This question is a doubt arising from the statement and the proof of proposition described below ( Book: Higher Order fourier analysis, by T. Tao) Suppose $\alpha \in \mathbf{T}^d$. Then,
Irrational :$\alpha$ is irrational, in the sense that $k\cdot \alpha\neq 0$ for any non-zero $k \in \mathbf{Z}^d$.
Rational: if $\alpha$ is rational in the sense that $m\alpha=0$ for some positive integer $m$, then the sequence $n\alpha +\beta$ is clearly periodic of period $m$.
Proposition 1.1.5 (Equidistribution for abelian linear sequences). Let $T$ be a torus, and let $x(n):=n \alpha+\beta$ for some $\alpha, \beta \in T$. Then there exists a decomposition $x=x^{\prime}+x^{\prime \prime}$, where $x^{\prime}(n):=n \alpha^{\prime}$ is totally asymptotically equidistributed on $\mathbf{Z}$ in a subtorus $T^{\prime}$ of $T$ (with $\alpha^{\prime} \in T^{\prime}$, of course), and $x^{\prime \prime}(n)=n \alpha^{\prime \prime}+\beta$ is periodic (or equivalently, that $\alpha^{\prime \prime} \in T$ is rational).
Proof. We induct on the dimension $d$ of the torus $T$. The claim is vacuous for $d=0$, so suppose that $d \geq 1$ and that the claim has already been proven for tori of smaller dimension. Without loss of generality we may identify $T$ with $\mathbf{T}^{d}$.
If $\alpha$ is irrational, then we know that $n\alpha + \beta$ is asymptotically equidistributed and hence we are done (from Ex 1.1.5) , so we may assume that $\alpha$ is not irrational; thus $k \cdot \alpha=0$ for some non-zero $k \in \mathbf{Z}^{d}$. We then write $k=m k^{\prime}$, where $m$ is a positive integer and $k^{\prime} \in \mathbf{Z}^{d}$ is irreducible (i.e., $k^{\prime}$ is not a proper multiple of any other element of $\mathbf{Z}^{d}$ ); thus $k^{\prime} \cdot \alpha$ is rational. We may thus write $\alpha=\alpha_{1}+\alpha_{2}$, where $\alpha_{2}$ is rational, and $k^{\prime} \cdot \alpha_{1}=0$. Thus, we can split $x=x_{1}+x_{2}$, where $x_{1}(n):=n \alpha_{1}$ and $x_{2}(n):=n \alpha_{2}+\beta$. Clearly $x_{2}$ is periodic, while $x_{1}$ takes values in the subtorus $T_{1}:=\left\{y \in T: k^{\prime} \cdot y=0\right\}$ of $T$. The claim now follows by applying the induction hypothesis to $T_{1}$ (and noting that the sum of two periodic sequences is again periodic).
Questions:
We are applying induction on the dimension $d$ of the torus. Then in the end of the proof, we have that $x_1$ takes values in the subtorus $T_1:=\{y\in T : k'\cdot y=0\}$ of $T$. But how are we able to apply induction hypothesis in $T_1$? How does one show that this subtorus $T_1$ has lower dimension than $T$? Otherwise, how come applying induction hypothesis makes sense?
In the statement of the proposition, we have that $x'(n)$ is totally asymptotically equidistributed on $\mathbf{Z}$ in a subtorus $T'$ of $T$. How is this subtorus defined?
In the proof of proposition we have defined the subtorus $T_1$ and then we apply induction hypothesis, but no where it is described how subtorus $T'$ is defined?
Thankyou!