -1/3, while leptons have charges 0 and -1. One gets the feeling, pondering this stuff, that there really is some truth to the SU(5) grand unified theory.

To give you just a little taste of what's going on, let me show you how the exterior algebra Λ(C5) corresponds to one generation of fermions and their antiparticles. For simplicity I'll use the first generation, since the other two work just the same:

Λ0(C5) ≅ <left-handed antineutrino>

Λ1(C5) ≅ <right-handed down quark> ⊕

<right-handed positron, right-handed antineutrino>

Λ2(C5) ≅ <left-handed up antiquark> ⊕

<left-handed up quark, left-handed down quark> ⊕

<left-handed positron>

Λ3(C5) ≅ <right-handed electron> ⊕

<right-handed up antiquark, right-handed down antiquark> ⊕

<right-handed up quark>

Λ4(C5) ≅ <left-handed up antiquark> ⊕

<left-handed electron, left-handed neutrino>

Λ5(C5) ≅ <right-handed neutrino>

All the quarks and antiquarks come in 3 colors, which I haven't bothered to list here. Each space Λp(C5) is an irreducible representation of SU(5), but most of these break up into several different irreducible representations of S(U(3) × U(2)), which are listed as separate rows in the chart above.

If you're curious how this "breaking up" works, let me explain - it's sort of pretty. We just use the splitting

C5 ≅ C3 ⊕ C2

to chop the spaces Λp(C5) into pieces.

To see how this works, remember that Λp(C5) is just the vector space analogue of the binomial coefficient "5 choose p". A basis of C5 consists of 5 things, and the p-element subsets give a basis for Λp(C5).

In our application to physics, these 5 things consist of 3 "colors" - red, green and blue - and 2 "isospins" - up and down. This gives various possible options.

For example, suppose we want a basis of Λ3(C5). Then we need to pick 3 things out of 5. We can do this in various ways:

We can pick 3 colors and no isospins - there's just one way to do that.

We can pick 2 colors and 1 isospin - there are six ways to do that.

Or, we can pick 1 color and 2 isospins - there are three ways to do that.

So, in terms of binomial coefficients, we have

(5 choose 3) = (3 choose 3)(2 choose 0) +

(3 choose 2)(2 choose 1) +

(3 choose 1)(2 choose 2)

= 1 + 6 + 3

= 10

In terms of vector spaces we have:

Λ3(C5) ≅ Λ3(C3) ⊗ Λ0(C2) ⊕

Λ2(C3) ⊗ Λ1(C2) ⊕

Λ1(C3) ⊗ Λ2(C2)

Taking dimensions of these vector spaces, we get 10 = 1 + 6 + 3. Finally, in terms of the SU(5) grand unified theory, we get this:

Λ3(C5) = <right-handed electron> ⊕

<right-handed up antiquark, right-handed down antiquark> ⊕

<right-handed up quark>

If we play this game for all the spaces Λp(C5), here's what we get:

Λ0(C5) ≅ Λ0(C3) ⊗ Λ0(C2)

Λ1(C5) ≅ Λ1(C3) ⊗ Λ0(C2) ⊕

Λ0(C3) ⊗ Λ1(C2)

Λ2(C5) ≅ Λ2(C3) ⊗ Λ0(C2) ⊕

Λ1(C3) ⊗ Λ1(C2) ⊕

Λ0(C3) ⊗ Λ2(C2)

Λ3(C5) ≅ Λ3(C3) ⊗ Λ0(C2) ⊕

Λ2(C3) ⊗ Λ1(C2) ⊕

Λ1(C3) ⊗ Λ2(C2)

Λ4(C5) ≅ Λ3(C3) ⊗ Λ1(C2) ⊕

Λ2(C2) ⊗ Λ2(C2)

Λ5(C5) ≅ Λ3(C3) ⊗ Λ2(C2)

If we interpret this in terms of physics, we get back our previous chart:

