Is the square root of a hyperbolic map hyperbolic? Suppose that we have an area preserving diffeomorphism of the $n$-dimensional torus,
$$f:\mathbb T^n \to \mathbb T^n,$$
such that the two-fold composition $f^2 := f \circ f$ is uniformly hyperbolic. That is, one has a splitting of the tangent bundle of $\mathbb T$
$$ T(\mathbb T^n) =  E_s \oplus E_u$$
and $\lambda \in (0,1)$ which satisfies the following, for all $x \in \mathbb T$,

*

*invariance under $f^2$: $\qquad T_xf^2(E_s(x)) = E_s(f^2(x))$  and  $T_xf^2(E_u(x)) = E_u(f^2(x))$.

*contraction on $E_s$: $\qquad\|D_xf^2|_{E_s(x)}\|_{op}<\lambda$.

*expansion on $E_u$: $\qquad \|D_xf^{-2}|_{E_u(x)}\|_{op}< \lambda.$
($\|\cdot\|_{op}$ denotes the operator norm, $T_xf^2$ is the tangent map of $f^2$ at $x$ and $E_s(x)$ denotes the fibre $T_x(\mathbb T^n)\cap E_s$, for example).
Question: Must $f$ be uniformly hyperbolic too?
 A: The answer is yes.

Let $M$ be  a compact $C^\infty$ manifold, $g:M\to M$ be a $C^1$ Anosov diffeomorphism. Thus there is a $(g,Tg)$-invariant topological splitting $TM=S(g)\oplus U(g)$ and for any $C^0$ fiberwise norm $|\bullet|$ on $M$ there are numbers $(C,\lambda)\in\mathbb{R}_{>0}\times ]0,1[$ such that for any $x\in M$ and any $n\in\mathbb{Z}_{\geq0}$ we have
\begin{align*}
S_x(g)=&\{\quad v\in T_xM\quad|\quad |T_xg^n v|\leq C\lambda^n|v|\quad\},\\
U_x(g)=S_x(g^{-1})=&\{\quad v\in T_xM\quad|\quad |T_xg^{-n}v|\leq C \lambda^{n}|v|\quad\}.
\end{align*}
Call $(C,\lambda)$ an Anosov pair for $g$ (with respect to $|\bullet|$) (see also What is the constant of hyperbolicity?). Note that $(C,\lambda)$ is an Anosov pair for $g^{-1}$ also.
For any $x\in M$ denote by $\mathcal{S}_x(g)$ and $\mathcal{U}_x(g)$ the global stable and unstable manifolds of $g$ at $x$, respectively. We have that $\mathcal{S}_x(g)$ and $\mathcal{U}_x(g)$ are injectively immersed $C^1$ submanifolds of $M$, $T_x\mathcal{S}_x(g)=S_x(g)$ and $T_x\mathcal{U}_x(g)=U_x(g)$, and they are uniquely defined by these properties. Further, if $d$ is the intrinsic distance function on $M$ defined by the fiberwise norm, then
\begin{align*}
\mathcal{S}_x(g)&=\left\{\quad y\in M\quad\left|\quad \lim_{n\to \infty} d(g^n(y),g^n(x))=0\quad\right\}\right.,\\ 
\mathcal{U}_x(g)=\mathcal{S}_x(g^{-1})&=\left\{\quad y\in M\quad \left|\quad \lim_{n\to \infty} d(g^{-n}(y),g^{-n}(x))=0\quad\right\}\right..
\end{align*}

Obs.: Let $f:M\to M$ be a $C^1$ diffeomorphism that commutes with a $C^1$ Anosov diffeomorphism $g:M\to M$, i.e. $f\circ g=g\circ f$. Then $f$ permutes the leaves of the stable foliation of $g$, that is,
$$\forall x\in M: f: \mathcal{S}_x(g)\to \mathcal{S}_{f(x)}(g).$$
Pf.: Let $x\in M, y\in\mathcal{S}_x(g)$. Then since $f$ is uniformly continuous
\begin{align*}
d(g^n(f(y)),g^n(f(x)))
=d(f(g^n(y)),f(g^n(x)))\xrightarrow{n\to\infty}0,
\end{align*}
that is, $f(y)\in \mathcal{S}_{f(x)}(g)$.
Applying the same argument to $g^{-1}$ we have that $f$ permutes the leaves of the unstable foliation of $g$ as well. Note that in this case we also have $T_xf : S_x(g)\to S_{f(x)}(g)$ and $T_xf: U_x(g)\to U_{f(x)}(g)$ by taking derivatives.
(For a general $g$ one can still define the global stable and unstable sets $\mathcal{S}_x(g)$ and $\mathcal{U}_x(g)$ as above, but they may fail to be injectively immersed submanifolds. In any event, if $f\circ g=g\circ f$, then again $f$ would permute the global stable sets of $g$ as well as the global unstable sets of $g$ (and vice versa).)

