Proof. Take a hyperbolic matrix $A \in \mathrm {M}_{2 \times 2}({\mathbb R})$ as in the hypothesis and let ${\varepsilon }>0$ . Let $\beta _{\unicode{x3bb} }$ be the function given by Lemma 4.1 for $b=1-\unicode{x3bb} $ and its corresponding function $\psi _{\unicode{x3bb} }$ , and let $\beta _{\mu }$ be the function given by Lemma 4.1 for $b=\mu -1$ and its corresponding function $\psi _{\mu }$ . Now, we define the function $g:{\mathbb R}^2\to {\mathbb R}^2$ by the equation:

(4.1) $$ \begin{align} g(x,y)=(\unicode{x3bb} x,\mu y) + (\beta_{\unicode{x3bb}}(r)x,-\beta_{\mu}(r)y), \end{align} $$

where $r=x^2+y^2$ . Notice that if $r\geq \epsilon $ , then $g=A$ . In particular, $Dg_{(x,y)}=A$ for every $(x,y)$ such that $x^2+y^2\geq \epsilon $ . In case $r<\epsilon $ , the differential is

$$ \begin{align*} Dg_{(x,y)}= \begin{bmatrix} \unicode{x3bb} +\beta_{\unicode{x3bb}}(r) +2x^2 \beta_{\unicode{x3bb}}'(r) & 2xy\beta_{\unicode{x3bb}}'(r) \\ -2xy\beta_{\mu}'(r) & \mu - \beta_{\mu}(r)-2y^2 \beta_{\mu}'(r) \end{bmatrix}. \end{align*} $$

In particular, we have that $Dg_{(0,0)}=\mathrm {Id}$ and, therefore, g is not hyperbolic. In case $r>0$ , we have that

$$ \begin{align*} \unicode{x3bb}+ \beta_{\unicode{x3bb}}(r)+2x^2\beta_{\unicode{x3bb}}'(r)< \unicode{x3bb} + \beta_{\unicode{x3bb}}(0)=1 \end{align*} $$

and

$$ \begin{align*} \mu -\beta_{\mu}(r)-2y^2\beta_{\mu}'(r)> \mu -\beta_{\mu}(0)=1. \end{align*} $$

Now, take the family of cones in ${\mathbb R}^2$ ,

$$ \begin{align*} {\mathcal C}^u(x,y)=\{ (a,b)\in {\mathbb R}^2: |a|\leq |b| \}. \end{align*} $$

We claim that this family of cones is $Dg$ -invariant. Therefore, we have to prove that if $(a,b)\in {\mathcal C}^u$ , then $Dg(a,b)=(a_1,b_1)\in {\mathcal C}^u$ , and this occurs if and only if $|a_1|\leq |b_1|$ . By the equations above, we have that

$$ \begin{align*} a_1 &= a ( \unicode{x3bb}+ \beta_{\unicode{x3bb}}(r)+2x^2\beta_{\unicode{x3bb}}'(r)) + b (2xy\beta_{\unicode{x3bb}}'(r)), \\ b_1 &= a (-2xy\beta_{\mu}'(r)) + b ( \mu - \beta_{\mu}(r)-2y^2 \beta_{\mu}'(r)). \end{align*} $$

Notice that if $r>\epsilon $ , the cones are $Dg$ -invariant since $g=A$ . However, for points close to zero (r small), this is not so clear. Therefore, we have to take a little more care with the choice of the functions $\beta _{\unicode{x3bb} }$ and $\beta _{\mu }$ , in particular, with the functions $\psi _{\unicode{x3bb} }$ and $\psi _{\mu }$ .

To find a suitable $C^{\infty }$ function $\psi _{\unicode{x3bb} }$ , we are going to approximate it by two $C^0$ functions $\psi _{\unicode{x3bb} }^1$ and $\psi _{\unicode{x3bb} }^0$ by above and below, which implies that $\beta _{\unicode{x3bb} }$ will be approximated by below and above by their corresponding $C^1$ functions $\beta _{\unicode{x3bb} }^1$ and $\beta _{\unicode{x3bb} }^0$ given by Lemma 4.1. Then, we use the density of $C^{\infty }$ functions to conclude.

