2 Eisenstein metrics

Let $\rho \colon \Gamma \to \operatorname {\mathrm {GL}}_{d}(\mathbf {C})$ be a representation of $\Gamma = \operatorname {\mathrm {SL}}_{2}(\mathbf {Z})$ . Let $\operatorname {\mathrm {Herm}}_{d}$ denote the real vector space of $d\times d$ Hermitian matrices, so that $M\in \operatorname {\mathrm {Herm}}_{d}$ means that $\bar M = M^{t}$ .

Definition 2.1 A metric for $\rho $ is a smooth function $H\colon \operatorname {\mathrm {\mathcal {H}}} \to \operatorname {\mathrm {Herm}}_{d}$ such that $H(\tau )$ is positive definite for all $\tau \in \operatorname {\mathrm {\mathcal {H}}}$ , and such that

(2.1) $$ \begin{align} \rho(\gamma)^{t} H(\gamma \tau)\overline{\rho(\gamma)} = H(\tau) \end{align} $$

holds for all $\gamma \in \Gamma $ .

Such functions can be used to define analogues of Petersson inner products on spaces of vector-valued modular forms associated to $\rho $ . For example, if F satisfies $F(\gamma \tau )=\rho (\gamma )F(\tau )$ for all $\gamma \in \Gamma $ , and similarly for G, then we can define an invariant scalar-valued form by the rule

$$ \begin{align*} \langle F,G\rangle_{\tau} := F(\tau)^{t}H(\tau)\overline{G(\tau)}. \end{align*} $$

Equation (2.1) implies that $\langle F,G\rangle _{\gamma \tau } = \langle F,G\rangle _{\tau }$ for all $\gamma \in \Gamma $ .

Let $h\colon \operatorname {\mathrm {\mathcal {H}}} \to \operatorname {\mathrm {Herm}}_{d}$ be a positive definite and smooth function that satisfies equation (2.1) for $\gamma =\pm T$ . More precisely, we assume that

(2.2) $$ \begin{align} \rho(T)^{t}h(\tau+1)\overline{\rho(T)} =& h(\tau), \end{align} $$

(2.3) $$ \begin{align} \rho(-I)^{t}h(\tau)\overline{\rho(-I)} =& h(\tau). \end{align} $$

If $\rho (-I)$ is a scalar matrix, then we must have $\rho (-I) = \pm I_{d}$ , and equation (2.3) is satisfied. This occurs, for example, if $\rho $ is irreducible.

Given an h as in the preceding paragraph, the corresponding Poincaré series is defined as usual by the formula

(2.4) $$ \begin{align} P(\rho,h,\tau) := \sum_{\gamma \in \langle \pm T\rangle\backslash \Gamma} \rho(\gamma)^{t}h(\gamma \tau)\overline{\rho(\gamma)}. \end{align} $$

This is well defined thanks to equations (2.2) and (2.3). Furthermore, if P converges absolutely, then we have

$$ \begin{align*} P(\rho,h,\alpha \tau) =&\sum_{\gamma \in \langle \pm T\rangle\backslash \Gamma} \rho(\gamma)^{t}h(\gamma \alpha\tau)\overline{\rho(\gamma)}\\ =&\sum_{\gamma \in \langle \pm T\rangle\backslash \Gamma} \rho(\gamma\alpha^{-1})^{t}h(\gamma\tau)\overline{\rho(\gamma\alpha^{-1})}\\ =& \rho(\alpha)^{-t}P(h,\tau)\overline{\rho(\alpha)}^{-1}, \end{align*} $$

for all $\alpha \in \Gamma $ . Thus, after moving the $\rho $ -terms to the other side, one sees that $P(h,\tau )$ satisfies equation (2.1) as a function of $\tau $ . If h is chosen so that $P(\rho ,h,\tau )$ converges absolutely to a smooth function, then the series $P(\rho ,h,\tau )$ defines a metric for $\rho $ .

Let $g \in \operatorname {\mathrm {GL}}_{d}(\mathbf {C})$ . Notice that if h satisfies equations (2.2) and (2.3) for $\rho $ , then $g^{-t}h\overline {g}^{-1}$ satisfies equations (2.2) and (2.3) for $g\rho g^{-1}$ .

Lemma 2.3 Assuming that both Poincaré series converge absolutely, then one has

$$ \begin{align*} P(g\rho g^{-1}, g^{-t}h\bar g^{-1},\tau) = g^{-t}P(\rho,h,\tau)\bar g^{-1}. \end{align*} $$

Proof The proof is a direct computation:

$$ \begin{align*} P(g\rho g^{-1}, g^{-t}h\bar g^{-1},\tau) =&\sum_{\gamma \in \langle \pm T\rangle\backslash \Gamma} (g\rho(\gamma)g^{-1})^{t}g^{-t}h(\gamma \alpha\tau)\bar g^{-1}\bar g\overline{\rho(\gamma)}\bar g^{-1}\\ =& g^{-t}P(\rho,h,\tau)\bar g^{-1}. \\[-32pt]\end{align*} $$

▪

Our next goal is to describe convenient choices of functions h satisfying equations (2.2) and (2.3). To this end, we introduce the notion of exponents. Define ${\mathbf {e}}(M) = e^{2\pi i M}$ for matrices M.

Since the matrix exponential is surjective, exponents always exist. They can be described explicitly in terms of a Jordan canonical form for $\rho (T)$ (cf. Theorem 3.7 of [Reference Candelori and Franc2]), and this description shows that if X commutes with $\rho (T)$ , then X also commutes with a choice of exponents.

Let L be a choice of exponents for $\rho $ . Since the matrix exponential satisfies ${\mathbf {e}}(X+Y) = {\mathbf {e}}(X)\,{\mathbf {e}}(Y)$ provided that $XY=YX$ , it follows that

$$ \begin{align*}{\mathbf{e}}(L(\tau+1)) = \rho(T)\,{\mathbf{e}}(L\tau)={\mathbf{e}}(L\tau)\rho(T).\end{align*} $$

Therefore, for any $h\in \operatorname {\mathrm {Herm}}_{d}$ , the function

(2.5) $$ \begin{align} h(\tau) = {\mathbf{e}}(-L\tau)^{t}h\overline{{\mathbf{e}}(-L\tau)} \end{align} $$

is Hermitian and satisfies equation (2.2). However, $h(\tau )$ need not satisfy equation (2.3), and it need not be positive definite in general. Therefore, we introduce a set of admissible choices for h.

Definition 2.5 Given a representation $\rho \colon \Gamma \to \operatorname {\mathrm {GL}}_{d}(\mathbf {C})$ , define

$$ \begin{align*} \operatorname{\mathrm{Pos}}(\rho) := \{h\in \operatorname{\mathrm{Herm}}_{d} \mid h \textrm{ is positive definite and } \rho(-I)^{t}h\overline{\rho(-I)}=h\}. \end{align*} $$

Definition 2.8 Let $\rho $ denote a representation of $\Gamma $ , and let L denote a choice of exponents for $\rho $ . Then, the associated Eisenstein metric is the infinite series

$$ \begin{align*} H(\rho,L,h,\tau,s) := \sum_{\gamma \in \langle \pm T\rangle\backslash \Gamma} \rho(\gamma)^{t}{\mathbf{e}}(-L^{t}(\gamma \cdot \tau))h\overline{{\mathbf{e}}(-L(\gamma \cdot \tau))}\overline{\rho(\gamma)}\operatorname{Im}(\gamma\cdot\tau)^{s}, \end{align*} $$

where $h \in \operatorname {\mathrm {Pos}}(\rho )$ and $\tau \in \operatorname {\mathrm {\mathcal {H}}}$ .

In the definition above, the dot in $\gamma \cdot \tau $ denotes the action of $\Gamma $ on $\operatorname {\mathrm {\mathcal {H}}}$ , not matrix multiplication. All other products in the expression defining Eisenstein metrics are ordinary matrix products. When $\rho $ is trivial, $L=0$ , and $h=1$ , then this definition gives the usual Eisenstein series.

The formation of $H(\rho ,L,h,\tau ,s)$ is linear in h. We will often write $H(h,\tau ,s)$ for these Eisenstein metrics when the dependence on $\rho $ and L is clear. Our next goal is to prove that $H(h,\tau ,s)$ converges absolutely if the real part of s is large enough. The proof is analogous to the proof of Theorem 3.2 in [Reference Knopp and Mason18]. We will show that one has convergence when $\operatorname {Re}(s)$ is large even if $(\rho ,L)$ varies in a family, as in the following definitions.

Definition 2.9 Let $U\subseteq \mathbf {C}^{m}$ denote an open subset. A family of representations for $\Gamma $ varying analytically over U consists of an analytic map

$$ \begin{align*} \rho \colon U \to \operatorname{\mathrm{Hom}}(\Gamma,\operatorname{\mathrm{GL}}_{n}(\mathbf{C})). \end{align*} $$

If $\rho $ is a family of representations, then we usually write $\rho _{u}$ for the value of $\rho $ at $u \in U$ . For each $\gamma \in \Gamma $ , we obtain a function $\rho (\gamma ) \colon U \to \operatorname {\mathrm {GL}}_{n}(\mathbf {C})$ , and the analyticity of $\rho $ consists of the analyticity of these maps. It suffices to test analyticity on a set of generators for $\Gamma $ . If $\mathcal {O}(U)$ denotes the ring of analytic functions on U, then a family of representations on U is the same thing as a homomorphism

$$ \begin{align*} \rho \colon \Gamma \to \operatorname{\mathrm{GL}}_{n}(\mathcal{O}(U)). \end{align*} $$

As above, we will sometimes write $L_{u}$ for the exponents evaluated at a point $u \in U$ .

Example 2.11 There exists an analytic family of representations on $\mathbf {C}^{\times }$ determined by

$$ \begin{align*} \rho(T) &= \left(\begin{matrix}u&u\\0&u^{-1}\end{matrix}\right), & \rho(S) &= \left(\begin{matrix}0&-u\\u^{-1}&0\end{matrix}\right). \end{align*} $$

The specializations $\rho _{u}$ are irreducible as long as $u^{4}-u^{2}+1 \neq 0$ . Since $\operatorname {\mathrm {Tr}}(\rho (T)) = u+u^{-1}$ , one sees that this family is nontrivial, in the sense that not all fibers are isomorphic representations, although one does have $\rho _{u} \cong \rho _{u^{-1}}$ . Modulo this identification, this describes one component of the universal family of irreducible representations of $\Gamma $ of rank two (cf. [Reference Mason22, Reference Tuba and Wenzl29]).

