Let $H$ be an infinite dimensional, separable complex Hilbert space with orthonormal basis $(e_n)_n$.
Let $T$ be the unilateral forward shift with $Te_n= t_n e_{n+1}$, where $(t_n) \in \ell^{\infty}$
Let $\mathcal{K}(H)$ denote the set of compact operators acting on $H$.
Let $\pi:\mathcal{B(H)}\rightarrow\mathcal{B(H)}/\mathcal{K(H)}$ be the canonical quotient map,
Note that $\pi(T^*)= \pi(T)^*$.
Find a necessary condition on the sequence $(t_n)$ so that $\pi(T^*)\pi(T)=\pi(T)\pi(T^*)$.
I suspect that the condition will have something to do we the normality and compactness of $T$. But I'm not sure about this.
Any ideas or help will be appreciated!
Since $\pi$ is multiplicative, the right point of view is to look at $\pi(T^*T)=\pi(TT^*)$. Even better, looking at $\pi(T^*T-TT^*)=0$, the condition required is that $T^*T-TT^*$ is compact.
In terms of the matrix units $\{E_{kj}\}$ coming from $\{e_n\}$, $$ T=\sum_k t_kE_{k+1,k},\qquad\qquad T^*=\sum_k\overline{t_k}\,E_{k,k+1}. $$ Then $$ T^*T=\sum_k|t_k|^2\,E_{kk},\qquad\qquad TT^*=\sum_k|t_k|^2\,E_{k+1,k+1}, $$ and $$ T^*T-TT^*=|t_1|^2\,E_{11}+\sum_{k>1}(|t_k|^2-|t_{k-1}|^2)\,E_{kk}. $$ This is a diagonal operator, and it is compact if and only if $\lim|t_k|^2-|t_{k-1}|^2=0$.
Edit: Working on the basis elements
$$ \langle T^*Te_n,e_m\rangle=\langle Te_n,Te_m\rangle=t_n\overline{t_m}\langle e_{n+1},e_{m+1}\rangle=\langle |t_n|^2e_n,e_m\rangle, $$ so $T^*Te_n=|t_n|^2e_n$. For $n,m>1$, $$ \langle TT^*e_n,e_m\rangle=\langle T\overline{t_{n-1}}e_{n-1},e_m\rangle=\langle |t_{n-1}|^2e_n,e_m\rangle. $$ So, $$ (T^*T-TT^*)e_1=T^*Te_1=|t_1|^2e_1, $$ and $$\tag{$*$} (T^*T-TT^*)e_n=(|t_n|^2-|t_{n-1}|^2)e_n,\qquad\qquad n>1. $$