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Stream: deprecated: algebraic geometry

Topic: Galois cohomology

John Baez (Apr 24 2020 at 23:39):

The subject header "Galois cohomology" sounds rather grand, but in this n-Cafe article I'm just trying to understand crossed homomorphisms to better understand an application of Galois cohomology:

• Crossed homomorphisms (part 1) .

I took a book I'm trying to read and extracted this result by generalizing what they actually said:

Theorem. Suppose we are given a group $G$ acting as not-necessarily-natural transformations of the identity functor on a category $X$, and an isomorphism $f \colon x \to y$ in $X$. Then we get an action of $G$ on the group $\mathrm{Aut}(x)$ and a crossed homomorphism $\phi \colon G \to \mathrm{Aut}(x)$. If $G$ acts as natural transformations, the crossed homomorphism is trivial.

Weird, huh? If you don't know what a "crossed homomorphism" is, don't worry, I explain it.

John Baez (Apr 25 2020 at 22:22):

I think I'm getting a better handle on the topology and category theory lurking behind the first cohomology of a group $G$ with coefficients in a (not necessarily abelian) group $A$ on which $G$ acts as automorphsms. I explained it in these two comments.

I'm still waiting for someone who really understands this stuff to weigh in. So far I've had two experts say they'd rather let me figure this out myself! I'm not sure if that means 1) it's trivial or 2) they don't really understand it themselves. :angry:

John Baez (Apr 27 2020 at 23:01):

I wrote up a much better explanation of the first cohomology of a group, how it comes up in the classification of "homotopy fixed points" of a group action on a category, and how this gets used in "Galois descent":

$\;$

• Group cohomology and homotopy fixed points.

Tim Hosgood (Apr 28 2020 at 10:41):

this is a really interesting story indeed!

John Baez (May 08 2020 at 17:53):

Thanks! It's just getting started.

John Baez (May 08 2020 at 21:03):

Okay, I think I'd like to talk about Galois cohomology and Galois descent a bit more.

John Baez (May 08 2020 at 21:10):

There's so much to say I have trouble knowing where to start... but here's something very nice and categorical.

Fix a field $k$. There's a bicategory with:

• finite-dimensional algebras over $k$ as objects
• bimodules as morphisms
• bimodule homomorphisms as 2-morphisms.

This is a monoidal bicategory since we can take the tensor product of finite-dimensional algebras, and everything else works nicely with that.

Given any monoidal bicategory we can take its core: that is, the sub-monoidal bicategory where we only keep invertible objects (invertible up to equivalence), and invertible morphisms (invertible up to iso-2-morphism), and invertible 2-morphisms.

The core is a monoidal bicategory where everything is invertible (in a suitably weakened sense) so it's called a 3-group.

The particular 3-group we get from our field $k$ is called the Brauer 3-group of $k$, $\mathbf{Br}(k)$.

It contains a lot of information that algebraists had already been studying before anyone knew about fields!

John Baez (May 08 2020 at 21:12):

It's discussed here, but they call it the "Picard 3-group":

https://ncatlab.org/nlab/show/Brauer+group#RelationToCatsOfModules

John Baez (May 08 2020 at 21:14):

Puzzle. Can you guess how a lot of what I just said can be abstracted a bit so it's not about commutative algebras over a field $k$?

John Baez (May 08 2020 at 21:20):

Anyone have any questions or comments? I guess my next step would be to talk in a bit more detail about the Brauer 3-group of a field and what information it contains, and then a bit how you can actually "compute" it using Galois theory.

John Baez (May 08 2020 at 21:21):

I'm into this because of a project on condensed matter physics... it's a long story... but it's probably easier to just focus on the algebra here.

John Baez (May 08 2020 at 21:40):

By the way, if anyone is intimidated by "monoidal bicategories", I can work with you on that.

