advanced-algebra

[subtle plaza]

I just adapted the hands on way of proving that Mod(T^2) = SL(2, Z)

I guess this ties back to what we were talking about earlier in terms of strong analogies between Out(F_n) and mapping class groups lol

[last talon]

Yeah lol

[subtle plaza]

I guess one point to be careful about here is orientation. In most texts you'll see people write that the mapping class group of a punctured torus is SL(2, Z). In this case the determinant just keeps track of what the self-homotopy equivalence does to orientation. So maybe you should compose with something to preserve orientation before invoking the Alexander trick

[limpid horizon]

For R noetherian (so that every ideal has a primary decomposition) Ass_R(R/I) is the same as the associated primes of I (from I’s primary decomposition) why is this?

[spice idol]

correspondence theorem + uniqueness

oh i think i misunderstood lol

[limpid horizon]

What did u think i asked

[spice idol]

i thought you were talking about the primes that appear in the decomposition of I/I in R/I for some reason

[limpid horizon]

Meaning like they would correspond to minimal primes containing I ?

[lone jacinth]

If you take the primary decomposition of I and take the intersection of all but one of the primary ideals you get something in R/I annihilated by p^n for p the radical of the last primary.

So you can find something with annihilator p from that

[digital parcel]

Why is e being used in the definition of character? This is from Kumar’s notes

I dont have a problem with it or anything just curious why it’s there

[limpid horizon]

Are u using A being noetherian here somewhere

p^n contained in rad or something

[lone jacinth]

For the existence of a primary decomposition of I

Anyway, my plane is taking off, bye bye

[digital parcel]

Enjoy your flight

Okay I’m looking in Demazure’s original paper and apparently there’s some map c_K: Z[M] -> K(X/B), and c_K(e^lambda) = cl( L(lambda))

Hmmm

[limpid horizon]

Thank u that was helpful

[last talon]

<@&268886789983436800>

[summer quest]

How else do you expect to write down characters

The exponential map is showing up naturally whenever you are translating between additive and multiplicative groups

[digital parcel]

Ah right, thanks

[near lantern]

Is it possible that this was supposed to be Out(ℤ^2)? Just an idle hope.

[last talon]

Nope

This is very specifically a short lecture series on Out(F_n)

Also Out(Z^2) is just Aut(Z^2) so it’s trivial
I wouldn’t’ve asked if it were

[pastel agate]

So I've just had the grothendieck group defined for me as some ridiculously large free abelian group consisting of isomorphism classes of finite length modules quotient a particular relation. It was emphasized in the lecture that before quotienting the free group we have is like absurdly large. From the looks of it it's so large that I'm having trouble even believing we're allowed to do this, it feels like there shouldn't even be a set of all finite length modules for us to be able to define all the other stuff on. The collection of all modules is a class, not a set, and I can find tons of subcollections of modules which also aren't sets. I don't expect to be able to understand fully why but could someone give any sort of indication as to why finite length modules are a set as opposed to other large collections of modules?

[summer quest]

This is a good question and there are several different answers which lead to slightly different definitions of K-theory

[pastel agate]

Interesting, haven't even heard of K-theory before (maybe in passing? idk)

[summer quest]

for any (commutative) ring R you can define the algebraic K-theory groups K_n(R) where K_0(R) is the Grothendieck group you're talking about

these are invariants you can extract from the category of finitely generated projective R-modules, or from perfect complexes

part of what is making this work is that although Mod_R is a large category, it is compactly generated by something much smaller

[fierce steeple]

(Do you really need commutative here – can't you use LMod_R)

Maybe I'm dumb tho

[summer quest]

I mean yes this is fine too

hence the parenthetical

[fierce steeple]

Ah lol I thought you meant it was an assumption you wanted to gloss over lol but yeah

Sorry this is a distraction lol

[summer quest]

yeah I only mention it keeping in mind how these things generalize to schemes

like the point is that Quilllen's K-theory for any scheme X comes from QCoh(X)=Ind(Perf(X)) and it suffices to define and compute with the much smaller category of perfect complexes Perf(X)

similar to how G-theory comes from IndCoh(X)=Ind(Coh(X))

[pastel agate]

Ok it seems this is probably a bit too out of reach for me

Thanks for the explanation though

[summer quest]

the upshot is that these large categories of modules are generated by smaller categories which make these potential size issues a nonissue

[pastel agate]

Ok that makes sense

[summer quest]

of course computing these things in practice is another issue entirely and this is very hard in general

[fierce steeple]

I guess with finite length it is even easier cause like every finite length module over a ring R is in particular finitely generated and hence has cardinality <= k := max{#R, aleph_0} or smth. So you could take a set of cardinality k and consider the set of finite length modules whose underlying set is a subset of k, and every finite length module be equivalent to one in this set

This is the sort of trick you can use to justify stuff if truly needed (slightly ad hoc tbh but ok)

[worldly zealot]

not really an answer, but if you have n isoclasses of simple modules, then by jordan holder the grothendieck group is isomorphic to Z^n

which would not be true without length

[fierce steeple]

How is the length of a module well-defined? You could write it out really big

[worldly zealot]

mooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooodule

cow doing algebra

[fierce steeple]

Lol

[near lantern]

I am actually quite curious about the details of this if you are willing to share (DMing if you don't feel like posting here).

[worldly zealot]

anyone have a good reading on the snapper polynomial?

[lone jacinth]

The isomorphism classes of finite length modules certainly form a set. This is even true for finitely generated modules.

To see why not that a finitely generated module is priceless one that's a quotient of R^n, so the number of isomorphism classes is bounded by the number of submodules of R^n as n ranges over natural numbers.

That being said, it's not hard to see that the group you're constructing just ends up being the free abelian group on the classes of simple modules, so you could just start with that if you want something more manageable.

[spice idol]

why do we have so much interest in K-theory?

[urban granite]

because theory named "K" sounds really chad and based /j

[spice idol]

https://tenor.com/view/ohh-i-under-black-guy-gif-1247944034220562367

[lone jacinth]

What even does the K stand for...

Just Klasse

That's boring

[spice idol]

what im getting from this is that the only reason to care about K-theory is because the name looks cool

got it

[ornate atlas]

It takes fun algtop and combines it with fun ring theory and I’m thinking that sounds pretty cool

[fierce steeple]

Lol I think there is even a quote from Grothendieck about wanting to use hhis mother tongue for this etc

[lone jacinth]

Using German is fine, but come up with some more interesting words will you

[spice idol]

i think the K should stand for kool

[young forge]

why are there sporatic groups, but for other similar group related structures the classifications are either simple or hoplessly complex(aka infinitely many "sporatic" seeming elements)

For example, finite simple moufang loops are all easily classified as being related to finite field octonions

or are finite simple groups

[spice idol]

"easily"

[young forge]

yeah valid

the paper doing so does seem nontrivial

[spice idol]

this seems easier because for moufang loops its the best we can do

while the classification of finite simple groups is very explicit

groups are kind of the perfect middle ground, and that is probably in part due to the existence of character theory

[last talon]

I don’t think it’s that unusual
Like for Coxeter groups, you get a classification with 4 infinite families, and 6 exceptional cases

[lone jacinth]

I mean, the classification of Dynkin diagrams/ semisimple lie algebras is similar with some infinite families and some "exceptional" examples

[last talon]

Probably also because of the amount of effort that’s been put into them (although that may well be a chicken and the egg)

[spice idol]

most likely

in any case classifying simple moufang loops includes classifying simple groups

[young forge]

I actually sorta see a relation here to chaos theory of all things.
In particular, there is sort of a "border" in dynamical systems. On one side being simple, pretty easy to predict and classify behavior. On the other side being things hoplelessly chaotic. In the middle, being the interesting structures that are neither trivial nor chaotic to the point of hopelessness.

Maybe this could be generalized to algebreic structures somehow? With groups being sort of on the edge of chaos so to speak. Cyclic groups would be an example of an "easy" structure to classify, and general magmas are an example of something "hard" to classify. With groups being somewhere on the edge of the two.

[wise sedge]

Finite simple Moufang loops have been classified

They are either associative or a Paige loop

[young forge]

yu[

yup

[spice idol]

alright now classify simple associative moufang loops and simple Paige loops

[young forge]

And yet, the classification of all finite simple loops, if such a thing can even be defined seems so hopeless i can't even find any results on it.

[spice idol]

because loops really arent nice objects at all

:P

[young forge]

And the question is, where does this border between "nice" and "not nice" really sit for general algebreic structures?

[spice idol]

i bet it has a lot to do with the fact that they cant nicely be thought of as sets of functions, unlike, say, semigroups

[wise sedge]

Are you still requiring finiteness?

[spice idol]

(implicitly)

[wise sedge]

Then https://en.wikipedia.org/wiki/Classification_of_finite_simple_groups in the former case

[young forge]

Simple associatvive moufang loops is just CFSG, paige loops are just finite field split octonions with norm 1 under multiplication

[spice idol]

tits group.

[spice idol]

"just"

[wise sedge]

You have to quotient by the center (which has order 2) in odd characteristic

[young forge]

interesting intermediary between moufang loops and general loops, has the classification of finite simple flexable loops been studied a all

i think that would include general caley dickson constructions over finite fields. So there would be a big class of order 2^n*p^m for integers n and m and for prime p

[wise sedge]

Are flexible loops necessarily power-associative?

[young forge]

yes

[wise sedge]

Oh yeah

Just set x=y

[young forge]

But do any sporatic flexable loops exist?

[wise sedge]

Probably

[young forge]

it perplexes me to no end that there are sporatic finite simple groups, yet i am not aware of a mere finite number of "sporatic" elements for other algebreic structures. There seem to be either a hopeless number of impossible to classify objects, or a few simple families in most I see.

[last talon]

Coxeter groups, semisimple Lie Algebras (for basically the same reasons) were given above

[wise sedge]

Finite real/complex/quaternionic reflection groups

There are 34 exceptional complex reflection groups (plus a single infinite family)

[last talon]

It would be nice to have a more “uniform” proof of CFSG where we could point to “there, that’s why these sporadic groups exist/the larger family (of possibly non-groups) they’re a part of”
But that seems like a pipe dream

[young forge]

lie algebras are another interesting case. The exceptional ones are all related to the octonions in some way.

At least thats one way to formulate them.

That makes me wonder, do the sporatic FSGs exist because of some other exceptional object(s) in some sort of vaguely unified way?

[wise sedge]

They aren’t really algebraic, but there are 73 exceptional finite planar uniform polyhedra

[young forge]

This also means that, large families of finite simple groups also exist due to the octonions

(The finite field analogues of the exceptional lie groups)

i don't know what it is, but the octonions are doing something important.

