advanced-algebra

[fierce steeple]

well actually idk how well-known the stuff i was gonna talk about is to rep theorists etc but i think it is familiar

(in the complex case)

[plucky arch]

sl_2(F_p)?

[fierce steeple]

ye

[lone jacinth]

Not really

[fierce steeple]

or sl_2(\bar{F_p}), doesn't matter for this

[plucky arch]

there's a whole part of the cambridge rep theory course dedicated to this

[fierce steeple]

nice

But yeah my question was like uh

[plucky arch]

https://www.dpmms.cam.ac.uk/~sjw47/2021RepTh.pdf

uh lemme cut that link down

[fierce steeple]

isn't that SL_2?

[plucky arch]

section 9 deals with the character table of GL_2(F_q)

oh rip i didn't realise this was the lie algebra 😭

[fierce steeple]

dww

[plucky arch]

my bad

[fierce steeple]

nws nws

Wadsey seems cool

[plucky arch]

i just had to suffer 3 lectures of this bs so i was looking forward to it finally being useful

[fierce steeple]

bruh bs

[plucky arch]

well idk the rep theory of GL_2(F_q) wasn't that fun

[lone jacinth]

Why did you have to suffer through it? Like because you need it for physics stuff?

[fierce steeple]

Sure I just mean like lol feels dismissive

but fair

[plucky arch]

this feels like a bit of a weird comment

[fierce steeple]

Idk just funny asking about something and it being called bs but it's okay lol

But ye i wondered like you can form $\Lambda^* \langle a_i, b_i \rangle$ on $2g$ generators $a_1,b_1,\dots,a_g, b_g$ and let $\omega = \sum_i a_i b_i$ and consider like multiplying by $\omega$ to get this operator $L$ and it's well-known that like you get this Lefschetz decomposition and stuff like $L^k$ inducing an iso $\Lambda^{g-k} \to \Lambda^{g + k}$ for $0 \le k \le g$

[plucky arch]

i was saying it more casually as an annoying part of the course

[broken turtle]

Prismatic Potato

[plucky arch]

idk why you interpreted that as me being dismissive

[fierce steeple]

Idk sorry

[fierce steeple]

but ye this all works beautifully in char 0 and you get this decomposition etc and the fact i say there

But i wasn't sure how this stuff works mod p - indeed it starts being weird lol

e.g. integrally, wedging by omega doesn't give isos here

But maybe this is just hard lol

[lone jacinth]

Hmm, do you have a good reference in char 0? Maybe one can step through the proof and see what is used

How does it fail to be iso integrally for example, is it still injective?

[fierce steeple]

Tbh if you don't know about it maybe dw aha but like the usual ingredients immediately break

[fierce steeple]

i think so. Like the source is free (as a Z-module) so I think that is immediate

But yeah like one way to prove this stuff is to use semisimplicity and that immediately breaks mod p lol

(that gives you a decomp as a rep of sl_2(C) basically and then you study that ^)

But for the fact about L^k inducing isos like a standard proof is to use a descending induction on k. But then the first step breaks because e.g. w^p = 0 mod p (for example)

[lone jacinth]

I see

[fierce steeple]

Basically what I was able to prove is that mod $p$, the dimensions of $\Lambda^* \langle a_i, b_i \rangle / \omega$ are as one expects until like degree $2p$ lol. The point is that if $g < p$ then $g!$ is invertible mod $p$ so you can just run the usual argument, and then an argument shows that this gives you $\omega \cdot -$ being injective up to degree $2p-1$

[broken turtle]

Prismatic Potato

[fierce steeple]

But ye dw

owo

[lone jacinth]

Sounds very possible that it's not iso for higher degrees then

Maybe it would be worthwhile to check for say p=2 if that's doable

[pastel agate]

Alright so I've sat with it for a bit and I think I understand free Algebras a lot better. I don't want to be annoying or anything but I do still have this one gap in my understanding of the universal property. So I understand that free structures are supposed to consist of all possible "combinations of operations" in your structure ala jagr's explanation. But I am still a bit perplexed that the universal property (and being clear now, I mean the property that a function from an alphabet to an algebra/group/module/etc extends uniquely to a homomorphism from a free structure over that alphabet to the algebra/group/etc) would induce this type of behaviour. What is it about this unique extension that forces free-whatevers to look like this? Maybe this is a question better asked once I have more experience in the field and I should just take it for granted for now but it really does bother me for some reason.

There's gonna be some category theory near the end of this course I'm taking so maybe it'll be explained better there, idk

[lone jacinth]

So the idea is that if something is generated by x and y you can determine a homomorphism by where it is mapped.

If you know f(x) and f(y) you can determine f(x^2) etc.

This gives you uniqueness, the value of x and y determines everything.

The tricky thing is existence in the first place. So how can existence fail?

That is, we fix f(x) and f(y) what could stop us from extending f to everything.

Well, if we just define f(x^2) as f(x)^2 and so on, everything works fine. So what could the problem be?

Well if for example x^3 equaled y^2, then we would need
f(x^3) = f(y^2).

But it could be that
f(x)^3 = f(y)^2.

So the thing that makes the free algebra special, is that we don't have any of these relations that stop us from extending f from {x, y}

[plucky arch]

there's actually quite a nice way to explain this with cat theory

[pastel agate]

I mean this does make sense to me as a consequence of being "generated by" something but pretty much everything I can find about this tends to like to emphasize the fact that the universal property is less of a property and more of a definition of a free thing. Like in the example of an algebra if I say that a free K-algebra in a set X is the set for which the universal property holds why would I immediately assume that it would just be this gigantic set consisting of words and linear combinations of those words etc. I understand that it is the case that this does happen but it's still odd to me that it does. Like I see exactly what you're saying about the necessity to be "relation-less" in a sense, but it's still not really clicking for me that this behaviour is induced by the property in question

I hope I'm not sounding pedantic lol I'm sure this is already crystal clear for most people here 😭

[plucky arch]

can i try to explain it?

[pastel agate]

Please

[plucky arch]

so the universal property of the free algebra can be stated as follows

[visual void]

Can someone help me with this

I’m in 8th grade in algebra 2 and I’m having a hard time with this

[plucky arch]

there's a natural correspondence between functions $X \to A$, and homomorphisms $K<X> \to A$

[broken turtle]

Pseudo (Cat theory #1 Fan)

[ornate atlas]

#prealg-and-algebra

[plucky arch]

in that - given any function $X \to A$, you can produce a homomorphism $K \langle X \rangle \to A$ - and given any homomorphism $K \langle X \rangle \to A$, youc an get a function $X \to A$

[broken turtle]

Pseudo (Cat theory #1 Fan)

[visual void]

Sorry thnak you

[plucky arch]

and these operations are inverses of each other

one direction is by "extending" the function to a homomorphism, the other direction is by "restricting" the homomorphism to its action on the generators

does that make sense so far?

[pastel agate]

Yeah that makes sense

[plucky arch]

next, we need the concept of the "double dual" map - i assume you've come across this from linear algebra

you can view a vector $v \in V$ as a way to evaluate a covector - it gives you a way to take in something of type $(V \to k)$, and output an element of $k$

[broken turtle]

Pseudo (Cat theory #1 Fan)

[pastel agate]

I actually have only seen this in the context of functional analysis, but I think it's fine yeah

[plucky arch]

hence the double dual map $V \to ((V \to k) \to k)$

[broken turtle]

Pseudo (Cat theory #1 Fan)

[plucky arch]

the thing is... we can generalise this to arbitrary vector spaces W

[lone jacinth]

So it's a fact that you might see when you learn more category theory or you can prove for yourself, that something satisfying a universal property is unique (up to unique isomorphism). So in that sense it functions as a definition.

I'm not sure exactly what you're point of confusion is, maybe Pseudo will say something.

But to spell out my thinking again.

k<X> should contain X and be such that a homomorphism k<X> -> A is the same as functions X -> A, by just restricting to X.

So k<X> contains the set X, hence also contains multiplications and sums of things in X. So the only question is if some of those sums and products are the same.

But if they were the same, then you couldn't map to an algebra A where they were different. So unless such expressions are equal for all algebras, they shouldn't be equal in k<X>

[plucky arch]

if you fix a particular $v_0 \in V$, then you obtain a way to "evaluate" any linear map $V \to W$ to produce an element of $W$

[broken turtle]

Pseudo (Cat theory #1 Fan)

[plucky arch]

via just $L \mapsto L(v_0)$

[broken turtle]

Pseudo (Cat theory #1 Fan)

[plucky arch]

this is something of type "forall W . (V -> W) -> W"

it's a kind of "generalised double dual", since we're not looking at just W = k, but all vector spaces W

does that track?

[fierce steeple]

Yee

[plucky arch]

if it's unclear please do say so, i will try my best to clarify

[pastel agate]

I'm reading up on duals for vector spaces, but I do follow I think

[plucky arch]

in the finite-dimensional case, the map $V \to V^{**}$ is an isomorphism, but this fails in the general case

[broken turtle]

Pseudo (Cat theory #1 Fan)

[plucky arch]

however, it turns out that for this "generalised" double dual, the map $V \to (\forall W . (V \to W) \to W)$ is an isomorphism!

[broken turtle]

Pseudo (Cat theory #1 Fan)

[plucky arch]

in other words, if you've got a way to take any vector space W and any linear map V -> W, and produce an element of W

i know for sure that, under the hood, you're just doing $L \mapsto L(v_0)$ for some $v_0 \in V$

[broken turtle]

Pseudo (Cat theory #1 Fan)

[plucky arch]

the point is that, if you're given no additional information about W other than it's a vector space (over the same field), and you're given a linear map V -> W

and you have to make a single formula that, no matter what, always gives back an element of W

your only option is basically to do L -> L(v_0)

(this is formalised in category theory by saying that the only natural transformations from Hom(V, -) to the identity/forgetful functor are maps of the form L -> L(v_0))

is that at all comprehensible? (note that i don't think it's obvious why this is the case, at least not yet)

[pastel agate]

So far I've understood what you've said, although not necessarily it's connection to my question (I do kinda get the feeling from the necessity of this map essentially being an evaluation, but not much more than that)

[plucky arch]

there's a quick motivation for why this is the case

since it works for all vector spaces W and all linear maps V -> W, it has to work for a special case

we choose W = V, and the map V -> V being the identity

now, $\textit{if}$ we knew that the map was $L \mapsto L(v_0)$ for some $v_0$, then using $L = \text{id}_V$ lets us recover what $v_0$ is

[broken turtle]

Pseudo (Cat theory #1 Fan)

[plucky arch]

because we'd have $\text{id}_V \mapsto \text{id}_V(v_0) = v_0$

[broken turtle]

Pseudo (Cat theory #1 Fan)

[plucky arch]

you can then argue that naturality/parametricity implies that, whatever v_0 it spits out for id_V, a general linear map L must spit out L(v_0)

anyway, all this is to say that you can embed any vector space in a kind of "generalised double dual", as $\forall W . (V \to W) \to W$

[broken turtle]

Pseudo (Cat theory #1 Fan)

[plucky arch]

but there's not really any point here where we used linearity specifically

this actually works for any kind of algebraic structure

you can embed an algebra $A$ in a kind of "generalised double dual", as $\forall B . (A \to B) \to B$

[broken turtle]

Pseudo (Cat theory #1 Fan)

[plucky arch]

if you've got a way that, for any algebra B and any algebra homomorphism A -> B, produces an element of B

i know that, under the hood, you're just doing f -> f(a_0) for some a_0 in A

[pastel agate]

I think I'm getting a lot more than I bargained for here but I am sort of following.

