[proofplan] Lift the pseudo-orbit from the torus to $\mathbb{R}^d$ so that the one-step errors become honest vectors of uniformly small Euclidean norm. The hyperbolicity of $A$ gives an invariant splitting $\mathbb{R}^d = E^s \oplus E^u$ and exponential estimates for forward iterates on $E^s$ and backward iterates on $E^u$. We solve the inhomogeneous recurrence $c_{k+1} = Ac_k + e_k$ by summing past stable errors and future unstable errors; subtracting this bounded correction from the lifted pseudo-orbit gives an exact lifted orbit. The uniform bound on the correction then projects to the desired $\varepsilon$-shadowing orbit on $\mathbb{T}^d$. [/proofplan] [step:Lift the pseudo-orbit to a sequence in $\mathbb{R}^d$ with small additive errors] Let $\pi: \mathbb{R}^d \to \mathbb{T}^d$ denote the quotient map, $\pi(y) = [y]$. Let $(x_k)_{k \in \mathbb{Z}}$ be a $\delta$-pseudo-orbit, where $\delta < 1/2$ will be fixed later. Choose any lift $y_0 \in \mathbb{R}^d$ with $\pi(y_0)=x_0$. We define inductively lifts $y_k \in \mathbb{R}^d$ for all $k \in \mathbb{Z}$ as follows. If $y_k$ is chosen, then the inequality \begin{align*} d_{\mathbb{T}^d}(\pi(Ay_k),x_{k+1}) < \delta \end{align*} means that there is a lift $y_{k+1} \in \mathbb{R}^d$ of $x_{k+1}$ such that \begin{align*} |y_{k+1}-Ay_k| < \delta. \end{align*} Similarly, moving backward, if $y_k$ is chosen, then the inequality \begin{align*} d_{\mathbb{T}^d}(f_A(x_{k-1}),x_k) < \delta \end{align*} means that there is a lift $w_{k-1} \in \mathbb{R}^d$ of $x_{k-1}$ and a vector $m \in \mathbb{Z}^d$ such that \begin{align*} |Aw_{k-1}-y_k-m| < \delta. \end{align*} Since $A \in GL(d,\mathbb{Z})$, we have $A^{-1}m \in \mathbb{Z}^d$. Define $y_{k-1}:=w_{k-1}-A^{-1}m$. Then $y_{k-1}$ is still a lift of $x_{k-1}$ and \begin{align*} |y_k-Ay_{k-1}| < \delta. \end{align*} Thus we obtain a bi-infinite lifted sequence $(y_k)_{k \in \mathbb{Z}}$ satisfying $\pi(y_k)=x_k$ for every $k \in \mathbb{Z}$. Define the error sequence $e: \mathbb{Z} \to \mathbb{R}^d$ by \begin{align*} e_k := y_{k+1}-Ay_k. \end{align*} Then \begin{align*} |e_k| < \delta \qquad \text{for every } k \in \mathbb{Z}. \end{align*} [guided] The torus point $x_k$ is a coset in $\mathbb{R}^d/\mathbb{Z}^d$, so before doing linear algebra we choose representatives in $\mathbb{R}^d$. Let $\pi: \mathbb{R}^d \to \mathbb{T}^d$ be the quotient map, $\pi(y)=[y]$. Choose one representative $y_0 \in \mathbb{R}^d$ with $\pi(y_0)=x_0$. The pseudo-orbit condition says that $x_{k+1}$ is close to $f_A(x_k)$. If $y_k$ represents $x_k$, then $Ay_k$ represents $f_A(x_k)$. Since \begin{align*} d_{\mathbb{T}^d}(\pi(Ay_k),x_{k+1}) < \delta, \end{align*} the definition of the quotient metric gives a representative $y_{k+1}$ of $x_{k+1}$ satisfying \begin{align*} |y_{k+1}-Ay_k| < \delta. \end{align*} This constructs the lifts forward in time. For negative indices, we use that $A \in GL(d,\mathbb{Z})$, so $A^{-1}\mathbb{Z}^d=\mathbb{Z}^d$. If $y_k$ is already chosen, the condition \begin{align*} d_{\mathbb{T}^d}(f_A(x_{k-1}),x_k) < \delta \end{align*} gives a lift $w_{k-1}$ of $x_{k-1}$ and a vector $m \in \mathbb{Z}^d$ such that $|Aw_{k-1}-y_k-m|<\delta$. Setting $y_{k-1}:=w_{k-1}-A^{-1}m$ keeps the same torus point because $A^{-1}m \in \mathbb{Z}^d$, and it gives \begin{align*} |y_k-Ay_{k-1}| < \delta. \end{align*} Repeating this gives a bi-infinite lift $(y_k)_{k \in \mathbb{Z}}$. Now define the additive error