Joe Moeller (May 09 2020 at 00:27):

So this core is like considering the monoidal bicategory as a tricategory and taking what I guess one might call the core ordinarily? Meaning, the maximal sub (3-)groupoid

Morgan Rogers (he/him) (May 09 2020 at 10:41):

John Baez said:

It's discussed here, but they call it the "Picard 3-group":

https://ncatlab.org/nlab/show/Brauer+group#RelationToCatsOfModules

Probably because to an algebraic geometer it looks like a Picard group! (group of invertible sheaves over a variety)
What Joe said sounds reasonable answer to your puzzle, although if so you gave the game away a little by explicitly pointing out that we were working with a monoidal bicategory :stuck_out_tongue_wink:

Simon Burton (May 09 2020 at 12:18):

John Baez said:

I'm into this because of a project on condensed matter physics... it's a long story... but it's probably easier to just focus on the algebra here.

I'm very interested in the cond-mat connection if you want to say a bit more about that...

John Baez (May 09 2020 at 22:59):

Joe Moeller said:

So this core is like considering the monoidal bicategory as a tricategory and taking what I guess one might call the core ordinarily? Meaning, the maximal sub (3-)groupoid

Right, exactly: the core of any sort of n-category is the biggest sub-thing that only contains weakly invertible morphisms, 2-morphisms etc. - so it's an n-groupoid.

We can take the core of a monoidal n-category, which is a one-object (n+1)-category, and we get a one-object (n+1)-groupoid, which is called an n-group.

John Baez (May 09 2020 at 23:02):

Morgan Rogers said:

John Baez said:

It's discussed here, but they call it the "Picard 3-group":

https://ncatlab.org/nlab/show/Brauer+group#RelationToCatsOfModules

Probably because to an algebraic geometer it looks like a Picard group! (group of invertible sheaves over a variety).

Well, Picard groups are sitting inside the Brauer 3-group, but so are Brauer groups.

John Baez (May 09 2020 at 23:02):

Let me say a bit more about how it works. I'll do it a bit more generally than before, because it's simpler. We fix a commutative ring $k$. (Before I was using a field.)

John Baez (May 09 2020 at 23:03):

There's a monoidal bicategory with:

• algebras over $k$ as objects
• bimodules as morphisms
• bimodule homomorphisms as 2-morphisms.

John Baez (May 09 2020 at 23:03):

Now let's think about the core of this.

John Baez (May 09 2020 at 23:04):

So, the objects will be algebras $A$ over $k$ that are invertible up to equivalence.

John Baez (May 09 2020 at 23:05):

What are those like?

John Baez (May 09 2020 at 23:05):

They're algebras $A$ that have an algebra $B$ with $A \otimes B \simeq k$ and $B \otimes A \simeq k$.

John Baez (May 09 2020 at 23:05):

Here the tensor product is the tensor product of $k$-algebras.

John Baez (May 09 2020 at 23:06):

The concept of equivalence is determined by the monoidal bicategory we're working in... I could say more about it, but it's called Morita equivalence.

John Baez (May 09 2020 at 23:07):

You can read about Morita equivalence for rings on Wikipedia:

• Morita equivalence.

but it works the same way for $k$-algebras. (A ring is a $k$-algebra with $k = \mathbb{Z}$.)

John Baez (May 09 2020 at 23:08):

So, objects in the core, i.e. objects in the Brauer 3-group, will be algebras $A$ that are "weakly invertible" in this sense: they have an algebra $B$ such that $A \otimes B$ and $B \otimes A$ are Morita equivalent to $k$.

John Baez (May 09 2020 at 23:10):

For example, suppose $k = \mathbb{R}$. Is the quaternions $\mathbb{H}$ a weakly invertible algebra?

John Baez (May 09 2020 at 23:10):

Yes, because it is its own inverse!

John Baez (May 09 2020 at 23:10):

$\mathbb{H} \otimes \mathbb{H}$ is isomorphic to the algebra of $4 \times 4$ real matrices.

John Baez (May 09 2020 at 23:11):

And it's a fact that a matrix algebra over any commutative ring $k$ is Morita equivalent to $k$ itself.

John Baez (May 09 2020 at 23:13):

Hmm, this is taking a long time! I should explain Morita equivalence - this is all a bit mysterious if one doesn't get that idea! But for now I'll just say that algebras that are weakly invertible in the sense I just described are called Azumaya algebras.

