Proof. Let $\rho :\pi _1(S_g)\to \operatorname {\mathrm {Sp}}(2n,\mathbb {R})$ be $\lbrace n\rbrace $ -Anosov. Let $(x,y,z)$ be a positively oriented triple in $\partial \pi _1(S_g)$ . Up to changing the symplectic basis, we can assume that for some $(\epsilon _i)\in \lbrace -1,1\rbrace $ :

$$ \begin{align*}\xi^n_\rho(x)=\langle x_1,x_2,\cdots, x_n\rangle,\end{align*} $$

$$ \begin{align*}\xi^n_\rho(y)=\langle x_1+\epsilon_1 y_1,x_2+ \epsilon_2 y_2,\cdots, x_n +\epsilon_ny_n\rangle,\end{align*} $$

$$ \begin{align*}\xi^n_\rho(z)=\langle y_1,y_2,\cdots, y_n\rangle.\end{align*} $$

Here the Maslov index of the triple $(\xi ^n_\rho (x),\xi ^n_\rho (y),\xi ^n_\rho (z))$ is equal to the sum of the $(\epsilon _i)$ .

Consider the following curve:

$$ \begin{align*}\tau_0:\theta\mapsto \langle \cos\left(\frac{\theta}{2}\right)x_1+\epsilon_1\sin\left(\frac{\theta}{2}\right)y_1,\cdots, \cos\left(\frac{\theta}{2}\right)x_n+ \epsilon_n\sin\left(\frac{\theta}{2}\right)y_n\rangle.\end{align*} $$

This loop is homotopic to the concatenation of the loops $\tau _i$ for $1\leq i\leq n$ :

$$ \begin{align*}\tau_i:\theta\mapsto \langle x_1, x_2,\cdots , \cos\left(\frac{\theta}{2}\right)x_i+ \epsilon_i\sin\left(\frac{\theta}{2}\right)y_i, \cdots , x_n\rangle.\end{align*} $$

These loops are homotopic to $\tau $ or its inverse depending on the sign of $\epsilon _i$ . The homotopy type of $\tau _0$ is hence equal to $(\epsilon _1+\epsilon _2+\cdots +\epsilon _n)[\tau ]$ . Moreover the set of Lagrangians transverse to a given Lagrangian is contractible, so one can homotope $\tau _0$ on the intervals $[0,\frac {\pi }{2}]$ , $[\frac {\pi }{2}, \pi ]$ and $[\pi ,2\pi ]$ to coincide with $\xi ^n_\rho $ . Hence the free homotopy class of $\xi ^n_\rho $ is equal to the one of $\tau _0$ , which is equal to $M(\xi ^n_\rho (x),\xi ^n_\rho (y),\xi ^n_\rho (z))[\tau ]$ . We therefore deduce that $\rho $ is maximal if and only if it is $\lbrace n\rbrace $ -Anosov and $[\xi ^n_\rho ]=n[\tau ]$ .

Proof. Fix $P\in \operatorname {\mathrm {Gr}}_d(\mathcal {Q})$ and suppose that there is a positive element $s\in E^\circ $ . We write s as a finite sum of positive rank one elements of the form $v_i\otimes v_i$ with $v_i\in V$ nonzero for $i\in I$ in a finite set. Indeed s defines a positive bilinear form on $V^*$ and hence is a sum of rank one symmetric positive bilinear forms. Let $q\in P$ be a semi-positive element. One has $q(s)=0$ since $s\in P^\circ $ but $q(v_i\otimes v_i)=q(v_i,v_i)\geq 0$ since q is positive and hence for all $i\in I$ , $q(v_i,v_i)=0$ .

Note that for all $v\in V$ nonzero we can choose a decomposition of s such that v is one of the $v_i$ , since $s-\epsilon v\otimes v$ is still positive for $\epsilon $ small enough.

We proved that $q(v,v)=0$ for all $v\in V$ so q is necessarily equal to $0$ . Hence P does not contain any nonzero semi-positive element.

Now suppose that $P^\circ $ is disjoint from the cone $S^2V^{>0}$ , so there must be a hyperplane H in $S^2V$ containing $P^\circ $ and disjoint from $S^2V^{>0}$ . This hyperplane corresponds to $\langle q\rangle ^\circ $ for some nonzero $q\in P$ . This element has the property that $q(s)\neq 0$ for all $s\in S^2V^{>0}$ . Up to exchanging q by $-q$ one can assume that $q(s)>0$ on $S^2V^{>0}$ , so $q(s)\geq 0$ on $S^2V^{\geq 0}$ . In particular for all $v\in V$ , one has $q(v\otimes v)=q(v,v)\geq 0$ , so q is semi-positive.

Proof. Let f be a face of $\mathbb {P}(S^2V^{\geq 0})$ , that is, the intersection of $\mathbb {P}(S^2V^{\geq 0})$ with a projective hyperplane corresponding to $[q]\in \mathbb {P}(S^2V^*)=\mathbb {P}(\mathcal {Q})$ not intersecting $\mathbb {P}(S^2V^{>0})$ trivially. It is the intersection of this convex with a projective hyperplane corresponding to $[q]\in \mathbb {P}(S^2V^*)=\mathbb {P}(\mathcal {Q})$ . The fact that this hyperplane intersects $\mathbb {P}(S^2V^{>0})$ trivially implies that there is a lift $q\in \mathcal {Q}$ that is a semi-positive element, by Proposition 3.2. Let $W\subset V$ be the vector subspace of isotropic vectors for q. The corresponding face is equal to $\mathbb {P}(S^2W^{\geq 0})$ .

Hence faces of $\mathbb {P}(S^2V^{\geq 0})$ are of the form $\mathbb {P}(S^2W^{\geq 0})$ for $W\subset V$ a linear subspace. This has dimension $0$ or at least $2$ , and therefore no face is a segment.

Suppose that for some general projective hyperplane, some extreme point $[s]$ of $\mathbb {P}(H\cap S^2V^{\geq 0})$ is not an extreme point of $\mathbb {P}(S^2V^{\geq 0})$ , then it belongs to the interior of a face of $\mathbb {P}(S^2V^{\geq 0})$ , that has dimension at least $2$ as stated previously. The intersection of this face with H contains therefore an open segment containing s, so s is not an extreme point of $H\cap \mathbb {P}(S^2V^{\geq 0})$ .

Proof of Proposition 3.4

Let us prove that $(i)$ implies $(ii)$ . The set $P_1^\circ \cap P_2^\circ $ is disjoint from $\mathbb {P}(S^2V^{\geq 0})$ if and only if there exists an element in $P_1+P_2$ that belongs to the dual of $\mathbb {P}(S^2V^{\geq 0})$ , that is, if there exists a positive bilinear form $q\in P_1+P_2$ . This form can be written as $q=q_1-q_2$ with $q_1\in P_1$ and $q_2\in P_2$ , therefore $(i)$ implies $(ii)$ .

Moreover $(ii)$ implies $(iii)$ . Indeed, if $q_2-q_1$ is positive then $\lbrace q_1\geq 0\rbrace \subset \lbrace q_2> 0\rbrace $ .

