Computation of $\Phi(T,[\gamma])$ in a paper of Poonen and Rodriguez-Villegas I am reading the paper "Lattice Polygons and the Number 12" by Bjorn Poonen and Fernando Rodriguez-Villegas (a copy of this paper can be found, e.g., on the first author's webpage). In it, the authors make use of the following weight $12$ cusp form for $SL(2,\mathbb{Z})$:
$$\Delta(z)=(2\pi)^{12}e^{2\pi iz}\prod_{n=1}^{\infty}(1-e^{2\pi inz})^{24}.$$
Let $\mathbb{H}$ denote the upper-half plane. The following is a quotation from the paper.

Also, $\mathbb{H}$ is simply connected, so we may fix once and for all a branch of $\log{\Delta(z)}$ on $\mathbb{H}.$ Then
  $$\log{\Delta(Mz)}-\log{\Delta(z)}=12\log(cz+d)+2\pi im$$
  for some integer $m$ depending on the branch of $\log(cz+d).$

Their idea is to choose a branch of $\log(cz+d)$ by choosing a path $\gamma$ in $\mathbb{R}^{2}\setminus\{(0,0)\}.$ The following is a quotation from the paper.

If $M=\begin{bmatrix} a & b \\ c & d \end{bmatrix}\in SL(2,\mathbb{R}),$ then having a path $\gamma$ from $(0,1)$ to $(c,d)$ in $\mathbb{R}^{2}\setminus\{(0,0)\}$ lets us make a canonical choice of branch of $\log(cz+d)$: for fixed $z\in\mathbb{H},$ we set $\log(0\cdot z + 1)=0$ and then make $\log(c^{\prime}z+d^{\prime})$ a continuous function of the path parameter, as $(c^{\prime},d^{\prime})$ moves from $(0,1)$ to $(c,d).$ Moreover this choice of branch depends only on the path-homotopy class of $\gamma;$ we call it $L(M,[\gamma];z).$ For $(M,[\gamma])\in\widetilde{SL(2,\mathbb{Z})},$
$$\log{\Delta(Mz)}-\log{\Delta(z)}=12L(M,[\gamma];z)+2\pi i\Phi(M,[\gamma])$$
  now defines a function $\Phi\colon \widetilde{SL(2,\mathbb{Z})}\to\mathbb{Z}.$

Here $\widetilde{SL(2,\mathbb{Z})}$ is a funky group, the definition of which is irrelevant to my question. Suffice it to say that the elements of $\widetilde{SL(2,\mathbb{Z})}$ look like $(M,[\gamma])$ for $M\in SL(2,\mathbb{Z})$ and $\gamma$ as above.
Let $T=\begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}$ and let $\gamma$ be the "trivial" (I understand this to mean "constant") path from $(0,1)$ to $(0,1)$ in $\mathbb{R}^{2}\setminus\{(0,0)\}.$ From the definition above, I compute that $\Phi(T,[\gamma])=0.$ Here's why:


*

*The left-hand side of the defining equation has to be $0,$ since, because $\Delta$ is a modular form of weight $12$ for $SL(2,\mathbb{Z}),$ we have $\Delta(Tz)=(0z+1)^{12}\Delta(z)=\Delta(z).$

*The right-hand side is just $2\pi i\Phi(T,[\gamma]),$ because $L(T,[\gamma];z)$ is the principal logarithm of $0z+1=1,$ which is $0.$
However, in the paper, the authors claim that $\Phi(T,[\gamma])=1,$ and in fact this is a crucial part of the proof of their main result. They compute the value a different way, which uses other ideas from the paper, but give a parenthetical remark saying that it could be calculated directly. I have gone through their calculation and have found no issues.

Question. What has gone wrong with my understanding of this definition of $\Phi$ that leads me to the (apparently incorrect) value $0$?

 A: Let $G = \{ M \in SL_2(\mathbb{Z}), c > 0\}$  then $\Im(cz+d)> 0$ and we can take everywhere the principal branch of $\log$. Since $\Im(z) > 0$ is simply connected and $\Delta$ doesn't vanish there is an holomorphic function $L$ such that $\Delta(z) = e^{L(z)}$.
For $M \in SL_2(\mathbb{Z})$ let $r(M)(z) = cz+d$. Then $\Delta \in M_{12}(G)$, $\Delta(M(z)) = r(M(z))^{12} \Delta(z)$ implies $$L(M(z)) = 12 \log(r(M)(z)) + L(z)+2i\pi n(M), \qquad n : SL_2(\mathbb{Z}) \to \mathbb{Z}$$
By direct computation $r(M_2M)(z) = r(M)(z) r(M_2)(M(z))$ so that  $$L(M_2 M(z)) = 12 \log r(M_2M)(z) + L(z)+2i\pi n(M_2M)\\=L(M_2 (M(z)))=12 \log r(M_2)(M(z)) + L(M(z))+2i\pi n(M_2)\\=12 \log r(M_2)(M(z)) +12 \log r(M)(z) + L(z)+2i\pi n(M)+2i\pi n(M_2)$$
$r(M_2M)(z) = r(M)(z) r(M_2)(M(z))$ implies $\log r(M_2M)(z) = \log r(M_2)(M(z)) + \log r(M)(z)+2i \pi k(M_2,M)$. 
For $M,M_2,M_2M \in G$ we know $\log r(M_2M)(z),\log r( M_2)(M(z)), \log r(M)(z)$ send $\Im(z)> 0$ to $\Im(z) \in (0,\pi)$ so $\log r(M_2)(M(z)) + \log r(M)(z)$ sends $\Im(z) > 0$ to $\Im(z) \in (0,2\pi)$ and hence $k(M_2,M) =0$.
Thus for $M,M_2,M_2M \in G$ $$n(M_2M) = n(M)+n(M_2)$$
The few remaining cases are $M$ or $M_2$ with $c=0$ and $M_2M$ with $c \le 0$ to find $n(M)$ for every $M \in SL_2(\mathbb{Z})$ and the transformation law of $L(z)$.
