Proof. Let $\hat H_0$ be a maximal strict subgroup of $\mathrm {SU}(2p+1)$ containing $H_0$ . Clearly the principal bundle $P_0=\mathrm {SU}(2p+2)\to S^{4p+3}$ also reduces to $\hat H_0$ . Then [Reference LeonardLeo71, Theorem 3] applied (with the notation of [Reference LeonardLeo71]) to $G_{2p + 1} := \mathrm {SU}(2p + 1)$ shows that $\hat H_0$ is a simple Lie group.

If $\hat H_0$ is a classical simple Lie group and $p\ge 2$ , by [Reference Čadek and CrabbČC06, Theorem 2.1, (D) and (E)] applied to $G=\hat H_0$ , we immediately obtain that the representation of $\hat H_0$ (and thus also that of H) on $\mathbb {C}^{2p+1}$ is reducible. The above result does not hold for $p=1$ (when the corresponding sphere $S^7$ is parallelizable), but it is easy to check that the only simple Lie group strictly contained in $\mathrm {SU}(3)$ whose representation on $\mathbb {C}^3$ is irreducible is $\mathrm {SO}(3)$ , embedded in $\mathrm {SU}(3)$ via the complexification of its standard representation on $\mathbb {R}^3$ . This corresponds to case (ii) of Theorem 3.1.

It remains to study the case where $\hat H_0$ is (a finite quotient of) one of the five exceptional simple compact Lie groups.

First of all, it suffices to look at complex irreducible representations of the exceptional Lie groups of odd dimension $2p+1$ . Moreover, by [Reference Čadek and CrabbČC06, Proposition 3.1], writing $4p+3=\dim \hat H_0 + k + 1$ for some integer k, there must exist at least k vector fields on the sphere $S^{4p+3}$ , so the Radon–Hurwitz number $\rho (n)$ (defined by the fact that $\rho (n) - 1$ is the maximal number of linearly independent vector fields on $S^{n-1}$ ) satisfies

$$ \begin{align*} \rho(4p+4)\ge 4p+3-\dim \hat H_0. \end{align*} $$

For all $p\ge 3$ , we have $\rho (4p+4)\leq 2p+3$ . Since no exceptional Lie group has an irreducible complex representation of dimension less than $7$ , it follows that $2p+1$ has to be the dimension of an irreducible complex representation of an exceptional Lie group $\hat H_0$ , with

(3.1) $$ \begin{align} 7\leq 2p+1\leq \dim \hat H_0+1. \end{align} $$

Denoting by $\widehat {\mathfrak {h}}$ the Lie algebra of $\hat H_0$ , it turns out that there is no complex odd-dimensional irreducible representation of an exceptional Lie group $\hat H_0$ satisfying equation (3.1) except in the following two cases.

Case 1. $\widehat {\mathfrak {h}}=\mathfrak {e}_6$ , $\dim \widehat {\mathfrak {h}}=78$ . There are two 27-dimensional irreducible representations of $\mathfrak {e}_6$ satisfying equation (3.1). This case corresponds to a (theoretical) reduction of the structure group $\mathrm {SU}(27)$ of $F_{\mathbb {C},\mathrm {SU}}S^{55}$ to a subgroup of $\mathrm {E}_6$ (case (iii) of Theorem 3.1).

Case 2. $\widehat {\mathfrak {h}}=\mathfrak {g}_2$ , $\dim \widehat {\mathfrak {h}}=14$ . The only complex odd-dimensional irreducible representation of $\mathfrak {g}_2$ satisfying equation (3.1) is the complexification of the real $7$ -dimensional representation $\rho _7:\mathrm {G}_2\to \mathrm {SO}(7)$ given by the embedding $\mathrm {G}_2\subset \mathrm {SO}(7)$ for $p=3$ . However, we will show that $F_{\mathbb {C},\mathrm {SU}}S^{15}$ does not admit any reduction to $\mathrm {G}_2$ .

Indeed, if such a reduction $P_{\mathrm {G}_2}$ exists, then the tangent bundle $TS^{15}$ is isomorphic to the direct sum $\mathbb {R}\oplus (\mathbb {C}\otimes F_7)$ , where $F_7:=P_{\mathrm {G}_2}\times _{\rho _7}\mathbb {R}^7$ is a real vector bundle of rank $7$ over $S^{15}$ .

Now, to each rank k real vector bundle E over $S^{15}$ , one can associate an element $\alpha (E)$ in the homotopy group $\pi _{14}(\mathrm {SO}(k))$ , namely, the homotopy class of its clutching function at the equator $S^{14} \hookrightarrow S^{15}$ . For $k<l$ , let $f_{k,l}:\mathrm {SO}(k)\to \mathrm {SO}(l)$ be the standard embedding and denote by $g_{k,l}:\pi _{14}(\mathrm {SO}(k))\to \pi _{14}(\mathrm {SO}(l))$ the group morphisms induced by $f_{k,l}$ in homotopy. If E and F have ranks k and l, respectively, clearly

$$ \begin{align*} \alpha(E\oplus F)=g_{k,k+l}(\alpha(E))+g_{l,k+l}(\alpha(F)). \end{align*} $$

In our situation, since $\mathbb {C}\otimes F_7$ is topologically isomorphic to $F_7\oplus F_7$ , $T S^{15}$ is isomorphic to $\mathbb {R}\oplus F_7\oplus F_7$ , so we can write

$$ \begin{align*} \alpha (T S^{15}) & =g_{14,15}(\alpha(F_7\oplus F_7)) \\ &=g_{14,15}(2g_{7,14}(\alpha(F_7)))=2g_{14,15}(g_{7,14}(\alpha(F_7)))=0, \end{align*} $$

because $\pi _{14}(\mathrm {SO}(15))=\mathbb {Z}_2$ . This is a contradiction since the tangent bundle of $S^{15}$ is non-trivial. Therefore, the case $\widehat {\mathfrak {h}}=\mathfrak {g}_2$ is impossible, thus finishing the proof.

Proof of Corollary 3.2

Up to replacing M by a finite covering if necessary, we can assume that the transitivity group $H \leqslant \mathrm {U}(m-1)$ is connected, see [Reference Cekić, Lefeuvre, Moroianu and SemmelmannCLMS21, Lemma 3.3]. In what follows, to keep notation simple, we will still denote this finite cover by M.

The representation $\rho : H \to \mathrm {U}(m-1)$ induces a representation $\det \rho : H \to \mathrm {U}(1)$ whose image is either $\mathrm {U}(1)$ or $\{1\}$ (by connectedness of H). Following an argument of Brin and Gromov [Reference Brin and GromovBG80], we first show that $(\det \rho )(H)= \mathrm {U}(1)$ .