vector \begin{align*} e_k := y_{k+1}-Ay_k. \end{align*} The construction gives the uniform estimate \begin{align*} |e_k| < \delta \qquad \text{for every } k \in \mathbb{Z}. \end{align*} This is the point of lifting: the pseudo-orbit condition on the quotient has become a linear recurrence in $\mathbb{R}^d$ with a small forcing term. [/guided] [/step] [step:Decompose the errors along the stable and unstable subspaces] Since $A$ has no eigenvalue of modulus $1$, the real [vector space](/page/Vector%20Space) $\mathbb{R}^d$ admits an $A$-invariant hyperbolic splitting \begin{align*} \mathbb{R}^d = E^s \oplus E^u, \end{align*} where $A|_{E^s}$ has spectral radius strictly less than $1$ and $A^{-1}|_{E^u}$ has spectral radius strictly less than $1$. Hence there exist constants $C_0 \ge 1$ and $\lambda \in (0,1)$ such that \begin{align*} |A^n v^s| \le C_0 \lambda^n |v^s| \qquad \text{for all } v^s \in E^s,\ n \in \mathbb{N} \cup \{0\}, \end{align*} and \begin{align*} |A^{-n} v^u| \le C_0 \lambda^n |v^u| \qquad \text{for all } v^u \in E^u,\ n \in \mathbb{N} \cup \{0\}. \end{align*} Let $P^s: \mathbb{R}^d \to E^s$ and $P^u: \mathbb{R}^d \to E^u$ be the linear projections associated to the direct sum $\mathbb{R}^d = E^s \oplus E^u$. Define \begin{align*} M := \max\{\|P^s\|_{\mathrm{op}},\|P^u\|_{\mathrm{op}}\}. \end{align*} For each $k \in \mathbb{Z}$ define \begin{align*} e_k^s := P^s e_k, \qquad e_k^u := P^u e_k. \end{align*} Then $e_k = e_k^s + e_k^u$, with $e_k^s \in E^s$, $e_k^u \in E^u$, and \begin{align*} |e_k^s| \le M\delta, \qquad |e_k^u| \le M\delta \end{align*} for every $k \in \mathbb{Z}$. [/step] [step:Construct a bounded correction sequence by summing stable past errors and unstable future errors] Define $c: \mathbb{Z} \to \mathbb{R}^d$ by \begin{align*} c_k := \sum_{j=-\infty}^{k-1} A^{k-1-j} e_j^s - \sum_{j=k}^{\infty} A^{k-1-j} e_j^u. \end{align*} The first series lies in $E^s$ and converges absolutely because \begin{align*} |A^{k-1-j} e_j^s| \le C_0 \lambda^{k-1-j} M\delta \qquad \text{for } j \le k-1. \end{align*} The second series lies in $E^u$ and converges absolutely because, for $j \ge k$, \begin{align*} A^{k-1-j} = A^{-(j-k+1)} \end{align*} on $E^u$, and therefore \begin{align*} |A^{k-1-j} e_j^u| \le C_0 \lambda^{j-k+1} M\delta. \end{align*} Consequently $c_k$ is well-defined for every $k \in \mathbb{Z}$ and satisfies \begin{align*} |c_k| \le \sum_{m=0}^{\infty} C_0\lambda^m M\delta + \sum_{m=1}^{\infty} C_0\lambda^m M\delta. \end{align*} Thus \begin{align*} |c_k| \le C_1 \delta \qquad \text{for every } k \in \mathbb{Z}, \end{align*} where \begin{align*} C_1 := C_0 M\left(\frac{1}{1-\lambda}+\frac{\lambda}{1-\lambda}\right) = C_0M\frac{1+\lambda}{1-\lambda}. \end{align*} [guided] We now solve the inhomogeneous recurrence \begin{align*} c_{k+1}=Ac_k+e_k. \end{align*} Why should the correction involve past stable errors and future unstable errors? A stable vector becomes small under positive powers of $A$, so a stable error made far in the past can be transported forward to time $k$ with a convergent sum. An unstable vector becomes small under negative powers of $A$, so an unstable error made far in the future can be transported backward to time $k$ with a convergent sum. Define \begin{align*} c_k := \sum_{j=-\infty}^{k-1} A^{k-1-j} e_j^s - \sum_{j=k}^{\infty} A^{k-1-j} e_j^u. \end{align*} The stable part is a sum over $j \le k-1$. For such $j$, the exponent $k-1-j$ is a