The Fried conjecture in small dimensions

Abstract

We study the twisted Ruelle zeta function $\zeta _X(s)$ for smooth Anosov vector fields X acting on flat vector bundles over smooth compact manifolds. In dimension 3, we prove the Fried conjecture, relating Reidemeister torsion and $\zeta _X(0)$. In higher dimensions, we show more generally that $\zeta _X(0)$ is locally constant with respect to the vector field X under a spectral condition. As a consequence, we also show the Fried conjecture for Anosov flows near the geodesic flow on the unit tangent bundle of hyperbolic 3-manifolds. This gives the first examples of non-analytic Anosov flows and geodesic flows in variable negative curvature where the Fried conjecture holds true.

Appendices

Appendix A: Proof of Lemma 3.2

1.1 A.1. Family of order functions

In this paragraph, we fix the aperture of the cones $\alpha _0>0$ small enough to ensure that $C^{ss}(\alpha _0)\cap C^u(\alpha _0)=\emptyset$ and we fix some small parameter $\delta >0$. We construct an order function for every X in a small enough neighborhood of $X_0$. For that purpose, we closely follow the lines of [21, Lemma 2.1]. We fix $T_{\alpha _0}'>1$$T_{\alpha _0}$ is given by Lemma 3.1. The time $T_{\alpha _0}'$ will be determined later on in a way that depends only on $\alpha _0$. For our construction, we also let $m_0(x,\xi )\in \mathcal {C}^{\infty }(S^*\mathcal {M},[0,1])$ to be equal to 1 on $C^{u}(\alpha _0)$ and to 0 on $C^{ss}(\alpha _0)$. Then, we set

$$\begin{aligned} m_X(x,\xi ):=\frac{1}{2T_{\alpha _0}'}\int _{-T_{\alpha _0}'}^{T_{\alpha _0}'}m_0\circ \tilde{\Phi }^{X}_t(x,\xi )dt. \end{aligned}$$

(A.1)

Note that $m_X$depends smoothly onX as we chose $T_{\alpha _0}'$ independently of X near $X_0$. First of all, we note that

$$\begin{aligned} {\tilde{X}}_Hm_X(x,\xi )=\frac{1}{2T_{\alpha _0}'}\left( m_0\circ \tilde{\Phi }^X_{T_{\alpha _0}'}(x,\xi )- m_0\circ \tilde{\Phi }^X_{-T_{\alpha _0}'}(x,\xi )\right) , \end{aligned}$$

(A.2)

where ${\tilde{X}}_H$ is the vector field of $\tilde{\Phi }^X_{t}.$ We also observe that, for every $(x,\xi )$ inside $S^*\mathcal {M}$, the set

$$\begin{aligned} \mathcal {I}_{X_0}(x,\xi ):=\left\{ t\in {\mathbb {R}}:\tilde{\Phi }^{X_0}_t(x,\xi )\in S^*\mathcal {M}\backslash \left( C^u(\alpha _0/2)\cup C^{ss}(\alpha _0/2)\right) \right\} \end{aligned}$$

is an interval whose length is bounded by some constant $T_{\alpha _0}''>0$. Fix now a point $(x,\xi )\in S^*{\tilde{M}}$ and a vector field which is close enough to $X_0$ (to be determined). If $\tilde{\Phi }^X_t(x,\xi )\in C^u(\alpha _0)$ for every $t\in {\mathbb {R}}$, then the set

$$\begin{aligned} \tilde{\mathcal {I}}_{X}(x,\xi ):=\left\{ t\in {\mathbb {R}}:\tilde{\Phi }^{X}_t(x,\xi )\in S^*\mathcal {M}\backslash \left( C^u(\alpha _0)\cup C^{ss}(\alpha _0)\right) \right\} \end{aligned}$$