I'll have to sit with it for a bit

[plucky arch]

yeah sorry 😅

i think the point is that you can always view elements of A as a kind of "evaluator"

they let you evaluate homomorphisms A -> B to produce an element of B

what's perhaps surprising is the reverse is true - if you have an "evaluator" that works for all B and all homomorphisms A -> B, it has to come from some element of A

[pastel agate]

Ok, and if they are to be an evaluator, then they need to be able to cover any possible "situation" you could find yourself in B

Am I getting that right?

[plucky arch]

yeah, exactly

they have to work for any B

which means you can't really use any specific properties of B itself

other than it being an algebra

[pastel agate]

Interesting

[plucky arch]

sometimes i like to imagine actually programming this out

[pastel agate]

I'm gonna do some practice problems and see if it sticks a bit better

[plucky arch]

if i had to write a function that took in a homomorphism A -> B and spit out an element of B

but i wasn't actually given access to what B was

[pastel agate]

My instructor really likes to point out how important certain things are before he proceeds to never mention them again so I'm gonna take him at his word and try to get this down

[plucky arch]

ah sorry, i'm almost done if you're still up for it

[pastel agate]

You can write it out and I'll revisit it later, for now I think I'm probably better off working through some of this myself

[plucky arch]

mhm mhm

[pastel agate]

If I have any questions I can always ask you later ig

[plucky arch]

so, this works in particular for the free algbera $K \langle X \rangle$

[broken turtle]

Pseudo (Cat theory #1 Fan)

[pastel agate]

Thanks though, I really do appreciate it 🙏

[plucky arch]

in other words, elements of $K \langle X \rangle$ correspond precisely to $\forall B . (K \langle X \rangle \to B) \to B$

[broken turtle]

Pseudo (Cat theory #1 Fan)

[pastel agate]

And to you too @lone jacinth 🙏

[plucky arch]

but this is where the universal property comes in!

because we know what $K \langle X \rangle \to B$ is! it corresponds precisely to functions $X \to B$

[broken turtle]

Pseudo (Cat theory #1 Fan)

[plucky arch]

so, the elements of $K \langle X \rangle$ correspond precisely to $\forall B . (X \to B) \to B$

[broken turtle]

Pseudo (Cat theory #1 Fan)

[plucky arch]

these are "evaluators" that take in a function from a set X to an algebra B, and spit out an element of B

hence why i said this earlier

there are a few ways i could implement such an evaluator, given f : X -> B and elements x_0, x_1, x_2... of X:

• f -> f(x_0)
• f -> f(x_0) + f(x_1)
• f -> f(x_1)^2 - 2 f(x_1) f(x_2) + f(x_3)^4
• f -> 0
• f -> f(x_1)^420

what's interesting is that, if you consider all ways to implement such an evaluator, these in themselves form an algebra

if you have a scalar s and an evaluator f -> ev(f), then you can form a new evaluator f -> s * ev(f)

if you have two evaluators ev_1 and ev_2, you can form a new evaluator f -> ev_1(f) + ev_2(f)

and you can also form f -> ev_1(f) * ev_2(f)

it's essentially "pointwise operations" but really souped-up

so, with the right category-theoretic machinery, you could define K < X > to be the algebra of evaluators "forall B . (X -> B) -> B"

the typical description of "linear combinations of noncommuting polynomials" is just a convenient way to encode the implementation of any evaluator

e.g. the evaluators i listed above would correspond to:

• x_0
• x_0 + x_1
• x_1^2 - 2 x_1 x_2 + x_3^4
• 0
• x_1^420

[past cove]

it's a complex that's used to detect when a sequence of elements is regular

or at least that's one interpretation

[limpid horizon]

Nice

[past cove]

it was defined for Lie algebra cohomology

but it's more general than that

it's essentially a very useful homological algebra tool for constructing complexes

or to be more precise

it's a very natural way to build a free resolution of a quotient ring

Tho I assume you're looking at it through a local cohomology lens right

tbh I'd really recommend you look at a homological algebra book like Weibel

you don't need everything from it

but it will make a lot of things clearer

[ornate atlas]

I think Miller and Strumfels also has quite a bit about the Koszul complex from a combinatorial lens

[worldly zealot]

miller :D

he taught me about koszul complex

[ornate atlas]

No way! Tell him he writes a mean book if you’re still there lol, really good author

[worldly zealot]

will do! yeah it’s a really great book. he’s an even better irl explainer

[pastel agate]

This sentence in particular made it click for me. Very insightful explanation, thank you very much!

[plucky arch]

Oh yay, I was worried I’d scared you off 😅

But yeah I really like this “evaluator” perspective on free objects

[pastel agate]

You definitely did yesterday but when I sat down again with it today I got a better perspective and it mostly clicked

[plucky arch]

There’s a quite pretty categorical reason for it (the forgetful functor is an algebra object in the category of set-valued functors)

[pastel agate]

I won't pretend and tell you I understood all the details but I do think I mostly got the intuition behind everything

[plucky arch]

Mhm mhm

[placid flare]

In here, Milne (in his CFT notes) defines the direct limit of sets to be the disjoint union (i.e. coproduct in Set) modulo an equivalence relation. He then says that if the A_i are abelian groups and the alpha_i homs of groups then the disjoint union has a natural group structure. Can this be? Wouldn't you want to use the coproduct in the cateogory of abelian groups instead when defining the direct limit?

[past cove]

this is just the explicit construction

wait

why would you wanna use the coproduct?

the coproduct in the category of abelian groups if the direct sum

but this is very different

so I don't understand your question

[spice idol]

it happens that the underlying set of the direct limit in any algebraic category (some category of algebraic structures) is precisely the direct limit of the underlying sets of the components

see if you can find a natural way to define an abelian group structure on this explicit construction of the direct limit

(hint: use the fact that for any sequence of elements a1, ..., an in the disjoint union, there is some component A_i where they all "meet" as you travel along the diagram)

[placid flare]

Oh I see now, the directness condition implies that for every two elements in the direct limit we can find an i such that both lie in A_i. Is the construction when you take the coproduct in Ab (i.e. direct sum) and then qoutient out by an apropriate subgroup a direct limit in the category Ab (i.e. satisfies the universal property)?

[spice idol]

yeah sure

there's a theorem that says every small colimit is a coequaliser of a coproduct

(and dually every small limit is an equaliser of a product)

but this construction is way nicer, especially on other categories (like e.g. groups or rings), as coproducts aren't as nice there

and, for example, as you can see here the forgetful functor F : Ab -> Set commutes with direct limits, while it does not commute with general colimits (coproduct in Set is disjoint union)

[placid flare]

I see, I wasn't aware that both construction yield the same thing. I can see that the one Milne uses is easer to work with. Thanks for the help

[spice idol]

ofc!!

[wise sedge]

Apparently GL(3,O) has an associative subloop isomorphic to F4

[vague flume]

Is there an elementary way to show that $\frac{k[X,Y]}{\mathfrak{m}}$ is a finite (hence algebraic) extension of $k$? (Elementary meaning no Nullstellensatz or Zariski's lemma)

[broken turtle]

Yuese

[lone jacinth]

So you have a field generated by x and y, and let's say m doesn't contain either x or y, because then it's trivial.

Think first about the subfield generated by k[x]. This is either finite or k(X). In the first case you'll have a finite extension of a finite extension and be done.

In the second case you have a field of the form k(X)[y] = k(X)[Y]/(f(Y)) for some monic polynomial with coefficients in k(X).

If you look at the denominators in the coefficients only finitely many of the primes in k[X] appear. Invert these primes to get the ring R.

Then your field is the field of fractions S := R[Y]/(f(Y)).

Now you can consider the homomorphism k[X, Y] -> S sending X to X and Y to Y. This is supposed to become surjective when passing to the field of fractions of S, but S is not a field. Contradiction.

Hence x satisfies some minimal polynomial and the whole extension is finite.

[unborn rampart]

I have a kinda stupid question. If the universal property of the exterior power Λ^k V is that every alternating multilinear from V^k factors uniquely through Λ^k V, then doesn't T^k V also satisfy this, since every multilinear map factors uniquely through T^k V? Clearly I'm missing something, because T^k V is too big. I'm guessing part of the universal property is that the canonical map ∧ from V^k to Λ^k V should be alternating? In which case it rules out T^k V. But I still feel like I'm missing something obvious

[past cove]

This is more strict

The tensor algebra also has a universal property like this but it's for any multilinear map

What this is saying is that for an alternating map you get a map from the tensor algebra but it factors through the exterior algebra!

So it's like a strengthening of the universal property for the tensor algebra

If you want to recover the map from the tensor algebra from this map just precompose this map with the natural map T^kV -> Λ^k V

[unborn rampart]

I don't quite understand what you're saying I know that alternating multilinear maps factor through the exterior power, but it seems to me that both Λ^k V and T^k V satisfy this universal property (which should be impossible, because objects defined by a universal property is unique up to iso)

[past cove]

They're technically different universal properties

[exotic hill]

can you guys teach what the symbols mean? I am Korea 6th grader and learning high school 2nd grader math in korea and we don’t learn that stuffs so…

High school 2nd grader math in korea includes calculus and these stuffs…

[past cove]

hold on

[fierce steeple]

in general

[unborn rampart]

wrong channel, try #prealg-and-algebra

[fierce steeple]

at least in general characteristics

[exotic hill]

oh, thanks for the guide😊

[digital parcel]

lol i think they're asking about the symbols in the commutative diagram

or maybe not

[exotic hill]

but the chat you suggested for are not my level though...

[exotic hill]

yup

[unborn rampart]

not sure, but let's say we leave out that requirement from the universal property, ie. \wedge : V^k \to T^k V can be any multilinear map. What object satisfy the resulting universal property? T^k V or Λ^k V?

[fierce steeple]

Well then you have changed the universal property

and now it's T^k V

But I am confused by how you have written wedge and then said tensor product

[unborn rampart]

hmm yeah... the universal propety is the same as for T^k V, but we still only want alternating multilinear maps to factor through it

[fierce steeple]

Idk what you mean lol

[unborn rampart]

yeah sorry, that was a mistake

[past cove]

he's saying that all alternating maps are a subset of multilinear maps so we should have some factorization involving T^k V

[digital parcel]

arrows just mean functions. so this one is phi (a greek letter) from M^l to P

[fierce steeple]

okay yeah sure

but what's the problem

[exotic hill]

than what does l mean ?

