I'm a big advocate for it becoming an unironic accessiblity feature
#groups-rings-fields
elder wave
I'm a big advocate for it becoming an unironic accessiblity feature
There’s a chrome extension for it
chilly ocean
Whats wrong with having solutions for self learning? In my analysis class my teacher gave us the instructors book
dim widget
Can you guys give me a chrome extension that plays subway surfer under y’all’s comments?
teal vessel
or I can make only certain letters bold to make some secret messages or something.
slim kayak
Wdym y'all can study without some obscure video essay as background noise
teal vessel
I've found that it's much better to not have solutions at least in the beginning, though much of my work is "prove X" at this point which makes it easy to know where I'm starting and where I'm ending.
delicate orchid
There’s a chrome extension for it
literally all of these extenstions sound like key loggers
I will do without
frigid lark
Wdym y'all can study without some obscure video essay as background noise
I think a parrot pecking at my laptop is enough distraction for me
dim widget
@delicate orchid embrace the surveillance state
delicate orchid
I'm such a NPCpilled psyopcel
chilly ocean
<@!233993116418441216> embrace the surveillance state
Say good bye to privacy
dim widget
say goodbye to piracy
chilly ocean
Lol
frigid lark
I think the book publishers are gonna stop piracy purely by making their pdf's too big
teal vessel
how can you read in all bolded words? I can't get past how lumpy and extremely jarring it is to read.
delicate orchid
it's not lumpy it's so much quicker and smoother
all these neurotyps in MY channel smh
slim kayak
Probs forces then to slowdown while reading while maintaining a regular reading pattern
Compared to whatever they usually do ig
dim widget
Hail galaxar 🫡
delicate orchid
no, it lets me read at the speed my brain goes at, it's faster
I feel like I'm in a twlight zone episode
teal vessel
literally learned a different language in order to use a logographic script instead of english so I don't have to read long words you wanna call me neurotyp again?
delicate orchid
yeah, neurotyp
frigid lark
no, it lets me read at the speed my brain goes at, it's faster
Standard textbook speed?
slim kayak
yeah, neurotyp
The worst insult in the abstract algebra channel
delicate orchid
oh and the miles davis' ones
frigid lark
don't know what that means
Slow
delicate orchid
badada ba bapppp [13 minute drum solo] bebababobo
teal vessel
that's the wrong color for miles davis
south patrol
As soon as I see that listing I assume chat gpt
dim widget
Bada bee bop bop badap bopp
south patrol
Lol
delicate orchid
Bada bee bop bop badap bopp
no no the one that goes be baba bap baba boop
south patrol
Blah blah blah.
That said, a list of blah is
1.
2.
3.
4.
Another paragraph reiterating the first one
delicate orchid
chat GPT uses the oxford comma
so... based...
south patrol
Where is the oxford comma lol
solemn garden
*literally learned a different language in order to use a logographic script ins...
Which one? Chinese?
delicate orchid
Where is the oxford comma lol
last paragraph
frigid lark
Yes, if K is a p-group, where every proper subgroup is contained in H and [K:H]=p, then we can say that K is cyclic.
To prove this, let's consider the case where K is not cyclic. Since K is a p-group, by the classification of finite abelian groups, we can write K as the direct product of cyclic groups of prime power order. Let's denote these cyclic factors as C_{p^{k_1}}, C_{p^{k_2}}, ..., C_{p^{k_r}}, where k_1 ≥ k_2 ≥ ... ≥ k_r > 0.
Now, let's consider the subgroup H of K such that every proper subgroup of K is contained in H. Since K is not cyclic, it means that at least two of the prime power factors, let's say C_{p^{k_i}} and C_{p^{k_j}}, are present in K. Without loss of generality, assume that k_i ≥ k_j.
Now, let's construct a subgroup M of K, which is isomorphic to C_{p^{k_i}}. Since C_{p^{k_i}} is a cyclic group, it has a unique subgroup of order p. Let's denote this subgroup as M. Since K is the direct product of cyclic groups, we can write K as K = M × N, where N is the remaining part of K after removing M.