Let $\rho>0$ be sufficiently small such that $3\rho <\epsilon $ and $2\rho ^2<\min \{1-\unicode{x3bb} , \mu -1\}$ , and let $m\in {\mathbb R}^+$ be such that $2m\rho ^2 <1-\unicode{x3bb} <2(m+1)\rho ^2$ . Now, for $i=0,1$ , we define the function $\psi _{\unicode{x3bb} }^i$ in the following way (see Figure 3):

$$ \begin{align*} \psi_{\unicode{x3bb}}^i(r)= \left\{\!\!\begin{array}{lll} (m+i)r & \text{if }\ r \in [0,\rho] , \\ (m+i)\rho & \text{if }\ r \in (\rho, 2\rho ) ,\\ (m+i)(-r+3\rho) & \text{if }\ r \in [2\rho,3\rho]. \end{array} \right.\end{align*} $$

Figure 3

Bump functions $\psi _{\unicode{x3bb} }^0, \psi _{\unicode{x3bb} }^1$ and $\psi _{\unicode{x3bb} }$ (colour online).

Notice that

$$ \begin{align*} \int_0^{\epsilon}\psi_{\unicode{x3bb}}^0(s)\,ds=2m\rho^2 < 1-\unicode{x3bb}<2(m+1)\rho^2=\int_0^{\epsilon}\psi_{\unicode{x3bb}}^1(s)\,ds. \end{align*} $$

Since $C^{\infty }$ functions are $C^0$ -dense, we can find a $C^{\infty }$ function $\psi _{\unicode{x3bb} }$ such that:

1. (1) supp $(\psi _{\unicode{x3bb} })\subseteq [0,\epsilon ]$ ;

2. (2) $\psi _{\unicode{x3bb} }^0(s) \leq \psi _{\unicode{x3bb} }(s) \leq \psi _{\unicode{x3bb} }^1(s)$ for every $s\in [0,\epsilon ]$ ;

3. (3) $\int _0^{\epsilon }\psi _{\unicode{x3bb} }(s)\,ds=1-\unicode{x3bb} $ .

Notice that point (2) implies that $\beta _{\unicode{x3bb} }^1(r)\leq \beta _{\unicode{x3bb} }(r)\leq \beta _{\unicode{x3bb} }^0(r)$ for every $r\in [0,\epsilon ]$ .

In the same way, we can find a suitable $C^{\infty }$ function $\psi _{\mu }$ by approximating it by two $C^0$ functions $\psi _{\mu }^1$ and $\psi _{\mu }^0$ by above and below, which implies that $\beta _{\mu }$ will be approximated by below and above by their corresponding $C^1$ functions $\beta _{\mu }^1$ and $\beta _{\mu }^0$ according to Lemma 4.1.

Recall that $2\rho ^2<\mu -1$ , then let $n\in {\mathbb R}^+$ be such that $2n\rho ^2<\mu -1<2(n+1)\rho ^2$ . Now, for $i=0,1$ , we define the function $\psi _{\mu }^i$ in the following way:

$$ \begin{align*}\psi_{\mu}^i(r)= \left\{\!\!\begin{array}{lll} (n+i)r & \text{if }\ r \in [0,\rho], \\ (n+i)\rho & \text{if }\ r \in (\rho, 2\rho), \\ (n+i)(-r+3\rho) & \text{if }\ r \in [2\rho,3\rho]. \end{array} \right.\end{align*} $$

In the same way as above, we observe that

$$ \begin{align*} \int_0^{\epsilon}\psi_{\mu}^0(s)\,ds=2n\rho^2<\mu-1<2(n+1)\rho^2=\int_0^{\epsilon}\psi_{\mu}^1(s)\,ds. \end{align*} $$

Since $C^{\infty }$ functions are $C^0$ -dense, we can find a $C^{\infty }$ function $\psi _{\mu }$ such that:

1. (1) supp $(\psi _{\mu })\subseteq [0,\epsilon ]$ ;

2. (2) $\psi _{\mu }^0(s) \leq \psi _{\mu }(s) \leq \psi _{\mu }^1(s)$ for every $s\in [0,\epsilon ]$ ;

3. (3) $\int _0^{\epsilon }\psi _{\mu }(s)\,ds=\mu -1$ .

