Proof. Indeed, if there were such a $\mathcal {L}$ , then, using the definition of $\phi ^u$ appearing before equation (5), we could define $\hat {\mathcal {L}}(\hat {x})$ on $\hat {\Sigma }$ by $ \hat {\mathcal {L}}(\hat {x}) = H^s_{\phi ^u(\hat {x}), \hat {x}} \mathcal {L}(x)$ so that

$$ \begin{align*} \hat{A}(\hat{x})_* \hat{\mathcal{L}}(\hat{x}) = H^s_{\phi^u(\hat{\sigma} \hat{x}), \hat{\sigma} \hat{x}} A(\hat{x})_* \mathcal{L}(x) = H^s_{\phi^u(\hat{\sigma} \hat{x}), \hat{\sigma} \hat{x}} \mathcal{L}(\sigma x) = \hat{\mathcal{L}}(\hat{\sigma} \hat{x}),\end{align*} $$

which yields invariance of $\hat {\mathcal {L}}$ . Moreover, since stable holonomies are continuous in the leaf and transverse direction, continuity of $\hat {\mathcal {L}}$ follows from continuity of $\mathcal {L}$ .

Proof of Proposition 3.3.

We take the continuous disintegration of m from Lemma 2.14. Suppose that $m_x(\mathcal {S})> 0$ for some $x \in \Sigma $ and some linear section $\mathcal {S}$ of $\mathrm {Gr}(k,d)$ . Let $d_0$ be the minimal dimension such that there exists a subspace S of $\Lambda ^k(\mathbb R^d)$ such that $m_x(\mathbb {P} S \cap \mathrm {Gr}(k,d))> 0 $ for some $x \in \Sigma $ . Let

(11) $$ \begin{align} \gamma_0=\sup_{x\in \Sigma, \mathbb{P}S: \dim S=d_0}m_x(\mathbb{P} S \cap \mathrm{Gr}(k,d)). \end{align} $$

We claim that by compactness $\gamma _0$ is realized. The space of all $\mathbb {P}S$ is compact and $m_x$ varies continuously in x in the weak* topology by Proposition 2.15. Hence, if we take any sequence $x_i, S_i$ such that

$$ \begin{align*} m_{x_i}(\mathbb{P}S_i\cap \mathrm{Gr}(k,d))\to \gamma_0, \end{align*} $$

we may pass to a subsequence so that we may assume that $x_i\to x'$ and $S_i\to S$ . Consider the set $B_{\epsilon }(S)$ ; then weak* convergence implies that for each $\epsilon $ as $B_{\epsilon }(S)$ is closed,

$$ \begin{align*}\begin{aligned} m_x(\mathbb{P}B_{\epsilon}(S)\cap \mathrm{Gr}(k,d))&\ge \limsup_{i \to \infty} m_{{x_i}}(\mathbb{P}B_{\epsilon}(S_i)\cap \mathrm{Gr}(k,d)) \\ &\ge \limsup_{i \to \infty} m_{x_i}(\mathbb{P}S_i\cap \mathrm{Gr}(k,d))=\gamma_0. \end{aligned}\end{align*} $$

Letting $\epsilon \to 0$ , we obtain that $m_{x'}(\mathbb {P}S\cap \mathrm {Gr}(k,d))=\gamma _0$ by continuity from above of the measures.

We now claim that for each $x \in \Sigma $ , there exists at least one subspace $S(x)$ of dimension $d_0$ such that $m_x(S(x) \cap \mathrm {Gr}(k,d)) = \gamma _0$ . The proof is along the same lines as in [Reference Avila and VianaAV07a, Lemma 5.2]. First, one shows that over every cylinder $[J]$ the supremum of $m_x(S\cap \mathrm {Gr}(k,d))$ is $\gamma _0$ . Then one can pick a sequence converging to any point $x'\in \Sigma $ as above. We are able to pass to this limit because the space of linear sections is compact.