I'd definitely say the octonions are an exceptional object of some kind

[spice idol]

minimal polynomial clones and Post's lattice are rather interesting

these both have very managable amounts of sporadic element

Post's lattice less, lol

[young forge]

alright i have no idea how to do this. How do I prove the multiplicitive loop of non zero divisor sedenonions over GF(3) is simple

well first how do i even prove that that forms a closed loop

sedentonions don' have norm like octonions do

[hard kite]

anyone have a good reference on representation theory over dedekind domains/rings of integers?

[young forge]

Regardless, by not trying to re-invent the wheel too much, AAG actually seems to work fine using octonions as a platform loop as long as you are careful about brackets

in fact though, the different bracketings seems to make it more secure than normal AAG

[young forge]

yup, AAG works over octonions!

Because every 2 element subalgebra is associative

and the final key computation involves a 2 element subalgebra

but, nonassociativity would heuristically hamper linear algbera attacks a lot and lead to combinatorial explosion way faster than normal aag

[sand falcon]

For the second question, is there not a square on the right hand side ?

[vague pawn]

where?

[sand falcon]

Sorry , when we express the discriminant as the product over the embeddings evaluated at alpha ?

[vague pawn]

ah yes we need a square there

[sand falcon]

Yes because we can express it as the square of a det of a vandermonde matrix right ?

Perfect thanks 🙂

[vague pawn]

ANT is amazing

have fun

[sand falcon]

Yeah i enjoy it very much , i followed it a first time last year but without any knowledge in Galois theory or localization i was so lost ah ah

[young forge]

Alright AAG over the octonions seems very very promising

in fact nonassociativity means public keys can seemingly be way smaller at the cost of slightly larger private keys

wait this is havin gissues

[young forge]

i might be cooking

[summer quest]

it's a real shame that octonion algebras over number fields do not enjoy the same deep arithmetic properties as quaternion algebras over number fields

[young forge]

Alright, I think I did it.

Generalized AAG for Moufang loops.

GENERALIZED AAG:

Public Parameters:
Some platform moufang Loop L, a number of elements n

Public key:
a list [a0, a1..., an] of elements in L, along with a companion element c such that (A^-1*x*A)*((A^-1*y*A)*c) = (A^-1*(xy)*A)*c for all x, y in L.

Private key:
some random word generated from (a0, a1, a2... an, a0^-1, a1^-1...an^-1)(could also have other powers here, but the goal of this is to leverage nonassociativity).

examples:

a4(a0(a1^-1a2^-1))
a0a1(a3(a4a0^-1))

make it generated long enough to reach whatever your desired security level is.

Note, for octonions over finite fields(their multiplicitive loop), c = k*A^-3 for some real scalar k. This means that the security of this algorithm is tired heavily to the size of the loop for octonion loops. But for a sufficiently large order base field we can just generate a random scalar k between 1 and |F|. Making this public likely isn't a big deal.

Key exchange:

Alice sends bob all the elements in his public key conjugated by A, along with c
[A^-1*b0*A, A^-1*b1*A...]

Bob does the same for alice.

Using this, alice computes B^-1*A*B, which she can do using the factorization of her private key A. along with bob's c.

For example, if her private key were (a0a1), she would compute B^-1a0B((B^-1a1B)B_c), which by the identities for B_c is equal to B^-1*A*B.
Unlike normal AAG, parenthesis matter A LOT here. In fact this leads to much greater security for shorter public keys. At the cost of more complex private keys. But private keys can be compressed anyways with PRNG tricks.
Due to A, B being 2 elements, they form an associative subalgebra. Thus Alice can compute A^-1(B^-1AB) without caring about parenthesis from this.

Bob can repeat the same steps to compute A^-1B^-1A (note, bob needs to invert his key symbolically, which key can do easily via the moufang identities. From this he can compute (A^-1B^-1A)B. Which is equal to alice's because A and B form a 2 element associative subalgebra.

this is informal. I'll probably ask one of my profs otmmorow to help write something a bit more formal

[young forge]

@broken wren what are your thoughts on this? I honestly have a feeling that this... has issues for octonions. But this seems like something very different than other primitives. Over an octonion loop, i'd set n=3 or n=4 because the automorphisms are linear. Heuristically that should be about a n^1/4 attack searching the kernal of the reconstructed linear map where n is the order of the field. n^1/8 on a quantum computer. The other attack would be to recover B^-3 from kB^-3. This is the more limiting attack and basically forces us to use a 256 bit prime field. with k being a random number between 1 and |F|, that should mean 128 bits of quantum security against that attack. Assuming recovering B from B^-3 is trivial in this case(it might not be for octonions, i'm not sure how many possible cube roots there are for a random element). For testing I will be using the prime field GF(2^256 - 189). The number of non unit norm octonions over a finite field is n^3(n^4-1)(n-1). the 8th root of this for n = 2^256-189 is huge. Near 2^256 itself. So an attack using grovers algorithm to recover kB^-3 seems to be the limiting factor.

Sorry for the ping if you do not care

this gives a public keysize of 10240 bits and a private keysize of 256 bits(compressing it and basically just regenerating the word as needed using the seed of a csprng)

[broken wren]

Are there octonions over finite fields?
Why not just start with the loop of octonions with Q coefficients? This fails to have key sizes, but worry about that later

[young forge]

Yeah your right that could have been a thing I started with

But generally using infinite fields like that tends to expose attacks to my knowledge

still, doing someting like this over an infinite field or ring could be something very much worth looking into

we might be able to get that keysize down to 8965 bits at the cost of a bit of security.

If we only use norm 1 octonions, we only need to publish 7 components and a sign

[young forge]

actually n=3.. is still very secure i think and leads to a smaller public keysize

for n=3 with uncompressed public octonions, that's 4*256*8 = 8192 bits for 128 bits of post quantum security

The main weakness is probably linear attacks on the public generators and also recovering A from kA^-3

in fact, if someone found a key exchange where we didn't have to publish that element, that would reduce keysize and increase performance by a tone

[young forge]

wait im not sure if my shitty protocol works

whatever i'll refine it a bit and speak to my professors

well it works but i think it might be like linear algebra attack levels of insecure

[hard kite]

.

[lone jacinth]

As in representations of those rings or modules/lattices of algebras/orders over those rings?

[hard kite]

Algebras over those rings. Specifically group representations.

[spice idol]

wdym by lattices of algebras/orders over those rings?

[lone jacinth]

We had a seminar at my uni not to long ago where we looked at the dvr case. Lots of interesting stuff was known there, so I can maybe dig up some reading material for that.

[lone jacinth]

Say you have an integral domain R, with field of fractions K.

Then an R-lattice A is an R-algebra that is fg free as R-module and such that K(x)A is a semisimple K-algebra.

Orders over a lattice are then modules that are fg free as R-modules.

These form an exact category that is a little more manageable than general A-modules

[spice idol]

hmm

is this related in any way to actual lattice orders?

[spice idol]

that is nice

[lone jacinth]

Well, a discrete subgroup of a vector space is called a lattice, so something like Z^n < Q^n is a lattice, and that's exactly what Z-lattices look like

I'm not sure where the word "order" comes from

[spice idol]

okay so
the lattice in "lattice of algebras" refers to that (obviously inferior) kind of lattice
but then they felt the need to use the terminology order??
this feels like explicit disrespect

[spice idol]

topos theory seems cool

[urban granite]

it is 🔥

[rose mirage]

of course it's Lawvere lol

[worldly zealot]

i think curtis and reiner's book is good for this?

[hard kite]

Yes please that would be great

Thanks

[hard kite]

Sweet

Ill look into it

[spice idol]

I mean at the end of the day it makes sense

[young forge]

Alright, in my work nonassociativity has yielded a few other promising ideas.(No concrete canidates though)

New hardness assumption.

Given a godawful loop element generated from a public list of generators, and a copy of this element with one random other unrelated element inserted somewhere in it's factorization, recover this other element

1 generator case

let L be some general loop(it need not be moufang)

say we have a public word p = a((aa)(a(aa)a)

given p, and the fact that it's generated with a, and the number of copies of a... this seems like a generalized version of discrete log for nonassociative structures

our core hardness assumpton is given p, a, and n, it's hard to recover the bracketing of a that lead to p

The problem is how do we let people publically mess with the bracketing of p in a public way without revealing it's bracketing?

my thought is, associator spam

in particular, we compose associators, commutators, and other junk on some other public elements b, c, along with publishign what band c are. and without knowing the structure of p it should be hard to factorize our public transform into commutators and associatos

so someone could predictably insert some word made of b and c into our bracketing of p. And us, knowing the bracketing of p, could recover it.

loops seem to be incredibly rich for cryptography actually

[young forge]

wait, this is a generalization of a scheme for nonabelian groups i've seen no one propose

that's related toc onjugcy

there seem to be very strange nonassociative analogues of conjugacy like problems that i can't quite crack yet

[limpid horizon]

If x is a regular element, can I say anything about the associated primes of M/xM if I know the associated primes of M?

I was wondering if that ses can help me, with how Ass gets transferred thru them

Like I know Ass(M) subset Ass(xM) union Ass(M/xM) i guess

ass moment

[hushed bone]

Hmm, I’m not really sure

[hushed bone]

So this is kinda pointless, if x is regular then M ≈ xM via multiplication by x, so then Ass M = Ass xM

[limpid horizon]

Ok yea i had a feeling the xM part would be redundant

[hushed bone]

It’s not really redundant, but look at what you’ve written

Ass M < Ass M U Ass M/xM

But is that actually what it is?

I thought it’s Ass xM U Ass M/xM < Ass M

No

I’m hallucinating, call me chat gpt

[young forge]

I have a slightly insane idea

public key encryption using a non moufang general loop

one that is not power associative

[limpid horizon]

Ma Boi what is moufang

[young forge]

a loop is a nonasociative group

a moufang loop is a group that satisfies some weaker versions of nonassociativity

it'd probably have to be a specally constructed loop

but the idea is that we public a public key k = some word made from 1 element\

which explodes combinatorially because oops n-nwo a-associtivuwuity :3

our private key is the exact bracketing of k

the idea is we construct some sort of way over our loop to embed elements into the bracketing of k without knowing the bracketing

we can give some public information allowing them to replace one of the a's with some arbitrary element k at some determined to the person with the bracketing of k but unknown to the public part of the key.

k is ideally chosen by the public key person

or it could be different parts of th ekey

the point is if you have the bracketing, you can recover k.