It only remains to show that $(iii)$ implies $(i)$ . Lemma 3.5 implies that:

$$ \begin{align*}\mathbb{P}(\langle q_1\rangle^\circ\cap S^2V^{\geq 0})=\text{Hull}(S^2\lbrace q_1=0\rbrace).\end{align*} $$

Hence $q_2\in P_2\subset \mathcal {Q}=S^2V^*$ is positive on the cone $\langle q_1\rangle ^\circ \cap S^2V^{\geq 0}\subset S^2V$ and therefore $\mathbb {P}(P_2^\circ )$ does not intersect $\mathbb {P}(\langle q_1\rangle ^\circ \cap S^2V^{\geq 0})$ .

Proof. We first show that $\operatorname {\mathrm {Ker}}(\mathrm {v}^\circ )=\operatorname {\mathrm {Im}}(\mathrm {v})^\circ $ , where we use the natural identification $(\mathcal {Q}/P)^*\simeq P^\circ $ .

We prove this by writing an equation that relates $\mathrm {v}$ and $\mathrm {v}^\circ $ . By definition, $\operatorname {\mathrm {Tr}}(qs)=0$ for $q\in P$ and $s\in P^\circ $ . Let us fix $q\in P$ and $s\in P^\circ $ and choose some representatives $\overline {\mathrm {v}(s)}\in S^2V$ and $\overline {\mathrm {v}^\circ (q)}\in \mathcal {Q}$ for $\mathrm {v}(s)\in S^2V/P^\circ $ and $\mathrm {v}^\circ (q)\in \mathcal {Q}/P$ . If $(P_t)$ is a smooth curve with $P_0=P$ and with derivative $\mathrm {v}$ at $t=0$ , $s+t\overline {\mathrm {v}^\circ (p)}+o(t)\in P_t$ and $q+t\overline {\mathrm {v}(q)}+o(t)\in P^\circ _t$ . Hence we get:

$$ \begin{align*}\operatorname{\mathrm{Tr}}\left(q \overline{\mathrm{v}^\circ(s)}+ \overline{\mathrm{v}(q)}s\right)=0.\end{align*} $$

An element $s\in P^\circ $ satisfies $s\in \operatorname {\mathrm {Ker}}(\mathrm {v}^\circ )$ if and only if $\operatorname {\mathrm {Tr}}(\overline {\mathrm {v}(q)}s)=0$ for all $q\in \mathcal {Q}$ , hence if and only if the corresponding linear form on $Q/P$ belongs to $\operatorname {\mathrm {Im}}(\mathrm {v})^\circ $ .

Now we prove the equivalence of the two definitions. If there exists $[q]\in \operatorname {\mathrm {Im}}(\mathrm {v})$ with q positive, then for any $s\in \operatorname {\mathrm {Ker}}(\mathrm {v}^\circ )$ , $\operatorname {\mathrm {Tr}}(q\circ s)=0$ . Hence s is not a positive tensor. Therefore $\operatorname {\mathrm {Ker}}(\mathrm {v}^\circ )\subset P^\circ $ intersects trivially $S^2V^{\geq 0}$ .

Conversely if $\operatorname {\mathrm {Ker}}(\mathrm {v}^\circ )$ intersects trivially $S^2V^{\geq 0}$ , there exists $q\in \mathcal {Q}$ that do not vanish on $S^2V^{\geq 0}$ , that is, q is positive. The class $[q]\in \mathcal {Q}/P$ belongs to $\operatorname {\mathrm {Im}}(\mathrm {v})$ .

Proof. Let us fix a complement H of $P^\circ $ in $S^2V$ . Let $N=\dim (\mathcal {Q})=\frac {n(n+1)}{2}$ . We identify a neighborhood of $P^\circ \subset \operatorname {\mathrm {Gr}}_{N-d}(\mathcal {Q})$ with $\operatorname {\mathrm {Hom}}(P^\circ ,H)$ . If there exists a nonzero element $s\in \operatorname {\mathrm {Ker}}(\mathrm {v}^\circ )\cap S^2V^{\geq 0}$ , one can consider the curve where $\gamma _t$ corresponds to

$$ \begin{align*}q\in P^\circ\mapsto t\mathrm{v}(q).\end{align*} $$

The nonzero element $s \in S^2V^{\geq 0}$ belongs to $P^\circ =\gamma _0^\circ $ and $\gamma _t^\circ $ hence the pair $(\gamma _0, \gamma _t)$ is not fitting for any $t>0$ .

In general the element corresponding to $\gamma _t$ is equal for t close to $0$ to the element of $\operatorname {\mathrm {Gr}}_{N-d}(\mathcal {Q})$ which is the graph of the map $P^\circ \to H$ :

$$ \begin{align*}q \mapsto t\mathrm{v}(q)+o(t).\end{align*} $$

The pair $(\gamma _o, \gamma _t)$ is fitting if and only if $\operatorname {\mathrm {Ker}}(\mathrm {v})\cap S^2V^{\geq 0}=\lbrace 0\rbrace $ , hence if $\mathrm {v}$ is fitting.

In this case the distance between $\mathbb {P}(\gamma _0^\circ \cap S^2V^{\geq 0})$ and $\mathbb {P}(\gamma _t^\circ \cap S^2V^{\geq 0})$ grows at least linearly in t for t close to $0$ .

Proof. Because of the second part of Definition 3.6 a vector is well-fitting if and only if it belongs to $C_{[q]}$ for some $[q]\in P$ . We just check that the sets $C_{[q]}$ are indeed open convex cones.

Let us fix a complement H of P in $\mathcal {Q}$ to identify $T_P\operatorname {\mathrm {Gr}}_d(\mathcal {Q})$ with $\operatorname {\mathrm {Hom}}(P,H)$ . If $\mathrm {v}_1,\mathrm {v}_2$ lie in $C_{[q]}$ and if $\lambda ,\mu \in \mathbb {R}_{>0}$ , then for some $q_1,q_2\in P$ one has $\mathrm {v}_1(q)+q_1$ and $\mathrm {v}_2(q)+q_2$ positive. Therefore $(\lambda \mathrm {v}_1+\mu \mathrm {v}_2)(q)+\lambda q_1+\mu q_2$ is positive so $\lambda \mathrm {v}_1+\mu \mathrm {v}_2$ belongs to $C_{[q]}$ .

Proof. Note that since M has dimension d and $u^\circ (x_0)$ has codimension d, the statement $(i)$ is equivalent to having for any Riemannian distance $d_R$ on $\mathbb {P}(S^2V)$ and $d_M$ on M, for some $\epsilon>0$ when x is close to $x_0$ :

$$ \begin{align*}d_R \left(\mathbb{P}\left(u^\circ(x)\cap S^2V^{\geq 0}\right),\mathbb{P}\left(u^\circ(x_0)\cap S^2V^{\geq 0}\right)\right)\geq \epsilon d_M(x,x_0).\end{align*} $$

Proposition 3.7 shows that the fitting condition is equivalent to having this distance growing linearly, hence the statements $(i)$ and $(ii)$ are equivalent.