[past cove]

here's the thing @unborn rampart it's a subtle thing

take any alternating map

call it f

[digital parcel]

just some number

[past cove]

by the universal property of T^k V, we get a map T^k V -> P

[exotic hill]

its like squares right?

[exotic hill]

oh sure thanks

[past cove]

but the universal property of Λ^k V tells us we also get a map Λ^k V -> P

now if you draw a commutative diagram

[unborn rampart]

I mean, let's say we are looking for an object X, such that every alternating multilinear map from V^k factors through X, and there's a canonical multilinear map f : V^k -> X. So basically, it's the universal property of the tensor power, but restricted to alternating maps. What would X be?

[past cove]

you'll see that the map T^k V -> P actually factors through Λ^k V -> P

so both universal properties are satisfied

[unborn rampart]

hmm, let me think

[proud mortar]

T^k V doesn't satisfy the universal property, because the map V^k -> T^k V isn't alternating

[unborn rampart]

yep, I got that, but I'm still curious about what the mystery object X is, when we remove that requirement

I'm guessing X would be something smaller than T^k V and bigger than Λ^k V

[proud mortar]

that property for X doesn't obviously look like a universal property to me, so I don't think it would be uniquely defined

[unborn rampart]

this makes sense 👍 I think my main confusion is cleared up at least, I was worried for a second that I found two non-isomorphic objects satisfying the same universal property

[unborn rampart]

I see

[plucky arch]

in this formulation it's important that the $\wedge$ map is alternating

[broken turtle]

Pseudo (Cat theory #1 Fan)

[plucky arch]

this is why i prefer the representability formulation here

the exterior power $\Lambda^k V$ $\textit{classifies}$ alternating multilinear maps, in the sense that linear maps $\Lambda^k V \to W$ represent alternating multilinear maps $V^k \to W$

[broken turtle]

Pseudo (Cat theory #1 Fan)

[plucky arch]

so any alternating multilinear map $V^k \to W$ produces a linear map $\Lambda^k V \to W$, and any linear map $\Lambda^k V \to W$ produces an alternating multilinear map $V^k \to W$

[broken turtle]

Pseudo (Cat theory #1 Fan)

[plucky arch]

and moreover, these operations are inverses of each other

if you try this with the tensor product $V^{\otimes k}$, then any alternating multilinear map $V^k \to W$ does indeed produce a linear map $V^{\otimes k} \to W$ - however, there's not an obvious way to go in reverse

[broken turtle]

Pseudo (Cat theory #1 Fan)

[plucky arch]

you could try something like the antisymmetrisation, but then that loses the operations being inverses of each other, since going linear -> multilinear -> linear is not the identity

another way to think about this is that the map multilinear -> linear is not surjective, so it can't be part of a bijection

[lone jacinth]

Remove what requirement? That the map is alternating?

Then there can't be a unique such map, since both the exterior power and the tensor power works.

In general you want "the universal X" to also be an X

[hushed bone]

A drawing isn’t a proof 😂

[past cove]

😂

Live, laugh, learn!

[rose mirage]

I'm gonna be sick

[worldly zealot]

@lone jacinth have you read keller's orbit categories paper? i tried to read it once but i dont understand dg categories enough to follow. if you have any good readings (on dg categories) that would be much appreciated

[lone jacinth]

I'm not sure Ive read that particular paper, but Keller himself has some papers on dg-categories like
https://arxiv.org/pdf/math/0601185

[unborn rampart]

Yep, that was my suspicion, so the universal property without \wedge being alternating isn't actually a universal property. I haven't really seen a precise definition of universal property yet (even Leinster seems to handwave the definition), but I think I can see that you need the map from V^k to \Lambda^k V to be alternating too, because otherwise the proof of uniqueness wouldn't go through

[unborn rampart]

That makes sense, thanks for the explanation 🙏

[plucky arch]

there are two "precise" definitions of universal property that i know of

one is "being initial or terminal in some category"

though i kind of dislike this one cause "some category" does feel a little vague

but it's still a good perspective to keep in mind

the other is "represents some functor on your category"

and i tend to prefer that a lot more

[restive pike]

Not sure if this is the right place to ask, but has anyone worked with semigroup applications to public key cryptography?

[slim dove]

probably the closest to it
would be very interested to hear about that as well

[jaunty storm]

I'm trying to understand this idea of 'formally' integrating a Lie algebra. The universal enveloping algebra of a Lie algebra has a comultiplication on it, turning it into a Hopf algebra; if you take the set of all group-like elements of this algebra, you get a group which people call the 'formal' Lie group integrating the Lie algebra. Is anyone familiar with this? How does this 'formal' Lie group relate to the actual Lie group (which there will be in a finite-dimensional setting) integrating the Lie algebra, e.g. how do you equip it with a topology, etc.? Does anyone know of a nice reference for this?

[restive pike]

Oh sure! It’s been the subject of study for my research at my university. The foundational paper that really launched the study of semigroup actions for public key cryptography is this one https://arxiv.org/pdf/cs/0501017

They show that the Diffie-Hellman protocol (which is based on finite cyclic groups) is really a special instance of a semigroup action

And so from here you can reformulate Diffie-Hellman (and the Discrete Log Problem) in terms of Semigroup actions (and the Semigroup Action Problem, respectively)

[unborn rampart]

We will not assume G has an identity. However, WLOG, we will always assume G has an identity
Is there something I don't know about semigroups? Are semigroups actually monoids? or am I just underestimating the power of WLOG?

[spice idol]

since you only care about semigroup actions you can always take the completion

it never hurts the semigroup to add an identity

(it just may get identified with an already existing identity, in which case it simply gets distinguished)

[lone jacinth]

You can always freely adjoin an identity to get a monoid, so presumably that doesn't change the argument here in any way

[unborn rampart]

Hmm, I see can you always freely adjoin an identity to a rng to get a ring?

[lone jacinth]

Yes

[spice idol]

will the natural Rng map be injective though?

[lone jacinth]

ZxR with multiplication
(n, r)(m, s) = (nm, ns + mr + rs)

[unborn rampart]

Wow WLOG every ring is unital

[lone jacinth]

I mean there might be loss in generality depending on what you're doing

Like the unitalization will get new ideals and so on

[spice idol]

where rings act like abelian groups, monoids act like sets

[lone jacinth]

Fun fact, the category of nonunital rings is equivalent to the over category (Ring, Z)

(i.e. homomorphisms from some ring to Z)

[forest turtle]

woag

[lone jacinth]

The equivalent is in one direction given by taking the kernel

And in the other with freely adjoining unit and the obvious map
ZxR -> Z

[forest turtle]

free-forgetful type beat?

[near lantern]

OK having understood this with your description I think I prefer to think of it as that the over category is equivalent to the category of rngs

[wise sedge]

Is GL(2,Q) a maximal subgroup of Sym_(Q^2/{0})?

[spice idol]

a matter of perspective

[near lantern]

A larger proper subgroup would be all permutations that preserve the relation of being proportional.

[near lantern]

This sounds very interesting - do you have any reference by any chance? Clearly someone must have considered this for there to be people who "call it the 'formal' Lie group".

[jaunty storm]

this is sort of the content of section 3 in https://arxiv.org/pdf/2404.09261

[slim dove]

i will say, for general semigroups, this sounds nasty difficult to reverse engineer

[spice idol]

is there already a good way for computers to deal with / compute in huge semigroups?

[slim dove]

not really

best I am aware of is the semigroup package in GAP

[spice idol]

hmm..

would the problem become harder if the semigroup in addition has unique factorisation, or easier?

[slim dove]

I would guess easier, but
not sure if this is very relevant? the usual idea to generate something hard would probably be getting some very funny semigroup presentation that turns out to be finite, and go from there

since semigroup presentations are just, really hard to deal with? i could easily expect brute force attacks to just be hard by virtue of calculating the entire semigroup being difficult, even if it's public

but i am bullshitting here, i'd need to read the article in more detail

[spice idol]

that's what you get when you don't have regular congruences..

[slim dove]

the article itself says it's not known how to adapt abelian group attacks for general semigroups

there is some chance I'd guess by going through krohn - rhodes? to do something, considering there seems to be a way to adapt these attacks to unions of cyclic (i think meaning monogenic) semigroups

(krohn-rhodes for context is a miracle theorem that basically generalises jordan - holder for groups, every finite semigroup is a homomorphic image of a subsemigroup of a wreath product of finite simple groups and finite aperiodic semigroups)

there's a version of this decomposition that seems to be relatively possible to calculate in GAP, and it seems to have applications by now

[worldly zealot]

whoa

[spice idol]

that's crazy

wtf

is there a classification of aperiodic semigroups yet?

[slim dove]

i don't think so

though

classifying any less nicely acting semigroup class sounds like hell

[spice idol]

forever glad that minimal algebras have a nice classifcation so we don't have to worry about that anymore 🙏

[spice idol]

mm yeah

[slim dove]

there is a way to do this if you decide to present things as subsemigroups of transformation monoids probably? aperiodic <--> H relation trivial, and cayley theorem analogue allows you to check the H-relation

like, the entire point of an aperiodic semigroup in that theorem is that they don't contain subgroups

[slim dove]

specifically, for context; H relation trivial iff no two elements have equal images and kernels at the same time

[spice idol]

https://en.wikipedia.org/wiki/Special_classes_of_semigroups
why are there so many lmaooo

[slim dove]

but actually getting a full description of every possible aperiodic semigroup sounds like hell

[slim dove]

atilla nagy and his obsession about finding lattice representations of every single little possible special semigroup type

[spice idol]

a fun problem to work on over the summer break :>

[slim dove]

tbf a lot of those do come up naturally

[spice idol]

lattice representations? :0

I gotta learn more about semigroup theory fr

[slim dove]

let me find what's the rough idea again

basically

probably one of the more better known types of semigroups is Clifford semigroups

you can describe them directly using their structure, so they are inverse semigroups (every element $x$ has a unique pseudoinverse $y$ such that $xyx = x$), that are completely regular (all elements are part of a subgroup, each subgroup being exactly an H-class)

alternatively, regular (every element has at least one pseudoinverse) and all idempotents are in the centre (defined as in groups)

[broken turtle]

JulieKitsune

[slim dove]

but the usual way of describing them is by a semilattice of groups

[restive pike]

the paper i linked was from 2007. here is a more recent paper from gnilke & zumbrägel (2023) in which they propose a generic algorithm for the semigroup action problem w/ lower + upper bounds https://arxiv.org/pdf/2301.01657

it is a little out of my depth at this moment since they introduce soem theory of computation to construct this abstract model of computation (section 3)

[slim dove]

by that, we mean a construction along the lines of: \

• A semilattice $L$ (commutative idempotent semigroup, or the usual order-theoretic definition of a poset with meets for all pairs of points) \
• Groups $G_\alpha$ where $\alpha \in L$ \
• Multiplication restricted by $G_\alpha G_\beta$ being a subset of $G_{\alpha\beta}$ when $\alpha, \beta$ multiplied in $L$

[broken turtle]

JulieKitsune

[slim dove]

you can describe a lot of semigroups by replacing the groups by some kind of semigroup instead

[spice idol]

ooo I see

[slim dove]

so like, you can often describe a global property by local properties\

[slim dove]

gonna read through that right now

[spice idol]

this seems like a general strife in algebra lol

find some way to look locally and conclude globally

I suppose the ide of looking locally is always to reduce a problem

[slim dove]

it's more a matter of this allowing you to prove something's a variety easier most of the time iirc

[slim dove]

gotta say, i really don't like the notation

[restive pike]

i have come across krohn-rhoes in passing during my rabbit hole of trying to understand possible attacks on SAP! i dont know exactly what a wreath product is. i'm going down a definition rabbit hole atm trying to stitch everything together

[restive pike]

oh YEAH tell me about it. the 2007 paper uses the letter G for a semigroup and the letter S for a set (imo this should be illegal)

[slim dove]

a wreath product is essentially a way to create a group/semigroup via having one group/semigroup act on a cross product of groups/semigroups by shifting around their indices

[restive pike]

when you say a cross product, is that different from something like the external direct product of groups?