[limpid horizon]

Lol

[young forge]

this problem gives np hard vibes ngl

over an arbitrary loop

but the big issue is... how the fuck do you even represent a general loop at all\

even if we can construct one

best solution, bite it and use 3 elements in a moufang loop instead.

but here, this might be better because we can probably come up with some sort of construction allowing someone to publically embed information in the bracketing using just the moufang identities

does anyone else have any other idea

s

And, we can actually construct symmetric key cryptography from this i think

[median nacelle]

oh wow this was a really good time to check in here for the first time in weeks as someone whose subfield is quasigroup theory lmao
some quick notes:

• the reason there is no literature on the classification of finite simple loops is because the vast majority of finite loops are simple, making the endeavor beyond hopeless
• re: "how to represent a general loop" uhhh if you mean in terms "how to construct a minimal presentation" that is a hard unsolved problem even for the finite case, if you mean "how to efficiently represent loops computationally as data structures" unless you go in with some kind of strong constraints you're basically going to have to just store the Cayley table
• there is a whole lot of literature on various approaches to cryptography based on quasigroups but that's not super my area so I'm unfortunately not super familiar with most of it

Also I'm kind of hazy on the details of your bracketing idea, how you chose to bracket a loop word isn't really a property of the loop itself, for sufficiently large cardinalities even if you fix a presentation for a specific loop it's often a nontrivial problem to determine if a given pair of loop words represent the same element

[spice idol]

this is slightly misleading: a loop is a unital magma with division. left and right inverses may not coincide

[spice idol]

loops dont have, unlike say quandle or racks or groups, a reduction algorithm for terms

[young forge]

i am too. But the idea is we can construct it in a special sort of loop, where someone can make subtle changes to bracketing and product of an element(with a known decomposition, they make changes to that known decompisiton) without revealing any of the bracketing structure. The public key would be what the bracketing evaluates to, the private key would be what the bracketing actually is.

this could be a nothingburger tho

for example pulic key:

w

private key, the fact that w = a(aa(a((aa)a))

public key:
also some way for someone to take some element b, which they randomly generate, and compute

a(aa(a((ba)a)) without revealing abything about the bracketing of w

private key:
the location of the b inserted into w

then, someone with the bracketing of w could reverse this bracketing to recover b

this sort of problem feels np hard

i.e, given some black box representation of an arbitrary non power associative loop, and an element w generated from 1 element a, and an element w' where 1 a is replaced with some element b(we don't know where or anything), recover b.

[swift cove]

Anybody has a proof for the following identity?

[median nacelle]

The situation is worse than that even - most loops don't have inverses period. Unless it's an inverse property loop, (ab)c=a doesn't imply (db)c=d

[spice idol]

lovely

I did a small study on isotopy actions on quasigroups a while back

that was pretty cool

or more accurately, row and column permutations on latin squares

but close enough

[median nacelle]

Nice!! Same idea lol. Love to see people interested in quasigroups

[spice idol]

Eventually got to a lovely result that an nxn reduced latin square (i.e. a loop) was the cayley table of a group (i.e. the loop is associative) if and only if its stabiliser wrt the row and column permutation action was of size n, in which case said stabiliser was isomorphic to the loop

[median nacelle]

Oh fascinating

[spice idol]

then you could get a ""closed form"" for the number of isotopies from such a latin square to itself, in terms of the size of the automorphism group

surprisingly combinatorical for what i usually do lmao

[spice idol]

its n!(n-1)!^2 / |Aut(G)| where n is the order of G

[golden osprey]

say I wanted to buckle down and properly learn commutative algebra

if I'm familar with say half of the stuff in AM (from a commutative algebra course + picking stuff up learning AG)

should I just go through AM fully or is it worth looking at say Matsumura instead

[median nacelle]

That's very cool, gotta check that out

[spice idol]

Bosch has a book that covers both AG and commutative algebra as a self contained package

though it certainly does not go in depth on AG as much as I'd like it to, the commutative algebra section is good

but this isnt really useful if youve already got an AG book lol

[golden osprey]

nah I know the AG

I really just need to learn commutative algebra properly

[spice idol]

okay lol ignore what i said then

[golden osprey]

rather than blackboxing things and picking stuff up piecemeal

[young forge]

I think it depends on semantics sometimes

i swear i've seen conflicting definitions

But a loop without inverses would just be a unital magma i think

[spice idol]

inverses =/= division

a right inverse of x is an element y such that (ax)y = a

[young forge]

I did see a key exchange actually which works over magmas(they need to be specially constructed to have special homomorphisms)

And to me, a moufang loop seems like a good canidate for a platform that's structured enough to have a.. sane key exchange. But not structured enough to make it easy to attack like a group

this wouldn't be bracketing based. Though AAG over a moufang loop is more secure for the same number of public elements due to bracketing combinations

[young forge]

What's the difference ooc

is division requiring both left and right inverses to exist/

[hushed bone]

Matsumura babyyyy

[spice idol]

a right inverse of x is an element y such that (ax)y = a for all a. Loops do not require this

Loops only require (ax)/x = a

[young forge]

isn't that just (ax)x^-1 with different notation?\

[spice idol]

for groups

but (ax)/x =/= (ax)(1/x)

for example

[golden osprey]

which one again, CA or CRT?

[spice idol]

or more explicitly: a/x =/= a(1/x)

[young forge]

oh wait division can be a different operation than the normal group one?

[spice idol]

well, no

but groups are associative

in which case it implies that a/x = ax^-1

(where we may define x^-1 as 1/x)

but loops dont usually form groups

[young forge]

would this explain why i've run into headaches trying to define conjugacy for moufang loops?

(you can do it but you need some sort of helper element which actually seems to give rise to it's own conjugacy like rpbolem)

[spice idol]

idk

[young forge]

in a moufang loop, (AxA^-1) (AyA^-1) = (Ax)(yA^-1) (Not what we want for something that preserves multiplication)

but i'm pretty sure for every A there exists an element w such that (AxA^-1)((AyA^-1)w) = (A(xy)A^-1)w

[spice idol]

but is the notion of A^-1 defined in moufang loops

[young forge]

inverses i think

moufang loops have strong enough conditions so that both left and right inverses exist and are the same for every element

i think

[spice idol]

prove it id say

seems pretty important to know for sure

[young forge]

from wikipedia

also idk if your familiar with them, but here are the conditions for a loop to be moufang

They aren't associative, but any 2 element generated subloop is

actually, sorta like how for a nonabelian group every subgroup generated by 1 element is abelian

[spice idol]

yeah i know

[young forge]

I think there might be analogues between studying noncommutative and nonassociative objects(especally alternative ones)

[spice idol]

not sure what kind of analogy youd be looking fkr

[young forge]

you can rearange

(AxA^-1)((AyA^-1)w) = (A(xy)A^-1)w
to get
((AxA^-1)((AyA^-1)w)w^-1) = (A(xy)A^-1)

This looks a lot like a weird generalized version of conjugacy to me

[hushed bone]

CRT

[spice idol]

"pretty sure"? how so

[young forge]

I loaded up an octonion algebra in sage over a finite field and it seemed to hold

[spice idol]

just one?

i can imagine that for moufang loops octonions are still relatively nice

[young forge]

octonions are actually special and related to the like... 1 family of finite simple moufang loops that aren't just groups that exist.(It suprises me genuinely that finite simple moufang loops are nowhere near as rich as fsgs. But i suppose allowing nonassociativity opens the door to far more complex ways for one loop to be a subloop of another)

but there are other families of moufang loops that seem unrelted to the octionions

1 sec i read a paper on this i'm not just pulling stuff out of my ass

https://www.cs.du.edu/~petr/data/papers/finite_simple_Moufang_loops.pdf

and here's a promising platform loop for cryptography that i need to read up on a bit more and understand better

https://arxiv.org/abs/1903.02748

octionions might not be optimal because they can be represented as zorn matricies and also the conjugation like action by one octonion on another is a linear map over the underlying 8d vector space. Which we odn't want

meaning we want either some sort of loop based crypto that isn't based on (generalized) conjugacy at all. Or a loop that's more nonlinear.

Loops are already nonlinear. But octionions don't feel nonlinear enough.(They technically don't have a faithful matrix representation, but both left and right multiplication can still be represented by a matrix vector product i think)

i'm going about this very heuristically right now. I need to lock in and teach myself formal abstract algebra

i think a good platform loop should have either trivial or way more constrained center/nucleus

than octonions have

wait maybe octonions over a characteristic 2 field could work?

the center is trivial there

i think

vaguely remembering the first paper

[spice idol]

maybe open your own thread here

[young forge]

oh sorry if i'm tking up space

Nonassociative/Moufang loop self study/cryptography

[zenith lichen]

what i just joined to see if i could get any help on loop theory and this is the first thing i see

[zenith lichen]

i've been trying to see what kind of loops the hyperbolic quaternions have

[barren obsidian]

hello guys I got preprint in zenodo https://zenodo.org/records/17580921 check it out

[barren obsidian]

hello check my preprint in osf https://osf.io/byqfw_v3/

[spice idol]

bro throwing his net out there

[limpid horizon]

So also I have M is CM btw, so by Thm2.1.3 in bruns herzog M/xM is CM too, so can I say something more about what Ass(M/xM) looks like in this case?

[spice idol]

oh its generatwd by an LLM

booooo

boooooooo

[hushed bone]

If it’s CM then I think there’s no embedded primes so I imagine it’s the subset of associated primes of M containing x?

[limpid horizon]

Btw, were you saying xM is isomorphic to M? Cause I thought xM is just a proper submodule

[hushed bone]

I told you how it’s isomorphic

[limpid horizon]

M -> M sending m to xm? since x is regular I thought that just tells us thats injective

[digital parcel]

booooo 🍅 🍅 🍅 🍅 🍅 🍅 🍅 🍅

[digital parcel]

the map is M -> xM

or if you prefer since the map M -> M is injective and its image is xM then M cong xM

[limpid horizon]

Oh .. right Lol

Umm I think I was unsure cause of that M neq IM condition when talking about reg sequences

there is certainly something probably silly im missing but idk what it is rn

We have M cong xM but M neq xM?

[plucky arch]

Is it normal for the snake lemma to feel opaque

[ornate atlas]

I guess if you haven’t seen it before yeah

I often find with a lot of diagram chasing stuff like that I don’t have a great feeling for why it should be true, and then you do the chase and you realise it just kinda has to be

[worldly zealot]

even after i believed it i found it opaque until i saw a theorem in a textbook with actual content outside of homalg which used the snake lemma

[ornate atlas]

I can’t think of any uses of it outside of the 3x3 lemma or the connecting homomorphism for LES in homology

But i guess those are big uses

[worldly zealot]

it was something about torsion pairs

[foggy galleon]

I mean it's fundamental for the long exact sequences that cohomology produces

[near lantern]

I think so.

[wary elbow]

imo it's normal for all of homological algebra to feel opaque

hence "abstract gibberish"

Statements of results in homological algebra are hard to parse but proofs are almost always easy. It's an inversion of the standard mathematical order

[young forge]

looking at the first 2 paragraphs of this, it already does not make sense

but i think this is already trivial. Just plug the number into the taylor expansion of e^x and/or ln(x) and ur good

[barren obsidian]

yeah that why it eliminate the use of transcendental e and infinite series to be purely algebraic

and in computation, especially with E^x I got more precise results, and no collapsing with catastrophic results under iteration

quaternion stability

with python

also relate it to hypercomplex cyclotomic and Galois theory, with use in hypercomplex to octonions with simpler calculations

[near lantern]

OK I got nerd sniped

[near lantern]

Lemma 2.1 is wrong: an element of a real algebra can generate a subalgebra of any dimension. Just consider x in ℝ[x]/(x^n - 1).

