@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.

monsoon0, to science
@monsoon0@mathstodon.xyz avatar

Does your work sit at the intersection of multiple fields of ? Imagine each field valued fundamental characteristics that the others do not. How do you build a career when those in one such field don't understand or care for the perspective of another?

johncarlosbaez,
@johncarlosbaez@mathstodon.xyz avatar

@monsoon0 - my solution, largely accidental, was to find a department that wasn't extremely competitive, and publish a lot, and teach well. That meant nobody was very upset by the fact that I kept switching from one subject to another. But they pegged me as an "analyst", so the only qualifier course I could teach was real analysis. Not too bad, but sort of funny.

mpiedrav, to random

I've grown mistrustful of the nature of multimedia. In many cases, the more mediatic a message is, the less important it might be, and the more misleading it could be. Really deep and insightful information is often only available in written media, and almost never as multimedia.

johncarlosbaez,
@johncarlosbaez@mathstodon.xyz avatar

@mpiedrav - yes, that's usually true. Also, if you're good at reading it's often faster to search for and find what you need to know in text.

But now I'm learning music theory, and it's great to be able to hear music while watching people play it and also see the score, so videos are great.

dpiponi, to random
@dpiponi@mathstodon.xyz avatar

I thought I'd see if there were any recent developments in the Bogdanoff affair and was surprised to find they both fell victim to covid last year

https://www.bbc.com/news/world-europe-59867046

johncarlosbaez,
@johncarlosbaez@mathstodon.xyz avatar

@dpiponi - Someone told me that. "Their friends said they were convinced their healthy lifestyle would protect them and they were admitted to hospital in mid-December."

MotivicKyle, to random
@MotivicKyle@mathstodon.xyz avatar

The entirety of de Rham cohomology hinges on partial derivatives of smooth functions commuting: ∂²f/(∂x∂y) = ∂²f/(∂y∂x).

What are other examples of mathematical apparatuses that would explode into Cantor dust if a small underlying fact were different?

johncarlosbaez,
@johncarlosbaez@mathstodon.xyz avatar

@MotivicKyle - all of math would collapse if you change any tiny fact... if you believe

𝐹 ⟹ 𝑃

But it might take a while to happen!

video/mp4

mc, to random
@mc@mathstodon.xyz avatar

This is the paper I needed to find! It lays out some basic facts about the structure of [0,oo] as an object of truth-values. It shows [0,oo] is a *-autonomous category, hence a model of classical linear logic (and more!)

https://www.emis.de/journals/JHRS/volumes/2007/n2a9/v2n2a9hl.pdf

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

@mc - the problem is that the "parallel" composition of circuits is not the tensor product in a monoidal category, it's an operation that takes two circuits with one input and one output

[ f , g : x \to x ]

and produces another such circuit like this:

[ m \circ (f \otimes g) \circ \Delta : x \to x ]

where (\Delta : x \to x \otimes x) "splits" one wire into two and (m: x \otimes x \to x) "joins" two wires into one. (\Delta) and (m) come from the basic object (x) being a Frobenius monoid.

I think this Frobenius stuff for electric circuits was worked out here:

https://arxiv.org/abs/1609.05382

before Sobocinski and collaborators did it. I've thought a lot about "harmonic sum" but it's still mysterious to me.

hallasurvivor, to science

Someone on mse asked a cute question last night. I answered it, but had a kind of weird experience doing so, and I'm curious if other people will have a similar experience.

The isn't so hard, so let me leave it here for a day so that people can have a chance to try it themselves. Then I'd love to use it to start a conversation about two approaches to problem solving in

Please post solutions as a reply to this, but put them in cw tags so that people can go spoiler-free if they want to! I'll @ everyone who replied when I post the followup sometime soon ^_^.

You and your friend play a game. Your friend gets to color each point on the unit circle either red or blue (she has very fine-tipped crayons) and you have to try and find 3 points on an equilateral triangle which get the same color.

You win if you can find such a triangle, and your friend wins if her coloring is able to stump you.

Can you always win? If not, how should your friend color the points to stop you from winning?

johncarlosbaez,
@johncarlosbaez@mathstodon.xyz avatar

@hallasurvivor - if the answer starts with something like "Assuming the axiom determinacy...", this is not the kind of game I want to play. 🙃

johncarlosbaez,
@johncarlosbaez@mathstodon.xyz avatar

@hallasurvivor -

This seems too easy. Iif she colors half the circle red and half blue, isn't that enough for her to win? An equilateral triangle inscribed in the circle always has at least one vertex in each half?