Let $\log $ denote the branch of the complex logarithm such that the imaginary parts of $\log (z)$ are contained in $[0,2\pi )$ . Then, a choice of exponents defined on $\mathbf {C} \setminus \mathbf {R}_{\geq 0}$ is

$$ \begin{align*} L = \frac{1}{2\pi i}\left(\begin{matrix}\log(u)&\frac{(\log(u)-\log(u^{-1}))u^{2}}{u^{2}-1}\\0&\log(u^{-1})\end{matrix}\right). \end{align*} $$

The apparent singularity at $u=-1$ is removable, whereas the singularity on the branch cut at $u=1$ is not. These values of u correspond to the specializations of $\rho $ where $\rho (T)$ is not diagonalizable.

Note that while each fiber $L_{u}$ of a choice of exponents can be put into Jordan canonical form, the existence of an analytic choice of change of basis matrix P putting L simultaneously into Jordan canonical form at all points of U is not guaranteed.

Definition 2.13 Let $\rho $ be a family of representations of $\Gamma $ on a set U. Define

$$ \begin{align*} \operatorname{\mathrm{Pos}}(\rho) := \bigcap_{u\in U} \operatorname{\mathrm{Pos}}(\rho_{u}). \end{align*} $$

Proof The proof is similar to the proof of Theorem 3.3 of [Reference Knopp and Mason17], although there are enough differences in the statements of these results that we repeat some details. If M is a matrix, then let $\left \lVert M\right \rVert $ denote the supremum norm on its entries. Then, we have

$$ \begin{align*} \left\lVert H(h,\tau,s)\right\rVert \leq 1+\left\lVert h\right\rVert\sum_{c=1}^{\infty}\sum_{\substack{d\in\mathbf{Z}\\ \gcd(c,d)=1}} \left\lVert \rho(\gamma)\right\rVert^{2}\left\lVert {\mathbf{e}}(-L(\gamma \cdot \tau))\right\rVert^{2}\frac{y^{s}}{\left\lvert c\tau+d\right\rvert^{2s}}, \end{align*} $$

where $\gamma = \left (\begin {smallmatrix}a&b\\c&d\end {smallmatrix}\right )$ for some choice of $a,b\in \mathbf {Z}$ . After possibly replacing $\gamma $ by $T^{n}\gamma $ , we may assume that $0\leq \operatorname {Re}(\gamma \cdot \tau ) < 1$ . We then have the basic estimate

$$ \begin{align*}\left\lvert \gamma \cdot \tau\right\rvert^{2} \leq 1+\operatorname{Im}(\gamma \cdot \tau)^{2}=1+\frac{y^{2}}{((cx+d)^{2}+c^{2}y^{2})^{2}}\leq 1+\frac{1}{c^{4}y^{2}}.\end{align*} $$

Use the existence of the Jordan canonical form on U to write $-L$ in the form $-L=P(D+N)P^{-1}$ where D is diagonal, $DN=ND$ , and $N^{d}=0$ , where $d=\dim \rho $ . We obtain the estimate

$$ \begin{align*} \left\lVert {\mathbf{e}}(-L(\gamma \cdot \tau))\right\rVert^{2} \leq &\left\lVert P\right\rVert^{2}\left\lVert P^{-1}\right\rVert^{2}\left\lVert {\mathbf{e}}(D(\gamma \cdot \tau))\right\rVert^{2}\left\lVert {\mathbf{e}}(N(\gamma \cdot \tau))\right\rVert^{2}\\ \leq &\left\lVert P\right\rVert^{2}\left\lVert P^{-1}\right\rVert^{2}\left\lVert {\mathbf{e}}(D(\gamma \cdot \tau))\right\rVert^{2}\sum_{k=0}^{d-1}\frac{(2\pi)^{k}}{k!}\left\lVert N\right\rVert^{2k}\left\lvert \gamma \cdot \tau\right\rvert^{2k}. \end{align*} $$

Thus, if we set $C_{1} = e\max (1,\left \lVert P\right \rVert ^{2},\left \lVert P^{-1}\right \rVert ^{2}, \left \lVert N\right \rVert ,\left \lVert N^{2}\right \rVert ,\ldots , \left \lVert N^{d-1}\right \rVert )$ , then we deduce that

$$ \begin{align*} \left\lVert {\mathbf{e}}(-L(\gamma \cdot \tau))\right\rVert^{2} \leq C_{1}\left\lVert {\mathbf{e}}(D(\gamma \cdot \tau))\right\rVert^{2}e^{\frac{2\pi}{c^{4}y^{2}}}. \end{align*} $$

To continue, write $D = U+iV$ where U and V are real diagonal matrices, so that in particular $UV=VU$ . Then,

$$ \begin{align*} \left\lVert {\mathbf{e}}(D(\gamma \cdot \tau))\right\rVert^{2} \leq &\left\lVert {\mathbf{e}}(U(\gamma \cdot \tau))\right\rVert^{2}\left\lVert {\mathbf{e}}(iV(\gamma \cdot \tau))\right\rVert^{2}\\ =&\left\lVert {\mathbf{e}}(iU\operatorname{Im}(\gamma \cdot \tau))\right\rVert^{2}\left\lVert {\mathbf{e}}(iV\operatorname{Re}(\gamma \cdot \tau))\right\rVert^{2}\\ =&\left\lVert e^{-\frac{2\pi yU}{\left\lvert c\tau+d\right\rvert^{2}}}\right\rVert^{2}\left\lVert e^{-2\pi V\operatorname{Re}(\gamma \cdot \tau)}\right\rVert^{2}\\ \leq & C_{2}\left\lVert e^{-\frac{2\pi yU}{\left\lvert c\tau+d\right\rvert^{2}}}\right\rVert^{2}, \end{align*} $$

where $C_{2} = \max (1,\left \lVert e^{-2\pi V}\right \rVert ^{2})$ . Putting these estimates together, we have shown that if we normalize our representatives $\gamma $ for cosets in $\langle \pm T\rangle \backslash \Gamma $ such that $0\leq \operatorname {Re}(\gamma \tau ) < 1$ , then

(2.6) $$ \begin{align} \left\lVert H(h,\tau,s)\right\rVert \leq 1+C_{1}C_{2}\left\lVert h\right\rVert y^{s}\sum_{c=1}^{\infty}\sum_{\substack{d\in\mathbf{Z}\\ \gcd(c,d)=1}} \left\lVert \rho(\gamma)\right\rVert^{2}\left\lVert e^{-\frac{2\pi yU}{\left\lvert c\tau+d\right\rvert^{2}}}\right\rVert^{2}\frac{e^{\frac{2\pi}{c^{4}y^{2}}}}{\left\lvert c\tau+d\right\rvert^{2s}}. \end{align} $$

It remains to estimate $\left \lVert \rho (\gamma )\right \rVert $ . For this, one can use Corollary 3.5 of [Reference Knopp and Mason20] to obtain an estimate for this term that is polynomial in $c^{2}+d^{2}$ , with constants and degree that depend only on $\rho $ . Since the exponential factors in equation (2.6) converge to $1$ as c grows, this estimate is sufficient to prove part (1) of the proposition. All constants in these various estimates can be chosen uniformly on K, so that (2) follows as well.▪

3 Tame harmonic metrics

Nonabelian Hodge theory describes a correspondence between categories of representations of fundamental groups, and categories of Higgs bundles on the underlying base manifold. A key ingredient for establishing such correspondences lies in proving the existence of metrics satisfying a nonlinear differential equation as in the following definition.

Definition 3.1 A Hermitian positive definite metric $H \colon \operatorname {\mathrm {\mathcal {H}}} \to M_{d}(\mathbf {C})$ for a representation $\rho $ is said to be harmonic provided that

$$ \begin{align*} \partial \bar \partial \log(H) = \tfrac 12 [\bar\partial \log(H),\partial \log(H)], \end{align*} $$

where $\partial \log (H) = H^{-1}\partial (H)$ , $\bar \partial \log (H) = H^{-1}\bar \partial (H)$ .

Proof First, since $\rho (\gamma )^{t}H(\gamma \tau )\overline {\rho (\gamma )} = H(\tau )$ , we find that

$$ \begin{align*} g^{t}H(\tau)\bar g &= g^{t}\rho(\gamma)^{t}H(\gamma \tau)\overline{\rho(\gamma)g}\\ &= (g^{-1}\rho(\gamma)g)^{t}g^{t}H(\gamma\tau)\bar g (\overline{g^{-1}\rho(\gamma)g}). \end{align*} $$

Next, notice that

$$ \begin{align*} \partial\log(g^{t}H\bar g) =(g^{t}H\bar g)^{-1}\partial(g^{t}H\bar g)=\bar g^{-1}H^{-1}g^{-t}g^{t}\partial(H)\bar g = \bar g^{-1}\partial \log(H) \bar g, \end{align*} $$

and similarly for $\bar \partial \log (g^{t}H\bar g)$ . Therefore,

$$ \begin{align*} \partial\bar\partial\log(g^{t}H\bar g ) &=\bar g^{-1}\partial\bar\partial \log(H) \bar g\\ &= \tfrac 12\bar g^{-1}[\bar\partial \log(H),\partial \log(H)] \bar g\\ &= \tfrac 12\bar [\bar g^{-1}\bar\partial \log(H)\bar g,\bar g^{-1}\partial \log(H) \bar g]\\ &=\tfrac 12[\bar\partial \log(g^{t}H\bar g),\partial \log(g^{t}H\bar g)].\\[-36pt] \end{align*} $$

▪

Definition 3.4 A metric $H \colon \operatorname {\mathrm {\mathcal {H}}} \to M_{d}(\mathbf {C})$ for a representation $\rho $ is said to be tame, or of slow growth, for a choice of exponents L provided that there exist constants C, N such that

$$ \begin{align*} \left\lVert {\mathbf{e}}(L^{t}\tau)H(\tau)\overline{{\mathbf{e}}(L\tau)}\right\rVert \leq Cy^{N}. \end{align*} $$

Tameness arises naturally in [Reference Simpson27] when considering correspondences between stable connections and stable Higgs bundles. In general, it can be difficult to write down explicit examples of harmonic metrics. Our interest in Eisenstein metrics is that they provide analytic families of metrics with appropriate invariance properties under the action of $\Gamma $ . The question is whether some specialization, residue, or some other metric derived from $H(\tau ,h,s)$ could satisfy the harmonicity and tameness conditions. Moreover, since the formation of Eisenstein metrics is well adapted to working with families of representations, one might obtain universal families of harmonic metrics living over moduli spaces of Higgs bundles. At present, no general results in this direction are known, although we discuss some preliminary examples below.