Notice that point (2) implies that $\beta _{\mu }^1(r)\leq \beta _{\mu }(r)\leq \beta _{\mu }^0(r)$ for every $r\in [0,\epsilon ]$ .

Recall that we want to prove that the cones $\mathcal {C}^u$ are $Dg$ -invariant. We are going to obtained the invariance of the cones for $C^1$ close diffeomorphisms $g^0$ and $g^1$ defined (in the same way as g) by equation (4.1), but using the functions $\beta _{\unicode{x3bb} }^0$ , $\beta _{\mu }^0$ and $\beta _{\unicode{x3bb} }^1$ , $\beta _{\mu }^1$ , respectively, instead of $\beta _{\unicode{x3bb} }$ and $\beta _{\mu }$ . Notice that for $i=0,1$ , the $C^1$ distance between g and $g^i$ can be taken as small as we want.

Now, for $i=0,1$ , to see the cones $\mathcal {C}^u$ are $Dg^i$ -invariant, we have to prove that if ${(a,b)\in {\mathcal C}^u}$ , then $Dg^i(a,b)=(a_1,b_1)\in {\mathcal C}^u$ , and this happens if and only if $|a_1|\leq |b_1|$ . We will obtain this inequality by studying different cases depending on r, which is the square of the distance of the point $(x,y)$ to the origin.

Case 1: $r\in [0,\rho ]$ . By our definitions above, we have

$$ \begin{align*}\beta_{\unicode{x3bb}}^i(r)=1-\unicode{x3bb} -(m+i)\frac{r^2}{2} \quad\text{and}\quad (\beta_{\unicode{x3bb}}^i)'(r)=-(m+i)r. \end{align*} $$

Then, we have that

$$ \begin{align*}\unicode{x3bb}+ \beta_{\unicode{x3bb}}^i(r)+2x^2(\beta_{\unicode{x3bb}}^i) '(r)=1-(m+i)\frac{r^2}{2}-2x^2 (m+i)r \end{align*} $$

and as a consequence, we obtain that $a_1$ is equal to

$$ \begin{align*}a_1 = a \bigg( 1-(m+i)\frac{r^2}{2}-2x^2 (m+i)r \bigg) + b (-2xy(m+i)r ). \end{align*} $$

Since $|a|\leq |b|$ , by taking the absolute value and applying the triangular inequality, we get

$$ \begin{align*} |a_1|&\leq |b|\bigg( 1-(m+i)\frac{r^2}{2}-2x^2 (m+i)r \bigg) + |b| ( 2|xy|(m+i)r) \\ &= |b|\bigg(1 -(m+i)\frac{r}{2}( r+4x^2 -4|xy| ) \bigg) \leq |b|, \end{align*} $$

where the last inequality holds as long as $r+4x^2-4|xy|=5x^2+y^2-4|xy|>0$ . We claim that this is always the case: if $|xy|=xy$ , we have to show that $5x^2+y^2-4xy>0$ . By solving the second degree equation in y, we obtain that

$$ \begin{align*}y=\frac{4x\pm \sqrt{16x^2-20x^2}}{2} \end{align*} $$

and this has no real roots. Since for $y=0$ we have $5x^2\geq 0$ , we obtain the desired inequality. The case where $|xy|=-xy$ is completely the same, since the discriminant in the equation above is the same. We conclude that $r+4x^2-4|xy|\geq 0$ and, moreover, $|a_1|<|b|$ if $r>0$ .

In the same way, we have that

$$ \begin{align*}\beta_{\mu}^i(r)=\mu-1 - ( n+i)\frac{r^2}{2}\quad\text{and}\quad (\beta_{\mu}^i)'(r)=-(n+i)r, \end{align*} $$

and then

$$ \begin{align*}\mu -\beta_{\mu}^i(r)-2y^2(\beta_{\mu}^i)'(r)=1+ (n+i)\frac{r^2}{2}+2y^2 (n+i )r. \end{align*} $$

We then have that $b_1$ is equal to

$$ \begin{align*} b_1=a (2xy( n+i)r)+ b \bigg( 1+ (n+i)\frac{r^2}{2}+2y^2 ( n+i)r\bigg). \end{align*} $$