Claim: Let $p\in\mathbb{Z}_{\geq1}$ and  $f,g:M\to M$ be $C^1$ diffeomorphisms such that $f^p=g$. Then $f$ is Anosov iff $g$ is Anosov.
Pf.: First note that $f$ and $g$ (as well as any powers of them) commute. It's clear that if ($f$ is Anosov and) $(C,\lambda)$ is an Anosov pair for $f$, then $(C,\lambda^p)$ is an Anosov pair for $g=f^p$.
Conversely, suppose ($g$ is Anosov and) $(C,\lambda)$ is an Anosov pair for $g$. Let $x\in M,n\in\mathbb{Z}_{\geq0}, v\in S_x(g)$. There are unique numbers $Q\in\mathbb{Z}_{\geq0}$ and $R\in\{0,1,...,p-1\}$ such that $n=pQ+R$, so that $f^n=g^Q\circ f^R$. By the above observation $T_x f^R:S_x(g)\to S_{f^R(x)}(g)$. We have $Q=\frac{n-R}{p}$, so that
$$\lambda^Q\leq (\lambda^{1/p})^n \lambda^{-\frac{p-1}{p}},$$
and also
$$\left\Vert T_xf^R\right\Vert\leq \max_{R\in\{0,1,...,p-1\}}\max_{x\in M}\left\Vert T_xf^R\right\Vert \leq \max_{R\in\mathbb{Z}_{|\bullet|<p}}\max_{x\in M}\left\Vert T_xf^R\right\Vert\in\mathbb{R}_{>0},$$
which does not depend on $R,n$ nor $x$.
(Here $\mathbb{Z}_{|\bullet|<p}=\{-p+1,-p+2,...,0,...,p-2,p-1\}$.)
Then we have
\begin{align*}
|T_xf^n v|
&=\left|T_x\left(g^Q\circ f^R\right)v\right|
=\left|T_{f^R(x)}g^Q\left( T_x f^Rv\right)\right|\\
&\leq C\lambda^Q \left|\left( T_x f^Rv\right)\right|
\leq C\lambda^Q \left\Vert T_x f^R\right\Vert |v|\\
&\leq \left(C\lambda^{-\frac{p-1}{p}} \max_{R\in\mathbb{Z}_{|\bullet|<p}}\max_{x\in M}\left\Vert T_xf^R\right\Vert\right) \,\, (\lambda^{1/p})^n |v|.
\end{align*}
Since $f^{-n}=g^{-Q}\circ f^{-R}$, for $v\in U_x(g)$ the appropriate estimate holds and we get that $\left(C\lambda^{-\frac{p-1}{p}} \max_{R\in\mathbb{Z}_{|\bullet|<p}}\max_{x\in M}\left\Vert T_xf^R\right\Vert,\lambda^{1/p}\right)$ is an Anosov pair for $f$, whence $f$ is Anosov (and $S_x(f)=S_x(g)$ and $U_x(f)=U_x(g)$ for any $x\in M$).
(An alternative argument is to use a uniform modulus of continuity for $\{f^R\,|\, R\in \{0,1,...,p-1\}\}$) to show that the global stable sets $\mathcal{S}_x(f)$ and $\mathcal{S}_x(g)$ coincide.)
Corollary: Let $M$ be a compact $C^\infty$ manifold, $f\in\operatorname{Diff}^1(M)$ be a $C^1$ diffeomorphism. Denote by $f^\mathbb{Z}=\{f^p\,|\, p\in\mathbb{Z}\}$ the subgroup of $\operatorname{Diff}^1(M)$ consisting of all iterates of $f$. Then $f^\mathbb{Z}$ contains an Anosov diffeomorphism iff $\forall p\in\mathbb{Z}\setminus0: f^p$ is an Anosov diffeomorphism.