We now show that the linear sections realizing the supremum form a continuous family. First, we show that there exists $\ell $ such that at every point $x\in \Sigma $ there exist exactly $\ell $ distinct linear $d_0$ -dimensional subspaces of $\Lambda ^k(\mathbb R^d)$ , $S_1(x),\ldots , S_{\ell }(x)$ such that $m_x(\mathbb {P} S_i(x) \cap \mathrm {Gr}(k,d))=\gamma _0$ . Take a point $x\in \Sigma $ that maximizes the number $\ell $ . Then by invariance of $m_x$ (that is, Proposition 2.16),

$$ \begin{align*}m_x = \sum_{z \in \sigma^{-k}x} \frac{1}{J\sigma^k(z)} (A^k(z))^{-1}_* m_z,\quad\text{with } \sum_{z \in \sigma^{-k}x} \frac{1}{J\sigma^k(z)} = 1,\end{align*} $$

it follows for each $z\in \sigma ^{-k}(x)$ that

$$ \begin{align*}m_{z}((A^k(z)^{-1})_*\mathbb{P} S_i(x) \cap \mathrm{Gr}(k,d)) = \gamma_0, \quad\text{for all }i = 1, \ldots, n.\end{align*} $$

Note that the preimages of x become dense in $\Sigma $ . Thus, for any point $q\in \Sigma $ , we may take a sequence of $z_n$ such that $z_n\to q$ and $m_{z_n}$ gives measure $\gamma _0$ to exactly $\ell $ subspaces $S_1(z_n),\ldots , S_\ell (z_n)$ . By the argument mentioned after equation (11), we may pass to a subsequence so that, for each i, $S_i(z_n)\to S_i'$ , where $S_i'$ are $d_0$ -dimensional subspaces of $\Lambda ^k(\mathbb R^d)$ . We claim that all the $S_i'$ are distinct. But this is clear: if two coincided, then the preceding argument implies that the limiting measure of some $S_i'$ would be at least $2\gamma _0$ , which is impossible as $\gamma _0$ is the maximum.

Now we show that these sections vary continuously as x varies. First observe that if $x_n\to x$ then for each n we must have that $S_i(x_n)$ is near some $S_j(x)$ ; otherwise, by taking a limit as before, we would obtain an additional section $S'$ of measure $\gamma _0$ for $m_x$ .

Finally, to see that the family $\{S_1(x),\ldots , S_\ell (x)\}_{x\in \Sigma }$ is invariant under the cocycle, note that this holds on the set of pre-images of z and that this set is dense. Thus, invariance holds by continuity. But we have now reached a contradiction to our assumption of irreducibility on $\mathrm {Gr}(k,d)$ , so we are done.

Proof. By the singular value decomposition, this reduces to the analogous claim for diagonal matrices, which is straightforward because every unit vector $Av$ uniformly transverse to $S_{d-k}$ has the length of its $U_k(A)$ component uniformly bounded below.

Proof. We begin with the following claim.

Now we return to the proof of Lemma 4.3. Because $\hat {q}$ is generic for $\hat {\mu }$ and by (1 $^\prime $ ) the set $G_l$ has positive measure there is a sequence of times $k_n \to \infty $ such that $\hat q^{k_n} \in G_l$ . Observe that for a given time $k_n$ , we can always take $\rho $ to be sufficiently large so that (1) holds, since $G_l \subseteq C_{2l}^u(\hat p)$ . We will now determine how large $k_n$ has to be for (2) and (3) to hold, and, thus increasing $\rho $ accordingly if necessary at the end, we can do so ensuring that (1) holds as well.

First, we will show that for $k_n$ sufficiently large, statements of (2) and (3) hold for $\hat y = \hat q^{k_n}$ and $\hat z = [\hat p, \hat q^{k_n} ]$ exactly. In order to do this, we use the hyperbolicity of $\hat q$ and Lemma 4.2. Then to conclude the proof we will later use continuity to show that (2) and (3) hold for any $\hat y \in \hat {\sigma }^{-k_n}(R_\rho (\hat q))$ .

Applying Lemma 4.2 with $\theta = \min \{\theta _0, \epsilon /2\}$ and $\gamma /2$ for the Lipschitz constant as in the statement of the lemma, we may obtain a constant K such that the conclusion of that lemma holds if $\sigma _k(A)>K\sigma _{k+1}(A)$ for a linear map $A\in \mathrm {SL}(d,\mathbb R)$ . Since $\hat q$ is generic for $\hat {\mu }$ and $\unicode{x3bb} _k(\hat {\mu })> \unicode{x3bb} _{k+1} (\hat {\mu })$ , for the K specified above if we choose $k_n$ large enough, then $\sigma _k(A^{k_n}(\hat {q}^{k_n}))>K\sigma _{k+1}(A^{k_n}(\hat {q}^{k_n}))$ . Hence, for any subspace U with $\angle (U,S_{d-k}(\hat {q}^{k_n}, k_n))> \theta _0$ , we have that

(12) $$ \begin{align} d_{\mathrm{Gr}}(A^{k_n}(\hat{q}^{k_n})(U),U_k(\hat{q}^{k_n}, k_n))<\epsilon/2 \end{align} $$

and $A^{k_n}$ is $\gamma /2$ -Lipschitz on the same set of U.