John Baez (May 09 2020 at 23:14):

It's clear from abstract general nonsense that Azumaya algebras taken up to Morita equivalence will form a group, with tensor product giving the multiplication.

John Baez (May 09 2020 at 23:15):

This is just because whenever you have an $n$-group, objects up to equivalence form a group.

John Baez (May 09 2020 at 23:15):

In the case at hand, though, this group is famous: it's called the Brauer group of our commutative ring $k$.

John Baez (May 09 2020 at 23:16):

So for example the Brauer group of $\mathbb{R}$ contains an element $[\mathbb{H}]$ that squares to the identity.

John Baez (May 09 2020 at 23:16):

So, the Brauer 3-group is a kind of high-powered way to understand the Brauer group.

John Baez (May 09 2020 at 23:18):

Here's a summary of what I've already said: to get the Brauer group of $k$, you take the monoidal bicategory of $k$-algebras, then you take its core to get the "Brauer 3-group", then you take objects up to equivalence and get the Brauer group.

John Baez (May 09 2020 at 23:18):

In this process we've thrown away a lot of interesting stuff, like the 2-morphisms and 3-morphisms.

John Baez (May 09 2020 at 23:19):

"Picard groups" are hiding in the morphisms of the Brauer 3-group.

John Baez (May 09 2020 at 23:21):

Simon Burton said:

John Baez said:

I'm into this because of a project on condensed matter physics... it's a long story... but it's probably easier to just focus on the algebra here.

I'm very interested in the cond-mat connection if you want to say a bit more about that...

This is a huge story unto itself, but luckily I've written something about it already:

$\;$

• The tenfold way.

You'll see that the "super-Brauer groups" of $\mathbb{R}$ and $\mathbb{C}$ show up here. These are used to classify matter in 10 kinds. These ten kinds have different properties with respect to time reversal (T) and switching particles and holes (C).

John Baez (May 09 2020 at 23:22):

I'm trying to understand this stuff better, so I'm digging deeper into Brauer groups.

John Baez (May 09 2020 at 23:29):

I feel like I'm sort of drowning in material I'd like to explain here.

Reid Barton (May 09 2020 at 23:31):

If I have, say, a roll of toilet paper, is it made up of one (or more) of these ten "kinds of matter"? Or is this a meaningless question?

John Baez (May 09 2020 at 23:34):

I don't know enough about the microstructure of toilet paper to easily answer that.

John Baez (May 09 2020 at 23:34):

Is the Hamiltonian of toilet paper time-reversal invariant? I don't know.

John Baez (May 09 2020 at 23:36):

Does it have a symmetry that interchanges particles and holes? I don't know.

John Baez (May 09 2020 at 23:36):

People usually focus on things like crystals of various kinds - things you can understand mathematically at the atomic level.

John Baez (May 09 2020 at 23:37):

So your question is not meaningless, but I've never read condensed matter theorists talking about the tenfold way for complicated materials like paper.

Reid Barton (May 09 2020 at 23:39):

I see. I thought maybe it would be an easier question to answer, like electrons are type 5 but muons are type 8 or something like that.

Reid Barton (May 09 2020 at 23:39):

But I guess this is at a larger scale than that?

John Baez (May 09 2020 at 23:41):

The tenfold way mainly comes up in condensed matter physics, though it could also be applied to particle physics.

David Michael Roberts (May 10 2020 at 01:43):

@Reid Barton yes, it's more like classifying crystal lattices, than fundamental particles. This relates to things like topological insulators, the fractional quantum Hall effect and so on. One can also use various flavours of K-theory of suitably periodic spaces (as in, tori, modelling objects with local translation symmetry) to classifying these. The relation of this to the Brauer-stuff @John Baez is talking about seems mysterious to me, though there's probably some spectrum/bundle of spectra-level way to look at it.

John Baez (May 10 2020 at 02:28):

This is the sort of thing I'm trying to get into. But I haven't even gotten into the twisted K-theory aspect of that paper by Freed and Moore you're alluding to! It should be connected, because the so-called bigger Brauer group classifies gerbes.

