# On the proof of four square theorems by modular forms

I am reading the usual computation of the number of ways of writing a number as sum of four squares (following these notes). At one point they use Cauchy theorem in order to integrate over the border of a fundamental domain of $\Gamma_0(2)$, and the multiplicities of zeroes are weighted as follows: $$\frac{1}{2}v_i(f) + v_\infty(f) + v_1(f) + \sum_{others \ p} v_p(f)$$

Why is there a half weight for $i$?

This is because the order of $i$ as an elliptic point is $2$, so that the arc angle of the little neighborhood on which you integrate if $\pi$, so $\frac{1}{2}$ after dividing by the factor in Cauchy's integral.
I believe there are some mistakes in the notes you linked: for the same reason, $v_1(f)$ should disappear (the arc angle is zero, look at the volume on a little neighborhood), and the others are also half-weighted. However, the others cancel out by periodicity of $f$, so only $v_i(f)$ remains, and it shows that every modular form of weight $2$ and level $2$ has $i$ as zero with order $1$.
• The arc length on $v_1$ is zero but a function vanishing there looks like $\exp(\pi i v/(\tau-1))$ so integrating over the horocycle gives the order of vanishing in the complexified space. – W. Cadegan-Schlieper May 9 '19 at 13:32
• Oops, that should say compactified, and for the theta function we're talking about, the vanishing order is nonzero at the cusp $1$, not $i$, as can be checked by checking the positivity of the terms $e^{\pi i n i}=e^{-\pi n}.$ – W. Cadegan-Schlieper May 9 '19 at 13:57