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This question is about euclidean lattices, regular arrays of points in $\mathbb R^N$.

Why are there 3 Gram matrices for the E8 lattice? They are not related by a similarity transformation; they have different eigenvalues. How do I simply prove that the three different gram matrices, with different eigenvalues, determine the same lattice?

Is there a concept for the degeneracy of the Gram matrix, which is 3 for E8, and what is the term for it? Are there any other examples?

The eigenvalues of the Gram matrices of other lattices seem unique, at least in the tables quoted below from Gabriele Nebe. I cannot find any other examples of this. E8 is the only lattice for which I can find different Gram matrices with different eigenvalues in Nebe's tables.

What property of E8 lattice is it that makes its Gram matrix eigenvalues undetermined?

What is the relationship among the 3 Gram matrices of E8? Do they actually determine the same lattice? I will check the depth of their holes by brute force, to see if they are different, but I have not done that yet.

https://www.math.rwth-aachen.de/~Gabriele.Nebe/LATTICES/E8.html

https://www.math.rwth-aachen.de/~Gabriele.Nebe/LATTICES/E8.b.html

https://www.math.rwth-aachen.de/~Gabriele.Nebe/LATTICES/E8_code.html

I think the 3 must be related by translations (and rotations). That would mean they correspond to a tricoloring of E8 & there are 3 different classes of points in E8, but I can't find that info. The chromatic number of E8 is 16, not 3.

Sikirić, Madore, Moustrou, and Vallentin - Coloring the Voronoi tessellation of lattices

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    $\begingroup$ If I am not misleaded, the Gram matrix of a lattice corresponds to a bilinear form. On the other hand, an Eigenvalue is a property of a transformation matrix, but not of a bilinear form. $\endgroup$ Sep 24 at 22:14
  • $\begingroup$ Thanks Martin mathoverflow.net/users/105705/martin-seysen. The matrix A that relates the two Gram matrices, computed by e.g. qfisom pari.math.u-bordeaux.fr/dochtml/html/… has a Jordan normal form, which seems to be the fundamental thing relating the two indexing conventions for the same points in space corresponding to the two Gram matrices. See Henry Cohn's mathoverflow.net/users/4720/henry-cohn answer below. The Jordan form of A seems to summarize the symmetry involved. Is there something more fundamental? $\endgroup$
    – Dan Haxton
    Sep 30 at 1:36
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The Gram matrix depends on the choice of a lattice basis (it's just the matrix of inner products of the basis vectors), which means all lattices in two or more dimensions have infinitely many Gram matrices, typically with very different eigenvalues. For example, for the integer lattice $\mathbb{Z}^2$, the basis $(1,0)$, $(0,1)$ has the identity matrix as its Gram matrix, while the basis $(2,3)$, $(3,4)$ has the Gram matrix $\begin{pmatrix} 13 & 18\\ 18 & 25 \end{pmatrix}$, which has eigenvalues $0.02633403\dots$ and $37.97366596\dots$. (To see that $(2,3)$ and $(3,4)$ span $\mathbb{Z}^2$, note that they certainly span a sublattice, and it can't be a proper sublattice since the basis matrix has determinant $-1$.) There's some regularity; for example, the determinant of the Gram matrix of a basis depends only on the lattice. However, the eigenvalues do not.

The Gram matrix uniquely determines the lattice basis, up to orthogonal transformations (i.e., isometries). That means the three given bases of $E_8$ are genuinely different from each other, but they all span the same lattice. This follows directly from known characterizations of $E_8$ (for example, it's the unique even unimodular lattice in $\mathbb{R}^8$), and it can also be checked by identifying the correct integer linear transformations to express each basis in terms of the others (see below for computational details).

So the issue here is that the Gram matrix is an invariant of the basis, not the lattice, and there are infinitely many different bases. I don't think the number $3$ has any particular significance here. It's just the number of bases given in Nebe and Sloane's catalog.

I don't know of any special reason why $E_8$ has more different bases listed by Nebe and Sloane than other lattices do. I'm not surprised that it has more than most lattices, because it's a particularly important lattice that can be constructed in a variety of different ways. However, I would have guessed that some other lattices might have as many.

To compute the change of basis matrices, one can use PARI/GP (see https://pari.math.u-bordeaux.fr). For example, two of the Gram matrices listed for $E_8$ by Nebe and Sloane are $$G_1 = \begin{pmatrix} 4 & -2 & 0 & 0 & 0 & 0 & 0 & 1\\ -2 & 2 & -1 & 0 & 0 & 0 & 0 & 0\\ 0 & -1 & 2 & -1 & 0 & 0 & 0 & 0\\ 0 & 0 & -1 & 2 & -1 & 0 & 0 & 0\\ 0 & 0 & 0 & -1 & 2 & -1 & 0 & 0\\ 0 & 0 & 0 & 0 & -1 & 2 & -1 & 0\\ 0 & 0 & 0 & 0 & 0 & -1 & 2 & 0\\ 1 & 0 & 0 & 0 & 0 & 0 & 0 & 2 \end{pmatrix}$$ and $$G_2 = \begin{pmatrix} 2 & 0 & -1 & 0 & 0 & 0 & 0 & 0\\ 0 & 2 & 0 & -1 & 0 & 0 & 0 & 0\\ -1 & 0 & 2 & -1 & 0 & 0 & 0 & 0\\ 0 & -1 & -1 & 2 & -1 & 0 & 0 & 0\\ 0 & 0 & 0 & -1 & 2 & -1 & 0 & 0\\ 0 & 0 & 0 & 0 & -1 & 2 & -1 & 0\\ 0 & 0 & 0 & 0 & 0 & -1 & 2 & -1\\ 0 & 0 & 0 & 0 & 0 & 0 & -1 & 2 \end{pmatrix}.$$ Suppose we give them to gp as variables G1 and G2. Then the command qfisom(G1,G2) will give us a matrix $$ A = \begin{pmatrix} 0&0&0&0&1&-2&0&-1\\ 0&0&0&1&0&-2&0&-1\\ 0&0&0&1&0&-3&0&-2\\ 0&0&0&2&0&-4&-1&-2\\ 0&0&0&2&0&-4&0&-1\\ 1&0&0&1&0&-3&0&0\\ 0&0&1&0&0&-2&0&0\\ -1&1&0&0&0&-1&0&0 \end{pmatrix} $$ such that $A^t G_2 A = G_1$. What this means is that $A$ is the change of basis matrix between these bases: the columns of $A$ tell which linear combinations of the basis vectors corresponding to $G_2$ give the basis vectors corresponding to $G_1$. In other words, if the basis matrix $B_i$ has the basis vectors as columns, then $G_i = B_i^t B_i$ and $B_1 = B_2 A$.

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  • $\begingroup$ 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. $\endgroup$
    – Dan Haxton
    Sep 26 at 3:59
  • $\begingroup$ 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? $\endgroup$
    – Dan Haxton
    Sep 26 at 3:59
  • $\begingroup$ 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. $\endgroup$
    – Dan Haxton
    Sep 26 at 4:03
  • $\begingroup$ 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. $\endgroup$
    – Henry Cohn
    Sep 26 at 4:03
  • $\begingroup$ 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. $\endgroup$
    – Henry Cohn
    Sep 26 at 4:05

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