[slim dove]

same thing, component-wise multiplication and all

i will say, krohn-rhodes is mainly an automata result and not necessarily a semigroup result, so most good explanations are in those terms instead

but automata are essentially subsemigroups of partial transformation semigroups, so you know, all comes down to that

[restive pike]

now thats quite remarkable

[worldly zealot]

wreath products are pretty confusing

[slim dove]

i'd honestly first look at semidirect products

[worldly zealot]

i understand it now but it took me a while to wrap my head around

[amber isle]

I have never seen this definition of a rank for an arbitrary subset of Z (can be extended to a Dedekind domain)

I’m guessing it should align with the rank of free Z-modules (therefore R-modules)

Anyone else seen this before?

[spice idol]

how do you see it defined usually?

[amber isle]

The rank that I'm familiar with is for R-modules. Specifically, I recall, if $M$ is an $R$-module, then
$\mathrm{rank}(M) = \mathrm{dim}_{\mathrm{Frac}(R)} M \otimes_R \mathrm{Frac}(R)$

[broken turtle]

The Big Show

[amber isle]

I've just never seen it in the context of an arbitary subset. Anything I'm pulling has to do with cumulative heirarchies, which I'm not familiar with

[limpid horizon]

i remember in dummit and foote i think they say its the maximal number of linearly independent elements. that is the same as that? I think i saw an argument for that at some point but i forget it

[spice idol]

maximal cardinality of linearly independent set would be my go to def too lol

[fierce steeple]

Is this chat

[amber isle]

Yes lol I've been criticizing chat because it got a lot of things wrong, and have been updating it

[limpid horizon]

lol ppl call it chat

[amber isle]

I'm wondering if this another thing it got wrong

[fierce steeple]

It should all be fine here cause we're over Z

[spice idol]

it will never be "chat" (voices in my head)

[fierce steeple]

It's just less clear what rank means if you're, say, over some random ring and not free

[limpid horizon]

it may not be well defined over non comm ring or something? or was that the size of a basis is not well defined

[fierce steeple]

Personally never seen rank of a set of elements

[fierce steeple]

Well the fact bases neednt always have the same size particular means rank shouldn't have a good meaning ig

[amber isle]

So if I let M instead be a set generated by its elements instead of just the elements themselves, then it should be a-okay in any ring?

Sounds kinda farce?

[limpid horizon]

but couldnt u say rank is just the bigger one lol

[fierce steeple]

Though you need R an integral domain here ofc for Frac R to make sense

[slim dove]

ok so the recursive method they have is extra clever and scares me lmao
essentially, there's a way to reduce the whole diffie-helman semigroup version to only checking on right principal ideals of specific elements (not a left zero, not invertible)
and this can be applied as much as needed before trying to find some stuff with invertible elements instead or probabilistic methods

[fierce steeple]

Maybe

[slim dove]

and yea I agree with their conclusion that uh
Picking a good semigroup to not be easily attacked is going to be difficult and crucial for this method of encrypting

[fierce steeple]

But I mean if you choose biggest one then I guess rank of a finitely generated module would usually be infinite

[amber isle]

Lemme walk through an example then.

[fierce steeple]

Since if R^m = R^(m + k) then this is = R^(m + nk) for any n

[amber isle]

Oh ew

[limpid horizon]

just curious why u wrote R^m cause i thought only free modules are like that

[spice idol]

I can remember commutative rings having invariant basis property, at least..

[amber isle]

Wait, if I want the M = set of primes, isn't <M> just Z since the primes have gcd = 1, so the rank of M is just 1?

[fierce steeple]

Ye this was for noncomm cause kiand said

[spice idol]

right okay didn't see lol mb

[fierce steeple]

Rank of any nonzero submodule of Z is 1

[amber isle]

Lame

[fierce steeple]

Lol

Why is maths hard.

[limpid horizon]

ikr

it shouldnt be hard and we should know everything immediately

[amber isle]

I guess there could be more interesting cases if R = O_K, ring of integers for a number field, so rank can be arbitrarily large

like K is cyclotomic

[limpid horizon]

wow i have not thought abt that stuff in a while

[limpid horizon]

i dont really get ur example, is it because u can have field extensions of arbitrarily high degree or something?

[amber isle]

Yes

I gave cyclotomic as an example since dim K = \phi(n), and \phi(n) can be made arbitrarily large if it has prime powers large enough

Hmm, I guess inductively adjoining an n-th root of a prime should just give dimension n.

Ah wait, scratch that, I'll need to sit down and verify, I'm missing somethign

[fierce steeple]

This is fine, just adjoin a root of x^n - 2

[amber isle]

Oh yeah lol

[foggy galleon]

Let f : R-->S be a ring homomorphism of integral domains. Let C=C_* be a complex of R-modules, so that, by virtue of f, C otimes_R S is a complex of S-modules. We can calculate the Euler characteristic chi_f(C)=chi(C otimes_R S). I'm interested in the problem of whether

chi(C)=chi_f(C)

[scarlet ermine]

what is your definition of euler characteristic here? alternating sum of free rank of modules?

[foggy galleon]

alternating sum of the ranks of the H_i viewed as R-modules

the rank of an R-module M is the dimension of M otimes_R k over the fraction field k of R

[scarlet ermine]

i see

I suspect you only have to think about fields here since you are tensoring over the field anyway

and localization is exact so it'll commute with taking homology of a complex

in more detail, H_i(C) otimes_R k is the same as H_i(C otimes_R k) because k is a flat R-module, since it is a localization. So, this tells you that chi(C) = chi(C otimes_R k).

(exact functors commute with homology. you can get a slightly more refined result for left/right exact functors, where you only get natural morphisms, see Vakil's FHHF theorem)

so now when you consider chi(C otimes_R S), this is the same as chi(C otimes_R S otimes_S Frac(S))

hmm I think this should be the same as $\chi(C \otimes_R \mathrm{Frac}(R) \otimes_{\mathrm{Frac}(R)} \mathrm{Frac}(S))$, which I think should let you that you can reduce to the case when $R$ and $S$ are fields, and $f$ is just the field extension (of the fraction fields)

[broken turtle]

joseph

[scarlet ermine]

so if i'm not screwing up, I think it should always be true that chi(C) = chi_f(C) because field extensions are also flat

[foggy galleon]

oh that's nice

[foggy galleon]

no I don't think this is true? Frac(S) could be like Z/pZ while Frac(R)=Q, then the tensor product is zero

[scarlet ermine]

aw yeah so i am screwing up i see

hmmm

like if f: Z-> Z/pZ is the quotient map then you definitely need not have same homology

[foggy galleon]

The case of field extensions is fine (probably, but haven't worked it out myself). So... we need only worry about R-->R/p where p is a prime ideal?

[scarlet ermine]

hmm

actually I think I change my mind about Z->Z/pZ I don't have a counterexample

[foggy galleon]

there are easy ones

[scarlet ermine]

what's your example?

[foggy galleon]

just take H^0

[scarlet ermine]

oh I agree about homology actually, I mean about the free rank/euler characteristic

[foggy galleon]

yeah you can take 0->M->0 then the Euler characteristic is just the rank/dimension of M

[scarlet ermine]

oh right, if you have 0->Z/2Z->0 as cx of Z-modules, you get euler char 0, but after you tensor over Z with Z/2Z along the projection Z->Z/2Z, then you get euler char 1

[foggy galleon]

I think what I asked for is true for R=Z and the case that C is a free complex

(free as in each term is a free R-module)

[scarlet ermine]

I would guess you want f:R->S to be flat or for C to be a complex of flat R-modules

[amber isle]

So apparently, every Dedekind domain finitely generated as a Z-algebra is just the ring of S-integers over a number field?

[scarlet ermine]

so for R=Z, a finitely generated module is flat iff free, so that checks out with your example

[amber isle]

Never heard of them

Gonna look it up

[foggy galleon]

maybe a "mod p flatness condition" is more interesting

for example if R=Z and the C_n are fg. with torsion coprime to p then it works for Z-->Z/pZ (I think what happens here is that you can compute the homology with the free part)

btw I don't know more obstructions than this one. So we can ask: Suppose rank(C_n)=rank(C_n otimes_R S) (the first is the rank as an R-module, the second as an S-module). Is it then true that chi(C)=chi_f(C)?

[scarlet ermine]

yeah I'm pretty sure that should be true

this is a classic thing where you split up your chain complex into lots of SESs describing homology. Let's use the following notation for the differential d_i: C_i -> C_{i-1}.

