Super-zeta functions and regularized determinants associated with cofinite Fuchsian groups with finite-dimensional unitary representations

Abstract

Let M be a finite volume, non-compact hyperbolic Riemann surface, possibly with elliptic fixed points, and let $\chi$ denote a finite dimensional unitary representation of the fundamental group of M. Let $\Delta$ denote the hyperbolic Laplacian which acts on smooth sections of the flat bundle over M associated with $\chi$. From the spectral theory of $\Delta$, there are three distinct sequences of numbers: the first coming from the eigenvalues of $L^{2}$ eigenfunctions, the second coming from resonances associated with the continuous spectrum, and the third being the set of negative integers. Using these sequences of spectral data, we employ the super-zeta approach to regularization and introduce two super-zeta functions, $\mathcal {Z}_-(s,z)$ and $\mathcal {Z}_+(s,z)$ that encode the spectrum of $\Delta$ in such a way that they can be used to define the regularized determinant of $\Delta -z(1-z)I$. The resulting formula for the regularized determinant of $\Delta -z(1-z)I$ in terms of the Selberg zeta function, see Theorem 5.3, encodes the symmetry $z\leftrightarrow 1-z$.

A Alternate expressions

The multiplicities $m_n$ of the trivial zeros of the Selberg zeta function carry important information related to the surface M and the character $\chi$. For this reason, we will deduce a different expression for $m_n$ (see Eq. (A.6)) from which it will be obvious that $m_n$ are nonnegative integers. Moreover, we will construct a different order-two entire function ${{\tilde{G}}}_1(s)$, given in terms of the gamma and the Barnes double gamma function (see Eq. (A.7)) and such that its null set coincides with the set of negative integers $-n$ with multiplicities $m_n$.

1.1 A.1 An alternate expression for the multiplicities $m_n$

We simplify (2.12) by combining it with (2.1) multiplied by $\frac{h}{2\pi }$ and (3.3) to get

$$\begin{aligned} m_n= & {} h\left( 2g-2 + \mathbf{c} + \mathbf{e} \right) (2n+1) \nonumber \\&- \sum _{j=1}^h \sum _{\{R\}_{\Gamma }} \frac{1}{d_R} \left( 2n+1 + \sum _{k=1}^{d_R-1} \left( \omega (R)^k_j \right) \frac{\sin \left( \frac{k\pi (2n+1)}{d_R}\right) }{\sin \left( \frac{k\pi }{d_R}\right) } \right) \end{aligned}$$

(A.1)

By (A.1), we can focus on the case of unitary characters rather than unitary representations.

Lemma A.1

Let $\omega$ be a unitary character of the finite cyclic group $\left\langle R \right\rangle$, where R is elliptic of order d. Further let $\omega (R)=\exp (2 \pi i q/d)$ for some integer q,  with $0 \le q \le d-1.$ Let $n \in \mathbb {N}= \{0,1,2,\dots \},$ then

$$\begin{aligned} \frac{1}{d}\left( 2n+1 + \sum _{k=1}^{d-1}\frac{\omega ^k(R)}{\sin \left( \frac{k\pi }{d}\right) }\sin \left( \frac{k\pi (2n+1)}{d}\right) \right) = \left| \{t \in \mathbb {Z}~|~ td \in \{-n+q,\dots n+q\} \} \right| , \end{aligned}$$

(A.2)

where |A| denotes the cardinality of the finite set A.

Proof

We apply the identity

$$\begin{aligned} \frac{\sin (k\pi (2n+1)/d)}{\sin (k\pi /d)}= \sum _{j=-n}^{n}\exp (2\pi i k j/d) \end{aligned}$$

and obtain

$$\begin{aligned}&\frac{1}{d}\left( 2n+1 + \sum _{k=1}^{d-1}\frac{\omega ^k(R)}{\sin \left( \frac{k\pi }{d}\right) }\sin \left( \frac{k\pi (2n+1)}{d}\right) \right) \nonumber \\&\quad = \frac{1}{d}(2n+1) + \sum _{k=1}^{d-1} \frac{1}{d} \exp (2 \pi i k q/d)\sum _{j=-n}^{n}\exp (2\pi i k j/d) \nonumber \\&\quad = \frac{1}{d}(2n+1) + \frac{1}{d} \sum _{j=-n}^{n} \sum _{k=1}^{d-1} \exp (2 \pi i kq/d) \exp (2\pi i k j/d) \nonumber \\&\quad = \frac{1}{d} \sum _{j=-n}^{n} \sum _{k=0}^{d-1} \exp (2 \pi i k (q+j)/d) \end{aligned}$$

