Timeline for Tri-coloring of E8 lattice? Why is the Gram matrix of E8 not unique?
Current License: CC BY-SA 4.0
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| Oct 2, 2021 at 18:20 | comment | added | Dan Haxton | In any case, the eigenvalues show that these 3 det 1 Gram matrices for E8 are all similarly "reduced" although the coding version is the "shortest". There is a symmetry, an isomorphism, of E8 revealed by the existence of these multiple gram matrices, involving the golden ratio, isn't there? (Icosahedron?) If the Gram matrices are brought into "correspondence" meaning that the matrix A returned by qfisom relating them is normal, then its eigenvalues would show the symmetry - rotations, reflections, stretching. If A cannot be normal, there is some translation in in the isomorphism, I suppose | |
| Sep 30, 2021 at 15:13 | comment | added | Dan Haxton | no the Jordan decomposition isn't quite it. By hand I can do qfisom on your example, with G1 = (1 0; 0 1) G2a = (13 18; 18 25) qfisom would return e.g. Aa=(2 3; 3 4) I suppose. But with G2b = (13 -18; -18 25), Ab = (-2 3; -3 4) you must flip the sign on one of the basis vectors in Ab before doing Jordan to see the same simple stretching isomorphism; otherwise Ab is not normal. I'll try all sign-flips & permutations in the A you gave for E8 to see whether I can bring it into normal form. But I think I need to search all relative orientations of the bases not just permutations & reflections | |
| Sep 27, 2021 at 20:43 | comment | added | Dan Haxton | There are not 8 eigenvalues. The Jordan normal form of A is -1 0 0 0 0 0 0 0 0 -P3 0 0 0 0 0 0 0 0 p3 0 0 0 0 0 0 0 0 1 1 0 0 0 0 0 0 0 1 1 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 1 where P3 is Phi^3 and p3 is Phi^-3. I think the Jordan normal form of A is the key thing that helps me understand the lattice isometry behind the different gram matrices. Too bad I don't have qfisom in matlab! | |
| Sep 27, 2021 at 18:28 | comment | added | Dan Haxton | Thanks again for your help Henry, and for the PARI/GP resource. I see that my original question, the idea of the tri-coloring was naive and mistaken. I understand that the matrix A you got with qfisom is a lattice isomorphism. If you permute the columns and compute the eigenvalues, eig(A(:,[5,4,6,7,8,1,3,2]), you get (1,1,1,1,1,-1,-4.236067977499789,0.236067977499790). You get the golden ratio cubed. I will investigate some more about golden ratio and E8. I would like to understand the geometric nature of this isomorphism, beyond the obvious. | |
| Sep 26, 2021 at 4:23 | history | edited | Henry Cohn | CC BY-SA 4.0 |
added more information
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| Sep 26, 2021 at 4:09 | comment | added | Henry Cohn | I'll add some information about how to compute this in the answer. | |
| Sep 26, 2021 at 4:09 | comment | added | Dan Haxton | Let us continue this discussion in chat. | |
| Sep 26, 2021 at 4:06 | comment | added | Henry Cohn | Then $G' = A^t G A$, which is a weird kind of relationship. It comes up quite a bit in number theory, but less often in linear algebra more broadly. | |
| Sep 26, 2021 at 4:05 | comment | added | Henry Cohn | Say $B$ is a basis matrix, and $G = B^t B$ is the corresponding Gram matrix. Then another basis matrix $B'$ will be of the form $B' = B A$, where $A$ has integer entries and determinant $\pm 1$; it tells which linear combinations you take to get the new basis. | |
| Sep 26, 2021 at 4:03 | comment | added | Henry Cohn | The normalized eigenvalues are typically distinct. For example, the bases I gave for $\mathbb{Z}^2$ give Gram matrices with different normalized eigenvalues. The relationship between them comes from change of basis. | |
| Sep 26, 2021 at 4:03 | comment | added | Dan Haxton | I thought that the lattice determined the spectrum of the normalized Gram matrix and vice versa. Because that is not true, I need to do more work to understand lattices. If I can't generate a lattice from its Gram matrix eigenvalues, then I don't understand lattices. I need to verify that the different E8 lattices that I can generate from the different Gram matrices are actually the same. | |
| Sep 26, 2021 at 3:59 | comment | added | Dan Haxton | So what relationship must the different Gram matrices satisfy? How can I tell that the lattices are the same, from the Gram matrix? E8 is the only example I can find of a lattice for which the normalized Gram matrix eigenvalues, which multiply to 1, can be different. Otherwise, I thought I understood lattices via the Gram matrix. I thought Gram matrices had to be related via similarity transformations, leaving the origin fixed. What is the relationship that the gram matrices must satisfy, if not that given by orthonormal transformations? | |
| Sep 26, 2021 at 3:59 | comment | added | Dan Haxton | Thanks Henry; let me explain better. Normalize the gram matrix to have determinant one; then take its eigenvalues, such that their product is 1. Seems that all lattices have only one such set of eigenvalues, except E8. If the normalized gram matrices have the same eigenvalues, then the bases are trivially related by a "similarity transformation", an orthonormal transformation, as you describe. But if the normalized gram matrices have different eigenvalues, then the relationship between the basis vectors is not a similarity transformation, an orthonormal transformation. | |
| Sep 26, 2021 at 3:19 | history | answered | Henry Cohn | CC BY-SA 4.0 |