Now, since H contains all proper subgroups of K, it must contain M. But this implies that the index [K:H] is greater than p because [K:H] = [M × N:H] = [M:H ∩ M] × [N:H ∩ N] = [M:H ∩ M] × [N:H]. Since H contains M, we have [M:H ∩ M] = 1, so [K:H] = [N:H]. But [N:H] > 1 because N is the remaining part of K after removing M, and it contains at least one cyclic factor C_{p^{k_j}}. This contradicts the assumption that [K:H] = p.
Therefore, our assumption that K is not cyclic leads to a contradiction. Hence, we can conclude that if K is a p-group, where every proper subgroup is contained in H and [K:H] = p, then K must be cyclic.
teal vessel
Which one? Chinese?
delicate orchid
oh lawddd the conlangers are here
south patrol
Yes, if K is a p-group, where every proper subgroup is contained in H and [K:H]=...
Lol
I sthis chat gpt
frigid lark
Yeah
delicate orchid
Yes, if K is a p-group, where every proper subgroup is contained in H and [K:H]=...
obvious. Consider the subgroup graph
dim widget
In general, ChatGPT's writing style does not rely heavily on listing items in an essay format. Instead, it tends to generate responses in a more conversational manner, with coherent paragraphs and sentences. However, depending on the context and prompts provided, ChatGPT may occasionally incorporate lists or bullet points to organize information or present a series of related points.
It's important to note that the specific writing style of ChatGPT can vary based on the training it has received and the patterns it has learned from the data it was trained on. While ChatGPT aims to provide informative and helpful responses, including organizing information into lists when appropriate, it is ultimately an AI model and its writing style may not perfectly align with typical human writing styles or preferences.
dim widget
The mods can’t help you now
teal vessel
the helps can't mod you now
chilly ocean
I wanna program a mathgpt
delicate orchid
shut up nerd
dim widget
@chilly ocean train it off of people who ask < 3 questions on this server
frigid lark
Surely irony defence works
chilly ocean
<@456226577798135808> train it off of people who ask < 3 questions on this serve...
That would make for a very good masters thesis
frigid lark
<@456226577798135808> train it off of people who ask < 3 questions on this serve...
furry gpt
elder wave
no chatgpt posting folks
dim widget
@elder wave are you trying to kill #groups-rings-fields
hidden haven
no chatgpt posting folks
I wasn't involved in this chat btw
elder wave
There was some context to it so whatever for now but don't do that
dim widget
We have to rise up against the mods. When they came for foundations I said nothing because idgaf…
teal vessel
oh lawddd the conlangers are here
I would throw a whole paragraph in sp in here but that's way too off topic for this chat, and also annoying to use a bot in another server for.
frigid lark
Algebra has words, conlang's have words
delicate orchid
Yes, if K is a p-group, where every proper subgroup is contained in H and [K:H]=...
what's the actual proof btw
teal vessel
algebra is the study of mathematical structure, it's just grammar but for math
frigid lark
what's the actual proof btw
Induction, centre of a p group is not trivial
Basically if you can get K abelian you are done
delicate orchid
swag
solemn garden
What’s it called
frigid lark
And K/Z(K) is a smaller p group
delicate orchid
(P is a p-group, [P:Q] = p, and H < P implies H <= Q) implies P cyclic
frigid lark
If K is a p-group, where every proper subgroup is contained in H neq K, show that K is cyclic
teal vessel
What’s it called
the good language. pronounced "toki pona"
south patrol
If K is a p-group, where every proper subgroup is contained in H neq K, show tha...
