advanced-algebra

[calm trellis]

got a stroke from the notation

[limpid horizon]

degree k terms like in what sense sorry

[calm trellis]

k-th term in your complex

[limpid horizon]

Ok

[calm trellis]

it's nothing that adds new information you just play with the indices

[limpid horizon]

so its really just relabelling

ok

[digital parcel]

Yeah

[calm trellis]

but seriously what is that lk_{st H_a} G_a notation

[limpid horizon]

lk is "link" an operation on simplicial complex

[calm trellis]

i see

[limpid horizon]

im trying to learn about Hochsters formula and reisners criterion

this is in chapter 5 of bruns and herzog

[quiet compass]

Say we have a filtration
$$0 = E_0 \subset \cdots \subset E_r = E$$
of $E$, then we get an associated graded object
$$G = \bigoplus_i \frac{E_i}{E_{i-1}},$$
and we can recover $E$ from $G$ by considering extensions of the summands. Is there a name for the condition where we ask for the extension
$$0 \to E_{i-1}/E_{i-2} \to E_i/E_{i-2} \to E_i/E_{i-2} \to 0$$
to be non-trivial for all $i$?

[broken turtle]

shingtaklam1324

[quiet compass]

If we think of the extension going from $G$ to $E$ as an upper triangular matrix $A$, this is the same as asking for $A_{i, i+1}$ to be non-zero for all $i$

[broken turtle]

shingtaklam1324

[quiet compass]

So this should be a condition on the filtration. Also the last term should say E_i/E_{i-1}

[lone jacinth]

The basis consist of the single basis vector [] so is rank 1

[ornate atlas]

Assuming my laptop feels functional when I get home I’ll send you some stuff, I don’t remember where but I remember finding a paper which was significantly less ass than Bruns and Herzog

My laptop feeling functional is currently a nontrivial assumption

[limpid horizon]

Oh, thank you, yeah i have been spending a long time on it its pretty annoying to parse through

[ornate atlas]

Yeah it’s certainly a book…

I don’t remember exactly where the exposition I found came from but I’d possibly check Miller and Strumfels, it could’ve been in there

There’s also a real possibility my laptop is just no longer working so I wouldn’t hold your breath, but there is definitely better stuff out there, I remember looking at all this for my UG thesis

[calm trellis]

book so ass the laptop gave up

[limpid horizon]

(for simplicial homology) it says if the complex is acyclic then its also split exact.. is that because the modules involved are all free modules?

[fierce steeple]

Sorry for the silly q but what precisely do you mean by split exact?

But yes this will very likely use the fact that the modules are free (it does if my interpretation is correct)

[fierce steeple]

Well this does not have the definition of split exact

[limpid horizon]

yeah thats just what it says

[fierce steeple]

oh lol

But okay one definition is that the canonical map 0 -> C is a chain homotopy equivalence

And then it'll be a special case of the fact that a quasi-iso of (homologically) bounded-below chain complexes which is degreewise free is a chain homotopy equivalence

[vapid axle]

Naive question, but what exactly is the purpose of forming rings and modules of fractions? Is it just for the purposes of localizing, which tells us information about the overall ring/module?

[fierce steeple]

I would view them as - analogously to forming quotients - simplifying your ring/module by cutting out some information. As a simple example, if you have some finitely generated abelian group $M$, then you can write $M$ as some direct sum $M = \mathbf Z^{\oplus m} \oplus \bigoplus_n \mathbf Z/n\mathbf Z$, and considering $M[1/p]$ will kill the bits where $p$ is not invertible (the $\mathbf Z/p^n$ bits). Or if you go all the way and form $M_{\mathbf Q} = M \otimes_{\mathbf Z} \mathbf Q$, then this kills all of the torsion and just leaves you with $\mathbf Q^{\oplus m}$. So this is much simpler and lets you e.g. work out $m$ more

[broken turtle]

Prismatic Potato

[fierce steeple]

You can also think about what it does to primes of a ring

[vapid axle]

Also, do we need Proposition 3.3 in order to see that S^{-1}M' is a submodule of S^{-1}M? It should be pretty clear without it, using the definitions (at least I hope, I did it in my head)

[vapid axle]

Oh interesting

[fierce steeple]

Yeah it isn't bad

i mean basically if m is zero (in either) then that is witnessed by some stuff in S which doesn't have anything to do with M beyond the span of m

[spice idol]

for example localising at a prime gives a local ring and those are nice to work with, and there are so-called local properties which hold iff they hold in all localisations at primes

[digital parcel]

If p is a prime ideal of a ring A, then primes of A/p correspond to primes of A containing p; primes of A_p correspond to primes of A contained in p

In this sense quotient and localization of primes are kind of complementary notions

[calm trellis]

complements?

[digital parcel]

Not like a real set theoretic complement

Like “complementary notions”

[calm trellis]

aah

[spice idol]

only with regards to primes

[digital parcel]

Yeah

Fixed for clarity

[vapid axle]

I like it when we make things easier to prove

[spice idol]

yes :3

[vapid axle]

wait localizations aren't always flat right

I know that S^{-1}A is flat whenever A is a ring

[digital parcel]

If $M$ isn’t flat then $S^{-1}M$ cant be flat, since $S^{-1}M \cong S^{-1}A \otimes_A M$

[broken turtle]

anamono

[digital parcel]

If that’s what youre asking

For M an A-module

[calm trellis]

you can cook up examples of this by looking at modules with torsion over integral domains

[fierce steeple]

This is false

e.g. Z/2 (+) Z is not flat over Z but is after tensoring up to Q or inverting 2

[digital parcel]

True

Whoops

[fierce steeple]

Or more dramatically dont even add the Z lol

[digital parcel]

I shouldve said S^-1M isnt always flat

[fierce steeple]

You can also just invert the multiplicative set {1}

[digital parcel]

Yeah I had some feeling the “can’t” part of my original message was off

[indigo lagoon]

What is R^m here given a presentation (X|Y)? In general, the submodule generated by the relations is not going to be free.

[scarlet ermine]

So the sub module generated by the relations is the image of the map R^m -> R^n, not R^m itself

the m here is the number of relations (the cardinality of Y), and you build the map by sending the basis vectors of R^m to relations

[indigo lagoon]

Ah, okay. That makes sense, thanks!

[limpid horizon]

When you do a tensor product of chain complexes, what happens to the differential maps?

Its fine ill just google a defn lol

[hushed bone]

You use the induced differential on the total complex of a bicomplex

[limpid horizon]

im in a situation where C is a fg free Z-module. k is a field, and we can consider HomZ(C,k) as a k-vector space right? Would it be the same rank as C?

Can we like write HomZ(C,k) = HomZ(Z (+) Z, k) = HomZ(Z,k) (+) HomZ(Z,k) like does that help (just in example where rank C = 2)

I guess so, since HomZ(Z,k) = k probably as k-modules

I havent studied dual vector spaces before but this must be relevant right?

[distant harness]

You don't really need duals, if you mix abstraction levels and say:
Let A be the set of generators of C. By abstract nonsense (free-forgetful adjunction in Zmod), HomZ(C,k) is in bijection with Set(A, k).
By the concrete construction of the free k-module with A generators, its elements are exactly all the functions in Set(A, k) that are nonzero at finitely many inputs. But since A is finite, that is everything in Set(A,k).

[digital parcel]

Iirc there’s a simple example for the tensor product of Koszul complex worked out in Serre’s Local Algebra book

Where he writes out the explicit differential

Hmm maybe I remember wrong

Anyway I think rotman also does it

[near lantern]

I suppose Hom_ℤ(C, k) = Hom_k(C (⨯)_ℤ k, k) (isomorphic as k-vector spaces) is the dual of the k-vector space C (⨯)_ℤ k,

[fierce steeple]

exactly this yeah, and note this is more canonical (i.e. does not require a choice of basis for C)

[lone jacinth]

Works for finite rank.

But countable rank turns into uncountable dimension and similarly for bigger cardinalities

[atomic crow]

If I have A,C two abelian group, P* a projective résolution of C, how can I create a bijection from H^1(Hom(P*,A)) to Ext(C,A) ?

[lone jacinth]

So you have
0 -> K -> P0 -> C -> 0
as the first part of P*
And H1(Hom(P*, A) is the cokernel of
Hom(P0, A) -> Hom(K, A)

The bijection is given by pushout of
0 -> K -> P0 -> C -> 0
along the map K -> A

It's an easy diagram chase to check that the pushouts splits iff K->A factors through P0, and you can show surjectivity using that P0 is projective

[quiet compass]

Suppose $E \to X$ is a complex vector bundle of rank $r$, and we have an $\mathbb C^$ action on $E$ fibrewise. In particular, we get representations $\mathbb C^ \to \mathrm{GL}(E_x)$ for all $x \in X$. Any such representation decomposes into weight spaces, so we have $\lambda_1(x), \dots, \lambda_r(x) \in \mathbb Z$ being the weights (not necessarily distinct). Is there anything we can say about how the weights vary as $x$ varies?

[broken turtle]

shingtaklam1324

[fierce steeple]

I strongly suspect the weights (with multiplicity) are locally constant, right?

[quiet compass]

That's what I would hope, and I feel like something like Peter-Weyl would imply that?

which would involve a non-trivial bit of analysis but that's fine with me

[fierce steeple]

Hm maybe, my suspicion is that it should be possible to generalise weight space decompositions to this setting

But idk a reference 😭

[vapid axle]

Very general question, but are there techniques to show that local properties are satisfied? (For example showing that a localization is zero)

Or is it usually much easier to show that the localization is zero for all prime ideals rather than showing the entire ring is zero

[digital parcel]

If S is a multiplicative set, A a ring, and M is an A-module, then S^{-1}M = 0 iff there exists s in S st sM=0

[fierce steeple]

I would say this is like too general to answer nicely

[vapid axle]

Okay what about local properties like injectivity lol

[fierce steeple]

Like local rings are ig nicer than just a random ring but it won't let you magically do what yoy want

[vapid axle]

Why are they nicer though, is it like you're cutting out information like you said earlier

[digital parcel]

Atiyah-Macdonald is littered with local property statements, if you’re curious

[vapid axle]

Yeah I'm going over local properties right now

I'm spoiled and am wondering if there's any way to make things easier 😂

i.e. anyway to reduce further

[fierce steeple]

For example some things like Nakayama become way easier to apply

And structurally much clearer

[digital parcel]

In general there’s a lot of nice statements for local rings that you get to apply after localizing at a prime ideal

[vapid axle]

I see

maybe there'll be some in the exercises

[digital parcel]

I’m pretty sure almost every exercise in chapter 3 of atiyah-macdonald is those kind of local property things

Or at least over half

[vapid axle]

have you gone through atiyah macdonald before

[digital parcel]

Mostly yeah

[vapid axle]

nice

I hope to finish it within this year so I can proceed with hartshorne 😊

[digital parcel]

I think a lot of ppl here have

Based

[vapid axle]

actually probably not going to happen within this year but whatever

[digital parcel]

But dont rush it. It’s very important stuff

[fierce steeple]

I am a local property.

[digital parcel]

Holy shit

Grok is this true

[vapid axle]

yeah for sure I'm averaging like 2-3 hours per exercise lol

[digital parcel]

Enjoy

It’s a good book

I should order a hard copy and throw away bruns herzog

That book so god damn boring I didnt even read it

I just use it as reference

[fierce steeple]

Atiyah–Macdonald had a farm

[digital parcel]

Eieio

I remember I bought a copy of bruns herzog and showed my prof

He said “as far as math books go, that’s one of the most boring ones ever. and we’re talking about math textbooks here”

(I forgot if i told that story before but i giggle every time i think abt it)

[ornate kindle]

For a infinite-dimension vector space $V$, we have that End($V) \not\cong V\otimes V^*$

[broken turtle]

NotABot

[ornate kindle]

Is there any way to express End(V) as the tensor product of two or more spaces?