(A.3)

The inner sum on the right is equal to d iff $d | (q+j).$ As j runs from $-n,-n+1,\dots ,n,$ the number of such j is equal to the number of multiples of d in the integer range $-n+q,\dots ,n+q.$ Recalling the factor $\tfrac{1}{d}$ out in front proves the lemma. $\square$

Lemma A.2

Let d, q, n be integers with $2 \le d, 0 \le q \le d-1,$ and $0 \le n.$

Then,

$$\begin{aligned} \left| \{t \in \mathbb {Z}~|~ td \in \{-n+q,\dots n+q\} \} \right| = \lfloor \frac{n+q}{d} \rfloor + \lfloor \frac{n+d-q}{d} \rfloor \end{aligned}$$

Here, $\lfloor x\rfloor$ denotes the floor function.

Proof

For $a \le b,$ both integers, define $f(a,b,d) = \left| \{t \in \mathbb {Z}~|~ td \in \{a,\dots b\} \} \right|$

Now, let a, b be integers with $a < 0$ and and $b > 0.$ Then,

$$\begin{aligned} f(a,b,d)= & {} f(-a,0,d) + f(0,b,d) - 1 = f(0,-a,d) + f(0,b,d) + 1\nonumber \\= & {} \lfloor \frac{d-a}{d} \rfloor + \lfloor \frac{b}{d} \rfloor \end{aligned}$$

(A.4)

To prove the lemma, we consider the following cases. The first case, $a = -n+q < 0,$ and $b=n+q > 0$ follows from (A.4).

In the case when $n=0,$ the lemma is trivial.

Next, consider the case where $0 < -n + q.$ Since $q < d,$ and $n \ge 0,$ it follows that $0<-n+q < d.$ If $n+q < d,$ then it follows that

$$\begin{aligned} f(-n+q,n+q,d) = 0 = \lfloor \frac{n+q}{d} \rfloor + \lfloor \frac{d + (n-q)}{d} \rfloor = 0 + 0, \end{aligned}$$

and the lemma is verified in this case.

Finally, we are left with the case $0<-n+q < d,$ and $d < n+q.$ We shift the integer interval $\{-n+q,\dots n+q\}$ to the left by d,  and obtain

$$\begin{aligned} \left| \{t \in \mathbb {Z}~|~ td \in \{-n+q,\dots n+q\} \} \right| = \left| \{t \in \mathbb {Z}~|~ td \in \{-n+q-d,\dots n+q-d\} \} \right| . \end{aligned}$$

We can apply (A.4) to the shifted interval, and we obtain

$$\begin{aligned} f(-n+q-d,n+q-d,d )= & {} \lfloor \frac{d-(-n+q-d)}{d} \rfloor + \lfloor \frac{n+q-d}{d} \rfloor \nonumber \\= & {} \lfloor 1 + \frac{n+d-q}{d} \rfloor + \lfloor -1 + \frac{n+q}{d} \rfloor \nonumber \\= & {} \lfloor \frac{n+d-q}{d} \rfloor + \lfloor \frac{n+q}{d} \rfloor , \end{aligned}$$

(A.5)

where the last equality follows because both $n+q$ and $n+d-q$ are positive. $\square$

One should note that Lemma A.2 is false for arbitrary n, q, d.

Combining (A.1), Lemma A.1, and Lemma A.2 we arrive at the following alternate expression for $m_n$:

$$\begin{aligned} m_n= & {} h\left( 2g-2 + \mathbf{c} + \mathbf{e} \right) (2n+1) \nonumber \\&- \sum _{\{R\}_{\Gamma }} \sum _{j=1}^h \left( \lfloor \frac{n+q(R)_j}{d_R} \rfloor + \lfloor \frac{n+d_R-q(R)_j}{d_R} \rfloor \right) \end{aligned}$$

(A.6)

1.2 A. 2 Double gamma function representation of the trivial zeros of the Selberg-zeta function

Recall that the Barnes double Gamma function, which is an entire order-two function defined by (2.4), has a zero of multiplicity n,  at each point $-n \in \{-1,-2,\dots \}.$ We start with the following lemma.

