Zariski closed sets in C^n are of measure 0 - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-21T03:20:53Z http://mathoverflow.net/feeds/question/25513 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/25513/zariski-closed-sets-in-cn-are-of-measure-0 Zariski closed sets in C^n are of measure 0 Akela 2010-05-21T17:48:31Z 2010-05-22T18:01:54Z <p>This is related to <a href="http://mathoverflow.net/questions/15569" rel="nofollow">another question</a> in which it is proved that Zariski open sets are dense in analytic topology.</p> <p>But it is intuitive that something more is true. Namely, that they are the sets where some polynomials vanish, and consideration of a few examples in $\mathbb R^n$ where they are of Lebesgue measure $0$, suggest strongly that the Zariski-closed sets(except the whole affine space) are of measure $0$ in $\mathbb C^n$ as well. This should be quite simple; but I am unable to prove it due to inexperience in measure theory.</p> <p>The nice thing about proving this is that once this is done, then we are able to claim safely that so-and-so statement is true almost everywhere, if it is true on a Zariski-open set.</p> <p>So, in a more measure theoretic formulation:</p> <blockquote> <p>Let $X$ be a set in $\mathbb C^n$ contained in the zero locus of some collection of polynomials. How to show that $X$ is of measure $0$?</p> </blockquote> <p>In fact my feeling is that more should be true, ie, we can replace polynomials by analytic functions at least, and get the same result.</p> http://mathoverflow.net/questions/25513/zariski-closed-sets-in-cn-are-of-measure-0/25516#25516 Answer by Henri for Zariski closed sets in C^n are of measure 0 Henri 2010-05-21T18:04:41Z 2010-05-21T18:04:41Z <p>Clearly, it is sufficient too show it for a closed set given by $f=0$ where $f$ is analytic. (write your set as included in a countable union of such described sets). Then, using the normal form of analytic germs as finite ramified coverings, you're done.</p> http://mathoverflow.net/questions/25513/zariski-closed-sets-in-cn-are-of-measure-0/25519#25519 Answer by Sergei Ivanov for Zariski closed sets in C^n are of measure 0 Sergei Ivanov 2010-05-21T18:16:55Z 2010-05-21T18:38:02Z <p>If a real analytic function $f:U\subset\mathbb R^n\to\mathbb R^m$ is zero on a set $Z$ of positive measure (and $U$ is connected), then $f\equiv 0$.</p> <p>Indeed, almost every point of $Z$ is a <a href="http://en.wikipedia.org/wiki/Lebesgue%2527s_density_theorem" rel="nofollow">density point</a>. It is easy to see that the derivative at a density point is zero. Therefore $df=0$ a.e. on $Z$. Applying the same argument to $df$, conclude that the second derivative vanishes a.e. on $Z$ too. And so on. Thus $f$ has zero Taylor expansion at some point, hence $f\equiv 0$.</p> http://mathoverflow.net/questions/25513/zariski-closed-sets-in-cn-are-of-measure-0/25586#25586 Answer by Robin Chapman for Zariski closed sets in C^n are of measure 0 Robin Chapman 2010-05-22T15:19:11Z 2010-05-22T15:19:11Z <p>There is a very naive argument for this. As Henri says, it reduces to a zero set of a polynomial $f$. Write $$f(z_1,\ldots,z_n)=\sum_{j=0}^d g_j(z_1,\ldots,z_{n-1})z_n^j.$$ where the polynomial $g_d$ is not identically zero. For each $(z_1,\ldots,z_{n-1})\in\mathbb{C}^n$, there are only finitely many $z_n$ with $f(z_1,\ldots,z_n)=0$ unless $g_d(z_1,\ldots,z_{n-1})=0$. Inductively these exceptional $(n-1)$-tuples form a set of measure zero in $\mathbb{C}^{n-1}$ and now the result follows from Fubini's theorem (regarding $\mathbb{C}^n$ as $\mathbb{R}^{2n}$ and going down two real dimensions).</p> http://mathoverflow.net/questions/25513/zariski-closed-sets-in-cn-are-of-measure-0/25594#25594 Answer by Michael Greenblatt for Zariski closed sets in C^n are of measure 0 Michael Greenblatt 2010-05-22T18:01:54Z 2010-05-22T18:01:54Z <p>And if you want to take this to an extreme... for a function on a domain in ${\bf R}^n$, it's enough to assume that at every $x$ there's a ball $B_x$ centered at $x$ and a multiindex $\alpha$ for which $\partial^{\alpha} f$ is nonzero and continuous on $B_x$. </p> <p>To see this, first note that it suffices to show that the zeroes of $f$ in a given $B_x$ have measure zero. This is proven by induction on $|\alpha|$. If $\alpha = 0$ it's trivial, and if $\partial^{\alpha'}f(x) \neq 0$ for any $\alpha '$ with $|\alpha '| &lt; |\alpha|$ it follows by the induction hypothesis, shrinking $B_x$ if necessary. Otherwise we can write $\partial^{\alpha} f = \partial_{x_i}\partial^{\beta} f$ for some $\beta$, where we can assume $\partial^{\beta} f(x) = 0$. By the implicit function theorem, if $B_x$ is small enough the zeroes of $\partial^{\beta} f$ in $B_x$ form a $C^1$ hypersurface with measure zero. For each $y$ off this surface, $\partial^{\beta} f$ is nonzero and then you can apply the inductive hypothesis to find an appropriate $B_y$. A simple compactness argument shows you need only countably many $B_y$.. so you're done. </p>