Theorem 3.1 is proved in exactly the same way it is classically, and the exp, cos, sin are also defined the same way as classically. To say in 3.4 that Euler's identity "fails" and give this as the correct replacement is a bit misleading since I think most mathematicians with some experience in algebra trying to write down a generalisation of Euler's identity would write down Theorem 3.1.

Also BTW (even in ℂ) the automorphism zeta |--> zeta^k (a) cannot be extended continuously to ℂ unless k = -1 (b) is not a rotation by a fixed angle.

[barren obsidian]

yeah in theorem 3.1 E isn't forcefully only imaginary i, it could be expression with hypercomplex elements that for example (e_i+e_j)^2=-1, that's why I think this is more useful and direct

[near lantern]

7 sections in I still haven't seen anything that doesn't follow almost immediately from identifying span{1, E} with ℂ.

[barren obsidian]

and that's why the results using python is more precise than normalized Euler, because you go two transcendental variables at least when you dig deeper pi and e or log, the numerical error get higher

[near lantern]

I don't think this can be considered a substantively new idea. That being said, I agree this is a useful way to calculate power series in E on a computer by directly using built-in functions on the real components like cos and sin.

[barren obsidian]

first thank you for reviewing seriously, but the key when you get not only one imaginary or hypercomplex element with power but an expression like like (j+k)^x or (e_1+e_5)^x, they are orthogonal to Real and C, but for only one element , it is just like C since we got the same multiplication table with e0

but the key here is when not only one element with exponent or with fractional power

or both

[near lantern]

e_1+e_5 is one element.

[barren obsidian]

yes chek example 5.2

[barren obsidian]

no e1 and e5 different element in octonion

e1 and e5 they are under addition not multiplication

[near lantern]

Sure, they're different elements. But e_1+e_5 is itself a single element.

[barren obsidian]

no it could never single element unless we multiply them

it is in hypercomplex numbers

[near lantern]

... I don't think I know what you mean by "single element".

[barren obsidian]

single element like 1, i, j, k they became later e0, e1, e2, e3

so like 1+i they arent same elements

[near lantern]

But why is this a distinction I should care about?

1, i, j, k, 1+i are all in the algebra.

[barren obsidian]

because when when make cayley-dickson construction from quaternion we get octonion and the notation start withe e0, and I know the notations get weird fast some keep use i,j, but for k is ij and e4 is l and some use graded notations for elements

from clifford algebra , it get weird fast

[near lantern]

Yes, but this is just a conventional choice of basis.

Switching between bases is a pretty foundational idea in anything with any linear algebra in it.

(Cf: diagonalisation.)

[barren obsidian]

yes I agree, I choose always e0 , e1, e2 notations because it is simple and direct

[near lantern]

I don't mean the notation. I mean the choice to write all the other elements as linear combinations of those elements, instead of some other elements.

[barren obsidian]

yes but here they are all the same things with different names like in "graded notation" the numbers of elements as binary here example https://arxiv.org/abs/2505.11747

he uses graded notations

[near lantern]

This doesn't visibly have anything to do with what I said.

[unborn rampart]

How much linear algebra have you actually studied? Are you familiar with isomorphisms, change of basis, etc?

[young forge]

i thought i sounded crank-ish

[limpid horizon]

Since x is M-regular its not in any associated prime of M

If p is in Ass(M/xM) tho then (x) is in p. So Ass(M) and Ass(M/xM) are disjoint?

[near lantern]

Sounds correct.

[limpid horizon]

The context is im trying to see if i can understand the elements in Ass(M/xM) if I know Ass(M). M and M/xM are also both cohen macaulay

[young forge]

also i hope i don't come off as like a crank or crackpot

i admit im an ameture. And I mean the fact that i'm even thinking about this as an undergrad shows i have potential i think. But still i souldn't get ahead of myself

[hushed bone]

Oh yeah tru

[thick oriole]

ass

[plucky arch]

What interpretations do people like to use for exact sequences

[spice idol]

i think of a long exact sequence as a bunch of short exact sequences smushed together, and a short exact sequence as an extension

[ornate atlas]

What do you mean interpretation? I just kinda think of them as a sequence of particularly “nice” maps

Nice in the sense that knowing about their kernels and images etc gives us a pretty good handle on how they behave, so perhaps well understood rather than nice

[spice idol]

great computational tool

because knowing some amount of terms with some amount of density often allows you to fill in the rest

or at least allows you to deduce properties about them

[distant harness]

I've never understood the choice of the word "exact". What would an approximate sequence be?

[worldly zealot]

a chain complex ig

[wary elbow]

I guess an "approximate" sequence would be an arbitrary complex, which need not be exact

[wary elbow]

haha nice

[worldly zealot]

lol

[distant harness]

But where does approximation or imprecision come into the mental picture?

[limpid horizon]

Image isnt the whole kernel but at least its part of it? Is what im assuming theyre getting at

[plucky arch]

Hm ok…

[plucky arch]

I guess I’m trying to find an interpretation that helps make homalg results more obvious

Currently trying to understand the five lemma

[ornate atlas]

This seems to be the most detailed historical answer I can find and it seems like no one knows for certain https://mathoverflow.net/a/160236

If I were to guess, and this may be ahistorical, it would be as opposed to chain complexes. So rather than the image of one map being simply contained in the kernel of the next, they are exactly the same. Possibly nonsense, but that’d be my best guess

(Edit: didn’t see people had already answered, train WiFi sucks)

[digital parcel]

I think exactness comes from diff top right?

In which case an exact form is like saying “this is exactly a differential”

[spice idol]

its all snake lemma...

[ornate atlas]

I had wondered if it came from closed and exact forms, but it doesn’t seem so, that’s why I looked up the etymology

[digital parcel]

Maybe it doesnt and im talking out my bum

[plucky arch]

That too

[digital parcel]

God dammit…

[spice idol]

the one time chicken and egg has a definitive answer

[ornate atlas]

I think for a lot of this stuff you just kinda have to “shut up and calculate” to build some intuition. That’s my feeling at least.

Like there’s a reason there’s jokes like “pick up any homalg textbook and prove every theorem without reading it”, a lot of these basic lemmas just have to be true, and if you do a bit of diagram chasing it falls out

Like to me that’s all diagram chasing even is, a lot of it is just unpacking definitions

[ornate atlas]

I mean I did like 5 minutes of research, it’s entirely possible, but I didn’t see a definitive answer in that direction and it was also my initial suspicion

[plucky arch]

ok, i guess i'll try doing some calculations and see if i can build intution that way

[spice idol]

exact sequences are very "rigid" in the sense that information on one end of the sequence can give information about further in the sequence and vice versa

esp when you have a commutative diagram involving exact sequences

[ornate atlas]

FWIW, I was saying the other night my (very basic) homalg was really rusty because I had just fallen out of practice with that kinda argument

But like for the 5 lemma say, there really is only one way to proceed (well really for the 4 lemma, 5 lemma is just apply the 4 lemma twice), like given the information you have you just pick an element and aim to use the fact that you have epis and monos (or I guess honest to god injections and surjections since I’m already picking elements) and it just kinda follows one step at a time

That’s not to say it’s easy,, but like at any one time you only have finitely many options of how to proceed. Do enough fumbling in the dark and you will get there

[digital parcel]

Hmm I used four lemma the other day on some vakil exercise

Cant remember which one I just had two exact rows and three vertical isomorphisms and one surjection and said “okay four lemma done”

[spice idol]

4 lemmas are about monos and epis

not about isomorphisms

right?

[ornate atlas]

Yeah

[digital parcel]

Isos are epis and monos

[spice idol]

right right

[barren obsidian]

please check hypercomplex numbers, and cayley-dickson algebra , it is very nich subject because we "lose nice properties" that we used in R or in C

[digital parcel]

Hmm anyway yesterday I was talking about how I need to learn leray spectral sequence

Then there were northern lights visible from my state and that was like well now I really gotta learn them

Then today we’re talking about exact sequences like damn i get the message

[spice idol]

spectral sequences are so scary

[digital parcel]

I need to learn demazure modules and characters

There’s some spectral sequence stuff in there

[spice idol]

oh damn that is cool

i probably need to learn them too considering im gonna be doing more hom alg pertaining to my research

but ive been putting it off

[urban granite]

they are very cool though

very much useful to calculate stuffs

[spice idol]

yes exactly but very daunting having scoured the wiki article

[urban granite]

I think it's fine if you learn spectral sequences on stuffs like Hodge-de Rham spectral sequences, Leray spectral sequences, or smth real stuffs rather than js doing it with full abstract from weibel chapter 5

[digital parcel]

My friend told me “the best way to learn Leray spectral sequences is by using them to prove Kunneth’s theorem”

[urban granite]

lmao good one

[digital parcel]

Not as a joke lol

[urban granite]

this one is also a good one

[urban granite]

i mean yes

[digital parcel]

Oh lol okay I misread what you said then

Mb

[fierce steeple]

I would argue they aren't too scary but have a reputation

[spice idol]

this is cool

right derived functors of the direct image functor

[fierce steeple]

I mean this is the best way to prove Künneth imo lol

It really makes clear where you use PID and where the Tor comes from

[urban granite]

yeah if you know what sheaf cohomology is

[fierce steeple]

I assumed you meant Künneth for just singular homology

[digital parcel]

I did

[fierce steeple]

Oh no they said Leray

[digital parcel]

Oh

Oops

[fierce steeple]

Well ok no it's fine

[digital parcel]

Hmmmm

[fierce steeple]

It still works

My point was there is also a Tor spectral sequence

Which gives you Künneth lol

[urban granite]

ah

that one

[digital parcel]

I see i see

What should I read to learn spectral sequences

[fierce steeple]

But I believe Künneth for sheaves reduces to the same fact anyway

[digital parcel]

Leray ones at least since I need those soon

[fierce steeple]

I think like using some examples is most illustrative if you can find good examples. Like I remember liking these little notes https://people.maths.ox.ac.uk/tanj/notes/sssguide.pdf

[digital parcel]

W thank you

[fierce steeple]

I liked uh Bott–Tu iirc

[fierce steeple]

But yeah this is tbh the sort of thing I learnt more from reading papers where spectral sequences are used a lot lol

[digital parcel]

So my friend mentioned that at some point, spectral sequences converge to the original cohomology. Is that total degree always finite? Do we have bounds?

