Proof. See [Reference Husemöller27, Ch. 6].

Proof. For $0\leq k\leq \infty $ , $\operatorname {\mathrm {Diff}}^k(M)$ is an infinite-dimensional separable Fréchet space [Reference Gorbatsevich, Onishchik and Vinberg17, §I.4.3]. Since all infinite-dimensional separable Fréchet spaces are homeomorphic to the Hilbert space $\ell ^2$ [Reference Anderson and Bing1], we see that $\operatorname {\mathrm {Diff}}^k(M)$ is homeomorphic to the Hilbert space $\ell ^2$ . Since $\ell ^2$ has the homotopy type of a CW-complex [Reference Smrekar and Yamashita40], we get that $\operatorname {\mathrm {Diff}}^k(M)$ is homotopy equivalent to a CW-complex [Reference Hatcher22]. Thus, by Whitehead’s theorem, to show that the inclusion $\iota :\operatorname {\mathrm {Diff}}^\infty (M)\hookrightarrow \operatorname {\mathrm {Diff}}^1(M)$ is a homotopy equivalence, it is sufficient to show that the induced map on homotopy groups $\iota _*:\pi _*(\operatorname {\mathrm {Diff}}^\infty (M))\to \pi _*(\operatorname {\mathrm {Diff}}^1(M))$ is an isomorphism.

To do this, we first recall that for any map in $\varphi \in C^1(M,M)$ , we can find a smooth map that is arbitrarily close to and homotopic to $\varphi $ and that the choice of this map depends continuously on our original map [Reference Lee28, Theorems 6.21 and 6.28]. This gives us a continuous map $\Phi :\operatorname {\mathrm {Diff}}^1(M)\to C^\infty (M,M)$ such that for any $\varphi \in \operatorname {\mathrm {Diff}}^1(M)$ , $\Phi (\varphi )\simeq \varphi $ and $d(\varphi ,\Phi (\varphi ))<\varepsilon $ . Since $\operatorname {\mathrm {Diff}}^\infty (M)\subset C^\infty (M,M)$ is open (by the inverse function theorem), by choosing $\varepsilon $ small enough, we get that $\Phi (\operatorname {\mathrm {Diff}}^1(M))\subset \operatorname {\mathrm {Diff}}^\infty (M)$ , so we can write $\Phi :\operatorname {\mathrm {Diff}}^1(M)\to \operatorname {\mathrm {Diff}}^\infty (M)$ .

Now, we show that the map induced by $\iota :\operatorname {\mathrm {Diff}}^\infty (M)\to \operatorname {\mathrm {Diff}}^1(M)$ on homotopy groups is an isomorphism. The map $\iota _*$ is surjective because any map $\phi :S^n\to \operatorname {\mathrm {Diff}}^1(M)$ is homotopic to the map $\Phi \circ \phi :S^n\to \operatorname {\mathrm {Diff}}^\infty (M)$ . To see that $\iota _*$ is injective, we take a map $\phi :S^n\to \operatorname {\mathrm {Diff}}^1(M)$ that is null-homotopic. Let $h_t:S^n\to \operatorname {\mathrm {Diff}}^1(M)$ be a null-homotopy for $\phi $ . Then the map $\Phi \circ \phi :S^n\to \operatorname {\mathrm {Diff}}^\infty (M)$ is homotopic to $\phi $ in $\operatorname {\mathrm {Diff}}^1(M)$ , and is null-homotopic in $\pi _n(M)$ via the homotopy $\Phi \circ h_t:S^n\to \operatorname {\mathrm {Diff}}^\infty (M)$ .

Proof. First, we note that M and $h^*M$ are isomorphic bundles. This is because, by assumption, M and $\tilde M$ are isomorphic, and since $h\sim {\text {id}}$ , the bundle $h^* \tilde M$ is isomorphic to the bundle ${\text {id}}^*\tilde M=\tilde M$ . Let $\phi :\tilde M\to h^*\tilde M$ be a bundle isomorphism, that is, the diagram in Figure 1a commutes. From the definition of $h^*\tilde M$ , we have that the commutative diagram in Figure 1b commutes. (Note that $\operatorname {\mathrm {proj}}_2:h^* \tilde M \to \tilde M$ is the projection onto the second coordinate from the definition of the pullback bundle $h^* \tilde M = \{ (b,x)\in B\times M: h(b)=\tilde \pi (x) \}$ .) Combining these diagrams gives us the commutative diagram from Figure 1c. So the continuous map

(3.1) $$ \begin{align} \tilde h:=\operatorname{\mathrm{proj}}_2\circ \phi:M\to \tilde M \end{align} $$

is a lift of h. Since $\tilde h$ is a continuous injection, by invariance of domain, $\tilde h:M\to \tilde M$ is a homeomorphism.

