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## Welcome to Our Blog & Errata Contest

Now that  Alice and Bob Meet Banach  is available,  it is time to bring this blog to life.

First, our plan is to write a few short entries on mathematical topics vaguely related to the subject of the book.  Next, we will spread gossip/announce events that may be of interest to the readers of the book.

Finally, this is also the place where you are encouraged to report typos/errors in the book, improvements, solutions to problems and the like.  To motivate submissions (via email or by commenting this entry), we offer a bottle of fine French wine (Châteauneuf-du-Pape) to the error-finder who will be most prolific over the next 365 days.

## PPT-inducing non-multiplicative channels

One success of the probabilistic method in quantum information theory is Hastings’s example of quantum channels, for which the capacity to transmit information is non-additive. The exegesis of that example has generated lot of activity in the community.

Here we consider a related question: for $p>1$, can we find two quantum channels $\Phi_1$ and $\Phi_2$ such that

$\|\Phi_1 \otimes \Phi_2\|_{1\to p} > \|\Phi_1\|_{1 \to p} \|\Phi_2\|_{1 \to p} \ ?$

This question is well-known to be much simpler than the one solved by Hastings (which corresponds to $p=1$).  However, one can make it a bit harder by demanding some extra properties from the channels $\Phi_1,\Phi_2$, for example of being PPT-inducing (meaning that applying the channel on one part on a bipartite system always outputs PPT states). Existence of such channels was proved by Moto Fukuda and Ion Nechita for $p>p_0=30.95$ by using tools from free probability. We noticed recently that one can push $p_0$ down to  $1$ if we use Asymptotic Geometric Analysis instead.

One motivation for such a question is to understand how PPT-inducing channels are different from entanglement-breaking channels (i.e., channels which output separable states), as the $1 \to p$ is always multiplicative for them. One famous open problem in this direction is the “PPT squared problem”: must the composition of two PPT-inducing channels be entanglement-breaking ?

As often in such questions, the example is obtained most efficiently by taking $\Phi_2 = \overline{\Phi_1}$, with $\Phi_1$ induced by a random isometric embedding $V : \mathbb{C}^m \to \mathbb{C}^k \otimes \mathbb{C}^d$. It remains to find the right dimensions: if we set $m=d^{\alpha}$ and $k=d^{\beta}$, one can check (applying Dvoretzky’s theorem for the Schatten $2p$-norm) that the “non-multiplicativity” holds provided $\beta-\alpha > 2/p-1$, while the resulting channel will be PPT-inducing provided $\alpha + \beta <1$ (this is essentially the “threshold theorem” for the PPT property of random states, since the PPT-inducing property can be checked on the Choi matrix, which here is random). And there is room for satisfying both inequalities whenever $p>1$. Some technical details have to enter the picture, but you don’t show these on blogs, and anyway here is a complete argument.

The apparent need for all the dimensions $k,m,d$ to go simultaneously to infinity makes the situation less feasible to analyze from the free probability point of view, while AGA handles that routinely.

And here is a challenge to our readers:  extend the above observation to  $p=1$, the context of Hastings’s result. At the first sight, the constraints on $k,m,d$ needed for additivity violations are fairly tight and appear to be incompatible with the PPT-inducing property being generic. But perhaps there is a trick…

## How optimal are random sphere coverings?

Suppose you would like to optimally place $N=2^{n}$ congruent spherical caps on the unit sphere in $\mathbb{R}^n, n>2$, each cap having volume $1/N$. (Here volume refers to the normalized surface area; the sphere has volume $1$). How much volume of the sphere can you cover?

Clearly the whole sphere cannot be covered: the caps must overlap if $N, n>2$ (which we assume). Moreover, it is not immediately obvious that a substantial proportion of the sphere can be covered this way. In fact, if we do insist that the caps (say, again, of volume $2^{-n}$) are disjoint, the proportion that can be covered is at most $(2/3)^{n}$ (Kabatianskii and Levenshtein, 1978, MR0514023).

This where the probabilistic method shows its effectiveness: placing the caps at random leaves an expected uncovered volume of $\left(1-\frac{1}{N}\right)^N \approx 1/e$. Thus one can cover a proportion $1 - \frac{1}{e} + o(1)$ of the sphere.

Can one do better? Can we hope to cover a proportion $1-o(1)$? Or even $\alpha+o(1)$ for some $\alpha>1-\frac{1}{e}$? The question was asked by Gil Kur,  a reader of our book, and we have no idea about the answer! There is nothing special about $N=2^n$,  you may replace it by other values.

There are few lower bounds on the density of spherical coverings (incidentally, the terms “overlap” or “redundancy” would be more logical). The general upper bound due to Rogers via random covering (Corollary 5.5 in our book) achieves a density of order $d \log d$. A lower bound of order $d$ holds for caps whose radii–in the geodesic metric–are either very close to $\pi/2$ (since $d+1$ caps are needed) or to $0$ (via the simplex bound of Coxeter-Few-Rogers), but little seems to be documented in the literature for the intermediate range. A conjectural spherical simplex bound would imply a lower bound of $d$ for all radii, see Conjecture 6.7.3 and the open Problem at the end of Section 6.8 in Károly Böröczky’s book Finite Packing and Covering.

## IHP Trimester “Analysis in Quantum Information” and more

Major ABMB-related events are taking place this fall all over the globe!

Both of us are co-organizing a trimester Analysis in Quantum Information Theory at the Institut Henri Poincaré in Paris. The trimester has been running successfully for 3 weeks already with plenty of stimulating activities.

If you cannot meet Alice and Bob in Paris, maybe try Santa Barbara! A trimester on Quantum Physics of Information is ran by the Kavli Institute for Theoretical Physics.

Still at the same time, many of Banach’s heirs are gathering at the MSRI for Geometric Functional Analysis and Applications.

Attending all these events may require that you violate the no-cloning theorem. But in the modern era science may spread via all kinds of channels: you may enjoy the playlist of the first IHP workshop, the  KITP podcasts or clicking on the film icons on the MSRI schedule.