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 $ST=TS?$

up vote 1 down vote accepted

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 $K$.

  • Very clear!Appreciate it. – mathrookie Sep 22 at 22:27

Your Answer

 

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.