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Corrected a minor error in the formula for eta, which does not affect the argument.
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Robert Bryant
Robert Bryant
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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 $$ \eta = A^{-1} \mathrm{d}A\, A^{-1} = (\eta^{ij}) = -\mathrm{d}(A^{-1}). $$$$ \eta = ({}^T\!\!A)^{-1} \mathrm{d}A\, ({}^T\!\!A)^{-1} = (\eta^{ij}). $$ Now, the framings $X$ and $Y$ are homotopic if and only if the coframings $\xi$ and $\eta$ are homotopic. Since $\xi = ({}^T\!\!A)^{-1} A\,\eta$$\eta = \xi A ({}^T\!\!A)^{-1}$, it follows that these framings are homotopic over $\mathrm{GL}(n,\mathbb{R})$ if and only if the map $\phi:GL(n,\mathbb{R})\to \mathrm{GL}(\mathfrak{gl}(n,\mathbb{R}))\simeq \mathrm{GL}(n^2,\mathbb{R})$ defined by $$ \phi(A) = L\bigl(({}^T\!\!A)^{-1} A\bigr) $$$$ \phi(A) = R\bigl(A({}^T\!\!A)^{-1}\bigr) $$ is null-homotopic, where $L(B)\in\mathrm{GL}(\mathfrak{gl}(n,\mathbb{R}))$$R(B)\in\mathrm{GL}(\mathfrak{gl}(n,\mathbb{R}))$ is leftright multiplication by $B$, i.e., $L(B)C = BC$$R(B)C = CB$. Note that $L(B)$$R(B)$ is simply $n$ copies of the natural leftright action of $B$ on $\mathbb{R}^n$, so $\phi$ is actually equivalent to the mapping $$ A\mapsto \bigl(({}^T\!\!A)^{-1} A,\ ({}^T\!\!A)^{-1} A, ({}^T\!\!A)^{-1} A\,,\ldots,\ ({}^T\!\!A)^{-1} A\bigr) $$$$ A\mapsto \bigl(A({}^T\!\!A)^{-1},\ A({}^T\!\!A)^{-1}, A({}^T\!\!A)^{-1}\,,\ldots,\ A({}^T\!\!A)^{-1}\bigr) $$ i.e., $n$ diagonal copies of the mapping $\psi:\mathrm{GL}(n,\mathbb{R})\to \mathrm{GL}(n,\mathbb{R})$ defined by $\psi(A) = ({}^T\!\!A)^{-1} A$$\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.

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 $$ \eta = A^{-1} \mathrm{d}A\, A^{-1} = (\eta^{ij}) = -\mathrm{d}(A^{-1}). $$ Now, the framings $X$ and $Y$ are homotopic if and only if the coframings $\xi$ and $\eta$ are homotopic. Since $\xi = ({}^T\!\!A)^{-1} A\,\eta$, 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 $$ \phi(A) = L\bigl(({}^T\!\!A)^{-1} A\bigr) $$ is null-homotopic, where $L(B)\in\mathrm{GL}(\mathfrak{gl}(n,\mathbb{R}))$ is left multiplication by $B$, i.e., $L(B)C = BC$. Note that $L(B)$ is simply $n$ copies of the natural left action of $B$ on $\mathbb{R}^n$, so $\phi$ is actually equivalent to the mapping $$ A\mapsto \bigl(({}^T\!\!A)^{-1} A,\ ({}^T\!\!A)^{-1} A, ({}^T\!\!A)^{-1} A\,,\ldots,\ ({}^T\!\!A)^{-1} A\bigr) $$ i.e., $n$ diagonal copies of the mapping $\psi:\mathrm{GL}(n,\mathbb{R})\to \mathrm{GL}(n,\mathbb{R})$ defined by $\psi(A) = ({}^T\!\!A)^{-1} A$.

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.

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.

Source Link
Robert Bryant
Robert Bryant
  • 108.4k
  • 8
  • 341
  • 453

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 $$ \eta = A^{-1} \mathrm{d}A\, A^{-1} = (\eta^{ij}) = -\mathrm{d}(A^{-1}). $$ Now, the framings $X$ and $Y$ are homotopic if and only if the coframings $\xi$ and $\eta$ are homotopic. Since $\xi = ({}^T\!\!A)^{-1} A\,\eta$, 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 $$ \phi(A) = L\bigl(({}^T\!\!A)^{-1} A\bigr) $$ is null-homotopic, where $L(B)\in\mathrm{GL}(\mathfrak{gl}(n,\mathbb{R}))$ is left multiplication by $B$, i.e., $L(B)C = BC$. Note that $L(B)$ is simply $n$ copies of the natural left action of $B$ on $\mathbb{R}^n$, so $\phi$ is actually equivalent to the mapping $$ A\mapsto \bigl(({}^T\!\!A)^{-1} A,\ ({}^T\!\!A)^{-1} A, ({}^T\!\!A)^{-1} A\,,\ldots,\ ({}^T\!\!A)^{-1} A\bigr) $$ i.e., $n$ diagonal copies of the mapping $\psi:\mathrm{GL}(n,\mathbb{R})\to \mathrm{GL}(n,\mathbb{R})$ defined by $\psi(A) = ({}^T\!\!A)^{-1} A$.

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.