johncarlosbaez

johncarlosbaez, 8 days ago

@JosephAdams - nice! There are lots of twice continously differentiable zig-zaggy functions that come close to making this zero.

johncarlosbaez, 8 days ago

@JosephAdams - right, that has one zig and one zag. But with more zigzags we can also make

[ \int_0^1 y^2 dx ]

as close to zero as we like, which is kind of cool.

johncarlosbaez, 9 days ago

@11011110 @gregeganSF - Great! What's Mahler's Last Conjecture?

My guess is that Hales will not move on until he's finished proving Reinhardt's conjecture. He wrote an NSF proposal saying

In 1934, Reinhardt considered the problem of determining the shape of the centrally symmetric convex disk in the plane whose densest packing has the lowest density. In informal terms, if a contract requires a miser to make payment with a tray of identical gold coins filling the tray as densely as possible, and if the contract stipulates the coins to be convex and centrally symmetric, then what shape of coin should the miser choose in order to part with as little gold as possible? Reinhardt conjectured that the shape of the coin should be a smoothed octagon. The smoothed octagon is constructed by taking a regular octagon and clipping the corners with hyperbolic arcs. The density of the smoothed octagon is approximately 90 per cent. Work by previous researchers on this conjecture has tended to focus on special cases. Research of the PI gives a general analysis of the problem. It introduces a variational problem on the special linear group in two variables that captures the structure of the Reinhardt conjecture. An interesting feature of this problem is that the conjectured solution is not analytic, but only satisfies a Lipschitz condition. A second noteworthy feature of this problem is the presence of a nonlinear optimization problem in a finite number of variables, relating smoothed polygons to the conjecturally optimal smoothed octagon. The PI has previously completed many calculations related to the proof of the Reinhardt conjecture and proposes to complete the proof of the Reinhardt conjecture.

johncarlosbaez, 9 days ago (edited 9 days ago)

@monsoon0 - because if you can't build the quantum computers that would run these algorithms, a mathematical proof that the algorithms could in theory offer large gains is not completely satisfying.

(Yes, the sentence didn't distinguish the algorithm and the quantum computer that would run it. The algorithms, being mathematical entities, are something you can prove theorems about. The computers that would run them effectively, being machines that nobody knows how to design yet, are not.)

johncarlosbaez, 9 days ago

@pschwahn - interesting question! But first, why do you think the subgroup of 𝐺₂ preserving the algebra of quaternions ℍ⊂𝕆 is isomorphic to SO(4)?

The group of automorphisms of ℍ is not SO(4), it's SO(3). Of course the group of automorphisms of ℍ may be different than the subgroup of 𝐺₂ preserving the algebra of quaternions ℍ⊂𝕆. But still, I'm not seeing how you get SO(4).

johncarlosbaez, 9 days ago (edited 9 days ago)

@pschwahn - thanks! It took me a while to reply because I didn't knew that the subgroup of Aut(𝕆) preserving a subalgebra isomorphic to ℍ is isomorphic to SO(4), and I wanted to understand this in a nice way! I'm not there yet but I'm close.

A 'basic triple' is an orthonormal triple of imaginary octonions, say e₁,e₂,e₃, where each is orthogonal to the algebra generated by the other two. Equivalently, it's a triple where each one squares to -1 and each one anticommutes with the other two and also the product of the other two.

A key fact is that Aut(𝕆) acts freely and transitively on basic triples. That is, for each pair of basic triples there exists a unique element of Aut(𝕆) mapping the first to the second. When you have a basic triple, any two generate a subalgebra isomorphic to ℍ while the third is orthogonal to this subalgebra.

Now, pick orthonormal imaginary octonions e₁,e₂. They generate a subalgebra isomorphic to ℍ. The orthogonal complement of this ℍ is 4-dimensional. Any unit vector in this ℝ⁴ can serve as the third element e₃ of a basic triple. Picking one, there is a unique element of Aut(𝕆) mapping this any other unit vector in this ℝ⁴ while holding e₁,e₂ and thus the whole algebra they generate pointwise fixed. So the subgroup of Aut(𝕆) pointwise fixing this ℍ is a 3-sphere, and thus SU(2).

johncarlosbaez, 8 days ago

@pschwahn - "In a nice way" just means "using facts I know by heart, with a minimum of calculations".

So yes, now I see in a nice way that the subgroup G of Aut(𝕆) preserving ℍ has an SU(2) subgroup fixing ℍ pointwise. There's also an obvious map G → Aut(ℍ). Once we know that map is onto, we get a short exact sequence

1→SU(2)→G→SO(3)→1.

This comes close to showing G is SO(4), but SU(2)×SO(3) could also fit into that exact sequence. To finish the job I bet it's enough to show the exact sequence doesn't split, but it's much more satisfying to do what you're doing: explicitly describe G! Your guesses look really plausible.

johncarlosbaez, 8 days ago (edited 8 days ago)

@pschwahn wrote: "Right now I'm still stuck at showing that my SO(4)-action above really preserves octonion multiplication..."