Since $|a|\leq |b|$ , taking the absolute value and by the triangular inequality, we have

$$ \begin{align*} |b_1|&\geq |b| \bigg( 1+ (n+i)\frac{r^2}{2}+2y^2 ( n+i)r \bigg)-|b| (2|xy|( n+i )r) \\ &= |b|\bigg( 1+\frac{r}{2} (n+i) ( r+4y^2-4|xy|)\bigg) \geq |b|, \end{align*} $$

where the last inequality holds as long as $r+4y^2-4|xy|=x^2+5y^2-4|xy|\geq 0$ . This is exactly the same equation we solve above, and thus we conclude that $|b_1|\geq |b|$ and, moreover, $|b_1|>|b|$ if $r>0$ . Then, we conclude that $|a_1|\leq |b|\leq |b_1|$ .

Case 2: $r\in [\rho ,2\rho ]$ . In this case, we obtain that $\beta _{\unicode{x3bb} }^i$ verifies

$$ \begin{align*}\beta_{\unicode{x3bb}}^i(r)=1-\unicode{x3bb}-(m+i)\frac{\rho^2}{2} -(m+i)\rho(r-\rho) \quad \text{and}\quad (\beta_{\unicode{x3bb}}^i)'(r)=-(m+i)\rho. \end{align*} $$

Then, we have that

$$ \begin{align*}\unicode{x3bb}+ \beta_{\unicode{x3bb}}^i(r)+2x^2(\beta_{\unicode{x3bb}}^i)'(r)=1-(m+i)\frac{\rho^2}{2} -(m+i)\rho(r-\rho) -2x^2(m+i)\rho; \end{align*} $$

therefore, $a_1$ is equal to

$$ \begin{align*}a_1 = a \bigg( 1-(m+i)\frac{\rho^2}{2} -(m+i)\rho(r-\rho) -2x^2(m+i)\rho \bigg) + b ( -2xy(m+i)\rho ). \end{align*} $$

Since $|a|\leq |b|$ , taking the absolute value and applying the triangular inequality, we get

$$ \begin{align*} |a_1|&\leq |b| \bigg( 1-(m+i)\frac{\rho^2}{2} -(m+i)\rho(r-\rho) -2x^2(m+i)\rho \bigg)\\ &\quad + |b| ( 2|xy|(m+i)\rho ) \\ &= |b|\bigg( 1-(m+i)\frac{\rho^2}{2} -(m+i)\rho(r-\rho) -2x^2(m+i)\rho + 2|xy|(m+i)\rho \bigg) \\ &= |b|\bigg(1-\bigg(\frac{(m+i)\rho}{2}\bigg)\bigg( 6x^2+2y^2-4|xy|-\rho \bigg) \bigg) \leq |b|, \end{align*} $$

where the last inequality holds as long as $6x^2+2y^2-4|xy|-\rho \geq 0$ . Notice that

$$ \begin{align*} 6x^2+2y^2-4|xy|-\rho=(x^2+y^2-\rho)+(5x^2+y^2-4|xy|). \end{align*} $$

The first term in the right expression is greater or equal to zero since $\rho \leq r$ . The second term $5x^2+y^2-4|xy|$ is exactly the same equation we solve in case 1.

In the same way, we have that

$$ \begin{align*} \beta_{\mu}^i(r)=\mu-1-\frac{\rho^2}{2}(n+i)-(r-\rho)\rho (n+i)\quad\text{and}\quad (\beta_{\mu}^i)'(r)= -(n+i)\rho, \end{align*} $$

and then we obtain

$$ \begin{align*} \mu-\beta_{\mu}^i(r)-2y^2(\beta_{\mu}^i)'(r)=1+ \frac{\rho^2}{2} (n+i) + (r-\rho)\rho(n+i)+2y^2 (n+i)\rho. \end{align*} $$

We then have that $b_1$ is equal to

$$ \begin{align*} b_1=a (2xy ( n+i) \rho) + b \bigg(1+\frac{\rho^2}{2} (n+i) + (r-\rho)\rho ( n+i) + 2y^2 (n+i)\rho \bigg). \end{align*} $$

