advanced-algebra

[spice idol]

ive kinda accepted ill never be able to learn modular forms and what theyre about

[steady lance]

weird question: if I have a Gorenstein R-module that is also an R-algebra, is it necessarily also a Gorenstein R-algebra?

I guess fundamentally this is a question about how injective dimension behaves on algebras vs modules

there's a paper where this is relevant and I'm wondering if this is just trivial or if I should spend some time rigorously proving it

[hushed bone]

The homological stuff for R-algebras is defined in terms of R-module structure unless you actually really need to use something about it being an algebra

That is to say, the answer is yes because everything involved in “Gorenstein” only makes use of the module structure

Whether or not this algebra is Gorenstein over itself is a different question though

[steady lance]

ahhhh perfect!

Thanks!

I could not find any citations for gorenstein algebras that didn't reduce to the module case so this makes a lot of sense. Thought I was taking crazy pills lol

I was thinking it was gonna be something like taking a finite injective resolution "as R-algebras", forgetting the algebra structure on the maps and then proving that injectivity as an R-algebra corresponded to injectivity as an R-module, but this makes that a hell of a lot easier lol

[hushed bone]

The category of algebras isn’t abelian so taking injectives there doesn’t really serve a point

The secret is, you can’t really do homological algebra on algebras because of this, which is why when you want to do things like take resolutions of rings you need to use for example simplicial techniques

[fierce steeple]

The humble simplicial ring

[hushed bone]

And then you start taking steps down to homotopical algebra

[steady lance]

this makes a lot of sense

[merry magnet]

are equivalence relation quotients the same idea as tagged unions in programming languages theory?

because with a set S and a relation R, the quotient S/R partitions S into disjoint subsets and each subset has elements that relate to each other in some way

since they are disjoint, given any element we can instantly figure out which partition It's in by finding its equivalnce class since forall a in [x], [a]=[x]

which means the elements carry the 'tag information' on their own by being disjoint

[plucky arch]

You can think of an equivalence relation as specifying a bijection between a set and a tagged union of sets

[merry magnet]

you mean an equivalence relation quotient?

[plucky arch]

yep

[merry magnet]

aight thanks

so using that interpretation/analogy, is G/H a tagged union where the tag type is G and the payload is the whole set H?

since we get a union of |G|-many disjoint H's?

[plucky arch]

well

the payload is H, yes

but the tag type is not G

you only get a union of |G|/|H|-many disjoint H's

[merry magnet]

ahh I didn't fully learn about full group quotients and tried to generalize 😅

don't really get it but ig I will when I get there

[elfin zephyr]

which book is best among herstein topics in algebra, abstract algebra dummit & foote & artin alegbra?

[broken wren]

Artin is the best to learn from

[steady lance]

ngl I'm a dummit and foote defender

I think it maintains the balance between rigor and readability quite well

[pastel shoal]

It's too long and has too many routine exercises

If someone is using it along with lectures and knows exactly what to read and which problems to solve it can be fine ig

[limpid horizon]

i vote dummit and foote for best intro text

[limpid horizon]

i disgaree about the exercises they are good for someone new to the subject

[toxic trout]

Everyone knows that the best intro algebra book is lang's algebra

[rose mirage]

it's actually Benson 1

[steady lance]

best intro algebra 101 text is just Hartshorne algebraic geometry. Just try to re-invent the algebra knowledge needed by scratch yourself as you read it

[toxic trout]

real

[steady lance]

tbh I don't see that as a bad thing necessarily

for an intro book I think it's good to have redundant exersizes. Nobody's gonna do every question but you slowly build intuition for what questions are worth doing - which builds an important skill in and of itself tbh

[ornate atlas]

I think routine exercises are great actually, and I get annoyed at books that don’t have them. I think one of my bigger issues with Hatcher is that it doesn’t really have many “easy” exercises. There’s some of course, but I like when every section starts with a couple of “just make sure you actually know the definitions and can directly apply big theorem”

[limpid horizon]

Yeah its nice

Its friendly

[broken wren]

Artin has exercises. Only the best exercises

[steady lance]

Not to bring up hartshorne again, but I also think that including unsolved problems as exercises is legitimately psychopathic and terrible pedagogy lol

[ornate atlas]

Nah I honestly kinda rate that, you’ll get someone who just doesn’t realise it’s supposed to be hard and then solves it

Though maybe the world will never know, who would be interested in their solution to exercise 3.28…

[steady lance]

there's like no chance that'll ever happen though, right?

much more likely it makes people waste time on an impossible task and then get demotivated

[ornate atlas]

I mean it’s unlikely, especially given how popular harthshorne is, but there are stories of people thinking some research problem was a homework question and knocking it out

[ornate atlas]

But yeah I think it makes more sense to indicate it’s an open problem included for interest. My complex analysis class had the Reimann Hypothesis on one homework but just worth 0 marks

Would’ve kinda sucked if it was for credit

[steady lance]

me solving the riemann hypothesis and then failing the class bc I didn't do the other problems

[broken wren]

Are any of those stories real? Like there’s a specific paper Milnor published as a freshman that was probably from calculus class, but someone asked if he really didn’t know that it was open and he said he didn’t want to ruin the legend
https://math.mit.edu/~hrm/papers/milnor-kinky.pdf

[ornate atlas]

Fields medal and millennium prize winner but you’re not allowed to graduate

[past cove]

idk why people shit on hartshorne genuinely

it is absolutely the best book for algebraic geometry out there

[ornate atlas]

There is one that seemed mostly true but I’m not sure I remember the specifics, like one of the more famous stories. I’m pretty sure it was somewhat exaggerated, and in any case would require like a once in a generation mathematician to come along with just the right background

[past cove]

(If you know commutative algebra properly)

like I'm not kidding when I say that I believe if you go through hartshorne you will know algebraic geometry better than any other book

it takes time and effort

but it makes you work and it pays off properly

actually makes you think and come up with original arguments vs. vakil absolutely spelling everything out for you like you're a child

[ornate atlas]

This just feels like the suffer through Rudin vs learn from Abbot debate again

I guess it’s a little different because it’s not your first introduction to maths, but still

[steady lance]

I go back and forth. Now that I have the commutative algebra background it seems great, but as someone who doesn't naturally think geometrically I initially found it rather difficult to get intuition on most of it

I think for those who are geometrically-inclined and have the necessary commutative algebra background it's probably great, but I prefer some pretty pictures

now that I'm thinking about it it probably makes sense that people who are more explicitly algebraic geometers tend to prefer hartshorne. But for me I kinda just use algebraic geometry as a way to justify the commutative algebra work I do. So I just want a quick geometric description of how to pretend my work matters

[hushed bone]

Hartshorne good

The longer and longer I am in my careeer the more I believe it

[muted sierra]

This, unironically.

[toxic trout]

I mean thats what I am currrently doing

[toxic trout]

this might happen if you use lang's complex analysis

[foggy galleon]

https://en.wikipedia.org/wiki/George_Dantzig chatgpt confirms. But I'm lazy to track the references

I had never heard the story of that paper recounted like that. This is what he (Milnor) said in the Abel prize interview

[distant harness]

Hmm, that sounds very far from the usual story told in the Wikipedia article (and elsewhere) -- it's not even in the same field, and the knot-theory one is 10 years later. Perhaps they're separate events?

Waitaminute -- Dantzig was born in 1914, so he would be 35 in 1949, not 18.

[foggy galleon]

sorry

the Abel prize interview is the one of Milnor

[distant harness]

Ah -- the interview above was Milnor not Dantzig.

[elfin zephyr]

@broken wren @steady lance @pastel shoal @limpid horizon thank you all for giving your suggesstions.. btw no one mentioned herstein's topic of algebra..

[steady lance]

np! I've never heard of Herstein ngl lol

[pastel shoal]

I haven't read it but my prof used to give exercises from Herstein, they were pretty good. But I've heard it's only good for groups and not rings and fields

[soft parcel]

Could someone help me understand this construction. If y_1, ..., y_n is the dual basis to x_1, ..., x_n, then beta is not even defined on (x_i, y_j), so what does the author mean here?

[pastel shoal]

It's dual relative to the bilinear form, it's not a basis of the dual space

[soft parcel]

Oh

Hmm, I see

And the existence then follows probably somehow from non-degeneracy?

[pastel shoal]

Yep

[soft parcel]

I see. That'll be a good exercise for me then

[pastel shoal]

It's similar to how you show it for the inner product

[soft parcel]

Great. Thank you!

[digital parcel]

does anyone have a pdf copy of M. DEMAZURE, Une nouvelle formule des caracteres (Bull. Sci. Math., T. 98, 1974, pp. 163-172).? I can't find anything online

the sciencedirect.com website for the journal only goes back to 1998

[weak lodge]

If you like, I can put in a request through my university's interlibrary loan system, though it's not guaranteed that they'll find it ofc. Let me know

[digital parcel]

yeah good idea, i'll try that first for my uni

[balmy rune]

Does anybody have a good "crash course" resource for a review of basic algebra (which I took like a decade ago, forgot a fair amount but remember the basics), rep theory, and maybe connections to topology? I'm trying to speedrun Artin's book right now but not sure if that's the best one.

[plucky arch]

how do schur functors interact with direct sums?

e.g. taking the antisymmetric square of $V \oplus W$

[broken turtle]

Pseudo (Cat theory #1 Fan)

[broken wren]

Artin is an introduction that doesn’t do anything systematically, so it’s not what you want. But I think people often want too much

[broken wren]

The total symmetric power and total alternating power turn sums into products
S^n(V+W)=Sum_a+b=n S^aV tensor S^b W
In general there is a coproduct on the ring of symmetric functions. (And a second coproduct for tensor product of representations)

[plucky arch]

interesting, so it's like a convolution?

how about the schur functors that aren't just symmetric or antisymmetric powers, though?

[broken wren]

How do you even describe other schur functors? As symmetric functions?

[plucky arch]

young diagrams, right?

[broken wren]

I guess. Turn it into a symmetric function and apply the coproduct

[plucky arch]

you mean with the young symmetriser...?

[broken wren]

[plucky arch]

otherwise what do you mean by "turn it into a symmetric function"

[rose mirage]

could be referring to the schur polynomials? idk

[near lantern]

By rep theory of finite groups (non-modular), there is a projector onto the isotypic component of a irrep in any representation; it is given by rho(P_chi) := 1/|G| ∑_{g in G} chi(g^{-1}) rho(g) for the isotypic component of the irrep with character chi in the representation rho. You could try applying these projections to the n^th tensor power of (V (+) W) and seeing what you get. In particular, for the "standard representation" of S_n, the character is chi(w) = #{fixed points of w} - 1. So very explicitly for S_3, the projector for the third Schur functor is (2 [1] - [(1 2 3)] - [(1 3 2)])/6, where [w] stands for the action of w.

ie we get Sch_3rd(V (+) W) = {2 x - (1 2 3)⋅x - (1 3 2)⋅x : x in (V (+) W)^{(⨯)3}} = ∑_{A = V (⨯) V (⨯) V, V (⨯) V (⨯) W, V (⨯) W (⨯) V, ..., W (⨯) W (⨯) W} {same expr for x in A}

The tensor power decomposes into parts with a given number of V's and W's and these are S_n-stable

So we can say Sch_chi(V (+) W) = P_chi((V (+) W)^{(⨯)n} = (+)_{k = 0}^n P_chi(direct sum of all tensor products of k V's and (n-k) W's in some order} =: (+)_{k = 0}^n Sch_{chi,k,n-k}(V, W) (introducing new notation)

[near lantern]

This sum is direct. Note that Sch_{chi,n,0}(V, W) = Sch_{chi}(V) and Sch_{chi,0,n}(V, W) = Sch_{chi}(W), so Sch_{chi}(V) and Sch_{chi}(W) do appear as the first and last terms in the formula.

Going back to the specific case of Sch_3rd for S_3, we can say the following: the action of (1 2 3) identifies VVW, WVV and VWV as isomorphic. So we can think of VVW (+) WVV (+) VWV as VVW (⨯) k[(1 2 3)] (latter is regular rep of <(1 2 3)> as a S_3-rep). Then Std sits inside the second factor and we get VVW (⨯) Std. But this is a very idiosyncratic description and I don't have anything more systematic for the middle terms yet.

[limpid horizon]

how should i think about in what sense is the module or ring extended

[fierce steeple]

Extension of scalars

[limpid horizon]

so the only sense in which a module is "extended" is through that

[fierce steeple]

Yee I think that is basicicalpy what they mean though a little more general as they allowed N

[limpid horizon]

Yea its like pretty general so im trying to figure out how it all fits lol

[limpid horizon]

later it says the extension R->Rp is flat. Does that mean Rp is a flat R-module?

[digital parcel]

yes

[limpid horizon]

Rp(x)R - is exact right?