A pair $(P,q_0)\in \operatorname {\mathrm {Gr}}_d(\mathcal {Q})\times \mathcal {Q}$ correspond to an element in $\mathcal {E}$ if and only if $q_0\in P$ . A pair $\left (\mathrm {v},\dot {q}\right )\in T_P\operatorname {\mathrm {Gr}}_d(\mathcal {Q})\times T_q\mathcal {Q}$ can be written as $\left (\mathrm {d}\operatorname {\mathrm {pr}}(\mathrm {w}),\mathrm {d}\pi (\mathrm {w})\right )$ for some $\mathrm {w}\in T\mathcal {E}$ if and only if $\dot {q}$ belongs to the class defined by $\mathrm {v}(q_0)$ .

Let us show this last claim. Let H be a complement of P in $\mathcal {Q}$ let $u:[0,1]\to \operatorname {\mathrm {Hom}}(P,H)$ and $q:[0,1]\to \mathcal {Q}$ be smooth curves with derivative $\mathrm {v}$ and $\dot {q}$ at $q=0$ and such that $q(t)$ belongs to the graph of $u(t)$ for $t\in [0,1]$ . For some smooth curve $\tilde {q}:[0,1]\to P$ with $\tilde {q}(0)=q_0$ and all $t\in [0,1]$ :

$$ \begin{align*}q(t)= \tilde{q}(t)+u(t)(\tilde{q}(t)).\end{align*} $$

Differentiating this at $t=0$ and we get exactly $\mathrm {v}(q_0)+\tilde {q}'(0)=\dot {q}$ , so $\dot {q}$ belongs to the class defined by $\mathrm {v}(q_0)$ , since $\tilde {q}'(0)\in P$ . Reciprocally if this holds, one can construct such a curve $\tilde {q}$ , so the pair corresponds to an element of $T\mathcal {E}$ .

We conclude that $(iii)$ is equivalent to the second characterization of fitting vectors in Definition 3.6: one can find such a positive lift $\mathrm {w}$ if and only if one can find a class in $\operatorname {\mathrm {Im}}(\mathrm {v})$ that contains a positive element.

Proof. Assume by contradiction that for some $t_0>0$ and $q\in \mathbb {S} u^*\mathcal {E}$ one has $u(x)=\operatorname {\mathrm {pr}}(q)=\operatorname {\mathrm {pr}}(\Phi _{t_0}(q))$ . The fact that the flow is fitting implies that for some $\lambda>0$ , $\lambda \pi (\Phi _{t_0}(q))-\pi (q)\in \mathcal {Q}$ is positive. This positive quadric belongs to $u(x)$ , contradicting the fact that $u(x)\in \operatorname {\mathrm {Gr}}_d^{{\operatorname {\mathrm {mix}}}}(\mathcal {Q})$ by Proposition 3.2.

Proof. Let $P=u(x)$ . Let $s_0\in P^\circ \subset S^2V$ be a positive tensor, which exists since the pencil P is assumed to be in $\operatorname {\mathrm {Gr}}^{{\operatorname {\mathrm {mix}}}}_d(\mathcal {Q})$ . Let $P'\in \operatorname {\mathrm {Gr}}_d(S^2V)$ be a supplement of $P^\circ $ . Since u is continuous for all y close enough to x the vector subspace $u^\circ (y)$ is transverse to $P'$ , therefore there exists a unique vector $\phi (y)\in P'$ such that:

$$ \begin{align*}s_0+\phi(y)\in (s_0+P') \cap u^\circ(y) \subset S^2V^{>0}.\end{align*} $$

This defines a continuous map $\phi $ from a neighborhood U of $x\in M$ to $P'$ .

Let $[q]\in \mathbb {S} u^*\mathcal {E}_x$ . For all t such that $s\circ \Phi _t(q)\in U$ , the linear form $\pi \left (\Phi _t(q)\right )\in \mathcal {Q}=S^2V^*$ vanishes on $\phi \left ( s\circ \Phi _t(q)\right )\in S^2V$ since this point belongs to $u^\circ (\operatorname {\mathrm {pr}}\circ \Phi _t(q))$ . Moreover $\pi \left (\Phi _t(q)\right )-\pi (q)$ is a positive bilinear form since $\Phi $ is a fitting flow.

In particular $\pi (q)\in \mathcal {Q}=S^2V^*$ is always negative on $\phi \left ( \operatorname {\mathrm {pr}}\circ \Phi _t(q)\right )\in S^2V$ . Hence for t small enough $[\phi \circ S_t]:\mathbb {S} u^*\mathcal {E}_x \to \mathbb {S} P'$ has the same degree as $[\pi ]: \mathbb {S} u^*\mathcal {E}_x \to \mathbb {S} P^{\prime *}\simeq \mathbb {S} P$ which associates to $[q]\in \mathbb {S} u^*\mathcal {E}_x $ the class $[\pi (q)]\in \mathbb {S} P$ . The map $[\pi ]$ is a diffeomorphism, so in particular $1=|\deg ([\pi ])|=|\deg (\phi )\deg (S_t)|$ . Hence $S_t$ is a generator of the homotopy group of $U\setminus \lbrace x\rbrace $ .

Proof. For all $[q]\in \mathbb {S} u^*\mathcal {E}$ we consider the convex $\mathbb {P}(\langle \pi (q)\rangle ^\circ \cap S^2V^{>0})$ .

We choose a continuous and $\rho $ -equivariant map $\sigma :\mathbb {S} u^*\mathcal {E}\to \mathbb {P}(S^2V^{>0})$ with the property that $\sigma ([q])\in \mathbb {P}(\langle \pi (q)\rangle ^\circ \cap S^2V^{>0})$ for all $[q]\in \mathbb {S} u^*\mathcal {E}$ . One can always find such a choice, as it amounts to finding a section of an open ball bundle over a compact set. Indeed $\mathbb {P}(\langle \pi (q)\rangle ^\circ \cap S^2V^{>0})$ is a nonempty convex set of codimension $1$ that forms an open ball for all $[q]\in \mathbb {S} u^*\mathcal {E}$ .

We fix a $\Gamma $ -invariant Riemannian metric g on $\mathbb {S} u^*\mathcal {E}$ with associated distance $d_g$ . We set $C_1$ to be the supremum of $d_H\left (\sigma ([q_1]),\sigma ([q_2])\right )$ for all $[q_1],[q_2]\in \mathbb {S} u^*\mathcal {E}$ such that $d_g([q_1],[q_2])\leq 1$ , which exists since $\Gamma $ acts cocompactly on $\widetilde {N}$ . The following inequality follows for all $[q_1],[q_2]\in \mathbb {S} u^*\mathcal {E}$ from the triangular inequality:

$$ \begin{align*}d_H\left(\sigma([q_1]),\sigma([q_2])\right)\leq C_1 d_g([q_1],[q_2])+C_1.\end{align*} $$

Let n be the unique integer $d_g([q_1],[q_2])\leq n< d_g([q_1],[q_2])+1$ . One can find elements $x_0=[q_1], x_1,\cdots , x_n=[q_2]$ in $\mathbb {S} u^*\mathcal {E}$ such that for all $1\leq i\leq n$ one has $d_g(x_{i-1},x_i)\leq 1$ . The triangular inequality for the Hilbert distance implies that:

$$ \begin{align*}d_H\left(\sigma([q_1]),\sigma([q_2])\right)\leq nC_1.\end{align*} $$