is empty and the same holds if $\tilde{\Phi }^X_t(x,\xi )\in C^{ss}(\alpha _0)$ for every $t\in {\mathbb {R}}$. Hence, it remains to bound the length of $\tilde{\mathcal {I}}_{X}(x,\xi )$ when the orbit of $(x,\xi )$ crosses $S^*\mathcal {M}\backslash \left( C^u(\alpha _0)\cup C^{ss}(\alpha _0)\right)$ and we may suppose without loss of generality that $(x,\xi )\in S^*\mathcal {M}\backslash \left( C^u(\alpha _0)\cup C^{ss}(\alpha _0)\right)$. Up to the fact that we may have to decrease a little bit the size of the set $\mathcal {U}_{\alpha _0}(X_0)$ appearing in Lemma 3.1, we have that $\tilde{\Phi }_{T_{\alpha _0}''}^{X}(x,\xi )$ belongs to $C^{u}(\alpha _0)$. Hence, thanks to Lemma 3.1, one finds that, for every $t\ge T_{\alpha _0}''+1$, one has $\tilde{\Phi }_t^X(x,\xi )\in C^u(\alpha _0)$. The same holds in backward times. Hence, the diameter of $\tilde{\mathcal {I}}_{X}(x,\xi )$ is uniformly bounded by $2(1+T_{\alpha _0}'')$ and we pick $T_{\alpha _0}'=\frac{1+T_{\alpha _0}''}{\delta }$ for $\delta <1$.

We set

$$\begin{aligned} \mathcal {O}^{u}(X)=\tilde{\Phi }^{X}_{T_{\alpha _0}'}(S^*\mathcal {M}\backslash C^{ss}(\alpha _0))\ \text {and}\ \mathcal {O}^{ss}(X)=\tilde{\Phi }^{X}_{-T_{\alpha _0}'} (S^*\mathcal {M}\backslash C^{u}(\alpha _0)). \end{aligned}$$

Let us now discuss the properties of $m_X$ for X belonging to $\mathcal {U}_{\alpha _0}(X_0)$:

• (1) If $(x,\xi )\in \mathcal {O}^u(X)$, then $\tilde{\Phi }^X_{-T_{\alpha _0}'}(x,\xi )\notin C^{ss}(\alpha _0)$. Hence, from the definition of $T_{\alpha _0}'$, one has $\tilde{\Phi }^X_{T_{\alpha _0}'}(x,\xi )\in C^{u}(\alpha _0)$ and, from (A.2), one deduce that ${\tilde{X}}_Hm_X\ge 0$ on $\mathcal {O}^u(X)$. Similarly, one has

  $$\begin{aligned} m_X(x,\xi )= & {} \frac{1}{2T_{\alpha _0}'}\left( \int _{-T_{\alpha _0}'}^{-T_{\alpha _0}'+2(T_{\alpha _0}+T_{\alpha _0}'')}m_0\circ \tilde{\Phi }^{X}_t(x,\xi )dt\right. \\&\left. + \int _{-T_{\alpha _0}'+2(T_{\alpha _0}+T_{\alpha _0}'')}^{T_{\alpha _0}'}m_0\circ \tilde{\Phi }^{X}_t(x,\xi )dt\right) , \end{aligned}$$

  from which one can infer

  $$\begin{aligned} \forall (x,\xi )\in \mathcal {O}^u(X),\quad m_{X}(x,\xi )\ge 1-\frac{T_{\alpha _0}+T_{\alpha _0}''}{T_{\alpha _0}'}=1-\delta . \end{aligned}$$