Indeed, assume that $(\det \rho )(H) = \{1\}$ . As $(\det \rho )(H)$ is the transitivity group of the frame flow of the complex line bundle $\Lambda ^{m-1,0} \mathcal {N}$ , we get by the non-Abelian Livšic Theorem 2.6 that $\Lambda ^{m-1,0} \mathcal {N}$ is trivial. Now, using that

$$ \begin{align*} \Lambda^{m-1,0} \mathcal{N} \to \Lambda^{m,0} \pi^*TM, \quad \omega \mapsto \omega \wedge (v-iJv) \end{align*} $$

is an isomorphism, the triviality of $\Lambda ^{m-1,0} \mathcal {N}$ implies that

$$ \begin{align*} c_1(\Lambda^{m,0} \pi^*TM) = \pi^* c_1(\Lambda^{m,0} TM) = 0 \in H^2(SM,\mathbb{Z}). \end{align*} $$

However, it can be easily checked using the Gysin sequence [Reference Bott and TuBT82, Proposition 14.33] (and the fact that the dimension of M is $n \geq 4$ ) that

$$ \begin{align*} \pi^* : H^2(M,\mathbb{Z}) \to H^2(SM,\mathbb{Z}) \end{align*} $$

is injective, so $c_1(\Lambda ^{m,0} TM) = 0 = -c_1(K_M)$ , where $K_M = \Lambda ^{m,0} T^*M$ is the canonical line bundle. This is impossible since $(M,g)$ has negative sectional curvature. Hence, $(\det \rho )(H)= \mathrm {U}(1)$ .

We will now show that the H-representation $\rho $ on $\mathbb {C}^{m-1}$ is reducible. Assume for a contradiction that $\rho $ is irreducible. By the Schur lemma, the center $C(H)$ of H is contained in the set $\mathrm {U}(1)$ of scalar matrices. However, the fact that $(\det \rho )(H) = \mathrm {U}(1)$ shows that H is not semi-simple, so its centre is at least one-dimensional. We thus obtain the equality $C(H)=\mathrm {U}(1)$ , that is, H contains the set of scalar matrices.

We now fix an arbitrary point $x_0 \in M$ , restrict the unitary frame bundle $F_{\mathbb {C}}M$ to a bundle over $S_{x_0}M$ and identify $S_{x_0}M \simeq S^{4p+3}$ . As the structure group of $F_{\mathbb {C}} M$ reduces to H by Theorem 2.5, we obtain by restriction to any fibre of $SM\to M$ that the structure group of $F_{\mathbb {C}}S^{4p+3}$ also reduces to H, that is, there exists a principal H-bundle $P_H \subset F_{\mathbb {C}}S^{4p+3}$ . We claim that $F_{\mathbb {C}}S^{4p+3}$ admits a further reduction to $H_0:=H\cap \mathrm {SU}(2p+1)$ . Indeed, the principal $\mathrm {SU}(2p+1)$ -bundle $F_{\mathbb {C},\mathrm {SU}}S^{4p+3}$ is already a reduction of $F_{\mathbb {C}}S^{4p+3}$ to $\mathrm {SU}(2p+1)$ and for every $v\in S^{4p+3}$ , the intersection of the fibres $F_{\mathbb {C},\mathrm {SU}}S^{4p+3}(v)\cap P_H(v)$ is non-empty: if $u\in P_H(v)\subset F_{\mathbb {C}}S^{4p+3}(v)$ , there exists $z\in \mathrm {U}(1)$ such that $uz\in F_{\mathbb {C},\mathrm {SU}}S^{4p+3}(v)$ , and since $\mathrm {U}(1)\subset H$ (that is, H contains scalar matrices), we also have $uz\in P_H(v)$ , so $uz\in F_{\mathbb {C},\mathrm {SU}}S^{4p+3}(v)\cap P_H(v)$ . It is then straightforward to check that $F_{\mathbb {C},\mathrm {SU}}S^{4p+3}\cap P_H$ is a principal bundle over $ S^{4p+3}$ with group $H_0$ .

As $m \neq 4, 28$ by assumption, we can apply case (i) of Theorem 3.1 to deduce that $H_0 \lneqq \mathrm {SU}(m-1)$ acts reducibly on $\mathbb {C}^{m-1}$ . However, as H was assumed to act irreducibly on $\mathbb {C}^{m-1}$ , $H_0 = H \cap \mathrm {SU}(m-1)$ also acts irreducibly on $\mathbb {C}^{m-1}$ and this is a contradiction. Therefore, H acts reducibly on $\mathbb {C}^{m-1}$ .

We can then conclude the following using the non-Abelian Livšic Theorem 2.6: by the remark after Theorem 2.6, there exists a smooth (non-zero) flow-invariant orthogonal projector $\pi _{\mathcal {V}'} \in C^\infty (SM,S^2 \mathcal {N})$ onto a flow-invariant smooth complex bundle $\mathcal {V}' \subset \mathcal {N}$ . Following [Reference Cekić, Lefeuvre, Moroianu and SemmelmannCLMS21, Lemma 3.10], one can find a (possibly different) smooth non-zero flow-invariant orthogonal projector $\pi _{\mathcal {V}} \in C^\infty (SM,S^2 \mathcal {N})$ of even Fourier degree onto a flow-invariant smooth complex bundle $\mathcal {V} \subset \mathcal {N}$ of complex rank $1 \leq r \leq m/2-1$ .

Proof. Corollary 3.5 follows immediately from Corollary 3.2 and the above discussion, except for the point that $k \geq 2$ which we now prove. If $k = 0$ , then $\pi _{\mathcal {V}} = u =u_0$ is of degree $0$ and thus $\pi _{\mathcal {V}}$ can be identified with a section in $C^\infty (M,S^2 TM)$ . However, we must also have $\iota _v \pi _{\mathcal {V}} = 0$ for all $v \in TM$ by equation (3.3), so $\pi _{\mathcal {V}} = 0$ , which contradicts the non-vanishing of $\pi _{\mathcal {V}}$ .

Proof. Using the upper bound in equation (2.14) on the sectional curvature from Lemma 2.3,

$$ \begin{align*} \langle R \nabla_{\mathbb{V}}^{E} f, \nabla_{\mathbb{V}}^{E} f \rangle_{L^2} & = \int_{M} \int_{S_xM}\sum_\alpha R(v,\nabla_{\mathbb{V}}^{E} f_\alpha,\nabla_{\mathbb{V}}^{E} f_\alpha,v) \\ & \leq - \frac{3\unicode{x3bb}-2}{4} \|\nabla_{\mathbb{V}}^{E} f\|^2_{L^2} - \frac{3\unicode{x3bb}}{4} \int_{M}\int_{S_xM}\sum_{\alpha} \langle v,J\nabla_{\mathbb{V}}f_\alpha \rangle^2 \\ & = - \frac{3\unicode{x3bb}-2}{4} \langle \Delta_{\mathbb{V}}^{E} f, f \rangle_{L^2} - \frac{3\unicode{x3bb}}{4} \int_{M}\int_{S_xM} \sum_{\alpha}\langle v,J\nabla_{\mathbb{V}}f_\alpha \rangle^2 \\ & = - \frac{3\unicode{x3bb}-2}{4}k(n+k-2)\|f\|^2_{L^2} - \frac{3\unicode{x3bb}}{4}\int_{M} \int_{S_xM}\sum_{\alpha} \langle v,J\nabla_{\mathbb{V}}f_\alpha \rangle^2. \end{align*} $$

Since $\alpha _{n,k} := k(n+k-2)$ , this completes the proof.