Since $|a|\leq |b|$ , taking the absolute value and by the triangular inequality, we have

$$ \begin{align*} |b_1|&\geq |b|\bigg(1+\frac{\rho^2}{2} (n+i) + (r-\rho)\rho ( n+i)+2y^2 (n+i)\rho\bigg)\\ &\quad- |b| ( 2|xy|( n+i) \rho ) \\ &= |b|\bigg( 1+ \frac{\rho^2}{2}( n+i)+ (r-\rho)\rho(n+i)+2y^2(n+i)\rho-2|xy|(n+i)\rho \bigg) \\ &= |b|\bigg( 1+ \bigg(\frac{(n+i)\rho}{2}\bigg)(2x^2+6y^2-\rho-4|xy|)\bigg) \geq |b| \end{align*} $$

by the same estimates as above. Moreover, we have that $|b_1|>|b|$ if $r>0$ . Then, we conclude that $|a_1|\leq |b| \leq |b_1|$ .

Case 3: $r\in [2\rho ,3\rho ]$ . This is the simplest case, since we are far enough to zero and so we omit the calculations.

To sum up, we have proved that for $i=0,1$ , the cones $\mathcal {C}^u$ are $Dg^i$ -invariant. In fact, we have proved that if $(a,b)\in {\mathcal C}^u$ , then $Dg^i(a,b)=(a_1,b_1)$ verifies $|a_1|\leq |b_1|$ . Now, just notice that the inequality $|a_1|\leq |b_1|$ depends on the values of the entries of $Dg^i$ . Since the entries of $Dg$ are bounded by above and below by the entries of $Dg^i$ , by taking suitable triangular inequalities, it is easy to see that the inequality $|a_1|\leq |b_1|$ also holds for $Dg$ , proving that the cones $\mathcal {C}^u$ are also $Dg$ -invariant.

Now, if we take the norm $\lVert (a,b)\rVert _1:=\max \{|a|,|b|\}$ in ${\mathbb R}^2$ , we have that vectors in ${\mathcal C}^u$ are expanded for the future: if $v=(a,b)\in {\mathcal C}^u$ , then $|a|\leq |b|$ and thus $\lVert v\rVert _1=|b|$ . Since $Dg(v)=(a_1,b_1) \in {\mathcal C}^u$ , this implies that $\lVert Dg(v)\rVert _1=|b_1|$ and we have just proved that $|b|\leq |b_1|$ . This implies that $\lVert Dg(v)\rVert _1 \geq \lVert v\rVert _1$ and, moreover, $\lVert Dg_{(x,y)}(v)\rVert _1> \lVert v\rVert _1$ if $(x,y)\neq (0,0)$ .

Next, we define the Lyapunov function $V:{\mathbb R}^2\times {\mathbb R}^2\to {\mathbb R}$ given by

$$ \begin{align*} V((x_1,y_1),(x_2,y_2))=(y_1-y_2)^2-(x_1-x_2)^2. \end{align*} $$

Let $P=(x_1,y_1)$ and $Q=(x_2,y_2)$ be two different points in ${\mathbb R}^2$ . Then, $V(P,Q)=(y_1-y_2)^2-(x_1-x_2)^2$ . Let us denote $g(P)=(x_1',y_1')$ and $g(Q)=(x_2',y_2')$ , then $V(g(P),g(Q))=(y_1'-y_2')^2-(x_1'-x_2')^2$ .

Now, take the function $\varphi :[0,1]\to {\mathbb R}$ given by $\varphi (t)=\Pi _2(g(tP+(1-t)Q))$ , where $\Pi _2:{\mathbb R}^2\to {\mathbb R}$ is the projection in the second coordinate. Notice that $\varphi (0)=y_2'$ , $\varphi (1)=y_1'$ and $\varphi $ is differentiable. Then, by the mean value theorem, we have

$$ \begin{align*}|y_2'-y_1'|&=|\varphi(0)-\varphi(1)|=|\varphi'(t_0)|=|\Pi_2 \circ Dg_{z_0}(P-Q)|\\&=|Dg_{z_0}(0,y_1-y_2)|>|y_1-y_2| \end{align*} $$

for some $z_0$ . In the same way, we get that $|x_1'-x_2'|<|x_1-x_2|$ , and this implies that $\Delta V(P,Q)>0$ . This ends the proof of the lemma.