• (2) Reasoning along similar lines, one also finds that, for every $(x,\xi )\in \mathcal {O}^{ss}(X)$, ${\tilde{X}}_Hm_X\ge 0$ and

  $$\begin{aligned} m_{X}(x,\xi )\le \delta . \end{aligned}$$

• (3) Let $(x,\xi )$ be an element of $S^*\mathcal {M}\backslash (\mathcal {O}^u(X)\cup \mathcal {O}^{ss}(X)).$ In that case, one has $\tilde{\Phi }^X_{-T_{\alpha _0}'}(x,\xi )\in C^{ss}(\alpha _0)$ and $\tilde{\Phi }^X_{T_{\alpha _0}'}(x,\xi )\in C^u(\alpha _0)$. Thus, one finds

  $$\begin{aligned} {\tilde{X}}_Hm_X(x,\xi )= & {} \frac{1}{2T_{\alpha _0}'}\left( m_0\circ \tilde{\Phi }^X_{T_{\alpha _0}'}(x,\xi )-m_0\circ \tilde{\Phi }^X_{-T_{\alpha _0}'}(x,\xi )\right) \nonumber \\= & {} \frac{1}{2T_{\alpha _0}'}>0. \end{aligned}$$

  (A.3)

• (4) Let now $(x,\xi )\in S^*\mathcal {M}\backslash C^{u}(\alpha _0)$. Write

  $$\begin{aligned} m_X(x,\xi )\le \frac{1}{2}+\frac{1}{2T_{\alpha _0}'} \int _{-T_{\alpha _0}'}^0m_0\circ \tilde{\Phi }^{X}_t(x,\xi )dt\le \frac{1+\delta }{2}. \end{aligned}$$

Let us conclude this construction with the following useful observation:

Lemma A.1

Let $\alpha _0>0$ be small enough to ensure that $C^{u}(\alpha _0)\cap C^{ss}(\alpha _0)=\emptyset$. Then, there exists $0<\alpha _1<\alpha _0$ and a neighborhood $\mathcal {U}_{\alpha _0}(X_0)$ of $X_0$ in $\mathcal {A}$ such that, for every $X\in \mathcal {U}_{\alpha _0}(X_0)$,

$$\begin{aligned} C^u(\alpha _1)\cap S^*\mathcal {M}\subset \mathcal {O}^u(X)\quad \text {and}\quad C^{ss}(\alpha _1)\cap S^*\mathcal {M}\subset \mathcal {O}^{ss}(X). \end{aligned}$$

Proof

First of all, we note that $S^*\mathcal {M}\cap C^u(0)$ is invariant by the flow $\tilde{\Phi }^{X_0}_t$ and it is disjoint from $C^{ss}(\alpha _0)\cap S^*\mathcal {M}$. Hence, by construction of $\mathcal {O}^u(X_0)$, one can find some small enough $\alpha _1>0$ such that $S^*\mathcal {M}\cap C^u(\alpha _1)$ is contained inside ${\mathcal {O}}^u(X_0)$. By continuity with respect to X, this property remains true for any X close enough to $X_0$, i.e. $S^*\mathcal {M}\cap C^u(\alpha _1)\subset {\mathcal {O}}^u(X)$ for any $X\in \mathcal {U}_{\alpha _0}(X_0)\cap \mathcal {A}.$ The same proof works for the second part of the Lemma. $\square$

Remark 11

In all the construction so far, we could have defined the cones $C^{uu}(\alpha )$ and $C^s(\alpha )$ (see paragraph 3.1) and a decaying order function ${\tilde{m}}_X(x,\xi )$ which is close to 0 on $C^{s}(\alpha )$ and close to 1 on $C^{uu}(\alpha )$.

1.2 A.2. Definition of the escape function

We start with the construction of the function $f(x,\xi )\in \mathcal {C}^{\infty }(T^*M,{\mathbb {R}}_+)$. For $\Vert \xi \Vert _x\ge 1$, it will be 1-homogeneous and equal to $\Vert \xi \Vert _x$ outside the cones $C^{uu}(\tilde{\alpha }_0)$ and $C^{ss}(\tilde{\alpha }_0)$ for $\tilde{\alpha }_0>0$ small enough (to be determined). Following the proof of [16, Lemma C.1] (see also [35, Lemma 2.2]), we set, for $(x,\xi )$ near $C^{ss}(\tilde{\alpha }_0/2)$ and $\Vert \xi \Vert _x\ge 1$,

$$\begin{aligned} f(x,\xi ):=\exp \left( \frac{1}{T_1}\int _{0}^{T_1}\ln \Vert (d\varphi ^{X_0}_{t}(x)^T)^{-1}\xi \Vert _{\varphi _{X_0}^t(x)}dt\right) . \end{aligned}$$