Since $\hat q^{k_n} \in G_l$ , by (2 $^\prime $ ) in Claim 4.4 we have $\angle (E^u( [ \hat p,\hat q^k_n]), E^s(\hat q^{k_n}))> 2 \theta _0$ . Thus, since $S_{d-k}(\hat {q}^{k_n},k_n)$ converges to $E^s(\hat {q}^{n_k})$ by (4 $^\prime $ ) in Claim 4.4, for $k_n$ sufficiently large we see that we may take $U = E^u( [ \hat p,\hat q^{k_n}])$ in (12) to obtain that $A^{k_n}(\hat q^{k_n})$ is $\gamma /2$ -Lipschitz on a $\theta _0$ -neighborhood of $E^u(\hat z)$ , verifying (3) for $\hat y = \hat q^{k_n}$ . Moreover, since $U_k(\hat {q}^{k_n}, k_n)$ converges to $E^u(\hat q)$ , (12) with $U = E^u(\hat z)$ gives

$$ \begin{align*} d_{\mathrm{Gr}}(A^{k_n}(\hat{q}^{k_n})(E^u(\hat z)),E^u(\hat q))<\epsilon/2 \end{align*} $$

for n sufficiently large. Thus, we have verified that (2) and (3) hold at exactly $\hat y = \hat q^{-k_n}$ as long as $k_n$ is sufficiently large, and we now let $k := k_n$ for any such large $k_n$ .

Finally, we verify that the statements of (2) and (3) hold for all $\hat y \in \sigma ^{-k}(R_\rho (\hat q))$ if $\rho $ is taken sufficiently large. The key observation is that $E^u(\hat z)$ , where $\hat z = [ \hat p,\hat y]$ , varies continuously on $\hat y$ . Since the cocycle for a fixed k also varies continuously, this implies that (2) and (3) are open conditions on $\hat y$ . Since (2) and (3) hold at $\hat q^{k}$ , increasing $\rho $ if necessary the same holds in the neighborhood $\sigma ^{-k}(R_\rho (\hat {q}))$ of $\hat q^{k}$ , with $\epsilon $ in equation (12) and $\gamma $ -Lipschitzness as required for (3).

Proof. Let $\hat {q}$ be a point such that $\hat {\mu }^s_{\hat q}$ -a.e. point in $W^s_{\mathrm {loc}}(\hat {q})$ is Oseledec regular and generic for $\hat {\mu }$ . We will need the following claim, which we will prove immediately after this proposition for the sake of clarity.

Claim 4.6. For any $\theta _0> 0$ sufficiently small, given any hyperplane section ${H\subseteq \mathrm {Gr}(k,d)}$ , there exists a subset $S \subseteq W^s_{\mathrm {loc}}(\hat q)$ of positive $\mu ^s_{\hat q}$ -measure such that for every $\hat z \in S$ ,

$$ \begin{align*} d_{\mathrm{Gr}(k,d)}(E^u(\hat z), H)> 2\theta_0. \end{align*} $$

Now let $\theta _0$ be any sufficiently small number such that Claim 4.6 and Lemma 4.3 hold. Observe that $\ell$ is independent of this small $\theta_0$ .

Fix a collection of hyperplane sections

$$ \begin{align*}\{H_1,\ldots,H_N\}\end{align*} $$

with the property that, given any other hyperplane section H, there exists some $1 \leq i \leq N$ such that the Hausdorff distance in $\mathrm {Gr}(k,d)$ between H and $H_i$ is less than $\theta _0/2$ . This is possible since the space of hyperplane sections is compact with the Hausdorff distance.

For each $1 \leq i \leq N$ , we apply Claim 4.6 with $H = H_i$ to obtain a positive $\mu ^s_{\hat q}$ -measure set $S_i \subseteq W^s_{\mathrm {loc}}(\hat q)$ of points $\hat z$ with

$$ \begin{align*}d_{\mathrm{Gr}(k,d)}(E^u(\hat z), H_i)> 2\theta_0.\end{align*} $$

For each $1 \leq i \leq N$ , we let $\hat q_i$ be any such point in $S_i$ which is, moreover, generic for and Oseledec regular for $\hat {\mu }$ . Let $0<\epsilon <\theta _0$ and apply Lemma 4.3 with this choice of $\epsilon ,\theta _0$ for each $\hat q_i$ to obtain a sequence of rectangles $R_{\rho _i}(\hat {q}_i)$ satisfying the conclusions of that lemma. Let $\rho $ be greater than the maximum of the $\rho _i$ and chosen large enough that $R_\rho (\hat q_i) \cap R_\rho (\hat q_j) = \varnothing $ , for $i \neq j$ , verifying (a). It is evident since the rectangles $R_\rho (\hat q_i)$ satisfy the conclusions of Lemma 4.3 that (b) and (d) also hold for the constructed rectangles. It remains to verify (c).