[broken turtle]

NotABot

[solemn urchin]

There is a natural isomorphism V \otimes V* ~ End(V)

[fierce steeple]

Only for finite dimensional V right

(Or at least, the canonical map one wants to write down is only an isomorphism in that case)

[fierce steeple]

oh lol this point has already been made

😭

[fierce steeple]

(so why this message?)

[solemn urchin]

Sorry lol, I was going to type a question myself and thought of answering the line above before but I didn't scroll enough apparently ahah

[fierce steeple]

I epxect it to be hard to formalise what you want -- you may want this to be a "natural" construction, but the assignment V |-> End(V) does not really lift to a functor

I guess one question would be whether there are spaces $A(V), B(V)$ (where $A$ is cotravariant and $B$ is covariant) such that $\mathrm{Hom}(V, W) \simeq A(V) \otimes B(W)$

[broken turtle]

Prismatic Potato

[solemn urchin]

Mm, that's an interesting question

Anyway, my problem instead was:
(I was cosidering linear algebra for a moment, but then opted for this channel)

Are real diagonizable matrices dense in M_n(R) ?

If n=2, observing the sign of the polynomial discriminant (under a perturbation) for the companion matrix of x^2+1 leads to a negative answer

Then, maybe I am missing something easy, but if n>2 I couldn't quite generalize the proof

So they are instead dense when n>=3?

[fierce steeple]

Oh wait R instead of C lol hm

Then this should be false for similar reasons: the characteristic polynomial depends continuously on the matrix coefficients, and the roots of a polynomial vary continuously with the coefficients

[quiet compass]

I asked ChatGPT and it gave me what looks like a correct answer lmao

[fierce steeple]

What did it say lol

I mean I also gave you what I think is a correct answer, just without proof

[quiet compass]

The weights are log of eigenvalues, which depend continuously on the base, but are also integers

[fierce steeple]

indeed

okay log of eigenvalues is good ig there

[solemn urchin]

That was my possible approach, but what we would like to happen is that, under any sufficiently small perturbation, a polynomial which doesn't split over R (it has an irreducible factor of degree 2) it still doesn't split

[fierce steeple]

Yeah but that will follow by what i said right

[solemn urchin]

For degree 2 polynomials the sign of the discriminant is a functional way to check this fact. But if the degree raise what ways do we have?

[fierce steeple]

One of the roots (over C) has non-zero imaginary part and this remains true under small perturbations of the coefficients

But I think this can also be done more explicitly by considering a block-diagonal matrix made up of, say, a 2x2 rotation matrix and the identity matrix

[solemn urchin]

Mm I see, that explaination seems nice. Although I am still a bit alarmed by this, so perturbating a polynomial doesn't mess at all with its factorization? I would have thought some wild behaviours could emerge, just instinctively

How do we make presice that roots depend continuously by the coefficients? Wouldn't this require the existence of a solving formula to express the roots for any degree?

That idea about the block matrix stopped me with this thought: perturbating a matrix doesn't leave invariant spaces still invariant right? So it should compromise it

[ornate kindle]

Well I'm not necessarily looking for something natural (although that'd be neat) I'm just wondering if it's possible at all

[ornate kindle]

Yeah that's more or less what I'm thinking

[fierce steeple]

I think the thing is that there are just gonna be somewhat trivial answers to this because vector spaces have so little structure

like End(V) is just a vector space of some infinite dimension and you can always write those "non-trivially" as tensor products in tons of different random ways

[ornate kindle]

Hmm that kinda makes sense

[quiet compass]

Also for infinite dimensional stuff you might want more structure? eg asking for V to be Banach, asking for continuous dual, using the tensor product of Banach spaces etc

[fierce steeple]

The answer to this should be no, like i believe you can just deduce that B(W) = W naturally and then you are trying to write Hom(V,W) = A(V) (x) W, but then also the assignment W |-> A(W) is meant to preserve limits and so it just takes duals

And then you will just wind up with Hom(V,W) = Hom(V, k) (x) W as the only option, via the usual map

And then ig you just check this is not always an iso

[fierce steeple]

Imagine working over R or C smh

[ornate kindle]

Hmm okay I think I get it

[solemn urchin]

@fierce steeple do you have any followup ?

[fierce steeple]

Hm well there are formulae for roots of polys i believe which are continuous, but you can also do ti more directly by like inverse function theorem i think

But also like e.g. if you consider the block diagonal matrix which is a rotation by 90 degrees in top 2x2 and 0 elsewhere, then the min poly is (x^2 + 1)x^n for some n. Thi scannot be very close to a polynomial which splits over R

Indeed note that $(x^2 + 1)x^n$ and $\prod_i (x-\lambda_i), \lambda_i \in \mathbf R$, must be far apart at $i$ because $|\prod_k (i-\lambda_k)| \ge 1$

[broken turtle]

Prismatic Potato

[solemn urchin]

Thank you! : ) It makes total sense.
I had given up too early on that path.
Very nice argument there

[fierce steeple]

thank, and np

[ruby otter]

i have a feeling these previous two exercises are supposed to help me prove that Q is not a free Z-module, but I don't see how exactly because Z is not a field?

maybe try to prove that there are supposed to be uncountably many elements to make a basis

yeah I don't see how. I did solve 1.13 and 1.14

[neon flare]

Arguing based on the size of the basis is a good path

Think about linear dependence

The other exercises are not necessary

[vapid axle]

The second claim in Exercise 15 seems to be false... for instance, if $\mu_2(x_2) = 0$, then $x_2 \coloneqq (0, x_2, 0, \dots) = n_1x_1 - n_1 \mu_{1, 7}(x_1) + n_{10} x_{10} - n_{10} \mu_{10, 18}(x_10)$, say. Writing $\mu_{1, 7}(x_1) = (0, 0, \dots, 0, x_1', 0, \dots)$ in the 7th coordinate and $\mu_{10, 18}(x_{10}) = (0, \dots, 0, x_{10}', 0, \dots)$ in the 18th coordinate gives $(0, x_2, 0, \dots) = (n_1 x_1, 0, \dots) - (0, \dots, 0, n_1 x_1', 0, \dots) + (0, \dots, 0, n_{10} x_{10}, 0, \dots) - (0, \dots, 0, n_{10} x_{10}', 0, \dots)$. Such an expression is impossible, since there are no elements of $\iota_2(M_2)$ on the right hand side. Or maybe this is the point?

[broken turtle]

okeyokay

[fierce steeple]

Considering Q is coumtable this should not be the case

[lone jacinth]

I'm not sure how to make a solution that encorporates the two other exercises, but it's not hard to show that any pair of elements in Q is linearly dependent. So if it was free it would need to be isomorphic to Z (which it is not)

[sly rune]

then you must have n₁ = n₁₀ = 0 so that j := i shows existence

hint: ||think about elements in D as "if you bring M_i to M_j then x_i - μ_ij(x_i) dies". then use the fact that you can represent x_i as a finite sum of elements of the form x_j - μ_jk(x_j)||

[ruby otter]

Thanks! Great!

[cerulean cove]

writing $H^n$ where H is quaternions space , as $C^{2n}$ then we can define determinant of matrix $A\in M_n(H)$ by defining 2n x 2n matrix as below but C can take 3 embeddings but why determinant is independent of embedding ?

[broken turtle]

Curvature

[summer quest]

that this is independent of the choice of embedding reduces to the case n=1 and it's a quick little computation with 2x2 complex matrices

[cerulean cove]

how it reduces to n=1 case

[summer quest]

you're writing a 2nx2n matrix in terms of nxn blocks and the determinant respects this block decomposition

e.g. for n=2 if A,B,C,D are 2x2 blocks then you have the following

$\mathrm{det}\begin{pmatrix} A & B\ C & D \end{pmatrix}=\mathrm{det}(AD-CB)$

[broken turtle]

nGroupoid

[cerulean cove]

ohk so this reduces it

[summer quest]

if you like you can check a similar thing involving nxn complex matrices written as 2nx2n real matrices

essentially the same argument

[cerulean cove]

we have to show for n=1 only by this

[summer quest]

yeah and then the n=1 computation is straightforward

this works for any embedding of C into H not just these three

[cerulean cove]

do u have proof of this

[cerulean cove]

does this formula always holds?

[atomic crow]

When I want to prove that is surjective, I use P0 is projective to obtain a morphism P0 —> E (E is my extension), then I take the pullback to complete my diagram an obtain a morphism f : ? —> A. How do I know f define a morphism from P1 to A ? (I’m really not familiar with pullback, pushout, sorry if it’s trivial)

In fact, I want f to be in H1(Hom(P*,A))

[cerulean cove]

@lone jacinth can u help me

[lone jacinth]

P0 -> E -> C has kernel K, so restricting P0-> E to K the image lands in A.

I.e. it restricts to a map K -> A, and K is a quotient of P1

[calm trellis]

don't just ping random people

[lone jacinth]

Help with what anyway?

[calm trellis]

it's just a lin alg fact you can probably find it on the wikipedia of block matrices somewhere

[cerulean cove]

he is helpful not random for me

[cerulean cove]

dont u think i tried it, he wrote the formula for AC=CA case

[cerulean cove]

see above

[lone jacinth]

Aren't these embeddings just conjugates of each other.

Seems like it to me without verifying anyway

[cerulean cove]

i am concerned about the aanswer given by groupoid above , the determinant formula only works for AC=CA

[atomic crow]

Yeah it’s true, look for Schur determinant formula

[lone jacinth]

Well you'll have to ask groupoid about that

[atomic crow]

Sorry I don’t get it. If you restrict a map from P0 to E, the image lands in E ? An why K is a quotient of P1 ?

[cerulean cove]

https://math.stackexchange.com/q/1905652/1166895

[lone jacinth]

You have a short exact sequence
0 -> A -> E -> C -> 0
then A is the kernel of E->C.

So since K -> P0 -> E -> C is 0, the image of K -> P0 -> E must be contained in A

[atomic crow]

Okk I agree, and why K will be a quotient of P1?

[lone jacinth]

That's how projective resolutions work. P0 is mapped to by P1, mapped to be P2 etc

[atomic crow]

So whenever you have a map P0->>C, you already know the kernel is a quotient of P1 ?

[lone jacinth]

Well it's not "whenever you have a map".

A projective resolution is an exact sequence
... -> P1 -> P0 -> C -> 0
So then the kernel of one map is the image of another

[atomic crow]

I agree with that, but here we’re taking the kernel of P0 -> E -> C -> 0, wich is not a morphism of our exact sequence (or I missing something)

[lone jacinth]

It is

That is how we chose P0 -> E

[atomic crow]

Thanks, I finally got it

[edgy pond]

Could someone explain why the eigenvalues must be integral linear combinations of L_i?

[cloud karma]

What is 12.2?

[edgy pond]

It is the eigenspace decomposition of an irreducible representation of sl3C

V = bigoplus V_alpha

That equation is 12.2

[tepid flower]

Don't you get integrality from the Cartan matrix?

[agile kraken]

For some Galois extensions K/R (it should equal C but we dont know that yet), write G = Gal(K/R). In some exercise, we are told "let H be a 2-Sylow subgroup of G [...]". Erm this might be dumb but how do we know it exists? (im really bad at group theory...)