To check that

(x,y) ↦ (q₁xq₁⁻¹, q₁yq₂⁻¹)

preserves octonion multiplication, I'd use the Cayley-Dickson formula expressing octonion multiplication as a way to multiply pairs of quaternions:

https://en.wikipedia.org/wiki/Cayley%E2%80%93Dickson_construction#Octonions

Then the verification involves only properties of quaternion addition, multiplication and conjugation, so it should straightforwardly succeed or fail.

I might also use this to check

"any f∈Aut(ℍ) can be extended to an automorphism of 𝕆 by setting f(x+ye):=f(x)+f(y)e for x,y∈ℍ"

if I wanted to be 100% sure of Mark Reeder's nice argument. But it's such a nice argument I'll probably just believe it!

Understanding SO(3)ᵢᵣᵣ seems harder, but it reminds me of how we can think of G₂ as the group of transformations of ℝ⁷ preserving a certain 'cross product'. We can define this cross product in terms of imaginary octonions by

x × y = (xy - yx)/2

So here's one way to pose your puzzle: you're trying to find a rep of SO(3) on imaginary octonions that's irreducible and preserves their cross product! This is extremely easy for imaginary 𝑞𝑢𝑎𝑡𝑒𝑟𝑛𝑖𝑜𝑛𝑠.

johncarlosbaez, 8 days ago

@pschwahn - oh, maybe I can guess how to get SO(3)ᵢᵣᵣ. Take a basic triple of octonions 𝑒₁, 𝑒₂, 𝑒₃. They span a copy of ℝ³ inside the imaginary octonions. If you can show that any rotation on this ℝ³ extends uniquely to an automorphism of the octonions, I think you're close to done since you get a representation of SO(3) on the imaginary octonions, which is 7-dimensional, and it should be easy to show it's irreducible. Why does any rotation on this ℝ³ extend to uniquely to an automorphism of the octonions? It suffices to show that it maps our basic triple to some other basic triple.

johncarlosbaez, 10 days ago

@highergeometer - "Does this role even have you considering a move to Adelaide to embark on your next career chapter?"

In spoken English you can turn any sentence into a question just by raising the pitch a bit at the end.

johncarlosbaez, 10 days ago

@TruthSandwich - hmm, how did that get called "Bazooka"?

johncarlosbaez, 10 days ago

@TruthSandwich - it seems strange to name some bubble gum after a portable anti-tank missile. What's next? RPG chewing gum? HIMARS candies?

The article doesn't explain this name choice except that it happened shortly after WW2 and the ad campaign featured one "Bazooka Joe". But Bazooka Joe wasn't a soldier: he was a kid with an eyepatch.

I'm left hoping that the original use of "bazoo" to mean "mouth" was still relevant....

johncarlosbaez, 11 days ago (edited 11 days ago)

To prove Dobiński's formula we can use combinatorial species and their generating functions.

There's a species (\mathrm{Part}), such that (\mathrm{Part}(n)) is the set of partitions of an (n)-element set into nonempty subsets. By definition its generating function is
[ \displaystyle{
|\mathrm{Part}|(x) =
\sum_{n \ge 0} \frac{|\mathrm{Part}(n)|}{n!} x^n =
\sum_{n \ge 0} \frac{B_n}{n!} x^n } ]
since we call the cardinality (|\mathrm{Part}(n)|) the (n)th Bell number.

To put a partition on a finite set amounts to chopping it into a finite set of nonempty finite sets, so using a cool fact about species, we have
[ |\mathrm{Part}| = |\mathrm{Exp}| \circ |\mathrm{NE}| ]
where (\mathrm{Exp}) is the species 'being a finite set' and (\mathrm{NE}) is the species 'being a nonempty finite set'. There's one way for an (n)-element set to be a finite set so
[ |\mathrm{Exp}|(x) = \sum_{n \ge 0} \frac{x^n}{n!} = e^x ]
and similarly
[ |\mathrm{NE}|(x) = \sum_{n \ge 1} \frac{x^n}{n!} = e^x - 1 ]
Thus we have
[ |\mathrm{Part}|(x) = e^{e^x - 1} ]
and thus
[ \sum_{n \ge 0} \frac{B_n}{n!} x^n = e^{e^x - 1} ]

Now let's use this to prove Dobiński's formula! We start by calculating the right hand side another way. We have
[ \displaystyle{ e^{e^x} = \sum_{k \ge 0} \frac{e^{kx}}{k!}
= \sum_{k \ge 0} \frac{1}{k!} \sum_{n \ge 0} \frac{(kx)^n}{n!} }]
so
[ \displaystyle{ e^{e^x - 1} =
\frac{1}{e} \sum_{k \ge 0} \frac{1}{k!} \sum_{n \ge 0} \frac{(kx)^n}{n!}} ]
The coefficient of (x^n) in this power series must be (B_n/n!), so Dobiński's formula follows:
[ \displaystyle{ B_n = \frac{1}{e} \sum_{k = 0}^\infty \frac{k^n}{k!} } ]

The moral: combinatorial species are cool!