Proof. The proof is the same as [Reference Cekić, Lefeuvre, Moroianu and SemmelmannCLMS21, Lemma 4.5] by inserting the bound in equation (2.17).

Proof. We start with the proof for $E = TM$ . Note that this equality is an integral equality over $SM$ . We will actually prove the integral equality over $S_xM$ for every $x \in M$ , and then it suffices to integrate over $x \in M$ to obtain the result. Recall that G is defined in equation (2.12). Using the expressions in equations (2.7) and (2.13),

(4.2) $$ \begin{align} 4 \sum_\alpha G(v, \nabla_{\mathbb{V}} f_\alpha, f, \operatorname{\mathrm{\mathbf{e}}}_\alpha) & = \sum_\alpha( \langle (v \wedge \nabla_{\mathbb{V}} f_\alpha) f, \operatorname{\mathrm{\mathbf{e}}}_\alpha \rangle \nonumber\\ &\quad + \langle (Jv \wedge J \nabla_{\mathbb{V}} f_\alpha) f, \operatorname{\mathrm{\mathbf{e}}}_\alpha \rangle + 2\langle{v, J \nabla_{\mathbb{V}} f_\alpha}\rangle\langle{f, J \operatorname{\mathrm{\mathbf{e}}}_\alpha}\rangle). \end{align} $$

The integral over $S_xM$ of the first term on the right-hand side can be immediately computed using [Reference Cekić, Lefeuvre, Moroianu and SemmelmannCLMS21, Lemma 4.6] (in the $\Lambda ^p$ case with $p=1$ ; we warn the reader that, in the notation of [Reference Cekić, Lefeuvre, Moroianu and SemmelmannCLMS21], G denotes the curvature tensor of the real hyperbolic space, that is, . The term $\sum _\alpha \int _{S_xM} \langle v,f\rangle \langle \nabla _{\mathbb {V}} f_\alpha , \operatorname {\mathrm {\mathbf {e}}}_\alpha \rangle - \langle v,\operatorname {\mathrm {\mathbf {e}}}_\alpha \rangle \langle \nabla _{\mathbb {V}} f_\alpha , f\rangle $ thus corresponds exactly to the term computed in [Reference Cekić, Lefeuvre, Moroianu and SemmelmannCLMS21, Equation after (4.14)] with $p=1$ ) since $\iota _v f$ is of degree $k-1$ and yields

$$ \begin{align*} \int_{S_xM} \sum_\alpha \langle (v \wedge \nabla_{\mathbb{V}} f_\alpha) f, \operatorname{\mathrm{\mathbf{e}}}_\alpha \rangle& = \int_{S_xM} \bigg(\sum_\alpha \langle v,f\rangle \langle \nabla_{\mathbb{V}} f_\alpha, \operatorname{\mathrm{\mathbf{e}}}_\alpha\rangle - \langle v,\operatorname{\mathrm{\mathbf{e}}}_\alpha\rangle \langle \nabla_{\mathbb{V}} f_\alpha, f\rangle\bigg) \\ &= (n+2k-4)\|\iota_v f\|^2_{L^2(S_xM)} + \|f\|^2_{L^2(S_xM)}, \end{align*} $$

where we use the notation

$$ \begin{align*} \|f\|^2_{L^2(S_xM)} = \int_{S_xM} g_{x}(f(v),f(v)), \quad \|\iota_v f\|^2_{L^2(S_xM)} = \int_{S_xM} |(\iota_v f)(v)|^2. \end{align*} $$

We claim that the integral over $S_xM$ of the second term in equation (4.2) is equal to $(n+2k-4)\|\iota _{Jv} f\|^2_{L^2(S_xM)} + \|f\|^2_{L^2(S_xM)}$ . Indeed, observe first that for each $\alpha $ ,

$$ \begin{align*}\langle{(Jv \wedge J\nabla_{\mathbb{V}} f_\alpha) f, \operatorname{\mathrm{\mathbf{e}}}_\alpha}\rangle = \langle{(v \wedge \nabla_{\mathbb{V}} f_\alpha)Jf, J\operatorname{\mathrm{\mathbf{e}}}_\alpha}\rangle.\end{align*} $$

Now, since $f_\alpha = \langle f,\operatorname {\mathrm {\mathbf {e}}}_\alpha \rangle = \langle Jf,J\operatorname {\mathrm {\mathbf {e}}}_\alpha \rangle $ , changing the basis $(\operatorname {\mathrm {\mathbf {e}}}_\alpha )$ by $(J\operatorname {\mathrm {\mathbf {e}}}_\alpha )$ , we see that the second term in equation (4.2) is the same as the first term with f replaced by $Jf$ . Using $\iota _v Jf = -\iota _{Jv} f$ , the result now follows from the previous computation. Inserting the previous equality in equation (4.2) gives the desired result (after integration over M) since $\delta _{n,k} = n+2k-4$ .

We now deal with the case $E = S^2TM$ . As before, using the expressions in equations (2.7), (2.11) and (2.13),

(4.3) $$ \begin{align} 4\! \sum_\alpha G_{S^2 TM}(v, \nabla_{\mathbb{V}} f_\alpha, f, \operatorname{\mathrm{\mathbf{e}}}_\alpha) &= {\kern-1pt}\sum_\alpha ( \langle [(v \wedge \nabla_{\mathbb{V}} f_\alpha), f], \operatorname{\mathrm{\mathbf{e}}}_\alpha \rangle \nonumber \\&\quad + \langle [(J v {\kern-1pt}\wedge{\kern-1pt} J \nabla_{\mathbb{V}} f_\alpha), f], \operatorname{\mathrm{\mathbf{e}}}_\alpha \rangle - 2 \langle{v, J \nabla_{\mathbb{V}} f_\alpha}\rangle.\langle{[J,{\kern-1pt}f], \operatorname{\mathrm{\mathbf{e}}}_\alpha}\rangle), \end{align} $$

where $(\operatorname {\mathrm {\mathbf {e}}}_\alpha )_{\alpha }$ is a local orthonormal basis of $S^2TM$ . The integral over $S_xM$ of the first term can be computed using [Reference Cekić, Lefeuvre, Moroianu and SemmelmannCLMS21, Lemma 4.6] (case of symmetric $2$ -tensors) since $\iota _v f$ is of degree $k-1$ and yields

(4.4) $$ \begin{align} \int_{S_xM} \sum_\alpha\langle [(v \wedge \nabla_{\mathbb{V}} f_\alpha), f], \operatorname{\mathrm{\mathbf{e}}}_\alpha \rangle = 2(n+2k-4)\|\iota_v f\|^2_{L^2(S_xM)} + 2\|f\|^2_{L^2(S_xM)}. \end{align} $$

The third term vanishes in equation (4.3) since $[J,f]=0$ by assumption.

