Question about characteristic classes
Xi Yin
flux quantization in M-theory.
1. For a spin manifold, the first Pontryagin class (of the tangent bundle)
is always divisble by 2 in the integer cohomology group. How to prove this?
What's the connection between the first Pontryagin class and the second
Stiefel-Whitney class?
2. For an E8 bundle V over a Riemannian manifold M of dim<16, why is the
homotopy type of V completely determined by the characteristic class
tr(F^F)/16pi^2? How is this related to the homotopy groups of E8?
3. Just for curiosity, can anyone sktech the idea of computing the homotopy
groups of E8?
Thanks,
Xi
Aaron Bergman
"Xi Yin" <xi...@fas.harvard.edu> wrote:
> These questions arises when I was trying to understand Witten's paper on
> flux quantization in M-theory.
>
> 1. For a spin manifold, the first Pontryagin class (of the tangent bundle)
> is always divisble by 2 in the integer cohomology group. How to prove this?
> What's the connection between the first Pontryagin class and the second
> Stiefel-Whitney class?
I'd guess that the argument goes something like this: recall that the
first Pontryagin class is the chern class of the complexification of the
canonical bundle. Now, two spinors include a vector (among other
things). Tensor with C, take the top exterior power of both sides and
then look at the first Chern class. I think that ought to work.
> 2. For an E8 bundle V over a Riemannian manifold M of dim<16, why is the
> homotopy type of V completely determined by the characteristic class
> tr(F^F)/16pi^2? How is this related to the homotopy groups of E8?
Witten works all this out in "TOPOLOGICAL TOOLS IN TEN-DIMENSIONAL
PHYSICS" Int.J.Mod.Phys.A1:39,1986
> 3. Just for curiosity, can anyone sktech the idea of computing the homotopy
> groups of E8?
Nope. A student here went on a long hunt to find this information
(including asking the man himself) and never really got an answer.
Aaron
--
Aaron Bergman
<http://www.princeton.edu/~abergman/>
<http://aleph.blogspot.com>
hyst...@mac.com
> 1. For a spin manifold, the first Pontryagin class (of the tangent bundle)
> is always divisble by 2 in the integer cohomology group. How to prove this?
> What's the connection between the first Pontryagin class and the second
> Stiefel-Whitney class?
This is in Borel-Hirzebruch, Characteristic Classes an Homogeneous Spaces
(Amer. J. Math. 81, 1959), p.376: B being the base manifold, Pontrjagin's
p_i \in H^2i(B,Z) and Stiefel-Whitney's w_2i \in H^2i(B,Z/2Z) are related
by
(p_i) modulo 2 = (w_2i)^2.
So, for a spin manifold (w_2 = 0) we get p_1 divisible by two.
Marc Nardmann
Aaron Bergman wrote:
> "Xi Yin" <xi...@fas.harvard.edu> wrote:
>
>>These questions arises when I was trying to understand Witten's paper on
>>flux quantization in M-theory.
>>
>>1. For a spin manifold, the first Pontryagin class (of the tangent bundle)
>>is always divisble by 2 in the integer cohomology group. How to prove this?
>>What's the connection between the first Pontryagin class and the second
>>Stiefel-Whitney class?
It's this: If M is any smooth manifold, if r:Z --> Z_2 is the mod 2
projection and (damn ACSII!) r_*: H^4(M;Z) --> H^4(M;Z_2) is the induced
group homomorphism, then the first Pontrjagin class p_1 \in H^4(M;Z) and the
second Stiefel/Whitney class w_2 \in H^2(M;Z_2) of the tangent bundle TM
satisfy the equation
r_*(p_1) = (w_2)^2 .
(This generalizes to a relation between p_k and w_{2k}.) If M is spin, then
w_2 = 0 and thus p_1 is divisible by 2.
(snip)
>>3. Just for curiosity, can anyone sktech the idea of computing the homotopy
>>groups of E8?
>
> Nope. A student here went on a long hunt to find this information
> (including asking the man himself) and never really got an answer.
I haven't seen Witten's paper, and I'm neither an expert in homotopy nor
in Lie group theory. But I guess the idea is to express E8 as the total
space of a fibration (e.g., take some exceptional Lie group G which
occurs as a subgroup of E8 and use the corresponding fibration G --> E8
--> E8/G) and then consider the homotopy exact sequence of this
fibration. You might have to consider several fibrations of this type
(with different Gs) and put this information together. Since you will
not know the homotopy groups of the fibre (probably another Lie group)
and the base space a priori, this approach might lead to a recursive
scheme where you go from bigger to smaller groups (E8 is quite big,
presumably an end in such a food chain).
As a toy model, think of the fibration SO(n) --> SO(n+1) --> S^n. When
you know some of the homotopy groups of SO(n) and S^n, you get a lot of
information on the corresponding homotopy groups of SO(n+1) from the
exact sequence of this fibration.
But don't trust me on these matters, ask your local homotopy theorist.
-- Marc Nardmann (to reply, remove the letter which doesn't belong
there three times from my e-mail address)
P.S.: From the MathSciNet database (I haven't looked at the paper but
it might contain some references to earlier results):
*****
Kachi, Hideyuki(J-SHINS); Mimura, Mamoru(J-OKAY)
Homotopy groups of compact exceptional Lie groups.
Proc. Japan Acad. Ser. A Math. Sci. 75 (1999), no. 4, 47--49.
Let $G$ be one of the simply-connected, compact exceptional Lie groups
$G_1,F_4,E_6,E_7$ or $E_8$. In this short note the authors show how to
calculate the 3-primary components of $\pi_i(G)$ in terms of the
homotopy groups of spheres. Explicit descriptions of the 3-primary
components of $\pi_i(G)$ are given for $i\leq 36$. The methods are
standard: killing homotopy classes and using the homotopy exact
sequences associated to various fibrations.
Reviewed by Jay A. Wood
*****
John Baez
Marc Nardmann <Marxc.N...@webx.de> wrote:
>Aaron Bergman wrote:
>> "Xi Yin" <xi...@fas.harvard.edu> wrote:
>>>3. Just for curiosity, can anyone sketch the idea of computing
>>>the homotopy groups of E8?
>> Nope. A student here went on a long hunt to find this information
>> (including asking the man himself) and never really got an answer.
The last time this question came up on sci.physics.research,
I told people to look at Mimura's review article on the homotopy
groups of simple Lie groups in the "Handbook of Algebraic Topology"
edited by I. M. James. This should at least have a reference to
the calculation. Also, the renowned homotopy theorist Frank Adams
was very fond of E8, and I think one may find some interesting
material on its homotopy groups in his "Collected Works".
Since I'm trying to become an expert on exceptional Lie groups,
I should learn how this stuff works. Unfortunately I haven't had
time to carefully read these references and post a summary.
>I haven't seen Witten's paper, and I'm neither an expert in homotopy nor
>in Lie group theory. But I guess the idea is to express E8 as the total
>space of a fibration (e.g., take some exceptional Lie group G which
>occurs as a subgroup of E8 and use the corresponding fibration G --> E8
>--> E8/G) and then consider the homotopy exact sequence of this
>fibration.
That's what I think too. And I have a strong hunch as to which fibration
to use! The Lie algebra e8 has so(16) as a maximal-rank Lie subalgebra,
and this gives us G = Spin(16)/(Z/2) as a subgroup of the simply-connected
form of E8 (which is also the centerless form). This lets us define a
fibration
G --> E8 --> E8/G
The space E8/G is truly remarkable: it's a 128-dimensional
Riemannian manifold called the "octooctonionic projective plane"
by Boris Rosenfeld, and it's a symmetric space whose isometry
group is exactly E8. Naively it has the right dimension to be
a projective plane over the "octooctonions", which is the
64-dimensional algebra
O tensor O,
O being the octonions. However, as far as I know, nobody has yet
been able to construct this space from the octooctonions without
first building E8 and then forming the quotient E8/G. One *can*
construct E8 using the octooctonions, and also the subgroup G, and
thus E8/G - but it would be nice to define the octooctonionic
projective plane *first*, and then *define* E8 as the isometry
group of this Riemannian manifold.
You see, it's annoying that nobody has ever come up with a nice
way to define E8 as the symmetry group of some structure that's
simpler than E8 itself! For example, the smallest nontrivial
representation of E8 is the representation on its own Lie algebra,
e8. Boris Rosenfeld has claimed the octooctonionic projective plane
is the answer to this puzzle. It's a tempting answer, but it will
only be satisfying to me if we can get our hands on the octooctonionic
projective plane without first building E8.
All this stuff is explained in much more detail here:
http://math.ucr.edu/home/baez/Octonions/node19.html
where I give 5 ways of constructing E8 from the octonions.
I urge all string theorists and anyone else fascinated by
E8 to read this webpage! Hopefully someone will use the
clues presented here to find a deeper understanding of E8.
And I'm really hoping there's some way to use this stuff to
prove that the homotopy groups of E8 vanish between the 3rd
and the 16th.
Aaron Bergman
> In article <3DECFD13...@webx.de>,
> Marc Nardmann <Marxc.N...@webx.de> wrote:
>>Aaron Bergman wrote:
>>> "Xi Yin" <xi...@fas.harvard.edu> wrote:
>>>>3. Just for curiosity, can anyone sketch the idea of computing
>>>>the homotopy groups of E8?
>>> Nope. A student here went on a long hunt to find this information
>>> (including asking the man himself) and never really got an answer.
> The last time this question came up on sci.physics.research,
> I told people to look at Mimura's review article on the homotopy
> groups of simple Lie groups in the "Handbook of Algebraic Topology"
> edited by I. M. James. This should at least have a reference to
> the calculation.
He looked there. Told me he couldn't find it.
Aaron
Thomas Larsson
John Baez <ba...@galaxy.ucr.edu> wrote in message news:aso9nn$pfh$1...@glue.ucr.edu...
> You see, it's annoying that nobody has ever come up with a nice
> way to define E8 as the symmetry group of some structure that's
> simpler than E8 itself!
The Lie algebra e8 can be realized as vector fields acting on a 57D space,
see http://www.arxiv.org/abs/hep-th/0008063. It looks like this
realization can be described in a rather straightforward way as contact
vector fields (i.e. as a subalgebra of the contact algebra k(57))
preserving extra structure. It will take a while to write this up, though.
Thomas Larsson
thomas....@hdd.se (Thomas Larsson) wrote in message news:<4b8cc0a6.02120...@posting.google.com>...
A while has now passed: http://www.arxiv.org/abs/math-ph/0301006.
Let me briefly describe e_8 as an algebra of vector fields. Consider a
57D space with coordinates t, x^ij = -x^ji, and y_ij = -y_ji, where i,j =
1,2,...,8. There are 8*7/2 = 28 x's, 28 y's, and one t, which gives us a
total of 57 indeterminates. We may think of x^ij as the coordinates of a
28D configuration space, of y_ij as the corresponding momenta, and of t
as time.