Proof. We begin by constructing principal H-bundles with the same transition data as $E_0$ and $E_1$ using Lemma 2.1. We will call these $q_0:Q_0\to B$ and $q_1:Q_1\to B$ . Let $f_0:B\to BH$ and $f_1:B\to BH$ be classifying maps for $Q_0$ and $Q_1$ . Note that since $E_0$ and $E_1$ (and therefore $Q_0$ and $Q_1$ ) are $C^k$ bundles, the maps $f_0$ and $f_1$ are $C^k$ .

Since $E_0$ and $E_1$ are isomorphic as continuous bundles, we get that there is a homotopy $f:B\times [0,1]\to BH$ from $f_0$ to $f_1$ . Since $f|_{B\times \{0,1\}}$ is $C^k$ , then by the Whitney approximation theorem (see [Reference Lee28, Theorem 6.26]), $f_t$ is homotopic to a $C^k$ map ${\overline {f}:B\times [0,1]\to BH}$ relative to $B\times \{0,1\}$ . So, we have a $C^k$ homotopy $\overline {f}:B\times [0,1]\to BH$ from $\overline {f_0}=f_0$ to $\overline {f_1}=f_1$ . If the classifying maps of two $C^k$ principal bundles are homotopic via a $C^k$ homotopy, then the bundles are isomorphic as $C^k$  bundles [Reference Husemöller27, Ch. 4.9]. (The argument in [Reference Husemöller27, Ch. 4.9] is only given for continuous bundles, but works for $C^k$ bundles.) Thus, we get that the pullback bundles $f_0^*EH$ and $f_1^*EH$ are isomorphic as $C^k$ principal H-bundles over B. Since $f_0$ and $f_1$ are the classifying maps for $Q_0$ and $Q_1$ , respectively, this means that $Q_0$ and $Q_1$ are isomorphic as $C^k$ principal H-bundles over B. Since $Q_0$ and $Q_1$ have the same transition functions as $E_0$ and $E_1$ , we get that $E_0$ and $E_1$ are isomorphic as $C^k$ bundles with structure group H.

From the definition of a $C^k$ isomorphism of F-bundles with structure group H, we see that this means that there is a $C^k$ isomorphism $\alpha :E_0\to E_1$ , such that, if ${(U_{0,i}, \phi _{0,i})}$ and $\{(U_{1,j},\phi _{1,j})\}$ are trivializing atlases for $E_0$ and $E_1$ , respectively, then there exists functions $d_{ij}:U_{0,i}\cap V_{1,j}\to H$ such that for $x\in U_{0,i}\cap U_{1,j}$ and $y\in F$ , we get that $\phi _{1,j}\circ \alpha \circ \phi _{0,i}^{-1}(x,y)=(x,d_{ij}(x)\cdot y)$ . Since H acts on F by isometries, we get that $\alpha $ is an isometry on fibers.

Proof. Since A lifts to a homeomorphism $\widehat {A}: M\to M$ , we can construct a (continuous) isomorphism $\theta : M\to A^*M$ of M and $A^*M$ as F-bundles with structure group H given by $\theta (z)=(\pi (z), \widehat {A}(z) )$ (see Figure 2a). By Lemma 4.1, there is a $C^\infty $ isomorphism ${\alpha :M\to A^*M}$ that is an isometry on fibers. We can then use $\alpha :M\to A^*M$ to define a $C^\infty $ diffeomorphism $g:M\to M$ by $g(z)=\operatorname {\mathrm {proj}}_2\circ \alpha (z)$ (see Figure 2b). Since $\alpha $ is an isometry on fibers, we get that g is an isometry on fibers.

Proof. We begin by constructing a Riemannian metric on M, with respect to which g is partially hyperbolic. This construction has three ingredients.

• • A smooth family $\langle \cdot ,\cdot \rangle ^F_x$ of Riemannian metrics on the fibers $\pi ^{-1}(x)$ such that ${Dg:T\pi ^{-1}(x)\to T\pi ^{-1}(f(x))}$ is an isometry for all $x\in B$ . (Such a family exists because g is an isometry on fibers of $\pi :M\to B$ .)