Then, we get a SES describing homology in our classic way,
0 -> im d_{i+1} -> ker d_i -> H_i -> 0
but then also SES describing these other pieces, for instance
0 -> ker d_{i+1} -> C_{i+1} -> im d_{i+1} -> 0

so then rank(H_i) = rank(ker d_i) - rank(im d_{i+1})

and then we can replace that image bit

rank(H_i) = rank(ker d_i) - rank(C_{i+1}) + rank(ker d_{i+1})

now as we do alternating sums of the rank(H_i), the ranks of kernels will all cancel out until we hit the edges of the complex (I'm assuming this is a bounded complex in order for this alternating sum to make sense)

so actually the alternating sum of rank(H_i) is the alternating sum of rank(C_i)

one thing that's neat about this argument---rank is not special, you could have repeated this argument for any assignment from R-modules to an abelian group that is alternating on exact sequences

this is pretty much the proof that in the grothendieck group K_0(R-mod), the alternating sum of homology equals the alternating sum of the objects in your complex

[foggy galleon]

I think maybe another condition is the following. Maybe consider the case R=Z. Given the complex C, we can either form the p-primary complex or the p-torsion complex or something like that. Then we can ask for this complex to be exact, and maybe this is what explains whether chi=chi_p or not

(just thinking outloud here)

[scarlet ermine]

yeah I've been doing a lot of thinking outloud too haha

I think it was bad of me to focus on homology when really this is more about rank, I like your criteria of just trying to check rank(C_n) = rank(C_n otimes_R S)

we're asking when is dim_Frac(R) (M otimes_R Frac(R)) = dim_Frac(S) (M otimes_R S otimes_S Frac(S))

where the second could be simplified to dim_Frac(S) (M otimes_R Frac(S)) if you'd like, but that's not much easier maybe

[foggy galleon]

I see. Yeah this is nice. The downside is that the H_i might have finite rank even if the C_i don't

[scarlet ermine]

oh that is true, I was assuming everything was maybe finite dimensional

[foggy galleon]

that case is interesting nonetheless, don't get me wrong

[digital parcel]

If the C_i in C are flat and f is injective, you should get some chain map like this, right? and each vertical arrow is injective

And each C_i is isomorphic to C_i \otimes_R R, so it's basically the same as giving a injective chain map phi from C_* to C_* \otimes_R S

Then I think you can get 0 -> C_* -> C_* \otimes_R S -> coker phi -> 0 an SES of bounded chain complexes

\chi is additive so you should get like \chi(C_* \otimes_R S) = \chi(C_*) + \chi(coker phi) if I have the direction right

well I guess I'd also need to show that the homologies of C_* \otimes_R S and coker phi are finite length

in order to use that \chi is additive

[foggy galleon]

In the case of fg abelian groups, we can write C=A oplus B with A having torsion coprime to p, and B like the p-primary part of C. Then chi(C)=chi(A)+chi(B) (with any coefficients). In general I don't think you have a direct sum. But nonetheless you can define the p-primary subcomplex C_p of C. Question: if chi(C_p)=0 (with coefficients in R/p) does it follow that chi(C)=chi_f(C) (for f: R-->R/p)

or even, is there a formula for chi_f(C) in terms of chi(C) and chi(C_p)? Maybe chi_f(C)=chi(C)+chi(C_p)? (When I write chi(C_p) I mean with coefficients in R/p ofc)

it isn't clear how finiteness of the homology of C relates to finiteness of the homology of C_p tho, but maybe just assume both of these complexes have finite homology

[digital parcel]

hmm i don't think the homologies of C_* \otimes_R S will be finite length without more assumptions

i think you need faithfully flat (e.g., the field extension that joseph mentioned above; field extensions are faithfully flat), maybe even flat local?

[scarlet ermine]

i'm going to make all the finiteness assumptions and say all the C_n's are finite rank, so it's just about asking when for an R-module M, whether rank(M) = rank(M otimes_R S).

I'm trying to interpret this geometrically (since I've been spending all day learning about flatness in AG to try to prep for my upcoming oral exam eek i need to grind).

Identify M with its quasicoherent sheaf F on Spec R, and denote the induced map Spec S -> Spec R by pi. Now,

• rank(M) is the dimension of the fiber of F at the generic point of Spec R
• rank(M otimes_R S) is the dimension of the fiber of pi^* M at the generic point of Spec S

I'm seeing that injectivity could potentially be nice, because that would be saying that pi maps generic point of Spec S to the generic point of Spec R, which is at least necessary for pi to be flat, but maybe we can get a more general sufficient condition by instead asserting that pi^* M is flat over Spec R, i.e.,

M otimes_R S is a flat S-module

which would maybe be saying something about p-torsion/primary stuff, for example in the case of R=Z, S=Z/pZ. Not sure haven't thought it through or whether this condition helps.

At the end of the day, I'm hoping that there's a nice formula that tells me something about the dimensions of the fibers of pi^* F (which over the generic point is going to be free rank)... So that I can see what the obstruction is or what conditions I should apply to M or f. I want a version of Hartshorne III.9.5 but instead about dimensions of fibers. so that's my essay (i'm no closer to understanding a nice sufficient condition but i'm having fun)

this seems like a fun problem but alas i am too eepy, maybe i will learn more flatness and find some nice commutative algebra to do this when we have good finiteness assumptions (everything finite free rank, maybe also assume R,S are Noetherian)

[foggy galleon]

Btw I think this is not a very important restriction. Every module is a limit of finitely generated ones. I believe it is possible to show that any complex is the limit of complexes of finitely generated modules. And the homology of the limit is the limit of the homologies

by which I mean that the kernels and the images and all the modules are fg

[lone jacinth]

Hard to define the Euler characteristic if stuff has infinite rank though

[foggy galleon]

why not just set chi=infty if any of the homologies is infinite rank?

[lone jacinth]

For the homologies sure, but you want to exploit that it's also the alternating sum of the ranks of Cn

[foggy galleon]

that's why I'm talking about limits

(although I haven't been super careful about my assertion yet)

[lone jacinth]

Anyway, if you're ring is Noetherianess and the complex is free, bounded above with finitely generated homology it should be homotopy equivalent to a complex of fg free modules

Might not be bounded on the left, but you can throw in R having finite global dimension for that.

Or I guess just have the complex be bounded to begin with

[foggy galleon]

yeah I'm fine with assuming the complex is bounded

[lone jacinth]

Anyway, then the Euler characteristic is preserved since it's just a question about ranks of free modules

[foggy galleon]

btw how do you prove this

I think if we want to understand torsion we should ask when is it true that rank(M)=rank(M otimes_R R/p) (the first as an R-module, the second as an R/p-module)

[lone jacinth]

Take a projective resolution of each homology group consisting of fg free modules. Lift maps between them and take total complex. This then induces a quasi-isomorphism to your original complex.

Quasi-isomorphism of bounded above complex of projectives is a homotopy equivalence (you can lift inductively starting at the top)

[foggy galleon]

in the torsion-free case I don't think you can do much better than assuming C is flat and writing the Künneth spectral sequence

[muted sierra]

I've been reading the nLab article on Schur functors since I woke up (an hour ago). I understood sections 1-3 well enough, I'm not sure about section 4, and I certainly didn't understand section 5. But I guess this is the first actual example I've seen of a reason to care about 2-categories. Is there a textbook that presents this kind of material, together with motivation?

[rose mirage]

do you just want to understand schur functors like 99% of people or are you really set on the categorical nonsense

[muted sierra]

What's the difference?

[rose mirage]

schur functors are usually very concrete combinatorial things

reading this article it just seems like they've found a way to characterise them as some specific 2-morphisms in what appears to be a tricategory (due to the mention of modifications)

[muted sierra]

I'll abbreviate X for “(Vect_C)-enriched category with a tensor product in which idempotents split”. What I understood so far is how Schur functors can be defined on any single fixed X. But the 2-categorical nonsense is about how Schur functors are actually “uniformly” defined over all X's simultaneously.

[rose mirage]

I guess I'm asking what you're learning schur functors for

[muted sierra]

Killing boredom, mainly.

[rose mirage]

either way there are a load of papers at the bottom of the nlab article

[muted sierra]

Yeah, but papers aren't textbooks.

[rose mirage]

there will not be a textbook on this

absolutely not

welcome to the real world!

there are textbooks which discuss schur functors combinatorially (i.e. sections 1-3ish of the nlab page), I think fulton-harris does this

doing some googling, maybe "Schur algebras and representations" By S.Martin will be good

[muted sierra]

I guess I should start with something more mundane, like how to express the Young symmetrizers in terms of the standard basis of the group algebra C[S_n], right?

Thanks.

[rose mirage]

yeah - you gotta remember half of what's on nlab is someone's pet research project lol

it's an advertisement board as much as it is a wiki

[muted sierra]

Advertisement for what?

[rose mirage]

their work, get people interested in the stuff they're doing

[digital parcel]

sometimes they're pretty damn good at it tbh

[rose mirage]

this is a very good thing to focus on and is a very cool, this is definitely covered in Fulton Harris

very tangible too, write down a young diagram and put some numbers in, then you get your symmeterizer out by just looking at the boxes

[jaunty sparrow]

Yes, good book and a good section in particular

[lime python]

Is the statement where $H = ...$ obvious? Also $\rho_x(f)g = f(gx)$

[broken turtle]

Funky_Funktor

[near lantern]

Not totally obvious, but not hard. If I is the vanishing ideal of the closed subvariety H of G, then (since rho_x is the pullback by the right-multiplication map r_x: G → G), rho_x(I) is the vanishing ideal of the closed subvariety r_x^{-1}(H) = Hx^{-1}. It is easily seen that this is equal to H iff x ∈ H (e.g. because if it's equal then x^{-1} ∈ Hx^{-1} = H).

[ornate atlas]

Does anyone have a good source for Verlinde categories and Tannakian reconstruction? (Together or separately) really no idea where to start with that

[foggy galleon]

is this true when considering R/I where I is not necessarily principal? Suppose R is an integral domain

[past cove]

Not really

Even when your ideal is f.g. you can have a complicated projective resolution for R/I

[lone jacinth]

The i=0 part is true

[lime python]

Thank you!

[lone jacinth]

I'm starting to think that this might even be equivalent to I being principal (for R integral domain)

[lone jacinth]

The vanishing higher tors means I is flat, so
0 -> I -> R
is a flat resolution.

I'm trying to check if it holds for
(2, 1+sq(-5)) over R=Z[sq(-5)]

[fierce steeple]

ignore lol

hmm

[lone jacinth]

Hmmm, okay. Starting to think it also holds here, even though it's not principal

[foggy galleon]

can't you consider M=Z[sqrt -5]/6, then 3 is killed by 2 but not by 1+sqrt -5 ?

and in general you have this thing

[lone jacinth]

Well, I would write the condition as x such that Ix=0

[mental escarp]

Is it possible to realize any univariate polynomial as a characteristic polynomial of some matrix?

[foggy galleon]

what is the definition of M[p]? It should be { x in M such that px=0}, no?

[lone jacinth]

Okay, I see what you're saying.

There is no m in M with Im = 0, but there is torsion

So then the dream lives on!

[lone jacinth]

Yes, any monic polynomial anyway

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

[foggy galleon]

I had another question. Let R be an integral domain, let p be a prime, and let M be a finitely generated flat R-module. Is it true that

rank M= rank M otimes_R R/p ?

Can one weaken the assumptions?

If R is noethy then you can localize I think and the module becomes free?

[worldly zealot]

okay so higher tor vanishes means pd R/I =< 1, so I is projective... and then should have rank 1

i think

[lone jacinth]

You just get a bound on flat dimension (which should equal projective dimension if I is finitely presented)

[foggy galleon]

oh I think we are still good if M is finitely presented?

by https://stacks.math.columbia.edu/tag/00NX

[lone jacinth]

Flat + finitely presented = fg projective

[foggy galleon]

every fg flat module is a limit of fg projectives, no? And limits commutes with ranks, no?

[lone jacinth]

A module is flat iff it is a direct limit of fg projectives.

[foggy galleon]

yeah just realized

[lone jacinth]

I'm not actually sure what you mean by limit committing with rank

[foggy galleon]

I mean that lim rank(M_i)=rank(lim M_i)

[lone jacinth]

so lim here meaning direct limit or just limit?

