Are these two structures isometric?

Question

Do the following Riemannian metrics on $GL(n,\mathbb{R})$ give us isometric structures?Do they generate the same volume forms? Is $O(n)$ a totally geodesic submanifold with respect to these metrics?

1. The metric with orthonormal frame $A\otimes A$ at each point $A\in GL(n,\mathbb{R})$

2. The metric with orthonormal frame $A\otimes A^{tr}$ for $A\in GL(n,\mathbb{R})$

Note that the tangent space $T_A GL(n,\mathbb{R})$ at each point $A$ is identified with $M_{n}(\mathbb{R})$ hence with $\mathbb{R}^{n^2}$.(Note that $GL(n,\mathbb{R})$ is counted as an open subset of Euclidean space hence has trivial tangent bundle so the above identification has an obvious desribtion). So at each point $A\in GL(n,\mathbb{R})$, each column of the tensor product matrix $A\otimes A$ (or $A\otimes A^{tr}$) can be considered as a tangent vector in $T_A GL(n,\mathbb{R})$. So $A\otimes A$ is obviously a fram, that is a base for the tangent space.This frame define a unique well defined Riemannian metric on $GL(n,\mathbb{R})$.

Edit: According to the comments on this question we clarify the identification $M_n(\mathbb{R})$ with $\mathbb{R}^{n^2}$: The identification is based on the lexicographic order on the index $i,j$ in $(a_{ij})$. For example $$ \begin{pmatrix} a_{11}&a_{12}\\ a_{21}&a_{22}\end{pmatrix}$$

is identified with $$(a_{11}, a_{12},a_{21}, a_{22})$$

Remark: For a homotopic version of this question please see this MO post.

Answer 1

Corrected answer: Well, technically, my answers to the OP's questions have not changed, but my argument has changed because I found a silly mistake in my original argument (which I will explain below).

First, computation shows that the first metric, described by giving the orthonormal frame as $A\otimes A$, can be written more conventionally as the symmetric quadratic differential form $$ ds^2_1 = \mathrm{tr}\bigl( (^{T}\!\!A^{-1}\mathrm{d}A\,A^{-1})\circ(^{T}\!A^{-1}\,{^T}\!(\mathrm{d}A)\,A^{-1})\bigr), $$ where $A:\mathrm{GL}(n,\mathbb{R})\hookrightarrow M_n(\mathbb{R})$ is the inclusion mapping. (N.B.: I use ${^T}\!\!B$ to mean the transpose of $B$.) Note that this metric is invariant under the left action of $\mathrm{O}(n)$ on $\mathrm{GL}(n,\mathbb{R})$ given by $R\cdot A = RAR^{-1}$, though it is not invariant under the action of $\mathrm{GL}(n,\mathbb{R})$ by either left or right multiplication. Direct computation shows that this metric is not flat. This is most easily carried out by making the substitution $A = B^{-1}$, for, in the $B$-coordinate, the metric becomes $$ ds^2_1 = \mathrm{tr}\bigl( (^{T}\!\!B\,B^{-1}\mathrm{d}B)\ \circ\ ^{T}\!(^{T}\!\!B\,B^{-1}\mathrm{d}B)\bigr), $$ which is somewhat simpler to work with. In fact, the $n^2$ components of the matrix $(^{T}\!\!B\,B^{-1}\mathrm{d}B)$ are an orthonormal basis for the $1$-forms for this metric on $\mathrm{GL}(n,\mathbb{R})$. Using this, it's not very hard to compute the curvature matrix and see that it is nonzero. Moreover, when $n=2$, one can compute the Ricci tensor of $ds^2_1$, and one finds that the four symmetric functions of its eigenvalues, $s_1$ (the scalar curvature), $s_2$, $s_3$, and $s_4$ have the property that $s_1$, $s_2$, and $s_3$ are independent functions on a dense open set in $\mathrm{GL}(2,\mathbb{R})$ while there is a nonzero polynomial $P_1$ of $4$ variables such that $P_1(s_1,s_2,s_3,s_4)$ vanishes identically. (It's not surprising that there would be at least one relation because there is a $1$-dimensional symmetry group of this metric, the $O(2)$-mentioned above.)

Meanwhile, the formula for the second metric, described by giving the orthonormal frame as $A\otimes {}^T\!\!A$, turns out to be more conventionally expressed in the form $$ ds^2_2 = \mathrm{tr}\bigl( (^{T}\!\!A^{-1}\mathrm{d}A\,^{T}\!\!A^{-1})\circ(A^{-1}\,{^T}\!(\mathrm{d}A)\,A^{-1})\bigr), $$ not as I originally had it, which was $$ \mathrm{tr}\bigl( (A^{-1}\mathrm{d}A\,A^{-1})\circ(^{T}\!\!A^{-1}\,{^T}\!(\mathrm{d}A)\,^{T}\!\!A^{-1})\bigr) =\mathrm{tr}\bigl( (\mathrm{d}A^{-1})\ \circ\ ^{T}(\mathrm{d}A^{-1})\bigr). $$ This latter metric, call it $ds^2_3$, is instead associated to the orthonormal frame field ${}^T\!\!A\otimes A$, and, as the formula above shows when one substitutes $A = B^{-1}$, this third metric is flat.

Unfortunately, $ds^2_2$ is not flat, so it is not immediately obvious whether $ds^2_1$ and $ds^2_2$ are isometric. (They certainly are not equal, but it's not immediately obvious that one can't be pulled back to the other by some complicated diffeomorphism.) However, in the case, $n=2$, one can, by a messy calculation using MAPLE, verify that, if $s_1'$, $s_2'$, $s_3'$ and $s_4'$ are the symmetric functions of the eigenvalues of the Ricci tensor of $ds^2_2$, then $P_1(s_1',s_2',s_3',s_4')$ does not vanish identically. Consequently, $ds^2_1$ and $ds^2_2$ are not isometric when $n=2$.

The two metrics do have the same volume form, but a more involved argument similar to the above works for all $n\ge 2$ to show that they are not isometric, even locally.

Finally, $\mathrm{O}(n)$ cannot be totally geodesic in either metric. The fact that both metrics are homogeneous of degree $-2$ under the scaling $A\mapsto \lambda A$ shows that there are no compact minimal submanifolds in $\mathrm{GL}(n,\mathbb{R})$ endowed with either metric, let alone totally geodesic ones.