@johncarlosbaez@mathstodon.xyz
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johncarlosbaez

@johncarlosbaez@mathstodon.xyz

I'm a mathematical physicist who likes explaining stuff. Sometimes I work at the Topos Institute. Check out my blog! I'm also a member of the n-Category Café, a group blog on math with an emphasis on category theory. I also have a YouTube channel, full of talks about math, physics and the future.

This profile is from a federated server and may be incomplete. Browse more on the original instance.

johncarlosbaez, (edited ) to random
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We were listening to bouzouki music, and the conversation naturally turned to bazookas. It turns out that there's also bazooka music!

The comedian Bob Burns invented a horn-like instrument in the 1910s - he's shown with it below. It actually caught on in jazz in the 1930s. Someone jokingly called it the "bazooka" after the word "bazoo", which was slang for "mouth".

Later, in World War II, "bazooka" became the name for a new American anti-tank weapon, because it looked like this instrument.

Ironically, the slang word "bazoo", for "mouth", probably came from the word "buisine", which was the name of a medieval trumpet! And that comes from "buccina", a brass horn used by the Roman army.

In case you're wondering, he word "bouzouki" is unrelated. It comes from the Turkish word "bozuk", meaning "broken" or "modified", which refers to a particular way of tuning a string instrument where the notes are not arranged from low to high.

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

johncarlosbaez, (edited ) to random
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Hardcore math puzzle:

Suppose raindrops are falling on your head, randomly and independently, at an average rate of one per minute. What's the average of the 𝑐𝑢𝑏𝑒 of the number of raindrops that fall on your head in one minute?

The probability that (k) raindrops fall on your head in a minute is given by the Poisson distribution of mean 1, so it's
[ \frac{1}{ek!} ]
I could explain this but let's move on. The puzzle asks us to compute the expected value of (k^3) for this probability distribution, which is
[ \sum_{k=0}^\infty \frac{k^3}{ek!} ]
The heart of the puzzle is to figure out this sum. It turns out that
[ \sum_{k = 0}^\infty \frac{k^n}{k!} = B_n e ]
where (B_n) is the (n)th 'Bell number': the number of partitions of an (n)-element set into nonempty subsets. This is called 'Dobiński's formula'. I'll prove it in my next post. Now let's just use it!

We're interested in the case (n = 3). There are 5 partitions of a 3-element set
[ {{1,2,3}}, ]
[ {{1,2}, {3}}, ; {{2,3}, {1}}, ; {{3,1}, {2}}, ]
[ {{1}, {2}, {3}} ]
so (B_3 = 5).

So, the average of the cube of the number of raindrops that fall on your head in one minute is 𝟓.

Wild, huh? From probability theory to combinatorics.

(1/3)

johncarlosbaez, (edited )
@johncarlosbaez@mathstodon.xyz avatar

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,
@johncarlosbaez@mathstodon.xyz avatar

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,
@johncarlosbaez@mathstodon.xyz avatar

@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,
@johncarlosbaez@mathstodon.xyz avatar

@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.

phonner, to math
@phonner@mathstodon.xyz avatar

Inspired by my brilliant student (https://mathstodon.xyz/deck/@phonner/112419322877058443) I've been playing around with (e)-like sums. Here's a fascinating one!

[ \sum_{n=0}^{\infty} {\frac{n^4}{n!}}=15e ]
This is strange enough to provoke wonder, but simple enough to serve as an entry-point to an interesting generalization.

johncarlosbaez, (edited )
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@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,
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@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

julesh, to random
@julesh@mathstodon.xyz avatar

I'm a city kid and I've never seen any astronomy before... but the northern lights were so bright we could see them in central Glasgow

johncarlosbaez,
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@julesh - I failed to look out our north-facing windows in Edinburgh. 😿

johncarlosbaez, to random
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The rise of book bans in Florida made Lauren Groff start a new bookstore there. It 𝑠𝑝𝑒𝑐𝑖𝑎𝑙𝑖𝑧𝑒𝑠 in banned books. It's called The Lynx.