Λ0(C5) ≅ <left-handed antineutrino>

Λ1(C5) ≅ <right-handed down quark> ⊕

<right-handed positron, right-handed antineutrino>

Λ2(C5) ≅ <left-handed up antiquark> ⊕

<left-handed up quark, left-handed down quark> ⊕

<left-handed positron>

Λ3(C5) ≅ <right-handed electron> ⊕

<right-handed up antiquark, right-handed down antiquark> ⊕

<right-handed up quark>

Λ4(C5) ≅ <left-handed up antiquark> ⊕

<left-handed electron, left-handed neutrino>

Λ5(C5) ≅ <right-handed neutrino>

Now, all this is really cool - but in fact, even before inventing the SU(5) theory, Georgi went a bit further, and unified all the left-handed fermions above into one irreducible representation of a somewhat bigger group: Spin(10). This is the double cover of the group SO(10), which describes rotations in 10 dimensions.

If you look at the chart above, you'll see the left-handed fermions live in the even grades of the exterior algebra of C5:

Λeven(C5) = Λ0(C5) ⊕ Λ2(C5) ⊕ Λ4(C5)

This big space forms something called the left-handed Weyl spinor representation of Spin(10). It's an irreducible representation.

Similarly, the right-handed fermions live in the odd grades:

Λodd(C5) = Λ1(C5) ⊕ Λ3(C5) ⊕ Λ5(C5)

and this big space forms the right-handed Weyl spinor representation of Spin(10). It's also irreducible.

I can't resist mentioning that there's also a gadget called the Hodge star operator that maps Λeven(C5) to Λodd(C5), and vice versa. In terms of physics, this sends left-handed particles into their right-handed antiparticles, and vice versa!

But if I get into digressions like these, it'll take forever to tackle the main question: how does this "Weyl spinor" stuff work?

It takes advantage of some very nice general facts. First, Cn is just another name for R2n equipped with the structure of a complex vector space. This makes SU(n) into a subgroup of SO(2n). So, it makes the Lie algebra su(n) into a Lie subalgebra of so(2n).

The group SU(n) acts on the exterior algebra Λ(Cn). So, its Lie algebra su(n) also acts on this space. The really cool part is that this action extends to all of so(2n). This is something you learn about when you study Clifford algebras, spinors and the like. I don't know how to explain it without writing down some formulas. So, for now, please take my word for it!

This business doesn't give a representation of SO(2n) on Λ(Cn), but it gives a representation of the double cover, Spin(2n). This is called the "Dirac spinor" representation. It breaks up into two irreducible parts:

Λ(Cn) = Λeven(Cn) ⊕ Λodd(Cn)

and these are called the left- and right-handed "Weyl spinor" representations.

Perhaps it's time for an executive summary of what I've said so far:

The Dirac spinor representation of Spin(10) neatly encodes everything about how one generation of fermions interacts with the gauge bosons in the Standard Model, as long as we remember how S(U(2) × U(3)) sits inside SO(10), which is double covered by Spin(10).

Of course, there's more to the Standard Model than this. There's also the Higgs boson, which spontaneously breaks electroweak symmetry and gives the fermions their masses. And, if we want to use this same trick to break the symmetry from Spin(10) down to S(U(3) × U(2)), we need to introduce more Higgs bosons. This is the ugly part of the story, it seems. Since I'm on vacation, I'll avoid it for now.

Next: how might exceptional Lie groups get involved in this game?

When Cartan classified compact simple Lie groups, he found 3 infinite families related to rotations in real, complex and quaternionic vector spaces: the SO(n)'s, SU(n)'s and Sp(n)'s. He also found 5 exceptions, which have the charming names G2, F4, E6, E7, and E8. These are all related to the octonions. G2 is just the automorphism group of the octonions. The other 4 are closely related to each other - thanks to the "magic square" of Rosenfeld, Freudenthal and Tits.