We now set $C_2$ to be the supremum of $d_H\left (\sigma ([q_1]),\sigma (\Phi _t([q]))\right )$ for all $[q_1]\in \mathbb {S} u^*\mathcal {E}$ and $0\leq t\leq 1$ . Similarly we get the following inequality for all $[q]\in \mathbb {S} u^*\mathcal {E}$ and $t\geq 0$ :

$$ \begin{align*}d_H\left(\sigma([q]),\sigma(\Phi_t([q]))\right)\leq C_2 t+C_2.\end{align*} $$

Let K be a compact fundamental domain for the action of $\Gamma $ on $\mathbb {S} u^*\mathcal {E}$ and let $\epsilon $ be the infimum of the Hilbert distance for any $[q]\in K$ between $\mathbb {P}\left (\langle \pi (q)\rangle ^\circ \cap S^2V^{>0}\right )$ and $\mathbb {P}\left (\langle \pi (\Phi _1(q))\rangle ^\circ \cap S^2V^{>0}\right )$ . Since the flow is fitting, the closures of these two sets in $\mathbb {P}(S^2V)$ are disjoint for any $[q]$ , and hence their Hilbert distance is positive. Since K is compact, the infimum $\epsilon $ is also positive.

The Hilbert distance between $\sigma (q)$ and $\sigma (\Phi _t(q))$ for $t>0$ and $q\in \mathbb {S} u^*\mathcal {E}$ is greater than $\epsilon (t-1)$ . Indeed for all integer $0\leq n\leq t$ the projective segment between $\sigma (q)$ and $\sigma (\Phi _t(q))$ , which is a geodesic for the Hilbert distance, intersects $\mathbb {P}\left (\langle \pi (\Phi _n(q))\rangle ^\circ \cap S^2V^{>0}\right ) $ in exactly one point $x_n$ . Hence the Hilbert distance between $x_n$ and $x_{n+1}$ for $0\leq n\leq t-1$ is at least $\epsilon $ .

Putting all of these inequalities together we get that for all $t\geq 0$ and $[q]\in \mathbb {S} u^*\mathcal {E}$ :

$$ \begin{align*}\frac{\epsilon}{C_1}(t-1)-1\leq d_g\left(\sigma([q]),\sigma(\Phi_t([q]))\right)\leq C_2 (t+1)\end{align*} $$

Hence the flow lines are quasi-isometric embeddings.

We now check that flow lines exist between any pair of points. Let $x\in M$ . Given $t\in \mathbb {R}$ we consider the d-sphere $S_t:q\in \mathbb {S} u^*\mathcal {E}_{|x}\mapsto \Phi _t(q)\in u^*\mathcal {E}$ . Suppose that some $y\in M$ , avoids the sphere $S_t$ for all $t>0$ . Consider a curve $\eta $ between x and y. The homological intersection between this segment and the spheres $S_t$ in $M\setminus \lbrace x,y\rbrace $ is constant, and is equal to zero for t large enough since the spheres $S_t$ are then uniformly far from x. However for t small enough, the homotopy class of $S_t$ is the one of any small sphere encircling x by Lemma 4.3. This leads to a contradiction since such a sphere will have homological intersection equal to $1$ or $-1$ with the curve $\eta $ . Hence there exists a flow line joining any pair of points.

Proof. Since there are flow lines between any pair of points in $\widetilde {N}$ , for every $x \neq y\in M$ one can find $q\in u(x)$ and $q'\in u(y)$ such that $q-q'$ is positive. Since the pencils $u(x)$ and $u(y)$ are mixed, one cannot have $q-q'\in u(x)=u(y)$ so $u(x)\neq u(y)$ . Furthermore the pair $(u(x),u(y))$ is a fitting pair for all $x \neq y\in M$ , hence u is a globally fitting map by Proposition 3.4.

For any $\Gamma $ -invariant Riemannian metric g on $\mathbb {S} \mathcal {E}$ , any map $\sigma $ as the proof of Proposition 4.5 is a quasi-isometric embedding. Indeed let $\delta $ be the maximum of $d_g(x,x')$ or $d_H(\sigma (x),\sigma (x'))$ for any $x,x'$ in the same fiber of $\operatorname {\mathrm {pr}}:\mathbb {S} \mathcal {E}\to \widetilde {N}$ . For every $x,y\in \mathbb {S} \mathcal {E}$ one can find $x',y'$ in the same fibers respectively as $x,y$ and in the same flow line for $\Phi $ . Hence for the constants $C,D$ from Proposition 4.5:

$$ \begin{align*}\frac{1}{C}d_g(x,y)-D-(\frac{2}{C}+2)\delta\leq d_H(\sigma(x),\sigma(y))\leq Cd_g(x,y)+D+(2C+2)\delta .\end{align*} $$

Hence $\rho $ is a quasi-isometric embedding.

Proof. Let us fix $x\in M$ . We construct a continuous map:

$$ \begin{align*}\phi:T_xM\setminus \lbrace 0\rbrace\to u^*\mathcal{E}_{x}\setminus \lbrace 0\rbrace.\end{align*} $$

We require that this map satisfies $\lambda \phi (v)=\phi (\lambda v)$ for all $v\in T_xM\setminus \lbrace 0\rbrace $ and $\lambda \in \mathbb {R}$ . Note that in particular $\phi $ defines an odd map:

$$ \begin{align*}\overline{\phi}:\mathbb{S} T_xM\to \mathbb{S} u^*\mathcal{E}_x.\end{align*} $$

We furthermore construct a lift:

$$ \begin{align*}\psi:T_xM\setminus \lbrace 0\rbrace\to Tu^*\mathcal{E}_.\end{align*} $$

In other words we assume that $\psi (v)\in T_{\phi (v)}(u^*\mathcal {E})$ for $v\in T_xM\setminus \lbrace 0\rbrace $ . We require that $\psi (\lambda v)=\lambda (\mathrm {d}m_\lambda )(\psi (v))$ for all $v\in T_xM\setminus \lbrace 0\rbrace $ and $\lambda \in \mathbb {R}$ where $m_\lambda :u^*\mathcal {E}\to u^*\mathcal {E}$ is the multiplication by $\lambda $ . We require $\mathrm {d}\pi (\psi (v))\in T\mathcal {Q}$ to be positive for all $v\in T_xM\setminus \lbrace 0\rbrace $ .

Finally we will make this construction so that in addition $\mathrm {d}\operatorname {\mathrm {pr}}\left ( \psi (v)\right )=v$ for $v\in T_xM\setminus \lbrace 0\rbrace $ , but this property will be only used during the construction.

The following diagram illustrates the situation.

If we can construct such continuous maps, the fact that $\overline {\phi }$ is an odd map between two spheres of the same dimension implies that it is homotopically nontrivial and therefore surjective. In particular for all $[q]\in \mathbb {S}\mathcal {E}_{u(x)}\simeq \mathbb {S} u(x)$ there exists a $v\in T_xM$ such that $\phi (v)=q$ . The element $\mathrm {w}=\psi (v)$ then satisfies the required conditions, so this finishes the proof.