Recall that, for every $\xi$ in $E_s^*(X_0,x)$, one has $\Vert (d\varphi ^{X_0}_{t}(x)^T)^{-1}\xi \Vert \le Ce^{-\beta t}\Vert \xi \Vert$ for every $t\ge 0$ (where $C,\beta$ are some uniform constants). Hence, if we set $T_1=2\frac{\ln C}{\beta }$, we find that, for every $(x,\xi )\in E_s^*(X_0)$ with $\Vert \xi \Vert _x\ge 1$, $X_{H_0}f(x,\xi )\le - f(x,\xi )\frac{\beta }{2}.$ Similarly, picking $T_1$ large enough, we set, for $(x,\xi )$ near $C^{uu}(\tilde{\alpha }_0/2)$ and $\Vert \xi \Vert _x\ge 1$,

$$\begin{aligned} f(x,\xi ):=\exp \left( \frac{1}{T_1}\int _{0}^{T_1}\ln \Vert (d\varphi ^{X_0}_{t}(x)^T)^{-1}\xi \Vert _{\varphi _{X_0}^t(x)}dt\right) , \end{aligned}$$

and we find that $X_{H_0}f(x,\xi )\ge f(x,\xi )\frac{\beta }{2}$ on $E_u^*(X_0)$. By continuity, we find that there exists some (small enough) $\tilde{\alpha }_0>0$ such that, for every $\Vert \xi \Vert _x\ge 1$,

$$\begin{aligned} (x,\xi )\in C^{ss}(\tilde{\alpha }_0/2)\Rightarrow X_{H_0}f(x,\xi )\le - f(x,\xi )\frac{\beta }{3}, \end{aligned}$$

(A.4)

and

$$\begin{aligned} (x,\xi )\in C^{uu}(\tilde{\alpha }_0/2)\Rightarrow X_{H_0}f(x,\xi )\ge f(x,\xi )\frac{\beta }{3}. \end{aligned}$$

(A.5)

As the function $f(x,\xi )$ is 1-homogeneous, we can find a neighborhood $\mathcal {U}(X_0)$ of $X_0$ in the $\mathcal {C}^{\infty }$-topology such that, for every X in $\mathcal {U}(X_0)$ and for every $\Vert \xi \Vert _x\ge 1$,

$$\begin{aligned} (x,\xi )\in C^{ss}(\tilde{\alpha }_0/2)\Rightarrow X_{H}f(x,\xi )\le - f(x,\xi )\frac{\beta }{4}, \end{aligned}$$

(A.6)

and

$$\begin{aligned} (x,\xi )\in C^{uu}(\tilde{\alpha }_0/2)\Rightarrow X_{H}f(x,\xi )\ge f(x,\xi )\frac{\beta }{4}. \end{aligned}$$

(A.7)

Finally, we note that there exists some uniform constant $C>0$ such that, for every X in $\mathcal {U}(X_0)$ and for $\Vert \xi \Vert _x\ge 1$,

$$\begin{aligned} -Cf(x,\xi )\le X_Hf(x,\xi )\le C f(x,\xi ) \end{aligned}$$

(A.8)