[fierce steeple]

So it kinda depends on ur interests

[digital parcel]

Ty I’ll look at that one too

[fierce steeple]

This isn't true in general. So like the zeroth question is whether the "E_oo" page gives you a complete and exhaustive filtration on what you started off with, but this. will be the case in most cases of interest. But generically to conclude that "E_n = E_oo" for some (finite) n depends on the shape of the spectral sequence

[urban granite]

some other places to learn spectral sequence would be Hodge-de Rham spectral sequence if you know some hodge stuffs

[fierce steeple]

In many good cases (e.g. if your spectral sequence only has terms in one quadrant) then each individual position stabilises at a finite stage. Then you can find a stage that works for all simultaneously if there are only finitely many terms in the spectral sequence

[spice idol]

big if

[fierce steeple]

And this can literally just be seen from deducing that the differentials cannot have source and target which are both nonzero aha

[fierce steeple]

I second this

[digital parcel]

Spectral sequences in more than one quadrant

[fierce steeple]

And you don't rly need to know Hodge stuffs ig

[urban granite]

i mean there aren't that much big stuffs to learn lol

[digital parcel]

Tyty

[fierce steeple]

Though maybe you do if you want to care more

A nice thing I liked which made me appreciate spectral sequences as even a conceptual tool is this like balancing tor and ext

You run two spectral sequences on the same thing and see you get the "two different notions"

[digital parcel]

Interesting

I wanna ask about the quadrant one

What’s the point of allowing negative (bi-)degrees?

In pretty much everything I’ve seen is just “if negative degree then it’s 0 by convention”

[spice idol]

homalg is shut up and calculate until you get heavier machinery so it becomes a corollary of a slightly shorter shut up and calculate

[past cove]

I sort of disagree mainly because for a lot of hom alg calcs you actually have to use noggin

[fierce steeple]

Disagree lol

[past cove]

especially when you use spectral sequences lol

god I still can't work with spectral sequences without a book open next to me to make sure the indices are correct

[fierce steeple]

I think hom alg has a weird reo just as the intro is mostly definitions and maybe a few computations

[fierce steeple]

Because they appear

[digital parcel]

[fierce steeple]

I mean like the common examples are one quadrant because you start with a complex concentrated in nonnegative or nonpositive homological degrees

But what if that isn't the case

[digital parcel]

I’ve heard of it but never really seen it

Where does this show up naturally?

[fierce steeple]

Often when stuff misaligns, like if you have a complex of sheaves in non-positive cohomological degrees (often happens) and then take sheaf cohomology which gives you positive cohomological degrees

But also just like it is very common more generally to have complexes which aren't just in non-negative/non-positive degrees

[digital parcel]

Hmm I see

[serene shoal]

I can feel the math pros here

I m 10th grader

[spice idol]

this just in: world is fucked everything sucks

at least positive bidegrees only seems way more nicely behaved

[fierce steeple]

Lol why

[digital parcel]

Another assumption added to my “list of assumptions to help me smile”

[fierce steeple]

If you don't allow nonnegative/positive degrees u can't even shift bro

[spice idol]

yes you can
may just have a little information loss

[fierce steeple]

Do some textbooks assume all chain complexes are like this

Maybe ig some topology books would

[digital parcel]

I think most hom alg textbooks allow negative degree lol I just always cover the left half with my hand

[spice idol]

well you get a nice equivalence between simplicial abelian groups and chain complexes

[fierce steeple]

Didn't realise

[urban granite]

[fierce steeple]

No I mean yeah good point like they are easier for some things

[distant harness]

All the algebra you will see at least until the end of high school is not the kind of algebra this channel is about. The #prealg-and-algebra channel will be more relevant for you, and perhaps eventually a small bit of #linear-algebra.

[fierce steeple]

Say non-negative tho smh lol

[spice idol]

and those are sufficient for e.g. derived functor stuff

[urban granite]

derived categories go brr

[spice idol]

need to learn that too

aaaaa

[digital parcel]

I tried watching some youtube lectures on them

[fierce steeple]

I wouldn't quite say so

[digital parcel]

Got lost somewhat quickly

[spice idol]

man....

[fierce steeple]

Like in AG for example many things are in negative homological degrees

Seems best to just work with unbounded stuff by default. And often "bounded above/below" is strong enough for the sort of difficulties that can arise

[serene shoal]

I hate straight lines

They ain't no straight no way

[distant harness]

Please stop making off-topic comments in the advanced channels.

[spice idol]

and thats a bad thing?

[spice idol]

i see

[serene shoal]

Good

Conic sections 🔥🔥

[spice idol]

#geometry-and-trigonometry

[serene shoal]

Ok

[urban granite]

moduli spaces of conic sections

[past cove]

Can you tell me how many plane conics are tangent to 5 other plane conics in general position?

[spice idol]

2.5

[digital parcel]

welcome back schubert

[past cove]

2 is the number of lines intersecting 4 given lines in general position in P^3

[spice idol]

6

[past cove]

It's good to be back

[urban granite]

3264?

idk somewhere i saw similar to this

[digital parcel]

they got a whole book on that dayum

[urban granite]

ah that book

lol

didnt notice

[past cove]

Yes! But the calculation is a bit complicated

Since the intersections aren't transversal you have an additional residual part of the intersection that you have to remove and only count the transverse one

Essentially what you want to do is like

You take the moduli space of smooth conics

Well

You take the moduli space of conics tangent to one of your five given conics

You do this through incidence correspondence

Now this moduli space is a hyperplane in P^5

And we wanna know what it's intersection looks like

Because those will be the conics tangent to all five conics

But these hyperplanes don't a priori intersect transversally (and the non transverse part isn't even finite)

If they did then you'd have 7776 such conics but this is false

You have to do a much more subtle analysis of the excess intersections where the intersections aren't transversal to get the correct number

And you can actually do this by blowing up the excess part to get something transversal

[urban granite]

so the 3264 and all that book builds up to prove this

[ornate atlas]

The introduction to 3264 and all that is great, I really love Eisenbuds writing

I will never remember the name of the book though, and I don’t care enough about AG to read it, but the intro is good

[past cove]

No it does a lot more

This is like midway through

[near lantern]

Let k[x1, ..., xn] be a polynomial ring over a field and A a graded subring. Suppose that under the specialisation x(r+1), ..., xn = 0, A is mapped isomorphically to a subring of k[x1, ..., xr] over which the latter is finitely generated (resp, finite-rank free) as a module. Is there a lift of k[x1, ..., xr] to a graded subring of k[x1, ..., xn] (i.e., a choice of linear lifts of xi to xi + ∑_{j > r} a_ij xj for i = 1, ..., r) which has the same property over A itself? Is such a lift the integral closure of A in k[x1, ..., xn]? If not, is it unique anyway?

[near lantern]

This is not a Lie algebra, but if I write w(x,y)c on the RHS it is, right?

[ornate kindle]

What does it mean to have a polynomial on a vector space? Is there something about the fact that V is a regular representation of G that gives V a multiplication?

[near lantern]

For V a vector space, the ring of "polynomials of V" is defined to be the symmetric algebra of V*.

[ornate kindle]

Ah okay I think I get that

[lone jacinth]

A polynomial is just a linear combination of products of linear maps

[near lantern]

Equivalently, given a basis e1, ..., en; let v = x1(v) e1 + ... + xn(v) en define the dual basis x1, ..., xn; then take the polynomial algebra in n variables and identify the generators with x1, ..., xn.

[ornate kindle]

Sure so rather than being built out of multiplying vectors in V, they're built out of multiplying elements of V* using standard function multiplication

That makes sense 👍

[lone jacinth]

This notation is kinda terrible.

What's the point of introducing this c thing if you're gonna write stuff like a(+)b

The notation was so bad the author forgot to use it

[near lantern]

Is what I said true though? 🥲

[fierce steeple]

Writing a (+) b is gross

[spice idol]

it is crazy 😭

[lone jacinth]

I mean, assuming c=1 the two would be the same

Like they just messed up their notation I think

I guess it should be a lie algebra either way though. Like it would just be rescaling the bracket

[near lantern]

No, I removed the mu lambda because it doesn't seem bilinear otherwise.

[lone jacinth]

Ah, I didn't even notice that.

Yeah you're right

[vague pawn]

@lone jacinth or any of the rep theory guys, is there something special about representations of p-adic groups?

or maybe its a question for nG?

[lone jacinth]

Sounds like number theory

[near lantern]

Maybe that they're not known.

[vague pawn]

forwarded to #advanced-number-theory

[vague pawn]

wdym?

[near lantern]

I feel like not knowing the representations of a thing is sufficient reason to want to know them

... and thus, representation theory.

[vague pawn]

yeah.... I was more asking what is there to say in that specific case

[near lantern]

Right, right. It was not a serious answer. Hence struck out.

[worldly zealot]

if struck out was struck out does that mean it was serious after all

[near lantern]

I thought about this as I typed it. I have decided to leave it as an exercise to the reader.

[near lantern]

This is actually really obviously false: take r = 1, n = 2 and the image of A to be k[x1^m]; then A is specified by a homogeneous polynomial of degree m of the form x1^m + a1 x1^(m-1) x2 + ... + am x2^m, and this to k[x1] extends iff it is an m^th power, i.e., iff (a1, ..., am) = (mCi lambda^i)_{i = 1}^m for some lambda.

[summer quest]

yes definitely

they are ubiquitous in number theory (particularly in the Langlands program) and a lot of aspects about their representation theory is in almost perfect parallel with what happens for real and complex Lie groups, though the p-adic situation is richer in many ways

although to be clear I'm talking about infinite dimensional representations of p-adic groups; the finite dimensional representations are not interesting

you have a reliable supply of such infinite dimensional representations induced from (finite dimensional!) representations of finite groups of Lie type, and these are much easier to understand

[vague pawn]

Thanks nG

[plucky arch]

Am trying to prove the snake lemma today

I really wanna understand how this long exact sequence in homology comes about

[limpid horizon]

in a CM ring is every permutation of a regular sequence remains regular?

[muted sierra]

Mmm... Is that true for the polynomial ring k[x1,...,xn]?

I'll go read Eisenbud.

[plucky arch]

Ok, continuing to try and understand homalg

I’ve come across the salamander lemma, which gives one a way to reduce the various diagram-chasing arguments in homalg to a single one

[muted sierra]

I recall the diagram chasing proof of the snake lemma being a tedious but mindless calculation.

[plucky arch]

Mhm, I agree

In my case I’m more looking to “understand” the proof

I’m quite confident I could parse a proof of the snake lemma

It’s more about trying to see what’s going on, so to speak

[muted sierra]

The two only things that aren't immediately obvious are

1. the boundary homomorphism is well-defined,
2. the resulting exact sequence is indeed exact at the source and target of the boundary homomorphism,
   right?

[plucky arch]

Mhm

[muted sierra]

But I'm pretty sure that the diagram chase that proves 1 is identical to the diagram chase that constructs the boundary homomorphism in the first place.

Which would leave us with just 2.

