Jtdylktuy

Under a modular transformation the Dedekind $eta$ function transforms as $$eta(-1/tau) = sqrt{-i tau}eta(tau) , .tag*{$(*)$}$$Siegel gives a proof in this paper here that uses complex analytic techniques. However, I am wondering how we can see it in the following fashion. The Dedekind $eta$ function can be represented in the form$$eta(tau) = q^{1/24} prod_{n = 1}^infty (1 - q^n) = sum_{n = -infty}^infty (-1)^n q^{{3over2}(n - 1/6)^2}.$$How can we use the Poisson resummation formula and this representation of the $eta$ function to verify the modular transformation $(*)$?

Let us start from one side of what we want to prove,$$eta(-1/tau) = r^{1/24} prod_{n = 1}^infty (1 - r^n), quad r = expleft(-{{2pi i}overtau}right).$$We were given Euler's (quite remarkable) pentagonal number theorem, so we might as well start by using it to transform the product into a sum$$eta(-1/tau) = r^{1/24} sum_{n in mathbb{Z}} (-1)^n r^{(3n^2 - n)/2}, quad r = expleft(-{{2pi i}overtau}right).$$The other hammer we were given is the Poisson summation, and this expression is beginning to look like the right nail: we just need to a find a function $f$ such that each term in the sum above is $f(n)$, and then Fourier transform it. Rewriting the above expression slightly as$$f(n) = expleft[pi ileft(-{1over{12tau}} + n - {{(3n^2 - n)}overtau}right)right],$$it becomes straightforward to extrapolate it to the reals by simply replacing $n$ with a real number $x$. We can now find the Fourier transform (note that $f$ does fall off "sufficiently fast" because $text{Im},tau > 0$),begin{align*}
tilde{f}(k) & = int_{-infty}^infty dx,expleft[pi ileft(-{1over{12tau}} + x - 2kx - {{(3x^2 - x)}overtau}right)right] \ & = sqrt{tauover{3i}} expleft[pi ileft({{tau(-2k + 1 + {1overtau})^2}over{12}} - {1over{12tau}}right)right] \ & = sqrt{{-itau}over3}expleft[pi ileft({{tau(2k - 1)^2}over{12}} - {{(2k - 1)}over6}right)right],end{align*}where in the first step we used the Gaussian integral$$int_{-infty}^infty dx,exp(-ax^2 + bx + c) = sqrt{piover a} expleft({{b^2}over{4a}} + cright),$$valid for $text{Re},a > 0$, which in our case translates to$$text{Re}left({{3pi i}overtau}right) = text{Im}left(-{{3pi}overtau}right) > 0.$$We can finally use the Poisson summation formula to obtain$$eta(-1/tau) = sqrt{{-itau}over3} sum_{k in mathbb{Z}} expleft[pi ileft({{tau(2k - 1)^2}over{12}} - {{(2k - 1)}over6}right)right].tag*{$(*)$}$$How close is this to the desired result? Let us use the pentagonal number theorem on $eta(tau)$ as well to transform it into a sum$$eta(tau) = q^{1/24} sum_{n in mathbb{Z}} (-1)^n q^{(3n^2 - n)/2} = sum_{n in mathbb{Z}} expleft[pi ileft({{tau(6n - 1)^2}over{12}} + nright)right].$$This is close! However, there is a factor of $sqrt{3}$, and the term proportional to $tau$ in the exponent has $6n - 1$ instead of $2k - 1$. To solve these assorted factors of $3$, we can split the sum in $(*)$ into three sums, with $k = 3l$, $3l + 1$, $3l + 2$. Then$$eta(-1/tau) = sqrt{{{-itau}over3}}sum_{l in mathbb{Z}} left[ expleft[pi ileft({{tau(6l - 1)^2}over{12}} - {{(6l - 1)}over6}right)right] + expleft[pi ileft({{tau(6l + 1)^2}over{12}} - {{(6l + 1)}over6}right)right] + expleft[pi ileft({{tau(6l + 3)^2}over{12}} - {{(6l + 3)}over6}right)right]right].$$The first term looks like the desired expression times $e^{pi i/6}$. In the second term, we can take $l to -l$ to transform it into the desired expression times $e^{-pi i/6}$. Since$$e^{pi i/6} + e^{-pi i/6} = 2cos pi/6 = sqrt{3},$$we have found the full answer and the last term must vanish!

We can rewrite it as a sum over odd integers,$$sum_{m in 2mathbb{Z} + 1} expleft[pi ileft({{3tau m^2}over4} - {mover2}right)right],$$and now separate the positive and negative $m$ parts to find$$sum_{m in 2mathbb{Z} + 1,,m > 0} expleft[pi i{{3tau m^2}over4}right]left(e^{pi im/2} + e^{-pi im/2}right),$$which vanishes since $e^{pi im} = -1$ for $m$ an odd integer.

Therefore, we finally have$$eta(-1/tau) = sqrt{-itau} sum_{l in mathbb{Z}} expleft[pi ileft({{tau(6l - 1)^2}over{12}} - lright)right] = sqrt{-itau}eta(tau),$$as we wanted to show.