[digital parcel]

yea

that's the definition of a flat module

[limpid horizon]

Ohh right lol

I was thinking something with localization

[digital parcel]

localization is always flat

[limpid horizon]

I dont know exactly how to interpret that but i know localizing is exact

[digital parcel]

If M is an R-module and S is a multiplicatively closed set in R, then S^{-1}M is the same thing as M \otimes_R S^{-1}R

In other words, doing - \otimes_R S^{-1}R is exact, i.e., S^{-1}R is a flat R-module. By "localization is always flat," I mean the localization map R -> S^{-1}R is a flat map

[limpid horizon]

Thank u

[limpid horizon]

M/xM \cong M(x)R R/(x)? x is M-regular

For the map M->M(x)R R/(x) I can see xM is in the kernel but not sure how to show the whole kernel is xM

[lone jacinth]

Consider the exact sequence
0 -> (x) -> R -> R/(x) -> 0

Tensor with M, use right exactness and that M = M(x)R

[limpid horizon]

Also (x) tensor M is xM?

[lone jacinth]

It is not

But what is the composite map
M(x)Rx -> M(x)R -> M?

[limpid horizon]

It has image xM?

[lone jacinth]

That is correct. But you should be able to just describe explicitly what the map is

[limpid horizon]

M(x) Rx -> M just multiplies the entries in the simple tensors together , is that what you mean?

Oh i see now why you wrote M (x) R in between now

Tensoring the exact sequence with M doesnt get multiply entry of tensors map but that isomorphism does

[plucky arch]

is the littlewood-richardson formula proof just a lot of computation

or is there, like, a nice reason to see why it works

[wary elbow]

I think the Littelmann path model is the "nice" reason

[plucky arch]

Ooh what’s this

[wary elbow]

It's a combinatorial framework for the decomposition of a tensor product of irreps for Kac-Moody algebras. This contains the Littlewood-Richardson formula as a special case (SL_n).

I'm not an expert, but Littleman has a survey paper here:
https://www.mi.uni-koeln.de/~littelma/papers/cambridge.pdf

Kashiwara has a related combinatorial construction called "crystals," but don't ask me what they are because unfortunately I don't know. Kashiwara has a lot of papers on crystals. Apparently they're related to D-modules and other geometric representation theory stuff.

Gaitsgory has some fancy looking papers about crystals that I'm hoping to understand someday

[plucky arch]

Looks quite beyond my level at the moment but perhaps I’ll return to it at some point

[past cove]

crystals are bascally qc sheaves that are rigid under infinitesimal deformation

that's why they're called "crystals" essentially because they "rigid" in this specific sense

[broken wren]

Crystals are sheaves on the crystalline site, closely related to D-modules. But the above is about crystal bases. They are bases, ie, subsets of a vector space, not objects in a category
https://en.wikipedia.org/wiki/Crystal_base

[wary elbow]

Whoops, yes, thanks for the correction. I was mixing up crystals and crystal bases.

[silver goblet]

if i had a solvable finite group G, is there a way to run a spectral sequence induced from the "filtration" coming from the composition series in hopes of calculating the group cohomology of G?

as in, we know the group cohomology of prime cyclic groups which are the "associated graded"

[broken wren]

Yes, I think you can roll several spectral sequences into one. It probably won’t help much compared to using them all individually

[silver goblet]

can you clarify what you mean by this

like which spectral sequences are you suggesting rolling into one

[broken wren]

If you have an extension of groups
N->G->Q
then there is a spectral sequence
H^q(Q;H^p(N)) => H^p+q(G)
Do this for all the extensions in the composition series

[silver goblet]

ah i see what you mean

i feel like i am missing something obvious but does the cochain complex for G does not get a filtration induced from the filtration on G?

[silver goblet]

ahhh nevermind, this makes sense now that i have thought some more

[weak spruce]

Hey! Can someone detail the difference between Isomorphism and Homomorphism?

[urban granite]

isomorphism is a homomorphism with an inverse (which is also a homomorphism)
homomorphism is a structure-preserving map between algebraic structures of fixed type (e.g. abelian groups, rings, modules)

[fierce steeple]

Remark is that for many algebraic structures with underlying sets, it is enough to check that the function is a bijection/invertible, as the set-theoretic inverse is automatically a homomorphism. So you may see it stated as like a bijective homomorphism or smth in some cases

[urban granite]

mfw topological abelian groups where R_{disc}\xrightarrow{x\mapsto x} R_{usual} is not an isomorphism of topological abelian groups because inverse function is not continuous while it is bijective

[fierce steeple]

Or sheaves of abelian groups where it does not make sense to ask the question of underlying sets

Well it does but in a different way

[weak spruce]

Thanks a lot!

[unborn rampart]

Not sure where to post this, exterior powers feel a bit too advanced for #linear-algebra. Anyways, to prove this, is it sufficient to first claim that $\Lambda^n V$ is 1-dimensional, and so $\alpha$ must be of the form $\alpha(w) = \lambda w$ for some constant $\lambda$. Then note that $\lambda$ is clearly an eigenvalue, and the determinant is the product of eigenvalues, so $det(\alpha) = \lambda$?

[broken turtle]

sheddow
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[last talon]

How does that last line work?

[unborn rampart]

The only eigenvalue of \alpha is \lambda, so det(\alpha) = \lambda?

[vague pawn]

Its more like when you write what alpha does to a basis (of size 1) of the exterior power you get that it multiplies by the determinant

By one of the definitions of the determinant

The one with permutations

[unborn rampart]

wdym? Does the proof I posted not work?

[vague pawn]

alpha:V-->V

So it doesn't have to be multiplication by a scalar

[unborn rampart]

oops, I meant \Lambda^n \alpha (w) = \lambda w

[vague pawn]

And then its not that direct to relate the determinants

[unborn rampart]

oh right, I see

[vague pawn]

Lambda^n V has basis
v1 ^ v2 ^ v3 ^ ... ^ vn

See how it acts on this

[unborn rampart]

I'll try, thanks for the hint

[unborn rampart]

do you mean the Leibniz formula? Or the characterization of the determinant as the unique alternating map such that det(I) = 1?

[vague pawn]

This

But probably any definition of determinant should be fine

[unborn rampart]

hmm, I'm struggling to see how to turn $\alpha(v_1) \wedge \dots \wedge \alpha(v_n)$ into something that looks like the formula above

[broken turtle]

sheddow

[lone jacinth]

Note that $\alpha(v_i) = \sum_j a_{ji}v_j$ where $a_{ji}$ is the matrix for $\alpha$

[broken turtle]

jagr2808

[lone jacinth]

So if you just write out the expression (and delete all the zero terms) you should get exactly this

I guess a simpler approach might be to just prove that it is true for elementary matrices. Then as all matrices are products of those you're done.

[unborn rampart]

thanks for the suggestions it's just that my brain shuts down whenever I see sums with multiple indices and matrices, and I have to figure out the correct order of the indices and which ones cancel and which ones don't etc. etc.

right now I wanna quit maths tbh, this is not fun at all

[lone jacinth]

Well that's why I'm suggesting elementary matrices. Then all the sums goes away

[unborn rampart]

yep, I'll try, but my brain is filled with non-elementary matrices right now, so I need a break to flush them out

[lone jacinth]

And I guess sort of general advice if you want to think about complicated sums. If you go through everything with like 2x2 matrices, then you can actually write out every term of the sum and see a pattern from there. Might help

[unborn rampart]

just proved it using this approach now, it was much easier, thanks I think I won't quit maths after all

[near lantern]

@plucky arch out of curiosity, did you try this and if so did it lead to anything? I would be very interested to see an answer to your question.

[plucky arch]

Hm I did not unfortunately

I did see in my notes something about splitting up the young diagram for the two factors?

[plucky arch]

Hm…

Suppose $V$ is an irreducible rep of a group $G$, and $W$ is another rep

[broken turtle]

Pseudo (Cat theory #1 Fan)

[plucky arch]

Can I extract the isotypic component corresponding to $V$ by doing $V \otimes_{k[G]} W$?

[broken turtle]

Pseudo (Cat theory #1 Fan)

[lone jacinth]

No, there are a few problems.

Number one is that $V \otimes_{kG} W$ is the dual of $Hom_{kG}(V, W^)$ so you would really be measuring the precense of $V$ in $W^$.

But the bigger problem is that $V \otimes_{kG} W$ isn't even a $kG$-module. It is just a vector space, but the dimension does measure how often $V$ appears in $W^*$

[broken turtle]

jagr2808

[lone jacinth]

The counit

$V \otimes_k Hom_{kG}(V, W) \to W$

is an embedding of the isotypic component though

[plucky arch]

hm ok

[broken turtle]

jagr2808

[wary elbow]

In general, branching problems are very challenging. There's no simple formula for obtaining the multiplicity of one irrep inside a tensor product (unless you would consider the Littlemann path formula "simple"). There are combinatorial techniques that work in theory, such as the Demazure character formula, but they become intractable in practical terms once you get above low dimensions.

I'm sure someone has implemented these in, say, Python, so if you have specific irreps you care about then you can probably do some computer calculations

[digital parcel]

speaking of which, my library was able to find the pdf via ILL, and here it is for posterity!

i don't think there are copyright concerns since it's a 1974 paper

there is a gap in one of the proofs of that paper as mentioned in andersen's paper:

[near lantern]

There is a natural map Hom_{k[G]}(V, W) (⨯)_{D} V → W, where D = End_{k[G]}(V) is the division ring of endomorphisms of the representation V. Its image is the V-isotypic component of W. (This works for modules over any ring.)

[near lantern]

Oh, it has been said. My bad.

[lone jacinth]

Yeah, I assumed k alg closed so D=k.

[graceful sluice]

Does a question about Schur polynomials belong here or in #advanced-analysis?

[vague pawn]

Here seems fine

[ornate atlas]

It’s probably best just to ask! Then people can see if they know or not lol

[graceful sluice]

Fair enough. Just didn't wanna start typing a bunch of stuff and then get redirected lol

So, the definition of Schur polynomials is giving me some trouble. I seem to understand most of the pieces, but I'm struggling with some of the notation.

Here's the definition from Wikipedia

Here's my current understanding of the relevant pieces:

Symmetric polynomial: a polynomial in n variables in which transposing x_i and x_j doesn't change the polynomial.
Alternating polynomial: a polynomial in n variables in which transposing x_i and x_j changes the sign of the polynomial.
Vandermonde polynomial: a polynomial in n variables which is the product of (x_j - x_i) for i < j.
All alternating polynomials are divisible by the Vandermonde polynomial because if x_i = x_j for some i and j, the only way for their transposition to change the sign is if the polynomial is 0.

What's primarily confusing to me is the indexing notation. I know broadly what a partition is, but not how to index something by integer partitions.
In particular, how to interpret \lambda_1, ..., \lambda_n, along with how to read this portion, continues to elude me.

For instance, can any alternating polynomial be written in this way? If so, since any alternating polynomial is divisible by the Vandermonde polynomial (and their quotient is a symmetric polynomial), does that mean any symmetric polynomial is a Schur polynomial? It says the degree d Schur polynomials in n variables are a linear basis for the space of homogeneous degree d symmetric polynomials in n variables, so I would imagine they're not the same thing, or they'd probably have the same name, but I'm not sure how to read this.

[jaunty sparrow]

The notation (λ_1,..., λ_n) usually denotes the partition of λ_1+...+λ_n into parts λ_1,..., λ_n.

[graceful sluice]

I'm not sure I follow.

[jaunty sparrow]

Each λ_i is a positive integer

[graceful sluice]

Yes.

[jaunty sparrow]

Let's say we partition 5 into (3,2)

Then λ_1=3 and λ_2=2

[jaunty sparrow]

The λ_i are usually in (not necessarily strictly, since parts can repeat) decreasing order.

[graceful sluice]

Okay, I think that seems alright.

[jaunty sparrow]

Is there still some confusion regarding the notation?

Hopefully not

[graceful sluice]

I'm fine with partitioning 5 into (3, 2).

What I'm not clear about is what it means for something to be indexed by such a partition. And moreover, what it is we're actually partitioning with this

[jaunty sparrow]

Like I said earlier, the partition (λ_1,... λ_n) is a partition of the positive integer λ_1+...+λ_n.

[graceful sluice]

I guess that makes sense. That's something I hadn't realized until now.

Why are we adding n-1, n-2, etc?

Are we now partitioning the sum of \lambda_i + (n^2 - n)/2?

Is it just so that we get something that's definitely strictly decreasing?

[jaunty sparrow]

Yes

[graceful sluice]

So, do the lambda even particularly matter at all? Would we have just as easily been able to say a_(strictly decreasing sequence of numbers)?

I don't understand what this indexing by partitions even has to do with what's being described.

[jaunty sparrow]

They clearly do matter, since they show up in the determinant

[graceful sluice]

I don't mean "does the chosen sequence matter"

[jaunty sparrow]

For one it shows a correspondence between the polynomials being described and a particular class of partitions.

[graceful sluice]

How do you mean?

So far I'm just seeing "a_{strictly decreasing sequence}(x_1, ..., x_n) is the determinant of a matrix where the columns are the variables raised to the elements of this sequence"

I'm getting all this stuff about semistandard Young tableaus, but I'm not seeing the deeper meaning beyond what I wrote a moment ago.