Let H be any hyperplane section of $\mathrm {Gr}(k,d)$ . By construction, there is some $1 \leq i \leq N$ such that the Hausdorff distance in $\mathrm {Gr}(k,d)$ between H and $H_i$ is less than $\theta _0/2$ . This, together with the fact that $\angle (E^u(\hat {q}_i),H_i)>2\theta _0$ , and

$$ \begin{align*}d_{\mathrm{Gr}}(A^{k_i}(E^u(\hat{z})), (E^u(\hat{q_i})))<\theta_0,\end{align*} $$

by (2) in Lemma 4.3 then gives (c) by the triangle inequality.

Proof of Claim 4.6.

Let $\hat {m}$ be a u-state over $\hat {\mu }$ on $\hat {\Sigma } \times \mathrm {Gr}(k,d)$ . For a hyperplane section H, let $H_\theta $ be its $\theta $ -closed neighborhood in $\mathrm {Gr}(k,d)$ . Since, by Proposition 3.3, $m_x(H) = 0$ for any hyperplane section H, by upper semi-continuity for each $1 \leq i \leq N$ there is some $\theta _i> 0$ sufficiently small to ensure that $m_{q}(H_{\theta _i}) < 1/2$ . By taking $\theta _0 = \min _{1 \leq i\leq N} \theta _i/3$ , we see that

(14) $$ \begin{align} \sup_{1 \leq i \leq N} m_q(H^i_{3\theta_0}) < \frac{1}{2}. \end{align} $$

Choose a set $\{H^i\}_{1 \leq i \leq N}$ of hyperplane sections of $\mathrm {Gr}(k,d)$ with the property that any other hyperplane section $H \subseteq H^i_{\theta _0/2}$ for some $1 \leq i \leq N$ . This is possible since, in the Hausdorff topology, the space of hyperplane sections is compact and the set of hyperplane sections contained in a neighborhood of a fixed hyperplane section contains an open set.

By the disintegration formula,

(15) $$ \begin{align} m_q &= \int_{W^s_{\mathrm{loc}}(\hat q)} \hat{m}_{\hat{x}}\, d\hat{\mu}^s_{\hat q}(\hat{x})\notag \nonumber\\ &= \int_{W^s_{\mathrm{loc}}(\hat q)} \delta_{E^u(\hat{x})}\, d\hat{\mu}^s_{\hat q}(\hat{x}),\end{align} $$

recalling that, by Proposition 3.5 for $\hat {\mu }^s_{\hat p}$ -a.e. $\hat {x}$ in $W^s_{\mathrm {loc}}(\hat q)$ , we have $\hat {m}_{\hat {x}} = \delta _{E^u(\hat {x})}$ .

Now, given an arbitrary hyperplane section H, by construction there is some ${1 \leq i \leq N}$ such that $H \subseteq H^i_{\theta _0/2}$ . Then by equation (14) and the disintegration formula (15), we see that set S of $\hat {x} \in W^s_{\mathrm {loc}}(\hat q)$ , with $E^u(\hat {x}) \notin H^i_{3\theta _0}$ , must have $\mu ^s_{\hat q}$ -measure at least $1/2$ . Finally, since $H \subseteq H^i_{\theta _0/2}$ , we conclude that for $\hat {x} \in S$ , we have $d_{\mathrm {Gr}}(E^u(\hat {x}), H)> 2\theta _0$ , as desired.

Proof. Clearly from local product structure it suffices to verify that $\hat {\nu }^s_{\hat {x}}(G\cap W^s_{\mathrm {loc}}(\hat {x}))=1$ for $\hat {\eta }$ -a.e. $\hat {x}$ . We will embed $W^s_{\mathrm {loc}}$ into Euclidean space and then make use of the following claim.