[silver goblet]

Every group G has Sylow p-subgroups for all p. They will be of order the p-adic valuation of |G|

[fierce steeple]

(In particular they might be trivial!)

[silver goblet]

Yes good point. The wikipedia article says n>0 but this is silly since the trivial thing satisfies all the properties of a Sylow p-subgroup

[fierce steeple]

I guess maybe this is a convention thing but it seems cleanest this way

[agile kraken]

ohhh i frgot to consider this subgroup

ty you two

nemeL

[agile kraken]

how do i get the algebraic closure of K((z)) ?

[lone jacinth]

https://en.m.wikipedia.org/wiki/Puiseux_series might help

In particular if K is characteristic 0, then the algebraic closure would be the Puiseux series over the closure of K

[agile kraken]

thanks !

[limpid horizon]

for a chain complex C, what does the complex C[-1] mean again ? C_i in C[-1] is C_i+1 in C?

[rare walrus]

I think it should be the other way around. The ith chain in C[n] should be the (i+n)th chain in C

This indexing always makes my eyes go crossed though so it's worth having someone else back it up

[limpid horizon]

what happens to the maps (or "differentials")?

[rare walrus]

...they also get shifted

The ith chain (resp. differential) in C[n] should be the (i+n)th chain (resp. differential) in C

[rare walrus]

Yeah indeed the stacks project agrees with me

https://stacks.math.columbia.edu/tag/0119

wait hold OOOOOOON

🙏 there is an extra thing happening to the maps

They get sent to (-1)^n times the (i+n)th chain map

that's a bit rogue

[limpid horizon]

it seems weird that C[-1] increases the index by 1 and doesnt decrease it

[lone jacinth]

The nth homology of C[-1] should be the (n-1)th homology of C. I find that pretty intuitive.

It's less clear to me whether this should be interpreted as increasing or decreasing the index.

But since you're writing C[-1] instead of C[1], maybe you're reading Weibel...?

Which I recently found out has the opposite convention, for some reason I cannot comprehend at all.

[limpid horizon]

im looking at something thats using simplicial homology, in a theorem they had to shift C to some C[-j-1] so i was needing to understand what that meant

[lone jacinth]

So I guess your saying that if C is concentrated in degree 0, then C[-(j+1)] is concentrated in degree j+1, which you would intuitively assume it was -(j+1)?

[fierce steeple]

The short answer is it depends on conventions lol

😭

Was talking to jagr about this all

But usually you can just infer the convention quickly from context

[lone jacinth]

Worst case just change notation to
$\Sigma C$ and $\Omega C$

[broken turtle]

Not jagr2808

[digital parcel]

Bruns-Herzog (if that's what you're still reading, kiand) uses the notation C[i]_n = C_{i + n}

I just checked lol, it's on page 32

[fierce steeple]

Real

[placid drift]

Example of transitive group action that is not faithful?

[weak lodge]

Action of $\bR$ on the unit circle $S^1$, namely $r\cdot e^{i\theta}=e^{i\theta r}$

[broken turtle]

harmacist

[placid drift]

Thankyou, what about the trivial group action , i.e gx=x for all g in G and x in X.

[stray grail]

I don’t think this action is transitive unless |X|=1. If there are two distinct points, there isn’t any group element which would get you from one to the other since the action is trivial.

[placid drift]

I get the kernel to be 1. Why is it not faithful?

[stray grail]

The kernel should be the whole group G

Since every element in the group is mapped to the identity map

[weak lodge]

The kernel is actually $2\pi\bZ$

[broken turtle]

harmacist

[stray grail]

Oops I didn’t see they were replying to your example… lol

[placid drift]

Understood, thanks 👍

[fierce steeple]

Let any non-trivial group act on a singleton 😄

But also like you can completely classify them: if G acts transitively on X then you can identify the action with the action of G on G/H

(Harmacist's example can then be understood as the fact "S^1 = R/2pi i Z")

[rare walrus]

Take any transitive group action of a group G on a set X. Choose any group H. Then GxH acts on X via (g, h).x = g.x. The kernel of this action is {1}xH.

[stray grail]

Isn’t the kernel not necessarily trivial (in particular ker (G->sym(X))? Like in Harmacists example? I think it should be ker(G-> sym(X)))xH.

But it does give a nice way to make an unfaithful action from a faithful action.

[rare walrus]

You're right, I mistakenly at the end assumed that the action of G was already faithful. But yes, my point is we can easily make new unfaithful actions, in fact "as unfaithful" as we like

[echo summit]

For I and J radical, is it true that rad(I+J)=rad(I)+rad(J)?

[digital parcel]

no

the sum of two radical ideals isn't always radical

but rad(I+J) = rad(rad I + rad J)

see Atiyah-Macdonald chapter 1

[digital parcel]

<@&268886789983436800>

[sour tartan]

lol that was strange

[sour tartan]

Is this a good book on ring theory ?

[void plank]

Is there a notion of a diagram of chain complexes commuting up to homotopy?

[digital parcel]

yes

[sour tartan]

Thanks for the rec

[digital parcel]

i'm not sure i understand what you're asking too well, maybe someone else does, but it sounds like you're talking about chain homotopy?

[void plank]

Well, a chain homotopy is an equivalence relation between two chain maps.

But I guess I'm asking if there exists a notion of a "map" between chain complexes that's weaker than a chain map, more specifically where the resulting diagram commutes up to homotopy?

[lone jacinth]

Just a commutative diagram in the homotopy category? Sure

[lone jacinth]

The question doesn't seem to quite make sense.

Homotopy is a relation between chain maps, the squares of a chain map can't commute up to homotopy

[void plank]

Ah sure.

On a related note, suppose I have a series of vertical maps connecting each degree of two chain complexes but that don't necessarily make the resulting diagram commute. Are there situations where this can be "deformed" into a proper chain map?

And are there any statements at all that can be made about what map in homology this series of maps induces (if that even makes sense)?

[lone jacinth]

I don't know about "deformed", but something that comes up is the Hom-complex whose degree i component is families of maps
An -> Bn+i
and where the differential takes
(fn) to
d_B o fn - fn+1 o d_A

The i-cycles of this complex are maps Hom(A, B[i]) and the homology are those maps up to homotopy

[lone jacinth]

You wouldn't really expect them to induce a map in homology

[void plank]

Ahh yes, I forgot about this example. This might be related to what I need in my practical application

Thanks!

[void plank]

I guess I want to reformulate my original question as follows: Can this idea be used to come up with a weaker notion of a chain map that still induces a well defined map in homology?

[lone jacinth]

So if you just look at what is needed to induce a map in homology:
Say you have f_i : Ai -> Bi
and you want f2 to induce a map in 2nd homology.

Then for every x in A2 with d(x) = 0 you would need d(f2(x)) = 0
and whenever x = d(y) there would need to be a z in B1 with f2(x) = d(z).

It's a little hard to imagine this as a condition on the fs in a natural way without just having z = f1(y). But I suppose they could anti-commute or commute up to some arbitrary automorphism

[void plank]

Sorry, which things are you saying anticommute?

[lone jacinth]

The
A1 -> A2
V V
B1 -> B2
square for example

And all the other squares

[void plank]

Ok cool, this should be sufficient for my purposes. Thanks a lot!

[limpid horizon]

If I have a complex whose modules are G-graded, to have a complex comprised of the a-th graded pieces, we need the original maps in the complex to be homogenous right?

Then in the a-th graded complex the maps there are just the restriction of d to that subgroup

[fierce steeple]

Sure, in other words you want it to be a complex of G-graded modules

[limpid horizon]

Oh yea like in the category of G-graded lodules

Modules

[cobalt sonnet]

is there a concrete example of an autonomous category having an object with a left dual which is not isomorphic to the right

[edgy pond]

Could someone explain the computation carried out here line by line? I am not able to follow anything past the first line :(

[weak lodge]

What are E_3,2 and E_2,1?

[mental escarp]

Matsumura is another good communist algebra book

[limpid horizon]

Upon the Witnessing

[spice idol]

what if you want to learn noncommutative rings

[tacit cipher]

i was looking for some thoughts on my publishment
https://divisionbyzeroaxiom.github.io/dbz-algebra-site, it's an algbera system.

[spice idol]

fix the latex 🙏

[tacit cipher]

wym?

[spice idol]

fix the latex, currently your latex has every special symbol written out

you probably need to add the amsmath and amssymb packages

[tacit cipher]

oh shoot, it wasnt like that before

when i’m home ill update it

[spice idol]

also, the fact that there is a parameter t here out of nowhere shouldnt be possible, any element in a commutative ring has at most one inverse

i dont get the premise either, a nilpotent element suggests an element that is "very small", in the sense that it multiplies to 0. So i dont know where the idea of x ⋅ κ = x/0 comes from because one should not expect 1/0 to behave like a nilpotent, because 1/0 should be treated like very big, more like an idempotent.

[tacit cipher]

yea there shouldnt be a T parameter it’s cuz I did /tfrac and it dropped it so it jus looked like T, mb

[spice idol]

whats tfrac?

$\tfrac ab$

$\frac ab$

hm

[tacit cipher]

setup:
work in the dual-number ring F[k]/(k^2). Let y = a + b*k with a ≠ 0.

find y^{-1} = α + βk so that (a + bk)(α + β*k) = 1.
multiply (using k^2 = 0):
aα + (aβ + bα)k = 1.

match coefficients:
aα = 1 -> α = 1/a
aβ + bα = 0 -> β = -b/a^2

so: (a + b*k)^{-1} = (1/a) + (-b/a^2)*k.

uniqueness: in any monoid, if u and v are both inverses of y, then
u = u*(yv) = (uy)v = 1v = v. So there’s no extra parameter.

the reason i’m using nilpotent is cuz im not making 0 invertible.

I define a total operator:
x / 0 := x * k
where k is central and nilpotent (k^2 = 0). It’s a bookkeeping tag that
carries the “remnant” (like a residue) through algebra.

reasons:
• keeps ring laws: associativity, commutativity, distributivity, and x0 = 0.
• linear: (x+y)/0 = x/0 + y/0.
• plays well with products: Res(xy) = Re(x)Res(y) + Re(y)Res(x).

this is different from frameworks that treat 1/0 as “very big” or idempotent;
those are wheels/transreals, a different semantics. I’m choosing a stable,
nilpotent totalization specifically to preserve ordinary algebra.

discord keeps fuckin up my syntax man

[spice idol]

still, a nilpotent element does not act like 1/0

nilpotents are more like infinitessimals

[forest turtle]

text fraction

[spice idol]

huh, cool

[forest turtle]

dfrac is display fraction

[spice idol]

FRACTION IS FRACTION

[tacit cipher]

i agree: a nilpotent does NOT “act like 1/0”. i’m not making 0 invertible.

what I’m doing is define a total operator:

x / 0  :=  x * k

inside the dual-number ring F[k]/(k^2) (k central, k^2=0). here “1/0” is just
shorthand for δ(1)=k, not an actual reciprocal of 0.

So this is bookkeeping,
not infinity.