Why do we index them the way that we do instead of just directly letting the labmdas be an arbitrary strictly decreasing sequence of nonnegative integers?

[jaunty sparrow]

On any other day I'd have been able to give a proper explanation, but I've just gotten done with my final exams lol. Hopefully some of the helpful people here will clear it all up.

[graceful sluice]

Moreover, why do we even care what order they're in? An alternating polynomial is gonna be the same in any order, up to a sign change. Does strictly decreasing guarantee positivity or something?

[graceful sluice]

No problem man! I really appreciate your help so far. Just understanding what the lambdas mean is already huge.

Honestly, we could just say a Schur polynomial is defined to be the determinant of any matrix whose entry in row j column k is x_j^{a_k} (divided by the classical Vandermonde determinant), then say if it happens to be zero then oh well, we just happened to have two x's or a's that were the same.

[near lantern]

Yeah, so you would get a lot of 0's and a lot of repetitions except for sign. If you want a possibly irredundant (including up to sign) list, you should not list the polynomials corresponding (a1, ..., an) such that some of the ai's are equal. Nor should you list the polynomials corresponding to different (ai)'s that are just permutations of each other. A natural way to achieve this is to restrict to (a1, ..., an)'s which are strictly decreasing.

[graceful sluice]

Sure, it seems like you'd get some trivial polynomials that are equal to zero, and you'd get some polynomials that are equivalent to others due to the alternating nature, but why define them with the lambdas and the ns when we could just refer to them in the other way? I

[near lantern]

Now as you have seen, there is a bijection between decreasing sequences (l1, ..., ln) and strictly decreasing sequences (a1, ..., an), given by (l1, ..., ln) <-> (l1 + n-1, ..., ln). So you could equally well index the set of Schur polynomials by the (li) corresponding to the (ai) used, instead of the (ai) directly.

[graceful sluice]

Why would we do that though instead of just indexing them by strictly decreasing sequences?

[near lantern]

The mathematical community has chosen to do the latter and there are probably good reasons for this.

[graceful sluice]

Or better yet, simply saying they're all Schur polynomials and the ones given by the formula are just in standard form?

It's not like we say x^3 + x + x^2 isn't a polynomial just because x and x^2 are out of order.

[near lantern]

(Off the top of my head, the degree of s_l is l1+...+ln instead of a1+...+an, because of the division by the Vandermonde determinant.)

[near lantern]

I mean, it's your opinion that that would be better. It's not an objective fact.

[near lantern]

Ah. So

[graceful sluice]

better meaning "more so", not strictly improved.

[near lantern]

The phrase "Schur polynomial" is not being used to describe some property of polynomials. It's just a specific family of polynomials with interesting properties.

Like "Legendre polynomials" or "Hermite polynomials" or even "rising/falling factorials" for that matter.

[graceful sluice]

Yes, I get that part. But why not refer to all of them as Schur polynomials and call the ones defined by the usual formula just Schur polynomials in standard form?

Since they're equivalent anyway.

Perhaps notated differently, but they'd still evaluate the same way, up to potentially a sign change.

[near lantern]

Why not refer to any scalar multiple of a Legendre polynomial as a Legendre polynomial?

[graceful sluice]

Like, if you transpose two pairs of lambda_i in the definition of a schur polynomial, we get the same polynomial.

[near lantern]

If you have a family of blah things a_i, you want "blah thing" to mean "a_i for some i".

[graceful sluice]

And if you just took a_{any decreasing sequence} divided by the Vandermonde determinant, you get a Schur polynomial. Just not the one indexed by standard partitions.

[near lantern]

You don't want it to mean "anything that is <closely related> to some a_i".

It's the more convenient way to have terminology work, for that use case.

[graceful sluice]

Is there some underlying property of Schur polynomials that makes the indexing especially important to the polynomials themselves?

I'm only familiar with them in the context of "determinant of generalized Vandermonde matrix divided by standard Vandermonde determinant"

[near lantern]

Yes, there should be.

[graceful sluice]

And don't really know anything else about them.

Like, I still don't understand why the partitions are even related to them.

[near lantern]

Well, they wouldn't be so widely studied if that's all they were.

[graceful sluice]

It feels like we could have used any sort of indexing to describe the same things.

[near lantern]

They turn out to be related to symmetric functions, representation theory, etc. in many ways.

[graceful sluice]

And just sorted them into equivalence classes.

[near lantern]

There the choice of indexing them that way does matter.

[graceful sluice]

I mean yeah, I can see why they're related to symmetric functions. They are symmetric functions, and apparently they form a basis for them

Ah, I don't really have any knowledge of representation theory. It's one of those phrases I sorta figured I'd have absorbed by now, but I've never actually done anything with it.

[near lantern]

If your problem is "I don't see why there's any benefit to doing the indexing and notation this particular way", when you've just seen the definition, you should probably keep reading.

In general.

[graceful sluice]

My main point of reference is that I'm studying which minors of DFT matrices can have determinant zero (or slightly stronger conditions), and they arise naturally as quotients of these determinants by the Vandermonde determinant.

Part of the difficulty is that each time I try to study them for their own sake, whenever I come across a new piece of information it carries with it several new terminologies I haven't seen before and suddenly I'm in a completely different field trying to figure out how any of what I'm reading is related to the task at hand.

Since I don't have a strong enough attention span to learn every piece of adjacent information, and I don't have the intuition/discernment with this particular field to know if what I am trying to read is going to be helpful or if I'm just in the weeds of a different branch of math and forget what I came for.

[near lantern]

I can appreciate that problem.

[graceful sluice]

I don't mean to make it other peoples' problem, but at the same time I don't know how to otherwise learn what I actually need to know without investing an unreasonable amount of time and effort into things that weren't as related as I hoped they would be.

[near lantern]

Sure. I think asking people is a good solution to figuring out what is related to and needed for what.

[graceful sluice]

I don't mean to come across as arrogant or dismissive when I ask the questions about why they're indexed the way that they are. If it turns out that the reason for this convention sheds some light on these generalized Vandermonde determinants, then I'm all ears for it. And if it turns out that it's super useful for something unrelated that won't have any bearing on what I'm doing, it's gonna be a lot easier for me to ignore the general convention and just treat them in the simpler, more intuitive way.

Because the actual object of interest for me isn't Schur polynomials, but minors of DFT matrices with determinant zero (more precisely, minors which admit full-support nullvectors). So really I'm only interested in studying Schur polynomials to the extent that they help me classify these matrices.

[near lantern]

But especially if you're reading many things outside your core focus because of their applications to it, it's not worth your time to get hung up on notational choices. People in a field are going to (try to) write in the way most suited to their field and not yours. It's sensible to translate that into a formulation that makes more sense to you and your application area if it helps you, but it's often kind of futile to expect or demand that they formulated it the same way (since it might not be the simplest formulation for how they use it).

[graceful sluice]

Oh for sure, I 100% agree. That's why I was asking the reason for the choice. It seems significantly more convoluted than the naive way to index them, which suggests there's some reason people have done so. I just wanted to know if that reason had any bearing on what I'm trying to do with them or if it's just something I can ignore for my own work and then translate whatever results I find into standard language when it comes time to publish.

[near lantern]

The only reason I can think of directly related to these determinants is that the degrees of the terms of s_l should be more directly related to l than a (because of the division by Vandermonde). It's probably safe for you to index by (ai) and switch later if you feel (li) might be better.

[graceful sluice]

Fair enough. Thanks for helping me understand!

To zoom all the way out, my work is essentially to generalize Tao's uncertainty principle as much and in as many ways as possible, and then to apply these generalizations as much as possible. So to attempt to classify spaces in which there are at most finitely many complex Hadamard matrices.

[vapid axle]

Can somebody please explain to me what exact sequences we should be taking in 6.3 to see this highlighted result? For a.c.c, we have the exact sequence 0 -> m1 -> A -> A/m1 -> 0 so it suffices to show that m1 and A/m1 are Noetherian. For m1, we have 0 -> m1m2 -> m1 -> m1/m1m2 -> 0 exact, so it suffices to show that m1m2 and m1/m1m2 are Noetherian. Continuing on this way we might be able to show that m1 is Noetherian. But I don't know how to show that A/m1 is Noetherian. I might be completely wrong

[worldly zealot]

A/m1 is a field/

?

[vapid axle]

m_1 is maximal

[digital parcel]

Ring mod maximal ideal is a field

Fields are noeth

Is what hk is getting at

[worldly zealot]

yez

amayesyes

[fierce steeple]

Does the 0 ring have a maximal ideal

More questions at 7.

I would say no fwiw

[digital parcel]

I sure hope not

[last talon]

The 0 ring is a lie

[fierce steeple]

No

I guess it is a Lie algebra yes with the trivial bracket

[last talon]

lies, not Lies

[worldly zealot]

i thought A-M assumes with unity

[ornate atlas]

Pull up the nLab page

[last talon]

0 ring has unity

It’s 0

[worldly zealot]

easter island emoji

octopus emoji

[near lantern]

I don't know much (or anything) about this topic myself, but I think it should be asked just to check: why wouldn't Tao have done this himself?

[fierce steeple]

And then do the opposite

[fierce steeple]

(Though this is also correct fr)

[ornate atlas]

Too simple to be simple

Clark went on a full rant about that in a topology lecture, shared that page with us all lol

[fierce steeple]

Lol

I like the idea but too preachy of a page iirc

According to the all-knowing nPOV...

[ornate atlas]

Yeah it really is

It’s not fundamentally wrong though

[fierce steeple]

One I do quite like is the negative thinking page

Lol

Have u seen this

[ornate atlas]

It would just be nice if it was more “this is a nice convention” rather than this is actual law

[ornate atlas]

I have not

[fierce steeple]

So like for example uh for n >= 0, a topological space X is called n-truncated if for all x in X and all k > n, the homotopy groups pi_k(X,x) vanish

This is standard

Okay so if you naively follow this, then (-1) truncated means that additiinally, for all x in X, the set pi_0(X) of path components vanishes

Which means X is either connected (and that component is contractible, so X is contractible) or empty

Lol

And (-2)-truncated by convention is then "contractible"

[ornate atlas]

I feel like I’m happy with that lol should that feel controversial

[fierce steeple]

For some reason this is occasionally useful lol

[ornate atlas]

Ok this is a bit more upsetting

[fierce steeple]

So like you can call a map of homotopy types n-truncated if like homotopy fibres are n-truncated, which corresponds to the map being injective/bijection on certain homotopy groups

Then a (-1)-truncated map is an inclusion of path components

And a (-2)-truncated map is an equivalence

I guess what I mean is here the negative numbers are super useful

Anyway this is kind of an example of negative thinking

Sometimes you can also do induction starting at like (-2)

[ornate atlas]

Is there a reason in particular beyond nice convention to define negative numbers for this? Like is there some sort of graded structure on homotopy groups or something

[fierce steeple]

Not rly like I wouldn't say negative homotopy groups of spaces have any meaning

[ornate atlas]

Yeah I guess it’s just nice convention then, I kinda expected that but you never know

[fierce steeple]

Or at least, are usually 0 by convention

[ornate atlas]

I need to start learning some homotopy theory soon

But I’ve got so much god forsaken lean to write

[fierce steeple]

Are you at Imperial

Jk I don't think u are lol

[ornate atlas]

I’m at Warwick, this is not Buzzard propaganda

He’s just managed to spread his disease

[fierce steeple]

Oop

[ornate atlas]

I don’t I’ve genuinely spent all night crashing out about this

I’ve done no maths all week because I’ve just been trying to get something to work in lean and I have not come close to succeeding

And id much rather just continue to do background reading for my diss

Anyway…

[ornate atlas]

As an actual maths question (though maybe more algtop appropriate), are you at all aware of “the squeezed model structure”? I’ve not been able to find much about it beyond a reference to a squeezed resolution

[fierce steeple]

I'm sorry eekies

No

Sorry lol

What context

[ornate atlas]

Yeah I guess lol

[fierce steeple]

Feel free to share more though

[ornate atlas]

My thesis title is “Loop space homology of classifying spaces and the squeezed model structure” and I don’t really know what that last part is

I’m guessing my advisor will give me some more details at some point but he’s currently being irritatingly aloof

[fierce steeple]

Ah yes a paper w essentially this title came up when I googled squeezed resolutions lol

[ornate atlas]

Ah maybe I’m just dumb and my Google fu is failing me lol

I found like one reference to the squeezed resolution but not much else

[ornate atlas]

Could you share the paper by any chance I genuinely cannot find this lol

[fierce steeple]

I mean this, where they are called \Omega-resolutions https://arxiv.org/pdf/1807.02353

I cannot find any reference to a model structure though

Though it seems reasonable one may exist

[ornate atlas]

I shall take a look at that tomorrow in between loosing my sanity marking and doing lean

[fierce steeple]

Glglgl

I need to learn material for a talk I am giving in a few days

https://tenor.com/view/muerto-gif-16250174877271582365

[ornate atlas]

Yeah it’s possible this is some new idea of my advisors he wants me to look into but as I said the guys being like unreasonabley hard to contact lol

[fierce steeple]

Ah ok lol bruh

[ornate atlas]

You’ve got it I believe

[fierce steeple]

Thank

[near lantern]

I want to believe this "diss" stands for "diss track", perhaps a Lean diss track.

