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Omar Antolín-Camarena
Omar Antolín-Camarena
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We can construct a homotopy retract of this to the set of diagonal matrices with nonzero diagonal entries by scaling nondiagonal entries to 0, and from there to a torus by scaling the diagonal entries to norm 1. Therefore, it is homotopy equivalent to $S_1^n$$(S^1)^n$, and is connected.

For real matrices, the retract above still works, and takes you to $\{-1,1\}^n$. The subset with positive determinant is the subset with product 1, and is not connected for $n>2$. I think the connected component you may want is the subset where all eigenvalues have positive real part, but have not checked.

We can construct a homotopy retract of this to the set of diagonal matrices with nonzero diagonal entries by scaling nondiagonal entries to 0, and from there to a torus by scaling the diagonal entries to norm 1. Therefore, it is homotopy equivalent to $S_1^n$, and is connected.

For real matrices, the retract above still works, and takes you to $\{-1,1\}^n$. The subset with positive determinant is the subset with product 1, and is not connected for $n>2$. I think the connected component you may want is the subset where all eigenvalues have positive real part, but have not checked.

We can construct a homotopy retract of this to the set of diagonal matrices with nonzero diagonal entries by scaling nondiagonal entries to 0, and from there to a torus by scaling the diagonal entries to norm 1. Therefore, it is homotopy equivalent to $(S^1)^n$, and is connected.

For real matrices, the retract above still works, and takes you to $\{-1,1\}^n$. The subset with positive determinant is the subset with product 1, and is not connected for $n>2$. I think the connected component you may want is the subset where all eigenvalues have positive real part, but have not checked.

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user44191
user44191
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We can construct a homotopy retract of this to the set of diagonal matrices with nonzero diagonal entries by scaling nondiagonal entries to 0, and from there to a torus by scaling the diagonal entries to norm 1. Therefore, it is homotopy equivalent to $S_1^n$, and is connected.

For real matrices, the retract above still works, and takes you to $\{-1,1\}^n$. The subset with positive determinant is the subset with product 1, and is not connected for $n>2$. I think the connected component you may want is the subset where all eigenvalues have positive real part, but have not checked.

We can construct a homotopy retract of this to the set of diagonal matrices with nonzero diagonal entries by scaling nondiagonal entries to 0, and from there to a torus by scaling the diagonal entries to norm 1. Therefore, it is homotopy equivalent to $S_1^n$, and is connected.

We can construct a homotopy retract of this to the set of diagonal matrices with nonzero diagonal entries by scaling nondiagonal entries to 0, and from there to a torus by scaling the diagonal entries to norm 1. Therefore, it is homotopy equivalent to $S_1^n$, and is connected.

For real matrices, the retract above still works, and takes you to $\{-1,1\}^n$. The subset with positive determinant is the subset with product 1, and is not connected for $n>2$. I think the connected component you may want is the subset where all eigenvalues have positive real part, but have not checked.

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user44191
user44191
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We can construct a homotopy retract of this to the set of diagonal matrices with nonzero diagonal entries by scaling nondiagonal entries to 0, and from there to a torus by scaling the diagonal entries to norm 1. Therefore, it is homotopy equivalent to $S_1^n$, and is connected.