and the

# McKay Correspondence

Dn corrrespondences

An correspondences

A-D-E series and their Fundamental Representations:

A0=D1 - A1=B1=C1 - A2 - A3=D3 - D4 - D5 - E6 - E7 - E8

### OY! Barry Simon has written YABOGR!

```The official title is:
Representations of Finite and Compact Groups (AMS 1996)
What is YABOGR? Read the Book!
```

### What does YABOGR do?

It, together with ideas of Onar Aam and Ben Goertzel about XOR and set theory, and work of J. Frank Adams on exceptional Lie algebras, inspired me to write THIS PAGE ABOUT SETS, CLIFFORD GROUPS AND ALGEBRAS, AND THE McKAY CORRESPONDENCE. (Any errors you see are due to me, not to Barry Simon, Onar Aam, Ben Goertzel, or J. Frank Adams.)

```
HOW TO BUILD CLIFFORD ALGEBRAS FROM SET THEORY:

Call it {e(1), ..., e(N)}.

Then consider the set 2^S(N) all of its 2^N subsets,
with a product on 2^S(N) defined as the symmetric set difference XOR.
Denote the elements of 2^S(N) by e(A) where A is in 2^S(N).

Then enlarge 2^S(N) to DCLG(N) by:
ordering the basis elements of S(N),
and then giving each element of 2^S(N) a sign,
either +1 or -1, so that DCLG(N) has 2^(N+1) elements.
(This amounts to orientation of the +/- unit basis of S(N).)

Then define a product on DCLG(N) by

(x1 e(A)) (x2 e(B)) = x3 e(A XOR B)

where A and B are in 2^S(N) with ordered elements,
and x1, x2, and x3 determine the signs.
For given x1 and x2,   x3 = x1 x2 x(A,B)
where x(A,B) is a complicated function that determines sign
by using the rules e(i)e(i) = +1          for   i   in S(N)
and                e(i)e(j) = - e(j)e(i)  for   i =/= j   in S(N),
then writing (A,B) as an ordered set of elements of S(n),
then using  e(i)e(j) = - e(j)e(i) to move each of the B-elements
to the left until it:
either
meets a similar element and then cancelling it
with the similar element by using e(i)e(i) = +1
or
it is in between two A-elements in the proper order.

DCLG(N) is a finite group of order 2^(N+1).
It is the Discrete Clifford Group of N signed
ordered basis elements.
(It is like a reflection group on the signed
ordered basis elements of S(N).)

Now, construct an N-dimensional vector space VN
with oriented orthonormal basis elements derived from
the signed ordered basis elements of S(N).

The discrete group DCLG(N) extends naturally
to the group CLG(N) acting on VN.

CLG(N) is the Clifford Group of N-dim Euclidean space.
(It is like the group of reflections through
the hyperplanes through the origin in VN.)

Then extend the Clifford Group CLG(N)
to the Clifford Algebra Cl(N) by
using the relations   e(i)e(j) + e(j)e(i) =  2  delta(i,j) 1
where 1 is the grade-0 scalar and delta(i,j) is the Kronecker delta.

The Clifford Group CLG(N) is then a subgoup
of the Clifford Algebra Cl(N).

There is a 1-1 correspondence between
the representations of Cl(N) and
those representations of DCLG(N) such that U(-1) = -1.

DCLG(N) has 2^N 1-dimensional representations,
each with U(-1) = +1.

The irreducible representations of DCLG(N)
with dimension greater than 1 have U(-1) = -1,
and are representations of CL(N).
If N is even,
there is one such representation, of degree 2^(N/2).
If N is odd,
there are two such representations, each of degree 2^((N-1)/2).

Roughly, all this means:

The CLIFFORD ALGEBRA is the GROUP ALGEBRA
of the DISCRETE CLIFFORD GROUP.

```
The Discrete Clifford Group contains many of the properties of the Clifford Algebra, including, as shown by Boya and Byrd, the 8-fold Periodicity Property.

