11
$\begingroup$

I'm reading some very old papers (by Birch et al) on quadratic forms and I don't get the following point:

If $f$ is a quadratic form in $X_1,X_2,\cdots,X_n$ over a finite field, then one can change variables such that $f$ can be written as $\sum_{i = 1}^s Y_{2i - 1}Y_{2i} + g$, where $g$ is a quadratic form which involves variables other than $Y_1,Y_2,\cdots,Y_{2s}$ and has order at most 2 (i.e. can be written using at most two linear forms).

So either this is a well-known result - but I don't find a reference - or either this is easy to see, but in that case I'm just missing the point. By the way, is this really true in characteristic 2?

(And in fact I'm not sure what role is played by the fact that the field is finite...)

Improve this question
$\endgroup$
7
  • $\begingroup$ "has order at most 2 (i.e. can be written using at most two linear forms)." What does this mean? $\endgroup$
    – darij grinberg
    Mar 4, 2010 at 16:39
  • $\begingroup$ Ok, I'll give a very precise definition - I thought it would be a well known term. If $f$ is a quadratic form over $k$ in $X_1,X_2,\cdots,X_n$, then we define $\gamma(f)$ as the number of variables which appear effectively in $f$, i.e. with a non-zero coefficient. For any $(n \times n)$-matrix $T$ over $k$, we have a quadratic form $f_T$ which maps $x \in k^n$ to $f(Tx)$. We then define the order $o(f)$ to be $\inf_T \gamma(f_T)$ where $T$ ranges over all non-singular transformations. So, "$f$ can be written using $o(f)$ linear forms in $X_1,X_2,\cdots,X_n$, but not fewer". $\endgroup$
    – Wanderer
    Mar 4, 2010 at 16:45
  • $\begingroup$ Ah, okay. I thought o(f) was called "rank" of the quadratic form. $\endgroup$
    – darij grinberg
    Mar 4, 2010 at 17:46
  • $\begingroup$ As for your question why the field should be finite: over the reals, $X_1^2+X_2^2+...+X_n^2$ would be a counterexample, so at least we need positive characteristic. $\endgroup$
    – darij grinberg
    Mar 4, 2010 at 17:46
  • 4
    $\begingroup$ For the case of characteristic not 2, see Serre, Course in arithmetic, section IV.1.7. I think you can find the case of characteristic 2 in the book Milnor & Husemoller, Symmetric bilinear forms. $\endgroup$
    – Bjorn Poonen
    Mar 4, 2010 at 18:20

2 Answers 2

Reset to default
10
$\begingroup$

I will sketch below a standard argument to show what you need, because I find it very neat!

Let $V$ be a finite dimensional vector space over a field $k$ and let $q \colon V \to k$ be a quadratic form on $V$. Denote by $b$ the symmetric bilinear form associated to $q$: thus for vectors $v,w \in V$ define $b(v,w) := q(v+w)-q(v)-q(w)$. Suppose that $v$ is a non-singular zero of $q$. Since $v$ is non-singular, it follows that there is a $w' \in W$ such that $\alpha := b (v , w') \neq 0$. Let $w := \frac{1}{\alpha^2} (\alpha w' - q(w') v)$; it is immediate to check that $q(w)=0$ and $b (v , w) = 1$. Observe that the ``orthogonal complement'' of $v,w$ with respect to the form $q$ has codimension two and does not contain the span of $v,w$. Thus, we conclude that we can find a basis of $V$ such that $q(x_1,\ldots,x_n) = x_1 x_2 + q'$, where $q'$ is a quadratic form over a space of dimension two less than the dimension of $V$.

The statement about finite fields follows at once, since over a finite field, any quadratic form in three or more variables admits a non-trivial zero. This is a consequence of the Chevalley-Warning Theorem. More generally, any field such that quadratic forms in three variables always admit a zero has the property you need, e.g. any $C_1$-field would work.

Improve this answer
$\endgroup$
5
  • $\begingroup$ Thanks. I guess I've seen you giving a talk about "grosses surfaces rationnelles" earlier this year in Paris, by the way. $\endgroup$
    – Wanderer
    Mar 4, 2010 at 20:01
  • $\begingroup$ Must've been me! Nice to meet you, d $\endgroup$
    – damiano
    Mar 4, 2010 at 20:24
  • $\begingroup$ By the way, don't you need a statement in your last paragraph which is slightly (but only a very little bit) stronger to make your argument work? I mean the fact that every quadratic form of order at least three has a non-singular zero (and not just a non-trivial zero). $\endgroup$
    – Wanderer
    Mar 4, 2010 at 20:51
  • $\begingroup$ It's ok, unless $k$ is imperfect of char 2. Let $R$ be the radical of $q$ and $W$ a complementary subspace. If $q|_R$ is non-zero, then ${\rm char}(k)=2$ and $q|_R=\sum a_ix_i^2$; if $k$ is perfect, this is the square of a linear form and hence $\dim(W)\geq o(q)-1$. Suppose that $k$ is a finite field and that $o(q)>2$. Then either $\dim(W)>2$, or $\dim(W)=2$ and $q$ is of the form $q'+ax^2$, with $a \in k^\times$ and $q'$ a form with $o(q')=2$. In either case there is a three dimensional subspace of $V$ where $q$ defines a smooth conic, and we conclude by the Chevalley-Warning Theorem. $\endgroup$
    – damiano
    Mar 5, 2010 at 10:27
  • $\begingroup$ Note that if $k$ has characteristic two and $a_1,\ldots,a_n$ are elements of $k^\times$ that are independent in $k^\times/(k^\times)^2$, then the quadratic form $\sum a_i x_i^2$ is not of the form that you mention as soon as $n \geq 3$. $\endgroup$
    – damiano
    Mar 5, 2010 at 10:50
8
$\begingroup$

Let me try a similar explanation with different words. (Note that my explanation does not cover characteristic $2$.)

A quadratic form is nondegenerate if any of its associated symmetric matrices has nonzero determinant. (Alternately, if the associated bilinear form $B(x,y) = q(x+y) - q(x) - q(y)$ is nondegenerate in the usual sense: $B(x,y) = 0 \ \forall y \in K \implies x = 0$.)

Let $K$ be a field of characteristic different from $2$. The hyperbolic plane is the special quadratic form

H(x,y) = xy.

(As with any quadratic form over $K$, it can be diagonalized: $\frac{1}{2} x^2 - \frac{1}{2} y^2$.)

A nondegenerate quadratic form $q(x_1,\ldots,x_n)$ is isotropic if there exist $a_1,\ldots,a_n \in K$, not all $0$, such that $q(a_1,\ldots,a_n) = 0$ and otherwise anisotropic.

Witt Decomposition Theorem: Any quadratic form $q$ can be written as an orthogonal direct sum of an identically zero quadratic form, an anistropic quadratic form, and some number of hyperbolic planes. In particular, any isotropic quadratic form $q(x_1,...,x_n)$ can be written, after a linear change of variables, as $x_1 x_2 + q(x_3,...,x_n)$.

For your purposes, you might as well assume your quadratic form is nondegenerate -- otherwise, it simply involves more variables than actually appear!

Now over a finite field, the Chevalley-Warning theorem implies that any nondegenerate quadratic form in at least three variables is isotropic, so that by Witt Decomposition, you can split off a hyperbolic plane. If you still have at least three variables, you can do this again. Repeated application gives your result.

References:

For Chevalley-Warning:

http://alpha.math.uga.edu/~pete/4400ChevalleyWarning.pdf

For Witt Decomposition:

http://alpha.math.uga.edu/~pete/quadraticforms.pdf

Improve this answer
$\endgroup$
0

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.