More here: https://archive.is/fD0uF

MartinEscardo, to random
@MartinEscardo@mathstodon.xyz avatar

It is just me? The following definition of category hurts my categorical instincts, because it uses object equality.

A category consists of

  1. A collection of objects.

  2. A collection of morphisms.

  3. Each morphism f has two assigned objects, its source s(f) and its target t(f).

  4. For each pair of morphisms f,g such that t(f)=s(g) there exists a specified morphism g ∘ f such that [it doesn't matter what]

  5. [Some axioms are satisfied.]

It is (4) that hurts my categorical instincts.

There is no reason to have "evilness" (in the categorical sense, rather than the emotional sense) built-in in the definition of category!

This definition is, for example, adopted by Freyd.

johncarlosbaez,
@johncarlosbaez@mathstodon.xyz avatar

@oantolin @MartinEscardo - indeed, most of us needed to learn any definition of category whatsoever - the nuances don't matter so much - before we could be attuned to the nuances of sameness (equality vs. isomorphism, etc.) and understand the problems of "evil", and appreciate why we'd want to avoid a definition of category that mentions equations between objects!

johncarlosbaez,
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@MartinEscardo @oantolin - I agree that it would be good to give people a nice definition of "category" the first time. But I don't see category theorists unwilling to give different styles of definition a fair chance. Maybe you're referring to non-category-theorists. I'd forgotten about them. 😳 I can easily imagine them sticking to the first definition of category they see, regarding it as sacrosanct.

johncarlosbaez, (edited )
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@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!

paulbalduf, to physics
@paulbalduf@mathstodon.xyz avatar

In , scattering amplitudes can be computed as sums of (very many) s. They contribute differently much, with most integrals contributing near the average (scaled to 1.0 in the plots), but a "long tail" of integrals that are larger by a significant factor.
We looked at patterns in these distributions, and one particularly striking one is that if instead of the Feynman integral P itself, you consider 1 divided by root of P, the distribution is almost Gaussian! To my knowledge, this is the first time anything like this has been observed. We only looked at one quantum field theory, the "phi^4 theory in 4 dimensions". It would be interesting to see if this is coincidence for this particular theory and class of Feynman integrals, or if it persists universally.
More background and relevant papers at https://paulbalduf.com/research/statistics-periods/

image/jpeg
image/jpeg

johncarlosbaez,
@johncarlosbaez@mathstodon.xyz avatar

@paulbalduf - that's a great observation! Have you written a paper about it? It's worthwhile, even if it's only true for the ϕ₄⁴ theory. Which class of Feynman diagrams were you considering? E.g. 4-point functions?

johncarlosbaez,
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@paulbalduf - thanks! I'll check out the paper and tell some people about it.

BartoszMilewski, to random
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I'm struggling with the definition of the category of elements--the direction of morphisms. Grothendieck worked with presheaves (C^{op} \to \mathbf{Set}), with a morphism ((a, x) \to (b, y)) being an an arrow (a \to b) in (C). The question is, what is it for co-presheaves? Is it (b \to a)? nLab defines it as (a \to b) and doesn't talk about presheaves. Emily Riehl defines both as (a \to b), which makes one wonder what it is for (𝐶ᵒᵖ)ᵒᵖ→𝐒𝐞𝐭 , not to mention (C^{op}\times C \to \mathbf{Set}).

johncarlosbaez,
@johncarlosbaez@mathstodon.xyz avatar

@BartoszMilewski - it's just a convention; either convention is possible. So in what sense are you struggling with it? Are you struggling to decide what you like best? You don't really need to pick a favorite.

johncarlosbaez,
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@BartoszMilewski - they both exist, neither is "right" by decree of god, so use the one that works for what you're doing... which may change next week.

johncarlosbaez, (edited ) to random
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Costa Talalaev gave me a fractal: a 3-dimensional Sierpinski gasket!