We now claim that the integral over $S_xM$ of the second term in equation (4.3) is also equal to $2(n+2k-4)\|\iota _v f\|^2_{L^2(S_xM)} + 2\|f\|^2_{L^2(S_xM)}$ , which will finish the proof. Indeed, observe that $JX \wedge JY = -J \circ (X \wedge Y) \circ J$ for any $X, Y \in TM$ , and therefore using also $[J, f] = 0$ , we get for each $\alpha $ ,

$$ \begin{align*}[Jv \wedge J \nabla_{\mathbb{V}} f_\alpha, f] = -J \circ [v \wedge \nabla_{\mathbb{V}} f_\alpha, f] \circ J.\end{align*} $$

Consequently, we can rewrite

$$ \begin{align*}\langle{[Jv \wedge J \nabla_{\mathbb{V}} f_\alpha, f], \operatorname{\mathrm{\mathbf{e}}}_\alpha}\rangle =\langle -J \circ [v \wedge \nabla_{\mathbb{V}} f_\alpha, f] \circ J, \operatorname{\mathrm{\mathbf{e}}}_\alpha\rangle= \langle{[v \wedge \nabla_{\mathbb{V}} f_\alpha, f],- J \circ \operatorname{\mathrm{\mathbf{e}}}_\alpha \circ J}\rangle.\end{align*} $$

Note that $(-J \circ \operatorname {\mathrm {\mathbf {e}}}_\alpha \circ J)_\alpha $ is also an orthonormal basis of $S^2TM$ , and since $f=-J \circ f \circ J =- \sum _\alpha f_\alpha J \circ \operatorname {\mathrm {\mathbf {e}}}_\alpha \circ J$ ,

$$ \begin{align*}f_\alpha=\langle f,\operatorname{\mathrm{\mathbf{e}}}_\alpha\rangle=\langle f,-J \circ \operatorname{\mathrm{\mathbf{e}}}_\alpha \circ J\rangle.\end{align*} $$

The claim thus follows from equation (4.4).

Proof. Using the twisted Pestov identity in equation (2.6) with $E =TM$ , and applying the bounds provided by Lemmas 4.1, 4.2 and 4.3,

$$ \begin{align*} &\frac{k(n+2k)}{k+1}\|\mathbf{X}_+f\|^2_{L^2}\\ &\quad\geq - \langle R \nabla_{\mathbb{V}}^{TM} f, \nabla_{\mathbb{V}}^{TM}f \rangle_{L^2} - \langle \mathcal{F}^{TM}f,\nabla_{\mathbb{V}}^{TM} f\rangle_{L^2} \\ &\quad\geq \frac{3\unicode{x3bb}-2}{4} \alpha_{n,k}\|f\|^2_{L^2} + \frac{3\unicode{x3bb}}{4} \int_M\int_{S_xM} \sum_{\alpha} \langle v,J\nabla_{\mathbb{V}}f_\alpha\rangle^2 -\frac{4(1-\unicode{x3bb})}{3}\beta_{n,k}\|f\|^2_{L^2} \\ & \qquad- \frac{1+\unicode{x3bb}}{2}\bigg( \dfrac{1}{4}\delta_{n,k}(\|\iota_v f\|^2_{L^2} + \|\iota_{Jv}f\|^2_{L^2}) + \frac{1}{2}\|f\|^2_{L^2}\\ &\qquad+ \frac{1}{2}\int_M \int_{S_xM} \sum_{\alpha} |\langle v,J\nabla_{\mathbb{V}}f_\alpha\rangle| |\langle f,J\operatorname{\mathrm{\mathbf{e}}}_\alpha \rangle| \bigg). \end{align*} $$

We now use the estimate

$$ \begin{align*} \frac{1}{2} \int_M \int_{S_xM} \sum_{\alpha} \langle v,J\nabla_{\mathbb{V}}f_\alpha\rangle \langle f,J\operatorname{\mathrm{\mathbf{e}}}_\alpha \rangle \leq \frac{1}{4\varepsilon} \|f\|^2_{L^2} + \frac{\varepsilon}{4}\int_M \int_{S_xM} \sum_\alpha \langle v,J\nabla_{\mathbb{V}}f_\alpha\rangle^2, \end{align*} $$

which holds for all $\varepsilon> 0$ , to deduce that

$$ \begin{align*} \frac{k(n+2k)}{k+1}\|\mathbf{X}_+f\|^2_{L^2} &\geq \frac{3\unicode{x3bb}-2}{4} \alpha_{n,k}\|f\|^2_{L^2} + \frac{3\unicode{x3bb}}{4}\int_M\int_{S_xM} \sum_{\alpha}\langle v,J\nabla_{\mathbb{V}}f_\alpha\rangle^2\\&\quad-\frac{4(1-\unicode{x3bb})}{3}\beta_{n,k}\|f\|^2_{L^2} \\&\quad - \frac{1+\unicode{x3bb}}{2}\bigg( \frac{1}{4}\delta_{n,k}(\|\iota_v f\|^2_{L^2} + \|\iota_{Jv}f\|^2_{L^2}) + \bigg(\frac{1}{2} +\frac{1}{4\varepsilon}\bigg)\|f\|^2_{L^2}\\&\quad+ \frac{\varepsilon}{4} \int_M \int_{S_xM}\sum_{\alpha} \langle v,J\nabla_{\mathbb{V}}f_\alpha\rangle^2 \bigg). \end{align*} $$

Taking the specific value $\varepsilon := 6\unicode{x3bb} /(1+\unicode{x3bb} )$ , we see that the coefficients in front of the term $\int _M \sum _{\alpha } \int _{S_xM} \langle v,J\nabla _{\mathbb {V}}f_\alpha \rangle ^2$ cancel out. Moreover, since by assumption $\unicode{x3bb} \in [2/3,1]$ , we have the lower bound $\varepsilon ={6\unicode{x3bb} }/({1 + \unicode{x3bb} }) \geq {12}/5$ and this eventually yields the result.

