# 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?

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

$$\left\Vert T_xf^R\right\Vert\leq \max_{R\in\{0,1,...,p-1\}}\max_{x\in M}\left\Vert T_xf^R\right\Vert \leq \max_{R\in\mathbb{Z}_{|\bullet|0},$$

which does not depend on $$R,n$$ nor $$x$$.

(Here $$\mathbb{Z}_{|\bullet|.)

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|

Since $$f^{-n}=g^{-Q}\circ f^{-R}$$, for $$v\in U_x(g)$$ the appropriate estimate holds and we get that $$\left(C\lambda^{-\frac{p-1}{p}} \max_{R\in\mathbb{Z}_{|\bullet| is an Anosov pair for $$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.