Lemma A.3

Let d, q, n be integers with $2 \le d, 0 \le q \le d-1,$ and $0 \le n.$ Set

$$\begin{aligned} g(n,q,d) = \lfloor \frac{n+q}{d} \rfloor + \lfloor \frac{n+d-q}{d} \rfloor . \end{aligned}$$

For $s \in \mathbb {C},$ define

$$\begin{aligned} G_{q,d}(s) = \prod _{m=0}^{d-1} G\left( \frac{s-q+m}{d} + 1 \right) G\left( \frac{s-(d-q)+m}{d} + 1 \right) . \end{aligned}$$

Then, $G_{q,d}(s)$ is an entire function with the set of zeros being the set of negative integers $\{-n:n = 1,2,\dots \}$ and each zero $s =-n$, $n = 1,2,\dots$ has multiplicity g(n, q, d).

Proof

We study $G_{q,d}(s)$ at $s = -n.$ Since $m \in \{0,\dots ,d-1\},$ the number $\frac{-n-q+m}{d}$ is a negative integer for exactly one value, say $m = m_1,$ in which case

$$\begin{aligned} \lfloor \frac{-n -q +m_1}{d} \rfloor = -\lfloor \frac{n+q}{d} \rfloor . \end{aligned}$$

Therefore, $\prod _{m=0}^{d-1} G\left( \frac{s-q+m}{d}+1\right)$ has a zero of order $\lfloor \frac{n+q}{d} \rfloor$ at $s = -n$ for any positive integer n. Similarly, $\prod _{m=0}^{d-1}G\left( \frac{s-(d-q)+m}{d} + 1 \right)$ has a zero of order $\lfloor \frac{n+q-d}{d} \rfloor$ at $s = -n.$

Moreover, $\frac{s-q+m}{d} = -k$ for some $k = 1,2,3,\ldots ,$ iff $s = -kd + q - m$ is a negative integer. This shows that there are no zeros of $\prod _{m=0}^{d-1} G\left( \frac{s-q+m}{d}+1\right)$ different from negative integers. A similar conclusion holds for $\prod _{m=0}^{d-1}G\left( \frac{s-(d-q)+m}{d} + 1 \right)$ and the proof is complete. $\square$

Recall that $\{R\}_{\Gamma }$ are the classes of elliptic elements of $\Gamma$ and that there are $\mathbf{e}$ of them. Further, recall the notation $\mathbf{c} ,h.$ Recall that $\omega (R)_j,$ for $j=1\dots h$ are the eigenvalues (and $d_R\mathrm {th}$-roots of unity) of $\chi (R)$, and that $q(R)_j,$ with $0 \le q(R)_j \le d_R-1$ is defined by (3.4).

Definition A.4

With the notation above, we define

$$\begin{aligned} G_E(s) = \prod _{\{R\}_{\Gamma }} \prod _{j=1}^{h} \prod _{m=0}^{d_R - 1} G\left( \frac{s-q(R)_j+m}{d_R} + 1 \right) G\left( \frac{s-\left( d_R-q(R)_j\right) +m}{d_R} + 1 \right) . \end{aligned}$$

and

$$\begin{aligned} {{\tilde{G}}}_1(s)= \left( G_E(s) \right) ^{-1}\left( \frac{(2\pi )^{-s} (G(s+1))^2}{\Gamma (s)}\right) ^{h(2g-2+\mathbf{c} + \mathbf{e} )} \end{aligned}$$

(A.7)

Lemma A.4

The function ${{\tilde{G}}}_{1}(s)$ is an entire of order two with zeros at points $-n \in -\mathbb {N}$ and corresponding multiplicities $m_n.$

Proof

The function $\left( \frac{(2\pi )^{-s} (G(s+1))^2}{\Gamma (s)}\right) ^{h(2g-2+\mathbf{c} + \mathbf{e} )}$ possesses zeros at points $s=-n$ with multiplicity $h(2g-2+\mathbf{c} + \mathbf{e} )(2n+1)$; hence, the statement follows by combining Eq. (A.1) with Lemmas A.1, A.2 and A.3. $\square$