[spice idol]

but why make it nilpotent then is my question

[tacit cipher]

why nilpotent? because it preserves ordinary algebra:

• (x+y)/0 = x/0 + y/0 (linearity)
• (xy)/0 = x*(y/0) + y*(x/0) (product rule for the k–coefficient)
• 0/0 = 0, and still x0 = 0
• multiplication: (a + bk)(c + d*k) = ac + (ad + bc)k

call Re(a+bk)=a and Res(a+bk)=b. Then Res(xy)=Re(x)Res(y)+Re(y)Res(x).

this lets me carry the “principal-part coefficient” (residue) through algebra
without breaking associativity/distributivity.

otherwise it breaks

Example in F[k]/(k^2):
5/0 = 5k = 0_5
(5/0) + (5/0) = 10k = 0_10
(5_3)*2 = (5+3k)*2 = 10 + 6k = 10_6
(5_3)*0 = 0 (and Res(5_3)=3; you read it with Res, not by *0)

[spice idol]

having it not be nilpotent preserves ordinary algebra too

or, if you want an algebra of dimension 2, any other quadratic polynomial besides x^2 would work just as well

[tacit cipher]

yep lots of tags would “preserve algebra”.

the reason I picked a nilpotent is that I want 3 extra properties at once:
1. keep the 2-slot form x = a + b*s (no higher powers),
2. Re(xy) = Re(x)Re(y), and
3. Res(xy) = Re(x)Res(y) + Re(y)Res(x) (the leibniz-style rule that matches simple pole calculus).

If we assume those, nilpotent drops out as a consequence:

work with a central symbol s and force closure in 2 slots:
(a + b s)(c + d s) = (ac) + (ad + bc) s + bd s^2.

since we only allow 1 and s, write s^2 = U + V s for some scalars U, V.
then
(a + b s)(c + d s) = (ac + U bd) + (ad + bc + V bd) s.

now impose the two desiderata:

(1) Re(xy) = Re(x)Re(y) for all x,y => U must be 0
(otherwise you'd get the extra term U bd in the real part).

(2) Res(xy) = Re(x)Res(y) + Re(y)Res(x) for all x,y => V must be 0
(otherwise you'd pick up + V bd).

Hence U = V = 0, i.e. s^2 = 0. that is exactly “nilpotent of order 2”.

[spice idol]

damn you type awfully fast

[tacit cipher]

TLDR
nilpotent isn’t about “acting like 1/0”. it’s the unique choice that (i) keeps the 2 slot numbers closed, (ii) makes Re a ring hom, and (iii) gives the clean product rule for Res that mirrors simple pole residue calculus.

[tacit cipher]

TLDR
lots of 2-D algebras “preserve algebra”, but if you also want
Re to be a hom and Res to satisfy the Leibniz-style product rule,
you are forced to the nilpotent case.

[tacit cipher]

162 monkeytype

[spice idol]

thats why i stopped writing, too many sweats

[spice idol]

huh, then maybe make that your introduction

your axioms are also a mix between definitions and immediate properties

[tacit cipher]

thsts a good suggestion thank you, ill be sure to add better clarifcation

[tacit cipher]

i’m new to writing axiom, feedback would be appreciated

[cobalt sonnet]

This definition is the same as saying it's a monoidal natural isomorphism plus 2.2.4 right?

[summer quest]

Yes

[cobalt sonnet]

anyone?

like i can't find any reason for why i should care about left and right duals

[lone jacinth]

Why is it called "division by 0", it doesn't appear to have anything to do with division by 0, it's just the dual numbers

[rocky falcon]

After looking it up I believe example 1 in section XIV.2 in Kassel is an example (page 347)
Take the category A-mod_f with A a Hopf algebra and he endows the left dual and the right dual with actions given by the antipode and it's inverse respectively

[cobalt sonnet]

i was looking at this decomposition result in wikipedia called pierce decomposition but when is this true? When can i write the identity as a sum of central orthogonal primitive idempotents?

[rare walrus]

Should be for any Noetherian ring

[velvet forge]

I'm a probabilist reading a course/paper on random walks on Lie groups (a theorem in it is necessary to prove smtg). Anyone here can explain what exactly is a "unitary representation" ? (I'm really unfamiliar with representation theory and have just basic knowledge on group theory and group action from my bachelor's)

Thanks in advance

[silver goblet]

it is an action of your group G on C^n, where G acts by unitary matrices (so distance-preserving)

also can/should be thought of as a group homomorphism G --> U(C^n), the group of n x n unitary complex matrices

[velvet forge]

thanks

[limpid horizon]

Basic question but im trying to make sense of the natural map like Rx1->Rx1x2
Where something like Rx1 is localization of R at the mult closed set generated by x1.

I guess im a little unsure because the denominators in Rx1x2 are of the form (x1x2)^n but sending r/x1 into Rx1x2 doesnt have a denominator of that form

[spice idol]

well the idea is that, if you can divide by x1x2 you can divide by x1

just take x2/x1x2

[limpid horizon]

Ok so, I think of localization as making inverses for the elements in S, but this is a case where we’re not only inverting elements of S but we’re also inverting others too

[spice idol]

so by the universal property of localisations (if there is a ring map f : R → R' where f(S') is a set of units in R', then we have a unique map from the localisation R_S to R'), there must exist a natural map R_x1 → R_x1x2, which looks like
a/x1^n ↦(a x2^n)/(x1x2)^n

[spice idol]

yes, that can happen

in particular, whenever fg ∈ S, then both f and g also get inverses in the localisation

so we can have different multiplicative sets giving the same localisation (up to canonical isomorphism)

[neon flare]

Does anybody know if there's a subring of R analogous to Z_p in Q_p? I've just noticed that it's very easy to justify the p-adic integers via the algebraic construction (as an inverse limit), but rather difficult via the analytic construction (completion of Q).

[spice idol]

i think the right way to view this is that, yes you invert elements in S, but that can have consequences being more invertible elements. just like how only asking for x to be 0 in the polynomial ring has as consequence that all polynomials collapse to their constant coefficient

[limpid horizon]

R->R’ in that case is R->Rx1x2 and that factors through a map Rx1->Rx1x2 by universal property? Are you using this to justify why there is a map from Rx1->Rx1x2 and that uni property also tells us what the map is?

[spice idol]

yis, basically

doesnt really help if you havent learned to think universal properties though

[limpid horizon]

Im getting a bit more used to them

Thanks for the explanation

[spice idol]

mull over it for a bit id suggest, commutative algebra can be hard

[digital parcel]

see atiyah-macdonald chapter 3 for universal property of localization

the explicit construction of the map is there

and i think they use it in some of the proofs of that section idr

[limpid horizon]

Yeah it factors through g(a)g(s)^-1 if g: R->R’ and g(S) are units in R’

[spice idol]

yis

[limpid horizon]

Yea sometimes just the little details can be weirdly confusing

[spice idol]

consider maybe Z_2 and Z_6.
if you have n/6, then surely 2 * 3/6 = = 1, which means that there must be an inverse of 2 in Z_6. As Z_2 is "the most general" ring with an inverse of 2, there must be a map from Z_2 to Z_6

[limpid horizon]

Yea that example is nice

[fierce steeple]

Please dont write Z_2 for this dawg 😭

[limpid horizon]

Lol

Z_(2)

[fierce steeple]

That is different too

[limpid horizon]

Dafoq

[fierce steeple]

Lol

[limpid horizon]

What is it

[fierce steeple]

Z[2^-1] = {a/2^i}

Z_(2) = { a/n where n is odd}

Z_2 = 2-adics

[digital parcel]

i remember in my commutative algebra course my prof said "i'm gonna write Z_n as Z/nZ" and a phd student shook his head and uttered "what?"

[limpid horizon]

O

[fierce steeple]

Though ofc some ||heathens|| use Z_2 = Z/2Z too

[fierce steeple]

Good

[digital parcel]

you just called my prof a heathen

[limpid horizon]

Early algebra stuff likes using Z_2 for Z/2Z

[fierce steeple]

But yeah the thing is like

[digital parcel]

better to not use bad notation from the get-go

[fierce steeple]

Often you write A_f for like inverting f and A_p for inverting away from a prime p

[limpid horizon]

Yea right

[fierce steeple]

For Z this gets really bad

Lmao

Like Z_(p) is standard and comes up for loclaising at the prime as mentioned but ye

[spice idol]

yeS i forgot the p-adics existed i may be stupid

or well, im just not a number theorist

god bless

[foggy galleon]

is using Z_2 really that bad? It's shorter than any other alternative

[spice idol]

<2>^-1 Z

where <2> denotes the submonoid of the multiplicative monoid of Z generated by 2

of course

[calm trellis]

I think people tend to overdramatize this

[fierce steeple]

Idk I geuss this is just me doing a field where i use the p-adics like very regularly lol

but perhaps yeah it is just number theorists + similra who care about this lol

[calm trellis]

it's a fair point sometimes but there's just always that one person who goes wah wah that's the p-adics when a prof uses Z_p for Z/pZ in a course where the p-adics won't come up a single time

here when talking about stuff with not that much context i agree that you should reserve Z_p for them

[limpid horizon]

The um akshually ppl in math are v annoying

[calm trellis]

it's not even a gotcha i can call the polynomial ring Z_p if i want to

p for polynomial

[spice idol]

horrid

[limpid horizon]

Its my
Polynomial i can call it Z_p if i want to 🎶

[spice idol]

Loc(R, S) for localisation of R by S

[limpid horizon]

Lol

[foggy galleon]

Z_n=Z/nZ. Z_p or Z_ell, p-adics or ell-adics.

[spice idol]

and for nontrivial prime powers Z_q means the localisation by q

[digital parcel]

do this and get the paper in annals and i'll personally give you one million usd

[calm trellis]

i personally denote prime numbers with n for non divisible

[spice idol]

i use S for algebraic Structure

[digital parcel]

i use F for sheaF

[spice idol]

R for gRoup

[digital parcel]

big things happening today

[spice idol]

G for rinG

given the group ring G[R]

[spice idol]

we are revolutionising notation

[forest turtle]

K for fieKld

[hard kite]

normalise Z/n

[hard kite]

this is awesome haha

[spice idol]

and of course Loc(R, 𝔭) means the localisation at 𝔭

[hard kite]

I still need to figure out a notation system where you can disinguish between localisation and homogenous localisation and p-adics and stuff but I cant figure out the right way to write it down

for some reason i really prefer M_f over M[1/f] and I can live with that conflicting with Z_p but now I cant figure out what to do with localisation of graded modules haha

[spice idol]

homogenous localisation? graded rings?

[hard kite]

yea like M is a graded module over a graded ring R and f is a homogenous element then M_f is m/f^e with m,f homogenous of the same degree

[spice idol]

mm i see

interesting

i wonder if graded structures form a universal algebra in some way

is the quotient of a graded module by a graded submodule (respecting the grading) also a graded module?

[hard kite]

yes

[spice idol]

oo

arbitrary products might pose a problem though

although..

wait no yeah they do

[digital parcel]

the category of graded R-modules forms an abelian category

[limpid horizon]

The direct sum when considering graded objects is like an “internal” direct sum right

[spice idol]

aint no variety though, only finite products 😔

[digital parcel]

doesn't internal direct sum mean you have an object and you take the direct sum of subobjects?

the direct sum of graded modules is just the typical direct sum

coproduct in the category

[limpid horizon]

Yea but arent the graded parts all like subgroups

[spice idol]

if you consider a graded ring as only its set of homogeneous components and multiplication • : R_n × R_m → R_n+m then this has been already UA-ized

by Birkhoff I think?

[lone jacinth]

Analogous in what way?

Z_p is the closure of Z inside Q_p, so for R I guess it would just be Z.

It's possible to cook up rings with field of fractions R, but they won't be very canonical (besides R itself).

