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Stream: community: general

Topic: group cohomology

John Baez (May 19 2020 at 18:57):

Today for some reason I need to compute the cohomology of the group $\mathbb{Z}/2$ with coefficients in the module $\mathbb{C}^\ast$ where $\mathbb{Z}/2$ acts on $\mathbb{C}^\ast$ nontrivially - namely, by complex conjugation.

I'm feeling humbled by the fact that I couldn't easily do this without Theorem 10.35 in Rotman's Homological Algebra, which looks like it gives a quick way to do this.

John Baez (May 19 2020 at 18:58):

Sometimes I feel I know my way around math and other times I feel like a real wimp.

John Baez (May 19 2020 at 18:59):

I guess everyone is like this.

Morgan Rogers (he/him) (May 19 2020 at 19:17):

On the contrary, it sounds like you've just bumped up against the fundamental problem of cohomology: even if you understand what cohomology is and can formally describe it in a bunch of ways, it's still hard work to calculate, even in simple cases. The rust builds up quickly.

John Baez (May 19 2020 at 19:28):

Okay, I feel that someone better than me would just reach into their pocket and pull out a spectral sequence...

Anyway, Rotman's theorem handles exactly the case I need, so I luck out!

Morgan Rogers (he/him) (May 19 2020 at 19:32):

What's the result? I'm curious but lazy

John Baez (May 19 2020 at 19:32):

If I haven't screwed up, the relevant cohomology gives $H^0 = \mathbb{R}^*$, and for higher even degrees $H^{\mathrm{even}} = \mathbb{Z}/2$, which is what I hoped for... but also $H^{\mathrm{odd}} = \mathrm{U}(1)$, which is surprising but cool. I'll have to check my calculation 5 times, especially that last one.

John Baez (May 19 2020 at 19:33):

Hmm, that last one could be wrong...

Morgan Rogers (he/him) (May 19 2020 at 19:34):

Interesting! Thanks for thinking out loud, it's reassuring to see someone else struggling a little with something they feel like they should understand (although the turnaround from not knowing to having completed the calculation was a little dizzying)

John Baez (May 19 2020 at 19:35):

Thank Rotman, he said how to do it...

John Baez (May 19 2020 at 19:36):

Okay, I was mixing up addition and multiplication at one point; straightening this out seems to give $H^{\mathrm{odd}} = \{0\}$, which is more believable.

John Baez (May 19 2020 at 19:36):

This is gonna be really fun to use.

Simon Pepin Lehalleur (May 19 2020 at 20:18):

The vanishing of H^1 is Hilbert 90 for the Galois extension C/R. One can do it by hand in this case because it is cyclic. I like Milne's notes on class field theory for these computations.

Simon Pepin Lehalleur (May 19 2020 at 20:23):

See II.1.20, II.1.22 and II.3.4.

John Baez (May 19 2020 at 20:27):

Right, I outlined a "conceptual" proof of Hilbert 90 here:

• Group cohomology and homotopy fixed points.

Right now I'm trying to understand the higher Galois cohomology groups for this very same Galois extension.

John Baez (May 19 2020 at 20:31):

Thanks for reminding me to read Milne's notes on class field theory. I tried these the other day and found them tough right at the start, perhaps because I'd forgotten a bunch of basic stuff.

Simon Pepin Lehalleur (May 20 2020 at 10:00):

John Baez said:

Right, I outlined a "conceptual" proof of Hilbert 90 here:

• Group cohomology and homotopy fixed points.

Right now I'm trying to understand the higher Galois cohomology groups for this very same Galois extension.

To compute $H^*(Gal(L/K),L^\times)$ in general is very difficult, I think.

For local and global fields, there are rather precise results. For instance, for an extension of local fields, the cohomological formulation of local class field theory due to Tate-Nakayama shows that the cup product with the fundamental class $\theta\in H^2(Gal(L/K),L^\times))$ (generator of this cyclic relative Brauer group)
$\theta\cup -: H^*_T(Gal(L/K),\mathbb{Z})\simeq H^{*+2}_T(Gal(L/K),L^\times)$ is an isomorphism.

In general, there is a closely related group which can be "computed" via the Bloch-Kato conjecture in terms of Milnor K-theory for a general finite Galois extension:
$H^m(Gal(L/K),\mu_n^{\otimes m})$ which when K contains all $n$-th roots of unity is closely related to the n-torsion in $H^m(Gal(L/K),L^\times)$.