[foggy galleon]

yeah I guess you should be careful. If M is fg flat, then we can find an increasing sequence of free modules N_i subset N_(i+1) subset ... subset M whose union is M (if I'm not mistaken). Then the question is whether rank(N_i)=rank(M) in the limit

[lone jacinth]

This is true, just by definition of rank as maximal linearly independent subset, but I'm not sure it helps

It doesn't use that M is fg at all

[foggy galleon]

Well we should inspect the RHS too. I.e., is rank N_i otimes_R R/p=rank M otimes_R R/p in the limit?

[lone jacinth]

Not for M=Q

The limit would just be
... -> Z/p -> Z/p ->...
with most maps 0, making the whole limit 0

[lone jacinth]

This MSE claim fg flat = projective over integral domain. So then you can just localize
https://math.stackexchange.com/q/1943418/306319

[foggy galleon]

good

thanks

[lone jacinth]

The result they're referring to is theorem 2 here
https://projecteuclid.org/journals/journal-of-the-mathematical-society-of-japan/volume-14/issue-3/On-flat-modules-over-commutative-rings/10.2969/jmsj/01430284.full

[fierce steeple]

This is cool

Thanks for sharing

[fierce steeple]

Or even fg frees ig right

[wary elbow]

Hey all! I'm wondering if there's a reference which describes (in terms of explicit matrix constructions) all real semisimple classical Lie algebras as subalgebras of GL(n, C) closed under hermitian transpose, i.e. as matrix algebras where the Cartan involution is negative hermitian transpose (which can always be done for real semisimple Lie algebras).

You can cobble together some examples using Knapp's Lie groups, and some are easy (compact algebras, split real forms of type A), but the problem in that book is that many algebras (eg the split real forms of the orthogonal groups) are conjugated to different matrix algebras whose Cartan involutions are not negative hermitian transpose. Fulton-Harris has the same issue: the Cartan involutions in some of their matrix descriptions are not negative hermitian transpose.

[wary elbow]

I'm aware of this thesis by S. Venkatesh, which contains an exposition on Verlinde categories:
https://dspace.mit.edu/handle/1721.1/122170
Milne has a book on Tannakian stuff:
http://jmilne.org/math/Books/tcdraft.pdf
(Disclaimer: I have not read either of these.)

[ornate atlas]

Perfect thank you! I was reading a paper by Venkatesh last night but it was quite dense, possibly the thesis will be a more explicit

[wary elbow]

Sure thing! I just recalled that there's also an introduction to Tannakian reconstructions in this paper:
https://arxiv.org/abs/1703.07288

[summer quest]

lovely notes

really interesting note on some subtleties about the Tannaka category appearing in geometric Satake among other things https://arxiv.org/abs/2004.10487

[last crane]

do you mean Lie algebras and Lie subalgebras of gl(n, C)?

[limpid horizon]

if f: C -> D is an isomorphism of complexes then you are pairing up the same indexed chain in the isomorphisms right

[lone jacinth]

An isomorphism of chain complexes is the same as a chain map such that
f_n: C_n -> D_n
is an isomorphism for all degrees n, yes

[wary elbow]

Yes, sorry, that should be lower case gl. I'm looking for subalgebras, not subgroups.

[last crane]

Interestingly, I am working with subgroups of GL(n, R) which are invariant under transpose: book Metric spaces of nonpositive curvature calls these groups reductive.

[wary elbow]

Yep, a reductive algebra is always the sum of an abelian algebra and a semisimple algebra, so the classification of semisimple algebras gets you all the reductives for free

It's easy to find lists of the real semisimple algebras but weirdly hard to find explicit matrix constructions

[urban granite]

<@&268886789983436800>

[spice idol]

<@&268886789983436800>

🙏

thats two in a row, lol

[lime saddle]

Very impressive

[shy cargo]

Algebra isn’t popular enough

[sly nymph]

Hello

[spice idol]

hello

[limpid horizon]

Greetings Earthlings

[spice idol]

all your base are belong to us

[digital parcel]

You must construct additional pylons

[pastel agate]

Does anyone know of a good proof/treatment of the structure theorem for modules over PIDs? My instructor tried his best but there just wasn't enough time to get through it and now everyone's confused

[lone jacinth]

There's a proof in Wikipedia that is pretty standard.

The proof is simply, a finitely generated module is the cokernel of a map
R^n -> R^m
When R is a pid every matrix can be put into Smith normal form

(See the algorithm here
https://en.wikipedia.org/wiki/Smith_normal_form )

[pastel agate]

Alright that makes sense

Gonna sit with this for a hot minute

Thanks

[unborn rampart]

If F is a linear map of k-tensors, is F(Alt(v)) = Alt(F(v)) for a k-tensor v?

hmm, no that would be too good to be true, that's only true for linear maps that's a scalar multiple of the identity IIRC

[fierce steeple]

By Alt(v) do you mean antisymmetrisation?

[unborn rampart]

yeah

[fierce steeple]

In degree k ig

[unborn rampart]

yeah, I was looking at f*(v \wedge w) = f*(v) \wedge f*(w) and wondering if any part of it generalizes in some way

[kindred maple]

Re-posting from #linear-algebra but here do we allow $V = V \oplus 0$?

[broken turtle]

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

[kindred maple]

I ask since there is an exercise which asks that if $\gamma^2 = 1$ then $\gamma$ can be made the grading operator for some decomposition of $V$. If $\gamma = \text{id}_V$ then we would need $V = V \oplus 0$

[broken turtle]

Khush

[spice idol]

yeah, V is the internal direct sum of itself with the 0 subspace and naturally isomorphic to the external direct sum V ⊕ 0

[kindred maple]

Ok bet, thank you

[kindred maple]

for this exercises can we take the generalized eigenspaces decomposition?

Its been a while since ive done JCF

because the minimal polynomial splits so the characteristic polynomial must also split

[void plank]

Suppose I'm working with chain complexes of vector spaces. Is there something analogous to tensor Hom adjunction in that category?

[lone jacinth]

Tensor product of complexes should be adjoint to the Hom-complex

Hom(A, B)_n = {fi: A_i -> B_i+n}
with differential
d(f) = dof - fod

(Possibly with some extra signs depending on conventions)

[void plank]

Ah very neat, thanks!

[lone jacinth]

https://stacks.math.columbia.edu/tag/0A8H

[candid igloo]

Let $I,J$ be ideals of a commutative ring $R$, let $f$ be a ring homomorphism from $R$ onto a commutative ring $R'$,
Prove that

$f(I \cap J) \subseteq f(I) \cap f(J)$

And equality holds if $ker(f) \subseteq I$ or $ker(f) \subseteq J$

I am asking is the last statement an if and only if condition?

That is, given the equality holds, does it necessarily imply $ker(f) \subseteq I$ or $ker(f) \subseteq J$ holds?

Can we get some counter example ? I have been stuck on this for weeks

[broken turtle]

Leno

[lone jacinth]

I mean you can have I=J without any relation to the kernel.

You can also have Z -> Z/2 and I = (3), J = (5) for example

[candid igloo]

Thanks

This was disgustingly simple
Can you please tell how you came up with it @lone jacinth

[dawn horizon]

just checking, f(I cap J) subset f(I) cap f(J) is just generally true for any sets I,J and function f right? that's not something that relies on any of the properties listed?

[candid igloo]

Yes it is a set theoretic property

[lone jacinth]

I mean, I thought what is the simplest example of a ring homomorphism I know. Checked that it worked

[candid igloo]

@dawn horizon and if for a map f from a set A to B, if for any to subsets C and D of A, if f distributes over intersections ie that Equality holds then f is injective
And vice versa ofc

[dawn horizon]

right

i think it was throwing me off that they asked to prove that given all the extra structure stuff but that makes sense

[spice idol]

f(ker f ∩ I) = f(ker f) ∩ f(I) = 0 ∩ f(I) = 0
So if equality were to hold for all I, J, then ker f ∩ I ⊆ ker f for all I. In other words, either ker f is 0 or R is subdirectly irreducible and ker f is its monolith (unique minimal nonzero ideal)

wait no that last part is not true

[lone jacinth]

Intersection something with ker f will always give a subset of ker f

[spice idol]

yes i noticed

[lone jacinth]

So an element
i + ker f
is in f(I) cap f(J) if there is a j with
i + ker f = j + ker f
i.e.
i + k = j
for some k in ker f.

Then to find a preimage in IcapJ you need to add something from the kernel so that the rhs is in J and the left hand side is in I. For the rhs to be in J you need to add something in JcapK (K being kerf).

So k needs to be able to be written as a sum of something in JcapK and IcapK. This leads to the condition
IcapK + JcapK > (I+J)capK.

So if either contains K both sums are K. And if for example both are relatively prime to K then both sides are (I+J)K

[spice idol]

f(I ∩ J) = f(I) ∩ f(J) => f^-1(f(I ∩ J)) = f^-1f(I) ∩ f^-1f(J)
=> (I ∩ J) + ker f = (I + ker f) ∩ (J + ker f)

conversely, if ker f distributes over intersection in this way then
f(I ∩ J) = f(f^-1f(I ∩ J)) = f((I ∩ J) + ker f) = f((I + ker f) ∩ (J + ker f)) = f(f^-1f(I) ∩ f^-1f(J)) = ff^-1(f(I) ∩ f(J)) = f(I) ∩ f(J)

so the equality holds for all ideals I, J if and only if ker f distributes over intersection

lmao i just noticed, the equality holding if ker f < I or ker f < J is just congruence-modularity for rings

[candid igloo]

I dont get it

[lone jacinth]

The basic conclusion is that it's kinda complicated whether the intersections are equal

[candid igloo]

Let $I,J$ be ideals of a commutative ring $R$, let $f$ be a ring homomorphism from $R$ onto a commutative ring $R'$,
Prove that

$f(I : J) \subseteq f(I): f(J)$
This is easy
And equality holds if $ker(f) \subseteq I$

I am asking is the last statement an if and only if condition?

That is, given the equality holds, does it necessarily imply $ker(f) \subseteq I$

Another one in same vein

Can we get some counter example ? I have been stuck on this for weeks

[broken turtle]

Leno

[candid igloo]

Hmmm I can't read it due to formatting
Can you do this one

[lone jacinth]

So, you say you've been stuck on this for weeks.

Have you tryed to compute some examples, which ones?

[candid igloo]

No
I tried (15) :(3) in Z with map to Z/2

[lone jacinth]

And then you noticed that answers your question...?

[candid igloo]

Yes
Are there more examples

[lone jacinth]

Should be plenty.

Seems to me it would be harder to find examples where they are not equal.

[lime python]

These are two separate questions. Can someone explain why both of these statements are true? Thank you

[limpid horizon]

“Funky Functor” lol

[sacred sentinel]

Phenomenal username

[spice idol]

she funky on my functor

[rare walrus]

For the first one, let's just focus on the semisimple part. Given R(G/R(G)), its preimage in G must be a normal subgroup containing R(G) by abstract group theory. I must admit I cannot quite remember why it must be connected, but it's definitely not hard to see that it must be solvable, since R(G/R(G)) being solvable means that the derived series of its preimage must end up in R(G), a solvable subgroup!

The unipotent statement comes from the fact that the map G -> G/R(G) preserves unipotence, we can talk about this if you like but the argument looks similar.

