I am having trouble parsing through this question about Nash's embedding theorem. My understanding of Riemannian geometry is unfortunately not strong enough to figure out an answer.
I am dealing with a problem involving a matrix Lie group $G$. Matrix Lie groups, being closed (in $GL(n)$) matrix subgroups, are clearly not closed in the ambient Euclidean space $\mathbb{R}^{n \times n}$. This is because $GL(n)$ is not closed.
I was hoping if there exists an isometric embedding $\iota: G \to \mathbb{R}^N$ such that the image $\iota(G)$ is closed. It seems that if $G$ is compact, then yes such an embedding exists (correct me if I am wrong). I was curious if this also holds for non-compact Lie groups?
A commonly used metric, and the one I using, with matrix Lie groups takes the form $\langle X_g , Y_g \rangle_g = \text{tr}\left[(g^{-1}X_g)^T (g^{-1}Y_g)\right]$, where $X_g, Y_g \in T_gG$.
I was hoping that due to the nice properties of Lie groups, there would exist such an embedding. Could someone help me find the answer to this question?
This paper states that if $(M,g)$ is a complete $n$-dimensional Riemannian manifold, then there is a closed isometric $C^\infty$-embedding of $(M,g)$ into some Euclidean space. Does this answer my question in the affirmative?
With help from Suárez-Álvarez, the answer to this question is in the affirmative. This paper proved if $(M,g)$ is a complete $n$-dimensional Riemannian manifold, then there is a closed isometric smooth embedding of $(M,g)$ into some Euclidean space $\mathbb{E}$. By closed, the authors means the image of any closed set in $G$ is closed in $\mathbb{E}$. Lie groups equipped with the metric I described in the post are always complete.