John Baez (May 10 2020 at 02:32):

So far I'm just trying to understand really clearly the relation between the Brauer group of a field $k$ and the group cohomology $H^2(\mathrm{Gal}(k), K^\times)$, where $K$ is the separable closure of $k$. This is old stuff. I think I get the basic idea but it's gradually getting more obvious.

John Baez (May 10 2020 at 02:34):

You can also think of $H^2(\mathrm{Gal}(k), K^\times)$ as classifying gerbes of some sort.

John Baez (May 10 2020 at 02:51):

The idea is to think of $G = \mathrm{Gal}(k)$ as the fundamental group of the space $BG$, and then look at $K^\times$-gerbes over this space.

Joe Moeller (May 11 2020 at 15:55):

John Baez said:

$\mathbb{H} \otimes \mathbb{H}$ is isomorphic to the algebra of $4 \times 4$ real matrices.

Why is this true?

John Baez (May 11 2020 at 16:26):

Joe Moeller said:

John Baez said:

$\mathbb{H} \otimes \mathbb{H}$ is isomorphic to the algebra of $4 \times 4$ real matrices.

Why is this true?

Finally, some questions about what I said!

Any algebra $A$ acts on itself by left and right multiplication, so you always get a representation of $A \otimes A^{\mathrm{op}}$ on $A$. The quaternions have $\mathbb{H} \cong \mathbb{H}^{\mathrm{op}}$, so $\mathbb{H} \otimes \mathbb{H}$ acts on $\mathrm{H}$. Since $\mathbb{H}$ is a real vector space of dimension 4, this means there's some homomorphism from $\mathbb{H} \otimes \mathbb{H}$ to the algebra of $4 \times 4$ matrices. You can check this is one-to-one and onto: in fact you just need to check one of these two things, since the dimension of $\mathbb{H} \otimes \mathbb{H}$ is 16, and so is the dimension of the space of $4 \times 4$ real matrices. So, $\mathbb{H} \otimes \mathbb{H}$ is isomorphic to the algebra of f $4 \times 4$ matrices.

That's a sketch of how it works. Basically I described a homomorphism from $\mathbb{H} \otimes \mathbb{H}$ to the algebra of $4 \times 4$ real matrices, and pointed out that they have the same dimension. It's just a calculation at this point to show they're isomorphic. Using facts about quaternions one can make this calculation pretty slick.

John Baez (May 11 2020 at 16:30):

Here's a super-slick way: $\mathbb{H}$ is a simple algebra whose center is just the ground field $\mathbb{R}$. It follows (using the theory of central simple algebras) that $\mathbb{H} \otimes \mathbb{H}$ is again a simple algebra. So it doesn't have any nontrivial ideals. So the homomorphism $\mathbb{H} \otimes \mathbb{H}$ from to the algebra of $4 \times 4$ matrices must be one-to-one... and therefore onto.

John Baez (May 11 2020 at 16:30):

There are other ways that use the geometry of the quaternions instead of fancy algebra facts. I guess I like these better.

John Baez (May 11 2020 at 16:34):

It's also fun to see why the quaternions are isomorphic to their op. That's not instantly obvious, since the quaternions aren't commutative!

Joe Moeller (May 11 2020 at 16:35):

John Baez said:

Here's a super-slick way: $\mathbb{H}$ is a simple algebra whose center is just the ground field $\mathbb{R}$. It follows (using the theory of central simple algebras) that $\mathbb{H} \otimes \mathbb{H}$ is again a simple algebra. So it doesn't have any nontrivial ideals. So the homomorphism $\mathbb{H} \otimes \mathbb{H}$ must be one-to-one.

I think you're missing the homomorphism

John Baez (May 11 2020 at 16:41):

I explained the homomorphism earlier, but yeah this sentence is broken. I meant:

Here's a super-slick way: $\mathbb{H}$ is a simple algebra whose center is just the ground field $\mathbb{R}$. It follows (using the theory of central simple algebras) that $\mathbb{H} \otimes \mathbb{H}$ is again a simple algebra. So it doesn't have any nontrivial ideals. So the homomorphism from $\mathbb{H} \otimes \mathbb{H}$ to 4x4 matrices must be one-to-one.

