Divisors, extensions of functions - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T17:14:28Z http://mathoverflow.net/feeds/question/4851 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/4851/divisors-extensions-of-functions Divisors, extensions of functions 1-- 2009-11-10T13:08:25Z 2010-01-02T07:04:37Z <p>This might be a silly/obvious question. I know that we have the removable singularity theorem (of Riemann) on the complex line, and we also have the generalization of this to algebraic curves. (Namely: if <img src="http://latex.mathoverflow.net/png?U" alt="U" title="" /> is an open subset of a Riemann surface and <img src="http://latex.mathoverflow.net/png?a%5Cin%20U" alt="a\in U" title="" />, and <img src="http://latex.mathoverflow.net/png?f%5Cin%5Cmathcal%7BO%7D%28U%2Da%29" alt="f\in\mathcal{O}(U-a)" title="" /> is bounded in some neighborhood of a, then f can be extended uniquely to a function <img src="http://latex.mathoverflow.net/png?F%5Cin%5Cmathcal%7BO%7D%28U%29" alt="F\in\mathcal{O}(U)" title="" />.) In particular, if a function on a curve vanishes over a divisor, we can extend over it uniquely.</p> <p>What do we know about extending of functions over divisors/hyperplanes in higher dimensions?</p> http://mathoverflow.net/questions/4851/divisors-extensions-of-functions/4853#4853 Answer by David Lehavi for Divisors, extensions of functions David Lehavi 2009-11-10T13:23:50Z 2009-11-10T13:23:50Z <p>I'm not sure I understand the question: extension is a local property, so if you can extend on C^n you can extend on any n-dimensional variety.</p> http://mathoverflow.net/questions/4851/divisors-extensions-of-functions/4858#4858 Answer by jvp for Divisors, extensions of functions jvp 2009-11-10T13:55:15Z 2009-11-10T13:55:15Z <p>The very same result holds in arbitrary dimensions: a locally bounded holomorphic function defined in the complement of a divisor extends. In many textbooks (like Gunning's) this is also called Riemmann's removable singularity Theorem. </p> <p>If you know more about your divisor you can do even better. For instance, if your ambient is a surface and you have a holomorphic function defined on the complement of a smooth and compact rational curve of self-intersection $-1$ then your function extends to the whole surface since the curve is contractible and after contraction you obtain a smooth surface. More generally, holomorphic functions defined on the complement of compact contractible divisors can also be extended if, after contraction, you obtain a normal variety. </p> http://mathoverflow.net/questions/4851/divisors-extensions-of-functions/4860#4860 Answer by David Speyer for Divisors, extensions of functions David Speyer 2009-11-10T14:04:52Z 2009-11-10T14:04:52Z <p>There is another very nice theorem: if $K \subset \mathbb{C}^n$ has complex codimension $\geq 2$ (real codimension $\geq 4$), and $f$ is an analytic function on $\mathbb{C}^n \setminus K$, then $f$ extends to $K$ -- no need for a boundedness hypothesis! I though this was called Hartog's theorem, but Wikipedia <a href="http://en.wikipedia.org/wiki/Hartogs%5Fextension%5Ftheorem" rel="nofollow">disagrees</a>.</p> http://mathoverflow.net/questions/4851/divisors-extensions-of-functions/10475#10475 Answer by Maharana for Divisors, extensions of functions Maharana 2010-01-02T07:04:37Z 2010-01-02T07:04:37Z <p>To add to the excellent answers of jvp and David Speyer above: </p> <p>Actually Hartog's theorem does even better: Given a domain $U\subset\mathbb{C}^{n{\geq{2}}}$ and a compact $K\subset{U}$ such that $U\setminus{K}$ is <strong>just connected</strong>, any holomorphic function on $U\setminus K$ extends holomorphically to $U$. Note that we don't need the codimension condition. This is actually a simple consequence of Cauchy's formula in several variables. </p> <p>For algebraic geometry (resp. complex geometry) and dimension $\geq{2}$: The analog of Hartog's theorem is that any regular (resp. holomorphic) function on the complement of an algebraic (resp. analytic) subset of codimension atleast $2$ in a normal algebraic(resp. analytic) variety, extends to the whole algebraic (resp. analytic) variety.</p> <p>I would also like to point out the important difference between the affine (resp. Stein) case and the projective case as perhaps alluded to by jvp above. If $X$ is a projective algebraic variety over $\mathbb{C}$ and you find a holomorphic function $f$ on the complement of a divisor $D$ then you will not be able to extend the function to all of $X$ however hard you try(!) since there are no global non-constant functions on projective varieties (similar statement for compact analytic vars). So to add to jvp's answer (which might seem like hair-splitting but is important IMHO): His second paragraph must refer to a non-projective neighborhood of the contractible curve that he discusses, otherwise the argument is still true but only vacuously! This is because there is no non-constant function to be found on a projective surface on the complement of a contractible curve and hence there is nothing to extend! </p> <p>Moreover, it will not be possible in general to extend holomorphic functions from the complement of divisors whose examples can be easily constructed. So there is no hope in this direction without any local boundedness hypothesis. </p>