Now let us construct the maps $\phi $ and $\psi $ . We first show that for all $v\in T_xM\setminus \lbrace 0\rbrace $ we can define $\phi (v)$ and $\psi (v)$ , and then we explain how to glue these maps together to get a continuous map.

Since u is a fitting immersion, the point $(iii)$ of Proposition 3.12 implies that given $v_0\in T_xM$ one can construct $\phi (v_0)\in u^*\mathcal {E}_{x}$ and $\psi (v_0)\in T_{\phi (v_0)} u^*\mathcal {E}$ such that $\mathrm {d}\pi (\psi (v_0))$ is positive and $\mathrm {d}\operatorname {\mathrm {pr}}\left ( \psi (v_0)\right )=v_0$ .

Note that the condition that $\mathrm {d}\pi (\psi (v))$ is positive is an open condition and the condition that $\mathrm {d}\operatorname {\mathrm {pr}}\left ( \psi (v)\right )=v$ requires that $\psi $ is a section of an affine sub-bundle. Hence for every $v\in T_xM\setminus \lbrace 0\rbrace $ we can find a small neighborhood $\mathcal {S}$ in a sphere in $T_xM$ containing $v_0$ on which we can define $\phi $ and a lift $\psi $ such that $\mathrm {d}\pi (\psi (v))$ is positive and $\mathrm {d}\operatorname {\mathrm {pr}}\left ( \psi (v)\right )=v$ for all $v\in \mathcal {S}$ . We take $\mathcal {S}$ small enough so that it does not contain any antipodal pair of points.

We define U to be the set of nonzero elements $\lambda v$ for all $\lambda \in \mathbb {R}$ and $v\in \mathcal {S}$ , and we extend $\phi $ and $\psi $ to U in a homogeneous way. We define $\phi $ on U so that $\phi _i(\lambda v)=\lambda \phi _i(v)$ for all $\lambda \in \mathbb {R}$ nonzero and $v\in \mathcal {S}$ . We set $\psi (\lambda v)=\lambda (\mathrm {d}m_\lambda ) \psi _i(v)$ for $\lambda \in \mathbb {R}$ , where $m_\lambda $ is the multiplication by $\lambda $ on $u^*\mathcal {E}_x$ . Note that $\mathrm {d}\pi \left (\psi _i(\lambda v)\right )=\lambda ^2\mathrm {d}\pi \left (\psi _i(v)\right )$ is positive and $\mathrm {d}\operatorname {\mathrm {pr}}\left ( \psi _i( \lambda v)\right )=\lambda v$ for all $\lambda v\in U$ . Indeed $\pi \circ m_\lambda =\lambda \pi $ so $\mathrm {d}\pi \circ \mathrm {d}m_\lambda =\lambda \mathrm {d}\pi $ and $p\circ m_\lambda =p$ so $\mathrm {d}\operatorname {\mathrm {pr}}\circ \mathrm {d}m_\lambda =\mathrm {d}\operatorname {\mathrm {pr}}$ .

We therefore can construct an open cover $\lbrace U_i\rbrace _{i\in I}$ of $T_xM$ and continuous maps $\phi _i:U_i\to \mathbb {S} u^*\mathcal {E}\subset u^*\mathcal {E}$ , and lifts $\psi _i:U_i\to Tu^*\mathcal {E}$ such that $\mathrm {d}\pi (\psi _i(v))$ is positive and $\mathrm {d}\operatorname {\mathrm {pr}}\left ( \psi _i(v)\right )=v$ for any $i\in I$ and $v\in U_i$ . The $U_i$ can be assumed invariant by scalar multiplication, and $\phi $ and $\psi $ satisfy the aforementioned homogeneity conditions.

We now glue these maps together. Let $\chi _i:U_i\to [0,1]$ for $i\in I$ be a family of functions that forms a locally finite partition of the unit. We define:

$$ \begin{align*}\phi:v\in \mathbb{S} T_xM \mapsto \sum_{i\in I}\chi_i(v)\phi_i(v)\in \left(u^*\mathcal{E}\right)_x.\end{align*} $$

Let us check that $\phi (v)$ is always nonzero. This is where we use that $\mathrm {d}\operatorname {\mathrm {pr}}\left ( \psi _i(v)\right )=v$ for $i\in I$ , and we also use that the fitting immersion u is defining a smooth fibration of the cone $S^2 V^{>0}$ . Let $\gamma :\mathbb {R} \to M$ be a curve such that $\gamma (0)=x$ and $\gamma '(0)=v$ . For all $t\in \mathbb {R}$ the intersection $ u^\circ (\gamma (t))\cap S^2V^{>0}$ is a nonempty convex set, so we can construct a section $s:t\in \mathbb {R} \mapsto S^2V^{>0}$ of the fibration, that is, such that for all $t\in \mathbb {R}$ , $s(t)\in u^\circ (\gamma (t))$ .

Let us fix $i\in I$ . Let $q:\mathbb {R}\to \mathbb {S} u^*\mathcal {E}$ be such that $q'(0)=\psi (v)$ , which implies that $p\circ q(t)$ is equal to $\gamma (t)$ at the first order around $t=0$ since $\gamma '(0)=\mathrm {d}\operatorname {\mathrm {pr}}\left (\psi (v)\right )=v$ . Since $s(t)\in u^\circ (\gamma (t))$ , one has $ \pi (q(t))\cdot s(t)=0$ at the first order around $t=0$ . Taking the first derivative at $t=0$ of this equation we get:

$$ \begin{align*}\pi(\phi_i(v))\cdot s'(0)+\mathrm{d}\pi(\psi_i(v) )\cdot s(0)=0.\end{align*} $$

Since $\mathrm {d}\pi (\psi _i(v) )$ is positive and $s(0)$ is a positive tensor:

$$ \begin{align*}\mathrm{d}\pi(\psi_i(v) )\cdot s(0)>0.\end{align*} $$

Hence for all $i\in I$ , $\pi (\phi _i(v))\cdot s'(0)<0$ and therefore $\pi (\phi (v))\cdot s'(0)<0$ . In particular $\phi (v)$ does not vanish.

In order to glue the $\psi _i$ we need to be careful since the vectors $\psi _i(v)$ do not belong to the same fiber of the tangent bundle $Tu^*\mathcal {E}$ . Let $\Sigma $ be the following map:

$$ \begin{align*}\Sigma: (q_i)_{i\in I}\in \left(u^*\mathcal{E}\right)_x^I\mapsto \sum_{i\in I}q_i\in \left(u^*\mathcal{E}\right)_x.\end{align*} $$

Given $v\in \mathbb {S} T_xM$ we set:

$$ \begin{align*}\psi(v)=\mathrm{d}\Sigma\left(\left(\chi_i(v) \psi_i(v)\right)_{i\in I}\right).\end{align*} $$

This combination still satisfies that $\mathrm {d}\pi (\psi )$ is positive, indeed:

$$ \begin{align*}\mathrm{d}\pi(\psi(v))=\sum_{i\in I}\chi_i(v)\mathrm{d}\pi(\psi_i(v)).\end{align*} $$

Note that we also get the following:

$$ \begin{align*}\mathrm{d}\operatorname{\mathrm{pr}}(\psi(v))=\sum_{i\in I}\chi_i(v)\mathrm{d}\operatorname{\mathrm{pr}}(\psi_i(v))=\left(\sum_{i\in I}\chi_i(v)\right)v=v.\end{align*} $$

This concludes the construction of $\phi $ and $\psi $ , and hence this concludes the proof.

