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Let $GL_n^+$ be the group of $n \times n$ real invertible matrices with positive determinant.

Let $g$ be the left-invariant Riemannian metric on $GL_n^+$ obtained by left translating the standard Euclidean inner product (Frobenius) on $T_IGL_n^+ \cong M_n \cong \mathbb{R}^{n^2}$. (i.e. for $X,Y \in M_n, \, \langle X,Y \rangle=\operatorname{tr}(X^TY) $).

$g$ is left-$GL_n$ invariant and right-$O_n$ invariant, where $O_n$ is the orthogonal group.

Question: Are there isometries of $(GL_n^+,g)$ which are not compositions of the above mentioned isometries? (i.e. do these symmetries generate the isometry group?).

Edit:

I think it might be useful to try to find the dimension of the group generated by left translations (matrix multiplications) and right orthogonal translations. Can we do that? I guess its dimension is less than $\dim O_n + \dim GL_n$.

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    $\begingroup$ I don't have an answer, but you may find this paper of Ravi Shankar of interest: www2.math.ou.edu/~shankar/papers/paper_thesis.pdf In it, he computes the full isometry group of several homogeneous spaces. Unfortunately, he only deals with compact things, so I'm not sure how much of what he does carries over to your case. $\endgroup$
    – Jason DeVito
    Mar 11, 2017 at 17:32
  • $\begingroup$ I guess the question should be about isometries of a connected component of $GL_n.$ Otherwise you can apply your known isometries on each component separately. $\endgroup$
    – Dap
    Jan 23, 2020 at 16:18
  • $\begingroup$ @Dap Thanks for your comment (and for the suggested bounty...). You are right of course in your remark. $\endgroup$
    – Asaf Shachar
    Jan 28, 2020 at 13:32

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