Joe Moeller (May 11 2020 at 21:12):

Maybe people talking about the physics side of stuff can split off into a different topic. I'd like to be able to read about the algebra without scrolling past a discussion about string theory.

Joe Moeller (May 11 2020 at 21:41):

There are two broader tricategories I can imagine that would include the ones at the beginning of this as subcategories. To get one, you consider the ones you gave as a family of monoidal bicategories indexed by fields or rings, deloop the fibres, then Grothendieck construction to give a tricategory with [rings, algebras, bimodules, bimod homs]. To get the other, you just Grothendieck the bicategories, and give the result a monoidal structure by the appropriate upgrade of monoidal Grothiendieck construction, then deloop, giving [*, ring algebra pairs, bimodules, bimod homs]. I wonder if either or both could be seen as useful in this story.

John Baez (May 11 2020 at 22:02):

Yes, let's talk about Galois cohomology and Brauer groups!

John Baez (May 11 2020 at 22:03):

I'm definitely interested in the tricategory

[rings, algebras, bimodules, bimodule homomorphisms].

It's easiest to get this "directly", not by focusing on one ring and then repairing that mistake by doing the Grothendieck construction.

John Baez (May 11 2020 at 22:05):

When we do focus on one ring $k$, of course we get a monoidal bicategory

[$k$-algebras, bimodules, bimodule homomorphisms].

But when $k$ is commutative - the case people think about most - this is actually a symmetric monoidal bicategory. For a lot of purposes I'm happy to live in here.

John Baez (May 11 2020 at 22:07):

I haven't been thinking about

[*, ring algebra pairs, bimodules, bimodule homomorphisms]

I guess I just don't want to treat a ring and an algebra as a "single thing" in this story.

John Baez (May 11 2020 at 22:07):

So here is the first big idea I'm aiming at in this story:

John Baez (May 11 2020 at 22:10):

If we take the "core" of the monoidal bicategory

[$k$-algebras, bimodules, bimodule homomorphisms]

we get the Brauer $3$-group of $k$, consisting of

[Azumaya algebras over $k$, Morita equivalences, bimodule isomorphisms]

I haven't said nearly enough about what Azumaya algebra and Morita equivalences are like! But anyway, any 3-group gives 3 groups called $\pi_1, \pi_2$ and $\pi_3$.

John Baez (May 11 2020 at 22:11):

In our example

$\pi_1$ = {equivalence classes of Azumaya algebras]

where "equivalence" the usual notion for objects in a bicategory.

Joe Moeller (May 11 2020 at 22:11):

I guess one way a category of ring module pairs can be useful is in defining modules of sheaves, as I talked about in the seminar two weeks ago.

John Baez (May 11 2020 at 22:12):

$\pi_1$ is called the Brauer group of $k$.

John Baez (May 11 2020 at 22:14):

I could explain $\pi_2$ and $\pi_3$, but let me get straight to the first really interesting fact: if $k$ is a field then

$\pi_1 \cong H^2(\mathrm{Gal}(K|k), K^\times)$

where $K$ is the separable closure of $K$ (kinda like the algebraic closure).

John Baez (May 11 2020 at 22:14):

The cohomology being used here is "group cohomology" - the cohomology of a group with coefficients in a module.

Joe Moeller (May 11 2020 at 22:15):

I'm sorta confused by your placement of Morita equivalences as the 1-morphisms.

John Baez (May 11 2020 at 22:15):

I've got a bicategory with Azumaya algebras as objects, Morita equivalences as 1-morphisms and bimodule isos between them as 2-morphisms.

John Baez (May 11 2020 at 22:16):

It's a monoidal bicategory.

John Baez (May 11 2020 at 22:16):

Of course there are lots of numbering-shift games you can play here.

John Baez (May 11 2020 at 22:16):

But I'm thinking of this as a 3-group, meaning a bicategorical analogue of a group, and the "group elements" are now objects in the bicategory.

John Baez (May 11 2020 at 22:17):

They're algebras of some sort, and you can multiply them (tensor them).