[ornate atlas]

Kevin Buzzard isn’t ready for the heat coming his way

[near lantern]

I'm guessing this is an instant corollary of Cohen's structure theorem or something similar, but for k a field and n a non-negative integer is there a classification of Artinian local rings with residue field k, of "order" n in the sense that the (n+1)^st but not n^th power of the maximal ideal m is 0, and whose tangent space (i.e., m/m^2) has dimension 1?

[lone jacinth]

If k has characteristic 0 then k[x]/(x^n+1) is the only option.

If k has characteristic p you can also have a Cohen ring modelo p^n+1. So then it's just Cohen structure theorem yes

[near lantern]

Skimming the statement of Cohen's structure theorem, it looks like formal power series cover the equicharacteristic case and Cohen rings (is W(k) the unique one-dimensional Cohen ring with residue field k?) cover the unramified mixed-characteristic case, there is still the ramified mixed-characteristic case.

[lone jacinth]

I'm not sure what you mean with ramified vs untamified case here.

Here is how you can construct your ring btw
https://stacks.math.columbia.edu/tag/03C3

[near lantern]

Will read in more detail later. By "unramified" I mean that p generates the maximal ideal rather than a higher power of it, in analogy with say the ramification of a finite extension of ℚ_p.

[lone jacinth]

That doesn't seem possible to me.

[near lantern]

ℤ/p^2ℤ and F_p?

Oh wait

No not that

[lone jacinth]

Wait you mean like p having a square root

I didn't think about that case

[near lantern]

Z_p[sqrt(p)] (or maybe that for p = 3 mod 4 and Z_p[(1+sqrt(p))/2] for p = 1 mod 4) its integral closure) should be a DVR with residue field F_p and valuation(p) = 2, so not a Cohen ring.

[lone jacinth]

Alright, so if x generates the maximal ideal then every ideal is off the form (x^n), in particular (p) is.

So you'll have a Cohen ring R and then take R[x]/(x^n - rp) for a unit r.

That should then cover everything

[lone jacinth]

Hmm, actually r could be a unit of the larger ring, not just of R. That would still just be r plus some multiple of x though. So
R[x]/(x^n - f(x)p)
where f has degree < n and invertible constant coefficient.

[hushed bone]

Yes,

The ramified case is always an Eisenstein extension of some W(k)[[some variables]]

And yes W(k) is the unique, but not up to unique isomorphism, cohen ring with residue field k

You can do some shit to lift some maps to show these guys are isomorphic, but there’s arbitrary choices made so there’s nothing functorial or nice about the maps

[broken wren]

Isn’t the only failure of unique isomorphism due to symmetries of k? So the map W(k)->k is unique up unique isomorphism of maps to k, ie, over the identity?

[hushed bone]

I don’t remember to that detail, I just remember that it wasn’t unique

[broken wren]

I think this is easy to see from Teichmueller lift:
k^* -> W(k)^*

[timber gazelle]

Let M be an finitely generated A-module with A being an integral domain. How does one prove that the kernel of the map f: M -> M^** given by f(m)(g) = g(m) is the torsion module T(M)? I can show that the torsion module sits inside ker(f), but proving that ker(f) sits inside the torsion module seems to be quite hard. If you pick m in the kernel of f, then for any g in M^* you get that g(m) = 0. Now somehow using this you should be able to construct a non-zero a in A such that am = 0. How does one cook up this a?

[broken wren]

Use the fact that it’s a domain. Use the field of fractions

[lone jacinth]

It's probably easier to show that any m not in T(M) is also not in the kernel.

I.e. show that M/T(M) can be embeded into A^n

[timber gazelle]

Why does the field of fractions make an appearance?

[past cove]

you can show that the dual is isomorphic {c\in K | cM \subseteq R}

but this doesn't feel very elementary to me and I was trying to think if there's a different way to prove this

[timber gazelle]

This kinda translates the problem to showing that instead of there existing an a in A with am = 0, one would need to exhibit a map h in M^* with h(m) = 0, but it still seems that one needs to actually construct either one of these

[lone jacinth]

I'm suggesting constructing a map with h(m) nonzero yes

Well "constructing". Showing it exists is really what I'm suggesting

[timber gazelle]

Are you claiming that this is for some reason easier than cooking up an element a that gives am = 0?

[lone jacinth]

At least I'm claiming that I know how to do that, but not really a way of finding this a.

It's probably possible to do it in a way where you show this a exist, I can try to think of a method if you prefer that

[past cove]

just construct h and dualize /j

[lone jacinth]

Like to construct this a you would need to use that h(m) = 0 for all h. But you don't really know very much about what all possible h is, so it seems much harder to deal with than just a single h.

I guess that's sort of the initiation for why one method is easier

[past cove]

maybe you can do a contradiction?

but ig that's what you're suggesting

[timber gazelle]

I'm really up for either one of these, but I would just like to understand the pieces of the argument. I cannot seem to be able to do this from first principles.

[lone jacinth]

Sort of the same idea would be let K be the field of fractions of A. Are you able to construct h from M to K such h(m) is nonzero?

(it might be easier to construct the map from M/T(M) and use induction)

[near lantern]

Ah, that's what I guessed. Neat.

What conditions should a cocomplete abelian category K satisfy so that for an object A of K, Hom(A, -) preserves arbitrary direct sums iff A is finitely generated (in the sense that if it is a directed union of subobjects it is equal to one of them)?

The following conditions are sufficient (which is more or less straightforward to show):
(i) a direct limit of subobjects is still a subobject (which holds in particular if directed limits are exact). In this case the direct sum is a subobject of the direct product.
(ii) The preimage by a map, as a homomorphism of subobject posets, preserves directed suprema.

Does (ii) follow if directed limits are exact?

[near lantern]

OK yes we can keep two objects the same in the pullback diagram and colimit over the last.

[broken wren]

In the category of modules over a ring that is not noetherian, is this true? It isn’t finite presentation?

[lone jacinth]

It is true.

Notice any particular element lands in a finite direct sum, so if you're finitely generated everything lands in a finite part.

Being finitely presented is equivalent to preserving direct limits. And being finitely generated is equivalent to preserving direct sums. This holds in ModR for any ring

[lone jacinth]

Hmm, so if you look at the sequence
0 -> K -> Sum -> Prod
and apply Hom(A, -) you get that Hom(A, K) = 0. So assuming the category is generated by fg objects it is at least necessary for the sum to be a subobject of the product.

[fierce steeple]

Is there a name for this property of an object more generally? Surprised I've not seen it

I mean there is this notion of being connected but it is kinda bad terminology here + the things are enriched homs

[lone jacinth]

I've only ever seen it in contexts where it is equivalent to fg (or compact). In which case people just call it that

[fierce steeple]

Ah ok

Sure

from what i could find online it seemed that "[this property] => fg" requires Noetherian but not conversely - is that right

Maybe there are silly counterexamples lol

[lone jacinth]

I'm not sure what you mean exactly. Requires Noetherian in what sense?

Oh, you mean in ModR for Noetherian R. Yeah then fg implies this, but not conversely in general

[lone jacinth]

In the triangulated setting this property is often called compact.

I don't know much about direct limits in triangulated category, so can't say if it's equivalent to the usual sense of the word compact, but it would imply "preserving" homotopy colimits at least.

[fierce steeple]

Compact is standard for this yeah with preserving filtered homotopy colimits

but that should be different to preserving just sums

[fierce steeple]

Yeah that's what I mean, sorry

(Of course if the module is Noetherian it is fg aha)

[broken wren]

Call it weakly compact

[lone jacinth]

I mean, in a triangulated category the hom-complex Hom(A, -) always takes triangles to triangles right?

So the only thing missing to preserve homotopy colimits is to preserve direct sums.

[fierce steeple]

true

[lone jacinth]

Here's nlab

[vapid axle]

Can I please get some help here

[fierce steeple]

Yeah it's just compact has some nonequivalent definitions from what i remember

The correct to me definitely seems to be preserving filtered colimits

[vapid axle]

Oh nvm

[north stream]

did you get it?

[lone jacinth]

I would prefer
finitely presented = preserves filtered colimits
compact = preserves direct sums

[broken wren]

Why? Preserving direct sums isn’t useful

[lone jacinth]

Why not

[vapid axle]

Yeah. If we assume a.c.c for each factor, then it suffices to show that m1 is Noeth, for 0 -> m1 -> A -> A/m1 -> 0 is exact (and A/m1 is Noeth). Then it suffices to show that m1m2 is Noeth, for 0 -> m1m2 -> m1 -> m1/m1m2 -> 0 is exact and m1/m1m2 is Noeth by assumption. Continuing on this way it suffices to show that m1m2...m_{n - 1} is Noeth, for which it suffices to show that m1m2...mn is Noeth (for the sequence 0 -> m1m2...mn -> m1m2...mn-1 -> m1...mn-1/m1...mn -> 0 ix exact). But m1m2...mn = 0 so it is Noetherian. That's at least one direction lol

[lone jacinth]

That's how compactly generated triangulated categories are defined

[broken wren]

Finite generation isn’t useful. You always want fp

Also, compact comes from topology, where direct sums are way too weak. You want to say that the sphere is compact

[broken wren]

But that’s only because weakly compact objects are truly compact in the triangular setting

[lone jacinth]

Truely compact meaning preserving filtered homotopy colimits or filtered colimits?

I just think the definition
"Map to direct sum factors through finite direct sum" matches well with "cover reduces to finite cover", and is the definition you use in practice, so why not let that have the name compact?

Especially when fp already has the name fp

[broken wren]

Maybe in the triangular setting, you don’t have filtered colimits or homotopy colimita, so you are forced to define by sums. But what’s really going on in the structured setting is preserving homotopy colimits

The word compact should be reserved for the useful concept. fp is useful. fg is useless

[lone jacinth]

I'm not really sure I see it. If you're forced to use a definition then surely that means the definition is useful.

So then what you're suggesting would just be using the same word for two different definitions depending on the context.

Which is fine I guess, but like... unnecessary

[broken wren]

The convention is to use compact for both categories and infinity categories to mean preserving filtered colimits, where homotopy is implicit

You propose to define it to mean something that is only useful in the stable setting where it implies the full property

Is there any setting in which you are forced to use fg modules, not fp modules?

[north stream]

And the same logic proves that is artinian.

[lone jacinth]

Well I'm more so thinking abelian vs triangulated categories than categories vs infinitely categories.

Sometimes you do want to think about filtered colimits in triangulated categories, so then redefining them to mean homotopy colimits seems like a problem.

I guess it doesn't matter to much, but in my circles compact objects are usually just talked about in triangulated categories in which case it's the sum definition.

[broken wren]

No, you never want to think about filtered colimits in triangulated categories as categories.
Never.

Yes, you define compact objects in triangulated categories by infinite sums, just as you define compact objects in noetherian categories as fg modules. But that doesn’t mean it’s the right definition

[fierce steeple]

If someone said filtered colimit in a triangulated category i would assume they meant homotopy lol. what's an example where they come up (or even always exist)

[fierce steeple]

No, you never want to think about filtered colimits in triangulated categories as categories. Fixed it

[broken wren]

I didn’t say that because there is a movement to redefine triangulated category to mean stable infinity category. You can see hints of this in terms like pre-triangulated

[silver goblet]

are there any triangulated categories that actually matter which aren't homotopy categories of a stable \infty-category?

[broken wren]

No

[muted sierra]

Why?

[latent crypt]

A triangulated structure is what is left over when you take the homotopy category of a stable infinity category. They are hard to work with whereas stable infinity categories are very nice. Triangulated structures are unnatural, clunky, archaic and reactive, whereas stable infinity categories are nice, natural, unbothered, moisturized and in their lane.

[muted sierra]

I'm pretty sure I've seen Urs Schreiber say this somewhere (with different words, obviously).

And I'm sure he and you must have your theoretical reasons why working with infinity categories is nicer. But I'm dumb, and I just want to do computations to solve specific problems. All this fancy schmancy theory goes way over my head.

[broken wren]

Computations with the octahedron axiom are a pain

[muted sierra]

Please correct me if I'm wrong, but IIUC, any functor from an infinity-category to an ordinary 1-category factors through the former's homotopy category, right?

So if I want to extract non-infinity-dimensional information about an infinity-category in a functorial way, then I'm unavoidably using the homotopy category anyway, right?

[fierce steeple]

Yes

[broken wren]

Sure, but when you use the triangulated structure, you’re using something other than the homotopy category

[fierce steeple]

Ig to me like even for working with the homotopy category ultimately and wanting to form some construction it can be more helpful to think in terms of universal properties etc

Or indeed near indispensable for some things

I should say my original comment was of course tongue in cheek

[muted sierra]

Ah, this is the Urs Schreiber post in question: https://golem.ph.utexas.edu/category/2010/02/intrinsic_naturalness.html

It reads very much like propaganda, though.