4 The inclusion representation

Suppose that $\rho \colon \Gamma \hookrightarrow \operatorname {\mathrm {GL}}_{2}(\mathbf {C})$ is the inclusion representation. In this case, a harmonic metric is known to be

(4.1) $$ \begin{align} K(\tau) = \frac{1}{y}\left(\begin{matrix}1&-x\\-x&x^{2}+y^{2}\end{matrix}\right), \end{align} $$

where $\tau = x+iy$ . This case is rather special, since $\rho $ in fact extends to the ambient Lie group $\operatorname {\mathrm {SL}}_{2}(\mathbf {R})$ . In fact, this metric is an example of a totally geodesic metric as in Example 4.4 of [Reference Franc and Rayan10], or Example 14.1.2 of [Reference Carlson, Müller-Stach and Peters3]. In this section, we show that this metric $K(\tau )$ arises as a residue of Eisenstein metrics.

Notice that since in this case $\rho (-I) = -I$ , the set $\operatorname {\mathrm {Pos}}(\rho )$ is the set of all positive definite matrices in $\operatorname {\mathrm {Herm}}_{2}$ . The possible exponent choices L take the form $2\pi i L = \left (\begin {smallmatrix}2\pi i n&1\\0&2\pi i n\end {smallmatrix}\right )$ for integers $n\in \mathbf {Z}$ . Fix $n \in \mathbf {Z}$ and observe that ${\mathbf {e}}(L\tau )=q^{n}\left (\begin {smallmatrix}1&\tau \\0&1\end {smallmatrix}\right )$ , where $q = {\mathbf {e}}(\tau )$ . Thus, if for our choice of $h\in \operatorname {\mathrm {Pos}}(\rho )$ we take $h=I$ , then

$$ \begin{align*} h(\tau) = \left\lvert q\right\rvert^{2n}\left(\begin{matrix}1&0\\-\tau&1\end{matrix}\right)\left(\begin{matrix}1&-\overline\tau\\0&1\end{matrix}\right)=e^{-4\pi ny}\left(\begin{matrix}1&-\bar\tau\\-\tau&1+\left\lvert \tau\right\rvert^{2}\end{matrix}\right). \end{align*} $$

Write $H(\tau ,s) = H(\rho ,L,I,\tau ,s)$ and compute

$$ \begin{align*} H(\tau,s) &= y^{s}h(\tau)+\sum_{c\geq 1}\sum_{\gcd(c,d) =1} \left(\begin{matrix}a&c\\b&d\end{matrix}\right)\left(\begin{matrix}1&0\\-\gamma \tau&1\end{matrix}\right)\left(\begin{matrix}1&-\gamma \bar\tau\\0&1\end{matrix}\right)\left(\begin{matrix}a&b\\c&d\end{matrix}\right)\\ &\quad \times e^{-4\pi n \frac{y}{\left\lvert c\tau+d\right\rvert^{2}}}\frac{y^{s}}{\left\lvert c\tau+d\right\rvert^{2s}}\\ &= y^{s}h(\tau)+\sum_{c\geq 1}\sum_{\gcd(c,d) =1} \left(\begin{matrix}a-c\gamma \tau&c\\b-d\gamma\tau&d\end{matrix}\right)\left(\begin{matrix}a-c\gamma\bar\tau&b-d\gamma\bar\tau\\c&d\end{matrix}\right)\\&\quad \times e^{-4\pi n \frac{y}{\left\lvert c\tau+d\right\rvert^{2}}}\frac{y^{s}}{\left\lvert c\tau+d\right\rvert^{2s}}\\ &= y^{s}h(\tau)+\sum_{c\geq 1}\sum_{\gcd(c,d) =1} \left(\begin{matrix}1&c\\-\tau&d\end{matrix}\right)\left(\begin{matrix}\left\lvert c\tau+d\right\rvert^{-2}&0\\0&1\end{matrix}\right)\left(\begin{matrix}1&-\bar\tau\\c&d\end{matrix}\right)\\&\quad \times e^{-4\pi n \frac{y}{\left\lvert c\tau+d\right\rvert^{2}}}\frac{y^{s}}{\left\lvert c\tau+d\right\rvert^{2s}}\\ &= y^{s}h(\tau)+\sum_{c\geq 1}\sum_{\gcd(c,d) =1} \left(\begin{matrix}\left\lvert c\tau+d\right\rvert^{-2}&c\\-\tfrac{\tau}{\left\lvert c\tau+d\right\rvert^{2}}&d\end{matrix}\right)\left(\begin{matrix}1&-\bar\tau\\c&d\end{matrix}\right)e^{-4\pi n \frac{y}{\left\lvert c\tau+d\right\rvert^{2}}}\frac{y^{s}}{\left\lvert c\tau+d\right\rvert^{2s}}\\ &= y^{s}h(\tau)+\sum_{c\geq 1}\sum_{\gcd(c,d) =1} \left(\begin{matrix}c^{2}+\frac{1}{\left\lvert c\tau+d\right\rvert^{2}}&cd-\tfrac{\bar\tau}{\left\lvert c\tau+d\right\rvert^{2}}\\cd-\tfrac{\tau}{\left\lvert c\tau+d\right\rvert^{2}}&d^{2}+\left\lvert \tfrac{\tau}{c\tau+d}\right\rvert^{2}\end{matrix}\right)e^{-4\pi n \frac{y}{\left\lvert c\tau+d\right\rvert^{2}}}\frac{y^{s}}{\left\lvert c\tau+d\right\rvert^{2s}}\\ &= y^{s}h(\tau)+\sum_{k\geq 0}\frac{(-4\pi n)^{k}}{k!}\sum_{c\geq 1}\sum_{\gcd(c,d) =1} \left(\begin{matrix}c^{2}+\frac{1}{\left\lvert c\tau+d\right\rvert^{2}}&cd-\tfrac{\bar\tau}{\left\lvert c\tau+d\right\rvert^{2}}\\cd-\tfrac{\tau}{\left\lvert c\tau+d\right\rvert^{2}}&d^{2}+\left\lvert \tfrac{\tau}{c\tau+d}\right\rvert^{2}\end{matrix}\right)\\&\quad \times \frac{y^{s+k}}{\left\lvert c\tau+d\right\rvert^{2(s+k)}}. \end{align*} $$

Notice that with $E(\tau ,s) = y^{s}+\sum _{c\geq 1}\sum _{\gcd (c,d)=1}\frac {y^{s}}{\left \lvert c\tau +d\right \rvert ^{2s}}$ equal to the usual real-analytic Eisenstein series, we have

(4.2) $$ \begin{align} \nonumber H(\tau,s) =& e^{-4\pi ny}y^{s}\left(\begin{matrix}0&0\\0&1\end{matrix}\right)+y^{-1}\sum_{k\geq 0}\frac{(-4\pi n)^{k}}{k!}E(\tau,s+k+1)\left(\begin{matrix}1&-\bar\tau\\-\tau&\left\lvert \tau\right\rvert^{2}\end{matrix}\right)+\\ &\sum_{k\geq 0}\frac{(-4\pi n)^{k}}{k!}\sum_{c\geq 1}\sum_{\gcd(c,d) =1} \left(\begin{matrix}c^{2}&cd\\cd&d^{2}\end{matrix}\right)\frac{y^{s+k}}{\left\lvert c\tau+d\right\rvert^{2(s+k)}}. \end{align} $$

We now focus on the final terms using a standard approach:

$$ \begin{align*} &\sum_{c\geq 1}\sum_{\gcd(c,d) =1} \left(\begin{matrix}c^{2}&cd\\cd&d^{2}\end{matrix}\right)\frac{y^{s}}{\left\lvert c\tau+d\right\rvert^{2s}}\\ =&\sum_{c\geq 1}c^{2-2s}\sum_{\substack{d=1\\ \gcd(c,d)=1}}^{c}\sum_{u\in\mathbf{Z}}\left(\begin{matrix}1&\tfrac dc + u\\\tfrac dc + u&(\tfrac dc+u)^{2}\end{matrix}\right)\frac{y^{s}}{\left\lvert \tau+\tfrac dc+u\right\rvert^{2s}}\\ =&\left(\begin{matrix}1&0\\-\tau&1\end{matrix}\right)\left(\sum_{c\geq 1}c^{2-2s}\sum_{\substack{d=1\\ \gcd(c,d)=1}}^{c}\sum_{u\in\mathbf{Z}} \left(\begin{matrix}1&\bar\tau+\tfrac dc + u\\\tau+\tfrac dc + u&\left\lvert \tau+\tfrac dc+u\right\rvert^{2}\end{matrix}\right)\frac{y^{s}}{\left\lvert \tau+\tfrac dc+u\right\rvert^{2s}}\!\right)\!\left(\begin{matrix}1&-\bar\tau\\0&1\end{matrix}\right). \end{align*} $$

If $f(\tau )$ denotes the sum over u in the previous line, then $f(\tau +1)=f(\tau )$ and the Poisson summation formula may be used. We must first compute the Fourier transform of the summand terms: if ${\mathbf {e}}(z) := e^{2\pi i z}$ , then the Fourier transform is

$$ \begin{align*} &y^{s}\int_{-\infty}^{\infty} \left(\begin{matrix}1&x-iy+\tfrac dc + t\\x+iy+\tfrac dc + t&\left\lvert x+iy+\tfrac dc+t\right\rvert^{2}\end{matrix}\right)\frac{{\mathbf{e}}(-mt)}{\left\lvert x+iy+\tfrac dc+t\right\rvert^{2s}}dt\\ =&y^{s}{\mathbf{e}}(mx + m\tfrac dc)\int_{-\infty}^{\infty} \left(\begin{matrix}1&r-iy\\r+iy&r^{2}+y^{2}\end{matrix}\right)\frac{{\mathbf{e}}(-mr)}{(r^{2}+y^{2})^{s}}dr. \end{align*} $$