## TheMcKayCorrespondence:

` `

### for the DN Lie Algebras,

```
consider some subgroups of DCLG(2N)
for an even-numbered basis of 2N elements.

Start with a set S(2N) with an even number, 2N, of elements.

There are 2^(2N) subsets of S(2N),
with the number of subsets in each grade
being determined by the binomial distribution.

Look at the 2N basis elements of the grade-1 subsets,
which are the elements of S(2N).

Choose one of the basis elements.

Choose a second basis element.
Let it, along with the first chosen element,

...

Choose a k-th basis element.
Let it, along with the preceding chosen elements,

...

Choose an (N-2)-nd basis element.
Let it, along with the preceding chosen elements,

Choose an (N-1)-st basis element.
Let it, along with the preceding chosen elements,

Choose an N-th basis element.
Let it, along with the preceding chosen elements,

Since, by the set theoretical counterpart of Hodge duality,
these N representatives characterize
the structure of DCLG(2N).

Note that
the fundamental representations of Spin(2N) that live
in exterior wedge-product multivector spaces
(as opposed to half-spinor spaces)

Consider those basis elements, of the set of 2N elements,
Let them form the vertices of an (N-2)-polygon.

Now consider the +/- signs of each of those N-2 basis elements.
They correspond to giving two faces to the (N-2)-polygon,
transforming it into an (N-2)-vertex dihedron.

The symmetry group of an (N-2)-vertex dihedron is
the binary dihedral group BD(N-2) is {2,2,N-2} of order 4(N-2).

Another way to look at it is:
The relevant part of the 2^2N graded sequence is
It has symmetry Cy(N-2) of order N-2 for cyclic permutations,
times 2 for Hodge duality (starting on the left or the right),
and
the +/- signs (or the two half-spinors)
have symmetry Cy(2) of order 2,
so
the total symmetry group is of order (N-2)x2x2 = 4(N-2),
the symmetry of BD(N-2).

BD(N-2) is a subgroup of DCLG(2N) that
represents the grades of the fundamental representations
of the bivector Lie Algebra Spin(2N)
of the Clifford Algebra Cl(2N)
that is the Group Algebra of DCLG(2N).

BD(N-2) corresponds, by the McKay correspondence,
to the DN Lie Algebra Spin(2N).

```

## For the AN correspondences:

```The 2^(N) sets can be graded into N+1 grades,
according to the binomial distribution.

Look at the N basis elements.

Choose one of the basis elements.

Choose a second basis element.
Let it, along with the first chosen element,

...

Choose a k-th basis element.
Let it, along with the preceding chosen elements,

...

Choose an (N-2)-nd basis element.
Let it, along with the preceding chosen elements,

Choose an (N-1)-st basis element.
Let it, along with the preceding chosen elements,

Choose an N-th basis element.
Let it, along with the preceding chosen elements,

For AN, do NOT use +/- signs for elements,
and do NOT use Hodge duality,
so that N grades correspond to representations.

Note that grade-N is the entire N-element set,
not a proper subset,
corresponding to the pseudoscalar trivial representation,
to the N-1 nontrivial fundamental representations of SU(N),
with
the form of its tensor product with grade-N,
forming the N^2-dimensional representation of U(N)
that includes
the (N^2 - 1)-dimensional adjoint representation of SU(N).

Now,
consider those basis elements, of the set of N elements,
Let them form the vertices of an (N-1)-polygon.

The symmetry group of an (N-1)-vertex polygon is
the cyclic group Cy(N-1) of order (N-1).

Another way to look at it is:
The relevant part of the 2^N graded sequence is
It has symmetry Cy(N-1) of order N-1 for cyclic permutations.
Note that Hodge duality takes the sequence into itself, so
that starting on the left is isomorphic to starting on the right,
and there are no +/- signs or spinors,
so
the total symmetry group is of order (N-1),
the symmetry of Cy(N-1).

Cy(N-1) corresponds, by the McKay correspondence,
to the A(N-1) Lie Algebra SU(N).

