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I'm assuming a Li-Yorke pair for $T$ is a pair $x,y$ such that $\limsup_{n\to \infty} d(T^n(x), T^n(y)) > 0 $ and $\liminf_{n\to \infty} d(T^n(x), T^n(y)) = 0$.

For toral automorphisms, I think the answer to your question is no, based on this:

Claim: A linear toral automorphism has a Li-Yorke pair if and only if it is ergodic.

Since ergodic linear toral automorphisms have positive entropy (see for instance here), the claim answers your question negatively.

Proof of the claim: The if part is easy, since a point with a dense orbit and a fixed point give a Li-Yorke pair. Now assume $T$ is not ergodic but has a Li-Yorke pair. Being non-ergodic implies that there is an eigenvalue which is a root of the unity. We may replace $T$ by some power of $T$ and assume that $1$ is an eigenvalue. Moreover you can find an integer eigenvector $v$ with eigenvalue $1$ (because the coefficients of the matrix corresponding to $T$ are integer), so passing to the quotient by $\mathbb{R}v$ you get a torus of dimension $n-1$. Since $T$ had a Li-Yorke pair, so does the automorphism induced by $T$ on the quotient, so you have reduced the dimension. The claim is trivial in dimension $1$, so the proof is completed by induction.

I'm assuming a Li-Yorke pair for $T$ is a pair $x,y$ such that $\limsup_{n\to \infty} d(T^n(x), T^n(y)) > 0 $ and $\liminf_{n\to \infty} d(T^n(x), T^n(y)) = 0$.

For toral automorphisms, I think the answer to your question is no. The idea is this: if $T$ is ergodic, then it has positive entropy (every ergodic linear toral automorphism does, see for instance here). If $T$ is not ergodic, you can find a factor of (some power of) $T$ on a torus of lower dimension which is ergodic, which means again that $T$ has positive entropy.

Indeed, being non-ergodic implies that there is an eigenvalue which is a root of the unity. We may replace $T$ by some power of $T$ and assume that $1$ is an eigenvalue. Moreover you can find an integer eigenvector $v$ with eigenvalue $1$ (because the coefficients of the matrix corresponding to $T$ are integer), so passing to the quotient by $\mathbb{R}v$ you should get a torus of dimension $n-1$. Since $T$ had a Li-Yorke pair, so does the automorphism induced by $T$ on the quotient, so you have reduced the dimension. If the new map is still not ergodic, we may repeat this reduction. But this process has to stop before reaching dimension $1$ since you cannot have a Li-Yorke pair in dimension $1$.

Note: I edited this answer since the previous one had a big mistake (the claim was false)

I'm assuming a Li-Yorke pair for $T$ is a pair $x,y$ such that $\limsup_{n\to \infty} d(T^n(x), T^n(y)) > 0 $ and $\liminf_{n\to \infty} d(T^n(x), T^n(y)) = 0$.

For toral automorphisms, I think the answer to your question is no, based on this:

Claim: A linear toral automorphism has a Li-Yorke pair if and only if it is ergodic.

Since ergodic linear toral automorphisms have positive entropy (see for instance here), the claim answers your question negatively.

Proof of the claim: The if part is easy, since a point with a dense orbit and a fixed point give a Li-Yorke pair. Now assume $T$ is not ergodic but has a Li-Yorke pair. Being non-ergodic implies that there is an eigenvalue which is a root of the unity. We may replace $T$ by some power of $T$ and assume that $1$ is an eigenvalue. Moreover you can find an integer eigenvector $v$ with eigenvalue $1$ (because the coefficients of the matrix corresponding to $T$ are integer), so passing to the quotient by $Rv$ you get a torus of dimension $n-1$. Since $T$ had a Li-Yorke pair, so does the automorphism induced by $T$ on the quotient, so you have reduced the dimension. The claim is trivial in dimension $1$, so the proof is completed by induction.