For the torus bit, let's say we have some normal unipotent subgroup of G x T. It's easy to see (I hope) that the unipotent elements of G x T are (u, 1) where u is a unipotent element of G. So really a unipotent subgroup of G x T is just a unipotent subgroup of G, since the torus component must be trivial. As for why it's not reductive, T embeds as a nontrivial connected sovable normal subgroup

[lime python]

Ahhhh, thank you. It seems to mostly make sense. Feel really stupid for not getting the last one

[rare walrus]

It's hard to picture these things, dw

You'll definitely get the hang of it

[limpid horizon]

Just noticed u also put the funk in functor. Sheer brilliance there

[worldly zealot]

@lone jacinth sorry that i ping you every time i have an AR/tau-tilting qestion lol. so my professor went to okinawa recently and met iyama and mousavand - apparently they were talking about bricks a lot (my professor knows nothing about this stuff). and you recently linked me a paper talking about bricks. so i get the sense that they are becoming important and people want to phrase representation theory statements in terms of bricks, but they still feel pretty foreign to me.

what motivates this? why do people think bricks are useful

[lone jacinth]

I was also recently in Okinawa on a conference Kaveh organized, so that's fun.

I guess the main thing is that there's a bijection between indecomposable tau-rigid modules and bricks, so one can translate some problems into the language of bricks which is hopefully easier.

Then it also connects to torsion classes, where a torsion class is determined by the bricks it contains, so you label irreducible inclusions of torsion classes by the one extra brick the bigger has.

I would imagine PersonX knows more about this than me actually, isn't he your advisor?

[worldly zealot]

nop, this is different professor lol

was for a TDA conference

[lone jacinth]

Right, but PersonX is your advisor.

[worldly zealot]

that is good to know though, i'll ask him

ya

xd

[digital parcel]

Oops in the first line where there's "semisimple, simply connected" ignore that

[near lantern]

Trying to read your question; just to confirm, does W_Q^min consist of the minimal elements in the left cosets of W_Q?

[digital parcel]

Yeah

The minimal length representatives of each wW_Q

[near lantern]

It seems like wQ = 1Q in this case. Maybe the question meant there is a bijection between W^min and T-fixed points, but didn't mean that only W^min give fixed points wQ?

Indeed, I think for w in W_Q, wQ = Q (at least for GL_n this is obvious). Therefore for general w, wQ only depends on the left coset of w in W/W_Q.

[digital parcel]

It was a statement they made, here it is verbatim (maybe I misunderstood):

Let $Q$ be a parabolic subgroup $Q \supset B$. Recall, $W = N(T)/T$ and $T \subset B \subset Q$ for any parabolic subgroup $Q$. Therefore, given any $w \in W$ there exists a well defined coset $wQ$ in $G/Q$ which we will denote by $e_{w, Q}$. Then the set of $T$-fixed points in $G/Q$ for the action given by left multiplication is precisely ${e_{w, Q} \mid w \in W^\text{min}_Q}$

[broken turtle]

anamono

[near lantern]

In fact for w in W, TwQ = wTQ = wQ (since w in N_G(T)). So wQ is a T-fixed point for all w in W. Presumably the ^min is just to eliminate redundancies

[digital parcel]

Ah yeah

It's that "precisely" part that's throwing me off though

Brion has a similar proposition in his notes as well

here F_w = wF, where F is the standard complete flag

[verbal panther]

does wheel theory have any applications?

[spice idol]

nice example of a universal algebra

[verbal panther]

can an operation be unital, self distributive, and associative?

[lone jacinth]

I did read a paper once that motivated wheels as a good alternative to fields, since the category is much more well behaved (which connects to universal algebra).

This then simplified some things in algebraic geometry and computer algebra.

It might have been this paper
https://www2.math.su.se/reports/2001/11/2001-11.pdf

[spice idol]

not sure about unital, but associative shelves exist

[lone jacinth]

Trivial group I guess

[spice idol]

a unital (left) self-distributive associative operation is equivalent to a unital associative operation satisfying yxy=yx

which is somewhat interesting

[verbal panther]

Truth nuke

[lone jacinth]

Multiplication in a boolean ring also works

[verbal panther]

Oh weird

[woven loom]

I mean, x = x1= x(11) = (x1)(x1) = x^2, so it’s gotta be like that

[spice idol]

so, for example, commutative idempotent monoids (i.e. semilattices bounded above)

yeah idempotency is guaranteed

it's like a.. skew lattice?

but semi

they can also be called "unital associative spindles"

fwiw

[spice idol]

https://arxiv.org/abs/1603.08590
lol

[unborn rampart]

love the title of the first chapter

[spice idol]

I'm surprised that Mohammed Elhamdadi isn't on this

aha

https://arxiv.org/abs/2301.12286
there we have him

[spice idol]

oh man this has a very nice free structure

on two generators at least

[vague pawn]

2001

Is jagr like 50yo or something

[fierce steeple]

Python

[fierce steeple]

Lol I am confused

Are we only allowed to read papers from the last 20 years

[vague pawn]

Oh readdddd

I read it as write

[fierce steeple]

Ah lol

He wrote it age 5 actually

[digital parcel]

wow...

[spice idol]

hmm...

now I am interested

[near lantern]

More generally, inf in a meet-semilattice (with top element).

[near lantern]

Oh we said the same thing

[spice idol]

hehe

xyzxy
is the longest element (up to permutation) you can form with three generators, and also the unique element up to permutation of this form. Then the rest (up to permutation) are:
xyzx
xyz
xy
x

I'm sensing a pattern here lol

[near lantern]

This is your result for parabolic = Borel (W_Q trivial).

[near lantern]

What it means or how to prove it?

[spice idol]

okay it gets complicated for >3 generators

someone can write a program to calculate the minimal representations of words in the free structure :P

[ornate atlas]

Whenever I need some obscure algebraic bullshit I know the exact person to come to, you have the most esoteric knowledge around (and yet also know a lot of the normal stuff)

[spice idol]

🙏 happy to help

[ornate atlas]

Do you just trawl the depths of arxiv looking for the most fucked objects you can find?

[vague pawn]

Why is xyxyxyxyxyxyxyxyxy... not considered longer

[ornate atlas]

Or are these of actual interest in UA?

[spice idol]

that is (xy)(xy)(xy)... which collapses to xy

[vague pawn]

Oh

[spice idol]

:3

[vague pawn]

Actually that is a really interesting question

[spice idol]

well they are just interesting structures in my opinion, because they behave fucky wucky

but I honestly just hear about them in passing, or see articles about them (usually from mohammed elhamdadi)

BCK-algebras, though, I heard from them from another universal algebraist I talked to (they're important to algebraic logic)

[ornate atlas]

I’m not saying they’re not interesting, I very much agree that they are, I’m more just curious where on earth one comes across spindles lol

[spice idol]

just a name people give them for completeness sake lol

[verbal panther]

That sounds fun

[spice idol]

well, idempotent stuff is always nice, and some big conjecture in constraint satisfaction problems has been reduced to the case of studying concrete clones (concrete clones are, on a high level, what universal algebra studies) where every operation is idempotent

[ornate atlas]

You should get into braces, they have a crazy structure

[verbal panther]

What are braces

[spice idol]

something something Yang-Baxter

[spice idol]

in fact, you can see (abstract) clones as generalisations of monoids, and the representation theory of clones is equivalent to universal algebra

[vague pawn]

This is exponential

[ornate atlas]

https://www.pure.ed.ac.uk/ws/portalfiles/portal/459312326/powerfulgroupsandbraces.pdf

[spice idol]

more generally, this is categorified to algebraic theories and Lawvere theories

I'd love to learn about braces it sucks that I'm knee deep in some UA stuff right now haha

[digital parcel]

What it means, since to me it reads like “wQ is T-fixed if and only if w in W^min”

[spice idol]

I can imagine

it is also important to note that with stuff you cannot be sure that taking the "greedy approach" will lead to a minimal representation of a term

[verbal panther]

Wait if aba=ab wouldn’t this just collapse to xyz

[fierce steeple]

Lmao

[vague pawn]

in the free strucute this doesn't happen

[verbal panther]

Ah

What is the free structure?

[spice idol]

the "most general" structure that satisfies the axioms

can be defined a couple of different ways; either categorically or via construction

[verbal panther]

So like a proper object?

[verbal panther]

What are the properties of the free structure

[spice idol]

I think the construction is the easiest to understand: take your language (in this case only consisting of the unit and a binary operation which I will write as juxtaposition). Then, take all expressions involving, in this case, the unit 1 and binary operation, and some elements of a set X, called the set of variables. You then quotient out by the axioms, essentially. It's a little more involved (you take a so-called fully invariant congruence), but that's the gist

so take the expression (informal, but as the operation is associative this is allowed)
xyxyxy...xy
we can use associativity to get the expression
(xy)(xy)(xy)...(xy)
and then use idempotency (which we have derived) to see that this collapses down to
xy

so it identifies two expressions precisely when you can "do the algebra" to turn one into the other

it turns out that this has many very nice properties, and you can prove a lot of nice stuff with them (ranging from stuff in universal algebraic geometry to general structure theorems, laying the link between algebra and logic, etc)

you encounter this in e.g. module theory or group theory with free modules and free groups

[verbal panther]

What would be the free structure of a group?

[spice idol]

the set of all reduced words in the set { x, x^-1, y, y^-1, ... } (where all the x, y, ... are your variables), where reduced means that xx is written as x^2, and xxx^-1 is written as x^2 etc.

[verbal panther]

Ok I think I get it

[spice idol]

nice!

okay actual crazy theorem

(a convex set is a set that, if a, b in the set, then all integers between a and b are in the set)

[verbal panther]

Is it like keeping the properties of a structure but replacing its elements with all unique permutations of its original elements?

[near lantern]

I think you should translate it as "for all w in W, wQ is a T-fixed point in G/Q, all T-fixed points are of this form, and w1 Q = w2 Q iff w1, w2 lie in the same coset in W/W_Q; as a corollary, since we can use W^min to canonically choose representatives for each coset, each T-fixed point is wQ for a unique w in W_Q".

[vague pawn]

not permuations but strings

and taking these strings and giving them an algebraic strucutre

for the free group, lets say you have X, then in any group containing X you must have XX, XXX, XXXX,XXXXXX,....

[digital parcel]

That makes more sense thank you!

[vague pawn]

but also X^-1 such that XX^-1 is the identity

[verbal panther]

You can “collapse” strings but only if the output of a collapse is the identity element?

[spice idol]

@ornate atlas if you thought "spindle" was weird

[near lantern]

"Groupoid with identity" is bad enough 🤡

[spice idol]

(algebraisations of Steiner systems btw)

[vague pawn]

you can collapse if you have an element times its inverse

[verbal panther]

But earlier they collapsed xyxy

[vague pawn]

the point of the free group is the each you want a group containing X, so you take all elements that must be in that group and make them distinct

[spice idol]

I suppose "collapsing" should be called "reducing"

[spice idol]

they also take their rings to be nonunital

[vague pawn]

there we also wanted idempotence

[verbal panther]

Does that include something like xyz where xy is the inverse of z

[vague pawn]

sure

[spice idol]

no, I am as an universal algebraist also not happy about this

[near lantern]

And even their groupoids? Their groups??