I talked about the magic square a bit in "week106" and "week145", and much more here:

3) John Baez, The magic square, http://math.ucr.edu/home/baez/octonions/node16.html

Instead of repeating all that, let me just summarize. The magic square gives vector space isomorphisms as follows:

f4 ≅ so(R ⊕ O) ⊕ (R ⊗ O)2

e6 ≅ so(C ⊕ O) ⊕ (C ⊗ O)2 ⊕ Im©

e7 ≅ so(H ⊕ O) ⊕ (H ⊗ O)2 ⊕ Im(H)

e8 ≅ so(O ⊕ O) ⊕ (O ⊗ O)2

Here f4, e6, e7 and e8 stand for the Lie algebras of the compact real forms of these exceptional Lie groups. R, C, H, and O are the usual suspects - the real numbers, complex numbers, quaternions and octonions. For any real inner product space V, so(V) stands for the Lie algebra of the rotation group of V. And, for each of the isomorphisms above, we must equip the vector space on the right side with a cleverly (but not perversely!) defined Lie bracket to get the Lie algebra on the left side.

Here's another way to say the same thing, which may ring more bells:

f4 ≅ so(9) ⊕ S9

e6 ≅ so(10) ⊕ S10 ⊕ u(1)

e7 ≅ so(12) ⊕ S12+ ⊕ su(2)

e8 ≅ so(16) ⊕ S16+

Here S9 and S10 are the unique irreducible real spinor representations of so(9) and so(10), respectively. In the other two cases, the little plus signs mean that there are two choices of irreducible real spinor representation, and we're taking the left-handed choice.

All this must seem like black magic of the foulest sort if you haven't wasted months thinking about the octonions and exceptional groups! Be grateful: I did it so you wouldn't have to.

Anyway: the case of E6 should remind you of something! After all, we've just been talking about so(10) and its left-handed spinor representation. These describe the gauge bosons and one generation of left-handed fermions in the Spin(10) grand unified theory. But now we're seeing this stuff neatly packed into the Lie algebra of E6!

More precisely, the Lie algebra of E6 can be chopped into 3 pieces in a noncanonical way:

so(10)

the unique irreducible real spinor representation of so(10), which by now we've given three different names:

S10 ≅ Λeven(C5) ≅ (C ⊗ O)2

u(1)

The first part contains all the gauge bosons in the SO(10) grand unified theory. The second contains one generation of left-handed fermions. But what about the third?

Well, S10 is defined to be a real representation of so(10). But, it just so happens that the action of so(10) preserves a complex structure on this space. This is just the obvious complex structure on (C ⊗ O)2, or if you prefer, Λeven(C5). So, there's an action of the unit complex numbers, U(1), on S10 which commutes with the action of so(10). Differentiating this, we get an action of the Lie algebra u(1):

u(1) ⊗ S10 → S10

And this map gives part of the cleverly defined Lie bracket operation in

e6 ≅ so(10) ⊕ S10 ⊕ u(1)

All this stuff is mysterious, but suggestive. It could be mere coincidence, or it could be the tip of an iceberg. It's more fun to assume the latter. So, let me say some more about it....

The copy of u(1) in here:

E6 ≅ so(10) ⊕ S10 ⊕ u(1)

is pretty amusing from a physics viewpoint. It's if besides the gauge bosons in so(10), there were one extra gauge boson whose sole role is to describe the fact that the fermions form a complex representation of so(10). This is funny, since one of the naive ideas you sometimes hear is that you can take the obvious U(1) symmetry every complex Hilbert space has and "gauge" it to get electromagnetism.

That's not really the right way to understand electromagnetism! There are lots of different irreducible representations of U(1), corresponding to different charges, and in physics we should think about all of these, not just the obvious one that we automatically get from any complex Hilbert space. If we only used the obvious one, all particles would have charge 1.