He made it using the 3d printer at the Hacklab, a makerspace here in Edinburgh. It was a bit hard to make since it's held together only at tiny spots. He had to build something with more plastic, a kind of scaffolding, and then tear that off. The end result is very light yet sturdy.

johncarlosbaez,
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@SamuelCraig - right, in the idealized limit it has zero volume!

johncarlosbaez,
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@bks - wow! Thanks!

johncarlosbaez,
@johncarlosbaez@mathstodon.xyz avatar

@codrusofathens - all this stuff is a lot of fun, indeed!

Unsolicited advice: fractals are just a tiny piece of the math landscape, so only do a math double major if you enjoy a lot of other math too. For example, to get serious about fractals you need to take advanced calculus and then real analysis, which are tough but to me utterly delightful subjects - and much bigger than fractals in the grand scheme of things.

johncarlosbaez,
@johncarlosbaez@mathstodon.xyz avatar

@codrusofathens - numerical analysis is a natural road to a job. Number theory is beautiful but there are few jobs in it and it's insanely competitive. But one great thing about a double major is that you can study something just for the love of it.

I'm studying a lot of number theory these days, and the layers of conceptual depth are absolutely enthralling.

johncarlosbaez, to random
@johncarlosbaez@mathstodon.xyz avatar

This is an ORC - an "odd radio cluster". It's a faint circle of radio emissions surrounding a distant galaxy. 5 ORCs have been found. This one is about a million light years in diameter, roughly 10 times the size of a galaxy like ours. So it was probably formed by some sort of explosion, and took a long time to get this big. But the details remain a mystery!

Astronomers have just discovered that one ORC is emitting X-rays. So at least in that one ORC, the diffuse gas must be hot: about 8 million Celsius.

Here are some theories of what an ORC might be:

• it's a spherical shock wave from a cataclysmic event in the host galaxy, such as a merger of two supermassive black holes

• it's the shock wave formed by a 'starburst wind' created by a burst of star formation in the host galaxy

• it's the jet produced by a supermassive black hole, seen head on.

• it's a wormhole

As often the case, the most exciting theory is the least likely unless the others get ruled out.

But we can still have some fun. So watch an animated gif of what the birth of an ORC might look like!

(1/2)

johncarlosbaez, (edited )
@johncarlosbaez@mathstodon.xyz avatar

Here's a possible scenario for the formation of an ORC: some sort of explosion in the host galaxy, producing a shock wave that expanded outwards for a billion years.

For more on ORCs, try these:

• Wikipedia, Odd radio cluster, https://en.wikipedia.org/wiki/Odd_radio_circle

• Ray P. Norris, Evan Crawford and Peter Macgregor, Odd radio circles and their environment, Galaxies, https://www.mdpi.com/2075-4434/9/4/83

• Ashley Bazar, X-ray satellite XMM-Newton sees ‘Space Clover’ in a new light, NASA, https://www.nasa.gov/missions/xmm-newton/x-ray-satellite-xmm-newton-sees-space-clover-in-a-new-light/

(2/2)

johncarlosbaez, (edited ) to random
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Tolstoy: "Happy families are all alike; every unhappy family is unhappy in its own way."

Mathematics: "Real tori are all alike; every complex torus is complex in its own way."

To be precise, a 'n-dimensional real torus' is a real manifold of the form V/Λ where V is an n-dimensional real vector space and Λ ⊆ V is a lattice of rank n in this vector space. They are all isomorphic.

An 'n-dimensional complex torus' is a complex manifold of the form V/Λ where V is an n-dimensional complex vector space and Λ ⊆ V is a lattice of rank 2n in this vector space. These are not all isomorphic, because there are different ways the lattice can get along with multiplication by i. For example we might have iΛ = Λ or we might not.

And so, it's possible to write a whole book - and indeed a fascinating one - on complex tori. For example a 1-dimensional complex torus is an elliptic curve, and there are whole books just about those.

johncarlosbaez,
@johncarlosbaez@mathstodon.xyz avatar

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

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