More specifically, she can chop the circle [0,2π) into the two halves' [0, π) and [π, 2π).

johncarlosbaez,
@johncarlosbaez@mathstodon.xyz avatar

@hallasurvivor - I came up with a solution, which alas was less clever than others, though it seems to work fine.

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

A minus sign can make a huge difference. Einstein discovered that the difference between space and time is all due to a minus sign.

Another amazing fact is that the difference between 'matter particles' (or more precisely fermions, like electrons, quarks, etc.) and 'force particles' (bosons, like photons, gluons, etc.) is mainly due to a minus sign:

When you switch two fermions, their quantum state gets multiplied by -1, while when you switch two bosons it get multiplied by 1.

This was discovered by Pauli, who realized that there must be some reason why the electrons in atoms go into 'shells' - why all the electrons in a big atom like iron don't all fall into the same lowest-energy state. The reason is that if two electrons were in the same state, switching them would do nothing but also multiply that state by -1: a contradiction. This rule, that fermions can't be in the same state, is called the Pauli exclusion principle.

Bose and Einstein realized that on the contrary, bosons actually like to be in the same lowest energy state at low temperatures! This is called Bose-Einstein condensation. Similarly, a laser beam has many photons in the same state.

Later people realized that if we replace vector spaces (like the Hilbert space of quantum states of some system) by super vector spaces, where every vector is a sum of a bosonic and fermionic part, we can impose a rule saying that switching two fermionic vectors should always introduce an extra minus sign.

It turns out that this rule is not arbitrary - it's mathematically very natural and it's lurking around all over in mathematics, even in contexts that superficially have nothing to do with bosons or fermions!

johncarlosbaez,
@johncarlosbaez@mathstodon.xyz avatar

@pgcd - great, thanks! I try to share the gems that many experts seem to prefer keeping hidden. 🙃

Seriously, I don't think they're actually trying to keep secrets - they just don't think hard about how to explain things.

johncarlosbaez,
@johncarlosbaez@mathstodon.xyz avatar

@j_bertolotti - Nice! But - nitpicky mathematical physicist speaking - I don't think it's reasonable to argue that the eigenvalues of P must be real, since there's no good reason to demand that switching corresponds to a self-adjoint operator. Sure it would be nice if this operator is an observable... but what it really needs to be is a unitary operator, snce a symmetry!

So, its eigenvalues need to be unit complex numbers, and again P² = 1 implies they must be 1 or -1.

But wait!

Even this argument is screwed up. In fact, not every self-adjoint operator or unitary operator has eigenvalues. They may have continuous spectrum.

Luckily, we don't even need to know that P is unitary! Here's the real argument:

As soon as we know any operator P has P² = 1, then every vector ψ is the sum of an eigenvector of P with eigenvalue 1, and an eigenvector of P with eigenvalue -1. Namely

(1+P)ψ/2

and

(1-P)ψ/2

johncarlosbaez,
@johncarlosbaez@mathstodon.xyz avatar

@jsdodge @j_bertolotti - the main thing about anyons is that switching them twice by moving them around does not get back to same state! Now we need representations of the braid group, not the symmetric group.

johncarlosbaez,
@johncarlosbaez@mathstodon.xyz avatar

@abuseofnotation - the stuff about space and time should be explained in any book on special relativity, but this video is a way to start:

https://www.youtube.com/watch?v=ZdrZf4lQTSg

Btw, people who think about special relativity often use units where the speed of light, c, is just 1. Then Sabine's formula simpifies and you see the difference between space and time is merely a minus sign!

johncarlosbaez,
@johncarlosbaez@mathstodon.xyz avatar

@bici - here's a quick explainer on how a minus sign makes fermions different from bosons:

https://www.youtube.com/watch?v=skFU7pmBOys

johncarlosbaez,
@johncarlosbaez@mathstodon.xyz avatar

@bici - Itchy brain? Just spread some hydrocortisone cream on it. 🙃

j_bertolotti, to random
@j_bertolotti@mathstodon.xyz avatar

Dear journals asking me to review a paper: what makes you think I am going to create an account on your system just to decline your invitation?

johncarlosbaez,
@johncarlosbaez@mathstodon.xyz avatar

@j_bertolotti - indeed! So what do you do?

JadeMasterMath, to random
@JadeMasterMath@mathstodon.xyz avatar

Question: Would you say that a preorder is a "depleted category"? Do you know a reference which uses depleted in this way?

johncarlosbaez,
@johncarlosbaez@mathstodon.xyz avatar

@JadeMasterMath - I've never seen that in writing. But people do really call a poset a "thin" category!

https://ncatlab.org/nlab/show/thin+category

MartinEscardo, to random
@MartinEscardo@mathstodon.xyz avatar

It appears that twitter is experiencing a rapid unscheduled disassembly.

johncarlosbaez,
@johncarlosbaez@mathstodon.xyz avatar

@MartinEscardo - what's today's outrage?