Let $g(m,y,s) = y^{s}\int _{-\infty }^{\infty } \left (\begin {smallmatrix}1&r-iy\\r+iy&r^{2}+y^{2}\end {smallmatrix}\right )\tfrac {{\mathbf {e}}(-mr)}{(r^{2}+y^{2})^{s}}dr$ . Then, putting everything together, we have shown that

(4.3) $$ \begin{align} \nonumber H(\tau,s) &= y^{s}e^{-4\pi ny}\left(\begin{matrix}0&0\\0&1\end{matrix}\right)+y^{-1}\sum_{k\geq 0}\frac{(-4\pi n)^{k}}{k!}E(\tau,s+k+1)\left(\begin{matrix}1&-\bar\tau\\-\tau&\left\lvert \tau\right\rvert^{2}\end{matrix}\right)+\left(\begin{matrix}1&0\\-\tau&1\end{matrix}\right)\\ & \quad \times \left(\sum_{k\geq 0}\frac{(-4\pi n)^{k}}{k!}\sum_{m\in\mathbf{Z}}{\mathbf{e}}(mx)g(m,y,s+k)\sum_{c\geq 1}c^{2-2(s+k)}\sum_{\substack{d=1\\ \gcd(c,d)=1}}^{c} {\mathbf{e}}(m\tfrac dc)\right)\\&\quad \times \left(\begin{matrix}1&-\bar\tau\\0&1\end{matrix}\right). \nonumber\end{align} $$

Recall that we have the following evaluation of the Ramanujan sum:

$$ \begin{align*} \sum_{\substack{d=1\\ \gcd(c,d)=1}}^{c} {\mathbf{e}}(m\tfrac dc)=\begin{cases} \sum_{g\mid \gcd(c,m)}\mu(\tfrac cg)g, & m\neq 0,\\ \phi(c), & m=0, \end{cases} \end{align*} $$

and therefore

$$ \begin{align*} \sum_{c\geq 1}c^{2-2(s+k)}\sum_{\substack{d=1\\ \gcd(c,d)=1}}^{c} {\mathbf{e}}(m\tfrac dc)=\begin{cases} \frac{\sigma_{3-2(s+k)}(\left\lvert m\right\rvert)}{\zeta(2(s+k)-2)}, & m\neq 0,\\ \frac{\zeta(2(s+k)-3)}{\zeta(2(s+k)-2)},&m=0. \end{cases} \end{align*} $$

That is, we have shown the following: $H(\tau ,s) = \left (\begin {smallmatrix}1&0\\-\tau &1\end {smallmatrix}\right ) \tilde H(\tau ,s) \left (\begin {smallmatrix}1&-\bar \tau \\0&1\end {smallmatrix}\right )$ where

(4.4) $$ \begin{align} \nonumber \tilde H(\tau,s) &= y^{s}e^{-4\pi ny}\left(\begin{matrix}0&0\\0&1\end{matrix}\right)+y^{-1}\sum_{k\geq 0}\frac{(-4\pi n)^{k}}{k!}E(\tau,s+k+1)\left(\begin{matrix}1&0\\0&0\end{matrix}\right)\\ &\quad + \sum_{k\geq 0}\frac{(-4\pi n)^{k}}{k!}\left(\frac{\zeta(2(s+k)-3)}{\zeta(2(s+k)-2)}g(0,y,s+k)\right.\nonumber\\&\quad + \left. \sum_{\substack{m\in\mathbf{Z}\\ m \neq 0}}\frac{{\mathbf{e}}(mx)\sigma_{3-2(s+k)}(\left\lvert m\right\rvert)g(m,y,s+k)}{\zeta(2(s+k)-2)}\right). \end{align} $$

To conclude the computation of the Fourier coefficients of $H(\tau ,s)$ , it now remains to give a more concrete expression for $g(n,y)$ . Equations (3.18) and (3.19) of [Reference Iwaniec15] yield

(4.5) $$ \begin{align} \int_{-\infty}^{\infty} \frac{{\mathbf{e}}(-mr)}{(r^{2}+y^{2})^{s}}dr = \begin{cases} \pi^{\frac 12}\frac{\Gamma(s-\frac 12)}{\Gamma(s)}y^{1-2s}, & m=0,\\ 2\pi^{s}\Gamma(s)^{-1}\left\lvert m\right\rvert^{s-\frac 12}y^{-s+\frac 12}K_{s-\frac 12}(2\pi \left\lvert m\right\rvert y), & m\neq 0. \end{cases} \end{align} $$

This allows us to evaluate the diagonal terms in $g(m,y,s)$ . It remains to treat the antidiagonal terms, and we first suppose $m\neq 0$ . By the product rule, and since $\operatorname {Re}(s) \gg 0$ ,

$$ \begin{align*} & \int_{-\infty}^{\infty} \frac{r{\mathbf{e}}(-mr)}{(r^{2}+y^{2})^{s}}dr \\ =&-\frac{1}{2\pi in}\lim_{N \to \infty}\left. \frac{r{\mathbf{e}}(-mr)}{(r^{2}+y^{2})^{s}}\right|^{N}_{-N}\!+\frac{1}{2\pi im}\!\int_{-\infty}^{\infty}\!\frac{((r^{2}+y^{2})^{s}-2sr^{2}(r^{2}+y^{2})^{s-1}){\mathbf{e}}(-mr)}{(r^{2}+y^{2})^{2s}}dr\\ =&\frac{1}{2\pi i m}\int_{-\infty}^{\infty}\frac{{\mathbf{e}}(-mr)}{(r^{2}+y^{2})^{s}}dr-\frac{2s}{2\pi i m}\int_{-\infty}^{\infty}\frac{{\mathbf{e}}(-mr)}{(r^{2}+y^{2})^{s}}dr+\frac{2sy^{2}}{2\pi i m}\int_{-\infty}^{\infty}\frac{{\mathbf{e}}(-mr)}{(r^{2}+y^{2})^{s+1}}dr. \end{align*} $$

That is, for $m\neq 0$ , we have shown that

$$ \begin{align*} &g(m,y,s)\\ &= y^{s}\!\!\!\!\int_{-\infty}^{\infty} \!\!\left(\!\begin{matrix}\frac{{\mathbf{e}}(-mr)}{(r^{2}+y^{2})^{s}}&\!\!\left(\!\frac{1-2s}{2\pi i m}-iy\!\right)\frac{{\mathbf{e}}(-mr)}{(r^{2}+y^{2})^{s}}+\frac{2sy^{2}}{2\pi i m}\frac{{\mathbf{e}}(-mr)}{(r^{2}+y^{2})^{s+1}}\\\left(\!\frac{1-2s}{2\pi i m}+iy\!\right)\frac{{\mathbf{e}}(-mr)}{(r^{2}+y^{2})^{s}}+\frac{2sy^{2}}{2\pi i m}\frac{{\mathbf{e}}(-mr)}{(r^{2}+y^{2})^{s+1}}&\!\!\frac{{\mathbf{e}}(-mr)}{(r^{2}+y^{2})^{s-1}}\end{matrix}\!\!\right)dr\\ &= y^{s}\left(\begin{matrix}1&\frac{1-2s}{2\pi i m}-iy\\\frac{1-2s}{2\pi i m}+iy&0\end{matrix}\right)2\pi^{s}\Gamma(s)^{-1}\left\lvert m\right\rvert^{s-\frac 12}y^{-s+\frac 12}K_{s-\frac 12}(2\pi \left\lvert m\right\rvert y)\\ &\quad + y^{s}\left(\begin{matrix}0&\tfrac{2sy^{2}}{2\pi i m}\\\tfrac{2sy^{2}}{2\pi i m}&0\end{matrix}\right)2\pi^{s+1}\Gamma(s+1)^{-1}\left\lvert m\right\rvert^{s+\frac 12}y^{-s-\frac 12}K_{s+\frac 12}(2\pi \left\lvert m\right\rvert y)\\ & \quad + y^{s}\left(\begin{matrix}0&0\\0&1\end{matrix}\right)2\pi^{s-1}\Gamma(s-1)^{-1}\left\lvert m\right\rvert^{s-\frac 32}y^{-s+\frac 32}K_{s-\frac 32}(2\pi \left\lvert m\right\rvert y). \end{align*} $$

This can be cleaned up somewhat using the functional equation $\Gamma (s+1)=s\Gamma (s)$ for the gamma function

$$ \begin{align*} g(m,y,s) &=\left(\begin{matrix}1&\frac{1-2s}{2\pi i m}-iy\\\frac{1-2s}{2\pi im}+iy&0\end{matrix}\right)2\pi^{s}\Gamma(s)^{-1}\left\lvert m\right\rvert^{s-\frac 12}y^{\frac 12}K_{s-\frac 12}(2\pi \left\lvert m\right\rvert y)\\ & \quad + \left(\begin{matrix}0&\tfrac{2y^{2}}{2\pi im}\\\tfrac{2y^{2}}{2\pi i m}&0\end{matrix}\right)2\pi^{s+1}\Gamma(s)^{-1}\left\lvert m\right\rvert^{s+\frac 12}y^{-\frac 12}K_{s+\frac 12}(2\pi \left\lvert m\right\rvert y)\\ & \quad + \left(\begin{matrix}0&0\\0&1\end{matrix}\right)2\pi^{s-1}(s-1)\Gamma(s)^{-1}\left\lvert m\right\rvert^{s-\frac 32}y^{\frac 32}K_{s-\frac 32}(2\pi \left\lvert m\right\rvert y). \end{align*} $$

