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The International Congress of Mathematicians (ICM 2022) will now be fully virtual. How best to utilise this format?

26 February, 2022 in non-technical, opinion | Tags: | by | 68 comments

[Note: while I am chair of the ICM Structure Committee, this blog post is not an official request from this committee, as events are still moving too rapidly to proceed at present via normal committee deliberations. We are however discussing these matters and may issue a more formal request in due course. -T.]

The International Mathematical Union has just made the following announcement concerning the International Congress of Mathematicians (ICM) that was previously scheduled to be held in St. Petersburg, Russia in July.

Decision of the Executive Committee of the IMU on the upcoming ICM 2022 and IMU General Assembly

On 26 February 2022, the Executive Committee of the International Mathematical Union (IMU) decided that:

1. The International Congress of Mathematicians (ICM) 2022 will take place as a fully virtual event, hosted outside Russia but following the original time schedule planned for Saint Petersburg.
2. Participation in the virtual ICM event will be free of charge.
3. The IMU General Assembly (GA) will take place as an in-person event outside Russia.
4. A prize ceremony will be held the day after the IMU GA, at the same venue as the IMU GA, for the awarding of the 2022 IMU prizes.
5. The dates for the ICM and the GA will remain unaltered.
6. We will return with further practical information regarding the two events.

An expanded version of the announcement can be found here. (See also this addendum.)

While I am not on the IMU Executive Committee and thus not privy to their deliberations, I have been in contact with several members of this committee and I support their final decision on these matters.

As we have all experienced during the COVID-19 pandemic, virtual conferences can be rather variable in quality, but there certainly are ways to make the experience more positive for both the speakers and participants. In the interest of maximizing the benefits that this meeting can still produce, I would like to invite readers of this blog to share any experiences they have had with very large virtual conferences, and any opinions on what types of virtual events were effective and engaging.

One idea that has been suggested to me has been to have (either unofficial, semi-official, or official) regional ICM hosting events at various places worldwide where mathematicians could gather in person to view ICM talks that would be streamed online (and perhaps some ICM speakers from that area could give talks in person in such locations). This would be very nonstandard, of course, but could be one way to salvage some of the physical ICM experience, and perhaps also a way to symbolically support the spirit of the Congress. I would be interested to get some feedback on this proposal.

Finally, I would like to request that comments to this post remain focused on the upcoming virtual ICM. Broader political issues are very much worth discussing at present, but there are other venues for such discussion, and as per my usual blog policy any off-topic comments may be subject to deletion.

Satz vom Nullprodukt

14 February, 2022 in diversions, math.RA | by | 56 comments

As I have mentioned in some recent posts, I am interested in exploring unconventional modalities for presenting mathematics, for instance using media with high production value. One such recent example of this I saw was a presentation of the fundamental zero product property (or domain property) of the real numbers – namely, that ab=0 implies a=0 or b=0 for real numbers a,b – expressed through the medium of German-language rap:

EDIT: and here is a lesson on fractions, expressed through the medium of a burger chain advertisement:

I’d be interested to know what further examples of this type are out there.

SECOND EDIT: The following two examples from Wired magazine are slightly more conventional in nature, but still worth mentioning, I think. Firstly, my colleague at UCLA, Amit Sahai, presents the concept of zero knowledge proofs at various levels of technicality:

Secondly, Moon Duchin answers math questions of all sorts from Twitter:

Perfectly packing a square by squares of nearly harmonic sidelength

8 February, 2022 in math.MG, paper | Tags: | by | 41 comments

I’ve just uploaded to the arXiv my preprint “Perfectly packing a square by squares of nearly harmonic sidelength“. This paper concerns a variant of an old problem of Meir and Moser, who asks whether it is possible to perfectly pack squares of sidelength {1/n} for {n \geq 2} into a single square or rectangle of area {\sum_{n=2}^\infty \frac{1}{n^2} = \frac{\pi^2}{6} - 1}. (The following variant problem, also posed by Meir and Moser and discussed for instance in this MathOverflow post, is perhaps even more well known: is it possible to perfectly pack rectangles of dimensions {1/n \times 1/(n+1)} for {n \geq 1} into a single square of area {\sum_{n=1}^\infty \frac{1}{n(n+1)} = 1}?) For the purposes of this paper, rectangles and squares are understood to have sides parallel to the axes, and a packing is perfect if it partitions the region being packed up to sets of measure zero. As one partial result towards these problems, it was shown by Paulhus that squares of sidelength {1/n} for {n \geq 2} can be packed (not quite perfectly) into a single rectangle of area {\frac{\pi^2}{6} - 1 + \frac{1}{1244918662}}, and rectangles of dimensions {1/n \times 1/n+1} for {n \geq 1} can be packed (again not quite perfectly) into a single square of area {1 + \frac{1}{10^9+1}}. (Paulhus’s paper had some gaps in it, but these were subsequently repaired by Grzegorek and Januszewski.)

Another direction in which partial progress has been made is to consider instead the problem of packing squares of sidelength {n^{-t}}, {n \geq 1} perfectly into a square or rectangle of total area {\sum_{n=1}^\infty \frac{1}{n^{2t}}}, for some fixed constant {t > 1/2} (this lower bound is needed to make the total area {\sum_{n=1}^\infty \frac{1}{n^{2t}}} finite), with the aim being to get {t} as close to {1} as possible. Prior to this paper, the most recent advance in this direction was by Januszewski and Zielonka last year, who achieved such a packing in the range {1/2 < t \leq 2/3}.

