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Ali Taghavi
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Put $M=GL(n,\mathbb{R})$Edit: According to comment of Prof.Assume that $n^2-n$ is not divided by $4$ otherwise GoodWillie we revise the answer to this question is obviuosely "yes" since the mapping.

Put $A\mapsto A^{tr}$ is homotopic to identity$M=GL(n,\mathbb{R})$.

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 $M$?

Put $M=GL(n,\mathbb{R})$.Assume that $n^2-n$ is not divided by $4$ otherwise the answer to this question is obviuosely "yes" since the mapping $A\mapsto A^{tr}$ is homotopic to identity.

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 $M$?

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 $M$?

added 123 characters in body
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Ali Taghavi
  • 356
  • 8
  • 31
  • 123

Put $M=GL(n,\mathbb{R})$.Assume that $n^2-n$ is not divided by $4$ otherwise the answer to this question is obviuosely "yes" since the mapping $A\mapsto A^{tr}$ is homotopic to identity. 

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 $M$?

Put $M=GL(n,\mathbb{R})$.Assume that $n^2-n$ is not divided by $4$. 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 $M$?

Put $M=GL(n,\mathbb{R})$.Assume that $n^2-n$ is not divided by $4$ otherwise the answer to this question is obviuosely "yes" since the mapping $A\mapsto A^{tr}$ is homotopic to identity. 

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 $M$?

added 23 characters in body
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Ali Taghavi
  • 356
  • 8
  • 31
  • 123

Put $M=GL(n,\mathbb{R})$.Assume that $n^2-n$ is not divided by $4$. 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 $M$?

Put $M=GL(n,\mathbb{R})$.Assume that $n^2-n$ is not divided by $4$. 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?

Put $M=GL(n,\mathbb{R})$.Assume that $n^2-n$ is not divided by $4$. 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 $M$?

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Ali Taghavi
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Ali Taghavi
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Ali Taghavi
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