[digital parcel]

yeah

[spice idol]

ugh cant find it but it was used to prove something related to Mal'cev conditions in a paper i read somewhere

[lone jacinth]

I guess something that might make sense would be to take the closure of the integral closure of Z inside R. I would guess this just equals R, but idk

[digital parcel]

hmm i guess you can look at the direct sum of graded R-modules as some form of internal direct sum

[spice idol]

there we go

[digital parcel]

like if I have two graded R-modules M and N, then you can look at the i-th graded piece M_i (+) N_i as a subobject of M (+) N as just modules (forgetting the grading)?

idk that's not really a fully formed thought

[fierce steeple]

Ye

[limpid horizon]

If I have M, N and L graded, is M (+) N (+) L graded with the decomposition of an element (m,n,l) into homogenous components as (m,0,0)+(0,n,0)+(0,0,l)?

[fierce steeple]

The usual geometric intuition is that Z_p should be like the closed ball in Q_p

[digital parcel]

yes

[limpid horizon]

That was assuming m n and l were homogenous already i guess

[lone jacinth]

How so?

[digital parcel]

if $M(i)$, for $i \in I$ some index set, is a collection of graded $R$-modules, then the direct sum $N = \oplus_I M(i)$ is given by the grading $N_j = \oplus_I M(i)_j$

[broken turtle]

anamono

[digital parcel]

"the degree j part of the sum is the sum of the degree j parts"

[lone jacinth]

You mean because it's not equal to the product of the underlying modules?

[limpid horizon]

Sometimes its confusing in direct sums when the sum + you can actually combine or just leave it as an expression with + if yanno what i mean

[spice idol]

that should work though

right

[fierce steeple]

I guess you can get sqrt(N) arbitrarily close to an integer, so the Z-linear span of square roots is dense

[spice idol]

the homogenous components would be M_j = ∏i M{j, i}

[lone jacinth]

Yeah, that's the product of graded modules

[spice idol]

oh i see how the underlying module would be different yeah

[fierce steeple]

Probably an easy way to do this is that like a Z-graded thing (except rings...) is a functor from discrete Z to the category of things

So basically everything nice is inherited

[spice idol]

this is about as general as you could have it

[lone jacinth]

I guess one would need to check that that is actually compatible with the module structure, but it is so....

[spice idol]

(without invoking category theory)

[fierce steeple]

Yeah i mean this is kinda clear from how everything is defined pointwise

[lone jacinth]

Well it's not exactly pointwise

You have maps R_i x M_j -> M_j+i going between the points

[fierce steeple]

Oh over a graded ring

Yeah that makes it different lol

(The categorical description is then that R is an algebra object in Fun(Z,Ab) and M is an R-module in that category)

Lol

[spice idol]

this might be a definition less general but closer to graded rings or modules:

Let F and F' be signatures, and G an F'-algebra. Then an F-algebra graded by G is a collection of F-algebras (A_g)_g∈G with n-homomorphisms of F-algebras or just functions idk either can work
f : A_g1 × ... × A_gn → A_f(g1, ..., gn)
for all gi ∈ G and f ∈ F'_n

[fierce steeple]

But still means it is very formally similar, replacing Ab by Fun(Z, Ab)

[limpid horizon]

what if m and n have the same degree, then (m,n,l) = (m,n,0) + (0,0,l) is another decomposition into homogenous components?

[spice idol]

yes

because then (m, 0, 0) and (0, n ,0) are in the same homogenous component, so their sum is too

[limpid horizon]

ok so the uniqueness of the decomposition is still satisfied

[spice idol]

yis

modules are just nice like that

🔥

[digital parcel]

yeh i guess modules and me got something in common

we both just kinda nice like that

B)

[limpid horizon]

damn 🔥

so real 💖

[spice idol]

youre ong just a chill guy

[digital parcel]

yea but i dont wanna go that far because im super humble and stuff

i aspire to be a chill guy

[spice idol]

i should probably be continuing doing my math but ong i am too lazy

[digital parcel]

i spent the last 8 hours or so doing stupid gre prep

im done for now

maybe after dinner i'll do some ag

[spice idol]

gre prep?

[digital parcel]

math subject gre

it's a dumb standardized exam for math grad school applications

[limpid horizon]

so like Rx1x2 does not have the element 1/x1 but like it actually does cause its x1/x1x2

like why does that make me feel weird

idk like i get it (i think) but something about it i feel like i dont understand and i also dont know how to describe what i dont understand about it

[spice idol]

it has x2/x1x2

just like 3/6 is the same as 1/2

[limpid horizon]

right

[spice idol]

this simplification can be done explicitly in a "larger" ring, that being Rx1x2 localised by x1 / 1

though that notation gets a little messy

[spice idol]

and just like 6/8 is the same as 3/4, the natural map Z_4 → Z_8 sends 3/4 to 6/8

it can be seen as a sort "unsimplification"

rather than simplifying the fraction, you make it more complicated so that the denominator matches

[jolly jetty]

Besides chris eigen's series on youtube for tensor algebra is there a good book for tensor algebra and calculus? Like I am not looking to derive everything from the first principle like pure math students do just get enough understanding for tensor decomposition and get comfortable with tensor calculus

[mental escarp]

What is tensor algebra and tensor decomposition? Sometimes you can factor a tensor into a symmetric and alternating part….

It sounds like you want to read lee’s smooth manifolds book. Although, if you want to learn about multilinear algebra I’d recommend multilinear algebra by greub yet it has nothing to do with calculus as far as I can tell

[lone jacinth]

Might be better to ask in #diff-geo-diff-top or perhaps a place with physicist. It's more of a differential geometry thing.

[jolly jetty]

Thanks

[kindred mauve]

I did a course on galois theory (we covered until kummer's theorem of abelian extensions), what's the next step if I wanna keep learning abt it?

[lone jacinth]

I'm not sure there's necessarily one natural next step, but there are also some classification theorems for inseperable extensions which might be interesting.

And there's infinite Galois theory and profinite groups.

Then I guess there are fields where Galois theory appears, like Galois cohomology, algebraic number theory, geometric invariant theory.

[kindred mauve]

Thanks. We talked about galois cohomology and used it to prove Kummer but I'm confused cause we introduced a new notion and barely used it, so I thought it would have been developed further

[ornate atlas]

Courses often do that, the very last thing in my noncom rings class was a definition

[lone jacinth]

What were they defining?

[ornate atlas]

But more commonly, in my experience, courses will pull together a few different ideas and maybe half explain a new tool to prove one final interesting result which brings together everything you’ve learned, not worrying too much if you don’t really get all of the tools, because it’s a satisfying place to end (generally)

[ornate atlas]

Holonomic modules, we did a bunch of stuff on Bernsteins inequality just before it

[kindred mauve]

Lol

[kindred mauve]

Thanks. I've looked something up and it still looks too advanced for now.

[ornate atlas]

The final peice of content lol

[lone jacinth]

Very analysisy for a ring theory class

[ornate atlas]

Yeah thats often the case, I remember my intro to diff geo course doing something similar. That could be a good direction to aim your studying now though!

[ornate atlas]

Yeah im not really sure why, that prof constantly joked about hating analysis but thats kinda where we ended. I think the main motivations were just that we studied the Weyl algebra quite a lot in the course, because its a "nice" noncom ring, and it lends its self naturally to talking about dimension, even if that means dipping your toes nto D-modules

The proof of Bernstein's even made us do like a week of comalg, the last 2 chapters of the course were just a bit odd, because this came after the noncom localisation week

[lone jacinth]

It's amazing all the math out there people be doing

[ornate atlas]

real

That lecturer does NCAG and enveloping algebra stuff, so I guess it does make an amount of sense, I think these things vaugley show up in all of that

But despite all the things I proved about universal enveloping algebras in that course, I still dont have the slightest clue what they actually are so I could be wrong, but I think its got something to do with differential algebra

[lone jacinth]

Universal enveloping algebra is usually an algebra with the same representation theory as something (often a lie algebra).

Not sure if that was what it meant to you

[ornate atlas]

It is, we proved some random results about universal enveloping algebras of Lie algebras but were never given more information than that these things were called that

[lone jacinth]

Makes sense, I think they usually have some interesting ring theoretic properties

[ornate atlas]

I still really want to learn representation theory of some kind, I did the tiniest amount in my analytic number theory class and that was it. I breifly looked at path algebras as part of that class and all that stuff seems fun, but I honestly have no understanding of what any of the very disperate fields of representation theory look like, or how it works

[lone jacinth]

One of us, one of us, one of us!

[ornate atlas]

Isnt that what you study? I just looked at them a little because one of the proposed topics for the final project in that class was "When are path algebras Noetherian" (iirc), which seemed like an interesting, but given the time contraints, too hard problem

[lone jacinth]

I study Artin algebras / finite dimensional algebras, and path algebras are a big part of that yeah.

Haven't really thought about when they're Noetherian, but I would guess when they have 1 or 0 oriented cycles.

Hmmm, no you can make it non-Noetherian with just one cycle can't you

[ornate atlas]

Ill hopefully have a bit more scope to look at these things in my masters but I think theyre much more commutative pilled unfortunatly, there is a class on Lie algebras though so ill hopefully have some fun with that and learn a little rep theory.

[ornate atlas]

Yeah it seems like an interesting question, but i was so busy at the time that it just seemed like too much background theory to learn before even starting the problem

[lone jacinth]

Hmmm, so no cycles -> finite dimensional, done.

One cycle x and an arrow y not in the cycle then
(y, yx, yx^2, yx^3 ...)
should be an infinitely generated left ideal.

A single cycle no other arrows, probably Noetherian. True for k[x] at least.

[ornate atlas]

I trust your wisdom, that sounds reasonable lol

[rich dock]

Guys, you all know that isomorphisms, equivalences, etc of categories are defined via the identity right?

Well the problem is that for C* algebras, approximate identity are useful rather than identity

So do we have like an alternative definition of them using approximate identity?

This seems really useful in study fields like functorial algebraic quantum field theory where there are a lot of categorical stuffs and C* algebras

[distant harness]

If f:A->B is "nice enough" and i is an approximate identity on A, will there always exist an approximate identity j on B such that fi=jf? (Or vice versa?) If so, you can define a category of nice maps modulo approximate identities and work with that.

[rich dock]

Wdym by A,B

[distant harness]

They're objects in your category. I guess probably C*-algebras.

[rich dock]

I don't know?

[rich dock]

I'm just curious about what I asked

We define categorical properties using identity but why not have them using approximate identity which comes even more handy

[distant harness]

That was what you already asked, wasn't it?

[rich dock]

It is

Talk about something on categorical stuffs duh?

[distant harness]

I was trying to suggest an answer, but since I don't know what you mean by "approximate identity", the suggestion had be a bit conditional on the properties of your "approximate identity" concept.

[rich dock]

Approximate identity are a well known concept in the study of C* algebras

[distant harness]

¯_(ツ)_/¯

[rich dock]

Bruh

[lone jacinth]

I don't know very much about this, but it is my understanding that the types of things of interest in QFT and such are categories of C*-algebras, not categorifications of C*-algebras, so then approximate identities wouldn't really be relevant to categorify.

But I guess the definition lends itself somewhat naturally to approximate inverses, so you could use that to define approximate isomorphisms I guess...

[cobalt sonnet]

I am looking into commutative FA and the author uses the terms anihilator algebra which is not standadrd so is an indecomposable algebra which has non zero radical (which is his def) same as a local algebra with a non zero radical?

[cobalt sonnet]

so my question is local the same as indecomposable

[lone jacinth]

So I'm assuming FA means finite dimensional algebras?

Anyway, indecomposable means not the product of two nontrivial things.

A commutative unital finite dimensional (or just artinian) algebra is indeed local, and any commutative unital artinian ring is a fintie product of local rings

So here anihhilator algebras would be local finite dimensional algebras that are not fields

[digital parcel]

Good books for Lie groups/algebras? even better if it has a bias towards AG

[void plank]

Let G be a finite group. Is there a condition on G that guarantees the 0th group homology and cohomology of G are isomorphic?

