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The $\ell^p$ unit sphere $\{x\in\Bbb R^n\mid |x_1|^p+\cdots+|x_n|^p=1\}$ with $p\in\Bbb N$ is a semi-algebraic set, and its polar dual is

$$(*)\quad \{x\in\Bbb R^n\mid |x_1|^q+\cdots +|x_n|^q=1\},$$

with $q=p/(p-1)$. Using Tarski-Seidenberg we see that the polar dual of a semi-algebraic set is semi-algebraic (see e.g. the discussion here). But if $p\ge 3$ then $q\not\in\Bbb N$, hence $(*)$ is not a semi-algebraic description.

Question: Is there a nice semi-algebraic description of $(*)$?

I am mostly interested in $n=2$.

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    $\begingroup$ Here is an easy, semi-algebraic, not-so-nice description: Let $y=\sum |x_i|^q$. Let basic terms be those of the form $\prod |x_i|^{a_i/p}$ where all $a_i$ are integers in $[0,p-1]$. Then for any integer $j>0$, the power $y^j$ is a $\mathbb{Z}[x_1,\ldots,x_n]$-linear combination of basic terms. Setting $y^j=1$ for $1\le j\le p^n$ gives $p^n$ equations in the basic terms. Solving them gives each basic term as an element of $\mathbb{Z}(x_1,\ldots,x_n)$, and thus gives $y$ in the same form. This way of writing $y$ makes $y \le 1$ a semialgebraic equation, as desired. $\endgroup$
    – user44143
    Commented Nov 25, 2022 at 22:08
  • $\begingroup$ @MattF. Great, thank you. I would also accept this as an answer since the procedure is nice (if no one comes up with a "nicer" descirption). $\endgroup$
    – M. Winter
    Commented Nov 25, 2022 at 22:28
  • $\begingroup$ In the above, I should have said $2\le j \le p^{n}+1$, since that should make $y\le 1$ a non-trivial semialgebraic equation. E.g. for $n=2$, $p=3$, $|x_1|^{3/2}+|x_2|^{3/2}\le 1$, the algorithm as in the previous comment gives the inequality $1\le 1$, and this adjustment gives an inequality which can be written as $4|x_1|^3 |x_2|^3\le (|x_1|^3+|x_2|^3-1)^2$. $\endgroup$
    – user44143
    Commented Apr 16, 2023 at 6:00

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