In the interest of pairing terms corresponding to $\pm m$ , notice that for $m> 0$

$$ \begin{align*} &{\mathbf{e}}(mx)g(m,y,s)+{\mathbf{e}}(-mx)g(-m,y,s) \\ &=\cos(2\pi mx)(g(m,y,s)+g(-m,y,s))+i\sin(2\pi mx)(g(m,y,s)-g(-m,y,s))\\ &= \frac{4y^{\frac 12}\pi^{s}m^{s-\frac 12}}{\Gamma(s)}\left(\cos(2\pi mx)\left(\begin{matrix}K_{s-\frac 12}(2\pi my)&-iyK_{s-\frac 12}(2\pi my)\\iyK_{s-\frac 12}(2\pi my)&\frac{(s-1)y}{\pi m}K_{s-\frac 32}(2\pi my)\end{matrix}\right) \right.\\ & \quad + \left.\sin(2\pi mx)\left(\frac{1-2s}{2\pi m}K_{s-\frac 12}(2\pi my)+yK_{s+\frac 12}(2\pi my)\right)\left(\begin{matrix}0&1\\1&0\end{matrix}\right)\right). \end{align*} $$

Finally, to evaluate $g(0,y)$ , it remains to observe that since $\tfrac {r}{(r^{2}+s^{2})^{s}}$ is an odd function of r, $\int _{-\infty }^{\infty } \frac {r}{(r^{2}+y^{2})^{s}}dr=0$ . Therefore, we find that

$$ \begin{align*} g(0,y,s) &= y^{s}\int_{-\infty}^{\infty} \left(\begin{smallmatrix}1&r-iy\\r+iy&r^{2}+y^{2}\end{smallmatrix}\right)\tfrac{1}{(r^{2}+y^{2})^{s}}dr\\ &=y^{s}\int_{-\infty}^{\infty} \left(\begin{smallmatrix}1&-iy\\iy&r^{2}+y^{2}\end{smallmatrix}\right)\tfrac{1}{(r^{2}+y^{2})^{s}}dr\\ &=y^{s}\left(\begin{matrix}\pi^{\frac 12}\frac{\Gamma(s-\frac 12)}{\Gamma(s)}y^{1-2s}&-iy\pi^{\frac 12}\frac{\Gamma(s-\frac 12)}{\Gamma(s)}y^{1-2s}\\iy\pi^{\frac 12}\frac{\Gamma(s-\frac 12)}{\Gamma(s)}y^{1-2s}&\pi^{\frac 12}\frac{\Gamma(s-\frac 32)}{\Gamma(s-1)}y^{3-2s}\end{matrix}\right), \end{align*} $$

so that

$$ \begin{align*} g(0,y,s) = \frac{\pi^{\frac 12}y^{1-s}\Gamma(s-\frac 12)}{\Gamma(s)}\left(\begin{matrix}1&-iy\\iy&\frac{2s-2}{2s-3}y^{2}\end{matrix}\right). \end{align*} $$

Finally, recall that the Fourier expansion for $E(\tau ,s+1)$ is

$$ \begin{align*} E(\tau,s+1) &= y^{s+1}+\frac{\pi^{2s+1}\Gamma(-s)\zeta(-2s)}{\Gamma(s+1)\zeta(2s+2)}y^{-s}\\ & \quad + \frac{4\pi^{s+1}y^{\frac 12}}{\Gamma(s+1)\zeta(2s+2)}\sum_{m=1}^{\infty} m^{-s-\frac 12}\sigma_{2s+1}(m)\cos(2\pi m x)K_{s+\frac 12}(2\pi m y). \end{align*} $$

Putting these computations together allows us to prove the following.

Theorem 4.1 Let $\rho $ denote the inclusion representation of $\Gamma $ , let $L = \left (\begin {smallmatrix}n&(2\pi i )^{-1}\\0&n\end {smallmatrix}\right )$ for $n \in \mathbf {Z}$ , and set $H(\tau ,s) := H(\rho ,L,I,\tau ,s)$ . Then, the Eisenstein metric $H(\tau ,s)$ has a Fourier expansion of the form

$$ \begin{align*}H(\tau,s) = \left(\begin{matrix}1&0\\-\tau&1\end{matrix}\right)\left(\sum_{m\geq 0} H_{m}(\tau,s)\right)\left(\begin{matrix}1&-\bar\tau\\0&1\end{matrix}\right),\end{align*} $$

where

$$ \begin{align*} H_{0}(\tau,s) &= y^{s}e^{-4\pi ny}\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\\ & \quad + \sum_{k\geq 0}\frac{(-4\pi n)^{k}}{k!}\frac{\pi^{2s+2k+1}\Gamma(-s-k)\zeta(-2s-2k)}{\Gamma(s+k+1)\zeta(2s+2k+2)}y^{-s-k-1}\left(\begin{matrix}1&0\\0&0\end{matrix}\right)\\ & \quad + \sum_{k\geq 0}\frac{(-4\pi n)^{k}}{k!}\frac{\pi^{\frac 12}y^{1-s-k}\Gamma(s+k-\frac 12)\zeta(2s+2k-3)}{\Gamma(s+k)\zeta(2s+2k-2)}\left(\begin{matrix}1&-iy\\iy&\frac{2s+2k-2}{2s+2k-3}y^{2}\end{matrix}\right), \end{align*} $$

and for $m\geq 1$ ,

$$ \begin{align*} & H_{m}(\tau,s) =4\pi^{s}\sum_{k\geq 0}\frac{(-4\pi^{2} n)^{k}}{k!}\left(\frac{\pi\sigma_{2s+2k+1}(m)\cos(2\pi m x)K_{s+k+\frac 12}(2\pi m y)}{m^{s+k+\frac 12}\Gamma(s+k+1)\zeta(2s+2k+2)y^{\frac 12}}\left(\begin{matrix}1&0\\0&0\end{matrix}\right) \right.\\ & + \frac{y^{\frac 12}m^{s+k-\frac 12}\sigma_{3-2s-2k}(m)}{\Gamma(s+k)\zeta(2s+2k-2)}\cos(2\pi mx)\left(\begin{matrix}K_{s+k-\frac 12}(2\pi my)&\!\!-iyK_{s+k-\frac 12}(2\pi my)\\iyK_{s+k-\frac 12}(2\pi my)&\!\!\frac{(s+k-1)y}{\pi m}K_{s+k-\frac 32}(2\pi my)\end{matrix}\right)\\ & + \left.\frac{y^{\frac 12}m^{s+k-\frac 12}\sigma_{3-2s-2k}(m)}{\Gamma(s+k)\zeta(2s+2k-2)}\sin(2\pi mx) \right. \\& \times \left. \left(\frac{1-2s-2k}{2\pi m}K_{s+k-\frac 12}(2\pi my)+yK_{s+k+\frac 12}(2\pi my)\right) \left(\begin{matrix}0&1\\1&0\end{matrix}\right)\right). \end{align*} $$

Moreover, $H(\tau ,s)$ admits meromorphic continuation to the region $\operatorname {Re}(s)> \tfrac 32$ , and its only pole in this region is simple and located at $s=2$ . This pole comes from the constant term $H_{0}(\tau ,s)$ , and the residue at $s=2$ is a tame harmonic metric for the inclusion representation

$$ \begin{align*} \operatorname{\mathrm{Res}}_{s=2}H(\tau,s) = \frac{3}{2\pi y}\left(\begin{matrix}1&-x\\-x&x^{2}+y^{2}\end{matrix}\right) = \frac{3}{2\pi}K(\tau). \end{align*} $$

In particular, the residue does not depend on the choice of exponent matrix L.

Proof The computation of the Fourier coefficients is accomplished by substituting our expressions for the Fourier transforms $g(m,y,s)$ into equation (4.4) and simplifying. For the meromorphic continuation, recall that $\Gamma (s/2)\zeta (s)$ has two simple poles, at $s=0$ and $s=1$ , but it is otherwise holomorphic. Furthermore, it is nonvanishing in a neighborhood of $s=1$ and for $\operatorname {Re}(s) \geq 1$ .

The first term in the expression for $H_{0}(\tau ,s)$ is holomorphic in s. For each summand in the second line of the expression for $H_{0}(\tau ,s)$ , consider the ratio

$$ \begin{align*} \frac{\Gamma(-s-k)\zeta(-2s-2k)}{\Gamma(s+k+1)\zeta(2s+2k+2)}, \end{align*} $$

which is $O(1)$ as a function of k. If $\operatorname {Re}(s)> \frac 32$ , then $\operatorname {Re}(-2s-2k) < -3-2k \leq -3$ . Therefore, the numerator of the lined formula above is holomorphic in this region. Likewise, $\operatorname {Re}(2s+2k+2)> 5+2k$ , so that the denominator is holomorphic and nonvanishing in this region. Therefore, the second line in the description of $H_{0}(\tau ,s)$ converges to a holomorphic function when $\operatorname {Re}(s)> \frac 32$ .

For the final set of terms in $H_{0}(\tau ,s)$ , we consider the expressions

$$ \begin{align*} \frac{\Gamma(s+k-\frac 12)\zeta(2s+2k-3)}{\Gamma(s+k)\zeta(2s+2k-2)}. \end{align*} $$

Here, if $\operatorname {Re}(s)> \tfrac 32$ , then $\operatorname {Re}(2s+2k-3)> 2k > 0$ , so that the only possible pole can come from solutions to $2s+2k-3=1$ , which is $s=2-k$ . In the region $\operatorname {Re}(s)>\tfrac 32$ , this can only occur if $k=0$ , in which case a pole occurs at $s=2$ . For the denominators, we have that $\operatorname {Re}(2s+2k-2)> 1+2k > 1$ , so that the denominator is holomorphic and nonvanishing. It follows that when $\operatorname {Re}(s)> \tfrac 32$ , the constant term $H_{0}(\tau ,s)$ is holomorphic save for a simple pole arising from the $k=0$ term in the last sum of its expression in Theorem 4.1.

The higher Fourier coefficients $H_{m}(\tau ,s)$ are holomorphic in the region $\operatorname {Re}(s)> \tfrac 32$ , and the proof is similar to the constant term. A new feature is that one must estimate sums such as

$$ \begin{align*}\sum_{k\geq 0}\frac{(4\pi^{2}n)^{k}\sigma_{2s+2k+1}(m)K_{s+k+\frac 12}(2\pi my)}{m^{k}k!\Gamma(s+k+1)\zeta(2s+2k+2)}.\end{align*} $$

When $\left \lvert \nu \right \rvert $ is large relative to x, one can estimate $K_{\nu }(x)$ via the first few terms of its Taylor expansion (see Appendix B above (B.35) of [Reference Iwaniec15]). More precisely, in the tail of the sum where $\left \lvert 2\pi m y\right \rvert \ll 1+\left \lvert s+k+\frac 12\right \rvert ^{1/2}$ , one can approximate

(4.6) $$ \begin{align} K_{s+k+\frac 12}(2\pi m y) \approx\tfrac{1}{2}\Gamma(s+k+\tfrac 12)(\pi m y)^{-s-k-\frac 12}. \end{align} $$

In this way, one can show that for large k, the Bessel terms are mollified by the $\Gamma $ -terms in the denominators. Standard and more elementary estimates for the remaining factors appearing in $H_{m}(\tau ,s)$ then allow one to deduce sufficiently fast convergence to show that these sums indeed yield a holomorphic function in the region $\operatorname {Re}(s)> \tfrac 32$ .