[foggy galleon]

what is the homology and cohomology of G?

[void plank]

Sorry, I mean group cohomology here

[foggy galleon]

don't you need a G-module?

[void plank]

Or was that a rhetorical question?

[foggy galleon]

or do you mean for all G-modules?

[void plank]

Yes, I have a G module in my situation

[void plank]

Yes, I believe so

I've seen a claim that this is true for a finite group of odd order

[foggy galleon]

aren't they always isomorphic?

[void plank]

Actually never mind, I need to think about my question some more. It doesn't make sense as stated

[fierce steeple]

Isn't this this just asking when M^G and M/G are isomorphic, which should be two quite different things

[void plank]

Yes, I was wondering about when the invariants and coinvariants are the same, but my situation is specific. Let me see if I can make my original question clearer now

Let V and W be vector spaces over field F. Let F[G] be a group algebra of a finite group G.

I think in my situation, I'm comparing the following situations

1. Suppose I also endow V and W with right (left) F[G] module structures. I then consider V (x)_F[G] W

2. I consider V (x)_F W and then endow this with a left F[G]-module structure by defining g*(v (x) w) = vg (x) (g^-1)w

I've seen a claim that if G is finite and of odd order, the invariants of M = V (x) W as a G-module (so H^0(G, M)) is isomorphic to V (x)_F[G] W under an averaging map where you take an element v (x)_F[G] w and send it to \sum_(g in G) vg (x) g^-1 w

[foggy galleon]

Chat gpt gives this example

[foggy galleon]

in (2), is this action well defined? I mean, is it associative?

shouldn't the actions be in the opposite direction?

[void plank]

Ah, I don't think it's associative

h(g v (x) w) =/= hg(v (x) w)

[foggy galleon]

you should define g*(v otimes w)=(gv) otimes (wg^(-1)), or act on the right

But I think what you are asking for is true. Essentially, let G be a finite group and A a G-module. Then you have a norm map A_G-->A^G and a conorm map A^G-->A_G. Their composition is the order of the group G, so if |G| is invertible in A you are good

edit: maybe you have to check the characteristic of F

[void plank]

Ah my bad, my vector spaces already have a F[G] module structure on them in both situations

In one situation, I'm taking the tensor product as F[G] modules and in the other situation, I'm taking the tensor product over F and then taking invariants

[foggy galleon]

what is R[G]?

oh

[void plank]

Fixed it, sorry

[void plank]

Ahh ok, my field is actually of characteristic 2 here so this makes sense

[foggy galleon]

lol I kinda forgot R and F were next to each other on the keyboard, otherwise I wouldn't have asked 😆

[void plank]

You're good, I mean to be working over a field F and not a ring R anyway 😁

[void plank]

But this doesn't coincide with the module structure I've defined on V and W right?

[foggy galleon]

fair point

why aren't you just defining g*(v otimes w)=(gv) otimes (gw)?

[foggy galleon]

this would make sense if W had a right action

and V a left action

[void plank]

This wouldn't make (V (x)_F W)^G isomorphic to (V (x)_F[G] W) right?

I guess one of these is invariants and the other coinvariants

[foggy galleon]

so is M= V otimes_F W? If the characteristic of F is coprime with |G| then H^0(G, M)=H_0(G, M). But where does V otimes_F[G] W come from?

F[G] is non-commutative, tensor products of modules over a non-commutative ring kill a lot of stuff, no?

[void plank]

Isn't V (x)_F[G] W = V (x) W/(vg (x) w - v (x) gw), the space of coinvariants?

[foggy galleon]

coinvariants meaning H_0?

[void plank]

Yes.

And yeah, so I suppose M = V (x) W here

[foggy galleon]

ah I see

yeah this is right

I think

[void plank]

Ok perfect, thanks!

And yes, I see that the composition of your co-norm and norm maps are basically |G| times the identity

[foggy galleon]

So. V has right action. W has left action. You want to give a left action on V (x) W=V (x)_F W, right? You have to set g(v x w)=(vg^{-1}) x (gw). Then the question is whether we have an equality of submodules (vg x w-v x gw)=(vg^{-1} x gw - v x w). This can be seen by a change of variables

right?

[void plank]

Oh that wasn't my question. My question essentially boiled down to "under what conditions on G do we have H^0(G, V (x) W) = H_0(G, V (x) W))" (where V and W have compatible F[G] module structures here)

And you said when |G| is coprime with char F, which I see

Unless you're making a separate point

But yes, I do see I have to be careful about what F[G] module structure I'm giving V (x)_F W

[foggy galleon]

right, I was commenting on this part. Yeah I think you are right, because there's no reason why (gv x gw-v x w)=(gv x w-v x gw)

(this is if V, W are left G-modules and define g (v x w)- (v x w), but your original set up is fine)

[void plank]

Ok cool!

It's strange, because the paper I'm reading defines g(v (x) w) = vg (x) g^-1 w, which doesn't seem associative

[foggy galleon]

maybe they mean to give V x W a right action?

[void plank]

So let (v (x) w) g := vg (x) g^-1 w

1. (v (x) w) (gh) = vgh (x) h^-1 g^-1 w

2. ((v (x) w) g)h = (v g (x) g^-1 w) h = vgh (x) h^-1 g^-1 w

Yeah, I think it's a typo in their paper

[smoky wasp]

whats this functor?

A and S are just two arbitrary C-algebras

where C is the complex numbers

i think its a functor form k algebras to groups?

maybe its S maps to S tensor A

because gamma has an action on S, so the function underneath makes some sense

[lone jacinth]

Since they first remark that A(x)S is an S-algebra, it would make sense if it was S-algebra automorphisms of A(x)S, but idk

[cobalt sonnet]

Sorry I should have specified i meant frobenius algebras

[smoky wasp]

ah i guess

[lone jacinth]

I see, well they are in particular finite dimensional algebras

[smoky wasp]

that would make more sense considering im trying to get that L is descended from this cocycle lol

thanks!

[cobalt sonnet]

in the last sentence you mean a commutative finite dimensional indecomposable algebra is local right?

[lone jacinth]

Yes

[cobalt sonnet]

and a commutative finite dimensional local algebra is indecomposable algebra

[lone jacinth]

Yes, though not necessarily Frobenius

[cobalt sonnet]

Initially, I was interpreting the last theorem incorrectly. It's only stating that a Frobenius algebra decomposes into local algebras which are either fields or they are not

[cobalt sonnet]

But the converse is not true?

[lone jacinth]

Converse of what?

[cobalt sonnet]

but do we have a characterization of when local algebras that are not fields are frobenius algebras?

[cobalt sonnet]

like of this decomposition

[lone jacinth]

I mean, it should be when they have simple socle for example

(having a unique simple ideal)

[cobalt sonnet]

simple ideal?

[lone jacinth]

One without any subideals

[cobalt sonnet]

So a unique, simple submodule

How do we prove that a finite-dimensional algebra can be written as a product of local algebras, though?

I haven't taken any advanced algebra so if you can give me some reference that would be great too

[lone jacinth]

The easiest way would be to mod out the radical, use Artin-Wedderburn, then lift idempotents

You might be able to do it more directly as well...

[cobalt sonnet]

So I can say any finite dimensional algebra is a product of indecomposables , modding out the indecomposable by the radical gives me a semisimple commutative algebra and i know these are direct product of fields by artin wedderburn but what next?

[lone jacinth]

Like an indecomposable algebra has only idempotents 0 and 1.

Consider any ideal J, then J^n get must eventually stabilize hence become idempotent. So unless J=A we must have J^n = 0.

So every nontrivial ideal is in the nilradical -> the nilradical is maximal -> you're local

[lone jacinth]

So A/J = k^n, since J is nil you can lift idempotents, i.e. find n idempotents in A that map to the ones in k^n.

So then A is a product of n algebras.

For the case A/J = k the algebra is local, because the radical is maximal

[cobalt sonnet]

So this directly proves that the indecomposable algebras in the decomposition of our fin dim algebra are local right?

But why is a local algebra with zero nilradical a field? i would have to prove that the nilradical consists of all the non units somehow?

[lone jacinth]

Same argument as above.

For any element consider (x^n)

This is prop 8 in your picture as well

[cobalt sonnet]

So I consider the ideal generated by x say J and it's a proper ideal unless x = 1, and thus J^n = 0. Hence, any non-unit is nilpotent?

[lone jacinth]

Yeah either (x) = (1) or x^n = 0

[cobalt sonnet]

yeah but i couldn't find the reference

So, this statement that every finite-dimensional commutative algebra decomposes into local algebras, which is either a field or not, is not very deep?

like from the direct proof you have shown me it doesn't seem like a non trivial thing to say

[cobalt sonnet]

here you mean product of n local algebras?

[lone jacinth]

Yes, well that remains to be shown I guess.

But product of n algebras with A/J = k

[lone jacinth]

It's easy to prove. But I'd say it's still important. And the methods can be used to make similar (though weaker) statements in the noncommutative case.

Maybe idk what you mean by deep

[cobalt sonnet]

Like, what can we say about the indecomposable factors of a noncommutative Frobenius algebra?

do they still have to be local?

[lone jacinth]

No they don't need to be

[cobalt sonnet]

I understand the direct proof you have shown me but i don't understand the other one in which we lift idempotents somehow

[lone jacinth]

So it's a general fact that if I is a nil-ideal, and
e in A/I is an idempotent, then there is an idempotent in A that maps onto e.

This is called lifting idempotents

[cobalt sonnet]

i guess the deep part that the author wanted to mention is the part about how the frobenius form on each factor determines the frobenius form on the whole

[lone jacinth]

So in general you might still get something like A/J = k^n, and you can still lift idempotents.

But in the noncommutative case those idempotents might not be central, so don't correspond to factors of the algebra. Instead they give the data of the vertecies of the Gabriel quiver, which you can use to understand the algebra

[cobalt sonnet]

and there is a one to one correspondence?

[cobalt sonnet]

Can I find this in Dummit and Foote?

[lone jacinth]

Not quite, there is in the commutative case, but you could set up some equivalence of idempotents to make it work.

[lone jacinth]

I haven't read D&F, but it seems plausible

[cobalt sonnet]

some other book then?

[lone jacinth]

It's definitely on stacks
https://stacks.math.columbia.edu/tag/00J9

The second proof is the same in the noncommutative case (though the uniqueness step doesn't work)

[fierce steeple]

stacks project for non-commutative rings pls

[cobalt sonnet]

So essentially two proofs

either fin dim comm algebra >> indecomposable factors are local >> local with no units is field >> fin dim comm algebra is sum of field and local non fields

or take fin comm algebra >> A/J semisimple=k^n >> lift idempotents which are in one to one correspondence >> decomposition into local algebras

[last talon]

What’s a proof of this that doesn’t use CFSG?
The proof the book gives uses CFSG to ensure that C has a group generated by 2 elements
C is an isomorphism-closed class of finite groups closed under quotients and subdirect products

Or is the use of a defn of “free pro-C group” where we have the universal property defined for groups G such that the image of X in it is topologically generating non-standard?

[foggy galleon]

What if you take |X|=2 and C be the smallest group formation containing a finite non-abelian simple group S? Every group G in C has S as a quotient (except the trivial group, of course), if I'm not mistaken. If iota is injective then this means that F is non-trivial, and therefore has a quotient in C, i.e., has S as a quotient. But F is generated by two elements, hence S is

but see https://mathoverflow.net/questions/59213/generating-finite-simple-groups-with-2-elements

@last talon

[last talon]

Ok yeah if you do it the way they do it you basically need the classification then

[last talon]

This seems nicer to me
For the reason of “you don’t need to invoke CFSG for such a simple lemma as this”

[foggy galleon]

What seem's nicer?

