TITLE: Homotopicity of two certain sections of frame bundle of $GL(n,\mathbb{R})$
QUESTION [7 upvotes]: Edit: According to comment of Prof. GoodWillie we revise the question.
Put $M=GL(n,\mathbb{R})$.
We identify $M_n(\mathbb{R})$ 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 $$(a_{11}, a_{12},a_{21}, a_{22})$$
So $M$ being an open subset of $\mathbb{R}^{n^2}$ has trivial tangent bundle and there is an obvious description for a tangent vector at a point $A\in M$.
The mapping $A\mapsto A\otimes A$ defines a section of fram bundle of $M$.Because each row of $A\otimes A$ is counted as a $n×n$ matrix via the above identification. So $A\otimes A$ actualy determines $n^2$ independent vectors in the tangent space to $M$ at $A$.
In a similar manner $A\otimes A^{tr}$ is another section of the frame bundle of manifold $M$ where $tr$ is transpose operator.

Are the above two sections $A\otimes A$ and $A\otimes A^{tr}$ homotopic sections of frame bundle of $M$?

REPLY [3 votes]: The two sections are not homotopic for large $n$. Here's why.
First, as you note, the tangent bundle of $GL_{n}$ is trivial, which means its frame bundle is as well. Next, as noted in the comments, we might as well replace $GL_{n}$ with $O(n)$, which is nicer because it is easier to typeset, and we can swap the transpose operator for the inverse operator.
The two sections you define are therefore maps $s,t:O(n)\rightarrow O(n)\times O(n^{2})$ given by $A\mapsto (A, A\otimes A)$ and $A\mapsto (A, A\otimes A^{-1})$. So to check homotopicity it's enough to check homotopicity of the maps $O(n)\rightarrow O(n^{2})$ that you get from the sections after projecting on to the second factor.
Let's check what the maps do on $\pi_{3}$. First note that $\pi_{3}O(n)=\mathbb{Z}$ for all $n>5$. Next, let's factor our maps $s=S\circ\Delta$, $t=T\circ\Delta$ where $\Delta:O(n)\rightarrow O(n)\times O(n)$ is the diagonal, $S(A,B)=A\otimes B$, and $T(A,B)=A\otimes B^{-1}$.
As Tom Goodwillie noted in the comments, the inverse map acts as $-1$ on all homotopy groups.
So the maps induced by $S$ and $T$ are two $1\times2$ integer matrices, and the second column of $T$ is the negative of the second column of $S$; i.e. $S=[a\ \ b]$ and $T=[a\ \ -b]$. If $s=t$ (on homotopy) then $a+b=a-b$ so $S=[a\ \ 0]$. But that clearly can't be, for symmetry reasons.