```

## For the A0=D1 correspondence:

```A0 = D1 = U(1) is Abelian and has no root vector diagram
and no Cartan matix.

```

## For the A1 correspondence:

```The A1=B1=C1 Coxeter-Dynkin diagram, in which
each vertex corresponds to a fundamental representation, is ```
`2`
```Click here to see the Cartan Matrix,
and here to see the Root Vector Diagram.

The 1 fundamental representation of A1 is:

The grade-1 spinor representation of A1=B1 is 2-dimensional.

A1 also has its trivial 1-dimensional scalar representation.

```

## For the A2 correspondence:

```The A2 Coxeter-Dynkin diagram, in which
each vertex corresponds to a fundamental representation, is ```
```3
|
3```
```Click here to see the Cartan matrix,
and here to see the Root Vector Diagram.

The 2 fundamental representations of A2 are:

The grade-1 vector representation of A2 is 3-dimensional.
The pseudo-vector grade-2 representation of A2 is 3-dimensional.

A2 also has its trivial 1-dimensional scalar representation.

```

## For the A3=D3 correspondence:

```The A3=D3 Coxeter-Dynkin diagram, in which
each vertex corresponds to a fundamental representation, is ```
```4
|
6-4```
```Click here to see the Cartan matrix,
and here to see the McKay Polytope and Root Vector Diagram.

The one vertex of the McKay Polytope corresponds to

The 3 fundamental representations of D3 are:

The grade-1 vector representation of D3 is 6-dimensional.
The +half-spinor representation of D3 is 4-dimensional.
The -half-spinor representation of D3 is 4-dimensional.

D3 also has its trivial 1-dimensional scalar representation.

```

## For the D4 correspondence:

```The D4 Coxeter-Dynkin diagram, in which
each vertex corresponds to a fundamental representation, is ```
```8
|
8-28-8```
```Click here to see the Cartan matrix,
and here to see the McKay Polytope and Root Vector Diagram.

The two vertices of the McKay Polytope correspond to
the 8-dimensional Vector representation and

Recall that, as in the D4-D5-E6-E7 physics model,
The 4 fundamental representations of D4 are:

The grade-1 vector representation of D4 is 8-dimensional.
The +half-spinor representation of D4 is 8-dimensional.
The -half-spinor representation of D4 is 8-dimensional.

D4 also has its trivial 1-dimensional scalar representation.

```

## For the D5 correspondence:

```The D5 Coxeter-Dynkin diagram, in which
each vertex corresponds to a fundamental representation, is ```
```    16
|
10 - 45 - 120 - 16```
```Click here to see the Cartan matrix,
and here to see the McKay Polytope and Root Vector Diagram.

The three vertices of the McKay Polytope correspond to
the 10-dimensional Vector representations,
the 45-dimensional Bivector Adjoint representations, and
the 120-dimensional Trivector representation.

Recall that, as in the D4-D5-E6-E7 physics model,
The 5 fundamental representations of D5 are:

The grade-1 vector representation of D5 is 10-dimensional.
The grade-3 representation of D5 is 120-dimensional,
where 16/\16 = 16x15/2 = 120 = 10/9x8/(2x3) = 10/\10/\10

The +half-spinor representation of D5 is 16-dimensional.
The -half-spinor representation of D5 is 16-dimensional.

D5 also has its trivial 1-dimensional scalar representation.

```

## For the E6 correspondence:

```The E6 Coxeter-Dynkin diagram, in which
each vertex corresponds to a fundamental representation, is ```
```78
|
27 - 351 - 2,925 - 351 - 27```
```

and here to see the McKay Polytope and Root Vector Diagram.

Recall that, as in the D4-D5-E6-E7 physics model,
E6 can be constructed from the representations of D5,
including the two D5 16-dimensional half-spinor representations.

The 5 fundamental representations of D5 are:

The grade-1 vector representation of D5 is 10-dimensional.
The grade-3 representation of D5 is 120-dimensional.
The +half-spinor representation of D5 is 16-dimensional.
The -half-spinor representation of D5 is 16-dimensional.

D5 also has its trivial 1-dimensional scalar representation.