John Baez (May 11 2020 at 22:18):

When we take $\pi_1$, we're crushing this 3-group down to a 1-group: we're decategorifying it twice.

John Baez (May 11 2020 at 22:19):

So the "Brauer 3-group" becomes the "Brauer group".

John Baez (May 11 2020 at 22:21):

Get it?

Joe Moeller (May 11 2020 at 22:25):

My problem was that I was thinking of "a Morita equivalence" as the equivalence between the categories of modules, but here it's a bimodule which represents that functor, right?

John Baez (May 11 2020 at 22:25):

Oh, good! There are two equivalent ways of thinking about Morita equivalence.

John Baez (May 11 2020 at 22:26):

Say $k$ is a field. Two $k$-algebras $A$ and $B$ are Morita equivalent iff they have equivalent categories of representations, where we look at their representations in $\mathrm{Vect}_k$.

John Baez (May 11 2020 at 22:28):

But they are also Morita equivalent iff there's an $A,B$-bimodule $M$ that has an "weak inverse": a $B,A$-bimodule $N$ such that $M \otimes N$ and $N \otimes M$ are isomorphic to the identity. (Here I'm tensoring over whatever makes sense!!!)

John Baez (May 11 2020 at 22:30):

Right now I'm mainly using the second viewpoint, since then a "Morita equivalence" is just an equivalence in the bicategory

[$k$-algebras, bimodules, bimodule homomorphisms].

John Baez (May 11 2020 at 22:31):

So, you don't need for now to understand why the two ways of thinking about Morita equivalence are equivalent!

John Baez (May 11 2020 at 22:32):

However, the equivalence of these two viewpoint is what makes Morita equivalence so beautiful.

Joe Moeller (May 11 2020 at 22:36):

ok, got it.

John Baez (May 11 2020 at 22:37):

Good. It's easy to see why the module definition implies the category-of-representations definition; the other way requires scratching my head a bit.

John Baez (May 11 2020 at 22:38):

But anyway, the cool part is that if $k$ is a field then its Brauer group is isomorphic to

$H^2(\mathrm{Gal}(K|k), K^\times)$

where $K$ is the separable closure of $K$ (kinda like the algebraic closure).

John Baez (May 11 2020 at 22:39):

The proof in Gille-Szaumely was hard for me to follow, but it turns out the reason was they weren't using enough category theory.

Joe Moeller (May 11 2020 at 22:54):

What's the Brauer group good for?

John Baez (May 11 2020 at 22:56):

Gille-Szaumely lists some applications in pure math, and Grothendieck wrote a 3-part paper on it so he must have thought it was interesting.

John Baez (May 11 2020 at 22:58):

I'm mainly interested in its applications to physics. The Brauer group of $\mathbb{R}$ is $\mathbb{Z}/2$, with its two elements being $[\mathbb{R}]$ and $[\mathbb{H}]$. The super-Brauer group of $\mathbb{R}$ is $\mathbb{Z}/8$, with its elements corresponding to the 8 kinds of Clifford algebras. All this is connected to Bott periodicity: complex K-theory has period 2 and real K-theory has period 8. These facts govern which kinds of physics can happen in which dimensions.

John Baez (May 11 2020 at 22:59):

I guess you could strip off the physics here and say it's about geometry and algebra, and how they work differently in different dimensions, with a strong period-8 thing going on.

John Baez (May 11 2020 at 23:00):

I have other things I'm trying to do but I'm not eager to reveal them before I do them.

John Baez (May 11 2020 at 23:01):

So, I don't think it's so much that the Brauer group is "good for" something - it's mainly just an incredibly interesting way to get a group that organizes our perception of algebras over a commutative ring.

John Baez (May 11 2020 at 23:05):

The fact that you can compute it using Galois cohomology is especially exciting, because naively you'd think Galois cohomology was about commutative algebra (field extensions), while the Brauer group is all about noncommutative algebra.

John Baez (May 11 2020 at 23:05):

I'm guessing this is why Hasse, Brauer and Noether teamed up to write a paper about it.

John Baez (May 11 2020 at 23:07):

This mysterious link between commutative algebra and noncommutative algebra makes sense when you understand "Galois descent".