We are now ready to construct our family of escape functions $G_X^{N_0,N_1}(x,\xi )$:

$$\begin{aligned} G_X^{N_0,N_1}(x,\xi ):=m_{X}^{N_0,N_1}(x,\xi )\ln (1+f(x,\xi )), \end{aligned}$$

with $m_{X}^{N_0,N_1}\in \mathcal {C}^{\infty }(T^*M,[-2N_0,2N_1])$ which is 0-homogeneous for $\Vert \xi \Vert _x\ge 1$. In order to construct this function, we will make use of the order functions defined in paragraph A.1 as in [21, p. 337-8]. Before doing that, let us observe that

$$\begin{aligned} X_HG_X^{N_0,N_1}(x,\xi )= & {} X_H(m_{X}^{N_0,N_1})(x,\xi )\ln (1+f(x,\xi ))\nonumber \\&+\,m_{X}^{N_0,N_1}(x,\xi )\frac{X_Hf(x,\xi )}{1+f(x,\xi )}. \end{aligned}$$

(A.9)

We now fix a small enough neighborhood $\mathcal {U}(X_0)$ of $X_0$ so that f enjoys (A.6) and (A.7) for all X in $\mathcal {U}(X_0)$ and so that we can apply the results of paragraph A.1. Following [21], we set, for $\Vert \xi \Vert _x\ge 1$,

$$\begin{aligned} m_X^{N_0,N_1}(x,\xi ):= & {} N_1\left( 2-m_X\left( x,\frac{\xi }{\Vert \xi \Vert _x}\right) -{\tilde{m}}_X\left( x,\frac{\xi }{\Vert \xi \Vert _x}\right) \right) \nonumber \\&-\,2N_0{\tilde{m}}_X\left( x,\frac{\xi }{\Vert \xi \Vert _x}\right) , \end{aligned}$$

(A.10)

where we used the conventions of paragraph A.1 and Remark 11. First, notice that, by construction, $X_H(m_{X}^{N_0,N_1})\le 0$ for $\Vert \xi \Vert _x\ge 1$. Recall that the order functions $m_X$ and ${\tilde{m}}_X$ depends on the parameters $\alpha _0>0$ and $\delta >0$ and that they depend smoothly on X. Now, we fix $0<\delta <\frac{1}{2}\min \{1,\min \{N_0,N_1\}/(N_0+N_1)\}$, $0<16N_0<N_1$ and $0<\alpha _0<\tilde{\alpha }_0/2$. We then find that $m_X^{N_0,N_1}(x,\xi /\Vert \xi \Vert _x)\ge N_1$ on $\mathcal {O}^{ss}(X)$ and $m_X^{N_0,N_1}(x,\xi /\Vert \xi \Vert _x)\le -N_0$ on $\mathcal {O}^{uu}(X)$. We also have that $m_X^{N_0,N_1}(x,\xi /\Vert \xi \Vert _x)\ge \frac{N_1}{4}-2N_0\ge N_1/8$ for $(x,\xi )$ outside $C^{uu}(\alpha _0)$ (as $N_1>16N_0$). We now fix $\alpha _1$ to be the aperture of the cone appearing in Lemma A.1. This allows to verify the first three requirements of $m_X^{N_0,N_1}$.

Remark 12

We could also have defined

$$\begin{aligned} {\tilde{m}}_X^{N_0,N_1}(x,\xi ):= N_1\left( 1-m_X\left( x,\frac{\xi }{\Vert \xi \Vert _x}\right) \right) -N_0{\tilde{m}}_X\left( x,\frac{\xi }{\Vert \xi \Vert _x}\right) . \end{aligned}$$

We still have ${\tilde{m}}_X^{N_0,N_1}(x,\xi )\ge N_1$ on $\mathcal {O}^{ss}(X)$, ${\tilde{m}}_X^{N_0,N_1}(x,\xi )\le \frac{N_1}{4}-N_0$ outside $C^{ss}(\alpha _0)$.

