Show that a Torus has an Isometry of Order4. If the fundamental region for a torus T is a square, show that T has an isometry of order 4 induced by a rotation of the plane.
Secondly, if the fundamental region is a rhombus with angles $\pi/3$ and $2\pi/3$, show that T has an isometry of order 6 induced by a rotation of the plane. 
We have defined the torus T as T=$R^2$$ /  \Gamma$, where $\Gamma$ is generated by two linearly independent translations, $t_1$ and $t_2$. The fundamental region of a T is a parallelogram, but in this special case we are assuming that this parallelogram is actually a square. I have no clue where to start with this problem, so any help is appreciated. 
 A: Suppose $R^2$ is generated by $u,v$ and $\Gamma$ is generated by the translations $t_u$ and $t_v$. The linear map $f$ defined by the matrix $A=\pmatrix{0 & -1\cr 1& 0}$ in $u,v$ is the matrix of the rotation of angle ${\pi\over 2}$.
Since $f\Gamma f^{-1}=\Gamma$, $f$ induces an automorphism of $T^2$. Its order is 4. 
A: You can evaluate this explicitly using complex algebra and roots of unity.
Since all that matters is the square's shape, we can WLOG assume $\Gamma$ is generated by $t(z)=z+1$ and $t_2(z)=z+i$. We wish to show the rotation $r(z)=iz$ induces an isometry of order $4$ in $\mathbb{T}=\mathbb{C}/\Gamma$.
Let $$z=a+bi,\quad u=c+di$$ be points in the fundamental square $S$. The distance between any two $\Gamma$-orbits of $z$ and $u$ is $$d(z',u')=\sqrt{(a+N_1-c-N_2)^2+(b+N_3-d-N_4)^2},\tag{*}$$ where $N_i\in\mathbb{Z}$. Under the rotation $r$ we get $$rz=-b+ai,\quad ru=-d+ci.$$ The distance between $\Gamma$-orbits of rotated points is $$d(rz',ru')=\sqrt{(b+N_3-d-N_4)^2+(a+N_1-c-N_2)^2}\tag{**}.$$ As the minimums of $*$ and $**$ are equal, $r$ induces an isometry of $\mathbb{T}$.
To show $r$ has order $4$, it suffices to follow the orbits of $z\in S$ under rotation. Note that, after a rotation of $\pi/2$ radians any point of $S$ will have shifted into the square on the left. Thus to work out repeated rotations in $\mathbb{T}$, each $r$ must be followed by a $t$, and the problem reduces to showing $$trtrtrtr(z)=z.$$ This is straightforward: $$tr(z)=iz+1$$ $$tr(iz+1)=-z+i+1$$ $$tr(-z+i+1)=-iz+i$$ $$tr(-iz+i)=z.$$
This setup also works for the case that the fundamental region is a rhombus $R$ with angles $\pi/3$ and $2\pi/3$. Some extra care is required this time, but not much, and it unearths a neat connection between tori and the roots of unity.
This time, let $$t(z)=z+\omega,\quad t_2(z)=z+\omega^2$$ where $\omega=e^{\frac{i2\pi}{3}}$ is a cube root of unity. We consider the rotation $r$ of $\pi/3$ radians about the origin. We will show $r$ maps $\Gamma$-orbits to $\Gamma$-orbits and thus is an isometry. Let $a=e^{\frac{i\pi}{3}}$.
For a rotated point $rz$ $$\Gamma rz=\{az+n\omega+m\omega^2:n,m\in\mathbb{Z}\}.$$ Let $z'\in\Gamma z$, then $$rz'=az+aN\omega+aM\omega^2$$ $$=az-N-\frac{M}{\omega^2}$$ $$=az+(\omega+\omega^2)N-M\omega $$ $$=az+(N-M)\omega+N\omega^2 .$$ Thus $rz'\in\Gamma rz$ and we conclude $r$ induces an isometry in $\mathbb{T}$.
So, we are left to show the order of $r$ in $\mathbb{T}$ is $6$.  Let $z$ be a point in the upper half of $R$, then the figure below shows that $tr^2(z)$ is a point in the upper half of $R$. Hence, we need only show $$tr^2tr^2tr^2(z)=z,$$ which follows once again from some simple properties of cube roots of unity: $$tr^2(z)=\omega z+\omega$$ $$tr^2(\omega z+\omega)=\omega^2z+\omega^2+\omega$$ $$tr^2(\omega^2z+\omega^2+\omega)=z+1+\omega+\omega^2=z.$$ The case for $z$ in the lower half being essentially the same, we conclude the result.

Clearly, Tsemo's proof via group actions is a slicker path to the result. There is something nice, however, about "seeing" the two geometries differ and connecting this to the roots of unity. I doubt this method would work for general $n^{th}$-roots of unity (maybe $6^{th}$), since really we were exploiting symmetries of regular tilings of the plane, of which there are only three (square, equilateral triangle, and hexagon). Basically, we used the  neat algebraic properties of the Gaussian and Eisenstein rings.
