Suppose $K=\mathcal{HS}(H)$,where$\mathcal{HS}(H)$ is the set of all Hilbert Schmidt operators on the Hilbert space $H$.I have two questions.

1.Can we construct a closed subspace $K_1$ of $K$ such that all elements in $K_1$ are commutative?

2.Does there exists a closed subspace $K_1$ of $K$ such that for any $T \in B(H),S \in K,$ we have $ST=TS?$


Hopefully I am not saying something stupid.

1) For each $(a_n) \in l^2$ define $$T(e_j)=a_je_j$$

Then the set of such operators is, I think a closed commutative subspace of $K$.

2) The answer is no. Indeed we follow the standard proof of a matrix is in the center of $\mathcal M_{n}$ if and only if is a scalar multiple of identity.

Pick $T(e_j)=a_je_j \forall j$ is a "diagonal" operator in $B(H)$ with $a_j \neq a_k$ for $j \neq k$. Such an operator can easily be made continuous.

Then, a simple computation shows that $TS=ST \Rightarrow S$ is diagonal. Indeed, for a fixd $j$

$$ a_j S(e_j)= ST(e_j)= TS(e_j) $$

So, if $S(e_j)= \sum_k b_ke_k$ then $$ \sum_k a_j b_ke_k =\sum_k a_k b_k ek \Rightarrow \sum_k (a_j-a_k) b_k e_k =0 \Rightarrow \\ b_k=0 \forall k \neq j \Rightarrow S(e_j)=b_j e_j$$

This gives that $K_1$ would need to consist only of diagonal operators. Pick such an operator : $S$ diagonal with $S(e_j)=b_je_j$

Next, if you use $T$ to be a transposition (fix $k \neq l$ and define $T(e_k)=e_l, T(e_l)=e_k, T(e_j)=e_j \forall j \neq k,l$) then $TS=ST$ if and only if $$b_k=b_l$$

This yields $S=b Id$ which is not in $K$.

  • $\begingroup$ Very clear!Appreciate it. $\endgroup$ – math112358 Sep 22 '18 at 22:27

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