Claim 4.11. For each $m\in {\mathbb N}$ , there exists $C_m>0$ such that if $\mathbb {P}$ is a probability measure in $\mathbb R^m$ with compact support S and $G\subseteq S$ is a set such that for $\mathbb {P}$ -a.e. $x\in S$ there exist arbitrarily small open balls $O(x)$ such that

$$ \begin{align*} \frac{\mathbb{P}(G\cap O(x))}{\mathbb{P}(O(x))}>\alpha>0, \end{align*} $$

then $\mu (G)\ge \alpha /C_m$ . In fact, if each $G\cap O(x)$ contains a closed subset K that is clopen in S such that

$$ \begin{align*} \frac{\mathbb{P}(G\cap K)}{\mathbb{P}(O(x))}>\alpha>0, \end{align*} $$

then G contains a closed subset of measure at least $\alpha /C_m$ that is clopen in S.

The set $W^s_{\mathrm {loc}}(\hat {x})$ is a Cantor set which we may embed Hölder continuously in $\{0,1\}^{{\mathbb N}}$ by coding the symbols used to define $\Sigma $ . The Cantor set $\{0,1\}^{{\mathbb N}}$ has a Hölder continuous embedding in $\mathbb R$ as the middle thirds Cantor set by sending each string to the appropriate real number with no $1$ s in its ternary expansion. In fact, the embedding into $\mathbb R$ carries balls in $\{0,1\}^{{\mathbb N}}$ into balls in $\mathbb R$ because the embedding is increasing in the obvious way. Let $\phi $ denote this particular embedding $\phi \colon W^s_{\mathrm {loc}}(\hat {x})\to \mathbb R$ .

Let $\mathbb {P} = \phi _* \hat \nu ^s_{\hat {x}}$ denote the pushforward of the measure $\hat {\nu }^s_{\hat {x}}$ on Euclidean space under the embedding $\phi $ and $G_\phi = \phi (G) \subseteq \Omega =\phi (W^s_{\mathrm {loc}}(\hat {x}))$ denote the pushforward of the set of good points. This set is still Borel because it is a Borel subset of a closed subspace. Moreover, as the embedding carries balls in the Cantor set into intervals in the real line, we may apply Claim 4.11, which requires a covering by balls. Thus, we see that there is a closed subset $K_1\subset G$ with $\mathbb {P}(K_1)\ge \alpha /C_1$ that is clopen in $\Omega $ . As $K_1$ is clopen in $\Omega $ , $\Omega _2=\Omega \setminus K_1$ is a closed set and a positive distance away from $K_1$ . Thus, $\Omega _2$ also has a covering by balls as in Claim 4.11. And, importantly, because $\Omega _2$ is a positive distance from $K_1$ , and the balls in Claim 4.11 are arbitrarily small, from Claim 4.11 we obtain another clopen subset $K_2\subseteq \Omega _2$ disjoint from $K_1$ with measure

$$ \begin{align*} \mathbb{P}(K_2)>(\alpha/C_1)\mathbb{P}(\Omega_2)\ge (\alpha/C_1)(1-\alpha/C_1). \end{align*} $$

We may then consider the set $\Omega _2\setminus K_2$ and continue inductively. The $\mathbb {P}$ -measure of the complement of $\sqcup _{i=1}^n K_i$ goes to zero exponentially quickly and the claim follows.

Proof of Proposition 4.9.

Let $X_{l}$ denote the set of l-good points, and let $R_1,\ldots , R_N$ be the rectangles guaranteed by Proposition 4.5 applied for the fixed $\theta _0>0$ and our choice of $l\in {\mathbb N}$ . We proceed by checking the criterion in Lemma 4.10 for $G=X_l$ . The point of the proof is that if a point $\hat {x}$ visits $R_i$ at some point $\hat {x}^n$ in its past, then there is some j such that the points in $W^s_{\mathrm {loc}}(\hat {x}^n)$ in $R_j$ are $\ell $ -good. This is because $A^n(\hat {x}^n)$ is constant on $W^s_{\mathrm {loc}}(\hat {x}^n)$ and this specific linear map is always l-good for the points in some $R_j$ .