Consider first the grade-1 10-dimensional D5 representation space.
Add to it the scalar 1-dimensional representation space
and one of the 16-dimensional half-spinor spaces,
to get
a 10+1+16 = 27-dimensional representation space.

Now, consider the antisymmetric exterior wedge algebra
of that 27-dimensional space.

The grade-1 part has dimension 27.
The grade-2 part has dimension 27/\27 = 351.
The grade-3 part has dimension 27/\27/\27 = 2,925.
As 27-2 = 25, the grade-25 part has dimension 351.
As 27-1 = 26, the grade-25 part has dimension 27.

These are 5 of the 6 fundamental representations of E6.

Since the two representations of dimension 27
are Hodge duals of each other,
as are the two representations of dimension 351,
only 3 of those 5 fundamental representations of E6
are independent:
the grade-1 part with dimension 27.
the grade-2 part with dimension 27/\27 = 351.
the grade-3 part with dimension 27/\27/\27 = 2,925.

They, like the D(N) and A(N) series constructions,
are all in the same exterior algebra (for E6, of /\27),
and so can be represented as the vertices of a triangle.

What about the 6th fundamental representation of E6?

Consider the grade-2 45-dimensional D5 representation space.
Add to it the scalar 1-dimensional representation space
and both of the 16-dimensional half-spinor spaces,
to get
a 45+1+16+16 = 78-dimensional representation space.

Now, consider the antisymmetric exterior wedge algebra
of that 78-dimensional space.

The grade-1 part has dimension 78.

It is the 6th fundamental representation of E6.
Since it is not in the same /\27 exterior algebra as
the other 3 independent fundamental representations,
it should not be a vertex in the same plane
as the triangle formed by them,
but should be a 4th vertex outside that plane,
with all 4 vertices forming a tetrahedron.

The binary tetrahedral group {2,3,3} is of order 24.

Another way to look at it is:

27  27/\27  27/\27/\27
has symmetry Cy(3) of order 3 for cyclic permutations,
times 2 for Hodge duality (starting on the left or the right).

78  78/\78
has symmetry Cy(2) of order 2 for cyclic permutation,
but since 78/\78 is fixed by its relation to 27/\27/\27,
do not use Hodge duality on the 78 graded sequence.

The +/- signs for the D5 half-spinors
have symmetry of order 2.

The "other" E6 sequence of 27/\27/\27  27/\27  27 is
the Hodge dual of the 27  27/\27  27/\27/\27
with Cy(3) symmetry first mentioned above,
so its symmetry it taken into account
by the Hodge duality of that sequence.

Therefore:
the total symmetry group is of order 3x2x2x2 = 24,
the symmetry of the binary tetrahedral group {2,3,3},
with 3 symmetries Cy(3), Cy(2), Cy(3).

It corresponds, by the McKay correspondence,
to the E6 Lie Algebra.

What are the relations between 2,925
and 27/\27/\27 and 78/\78 ?

2,925 = 27/\27/\27

2,925 = 78/\78 - 78 =
= 3,003  - 78

```

## For the E7 correspondence:

```The E7 Coxeter-Dynkin diagram, in which
each vertex corresponds to a fundamental representation, is ```
```        912
|
56 - 1,539 - 27,664 - 365,750 - 8,645 - 133```
```
and here to see the McKay Polytope and Root Vector Diagram.

Note that E7 can be constructed
from the representations of E6 and D6,
including
the two D6 32-dimensional half-spinor representations.

The grade-1 vector representation of D6 is 12-dimensional.

D6 and E6 both have trivial 1-dimensional scalar representations.

E6 has 27-dimensional and 78-dimensional representations.

Consider first the 27-dimensional E6 representation space.
Take two copies of them, and
add two copies of the scalar 1-dimensional representation space
to get
a 27+27+1+1 = 56-dimensional representation space.

Now, consider the antisymmetric exterior wedge algebra
of that 56-dimensional space.

The grade-1 part has dimension 56.
The grade-2 part has dimension 56/\56 = 1,540.
The grade-3 part has dimension 56/\56/\56 = 27,720.
The grade-4 part has dimension 56/\56/\56/\56 = 367,290.