Proof. Let $A\in \mathrm {SL}(2,{\mathbb Z})$ be as in the hypothesis and suppose the eigenvalues of A are $\unicode{x3bb} $ and $\mu $ with $0<\unicode{x3bb} <1<\mu $ . Let $E^s$ be the eigenspace associated to $\unicode{x3bb} $ , and let $E^u$ be the eigenspace associated to $\mu $ . Then, we have that ${\mathbb R}^2=E^s\oplus E^u$ . Now, for ${\varepsilon }>0$ and identifying ${\mathbb R}^2$ with $E^s\oplus E^u$ , we take the map $g:{\mathbb R}^2\to {\mathbb R}^2$ given by Lemma 4.4. We can take $\epsilon $ small enough so we can send the map g to the quotient ${\mathbb T}^2={\mathbb R}^2/{\mathbb Z}^2$ (notice that we are making an abuse of notation between g and its lift to ${\mathbb R}^2$ ).

We know that the map g has a Lyapunov function $V:{\mathbb R}^2\times {\mathbb R}^2\to {\mathbb R}$ given by

$$ \begin{align*} V((x_1,y_1),(x_2,y_2))=(y_1-y_2)^2-(x_1-x_2)^2, \end{align*} $$

which verifies that $\Delta V((x_1,y_1),(x_2,y_2))>0$ for every pair of different (and sufficiently close) points $(x_1,y_1),(x_2,y_2) \in {\mathbb R}^2$ . Then, by Proposition 4.3, the map g is expansive. Notice that we are only interested in what happens in a fundamental domain, so the non-compactness of ${\mathbb R}^2$ is not a problem.

Finally, since g fails to be hyperbolic only at $(0,0)$ , we know that g is in the $C^1$ boundary of Anosov diffeomorphisms. Then, by [Reference Lewowicz13, Corollary 6.2], g is conjugated to A. Moreover, this implies g is infinite expansive in ${\mathbb R}^2$ and, in particular, the semiconjugacy $H_g$ is in fact injective and $\Lambda (g)=0$ .

Proof. Let $A\in \mathrm {SL}(4,{\mathbb Z})$ be a matrix with four eigenvalues $\unicode{x3bb} ^{ss},\unicode{x3bb} ,\mu ,\mu ^{uu}$ such that

$$ \begin{align*} 0<\unicode{x3bb}^{ss}<\unicode{x3bb}<1<\mu<\mu^{uu}. \end{align*} $$

We can assume that in the basis given by the eigenspaces associated to the eigenvalues, we have that $A(x,y,z,t)=(\unicode{x3bb} x, \mu y, \unicode{x3bb} ^{ss}z,\mu ^{uu}t)$ . Take the same functions $\beta _{\unicode{x3bb} }$ and $\beta _{\mu }$ as in Lemma 4.4 and define the map $f:{\mathbb R}^4\to {\mathbb R}^4$ by

$$ \begin{align*}f(x,y,z,t)=(\unicode{x3bb} x,\mu y, \unicode{x3bb}^{ss} z, \mu^{uu} t) + \rho(w)(\beta_{\unicode{x3bb}}(r)x,-\beta_{\mu}(r)y,0,0), \end{align*} $$

where $\rho $ is a bump function supported in $[0,{\varepsilon }]$ and $w=z^2+t^2$ . We can take $\epsilon $ small enough so we can send the map f to the quotient ${\mathbb T}^4={\mathbb R}^4/{\mathbb Z}^4$ .

If $\lVert (x,y,z,t)\rVert \geq {\varepsilon }$ , we have that $f=A$ . For points with $\lVert (x,y,z,t)\rVert < {\varepsilon }$ , the differential of f at a point $(x,y,z,t)$ is

$$ \begin{align*} \begin{bmatrix} \unicode{x3bb} +\rho(w) ( \beta_{\unicode{x3bb}}(r) +2x^2 \beta_{\unicode{x3bb}}'(r)) & \rho(w) (2xy\beta_{\unicode{x3bb}}'(r)) &2xz\rho'(w)\beta_{\unicode{x3bb}}(r) & 2tz\rho'(w)\beta_{\unicode{x3bb}}(r) \\ -\rho(w) (2xy\beta_{\mu}'(r)) & \mu -\rho(w)( \beta_{\mu}(r)-2y^2 \beta_{\mu}'(r) ) &-2yz\rho'(w)\beta_{\mu}(r) & 2yt\rho'(w)\beta_{\mu}(r) \\ 0 & 0 & \unicode{x3bb}^{ss} & 0 \\ 0 & 0 & 0 & \mu^{uu} \end{bmatrix}\!. \end{align*} $$

