MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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 $x-\omega^{-1}y=0$. – Anton Fonarev Nov 18 '11 at 0:13
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. – Anton Fonarev Nov 18 '11 at 0:18
The shadow of that point seems entirely well-behaved: it's a point, $[0,1]$ since $[\omega,\omega^2]=[1/\omega,1]$ with shadow $[0,1]$. – Will Sawin Nov 18 '11 at 1:55
Sorry, I was wrong in my first comment. Indeed, one should divide every equation by the biggest coefficient. – Anton Fonarev Nov 18 '11 at 8:53

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 $\vec u$.

share|cite|improve this answer
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. – Daryl Cooper Nov 21 '11 at 22:03

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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