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For $n$ an integer divisible by $8$, let me denote by $E_n$ the "usual" even non-degenerate positive definite integral symmetric bilinear form over $\mathbf Z^n$.

It is well known that in dimension 16, the only even non-degenerate positive definite integral symmetric bilinear form are (up to isometry) $E_8 \oplus E_8$ and $E_{16}$. Moreover, it is known that these two have same theta series (i.e. their quadratic forms represent the same integers, the same amount of times).

My question is thus: given a dimension 16 non-degenerate positive definite integral symmetric bilinear form, what is the easiest test to determine whether it is isometric to $E_8 \oplus E_8$ or to $E_{16}$?

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    $\begingroup$ It depends on what tools you have ... you could compute the 4th Siegel gamma which distinguishes between these two lattices, or study the graphs of roots (one vertex per root, one edge between two vertices if the corresponding roots are not orthogonal), or compute the automorphism groups, etc... $\endgroup$
    – few_reps
    Commented Jun 24, 2014 at 9:52
  • $\begingroup$ Compute some Coxeter diagrams? $\endgroup$
    – S. Carnahan ♦
    Commented Jun 24, 2014 at 11:00
  • $\begingroup$ All this seems like a "rather complicated task". I mean: I am in front of a $16 \times 16$ matrix, with not so small integers. Pari/gp can list all 480 minimal length vectors, but analyzing this data will be tedious. I would have hoped for a trick. $\endgroup$
    – Oblomov
    Commented Jun 24, 2014 at 12:18
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    $\begingroup$ Well ... if you have a concrete Gram matrix M, ask Magma ( just enter M and type IsIsometric(LatticeWithGram(M),Lattice("E", 8)+Lattice("E", 8)), the online calculator will do it in few seconds. $\endgroup$
    – few_reps
    Commented Jun 24, 2014 at 14:18
  • $\begingroup$ I think it likely they will have different covering radius. Magma calculates the square of the covering radius, algorithm by G. Nebe. Squares of covering radii add for orthogonal sum, so the radius for $E_8 + E_8$ is just double that for $E_8.$ I will see if this is listed on the Catalogue of Lattices. $\endgroup$
    – Will Jagy
    Commented Jun 25, 2014 at 3:01

2 Answers 2

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One of the possible tricks I was after was the following: the lattice generated by the roots (vectors of length $2$) of $E_8 \oplus E_8$ is $E_8 \oplus E_8$, whereas the lattice generated by the roots of $E_{16}$ is $D_{16}$ (which is of index $2$ in $E_{16}$).

Pari/gp has a very efficient way to list the roots. Putting the resulting matrix in Hermite normal form lets me see whether the index is $1$ or $2$.

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In sage, you can use the command .is_globally_equivalent_to(), which allows you to compare two quadratic forms. See the Quadratic Forms Overview.

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