nonnegative integer, and the stable estimate gives \begin{align*} |A^{k-1-j} e_j^s| \le C_0 \lambda^{k-1-j} |e_j^s|. \end{align*} Since $|e_j^s| \le M\delta$, we obtain \begin{align*} |A^{k-1-j} e_j^s| \le C_0 \lambda^{k-1-j} M\delta. \end{align*} The geometric series $\sum_{m=0}^{\infty}\lambda^m$ converges because $0<\lambda<1$, so the stable series converges absolutely. For the unstable part, $j \ge k$, so \begin{align*} A^{k-1-j}=A^{-(j-k+1)} \end{align*} on $E^u$. The unstable estimate is precisely an estimate for negative iterates on $E^u$, hence \begin{align*} |A^{k-1-j}e_j^u| \le C_0\lambda^{j-k+1}|e_j^u| \le C_0\lambda^{j-k+1}M\delta. \end{align*} Again this is bounded by a convergent geometric series. Combining the two estimates gives \begin{align*} |c_k| \le \sum_{m=0}^{\infty} C_0\lambda^m M\delta + \sum_{m=1}^{\infty} C_0\lambda^m M\delta. \end{align*} Evaluating the geometric sums, \begin{align*} \sum_{m=0}^{\infty}\lambda^m = \frac{1}{1-\lambda} \end{align*} and \begin{align*} \sum_{m=1}^{\infty}\lambda^m = \frac{\lambda}{1-\lambda}. \end{align*} Therefore \begin{align*} |c_k| \le C_0M\left(\frac{1}{1-\lambda}+\frac{\lambda}{1-\lambda}\right)\delta = C_0M\frac{1+\lambda}{1-\lambda}\delta. \end{align*} Thus the correction sequence is uniformly bounded. This uniform bound is what will become the shadowing estimate on the torus. [/guided] [/step] [step:Subtract the correction to obtain an exact lifted orbit] We claim that the correction sequence satisfies \begin{align*} c_{k+1}=Ac_k+e_k \qquad \text{for every } k \in \mathbb{Z}. \end{align*} Indeed, using absolute convergence to reindex the stable series, \begin{align*} c_{k+1}^s := \sum_{j=-\infty}^{k} A^{k-j}e_j^s = A\sum_{j=-\infty}^{k-1} A^{k-1-j}e_j^s + e_k^s. \end{align*} For the unstable series, \begin{align*} c_{k+1}^u := - \sum_{j=k+1}^{\infty} A^{k-j}e_j^u = A\left(-\sum_{j=k}^{\infty}A^{k-1-j}e_j^u\right)+e_k^u. \end{align*} Adding the stable and unstable identities gives \begin{align*} c_{k+1}=Ac_k+e_k. \end{align*} Define $z: \mathbb{Z} \to \mathbb{R}^d$ by \begin{align*} z_k := y_k-c_k. \end{align*} Then \begin{align*} z_{k+1} = y_{k+1}-c_{k+1} = Ay_k+e_k-(Ac_k+e_k) = A(y_k-c_k) = Az_k. \end{align*} Thus $(z_k)_{k \in \mathbb{Z}}$ is an exact orbit of the [linear map](/page/Linear%20Map) $A$ on $\mathbb{R}^d$, and in particular \begin{align*} z_k=A^k z_0 \qquad \text{for every } k \in \mathbb{Z}. \end{align*} [/step] [step:Choose $\delta$ so the exact orbit shadows after projection to the torus] Let $\varepsilon>0$ be sufficiently small. Choose \begin{align*} \delta := \min\left\{\frac{1}{2},\frac{\varepsilon}{2C_1}\right\}. \end{align*} For each $k \in \mathbb{Z}$, the quotient metric is bounded above by Euclidean distance between any chosen lifts, hence \begin{align*} d_{\mathbb{T}^d}(\pi(z_k),\pi(y_k)) \le |z_k-y_k| = |c_k| \le C_1\delta \le \frac{\varepsilon}{2} < \varepsilon. \end{align*} Set \begin{align*} x := \pi(z_0) \in \mathbb{T}^d. \end{align*} Since $z_k=A^kz_0$, we have \begin{align*} f_A^k(x)=\pi(A^kz_0)=\pi(z_k) \qquad \text{for every } k \in \mathbb{Z}. \end{align*} Also $\pi(y_k)=x_k$ by construction. Therefore \begin{align*} d_{\mathbb{T}^d}(f_A^k(x),x_k) = d_{\mathbb{T}^d}(\pi(z_k),\pi(y_k)) < \varepsilon \qquad \text{for every } k \in \mathbb{Z}. \end{align*} Thus the true orbit of $x$ under $f_A$ $\varepsilon$-shadows the given bi-infinite $\delta$-pseudo-orbit. This proves the theorem. [/step]