Proof of Proposition 4.8

We construct a continuous vector field $\mathrm {W}$ over $u^*\mathcal {E}$ , except the zero section that is homogeneous, that is, such that for all $x\in M$ and non zero $q\in u(x)$ one has $W_{\lambda q}=\mathrm {d}(m_\lambda )W_q$ for all $\lambda \in \mathbb {R}^{> 0}$ , where $m_\lambda $ is the multiplication by $\lambda $ on $\mathcal {E}$ . Such a vector field defines a vector field $\overline {W}$ on $\mathbb {S} u^*\mathcal {E}$ . We require moreover that $\mathrm {d}\pi (W)$ is always positive.

Given any nonzero $q_0\in u^*\mathcal {E}$ , Lemma 4.9 provides the existence of some $\mathrm {w}_0\in T_{(u(x),q)}\mathcal {E}$ for all $(x,q)\in u^*\mathcal {E}$ such that:

1. (i) $\mathrm {d}\operatorname {\mathrm {pr}}(\mathrm {w}_0)\in \mathrm {d}u(T_xM)$ ,

2. (ii) $\mathrm {d}\pi (\mathrm {w}_0)$ is positive.

Property $(i)$ implies that $\mathrm {w}_0$ defines a vector in $T_{x,q_0}u^*\mathcal {E}$ . For each such $q_0$ one can find a neighborhood U of it in $u^*\mathcal {E}$ that is invariant by the $\mathbb {R}^{> 0}$ -action, and on which one can define a map W satisfying $(i)$ as well as the homogeneity condition $W_{\lambda q}=\mathrm {d}(m_\lambda )W_q$ . Condition $(ii)$ being an open condition invariant by the action of $m_\lambda $ it is satisfied automatically for U small enough.

Using a partition of the unit as in Lemma 4.9 we construct the desired vector field on $u^*\mathcal {E}$ . Note that both property $(i)$ and the homogeneity are preserved by linear combinations, and $(ii)$ is preserved by positive combinations.

Finally note that the collection $U_i$ as well as the partition of the unity can be chosen to be $\rho $ -equivariant, so that the vector field W is also $\rho $ -equivariant, and hence also the fitting flow $\Phi $ .

Proof. First note that the assumption $q_2-q_1$ is positive implies that $\lbrace q_1\geq 0 \rbrace \subset \lbrace q_2>0\rbrace $ and $\lbrace q_2\leq 0 \rbrace \subset \lbrace q_1<0\rbrace $ . If L intersects $\lbrace q_1>0\rbrace $ and $\lbrace q_1< 0 \rbrace $ then the quadratic forms $q_1$ and $q_2$ are both positive on some points and negative on some points of L, and hence they both admit exactly two zeroes on L. These zeroes, denoted respectively $\ell _1$ , $\ell _1'$ for $q_1$ and $\ell _2$ , $\ell _2'$ for $q_2$ are in the following cyclic order on L up to exchanging $\ell _1$ and $\ell _1'$ : $(\ell _1,\ell _2,\ell _2',\ell _1')$ . One can refer to Figure 1 where $\lbrace q_i>0\rbrace $ is the interior of the corresponding ellipse.

Figure 1 Illustration of Proposition 5.2.

Consider the compact space X of projective lines L intersecting $\lbrace q_1\geq 0\rbrace $ and $\lbrace q_2\leq 0\rbrace $ . For every such line we can define the lines $\ell _1,\ell _1'$ and $\ell _2, \ell _2'$ as in previously where $\ell _1,\ell _1'\neq \ell _2,\ell _2'$ but eventually $\ell _1=\ell _1'$ or $\ell _2=\ell _2'$ which occur respectively exactly if L intersects trivially $\lbrace q_1>0\rbrace $ or $\lbrace q_2>0\rbrace $ . Indeed in these cases the quadratic forms $q_i$ are semi-positive or semi negative on L and hence have a double zero. By looking at the formula (2) on an affine chart such that $x_1'< x_2'\leq x_2< x_1$ we see the cross ratio $[\ell _1,\ell _2,\ell _2',\ell _1']$ is in $(1,\infty ]$ . This ratio is finite if and only if $\ell _1\neq \ell _1'$ and $\ell _2\neq \ell _2'$ which occurs exactly when L belongs to the interior of X, that is, when L intersects $\lbrace q_1> 0\rbrace $ and $\lbrace q_2< 0\rbrace $ . Finally note that this ratio depends continuously on $L\in X$ and that the interior of X is nonempty since $q_1$ and $q_2$ are mixed. Therefore the minimum on X is well defined in $ (1,\infty )$ and is reached on the interior of X.

Proof. Pick any projective line that crosses $\lbrace q_1>0\rbrace $ and $\lbrace q_3<0\rbrace $ . Let the intersections of L with the zeroes of $q_1$ , $q_2$ and $q_3$ be respectively $(\ell _1,\ell _2,\ell _3,\ell _3',\ell _2',\ell _1')$ , counted with multiplicity and cyclically ordered. We fix an affine chart for L such that this tuple corresponds to the tuple $x_3'< x_2'<x_1'\leq x_1<x_2<x_3$ of real numbers. This yields the following:

$$ \begin{align*}[\ell_1,\ell_3,\ell_3',\ell_1']& =\frac{x_3-x_1'}{x_3-x_3'}\times\frac{x_1-x_3'}{x_1-x_1'} .\\ [\ell_1,\ell_3,\ell_3',\ell_1']& = \left( \frac{x_3-x_1'}{x_3-x_2'}\times\frac{x_1-x_2'}{x_1-x_1'}\right)\times \left(\frac{x_3-x_2'}{x_3-x_3'}\times\frac{x_1-x_3'}{x_1-x_2'}\right) .\end{align*} $$

Note that if $B>A>0$ , for all $c>0$ one has $\frac {A}{B}\leq \frac {A+c}{B+c}$ and respectively $\frac {B+c}{A+c}\leq \frac {B}{A}$ . Hence, due to the ordering of the tuple, we have the following inequalities by setting respectively $A=x_2-x_1'$ , $B=x_2-x_2'$ , $c=x_3-x_2$ and $A=x_1-x_2'$ , $B=x_1-x_3'$ , $c=x_2-x_1$ :

$$ \begin{align*}\frac{x_2-x_1'}{x_2-x_2'}\leq \frac{x_3-x_1'}{x_3-x_2'}\;,\;\frac{x_2-x_3'}{x_2-x_2'} \leq \frac{x_1-x_3'}{x_1-x_2'}.\end{align*} $$

Hence we obtain:

$$ \begin{align*}\text{cr}([q_2],[q_1])\leq [\ell_1,\ell_2,\ell_2',\ell_1']&=\frac{x_2-x_1'}{x_2-x_2'}\times\frac{x_1-x_2'}{x_1-x_1'} \leq \frac{x_3-x_1'}{x_3-x_2'}\times\frac{x_1-x_2'}{x_1-x_1'} .\\\text{cr}([q_3],[q_2])\leq [\ell_2,\ell_3,\ell_3',\ell_2']&=\frac{x_3-x_2'}{x_3-x_3'}\times\frac{x_2-x_3'}{x_2-x_2'} \leq \frac{x_3-x_2'}{x_3-x_3'}\times\frac{x_1-x_3'}{x_1-x_2'}.\end{align*} $$

Hence one has $[\ell _1,\ell _3,\ell _3',\ell _1']\geq \text {cr}([q_2],[q_1])\text {cr}([q_3],[q_2])$ for every such projective line L.