Let's draw the diagram for concreteness:
$$
\newcommand \coker {\mathrm{coker}}
\begin{tikzcd}
& \ker(f') \rar \dar & \ker(f) \rar \dar & \ker(f'') \dar & \
& M' \rar{\iota} \dar{f'} & M \rar{\pi} \dar{f} & M'' \rar \dar{f''} & 0 \
0 \rar & N' \rar{\iota'} \dar & N \rar{\pi'} \dar & N'' \dar & \
& \coker(f') \rar & \coker(f) \rar & \coker(f'')
\end{tikzcd}
$$
We both know where the boundary homomorphism $\partial : \ker(f'') \to \mathrm{coker}(f')$ should go, and I can't be bothered to typeset it correctly.

[broken turtle]

Eduardo León

[plucky arch]

Dw I think I am now happy with the snake lemma

I read up on the salamander lemma and it was quite enlightening

[spice idol]

you have the amazingly strange quirk of, the more high level it is, the easier it is for you to understand

(/pos)

[plucky arch]

Wait what do you mean

[plucky arch]

Also what is this lol

[muted sierra]

Normal people find concrete stuff more accessible.

[spice idol]

its meant positively, like a compliment of sorts

[plucky arch]

I mean so do I

I’m a physicist

I love concreteness

[spice idol]

well, i wouldnt call the salamamder lemma more concrete than the snake lemma

[plucky arch]

Why not?

[muted sierra]

Because you're trying to come to terms with ordinary chain complexes and then suddenly you bring double complexes into the picture?

[plucky arch]

I suppose the way I’d explain it is

You can extend a finite rectangular diagram to a double complex, for one

It’s similar to extending an N-graded complex to a Z-graded one

Also, the main lemmas of homalg all seem to involve finite segments of double complexes

[muted sierra]

Working with Z-graded complexes actually complicates things in unexpected ways, e.g., the construction of tensor products.

[plucky arch]

Right, sure

i think for the purposes of the salamander lemma it just gives you enough room to move around

[muted sierra]

You “reduced” proving the snake lemma to proving the salamander lemma.

[plucky arch]

Mhm, exactly

[muted sierra]

Now you have to argue that the proof of the salamander lemma is simpler than the direct diagram chase proof of the snake lemma.

Well, of course it all boils down to the same thing, in the end.

[plucky arch]

For that I think

I find it “conceptually” simpler than the diagram chasing proof of the snake lemma

If I was focused on shortest proof length, I could just explicitly write down the connecting homomorphism and verify exactness

[muted sierra]

Does your “shortest proof length” include the length of the proof of the salamander lemma?

[plucky arch]

But for me, this only convinces me “that” the snake lemma is true

And doesn’t necessarily tell me “why” the snake lemma is true

[muted sierra]

If yes, then good. If not, then that's cheating.

[plucky arch]

On the other hand, decomposing it into salamanders helps me a lot more with understanding “why” the snake lemma is true, since it’s significantly easier for me to see why the salamander lemma is true

In other words, I am happy to sacrifice some amount of proof efficiency if it gives me conceptual clarity

[muted sierra]

Okay, sure.

[plucky arch]

Yes it would

[muted sierra]

\textbf{Construction of $\partial$:} Let $x'' \in M''$ such that $f''(x'') = 0$. By exactness at $M''$, there exists $x \in M$ such that $\pi(x) = x''$. Then $\pi'(f(x)) = f''(\pi(x)) = f''(x'') = 0$. By exactness at $N$, there exists $y' \in N'$ such that $\iota'(y') = f(x)$. By exactness at $N'$, this $y'$ is unique. We let $\partial(x'')$ be the image of this $y'$ in $\mathrm{coker}(f')$.

\textbf{Well-definedness of $\partial$:} In the original construction, suppose we started with $x'' = 0$. By exactness at $M$, there exists $x' \in M$ such that $\iota(x') = x$. Then $\iota'(f'(x')) = f(\iota(x')) = f(x)$. Then $y' = f'(x')$ and its image is $\partial(x'') = 0$.

\textbf{Exactness at $\ker(f'')$:} In the original construction, suppose that $\partial(x'') = 0$. Then there exists $x' \in M'$ such that $f'(x') = y'$. Then $f(\iota(x')) = \iota'(f'(x')) = y = f(x)$. Then $h = x - \iota'(x) \in \ker(f)$, and by construction $\pi(h) = \pi(x) = x''$.

\textbf{Exactness at $\mathrm{coker}(f')$:} Let $y' \in N'$ such that the image of $y = \iota'(y')$ in $\mathrm{coker}(f)$ is zero. There exists $x \in N$ such that $f(x) = y$. We can take $x'' = \pi(x)$ and repeat the original construction, and we'll find that $\partial(x'')$ is the image of $y'$ in $\mathrm{coker}(f')$.

Is the proof of the salamander lemma really shorter than this?

[broken turtle]

Eduardo León

[lone jacinth]

Pseudo said that using the salamander lemma would not give a shorter proof

[plucky arch]

Yeah I definitely don’t think it does

I think this corollary in particular is helpful for me, since it tells me under what conditions I can go “against the arrows”

[muted sierra]

Just reading the definitions is giving me a headache. I only have so much working memory.

[plucky arch]

Which definitions

[muted sierra]

2.2

I'm exaggerating, though. I can deal with it just fine if I do it slowly.

[plucky arch]

Ah I see I see

I think the main new things to understand are these receptor and donor objects

[muted sierra]

It's not thaaat far from elementary. But the direct diagram chases you can actually teach to undergraduates.

It's cool that there's a single homological algebra result that subsumes so many others, though.

[plucky arch]

Hm interesting

Would you say the salamander lemma would be too advanced for undergrads?

I never took algtop formally as a course so I don’t know the level at which it’s taught very well

[muted sierra]

Not so much “too advanced” as it is “too long”.

[plucky arch]

Ah right

[muted sierra]

The “working memory” issue I mentioned earlier.

Algebraic topology can be done by jumping directly into abstract categorical nonsense à la tom Dieck, but it can also be done in a more intuitive, geometrically minded way.

[idle copper]

why every element of the commutator sub group is a product of commutators

[plucky arch]

Hm isn’t the commutator subgroup defined to be the subgroup generated by the commutators?

[spice idol]

yes

[spice idol]

and do try to refrain from crossposting

[idle copper]

yes ty i ll keep that in mind

it is just that i can't find why, that every element is the product of commutators, it is obvious why the product if commutators is in there to achieve stability
but i am paranoid what if there is a needed element that can't be generated by any finite product needed to achieve the structure of a subgroup

[plucky arch]

I guess the main thing you’d need to show is that the inverse of a commutator is a commutator

Because the subgroup generated by S consists of products of elements of S and their inverses

There’s also a linear algebra version of this statement which I found interesting

The set of all commutators of matrices forms an abelian group under matrix addition

[idle copper]

can you explain to me the link between

[plucky arch]

Hm I don’t know of a direct link

[idle copper]

i find it interesting as set of matrices already is an abelian group under addition

unless we speaking about matrices defined on a non abelian structures

[plucky arch]

The main surprising fact is that a sum of two matrix commutators is a matrix commutator

[idle copper]

oooh

[plucky arch]

Proving this takes some effort

[plucky arch]

This isn’t true in the group case

A product of group commutators is not in general a group commutator

[idle copper]

yes

[idle copper]

i ssee so it became pure sub group definition

<S> ={ all finite products of elements of S and ther inverses } is a subgroup

interesting how the i was paranoid about an element that shouldn't exist

ty for the answer

[spice idol]

many such cases, don't worry

often a lot of the hard work comes in proving that such an element (or whatever you're studying) does not exist

[idle copper]

yeah true , i was just teaching groups and got this part where i have to prove that commutator subgroup is normal

and i wanted my students to be careful with generated subgroups that they may include lot of extra things

[surreal isle]

how's everyone doing do yall have a good teacher for math on youtube

[spice idol]

depends on what you want to learn

at a certain point you'll start needing to learn from textbooks though

[idle copper]

yes, i prefer just following a solid textbook, and may be look specific stuff on youtube

[surreal isle]

well right now I'm thinking of starting from ground zero

[surreal isle]

alright.

[spice idol]

so like, pre algebra and calculus? or proofs?

those are two different ground zeros

[surreal isle]

correct

[spice idol]

that was a question. Either of the two

[surreal isle]

pre algebra and calculus

[spice idol]

then I shall refer you to #prealg-and-algebra , they will probably have better answers for you

this is a channel for abstract algebra, a collection of subjects from university and beyond

[surreal isle]

thank you

[idle copper]

@spice idol can i ask you something about constructive proofs in the #math-pedagogy it is about teaching advanced algebra

[spice idol]

I'm not any good a teacher I'm afraid

[idle copper]

anyway, i ll post there ty so much for the replays 🙏

[young forge]

I believe i may have found some more moufang loop key exchanges

See my thread for more

[wise sedge]

Is the subfield of F_2(x) fixed by all automorphisms itself isomorphic to F_2(x)?

[mild imp]

Does Advanced Algebra also include Abstract Algebra?

[lone jacinth]

This channel covers the more advanced parts of abstract algabra yes. See the channel description

[mild imp]

Awesome

[lone jacinth]

See also #groups-rings-fields , the other abstract algabra channel

[lone jacinth]

It seems that way to me.

Like let r be the automorphism r(x) = 1/(x+1). Then the fixed field of r is generated by the coefficients of
(t - x)(t - r(x))(t - rr(x))
and it seems to be
t^3 + ft^2 + (f+1)t + 1
where f = (x^3 + x + 1)/x(x+1). So this fixed field is F2(f). Letting s be s(x) = x+1 the fixed field by r and s is the fixed field by s in F(f). Which is generated by the coefficients of
(t - f)(t - s(f)) = t^2 + t + f(f+1)

So the fixed field is equal to F2(f(f+1)) ~= F2(x).

[onyx imp]

Could anyone help me check if the following is true? Let $F$ be a field with characteristic $p$ and $G$ be a group, Suppose we have an indecomposable module $FG$-module $M$ such that $p \mid \dim M$, then for any module $N$, the indecomposable modules $M_i$ of $M\otimes N$ will also satisfy $p \mid \dim M_i$

[broken turtle]

somethingwrong

[lone jacinth]

Doesn't seem true to me. Letting F be F2, G=C3 and M=N the 2d indecomposable

But maybe you want F alg closed...

[onyx imp]

yes sorry in my context F is alg closed and i might only want p | |G|

[plucky arch]

how should i actually think about clifford algebras, pin groups, spin groups and the like

I’ve seen elements of this story from the physicist point of view

[foggy galleon]

You will probably get better answers in #diff-geo-diff-top .

It's probably useful to read about Clifford's intuition and classical geometric algebra. Wikipedia says spinors were first introduced by Cartan, idk why. It's genuine motivation to think of dirac operators as "square roots" of the Laplacian. Why should that be a useful notion, mathematically? I have no idea, and I'd like to hear more.