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

Yesterday I said a bit about representations of Clifford algebras and raised a puzzle, which @4raylee answered. This chart summarizes what's going on.

For example we get Cliff₂ by taking the real numbers and throw in 2 anticommuting square roots of -1, called i and j. Their product gives another, called k. So we get the quaternions! Thus the category of representations of Cliff₂, Rep(Cliff₂), is the category of quaternionic vector spaces. And so on.... the pattern repeats mod 8.

The puzzle was: what's the visible pattern in this chart? And the answer is: it has bilateral symmetry across the diagonal line from Cliff₃ to Cliff₇.

Why such a weird diagonal line? It's because we're doing things a bit wrong! The Clifford algebras aren't just algebras: they are 'superalgebras', meaning that every element is a sum of two parts, which we can jazzily call the bosonic and the fermionic part, and multiplication obeys these rules:

bosonic × bosonic = bosonic
bosonic × fermionic = fermionic
fermionic × bosonic = fermonic
fermionic × fermionic = bosonic

These rules are motivated by pure math and also what happens in nature when you combine bosons and fermions into bigger particles.

How do we make the Clifford algebras into superalgebras? We just decree that the square roots of -1 we throw in are fermionic. In Cliff₂ this means that i and j are fermionic and k is bosonic. That may seem weird, but that's because we're getting the quaternions from studying 2d space, which is also a bit weird.

Believe it or not, working with superalgebras and their super-representations takes our chart and rotates it a bit, so the weird diagonal line becomes a vertical line!

(1/2)

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

It's super time!

A 'super vector space' is a vector space where each element is a sum of a bosonic and a fermionic part. And superalgebra can have a super-representation on a super vector space. It's just like an ordinary representation except that when we let guys in our superalgebra act on guys in our superalgebra, we require these rules:

bosonic × bosonic = bosonic
bosonic × fermionic = fermionic
fermionic × bosonic = fermonic
fermionic × fermionic = bosonic

This chart shows the categories of super-representations of the Clifford algebras. Amazingly, it's just like the previous chart rotated an eighth of a turn clockwise! Now the axis of symmetry is the vertical line!

This doesn't explain yet why the symmetry exists in the first place, but it's a step in the right direction.

As a quick sanity check, think about the category of super-representations of Cliff₀. Cliff₀ is just ℝ, with every element bosonic. So a super-representation of Cliff₀ is just a super-vector space where you can multiply by real numbers... which is just a super-vector space. But a super-vector space is just a vector space V split as a direct sum of two parts:

V = V₀ ⊕ V₁

where to act cool we are calling V₀ the bosonic part and V₁ the fermionic part. So yes, the category of super-representations of Cliff₀ is just the category of real vector spaces split into two parts in this way!

So at least that case checks out.

If you missed my more elementary post about this stuff yesterday, you can see it here:

https://mathstodon.xyz/@johncarlosbaez/110255214021130500

And by the way, if you're scared to ask questions, please know that I'm really just trying to make conversation (in my own nerdy and inept way).

(2/2)

johncarlosbaez,
@johncarlosbaez@mathstodon.xyz avatar

@leemph - it's an amazing fact that the difference between "matter particles" (fermions, like electrons, quarks, etc.) and "force particles" (bosons, like photons, gluons, etc.) is mainly due to the fact that when you switch two matter particles, their quantum state gets multiplied by -1, while when you switch two identical force particles it get multiplied by 1.

This was discovered by Pauli, who realized that there must be some reason why the electrons in atoms go into "shells" - why all the electrons in a big atom like iron don't all go to the same lowest-energy state. The reason is that if two electrons were in the same state, switching them would do nothing but also multiply that state by -1: a contradiction.

Bose and Einstein realized that on the contrary, bosons can be in the same lowest energy state, and indeed that's what they do at low temperatures. This is called a Bose-Einstein condensate.

Later people realized that if we replace vector spaces (like the space of quantum states of some system) by super vector spaces, we can build in this "switching rule": switching two fermionic vectors should always introduce an extra minus sign.

Later still it was realized that this rule is not arbitrary - it's mathematically very natural and it's lurking around all over in mathematics, even in contexts that superficially have nothing to do with bosons or fermions!

That's a short version of the story. But there's a lot more to it. It's one of the most amazing discoveries of the 20th century.

