Misha Verbitsky Голосов: 1 Адрес блога: http://lj.rossia.org/users/tiphareth/ Добавлен: 2008-01-02 18:18:22 блограйдером Robin_Bad

степень формы рациональна

2014-08-18 12:32:27 (читать в оригинале)

Доказал такую простую штуку: если какая-то степень
симметрической или симплектической формы рациональна,
то она рациональна. Положу сюда, чтобы не потерялось.

\subsection{Rational symmetric and skew-symmetric forms}

For the purpose of this paper,
{\bf a rational tensor} on $\R^n$ is a tensor $\Theta$ such that
$c\Theta$ is rational, for some $c\in \R$.
Further on, we shall need the following two
propositions.

\hfill

\proposition
Let $V=\R^n$ be a vector space, and $g\in \Sym^2 V^*$ a bilinear symmetric
form on $V$ such that $g^k\in \Sym^{2k}V^*$ is rational. Then
$g$ is rational.

\hfill

{\bf Proof:} Let $x\in V$ be a vector such that $g(x,x)\neq 0$.
Consider the map $R_g:\; V \arrow \R$ mapping $y$ to $g^k(x,x,x, ..., x, y)$.
Clearly, $x^\bot=\ker R_g$. Therefore, for each rational $x$,
its orthogonal complement $x^\bot$ is also rational.
Gramm-Schmidt orthogonalization method allows one to
construct a rational orthogonal basis $x_i$ for $g$.
Modifying $g$ by a constant multiplier, we may assume that
$g(x_1, x_1)=\pm 1$. Then $g^k(x_i, x_i, x_1, ..., x_1)=g(x_i, x_i)$,
and this number must be rational, because $g^k(x_1, ..., x_1)$
is rational. \endproof

\hfill

\proposition
Let $V=\R^{2n}$ be a vector space, and $g\in \Lambda^2 V^*$
a bilinear, non-degenerate skew-symmetric
form on $V$ such that $\omega^k\in \Lambda^{2k}V^*$ is rational,
for some $k<n$. Then $\omega$ is rational. \hfill {\bf Proof. Step 1:} Assume that $k= n-1$. Then the map $v, v' \arrow \frac{\omega^k\wedge v \wedge v'}{\omega^n}$ defines a non-degenerate form on $V^*$, dual to $\omega$. Since this form is rational, $\omega$ is also rational. {\bf Step 2:} Now assume that $1< k< n-1$. Fix two rational vectors $x, x'$ such that $\omega(x, x')\neq 0$. Rescaling $\omega$, we may assume that $\omega(x, x')=1$. Then for any $k+1$-dimensional subspace $W\subset V$ containing $x, x'$ and such that $\omega\restrict W$ is non-degenerate, $\omega\retsrict W$ is rational and takes rational values on all rational vectors. Since any two vectors $y, y'\in V$ are contained in appropriate $W$, this implies that $\omega(y, y')$ is rational. \endproof

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Тэги: math