Now consider the algebra of vector fields vect(57). As is well known,
the subalgebra of contact vector fields k(57) preserves the contact
one-form
alpha = dt + x^ij dy_ij - y_ij dx^ij
up to a function. We can also write down the bilinear forms
beta^ijkl = dx^[ij dx^kl] + eps^ijklmnpq dy_mn dy_pq.
The brackets in the first term indicate total anti-symmetrization, i.e.
dx^[ij dx^kl] = dx^ij dx^kl - dx^ji dx^kl - dx^kj dx^il + ...,
and eps^ijklmnpq is the totally anti-symmetric constant tensor in 8D.
Now I claim that e_8 is the subalgebra of k(57) which additionally
preserves beta^ijkl up to functions. In other words, X is in e_8 if
L_X alpha = f alpha,
L_X beta^ijkl = g^ijkl_mnpq beta^mnpq
for some X-dependent functions f and g^ijkl_mnpq.
Why do I claim that these vector fields generate an algebra isomorphic
to e_8? Well, first of all the explicit realization of e_8 on 57D space
was written down by M. Gunaydin, K. Koepsell and H. Nicolai in
http://www.arxiv.org/abs/hep-th/0008063. One sees that e_8 is a graded
Lie algebra of depth 2 (5-graded structure),
e_8 = g_-2 + g_-1 + g_0 + g_1 + g_2.
The grading is defined by deg t = 2, deg x^ij = deg y_ij = 1, which
makes alpha and beta^ijkl homogeneous of degree 2.
The subspaces g_1 and g_2 are realized non-linearly which makes this
realization quite opaque. Fortunately, it ususally suffices to know the
structures preserved by g_-2 + g_-1 + g_0, and one verifies that this
subalgebra indeed preserves alpha and beta^ijkl up to functions.
Moreover, the vector fields which preserve some structure automatically
generate a closed subalgebra, and alpha and beta^ijkl are the only
natural structures that can be written down. Finally, we know that this
subalgebra is not empty because g_-2 + g_-1 + g_0 belongs to it. Note
that beta^ijkl is an exceptional structure because eps^ijklmnpq only
exists in 8D.
All five exceptional Lie algebras can be described in the same way: as
vector fields that preserve the contact one-form and exceptional bilinear
forms up to functions. Moreover, e_6 and e_7 can alternatively be
described as vector fields preserving the length element and exceptional
bilinear forms up to functions; e_8, f_4 and g_2 do not possess gradings
of depth 1.
zirkus
thomas....@hdd.se (Thomas Larsson) wrote in message news:
> A while has now passed: http://www.arxiv.org/abs/math-ph/0301006.
I have not yet read this paper and before I do I am hoping that you
might be able to address this question:
The exceptional Lie algebras are related to the octonions and
exceptional Jordan algebra as discussed in the paper [1] by J. Baez.
Would your approach be related to Jordan superalgebras including the
exceptional Jordan superalgebras discussed starting in Section 2 of a
paper [2] by D. Leites?
John Baez
zirkus <zir...@hotmail.com> wrote:
>thomas....@hdd.se (Thomas Larsson) wrote in message news:
>> A while has now passed: http://www.arxiv.org/abs/math-ph/0301006.
>I have not yet read this paper and before I do I am hoping that you
>might be able to address this question:
>The exceptional Lie algebras are related to the octonions and
>exceptional Jordan algebra as discussed in the paper [1] by J. Baez.
Yup.
>Would your approach be related to Jordan superalgebras including the
>exceptional Jordan superalgebras discussed starting in Section 2 of a
>paper [2] by D. Leites?
Everything is related - some things more than others.
Thanks for pointing out that paper of Leites! Larsson's work
is indeed related to that. But as Larsson points out, his work
is even *more* related to this paper:
M. Gunaydin, K. Koepsell and H. Nicolai,
Conformal and Quasiconformal Realizations of Exceptional Lie Groups,
http://www.arxiv.org/abs/hep-th/0008063.
That was obvious to me as soon as Larsson mentioned that he'd gotten
E8 to act on a 57-dimensional manifold, since the main thing
about the Gunaydin et al paper is that they got E8 to act on a
57-dimensional manifold.
The appearance of the number 57 here is fascinating.
And this is not because Heinz claims to make 57 flavors of ketchup!
Rather, it's because the smallest nontrivial rep of the Lie group E7
is 56-dimensional. The intrigue is heightened by the fact that
Gunaydin et al get E7 to act on a 27-dimensional manifold, and the
smallest nontrivial rep of E6 is 27-dimensional!
What's the pattern, if any?
Gaze at this:
21
|
|
|
E6 16------25------29------25------16
42
|
|
|
E7 33------47------53------50------42------27
92
|
|
|
E8 78------98-----106-----104------99------83------57
These are the Dynkin diagrams of the exceptional simple Lie groups E6, E7
and E8. As explained in "week181", for each dot of the Dynkin diagram
of a simple Lie group we get a space on which the group acts. The
numbers above are the dimensions of these spaces.
This stuff is not new. So, it was obvious long before Gunaydin & Co.
came along that E8 acts on a 57-dimensional manifold, and E7 acts on
a 27-dimensional manifold, and E6 acts on two different 16-dimensional
manifolds (the octonionic projective plane, and the space of lines in
the octonionic projective plane). What they did is come up with a new
way of thinking about this. Larsson has come up with yet another.
But: why 57? Well, you can get a pretty good explanation from the
paper by Gunaydin et al, who by the way are eager to relate all this
to M-theory. It's too much work for me to explain it, and hard enough
for me just to understand it, but the first big idea is to use the
fact that E7 sits inside E8. If we decompose E8 as a rep of E7,
it breaks up like this:
248 = 1 + 56 + (133+1) + 56 + 1
where 1 is the trivial rep of E7, 56 is the smallest nontrivial
rep of E7, and 133 is E7 itself. This way of chopping up E8
in fact gives it a "5-grading":
248 = 1 + 56 + (133+1) + 56 + 1
E8 = G(-2) + G(-1) + G(0) + G(1) + G(2)
If you take the bracket of something in G(i) and something in
G(j), you get something in G(i+j).
Using this we can get E8 to act on a space of dimension 56+1 = 57.
A similar trick gets E7 to act on something 27-dimensional, since
decomposition E7 as a rep of its Lie subalgebra E6 gives it a
3-grading:
133 = 27 + (78+1) + 27*
E7 = G(-1) + G(0) + G(1)
where 1 is the trivial rep of E6, 27 and 27* are the two smallest
nontrivial reps of E6, and 78 is E7 itself.
Thomas Larsson
The answer is no, because my paper is not super; I describe structures
preserved by non-super exceptional Lie algebras. However, there is a
relation to the exceptional Jordan (non-super) algebra, and thus to
octonions.
I recently stumbled across a connection between graded Lie algebras and
Jordan algebras, which seems to be well known to algebraists but was
news to me. Given a graded Lie algebra of depth 1 and a fixed element K
of degree +1, the product
XoY = [X, [K, Y]]
makes the subspace of degree -1 into a Jordan algebra. This strengthens
my belief that all interesting algebraic structures are associative or
closely related to associativity, like the Lie bracket and and Jordan
product.
Let me elaborate. Since deg X = deg Y = -1 and deg K = +1,
deg [K, Y] = 0 and deg [X, [K, Y]] = -1, so o defines a closed product.
Using that [X,Y] = 0 (deg [X,Y] = -2 but there are no such elements
because depth is 1) and the Jacobi identity one now proves that the
product defines a Jordan algebra, i.e. Xo(Yo(XoX)) = (XoY)o(XoX).
Actually, I have been too lazy to check this, but I have read it, so
it should be true.
Now, if you plug in the depth 1 realization of e_6 or e_7 into the
formula for the Jordan product, you should get a realization of the
exceptional Jordan algebra. The other exceptions only admit gradings of
depth 2, which do not give rise to Jordan algebras.
We also get a straightforward strategy for proving the classification of
simple Jordan algebras:
1. Prove that the product o defines a Jordan algebra.
2. Prove that every simple Jordan algebra comes from a simple Lie
algebra of depth 1 in this way. This may, or may not, be the hard step.
3. Apply this idea to the known list of simple Lie algebras of depth 1,
i.e. sl(n), so(n), e_6, e_7 and perhaps also the infinite-dimensional
ones vect(n), svect(n) and h(n).
4. Remove duplicates; different Lie algebras may give rise to isomorphic
Jordan algebras.
Now, if you start with a Lie superalgebra instead, the product o gives
you a Jordan superalgebra, which is like a Jordan algebra except
that it graded-commutes. In fact, I believe that Kac used this strategy
to classify simple Jordan superalgebras. So in order to obtain a
realization of the exceptional Jordan superalgebras, you should start
with the exceptional Lie superalgebras of depth 1. There are five
infinite-dimensional exceptions admitting totally 15 gradings, some of
which are of depth 1. There are also two finite-dimensional exceptions
F(4) and G(3) and the family D(2,1;alpha), but probably their gradings
have depth 2.
Anyway, I am looking into the 15 infinite-dimensional exceptions and
their preserved structures, which should then result in realizations of
the exceptional Jordan superalgebras as well.
Thomas Larsson
> > exceptional Jordan superalgebras discussed starting in Section 2 of a
> > paper [2] by D. Leites?
> >
>
> The answer is no, because my paper is not super; I describe structures
> preserved by non-super exceptional Lie algebras. However, there is a
> relation to the exceptional Jordan (non-super) algebra, and thus to
> octonions.
This grading stuff might be confusing, because it can mean two different
things. A superalgebra has a Z_2 grading; bosons have degree 0 and
fermions degree 1 modulo 2. However, when I talk about a grading of
depth 1 I mean a Z-grading of the form
g = g_-1 + g_0 + g_1
in finite dimension, or
g = g_-1 + g_0 + g_1 + g_2 + g_3 + ...
in infinite dimension. A well-known example is the conformal algebra
so(n+2), which is graded in the following way:
-1: translations
0: rotations, dilations
1: conformal boosts.
> I recently stumbled across a connection between graded Lie algebras and
> Jordan algebras, which seems to be well known to algebraists but was
> news to me. Given a graded Lie algebra of depth 1 and a fixed element K
> of degree +1, the product
>
> XoY = [X, [K, Y]]
>
> makes the subspace of degree -1 into a Jordan algebra.
...
> Actually, I have been too lazy to check this, but I have read it, so
> it should be true.
Now I realize where I read it - in the Leites paper you referred to!
> We also get a straightforward strategy for proving the classification of
> simple Jordan algebras:
>
> 1. Prove that the product o defines a Jordan algebra.
>
> 2. Prove that every simple Jordan algebra comes from a simple Lie
> algebra of depth 1 in this way. This may, or may not, be the hard step.