It remains to compute the residue at $s=2$ , and by the preceding analysis, we have

$$ \begin{align*} \operatorname{\mathrm{Res}}_{s=2} H(\tau,s) =& \left(\begin{matrix}1&0\\-\tau&1\end{matrix}\right)\operatorname{\mathrm{Res}}_{s=2}H_{0}(\tau,s)\left(\begin{matrix}1&-\bar\tau\\0&1\end{matrix}\right)\\ =&\left(\begin{matrix}1&0\\-\tau&1\end{matrix}\right)\operatorname{\mathrm{Res}}_{s=2}\frac{\pi^{\frac 12}y^{1-s}\Gamma(s-\frac 12)\zeta(2s-3)}{\Gamma(s)\zeta(2s-2)}\left(\begin{matrix}1&-iy\\iy&\frac{2s-2}{2s-3}y^{2}\end{matrix}\right)\left(\begin{matrix}1&-\bar\tau\\0&1\end{matrix}\right)\\ =&\frac{\pi^{\frac 12}\Gamma(\frac 32)\operatorname{\mathrm{Res}}_{s=2}\zeta(2s-3)}{y\Gamma(2)\zeta(2)}\left(\begin{matrix}1&0\\-\tau&1\end{matrix}\right)\left(\begin{matrix}1&-iy\\iy&2y^{2}\end{matrix}\right)\left(\begin{matrix}1&-\bar\tau\\0&1\end{matrix}\right)\\ =&\frac{3}{2\pi y}\left(\begin{matrix}1&-x\\-x&x^{2}+y^{2}\end{matrix}\right). \end{align*} $$

This concludes the proof of Theorem 4.1.▪

5 Unitary monodromy at the cusp

Now, let $\rho $ be a representation with $\rho (T)$ unitary. Then, we can write $\rho (T) = e^{2\pi i L}$ where L is Hermitian and real. In fact, we may and shall suppose that $\rho (T)$ and L are both diagonal.

Lemma 5.1 Suppose that $\rho (T)$ and L are both diagonal. If $h\in \operatorname {\mathrm {Pos}}(\rho )$ , then h satisfies

$$ \begin{align*} {\mathbf{e}}(-Lz)h{\mathbf{e}}(Lz)= h, \end{align*} $$

for all $z\in \mathbf {C}$ .

Write $\tau =x+iy$ , so that for $h\in \operatorname {\mathrm {Pos}}(\rho )$ , the preceding lemma implies that

$$ \begin{align*} h(\tau) = e^{-2\pi i L(x+iy)}h e^{2\pi i L(x-iy)} = e^{4\pi Ly}h. \end{align*} $$

By abuse of notation below, $a,b$ denote integers chosen, so that the corresponding matrix is unimodular; in particular, b changes in the last line of the following computation, but the result is independent of this choice, which justifies our abuse of notation. With this point made, we compute

$$ \begin{align*} & H(h,\tau,s)\\ &= e^{4\pi Ly}y^{s}h + \sum_{c\geq 1}\sum_{\gcd(c,d)=1} \rho\left(\begin{smallmatrix}a&b\\c&d\end{smallmatrix}\right)^{t} e^{4\pi L \frac{y}{\left\lvert c\tau+d\right\rvert^{2}}}h\overline{\rho\left(\begin{smallmatrix}a&b\\c&d\end{smallmatrix}\right)}\frac{y^{s}}{\left\lvert c\tau+d\right\rvert^{2s}}\\ &= e^{4\pi Ly}y^{s}h + \sum_{c\geq 1}\sum_{\substack{d=1\\\gcd(c,d)=1}}^{c}\sum_{r\in\mathbf{Z}} \rho\left(\begin{smallmatrix}a&b\\c&cr+d\end{smallmatrix}\right)^{t}e^{4\pi L \frac{y}{\left\lvert c(\tau+r)+d\right\rvert^{2}}}h\overline{\rho\left(\begin{smallmatrix}a&b\\c&cr+d\end{smallmatrix}\right)}\frac{y^{s}}{\left\lvert c(\tau+r)+d\right\rvert^{2s}}\\ &=e^{4\pi Ly}y^{s}h + \sum_{c\geq 1}\sum_{\substack{d=1\\\gcd(c,d)=1}}^{c}\sum_{r\in\mathbf{Z}}\sum_{k\geq 0} \frac{(4\pi)^{k}}{k!}\rho\left(\begin{smallmatrix}a&b\\c&cr+d\end{smallmatrix}\right)^{t} L^{k}h\overline{\rho\left(\begin{smallmatrix}a&b\\c&cr+d\end{smallmatrix}\right)}\frac{y^{s+k}}{\left\lvert c(\tau+r)+d\right\rvert^{2(s+k)}}\\ &=e^{4\pi Ly}y^{s}h + y^{s}\sum_{k\geq 0} \frac{(4\pi y)^{k}}{k!}\sum_{c\geq 1}\frac{1}{c^{2(s+k)}}\sum_{\substack{d=1\\\gcd(c,d)=1}}^{c}\\&\quad \times \sum_{r\in\mathbf{Z}}\frac{\rho(T^{r})\rho\left(\begin{smallmatrix}a&b\\c&d\end{smallmatrix}\right)^{t} L^{k}h\overline{\rho\left(\begin{smallmatrix}a&b\\c&d\end{smallmatrix}\right)}\rho(T^{-r})}{\left\lvert \tau+r+\frac{d}{c}\right\rvert^{2(s+k)}}, \end{align*} $$

where in the last line we have used that $\rho (T)$ is diagonal to drop the transpose. Likewise, since L is real, this means $\rho (T)$ is unitary and we have used the identity $\overline {\rho (T^{r})} = \rho (T^{-r})$ .

Let $G(\tau )$ denote the sum over r above. Notice that $G(\tau +1) = \rho (T)^{-1}G(\tau )\rho (T)$ , so that if we write

$$ \begin{align*} \tilde G(\tau) = {\mathbf{e}}(Lx)G(\tau)\,{\mathbf{e}}(-Lx), \end{align*} $$

then $\tilde G(\tau +1)=\tilde G(\tau )$ . For simplicity, to study $\tilde G$ , we momentarily write

$$ \begin{align*} M =& \rho\left(\begin{smallmatrix}a&b\\c&d\end{smallmatrix}\right)^{t} L^{k}h\overline{\rho\left(\begin{smallmatrix}a&b\\c&d\end{smallmatrix}\right)},\\ L =& \operatorname{\mathrm{diag}}(e_{1},\ldots, e_{n}), \end{align*} $$

for real exponents $e_{j}$ . Then, the $(i,j)$ -entry of $\tilde G$ is

$$ \begin{align*} \tilde G_{ij} =& M_{ij}\sum_{r\in\mathbf{Z}}{\mathbf{e}}((e_{i}-e_{j})(x+r))\frac{1}{((x+r+\frac dc)^{2}+y^{2})^{s+k}}\\ =& {\mathbf{e}}\left((e_{j}-e_{i})\frac dc\right) M_{ij}\sum_{r\in\mathbf{Z}}{\mathbf{e}}((e_{i}-e_{j})(X+r))\frac{1}{((X+r)^{2}+y^{2})^{s+k}}, \end{align*} $$

where $X = x+\frac dc$ . We can evaluate this last sum using Poisson summation: with the Fourier transform,

$$ \begin{align*} f(u) =& \int_{-\infty}^{\infty} {\mathbf{e}}((e_{i}-e_{j})(X+r))\frac{1}{((X+r)^{2}+y^{2})^{s+k}}{\mathbf{e}}(-ur)dr\\ =&{\mathbf{e}}(Xu)\int_{-\infty}^{\infty} {\mathbf{e}}((e_{i}-e_{j}-u)r)\frac{1}{(r^{2}+y^{2})^{s+k}}dr. \end{align*} $$

Poisson summation gives

$$ \begin{align*} \tilde G_{ij} ={\mathbf{e}}\left((e_{j}-e_{i})\frac dc\right) M_{ij}\sum_{u\in\mathbf{Z}} f(u). \end{align*} $$

Formulas (3.18) and (3.19) of [Reference Iwaniec15] yield expressions

(5.1) $$ \begin{align} f(u) = \begin{cases} \pi^{\frac 12}{\mathbf{e}}((x+\frac dc)u)\frac{\Gamma(s+k-\frac 12)}{\Gamma(s+k)}y^{1-2(s+k)}, & e_{i}-e_{j}=u,\\ 2\frac{\pi^{s+k}{\mathbf{e}}((x+\frac dc)u)\left\lvert u+e_{j}-e_{i}\right\rvert^{s+k-\frac 12}}{y^{s+k-\frac 12}\Gamma(s+k)}K_{s+k-\frac 12}(2\pi \left\lvert u+e_{j}-e_{i}\right\rvert y),&e_{i}-e_{j} \neq u. \end{cases} \end{align} $$

Thus, if

$$ \begin{align*} z_{ij} = \begin{cases} 1, & e_{i}-e_{j} \in \mathbf{Z},\\ 0, & e_{i}-e_{j} \not \in \mathbf{Z}, \end{cases} \end{align*} $$

then we deduce that

$$ \begin{align*} \tilde G_{ij} &= \pi^{\frac 12}{\mathbf{e}}((e_{i}-e_{j})x)\frac{\Gamma(s+k-\frac 12)}{\Gamma(s+k)}y^{1-2(s+k)}M_{ij}z_{ij}\\ &\quad +\frac{2\pi^{s+k}{\mathbf{e}}\left((e_{j}-e_{i})\frac dc\right) M_{ij}}{y^{s+k-\frac 12}\Gamma(s+k)}\\&\quad \times \sum_{\substack{u\in\mathbf{Z}\\ u\neq e_{i}-e_{j}}} {\mathbf{e}}((x+\tfrac dc)u)\left\lvert u+e_{j}-e_{i}\right\rvert^{s+k-\frac 12}K_{s+k-\frac 12}(2\pi \left\lvert u+e_{j}-e_{i}\right\rvert y). \end{align*} $$

