Root systems and sums of squares

Question

It is easy to see that the quadratic form for the root system $A_n$ is a sum of $n+1$ squares of integral linear forms:

$$q_{A_n} = 2 \sum_{i=1}^n x_i^2 - 2 \sum_{i=1}^{n-1} x_i x_{i+1} = x_1^2 + (x_1-x_2)^2 + \dots (x_{n-1}-x_n)^2 + x_n^2$$

It is equally easy to see that $q_{D_n}$ is a sum of $n$ squares. It is a little harder to see but I think is true that $q_{E_n}$ ($n=6,7,8$) is not a sum of $\ge n$ squares of integral forms.

Question: is this a standard fact, well-known to experts? Is there a standard reference? (I hate to reinvent a bycicle.) And has this fact been used for something interesting? (I have an interesting application in mind, so I am looking for connections...)

Answer 1

If a quadratic form in $n$ variables is the sum of the squares of $n$ integer linear forms, it's the sum of the squares of $n$ rational linear forms. Thus it's equivalent as a rational quadratic form to $x_1^2+\cdots+x_n^2$. In particular its discriminant is a square. This rules out $E_6$ and $E_7$.

The case of $E_8$ is trickier, as it is equivalent over the rationals to $x_1^2+\cdots+x_8^2$. This time you have to show there's no equivalence over the integers, but one form takes solely even values and the other doesn't.

Added The first time round I didn't clock the $\ge n$ condition. But there's certainly a way to put an upper bound on the number $m$ of linear forms one needs.

I'll stick to the $E_8$ form. One can think of this as describing a lattice $L$ in Euclidean space. This lattice $L$ is self-dual, unimodular and even. Its shortest vectors form the set of 240 roots $R$. This set of roots has the nice property (I think it may be called the eutactic property or the perfect property; all this is in Martinet's book on lattices) that $x\mapsto\sum_{y\in R}(x\cdot y)^2$ is proportional to the quadratic form $x\mapsto x\cdot x$. I think actually $\sum_{y\in R}(x\cdot y)^2=30x\cdot x$ as $30=240/8$. Now an integer linear form is a linear form taking integer values on the lattice $L$ and so is $x\mapsto x\cdot z$ for some $z$ in the dual of $L$, so here $z\in L$. If $x\cdot x=\sum_{j=1}^m(x\cdot z_j)^2$ for $z_j\in L$ then $$480=2|R|=\sum_{y\in R}y\cdot y=\sum_{j=1}^m\sum_{y\in R}(y\cdot z_j)^2 =30\sum_{j=1}^m z_j\cdot z_j.$$ Each $z_j\cdot z_j\ge 2$ so we must have $m=8$ and each $z_j\cdot z_j=2$.

You should check my numbers... For $E_6/E_7$ the dual lattice is different from the original but I'm sure they still have the eutactic(?) property. Any way one can get an effective upper bound on $m$, probably not much bigger than $n$.

Answer 2

Just saw this thanks to a "Related" link from Question 154928. Yes, it is known that the $E_6$ form cannot be written as a sum of integral squares, and thus (by specialization) that the same is true of $E_7$ and $E_8$; moreover the representations of $D_n$ ($n>2$) and $A_n$ as the sum of $n$ and $n+1$ squares respectively are the only ways (up to isomorphism) to write these forms as the sum of any number of nonzero squares, with the exception of $D_3 \cong A_3$ which has both a three-square and a four-square representation. Or at least it is known once one makes Will Jagy's key observation that writing a form as a sum of $m$ integral squares is tantamount to embedding the corresponding lattice into ${\bf Z}^m$. (As it happens I was just asked a few days ago whether any integral positive-definite lattice can be embedded in some ${\bf Z}^m$, so this question was very familiar.)

I don't know a reference, but the result is not hard starting from the Coxeter diagrams and the fact that the vectors of norm $2$ in ${\bf Z}^m$ are exactly the vectors $e+e'$ for some $\pm$ unit vectors $e,e'$ with $e' \neq \pm e$.

For $A_n$ we need a sequence of $n$ such vectors any two of which are orthogonal except that consecutive vectors have inner product $-1$. The first two must be $e'-e, \, e''-e'$ with $e,e',e''$ orthogonal unit vectors. The third could be either $e'''-e''$ or $e+e'$. The latter choice does not extend to $A_4$, and the former extends uniquely to $e''''-e'''$, and then by induction to $\{ e^{(i)} - e^{(i-1)} \}_{i=1}^n$ with all $n+1$ unit vectors orthogonal.

For $D_n$ we need an $A_{n-1}$ configuration together with a norm-$2$ vector orthogonal to all but the second vector, with which it has inner product $-1$. For $n=3$ we've done this already because the $D_3$ and $A_3$ diagrams are isomorphic. For $n=4$ either of our two $n=3$ solutions extends uniquely and both give $e'-e, e''-e', e'''-e'', e+e'$. For all $n > 4$ the unique $A_{n-1}$ diagram extends uniquely, again with extra vector $e+e'$.

We can now obtain the impossibilty of the $E_6$ configuration by trying to extend either $D_5$ at a short end or $A_5$ at the middle vertex (or by trying to overlap $D_5$ with $A_5$ or with another $D_5$). Since $E_7$ and $E_8$ contain $E_6$, they are impossible too.

Hence none of the $E_n$ lattices are contained in any ${\bf Z}^m$, whence the corresponding quadratic forms are not sums of integral squares, QED.

Answer 3

If a quadratic form $x^TAx$ (for lattice $L$) in $n$ variables is expressible as a sum of $m (\ge n)$ squares of integral linear forms
then $x^TAx=||Fx||^2$, where $F$ is an $m \times n$ integral matrix and the columns of $F$
gives an explicit embedding of $L$ as a sublattice of $\mathbb{Z}^m$. For $E_8$ and $m=8$, this will mean
$E_8=\mathbb Z^8$ since both have the same volume, which is not possible. Also from $A=F^TF$ and the Cauchy-Binet formula,
$\det A$ is a sum of $m \choose n$ integral squares which are squared volume of the projections.
Since $\det E_8=1$, there can only be a single term 1 and the rest are zero and this means $E_8$ is embedded inside some
$\mathbb Z^8$ inside $\mathbb Z^m$ so this reduces to the case $m=8$. For $E_6,E_7$, the case $m=n$ can be ruled out since $\det A$ is
not a square as Robin Chapman noted. For $m>n$, since $2=1+1,6=1+1+4=1+1+1+1+1+1$ are the only partition of $\det A$ a a sum of squares, many projections have to be zero which means large $m$ can be reduced to smaller $m$. Can this be used to reduce to the case $m=n$ ?

Answer 4

Fairly recent work by Ellenberg and Venkatesh, later improved by Schulze-Pillot, show that it is reasonable to hope that $n+3$ squares of linear forms suffice. See Theorem 11, bottom of pdf page 8

http://arxiv.org/abs/0804.2158

Of the hypotheses involved, the more serious is that of sufficiently large minimum, as your quadratic forms have very small minima. Well, if you have favorite expressions for the $E_n$ quadratic forms, let me know, I can probably program something definitive up to $n+3.$