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 $M$?

  • $\begingroup$ Since the tangent bundle is trivial, we have $TGL_n(\mathbb{R}) = GL_n(\mathbb{R}) \times M_n(\mathbb{R})$. So do you mean the sections $A \mapsto (A,A)$ and $A \mapsto (A,A^{tr})$? $\endgroup$ Jan 3, 2020 at 21:17
  • $\begingroup$ @UlrichPennig I am not talking about a single section for tangent bundle but I am talking about a section for frame bundle so actualy it consist of $n^2$ sections for the tangent bundles. They are liying at rows of $A\otimes A$ or $A\otimes A^{tr}$. $\endgroup$ Jan 3, 2020 at 21:28
  • $\begingroup$ We remove the base point from the first component of sections that is we remove $x$ from $(x, f(x)$. $\endgroup$ Jan 3, 2020 at 21:30
  • 2
    $\begingroup$ So, in the case of $n = 2$, the frame would consist of the vectors $$ e_{11} = (A_{11}A_{11}, A_{11}A_{12}, A_{12}A_{11}, A_{12}A_{12})$$ $$e_{12} = (A_{11}A_{21}, A_{11}A_{22}, A_{12}A_{21}, A_{12}A_{22}) $$ $$ e_{21} = (A_{21}A_{11}, A_{21},A_{12}, A_{22}A_{11}, A_{22}A_{12}) $$ $$e_{22} = (A_{21}A_{21}, A_{21}A_{22}, A_{22}A_{21}, A_{22}A_{22}) $$. Is that correct? $\endgroup$
    – Deane Yang
    Jan 3, 2020 at 21:51
  • 1
    $\begingroup$ The transpose operator on n by n matrices is homotopic (through linear maps) to the identity in that case, but the homotopy cannot be chosen to take invertible matrices to invertible matrices. $\endgroup$ Jan 4, 2020 at 19:51

2 Answers 2


They are homotopic when $n=2$, but not when $n>2$. Here is the argument:

Let $A=(a_{ij})$. Then, the definitions of the two framings can be made more explicit as follows: For the first frame field, each vector field, say, $X_{ij}$, can be thought of as an $n$-by-$n$ matrix, and the formula for the $kl$ entry of $X_{ij}$ is $$ (X_{ij})_{kl} = a_{ik}a_{jl}\,, $$ while for the second frame field, each vector field, say $Y_{ij}$, can be thought of as an $n$-by-$n$ matrix, and the formula for the $kl$ entry of $Y_{ij}$ is $$ (Y_{ij})_{kl} = a_{ik}a_{lj}\,. $$ The $1$-forms $\xi^{ij}$ of the coframing dual to the framing $X_{ij}$ are then seen to be the components of the $n$-by-$n$ matrix-valued $1$-form $$ \xi = ({}^T\!\!A)^{-1} \mathrm{d}A\, A^{-1} = (\xi^{ij}) $$ while the $1$-forms $\eta^{ij}$ of the coframing dual to the framing $Y_{ij}$ are then seen to be the components of the $n$-by-$n$ matrix-valued $1$-form $$ \eta = ({}^T\!\!A)^{-1} \mathrm{d}A\, ({}^T\!\!A)^{-1} = (\eta^{ij}). $$ Now, the framings $X$ and $Y$ are homotopic if and only if the coframings $\xi$ and $\eta$ are homotopic. Since $\eta = \xi A ({}^T\!\!A)^{-1}$, it follows that these framings are homotopic over $\mathrm{GL}(n,\mathbb{R})$ if and only if the map $\phi:GL(n,\mathbb{R})\to \mathrm{GL}(\mathfrak{gl}(n,\mathbb{R}))\simeq \mathrm{GL}(n^2,\mathbb{R})$ defined by $$ \phi(A) = R\bigl(A({}^T\!\!A)^{-1}\bigr) $$ is null-homotopic, where $R(B)\in\mathrm{GL}(\mathfrak{gl}(n,\mathbb{R}))$ is right multiplication by $B$, i.e., $R(B)C = CB$. Note that $R(B)$ is simply $n$ copies of the natural right action of $B$ on $\mathbb{R}^n$, so $\phi$ is actually equivalent to the mapping $$ A\mapsto \bigl(A({}^T\!\!A)^{-1},\ A({}^T\!\!A)^{-1}, A({}^T\!\!A)^{-1}\,,\ldots,\ A({}^T\!\!A)^{-1}\bigr) $$ i.e., $n$ diagonal copies of the mapping $\psi:\mathrm{GL}(n,\mathbb{R})\to \mathrm{GL}(n,\mathbb{R})$ defined by $\psi(A) = A({}^T\!\!A)^{-1}$.

Now, as is well known, $\mathrm{GL}(n,\mathbb{R})$ is diffeomorphic to $S^+_n\times \mathrm{O}(n)$, where $S^+_n$, which is contractible, is the space of $n$-by-$n$ positive definite matrices. This is the famous $QR$-decomposition, i.e., $A = QR$ where $Q$ is symmetric positive definite and $R$ is orthogonal. Since $S^+_n$ is contractible, we can answer the homotopy question by restricting $\phi$ to $O(n)$, i.e., we set $Q=I_n$. On $\mathrm{O}(n)$ the mapping becomes $$ \phi(R) = (R^2,\ R^2,\ ,\ldots,\ R^2)\in \mathrm{SO}(n^2) $$ Note that the image goes diagonally into $\mathrm{SO}(n)\times\cdots\times\mathrm{SO}(n)\subset \mathrm{SO}(n^2)$.

Now, when $n=2$, $\mathrm{SO}(2)\simeq S^1$, and $\pi_1\bigl(\mathrm{SO}(2)\bigr)\simeq\mathbb{Z}$. Meanwhile, $\pi_1\bigl(\mathrm{SO}(4)\bigr)\simeq\mathbb{Z}_2$, and $\phi$ takes a generator of $\pi_1\bigl(\mathrm{SO}(2)\bigr)$ to $4$ times a generator of $\pi_1\bigl(\mathrm{SO}(4)\bigr)$, i.e., to zero. Thus, when $n=2$, the mapping $\phi$ is null-homotopic, so the two framings $X$ and $Y$ are homotopic when $n=2$.

However, when $n>2$, we have $\pi_3\bigl(\mathrm{SO}(n)\bigr)\simeq\mathbb{Z}$ (except when $n=4$, when it equals $\mathbb{Z}\oplus\mathbb{Z}$). Moreover, it is easy to see that, when $n\not=3$, the mapping $\phi$ takes a generator of $\pi_3\bigl(\mathrm{SO}(n)\bigr)$ to $2n$ times a generator of $\pi_3\bigl(\mathrm{SO}(n^2)\bigr)$, which is nontrivial. Thus, $\phi$ is not homotopically trivial in these cases. A similar argument applies in the case $n=4$, to show that $\phi$ is not homotopically trivial in this case either. Thus, when $n>2$, the two framings are not homotopic.

  • $\begingroup$ Thank you very much for your answer to this question and its geometric version. I effort to understand their details. Regarding the geometric version we realize that both frames consiste of mutually orthonormal (wrt the metric) matrice but no matric of this frame is invertible. So one can be wonder if there is an orthogonal section section of this frame whose all matrices ate onvertible? $\endgroup$ Jan 8, 2020 at 12:09

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.

  • $\begingroup$ Thank you very much for your answer I try to understand its detail. $\endgroup$ Jan 8, 2020 at 12:10

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