I tried to show that Lemma 3.3.1 (together with existence, which is the next result and elementary) implies that every finite simple group is generated by 2 elements

[last talon]

Defining the universal property as you would normally (ie for all groups, not just those t.g by X)

[foggy galleon]

I mean the normal definition has an X which is discrete

[last talon]

For all profinite groups I meant

[foggy galleon]

Their definition is standard tho. There are certain situations where you would like to consider procyclic, proabelian, prosolvable, pro-p, etc.

In the discrete case, free abelian groups are also considered

But you are right that in this cases you dont need CFSG. The point is simply that Lemma 3.3.1 is equivalent to the question of whether the class C contains a group generated by exactly 2 elements

[last talon]

Yeah
And shouldn’t that imply every finite simple group is 2-generated (by taking a simple quotient of such a group, for C being those groups whose composition series is just copies of S)?

[foggy galleon]

yeah, that's where I was going

[last talon]

And then by the link you sent, there doesn’t seem to be a known proof of that without CFSG
So it’s not equivalent, but there’s no known proof without it

[foggy galleon]

yeah, this is rather interesting imo

[last talon]

Which seems like
3.3.1 seems like it shouldn’t need CFSG

[foggy galleon]

why not?

simple statements with incredibly difficult proofs abound in math

[last talon]

Mostly the idea of “it feels simple enough to have been proven before CFSG”

[smoky wasp]

i reread all the notes up to here turns out its R algebras i think

since what i need is a map from Gamma to Aut_R((A x R) x S)

[cobalt sonnet]

does anyone know a reference for this?

[cobalt sonnet]

can you tell me how to prove it?

[lone jacinth]

So say A is your local algebra and S is the simple socle. Pick v non-zero in S and extend to a basis for A.

Then the projection to kv defines a non-degenerate bilinear form.

It's non-degenerate because there's always a smallest n with J^n x in the socle, hence equal to S. So you can always find y with xy = v

Conversely, if A is Frobenius then A is isomorphic to DA, but if A is local it has simple top. And the top of DA is the dual of the socle, so A has simple socle

[cobalt sonnet]

top?

[lone jacinth]

The top of a module is what you get by modding out the radical

[cobalt sonnet]

projection to kv means exactly?

[lone jacinth]

Like say the basis was {v, w}, then you take
av + bw to a

[cobalt sonnet]

ah ok

[cobalt sonnet]

what is J here ?

"It's non-degenerate because there's always a smallest n with J^n x in the socle, hence equal to S. So you can always find y with xy = v"

[lone jacinth]

The radical, sorry

[cobalt sonnet]

why is that? "because there's always a smallest n with J^n x in the socle"

[lone jacinth]

x is in the socle iff Jx = 0.

So say J^n+1 x = 0, then J^n x is in the socle

[cobalt sonnet]

in this case we can just say x in socle iff mx=0 right?

since it's local

[cobalt sonnet]

Why must it eventually go to zero?

[lone jacinth]

Yeah m=J

Because J^n is eventually 0

Just cuz A is finite dimensional

[cobalt sonnet]

i know nilradical consists of nilpotents

but it is itself nilpotent for fin dim?

how

[lone jacinth]

Several ways to see it, but easiest is maybe just to think about
J > J^2 > J^3 > ...
since everything is finite dimensional this can't decreases forever

Alternativenely pick a basis for J. Then a basis for J^n consists of all possible products of n of these vectors.

If there are d of them, then one must appear at least n/d times.

So just pick n big enough so that x^(n/d) = 0 for all of them

[cobalt sonnet]

"
It's non-degenerate because there's always a smallest n with J^n x in the socle, hence equal to S. So you can always find y with xy = v"

what definition are using in proving non degeneracy?

we should prove that xy=0 for all y in A implies x=0 right?

[lone jacinth]

For every nonzero x, there exists a y such that f(yx) = 1

[cobalt sonnet]

so surjectivity of A to DA then

[lone jacinth]

It would correspond to injectivity actually, but for finite dimensional injective, surjective and bijective are all equivalent

[cobalt sonnet]

no yeah you are right sorry

"there's always a smallest n with J^n x in the socle, hence equal to S." being in socle means equal to socle??

why can't it be less

[lone jacinth]

The socle was simple by assumption

So nothing smaller

[cobalt sonnet]

right nvm

[cobalt sonnet]

This is the proof i was looking at and I couldn't understand it

how does nakayama lemma help here?

[cobalt sonnet]

this is the form of nakyama lemma i am familiar with

[cobalt sonnet]

your proof doesn't seem to need it right?

[lone jacinth]

Well, you would need it to show that m is nilpotent

[lone jacinth]

Nakyamas lemma is used where they reference it to establish that m(x) is a proper submodule of (x)

[cobalt sonnet]

i can prove it directly without nakayama i think

if (xy)=(x) implies xry=rx implies x(1-yr)=0 but yr in m and A local implies 1-yr is a unit and thus x=0 right?

[lone jacinth]

I mean 1-yr being a unit is basically the proof of Nakayamas lemma

But yeah you can just inline it like that

[cobalt sonnet]

but they are writing xy not mx

[lone jacinth]

y is in m

[cobalt sonnet]

But this could be smaller no?

[cobalt sonnet]

(xy)=x is same as xm=x?

[lone jacinth]

Well (xy) <= m(x)

[cobalt sonnet]

so i can't directly use nakayamma?

i am sure it must be trivial i am just bad at this

[lone jacinth]

(xy) < m(x) < (x)

[cobalt sonnet]

and we use that artinian implies noetherian to use nakyama here right?

like (x) is finitely generated so it applies

[cobalt sonnet]

oh right by artinian mm^n=m^n and thus m^n=0

Is it obvious that the following examples are local algebras?

how does one quickly check that?

[lone jacinth]

(x, y, z) is a maximal ideal, and it's nilpotent, so must be the unique one

[cobalt sonnet]

if a maximal ideal is nilpotent it's unique?

so becasue (x,y) (x) (x,y) (x,y,z)are nilpotent these are unique maximal?

[cobalt sonnet]

nvm this is easy

[cobalt sonnet]

and how do quickly find the socle?

[lone jacinth]

I mean, it's just the stuff that is annihilated by both x, y and z

[cobalt sonnet]

like perhaps i could have thought that xz is the anihilator

how do i know when i am done

[lone jacinth]

Like just look at what x annihilates, what y does and what z does, then take the intersection

x annihilates (x, y), y annihilates (x, y) and z annihilates (z)

Intersection is (xz, yz)

[cobalt sonnet]

I get that the top of A is iso to the dual of the socle of A

how do we get that socle is simple?

[lone jacinth]

Because the top is simple

[cobalt sonnet]

so dual of socle is simple

why must the socle be?

[lone jacinth]

Well say the socle had a submodule, then the dual would aswell, so not simple

If N < M is a submodule then the functions that vanish on N is a submodule of DM (equal to D(M/N))

[cobalt sonnet]

now the only part i am unclear on is why the top of DA is dual of the socle

[lone jacinth]

The basic idea is that quotient modules of A give submodules of DA and vica versa.

So a maximal submodule of DA will look like D(A/S) for a minimal submodule S.

Then the intersection of maximal submodules will correspond to the sum of minimal submodules i.e. the socle

[cobalt sonnet]

so Top of DA=DA/ intersection of maximal submodules of DA= D(sum of minimals of A)?

[cobalt sonnet]

and is the residue field of a k algebra always k?

[lone jacinth]

Not necessarily no

I mean you can just take any field extension as your k algebra

[cobalt sonnet]

Then, checking the socle is simple is same as being one dimensional but only when we are looking at it over the residue field right?

[lone jacinth]

Yeah that's true

In general you can just check if it has the same dimension as the residue field I guess

[cobalt sonnet]

this condition seems odd tho like when does an algebra not contain its residue field

also this means classifying commutative frobenius algebras is same as classifying local fin dim gorenstein algebras right?

do we have anything like that?

[lone jacinth]

Z/4 is a local ring with residue field Z/2 for example

[cobalt sonnet]

are fin dim local Gorenstein algebras frobenius as well? i think all our arguments generalize

[lone jacinth]

The commutative ones are at least

[cobalt sonnet]

yeah but the non commutative ones have anything to do with Gorenstein?

[lone jacinth]

Well a finite dimensional algebra is Frobenius if
inj-dim A = 0
and Gorenstein if
(left)-inj-dim A = (right)-inj-dim A < infinity

So the notions are related

(if the left injective dimension is 0 then the right one is also 0, but it's an open problem if the left and right injective dimension are always equal)

Called the Gorenstein symmetry conjecture.

If they're both finite they're equal, and your algebra is Gorenstein. But somewhere out there, maybe there's an algebra with the left one finite and the right one infinite

Let me know if you find one

[cobalt sonnet]

Checking Gorenstein seemed to be the only way to disprove that an algebra is not Frobenius-like in these examples

in general i only know of using the structure constants of the algebra

(to make the matrix of A iso to A star and then show no such matrix is possible)

but is there some other way to do that's not so computational? like i want to prove that upper triangular 2 by 2 don't have a frobenius structure

how do i do that

[lone jacinth]

Guess it depends what tools you have in your belt.

Showing that A isn't injective by making a non-split injective map between the projectives is not too bad

[cobalt sonnet]

any other way?

[lone jacinth]

I'm sure there are loads of ways

Could also just try to think about what this bilinear form could be.

Like you have the matrices
E11, E12, E22
as a basis for A.
AE12 is just the span of E12 so f(E12) would need to be non-zero (might as well assume it's 1).

Similarly AE11 is also just the span of E11, so say f(E11) = x =/= 0.

Then f(E11 - xE12) = 0, and multiplying by anything in A just scales this, so the form can't be non-degenerate.

[cobalt sonnet]

is this appraoch fast too? i have no idea what these terms are but i can look up the definitions

this makes me want to learn homological algebra

[lone jacinth]

Pretty fast yeah.

You have two projectives, corresponding to the two columns. And you have an inclusion map from one to the other.

It is indeed some basic homological algebra

It is in some sense very analogous to what's written above, since the projectives are
AE11 and AE22 and the embedding takes E11 to E12

So you can sort of see a shadow of what happens in the other version

[cobalt sonnet]

what's the inclusion map

[lone jacinth]

The one taking E11 to E12

[cobalt sonnet]

you are talking about E11 and E22 as the projectives

[lone jacinth]

AE11 and AE22 are the indecomposable projective A-modules yeah.

And you have an embedding of AE11 into AE22

[cobalt sonnet]

the embedding just takes the AE11 span E11 to AE22=span of E22

[cobalt sonnet]

simply by mapping e11 to e22 you mean right

[lone jacinth]

No that's not a homomorphism

e11 * e11 = e11, but e11 * e22 = 0

However e11 * e12 = e12

[lone jacinth]

AE22 is the span of E12 and E22

Since E12 * E22 = E12

[cobalt sonnet]

Thanks got it! This is slick!