But in the Spin(10) grand unified theory, the electromagnetic u(1) Lie algebra is sitting inside so(10); it's not the u(1) you see above. The u(1) you see above is the "obvious" one that the spinor representation S10 gets merely from being a complex Hilbert space.

The splitting

e6 = so(10) ⊕ S10 ⊕ u(1)

also hints at a weird unification of bosons and fermions, something different from supersymmetry. We're seeing e6 as a Z/2-graded Lie algebra with so(10) ⊕ u(1) as its "bosonic" part and S10 as its "fermionic" part. But, this is not a Lie superalgebra, just an ordinary Lie algebra with a Z/2 grading!

Furthermore, an ordinary Lie algebra with a Z/2 grading is precisely what we need to build a "symmetric space". This is really cool, since it explains what I meant by saying that the split of e6 into bosonic and fermionic parts is "noncanonical". We'll get a space, and each point in this space will give a different way of splitting e6 as

e6 = so(10) ⊕ S10 ⊕ u(1)

It's also cool because it gives me an excuse to talk about symmetric spaces... a topic that deserves a whole week of its own!

Symmetric spaces are the epitome of symmetry. A "homogeneous space" is a manifold with enough symmetry that any point looks like any other. A symmetric space is a homogeneous space with an extra property: the view from any point in any direction is the same as the view in the opposite direction!

Euclidean spaces and spheres are the most famous examples of symmetric spaces. If an ant decides to set up residence on a sphere, any point is just as good any other. And, if sits anywhere and looks in any direction, the view is the same as the view in the opposite direction.

The symmetric space we get from the above Z/2-graded Lie algebra is similar, but more exotic: it's the complexified version of the octonionic projective plane!

But let's start with the basics:

Suppose someone hands you a Lie algebra g with a Lie subalgebra h. Then you can form the simply-connected Lie group G whose Lie algebra is g. Sitting inside G, there's a connected Lie group H whose Lie algebra is h. The space

G/H

is called a "homogeneous space". Such things are studied in Klein geometry, and I've been talking about them a lot lately.

But now, suppose g is a Z/2-graded Lie algebra. Its even part will be a Lie subalgebra; call this h. This gives a specially nice sort of homogeneous space G/H, called a "symmetric space". This is better than your average homogeneous space.

Why? Well, first of all, for each point p in G/H there's a map from G/H to itself called "reflection through p", which fixes the point p and acts as -1 on the tangent space of p. When our point p comes from the identity element of G, this reflection map corresponds to the Z/2 grading of the Lie algebra, which fixes the even part and acts as -1 on the odd part.

This is what I meant by saying that in a symmetric space, "the view in any direction is the same as the view in the opposite direction".

Second, these reflection maps satisfy some nice equations. Write p>q for the the result of reflecting q through p. Then we have:

p>(p>q) = q

p>p = p

and

p>(q>r) = (p>q) > (p>r)

A set with an operation satisfying these equations is called an "involutory quandle". Quandles are famous in knot theory. Now we're seeing them in another role.

Let me summarize with a few theorems - I hope they're all true, because I don't know a book containing all this stuff. We can define a "symmetric space" to be an involutory quandle that's a manifold, where the operation > is smooth and the reflection map

x |→ p>x

has derivative -1 at p. Any Z/2-graded Lie algebra gives a symmetric space. Conversely, any symmetric space has a universal cover that's a symmetric space coming from a Z/2-graded Lie algebra!

Using this correspondence, the Lie algebra e6 with the Z/2-grading I described gives a symmetric space, roughly:

E6/(Spin(10) × U(1))

But, this guy is a lot better than your average symmetric space!

For starters, it's a "Riemannian symmetric space". This is a symmetric space with a Riemannian metric that's preserved by all the operations of reflection through points.