Let $R=\sqcup _{1\le i\le N} R_i$ . The following statement is immediate from the continuous product structure. There exists $K>1$ such that for any $\hat {x}\in R$ and $1\le i\le N$ ,

(16) $$ \begin{align} K^{-1}<\hat{\eta}^s_{\hat{x}}(R_i\cap W^s_{\mathrm{loc}}(\hat{x}))<K. \end{align} $$

For $\hat {\eta }$ -a.e. $\hat {x}\in \hat {\Sigma }$ , there exist infinitely many $n\ge l$ such that $\hat {\sigma }^{-n}(\hat {x})\in R$ . Consider now the matrix $A^n(\sigma ^{-n}(\hat {x}))$ . Note that as the cocycle is constant on local stable leaves, for any $\hat {x}'\in W^s_{\mathrm {loc}}(\hat {\sigma }^{-n}(\hat {x}))$ we have that $A^n(\hat {x}')=A^n(\hat {\sigma }^{-n}(\hat {x}))$ . By Proposition 4.5, there exists some $i\in \{1,\ldots , N\}$ such that if we take $H=S_{d-d^u}(A^n(\hat {\sigma }^{-n}(\hat {x})))$ , and any $\hat {y}$ in $R_i\cap W^s_{\mathrm {loc}}(\hat {x}')$ , then (13) holds for that $\hat {y}$ . This means that $\hat {\sigma }^n(\hat {y})$ is $\ell $ -good at time n.

In particular, we see that all points in $\hat {\sigma }^n(R_i\cap W^s_{\mathrm {loc}}(\hat {\sigma }^{-n}(\hat {x})))$ are in $X_l$ . Call this set $G_{n,\hat {x}}$ . In order to conclude, it suffices to show that there exists some $D>0$ independent of $\hat {x}$ such that

$$ \begin{align*} \frac{\hat{\eta}^s_{\hat{x}}(G_{n,\hat{x}})}{\hat{\eta}^s_{\hat{x}}(\hat{\sigma}^n(W^s_{\mathrm{loc}}(\hat{\sigma}^{-n}(\hat{x})))}>D>0. \end{align*} $$

But by equation (16), this is immediate from Lemma 2.6. We are now done because we verified the criterion in Lemma 4.10.

Proof. Let $A \in \mathrm {GL}(d,\mathbb R)$ . Since the metric on the Grassmannian is invariant by isometries of $\mathbb R^d$ , by the singular value decomposition theorem we may assume that A is of the form $A = \text {diag}(a_1, \ldots , a_d)$ , with $a_1 \geq \cdots \geq a_d> 0$ . Without loss of generality we may also assume that H is the hyperplane section defined by $e_{k+1} \wedge \cdots \wedge e_d$ .

Thus, we can describe elements of $\{\xi \in \mathrm {Gr}(k,d): d_{\mathrm {Gr}(k,d)}(\xi , H)> \theta _0\}$ as graphs with bounded slopes over the first k coordinates, that is, we may embed the preceding set into the set $\text {End}_M(\mathbb R^k, \mathbb R^{d-k})$ of linear maps from the first k coordinates to the last $d-k$ coordinates of $\mathbb R^d$ with a norm bounded by $M> 0$ , where M only depends on $\theta _0$ . Moreover, it is not hard to see that in this set $d_{\mathrm {Gr}(k,d)}$ is equivalent to the distance induced by the operator norm on $\text {End}_M(\mathbb R^k, \mathbb R^{d-k})$ . Since all matrix norms are equivalent we endow $\text {End}_M(\mathbb R^k, \mathbb R^{d-k})$ with the max norm, that is, for $\phi \in \text {End}_M(\mathbb R^k, \mathbb R^{d-k})$ given in a basis by $\phi _{ij}$ , we let $\|\phi \| = \max _{i,j} |\phi _{ij}|$ .

Let $\phi _1, \phi _2 \in \text {End}_M(\mathbb R^k, \mathbb R^{d-k})$ correspond to $\xi _1, \xi _2 \in \mathrm {Gr}(k,d)$ as in the hypotheses of the theorem and $A\phi _1, A\phi _2 \in \text {End}_M(\mathbb R^k, \mathbb R^{d-k})$ correspond to $A\xi _1, A\xi _2 \in \mathrm {Gr}(k,d)$ . To conclude, it suffices to find a uniform bound for $|(A\phi _1-A\phi _2)_{ij}|$ in terms of $|(\phi _1-\phi _2)_{ij}|$ . It is clear that for any $\phi \in \text {End}_M(\mathbb R^k, \mathbb R^{d-k})$ , we have $|(A\phi )_{ij}| \leq (a_{k+1}/a_k) \dot |\phi _{ij}|$ , since the action of A expands the first k coordinates by at least $a_k$ and the last $d-k$ by at most $a_{k+1}$ . Since $a_{k+1}/a_k \leq 1$ the proof is complete.

Proof of Proposition 4.1.