Reid Barton (May 20 2020 at 11:15):

(You can use ```latex ... ``` for displays.)

John Baez (May 20 2020 at 14:55):

Thanks, Simon, that's very helpful!

Joe Moeller (May 20 2020 at 18:09):

This is 10.35 in my Rotman: image.png

Joe Moeller (May 20 2020 at 18:09):

are we looking at different editions?

Sahil Imtiyaz (Apr 11 2022 at 09:51):

I am working on computational aspect of Brauer monoid. At first, it looked a bit easy as a generalization of symmetry group and working with tangles. But it seems I am not very far from understanding the real elegance of this algebraic structure. I realized this after reading papers from Swedler, Haige and Novikov etc. It has also some connections with Galois cohomology. I am not a mathematician but I like to exercise my mind in abstract math thing. Can anyone help me in understanding what makes a monoid a Brauer monoid? If possible, assume I am not an expert or a good student.

John Baez (Apr 26 2022 at 18:28):

@Sahil Imtiyaz - I know what the Brauer group of a commutative ring is, and I've written about the Brauer 3-group of a commutative ring. I may have even defined something called the Brauer monoid of a commutative ring - but I'm not sure that's what you're talking about.

Eric M Downes (Mar 16 2024 at 08:08):

Sahil Imtiyaz said:

Can anyone help me in understanding what makes a monoid a Brauer monoid? If possible, assume I am not an expert or a good student.

This is probably too late to be useful, but maybe somebody else will find it helpful.

Daniel Tubbenhauer (no doubt hearing your question) made a nice video answering visually. https://youtu.be/5UeAjqzNdWk?si=PJeR3VHE2IKcdM8k

For the impatient:

Familiar thing. You can realize the symmetric group $S_n, Aut(n)$ on a set as a bunch of strings swapping places from say, left to right: $(1324)\in S_4$ takes string 1 to position 3, string 3 to position 2, 2 to 4, and 4 to 1. Strings are always going in one direction; they never start and end on the "same side" and they never join. (Also the "strings" can pass through one another.)

You can get to the full endomorphism monoid ($T_n$, $End(n)$) on the same set by allowing pictures in which strings do join on the right (many to one maps), but the strings still start on one side and end on the other, because its a set function. So here people use list notation $[1132]$ takes string 1 to pos. 1, 2 joins 1, and string 3 goes to pos. 3, 4 to 2; (we're using 1 indexing).

A different generalization: you don't allow strings to join but you do allow strings to start and end on the same side, which is not a function or even a relation on that set. This is the Brauer monoid "$B_n$". It's maybe better to think of these things topologically, as cobordisms, which is what Daniel does, though there are also combinatorial interpretations in terms of set partitions.

The set partition picture: maybe you feel of the $S_n$ pictures I was drawing in para. 1, that its inefficient to represent position $k$ twice. Ok, for $S_{2n}$, you can represent a permutation+inverse pair by labelling $1..n/2$ on the top and $n/2+1..n$ on the bottom; a partition whose parts are all of size 2. e.g. $\{1,7\}\{3,6\}\{2,8\}\{4,5\}$ achieves the same picture as the permutation in para 1.

What makes the Brauer monoid a monoid is that there are the same number of starting and ending positions; its an endomorphism. What makes it "Brauer" seems to be a convention to do with isomorphism classes. The Brauer monoid on a set is in terms of equivalence classes of elements. The Brauer monoid of a ring involves isomorphism classes of certain algebras over that ring. I don't think its a categorical adjective, though.

John Baez (Mar 16 2024 at 15:56):

I get the impression people usually take formal linear combinations of elements in the monoid you just described, getting an algebra called the Brauer algebra. At least there's a Wikipedia article on the Brauer algebra and not an article on the monoid!

John Baez (Mar 16 2024 at 15:57):

However, I just found a paper that mentions the Brauer monoid you're talking about.

John Baez (Mar 16 2024 at 15:59):

Warning: all this is completely unrelated to the [[Brauer group]] of a field and my generalization which I called the Brauer monoid of a field! This is connected to Galois cohomology.

John Baez (Mar 16 2024 at 16:00):

I don't know which kind of Brauer monoid @Sahil Imtiyaz was talking about.

Eric M Downes (Mar 17 2024 at 01:19):

I suppose I should start saying "one of the X" instead of "the X", when I lack a uniqueness proof. Yet more vindication for the Arkor Principle! At least its not as bad as some algebraists calling magmas "groupoids"!