Existence of spectral gap

Question

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?

Answer 1

The answer to question (i) is negative. First of all, I'd like to remark that the shouldn't be an $i$ in the definition of $S_T$ therefore for me $S_T^2(H)$ will be $[T,[T,H]]$.

Let us first consider the case in which $T=\textrm{diag}(c_i)$ is a diagonal operator. Then $\textrm{Tr}(S_T^2(H) H^{\ast}) = \sum_{i,j} (c_i-c_j)^2 |h_{ij}|^2$, where $H=(h_{ij})$. Now $H=e_{ii}$ is in the kernel of $T$, so there is no spectral gap.

To prove it in the general case, assume that $T$ is self-adjoint contraction such that $S_T^2$ has a spectral gap $\lambda>0$, which means that for any $H$ with $\textrm{Tr}(H H^{\ast})=1$ we have $\textrm{Tr}(S_T^2(H) H^{\ast}) \geqslant \lambda$. Now, by the Weyl-von Neumann theorem, we can write (up to unitary conjugacy which is harmless) $T= D + \varepsilon$, where $D=\textrm{diag}(d_i)$ is diagonal and $\varepsilon$ is Hilbert-Schmidt with arbitrarily small Hilbert-Schmidt norm. We can compute $S_T^2(H) = [D,[D,H]] + [\varepsilon,[D,H]]+[D,[\varepsilon,H]] + [\varepsilon,[\varepsilon,H]]$. By our assumption on $H$ and $T$, the norm of $D$ is at most slightly above, and $H$ has Hilbert-Schmidt norm equal to $1$, so the Hilbert-Schmidt norm of the last three terms can be made arbitrarily small. If we assume that the Hilbert-Schmidt norm of $\varepsilon$ is small enough, this would force $S_D^2$ to have a spectral gap, which we excluded earlier.

Answer 2

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.