Finally, combining $X_H(m_{X}^{N_0,N_1})\le 0$ with (A.9) for $||\xi ||\ge 1$, we immediately get the upper bound (3.5). It now remains to verify the decay property (3.3). For that purpose, we shall use the conventions of paragraph A.1 and set, for every $X\in \mathcal {U}(X_0)$,

$$\begin{aligned}&\tilde{\mathcal {O}}^{uu}(X)=\mathcal {O}^{uu}(X)\cap \mathcal {O}^u(X),\ \tilde{\mathcal {O}}^{0}(X)=\mathcal {O}^{s}(X)\cap \mathcal {O}^u(X),\ \text {and}\ \tilde{\mathcal {O}}^{ss}(X)\\&\quad = \mathcal {O}^{ss}(X)\cap \mathcal {O}^s(X), \end{aligned}$$

which contains respectively $C^{uu}(\alpha _1)$, $C^u(\alpha _1)\cap C^s(\alpha _1)$ and $C^{ss}(\alpha _1)$ for $\alpha _1>0$ small enough (see Lemma A.1). Note also that $\tilde{\mathcal {O}}^{0}(X)$ is contained inside $C^u(\alpha _0)\cap C^s(\alpha _0)$ which is a small vicinity of $E_0^*(X_0)$. Based on (A.9), we can now establish (3.3) except in this small cone around the flow direction. Outside $\tilde{\mathcal {O}}^{uu}(X)\cup \tilde{\mathcal {O}}^{0}(X)\cup \tilde{\mathcal {O}}^{ss}(X)$, it follows from (A.3) and (A.9). Inside $\tilde{\mathcal {O}}^{uu}(X)$ and $\tilde{\mathcal {O}}^{ss}(X)$, it follows from (A.6), (A.7) and (A.9).

Appendix B: Selberg zeta function on trace-free symmetric tensors

Proposition B.1

Let n be even and $M=\Gamma \backslash {\mathbb {H}}^{n+1}$ be a compact hyperbolic manifold. Let $\rho :\pi _1(M)\rightarrow U(V_\rho )$ be a finite dimensional unitary representation and let $\sigma _m$ be the irreducible unitary representation of $\mathrm{SO}(n)$ into the space $S_0^m{\mathbb {R}}^n$ of trace-free symmetric tensors of order $m\ge 1$ on ${\mathbb {R}}^n$. Then the Selberg zeta function $Z_{S,\sigma _m}(s)$ on M associated to $\sigma _p$ and $\rho$ is holomorphic and the order of its zeros are given by

$$\begin{aligned}&\mathrm{ord}_{s_0}Z_{S,\sigma _m}(s)\\&\quad =\left\{ \begin{array}{ll} \dim \ker (\nabla ^*\nabla -n^2/4-m+(s_0-n/2)^2)\cap \ker D^* &{} \text { if }s_0\not =n/2\\ 2\dim \ker (\nabla ^*\nabla -n^2/4-m)\cap \ker D^* &{} \text { if }s_0=n/2 \end{array}\right. \end{aligned}$$

where $\nabla$ is the twisted Levi-Civita covariant derivative on $S_0^mT^*M\otimes E$, $E\rightarrow M$ being the flat bundle over M obtained from the representation $\rho$, and $D^*=-\mathrm{Tr}\circ \nabla$ is the divergence operator.

Proof

We follow [6, Theorem 3.15]. First we need to view $\sigma _m$ as the restriction of a sum of irreducibles representations of $\mathrm{SO}(n+1)$ as in Section 1.1.2 [6]: it is not difficult to check that

$$\begin{aligned} \sigma _m=(\Sigma _m-\Sigma _{m-1})|_{\mathrm{SO}(n)} \end{aligned}$$

