Fdtxjr

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 $SL(2,mathbbZ)$:
$$Delta(z)=(2pi)^12e^2pi izprod_n=1^infty(1-e^2pi inz)^24.$$
Let $mathbbH$ denote the upper-half plane. The following is a quotation from the paper.

Also, $mathbbH$ is simply connected, so we may fix once and for all a branch of $logDelta(z)$ on $mathbbH.$ Then
$$logDelta(Mz)-logDelta(z)=12log(cz+d)+2pi im$$
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 $mathbbR^2setminus(0,0).$ The following is a quotation from the paper.

If $M=beginbmatrix a & b \ c & d endbmatrixin SL(2,mathbbR),$ then having a path $gamma$ from $(0,1)$ to $(c,d)$ in $mathbbR^2setminus(0,0)$ lets us make a canonical choice of branch of $log(cz+d)$: for fixed $zinmathbbH,$ we set $log(0cdot z + 1)=0$ and then make $log(c^primez+d^prime)$ 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 $(M,[gamma])inwidetildeSL(2,mathbbZ),$
$$logDelta(Mz)-logDelta(z)=12L(M,[gamma];z)+2pi iPhi(M,[gamma])$$
now defines a function $Phicolon widetildeSL(2,mathbbZ)tomathbbZ.$

Here $widetildeSL(2,mathbbZ)$ is a funky group, the definition of which is irrelevant to my question. Suffice it to say that the elements of $widetildeSL(2,mathbbZ)$ look like $(M,[gamma])$ for $Min SL(2,mathbbZ)$ and $gamma$ as above.

Let $T=beginbmatrix 1 & 1 \ 0 & 1 endbmatrix$ and let $gamma$ be the "trivial" (I understand this to mean "constant") path from $(0,1)$ to $(0,1)$ in $mathbbR^2setminus(0,0).$ 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 $SL(2,mathbbZ),$ we have $Delta(Tz)=(0z+1)^12Delta(z)=Delta(z).$
  


• The right-hand side is just $2pi iPhi(T,[gamma]),$ 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 $0$?

Let $G = M in SL_2(mathbbZ), c > 0$ 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 $M in SL_2(mathbbZ)$ let $r(M)(z) = cz+d$. Then $Delta in M_12(G)$, $Delta(M(z)) = r(M(z))^12 Delta(z)$ implies $$L(M(z)) = 12 log(r(M)(z)) + L(z)+2ipi n(M), qquad n : SL_2(mathbbZ) to mathbbZ$$

By direct computation $r(M_2M)(z) = r(M)(z) r(M_2)(M(z))$ so that $$L(M_2 M(z)) = 12 log r(M_2M)(z) + L(z)+2ipi n(M_2M)\=L(M_2 (M(z)))=12 log r(M_2)(M(z)) + L(M(z))+2ipi n(M_2)\=12 log r(M_2)(M(z)) +12 log r(M)(z) + L(z)+2ipi n(M)+2ipi n(M_2)$$

$r(M_2M)(z) = r(M)(z) r(M_2)(M(z))$ 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 $Im(z) in (0,2pi)$ 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 $M in SL_2(mathbbZ)$ and the transformation law of $L(z)$.