>
> 3. Apply this idea to the known list of simple Lie algebras of depth 1,
> i.e. sl(n), so(n), e_6, e_7 and perhaps also the infinite-dimensional
> ones vect(n), svect(n) and h(n).
>
> 4. Remove duplicates; different Lie algebras may give rise to isomorphic
> Jordan algebras.
>
Apparently this approach goes back to Kantor in the 1960s. When I look
at Table 2 at the end of Leites' paper, I notice only finite-dimensional
Lie superalgebras. This makes me somewhat worried, since there are four
series of simple Jordan algebras but only three series of simple Lie
algebras of depth 1: sl(n), so(2n), so(2n+1). AFAIU, sp(2n) does not
admit a grading of depth 1 - the dimensions just don't match.
In fact, a simple argument explains why the infinite-dimensional Lie
algebras are irrelevant here. The only part degree 1 that enters is the
fixed element K, and since [K,K] = 0, elements of degree >= 2 never enter.
> Now, if you start with a Lie superalgebra instead, the product o gives
> you a Jordan superalgebra, which is like a Jordan algebra except
> that it graded-commutes. In fact, I believe that Kac used this strategy
> to classify simple Jordan superalgebras.
This classification was made in 1977, with a correction in 1983. At that
time, only finite-dimensional simple Lie superalgebras were classified,
which again indicates that finite dimension is enough here.
> There are also two finite-dimensional exceptions
> F(4) and G(3) and the family D(2,1;alpha), but probably their gradings
> have depth 2.
This is wrong, which is clear from Leites' Table 2, the two bottom lines;
osp(4,2;alpha) = D(2,1;alpha) and ab(3) have depth 1. I suspect that
ab(3) is the same as F(4), but it could in principle be G(3).
The last paragraph illustrates the confusing fact that the naming systems
used by the Kac and Leites camps are incompatible, e.g. E(3|8) = mb(3|8).
That Kac and Leites dislike each other does not help.
Thomas Larsson
> Jordan algebras, which seems to be well known to algebraists but was
> news to me. Given a graded Lie algebra of depth 1 and a fixed element K
> of degree +1, the product
>
> XoY = [X, [K, Y]]
>
> makes the subspace of degree -1 into a Jordan algebra.
> Apparently this approach goes back to Kantor in the 1960s. When I look
> at Table 2 at the end of Leites' paper, I notice only finite-dimensional
> Lie superalgebras. This makes me somewhat worried, since there are four
> series of simple Jordan algebras but only three series of simple Lie
> algebras of depth 1: sl(n), so(2n), so(2n+1). AFAIU, sp(2n) does not
> admit a grading of depth 1 - the dimensions just don't match.
There were reasons to worry; I had missed two series.
sp(2n) admits a three-grading of the form
n(n+1)/2 + gl(n) + n(n+1)/2.
Since dim gl(n) = n^2, we get dim sp(2n) = 2n(2n+1)/2 as we should.
The n(n+1)/2 - dimensional coordinates are x^ij = + x^ji, where
i,j are n-dimensional indices. The preserved form must single out
gl(n) from gl(n(n+1)/2), which means that we must use the n-dimensional
epsilon symbol eps_{ij..k}. The only possibility is that sp(2n)
preserves
eps_{ij..k} eps_{lm..n} dx^il dx^jm ... dx^kn
up to a function.
so(2n) admits a three-grading of the form
n(n-1)/2 + gl(n) + n(n-1)/2.
We check that dim so(2n) = 2n(2n-1)/2, as it should. The difference
is that the coordinates now obey x^ij = - x^ji.
Hence we have four infinite series of simple Lie algebras of depth 1,
sl(n+1), so(n+2), sp(2n) and so(2n), which correspond to the four
infinite series of simple Jordan algebras. Note that although the
second and fourth series contain isomorphic abstract algebras
(grading forgotten), they are distinct as graded algebras. We also
have two exceptions e_6 and e_7, which must both correspond to the
same exceptional Jordan algebra.
Some people will be amused to know that I learned about my omissions
when trying to understand M-theory. Namely, sp(2n) is the bosonic
part of the so-called conformal M-superalgebra [West, Ferrara,
Toppan, and others]. In particular, M theorists are especially
intested in the case n = 32, because the number of spinor components
in 11D is 2^[11/2] = 32. For opaque reasons, they like to restrict
the degree 0 subalgebra gl(32) to so(32). This implies that the
degree 1 subspace vanishes as well, and we are left with a
Poincare-like algebra known as M-algebra or M-superalgebra; AFAIU, it
is of no particular mathematical interest. Anyway, under this
restriction, the degree -1 subspace decomposes into so(32) irreps:
528 = 11 + 55 + 462
= (11 choose 1) + (11 choose 2) + (11 choose 5).
The first term is translations in 11D, and the two others seem to be
the reason why M theorists talk about 2-branes and 5-branes.
What I don't understand is the special role of 11D or n = 32. Long
ago Haag-Luposzanski-Sohnius proved that susy only exists in <= 11D,
but M-algebra evidently relaxes some overly restrictive assumption of
theirs; this is why Toppan call these kinds of algebras "generalized
spacetime supersymmetries".
zirkus
> Using this we can get E8 to act on a space of dimension 56+1 = 57.
In a subsequent paper [1], the 57 variable is eliminated so that E8 is
realized on R^56.
Btw, Gunaydin also discusses Jordan superalgebras in Section 6 on page
13 of [2].
Speaking of split octonions, here is the abstract of a new paper [3]
which might be interesting:
"Observable Algebra" by M. Gogberashvili
A physical applicability of normed split-algebras, such as hyperbolic
numbers, split-quaternions
and split-octonions is considered. We argue that the
observable geometry can be described by the
algebra of split-octonions, which is naturally equipped by
zero divisors (the elements of split-
algebras corresponding to zero norm). In such a picture
physical phenomena are described by the
ordinary elements of chosen algebra, while the zero divisors
give raise the coordinatization and
two fundamental constants, namely velocity of light and
Planck constant. It turns to be possible
that uncertainty principle appears from the condition of
positively defined norm, and has the
same geometrical meaning as the existence of the maximal
value of speed. The property of
non-associativity of octonions could correspond to the
appearance of fundamental probabilities
in physics. Grassmann elements and a non-commutativity of
space coordinates, which are widely
used in various physical theories, appear naturally in our
approach.
[1] http://arxiv.org/abs/hep-th/0205235
Thomas Larsson
First, I made two more omissions.
1. sl(2n) admits a grading of the form
n^2 + (sl(n) + sl(n) + gl(n)) + n^2.
The n^2 indeterminates are x^ia, where i and a are two different sets of
n-dimensional indices, and let eps_{ij..k} and eps_{ab..c} be the
corresponding epsilon symbols. sl(2n) is the subalgebra of gl(n^2) that
preserves
eps_{ij..k} eps_{ab..c} dx^ia dx^jb ... dx^kc
up to a function.
2. so(4n) admits another grading of the form
(2n^2-2) + (sl(2n) + gl(1)) + (2n^2-2).
The 2n^2-2 indeterminates are x^ij = -x^ji, where i,j are 2n-dimensional
indices, and eps_{ijkl..mn} is the 2n-dimensional epsilon. so(4n) preserves
eps_{ijkl..mn} dx^ij dx^kl ... dx^mn
up to a function.
The Jordan algebras are described in Baez' octonion paper. Let h_n(K) be
the set of nxn symmetric matrices with values in the field K. Such a
matrix is specified by K-numbers in the lower triangle and reals in the
diagonal, so
dim h_n(K) = dim K n(n-1)/2 + n,
where dim K = 1,2,4,8 for K = R,C,H,O. The dimensions of the simple
Jordan algebras are thus
R^n + R n+1
h_n(R) n(n+1)/2
h_n(C) n^2
h_n(H) 2n^2 - n
h_3(O) 27
A simple Lie algebra of depth 1 has a three-grading
g = g_-1 + g_0 + g_1.
g_0 always contains the grading operator (dilatation) generating gl(1),
so g_0 = g'_0 + gl(1), where g'_0 is usually simple. g_-1 and g_1 are
isomorphic g_0 modules, described by their dimension. Since the formula
XoY = [X, [K,Y]] equips the degree -1 subspace with a Jordan structure,
dim g_-1 equals the dimension of the corresponding Jordan algebra.
Therefore, the Lie-Jordan correspondence must be given by the following
table:
Lie g'_0 g_-1 Jordan
sl(n+1) sl(n) n R^{n-1} + R
so(n+2) so(n) n R^{n-1} + R
sp(2n) sl(n) (n^2+n)/2 h_n(R)
so(2n) sl(n) (n^2-n)/2 h_{n-1}(R)
sl(2n) sl(n)+sl(n) n^2 h_n(C)
so(4n) sl(2n) 2n^2-n h_n(H)
e_7 e_6 27 h_3(O)
e_6 so(10) 16 h_4(C)
This should complete the story. It is of course possible that I have
missed yet another grading, but I don't think so since I have enough
to get all Jordan algebras.
BTW, does anyone know where to find lists of graded simple Lie algebras
and superalgebras? I'm pretty sure that the stuff above was done by Kantor
(the grading, not the preserved structures), but I don't have easy access
to a library and I don't read Russian, so finding it on the net is much better.
> the degree 0 subalgebra gl(32) to so(32). This implies that the
^^^^^^
> restriction, the degree -1 subspace decomposes into so(32) irreps:
^^^^^^
Here I mean so(11), of course. so(11) sits naturally inside gl(32) since
the spinors have 32 components.
Tony Smith
John Baez said:
"... E8 ...[has]... a "5-grading":
248 = 1 + 56 + (133+1) + 56 + 1
E8 = G(-2) + G(-1) + G(0) + G(1) + G(2)
... A similar ...decomposition E7 ... gives it a 3-grading:
133 = 27 + (78+1) + 27*
E7 = G(-1) + G(0) + G(1) ...".
Soji Kaneyuki has a nice chapter entitiled
Graded Lie Algebras, Related Geometric Structures,
and Pseudo-hermitian Symmetric Spaces,
which is Part II of the book
Analysis and Geometry on Complex Homogeneous Domains,
by Faraut, Kaneyuki, Koranyi, Lu, and Roos (Birkhauser 2000).
In that chapter, at pages 119-126, Kaneyuki has
Tables of the Simple GLAs (Graded Lie Algebras).
