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Real semialgebraic sets are sets definable in the language of the reals: $(\mathbb{R},0,1,+,\cdot)$, which has as a definitional extension $(\mathbb{R},0,1,+,\cdot,\leq)$ by the useful fact that $a< b$ iff there exists $t\in\mathbb{R}$ such that $(b-a)t^2=1$.

For better or worse, the same trick does not work for the language of the complex numbers. Is there a name for the sorts of sets and functions definable over the complex numbers equipped with the absolute value symbol? I'm imagining something like definable sets over local fields, where the logical structure provides you a valuation symbol and an ordering on the valuation group. Have the complex version of these been studied at all?

KReiser
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    We can identify $\Bbb C^n$ with $\Bbb R^{2n}$. So, are your definable sets in $\Bbb C^n$ just the same as the semialgebraic sets in $\Bbb R^{2n}$? I think they might be, which would explain why they apparently haven't been studied. – Angina Seng Jun 07 '17 at 17:58

1 Answers1

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If you add $|\cdot|$ to the language of $\Bbb{C}$ and interpret it as absolute value, then the real line $\Bbb{R}$ becomes definable as $\{z \mid |z| = z \lor |z| = -z\}$ and you are looking at $\Bbb{C}$ as an $\Bbb{R}$-algebra. The definable sets in $\Bbb{C}^n$ then coincide with the semialgebraic subsets of $\Bbb{R}^{2n}$ (as Lord Shark surmised in his comment).

Rob Arthan
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    This works if $|\cdot|$ is a function $\mathbb{C}\to\mathbb{C}$ which always takes values in $\mathbb{R}$. But what if we look at a $2$-sorted structure $(\mathbb{C},\mathbb{R})$, equipped with the field structures on $\mathbb{C}$ and $\mathbb{R}$ and a function $|\cdot|\colon \mathbb{C}\to \mathbb{R}$? Can we still define $\mathbb{R}$ as a subset of $\mathbb{C}$? This seems to be closer to what the OP was interested in, since they mention valued fields, which are typically presented as two-sorted structures. – Alex Kruckman Jun 07 '17 at 19:56
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    Ah, yes: $\mathbb{R}\subseteq \mathbb{C}$ is defined by the formula $(|1 + x| = 1 + |x|) \lor (|1-x| = 1 + |x|)$. – Alex Kruckman Jun 07 '17 at 20:08
  • I was indeed imagining the scenario presented by @Alex Kruckman. Finding $\Bbb R$ in the way presented leads me to suspect that this answer is correct. I'd prefer to accept an edited version of this answer using the 2-sorts language, but if the two are equivalent, perhaps it doesn't matter that much. – KReiser Jun 07 '17 at 23:25
  • In the two-sorted language appropriate for $\Bbb{C}$ viewed as an algebra over $\Bbb{R}$, the real line is already definable as ${z : \Bbb{C} \mid \exists x : \Bbb{R}, z = x\cdot1}$. Or am I misunderstanding what you have in the signature for your two-sorted system? – Rob Arthan Jun 08 '17 at 17:40
  • @RobArthan in the two-sorted language I am envisioning, $\Bbb C$ is equipped with an absolute value in to $\Bbb R$, but there's no interaction of the two parts- you don't get to ask about $z\in\Bbb C$ such that $z=|z|$, for instance, since the element on the LHS is in $\Bbb C$ but the RHS is not. One other brief remark: is it clear how to define something like $p(Im(z),Re(z))\geq 0$ for $p$ a polynomial? I admit that finding $\Bbb R$ is possible, but I'm struggling to see how to pass to that- I'm probably missing something easy. – KReiser Jun 08 '17 at 22:22