Dn corrrespondences

A-D-E series and their **FundamentalRepresentations**:

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

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

It, together with ideas of OnarAam and Ben Goertzel about XORand set theory, and work of J. Frank Adams on exceptional Liealgebras, inspired me to write THIS PAGE ABOUT SETS,CLIFFORD GROUPS AND ALGEBRAS, AND THE McKAYCORRESPONDENCE. (Any errors you see are due to me, not to BarrySimon, Onar Aam, Ben Goertzel, or J. Frank Adams.)

HOW TO BUILD CLIFFORD ALGEBRAS FROM SET THEORY: You can start with a set S(N) of N elements. 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 VNwith 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.

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), which subsets can be graded into 2N+1 grades, 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. Let it represent grade-1. Choose a second basis element. Let it, along with the first chosen element, represent grade-2. ... Choose a k-th basis element. Let it, along with the preceding chosen elements, represent grade-k. ... Choose an (N-2)-nd basis element. Let it, along with the preceding chosen elements, represent grade-(N-2). Choose an (N-1)-st basis element. Let it, along with the preceding chosen elements, represent grade-(N-1). Choose an N-th basis element. Let it, along with the preceding chosen elements, represent grade-N. Since, by the set theoretical counterpart of Hodge duality, grade-k is isomorphic with grade-(N-k), 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) are in spaces of grade-1, grade-2, ... , grade-(N-2). Consider those basis elements, of the set of 2N elements, that represent grade-1, grade-2, ... , grade-(N-2). 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 N N/\N ... N/\grade-(N-2)/\N 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).

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. Let it represent grade-1. Choose a second basis element. Let it, along with the first chosen element, represent grade-2. ... Choose a k-th basis element. Let it, along with the preceding chosen elements, represent grade-k. ... Choose an (N-2)-nd basis element. Let it, along with the preceding chosen elements, represent grade-(N-2). Choose an (N-1)-st basis element. Let it, along with the preceding chosen elements, represent grade-(N-1). Choose an N-th basis element. Let it, along with the preceding chosen elements, represent grade-N. 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, so that the N-1 grades grade-1 through grade-(N-1) correspond to the N-1 nontrivial fundamental representations of SU(N), with grade-(N-1) being used for the adjoint representation in 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, that represent grade-1, grade-2, ... , grade-(N-1). 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 N N/\N ... N/\grade-(N-2)/\N N/\grade-(N-1)/\N 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).

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

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.

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.

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 6-dimensional Adjoint representation. 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.

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 the 28-dimensional Adjoint representation. 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 grade-2 adjoint representation of D4 is 28-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.

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 tothe 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-2 adjoint representation of D5 is 45-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.

The E6 Coxeter-Dynkin diagram, in which each vertex corresponds to a fundamental representation, is

78|27 - 351 - 2,925 - 351 - 27

Click here to see the Cartan matrix, 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-2 adjoint representation of D5 is 45-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: The graded sequence 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). The graded sequence 78 78/\78has 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

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

Click here to see the Cartan matrix, 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. The grade-2 adjoint representation of D6 is 66-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: The graded sequence 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. The 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. The graded sequence 912 912/\912has 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.

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 Click here to see the Cartan matrix, 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. The grade-3 part has dimension 248/\248/\248 = 2,511,496. The grade-4 part has dimension 248/\248/\248/\248 = 153,829,130. The grade-5 part has dimension 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: The graded sequence 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. The 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. The 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 - 3875In 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 b147250 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. J. F. Adams says:"... we can allow ourselves to construct new representationsfrom old by taking exterior powers, as well as tensor productsand 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 aneigenvector 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 eigenvectorcorresponding 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.

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