Among the Exceptional Simple GLAs of the Second Kind are:
(e16) g = E8(8) , II = E8 , II1 = {y8}
g(0) = E7(7) + R , dimR g(-1) 56 , dimR g(-2) = 1
(e18) g = E8(8) , II = F4 , II1 = {y1}
g(0) = E7(-25) + R , dimR g(-1) 56 , dimR g(-2) = 1
(e30) g = E8C , II = E8 , II1 = {y8}
g(0) = E7C + C , dimC g(-1) 56 , dimC g(-2) = 1
and among the Simple GLAs of the First Kind are:
(I11) g = E7(7) , II = E7 , II1 = {a7}
g(0) = E6(6) + R , g(-1) = H3(O')
(I12) g = E7(-25) , II = C3 , II1 = {a3}
g(0) = E6(-26) + R , g(-1) = H3(O)
(I18) g = E7C , II = E7 , II1 = {a7}
g(0) = E6C + C , g(-1) = H3(O)C
which correspond to the gradings described above by John Baez.
As to notation,
H3(O) is the vector space of Octonion-hermitian matrices of degree 3,
M1,2(O) is the vector space of 1x2 matrices with Octonion entries,
II is a restricted fundamental root system of g,
II1 is the part of II corresponding to grade g(1),
R, C, H, and O are division algebras, O' is the split octonions,
and I use a for alpha and y for gamma.
For more details, see the book.
As to E6, among the Simple GLAs of the First Kind are:
(I9) g = E6(6) , II = E6 , II1 = {a7}
g(0) = so(5,5) + R , g(-1) = M1,2(O')
(I10) g = E6(-26) , II = A2 , II1 = {a1}
g(0) = so(1,9) + R , g(-1) = M1,2(O)
(I17) g = E6C , II = E6 , II1 = {a1}
g(0) = so(10)C + C , g(-1) = M1,2(O)C
Going down to structures related to so(1,9) etc, there
are among the Simple GLAs of the First Kind are:
(I6) g = so(p+1,q+1) , 0 <= p < q or 3 <= p = q
II = B(p+1) for p < q or D(p+1) for p = q, II1 = {a1}
g(0) = so(p,q) + R , g(-1) = M1,(p+q)(R)
(I15) g = so(n+2)C , II = B[(n+2)/2] or D[(n+2)/2] , II1 = {a1}
g(0) = so(n)C + C , g(-1) = M1,n(C)
Note that GLAs of type I6 and I15 are related
to the conformal Lie sphere geometry
of symmetric spaces like SO(2,4) / SO(1,3)xU(1).
As Kaneyuki says on pages 105-106: "... The purpose of these notes
is to give an introduction and survey of recent results on
semisimple pseudo-Hermitian symmetric spaces. ...
By a symmetric R-s-pace we mean a compact irreducible Hermitian
symmetric space or a real form ... of it.
The class of symmetric R-spaces contains the Shilov boundaries
of symmetric bounded domains of tube type. ...".
Motivated by work of Armand Wyler back in the 1960s-70s,
I have found such Shilov boundary structures to be useful
in building physics models. I should say that such models
are unconventional, but in my opinion they are useful models.
Tony Smith 22 Jan 2003
Thomas Larsson
Tony Smith has commented on this thread by email, but I am for some reason
unable to get my responses through. Maybe you have some filter that blocks
my email address or something. So it seems that spr is my only means to
answer. This might be a good thing, though, since your comments should be of
general interest. Sorry that I have not asked for your consent to reproduce
your email, but as I said I am unable to do so.
[Moderator's note: Most or all of this did actually appear in SPR,
although perhaps not by the time that Thomas Larsson wrote this. -- KS]
Your web page http://www.innerx.net/personal/tsmith/GLA.html
is really wonderful, with all simple graded Lie algebras are
C as my base field, only cases I7 and I13-I18 apply, whereas the others are
different real forms of these complex algebras. My only omission is in case
I13, where I only considered the case p = n-p, i.e. n = 2p.
In depth 2 I missed several cases, though. However, with this list at hand,
it should be straightforward to write down the correct preserved structures.
In your first mail, you observed that I made some mistakes, in particular
that I believed that e_6 and e_7 give rise to the same Jordan algebra. I
realized that myself when comparing the dimensions of g_-1 and the Jordan
algebras, see my latest post news:<4b8cc0a6.03012...@posting.google.com>.
It took me a week to get the connection between the graded Lie algebras and
the Jordan algebras right, and it was probably stupid to display this
process in public.
On den 27 januari 2003 12:22, Tony Smith [SMTP:tsm...@innerx.net] wrote:
> On 26 January 2003 Thomas Larsson asked in a post to spr:
>
> "... does anyone know where to find lists of graded simple Lie algebras
> and superalgebras? ...".
>
>
> As to simple Graded Lie Algebras, yes.
> I sent you a message on Friday 24 January 2003 that gave such
> a reference. In case that reference is hard to find in a library,
> I have put a copy of the table on the web at
>
> http://www.innerx.net/personal/tsmith/GLA.html
>
> as you can see, the list includes the following 3-level gradings:
>
>
> (I10) g = E6(-26) ,
> II = A2 ,
> II1 = {a1},
> g(0) = so(1,9) + R ,
> g(-1) = M1,2(O).
>
> and
>
>
> (I12) g = E7(-25) ,
> II = C3 ,
> II1 = {a3},
> g(0) = E6(-26) + R ,
> g(-1) = H3(O).
>
> where M1,2(O) is the vector space of 1x2 octonion matrices
> and H3(O) is the 27-dim vector space of the exceptional Jordan algebra.
>
>
> Maybe the internet worm that attacked at aaround last Friday prevented
> you from receiving that message, so here is a copy of it:
>
> =================================================================
>
>
> Delivered-To: tsm...@innerx.net
> X-Sender: tsm...@pop3.innerx.net
> Mime-Version: 1.0
> Date: Fri, 24 Jan 2003 20:58:51 -0500
> To: ba...@math.ucr.edu, thomas....@hdd.se
> From: Tony Smith <tsm...@innerx.net>
> Subject: Re: Structures preserved by e_8
> Cc: tsm...@innerx.net
>
> Thomas Larsson said, in a sci.physics.research post
> on the thread Re: Structures preserved by e_8
>
> "... We also have two exceptions e_6 and e_7,
> which must both correspond to the same exceptional Jordan algebra. ...".
>
> -----------------------------------------------------------
>
> Soji Kaneyuki has a nice chapter entitiled
> Graded Lie Algebras, Related Geometric Structures,
> and Pseudo-hermitian Symmetric Spaces,
> which is Part II of the book
> Analysis and Geometry on Complex Homogeneous Domains,
> by Faraut, Kaneyuki, Koranyi, Lu, and Roos (Birkhauser 2000).
>
> In that chapter, at pages 119-126, Kaneyuki has
> Tables of the Simple GLAs (Graded Lie Algebras).
>
> Please note that E6 and E7 do have depth-1 gradings,
> but they do NOT correspond to the same Jordan algebra.
>
> The E6 grading corresponds to the 1x2 matrix algebra with octonion,
> or split octonion, entries. Its g(0) grade element is (up to
> signature) the D5 Lie algebra so(10).
>
> The E7 grading corresponds to the 3x3 Hermitian matrix algebra with
> octonion, or split octonion, entries. The g(0) grade element is (up to
> signature) the E6 Lie algebra plus a 1-dimensional factor that is R or,
> in the complexified version, C, that corresponds to a generator of U(1)
> or its complexification.
>
> Although there is no depth-1 grading for E8, there are 5-element
> depth-2 gradings, including the following from Kaneyuki's Tables:
>
> (e16) g = E8(8) , II = E8 , II1 = {y8}
> g(0) = E7(7) + R , dimR g(-1) 56 , dimR g(-2) = 1
>
>
> (e18) g = E8(8) , II = F4 , II1 = {y1}
> g(0) = E7(-25) + R , dimR g(-1) 56 , dimR g(-2) = 1
>
>
> (e30) g = E8C , II = E8 , II1 = {y8}
> g(0) = E7C + C , dimC g(-1) 56 , dimC g(-2) = 1
>
>
> Note that if you combine all the even grade things,
> you get g(-2) + g(0) + g(2) = E7 + SU(2) (or complexifications)
> (by noting that 3-dim SU(2) is generated by R of g(0) plus
> the two 1-dim elements of g(-2) and g(2)).
>
> Then you have left over the 56-dim g(-1) and 56-dim g(1),
> and
> they seem to correspond to the 56-dimensional Freudenthal algebra.
> If you look at the g(-2) + g(-1) + g(1) + g(2) elements,
> you see two 57-dim things, which seem to correspond to
> the 57-dim realization of E8 that is described by Murat Gunaydin at
> http://xxx.lanl.gov/abs/hep-th/0008063
>
>
>
> Note that the 56-dim Freudenthal algebra is similar to
> the 27-dim exceptional Jordan algebra H3(O),
> in that:
> F4 is the automorphism group of 27-dim H3(O)
> and
> E6 is the automorphism group of the 56-dim Freudenthal algebra.
>
> As to the algebras of which E7 and E8 might be automorphism groups,
> and related topics,
> there are some interesting resouces, such as:
>
> The web site of R. Skip Garibaldi and his papers Structurable Algebras
> and Groups of Types E6 and E7,
> http://xxx.lanl.gov/abs/math.RA/9811035
> and
> Groups of Type E7 Over Arbitrary Fields,
> http://xxx.lanl.gov/abs/math.AG/9811056
>
> A Taste of Jordan Algebras, by Kevin McCrimmon,
> to be published (according to Amazon) in January 2003.
>
> The Book of Involutions, by M.-A. Knus, A. Merkurjev, M. Rost,
> and J.-P. Tignol (with a preface by J. Tits), American Mathematical
> Society Colloquium Publications, vol. 44 (American Mathematical
> Society,Providence, RI, 1998); also see
> http://www.math.ohio-state.edu/~rost/BoI.html
>
> Tony Smith 24 Jan 2003
Thomas Larsson
Tony Smith <tsm...@spamfree.com> wrote in message news:<l03102800ba53dc472ef9@[209.246.179.99]>...
>
> Soji Kaneyuki has a nice chapter entitiled
> Graded Lie Algebras, Related Geometric Structures,
> and Pseudo-hermitian Symmetric Spaces,
> which is Part II of the book
> Analysis and Geometry on Complex Homogeneous Domains,
> by Faraut, Kaneyuki, Koranyi, Lu, and Roos (Birkhauser 2000).
>
> In that chapter, at pages 119-126, Kaneyuki has
> Tables of the Simple GLAs (Graded Lie Algebras).
>
As I mentioned on a post that has not yet appeared, I am unable to
respond to your emails; after a while, I get an unknown recipient message.
It might be worth pointing out that most information about realizations
can be read off immediately from the grading. Assume that the Lie algebra
g has a 3-grading, g = g_-1 + g_0 + g_1, and that it admits a realization
on an C^n with coordinates x^i, i = 1,2,...,n. Then the subspace g_-1
is spanned by the translations d/dx^i, and in particular dim g_-1 = n.
So given g_-1 we can immediately read off the dimension of the space
on which g is realized.