This is actually true
hm
delicate orchid
yeah I mean I can see it's absolutely obvious by just pointing at subgroup lattices
south patrol
Why?
south patrol
Why can't it be a mess lol
delicate orchid
contains at least one C_p \times C_p right
hmm
ok so that does it for the p^3 case
solemn garden
~~the~~ good language. pronounced "toki pona"
oh I’ve heard of that, how do you deal with the word for “simple” being the same as the one for “good”
frigid lark
So K/Z(K) is a smaller p group, and every proper subgroup of K/Z(K) is contained in H/Z(K) by correspondence theorem
frigid lark
Then you do induction
south patrol
Oh okay that's nice
frigid lark
Also if K is abelian, K is cyclic
frigid lark
By finitely venerated abelian grapes
delicate orchid
actually does it do it for p_-^{1+2}? that has exponent p^2
south patrol
I was wondering how you'd use induction but using the centre makes sense
delicate orchid
hmmmm
south patrol
I guess the idea is just you want to pick smth liek canonical to reduce the size and for p-groups the centre works?
teal vessel
oh I’ve heard of that, how do you deal with the word for “simple” being the same...
context, primarily. We should probably take this to #chill if you wanna talk more.
delicate orchid
oh I'm not using the centre at all, I have no idea how that would go
dim widget
Can’t you prove this directly?
Because if g is an element which maps to a generator of G/Q then <g> = G
delicate orchid
actually does it do it for p_-^{1+2}? that has exponent p^2
nah it still works :iwon:
dim widget
It actually seems to have nothing to do with being a p-group..
frigid lark
Hmmm, nice
dim widget
It’s a much weaker condition than being a path
delicate orchid
that reformulation only holds in the p-group case
dim widget
The point is that you can’t have a single maximal nontrivial subgroup in a group.
delicate orchid
it's like having a little node at the top and then the rest of the lattice exploding off of the one beneath
frigid lark
It actually seems to have nothing to do with being a p-group..
I do think the restrictions force it to be a p group, cause if it's not I think Sylow will make it so such a Q doesn't exist
In the finite case
dim widget
@frigid lark that’s true
delicate orchid
index p p-subgroup => p-group yurrr
dim widget
@delicate orchid we’re not assuming Q is a p-group
But it will have to be if it satisfies the condition in the theorem
delicate orchid
I'll just let you two get on with it 
ok no I won't actually
I've re-read this like three times now
delicate orchid
Then you do induction
what does this show?
delicate orchid
Also if K is abelian, K is cyclic
why?
delicate orchid
Because if g is an element which maps to a generator of G/Q then <g> = G
what?
thanks for coming to my ted talk. I will now enjoy my ham and cheese toastie
delicate orchid
quite a witty quip
dim widget
white a quitty whip
delicate orchid
it's fine dw I will have long since forgotten by then
frigid lark
But Topos' proof is league's better
frail summit
Shouldn't this say "where p'(x) is a product of irreducibles in both R[x] and F[x]"?
dim widget
@frail summit yes.
frail summit
Oki, thx
glossy crag
If A and B are similar CSAs (there is a division algebra with A\cong M_n(D) and B\cong M_m(D)) and C is a simple central artinian algebra (not necessarily finite-dimensional), are A\otimes C and B\otimes C similar? I know this holds if C is finite-dimensional, i.e. a CSA.
glossy crag
If A and B are similar CSAs (there is a division algebra with A\cong M_n(D) and ...
NVM, D is necessarily finite-dimensional, so D\otimes C is artinian.
glossy crag
Is it correct to say that if K-algebras A and B are isomorphic as rings, then they're isomorphic as algebras? K is a subring of either centre and A\cong B\implies Z(A)\cong Z(B), so the K-actions should be the same, no?
oblique river
is that true? let's say that you have a field K and a (strict) subfield L which happens to be abstractly isomorphic to K. Then, you can consider K as a K-algebra in two ways: it acts on itself by the identity map K --> K, and it also acts on itself via the composition K \cong L \subset K
K is isomorphic to itself as a ring but I don't think these are the same K-algebra structure
night onyx
Yeah tl;dr an algebra structure on a ring is given by the ring structure together with an embedding of the field into the center of the ring, and you can have different embeddings of K into Z(A) which produce non-isomorphic algebras
glossy crag
is that true? let's say that you have a field K and a (strict) subfield L which ...