I'm assuming a Li-Yorke pair for $T$ is a pair $x,y$ such that $\limsup_{n\to \infty} d(T^n(x), T^n(y)) > 0 $ and $\liminf_{n\to \infty} d(T^n(x), T^n(y)) = 0$.

For toral automorphisms, I think the answer to your question is no, based on this:

Claim: A linear toral automorphism has a Li-Yorke pair if and only if it is ergodic.

Since ergodic linear toral automorphisms have positive entropy (see for instance here), the claim answers your question negatively.

Proof of the claim: The if part is easy, since a point with a dense orbit and a fixed point give a Li-Yorke pair. Now assume $T$ is not ergodic but has a Li-Yorke pair. Being non-ergodic implies that there is an eigenvalue which is a root of the unity. We may replace $T$ by some power of $T$ and assume that $1$ is an eigenvalue. Moreover you can find an integer eigenvector $v$ with eigenvalue $1$ (because the coefficients of the matrix corresponding to $T$ are integer), so passing to the quotient by $\mathbb{R}v$ you get a torus of dimension $n-1$. Since $T$ had a Li-Yorke pair, so does the automorphism induced by $T$ on the quotient, so you have reduced the dimension. The claim is trivial in dimension $1$, so the proof is completed by induction.

I'm assuming a Li-Yorke pair for $T$ is a pair $x,y$ such that $\limsup_{n\to \infty} d(T^n(x), T^n(y)) > 0 $ and $\liminf_{n\to \infty} d(T^n(x), T^n(y)) = 0$.

For toral automorphisms, I think the answer to your question is no, based on this:

Claim: A linear toral automorphism has a Li-Yorke pair if and only if it is ergodic.

Since ergodic linear toral automorphisms have positive entropy (see for instance here), the claim answers your question negatively.

Proof of the claim: The if part is easy, since a point with a dense orbit and a fixed point give a Li-Yorke pair. Now assume $T$ is not ergodic but has a Li-Yorke pair. Being non-ergodic implies that there is an eigenvalue which is a root of the unity. We may replace $T$ by some power of $T$ and assume that $1$ is an eigenvalue. This implies that the dynamics fibers over a circle: you can find an integer eigenvector with eigenvalue $1$, and after a linear change of coordinates you may assume the dynamics is $(x,y) \mapsto (x, T'y)$ for $(x,y)\in \mathbb{T}\times \mathbb{T}^{d-1}$. Clearly if you have a Li-Yorke pair, both points must belong to the same fiber, and the automorphism $T'$ of $\mathbb{T}^{d-1}$ also has a Li-Yorke pair. So you have reduced the dimension. The claim is trivial in dimension $1$, so the proof is completed by induction.

I'm assuming a Li-Yorke pair for $T$ is a pair $x,y$ such that $\limsup_{n\to \infty} d(T^n(x), T^n(y)) > 0 $ and $\liminf_{n\to \infty} d(T^n(x), T^n(y)) = 0$.

For toral automorphisms, I think the answer to your question is no, based on this:

Claim: A linear toral automorphism has a Li-Yorke pair if and only if it is ergodic.

Since ergodic linear toral automorphisms have positive entropy (see for instance here), the claim answers your question negatively.

Proof of the claim: The if part is easy, since a point with a dense orbit and a fixed point give a Li-Yorke pair. Now assume $T$ is not ergodic but has a Li-Yorke pair. Being non-ergodic implies that there is an eigenvalue which is a root of the unity. We may replace $T$ by some power of $T$ and assume that $1$ is an eigenvalue. Moreover you can find an integer eigenvector $v$ with eigenvalue $1$ (because the coefficients of the matrix corresponding to $T$ are integer), so passing to the quotient by $Rv$ you get a torus of dimension $n-1$. Since $T$ had a Li-Yorke pair, so does the automorphism induced by $T$ on the quotient, so you have reduced the dimension. The claim is trivial in dimension $1$, so the proof is completed by induction.