Notice that

$$ \begin{align*} G_{ij} = ({\mathbf{e}}(-Lx)\tilde G {\mathbf{e}}(Lx))_{ij} ={\mathbf{e}}((e_{j}-e_{i})x)\tilde G_{ij}. \end{align*} $$

Therefore, putting all of this together, we have shown that

$$ \begin{align*} H_{ij} &= e^{4\pi e_{i}y}y^{s}h_{ij}+\pi^{\frac 12}y^{1-s}\sum_{k\geq 0} \frac{(4\pi)^{k}}{k!}\sum_{c\geq 1}\frac{1}{c^{2(s+k)}}\sum_{\substack{d=1\\\gcd(c,d)=1}}^{c}\frac{\Gamma(s+k-\frac 12)}{y^{k}\Gamma(s+k)}M_{ij}z_{ij}\\ & \quad + y^{\frac 12}\sum_{k\geq 0} \frac{(4\pi)^{k}}{k!}\sum_{c\geq 1}\frac{1}{c^{2(s+k)}}\sum_{\substack{d=1\\\gcd(c,d)=1}}^{c}\frac{2\pi^{s+k} M_{ij}}{\Gamma(s+k)} \\ & \quad \times \sum_{\substack{u\in\mathbf{Z}\\ u\neq e_{i}-e_{j}}} {\mathbf{e}}((x+\tfrac dc)(u+e_{j}-e_{i}))\left\lvert u+e_{j}-e_{i}\right\rvert^{s+k-\frac 12}K_{s+k-\frac 12}(2\pi \left\lvert u+e_{j}-e_{i}\right\rvert y). \end{align*} $$

Rearranging terms yields the following Fourier expansion for the $(i,j)$ -entry of $H(h,\tau ,s)$ :

(5.2) $$ \begin{align} \nonumber H_{ij} &= e^{4\pi e_{i}y}y^{s}h_{ij}+\pi^{\frac 12}y^{1-s}\frac{\Gamma(s-\frac 12)}{\Gamma(s)}\nonumber\\&\quad \sum_{c\geq 1}\frac{1}{c^{2s}}\left(\sum_{\substack{d=1\\\gcd(c,d)=1}}^{c}\rho\left(\begin{smallmatrix}a&b\\c&d\end{smallmatrix}\right)^{t}{}_{1}F_{1}(s-\tfrac 12,s,\tfrac{4\pi L}{c^{2}y})h\overline{\rho\left(\begin{smallmatrix}a&b\\c&d\end{smallmatrix}\right)}\right)_{ij}z_{ij}\nonumber\\ &\quad +2y^{\frac 12}\pi^{s}\sum_{\substack{u\in\mathbf{Z}\\ u\neq e_{i}-e_{j}}}\left\lvert u+e_{j}-e_{i}\right\rvert^{s-\frac 12}{\mathbf{e}}((u+e_{j}-e_{i})x)\nonumber\\&\quad \times \sum_{c\geq 1}\frac{1}{c^{2s}}\sum_{\substack{d=1\\\gcd(c,d)=1}}^{c}{\mathbf{e}}((u+e_{j}-e_{i})\tfrac dc)\\ \nonumber & \left(\rho\left(\begin{smallmatrix}a&b\\c&d\end{smallmatrix}\right)^{t} \sum_{k\geq 0}\left(\frac{K_{s+k-\frac 12}(2\pi \left\lvert u+e_{j}-e_{i}\right\rvert y)}{k!\Gamma(s+k)}\left(\frac{4\pi^{2}\left\lvert u+e_{j}-e_{i}\right\rvert L}{c^{2}}\right)^{k}\right)h\overline{\rho\left(\begin{smallmatrix}a&b\\c&d\end{smallmatrix}\right)}\right)_{ij}. \end{align} $$

Note that the convergence of the sum on k is deduced as in equation (4.6) in the proof of Theorem 4.1, which uses the expansion of $K_{s+k-\frac 12}$ near zero when k is large.

The first line in equation (5.2) gives the constant term of $H(\tau ,s)$ , which in particular is independent of x, unlike for the inclusion representation in Section 4. Inspired by the discussion in Section 4, it is natural to consider the rightmost pole of this constant term (if such a pole exists!) and its corresponding residue. Restrict to the case where $e_{i}-e_{j}\not \in \mathbf {Z}$ unless $i=j$ , which is a familiar condition from the study of ordinary differential equations. This condition implies that $z_{ij} = \delta _{ij}$ . Since the term $e^{4\pi Ly}y^{s}h$ is entire as a function of s, we are interested in the diagonal terms of the matrix valued function

$$ \begin{align*} C(\tau,s)=\pi^{\frac 12}y^{1-s}\frac{\Gamma(s-\frac 12)}{\Gamma(s)}\sum_{c\geq 1}\frac{1}{c^{2s}}\left(\sum_{\substack{d=1\\\gcd(c,d)=1}}^{c}\rho\left(\begin{smallmatrix}a&b\\c&d\end{smallmatrix}\right)^{t}{}_{1}F_{1}(s-\tfrac 12,s,\tfrac{4\pi L}{c^{2}y})h\overline{\rho\left(\begin{smallmatrix}a&b\\c&d\end{smallmatrix}\right)}\right). \end{align*} $$

Unfortunately, the Kloosterman sums appearing above are somewhat unwieldy to handle via a direct approach in general. However, basic estimates show that the rightmost pole arises from the constant term of ${}_{1}F_{1}(s-\tfrac 12,s,\tfrac {4\pi L}{c^{2}y})$ in its Taylor expansion in $4\pi L/c^{2}y$ , so that one is really interested in the analytic properties of the diagonal terms of

$$ \begin{align*} C_{0}(\tau,s) = \pi^{\frac 12}y^{1-s}\frac{\Gamma(s-\frac 12)}{\Gamma(s)}\sum_{c\geq 1}\frac{1}{c^{2s}}\left(\sum_{\substack{d=1\\\gcd(c,d)=1}}^{c}\rho\left(\begin{smallmatrix}a&b\\c&d\end{smallmatrix}\right)^{t}h\overline{\rho\left(\begin{smallmatrix}a&b\\c&d\end{smallmatrix}\right)}\right). \end{align*} $$

This expression is a little more manageable. For example, suppose that $h=I_{d}$ and $\rho $ is unitary, so that $\rho (\gamma )^{t}\overline {\rho (\gamma )}=I_{d}$ . In this case,

$$ \begin{align*} C_{0}(\tau,s) = \pi^{\frac 12}y^{1-s}\frac{\Gamma(s-\frac 12)}{\Gamma(s)}\sum_{c\geq 1}\frac{\phi(c)}{c^{2s}}I_{d}=\pi^{\frac 12}y^{1-s}\frac{\Gamma(s-\frac 12)\zeta(2s-1)}{\Gamma(s)\zeta(2s)}I_{d}. \end{align*} $$

It follows that the rightmost pole occurs at $s=1$ , and the residue is a multiple of the identity matrix. Therefore, this shows that when $\rho $ is unitary, the rightmost pole of $H(\tau ,s)$ for $h=I_{d}$ occurs at $s=1$ and the residue is a multiple of the Petersson inner product, which is a harmonic metric for trivial reasons. It is unclear how general this phenomenon is. In the next section, we discuss an example where $\rho (T)$ is unitary, but $\rho $ is not unitarizable, to indicate some of the difficulties of analyzing expressions like $C_{0}(\tau ,s)$ in more general circumstances.

Remark 5.2 Since L is diagonal, the two matrix sums

$$ \begin{align*} &\sum_{\substack{d=1\\\gcd(c,d)=1}}^{c}\rho\left(\begin{smallmatrix}a&b\\c&d\end{smallmatrix}\right)^{t}h\overline{\rho\left(\begin{smallmatrix}a&b\\c&d\end{smallmatrix}\right)},&& \sum_{\substack{d=1\\\gcd(c,d)=1}}^{c}{\mathbf{e}}(-L\tfrac dc)\rho\left(\begin{smallmatrix}a&b\\c&d\end{smallmatrix}\right)^{t}h\overline{\rho\left(\begin{smallmatrix}a&b\\c&d\end{smallmatrix}\right)}{\mathbf{e}}(L\tfrac dc) \end{align*} $$

have the same diagonal entries. The matrix on the right, however, only depends on $d \mod {c}$ (this observation uses that L is real), whereas the matrix expression on the left undergoes a monodromy transformation after changing the values of d mod c. Thus, the right sum involving ${\mathbf {e}}(\pm L\tfrac dc)$ could be used in the definition of $C_{0}(\tau ,s)$ , giving a more natural expression. This would also give an expression for the constant term of $H(\tau ,s)$ that is more uniform with the higher Fourier coefficients, which already incorporate such exponential factors. In the following section, we consistently work with Kloosterman sums that include these additional exponential factors.

6 An indecomposable family

Let $\chi $ be the character of the modular form $\eta ^{4}$ , where $\eta $ is the Dedekind eta function. For $z \in \operatorname {\mathrm {\mathcal {H}}}$ and $\alpha \in \mathbf {C}$ , define a $\mathbf {C}$ -valued group cocycle on $\Gamma $ by the integral

$$ \begin{align*} \kappa(\gamma) = \int_{z}^{\gamma z}\alpha \eta^{4}(\tau)d\tau. \end{align*} $$

This satisfies the cocycle identity

$$ \begin{align*} \kappa(\gamma_{1}\gamma_{2}) = \chi(\gamma_{1})\kappa(\gamma_{2})+\kappa(\gamma_{1}), \end{align*} $$

for all $\gamma _{1},\gamma _{2} \in \Gamma $ . Changing z adjusts $\kappa $ by a coboundary. We can use this cocycle to define a representation of $\Gamma $ that contains a nontrivial subrepresentation, but which is not completely reducible into a direct sum of irreducible representations:

$$ \begin{align*} \rho(\gamma) = \left(\begin{matrix}\chi(\gamma)&\kappa(\gamma)\\0&1\end{matrix}\right). \end{align*} $$

This defines a family of representations in the parameters z and $\alpha $ defining $\kappa $ . If $\zeta = e^{2\pi i/6}$ , then

$$ \begin{align*} \rho(T) &= \left(\begin{matrix}\zeta&\kappa(T)\\0&1\end{matrix}\right), & \rho(S) &= \left(\begin{matrix}-1&\kappa(S)\\0&1\end{matrix}\right), & \rho(-1) &= \left(\begin{matrix}1&0\\0&1\end{matrix}\right). \end{align*} $$

Observe that $\rho (T)^{6}=I$ , so that $\rho (T)$ is diagonalizable. However, it is not possible to diagonalize $\rho (T)$ while keeping $\rho (S)$ diagonal. From this, one sees that $\rho $ contains a nontrivial subrepresentation, but it is not completely reducible for generic values of $\alpha $ and z.

