Identification of points on a Euclidean torus Consider the following paragraph from page 214-215 of Thomas Hartman's notes on Quantum Gravity and Black Holes.

Consider a $2$d QFT (not necessarily conformal yet) on a Euclidean torus. The most
  general torus is specified by two lattice vectors $\vec{v}_{1}$, $\vec{v}_{2}$ on the $(t_{E},\phi)$ plane, meaning that we identify all points related by
  $$(t_{E},\phi) \sim (t_{E},\phi) + m\vec{v}_{1} + n\vec{v}_{2}, \qquad m,n \in \mathbb{Z}.$$
  If the theory is rotationally invariant (i.e., if its Lorentzian counterpart is Lorentz invariant), then we may w.l.o.g. rotate $\vec{v}_{1}$ to lie on the $t_E$ axis. If the theory is scale invariant, then we can also w.l.o.g. set its length to $\vec{v}_{2} = (0, 2\pi)$. Thus in a conformal field theory, we are led to consider the theory on the torus
  $$(t_{E},\phi) \sim (t_{E} + \beta, \phi + \theta) \sim (t_{E},\phi + 2\pi),$$
  where $\phi$ and $\theta$ are arbitrary real numbers.

Radius of the $\phi$-circle:
I understand that the coordinates $(t_{E},\phi + 2\pi)$ are related to the coordinates $(t_{E},\phi)$ by a simple translation in the $\phi$-direction. This means that the circumference of the $\phi$-circle is $2\pi$.
Radius of the $t_{E}$-circle:
I also understand that the circumference of the $t_E$-circle is $\beta$. But I am not sure how this follows from $(t_{E},\phi) \sim (t_{E} + \beta, \phi + \theta)$. In other words, what is the meaning of the parameter $\theta$ in the identification $(t_{E},\phi) \sim (t_{E} + \beta, \phi + \theta)$?
How is the parameter $\theta$ related to the circumference $L=2\pi$ of the $\phi$-circle?
 A: In declaring that we're working on a torus, we defined two lattice vectors, and we require, on any torus, that displacements equal to an integer number of lengths of those lattice vectors bring us back to our starting point. Since the theory is scale invariant, these lattice vectors can have any length. Let $\beta$ be the arbitrary length of $\vec{v}_1$ and $\theta$ be the arbitrary length of $\vec{v}_2$. Then displacements of $\beta$ along $\vec{v}_1$ and $\theta$ along $\vec{v}_2$ bring us back to the same point. Since the theory is rotationally invariant, we can choose the orientation of the two lattice vectors. Therefore, we choose that $\vec{v}_1$ points in the $t_E$-direction and $\vec{v}_2$ points in the $\phi$-direction. Applying a displacement of $\beta$ along $\vec{v}_1$ and $\theta$ along $\vec{v}_2$, we have that 
$$(t_E,\phi)\sim(t_E+\beta,\phi+\theta)$$
In order to make this look more like a conventional torus, we can set the length of $\vec{v}_2$ (the lattice vector pointing in the $\phi$-direction) to be $2\pi$. Due to scale invariance, this doesn't change anything. Under these conditions, a displacement of $2\pi$ in the $\phi$ direction brings us back to the starting point. In other words, under these conditions,
$$(t_E,\phi)\sim(t_E,\phi+2\pi)$$