Now:

Keep the grade-1 part of dimension 56.

Subtract off 1
from 56/\56 = 1,540 to get 1,539.

Subtract off 56
from 56/\56/\56 = 27,720 to get 27,664.

Subtract off 56/\56 = 1,540
from 56/\56/\56/\56 = 367,290 to get 365,750.

These are 4 of the 7 fundamental representations of E7.

They, like the D(N) and A(N) series constructions,
are all in the same exterior algebra (of /\56),
and so can be represented as the vertices of a square.

A 5th representation of E7, 912-dimensional,
can be represented as 16x(56+1) = 912,
and so is not independent of the plane of the /\56 square.

What about the 6th and 7th fundamental representations of E7?

Consider the 78-dimensional E6 representation space.
Add two copies of the 27-dimensional E6 representation space
and one copy of the scalar 1-dimensional representation space
to get
a 78+27+27+1 = 133-dimensional representation space.

Now, consider the antisymmetric exterior wedge algebra
of that 133-dimensional space.

The grade-1 part has dimension 133.
The grade-2 part has dimension 133/\133 = 8,778.
The grade-3 part has dimension 133/\133/\133 = 383,306.

Now:

Keep the grade-1 part of dimension 133.

Subtract off 133
from 133/\133 = 8,778 to get 8,645.

They are the 6th and 7th fundamental representations of E7.
Since they are not in the same /\56 exterior algebra as
the 4 square-vertex fundamental representations of E7,
they should not be vertices in the same plane as the square.
However, since they are in the same /\133 exterior algebra,
they should be collinear, one above and one below the square,
thus forming a square bipyramid, or octahedron.

The binary octahedral group {2,3,4} is of order 48.

Another way to look at it is:

56  56/\56  56/\56/\56  56/\56/\56/\56
has symmetry Cy(4) of order 4 for cyclic permutations,
but since 56/\56/\56/\56 is fixed by its relation to 133/\133/\133,
do not use Hodge duality on the 133 graded sequence.

133  133/\133  133/\133/\133
has symmetry Cy(3) of order 3 for cyclic permutations,
but since 133/\133/\133 is fixed by its relation to 56/\56/\56/\56,
do not use Hodge duality on the 133 graded sequence.

The +/- signs for the D5 half-spinors inherited from E6
have symmetry of order 2.

912  912/\912
has symmetry Cy(2) of order 2 for cyclic permutation,
but since 912/\912 is fixed by its relation to 56/\56/\56/\56,
do not use Hodge duality on the 78 graded sequence.

Therefore:
the total symmetry group is of order 4x3x2x2 = 48,
the symmetry of the binary octahedral group {2,3,4},
with 3 symmetries Cy(4), Cy(3), Cy(2).

It corresponds, by the McKay correspondence,
to the E7 Lie Algebra.

What are the relations between 365,750
and 56/\56/\56/\56, 133/\133/\133, and 912/\912 ?

365,750 = 56/\56/\56/\56 - 56/\56 =
=    367,290     - 1,540  = 365,750

365,750 = 133/\133/\133 - 2 x 133/\133 =
=    383,306     - 2 x 8,778   =
=    383,306     -  17,556     = 365,750

365,750 = 912/\912 - 5 x 133/\133 - 3 x 56/\56
- 8 x 133 - 56 - 28 - 8       =
=  415,416 - 5 x 8,778 - 3 x 1,540
- 8 x 133 - 56 - 28 - 8       =
=  415,416 - 43,890 - 4,620 - 1,064 - 92 = 365,750

In the 912/\912 relations,
the 28 and 8 dimensional representations of D4 were also used.

Compare the above relations, particularly
those involving 912, with
the paper of J. F. Adams entitled
The Fundamental Representations of E8.

```

## For the E8 correspondence:

```
The E8 Coxeter-Dynkin diagram, in which
each vertex corresponds to a fundamental representation, is

147,250
|
248 - 30,380 - 2,450,240 - 146,325,270 - 6,899,079,264 - 6,696,000 - 3,875

and here to see the McKay Polytope and Root Vector Diagram.

