Let $H$ be a Hilbert space and $B(H)$ be the space of all bounded operators on $H$.


The Wold decomposition says that: an operator $x$ in $B(H)$ is an isometry if and only if $x=x\_u\oplus x\_s$ where $x\_u$ and $x\_s$ are unitary and unilateral shift respectively. Indeed $x\_u$ is the restriction of $x$ to $H\_u=\bigcap x^nH$ and $x\_s$ is the restriction of $x$ to $H\_s$ where,  

$$H\_s:=H\ominus H\_u=\bigoplus\_{n\geq0} (x^nH\ominus x^{n+1}H)$$


For $x\in B(H)$, let us denote $e\_x$ by the projection onto $xH$.


Q. I am looking for a commuting pair of isometries $(x\_1,x\_2)$ in $B(H)$ such that all the following hold:


i) $e\_y$ commutes with all projections $e\_{x^n}$ for $n\geq1$.


ii) $H\_s^{(x)}:=H\ominus (\cap x^nH)$ is not $y$-invariant.



A commuting pair of isometries
Such a pair $(X,Y)$ is constructed as follows. Consider a Hilbert space $M$ with an orthonormal basis $\{e\_n:n\in\mathbb Z\}$ and the bilateral shift $U$ on $M$ such that $Ue\_n=e\_{n+1}$. Denote by $S$ the restriction of $U$ to the space $N$ generated by $\{e\_n:n\geq 0\}$, and construct the space $H=N\oplus M\oplus M\oplus\cdots$, that is, the space of square summable sequences $(m\_0,m\_1,\dots)$ such that $m\_0\in N$ and $m\_j\in M$ for $j>0$. Now we define $X,Y$ by setting
$$X(m\_0,m\_1,\dots)=(Sm\_0,Um\_1,Um\_2,\dots),$$ and
$$Y(m\_0,m\_1,\dots)=(0,m\_0,m\_1,\dots).$$
The range projections are easily calculated: $$YY^\*(m\_0,m\_1,\dots)=(0,P\_Nm\_1,m\_2,\dots),$$ and
$$X^nX^{\*n}(m\_0,m\_1,\dots)=(S^nS^{\*n}m\_0,m\_1,\dots).$$
It is trivial to verify that $YY^\*$ commutes with $X^nX^{\*n}$ for every $n$, but $Y$ does not leave the space
$$H\ominus\left(\bigcap\_n X^nH\right)=\{(m\_0,0,0,\dots):m\_0\in N\}$$
invariant.


This construction is based on an idea suggested by Ronald Douglas and formalized in *Canonical models for bi-isometries* ([actual paper](https://doi.org/10.1007/978-3-0348-0221-5_7), [MR review](https://mathscinet.ams.org/mathscinet/search/publdoc.html?arg3=&co4=AND&co5=AND&co6=AND&co7=AND&dr=all&pg4=AUCN&pg5=TI&pg6=PC&pg7=JOUR&pg8=ET&review_format=html&s4=bercovici%20and%20douglas&s5=&s6=&s7=&s8=All&sort=Newest&vfpref=html&yearRangeFirst=&yearRangeSecond=&yrop=eq&r=1&mx-pid=2931928)).


