Timeline for Complex semi-algebraic sets
Current License: CC BY-SA 4.0
32 events
| when toggle format | what | by | license | comment | |
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| Aug 11, 2019 at 18:44 | answer | added | Todd Trimble | timeline score: 1 | |
| Aug 11, 2019 at 17:13 | comment | added | Pietro Paparella | @ToddTrimble: I’m sure those are great questions you’re asking, but, alas, I don’t have the background in semi-algebraic geometry/logic to answer them. | |
| Aug 11, 2019 at 17:07 | comment | added | Todd Trimble | There are still some loose threads which have not been entirely sewn up. Following up on YCor's question on definability constraints: the collection of semi-algebraic sets can also be defined so that by definition it is closed under taking images along projection maps (and then Tarski-Seidenberg assures us we could do with less). Should I assume you don't allow that, i.e., "definability" here excludes existential quantification? Also, would you allow the modulus map to be seen as a function $\mathbb{C} \to \mathbb{C}$, i.e., as the interpretation of an unary function symbol in the signature? | |
| Aug 10, 2019 at 19:32 | comment | added | Dima Pasechnik | please see my answer regarding MR2399570 below. Hope it clarifies. | |
| Aug 10, 2019 at 19:31 | answer | added | Dima Pasechnik | timeline score: 1 | |
| Aug 10, 2019 at 14:28 | comment | added | Pietro Paparella | @DimaPasechnik: 1) as quoted above, Bharali ad Holtz asset that $\mathbb{L}^n$ is “semi-algebraic” in the real sense; 2) they never defined complex semi-algebraic set in the manner in which I did above; 3) the conclusion about polynomial inequalities is, given the example I gave above, at the very least problematic; 4) their proof sketch implies my definition and therefore the problems in the three items above abound. | |
| Aug 10, 2019 at 12:28 | comment | added | Dima Pasechnik | you never explained what in your opinion is wrong with MR2399570. It is certainly true that $\mathbb{L}^n=\mathbf{T}^{-1}(S)$ for $S$ a semialgebraic set. | |
| Aug 10, 2019 at 7:08 | comment | added | Pietro Paparella | @DimaPasechnik: yes. I work on the NIEP and am one of the world’s experts on the problem. Hence my original post. | |
| Aug 10, 2019 at 6:20 | comment | added | Dima Pasechnik | One further point regarding NIEP is that there the complex semialgebraic set in question is invariant under conjugation. | |
| Aug 9, 2019 at 21:38 | comment | added | Pietro Paparella | @DimaPasechnik: your comment hints at points to the crux of my question: what do polynomial inequalities in $\Re x_k$, $\Im x_k$ ($1 \le k \le n$) imply about the complex variables $x_1,\dots,x_n$? | |
| Aug 9, 2019 at 21:14 | comment | added | Dima Pasechnik | What's wrong with inequalities in $\Re x_k, \Im x_k$ ? They do describe the $x_k$'s. And showing that the solutions to NIEP form a semialgebraic set is easy, one needs to recall that a projection of a semialgebraic set is semialgebraic. | |
| Aug 9, 2019 at 15:59 | comment | added | Pietro Paparella | @DimaPasechnik: it may be possible that their conclusion is correct, but their argument that leads to the conclusion is not. It is clear that there must be polynomial inequalities in $\Re x_1, \Im x_1,\dots,\Re x_n, \Im x_n$, but what is desired are equations and/or inequalities in $x_1, \dots, x_n$. | |
| Aug 9, 2019 at 9:32 | comment | added | Dima Pasechnik | Do you disagree with what [MR2399570] says in the place you quoted in boldface? I suppose they mean the usual identification of $\mathbb{C}$ as $\mathbb{R}\oplus \sqrt{-1}\mathbb{R}$, and everything works as claimed. | |
| Aug 9, 2019 at 7:43 | comment | added | Ben McKay | @PietroPaparella: I don't know if such a notion has been defined before; I was using the term "complex semi-algebraic set" as you defined it in your question. | |
| Aug 8, 2019 at 22:46 | history | edited | Pietro Paparella | CC BY-SA 4.0 |
