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A subset $X \subseteq \mathbb{R}^n$ is called semialgebraic if it is of the form $$ X = \bigcup_{finite} \bigcap_{finite} \{ x \in \mathbb{R}^n \colon f_{i,j}(x) \star 0 \} $$ where $\star$ represents any of the symbols $=, \leq , \geq, <, >$ and $f_{i,j} \in \mathbb{R}[X_{11}, \ldots , X_{nn}]$.

A subgroup $G < \operatorname{GL}(n,\mathbb{R}) \subseteq \mathbb{R}^{n\times n}$ is called a semialgebraic group if $G$ is a semialgebraic subset of $\mathbb{R}^{n\times n}$.

Is the following true?

Conjecture: A subgroup $G < \operatorname{GL}(n,\mathbb{R})$ is semialgebraic if and only if for every $p \in \mathbb{R}^n$, the orbit $G.p \subseteq \mathbb{R}^n$ is semialgebraic.

One direction follows from the Tarski-Seidenberg transfer principle: If $G$ is semialgebraic, orbits $$ G.p = \{x \in \mathbb{R}^n \colon \exists g \in G , x=g.p\} $$ are semialgebraic. What about the other direction?

Strichcoder
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1 Answers1

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Here is a counter-example. Consider the subgroup of $GL(2,{\mathbb R})$ consisting of matrices with rational determinants.

Another example is the following subgroup of $SO(4)$: Unitary matrices (elements of $U(2)$) whose complex determinants are roots of unity.

Conjecture. Suppose that $G<GL(n, {\mathbb R})$ is a subgroup with semialgebraic orbits. Then $G$ contains a semialgebraic subgroup with exactly the same orbits as $G$.

Moishe Kohan
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