For an invertible linear transformation \(f: \mathbb{R}^n \to \mathbb{R}^n\) show that the induced map on \(H_n(\mathbb{R}^n, \mathbb{R}^n - \{0\}) \approx \tilde{H}_{n-1}(\mathbb{R}^n - \{0\}) \approx \mathbb{Z}\) is \(\mathbb{1}\) or \(-\mathbb{1}\) according to whether the determinant of \(f\) is positive or negative. [Use Gaussian elimination to show that the matrix of \(f\) can be joined by a path of invertible matrices to a diagonal matrix with \(\pm 1\)’s on the diagonal.]
Solution
Let \(f:\mathbb R^n\to\mathbb R^n\) be invertible. The induced automorphism of \[ H_n(\mathbb R^n,\mathbb R^n-\{0\})\cong\mathbb Z \] depends only on the path component of \(f\) in \(GL_n(\mathbb R)\).
1. If \(f_t\) is a path in \(GL_n(\mathbb R)\), then all \((f_t)_*\) on local homology are equal.
The map \[ H(x,t)=f_t(x) \] is a homotopy of maps of pairs \[ (\mathbb R^n,\mathbb R^n-\{0\})\to(\mathbb R^n,\mathbb R^n-\{0\}), \] since every \(f_t\) is invertible and hence takes nonzero vectors to nonzero vectors. Homotopy invariance of relative homology gives equality of the induced maps.
2. Every matrix with positive determinant is path connected in \(GL_n(\mathbb R)\) to the identity, while every matrix with negative determinant is path connected to\[ r=\operatorname{diag}(-1,1,\dots,1). \]
Gaussian elimination through invertible elementary matrices reduces any matrix to a diagonal one without crossing determinant zero. Positive diagonal entries can be deformed to \(1\), and negative entries can be paired and deformed through an invertible \(2\times2\) block to two positive entries. Thus the positive-determinant component contains \(I\) and the negative-determinant component contains \(r\). Equivalently, \(GL_n(\mathbb R)\) has exactly the two components distinguished by the sign of the determinant.
3. The identity induces \(+1\) and the reflection \(r\) induces \(-1\) on local homology.
Under the boundary isomorphism \[ H_n(\mathbb R^n,\mathbb R^n-\{0\}) \cong \widetilde H_{n-1}(\mathbb R^n-\{0\}) \cong \widetilde H_{n-1}(S^{n-1}), \] the map induced by \(r\) is the usual reflection of \(S^{n-1}\). A reflection reverses orientation and therefore has degree \(-1\). The identity has degree \(+1\).
Combining <1>1–<1>3, \[ \boxed{ f_*= \begin{cases} +1,&\det f>0,\\ -1,&\det f<0. \end{cases}} \]