Proof of Lemma 4.4

Using the equality $\mathbf {X}(\iota _v u) = \iota _v \mathbf {X} u = \iota _v \mathbf {X}_- u$ and the fact that $\iota _{Jv}\mathbf {X}_-u=J(\iota _{v}\mathbf {X}_-u)$ ,

$$ \begin{align*} \|\mathbf{X}_-u\|^2_{L^2} & \geq \|\iota_v \mathbf{X}_- u\|^2_{L^2} + \|\iota_{Jv}\mathbf{X}_-u\|^2_{L^2} =2\|\iota_v \mathbf{X}_- u\|^2_{L^2} \\ & =2 \|\mathbf{X}(\iota_vu)\|^2_{L^2} =2 \|\mathbf{X}_+(\iota_v u)\|^2_{L^2} +2 \|\mathbf{X}_-(\iota_v u)\|^2_{L^2} \\ & \geq 2\|\mathbf{X}_+(\iota_v u)\|^2_{L^2} , \end{align*} $$

where in the second line, we used that $\iota _v u$ and $\iota _{Jv} u$ are of degree $k - 1$ and the mapping property in equation (2.5). By assumption, u is a complex normal twisted conformal Killing tensor, so this implies that $\iota _v \iota _{Jv} u = \iota _{Jv} \iota _v u = 0$ and $\iota _v \iota _v u= \iota _{Jv}\iota _{Jv}u$ is of degree $k-2$ . As a consequence, we can apply Lemma 4.5 with $f:=\iota _v u$ (which is of degree $k-1$ ). Using the fact that $\iota _{Jv} f=0$ , together with the fact that

$$ \begin{align*}\|\iota_v f\|^2_{L^2}=\|\iota_v \iota_v u\|^2_{L^2}\leq\|\iota_v u\|^2_{L^2}, \end{align*} $$

we obtain the announced result.

Proof. Straightforward computation, inserting in the twisted Pestov identity in equation (2.6) the lower bound for $\|\mathbf {X}_-u\|^2$ (Lemma 4.4), the upper bounds for the terms on the right-hand side (Lemmas 4.1 and 4.2 applied to $f=u$ , and Lemma 4.3(ii)), as well as using the facts that $\mathbf {X}_+ u = 0$ (by the definition in equation (3.3)) and $\|Z(u)\|_{L^2}^2 \geq 0$ .

Proof. If ${C}_{n,k}(\unicode{x3bb} ) \geq 0$ , this is immediate from the first part of equation (5.4). If ${C}_{n,k}(\unicode{x3bb} ) <0$ , using that $ \|u\|^2_{L^2}\geq \|\iota _vu\|^2_{L^2}+\|\iota _{Jv}u\|^2_{L^2}=2\|\iota _vu\|^2_{L^2}$ , we get by equation (5.1):

$$ \begin{align*} (B_{n,k}(\unicode{x3bb}) + \tfrac12{C}_{n,k}(\unicode{x3bb}))\|u\|^2_{L^2} \leq 0, \end{align*} $$

so $u \equiv 0$ by the second part of equation (5.4).

Proof. Follows immediately from equations (5.2) and (5.3), as the expressions of $B_{n,k}$ and $C_{n,k}$ are affine functions in $\unicode{x3bb} $ .

Proof. It is straightforward to check that both sequences are positive for $k\ge 2$ and $n\ge 4$ . Using that $\alpha _{n,k}={\beta _{n,k}^2}/({n-1})$ , we can write

Since $\alpha _{n,k}$ and $\beta _{n,k}$ are positive and increasing in k, the numerator in the right-hand term is increasing in k, whereas the denominator is positive and decreasing in k. Thus, $k \mapsto \unicode{x3bb} _{1}(n,k)$ is decreasing.

Consider now the expression $\unicode{x3bb} _{2}(n,k)$ . Again it is easy to check that $\unicode{x3bb} _{2}(n,k)>0$ for $k\ge 2$ and $n\ge 4$ , and

Let us denote this last expression by ${E_{n,k}}/{F_{n,k}}$ . We claim that $E_{n,k}$ is increasing in k and $F_{n,k}$ is decreasing in k.

Using the fact that ${\gamma _{n,k}}/{\alpha _{n,k}}=({n+2k-4})/({(n+2k-2)\alpha _{n,k-1}})$ ,

$$ \begin{align*}E_{n,k}&=3+3\frac{\gamma_{n,k}}{\alpha_{n,k}}\alpha_{n,k-1}-\frac{12+6\delta_{n,k}}{\alpha_{n,k}}-\frac{29}{4}\frac{\gamma_{n,k}}{\alpha_{n,k}}-3\frac{\gamma_{n,k}}{\alpha_{n,k}}\delta_{n,k-1}\\ &=6\frac{n+2k-3}{n+2k-2}-6\frac{n+2k-2}{\alpha_{n,k}}-\frac{29}{4}\frac{\gamma_{n,k}}{\alpha_{n,k}}-3\frac{\gamma_{n,k}}{\alpha_{n,k}}\delta_{n,k-1}\\ &=6-\frac6{n+2k-2}-\frac6k-\frac6{n+k-2}-3\frac{\gamma_{n,k}}{\alpha_{n,k}}\bigg((n+2k-4)+\frac5{12}\bigg). \end{align*} $$

To express $F_{n,k}$ , we remark that

$$ \begin{align*}\frac{\gamma_{n,k}}{\alpha_{n,k}}\beta_{n,k-1}=\frac{n+2k-4}{n+2k-2}\frac{\beta_{n,k-1}}{\alpha_{n,k-1}}=\frac{s_{n,k-1}}{n+2k-2},\end{align*} $$

where we denote $s_{n,k}:={(n+2k-2)\beta _{n,k}}/{\alpha _{n,k}}.$ A straightforward computation similar to that for $E_{n,k}$ shows that

$$ \begin{align*}F_{n,k}&=6+3\frac{n+2k-2}{\alpha_{n,k}}+6\frac{n+2k-4}{n+2k-2}+\frac{3}{2}\frac{\gamma_{n,k}}{\alpha_{n,k}}\bigg((n+2k-4)+\frac5{12}\bigg)\\ &\quad+32\frac{s_{n,k}}{n+2k-2}+16\frac{s_{n,k-1}}{n+2k-2}.\end{align*} $$

To prove our claim, it is thus enough to remark that

$$ \begin{align*}(n+2k-4)\frac{\gamma_{n,k}}{\alpha_{n,k}}&=\frac{(n+2k-4)^2}{(k-1)(n+k-3)(n+2k-2)}\\ &=\frac{n-4}{n-2}\cdot\frac1{k-1}+\frac{n}{n-2}\cdot\frac1{n+k-3}+\frac{4}{(k-1)(n+k-3)(n+2k-2)}\end{align*} $$

is decreasing in k (and thus ${\gamma _{n,k}}/{\alpha _{n,k}}$ is decreasing in k too);

$$ \begin{align*}s_{n,k}=(n+2k-2)\frac{\sqrt{(n-1)k(n+k-2)}}{k(n+k-2)}=\sqrt{(n-1)\bigg(4+\frac{(n-2)^2}{k(n+k-2)}\bigg)}\end{align*} $$

is decreasing in k, and

$$ \begin{align*}\frac{n+2k-2}{\alpha_{n,k}}+2\frac{n+2k-4}{n+2k-2}&=2+\frac{n+2k-2}{k(n+k-2)}-\frac4{n+2k-2}\\&=2+\frac{(n-2)^2}{k(n+k-2)(n+2k-2)}\end{align*} $$

is decreasing in k.