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

@leemph "To me, as a math fan, would be like discovering that the basic ontology of the universe comes in two radically different kind of sorts, not just one, eg. like a two sorted theory."

Exactly! People call this insight "supermathematics".

"Then is switching is the practical action of moving one in the place of the other and vice versa, or is more like an instantaneous mental experiment of switching, like writing ab=(-1)ba?"

Both! Mathematically we can specify a state

ψ = "electron 1 is in state a and electron 2 is in state b"

and a state

ϕ = "electron 1 is in state a and electron 2 is in state b"

Then the surprise is that when we describe ψ and ϕ as vectors in a vector space we have

ψ = -ϕ.

But physically, one way to turn state ψ into state ϕ is by grabbing the two electrons somehow and moving them so they trade places. (This is harder than it sounds,, but doable.) Then you can show experimentally that this process actually turned state ψ into state -ψ.

"what happens if I stop that action in the middle of it? Does the state get multiplied by (-1)^(1/2)?"

No, you just get two electrons in some other locations, like when you're trying to switch chairs between two apartments and you pull over to take a lunch break halfway thorugh the job!

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

In the Clifford algebra Cliffₙ we start with the real numbers and then throw in n anticommuting square roots of -1. For example:

Cliff₀ = ℝ, Cliff₁ = ℂ, Cliff₂ = ℍ (the quaternions), etc.

With the quaternions, once you throw in i and j with i² = j² = -1 and make them anticommute (ij = -ji) you get k = ij for free.

Each Clifford algebra has 'representations': roughly, real vector spaces where elements of the Clifford algebra act as linear operators. The most famous representations of Cliffₙ are the 'pinor' representations, which describe spin-1/2 particles in n-dimensional space along with how reflections act on these particles. You get all the other representations by taking direct sums of pinor representations.

For example, in 2d space Cliff₂ = ℍ has a representation on itself, and this is the only pinor representation in 2d space. All other representations of Cliff₂ are direct sums of this one - so its category of representations is the category of quaternionic vector spaces!

This chart shows the categories of representations of the Clifford algebras up to dimension 7. After that they repeat.

The symbol ≃ means that two algebras have equivalent categories of representations. For example, Cliff₆ is the category of 8×8 real matrices! So it's not isomorphic to ℝ, but you can show they have equivalent categories of representations.

One last thing: Cliff₃ is isomorphic to ℍ⊕ℍ, meaning an element is a pair of quaternions. So a representation is a pair of quaternionic vector spaces - or equivalently, a quaternionic vector space that's been 'split' as the direct sum of two pieces. That's what I mean by 'split' here.

Do you see the surprising pattern in this chart? Very visible.

johncarlosbaez,
@johncarlosbaez@mathstodon.xyz avatar

@4raylee - the octonions are not Clifford algebras because they're not associative, and Clifford algebras are associative. Indeed in my post "algebra" meant "associative algebra" - a common usage!

Still, there is a profound connection between octonions and Clifford algebras. But that's another story for another day.

"Other than that the only other surprising thing to me (so far) is the progression R -> C -> H is reversed on the other side of the split."

That's the surprising pattern! Can you describe it as a symmetry of the picture?

"And no C ⨁ C?"

That's a good observation too! Here I've been talking about real Clifford algebras, where you take the real number and throw in a bunch of anticommuting square roots of -1. C ⨁ C shows up when we look at complex Clifford algebras.

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

@4raylee - right! There's a visually evident symmetry is across the diagonal line from Cliff₃ to Cliff₇. And this is tantalizing and also weird, because it might seem more natural to have the symmetry axis to be vertical (since Cliff₀ is where the whole story starts).

I think there's an explanation of this, but it's too long to fit into this toot. 🙃

BartoszMilewski, to random
@BartoszMilewski@mathstodon.xyz avatar

Does a point in space exist if there's nobody around to watch it? Einstein spent three years pondering this question. It turns out that, if there is a hole in space, devoid of any matter, the solution to Einstein's equations is not uniquely determined. You can take one solution for the metric tensor and "slide it around" the hole, like a patch of loose skin, to get another solution. The problem is that this new solution gives different values for physical curvature. So if there is a bump in curvature, say, at the center of the hole, you can slide it off center, and it still satisfies Einstein's equations. The very idea of the "center" of the hole makes no sense in the absence of matter. You have to put a material observer in it to pin the field down to a given location. https://en.wikipedia.org/wiki/Hole_argument

johncarlosbaez,
@johncarlosbaez@mathstodon.xyz avatar

@BartoszMilewski - in loop quantum gravity we (meaning mainly Rovelli and Smolin) tried to eliminate spacetime and make locations relative to matter (and the gravitational field).

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