TITLE: Existence of spectral gap
QUESTION [7 upvotes]: I would like to start by saying that any comment or idea is highly appreciated.
Let us observe that for Hilbert-Schmidt operators $H_1,H_2$ on an infinite-dimensional separable complex Hilbert space $H$ and a bounded self-adjoint operator $T: H \rightarrow H$
$$\operatorname{Tr} \left(H_1 ([T,H_2])^*\right) = \operatorname{Tr} \left([T,H_1] H_2^* \right).$$ 
In other words, the operator $S_T(H_1):=[T,H_1]$ is self-adjoint with respect to the Hilbert-Schmidt inner product.
Then, we can look at $0 \le S_T^2(H_1) = [T,[T,H]]$
and see that this operator is positive, as it is a square.
Now, I would like to ask the following questions: 
(i)Is there a self-adjoint bounded $T$ such that $S_T^2$ has a spectral gap? (FALSE! See answer by Mateusz Wasilewski.)
(ii)Is there a finite collection of self-adjoint bounded $T_1,...,T_n$ such that $S_{T_1}^2 + ... + S_{T_n}^2$ has a spectral gap?
(iii) Is there an infinite collection of positive bounded $(T_n)$ such that $\sum_n T_n$ converges pointwise such that $\sum_n S_{T_n}^2$ has a spectral gap?
I say $S_T^2$ has a spectral gap if
$$\operatorname{Tr}(S_T^2(H_1)H_1^*) \ge \lambda \operatorname{Tr}(H_1H_1^*)$$ 
for some positive $\lambda>0$.
It looks to me that somebody with a operator algebra background may be more familiar with this problem. I assume the three questions are increasingly likely to be true?

REPLY [7 votes]: The answer to (ii) is positive. Here is a construction, which relies on graphs with spectral gap. For simplicity I write things for trees, but one can probably do the same for more general graphs with spectral gap. 
Start with $X_n$ an $n \geq 3$-regular tree, and label each edge with a label in $\{1,\dots,n\}$ in such a way that no edge with the same label share a vertex. Define $\sigma_i$ the permutation of $X_n$ permuting the endpoints of every edge labelled $i$, and $T_i \colon \xi \in \ell_2(X_n) \mapsto \xi_i \circ \sigma_i$. If you prefer groups, $X_n$ is the Cayley graph of $G_n$, the free product of $n$ copies of $(\mathbf Z/2\mathbf Z)$, $\sigma_i$ are the standard generators of $G_n$ and $T_i =\lambda(\sigma_i)$ for the left-regular representation.
Then $S_{T_i}^2(H) = 2(H-T_i H T_i)$ and $\sum_{i=1}^n S_{T_i}^2(H) = 2n(H-\frac{1}{n} \sum_{i}^n T_i H T_i^*)$ (I write $T_i^*$ eventhough $T_i$ is self-adjoint). So it suffices to show that the map $H \mapsto \frac{1}{n} \sum_{i}^n T_i H T_i^*$ has norm $<1$. But identifying the space of Hilbert-Schmidt operators on $H$ with $H \otimes \overline H$, this operators is $\frac{1}{n} \sum_{i}^n T_i \otimes T_i^*$, which (by Fell's absorption principle) has the same norm as $\frac{1}{n} \sum_{i}^n T_i$, the operator of the Random walk on $X_n$. By a famous computation by Kesten, this norm is $\frac{2 \sqrt{n-1}}{n}$. This gives spectral gap $\lambda = 2n-4\sqrt{n-1}$.
A last remark: this method cannot be applied to $n=2$ (because it constructs  self-adjoint unitaries $T_1,\dots,T_n$, and for $2$ unitaries there cannot be spectral gap). I am sure that (ii) has already positive answer for $n=2$, but this would require a different construction.