${\mathbb{R}}^n⋊G(n)$ is Algebraically Determined

Let $G$ be a Polish (i.e., complete separable metric topological) group. Define $G$ to be an algebraically determined Polish group if for any Polish group $L$ and algebraic isomorphism $\phi :L\mapsto G$, we have that $\phi$ is a topological isomorphism. Let $M(n,\mathbb{R})$ be the set of $n \times n$matriceswithrealcoefficientsandletthegroup$G$ in the above definition be the natural semidirect product ${\mathbb{R}}^n⋊G(n)$, where $n \ge 2$and$G(n)$ is one of the following groups: either the general linear group $GL(n,\mathbb{R})=\{A\in M(n,\mathbb{R})\mathrm{\mid }\det (A)\mathrm{\ne }0\}$, or the special linear group $SL(n,\R) = \left{ A \in GL(n,\R) \ | \ \det(A) = 1 \right}$,or$|SL(n,\R)| = \left{ A \in GL(n,\R) \ | \ |\det(A)| = 1 \right}$or$GL^{+}(n,\R) = \left{ A \in GL(n,\R) \ | \ \det(A) > 0 \right}$. These groups are of fundamental importance for linear algebra and geometry. The purpose of this paper is to prove that the natural semidirect product ${\mathbb{R}}^n⋊G(n)$ is an algebraically determined Polish group. Such a result is not true for ${\text{complexes}}^n⋊GL(n,\text{complexes})$ nor even for ${\mathbb{R}}^3⋊SO(3,\mathbb{R})$. The proof of this result is done in a sequence of steps designed to verify the hypotheses of the road map Theorem 2. A key intermediate result is that ${\phi }^{-1}(SO(n,\mathbb{R}))$ is an analytic subgroup of $L$ for every $n\ge 2$.