In this case, the subspace $E^c_f=\{(x,y,0,0)\}$ is $Df$ invariant, and it is quite direct to see that $Df|_{E^c_f}$ is basically $Dg$ like above (we have to deal with the function $\rho $ , but this is not a problem). In particular, we have that

$$ \begin{align*} Df_0= \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & \unicode{x3bb}^{ss} & 0 \\ 0 & 0 & 0 & \mu^{uu} \end{bmatrix}. \end{align*} $$

The strong bundles are not going to be the canonical ones, but if we ask for the strong eigenvalues $\unicode{x3bb} ^{ss}$ and $\mu ^{uu}$ to be sufficiently far away from 1 (and we can do this by iterating the matrix), the same strong cones for the matrix A are going to be $Df$ -invariant.

Finally, in the same way as in Lemma 4.4 and Corollary 4.6, by taking the Lyapunov function $V:{\mathbb R}^4 \times {\mathbb R}^4 \to {\mathbb R}$ given by

$$ \begin{align*} V((x_1,y_1,z_1,t_1),(x_2,y_2,z_2,t_2))=(y_1-y_2)^2 -(x_1-x_2)^2 + (t_1-t_2)^2-(z_1-z_2)^2, \end{align*} $$

it is easy to see that $\Delta V((x_1,y_1,z_1,t_1),(x_2,y_2,z_2,t_2))>0$ for every pair of different (and sufficiently close) points $(x_1,y_1,z_1,t_1),(x_2,y_2,z_2,t_2) \in {\mathbb R}^4$ . Then, by Proposition 4.3, we conclude that $f:{\mathbb T}^4\to {\mathbb T}^4$ is an expansive diffeomorphism. Moreover, f is in the $C^1$ boundary of Anosov diffeomorphisms and by [Reference Lewowicz13, Corollary 6.2], f is conjugated to A (which implies $\Lambda (f)=0$ ).

Proof. Take a matrix $A\in M_{k\times k}({\mathbb Z})$ as above. Then the eigenvalues of A verify that

Fix ${\varepsilon }>0$ small and take $\unicode{x3bb} $ such that $\unicode{x3bb} _k < \unicode{x3bb} < 1$ . In particular, $\unicode{x3bb}> \unicode{x3bb} _j$ for every $j=1,\ldots ,k$ . Now take a number $c\in (0,{\varepsilon })$ such that $c< ({1-\unicode{x3bb} })/({k(\unicode{x3bb} -\unicode{x3bb} _1)})$ . Now, for this $c>0$ , take the function $\beta $ given by Lemma 4.1 for $b=1$ . In particular, the function $\beta $ verifies:

• • $\beta $ is $C^{\infty }$ and decreasing;

• • $\beta $ is supported in $[0,{\varepsilon }]$ ;

• • $\beta (0)=1$ .

Moreover, we can ask for $\beta $ to be equal to 1 in a small interval $[0,\delta ]$ . Now, we can define the map $g_1:{\mathbb R}^k \to {\mathbb R}^k$ by

where $r=x_1^2+\cdots +x_k^2$ . Since supp $(\beta )\subseteq [0,{\varepsilon }]$ , we have that if $\lVert x\rVert>{\varepsilon }$ , then $g_1=A$ . The differential of $g_1$ in a point x is

where $M(x)$ is the matrix given by

In particular, since $\beta (0)=1$ , we have that $D(g_1)_0=A+\beta (0)(\unicode{x3bb} \mathrm {Id} -A)+M(0)=\unicode{x3bb} \mathrm {Id}$ . Now, take a point $x\in {\mathbb R}^k$ and a vector $v\in {\mathbb R}^k$ , then we have that

$$ \begin{align*}D(g_1)_x(v)=Av+\beta(r)(\unicode{x3bb} \mathrm{Id}-A)v+M(x)v. \end{align*} $$

