0
$\begingroup$

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?

$\endgroup$
2
  • $\begingroup$ 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

3
$\begingroup$

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.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.