Consider the representation of $\Gamma $ on the real vector space $\operatorname {\mathrm {Herm}}_{2}$ defined by

$$ \begin{align*} M\cdot \gamma = \rho(\gamma)^{t}M\overline{\rho(\gamma)}. \end{align*} $$

Let $U=\operatorname {\mathrm {Herm}}_{2}^{\Gamma }$ denote the invariants for this action. A simple computation shows that generically U is spanned by $\left (\begin {smallmatrix}0&0\\0&1\end {smallmatrix}\right )$ . In particular, U does not contain any positive definite matrices, so that there does not exist a harmonic metric for $\rho $ that is constant as a function of $\tau $ for generic choices of $\kappa $ .

To apply the material of Section 5 in our study of metrics for this representation, it will be necessary to change basis, so that the T-matrix is diagonal. If we set $\zeta = e^{2\pi i/6}$ , $P=\left (\begin {smallmatrix}0&\zeta \\1&-\zeta \kappa (T)\end {smallmatrix}\right )$ , and $\psi = P\rho P^{-1}$ , then one checks that

$$ \begin{align*} \psi(T) &= \left(\begin{matrix}1&0\\0&\zeta\end{matrix}\right), &\psi(S) &= \left(\begin{matrix}1&0\\ (1-\zeta)\kappa(S)-2\kappa(T)&-1\end{matrix}\right), \\[-10pt]\end{align*} $$

and more generally

$$ \begin{align*} \psi(\gamma) = \left(\begin{matrix}1&0\\ (1-\zeta)\kappa(\gamma)+(\chi(\gamma)-1)\kappa(T)&\chi(\gamma)\end{matrix}\right).\\[-10pt] \end{align*} $$

The identity

(6.1) $$ \begin{align} \frac{d\kappa}{dz}(\gamma) = (\chi(\gamma)-1)\alpha\eta^{4}(z)\\[-10pt]\nonumber \end{align} $$

implies that $\frac {d\psi }{dz}=0$ , so that the change of basis has made $\psi $ independent of z. It is thus a one-parameter family of representations determined by the choice of $\alpha \in \mathbf {C}$ in the definition of $\kappa $ . The specializations of this family are not completely reducible unless $\alpha =0$ .

Turning to the associated Eisenstein metrics, we take for our exponent matrix $L = \left (\begin {smallmatrix}0&0\\0&\frac 16\end {smallmatrix}\right )$ . Observe that since $\psi (T)$ is diagonal with distinct eigenvalues, $\operatorname {\mathrm {Herm}}_{2}^{\psi (T)}$ consists of real diagonal matrices. Therefore, $\operatorname {\mathrm {Pos}}(\psi )$ consists of positive real diagonal matrices, and since the choice of $h \in \operatorname {\mathrm {Pos}}(\psi )$ only really depends on h up to scaling, we can write

$$ \begin{align*}h = \left(\begin{matrix}1&0\\0&A\end{matrix}\right),\\[-10pt]\end{align*} $$

for $A \in \mathbf {R}_{>0}$ . With these choices of parameters, we write $H(\tau ,s) = H(\psi ,L,h,\tau ,s)$ , whose Fourier expansion is given by equation (5.2).

As usual, much of the difficulty in studying the Fourier expansion of $H(\tau ,s)$ lies in understanding the Kloosterman sumsFootnote ¹ and their associated generating series:

$$ \begin{align*} \operatorname{\mathrm{Kl}}(c) :=& \sum_{\substack{d=1\\\gcd(c,d)=1}}^{c}{\mathbf{e}}(-L\tfrac{d}{c})\psi\left(\begin{matrix}a&b\\c&d\end{matrix}\right)^{t}\left(\begin{matrix}1&0\\0&A\end{matrix}\right)\overline{\psi\left(\begin{matrix}a&b\\c&d\end{matrix}\right)}{\mathbf{e}}(L\tfrac{d}{c}),\\ D(s) :=& \sum_{c\geq 1}\frac{\operatorname{\mathrm{Kl}}(c)}{c^{s}}. \end{align*} $$

These families of Kloosterman sums admit a second-order Taylor expansion centered on the reducible specializations of $\psi $ satisfying $\kappa (S)=2\zeta \kappa (T)$ (which we have seen is equivalent to the condition $\alpha =\kappa =0$ in the definition of $\kappa $ ).

Proposition 6.1 There exist sequences $a_{c}\in \mathbf {Z}_{\geq 0}$ and $b_{c}\in \mathbf {Z}[e^{2\pi i/6c}]$ , independent of $\kappa $ , such that

$$ \begin{align*} \operatorname{\mathrm{Kl}}(c) = \left(\begin{matrix}\kappa(S)-2\zeta\kappa(T)&0\\0&1\end{matrix}\right)\left(\begin{matrix}a_{c}&b_{c}\\\overline{b_{c}}&0\end{matrix}\right)\left(\begin{matrix}\overline{\kappa(S)-2\zeta\kappa(T)}&0\\0&1\end{matrix}\right)A+\phi(c)\left(\begin{matrix}1&0\\0&A\end{matrix}\right), \end{align*} $$

for all $c\geq 1$ .

Proof Write $\gamma = \left (\begin {smallmatrix}a&b\\c&d\end {smallmatrix}\right ) \in \Gamma $ , and observe that the lower-triangular forms of $\psi (T)$ and $\psi (S)$ show that after writing $\gamma $ as a word in S and T, we have

$$ \begin{align*} \psi(\gamma) = \left(\begin{matrix}1&0\\\lambda a(\gamma)&\chi(\gamma)\end{matrix}\right), \end{align*} $$

where $\lambda = \kappa (S)-2\zeta \kappa (T)$ and $a(\gamma ) \in \mathbf {Z}[\zeta ]$ . Furthermore, $a(\gamma )$ is independent of $\kappa $ , as it only depends on how one writes $\gamma $ as a word in S and T. Equivalently, this independence can be seen by writing $a(\gamma )$ as a ratio of linear combinations of values of $\kappa $ . Then, $a(\gamma )$ is seen to be independent of z by differentiation, via equation (6.1). The possible dependence of $a(\gamma )$ on $\alpha $ cancels in the ratio defining $a(\gamma )$ , so that it is indeed entirely independent of the choice of $\kappa $ .

If we write $\omega = e^{2\pi i /c}$ , then the general term in the sum defining $\operatorname {\mathrm {Kl}}(c)$ takes the form

$$ \begin{align*} &{\mathbf{e}}(-L\tfrac{d}{c})\psi\left(\begin{matrix}a&b\\c&d\end{matrix}\right)^{t}\left(\begin{matrix}1&0\\0&A\end{matrix}\right)\overline{\psi\left(\begin{matrix}a&b\\c&d\end{matrix}\right)}{\mathbf{e}}(L\tfrac{d}{c})\\ =&\left(\begin{matrix}1&0\\0&w^{-d}\end{matrix}\right)\left(\begin{matrix}1&\lambda a(\gamma)\\0&\chi(\gamma)\end{matrix}\right)\left(\begin{matrix}1&0\\0&A\end{matrix}\right)\left(\begin{matrix}1&0\\\overline{\lambda a(\gamma)}&\overline{\chi(\gamma)}\end{matrix}\right)\left(\begin{matrix}1&0\\0&\omega^{d}\end{matrix}\right)\\ =&\left(\begin{matrix}1&A\lambda a(\gamma)\\0&A\omega^{-d}\chi(\gamma)\end{matrix}\right)\left(\begin{matrix}1&0\\\overline{\lambda a(\gamma)}&\overline{\chi(\gamma)}\omega^{d}\end{matrix}\right)\\ =&\left(\begin{matrix}1+A\left\lvert \lambda a(\gamma)\right\rvert^{2}&A\lambda a(\gamma)\overline{\chi(\gamma)}\omega^{d}\\A\overline{\lambda a(\gamma)}\chi(\gamma)\omega^{-d}&A\end{matrix}\right). \end{align*} $$

Therefore, from this expression, we see that the Proposition holds with

$$ \begin{align*} a_{c} &= \sum_{\substack{d=1\\\gcd(c,d)=1}}^{c} \left\lvert a(\gamma)\right\rvert^{2}, & b_{c}&= \sum_{\substack{d=1\\\gcd(c,d)=1}}^{c}a(\gamma)\overline{\chi(\gamma)}\omega^{d}.\\[-46pt] \end{align*} $$

▪

Values of $a_{c}$ and $b_{c}$ from Proposition 6.1 are listed in Table 1, and plots of the values $a_{c}/\phi (c)$ and $\left \lvert b_{c}\right \rvert /\phi (c)$ are contained in Figure 1. Polynomial growth estimates can be obtained for both $a_{c}$ and $b_{c}$ using Eichler length estimates as in Corollary 3.5 of [Reference Knopp and Mason20], although establishing an exact abscissa of convergence for $D(s)$ , and the computation of the corresponding residue, would require a more detailed analysis. Completion of such an analysis would likely enable one to establish analytic continuation of $H(\tau ,s)$ around its rightmost pole and compute the corresponding residue. The analysis of Section 5 shows that it is really the diagonal terms of $D(s)$ that intervene in this residue computation, so that it is the sequence $a_{c}$ that is most important for this analysis. We leave this computation, and whether the resulting computation produces a harmonic metric for these nonunitary representations, as an open question for future investigation.