Witten gave a quite simple proof of the positive mass theorem (this can be understood purely mathematically, and plays a role in the Yamabe problem). It's parseable, but I have no intuition on why Dirac operators should simplify things. Witten's intuition is physical and comes from quantum gravity, as explained in the paper.

Lawson-Michelsohn give a construction of linearly independent vector fields on spheres via Clifford algebras (in fact, the maximum number, by Adams).

Over Riemann surfaces, spin structures are related to theta characteristics (square roots of the canonical bundle -- which are related to theta functions)

[foggy galleon]

oh also I believe for the Atiyah-Singer index theorem it suffices to prove it for Dirac operators and the rest is quite topological (I'm not completely sure). So Dirac operators are a quite important class of operators among all elliptic operators. An example of a dirac operator comes from the Cauchy-Riemann equations

[summer quest]

Pin groups are sort of straightforward once you understand Spin groups, the main thing to understand is how Spin groups are related to Clifford algebras and what makes these things natural

although it is not necessary, it is very helpful to understand the Clifford algebra and Spin group story through their relation to topological K-theory (and this is closer to necessary if you want to understand how these things are related to things like the Atiyah-Singer index theorem for example)

Pin groups and Spin groups naturally fall out of Clifford algebras in the following way

if you have a vector space V equipped with a symmetric bilinear form <-,-> or equivalently a quadratic form q, then you can construct a Clifford algebra Cl(V,q) as the quotient T(V)/I(V,q) of the tensor algebra T(V) by the ideal I(V,q) generated by elements of the form v\otimes v-q(v)

why this is a natural thing to consider has various explanations, one of which is in terms of quantization but that's also not such a necessary story (although it is a good justification from the point of view of physics)

Both Pin(V,q) and Spin(V,q) are subgroups of the group of units Cl(V,q)*

elements of Pin(V,q) are multiples v_1...v_n of elements v_i in V with q(v_i)=1

elements of Spin(V,q) are multiples v_1...v_2n of elements v_i in V with q(v_i)=1

one natural justification for considering Spin(n) (that is the Spin group constructed from V=R^n with the usual Euclidean inner product) is through its relation to SO(n), similar to the relation between SO(n) and O(n)

if you have a smooth manifold X then the tangent bundle gives you a map X->BO(n), a choice of orientation lifts this to a map X->BSO(n), and a choice of spin structure lifts this to a map X->BSpin(n)

This is the start of the Whitehead tower ...->BSpin(n)->BSO(n)->BO(n)

you can sort of discover Spin(n) just from this Whitehead tower alone but this gives you just the case of Euclidean signature and there are Spin groups of other signature as well which is what the Clifford algebra picture explains most cleanly

[muted sierra]

Other signatures as in bilinear forms represented (in some basis) by the matrix diag(1,...,1,-1,...,-1) for some number of 1's and some number of -1's?

[summer quest]

yes

you can talk about Spin(p,q) coming from R^n equipped with a symmetric bilinear form of signature (p,q) for p+q=n

it's a good exercise in understanding Pin groups and Spin groups to check that Spin(n,0) and Spin(0,n) are the same (and both are usually just written as Spin(n) under this identification) whereas Pin(n,0) and Pin(0,n) are typically not the same (and are usually written as Pin^+(n) and Pin^-(n) to distinguish them)

there is of course a similar verison of this for complex vector spaces rather than real vector spaces but then you don't really have any interesting distinctions in signature

there is also a quaternionic version of this but the quaternionic Clifford algebras end up being essentially equivalent to the real Clifford algebras after some reindexing

the real, complex, and quaternionic versions of this story around Clifford algebras and Spin groups corresponds to the real, complex, and quaternionic versions of topological K-theory KO, KU, and KSp

in fact you can construct these respective K-theory spectra in terms of these respective versions of Clifford algebras, with Bott periodicity in K-theory matching periodicity at the level of Clifford algebras

[muted sierra]

#advanced-algebra message
#advanced-algebra message
These descriptions only work when q is positive-definite, right?

[plucky arch]

Ooh what’s the quantization pov?

[summer quest]

the quantization perspective on Clifford algebras and Weyl algebras is in terms of CAR algebras (Canonical Anticommuting Relation) and CCR algebras (Canonical Commuting Relations) respectively

[plucky arch]

I’m familiar with these from physics actually

[summer quest]

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

basically if you have an orthogonal vector space (a vector space with symmetric bilinear form) then the unital C*-algebra generated by elements of V with CAR relations is a canonical quantization and this is identified with the corresponding Clifford algebra

[plucky arch]

Also a silly Q but - is there a relation between the group algebra of the spin group and the Clifford algebra it comes from?

[summer quest]

similarly if you have a symplectic vector space (a vector space with antisymmetric bilinear form) then the unital C*-algebra generated by elements of V with CCR relations is a canonical quantization and this is identified with the corresponding Weyl algebra

[ornate atlas]

Weyl algebras in the sense of A_1 =K<x,y>/(xy-1-yx)?

we were always told these are related to physics in some way in our non com class, and like I see some kinda relation to them through commutation relations in QM, but what specifically do the Weyl algebras arise from?

[summer quest]

so in the same way that the CAR quantization gives rise to Clifford algebras and Spin groups, the CCR quantization gives rise to Weyl algebras and Heisenberg/metaplectic groups

[plucky arch]

Oh interesting

[summer quest]

in particular these Heisenberg/metaplectic groups are the symplectic version of Spin groups

what is interesting about this is that the Clifford algebras are finite dimensional and the corresponding Spin groups have plenty of interesting finite dimensional representations giving rise to spinors, whereas the Weyl algebras are very much infinite dimensional and the corresponding metaplectic groups only have interesting infinite dimensional representations

[summer quest]

yes, these are noncommutative deformations of polynomial algebras; one way to say this is that you're deforming the polynomial algebra C[x_1,...,x_n,y_1,...,y_n] of functions on the cotangent space T*A^n to a noncommutative algebra

[plucky arch]

I remember learning this in my linear algebra course implicitly

The issue is trace, right?

Trace of a commutator is zero in finite dimensions

[summer quest]

the Weyl algebra attached to a polynomial algebra C[x_1,...,x_n] is naturally identified with the Weyl algebra of polynomial differential operators C<x_1,...,x_n,y_1,...,y_n>/(Weyl relations) where you can think of y_i as \partial x_i

[summer quest]

well the issue is that the metaplectic double cover Mp(V) of the symplectic group Sp(V) has no finite dimensional faithful representations, in particular it cannot be identified with a group of matrices acting on a finite dimensional vector space

whereas the Spin double cover Spin(V) of the special orthogonal group SO(V) certainly has finite dimensional faithful representations since this can be identified with a group of matrices acting on the corresponding Clifford algebra

[plucky arch]

Essentially I was wondering whether like

The direction I’m familiar with is starting from a group and obtaining the group algebra

Are spin groups kind of the opposite

You start with the group algebra and obtain the group

[summer quest]

sort of in the sense that I think it's natural to view the Spin groups as arising from Clifford algebras and not so much the other way around, but I'm not sure that this is exactly the same as constructing group algebras

[plucky arch]

Like I was thinking

You take the Clifford algebra

And then the even subalgebra

And then maybe the spin group from that?

[summer quest]

it is coming from group algebras in the sense that for both CAR and CCR algebras you're looking at group C* algebras C*(G) whose spectrum is the Pontryagin dual of G

https://en.wikipedia.org/wiki/Stone–von_Neumann_theorem

this is worth reading about, the essential uniqueness of the Schrodinger oscillator representation of metaplectic groups that Stone-Von-Neumann describes is the main structural theorem in the CCR situation

usually when the Schrodinger oscillator representation is stated in quantum mechanics it's viewed as a projective unitary representation of the symplectic group (not least because in quantum mechanics you're looking at projective symmetries of some Hilbert space of states)

the Schrodinger oscillator representation is only a projective representation of the symplectic group, it does not lift to a genuine representation of the symplectic group, only of the corresponding metaplectic double cover

[ornate atlas]

Yeah my non com class spent a lot of time on them (we defined like holonomic modules and stuff) but what is the physical motivation for them? I know you can think of it as a ring of differential operators but apparently these came from Weyl considering something in physics and I was wondering what that was (if you or pseudo knows)

[summer quest]

well the oscillator representation was originally discovered by Heisenberg and Schrodinger in the simplest case of the quantum harmonic oscillator (which is why this is called the oscillator representation)

in the usual picture of quantization you are starting out with some symplectic manifold as a phase space, and quantizing this to some algebra of operators acting on some Hilbert space of states

the simplest case of the quantum harmonic oscillator comes from viewing the cotangent space T*R^n as a symplectic vector space of dimension 2n splitting into position and momentum coordinates

upon quantization you get a Hilbert space of states L^2(R^n) for the quantum harmonic oscillator with the x_1,...,x_n acting by position operators and with the y_1,...,y_n acting by momentum operators

[muted sierra]

So you're passing from a vector space to some kind of operator algebra generated by it. But how would that work when you start with an arbitrary symplectic manifold, rather than T*R^n?

[summer quest]

in general the quantization procedure produces a Hilbert space of states by geometric quantization, and produces a (noncommutative) algebra of operators by deformation quantization, which acts on this Hilbert space of states

the geometric quantization comes from looking at the space of L^2 sections of certain vector bundles over your symplectic manifold

the deformation quantization essentially comes from the Moyal *-algebra quantization procedure

[muted sierra]

Which vector bundles? And how much do spaces of L^2 sections of those vector bundles remember about the original symplectic manifold?

I guess I'm having the wrong intuition from algebraic geometry, where vector bundles can fail very badly to have nonzero global sections at all.

[summer quest]

well so the input for geometric quantization is to start with some symplectic form \omega. Before quantizing this, you are meant to prequantize this by replacing \omega with a U(1)-principal bundle with connection (L,\nabla) with curvature \nabla

[muted sierra]

Ah.

[summer quest]

the inverse Planck constant and the fact that things are quantized in terms of multiples of this comes from the first Chern class of this circle bundle, and the charge quantization condition demands that curvatures and Chern classes should define integer cohomology classes

to quantize this you then want to fix a polarization and then consider polarized L^2 sections

see the section on geometric quantization here https://ncatlab.org/nlab/show/geometric+quantization

[muted sierra]

Let's see. Thanks!

[summer quest]

one nice perspective on this geometric quantization procedure explained on the same page is that by fixing such a polarization you are basically fixing a Spin^c structure on your symplectic manifold, which gives you a Spin^c Dirac operator, whose index is exactly this space of states

there is a nice realization of topological K-theory in terms of spaces of Fredholm operators, which makes the connection between such Dirac operators and classes in topological K-theory more transparent. From this point of view, the geometric quantization is computed by pushforward to topological K-theory of a point

[muted sierra]

What I'm reading looks like some unholy marriage of differential geometry and functional analysis.