Proof of ergodicity in Theorem 1.1

We need to show that under the pinching condition $\unicode{x3bb}> \unicode{x3bb} (m)$ of Theorem 1.1, the unitary frame flow is ergodic. If the frame flow is not ergodic, and $m \neq 4,28$ , we know by Corollary 3.2 that there exists a flow-invariant orthogonal projector $\pi _{\mathcal {V}} \in C^\infty (SM,S^2 \mathcal {N})$ of even Fourier degree onto a complex vector bundle $\mathcal {V} \subset \mathcal {N}$ of rank $1 \leq r \leq m/2-1$ .

By Corollary 3.5 and Lemma 5.1, this yields the existence of a non-zero complex normal twisted conformal Killing tensor $u \in C^\infty (M,\Omega _k \otimes S^2 T^*M)$ of even degree $k \geq 2$ which satisfies the inequality in equation (5.1). We distinguish two cases.

Case 1. If $k \geq 4$ , Lemmas 5.3 and 5.4 show that if the holomorphic pinching $\unicode{x3bb} $ satisfies $\unicode{x3bb}> \max (\unicode{x3bb} _1(n,4),\unicode{x3bb} _2(n,4))$ , we then have $B_{n,k}> 0$ and $B_{n,k}+\frac 12C_{n,k}> 0$ which by Lemma 5.2 implies that $u \equiv 0$ .

Case 2. If $k=2$ , let $\pi _{\mathcal {V}}$ be the projector given by Corollary 3.2, whose complex rank r satisfies $1\le r\le m/2-1$ . By [Reference Cekić, Lefeuvre, Moroianu and SemmelmannCLMS21, Lemma 4.2], one can then show that $\pi _{\mathcal {V}}$ must be of the form

with $u \in C^\infty (M,\Omega _2 \otimes S^2 T^*M)$ as above. In particular, since is parallel, this implies $\mathbf {X} u = 0$ , that is, $\mathbf {X}_\pm u = 0$ . Moreover, a quick algebraic computation (see [Reference Cekić, Lefeuvre, Moroianu and SemmelmannCLMS21, p. 38]) gives

(5.5) $$ \begin{align} \|\iota_v u\|^2_{L^2} = \frac{2r}{n(n-2r)}\|u\|^2_{L^2}, \end{align} $$

and since $2r\le m-2=n/2-2$ ,

(5.6) $$ \begin{align} \|\iota_v u\|^2_{L^2} \le\frac{n-4}{n(n+4)}\|u\|^2_{L^2}. \end{align} $$

Applying the twisted Pestov identity in equation (2.6) to u for $k=2$ , using that $\mathbf {X}_\pm u = 0$ (the bound of Lemma 4.4 is therefore useless), and the upper bounds of Lemmas 4.1, 4.2 and 4.3, we find that

$$ \begin{align*} \bigg(\frac{3\unicode{x3bb}-2}{2}n-\frac{8(1-\unicode{x3bb})}{3}(2n(n-1))^{1/2}-\frac{1+\unicode{x3bb}}{2}\bigg)\|u\|^2_{L^2} - \frac{1+\unicode{x3bb}}{2} n \|\iota_vu\|^2_{L^2} \leq 0. \end{align*} $$

Inserting equation (5.5) in the previous inequality, then

$$ \begin{align*} \bigg(\frac{3\unicode{x3bb}-2}{2}n-\frac{8(1-\unicode{x3bb})}{3}(2n(n-1))^{1/2}-\frac{1+\unicode{x3bb}}{2} \frac{2n}{n+4} \bigg)\|u\|^2_{L^2} \leq 0. \end{align*} $$

This shows that $u \equiv 0$ when $k=2$ , as soon as the holomorphic pinching $\unicode{x3bb} $ satisfies

Summarizing the two cases $k\ge 4$ and $k=2$ , we have obtained that the the unitary frame flow is ergodic as soon as the holomorphic pinching $\unicode{x3bb} $ satisfies

Using a formal computing tool, it can be easily checked that $ \unicode{x3bb} _0(m)=\unicode{x3bb} _2(2m)$ for $m \geq 6$ . However, for practical reasons, we will give a bound which is slightly less accurate, but much easier to compute.

Namely, using the obvious inequality $((n+2)(n-1))^{1/2}<n+1/2$ , we can write

Similarly, since $16\sqrt 2<68/3$ ,

Finally, using the calculation in the proof of Lemma 5.4 together with the obvious inequalities $64/\sqrt 3<37$ and

$$ \begin{align*}\dfrac43>\gamma_{n,4}= \dfrac{4(n+2)(n+4)}{3(n+1)(n+6)}> \dfrac{4n}{3(n+1)}\quad\text{for all } n\ge3,\end{align*} $$

we obtain

It is straightforward to check that

$$ \begin{align*}\dfrac{6n+6}{44n+43}>\dfrac{9n}{86 n+362}>\dfrac{14n-26}{154n+131}\quad\text{for all } n\ge 0,\end{align*} $$

which implies $1/{\unicode{x3bb} _0(m)}-1>({14n-26})/({154n+131})$ , so eventually

thus proving the ergodicity statement of Theorem 1.1.