Compact Riemannian symmetric spaces were classified by Cartan, and you can see the classification here, in a big chart:

4) Riemannian symmetric spaces, Wikipedia, http://en.wikipedia.org/wiki/Riemannian_symmetric_space

As you'll see, there are 7 infinite families and 12 exceptional cases. The symmetric space I'm talking about now, namely E6/(Spin(10) × U(1)), is called EIII - it's the third exceptional case. And, as you can see from the chart in this article, it's the complexified version of the octonionic projective plane! For this reason, I sometimes call it

(C ⊗ O)P2

In fact, this space is better than your average Riemannian symmetric space. It's a Kähler manifold, thanks to that copy of U(1), which makes each tangent space complex. Moreover, the Kähler structure is preserved by all the operations of reflection through points. So, it's a "hermitian symmetric space".

You're probably drowning under all this terminology unless you already know this stuff. I guess it's time for another executive summary:

Each point in the complexified octonionic projective plane gives a different way of splitting the Lie algebra of E6 into a bosonic part and a fermionic part. The fermionic part is just what we need to describe one generation of left-handed Standard Model fermions. The bosonic part is just what we need for the gauge bosons of the Spin(10) grand unified theory, together with a copy of u(1), which describes the complex structure of the left-handed Standard Model fermions.

Another nice fact is that (C ⊗ O)P2 is one of the Grassmannians for E6. I explained this general notion of "Grassmannian" back in "week181", and you can see this 16-dimensional one in the list near the end of that Week.

Even better, if you geometrically quantize this Grassmannian using the smallest possible symplectic structure, you get the 27-dimensional representation of E6 on the exceptional Jordan algebra!

So, there's a lot of seriously cool math going on here... but since the basic idea of relating the Standard model to E6 is only half-baked, all the ideas I'm mentioning now are at best quarter-baked. They're mathematically correct, but I can't tell if they're leading somewhere interesting.

In fact, I would have kept them in the oven longer had not Garrett Lisi brought E6's big brother E8 into the game in a tantalizing way. I'll conclude by summarizing this... and you can look at his website for more details. But first, let me emphasize that this E8 business is the most recent and most speculative thing Garrett has done. So, if you think the following idea is nuts, please don't jump to conclusions and decide everything he's doing is nuts!

Briefly, his idea involves taking the description of e8 I already mentioned:

e8 = so(O ⊕ O) ⊕ (O ⊗ O)2

and writing the linear transformations in so(O ⊕ O) as two 8×8 blocks living in so(O), together with an off-diagonal block living in O ⊗ O. This gives

e8 = so(O) ⊕ so(O) ⊕ (O ⊗ O)3

Then, he wants to use each of the three copies of O ⊗ O to describe one of the three generations of fermions, while using the so(O) ⊕ so(O) stuff to describe bosons (including gravity).

For this, he builds on some earlier work where he sought to combine gravity, the Standard Model gauge bosons, the Higgs and one generation of Standard Model fermions in an so(7,1) version of MacDowell-Mansouri gravity.

If I were really being responsible, I would describe and assess this earlier work. But, it's summer and I just want to have fun....

In fact, the above alternate description of E8 is the one Bertram Kostant told me about back in 1996. He said it a different way, which is equivalent:

E8 = so(8) ⊕ so(8) ⊕ End(V8) ⊕ End(S8+) ⊕ End(S8-)

Here V8, S8+ and S8- are the vector, left-handed spinor, and right-handed spinor representations of Spin(8). All three are 8-dimensional, and all are related by outer automorphisms of Spin(8). That's what "triality" is all about. You can see more details in "week90".

The idea of relating the three generations to triality is cute. Of course, even if it worked, you'd need something to give the fermions in different generations different masses - which is what happens already in the Standard Model, thanks to the Higgs boson. It's the bane of all post-Standard Model physics: symmetry looks nice, but the more symmetry your model has, the more symmetries you need to explain away! As the White Knight said to Alice:

But I was thinking of a plan

To dye one's whiskers green,

And always use so large a fan

That they could not be seen.

