Let $H$ be a separable infinite-dimensional Hilbert space and $S$ be the left shift operator. Is there any operator $T$ in $B(H)$ such that $TS=ST$ except $\mathbb{C}\textbf{1}$ or $0$?
Asked
Active
Viewed 244 times
1
-
I may be mistaken, but I believe $S$ is an operator in $B(H)$ and does commute with itself, right? – Matthias Klupsch Sep 14 '16 at 07:47
-
@Matthias Klupsch Sorry！so, thx, i will keep thinking – user92646 Sep 14 '16 at 08:12
-
(assuming $S$ is bounded) it commutes with the convolution operators $T = \sum_{k=a}^b c_k S^k$ or $T = \int_a^b c(x) S^x dx$ – reuns Sep 14 '16 at 08:24
-
@usee1952009 THX！ it does help！ – user92646 Sep 14 '16 at 08:32
-
What do you mean by the left shift operator on a general separable infinite-dimensional Hilbert space? – Christian Sep 14 '16 at 08:40
-
@Christian it is not accurate should be l_2 – user92646 Sep 14 '16 at 09:05
-
see also this question http://math.stackexchange.com/questions/647866/in-ellp-if-an-operator-commutes-with-left-shift-it-is-continuous – daw Sep 14 '16 at 09:17
-
It might be worth mentioning that your guess is correct if you require $T$ to commute with both $S$ and $S^$. That is, if $TS = ST$ and $TS^ = S^*T$, then $T \in \mathbb{C}1$. See, for example, the remark after Theorem 4.1.12 here. – Josh Keneda Sep 14 '16 at 15:21
-
@JoshKeneda wOO, it could be useful! Thx! – user92646 Sep 15 '16 at 01:22
-
Are you talking about a shift on $\ell^2(\mathbb{N})$ or on $\ell^2(\mathbb{Z})$? It makes a difference because the first $S$ is not normal, while the second is unitary. – Disintegrating By Parts Sep 15 '16 at 02:24
-
@TrialAndError Actually, I just wanna find a case such that my guess holds true. If $l^2(\mathbb{N})$ holds, I will take it, if not, I will take the other one. – user92646 Sep 15 '16 at 04:26