In the study of superconformal indices for certain quantum field theories, one encounters the elliptic $\Gamma$ function, which can be expressed as: $$ \log \Gamma(z;\tau,\sigma)=\sum^{\infty}_{l=1}\frac{1}{l}\frac{x^l-(x^{-1}pq)^l}{(1-p^l)(1-q^l)}, \quad q=e^{2\pi i\tau},p=e^{2\pi i\sigma},x=e^{2\pi i z}$$ I am interested in a certain limit of this function, where the imaginary parts of $\tau$ and $\sigma$ are taken to zero. Now, I have two methods at my disposal to compute this limit, which have differing domains of validity in $z$ space, but agree on the overlap. I have tried to abstract the problem in a way that does not refer to the specific details mentioned above.

Suppose $f(x+1)=f(x)$ is real analytic for $x\in \mathbb{R}\setminus \mathbb{Z}$. Moreover, it is once-differentiable at $x\in \mathbb{Z}$.

In addition, suppose that on the interval $(0,1)$, we know the analytic continuation into the complex plane. Let us call this function $g(z)$, where $0<\mathrm{Re}(z)<1$. Finally, we also know that $g(z+n)=g(z)$ for some integer $n>1$. So really, we know the analytic continuation of $f(x)$ for $\mathrm{Re}(z)\in (mn,mn+1)$ for $m\in\mathbb{Z}$.

I want to find the analytic continuation valid for any $\mathrm{Re}(z)\in \mathbb{R}\setminus \mathbb{Z}$. An obvious continuation is that one takes $g([z])$, where the fractional part is defined as: $$ \text{$[z]=z+k$, $k\in\mathbb{Z}$ such that $0<\mathrm{Re}(z)+k<1$}.$$ This function agrees with $f(x)$ on $\mathbb{R}\setminus \mathbb{Z}$. For the specific function at hand, it turns out that $g(z)$ at $z=0,1$ is once differentiable, even though the fractional part is not defined there. This matches with the behaviour of $f(x)$.

My question: is this continuation unique?

  • $\begingroup$ You mean $\lfloor z \rfloor = z + k$, $k \in \mathbb Z$ such that $0 < \text{Re}(z)+k < 1$. $\endgroup$ Apr 1, 2021 at 14:29
  • $\begingroup$ o yes corrected, thanks. $\endgroup$
    – sam
    Apr 1, 2021 at 14:53

1 Answer 1


Yes, of course. Since $g(\lfloor z \rfloor)$ is analytic on $k < \text{Re}(z) < k+1$, and agrees with $f(z)$ for $z \in (k,k+1)$, it is the unique analytic continuation of $f$ from $(k,k+1)$ to the strip $k < \text{Re}(z) < k+1$ by the identity theorem for analytic functions.

  • $\begingroup$ I see, ok thank you for the clear answer. $\endgroup$
    – sam
    Apr 1, 2021 at 14:55

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