[wary elbow]

lol I've heard so much about how infinity categories are The New Hotness over the years that I am genuinely unable to tell what is a parody of category theorists vs the real thing

[fierce steeple]

Ah lol

[fierce steeple]

Tbh I would say "people using infinity categories" and "category theorists" are often two quite different things

[latent crypt]

This is a good point. In triangulated categories, the distinguished triangles are additional data and there's nothing special about them in the sense of universal properties. In the stable infinity category lying over the triangulated category, extending a map to a distinguished triangle on the right is just taking a cokernel and to the left is taking a kernel. You can then use the infinity categorical universal properties of the kernel and cokernel to work with maps in and out of different parts of the distinguished triangle.

The definition of a stable infinity category is that it is an infinity category in which certain limits and colimits exist and certain diagrams are both limit and colimits diagrams. It is a property rather than extra structure.

[lone jacinth]

Tbf being an infinity category is quite a lot of extra structure

[latent crypt]

Shush

It's structure that should've been there in the first place

[plucky arch]

Is it at all sensible to think of chain complexes as tangent vectors

[ornate atlas]

What inspired this thought?

I don’t immediately see a reasonable way to think of that, but I’m by no means an expert and haven’t considered this before

[muted sierra]

Tangent to what anyway?

Chain complexes are just chain complexes. Of course, specific chain complexes might be built with specific geometric motivation, but why would that be a good way to think of all chain complexes in general?

[broken wren]

Goodwillie sees the category of chain complexes as tangent vectors to the category of derived rings

[past cove]

Goodwillie should've been told that some thoughts are meant to remain on the inside

[past cove]

idrg what this type of thinking gets you in practice
Chain complexes are sort of well behaved enough where I don't see a reason to try and give some weird geometric picture for them

and I'd consider chain complexes to be one of those fundamental ideas in math that can't really be massively simplified
It's already basically fundamental to so many things that it's just another building block instead of it being an example of something already out there imho

[past cove]

de Rham cohomology I'd maybe assume?

[ornate atlas]

This was my initial thought but then are you really thinking of the chain complex as a tangent vector?

My feeling is the same as yours that a chain complex already isn’t like that much structure* I’m not sure you can simply it much more than that

[scarlet ermine]

maybe a reason psuedo is guessing this is because d^2 = 0 looks like how a (k-valued) tangent vector on a scheme is a map from k[e]/e^2, where this works out because you are in some sense asserting “epsilon squared is really small”

or i’m guessing (maybe psuedo has deeper reasons)

[silver goblet]

does this have something to do with the theory of cotangent complexes of E_\infty rings? I know basically nothing about this but just curious

[scarlet ermine]

https://mathoverflow.net/a/68635/525690
this seems neat, would like to understand this

[muted sierra]

That sounds more like philosophy than anything else...

[scarlet ermine]

i mean sure yeah I just think it seems neat

[broken wren]

The Goodwillie derivative applied to E_oo rings or simplicial commutative rings gives the cotangent complex. Applied to DGAs it gives Hochschild homology

[muted sierra]

I mean, the MO post, not what you said earlier about tangent vectors being maps from k[e] / (e^2).

[silver goblet]

huh that is funky. i think this just means i should start learning about goodwillie calculus haha

[plucky arch]

Some comment earlier in this channel about chain complexes being functors from Z[x]/(x^2)

Sorry a module

#advanced-algebra message

[plucky arch]

That’s strange to me because the chain complex def seems fairly complicated for a “fundamental” concept

[plucky arch]

Yes exactly

[fierce steeple]

I am not sure this quite works for simplicial commutative rings? Maybe I am wrong

[plucky arch]

oh it's popato

[fierce steeple]

My understanding was that there is one notion of cotangent complex for simplicial commutative rings where you stabilise etc but it is distinct from what you get by deriving the Kähler differentials

[muted sierra]

Is this cotangent complex business in any way related to the associated graded ring of a local ring (A,m), which is the coordinate ring of the tangent cone of Spec(A) at m?

[broken wren]

They are orthogonal

[muted sierra]

Orthogonal?

[broken wren]

Maybe independent would be a better word

[plucky arch]

i'm trying to understand the conditions under which it makes sense to think about complexifications of lie algebras

the lie algebra obtained from a lie group is always real, i think

however, if we're considering complex representations of the lie algebra, then it makes sense to take the complexification...?

[broken wren]

For any real Lie algebra you can tensor with C to get a complex Lie algebra. The complex representations of the original are the same as the complex representations of the complexification

If your Lie group has a complex structure, then its Lie algebra has a natural complex structure already. You can forget the complex structure and complexify again

[plucky arch]

gotcha, this makes sense

[wary elbow]

As a concrete example: if you want a unitary representation of a real algebra, it can sometimes be more straightforward to construct a representation of the complexification which is unitary when restricted to the real algebra. This is one way of constructing the Fock representation of the Heisenberg group.

This is also hugely important for the classification of real semisimple Lie algebras and groups. The classification relies fundamentally on embedding the real algebra in its complexification, which is a complex semisimple Lie algebra, and then using constructions in the complex algebra.

[plucky arch]

currently trying to understand lie's theorem on solvable lie algebras

the proof seems fairly involved, idk if there's a particularly nice way to do it?

[broken wren]

There’s an easy proof for the nilpotent case. For the solvable case you need characteristic zero, so you can’t expect anything too nice

[plucky arch]

hm ok

[pastel agate]

I'm tentatively interested in taking a reading course at my university next semester on semigroups. I've been told that prerequisites are minimal, needing only basic algebraic structures and some combinatorics. I'm confident on the algebra side of things but I haven't ever taken a course in combinatorics, does anyone know the extent to which I need to be comfortable with it in order to study semigroups?

My university does offer a course in combinatorics but I simply never had the opportunity to take it unfortunately

[distant harness]

"Some combinatorics" sounds fairly minimal; it's not necessarily more than you can investigate in half an hour and go "yes, of course" when you come to it.

If you can wing it through problems like "count the number of possible poker hands; how many of them are two pairs?" that's most likely enough.

[pastel agate]

Alright that's fair, thanks

[plucky arch]

This is a bit more of a philosophical Q, but

In algebra, when should I expect there to be an explanation for “why” something is true

For an example of what I mean, I remember quite early on in my group theory course doing the classification of all groups of order 8

[distant harness]

As somehow distinct from a proof?

[plucky arch]

I don’t remember all the details, but it felt like a lot of case-bashing

[plucky arch]

Yes so what I mean is something like - there are some proofs that tell me that a certain result is true, but (subjectively) don’t tell me why a certain result is true

[plucky arch]

For this, I’m not sure I have a good sense of “why” there are only 5 groups of order 8 up to iso, even though I could tell you “that” this statement was true and prove it

Idk if I’m making sense with that

Maybe a simpler example would be associativity of matrix multiplication?

[distant harness]

I'm not sure there is any better "why" than the fact that those happen to be the groups that turn up if you do an exhaustive set for group operations on a set with 8 elements, and the case-bashing proof amounts to a lot of shortcuts in that exhaustive search.

[plucky arch]

One approach to proving this is a brute force computation

Another approach is to show that it’s a conjugation of linear map composition, and hence inherits associativity

Both approaches tell you “that” matrix multiplication is associative, but somehow I feel only the second approach tells you “why” it’s associative

[plucky arch]

I see

[ornate atlas]

Yeah I agree with tropo here in that I view this as basically a combinatorial result. If you only have 8 elements, there’s only so many choices to make, and it’s small enough that working out all the cases isn’t so unreasonable.

For the matrix multiplication thing, I think I view that as “well matrices are supposed to represent linear maps, and I generally expect maps to be associative”. I’m not sure if that’s much of a “why?” It’s more why morally we want/expect it to be so but yeah

I’m not sure there’s any great answer to give here, or at least one specific to algebra

[worldly river]

when constructing the algebraic closure of a field, in point ii), rotman says that "It now follows from induction on deg f that f(x) splits on K" I don't see how this induction works, as you just showed that f has a root in K[x] but then you don't know that it can be factored in k[x]. Also, in the next paragraph, he just says "copy the construction so that every polynomial has a root", so maybe this is a mistake? Otherwise I don't really understand the proof

I read online that he had a mistake in the proof so maybe he forgot to edit that part out?

[lone jacinth]

So it actually is true that every nonconstant polynomial splits in K. But it does not follow by induction, the proof is somewhat complicated.

However it is true that you can finish the proof just by repeating the construction, so you don't need this complicated part

[hushed bone]

So I looked it up and see that this is proven by Gilmer

But does that automatically mean that K is algebraically closed? I'm pretty sure the answer is yes, but you'd need to show every polynomial over K has a root in K, but a priori you only know that every polynomial over k splits completely in K

how do you go from that to polynomials over K?

is it because all the coefficients are alg over k and then you do some bs algebra?

[lone jacinth]

Yes, for a given polynomial with coefficients in K just take the product with all its Galois conjugates (counting itself several times in the non-seperable case) to get a polynomial over k

[hushed bone]

okay yah

cool

wait uh

[lone jacinth]

Or I guess possibly simpler, just adjoin a root over K and take it's minimal polynomial over k

[hushed bone]

Sorry, there might be infinitely many automorphisms

So do you look at the finitely many ones which do anything to at least one coefficient

[lone jacinth]

There won't be because the extension generated by the coefficients is finite

[hushed bone]

and then take the product over that?

[lone jacinth]

Anyhow, whit this approach you don't need Galois theory

[hushed bone]

yeh

cool

[lone jacinth]

If k is perfect you can use the primitive element theorem to show that K is alg closed.

But the imperfect case it's quite complicated as I remember

[hushed bone]

it looked like what you did was adjoin all purely inseparable elements or something

[lone jacinth]

Yeah, there's something like that to reduce to the perfect case, but I can't remember exactly how it goes

[hushed bone]

surfe

[lone jacinth]

I have read this proof before
#groups-rings-fields message

[candid igloo]

Please tell me how to do the last step

[limpid horizon]

Question meant to say only finitely many max ideals containing a nonzero ideal in R?

[candid igloo]

Yes, because every element is contained in a finitely many maximal ideals

[limpid horizon]

how does "only finitely many maximal ideals" contain any r in R mean only finitely many maximal ideals contain an ideal I?

[candid igloo]

If infinite maximal ideals contain I then every element of I is contained in infinite maximal ideals contradicting problem statement

[limpid horizon]

ty

[candid igloo]

Can you solve it please @limpid horizon

[limpid horizon]

well im looking at it rn

[still ledge]

😭 idk

[past cove]

this isn't algebra

[still ledge]

idk 😭 👍

[restive vale]

I've known about Frobenius reciprocity for a while, even when G is infinite. Recently, I have been asked about some geometric interpretations of rep theory as a whole (using G-structures etc.) and kinda feel like I don't know anything about this. Is there a nice place to read about G-structures for the sake of rep theory?

[wary elbow]

What do you mean by G-structure? Do you mean a G-action on a variety or scheme? If so, Jantzen has a thorough overview in Representations of Algebraic Groups, including connections to Frobenius reciprocity.

[forest turtle]

bruh why are tensor products of chain complexes/maps so hard to work with

[latent crypt]

To torment you specifically

[scarlet ermine]

This doesn’t change the fact that they are very annoying to work with and maybe you don’t need/want a paragraph, but here’s the way I like to think about it! I’ll consider a chain complex as a graded abelian group A with a degree 1 morphism d: A->A(1). Then, given two chain complexes A,B, you form the tensor product graded abelian group, and then to get a degree 1 map you try to do d (x) 1 + 1 (x) d.

The really annoying part is keeping track of signs for the graded tensor product of graded maps… but I very recently figured out how to remember lol, maybe everyone else knew this but I was just memorizing the signs until now—it’s about counting how many times you move symbols past each other, where the degree is the “number of symbols”.

So if A,B,C,D are graded abelian groups, and f:A->C and g:B->D are graded morphisms, then

(f (x) g) (a (x) b) = f(a) (x) g(b)

BUT you throw in a sign for swapping the symbols g and a, (ie, we went from fgab to fagb if I drop all the parentheses and tensors) where you think of g as |g| symbols, and a as |a| symbols, so you could accomplish this with |g||a| transpositions, incurring you with a sign of

(-1)^{ |g| |a| }

[round ocean]

can this channel be used for asking for abstract algebra help?

[vague pawn]

Yeah, depending on what you want to ask #groups-rings-fields might be better

see channel descriptions

[round ocean]

Alright thanks

[late marten]

Hello. Sorry for interrupting discussions like this, but someone from this server sent me a bunch of exercises and solutions on properties of direct and inverse limits. I unfortunately lost him from my dms ? Can you kindly send it again or atleast ping me up ? Thank you !

[urban granite]

I think this is false? let R=k[x_1,x_2,...] and take I=(x_1,x_2,...) and J=(x_2,x_3,...). Then I\cong J since we can just shift variables by 1 but there does not exist a,b\in R\setminus 0 such that aI=bJ?