Someday we may think of a way around this problem. But for now, I've got a more pressing worry. This splitting of E6:

E6 = so(10) ⊕ S10+ ⊕ u(1)

corresponds to a Z/2-grading where so(10) ⊕ u(1) is the "bosonic" or "even" part and S10+ is the "fermionic" or "odd" part. This nicely matches the way so(10) describes gauge bosons and S10+ describes fermions in Georgi's grand unified theory. But, this splitting of E8:

E8 = so(8) ⊕ so(8) ⊕ End(V8) ⊕ End(S8+) ⊕ End(S8-)

does not correspond to any Z/2-grading where so(8) ⊕ so(8) is the bosonic part and End(V) ⊕ End(S+) ⊕ End(S-) is the fermionic part. There is a closely related Z/2-grading of E8, but it's this:

E8 = so(16) ⊕ S16+

So, right now I don't feel it's mathematically natural to use this method to combine bosons and fermions.

But, only time will tell.

Here are some more references. The SU(5) grand unified theory was published here:

5) Howard Georgi and Sheldon Glashow, Unity of all elementary-particle forces, Phys. Rev. Lett. 32 (1974), 438.

For a great introduction to the Spin(10) grand unified theory - which is usually called the SO(10) GUT - try this:

6) Anthony Zee, Quantum Field Theory in a Nutshell, Chapter VII: SO(10) unification, Princeton U. Press, Princeton, 2003.

Then, try these more advanced review articles:

7) Jogesh C. Pati, Proton decay: a must for theory, a challenge for experiment, available as hep-ph/0005095.

8) Jogesh C. Pati, Probing grand unification through neutrino oscillations, leptogenesis, and proton decay, available as hep-ph/0305221.

The last two also consider the gauge group "G(224)", meaning SU(2) × SU(2) × SU(4).

By the way, there's also a cute relation between the SO(10) grand unified theory and 10-dimensional Calabi-Yau manifolds, discussed here:

9) John Baez, Calabi-Yau manifolds and the Standard Model, available as hep-th/0511086

This is an easy consequence of the stuff I've explained this week.

To see what string theorists are doing to understand the Standard Model these days, see the following papers. Amusingly, they also use E8 - but in a quite different way:

10) Volker Braun, Yang-Hui He, Burt A. Ovrut and Tony Pantev, A heterotic Standard Model, available as hep-th/0501070.

A Standard Model from the E8 × E8 heterotic superstring, hep-th/0502155.

Vector bundle extensions, sheaf cohomology, and the heterotic Standard Model, available as hep-th/0505041.

Heterotic Standard Model moduli, available as hep-th/0509051.

The exact MSSM spectrum from string theory, available as hep-th/0512177.

All this stuff is really cool - but alas, they get the "minimal supersymmetric Standard Model", or MSSM, which has a lot more particles than the Standard Model, and a lot more undetermined parameters. Of course, these flaws could become advantages if the next big particle accelerator, the Large Hadron Collider, sees signs of supersymmetry.

For more on symmetric spaces, try these:

11) Sigurdur Helgason, Differential Geometry, Lie Groups, and Symmetric Spaces, AMS, Providence, Rhode Island, 2001.

12) Audrey Terras, Harmonic Analysis on Symmetric Spaces and Applications I, Springer, Berlin, 1985. Harmonic Analysis on Symmetric Spaces and Applications II, Springer, Berlin, 1988.

13) Arthur Besse, Einstein Manifolds, Springer, Berlin, 1986.

They're all classics. Helgason's book will teach you differential geometry and Lie groups before doing Cartan's classification of symmetric spaces. Terras' books are full of fun connections to other branches of math. Besse's book has lots of nice charts, and goes much deeper into the Riemannian geometry of symmetric spaces.

These dig deeper into the algebraic aspects of symmetric spaces:

14) W. Bertram, The Geometry of Jordan and Lie structures, Lecture Notes in Mathematics 1754, Springer, Berlin, 2001.