• • A Riemannian metric $\langle \cdot , \cdot \rangle ^B$ on B that is adapted to the Anosov diffeomorphism ${f:B\to B}$ .

• • An Ehresmann connection H on M, that is, H is a smooth subbundle of the tangent bundle $TM$ such that for all $p\in M$ , $T_pM=H_p\oplus \ker (D_p\pi )$ . Note that from the definition of an Ehresmann connection, we know that $D_p\pi |_{H_p}:H_p\subset T_pM \to T_{\pi (p)}B$ is an isomorphism and the map $p\mapsto H_p$ is smooth.

We define a Riemannian metric $\langle \cdot , \cdot \rangle $ on M by letting, for all $p\in M$ :

• • $\langle v,v'\rangle =\langle D_p\pi (v), D_p\pi (v')\rangle ^B$ for $v,v'\in H_p$ ;

• • $\langle v,v'\rangle = \langle v,v'\rangle _{\pi (p)}^F$ for $v,v'\in \ker (D_p\pi )$ ; and

• • $H_p\perp \ker (D_p \pi )$ .

Now, we need to show that g is partially hyperbolic with respect to the metric $\langle \cdot , \cdot \rangle $ . To do this, we construct a dominated splitting $TM=E^s\oplus E^c\oplus E^u$ such that g is uniformly contracting on $E^s$ and uniformly expanding on $E^u$ . We begin by letting $E^c=\ker (D\pi )$ .

Next, we construct the unstable bundle $E^u$ using a graph transform argument. We begin by lifting the unstable bundle $E^u_f\subset TB$ for the Anosov diffeomorphism $f:B\to B$ to the bundle $\hat E^u\subset TM$ given by $\hat E^u_p:= H_p\cap D_p\pi ^{-1}(E^u_f(\pi (p)))$ for $p\in M$ . Note that $D_p\pi ^{-1}(E^u_p(\pi (p)))=\hat E^u_p \oplus \ker (D_p\pi )$ , and that $Dg$ preserves $\hat E^u\oplus E^c$ because $Df$ preserves $E^u_f$ and g covers f (so $D_{\pi (p)}f\circ D_p\pi =D_{g(p)}\pi \circ D_pg$ ).

Let

$$ \begin{align*} \Sigma&=\{\sigma:\hat E^u\to E^c: \ \sigma \text{ is fiber preserving over id},\\&\hspace{7pt}\quad \text{and } \sigma_p:\hat E^u(p)\to E^c(p) \text{ is linear } \text{for all } p\in M \}. \end{align*} $$

We put the norm $\|\cdot \|_\Sigma $ on $\Sigma $ given by

$$ \begin{align*} \|\sigma\|_\Sigma=\sup_{p\in M} \|\sigma_p\|, \end{align*} $$

where $\|\sigma _p\|$ is the operator norm. Note that this norm makes $\Sigma $ a Banach space.

Now, we want to define a map $\Gamma :\Sigma \to \Sigma $ , called the linear graph transform covering g, so that $D_pg(\operatorname {\mathrm {graph}}(\sigma _p))=\operatorname {\mathrm {graph}}(\Gamma (\sigma _p))$ . We now give $\Gamma :\Sigma \to \Sigma $ explicitly. Since by assumption, $Dg$ preserves $E^c=\ker (D\pi )$ (and $Dg$ preserves $\hat E^u\oplus E^c$ ), we can write for each $p\in M$ ,

$$ \begin{align*} D_pg=\begin{pmatrix} A_p & 0\\ C_p & K_p \end{pmatrix}:\hat E^u(p)\oplus E^c(p) \to \hat E^u(g(p))\oplus E^c(g(p)), \end{align*} $$

where

$$ \begin{align*} A_p:\hat E^u(p)\to \hat E^u(g(p)), \quad C_p:\hat E^u(p)\to E^c(g(p)), \quad K_p:E^c(p)\to E^c(g(p)) \end{align*} $$

are all linear. Also note that since $D_pg$ is invertible, both $A_p$ and $K_p$ are invertible.

