In this question $\mathbb F$ is a field and $P({\mathbb F}^{n+1})$ is the projective space of dimension $n$ over $\mathbb F$. The term algebraic variety means a subset of $P({\mathbb F}^{n+1})$ which is the set of zeroes of finitely many homogeneous polynomials with coefficients in $\mathbb F$. The field $\mathbb K$ is the field of non-standard complex numbers (real and imaginary parts are hyperreal) and $\mathbb C$ is the subfield of complex numbers. If V is a variety in $P({\mathbb K}^{n+1})$, is the shadow of V a variety in $P({\mathbb C}^{n+1})$ ? A point is in the shadow of V means it has some projective coordinates in ${\mathbb C}^{n+1}$ infinitesimally close to some projective coordinates in ${\mathbb K}^{n+1}$ of some point in V.
-
$\begingroup$ It is not always true: take some infinite $\omega\in\mathrm{F}$ (a hyperreal number, s.t. $n<\omega$ for any $n\in\mathbb{N}$). And the point [$\omega\ $:$\omega^2$] defined in $\mathbb{FP}^2$ by the obvious equation $x-\omega^{-1}y=0$. $\endgroup$– Anton FonarevCommented Nov 18, 2011 at 0:13
-
$\begingroup$ I'd say that the proper question is whether such a projective variety is a shadow in the case where all the coefficients are finite and not infinitesimally small. Then the answer is affirmative. $\endgroup$– Anton FonarevCommented Nov 18, 2011 at 0:18
-
2$\begingroup$ The shadow of that point seems entirely well-behaved: it's a point, $[0,1]$ since $[\omega,\omega^2]=[1/\omega,1]$ with shadow $[0,1]$. $\endgroup$– Will SawinCommented Nov 18, 2011 at 1:55
-
$\begingroup$ Sorry, I was wrong in my first comment. Indeed, one should divide every equation by the biggest coefficient. $\endgroup$– Anton FonarevCommented Nov 18, 2011 at 8:53
1 Answer
The answer is "yes" if $V$ is the projective zero-set $Z(f)$ of a single homogeneous polynomial $f$. Perhaps an expert in several complex variables will provide a more general answer.
Without loss of generality, every coefficient of $f$ is finite (i.e., they all have modulus less than some standard $R$) and at least one coefficient is not infinitesimal. Let $st(f)$ be the "standard part" of $f$ obtained by replacing every coefficient with its standard part. Let $W$ be the shadow of $V$. I claim that $Z(st(f))=W$ where $Z(st(f))$ is the set of standard projective zeroes of $st(f)$.
The easy half, which generalizes to all algebraic varieties, is that $W\subseteq Z(st(f))$. Let $x\approx y$ iff $\lvert x-y\rvert$ is infinitesimal. If $[\vec w]\in W$ and (without loss of generality) all coordinates of $\vec w$ are finite, then $f(w)\approx 0$ by continuity of $f$, so $st(f)(\vec w)\approx 0$, so $st(f)(\vec w)=0$ because $st(f)(\vec w)$ is standard.
To prove the other half, suppose $[\vec u]\in Z(st(f))$. Choose a standard unit vector $\vec e$ such that $h(z)=st(f)(\vec u+z\vec e)$ is not constant on any open disc $D_r=\{z\in\mathbb{K}:\lvert z\rvert< r\}$ where $r>0$ is standard. Since $h$ has only finitely many roots, we may choose an arbitrarily small standard $r$ such that $h$ has no root on $\partial D_r$. Let $M$ be the (necessarily standard) minimum modulus of the $h$-image of $\partial D_r$. Setting $p(z)=f(\vec u+z\vec e)$, we have $p(z)\approx h(z)$ for all $z\in\mathbb{K}$. Hence, $\lvert p(z)-h(z)\rvert< M\leq\lvert h(z)\rvert$ for all $z\in\partial D_r$. By Rouche's Theorem, $h$ and $p$ have the same number of roots (counting multiplicities) in $D_r$. By overspill, $h$ and $p$ have the same number of roots in $D_\delta$ for some positive infinitesimal $\delta$. In particular, $p$ has an infinitesimal root, so $f$ has a root infinitely close to $\vec u$.
-
$\begingroup$ I agree. It is also true for subvarieties defined by linear equations: projective subspaces. Maybe the general case follows from Bishop's theorem (MR0168801) "Conditions for the analyticity of certain sets" I am hoping someone out there knows the answer. $\endgroup$ Commented Nov 21, 2011 at 22:03