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](https://mathoverflow.net/questions/349671/homotopicity-of-two-certain-sections-of-frame-bundle-of-gln-mathbbr).



Are these two structures isometric?
**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.