Note that we can write a point in the graph of $\sigma _p$ as $(v,\sigma _pv)\in \operatorname {\mathrm {graph}}(\sigma _p)\subset \hat E^u(p)\oplus E^c(p)$ . Applying $D_pg$ to this point gives us

$$ \begin{align*} D_pg(v,\sigma_p v) = \begin{pmatrix} A_p&0\\C_p&K_p \end{pmatrix} \begin{pmatrix} v \\ \sigma_pv \end{pmatrix} = \begin{pmatrix} A_pv\\ C_p v +K_p\sigma_p v \end{pmatrix}. \end{align*} $$

So, we can write

$$ \begin{align*} D_pg(\operatorname{\mathrm{graph}}(\sigma_p))&= \left\{ \begin{pmatrix} A_pv\\ C_p v +K_p\sigma_p v \end{pmatrix} : \ v\in \hat E^u(p) \right\}\\& = \left\{ \begin{pmatrix} w\\ (C_p +K_p\sigma_p )\circ A_p^{-1}w \end{pmatrix} : \ w\in \hat E^u(g(p)) \right\} \end{align*} $$

by reparameterizing. Thus, the requirement that $D_pg(\operatorname {\mathrm {graph}}(\sigma _p))=\operatorname {\mathrm {graph}}(\Gamma (\sigma _p))$ is equivalent to saying that

$$ \begin{align*} \begin{pmatrix} w\\ (C_p +K_p\sigma_p )\circ A_p^{-1}w \end{pmatrix} = \begin{pmatrix} w \\ \Gamma\sigma_p w \end{pmatrix} \end{align*} $$

for all $w\in \hat E^u(g(p))$ . This gives us an equation for $\Gamma $ in terms of $C, \ K,$ and A:

$$ \begin{align*} \Gamma\sigma_p =(C_p +K_p\sigma_p )\circ A_p^{-1}:\hat E^u(g(p))\to E^c(g(p)) \end{align*} $$

for all $\sigma \in \Sigma $ , $p\in M$ , and $v\in \hat E^u(p)$ . Omitting base points, we get that

$$ \begin{align*} \Gamma \sigma = (C +K\sigma )\circ A^{-1}. \end{align*} $$

Now, our goal is to find an invariant section for $\Gamma $ (the graph of which we will then show is the unstable bundle $E^u$ for g). To do this, it suffices to show that $\Gamma :\Sigma \to \Sigma $ is a contraction. Take $\sigma , \sigma '\in \Sigma $ . For $p\in M$ , we have

(4.1) $$ \begin{align} \|\Gamma \sigma_p - \Gamma \sigma_p'\| &= \| ( C_p+K_p \sigma_p )\circ A_p^{-1} - ( C_p+K_p \sigma_p' )\circ A_p^{-1} \|\nonumber\\ &= \| ( C_p+K_p \sigma_p-C_p-K_p\sigma^{\prime}_p )\circ A_p^{-1}\|\nonumber\\ &= \| (K_p \sigma_p-K_p\sigma^{\prime}_p )\circ A_p^{-1}\|\nonumber\\ &= \| K_p \circ (\sigma_p-\sigma^{\prime}_p )\circ A_p^{-1}\|\nonumber\\ &\leq \|K_p\| \|\sigma_p-\sigma^{\prime}_p\|\|A_p^{-1}\|. \end{align} $$

We now bound $\|K_p\|$ and $\|A_p^{-1}\|$ . Since g is an isometry on fibers of $\pi :M\to B$ and $E^c=\ker (D\pi )$ , we have that $D_pg|_{E^c(p)}=K_p:E^c(p)\to E^c(g(p))$ is an isometry. Thus, $\|K_p\|=1$ . Now, we bound $\|A_p^{-1}\|$ by relating the norm of A to the norm of $Df$ on $E^u_f$ . Since $E^u_f$ is the unstable bundle for the Anosov diffeomorphism $f:B\to B$ and the norm $\|\cdot \|^B$ is adapted to f, we know that there exists a constant $\unicode{x3bb}>1$ such that for all $w\in E^u_f$ , $\|Df(w)\|^B\geq \unicode{x3bb} \|w\|^B$ . Take $v\in \hat E^u(p)$ . Since $D_p\pi (v)\in E^u_f(\pi (p))$ , we therefore have that

$$ \begin{align*} \|D_{\pi(p)}f(D_p\pi(v))\|^B\geq \unicode{x3bb} \|D_p(v)\|^B. \end{align*} $$

Since g covers f, we see that $D_{\pi (p)}f( D_p\pi (v))=D_{g(p)}\pi (D_pg(v))$ . This along with the fact that $E^c=\ker (D\pi )$ and $C_p(v)\in E^c(g(p))$ gives that

$$ \begin{align*} D_{\pi(p)}f(D_p\pi(v)) = D_{g(p)}\pi(D_pg(v)) = D_{g(p)}\pi(A_p(v)+C_p(v)) = D_{g(p)}\pi(A_p(v)). \end{align*} $$