Therefore $\text {cr}([q_3],[q_1])\geq \text {cr}([q_3],[q_2])\text {cr}([q_2],[q_1])$ .

Proof. The intersection $I=\bigcap _{n\in \mathbb {N}} \lbrace q_n\leq 0\rbrace $ is a compact nonempty subset. Let $x\neq y\in I$ and let L be the projective line from x to y. Suppose that there exists $z\in L$ such that $z\notin I$ . Without any loss of generality one can assume that the open interval $S\subset L$ bounded by $x,y$ and containing z does not intersect I. Indeed one can otherwise replace $x,y$ by the points on $L\cap I$ closest to z on both sides.

Since $z\notin I$ there exist $n_0\in \mathbb {N}$ such that $q_{n_0}$ is positive on z. For $n\geq n_0$ , let $z_1,z_2$ be the two intersections of L with the zeroes of $q_n$ so that the points $(x,z_n^-,z,z_n^+,y)$ are cyclically ordered. The sequences $(z_n^-)$ and $(z_n^+)$ are monotonic in S and must converge to x and y since $S\cap I=\emptyset $ . The value of $\text {cr}([q_{n_0}],[q_n])$ is bounded from above by the cross ratio $[z_{n_0}^+,z_n^+,z_n^-,z_{n_0}^-]$ , which in turn converges to the cross ratio $[z_{n_0}^+,x,y,z_{n_0}^-]<\infty $ when n goes to $+\infty $ . This contradicts the fact that $\text {cr}([q_{n}],[q_0])$ goes to $+\infty $ . Hence for every pair of points in I, the associated projective line is contained in I. In particular I is a projective subspace.

Proof. The flat V-bundle over $\mathbb {S} u^*\mathcal {E}$ associated with $\rho $ admits a continuous splitting $E\oplus F$ where for $q\in \mathbb {S} u^*\mathcal {E}$ :

$$ \begin{align*}\mathbb{P}\left(E_q\right) &=\bigcap_{t\geq 0} \lbrace \pi(\Phi_{-t}(q))\geq 0\rbrace,\\\mathbb{P}\left(F_q\right) &=\bigcap_{t\geq 0} \lbrace \pi(\Phi_{t}(q))\leq 0\rbrace.\end{align*} $$

This defines transverse vector subspaces by Proposition 5.3 since $\text {cr}(\Phi _t(q),q)$ and $\text {cr}(-\Phi _{-t}(q),-q)$ go to $+\infty $ when t goes to $+\infty $ . Moreover the quadrics in the pencils in the image of u are of signature $(n,n)$ , one must have $\dim (E_q)=\dim (F_q)=n$ so this splitting is well-defined. This splitting is preserved by $\Phi $ .

We now construct a metric h on this flat V-bundle over $\mathbb {S} u^*\mathcal {E}$ . Given $q\in \mathbb {S} u^*\mathcal {E}$ we define the symmetric bilinear form $h_q$ on $V=E_q\oplus F_q$ so that this sum is orthogonal and $h_q$ is equal to $\pi (q)\in \mathcal {Q}$ on $E_q$ and $-\pi (q)\in \mathcal {Q}$ on $F_q$ . Note that by definition of $E_q$ and $F_q$ , $h_q$ is positive.

We also introduce an auxiliary symmetric bilinear form $\bar {h}$ of signature $(n,n)$ on this flat V-bundle over $\mathbb {S} u^*\mathcal {E}$ so that the sum $E_q\oplus F_q$ is orthogonal and $\bar {h}_q$ is equal to $\pi (q)\in \mathcal {Q}$ on $E_q$ and on $F_q$ .

Our first step is to compare the quadric $\bar {h}_q$ with $\pi (q)$ . Let L be a projective line intersecting $\mathbb {P}(E_q)$ at some e and $\mathbb {P}(F_q)$ at some f. In Figure 3, we illustrate some $\mathbb {RP}^2\subset \mathbb {RP}^{2n-1}$ containing the projective line L. Let $\ell _1,\ell ^{\prime }_1$ be the zeroes of $\pi (q)$ on L and $\ell _2,\ell ^{\prime }_2$ be the zeroes of $\bar {h}_q$ on L, so that $\ell _1,\ell _2$ lie in the same connected component of $L\setminus \lbrace e,f\rbrace $ . Since N is compact, there exists a maximum $\delta <\infty $ for all $q\in \mathbb {S} u^*\mathcal {E}$ and all such projective line L of the quantity $|\log \left ([\ell _1,\ell _2,f,e]\right )|$ .

Now we turn our attention to the contraction properties of $\Phi $ . Let $t>0$ be a real number and let $q\in \mathbb {S} u^*\mathcal {E}$ . Let $v\in E_q$ and $w\in F_q$ . We are interested in the following ratio:

$$ \begin{align*}R=\frac{h_{\Phi_t(q)}(v)h_q(w)}{h_{\Phi_t(q)}(w)h_q(v)}.\end{align*} $$

Let $e,f$ be the lines generated by $v,w$ and L be the projective line joining them. Let $\ell _1,\ell ^{\prime }_1$ be the zeroes of $\pi (q)$ on L, $\ell _2,\ell ^{\prime }_2$ the zeroes of $\bar {h}_q$ on L, $\ell _3,\ell ^{\prime }_3$ be the zeroes of $\pi (\Phi _t(q))$ on L and finally $\ell _4,\ell ^{\prime }_4$ the zeroes of $h_{\Phi _t(q)}$ on L. We assume that $\ell _1,\ell _2,\ell _3,\ell _4$ all lie on the same component of $L\setminus \lbrace e,f\rbrace $ .