15) Ottmar Loos, Jordan triple systems, R-spaces and bounded symmetric domains, Bull. AMS 77 (1971), 558-561.

16) 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.

Finally, an obnoxious little technical note. The complexification of the octonionic projective plane is not really E6/(Spin(10) × U(1)); it's

E6/((Spin(10) × U(1))/(Z/4))

This is worked out here:

17) John Frank Adams, Lectures on Exceptional Lie Groups, eds. Zafer Mahmud and Mamoru Mimura, University of Chicago Press, Chicago, 1996.

Addendum: Joseph Hucks points out his paper describing the 13 different groups with Lie algebra su(3) ⊕ su(2) ⊕ u(1), and their implications for physics:

18) Joseph Hucks, Global structure of the standard model, anomalies, and charge quantization, Phys. Rev. D 43 (1991), 2709-2717.

Using S(U(3) × U(2)) and demanding anomaly cancellation, we automatically get a lot of the features of the Standard Model fermions.

Toby Bartels wisely points out that my basic examples of symmetric spaces - Euclidean spaces and sphere - are actually a bit misleading. I'd written:

Euclidean spaces and spheres are the most famous examples of symmetric spaces. If an ant decides to set up residence on a sphere, any point is just as good any other. And, if sits anywhere and looks in any direction, the view is the same as the view in the opposite direction.

But in these particular examples, the view in any direction is the same as the view in any other direction! These spaces are more symmetrical than your average symmetric space: they're isotropic.

So, it's good to see some other examples, like a torus formed as the product of two circles of different radii. Any product of symmetric spaces is a symmetric space, so this is definitely a symmetric space. And, if you think about it, the ant's-eye view in any direction is just the same as the view in the opposite direction. But, this space is not isotropic: there are special directions, corresponding to "the short way around the torus" and "the long way around the torus".

The octonionic projective plane OP2 is not only a symmetric space: it's isotropic! But according to Tony Smith, the complexified version (C ⊗ O)P2 is not isotropic:

In Spaces of Constant Curvature, Fifth Edition (Publish or Perish 1984), Joseph A. Wolf says (pages 293-294):

... M is called isotropic at x if I(M)x is transitive on

the unit sphere in Mx;

it is isotropic if it is isotropic at every point.

...

M is isotropic if and only if it is two point homogeneous.

...

Let M be a riemannian symmetric space.

Then the following conditions are equivalent.

(i) M is two point homogeneous.

(ii) Either M is a euclidean space or M is irreducible and

of rank 1.

Since (C ⊗ O)P2 = E6/(SO(10) × SO(2)) is rank 2, it is NOT isotropic.

In the quote by Wolf, I can only guess that I(M)x is the group of isometries of M that fix the point x, while Mx is the tangent space of M at x. Similarly, I guess that "two point homogeneous" means that for any D ≥ 0, the isometry group of M acts transitively on the set of pairs of points in M whose distance from each other is D.

I also thank Tony for correcting some errors involving spinors. There's some quite subtle stuff going on here. For example, above it says that:

Well, S10 is defined to be a real representation of so(10). But, it just so happens that the action of so(10) preserves a complex structure on this space. This is just the obvious complex structure on (C ⊗ O)2, or if you prefer, Λeven(C5). So, there's an action of the unit complex numbers, U(1), on S10+ which commutes with the action of so(10).

But in fact, so(10) preserves two complex structures on S10. This is how it always works: if some complex structure J is preserved by some group or Lie algebra action, so is -J. In the case at hand, one of these makes the representation of so(10) into a complex representation isomorphic to its left-handed Weyl spinor representation on Λeven (C5). The other gives the right-handed Weyl spinor representation on Λodd(C5). Neither one of these is "more correct" than the other. So, whenever I talk about e6 as being related to one generation of left-handed Standard Model fermions, I could equally well say "right-handed Standard Model fermions". It just depends on which complex structure we choose!

