# construct a closed subspace of $\mathcal{HS}(H)$ such that all elements in the subspace are commutative.

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 $$K$$.

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