[lone jacinth]

If I'm interpreting what "shift variables" means correctly, then this is not an isomorphism of R-modules

(a map taking x1 to x2 and x2 to x3 is not R-linear)

[urban granite]

ah

nvm (it was a trivial exercise i was just overthinking about it )

[eager rock]

There are many conditions of the form “A commutative ring is Artinian if and only if it’s Noetherian and X.” But what about conditions that don’t reference Noetherian?

[ornate atlas]

Any Artinian ring is Noetherian

This is a theorem of Hopkins and Levitsky

[distant harness]

There could still be conditions that don't explicitly say "Noetherian".

[digital parcel]

A commutative ring is artinian iff it satisfies acc on ideals and is dimension 0

Amen

I think there’s an iff with finite length

Like A is artinian iff it’s a finite length A-module

Yeah https://stacks.math.columbia.edu/tag/00JB

[lone jacinth]

A ring is artinian if A/J is semisimple and J is finitely generated, where J is the nilradical

[worldly zealot]

iff?

[lone jacinth]

Finite length is also equivalent to artinian+Noetherian in general

[digital parcel]

Yeah I was just saying one that avoids explicitly saying “Noetherian”

[digital parcel]

This is cool didnt know this one

[lone jacinth]

Not avoiding Noetherian, but artinian rings are exactly Noetherian perfect rings.

[eager rock]

Do you have a reference for that?

[lone jacinth]

I might be able to dig up a reference, but it's not too hard to prove yourself.

It's also how the standard proof that artinian rings are Noetherian goes

J is fg + nil so nilpotent.

So A is filtered by A/J, J/J^2, ... which are all finitely generated over the semisimple ring A/J, so have finite length. Conclusion A has finite length.

[lone jacinth]

Conversely let A be artinian and J the Jacobson radical (which we will prove equals the nilradical).

Then as A is artinian we can write J as the intersection of finitely many maximal ideals. Then by CRT A/J is semisimple.

Again by artinian we know J^n = J^n+1 for some n. If J^n consider the set of ideals I in J^n such that J^n I is nonzero. This must have a minimal element by artinian. And a minimal element is necessarily principal, so in particular finitely generated. But then by Nakayamas lemma J^n I is a smaller ideal contradicting minimality.

Hence J is nilpotent (so must be the nilradical). And as A/J, J/J^2, J^2/J^3, ... are all semisimple artinian they must have finite length. So A has finite length, hence A is Noetherian.

[lone jacinth]

Here's a fun one I thought up:

Every nonzero fg module has a nonzero fg socle.

Why this implies artinian:
Let A be non-artinian. Then there exists a decreasing sequence of ideals I1 > I2 > ... Let I be the intersection. Then A/I is fg, let's assume it still has fg nonzero socle.

As the socle is semisimple and fg it must have finite length. Intersecting In with the socle we get a sequence with intersection 0, so there must be an In that doesn't intersect the socle.

But if In/I is nonzero it has a nonzero fg submodules, which must have a socle, which would then also be a part of the socle of A/I. So In = I and A is artinian.

Conversely if A is artinian every fg module has finite length, so easy peasy

[steady lance]

I like this a lot lol

I like the idea of someone who's uncomfortable thinking about noetherianness but somehow inexplicably has intuition about socles

[lone jacinth]

And I guess for one that actually used commutativity: spectrum is a finite discrete space and nilradical fg

[lone jacinth]

I mean it's not really that weird. Socles are sort of a big part of artinianess so if you're doing something like rep theory of artinian rings you might be exactly such a person.

The proof that artinian implies Noetherian is built around similar ideas, so someone learning that proof might be that person

[steady lance]

yeah I guess you're right

[limpid horizon]

but Jagr... Jagr is every kind of person

[lone jacinth]

🎵 we are the world 🎶

🎶 we are the people not comfortable with noetherianness but inexplicably comfortable with socles 🎶

We are the ones who make a brighter day

So let's start mathin'

[steady lance]

My personal math playground just always assumes noetherianness and is characteristic-invariant

I refuse to consider anything beyond that

[lone jacinth]

My playground always assumes finite length, so specifying noetherian confuses me

[steady lance]

so based

[lone jacinth]

Another I was thinking about, but I wasn't able to convince myself implies artinian was that the category of finite length modules has a projective generator

[limpid horizon]

W

[olive schooner]

the best playground (guessing reps of artin algebras?)

[eager rock]

For the record I’m not uncomfortable thinking about Noetherianness, it’s just that for the work I’m doing a Noetherian characterization won’t suit my purposes.

[lone jacinth]

Ooo, what's your purpose?

[eager rock]

Mathematical logic

@lone jacinth Noetherian is a second-order property of rings, and I’m trying to see if Artinian can be made into a first-order property of rings.

[fierce steeple]

Based lol

[lone jacinth]

I see. I would guess they can't, but maybe there is a good first order "approximation"

[eager rock]

I would guess they can’t too.

[steady lance]

oh yeah I didn't see what the broader conversation was about. Just saw that one message and was amused. Hope it wasn't taken as an insult haha

[ornate atlas]

Interesting. I admittedly know marginally more than nothing about logic, but is there a reason you might suspect something could work for Artinianness but not Noetherianity? To me they exist as very similar properties so I guess I’d be surprised if they were different in that way

But I also know so little about logic that that could be a very uninformed opinion. Well I guess it is lol.

[lone jacinth]

How about this:

Say Pn is the statement that there exists
x1, ..., xn such that
x1 is nonzero, x[i+1] is not in the span of x1, ..., xi.

Then for every artinian ring there will be an n such that it doesn't satisfy Pn. Hence a non-artinian ring can never be elementary equivalent to an artinian ring.

That's something at least

[lone jacinth]

I mean being artinian is a much stronger property

[ornate atlas]

I suppose yeah, all superficial similarities aside

[lone jacinth]

I feel like there should be some kind of argument like look at k[x1, ...]/(x1, ...)^2 and any first order formula. Then something something finitely many variables in the formula means you can reduce to a finitely generated subring which is artinian.

Not sure about the something something part though

[lone jacinth]

What's the proof that Noetherian is not first order?

[unborn rampart]

btw, I just found out that for modules Artinian doesn't imply Noetherian, unlike the case for rings is there any intuition for this discrepancy? I had the impression that rings and modules were kinda two sides of the same coin

[limpid horizon]

upon the witnessing ....

[lone jacinth]

Well, a ring A being artinian is just the module A being artinian. But the module A is a lot nicer than just any old module.

For one it is finitely generated

[vast estuary]

Well think in terms of fields and vector spaces maybe.
Just because there is some property that holds for fields, doesn't mean it should hold for vector spaces

[unborn rampart]

Hmm yeah, good points

[lone jacinth]

For a commutative ring at least artinian+finitely generated implies Noetherian

[eager rock]

Basically you can encode any tree in the set of finite sequences of natural numbers into an integral domain. Whether such a tree has an infinite path is equivalent to whether the resultant integral domain is Noetherian. So if Noetherian was a first order property, then whether such trees have infinite paths would also be a first order property. But it’s a theorem of mathematical logic that trees having infinite paths is a second-order property.

[lone jacinth]

Hmm, I see. Maybe there is a similar construction for perfect rings or something then

[vast estuary]

(ams) \def\bZ{\mathbb Z}\def\U{\mathcal U}

@eager rock okay here's an idea.
If the theory of artinian rings were first order definable, its class of models would be closed under ultraproducts.

Now, take a field $k$ and consider the algebras
$$ A_n = k[\bZ/n\bZ] $$
(Ie the group algebra of the cyclic group).
These are all finite-dimensional $k$ algebras, and as such are Artinian.

Now let $\U$ be a non-principal ultra filter on $\omega$, and let $A$ be the ultraproduct of $A_n$s.
If I'm not mistaken, $A$ should be isomorphic to $k[\bZ]$, which is not Artinian.

[broken turtle]

ΡΟΟΡ ΡΙΡΕ ΒΟΜΒ

[vast estuary]

I haven't fully checked everything about this, so I could be totally wrong

[lone jacinth]

Cool idea. Perhaps a simpler example:

Take An = k[x]/(x^n).

Then the element that is x for each n satisfies 1 - rx being invertible, so is in the Jacobson radical. But it's not nilpotent since for any m, x^m is only zero for finitely many n. Hence the ultra product can't be artinian.

[vast estuary]

Ooo nice

[lone jacinth]

It feels to me that this ultraproduct should be more complicated than kZ, but I don't really have a good reason.

Did you have (part of) an argument in mind?

[vast estuary]

Oh yeah oops I don't think it's kZ. Made a silly mistake

[lone jacinth]

I guess an easy argument is that having zero divisors is a first order property

And kZ does not

[lone jacinth]

This ring must be pretty horrible then I guess

[vast estuary]

Ah true

[vast estuary]

I still think it works as a counterexample

Ugly or not

[lone jacinth]

Yeah the argument seems sound

[analog abyss]

What is a good place to learn about the representation theory of infinite discrete groups, especially free groups and free products of finite groups?

[woven loom]

Easier than that is just take an ultraproduct of Z[x1, …, xn] along n, which should have a copy of Z[x1, …] in it?

I think that should kill that anyway since you should have those ideal issues

Now, important to note is there’s some infinitary axioms to give you conditions like PID and noetherian and such iirc

At least especially in commutative settings for noetherian anyway

Also some related properties, which I believe some of this is in an exercise in Hodges Model Theory, chapter 2

[woven loom]

Anyhow, see also Prest’s Model Theory and Modules, which has some extra tidbits on rings <-> first order stuff on modules, and other work by Prest around purity and spectra and definable category things

[woven loom]

This I think is curious around things being essentially second order, encodings in theories, etc, but that’s neither here no there

Ideals in subrings are a tad off so maybe something to be said around locality of artinian, as in closure in ultrapowers

Since what jagr said about elementary equivalence and all

[unkempt crane]

bumping this up

[torn harbor]

Tbh I'm not aware of any focused books, but some representation theory of discrete groups shows up in some ergotic theory books and in certain subareas of functional analysis at least for the unitary ones

Representations of discrete groups iirc are just very complicated

I think lattices within lie groups are a bit better understood so you might also be able to look in that direction

I'm not really an expert but those are maybe some places you could look further

[analog abyss]

Thank you! I'm contemplating looking into Einsiedler-Ward's new book on Unitary Representation Theory for these purposes, though that is a bit more focused on Lie Theory.

[elfin ice]

This might be dumb but
How should I verify that, a scalar multiple of the Killing form on sl,so,sp, which is tr(XY)=B(X,Y)*t, is invariant under lie algebra automorphisms

After its verification, I may proceed to calculate the constant coefficient for these three lie algebras

[summer quest]

Trace is independent of basis

[nimble orchid]

Hi, is there anyone here expert in Lie groups theory and representations? I have to prove that in the Poincarè group S^(-1)(a)=S(a^(-1)), where a is an element of the group for a field psi(x) (satisfying a covariant equation) and a is the related transformation for x (a general isometry in the Minkowski space). What can I do? The problem is that in my QFT book the property Is given without proof. It seems It Is due to the fact S(a1•a2)=S(a1)•S(a2) () because if so aa^(-1)=1 and S(a)S(a^(-1))=1 because S(e)=1, but why the omomorphism () exist?. If you tell me how to write here in latex, I' ll put the question in a formal way.

[elfin ice]

I understand that for B(X,Y)
I can't make sure of that for tr(XY)

[plucky arch]

Why is it the case that “abelianisation” is often considered an analog of “linearisation”?

As a physicist i use linearisation all the time for example

But I haven’t yet appreciated in what way these concepts are similar

[summer quest]

this can mean a lot of different things, but one way in which this is usually meant is that one typically likes to replace more complicated "nonlinear" objects or categories of such (e.g. topological spaces) with less complicated "linear" approximations of these objects (e.g. Abelian groups or modules, and chain complexes of such)

[plucky arch]

Yeah the way I’ve seen it used, “linear” seems to refer to abelian categories, and “nonlinear” seems to refer to the rest

[summer quest]

one way in which the latter are more "linear" is that objects like Abelian groups and modules form Abelian categories and those are categories in which you can do linear algebra so to speak

[plucky arch]

I’d be interested to hear what you mean by “do linear algebra”

I don’t find myself computing determinants or eigenvalues or eigenvectors or traces much when working in abelian categories

But maybe I haven’t done enough work with them

[summer quest]

yes but you can state all of those constructions in such categories that's the point

[plucky arch]

Hm are you saying that the internal language of an abelian category is sufficiently powerful to express linear algebra?

