Irrational rotation and recurrence time Let $\mathbb{T}=\mathbb{R}/\mathbb{Z}$ be the torus, and $\alpha\in(0,1)$ be an irrational number, then the transformation $T$ defined by $Tx=x+\alpha$ is the irrational rotation on $\mathbb{T}$. 
Now for a neighborhood $(-\epsilon,\epsilon)\pmod 1$ of the point $0$ we consider the set of the recurrence time $$A = \{n:T^n(0)\in (-\epsilon,\epsilon)\pmod 1 \} = \{n:n\alpha\in(-\epsilon,\epsilon)\pmod 1 \} \;.$$ For a quadratic polynomial, for example $n^2$, I want to ask whether we can find two numbers $\beta,\delta\in (0,1)$,such that the set of polynomial recurrence time $\{n:n^2\beta\in(-\delta,\delta)\pmod 1\}$ is contained in the set $A$.

No, because for $\epsilon$ small enough, we can put a lower bound in the differences of the elements in $A$, while in the set of polynomial recurrence time with $n^2 \beta$, there always are some consecutive numbers :


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*If $n_1 \alpha$ and $n_2 \alpha \pmod 1$ are in $(- \epsilon, \epsilon)$, then $(n_2-n_1)\alpha \in (- 2\epsilon, 2\epsilon) \pmod 1$. If $N$ is an integer, then since $\alpha$ is irrational, $\min \{d(k\alpha,\mathbb Z)$ where $0\le k<N\} > 0$, and by picking $\epsilon$ smaller than this number, we force the elements of $A$ to always be at least $N$ apart.

*For any irrational $\beta$, and any $\delta > 0$, since $\{(n\beta,n^2\beta)\}$ is equidistributed, there are infinitely many $n$ such that $n\beta \in (- \delta/4-\beta, \delta/4+\beta)$ and $n^2\beta \in (- \delta/2, \delta/2)$, and in this case, $(n+1)^2\beta = n^2\beta + 2n\beta + \beta \in (-\delta,\delta)$, so both $n$ and $n+1$ are in the set of recurrent times. 
Therefore, in most cases the second set will not be included in the first set.
(and of course if you pick $\epsilon > 1/2$, then $A = \mathbb N$ so it will contain anything)