added 2 characters in body
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| Aug 8, 2019 at 22:14 | comment | added | Pietro Paparella | @DimaPasechnik: I am not forbidding anything; I simply want to know, with the definition above, how one can define a complex semi-algebraic set without appealing to the mapping above. | |
| Aug 8, 2019 at 22:01 | comment | added | Pietro Paparella | @BenMcKay: your question pre-supposes the concept of a complex semi-algebraic set – has such a notion been defined already? | |
| Aug 8, 2019 at 17:29 | history | edited | Pietro Paparella | CC BY-SA 4.0 |
edited definition
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| Aug 8, 2019 at 17:20 | history | edited | Pietro Paparella | CC BY-SA 4.0 |
added context to the post
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| Aug 8, 2019 at 9:19 | comment | added | Ben McKay | Are you asking if every complex semialgebraic set is the preimage of a real semialgebraic set via the map $z\mapsto(|z_1|,\dots,|z_n|)$? | |
| Aug 8, 2019 at 6:31 | comment | added | Dima Pasechnik | Are you allowing equations, e.g. $|z|-z=0$, too? Anyway, I have trouble trying to describe the positive ortant in $\mathbb{C}$ using the norm. | |
| Aug 8, 2019 at 3:44 | comment | added | Pietro Paparella | @ZachTeitler: no, this was an oversight on my part. I have edited the post to reflect your comment. | |
| Aug 8, 2019 at 3:43 | history | edited | Pietro Paparella | CC BY-SA 4.0 |
better definition for SA sets
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| Aug 8, 2019 at 2:18 | comment | added | Zach Teitler | That's not the usual definition of semialgebraic set. Usually they are taken to be a finite union of sets $P_i$ which are each defined by one or more polynomial equalities and inequalities. Are you restricting to semialgebraic sets of a special form? | |
| Aug 8, 2019 at 1:29 | comment | added | Pietro Paparella | @Qfwfq: pertaining to your third comment, I would accept that definition so long as it agrees with the definition above, but this seems difficult to prove. | |
| Aug 8, 2019 at 1:27 | comment | added | Pietro Paparella | @Qfwfq: pertaining to your first comment, the set of real numbers is not a subset of the set of complex numbers. One may "view" $\mathbb{R}$ as a subset of $\mathbb{C}$ given that the mapping $x \in \mathbb{R} \longmapsto x+0i$ is a field isomorphism. Thus, it is more accurate to say that, under the definition above, the real-axis in the complex plane is complex algebraic since the x-axis is semi-algebraic in $\mathbb{R}^2$. Thus, it is inaccurate to conclude that "$\mathbb{R}$ is complex algebraic" because it is not a subset of $\mathbb{C}$. | |
| Aug 8, 2019 at 0:46 | comment | added | Qfwfq | By the way (and sorry for the many questions), would it be interesting to define a notion of complex semi-algebraic set by some $|p_i(z)|\geq c$ for complex $p_i$ and real $c$, instead? | |
| Aug 8, 2019 at 0:43 | comment | added | Qfwfq | Also, you obtain all the (real) semi-algebraic subsets of $\mathbb{R}^{2n}$ in this way. So you're asking if every real semi-algebraic subset of $\mathbb{R}^{2n}$ can be expressed in some way in terms of the norm on $\mathbb{C}^n$, correct? | |
| Aug 8, 2019 at 0:36 | comment | added | Qfwfq | Then $\mathbb{R}\subset\mathbb{C}$ would be a "complex semi-algebraic" subset according to your definition, and also of course a real algebraic subset of odd dimension. Are you ok with allowing this? | |
| Aug 7, 2019 at 20:57 | comment | added | Pietro Paparella | @YCor: I make no requirements on the inequalities; I mentioned modulus because inequalities involving complex numbers are not possible without it. | |
| Aug 7, 2019 at 20:55 | comment | added | YCor | Could you be more precise (than "involving the modulus") on what you allow and what you don't allow to define subsets? | |
| Aug 7, 2019 at 20:04 | history | asked | Pietro Paparella | CC BY-SA 4.0 |