[summer quest]

it's like how toposes are the kinds of categories with the kinds of structures and properties necessary to interpret set theory in some sense yes

[plucky arch]

Mhm mhm

[summer quest]

of course the kind of set theory that makes sense in this level of generality is much more general than the initial examples which motivate the definition, the same is true for the kind of linear algebra I am talking about in Abelian categories

[plucky arch]

Sure sure

[summer quest]

I guess Abelian categories are also not the only setting that you might call linear, various stable infinity categories are like this too

[plucky arch]

Hm yeah I have heard stabilisation also referred to as “linearisation”

[summer quest]

one thing that makes these situations behave more linearly so to speak is that finite products and finite coproducts coincide, and kernels and cokernels satisfy certain properties similar to what you see from linear algebra in a certain level of generality

when finite products and finite coproducts agree like this then you automatically have a good notion of matrices

[plucky arch]

Right, because you can decompose morphisms into matrices

And the way they compose is exactly matrix multiplication

[summer quest]

right exactly, and this fails quite badly without this kind of property involving biproducts

[plucky arch]

Yeah in general you can represent a morphism by a “matrix” if you’re going from a (finite) coproduct to a (finite) product I think

But the issue is that you can’t describe composition in terms of matrix multiplication in a nice way

[summer quest]

right something like that

you often need a bit more structure to properly interpret things like traces, like you need a categorical notion of duals and so on

there is a lot of well developed technology for this sort of thing

[plucky arch]

Yeah I’ve come across constructions like traced monoidal categories

Which axiomatise the notion of a partial trace operation

[summer quest]

yeah categorical traces are quite hot these days

[plucky arch]

oh?

[summer quest]

yeah I mean you can categorify these things pretty heavily in a lot of contexts

this is very popular in geometric representation theory for example, where one often encounters categorified versions of traces and determinants in various contexts

[plucky arch]

That’s awesome actually

Common linear algebra W

[summer quest]

Hochschild (co)homology gives a pretty general formalism for talking about this sort of thing and it's very rich

also very very intimately related to TQFTs and so on

[plucky arch]

Oh I’ve heard that term so many times

[summer quest]

that's the other big motivation for that

[plucky arch]

Cool!!!

The other main way I’ve come across to categorify matrices is with profunctors

The coend you use for profunctor composition is very analogous to matrix multiplication

[summer quest]

yeah there is some amount of profunctor/(co)end calculus that helps with this, sometimes it's very useful other times it's mostly just fancy windowdressing

[plucky arch]

Mhm

[summer quest]

a really nice motivating example of categorical traces is to take the category Rep(G) of (C-linear) representations of some finite group G (or more generally some reductive algebraic group)

you can take the categorical trace of the identity functor

[plucky arch]

Yes I’ve come across this in a few seminars I’ve been to

[summer quest]

Tr(Rep(G))=\bigoplus_Irr(G) C=O(G)^G

[plucky arch]

It’s often used as a model of “generalised symmetry”

[summer quest]

https://chenhi.github.io/notes/traces.pdf

yeah these notes are super good

I love these kinds of pictures

[fierce steeple]

Ah classic

[fresh crest]

Guys what is the answer to my name?
(√π to the power of 9 or am I typing it incorrectly)

[lethal gulch]

Let K be a finite field extension of F. Let L be the field of purely inseparable elements of K over F. Is it true that K is a separable extension of L? If so why?

[fierce steeple]

If an element of K is inseparable over L then it is inseparable over F and hence in L

[lethal gulch]

@fierce steeple But, L is the field of purely inseparable elements. So L may not contain all inseparable elements of K over F, right?

[fierce steeple]

Oh wait lol this is the other way round to what I thought it was

Usually you do separable stuff first and then are left with purely inseparable

Mb

[lethal gulch]

@fierce steeple Yes. My question is whether the other way around is also true?

[fierce steeple]

Then no - there are non-separable extensions without any purely inseparable elements https://math.stackexchange.com/questions/1275070/does-an-inseparable-extension-have-a-purely-inseparable-element (except the ground field lol)

[lethal gulch]

@fierce steeple Thanks a lot

[broken wren]

Can you understand this example by generalizing Galois groups to algebraic groups? The connected component is always a normal subgroup, which tells you that you can split an extension by doing a separable extension first. And if you actually computed this example, you could see that it doesn’t split?

[lone jacinth]

Like you tensor the extension with a ring and take the automorphism group or what is the algebraic group in question?

[north stream]

But there is not a Galois group to work with. Or what do you mean exactly?

[pastel agate]

Is there a classification of rings where all epimorphisms are surjections?

[broken wren]

That’s why you have to generalize the theory

[pastel agate]

Or at least some nice characterization?

[lone jacinth]

I don't about a classification, but an epimorphism to a commutative artinian ring should be surjective.

Something like rings with unit group {±1} would also make sense, because they can't be made from a (non-surjective) localization.

But epimorphisms can be even more wacky than that, so probably needs more restricting

[pastel agate]

I've seen a stack exchange post that says finite epis are surjections but the proof seems out of my reach

[north stream]

I think you can archieve this if you restric over some "nice" f.g. k-algebras.

[pastel agate]

Ok fair enough this seems a bit complicated for me lol

[lone jacinth]

Allright, well then finite morphism would be a nice characterization of when epi implies surjective

[north stream]

Isn't it an application of noether normalization lemma?

[pastel agate]

No idea, but here's the stackexchange post in case anyone's interested

https://math.stackexchange.com/questions/3847521/proofs-that-a-morphism-of-rings-is-surjective-if-and-only-if-it-is-a-finite-epim

[lone jacinth]

Doesn't really seem that complicated to me.

They use Nakayamas lemma otherwise not much hard hitting

[pastel agate]

I learned what an epimorphism was 2 days ago

For me it's a bit much right now but I'm sure it's doable with some work

Just not something I want to put too much time into when I'm just learning the basics

[lone jacinth]

They use the going up theorem I guess is the most complicated part

So that if R -> S finite any maximal ideal in S restricts to a maximal ideal in R.

[pastel agate]

I'm gonna take some time out of my day to learn this now just because of the name

[north stream]

This is even more general

[lone jacinth]

More general than what?

[pastel agate]

^

[north stream]

Than what I had in my mind haha

[lone jacinth]

Ah, the fg k-algebras thing

[north stream]

I'm not being productive, I'm sorry

[lone jacinth]

I mean, if your thing is easier to prove it might be productive

[north stream]

I don't think it is. Since Nakayama lemma and the going up theorem are easier to prove in this case

[lone jacinth]

I guess you don't need the full going up either. Just that if R -> S is finite and m maximal in S then m' := R\cap m is maximal in R.

For any x in R/m', 1/x in S/m is integral so satisfies a monic polynomial. Then just multiplying by x^n and rearranging the equation you get
1 = x f(x)
for some f, so x is a unit and R/m' is a field

[vast estuary]

What is a finite epi?

Ah found a defn

Hmm actually
Okay think I got it

[lone jacinth]

Epimorphism that is finite

[vast estuary]

f:R->S is finite if the R-module structure given to S by f is finitely generated?

[north stream]

The finite part refers to S being a finite R-module

[vast estuary]

What is the stacks project? This is the first I've heard of it I think

Seems pretty large

[lone jacinth]

It's a wikisite that functions as a ag textbook

[digital parcel]

No way uve never heard of stacks project before

Top 10 things on the internet

[north stream]

Is the bible for nerds

[vast estuary]

Scary what it considers as preliminaries holy shit

[lone jacinth]

Same as nlab is to homotopy theory basically

[digital parcel]

Look up anything and it’s there in full generality

[north stream]

It looks more scary than it actually is.

[vast estuary]

I've tried to keep it secret but I don't actually do that much math 🤫

[digital parcel]

Wtf

[lone jacinth]

It's very useful for looking up proofs of standard ag/comalg stuff

[vast estuary]

Oh cool

Good to know

[vast estuary]

How?

😭

[lone jacinth]

I mean, you probably shouldn't actually follow it like a text book

[vast estuary]

I'm sure

It seems much too large to do that

[lone jacinth]

Like you don't have to actually venture outside the prelims

[north stream]

I mean, its porpuse is like those books in a airplanes the pilots uses it every now and then. It won't teach you how to fly but it you have any doubts you can quickly search for exactly what you ate looking for.

[vast estuary]

Gotcha makes sense

[north stream]

It does its job really well.

[restive pike]

Not sure what the best channel to ask this question in is.

I’m trying to show that a given distance matrix D is a Euclidean distance matrix if and only if the corresponding centered inner product matrix B is positive semi definite.

I know that each entry of the distance matrix D is given by ||x_i - x_j|| = <x_i - x_j, x_i - x_j>. Furthermore, the matrix B = HAH, where H is the centering matrix and A = (-1/2)D^2 (element-wise multiplication). I also know that B is PSD if and only if it’s eigenvalues are all greater than or equal or zero.

I’m confused on where to start the forward and backwards direction of the if and only if statement.

[thick oriole]

is this where discussion about abstract algebra would take place, or a different channel?

ah nvm, #groups-rings-fields

[near lantern]

Is it typically practical to compute the (inverse) Nakayama and (inverse) Auslander-Reiten translation functors for a finite acyclic quiver directly using a projective (resp. injective) resplution? Or are they usually computed using more machinery?

[lone jacinth]

This is a fairly efficient way to compute it yes, and I don't think there's much you can do to speed it up.

QPA computes it in this way.

[lone jacinth]

Okay, if you're quiver is without relations and you don't have projective/injective summands you could use the Cartan matrix to compute the dimension vector of the AR translate.

In the rep finite case the dimension vector determines the module, and more generally it contains most of the relevant information (at least if you know the module is indecomposable). But it might not count as "computing the module".

[void plank]

Let P be a complex of flat R-modules

Is there a sense in which tensoring another complex of R-modules by P and taking the resulting total complex is an exact functor?

If so, I would appreciate a reference where I can read a proof of this

[lone jacinth]

I doubt there is such a sense at with kinda go against the main deal with complexes.

You wouldn't expect a complex to be flat just because the individual modules making it up are

[fierce steeple]

Well I guess there is a sense, in that this tensor product computes the derived tensor product

[lone jacinth]

Okay, I guess exact as meaning takes quasi-isomorphism to quasi-isomorphism.

[void plank]

Yes, is there anywhere I can look up a proof that tensoring by such a complex takes quasi-iso to quasi-iso?

[lone jacinth]

https://stacks.math.columbia.edu/tag/06XY

[void plank]

I'm guessing you're referring to Lemma 15.59.2 on that page right? If so, I'm assuming K(R) there refers to the homotopy category of chain complexes in R-mod?

[lone jacinth]

Lemma 15.59.7 is the most relevant I guess

Though 15.59.2 aswell yeah

[void plank]

I love following the rabbit hole of stacks hyperlinks indefinitely

[worldly zealot]

im enjoying this arc

[limpid horizon]

Raghuram always be like: ❔

[near lantern]

Is there a way to recover the relations of a bound quiver from its category of (finite-dimensional) representations?

[worldly zealot]

not uniquely

well idk. you can reconstruct the quiver Q, vertices correspond to isoclasses of simples and arrows correspond to dimension of Ext^1(S_i, S_j)

i think you can also get the algebra A itself up to isomorphism as End(P) (maybe opposite idk) where P is a basic progenerator such as the sum of all indecomposable projectives

so then once you fix your basis of each e_irad(A)e_j/e_irad^2(A)e_j then maybe you can read off I = ker(kQ ->> A) which gives you some presentation of the algebra

for stupid reasons you can like quotient by scalar multiples of an arrow and not change the algebra

[lone jacinth]

So like hk says the path algebra itself is recoverable as an endomorphism ring, but one useful relationship is that
dim Ext^2(Si, Sj) is the number of minimal relations between i and j.

[quiet compass]

Is there a name for this condition: Let $R$ be a ring, $S$ a subring. Then for every $r \in R$, there exists a unit $u \in R$ such that $ur \in S$

[broken turtle]

shingtaklam1324

[quiet compass]

context: Weierstrass preparation theorem

[broken wren]

Does this imply that R is a localization of S?
A localization is where you force some elements to be invertible. (This is stronger than localization, because localization does not imply injective)

[lone jacinth]

I mean Z inside any field of characteristic 0 satisfies this

[broken wren]

Oh, yeah

[hushed bone]

Isn’t this true for any subring of any field? Just let u be r^-1

[limpid horizon]

Idk if im interpreting this correctly

Ya i guess i was thinking like chmonkey

[ornate atlas]

And also don’t cross post please

[limpid horizon]

I dont really understand this definition for localization

Why consider C(S)..?

Oh cause its the solution in C(S) to that “universal mapping problem” i guess

[spice idol]

i find "A a commutative R-algebra and φ: R → A an R-algebra map" a little silly as there is only a unique such map for any R-algebra A, being the structure map

the idea of this definition is that S^-1R is the universal R-algebra where the elements in S (interpreted in the R-algebra using the structure map) have an inverse

[limpid horizon]

Oh yeah

The way this book presents localization i have never seen before

Proof of existence is weird

[spice idol]

how do they prove it?

not via the usual construction of RxS/~?

[limpid horizon]

No

[spice idol]

lmao

that sure is the obvious way to do it

[limpid horizon]

The page before argued this is actually the less tedious way lol

[spice idol]

well, they are correct in some sense

[limpid horizon]

Yea

[spice idol]

the proof falls directly out of elementary properties of inverses and the universal properties of quotients and polynomial rings

[limpid horizon]

Why the emphasis on R-algebra btw?