Proof. Mixing implies ergodicity so it remains to show that ergodicity implies mixing in this setting. By [Reference LefeuvreLef23, Proof of Lemma 3.7], it suffices to show that the equation

(5.7) $$ \begin{align} X_{F_{\mathbb{C}}M} u = i \unicode{x3bb} u, \quad \unicode{x3bb} \in \mathbb{R} \setminus \{0\},\ u \in L^2(F_{\mathbb{C}}M) \end{align} $$

implies that $u \equiv 0$ . Let $u \in L^2(F_{\mathbb {C}}M)$ be a solution to equation (5.7). Since $F_{\mathbb {C}}M \to SM$ is a principal $\mathrm {U}(m-1)$ -bundle over $SM$ , the space $L^2(F_{\mathbb {C}}M)$ splits as

(5.8) $$ \begin{align} L^2(F_{\mathbb{C}}M) = \bigoplus_{k=0}^{+\infty} L^2(SM, \Theta_k), \end{align} $$

where $\Theta _k$ is the vector bundle over $SM$ whose fibre over $v\in SM$ is the eigenspace of the Casimir operator of $\mathrm {U}(m-1)$ acting on functions of the fibre $(F_{\mathbb {C}}M)_v$ , associated to the eigenvalue $\mu _k \geq 0$ (and $\mu _k \neq \mu _j$ for $k \neq j$ ). Since $X_{F_{\mathbb {C}}M}$ preserves the splitting equation (5.8) (because the frame flow $(\Phi _t)_{t \in \mathbb {R}}$ commutes with the right-action of $\mathrm {U}(m-1)$ ), we deduce that for each $k\ge 0$ , the component $u_k\in L^2(SM,\Theta _k)$ of some function $u = \sum _{k} u_k \in \oplus _{k=0}^{+\infty }L^2(SM,\Theta _k)$ satisfying equation (5.7) must satisfy

$$ \begin{align*} X_{F_{\mathbb{C}}M} u_k = i \unicode{x3bb} u_k. \end{align*} $$

Since $\Theta _0 = \mathbb {C}$ is trivial over $SM$ , the equation $X_{F_{\mathbb {C}}M} u_0 = i\unicode{x3bb} u_0$ reads $X u_0 = i \unicode{x3bb} u_0$ , where $u_0 \in L^2(SM)$ and X is the geodesic vector field. However, the geodesic flow is mixing as $(M,g)$ has negative curvature, so we deduce that $u_0 = 0$ .

For $k \neq 0$ , [Reference LefeuvreLef23, Proposition 2.7] based on microlocal analysis shows that $u_k \in ~L^2 (SM,\Theta _k)$ is actually smooth, that is, $u_k \in C^\infty (SM,\Theta _k)$ . Moreover, $X_{F_{\mathbb {C}}M}|u_k|^2 = 0$ , so using the ergodicity assumption on the frame flow $(\Phi _t)_{t \in \mathbb {R}}$ , we get that $|u_k|$ is constant on $F_{\mathbb {C}}M$ . If $u_k \neq 0$ , then up to rescaling, we obtain that $u_k : F_{\mathbb {C}} M \to \mathrm {U}(1)$ is a well-defined smooth map. To simplify the notation, we will drop the index k from now on and simply write u instead of $u_k$ , and $\mu $ instead of $\mu _k$ .

Denote by P the unit circle bundle of the complex line bundle $\Lambda ^{m-1,0}\mathcal {N} \to SM$ . Now, observe that there is a natural surjective bundle map $\psi : F_{\mathbb {C}}M \to P$ given by

$$ \begin{align*} \psi : (v, \operatorname{\mathrm{\mathbf{e}}}_2, \ldots,\operatorname{\mathrm{\mathbf{e}}}_m) \mapsto \bigg(v,\frac{1}{\sqrt{2}} (\operatorname{\mathrm{\mathbf{e}}}_2-i J\operatorname{\mathrm{\mathbf{e}}}_2) \wedge \ldots\wedge \frac{1}{\sqrt{2}} (\operatorname{\mathrm{\mathbf{e}}}_m-iJ\operatorname{\mathrm{\mathbf{e}}}_m)\bigg). \end{align*} $$

There is a natural unitary parallel transport along geodesic flow-lines of sections of $\Lambda ^{m-1,0}\mathcal {N}$ , so there is a flow $(\Phi _t^P)_{t \in \mathbb {R}}$ on P with generator $X_P$ extending the geodesic flow $(\varphi _t)_{t\in \mathbb {R}}$ as in §2.3, that is, writing $\pi : P \to SM$ for the projection map, one has $\pi \circ \Phi _t^P = \varphi _t \circ \pi $ for all $t \in \mathbb {R}$ . Moreover, $\psi $ intertwines the frame flow on $F_{\mathbb {C}}M$ and the flow on P, that is

(5.9) $$ \begin{align} \Phi_t^P \circ \psi = \psi \circ \Phi_t. \end{align} $$

We claim that the following holds.

Note that, once we know that $u = \psi ^*w$ for some function w, the relation $X_P w = i \unicode{x3bb} w$ is immediate using equation (5.9) and $X_{F_{\mathbb {C}}M} u = i \unicode{x3bb} u$ .

Proof. We show that $u = \psi ^*w$ for some $w \in C^\infty (P)$ . We fix an arbitrary point $v_0 \in SM$ and $w_0 \in (F_{\mathbb {C}}M)_{v_0}$ . There is then a commutative diagram

(5.10)

where the downward arrows are isometries. Hence, by restricting u to the fibre $(F_{\mathbb {C}}M)_{v_0}$ and identifying isometrically $(F_{\mathbb {C}}M)_{v_0} \simeq \mathrm {U}(m-1)$ , we get a map $F := u(v_0)$ such that

$$ \begin{align*} F : F_{\mathbb{C}}M \simeq \mathrm{U}(m-1) \to \mathrm{U}(1), \end{align*} $$

and $\Delta _{\mathrm {U}(m-1)}F = \mu F$ with $\mu \neq 0$ since u takes values in $\Theta _k$ for $k \neq 0$ . In other words, F is an eigenfunction of the Casimir operator on $\mathrm {U}(m-1)$ associated to the eigenvalue $\mu \neq 0$ and of constant modulus.

Now, recall that $\mathrm {U}(m-1)$ is a split group extension of the circle group $\mathrm {U}(1)$ by the special unitary group $\mathrm {SU}(m-1)$ , that is, $\mathrm {det} : \mathrm {U}(m-1) \to \mathrm {U}(1)$ is a fibre bundle with fibres isometric to $\mathrm {SU}(m-1)$ . We claim that such a function F must then necessarily be constant on the $\mathrm {SU}(m-1)$ -fibres of the bundle $\mathrm {U}(m-1)$ , that is, $F = \mathrm {det}^* f$ for some $f \in C^\infty (\mathrm {U}(1))$ , $\mu = j^2$ for some integer $j \geq 0$ and f is an eigenfunction of $\Delta _{\mathrm {U}(1)}$ with eigenvalue $j^2$ .

