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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?

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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.

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  • $\begingroup$ thank you very much, also for the correction. Do you have any conjectures on the general case as well? $\endgroup$ – Dixmier Mar 23 '18 at 16:55
  • $\begingroup$ Well, I am not sure. I assume that you want your Hilbert space to be infinite dimensional? Otherwise the answer to question (ii) has to be positive for finite matrices. It also feels that the answer to (iii) should be positive but it somehow feels that (ii) is more interesting. $\endgroup$ – Mateusz Wasilewski Mar 23 '18 at 17:01
  • $\begingroup$ yes, I had complex infinite-dimensional and separable Hilbert spaces in mind. I added this now as an additional assumption. $\endgroup$ – Dixmier Mar 23 '18 at 17:20
  • $\begingroup$ I don't understand your argument for the general case... why can you assume that $T$ is a self-adjoint contraction with a spectral gap? And why does this imply the estimate you state (wouldn't that be the claim?). Also, how is this assumption used later on? $\endgroup$ – Matthias Ludewig Mar 26 '18 at 3:10
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    $\begingroup$ @MatthiasLudewig, I assume that $T$ is a contraction just to normalise things so that I don't have to worry about various parameters. The assumption $\textrm{Tr}(H H^{\ast})=1$ is also just a normalisation, which can be done, as both sides in the definition of spectral gap have the homogeneity in $H$. It is not visible that I use these assumptions because I did not write explicitly the estimates that yield the contradiction; if I did, it would be clear that everything depends only on the Hilbert-Schmidt norm of $\varepsilon$ and the assumed spectral gap $\lambda$. $\endgroup$ – Mateusz Wasilewski Mar 26 '18 at 15:06
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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.

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  • $\begingroup$ that's a great point of view. $\endgroup$ – Sascha Mar 26 '18 at 9:56

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