Note that E8 can be constructed
from the representations of E6 and D8,
including
the two D8 128-dimensional half-spinor representations.

The grade-1 vector representation of D8 is 120-dimensional.
The half-spinor representation of D8 is 128-dimensional.

The adjoint representation of E8 is 120 + 128 = 248-dimensional.

D8 and E6 both have trivial 1-dimensional scalar representations.

E6 has 27-dimensional and 78-dimensional representations.

The grade-1 part has dimension 248.
The grade-2 part has dimension 248/\248 = 30,628.
248/\248/\248 = 2,511,496.
248/\248/\248/\248 = 153,829,130.
248/\248/\248/\248/\248 = 7,506,861,544.

Now:

Keep the grade-1 part of dimension 248.

Subtract off 248
from 248/\248 = 30,628 to get 30,380.

Subtract off 2 x 248/\248 = 2 x 30,628
from 248/\248/\248 = 2,511,496 to get 2,450,240.

Subtract off 2 x 2,511,496 and 2,450,240 and 30,628
from 248/\248/\248/\248 = 153,829,130 to get 146,325,270.

Subtract off 2 x 153,829,130 and 2 x 146,325,270
and 2 x 2,511,496 and 2,450,240 and 248
from 248/\248/\248/\248/\248 = 7,506,861,544 to get 6,899,079,264.

These are 5 of the 8 fundamental representations of E8.

They, like the D(N) and A(N) series constructions,
are all in the same exterior algebra (of /\248),
and so can be represented as the vertices of a pentagon

What about the 6th and 7th fundamental representations of E8?

Consider the 27-dimensional E6 representation space.
Add 32 copies of the 128-dimensional D8 half-spinor space,
and subtract off
one copy of the 248-dimensional E8 representation space
to get
a 27 + 32 x 128 - 248 = 3,875-dimensional representation space.

Now, consider the antisymmetric exterior wedge algebra
of that 3,875-dimensional space.

The grade-1 part has dimension 3,875.
The grade-2 part has dimension 3,875/\3,875 = 7,505,875.

Now:

Keep the grade-1 part of dimension 3,875.

Subtract off 5 x 147,250 and 2 x 30,628
and 3 x 3,875 and 3 x 248
from 3,875/\3,875 = 7,505,875 to get 6,696,000.

They are the 6th and 7th fundamental representations of E8.
Since they are not in the same /\248 exterior algebra as
the 5 pentagon-vertex fundamental representations of E8,
they should not be vertices in the same plane as the pentagon.
However, since they are in the same /\3,875 exterior algebra,
they should be collinear, one above and one below the pentagon,
thus forming a pentagonal bipyramid.

What about the 8th fundamental representations of E8?

Consider 2 x 24x24 - 1 = 2 x 576 - 1 = 1,151 copies of
the 128-dimensional D8 half-spinor space,
and subtract off
one copy of the 78-dimensional E6 representation space
to get a representation space of dimension
1,151 x 128 - 78 = 147,328 - 78 = 147,250.

Now, consider the antisymmetric exterior wedge algebra
of that 147,250-dimensional space.
The grade-1 part has dimension 147,250.

It is the 8th fundamental representation of E8.
Since it is not in the same /\248 exterior algebra as
the 5 pentagon-vertex fundamental representations of E8,
it should not be a vertex in the same plane as the pentagon.
Also, since it is not in the same /\3,875 exterior algebra as
the two bipyramid-peak-vertex fundamental representations of E8,
it should not be a vertex on the same line as
the pentagonal bipyramid axis.
It should represent a vertex creating a triangle
whose base is one of the sides of the pentagon
and whose top is near one of the bipyramid-peak-vertices,
to which it is connected by a line.
To produce a symmetric figure,
the vertex must be reproduced in 5 copies, one over
each of the 5 sides of the pentagon.
Then, for the entire figure to be symmetric,
it must form an icosahedron.
The binary icosahedral group {2,3,5} is of order 120.