[warped cape]

Hello so this is probably going to be an uncommon question but let me explain the context. So I recently did a research internship in theorical computer science about the complexity of a particular case of the vertex enumeration problem (it doesn’t really matter for the question but it’s for context purpose, but it’s the computational problem of given the face representation of a polyhedra, outputing its extreme vectors). Now that it’s finished I still thought about it and had an idea of a generalization of it (I would like to talk about it with my supervisor but he is in holidays and I keep thinking about it so I must share it somewhere), so here it is. Let [ n \in \mathbb{N}, E_1,…,E_n] be any sets. Let F be a set of functions of the form [ f : E_1 x … x E_n \longrightarrow E_1 x … x E_n ], and [(x_1,…,x_n) \in E_1 x … x E_n ] the problem is finding the set S of every tuple of the form [ (f_1,…,f_k), k \in \mathbb{N} ] such that [ (f_1 \circ … \circ f_n)((x_1,…,x_n)) = (x_1,…,x_n) ]. So I know said like this it is too general to say anything about S, but the idea would be to find some properties on S with restrictions on F or the E_i’s. The only result that I could think of at the moment (which is not really impressive) is that if F is finite and its functions comute then there is a finite set of "generators" of S (i.e elements of S which are not concatenations/compositions of 2 or more elements of S). So I have several questions about this problem. First, do you know if anything like this already exists and has already been studied ? Secondly, do you have any idea of things that could be interesting to search for (like do you have any intuition of a result or which restrictions could be interesting or maybe what other problems are particular cases of this one) ? And finally, do you think this problem is actually interesting to study or is it too vague or just not useful or idk ?

Please tell me if you think this message is in the wrong channel so I can delete and repost in the right one, thanks for reading and don’t hesitate if you need more explanation on something !

[uneven nimbus]

I don't think splitting the (co)domain of the functions into a product like that actually does anything, the way you've stated the problem?

if we call the product X, then (the composition closure of) F is a subset of the transformation monoid T_X, so we're looking for words in this monoid semigroup equivalent to 1;
if X is finite, then all of the viable f_i must be permutations and this reduces to an easier combinatorics problem

for infinite X and F, it feels like this is an instance of the word problem, so some theory there might be helpful

[warped cape]

Oh yes you are right about the splitting the domain thanks

And thank you for the two directions !!! I look them up

[warped cape]

I don’t think the "we’re looking for words equivalent to 1" is true as I stated the problem, because we search to let one element invariant only not to get identity (but it’s still an interesting thing it’s just to clarify the problem)

[uneven nimbus]

oh, is the point fixed?

this might be relevant then? https://en.wikipedia.org/wiki/Group_action#Fixed_points_and_stabilizer_subgroups

you're looking for the stabilizer subgroup of your given point under the action of the transformation semigroup/monoid generated by F

[warped cape]

Yes but both problems are important/interesting I just tried to generalize as much as I could

Ok thanks

[uneven nimbus]

I think whether such a tuple exists is undecidable in general; I'll do a reduction from PCP (post correspondence):

Let $\Sigma$ be some finite alphabet with at least $2$ symbols, and let $\alpha_1,\ldots,\alpha_N$ and $\beta_1,\ldots,\beta_N$ be two given lists of words over $\Sigma$. Let $X=\Sigma^\ast\times\Sigma^\ast/\sim$, where $(a,a)\sim(b,b)$ for all words $a,b\in\Sigma^\ast$. Now, define $F$ to be the set of functions $f_i:X\rightarrow X$ defined by $f_i(x,y)=(x\alpha_i,y\beta_i)$. Finally, pick $x=(\varepsilon,\varepsilon)$.

So, applying a chain of $f_i$s to $(\varepsilon,\varepsilon)$ yields something like:
\begin{equation*}
f_{i_1}\circ\cdots\circ f_{i_K}(\varepsilon,\varepsilon)=(\alpha_{i_1}\cdots\alpha_{i_K},\beta_{i_1}\cdots\beta_{i_K})
\end{equation*}
Thus, there exists a sequence of functions $f_{i_1}\circ\cdots\circ f_{i_K}$ fixing $(\varepsilon,\varepsilon)$ if and only if this instance of PCP has a solution.

[broken turtle]

Desync

[uneven nimbus]

I haven't done reductions in a while, so do check this over, but it definitely feels like the general case is not going to be tractable at all

edit: I have also just spotted a problem with the above reduction in that we don't disallow empty compositions; easy fix though, replace $\Sigma$ by $\hat\Sigma:=\Sigma\cup{\bullet}$, where $\bullet\notin\Sigma$. Now, add an extra map $g:X\rightarrow X$ defined by
g(a,b)=\begin{cases}
(\varepsilon,\varepsilon) & a = b \text{ and }(a,b)\neq(\varepsilon,\varepsilon)\
(\bullet,\bullet) & a=b=\varepsilon\
(a,b) & \text{else}
\end{cases}
so $g$ moves every non-empty equal pair to $x$ (and moving $x$ itself to a dead state instead), forcing the composition to be non-empty. i.e., we move the earlier quotient into the map $g$ and force it to be used.

[warped cape]

Wow thank you so much yes I imagine that we can’t say much about the general case but maybe about particular cases. The problem I am working on is "can we enumerate all the "generators" of S in polynomial total time (i.e polynomial time in the size of input + output) in the particular case where there are a finite number of f_i all of the form f_i : x —> x + k_i where k_i is in Z x Z x … x Z (or Q x … x Q) and (x_1,…,x_n) = (0,…,0) (so it is the same as searching to obtain 1 or identity)" just to give context and to help with the results it would be ideal to have but any idea is cool like your reduction thanks a lot it’s very late in the night where i live so I will look at it more closely tomorrow but thanks again

[cerulean cove]

What does this symbol mean

[foggy galleon]

coproduct

[unreal epoch]

does anyone have any ideas on how to do the reverse direction "generally"? this is all fine when K is "nice" i.e. alg closed and char 0 because this collapses to understanding the central character/turns into character theory. but otherwise i have no idea how to proceed. to make things worse im reverse engineering the qual syllabus from the problem list so i dont even know what i'm supposed to know

[urban granite]

product with arrows reversed

[cobalt sonnet]

btw do you know how to prove that the generator of socle of a semisimple frobenius algebra must be a unit?

[lone jacinth]

If it's semisimple then the socle is everything

[cobalt sonnet]

Oh right sorry, what I meant was how to prove that the Socle is always a principal ideal

[lone jacinth]

I mean if it's equal to (1) it's a principal ideal

[cobalt sonnet]

but that's just one side though

[lone jacinth]

I'm not sure I understand what you're asking

[cobalt sonnet]

For commutative Frobenius algebras there's this result but i don't understand the proof

[cobalt sonnet]

The distinguished element in this means the number specifying comultiplication, followed by multiplication

I said socle here because he also proves that this element generates the socle.

[lone jacinth]

So you're asking why the socle of an annihilator algebra is principal?

This is because it is simple

That's what we talked about yesterday

[cobalt sonnet]

but the author claims that this is true for any Frobenius algebra not just a local one

[lone jacinth]

Where do they claim that?

[cobalt sonnet]

Commutative*

[lone jacinth]

The assumption here seems to be that A is an annihilator algebra, since they're talking about N and U

Or indecomposable at least

What is the preceding text where they presumably define S and A?

[cobalt sonnet]

how do you deduce that?

[lone jacinth]

They're using the fact that N consists of all non-units for example

[cobalt sonnet]

i think i get it it's because when it's indecomposable then we it splits into units and non units

[smoky wasp]

what does this mean

Woul dit just end up being like the morphism that completes the diagram

[digital parcel]

should just be the map
[ N \otimes R' \xrightarrow{\varphi'} N \otimes R' \xrightarrow{\varphi' \otimes \text{id}_{R''}} N \otimes R' \otimes R'' \cong N \otimes R'' ]

[broken turtle]

anamono

[digital parcel]

the isomorphism on the right comes from your homomorphism alpha

[digital parcel]

oh wait there's a typo in my tex lol just realized

lemme fix it

hmm yeah im not sure actually ignore what i said above lol

sorry

[ornate kindle]

Regarding (iii) in 1.3.3, what does "isotropic" mean in this context?

Is it maybe like

[broken turtle]

NotABot

[ornate kindle]

to match (16) in the proof of 1.3.4?

[dusty warren]

i think its in this sense https://ncatlab.org/nlab/show/isotropic+subspace

specifically that ( , ) vanishes on frak p_+ timex frak p_+ and on p_- too

[ornate kindle]

$W\subseteq W^\perp$

[broken turtle]

NotABot

[ornate kindle]

Yeah that looks like what I was thinking, thanks!

[dusty warren]

the place ive heard this word before is in the solution to eventown

ymmv

[pallid pendant]

Can anyone recommend a book that roughly(at least) covers the first half of Atiyah - Intro to commutative algebra. We are going to cover the first 5 chapters of this book "religiously" - Rings and Ideals, Modules, Localization, Primary decomposition, Integral extensions.
I am mostly self studying.
So far, I have covered all the relevant topics from Aluffi - Notes from the underground like Rings and Ideals, Modules. I really really liked his style of being more hand wavy , approachable and far from being terse.
So, I am looking for a book of similar style that covers rest of the topics in a gentle manner.

[ornate atlas]

Is there a reason you can’t just use Atiyah MacDonald?

[tulip flicker]

Tersity if i'm reading correctly

[pallid pendant]

I don't find it very illuminating so I need to supplement it with a more elaborate text

Very helpful. Thanks.

[ornate atlas]

Reid’s undergraduate commutative algebra may be of interest then

[pallid pendant]

Looking at it, I think it is best suited for my requirements since it covers the exact same topics and seems more elaborate.

Thanks

[cerulean cove]

brooo my account is hack???

nooooo way

[vapid axle]

Is 10 exercises per chapter of Atiyah Macdonald enough

Lowkey I'd like to do more but I've been on this chapter for like 1.5 months now

[ornate atlas]

I would say move on

You can always come back and do more if you need to, but 1.5 months per chapter is likely more than enough time

I also think excercise choice and what you take away from them is more important than the raw number of them, I think as soon as you feel you have a pretty reasonable grasp on the content its more than fair to move on

[vapid axle]

Yeah basically half of chapter 2's questions are on direct limits, and I'm skipping those

[ornate atlas]

And youre not banned from coming back to it, its a good thing to do in fact

Make some progress, check back in and do some more in a few months is my advice

[vapid axle]

Okay thanks

[ornate kindle]

$a$ is a $\mathfrak{g}$-invariant element of $\mathfrak{g}\otimes\mathfrak{g}$

[broken turtle]

NotABot

[ornate kindle]

This means that all elements of g act on a by scalar multiplication yeah?

[lethal scarab]

for b and c
intuitively i feel its some sort of conjugacy class/orbit type argument but im stuck

[limpid horizon]

are these isomorphisms as k-modules?

[spice idol]

"k-modules"

[limpid horizon]

lol

[lone jacinth]

The second isomorphism should only be true with some finiteness assumption though.

Maybe that's part of your definition of simplicial complex...

[limpid horizon]

yeah

idrk how to do the first one tbh

[lone jacinth]

Just writing out the definition of cohomology goes a long way

[fierce steeple]

Saying k-modules is based

[fierce steeple]

First find like a natural map between the two

I mean maybe that is a dumb hint sorry lol

[vague pawn]

btw speaking of Cohomology are there any tips for how to study the basics of it?

[fierce steeple]

But tbh i do not think this is 100% obvious without like other facts

[lone jacinth]

I mean like with anything I guess pick up a book or take a class on it.

Rotmann has an intro to homalg book for example

Might be beneficial to know some algebraic topology or representation theory first though, to have something to apply it to

[fierce steeple]

Also unclear what is meant by "cohomology"

Can mean many things

[vague pawn]

Is Class Field Theory good for applying it to?