Similarly, if g = g_-2 + g_-1 + g_0 + g_1 + g_2 has a 5-grading, we have
a realization on C^(m+n) with two types of coordinates: the m coordinates
t^a and the n coordinates x^i. The negative part is realized by the
vector fields
g_-2: d/dt^a
g_-1: d/dx^i + c^ib_j x^j d/dt^b
for some structure constants c^ib_j. So again we immediately get the
coordinates from g_-2 and g_-1, and in particular dim g_-2 = m and
dim g_-1 = n.
The subspace g_0 is even a subalgebra. Almost always, it is of the
form g_0 = g'_0 + gl(1), where the second term is generated by the
grading operator Z. In depth 1, this is simply the dilatation,
Z = x^i d/dx^i, and in depth 2
Z = 2 t^a d/dt^a + x^i d/dx^i.
When we have come this far, we can usually determine the geometry
preserved by g by inspection. If Z is part of g_0, g can only preserve
some multilinear form conformally, i.e. up to functions, because Z
has this property; Z w = k w for a k-linear form w. Then we look at
the rest g_0 to see what structure constants we may use.
If g_0 contains sl(n), we may only use the epsilon symbols eps_{ij..k}
and eps^{ij..k}.
If g_0 contains so(n), we may also use the metric g_ij = g_ji.
If g_0 contains sp(n), we may also use the symplectic metric w_ij = -w_ji.
So given the lists of simple GLAs in Kaneyuki paper and on your webpage,
we get almost directly the list of all possible preserved structures.
There is a lesson to be learned from this. Knowing the Cartan
classification of abstract Lie algebras is not enough; we really need
the classification of graded Lie algebras. In physics, spacetime and
phase-space symmetries are always graded - Poincare, conformal,
super-Poincare, superconformal, Virasoro, super-Virasoro, canonical
transformations, contact transformations - because we need to know the
dimension on the spacetime on which it is realized. Internal symmetries
are usually considered only abstractly, but that does not mean that
understanding the gradings would be helpful.
Another lesson is that the division between finite- and infinite-dimensional
algebras is somewhat artificial. Both types can be described as the
subalgebra of vect(n) preserving some geometry, usually a multi-linear
form conformally.
Finally, let me point out that everything can be superized. However,
something new happens for infinite Lie superalgebras: depth 3. The
grading takes the form
g = g_-3 + g_-2 + g_-1 + g_0 + g_1 + g_2 + g_3 + g_4 + ...
A priori, depth 3 is unlikely to occur, because any element in g_-3 can
be written as [A, [B, C]] for A, B, C in g_-1. Now, if [A, [B, C]],
[B, [C, A]] and [C, [A, B]] are all linearly independent, they must all
vanish because of the Jacobi identity. We must thus have dim g_-3 <= 2,
and indeed we get equality. Since sl(2) is the only Lie algebra with
2-dim reps, g_0 must contain an sl(2) term, as well as the gl(1) grading
operator. When things are worked out properly, we see that in fact
g_0 = sl(3) + sl(2) + gl(1). A real form of this algebra is
g_0 = su(3) + su(2) + u(1), which is precisely the symmetries of the
Standard Model in particle physics.
I find it remarkable that the postulate of maximal depth immediately and
unambigously selects the correct symmetries of the Standard Model,
something that string theorists have failed to do for 20,000 man-years.
Alas, as A J Tolland pointed out, physicists are not interested in
learning about new algebraic structures, since they already know the
correct Theory of Everything. However, I am somewhat surprised that
people working on supersymmetry don't want to know about progress in
superalgebras.
Tony Smith
on the thread Re: Structures preserved by e_8
"... We also have two exceptions e_6 and e_7,
which must both correspond to the same exceptional Jordan algebra. ...".
-----------------------------------------------------------
Soji Kaneyuki has a nice chapter entitiled
Graded Lie Algebras, Related Geometric Structures,
and Pseudo-hermitian Symmetric Spaces,
which is Part II of the book
Analysis and Geometry on Complex Homogeneous Domains,
by Faraut, Kaneyuki, Koranyi, Lu, and Roos (Birkhauser 2000).
In that chapter, at pages 119-126, Kaneyuki has
Tables of the Simple GLAs (Graded Lie Algebras).
Please note that E6 and E7 do have depth-1 gradings,
John Baez
anything terribly intelligent to say about it. But I don't
want to give the impression I don't care, so I'll just go
ahead and say something unintelligent!
I like this business about getting Jordan algebras from
depth-1 gradings of Lie algebras, and I'm inclined to agree
that this trick explains a lot about the deep inner meaning of
Jordan algebras - and also their relation to Lie triple systems,
since we get a Lie triple system from an involution on a Lie algebra,
and I think lots of these depth-1 gradings come with involutions
that "flip the grading".
In article <4b8cc0a6.03012...@posting.google.com>,
Thomas Larsson <thomas....@hdd.se> wrote:
>This grading stuff might be confusing, because it can mean two different
>things. A superalgebra has a Z_2 grading; bosons have degree 0 and
>fermions degree 1 modulo 2. However, when I talk about a grading of
>depth 1 I mean a Z-grading of the form
>
> g = g_-1 + g_0 + g_1
>
>in finite dimension, or
>
> G = g_-1 + g_0 + g_1 + g_2 + g_3 + ...
>
>in infinite dimension. A well-known example is the conformal algebra
>so(n+2), which is graded in the following way:
>
>-1: translations
> 0: rotations, dilations
> 1: conformal boosts.
Right. Here's a related puzzle that Jim Dolan emailed me:
> here's a little practice quiz. here's the footprint of the simple
> lie algebra b2:
>
> . . .
> . . .
> . . .
>
>
> now, do you get one of those tri-gradings by taking the short root
> copy of a1 to be grade zero?
[In case anyone has trouble understanding the nonstandard jargon
here: I think the "footprint" consists of the roots of this Lie
algebra together with the origin, and I know a "tri-grading" is
just another name for what Larsson is caling a "grading of depth
one".]
Anyway, it seems like we get a tri-grading by taking the short root
copy to have grade zero. Here are the roots (and origin) divided
up and labelled by their grades:
| |
. | . | .
. | . | .
. | . | .
| |
-1 0 1
To check that this is a tri-grading we need to use the formula
for the bracket of root vectors:
[e_a, e_b] = const. e_{a+b}
where a is a root and e_a is the corresponding root vector.
We want the bracket of two root vectors of grade 1 to equal zero, so
adding two of the x-marked roots:
. | . | x
. | . | x
. | . | x
should give a dot that "falls off the footprint", where the
dot in the very middle counts as the origin. It works!
For example, this root
. | . | x
. | . | .
. | . | .
plus this root
. | . | .
. | . | x
. | . | .
gives this:
. | . | .
. | . | . x
. | . | .
which "falls off the footprint".
We also want the bracket of two root vectors of grade -1 to equal zero,
so when we add any two of these marked roots:
x | . | .
x | . | .
x | . | .
the result should again "fall off the footprint". It does!
Finally, we want the bracket of a root vector of grade 1 and a root
vector of grade -1 to give something of grade 0, so when we add, say,
this root
. | . | x
. | . | .
. | . | .
with this root
. | . | .
x | . | .
. | . | .
we want to get one of these:
. | x | .
. | x | .
. | x | .
and we do!
So, unless I'm mixed up, this gives a tri-grading of b2 = so(5),
and I bet it's the same as the one Larsson is talking about.
> how about the long root copy?
This appears not to work. To draw the roots of grade -1, 0, 1
in this case, it's easier to use another style of picture:
0 1 1
-1 0 1
-1 -1 0
I've just numbered them by their grade. Here the problem is that
the sum of two roots of grade 1 need not "fall off the footprint":
adding this
. x .
. . .
. . .
and this
. . .
. . x
. . .
gives this:
. . x
. . .
. . .
So, it looks like there's just one tri-grading, which is why
I guess it's the same as the one Larsson is talking about.
It's easy to generalize this discussion to any simple Lie algebra.
It's a bit too tiring for me to use this technique to work out all
the tri-gradings of all the simple Lie algebras... but if I didn't
know that someone had already done it, I can imagine doing it.
John Baez
>thomas....@hdd.se (Thomas Larsson) wrote in message
>news:<4b8cc0a6.03012...@posting.google.com>...
>> I recently stumbled across a connection between graded Lie algebras and
>> Jordan algebras, which seems to be well known to algebraists but was
>> news to me. Given a graded Lie algebra of depth 1 and a fixed element K
>> of degree +1, the product
>>
>> XoY = [X, [K, Y]]
>>
>> makes the subspace of degree -1 into a Jordan algebra.
I forget if this is in McCrimmon's wonderful review article
on Jordan algebras, which you should definitely read if you
haven't yet, even if you know most of the stuff in it by now,
because it's quite well-written and fun:
Kevin McCrimmon, Jordan algebras and their applications,
Bull. Amer. Math. Soc. 84 (1978), 612-627.
>Hence we have four infinite series of simple Lie algebras of depth 1,
>sl(n+1), so(n+2), sp(2n) and so(2n), which correspond to the four
>infinite series of simple Jordan algebras. Note that although the
>second and fourth series contain isomorphic abstract algebras
>(grading forgotten), they are distinct as graded algebras. We also
>have two exceptions e_6 and e_7, which must both correspond to the
>same exceptional Jordan algebra.
I guess one of these is the 3-grading of e_7 which I mentioned
earlier: the one coming from the decomposition of e_7 as a rep of
its Lie subalgebra e_6:
e_7 = g(-1) + g_0 + g(1)
133 = 27 + (78+1) + 27*
where 1 is the trivial rep of e_6, 27 and 27* are the representations
e_6 on the exceptional Jordan algebra and its dual, and 78 is the
adjoint representation of e_6 on itself.
As for the 3-gradings of e_6, Tony Smith has listed (some of?) these,
but I'm a little confused because instead of getting g(-1) to be
a Jordan algebra, he seems to be getting it to be part of a Jordan
pair - not 3x3 octonion matrices, but 1x2 matrices of various sorts:
>(I9) g = E6(6) , II = E6 , II1 = {a7}
> g(0) = so(5,5) + R , g(-1) = M1,2(O')
>(I10) g = E6(-26) , II = A2 , II1 = {a1}
> g(0) = so(1,9) + R , g(-1) = M1,2(O)
>(I17) g = E6C , II = E6 , II1 = {a1}
> g(0) = so(10)C + C , g(-1) = M1,2(O)C
>As to notation,
>H3(O) is the vector space of Octonion-hermitian matrices of degree 3,
>M1,2(O) is the vector space of 1x2 matrices with Octonion entries,
>II is a restricted fundamental root system of g,
>II1 is the part of II corresponding to grade g(1),
>R, C, H, and O are division algebras, O' is the split octonions,
>and I use a for alpha and y for gamma.
So is there a 3-grading of e_6 for which g(-1) is the exceptional
Jordan algebra, or not???
Inquiring minds want to know!