[spice idol]

apparently...

well no not groups, obviously

we're universal algebraists

every. operation. has. to. be. included. in. the. signature.

[verbal panther]

universal algebra is cool

[verbal panther]

What’s idempotence

[spice idol]

I'm sadly one of (if not the only) one studying it in this server

[spice idol]

xx=x

[spice idol]

at least, who is active here

[vague pawn]

sadly I tend towards wanting to learn NT more

[verbal panther]

What’s nt

[vague pawn]

number theory

[spice idol]

number theory ewwwwwwwww /j

[vague pawn]

I need to get my shit together and start reading class field theory for real

but its really hard for me

but now I am trying to build some background in AG

@spice idol can I have some motivation (like cheering) to study AG tomorrow

will start the chapter on algebraic variates

[spice idol]

oh sure!

what're you reading?

[vague pawn]

milne

[spice idol]

hmm don't know much about that

I was reading Liu but got sidetracked doing UA

[vague pawn]

I wanted motivation like cheer me up xD

[digital parcel]

wait have you not been doing algebraic varieties lol

[vague pawn]

like not affine

[digital parcel]

ah

[vague pawn]

next chapter after this are projective varieties

[verbal panther]

Who doesn’t love sticking properties together and seeing what freak of nature pops out

[spice idol]

yurr, however I'm working on a pretty high level at the moment

much less individual algebras / algebraic structures

[verbal panther]

Also is there a notion of identity for higher order operations

[spice idol]

eh

[verbal panther]

Sounds fun

[spice idol]

multiple

f(1, ..., x, ..., 1) = x for all positions of x

is the most common one

[fierce steeple]

f(1...,1, x, 1, ... 1) ig lol

[spice idol]

yeah

idk it just doesn't really come up often / is that useful

idempotent varieties are more important

[verbal panther]

How would that work for higher orders?

[spice idol]

just f(x, ..., x) = x

[verbal panther]

Ah

[spice idol]

okay now we're just making up shit
"demi-semi-primal"

pack it up...

[verbal panther]

Real math done by real mathematicians

[spice idol]

"If U is a locally <a, b>-minimal neighborhood and V is minimal wrt a|V ≠ b|V, then there is a polynomial isomorphism between U and V. Consequently, V is also locally <a, b>-minimal"

this ts i have to prove

💔

[verbal panther]

Is abstract algebra basically intro to universal algebra

[woven loom]

Boy do I know the guy to find those

[spice idol]

universal algebra isn't done much right now

so I wouldn't say that necessarily (mainly because lattice theory isn't standard)

[verbal panther]

Lattice theory?

[spice idol]

posets with finite sups and infs

[spice idol]

do say I want to know more fuckass algebras

[woven loom]

I’m referencing how every time I bring up some question I get told I study the worst questions

[spice idol]

😭

rude

[woven loom]

But I mean I did eventually prove those terrible groups are often closed under finite products

With a gorillion technical assumptions on structure or analysis

[spice idol]

I vaguely remember you asking that question

what were they again

[woven loom]

Group stability stuff

[spice idol]

ah, model theory stuff

[spice idol]

lattice theory is in some sense the "language" of much of universal algebra

[woven loom]

Indeed

[spice idol]

though lately category theory has been creeping its way into my research

@.@ you cannot escape it, truly

[verbal panther]

Category theory :3

[woven loom]

Stinky

[spice idol]

well, it is a natural setting for duality stuff, which pops up a lot in universal algebraic geometry

[woven loom]

Fair

[spice idol]

and I'm slowly coming to the opinion that it is also the natural setting for tame congruence theory

[woven loom]

Wild congruence theory:

[spice idol]

they dont tell you much about the algebra...

[verbal panther]

I love self distributivity now such a fucked up property

[verbal panther]

What’s that

[spice idol]

a joke, lol

[verbal panther]

So it is

What is tame congruence theory

[spice idol]

you look at properties of some algebra by looking "locally" at how the operations behave on "nice" subsets

these "nice" subsets are called neighborhoods and correspond to splittings equivalence classes of splittings of some algebraic structure. Currently I am in the process of trying to rewrite tame congruence theory purely in terms of these splittings

a splitting of an object Y is a diagram X -s-> Y -r-> X with rs = id_X btw

[verbal panther]

Neat

[spice idol]

and there's this really nice interplay between "local" and "global" properties

that's where the name "tame" comes from; a quotient R < S is tame whenever you can do this local/global stuff

[verbal panther]

Subsets of what?

[spice idol]

subsets of your algebra

[verbal panther]

Ah

[digital parcel]

Let $G$ be an algebraic group, $B$ a Borel subgroup, $W$ its Weyl group, $R^+$ the set of positive roots. For $w \in W$, let $B_w = B \cap wBw^{-1}$ and $X(w)$ be the Zariski closure of $BwB$ in $G/B$. Then we have an isomorphism $B/B_w \cong \prod_{\beta \in N(w)} U_\beta$, where $N(w) = {\beta \in R^+ \mid w^{-1}(\beta) < 0}$ and $U_\beta$ is the root subgroup corresponding to $\beta$. Why does it follow then that $\dim X(w) = # N(w)$? I can understand that $\dim B/B_w = # N(w)$, since each $U_\beta$ is one-dimensional, but I'm not sure how they got $\dim X(w) = \dim B/B_w$

[broken turtle]

anamono

[digital parcel]

Maybe there's some orbit-stabilizer theorem thing here

Oh I guess you have an action on B and B_w is the stabilizer so BwB is isomorphic to B/B_w?

[near lantern]

If you mean that there is an action of B on G/B and B_w is the stabiliser of the point wB, then yes.

[near lantern]

Also BwB/B is open in X(w) (orbits of algebraic group actions are locally closed) and dense too (BwB/B is irreducible (image of B under the map b ↦ b⋅wB = bwB, and B is connected hence irreducible), so X(w) is irreducible, and its non-empty open subset BwB/B is dense). Hence dim X(w) = dim BwB/B.

[digital parcel]

I see, thanks

[verbal panther]

I’m pretty sure you can only have a finite amount of members given a finite amount of generators

Now im curious what’s the minimum amount of generators required for a structure with that operation and n members

Does distributivity require a(bcd…)=abacad… or just a(bc)=abac

[lone jacinth]

The latter implies the former

[verbal panther]

Oh yeah you’re right

[verbal panther]

@spice idol would you know about this?

[spice idol]

depends on your variety

its terribly hopeless to answer it in general though lol

i cant say ive got a good feel for these structures yet though so i cant say

(the associative unital shelves)

[spice idol]

yes, a letter can only occur twice if it is at the far right of the word (assuming left-distributivity)

wait no thats not even true

letters cannot occur twice in any circumstance

[verbal panther]

abca?

Oh

Huh you’re right

[spice idol]

yxy = yx for all x, y

shit's fucked

[weak lodge]

~~actually all unital shelves are associative; see Cor. 1.6 of https://arxiv.org/abs/1603.08590 ~~

[worldly zealot]

shelves

[verbal panther]

So given n generators you could have at most sum up to n choose n elements?

[digital parcel]

lol i didn't even notice i read it as sheaves

[spice idol]

ooo

so we are just working with unital shelves

self distributivity is like

weirdly poweful

combined with unitalness

[spice idol]

ye

but thats an upper bound

i think its closer to n! + (n-1)! + ... + 2! + 1 + 1

no thats not right

uhh

the sum of k!(n choose k) as 0 ≤ k ≤ n

Consider the free idempotent monoid F(X). The reduction r(w) of a word w = x1...xn is the largest word w' = x1...xk with k ≤ n where no xi occurs more than once.

We say w ~ w' iff r(w) = r(w')

show that this is a congruence and that F(X)/~ is a unital shelf

then necessarily, it must be fully invariant and so the free unital shelf because of reasons

that its a congruence shouldnt be too hard to see

[verbal panther]

I love it it’s so fucked

[spice idol]

Let w, w' be words. Then clearly r(ww'w) = r(ww'), as everything from the end of ww' appears in the beginning, so indeed F(X)/~ is an associative idempotent magma with aba = ab, i.e. a unital shelf

so this means there is a really simple way to determine whenever to terms are the same after doing "the algebra" in associative shelves!

just throw away all parentheses, remove repeats, and trim off the right until there are no repeats!

[spice idol]

i.e. this gives us the exact amount of elements!

thats kinda crazy

[verbal panther]

Are you sure it wouldn’t just be n choose k?

I’m pretty sure when order doesn’t matter it’s n choose k / k! When order does matter ie commutative structure and just n choose k when it does

[spice idol]

order matters

our structure is not commutative

[verbal panther]

Yeah

[spice idol]

n choose k chooses k elements from a set of n elements, regardless of order

so you multiply by k! to account for the k! different orderings

[verbal panther]

Ah I was just misremembering the formula then

[spice idol]

hehe

[verbal panther]

Kind of crazy how much adding an identity changes it

[spice idol]

yeah its power is unexpected

but on the other hand, the existence of the identity in rings is very important too

[verbal panther]

Self distributivity is awesome

[verbal panther]

Oh that’s wild

[spice idol]

look this is cool but

i dont see any reason to care about this lol

as is with most papers in this part of algebra

i love me a cool homology theory but never (except in like 2 or 3 cases) do they actually give a reason to care about the homology lol

like, what does the homology actually tell us about these structures

the problem i think is that were doing homology instead of cohomology

[digital parcel]

I'm a little confused on the notation between $BwB$ and $BwB/B$. My book denotes $e_w \coloneqq wB$. Then there's a Bruhat decomposition $G/B = \cup_{w \in W^B} Be_w$. So if I plug in $wB$ for $e_w$, I should get that $G/B = \cup BwB$. But the correct bruhat decomposition should be $G/B = \cup BwB/B$. So what's the difference between the two

[broken turtle]

anamono

[rose mirage]

This was perfect bedtime reading, thank you for sharing

[rose mirage]

The computations in the appendix make my neurons activate. Especially the “two term” H^0

[near lantern]

I think they use BwB for what I call BwB/B. I feel that it's not quite correct to use BwB to denote a subset of G/B since it's already a subset of G.

That's probably all that happened.

[spice idol]

yeah thats very nice, i wonder for what associative shelves the rack homology is H_n = Z^{k^n-1} if H_0 = Z^k

[digital parcel]

Ahh okay thanks yeah that makes sense

[verbal panther]

Are associative shelves necessarily idempotent, or just unital ones?

aaa=aaaa at least

Are there any useful non transitive structures

[spice idol]

not necessarily

only if theyre unital

[spice idol]

wdym by transitive

[verbal panther]

If a=b and b=c a=c

If for every a and b in a structure there’s an n such that an=b is the operation invertible

[spice idol]

? equality is transitive thats literally in the definition

[spice idol]

no, the n may not be unique