In this paper we are able to get {t} arbitrarily close to {1} (which turns out to be a “critical” value of this parameter), but at the expense of deleting the first few tiles:

Theorem 1 If {1/2 < t < 1}, and {n_0} is sufficiently large depending on {t}, then one can pack squares of sidelength {n^{-t}}, {n \geq n_0} perfectly into a square of area {\sum_{n=n_0}^\infty \frac{1}{n^{2t}}}.

As in previous works, the general strategy is to execute a greedy algorithm, which can be described somewhat incompletely as follows.

  • Step 1: Suppose that one has already managed to perfectly pack a square {S} of area {\sum_{n=n_0}^\infty \frac{1}{n^{2t}}} by squares of sidelength {n^{-t}} for {n_0 \leq n < n_1}, together with a further finite collection {{\mathcal R}} of rectangles with disjoint interiors. (Initially, we would have {n_1=n_0} and {{\mathcal R} = \{S\}}, but these parameter will change over the course of the algorithm.)
  • Step 2: Amongst all the rectangles in {{\mathcal R}}, locate the rectangle {R} of the largest width (defined as the shorter of the two sidelengths of {R}).
  • Step 3: Pack (as efficiently as one can) squares of sidelength {n^{-t}} for {n_1 \leq n < n_2} into {R} for some {n_2>n_1}, and decompose the portion of {R} not covered by this packing into rectangles {{\mathcal R}'}.
  • Step 4: Replace {n_1} by {n_2}, replace {{\mathcal R}} by {({\mathcal R} \backslash \{R\}) \cup {\mathcal R}'}, and return to Step 1.

The main innovation of this paper is to perform Step 3 somewhat more efficiently than in previous papers.

The above algorithm can get stuck if one reaches a point where one has already packed squares of sidelength {1/n^t} for {n_0 \leq n < n_1}, but that all remaining rectangles {R} in {{\mathcal R}} have width less than {n_1^{-t}}, in which case there is no obvious way to fit in the next square. If we let {w(R)} and {h(R)} denote the width and height of these rectangles {R}, then the total area of the rectangles must be

\displaystyle \sum_{R \in {\mathcal R}} w(R) h(R) = \sum_{n=n_0}^\infty \frac{1}{n^{2t}} - \sum_{n=n_0}^{n_1-1} \frac{1}{n^{2t}} \asymp n_1^{1-2t}

and the total perimeter {\mathrm{perim}({\mathcal R})} of these rectangles is

\displaystyle \mathrm{perim}({\mathcal R}) = \sum_{R \in {\mathcal R}} 2(w(R)+h(R)) \asymp \sum_{R \in {\mathcal R}} h(R).

Thus we have

\displaystyle n_1^{1-2t} \ll \mathrm{perim}({\mathcal R}) \sup_{R \in {\mathcal R}} w(R)

and so to ensure that there is at least one rectangle {R} with {w(R) \geq n_1^{-t}} it would be enough to have the perimeter bound

\displaystyle \mathrm{perim}({\mathcal R}) \leq c n_1^{1-t}

for a sufficiently small constant {c>0}. It is here that we now see the critical nature of the exponent {t=1}: for {t<1}, the amount of perimeter we are permitted to have in the remaining rectangles increases as one progresses with the packing, but for {t=1} the amount of perimeter one is “budgeted” for stays constant (and for {t>1} the situation is even worse, in that the remaining rectangles {{\mathcal R}} should steadily decrease in total perimeter).

In comparison, the perimeter of the squares that one has already packed is equal to

\displaystyle \sum_{n=n_0}^{n_1-1} 4 n^{-t}

which is comparable to {n_1^{1-t}} for {n_1} large (with the constants blowing up as {t} approaches the critical value of {1}). In previous algorithms, the total perimeter of the remainder rectangles {{\mathcal R}} was basically comparable to the perimeter of the squares already packed, and this is the main reason why the results only worked when {t} was sufficiently far away from {1}. In my paper, I am able to get the perimeter of {{\mathcal R}} significantly smaller than the perimeter of the squares already packed, by grouping those squares into lattice-like clusters (of about {M^2} squares arranged in an {M \times M} pattern), and sliding the squares in each cluster together to almost entirely eliminate the wasted space between each square, leaving only the space around the cluster as the main source of residual perimeter, which will be comparable to about {M n_1^{-t}} per cluster, as compared to the total perimeter of the squares in the cluster which is comparable to {M^2 n_1^{-t}}. This strategy is perhaps easiest to illustrate with a picture, in which {3 \times 4} squares {S_{i,j}} of slowly decreasing sidelength are packed together with relatively little wasted space:

By choosing the parameter {M} suitably large (and taking {n_0} sufficiently large depending on {M}), one can then prove the theorem. (In order to do some technical bookkeeping and to allow one to close an induction in the verification of the algorithm’s correctness, it is convenient to replace the perimeter {\sum_{R \in {\mathcal R}} 2(w(R)+h(R))} by a slightly weighted variant {\sum_{R \in {\mathcal R}} w(R)^\delta h(R)} for a small exponent {\delta}, but this is a somewhat artificial device that somewhat obscures the main ideas.)

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