Furthermore, when we complexify the real Lie algebra e6, we get

C ⊗ e6 ≅ so(10,C) ⊕ Λ(C5) ⊕ C

where Λ(C5) is the Dirac spinor representation of so(10,C), describing both left- and right-handed fermions.

For more discussion, go to the n-Category Café.

The Big Crunch was her nickname for the mythical result that the Niah had aspired to reach: a unification of every field of mathematics that they considered significant. - Greg Egan, Glory

© 2007 John Baez

baez@math.removethis.ucr.andthis.edu

Chào Kaka, em biên soạn cẩn thận hơn nhất là các biểu thức toán. Em đưa nguyên xi các bài viết theo tuần của John Baez thật khó gây cảm hứng cho người đọc, có khi còn bị khó chịu. Em hãy viết theo ý em và nếu cần thì trích dẫn những đoạn thật súc tích, tốt nhất là có hình ảnh và số liệu tính toán minh họa.

Tiếp thị thế này có nguy cơ bị phá sản mất...

Thân mến

Muốn tiếp thị thì phải viết kiểu này vậy

http://aimath.org/E8/

Lời giải câu đố 120 năm phủ kín đảo Manhattan

Các nhà nhà toán học và chuyên gia máy tính quốc tế đã sắp xếp được cấu trúc bên trong của một trong những vật thể phức tạp nhất mà con người từng biết đến với 248 chiều.

Từ cách đây 4 năm, 18 nhà nghiên cứu đã được Viện toán học Mỹ (AIM), trụ sở tại Palo Alto, mời tham gia dự án mô hình hóa vật thể có tên "Lie group E8".

Những nhóm Lie này do nhà toán học người Na Uy Sophus Lie phát triển từ thế kỷ 19 khi nghiên cứu các cấu trúc đối xứng như hình cầu và phép vi phân. Trong đó, nhóm E8, ra đời năm 1887, là nhóm Lie phức tạp đến mức người ta cho là không thể tìm ra lời giải.

"Để nói chính xác E8 là cái gì thì ngay cả nhiều chuyên gia toán học kỳ cựu cũng không thể hiểu nổi", Jeffrey Adams, trưởng dự án và là nhà toán học tại Đại học Maryland (Mỹ), cho hay.

Sau khi được nén lại để lưu trong ổ cứng, lời giải cho câu đố E8 vẫn có dung lượng lớn tương đương một file nhạc MP3 được phát liên tục 45 ngày. "Nó giống như đỉnh Everest trong toán học mà chúng ta vừa chinh phục và sẽ tạo ra bước tiến lớn trong lĩnh vực hình học và vật lý", Brian Conrey, Giám đốc viện AIM, phát biểu.

Bạn đưa lên đây thì chú thích giùm cho tiếng Việt, bạn thông cảm, học ÂDNH đã mệt giờ lại phải đọc bằng Anh ngữ thì khổ quá. Đọc tiếng Tàu còn lọ mọ được chút chút :lol:. Không lẽ đọc một hồi vã mồ hôi ra mới hiểu Mr Khong = Khổng tử, Mr Giong = Thánh Gióng.

Bạn đưa lên đây thì chú thích giùm cho tiếng Việt, bạn thông cảm, học ÂDNH đã mệt giờ lại phải đọc bằng Anh ngữ thì khổ quá. Đọc tiếng Tàu còn lọ mọ được chút chút :lol:. Không lẽ đọc một hồi vã mồ hôi ra mới hiểu Mr Khong = Khổng tử, Mr Giong = Thánh Gióng.

Bạn hỏi thế thì khó quá. Tiếng anh dài thế dịch sao cho hết được. Để tôi nghĩ cách khác. tài liệu tiếng việt thì khó quá, vì VN minh không mạnh về khoa học.