Another way to look at it is:

248
248/\248
248/\248/\248
248/\248/\248/\248
248/\248/\248/\248/\248
has symmetry Cy(5) of order 5 for cyclic permutations,
but since 248/\248/\248/\248/\248 is fixed
by its relation to 3,875/\3,875/\3,875,
do not use Hodge duality on the 248 graded sequence.

3,875
3,875/\3,875
3,875/\3,875/\3,875
has symmetry Cy(3) of order 3 for cyclic permutations,
but since 3,875/\3,875/\3,875 is fixed by
its relation to 248/\248/\248/\248/\248,
do not use Hodge duality on the 3,875 graded sequence.

147,250
147,250/\147,250
has symmetry Cy(2) of order 2 for cyclic permutations,
but since 147,250/\147,250 is fixed by
its relation to 248/\248/\248/\248/\248,
do not use Hodge duality on the 147,250 graded sequence.

The +/- signs for the D5 half-spinors inherited
from E6 through E7
have symmetry of order 2.

Since the dimension of E8 is 248 = 120 +128,
the sum of the 120-dimensional adjoint representation of D8
plus
ONE of the 128-dimensional half-spinor representations of D8,
there is a choice to be made as to
which of the two half-spinor representations of D8 are used.
As they are mirror images of each other,
that choice has a symmetry of order 2.

Therefore:
the total symmetry group is of order 5x3x2x2x2 = 120,
the symmetry of the binary icosahedral group {2,3,5},
with 3 symmetries Cy(5), Cy(3), Cy(2).

It corresponds, by the McKay correspondence,
to the E8 Lie Algebra.

What are the relations between 6,899,079,264
and 248/\248/\248/\248/\248, 3,875/\3,875/\3,875,
and 147,250/\147,250 ?

6,899,079,264 = 248/\248/\248/\248/\248 -
- 2 x 248/\248/\248/\248 -
- 2 x (248/\248/\248/\248 - 2 x 248/\248/\248 -
- (248/\248/\248 - 2 x 248/\248) - 248/\248 ) -
- 2 x 248/\248/\248
-  (248/\248/\248 - 2 x 248/\248) - 248 =

=      248/\248/\248/\248/\248 -
- 4 x      248/\248/\248/\248 -
+ 3 x           248/\248/\248 -
-                         248 =

=      7,506,861,544 -
- 4 x   153,829,130 +
+ 3 x     2,511,496 -
-               248 =

=      7,506,861,544 -
-       615,316,520 +
+         7,534,488 -
-               248 = 6,899,079,264

6,899,079,264 = 3,875/\3,875/\3,875 - 18 x 248/\248/\248/\248 -
- 3 x 3,875/\3,875 +
+ 3 x 147,250 -
- 20 x 248 -
- 6 x 27 - 3 x 8 =
= 9,690,084,625 - 18 x 153,829,130 -
- 3 x 7,505,875 +
+ 441,750 -
- 4,960 -
- 162 - 24 =
= 9,690,084,625 - 2,768,924,340 -
- 22,517,625 +
+ 441,750 -
- 4,960 -
- 162 - 24 = 6,899,079,264

6,899,079,264 = 147,250/\147,250 - 25 x 248/\248/\248/\248 -
- 38 x 248/\248/\248 -
- 31 x 248/\248 -
- 55 x 248 - 128 -27 =
=  10,841,207,625 - 25 x 153,829,130 -
- 38 x 2,511,496 -
- 31 x 30,628 -
- 55 x 248 - 128 - 27 =
=  10,841,207,625 - 3,845,728,250 -
- 95,436,848 -
- 949,468 -
- 13,640 -
- 155 = 6,899,079,264

In the relations,
the 8 dimensional representation of D4 was also used.