Thus,

$$ \begin{align*} \unicode{x3bb} \|D_p(v)\|^B \leq \|D_{\pi(p)}f(D_p\pi(v))\|^B = \| D_{g(p)}\pi(A_p(v)) \|^B. \end{align*} $$

Finally, note that since $v, A_p(v)\in \hat E^u(P)\subset H_p$ , by our definition of the norm on M, we get that

$$ \begin{align*} \|D_p\pi(v)\|^B=\|v\| \quad \text{and} \quad \| D_{g(p)}\pi(A_p(v)) \|^B=\|A_p(v)\|. \end{align*} $$

We have therefore shown that $\unicode{x3bb} \|v\|\leq \|A_p(v)\|$ , which implies that $\|A_p^{-1}\|\leq \unicode{x3bb} ^{-1}$ .

Combining our estimates for the norms of $\|K_p\|$ and $\|A_p^{-1}\|$ with equation (4.1) gives that

$$ \begin{align*} \|\Gamma \sigma_p - \Gamma \sigma_p'\|\leq \unicode{x3bb}^{-1}\|\sigma_p-\sigma^{\prime}_p\|. \end{align*} $$

We have therefore shown that $\Gamma $ is a contraction map. Then, by the contraction mapping principle, we get a $\Gamma $ -invariant section $\sigma ^u\in \Sigma $ . We now define a bundle $E^u\subset TM$ by letting $E^u(p):=\operatorname {\mathrm {graph}}(\sigma ^u_p)$ . Note that $E^u$ is $Dg$ invariant since $\sigma ^u$ is $\Gamma $ invariant and $D_pg(\operatorname {\mathrm {graph}}(\sigma _p))=\operatorname {\mathrm {graph}}(\Gamma (\sigma _p))=\operatorname {\mathrm {graph}}(\sigma _{g(p)})$ .

The construction of the bundle $E^s$ is analogous.

We now have a $Dg$ -invariant splitting, $TM=E^s\oplus E^c\oplus E^u$ . We now need to show that this splitting is partially hyperbolic. To do this, we construct a new metric $\langle \cdot , \cdot \rangle '$ on M by letting for all $p\in M$ :

• • $\langle v,v'\rangle ' = \langle D_p\pi (v),D_p\pi (v')\rangle ^B$ for $v,v'\in E^s$ ;

• • $\langle v,v'\rangle ' = \langle D_p\pi (v),D_p\pi (v')\rangle ^B$ for $v,v'\in E^u$ ;

• • $\langle v,v'\rangle ' = \langle v,v' \rangle ^F_{\pi (p)}$ for $v,v'\in E^c$ ; and

• • $E^s$ , $E^c$ , and $E^u$ be pairwise orthogonal.

From our construction of $E^s$ and $E^u$ , we get that $Dg$ is uniformly expanding on $E^u$ and uniformly contracting on $E^c$ with respect to this new metric. Finally, the splitting is dominated because $Dg$ restricted to $E^c$ is an isometry.

Proof of Proposition B

The setup of Proposition B is that we are given a fibered partially hyperbolic system $\hat f:M\to M$ with quotient a nilmanifold B and $C^1$ fibers F. We assume that the F-bundle M is trivial. Note that this means that the structure group of the bundle $\pi :M\to B$ is the trivial group. We can then proceed with an analogous argument to that given in the proof of Theorem A to get the conjugacy $h:B\to B$ between the Anosov homeomorphism $f:B\to B$ induced by $\hat f:M\to M$ and a hyperbolic nilmanifold automorphism $A:B\to B$ , and to get a smooth bundle $\widehat \pi : \widehat M \to B$ with trivial structure group that is isomorphic to the original bundle $\pi : M\to B$ . This means that identifying the smooth F-bundle $\widehat \pi : \widehat M \to B$ with the smooth bundle ${\operatorname {\mathrm {proj}}_1: B\times F \to B}$ (that is, projection onto the first coordinate), we can smoothly lift $A:B\to B$ to the fibered partially hyperbolic diffeomorphism $g:\widehat M\cong B\times F\to \widehat M\cong B\times F$ given by $g: (x,y)\mapsto (Ax,y)$ for $(x,y)\in B\times F$ .