Timeline for Determining if two algebraic sets are homeomorphic
Current License: CC BY-SA 2.5
18 events
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| May 22, 2010 at 2:36 | comment | added | Joel David Hamkins | Probably one can trace back to how the halting problem is encoded into the problem. The basic situation is that one can write down a specific Turing machine program $p$, such that ZFC proves that $p$ halts on input $0$ if and only if ZFC is inconsistent. Thus, if ZFC is consistent, then $p$ does not halt, but ZFC cannot prove this. This situation would get embedded into your homeomorphism problem via the undecidability result, and I expect one might extract a specific example this way. | |
| May 21, 2010 at 20:10 | vote | accept | Guy Katriel | ||
| May 21, 2010 at 20:09 | comment | added | Guy Katriel | Thanks Joel! So we know there exists a pair of polynomials for which we could never know whether their zero sets are homeomorphic! Is there any chance of finding such a pair and proving that the question of the homeomorphism of their zero sets is undecidable? | |
| May 21, 2010 at 19:05 | comment | added | Joel David Hamkins | Guy, regarding your last point. If all actual instances of pairs of rational polynomials either had a proof (in whatever formal system) that the solutions sets were homeomorphic or that they weren't homeomorphic, then there would be a decision procedure, namely, go and look for a proof one way or the other and output the corresponding answer. That is, if the problem is undecidable, then there is a pair of rational polynomials such that it is neither provable nor refutable in ZFC that the solution sets are homeomorphic. (This would be an existence proof that there is an explicit example.) | |
| May 21, 2010 at 18:54 | comment | added | Guy Katriel | Thank you for the enlightening responses. Assuming, as it seems, that the problem I stated is undecidable, does this mean that one could find two explicit polynomials with rational coefficients for which the question whether their zero sets are homeomorphic is an unsolvable one within the standard axiomatic framework of mathematics? | |
| May 21, 2010 at 18:33 | comment | added | David E Speyer | @Charles: Good point! I missed that example because I was looking to cut out the right radical ideal, at least locally. But, yes, on the level of sets of real points, $(0,0)$ is the "hypersurface" defined by $x^2+y^2=0$. | |
| May 21, 2010 at 18:11 | answer | added | algori | timeline score: 3 | |
| May 21, 2010 at 18:06 | comment | added | Charles Siegel | Oh, and regarding being a hypersurface, that's completely meaningless over a nonalgebraically closed field, because I can always make a polynomial $f(x_1,\ldots,x_k)$ such that for any $g_1,g_k$ in $t_1,\ldots,t_n$, we have $f(g_1(t),\ldots,g_k(t))=0$ if and only if all the $g$'s vanish, so every real variety is a "hypersurface"...also over any other nonclosed field. | |
| May 21, 2010 at 18:04 | comment | added | Charles Siegel | @Andy and David, by a version of the Weierstrass approximation theory, any manifold is homeomorphic to a real variety. Thus, real varieties can have any fundamental group a manifold can have, and are not restricted like Kahler groups are. | |
| May 21, 2010 at 17:58 | comment | added | David E Speyer | @Andy I don't claim to have a rigorous argument. However, real varieties can have much more flexible geometry than complex varieties. On the other hand, being hypersurfaces might impose all sorts of nontrivial conditions. | |
| May 21, 2010 at 16:47 | comment | added | Andy Putman | @David : I'm not sure that you could get any group like this. It is definitely known that there are strong restrictions on the fundamental groups of smooth projective varieties over C. Determining which groups can be fundamental groups of compact Kahler manifolds is an active field of study. | |
| May 21, 2010 at 16:26 | comment | added | Dan Piponi | @Guy To make sense of this problem for reals you can represent the real $x$ as a function acting as a Cauchy sequence $f:\mathbb{N}\rightarrow\mathbb{Q}$ such that $|f(i)-x|<2^{-i}$. With this representation, equality of reals is undecidable, so your problem certainly is. I'd stick with rational coefficients. | |
| May 21, 2010 at 14:05 | comment | added | David E Speyer | I should make a more basic point: if the coefficients of your polynomials are given in a computable form, such as rational numbers, then there is an algorithm to find a simplicial complex homeomorphic to your zero locus. In other words, we "know" what the geometry is. Unfortunately, deciding whether two simplicial complexes are homeomorphic is not computable. | |
| May 21, 2010 at 14:03 | comment | added | David E Speyer | My intuition is that the answer should be no, for the same reason that there is no algorithm to decide whether two manifolds are homeomorphic: I think I should be able to encode any finitely presented group as the fundamental group of a real algebraic variety, and come up with some trick so that, if the groups are isomorphic, the varieties are homeomorphic. (I don't know exactly what this trick would be.) But you make the problem harder by requiring me to use hypersurfaces, which is very restrictive. | |
| May 21, 2010 at 13:21 | answer | added | Joel David Hamkins | timeline score: 2 | |
| May 20, 2010 at 21:23 | comment | added | Guy Katriel | My question is about computability in principle. You're right about real numbers containing infinite data, so I don't mind restricting the question to polynomials with rational coefficients, and the model of computability being (say) a Turing machine. But perhaps the question with real coefficients can also make sense, for example maybe for polynomials of a given maximal degree of $p,q$ one can define some polynomial expression in the coefficients of $p,q$ which will be non-negative if and only if the zero sets of $p,q$ are homeomorphic. | |
| May 20, 2010 at 20:58 | comment | added | Joel David Hamkins | Do you seek a useful algorithm, or are you asking whether the problem is computable in principle? If the latter, then could you clarify what model of computability you have in mind? After all, the input should presumably include the real coefficients, an infinite object, and probably you don't expect to get the answer in finitely many steps of an oracle Turing machine. | |
| May 20, 2010 at 19:59 | history | asked | Guy Katriel | CC BY-SA 2.5 |