(2/3)

johncarlosbaez, 11 days ago

To learn about combinatorial species, try any of the 3 free books linked to here:

https://math.ucr.edu/home/baez/permutations/

Or if you're really brave, just dive in and see how I apply them to random permutations. In learning about these, I learned something really surprising to me: random permutations are largely governed by the Poisson distribution. That got me interested in combinatorial proofs of Dobiński's formula. Besides the proof I just gave you, there's a nice proof due to Rota that uses Stirling numbers:

https://math.ucr.edu/home/baez/permutations/permutations_8.html

There's a lot of category theory in this business, so it's a fun mix of ideas.

(3/3)

johncarlosbaez, 10 days ago

@dougmerritt - I'm glad you know 𝐺𝑒𝑛𝑒𝑟𝑎𝑡𝑖𝑛𝑔𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛𝑜𝑙𝑜𝑔𝑦 - it's a blast. Flajolet's book is also lots of fun, full of concrete examples of how to use generating functions. I think it gently brings in the fact that what you're computing an (exponential) generating function 𝑜𝑓 is actually a species: a functor from the groupoid of finite sets to Set, sending each finite set to some set of structures you can put on it. The book by Bergeron, Labelle, and Leroux digs a bit deeper into the functorial approach to species. But none of these gets anywhere near the full glory of that viewpoint.... you'll probably be relieved to know. I got into that more deeply in this course:

https://math.ucr.edu/home/baez/qg-fall2019/

which someday should become a book.

johncarlosbaez, 10 days ago

@uxor - nice, I don't know the Wyman Moser asymptotic formula, but it sounds like a fun example of approximating something by a Gaussian, and I bet there is a lot of deep combinatorial significance lurking beneath it.

johncarlosbaez, 11 days ago (edited 11 days ago)

@paulmasson @phonner - as you probably know, Dobiński's formula says

[ \displaystyle{ \frac{1}{e} \sum_{k = 0}^\infty \frac{k^n}{k!} = B_n } ]

where (B_n) is the nth Bell number. Dobiński published a paper about it in 1877, but it seems that he only proved some special cases. Here I give a combinatorial proof following Gian-Carlo Rota:

https://math.ucr.edu/home/baez/permutations/permutations_8.html

johncarlosbaez, 11 days ago

@svat @phonner @paulmasson - I expand on that proof of Dobiński's formula using Stirling numbers here:

https://math.ucr.edu/home/baez/permutations/permutations_8.html

and I just gave another proof here on Mathstodon, using a lot of LaTeX:

https://mathstodon.xyz/@johncarlosbaez/112427042889044371

johncarlosbaez, 12 days ago (edited 11 days ago)

@antoinechambertloir @MartinEscardo @oantolin - There is a rather elaborate and carefully worked out esthetic of formulating definitions in categorical logic, aimed to ensure that everything works out as smoothly as possible, e.g. that everything you say about a functor is invariant under natural isomorphisms, and everything you say about a category is invariant under equivalence. Equations between objects can easily break these principles so we call them 'evil'; this then pressures us to take an approach where we don't need to check that the source of one morphism equals the target of the next.

A common approach is to make morphisms 'dependently typed', so that for each pair of objects (a,b) you have a set of morphisms hom(a,b). You never talk about the set of all morphisms, so you never mention source and target maps. Composition is not a single partially defined function, but instead a bunch of functions hom(a,b) × hom(b,c) → hom(a,c). So, you never need to check that the source of one morphism equals the target of another: it's impossible to even dream of composing morphisms unless you already know you can do it!

johncarlosbaez, 8 days ago (edited 8 days ago)

@madnight @BartoszMilewski - I doubt anyone can explain in simple terms why the derived category of coherent sheaves on one Calabi–Yau manifold should be equivalent to the Fukaya category of some other Calabi–Yau manifold.

This is an unproved conjecture due to Kontsevich called "homological mirror symmetry". You can read an elementary introduction to it on Wikipedia:

https://en.wikipedia.org/wiki/Homological_mirror_symmetry

but one thing you won't find is anything about why it should be true! They say there was a year-long program on it at the Institute for Advanced Studies, but "only in a few examples have mathematicians been able to verify the conjecture."

The nLab article

https://ncatlab.org/nlab/show/mirror+symmetry

gives a bit of an explanation: physicists believe for every 3d complex Calabi-Yau variety (X) there are associated two field theories, and at least some X have a "mirror partner" (\hat{X}) such that the first field theory built using (X) is equivalent to the second one build using (\hat{X}). But this has not been proved - in fact, these field theories have only been fully constructed in some cases!

I would hope some good string theorists could tell a good simplified story about why they believe this stuff, but I haven't seen it.

When you get mathematicians involved in a difficult unsolved problem like this, things tend to become technical. This supposedly introductory paper:

https://arxiv.org/abs/0801.2014

says

"Part of the difficulty in dealing with homological mirror symmetry is the breadth of knowledge required for a proper formulation."

It doesn't give a good story about why the homological mirror symmetry conjecture should be true.

If you want to study this stuff, learn lots and lots of math first.

johncarlosbaez, 10 days ago

@battaglia01 - great question! Let me get back to you on this... it's a good excuse to study some stuff.

johncarlosbaez, 9 days ago

@j_bertolotti - sounds like a fun book! Chapter 3 explains why Chapters 1 and 2 are wrong. 😜