Let \(x_0\) be the image of \(0\), show that there is an order 6 homeomorphism \(f: T \to T\) fixing \(x_0\). Find a representation of \(f_*\) as a matrix, and find its determinant.
Solution
1. Identify \(T\) with \(\RR^2/\ZZ^2\), with \(x_0\) the image of \(0\).
This is the standard linear model of the torus.
2. Let\[ B=\begin{pmatrix}0&-1\\1&1\end{pmatrix}\in SL_2(\ZZ) \]and define\[ f([x,y])=[-y,x+y]. \]
Because \(B\ZZ^2=\ZZ^2\), the integral linear automorphism \(B:\RR^2\to\RR^2\) descends to a homeomorphism of \(\RR^2/\ZZ^2\).
3. The map \(f\) fixes \(x_0\).
\(B(0,0)=(0,0)\).
4. One has\[ B^2=\begin{pmatrix}-1&-1\\1&0\end{pmatrix}, \qquad B^3=-I, \qquad B^6=I. \]
These identities follow by direct matrix multiplication.
5. No positive power \(B^j\) with \(1\le j<6\) is the identity, so \(f\) has order exactly \(6\).
The matrices \(B\) and \(B^2\) displayed above are not \(I\), while \(B^3=-I\ne I\). Then \(B^4=-B\ne I\) and \(B^5=-B^2\ne I\). Together with \(B^6=I\), this proves that the order is \(6\).
6. In the standard basis of \(H_1(T;\ZZ)\cong\ZZ^2\), the induced map is represented by\[ \boxed{f_*=B=\begin{pmatrix}0&-1\\1&1\end{pmatrix}}. \]
The standard generators of \(H_1(T;\ZZ)\) are the images of the coordinate vectors in \(\ZZ^2\), and the induced map on these generators is exactly the lattice automorphism \(B\).
7. Its determinant is\[\boxed{\det f_*=1}.\]
\(\det B=0\cdot1-(-1)\cdot1=1\).