I guess… cause it is one

[spice idol]

natural setting for when you work with maps out of R

[lone jacinth]

I mean, an R-algebra is the same as a (commutative) ring with a ring homomorphism R -> A, and this homomorphisms is kind of key.

[limpid horizon]

Ok thats fair

[spice idol]

its more a happy coincidence that R-algebras are the same as homomorphisms out of R

[limpid horizon]

Yea i think the other day i was trying to show how they imply each other

[lone jacinth]

Now it's kind of bag in a purse to specify that A is an R-algebra and phi is R-linear, but doesn't hurt I guess

[spice idol]

i guess formally the under category and category of R-algebras are different lol

[lone jacinth]

Depending on whose doing the defining they might formally be the same

[spice idol]

secret third definition where you add the elements of R as constants

[lone jacinth]

Weird that Norwegian has two expressions "bag in a purse" and "butter on ham" that both mean something is redundant

One of these expressions are redundant!

[spice idol]

maybe they are both redundant until one is removed

[lone jacinth]

Is there an English version of this?

There's having your cake and eating it too, but that's not quite the same

[spice idol]

i actually don't know any expressions for this in dutch

id say it was a pleonasm or tautology, depending on context

[limpid horizon]

Im not sure lol

[ornate atlas]

I was about to Google bag in a purse to clarify what that meant lol

There’s a saying in the UK, though admittedly not a very common one, not anymore anyway, of carrying coal to Newcastle which would be the same I guess

There’s others in a similar vein of [product] in [place that makes product] but I’m blanking on examples

[wary elbow]

https://en.wiktionary.org/wiki/put_a_hat_on_a_hat

[lone jacinth]

Now that's the expression I'm looking for

[digital parcel]

oh yeah in chinese we have 脱裤子放屁 which means "to take off your pants to fart"

[worldly zealot]

i do that

[hushed bone]

More like amanono

[tribal viper]

is there a nice (layperson-ish) explanation of why the cayley-dickson constructions of complex, quaternion, octonion, etc lose their relevant properties one at a time in that specific order?

like why does R->C lose ordering and not, say, associativity?

[arctic pond]

Hello everyone, I wonder if there is any academic community/group related to representation theory?

[lone jacinth]

Besides this one you mean?

[lone jacinth]

I don't know how satisfying this is as an answer, but if you look at the way multiplication is defined since the involution appears somewhat assymetrically you need * to be trivial for commutative to carry over.

Similarly since the operation involves some switching of the order you need commutativity for associativity to carry over.

Ordering is more complicated in how it relates to the multiplication, but something true is that if you have a square root of -1 you can't be ordered. And I guess the main thing the construction does is add lots of square roots to -1

[arctic pond]

yes

[lone jacinth]

Allright, let me know if you find one

[tribal viper]

I don't know how satisfying this is as an answer, but if you look at the way multiplication is defined since the involution appears somewhat assymetrically you need * to be trivial for commutative to carry over.
sure, but then why does it only start with the quaternions

Ordering is more complicated in how it relates to the multiplication, but something true is that if you have a square root of -1 you can't be ordered
again, but why? I mean I can obviously see why you cannot order complex numbers but I'm curious as to how it gets lost

[lone jacinth]

I mean it starts with quaternions because * is nontrivial for C.

Like for R * is trivial, hence commutative carries over to C, but * is no longer trivial. Now C is commutative, so associativity carries over to H, but since * is nontrivial commutative does not carry over. Then as H isn't commutative, associativity doesn't carry over to O.

As for the ordering. It's not so hard to show that squares are positive. So if you have a square root of -1 you can't be ordered.

In the construction (0, 1) is always a square root of -1, so nothing produced by the construction can ever be ordered.

[tribal viper]

I mean it starts with quaternions because * is nontrivial for C. Like for R * is trivial..
ah gotcha
Now C is commutative, so associativity carries over to H, but since * is nontrivial commutative does not carry over. Then as H isn't commutative, associativity doesn't carry over to O.
also makes sense, thanks
As for the ordering. It's not so hard to show that squares are positive.
sure, got it
So if you have a square root of -1 you can't be ordered.
not following on this bit - okay, so we've lost the property that a square is always positive, but how exactly does that relate to ordering? I imagine it's something to do with going from x > y, x y > 0 => x^2 > y^2...?

[lone jacinth]

So if x > 0, then multiplying by x gives
x^2 > x*0 = 0.

If x is negative, then just replace x by (-x). Then (-x)^2 = x^2 is positive

So all squares are positive

But 1 and -1 can't both be positive

[tribal viper]

yes yes all squares are positive (in R, not in C) but I'm just not sure why that leads to "if not all squares are positive, we cannot define an ordering"

or we cannot define a...consistent? total? not sure. a good ordering.

[lone jacinth]

So you cannot define an ordering in the sense of an ordered ring.

I.e. a total ordering such that
x > y implies x+z > y+z and
x > y and z > 0 implies xz > yz.

This is because in such an ordering we just saw that all squares must be positive.

But if 1 and -1 are both positive, if I take
1 > 0 and add -1 to both sides I get
0 > -1

[tribal viper]

ooh yes it's clicking now

thank you kindly

[near lantern]

(I kind of figured this out while typing it out, but could use confirmation.)

I think for this to work one needs to define the morphism kQ -> A. I'm also trying to think about how to show any fd algebra is Morita equivalent to a bound quiver, so I really want to figure out the map in that context. So let's work in the context that A is a basic fd algebra whose simple modules are 1-dimensional and Q is the quiver with vertices correponding to the projective summands of A (which are distinct by the basic hypothesis), and number of edges from i to j equal to the dimension of e_j rad(A)/rad(A)^2 e_i.

Now, obviously the vertex idempotents in kQ map to the idempotents projecting onto the corresponding projectives in A. The arrows from i to j should map to elements of e_j A e_i which form a basis of e_j rad(A)/rad(A)^2 e_i. I now believe we can choose these elements arbitrarily. Once we do this, we should get that the kernel I is an admissible ideal: the vertex idempotents and of all the edges respectively form bases of A/rad(A) and rad(A)/rad(A)^2 under the map, and they also form bases of kQ/rad(kQ) and rad(kQ)/rad(kQ)^2. This implies that I ⊆ rad(kQ)^2. On the other hand, since rad(A) is nilpotent, say rad(A)^n = 0, rad(kQ/I)^n = 0 as well as kQ/I ⊆ A. So I ⊇ rad(kQ)^n.

If we want to show End((+) projectives) = kQ/I i#n the setting where we start with a quiver, or that any basic fd algebra is isomorphic to a bound quiver path algebra, then we also need to show that the map is surjective. This follows because the map f surjects onto A/rad(A) and rad(A)/rad(A)^2: we can use a Hensel's Lemma style argument to show it surjects onto rad(A)^n/rad(A)^{n+1} (because mult (rad(A)/rad(A)^2)^{(⨯)n} → rad(A)^n/rad(A)^{n+1} is surjective) and then A/rad(A)^{n+1} for any n. Explicitly, given b in A, we can find a0 in kQ such that b-f(a0) in rad(A), then a1 such that b-f(a0+a1) in rad(A)^2, then a2 = ∑ a1_i a1'_i such that b-f(a0+a1+a2) in rad(A)^3, etc.

Is this correct?

[near lantern]

OK, now why do we get zero kernel if A is hereditary?

(Do we?)

[lone jacinth]

So A = kQ/I with I admissible ideal. Then what you need to show is that I is 0 iff A is hereditary.

Let Pi be the indecomposable projective Aei, and let Si be it's top.

Then a minimal projective presentation of Si is
Sum[arrow i -> j] Pj -> Pi
Now A is hereditary iff these maps are always injective. Since then they give the minimal projective resolution of Si.

Now if Iei is nonzero, then there is combination of paths that is 0. So we can consider the element in Sum Pj where we split the combination up according to their last arrow.

So if this is injective each of these split up combinations must be 0. And then you can do induction on for example the length of the longest path.

Said another way, minimal relations in ej I ei correspond to elements of Ext^2(Si, Sj)

[spice idol]

back to this apparently we have the saying "balken naar Noorwegen dragen" which roughly translates to "carrying beams to Norway"

no clue where it comes from but apparently Norway has enough beams already

[lone jacinth]

I mean, Norway was pretty big in lumber production in like 17 and 18 hundreds (and before I guess)

[spice idol]

that makes sense

lol we've also got "carrying water into the Rijn"

[near lantern]

I believe A is hereditary iff for all i the image of the map, which is the unique maximal proper submodule of Pi, is projective (because we can then show any submodule of a projective is projective by some dimension induction). Thus, it is sufficient for the map to be injective. Why is it necessary?

[lone jacinth]

So I will just write P for the big sum.
Notice that
P/radP -> radPi/rad^2Pi
is an isomorphism, so the kernel of the map is in the radical and hence not a direct summand.

But then the image is not projective

(unless the kernel is 0)

[near lantern]

Also, I think I understand the core of the argument detailed in your message to be as follows:

The map is injective for all i iff I = 0. Indeed, the backwards direction is a basic fact in the unbound theory. For the forward direction:

First observe that for a given i the map is injective iff (i) the spaces P_e of paths (in P_i, so mod relations of paths starting at i) starting with e as e ranges over all edges out of i are linearly independent of each other (ii) for each edge e: i → j, precomposing paths by e is injective mod relations, i.e., a relation cannot hold (be in I) after prefixing e to all paths unless it holds to begin with.
((i) says the kernel is a "homogeneous subspace": a direct sum of the kernels on P_j for each edge e: i → j. (ii) says those kernels are 0.)

So if these conditions (i) and (ii) hold for all i, given any relation ∑ paths = 0 then for any k we can, by induction on k, group the terms by the first k edges of the path and each group will also be 0, without the prefix. But of course for k the longest length of a path in the relation the groups become single terms and removing the prefix gives the relation 1 = 0. The only way out is to have no terms.

Oh, you did say induct on the longest path and everything 😅. I think I understood that correctly then.

[near lantern]

Got it, thanks! How did you come up with that? Is there some picture in your head that you got that argument from or was it just trying things until they worked or is this just something you remember?

[cerulean cove]

I wanted to ask:
Let's say R is a noetherian ring and P is a prime ideal such that for a maximal ideal containing P,
R_m/P_m is a regular local ring. Now that means there's a regular sequence {y_i} in R_m/P_m generating the maximal ideal mR_m/PR_m. We lift these y_i's to x_i's in R_m such that their image lands at y_i's. It says that {x_i}'s are M_i/M_{i+1} and M/M_i regular sequences where M is a finitely generated R_m-module defined as: M_i = P^i_mJ_m/P^{i+1}_mJ_m where J is a finitely generated R-module. I want to know why they lift to regular sequences.

PS. the question is easy really to understand, if you feel confused please ask me again.

[lone jacinth]

Well, it's a fact I know quite well. So I guess I remembered it.

Like I said before the intuitive idea is that Ext^1(Si, Sj) corresponds to arrows j->i and that Ext^2(Si, Sj) corresponds to relations.

In general global dimension (of finite length modules) can be computed at the simples. So having global dimension less than n just amounts to showing Ext^n(Si, Sj) = 0. So from there it's just take the projective resolution of Si and check that it stops at the second step.

[dull cedar]

If the P‑adic filtration is nice (good/I-stable) then lifting a regular sequence from Rm/Pm to Rm gives a regular sequence on Mi/Mi+1 and M/Mi otherwise it may not

[cerulean cove]

okay! thanks for help. the only extra information I have is that M_i/M_(i+1) is a free-(R/p)_m module. Is that nice enough?

[dull cedar]

Yes, that’s enough If each Mi/Mi+1 is free over (R/P)m then the lifts xi act like yi on a free module so they form a regular sequence on Mi/Mi+1 and on M/Mi

[cerulean cove]

thank you for your help, kind person, i'll try to prove it.

[dull cedar]

though technically some extra conditions on the filtration may still be needed in full generality

No worries

[cerulean cove]

i forgot to mention that Ass(M_m) = {P_m} 😭

[dull cedar]

Oh lmfao

That clears things

[cerulean cove]

hi sorry to disturb, can you give me a proof for it? like why it lifts to regular sequence on (M/M_i)_m?... thanks, even if you don't, kind man.

[hushed bone]

Okay this kinda was pmo the whole time but all of this is happening localized at m so you should just reformulate the question to be about a local ring so you can quit writing _m everywhere lol

Anyway

@cerulean cove I’m localizing everything at m always

You know that y_i is regular on R/P, and x_i are lifts thereof. You said that M_i/M_i+1 is a free R/P module, so it has the same associated primes, aka anything regular on R/P is regular on M_i/M_i+1. Because M_i/M_i+1 is naturally an R/P module (as in P kills it), elements x in R act on M_i/M_i+1 the same way their reduction mod P does.

By construction of the x_i they act on M_i/M_i+1 the same way the y_i do, which were a regular sequence on R/P and therefore a regular sequence on M_i/M_i+1