Tony Smith
John Baez says:
"... As for the 3-gradings of e_6, Tony Smith has listed (some of?) these,
but I'm a little confused because instead of getting g(-1) to be
a Jordan algebra, he seems to be getting it to be part of a Jordan
pair - not 3x3 octonion matrices, but 1x2 matrices of various sorts
...
So is there a 3-grading of e_6 for which g(-1) is the exceptional
Jordan algebra, or not??? Inquiring minds want to know! ...".
I should say that to attribute the list to Tony Smith is an error.
The list is (as I think I stated in my post) due to Soji Kaneyuki
and is taken from the book
Analysis and Geometry on Complex Homogeneous Domains,
by Jacques Faraut, Soji Kaneyuki, Adam Koranyi, Qi-keng Lu,
and Guy Roos (Birkhauser 2000).
I think that Soji Kaneyuki has proven his table to be complete,
which does indicated that E6 does NOT have
a 3-grading for which g(-1) is the exceptional Jordan algebra J3(O),
although the 1x2 octonion matrices are related to J3(O).
Therefore,
you have to go to E7 to get a 3-grading with J3(O) as g(-1).
Since Soji Kaneyuki did the proofs, and I did not,
please read the book if you want details. I would most
likely botch up any attempt I might make at giving such details.
I will say that I tried to copy the tables, which I put on the web at
http://www.innerx.net/personal/tsmith/GLA.html
accurately, but it is always possible that I could have made
a typo or mental error. However, in checking them over I do
think that in fact there is no 3-grading of E6 with g(-1) = J3(O).
On a more general level, it is interesting to me that,
even though E6 is not the automorphism group of J3(O)
and does not have J3(O) as a 3-grading g(-1),
E6 sort of "sits in between" the two groups:
F4 - automorphism group of J3(O)
and
E7 - J3(O) as a 3-grading g(-1),
because (ignoring some technicalities about real forms etc)
E6 = F4 + 26-dim traceless part of J3(O)
and
E7 3-grading has g(0) = E6 + R,
so that
E7 = E6 + R + two copies of J3(O).
As has been discussed earlier, E6 is the automorphism group of
the Freudenthal algebra, which is "sort of like" a complexification
of J3(O).
As to what the 1x2 octonion matrices that
are g(-1) of 3-grading of E6 with g(0) = so(1,9) + R
here is how I picture that E6 3-grading in my mind:
Denote so(1.9) by D5.
E6 = ( adjoint of D5 + R ) + two copies of spinor of D5
so that the 1x2 octonion matrices would correspond
to the 16-dimensional spinor representation of D5 = so(1,9).
That is, you can go from D5 to E6 by "adding" two D5 spinors plus R or U(1).
When you go up to E7 from E6,
you are also "addding" two "things" plus R or U(1),
but here the "thing" is more complicated than a spinor,
and is in fact the vector space of the J3(O) Jordan algebra.
It is interesting that the two copies of J3(O) plus a U(1) might
be regarded as a Freudenthal algebra (no, I haven't proven it,
it just looks likely to me), so that, IF that view is correct:
E7 = E6 plus (the algebra of which E6 is the automorphism algebra).
Going to E8 from E7 is sort of similar, but you have to
use 5-gradings because (in my opinion but not proven by me)
going to E8 from E7 is a quaternification, which requires
an SU(2) symmetry instead of
the U(1) that can be put in the g(0) going to E7 from E6
and
only the U(1) subgroup of the SU(2) can be put in the g(0) for E8 5-grading
and
the other two generators of the SU(2) have to be put outside
the g(0) and g(-1) and g(1) grades,
so one of them is put in each of g(-2) and g(2) (which are in
fact 1-dimensional).
Tony 17 Feb 2003
Thomas Larsson
> In article <4b8cc0a6.03012...@posting.google.com>,
> So is there a 3-grading of e_6 for which g(-1) is the exceptional
> Jordan algebra, or not???
>
> Inquiring minds want to know!
The answer is no. The statement I made in my post was simply wrong,
which Tony Smith also pointed out. In two later posts I corrected this
error. Zirkus' post prompted me to explain the connection between Jordan
algebras and simple GLAs of depth 1, but I didn't understand it very well
myself. So I posted on spr in parallel with learning the stuff myself,
which led to some incorrect statements. In particular, once one thinks
about the dimensions of the Jordan algebras is becomes obvious that e_6
does not give rise to h_3(O).
Another misconception of mine, that I fortunately did not put into my
e-print, is that I believed that the gradings given by
Gunaydin-Koepsell-Nicolai exhausted the possibilities. Looking into
Kaneyuki's list, it seems that they covered all cases with dim g_-2 = 1,
which apparently is necessary to obtain a connection with Freudental
triple systems, but there are also cases with dim g_-2 > 1. In this
case, there is a connection to something more general called Kantor
triple systems (which I don't know anything about).
Unlike me, Kaneyuki has clearly thought seriously about these gradings
for a long time, so you should trust his list. In particular, e_8 has
two gradings of depth 2, the one I described
1 + 56 + (e_7 + 1) + 56 + 1
and
14 + 64 + (so(14) + 1) + 64 + 14
(and various real forms of these). The second can be more explicitly
described as follows. g_-1 is the 64D spinor rep S and g_-2 the 14D
vector rep V. Let the spinor Q_a form a basis for S and the vector P_i
a basis for V. Now, there must better exist a map c: S /\ S --> V with
structure constants c^i_ab. The nilpotent algebra g_-2 + g_-1 becomes
[Q_a, Q_b] = c^i_ab P_i,
[P_i, Q_a] = [P_i, P_j] = 0.
This is very reminiscent of a supersymmetry algebra except that the
spinor Q_a is bosonic. An obvious candidate for a conformally preserved
structure is the one-form
dt^i + c^i_ab x^a dx^b,
where t^i and x^a are the coordinates of C^78 = C^(14+64).
However, I am a little worried about two things. The vector fields that
preserve the one-form above almost certainly generate an infinite-
dimensional Lie algebra, so there must be some other preserved structure.
Second, I don't see what's exceptional about this. If it works for so(14),
it should work for so(6+8n) by Bott periodicity.
What surprised me is that e_8 also seems to admit a 7-grading,
g = g_-3 + g_-2 + g_-1 + g_0 + g_1 + g_2 + g_3,
of the form
e_8 = 8 + 28* + 56 + (sl(8) + 1) + 56* + 28 + 8*.
Kaneyuki does not mention anything about this, because from his point of
view 3- and 5-gradings are more interesting. Incidentally, this grading
refutes my claim that mb(3|8) is deeper than anything seen in string
theory, since now e_8 also admits a grading of depth 3 and I learned
about it in an M theory paper:
P West, E_11 and M theory, hep-th/0104081, eqs (3.2) - (3.8).
OTOH, the above god-given 7-grading of e_8 is not really useful in M
theory, because g_-3 is identified with spacetime translations and one
would therefore get that spacetime has 8 dimensions rather than 11. For
this reason, West tries to tweak this construction into a similar
realization of the hyperbolic Kac-Moody algebra e_11. It is unclear to
me if he succeeds or indeed why he believes that this may be possible,
since e_11 is of exponential growth and thus cannot be realized as an
algebra of vector fields acting on a finite-dimensional manifold.
Tony Smith
Thomas Larsson says
"... e_8 also seems to admit a 7-grading,
g = g_-3 + g_-2 + g_-1 + g_0 + g_1 + g_2 + g_3,
of the form
e_8 = 8 + 28* + 56 + (sl(8) + 1) + 56* + 28 + 8*.
Kaneyuki does not mention anything about this, because from his point of
view 3- and 5-gradings are more interesting. Incidentally, this grading
refutes my claim that mb(3|8) is deeper than anything seen in string
theory, since now e_8 also admits a grading of depth 3 and I learned
about it in an M theory paper:
P West, E_11 and M theory, hep-th/0104081, eqs (3.2) - (3.8).
OTOH,
the above god-given 7-grading of e_8 is not really useful in M theory,
because g_-3 is identified with spacetime translations and
one would therefore get that spacetime has 8 dimensions rather than 11. ...".
To me, that structure is interesting because it shows the
relationship between 248-dim E8
and
the 256-dim graded exterior algebra /\(8) with graded structure
1 8 28 56 70 56* 28* 8* 1* (the 70 is self-dual)
That is, E8 = 8 + 28 + 56 + 64 + 56* + 28* + 8*
and
you get E8 by dropping the 1 and 1* and 70, and adding a 64.
If you look at 64 in Lie algebra terms,
it is natural to think of it as 8 x 8* which
is in compact terms the adjoint rep of U(8) = SU(8) x U(1).
However,
because of connections with my (admittedly unconventional) physics
model,
I like to think of the 64 as the space of the Clifford algebra Cl(6)
of 8x8 real matrices, or
if you look at signature Cl(2,4) as 4x4 quaternionic matrices.
(I like the quaternionic better, because I like to think of
the Clifford algebra of physical spacetime as quaternionic Cl(1,3)
(using the signature convention of Porteous), but that is another
spr thread with post by John Baez where he said
"... fermions are quaternionic and bosons are real ...".)
>From that point of view the 7-grading of E8 looks like
E8 = 8 + 28 + 56 + Cl(2,4) + 56* + 28* + 8*
If you do the natural thing and let the 28 be the D4 Lie algebra,
you have
E8 = 8 + D4 + 56 + Cl(2,4) + 56* + D4* + 8*
and if you regard the 8 as the root vector space of E8 you have
Vector Lie alg Clif alg Lie alg* vect*
E8 = 8 + D4 + 56 + Cl(2,4) + 56* + D4* + 8*
As to the 56, recall that the Lie Group E6 is
the Automorphism Group of
the 56-dimensional Freudenthal Algebra Fr3(O)
of 2x2 Zorn-type vector-matrices
a X
Y b
where a and b are real numbers and
X and Y are elements of the 27-dimensional Jordan Algebra J3(O).
Vector Lie alg Fr alg Clif alg Fr alg* Lie alg* vect*
E8 = 8 + D4 + Fr3(O) + Cl(2,4) + Fr3(O)* + D4* + 8*
I am NOT saying that the E8 Lie multiplication is the same as
the Lie, Freudenthal, and Clifford multiplications,
but
it does seem that E8 has graded structure such that
the vector spaces of the graded elements are the same
as vector spaces of interesting Lie, Freudenthal, and Clifford algebras.
Soji Kaneyuki makes that clear when he says that
"... Hn(F) the vector space of F-hermitian matrices of degree n ...",
so that in his notation H3(O) is NOT the Jordan algebra J3(O),
but
is only the vector space on which you can put the Jordan product
if you want to make J3(O) from it.
Since the Freudenthal algebra Fr3(O) is closely related to the Jordan
algebra J3(O), you can say that the 7-grading structure shows that
E8 looks like a combination of Vector, Lie, Jordan, and Clifford things.
