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The answer is no. This is a good illustration of a reasoning principle identified explicitly in Gowers, W. T., The two cultures of mathematics, Arnold, V. (ed.) et al., Mathematics: Frontiers and pers …

This appears to be the case, but I was forced to rely on a somewhat complicated inequality on two real variables that looks quite plausible numerically, though I do not have a 100% rigorous proof of i …

This is an infinite commutative diagram on $M$ (viewed as a category with a single object $\bullet$). $\require{AMScd}$ \begin{CD} \vdots @. \vdots @. \vdots\\ @VVR_y(0,2)V @VVR_y(1,2)V @VVR_y(2,2)V …

One can get a certain way towards this goal via a sort of "dimensional analysis". This isn't a completely satisfying heuristic argument - in particular, it only partially specifies what compression m …

There is certainly no asymptotic independence between $\det M_{11}, \det M_{22}$. From the base times height formula for parallelepipeds we see that \begin{align*} \frac{|\det M_{12}|}{|\det M_{22}|} …

Shorn of probabilistic language, this inequality follows from the assertion that $|x+y|-|x-y|$ is a positive semi-definite kernel, and is therefore the sum (or integral) of squares. Your Fourier-anal …

The continuous comparison method of Knowles and Yin, Knowles, Antti; Yin, Jun, Anisotropic local laws for random matrices, Probab. Theory Relat. Fields 169, No. 1-2, 257-352 (2017). ZBL1382.15051. fol …

The theory of classical independence and classical convolution can be generalised to noncommutative settings in several ways. The most famous one is that of free independence and free convolution (int …

Using the Newton identities, one can (in the high characteristic regime $p>k$) express the elementary symmetric polynomial $\sum_{i_1 < \dots < i_k} a_{i_1} \dots a_{i_k}$ in terms of the moments $\su …

One heuristic is to replace the $n^{th}$ roots of unity by $n$ iid elements $\zeta_1,\dots,\zeta_n$ of the unit circle, drawn uniformly at random. For any sum $\zeta_{i_1} + \dots + \zeta_{i_k}$ of $ …

We have $$ C(W) = 2 A \circ A + v v^\top$$ where $v$ is the vector with entries $\|w_i\|^2$, $A$ is the Wishart matrix with entries $w_i^\top w_j$, and $\circ$ is the Hadamard product. From the Schur …

Using (say) decimal notation, ASCII encoding, and a delimiter symbol such as a space or comma, as well as the law of large numbers, one can almost surely encode $N$ independent copies of $\lfloor X \r …

By the Chernoff bound, we see that for each $1 \leq i < j \leq m$, one has $u_i \cdot u_j = O(\sqrt{n})$ with probability at least $1-\frac{1}{10m^2}$ (say), where implied constants are allowed to dep …

After chasing down references relating to the paper of Shelah mentioned by Will Brian, I now have a satisfactory answer to the question. It all hinges on whether there is a splitting of the quotient …

To complement my other answer, I will show Proposition 1 Let $\xi_k$ be a finite number of iid Bernoulli random variables of expectation $p > 1/2$, and let $a_k > 0$ be real numbers. Then ${\bf …