Assume the vector v is equal to $v=(a,\ldots ,a)\in {\mathbb R}^k$ for a given $a\in {\mathbb R}$ and denote $D(g_1)_x(v)=(a_1,\ldots ,a_k)$ . If we prove that $|a_j|<|a|$ for every $j=1,\ldots ,k$ , we obtain that $D(g_1)_x$ is a contraction (by taking the norm of the maximum). Let us take a look at the first coordinate $a_1$ :

By taking the absolute value and applying the triangular inequality, we obtain

Notice that $0\leq (x_i+x_j)^2=x_i^2+x_j^2+2x_ix_j$ and this implies $2|x_ix_j|\leq x_i^2+x_j^2\leq r$ . Now, recall that $|\beta '(r)|\leq ({c}/{r}) < ({1-\unicode{x3bb} })/({k(\unicode{x3bb} -\unicode{x3bb} _1)r}) $ and, in particular, we have that

This implies that

Since $\beta $ is a decreasing function, we have that $1=\beta (0)\geq \beta (r)$ and then

The exact same calculation shows that $|a_j|<|a|$ for every $j=1,\ldots ,k$ . This shows that $D(g_1)_x$ is a contraction (for the norm of the maximum) for every $x\in {\mathbb R}^k$ and, in particular, $g_1$ is expansive. Notice that since $\beta (r)=1$ for every $r\in [0,\delta ]$ , we have that $g_1(x)=\unicode{x3bb} x$ for every $x\in B(0,\delta )$ . Now, take the function $h:{\mathbb R}^k\to {\mathbb R}^k$ given by $h(x)=(1-r)x$ , where $r=\lVert x\rVert ^2$ and consider a bump function $\rho :[0,+\infty ) \to {\mathbb R}$ such that:

• • $\rho (t)=1$ for every $t\in [0,\delta /2]$ ;

• • $\rho (t)=0$ for every $t\geq \delta $ .

Now, we define $g:{\mathbb R}^k\to {\mathbb R}^k$ given by the equation:

$$ \begin{align*} g(x)=\rho(r)h(x)+(1-\rho(r))g_1(x), \end{align*} $$

where $r=\lVert x\rVert ^2$ . The first observation is that if $\lVert x\rVert \geq \delta $ , then $g(x)=g_1(x)=Ax$ . However, if $r=\lVert x\rVert ^2\leq \delta $ , then we have that $g_1(x)=\unicode{x3bb} x$ and, therefore,

$$ \begin{align*}g(x)=\rho(r)h(x)+(1-\rho(r))\unicode{x3bb} x=[\rho(r)(1-r)+(1-\rho(r))\unicode{x3bb}]x\end{align*} $$

and the function g is radial. Denote $\alpha (r):=\rho (r)(1-r)+(1-\rho (r))\unicode{x3bb} $ , then it is direct to see that

$$ \begin{align*} \alpha(r)=\rho(r)(1-r)+(1-\rho(r))\unicode{x3bb}=\rho(r)(1-r-\unicode{x3bb})+\unicode{x3bb} \leq 1-r.\end{align*} $$

As a result, g sends every sphere of radius R to a sphere of radius $\alpha (R)R$ which is strictly smaller than R. This implies that $Dg_x$ is a contraction for every $x\in B(0,\delta )$ . To see this, just notice that given $x\in {\mathbb R}^k$ , we have that $T_x{\mathbb R}^k=T_x S_{\lVert x\rVert }+ \langle x \rangle $ , where $S_{\lVert x\rVert }$ is the sphere centred at 0 of radius $\lVert x\rVert $ . The same happens with $g(x)$ , that is, $T_{g(x)}{\mathbb R}^k=T_{g(x)} S_{\alpha (\lVert x\rVert )\lVert x\rVert }+ \langle x \rangle $ and the differential of g at x restricted to this subspace is exactly

$$ \begin{align*}Dg_x|_{T_xS_{\lVert x\rVert}}=\alpha(\lVert x\rVert)\mathrm{Id},\end{align*} $$

which is a contraction. The other direction $\langle x \rangle $ is exactly the same, and, therefore, $Dg_x$ is a contraction. To finish the proof, just observe that if $r=\lVert x\rVert ^2<\delta /2$ , we have that $g(x)=(1-r)x$ and, in particular, $Dg_0=\mathrm {Id}$ .