I very much like that 7-grading because in my physics model I use:
the Vector part as my 8-dim spacetime (prior to dimensional reduction);
the Lie part as my 28-dim gauge group (prior to dimensional reduction); and
the Clifford part to get Spin(2,4) = SU(2,2) conformal group (after
dimensional reduction) to give gravity etc by the MacDowell-Mansouri
mechanism.
As to the Freudenthal (Jordan) part, its E6 automorphism group is
where my 8 first-generation fermion particles
and 8 first-generation fermion antiparticles live.
In other words, unlike West,
I am happy with the 7-grading as is,
because my model (unlike M-theory) has a physical interpretation
for each term in the 7-grading.
Tony 19 Feb 2003
John Baez
be a theory of everything, examines an 8-dimensional gemstone with
240 vertices, turning it until he finds that the vertices line
up to form 5 parallel planes.
In article <4b8cc0a6.03021...@posting.google.com>,
Thomas Larsson <thomas....@hdd.se> wrote:
>ba...@galaxy.ucr.edu (John Baez) wrote in message
>news:<b2ps9f$kbl$1...@glue.ucr.edu>...
>> In article <4b8cc0a6.03012...@posting.google.com>,
>> So is there a 3-grading of e_6 for which g(-1) is the exceptional
>> Jordan algebra, or not???
>> Inquiring minds want to know!
>The answer is no.
Okay, that's sort of a relief.
>The statement I made in my post was simply wrong,
>which Tony Smith also pointed out. In two later posts I corrected this
>error.
Sorry, I'm still sort of catching up on this stuff.
>Zirkus' post prompted me to explain the connection between Jordan
>algebras and simple GLAs of depth 1, but I didn't understand it very well
>myself. So I posted on spr in parallel with learning the stuff myself,
>which led to some incorrect statements.
That's commendable - I like to "learn by posting" myself, and as
a result I screw up a fair amount.
>In particular, once one thinks
>about the dimensions of the Jordan algebras is becomes obvious that e_6
>does not give rise to h_3(O).
But as you undoubtedly know, h_3(O) goes give rise to e_6. :-)
More precisely, the group E6 has a real form called E6(-26)
which is the group of all transformations of h_3(O) that preserve
the "determinant". This is why some people define E6 as the
group of all transformations of a certain 27-dimensional space
preserving a cubic form.
Equivalently, E6(-26) is the group of all transformations of
the octonionic projective plane that map lines to lines.
>Another misconception of mine, that I fortunately did not put into my
>e-print, is that I believed that the gradings given by
>Gunaydin-Koepsell-Nicolai exhausted the possibilities. Looking into
>Kaneyuki's list, it seems that they covered all cases with dim g_-2 = 1,
>which apparently is necessary to obtain a connection with Freudenthal
>triple systems, but there are also cases with dim g_-2 > 1.
Hmm, interesting!
>In this
>case, there is a connection to something more general called Kantor
>triple systems (which I don't know anything about).
I guess everyone gets their own "triple system": Lie, Jordan, Freudenthal,
Kantor... with the later people getting increasingly weird ones.
>Unlike me, Kaneyuki has clearly thought seriously about these gradings
>for a long time, so you should trust his list. In particular, e_8 has
>two gradings of depth 2, the one I described
>
> 1 + 56 + (e_7 + 1) + 56 + 1
>
>and
>
> 14 + 64 + (so(14) + 1) + 64 + 14
>
>(and various real forms of these).
>What surprised me is that e_8 also seems to admit a 7-grading,
>
> g = g_-3 + g_-2 + g_-1 + g_0 + g_1 + g_2 + g_3,
>
>of the form
>
> e_8 = 8 + 28* + 56 + (sl(8) + 1) + 56* + 28 + 8*.
I would like to understand all the gradings of E8 in the following
way.
The E8 lattice consists of all 8-tuples (x_1,...,x_8) of real numbers
such that the x_i are either all integers or all half-integers
(a half-integer being an integer plus 1/2), and they satisfy
x1 + ... + x8 is even.
The nearest neighbors of the origin are called the "roots" of E8.
They all have length equal to sqrt(2), and here they are:
(1,1, 0,0,0,0,0,0) and all permutations:
there are 8 choose 2 = 28 of these
(-1,-1, 0,0,0,0,0,0) and all permutations:
there are 8 choose 2 = 28 of these
(1,-1, 0,0,0,0,0,0) and all permutations:
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
(-1/2,-1/2,-1/2,-1/2,-1/2,-1/2,-1/2,-1/2)
there is 1 of these
So, there are a total of
28 x 6 + 70 + 2 = 168 + 72 = 240
roots. I wrote out this calculation because it's one of the
strangest ways I've seen so far of counting the 240 roots of E8;
for example, the number 168 is the size of the symmetry group of
the Fano plane! One can construct the octonions starting with the
Fano plane, and E8 from the octonions... hmm....
Anyway, by James Dolan's little quiz which I answered earlier
in this thread, it seems evident that we get gradings of E8 by
finding integer-valued linear functionals L on the E8 lattice;
the value of L on a given root is its "grade". We say we have
an "n-grading" if L takes exactly n values on the roots and
the origin.
I could be wrong, but that's how I think it works.
So, let's find some gradings.
This sounds very technical, but you should really imagine it
like this: you're a jeweler holding up a precious 8-dimensional
gemstone cut in a shape with 240 vertices, and you want to
rotate it around and see how the vertices (and the center of the gem)
can line up to lie on parallel planes.
If you can get them all to lie on just 3 planes, you're very lucky:
your gem gives you a Jordan algebra! If you can get them to
lie on 5, I guess you get a Kantor triple system.
I'll just try a simple method: I'll define L to equal the first
component of our vector. This can be 1, 1/2, 0, -1/2, or -1, so
we get a 5-grading. Let's figure out the dimension of each grade,
to guess if it's one of the 5-gradings we've already seen above.
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 "-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
The rest of the roots have a "0" as the first component; there are
240 - 14 - 64 - 64 - 14 = 84
We also have to count the Cartan subalgebra (corresponding to
the origin); this gives 8 more dimensions in this grade, for a total of 92.
So we get a 5-grading of this sort:
14 + 64 + 92 + 64 + 14
Unless the gods are playing pranks, this must be Larsson's
> 14 + 64 + (so(14) + 1) + 64 + 14
So, while not terribly elegant, I think this approach can work.
The next obvious grading might come from taking L to be the sum
of the first two components of the vector. This seems to give
a 7-grading.
I should mention that while I was doing these calculations, I noticed
all sorts of strange things. For example, 64 + 14 = 78, the dimension
of e_6. And the elements of grade 2 and 1 in the above grading of e_8
do form a 78-dimensional Lie algebra, right? But unless I'm mixed
up it must be nilpotent Lie algebra, not e_6.
Thomas Larsson
> But as you undoubtedly know, h_3(O) goes give rise to e_6. :-)
>
> More precisely, the group E6 has a real form called E6(-26)
> which is the group of all transformations of h_3(O) that preserve
> the "determinant". This is why some people define E6 as the
> group of all transformations of a certain 27-dimensional space
> preserving a cubic form.
Hmm. There is no grading corresponding to this. Kaneyuki lists four
different gradings for e_6:
i17. e_6 = 16 + (so(10) + C) + 16
e24. e_6 = 5 + 20 + (sl(5) + sl(2) + C) + 20 + 5
e25. e_6 = 1 + 20 + (sl(6) + C) + 20 + 1
e26. e_6 = 8 + 16 + (so(8) + C + C) + 16 + 8
which act on spaces of dimension 16, 25, 21, 24 but not 27.
>
> I would like to understand all the gradings of E8 in the following
> way.
...
> Anyway, by James Dolan's little quiz which I answered earlier
> in this thread, it seems evident that we get gradings of E8 by
> finding integer-valued linear functionals L on the E8 lattice;
> the value of L on a given root is its "grade". We say we have
> an "n-grading" if L takes exactly n values on the roots and
> the origin.
>
Let me see if I understand what you do. You slice the jewel into
hyperplanes. The degree 0 subalgebra becomes g_0 = h_0 + C, where the h_0
roots lie inside the middle hyperplane and C gives a root perpendicular to
it. The latter corresponds to the grading operator, which is simply
dilatations in depth 1. Since the root space for h_0 lies in a hyperplane
of codimension 1, we must have rank g = rank g_0 = rank h_0 + 1. E.g.
rank e_6 = rank so(10) + 1 = rank sl(5) + rank sl(2) + 1
= rank sl(6) + 1 = rank so(8) + 2 = 6.
Similarly, for the e_8 gradings in my previous post,
rank e_8 = rank e_7 + 1 = rank so(14) + 1 = rank sl(8) + 1 = 8.
Another way to see that g and g_0 must have the same rank is to use the
1-1 correspondence between g and g_0 irreps. Given a g_0 irrep R, we can
construct the induced g rep Ind R = R @_U(g_0) U(g). This just means that
Ind R is obtained from the universal envelope U(g) by factoring out the
relations in g_0 defined by R. For each R there is a g irrep Irr R which
is a subrepresentation in Ind R.
This correspondence is not terribly useful for finite-dimensional
algebras, since Irr R might be hidden deep inside Ind R and it might be
difficult to disentangle it, but it follows immediately that g and g_0
must have the same rank. However, infinite-dimensional Lie algebras of
vector fields are also graded Lie algebras, although the grading extends
to plus infinity. In this case the correspondence is really useful because
Ind R is often irreducible by itself. The canonical example is vect(n).
Given a gl(n) irrep R, Ind R is equivalent to (contragredient to) the
representation on tensor fields of type R. This is irreducible except for
differential forms, which have a subrepresentation on closed forms.
Induced representations are useful even if g is non-simple, such as the
Poincare algebra. Here the representations induced from the so(n)
subalgebra are flat tensor and spinor fields.
> I should mention that while I was doing these calculations, I noticed
> all sorts of strange things. For example, 64 + 14 = 78, the dimension
> of e_6. And the elements of grade 2 and 1 in the above grading of e_8
> do form a 78-dimensional Lie algebra, right? But unless I'm mixed
> up it must be nilpotent Lie algebra, not e_6.
It is the nilpotent "supersymmetry" algebra with a bosonic spinor that I
described in my previous post, i.e.
Tony Smith
John Baez says:
"... we get a 5-grading [of e_8] of this sort:
14 + 64 + 92 + 64 + 14
And the elements of grade 2 and 1 in the above grading of e_8
do form a 78-dimensional Lie algebra, right?
But unless I'm mixed up it must be nilpotent Lie algebra, not e_6. ...".
Actually,
as Kaneyuki makes clear in his book cited earlier in this thread,
the non-zero gradings do not constitute sub-algebras of
the Lie algebra that is being graded,
but instead
are vector spaces (some of which vector spaces could be
given a different product to form Jordan or Lie algebras).