[summer quest]

many such cases

[mild imp]

What ai model is best for help on Abstract Algebra problems?

[spice idol]

dummit&footeGPT

[mild imp]

I’ve heard Wolfram Alpha Pro is pretty good

You can feed it images of abstract algebra problems and it’ll provide step by step solutions

[forest turtle]

!nogpt

[upper beacon]

Please do not trust ChatGPT or similar AI tools for mathematical tasks, as they often generate output which "sounds correct" but has numerous factual or logical errors. Use of these AI tools to answer other people's help questions is strictly against server rules (see #rules).

[forest turtle]

better to just ask here honestly

[mild imp]

I agree

Although I’ve also noticed humans are prone to make mistakes too

Wouldn’t it be of value to use AI as a supportive tool to aid in learning, all the while verifying answers with actual human beings?

Basically treat AI as a potentially helpful tool in certain contexts but not relying on their answers as the true solution but rather using their solutions to spur more thought?

[forest turtle]

my opinions on this are going to be pretty antithetical to yours. i mostly avoid ai

[muted sierra]

The problem with AI is... How are you going to judge the quality of the AI's answers, when you don't know the topic yet?

Even conceding that AI can be a time saver when you can check by yourself the quality of its output.

[plucky arch]

Does the category of chain complexes in an abelian category also form an abelian category?

And - are the relevant limits/colimits and additive structure all degreewise?

[muted sierra]

To the first question, yes.

[lone jacinth]

Same for the second

[fierce steeple]

Trying to think like what is the best way to see this lol

In many cases I think a cute way to argue this is that like a chain complex should be a module (in graded modules) over Z[x]/x^2 where |x| = -1

[plucky arch]

hm hang on

that looks suspiciously tangent vectory

[mild imp]

It breaks every problem down in detailed steps which makes it much easier to assess validity. Then you can cross check it with humans. I just got Chat GPT 5 Plus and it is amazing! Already learning tons from it. It certainly is speeding learning process.

[fierce steeple]

I am kinda unsure there is a link. An interesting link though is that this exterior algebra is the cohomology ring of a circle lol

[scarlet ermine]

I like doing this by proving that (possibly additive) functors valued in an abelian category form an abelian category, and then chain complexes are additive functors out of the walking chain complex category https://math.stackexchange.com/questions/2139250/can-the-category-of-chain-complexes-be-realized-as-a-functor-category

[fierce steeple]

Sounds good, thanks

I like this description as the idempotent completion of the linearisation of the (opposite of) the simplex category, too lol

[scarlet ermine]

haha nice

[hushed bone]

Nerd.

[hushed bone]

U go to Utah now?

R u tryna work with Hacon?

[fierce steeple]

I started a videogame where you have to become a fashion icon as quickly as possible. It's called the minimal model program

[scarlet ermine]

hehe nah I’m working with Bertram, trying to do some derived cats, maybe some bridgeland stability

[fierce steeple]

Nice

[digital parcel]

W

[fierce steeple]

Thanks

[digital parcel]

No problem

[young forge]

wut
DImension of the monster: 196883
Densest 24d sphere packing(related to the leech lattice):196560

Why are these so close? Normally i'd dismiss this as a coincidence, but monsterous moonshine exists so

[fierce steeple]

Have you seen this https://mathoverflow.net/q/429505/510603

[torn harbor]

both are closely related to modular forms, I'd imagine its something to do with that

[fierce steeple]

V cool

[torn harbor]

whenever something doesn't make sense in math the answer is probably modular forms

[fierce steeple]

Tl;dr can explain the two being close w modular forms lol

[young forge]

hmm

interesting

[broken wren]

The MO answer computes that it is correct via modular forms, but I doing think it explains anything

The posed question says that it makes sense in dimension 8, but it sounds like numerology to me. Presumably it’s the hint of a geometric construction, but I don’t see it

[torn harbor]

to be honest its really going to be impossible to explain it I think

in a satisfying way

[past kestrel]

Hello! I need some feedback on my mathematical paper about indeterminate forms. If you are interested, here is my link to the post on reddit: https://www.reddit.com/r/Arsammath/comments/1oy2te9/i_found_a_way_to_solve_indeterminate_forms_by/?utm_source=share&utm_medium=web3x&utm_name=web3xcss&utm_term=1&utm_content=share_button

[spice idol]

this is complete nonsensw lol

until you write it down formally in the currently existing mathematical framework, no one will take you seriously

this probably starts with taking an intro to proofs class at a university

[past cove]

sorry mister ramaswamy

but you're a crank

[forest turtle]

i think its good that you are thinking about math. but your first few pages loosely resemble asymptotic analysis and big O manipulation. maybe you should read more about those.

in general, there is no hierarchy that you speak of; functions on R are somewhat messy and you cant meaningfully linearly (or well) order them

[past kestrel]

Yeah about that. I had made this system before knowing about BIG-O-notation. Well the hierarchy is complicated and infinitely long, the only way to distinguish infinity levels is from the growth rates. I take tools from well know fields and implement them with a hierarchy of infinities.

This a concept not a proof guys. I need help to prove it of course but i wanted the idea to become public.

[past kestrel]

they called Ramanujan a crank

[ornate atlas]

You have been given well formed, respectful and helpful feedback and you chose to ignore it. If youre not willing to engage with it, then im not sure what youre looking for. In any case, this isnt the channel to discuss it

[gilded sleet]

Aint no way bro wrote a paper on reddit

[forest turtle]

what is the purpose of this comment?

[distant harness]

I'm not sure Irony's response in particular should be characterized as "respectful". Even though Hastysnail may well be a crank, calling him that to his face doesn't feel productive.

[past kestrel]

i mean that was respectful feedback, anyways Nope i was given feedback and i am grateful. But most were not related to the idea, but at minor misunderstandings.

[ornate atlas]

No im not refering to Ironys response, this has been going on a while, mostly in #discussion where there was genuine, helpful, respectful feedback about where the mistakes where, and further sources which could help explain them

I was reffering to a response from, IIRC, sheddow

[past kestrel]

Yeah there was some that actually enlightened me. But why cant i talk about the idea here? I used advanced algebra to solve indeterminate forms. I am not sending links anymore, only answering and debating.

[gilded sleet]

All standard algebraic rules apply within Arsam’s Calculus, except in cases where
asymptotic dominance causes a change in the infinity level of the resulting expression. If
uncertainty arises regarding which term dominates, Big O notation may be used as an
asymptotic verification tool to determine the dominant level.
This ensures that Arsam’s Calculus remains consistent with traditional asymptotic
analysis, while extending it to permit formal algebraic manipulation of infinity levels
beyond conventional limit -based approaches.

I'm having problems understanding this part. NOT because it is hard to understand or anything but im wondering why are you speaking in third person?

[past kestrel]

😂 its on zenodo, i was using reddit to reach out to people. I have contacted countless university but hey said that they cant help me.

[ornate atlas]

Please see the channel description, there is not any advanced algebra used in your writing, as far as the scope of this channel is concerned

[gilded sleet]

Im not trynna be rude but this is a nothing burger man im sorry

It mightve been written partly by ai too

[past kestrel]

bro its my first paper ever, i am a high school student talking about university level mathematics. I am not good at explaining, only asking and working on questions.

[gilded sleet]

Ask better questions then. Study the field you are going to touch on. Using AI is just gonna give you false hopes

[ornate atlas]

Im not sure what you want then. You admit you dont havre much background, yet dont take on the feedback from people who know more, and when you have been directed to sources which might improve your work you have refused to look.

[last talon]

Half of maths is being able to communicate your results in a way that makes it as easy as reasonable for others to understand

[gilded sleet]

I understand where you are coming from but high school maths is far from advanced maths

@past kestrel If you are going to progress in this field, put something in your burger before serving it

[past kestrel]

I want to see if my idea holds up before i commit fully to it. I am going to take the advice but i could easily make it easier to read and understand. What is not easy is to see if this idea or concept holds up.

[past kestrel]

Mathematics is not set in stone, but an ever-growing and evolving body of knowledge. Many great insights and discoveries have been made just because somebody stopped taking “you can’t do that” as an answer, and instead wondered, “what if you could?”

What assumptions will you question?

[gilded sleet]

This is not even a concept to hold up. What is Arsams calculus????? It just makes you sound like a big narcissist or just AI slop consumer.

[gilded sleet]

This is AI too I think

[past kestrel]

no its from the link Nope sent

[ornate atlas]

Ok, im just going to redirect this because its going in circles and not on topic. If you want to discuss this further, there is #math-discussion , but until you make it more clear what youre looking for and actually engage with people in a productive way, im not sure what you expect to get from this.

[past kestrel]

ok, milo if you want to talk, talk in math discussion.

i am going there

[gilded sleet]

True. Ngl Im tired too. Just wanted to give constructive criticism but the guy is a big time narcissist

[ornate atlas]

Insulting people is generally a poor way to get them to take your point

[gilded sleet]

Then you convince him. Also, Im sorry for that.

[young forge]

I mean, ramanujan
(we shouldn't dismiss someone just because they don't use formal notation)
But still

[plucky arch]

can i always realise an (associative) algebra as a quotient of a tensor algebra?

in quantum mechanics we often specify "algebras" by giving a list of generators, together with commutation (or anticommutation) relations

[young forge]

to me, a big sign on if someone like that is a crank or someone just not using familiar notation is how they respond to criticism

[spice idol]

iirc the free part of the adjunction R-Mod and R-Alg is given by the symmetric tensor product

so yes

[plucky arch]

hm, i see

[spice idol]

or whatever its called

[plucky arch]

wait, symmetric?

[spice idol]

wait thinking about commutative algebras

too AG-pilled

[distant harness]

If you're okay with infinitary tensor products, you can always take a copy of the underlying ring for each element in the algebra you want.

[plucky arch]

hm i see

[spice idol]

there is a natural projection of R-algebras T(A) → A

by sending a1 (x) a2 (x) ... (x) an to a1a2...an

[plucky arch]

What I mean is something like the construction of the universal enveloping algebra

[spice idol]

that is the counit of the adjunction of R-Mod and R-Alg

[young forge]

yeah nvm this guy is a crank

[spice idol]

or are we talking about a different tensor algebra

[young forge]

1000%

[lone jacinth]

The tensor algabra is the free algebras yes

[muted sierra]

Honest question, what's a reason to care about modular forms?

[torn harbor]

I'm not an expert but for example they feature heavily in number theory (the proof of fermats last theorem for instance)

[muted sierra]

I'm too dumb for number theory.

[torn harbor]

they are also crucial to studying sphere packings and lattices, as well as simple groups if you look at the comments above

They have a habit of showing up in a lot of places

Where you wouldn't expect

[muted sierra]

Okay, sphere packings sounds like a better reason.

[torn harbor]

The proof of the optimal sphere packings in 8 and 24 dimensions features modular forms heavily