9
$\begingroup$

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...)

$\endgroup$
1
  • $\begingroup$ In fact, I think $E_n$ is not a sum of ANY number of squares of integral linear forms. $\endgroup$
    – VA.
    Sep 30, 2010 at 16:05

4 Answers 4

9
$\begingroup$

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 $n$.

$\endgroup$
4
  • 1
    $\begingroup$ This does not rule out that $E_n$ is a sum of $>n$ squares. But it's a good observation. $\endgroup$
    – VA.
    Sep 30, 2010 at 16:09
  • $\begingroup$ To amplify VA's remark, this argument "rules out" most of $A_n$ as well, but of course the corresponding quadratic forms can be represented as sums of $n+1$ squares, per the original question. $\endgroup$ Sep 30, 2010 at 16:39
  • $\begingroup$ For simply laced root systems the "eutactic" property you need is the corollary to Theorem 1 of paragraph 6 of chapter 5 of Bourbaki's "Lie Groups and Lie Algebras". The 30 that appears is, of course, the Coxeter number of $E_8$. $\endgroup$ Oct 1, 2010 at 9:12
  • $\begingroup$ Robin, thank you for your answer. I think I can prove all my statements quite easily by a direct combinatorial argument, working with the Dynkin graph. But you give a nice argument. $\endgroup$
    – VA.
    Oct 1, 2010 at 15:22
5
$\begingroup$

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.

$\endgroup$
1
  • 1
    $\begingroup$ I just came across this question. It was Mordell who proved that $E_6$ is not represented by any sum of squares. See "The representation of a definite quadratic form as a sum of two others’, Ann. Math. 38 (1937), 751–757. $\endgroup$
    – WKC
    May 31, 2015 at 6:10
4
$\begingroup$

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 $m=n$ ?

$\endgroup$
1
  • $\begingroup$ The argument shows that an unimodular lattice not a $\ZZ^n$ cannot be embedded in any $\ZZ^m$ ie. its quadratic form cannot be expressed as a sum of squares of any number of linear forms $\endgroup$
    – Chua KS
    Jul 18, 2011 at 2:27
3
$\begingroup$

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 $n+3.$

$\endgroup$
2
  • $\begingroup$ Although the paper of Ellenberg-Venkatesh you quoted is really interesting, I don't see how to easily apply it: it only applies to forms with large minima, plus there are local conditions to check. I actually think it does not apply at all, since neither of $A_n,D_n,E_n$ is a sum of $\ge n+2$ squares. In any case, I can prove the statement easily, so my questions were (1) standard reference or a 2-line proof? and (2) did this simple characterization of $A_n,D_n,E_n$ show up in relation with something interesting? $\endgroup$
    – VA.
    Sep 30, 2010 at 20:09
  • 1
    $\begingroup$ Dear VA, I see what you mean, about $\geq n+2.$ I found nice Gram matrices for $E_6, E_7, E_8$ on Sloane's website, I am beginning to see, by hand, that they may not be sums of squares by pigeonhole arguments, anyway math.rwth-aachen.de/~Gabriele.Nebe/LATTICES/E6.html math.rwth-aachen.de/~Gabriele.Nebe/LATTICES/E7.html math.rwth-aachen.de/~Gabriele.Nebe/LATTICES/E8.b.html $\endgroup$
    – Will Jagy
    Sep 30, 2010 at 20:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.