In the above case,
the 78-dimensional grade 2 plus grade 1 vector space
could indeed be given a product to make it into e_6
but
the product structure directly inherited from e_8 would
not be the e_6 product structure.
For example, in the grading of e_7
H3(O) + ( e_6 + R) + H3(O)
H3(O) is as Kaneyuki says
"... the vector space of O-hermitian matrices of degree 3 ..."
and
although it can be given (in a way that seems natural to me)
a product such that it would be the Jordan algebra J3(O)
in fact
as a grade-1 part of 3-graded e_7 it is only a vector space.
The way I see it geometrically is that
only the grade-0 part of those gradings is actually a sub-algebra
and
the grade-1, grade-2 etc parts are vector spaces,
with the group structure analogy being symmetric spaces
such as
E7 / E6xU(1) = 27+27=54-dim homogeneous space that is only a
manifold, with no Lie or Jordan product directly inherited
and
only the E6xU(1) having Lie group structure directly inherited
as a subgroup of E7.
Tony 20 Feb 2003
John Baez
In article <b34cr7$f2r$1...@panther.uwo.ca>,
Tony Smith <tsm...@spamfree.com> wrote:
>John Baez says:
>
>"... we get a 5-grading [of e_8] of this sort:
>
>14 + 64 + 92 + 64 + 14
>
>And the elements of grade 2 and 1 in the above grading of e_8
>do form a 78-dimensional Lie algebra, right?
>But unless I'm mixed up it must be nilpotent Lie algebra, not e_6. ...".
>Actually,
>as Kaneyuki makes clear in his book cited earlier in this thread,
>the non-zero gradings do not constitute sub-algebras of
>the Lie algebra that is being graded,
>but instead
>are vector spaces (some of which vector spaces could be
>given a different product to form Jordan or Lie algebras).
Well...
In a graded Lie algebra, if x has grade n and y has grade m,
then [x,y] has grade n+m. So, the guys of grade n will
typically be closed under bracketing only when n = 0.
So far I'm agreeing with you. But...
We can also take all the guys of grade n or more, and if
n is nonnegative, they will be closed under bracketing.
So we get a Lie subalgebra consisting of all guys of
degree n or more when n is nonnegative. If n is positive
and our Lie algebra is finite-dimensional, this subalgebra
must be nilpotent, because repeated bracketing keeps raising
the grade until we "take a long walk off a short pier",
going above the highest possible grade and therefore getting zero.
That's what I was doing above. I was taking the guys in
e_8 of grade 1 or more, and getting a nilpotent Lie subalgebra of
dimension 78.
I'm quite confident that this is correct. My querulousness
was due only to the fact that e_8 also has a very famous
NON-nilpotent Lie subalgebra of dimension 78, namely e_6.
It just seemed too weird. When I think "78-dimensional Lie
subalgebra of e_8", I think e_6!
But, a new post by Thomas Larsson has reassured me that it's true.
>In the above case,
>the 78-dimensional grade 2 plus grade 1 vector space
>could indeed be given a product to make it into e_6
>but
>the product structure directly inherited from e_8 would
>not be the e_6 product structure.
Well, I claim it inherits a Lie bracket directly from e_8,
and that this makes it into a Lie algebra that's not e_6.
>For example, in the grading of e_7
>
>H3(O) + ( e_6 + R) + H3(O)
>
>H3(O) is as Kaneyuki says
>"... the vector space of O-hermitian matrices of degree 3 ..."
>and
>although it can be given (in a way that seems natural to me)
>a product such that it would be the Jordan algebra J3(O)
>in fact
>as a grade-1 part of 3-graded e_7 it is only a vector space.
The grade-1 part of a 3-graded Lie algebra won't be a Lie
subalgebra. But it will be more than a mere vector space:
it will be a Jordan algebra! That's what Larsson and I have
been rhapsodizing about. And in this case you get the Jordan
algebra J3(0).
Admittedly, to make the grade-1 part into a Jordan algebra
you need to make an arbitrary choice: you need to pick an
element p of grade -1. Then you can define this product
on the elements of grade 1:
x o y = [x,[p,y]]
and get a Jordan algebra! Like Larsson, I haven't checked
that this product satisfies the crucial Jordan identity:
x o ((x o x) o y)) = (x o x) o (x o y),
But people say it's true. The calculation seems quite
obnoxious, so I will wait for some day when I am very
bored before checking it.
John Baez
Thomas Larsson <thomas....@hdd.se> wrote:
>John Baez <ba...@galaxy.ucr.edu> wrote in message
>news:b31qct$1bd$1...@glue.ucr.edu...
>> But as you undoubtedly know, h_3(O) goes give rise to e_6. :-)
>> More precisely, the group E6 has a real form called E6(-26)
>> which is the group of all transformations of h_3(O) that preserve
>> the "determinant". This is why some people define E6 as the
>> group of all transformations of a certain 27-dimensional space
>> preserving a cubic form.
>Hmm. There is no grading corresponding to this.
And there'd darn well better not be.
>Kaneyuki lists four different gradings for e_6:
>
>i17. e_6 = 16 + (so(10) + C) + 16
>
>e24. e_6 = 5 + 20 + (sl(5) + sl(2) + C) + 20 + 5
>
>e25. e_6 = 1 + 20 + (sl(6) + C) + 20 + 1
>
>e26. e_6 = 8 + 16 + (so(8) + C + C) + 16 + 8
>
>which act on spaces of dimension 16, 25, 21, 24 but not 27.
Here what is acting on the spaces
16, 25, 21 and 24
is not e_6 but its degree-0 subalgebra, namely
so(10)+C, sl(5)+sl(2)+C, sl(6)+C and so(8)+C+C
This is because the degree-0 subalgebra acts by bracketing
on the subspace of degree > 0, but the whole Lie algebra does not.
So, you shouldn't find the 27-dimensional rep of e_6 on *this* list.
(Instead, we see things like the famous 16-dimensional rep
of so(10)+C - famous because the spinor rep of so(10) is
16-dimensional, and this spinor rep is complex, so it becomes
a rep of so(10)+u(1), but you're complexifying everything,
and complexifying u(1) turns it into C.)
But, you might find the 27-dimensional rep of e_6 showing up
if there were a similar list where you had a graded Lie algebra
whose degree-0 subalgebra is e_6.
And there is, and you do:
e_7 = 27 + (e_6 + C) + 27
So, I think everything is fine and dandy! The complex form of e_7
admits a 3-grading where the degree-zero subalgebra is e_6 + C.
This acts by bracketing on the 27-dimensional degree-1 part.
As usual, this degree-1 part becomes a Jordan algebra if we
pick a nonzero element of the degree-(-1) part. This Jordan
algebra is the complexification of the exceptional Jordan algebra.
The complex form of E6 acts as determinant-preserving transformations
of this Jordan algebra, and the Lie group corresponding to C acts
on it in some way that I don't understand... but something
boring, like trivially or multiplication by scalars.
(I would like to understand this annoying little C better.)
[See below, where I actually do.]
By the way, the complexification of the exceptional Jordan
algebra is again a simple Jordan algebra over the real numbers.
One tends to forget this one, but it's important.
>> I would like to understand all the gradings of E8 in the following
>> way.
>...
>> Anyway, by James Dolan's little quiz which I answered earlier
>> in this thread, it seems evident that we get gradings of E8 by
>> finding integer-valued linear functionals L on the E8 lattice;
>> the value of L on a given root is its "grade". We say we have
>> an "n-grading" if L takes exactly n values on the roots and
>> the origin.
>Let me see if I understand what you do. You slice the jewel into
>hyperplanes. The degree 0 subalgebra becomes g_0 = h_0 + C, where the h_0
>roots lie inside the middle hyperplane and C gives a root perpendicular to
>it.
I'd say: the degree 0 subalgebra is g_0 = h_0 + H, where the h_0
roots lie inside the middle hyperplane and H is the Cartan subalgebra
of the Lie algebra you started with.
(Btw, this means that this "jewel" trick only gives those gradings where
the degree 0 subalgebra contains the Cartan. I believe it gives all
gradings of this type.)
>The latter corresponds to the grading operator, which is simply
>dilatations in depth 1.
There might always be an operator in g_0 that corresponds to the
grading operator, but if so that's because some theorem makes it happen,
not because we go and "throw it in".
A simple Lie algebra is spanned by one vector for every root, together
with the Cartan subalgebra. So, when we use this "jewel" trick to
put a grading on our Lie algebra, we need to assign a grade to each
root, and to the elements of the Cartan. It makes sense to assign
grades to roots using a linear functional on the root lattice, since
this automatically implies
[g_n , g_m] is contained in g_{n+m}
It then makes sense to assign grade zero to everyone in the
Cartan subalgebra H, because it's a fact that
[H , g_n] is contained in g_n
Now I think I see why there will be an element of degree zero that
serves as a "grading operator" in your sense.
I'm too hungry to explain this well, or to check that it *always*
happens. But maybe I can illustrate it in an example, since
you've done most of the work. In fact, below you seem to agree
with almost everything I just said... so we may just be suffering
from some slight difference in terminology:
>Since the root space for h_0 lies in a hyperplane
>of codimension 1, we must have rank g = rank g_0 = rank h_0 + 1. E.g.
>
> rank e_6 = rank so(10) + 1
Let me say how I'd describe what's going on:
Here g_0 will be spanned by the roots lying in the middle hyperplane
together with a basis of the Cartan subalgebra of e_8. However,
the roots lying in the middle hyperplane look just like the roots
of so(10). Also, the Cartan of so(10) is *one less* than that of e_8.
So, g_0 will be spanned by so(10) and *one more* vector. This
serves as the "grading operator".
Nice!
However, there might be other cases where the roots lying
in the middle hyperplane aren't themselves the roots of a simple
Lie algebra. In such cases things wouldn't need to work so prettily.
But for some reason, it seems to works in all the cases you've shown me.
Here's a guess as to why: to get these other cases, we'd need to
slice our jewel at a pretty weird angle, which would produce an
n-grading with fairly large n.
>> I should mention that while I was doing these calculations, I noticed
>> all sorts of strange things. For example, 64 + 14 = 78, the dimension
>> of e_6. And the elements of grade 2 and 1 in the above grading of e_8
>> do form a 78-dimensional Lie algebra, right? But unless I'm mixed
>> up it must be nilpotent Lie algebra, not e_6.
>It is the nilpotent "supersymmetry" algebra with a bosonic spinor that I
>described in my previous post, i.e.
>
>[Q_a, Q_b] = c^i_ab P_i,
>
>[P_i, Q_a] = [P_i, P_j] = 0.
Hmm. I'll have to reread this, and this time I'll have to
actually *understand* it! But I'm pleased to hear that I'm
not just insane.
I'm really enjoying thinking about this stuff with you.