J. F. Adams has written a paper entitled```

The Fundamental Representations of E8

```published in Contemporary Mathematics, Volume 37, 1985, 1-10,
and reprinted in The Selected Works of J. Frank Adams, Volume 2,
edited by J. P. May and C. B. Thomas, Cambridge 1992, pp. 254-263.

The 8 fundamental E8 representations are

147250
|
248 - 30380 - 2450240 - 146325270 - 6899079264 - 6696000 - 3875

In his paper, Adams denotes the three at the ends as follows:

248 is denoted by alpha, which I will write here as   a

3875 is denoted by beta, which I will write here as   b

147250 is denoted by gamma, which I will write here as c.

Adams denotes the kth exterior power by lambda^k
which I will write here as /\k
and he denotes the kth symmetric power by sigma^k
which I will write here as Sk .
Also, here I write (x) for tensor product.

"... we can allow ourselves to construct new representations
from old by taking exterior powers, as well as tensor products
and Z-linear combinations ...

... seven of the eight generators for the polynomial ring R(E8)
may be taken as

a, /\2 a , /\3 a , /\4 a , b , /\2 b , c .

The eighth may be taken either as  /\5 a , or as  /\3 b , or as /\2 c .

The corresponding argument for Dn would say that one should begin
by understanding three representations of Dn = Spin(2n) ,
namely the usual representation on R^2n and the two half-spinor
representations delta+ , delta- .
This we believe, so perhaps we can accept the analogue for E8 .

...

we must get back to business and construct b and c .

...

To give explicit formulae we must agree on a coordinate system.
The group E8 contains a subgroup of type A8 .
...
As a representation of A8 , the Lie lagebra L(E8) splits to give

L(E8) = L(A8) + /\3 + /\3* .

Let e1, e2, ... , e9 be the standard basis in the vector
space V = C9 on which A8 acts
...
We now introduce the element

v_ik = SUM(j) ( ei* ej* ek* (x) ej ek* + ej ek* (x) ei* ej* ek* )

in the symmetric square S2(a) in a (x) a , where  a = L(E8) .
...
THEOREM 4.
(a) The representation S2(a) of E8 contains a unique copy of b .
(b) This copy of  b  contains the elements  v_ik .
(c) For i =/= k the elements v_ik are eigenvectors corresponding
to extreme weight of 8 .
(d) In particular (with our choice of details) v_gl is an
eigenvector corresponding to the highest weight of b .
...
Theorem 4 allows us to realize  b  as the E8-submodule of S2(a)
generated by v_91 (or any other v_ik with i =/= k . )
...
In fact, S2(a) contains a trivial summand ... and also an
irreducible summand of highest weight ...
It turns out that the remaining summand has dimension 3,875,
which is precisely the dimension of  b  .

We now introduce the element

w_k = SUM(i) ei ek* (x) v_ik

= SUM(i,j) ei ek* (x) (ei* ej* ek* (x) ej ek* + ej ek* (x) ei* ej* ek*)

in  a (x) b  in   a (x) S2(a)  in  a (x) a (x) a .

...

THEOREM 5.
(a) The representation  a (x) b  of E8 contains a unique copy of  c .
(b) This copy of  c  contains the elements w_k .
(c) The elements w_k are eigenvectors corresponding to the extreme
weights of  c .
(d) In particular (with our choice of details) w_1 is an eigenvector
corresponding to the highest weight of  c .
...
we may realize  c  as the E8-submodule of  a (x) b  generated
by w_1 (or any other w_k).
...
In fact,  a (x) b  contains an irreducible summand of highest weight
... and the remaining summand has too small a dimension to contain
two copies of  c .   ...".

Adams then gives proofs of the theorems,
involving such things as looking at the Lie algebra L(E8) as
a representation of D8 as which it splits to give
L(E8) = L(D8) + delta-
where delta- is a half-spinor representation of D8.

One of the things that I think that I get out of Adams's paper is
that it seems to me that in order to get 3875 and 147250
you have to look not only at the 248 representation of E8 but
also at representations of some subgroups of E8
such as D8 etc.

Also, I think that Adams's methods may give
an interesting structure of the 912 of E7,
perhaps by looking at the Lie algebra L(E7) as a
representation of some subgroup (perhaps D4 and/or D6),
but Adams's paper is restricted to E8.

It makes me very sad that Adams was killed in a car accident
some years ago,
and is not around to explain more about such stuff.

```