Since the countable intersection of full measure sets has full measure, by Proposition 4.9, we see that $\hat \eta $ -a.e. $\hat {x}$ is l-good for every l. Recall that for $\hat \eta $ -a.e. $\hat {x}$ by Proposition 2.13,

$$ \begin{align*}\hat{m}^{\eta}_{\hat{x}} = \lim_{n \to \infty} A^n(\hat{x}^n)_* m^{\eta}_{ P(\hat{x}^n)}.\end{align*} $$

Hence, there is a full $\hat \eta $ -measure set $S \subseteq \hat {\Sigma }$ such that the above holds and every point in S is l-good for every $l> 0$ . Fix any such $\hat {x}$ in S. For all $l> 0$ fix $n_l> l$ to be a time guaranteed by the definition of l-good for $\hat {x}$ . We say that a probability measure m on a metric space is $\varepsilon $ -concentrated at a point z if $m(B_{\epsilon }(z))\ge 1-\varepsilon $ , where $B_{\epsilon }(z)$ is the open ball of radius $\epsilon $ about z.

Claim 4.14. Let $\varepsilon> 0$ . Then for all $l> 0$ sufficiently large there is an $i \in \{1, \ldots ,N\}$ such that for $k_i = k_i(l)$ as in Proposition 4.5, the measure

$$ \begin{align*}A^{n_l+k_i+l}(\hat{x}^{n_l+k_i+l})_* m^{\eta}_{ P(\hat{x}^{n_l+k_i+l})}\end{align*} $$

is $\varepsilon $ -concentrated.

We will now show how to conclude given the claim. Since $n_l> l$ , by passing to a subsequence of $n_l$ we may assume that $n_{l+1}> n_l+k_i(l)+l$ , where $k_i(l)$ is as in the claim, and that the point around which the masses $\varepsilon $ -concentrate converges to some $\xi ^u(\hat {x}) \in \mathrm {Gr}(k, d)$ , by compactness of the Grassmannian, from which we conclude that

$$ \begin{align*}\lim_{l \to \infty} A^{n_l+k_i+l}(\hat{x}^{n_l+k_i+l})_* m^{\eta}_{ P(\hat{x}^{n_l+k_i+l})} = \delta_{\xi^u(\hat{x})},\end{align*} $$

and the proof is complete. Hence, we are done once we prove the claim about $\varepsilon $ -concentration of masses.

Proof of Claim 4.14.

We begin my fixing some constants. Fix some small $\varepsilon>0$ . From Proposition 4.5, there exists a fixed $\theta _0>0$ such that the conclusions of that proposition hold. For each l, we let $R_i^{l}$ denote the collection of rectangles obtained from Proposition 4.5 applied with that fixed choice of $\theta _0>0$ . By definition of $n_l$ , $\hat {x}^{n_l} \in R_i^l$ for some $1 \leq i \leq N$ and a rectangle obtained from Proposition 4.5. Hence, $\hat {x}^{n_l + k_i} \in C^u_l(\hat p)$ , where $k_i$ is also given by Proposition 4.5.

By Lemma 4.12, given this choice of $\varepsilon>0$ and $\theta _0>0$ , there exists some $\delta _1>0$ such that if m is a measure on $\mathrm {Gr}(k,d)$ , $A\in \mathrm {GL}(d,\mathbb R)$ , and m is $\delta _1$ concentrated at a point $V\in \mathrm {Gr}(k,d)$ that is $\theta _0$ away from H, a geometric hyperplane defined by a subspace containing the most contracting subspace $S_q(A)$ of A as defined in Lemma 4.12, then $A_*m$ is $\varepsilon $ -concentrated. Note that this $\delta _1$ is not literally the $\delta $ from that lemma but may be smaller to ensure that the conclusion of the lemma applies to the set where m is concentrated.

Let $\alpha =\min \{\theta _0/2,\delta _1\}>0$ .

We will show $\varepsilon $ -concentration by breaking down

$$ \begin{align*} A^{n_l+k_i+l}(\hat{x}^{n_l+k_i+l})_* m^{\eta}_{ P(\hat{x}^{ n_l+k_i+l} )} \end{align*} $$

into

$$ \begin{align*} A^{n_l}(\hat{x}^{n_l})_*A^{k_i}(\hat{x}^{n_l+k_i})_*A^{l}(\hat{x}^{n_l+k_i+l})_* m^{\eta}_{ P(\hat{x}^{n_l+k_i+l} )}, \end{align*} $$

and pushing the measure through each of these terms in order. We give each of these terms names:

(17) $$ \begin{align} m_1&=A^l(\hat{x}^{n_l+k_i+l})_*m^{\eta}_{ P(\hat{x}^{n_l+k_i+l} )}, \end{align} $$

(18) $$ \begin{align} m_2&=A^{k_i}(\hat{x}^{n_l+k_i})_*m_1,\end{align} $$

(19) $$ \begin{align} m_3&=A^{n_l}(\hat{x}^{n_l})_*m_2. \end{align} $$

The proof then breaks down into a sequence of three claims, which follow from our choice of constants at the beginning of the proof.