The cross ratio $[\ell _2,\ell _4,f,e]$ is equal to $R^{\frac {1}{2}}$ . Indeed $\ell _2$ is generated by $h^{\frac {1}{2}}_q(w)v+h^{\frac {1}{2}}_q(v)w$ and $\ell _4$ is generated by $h^{\frac {1}{2}}_{\Phi _t(q)}(w)v+h^{\frac {1}{2}}_{\Phi _t(q)}(v)w$ , up to changing $\ell _i$ by $\ell _i'$ for $1\leq i\leq 4$ . Hence:

$$ \begin{align*}[\ell_2,\ell_4,f,e]=\frac{\left( h^{\frac{1}{2}}_{q}(w)v+h^{\frac{1}{2}}_{q}(v)w\right)\wedge w}{\left( h^{\frac{1}{2}}_{q}(w)v+h^{\frac{1}{2}}_{q}(v)w\right)\wedge v}\times \frac{\left( h^{\frac{1}{2}}_{\Phi_t(q)}(w)v+h^{\frac{1}{2}}_{\Phi_t(q)}(v)w\right)\wedge v}{\left( h^{\frac{1}{2}}_{\Phi_t(q)}(w)v+h^{\frac{1}{2}}_{\Phi_t(q)}(v)w\right)\wedge w}=R^{\frac{1}{2}}.\end{align*} $$

However due to our comparison of $\pi (q)$ and $\bar {h}_q$ one has:

$$ \begin{align*}e^{-\delta}\leq [\ell_2,\ell_4,f,e],[\ell_1,\ell_3,f,e]\leq e^{\delta}.\end{align*} $$

Therefore $[\ell _2,\ell _4,f,e]/[\ell _1,\ell _3,f,e]\geq e^{-2\delta }$ . Hence $R^{\frac {1}{2}}>e^{-2\delta }[\ell _1,\ell _3,f,e]$ . This last cross ratio is larger than:

$$ \begin{align*}[\ell_1,\ell_3,\ell_3',\ell_1']\geq \text{cr}\left(\pi(\Phi_t(q)),\pi(q),\right).\end{align*} $$

Since $\Phi $ is a fitting flow and since N is compact, there exist $\alpha>0$ such that $\text {cr}\left (\pi (\Phi _1(q)),\pi (q)\right )\geq e^\alpha $ for all $q\in \mathbb {S} u^*\mathcal {E}$ . Hence by the triangular inequality from Proposition 5.2, for all $t>0$ :

$$ \begin{align*}\text{cr}\left(\pi(\Phi_t(q)),\pi(q)\right)\geq e^{\alpha (t-1)}.\end{align*} $$

Hence we get the following estimate:

$$ \begin{align*}\frac{h_{\Phi_t(q)}(v)h_q(w)}{h_{\Phi_t(q)}(w)h_q(v)}\geq e^{2\alpha t-4\delta-2\alpha} .\end{align*} $$

This implies that the splitting $V=E_q\oplus F_q$ is $\lbrace n\rbrace $ -contracting in the sense of [Reference BochiBPS19] with respect to the flow $\Phi $ for the metric h. Moreover $\Gamma $ acts cocompactly on $\mathbb {S} u^*\mathcal {E}$ and every geodesic in $\Gamma $ is at uniform distance from a flow line of $\Phi $ . The domination of this splitting implies an exponential gap for the singular values [Reference BochiBPS19, Theorem 2.2], which implies that $\rho $ is $\lbrace n\rbrace $ -Anosov. The vector subspace $\xi _\rho ^n(\zeta )$ is the contracted subspace $F_q$ .

Proof. Let us first prove that all of $\mathbb {P}(S^2V^{>0})$ is covered by the union of $\mathbb {P}(u^\circ (x)\cap S^2V^{>0})$ for $x\in \widetilde {N}$ . We fix a Riemannian metric on N that defines a Riemannian metric g on $\widetilde {N}$ , with associated distance $d_g$ . Since u is globally fitting, see Remark 4.6, for all $x\in \widetilde {N}$ there exists a neighborhood U of $\mathbb {P}(u^\circ (x)\cap S^2V^{\geq 0})$ in $\mathbb {P}(S^2V^{\geq 0})$ that is covered by the manifolds $\mathbb {P}(u^\circ (y)\cap S^2V^{\geq 0})$ for y in the ball for $d_g$ of radius $1$ centered at $0$ , by Proposition 3.12, part (i).

This neighborhood contains an $\epsilon $ -neighborhood of $\mathbb {P}(u^\circ (x)\cap S^2V^{> 0})$ for the Hilbert metric on $\mathbb {P}(S^2V^{>0})$ for some $\epsilon>0$ . Since N is compact, this $\epsilon>0$ can be chosen independently of x.

Now let us fix some $x_0\in \widetilde {N}$ and some $[s_0]\in \mathbb {P}(u^\circ (x_0)\cap S^2V^{> 0})$ . Given any $[s']\in \mathbb {P}(S^2V^{> 0})$ one can find a finite sequence $s_0,s_1,\cdots , s_k=s'$ in $S^2V^{>0}$ so that the Hilbert distance between $[s_i]$ and $[s_{i+1}]$ is less than $\epsilon $ for all $0\leq i< k$ . By induction, and since the Riemannian metric $d_g$ is complete, one can construct for all $1\leq i\leq k$ a point $x_i\in \widetilde {N}$ such that $[s_i]\in \mathbb {P}(u^\circ (x_i)\cap S^2V^{> 0})$ and $d_H(x_{i-1},x_i)\leq 1$ . Therefore the manifolds $\mathbb {P}(u^\circ (x)\cap S^2V^{> 0})$ cover all of $\mathbb {P}(S^2V^{>0})$ .

Now let us consider the fibered domain in projective space. We first prove that the fibers are contained in the domain $S^2\Omega $ . Consider a rank one line $[s]\in S^2\mathbb {P}(\xi ^n_\rho (x))$ for some $\zeta \in \partial \Gamma $ . Suppose that p belongs to $u^\circ (x)$ for some $x\in \widetilde {N}$ . There exists a flow line $(\Phi _t([q]))_{t\geq 0}$ starting at $\operatorname {\mathrm {pr}}([q])=x$ and converging to $\zeta \in \partial \Gamma \simeq \partial \widetilde {N}$ by Proposition 4.5. Theorem 5.4 implies that $\pi (q)\in \mathcal {Q}$ must be negative on $\xi ^2_\rho (\zeta )$ and hence p cannot belong to $u^\circ (x)$ .

Conversely let us prove that every point in $S^2\Omega $ is covered by some fiber. Fix a point $x\in \widetilde {N}$ and take any rank one point $[s]\in S^2\Omega $ in the Guichard-Wienhard domain of discontinuity. There exists a sequence $(x_n)$ such that $[s]$ belongs to the limit of $\mathbb {P}(u^\circ (x_n)\cap S^2V^{\geq 0})$ , since these manifolds cover $\mathbb {P}(S^2V^{>0})$ . We consider some $[q_n]\in \mathbb {S}\mathcal {E}_x$ such that $\Phi _t([q_n])\in \mathbb {S}\mathcal {E}_{x_n}$ for some $t_n>0$ , which exist by Proposition 4.5. If $t_n$ diverges when n varies, then the set $\mathbb {P}(u^\circ (x_n)\cap S^2\mathbb {P}(V))$ becomes arbitrarily close to $S^2\mathbb {P}(\xi ^n_\rho (\zeta _n))$ where $\zeta _n$ is the limit when t goes to $+\infty $ of $\Phi _t([q_n])$ . This would contradict the fact that $[s]\in S^2\Omega $ , as in this case $[s]\in \mathbb {P}(\xi _\rho ^n(\zeta ))$ where $\zeta $ is a limit point of $(\zeta _n)$ . Hence the sequence $(x_n)$ is bounded and therefore converges up to subsequence to some $x_\infty \in \widetilde {N}$ , and $[s]\in \mathbb {P}(u^\circ (x_\infty ))$ .