Indeed, since $\mathrm {U}(m-1) = \mathrm {SU}(m-1) \times _{\mathbb {Z}_{m-1}} \mathrm {U}(1)$ , there is a $\mathbb {Z}_{m-1}$ -bundle map $p : \mathrm {SU}(m-1) \times \mathrm {U}(1) \to \mathrm {U}(m-1)$ , $p(w, z) = wz$ , and this map is locally a Riemannian isometry. (The $\mathbb {Z}_{m-1}$ -action on $\mathrm {SU}(m-1) \times \mathrm {U}(1)$ is simply given by $(w,z) \mapsto (w \omega ^{-k}, \omega ^k z)$ for $(w,z) \in \mathrm {SU}(m-1) \times \mathrm {U}(1)$ , $k \in \{0,\ldots ,m-2\}$ and $\omega := e^{2i\pi /{(m-1)}}$ .) As a consequence, we get that $\Delta _{\mathrm {SU}(m-1) \times \mathrm {U}(1)} p^*F = \mu ~ p^* F$ . As $\mathrm {SU}(m-1) \times \mathrm {U}(1)$ is a Riemannian product, its eigenfunctions are obtained as sums of products of eigenfunctions on each factor and the eigenvalues are sums of the eigenvalues on each factor. Hence, we can write

$$ \begin{align*} p^*F(w,z) = \sum_{j=1}^N a_j(w) b_j(z) = \sum_{j=1}^N a_j(w) z^{k_j} \end{align*} $$

for some finite number $N> 0$ , where $(w,z) \in \mathrm {SU}(m-1) \times \mathrm {U}(1)$ , $k_j \in \mathbb {Z}$ and $k_{j_1} \neq k_{j_2}$ for $j_1 \neq j_2$ , $a_j \neq 0$ is an eigenfunction of $\Delta _{\mathrm {SU}(m-1)}$ associated to the eigenvalue $\unicode{x3bb} _j$ and $\unicode{x3bb} _j + k^2_j = \mu> 0$ . However, we also know that $p^*F$ has constant modulus (equal to $1$ ). Freezing an arbitrary point $w \in \mathrm {SU}(m - 1)$ and looking at $|p^*F|^2(w, z)$ as a function of $z \in \mathbb {S}^1$ , we easily get that N must be equal to $1$ , that is, $p^*F(w,z) = a(w) z^{k}$ for some $k\in \mathbb {Z}$ . Hence, $a \in C^\infty (\mathrm {SU}(m-1))$ satisfies $|a|=1$ and $\Delta _{\mathrm {SU}(m-1)} a = \unicode{x3bb} a$ for some $\unicode{x3bb} \geq 0$ (with $k^2+\unicode{x3bb} =\mu $ ).

We claim that $\unicode{x3bb} = 0$ , that is, a is constant, and F is thus constant along the $\mathrm {SU}(m-1)$ -fibres. Indeed, if $\unicode{x3bb} \neq 0$ , using

$$ \begin{align*} 0&=\Delta_{\mathrm{SU}(m-1)} |a|^2 = (\Delta_{\mathrm{SU}(m-1)} a) \overline{a} + a (\Delta_{\mathrm{SU}(m-1)} \overline{a}) - 2 \nabla a \cdot \nabla \overline{a} \\ & = 2 \unicode{x3bb} - 2 |\nabla a|^2, \end{align*} $$

we get that $|\nabla a|$ is a non-zero constant. Since $\mathrm {SU}(m-1)$ is simply connected, the map a lifts to the universal cover of $\mathrm {U}(1)$ , so there exists a map $\theta :\mathrm {SU}(m-1)\to \mathbb {R}$ such that $a=e^{i\theta }$ . At a critical point of $\theta $ , we thus get $\nabla a=0$ , which is absurd.

As a consequence, $\unicode{x3bb} = 0$ and this shows that F is constant along $\mathrm {SU}(m-1)$ -fibres of the bundle map $\mathrm {det} : \mathrm {U}(m-1) \to \mathrm {U}(1)$ . In turn, as equation (5.10) commutes, we get that u is constant on the preimages of the map $\psi : F_{\mathbb {C}} M \to P$ so $u = \psi ^* w$ for some smooth $w \in C^\infty (P)$ .

We thus have just shown that if there is a non-zero solution to equation (5.7), then there is also a non-zero solution to

(5.11) $$ \begin{align} X_P w = i \unicode{x3bb} w, \quad \unicode{x3bb} \in \mathbb{R} \setminus \{0\}, w \in C^\infty(P). \end{align} $$

Hence, it remains to show that equation (5.11) has no non-zero solutions, and we will actually show it for $w \in L^2(P)$ .

Assume that $w \in L^2(P)$ is a solution to equation (5.11). As above, using that P is a principal $\mathrm {U}(1)$ -bundle, we can decompose

(5.12) $$ \begin{align} L^2(P) = \bigoplus_{k \in \mathbb{Z}} L^2(SM, L_k), \end{align} $$

where the sections of $L_k$ are given by functions $f \in L^2(P)$ satisfying $V f = ik f$ , where V is the infinitesimal generator of the $\mathrm {U}(1)$ -action on the fibres of P. Observe that $L_k = L^{\otimes k}$ for $k \in \mathbb {Z}$ where $L := L_1$ and $L_1 \simeq \Lambda ^{m-1,0} \mathcal {N}$ . Now, using the splitting equation (5.12), the equation $X_P w = i \unicode{x3bb} w$ reads $X_P w_k = i \unicode{x3bb} w_k$ for all $k \in \mathbb {Z}$ , where $w_k \in L^2(SM,L_k)$ (similarly to equation (5.8) and the subsequent argument). By [Reference LefeuvreLef23, Proposition 2.7], we also have that $w_k$ is smooth, that is, $w_k \in C^\infty (SM,L_k)$ . Moreover, $X|w_k|^2 = 0$ , where $|w_k|^2(v) := \int _{P_v} |w_k(v, u)|^2\, du$ and $u \in P_v$ is the variable in the $\mathrm {U}(1)$ -fibre, so $|w_k|$ is constant by ergodicity of the flow $(\varphi _t)_{t \in \mathbb {R}}$ . If $k=0$ , $L_0 = \mathbb {C}$ is trivial over $SM$ so the mixing of the geodesic flow $(\varphi _t)_{t \in \mathbb {R}}$ then implies that $w_0 \equiv 0$ . If $k \neq 0$ and $w_k \neq 0$ , then we obtain a smooth nowhere vanishing section $w_k \in C^\infty (SM,L_k)$ , so $L_k$ is trivial. However, in turn, this implies that the first Chern class of $L_1 = \Lambda ^{m-1,0}\mathcal {N}$ is torsion. From the proof of Corollary 3.2, we obtain that the canonical bundle $K_M$ of M has torsion first Chern class so the image of $c_1(K_M)$ in $H^2(M,\mathbb {R})$ vanishes. This is a contradiction since this image is equal to $[-1/({2\pi })\rho ]$ where $\rho $ is the Ricci form of M, which is not exact since M has negative definite Ricci tensor. Hence, any solution to equation (5.11) is trivial, and thus, so is any solution to equation (5.7). This finishes the proof of Proposition 5.6.