Subclaim 1. The measure $m_1$ is $\alpha $ -concentrated around $E^u(\hat z)$ , where $\hat z=W^s_{\mathrm {loc}}(\hat {x}^{n_l+k_i})\cap W^u_{\mathrm {loc}}(\hat p)$ .

Proof. By definition $\hat {z}\in W^s_{\mathrm {loc}}(\hat {x}^{n_l+k_i})$ . But note that as $n_l$ is an l-good time, $\hat {x}^{n_l+k_i}\in C^u_l(\hat p)$ . This implies that $\hat {x}^{n_l+k_i+l}$ and $\hat {z}^l$ lie in the same local stable leaf. Thus, as the cocycle is constant along local stable leaves,

(20) $$ \begin{align} A^{l}(\hat{x}^{n_l+k_i+l})_* m^{\eta}_{ P(\hat{x}^{n_l+k_i+l} )} = A^{l}(\hat z^l)_* m^{\eta}_{ P( \hat z^l )}, \end{align} $$

hence it suffices to show that the right-hand side of (20) is $\alpha $ -concentrated around $E^u(\hat {z})$ . To see this, note that that

$$ \begin{align*} A^{l}(\hat z^l)_* m^{\eta}_{ P(\hat z^l )} = (A^{-l}(\hat z)^{-1} A^{-l}(\hat p))_* A^{l}(\hat p)_* m^{\eta}_{ P(\hat z^l )}. \end{align*} $$

The expression in parentheses converges to $H^u_{\hat p \hat z}$ as $l \to \infty $ . Moreover, $A^{l}(\hat p)_* m^{\eta }_{ P(\hat z^{ l})}$ converges to $\delta _{E^u(\hat p)}$ since $m^{\eta }_{ P(\hat z^{l})}$ converges weak* to $m^{\eta }_{\hat p}$ and the latter gives 0 measure to hyperplane sections. Hence, by Lemma 2.10, we see that the right-hand side of (20) converges to $\delta _{H^u_{\hat p\hat z}E^u(\hat p)}$ .

Therefore, for l sufficiently large it follows that

$$ \begin{align*} m_1 := A^{l}(\hat z^l)_* m^{\eta}_{ P( \hat z^l)} \end{align*} $$

is $\alpha $ -concentrated around $H^u_{\hat p \hat z} E^u(\hat p) = E^u(\hat z)$ (Lemma 2.3), as desired.

Subclaim 2. The measure $m_2$ is $\delta _1$ -concentrated at a point at least $\theta _0$ away from $H=H_{\hat {x}^{n_l}}$ , where H is defined as in the definition of $\ell $ -good.

Proof. From Subclaim 1, $m_1$ is $\alpha $ -concentrated at $E^u(\hat z)$ . The application of Proposition 4.4 gives that

$$ \begin{align*} d_{\mathrm{Gr}}(A^{k_i}(\hat{x}^{n_l+k_i})E^u(\hat z),H_{\hat{x}^{n_l}})>\theta_0, \end{align*} $$

and that $A^{k_i}(\hat {x}^{n_l+k_i})$ is $1$ -Lipschitz in a $\theta _0$ -neighborhood of $E^u(\hat z)$ . Thus, as $\alpha <\theta _0$ , by the $1$ -Lipschitzness $m_2$ is $\alpha $ -concentrated at a point $\theta _0$ away from H. As $\alpha \le \delta _1$ we are done.

Subclaim 3. The measure $m_3$ is $\varepsilon $ -concentrated.

Proof. From the conclusion of Subclaim 2, H is a geometric hyperplane defined by a subspace containing the appropriate most contracting subspace of $A^{n_l}(\hat {x}^{n_l})$ specified by Lemma 4.12. Thus, by the choice of $\delta_1$ in the second paragraph of the proof of Claim 4.14, as $m_2$ is $\delta _1$ -concentrated $\theta _0$ away from H, it follows that $m_3$ is $\varepsilon $ -concentrated.

We are now done with the proof of Claim 4.14 as Subclaim 3 is the conclusion needed.

Having finished the proof of Claim 4.14, we obtain the proof of Proposition 4.1 by the discussion after the statement of that claim.