where $\Sigma _m$ denotes the irreducible unitary representation of $\mathrm{SO}(n+1)$ into the space $S_0^m{\mathbb {R}}^{n+1}$. By Section 1.1.3 of [6], there is a ${\mathbb {Z}}^2$-graded homogeneous vector bundle $V_{\sigma _m}=V_{\Sigma _m}^+\oplus V_{\Sigma _m}^-$ over ${\mathbb {H}}^{n+1}$ with $V_{\Sigma _m}^+=S_0^{m}{\mathbb {R}}^{n+1}$ and $V_{\Sigma _m}^-=S_0^{m-1}{\mathbb {R}}^{n+1}$, and we define the bundle $V_{M,\rho \otimes \sigma _m}=\Gamma \backslash (V_\rho \otimes V_{\sigma _m})$ over M. Denoting $E\rightarrow M$ the bundle over M obtained from $V_\rho$ by quotienting by $\Gamma$ and $S^m_0T^*M$ the bundle of trace-free symmetric tensors of order m on M, the bundle $V_{M,\rho \otimes \sigma _m}$ is isomorphic to the bundle $\mathcal {E}:=(S^m_0T^*M\oplus S^{m-1}_0T^*M)\otimes E$. There is a differential operator $A^2_{\sigma _m}$ on $\mathcal {E}$ constructed from the Casimir operator that has eigenvalues in correspondence with the zeros/poles of $Z_{S,\sigma _m}(s)$, it is given $A^2_{\sigma _m}=-\Omega -c(\sigma _m)$ where $\Omega$ is the Casimir operator and $c(\sigma )=n^2/4-|\mu (\sigma _m)|^2-2\mu (\sigma ).\rho _{\mathrm{so(n)}}$ with $\mu (\sigma _m)$ the highest weight of $\sigma$ and $\rho _\mathrm{so(n)}=(\tfrac{n}{2}-1,\tfrac{n}{2}-2,\dots ,0)$. Here we have $\mu (\sigma _m)=(m,0,\dots ,0)$ thus

$$\begin{aligned} c(\sigma _m)=\frac{n^2}{4}-m(m+n-2). \end{aligned}$$

We then obtain the formula

$$\begin{aligned} A^2_{\sigma _m}=(\Delta _m-c(\sigma _m))\oplus ( \Delta _{m-1}-c(\sigma _m)) \end{aligned}$$

where $\Delta _m=\nabla ^*\nabla -m(m+n-1)$ is the Lichnerowicz Laplacian on (twisted) trace-free symmetric tensors of order m on M (see for instance [38, Section 5]). Now we have by [38, Lemma 5.2] that $D^*\Delta _m=\Delta _{m-1}D^*$ if $D^*$ is the divergence operator defined by $D^*u=-{{{\,\mathrm{Tr}\,}}}(\nabla u)$, and whose adjoint is $D=\mathcal {S}\nabla$ is the symmetrised covariant derivative. This gives $\Delta _mD=D\Delta _{m-1}$, but since D is elliptic with no kernel by [40, Proposition 6.6], it has closed range and D gives an isomorphism

$$\begin{aligned} D: \ker (\Delta _{m-1}-c(\sigma _m)-s)\rightarrow \ker (\Delta _{m}-c(\sigma _m)-s)\cap (\ker D^*)^\perp \end{aligned}$$

for each $s\in {\mathbb {R}}$. In particular, one obtains that for each $s\in {\mathbb {R}}$

$$\begin{aligned}&\dim \ker (\Delta _{m}-c(\sigma _m)-s)-\dim \ker (\Delta _{m-1}-c(\sigma _m)-s)\\&\quad =\dim ( \ker (\Delta _{m}-c(\sigma _m)-s)\cap \ker D^*). \end{aligned}$$

Now by [6, Theorem 3.15], the function $Z_{S,\sigma _m}(s)$ has a zero at s of order

$$\begin{aligned} \begin{aligned}&2 \dim (\dim ( \ker (\Delta _{m}-c(\sigma _m)\cap \ker D^*)) \text { if }s=\tfrac{n}{2}\\&\dim (\dim ( \ker (\Delta _{m}-c(\sigma _m)\cap \ker D^*)) \text { if }s\not =\tfrac{n}{2}. \end{aligned} \end{aligned}$$

$\square$