What's an example of such K and L?
Something to do with transcendence bases?
oblique river
you can do that yeah
glossy crag
Something to do with transcendence bases?
Because i can't think of any number fields like this
oblique river
like K = C(x1, x2, x3, ...) and L = C(x2, x3, ...)
you can even take K = C which happens to be isomorphic to a strict subfield of itself
this wont work for anything finite dimensional over Q
night onyx
I think R is a Q(root(2)) algebra with root(2) -> +-root(2), and the algebra structures aren't the same (though I don't remember why lol)
glossy crag
you can even take K = C which happens to be isomorphic to a strict subfield of i...
how do you figure that?
oblique river
there is a theorem that C is teh only algebraically closed field with characteristic 0 and cardinality of the continuum
the algebraic closure of the field C(x) also has both of those properties
so it's isomorphic to C
so you can consider the inclusions
C --> C(x) --> \bar{C(x)} \cong C
dim widget
It’s easy to find ring isomorphisms which are not K-algebra isomorphisms as long as K is not Q. But if A, B are CSA’s and they are ring isomorphic then they are morita equivalent thus equivalent in the usual sense
Then the theory of morita equivalence for CSA’s tells you that there is a K-algebra isomorphism
delicate orchid
I think R is a Q(root(2)) algebra with root(2) -> +-root(2), and the algebra str...
R is a Q(sqrt(2)) algebra with all the operators you'd expect because Q(sqrt(2)) is a subfield of R. I don't know what you meant by root(2) -> +- root(2)
dim widget
@glossy crag wouldn’t it be the Swiss guard bursting in?
oblique river
i think they meant Q(sqrt(2)) can act on R where sqrt(2) acts by multiplication by -sqrt(2)
glossy crag
<@233934386863669248> wouldn’t it be the Swiss guard bursting in?
i'm too inept to edit Cantor's face in. why Swiss?
night onyx
yeah exactly, basically two different embeddings of the field Q(root(2)) into the ring R
delicate orchid
i think they meant Q(sqrt(2)) can act on R where sqrt(2) acts by multiplication ...
oh I see I got it backways
my apologies
night onyx
np
dim widget
@glossy crag because a cardinality is the domain of influence of a cardinal
glossy crag
<@233934386863669248> because a cardinality is the domain of influence of a card...
Ahahahaha, good one, didn't click at first
Swiss guard didn't register as the papal guard and I thought your implication was Cantor is Swiss (which he isn't)
south patrol
bump lol
rocky cloak
I think R is a Q(root(2)) algebra with root(2) -> +-root(2), and the algebra str...
Well in one sqrt(2) has a square root and in the other it doesn't. So that's why they're different.
I'm guessing it's supposed to say sigma(1) = i and sigma (2) = j ...?
dim widget
@south patrol what is your question?
glossy crag
If R and S have are rings of lengths n and m (as left modules over themselves), then R\times S is of length n+m, right?
dim widget
Yes
delicate orchid
yeah you just take the left doodad x S and then the right doodad
glossy crag
Yeah, just checking.
glossy crag
Do you concur, @delicate orchid?
south patrol
I'm guessing it's supposed to say sigma(1) = i and sigma (2) = j ...?
Ah I assume so
Thank
white oxide
what do they mean by indeterminate over F?
rocky cloak
what do they mean by indeterminate over F?
Freely adjoined variable. I.e. x is transcendental over F.
white oxide
Freely adjoined variable. I.e. x is transcendental over F.
ah ok, thanks
hollow fjord
Hi guys! I have the following problem. Let $G$ be a compact group, $V$ a complex vector space (finite dimensional), if $f: G \times V \to V$ is a group action that is compatible with the structure of $V$ and $\rho: G \to GL(V), g \mapsto \rho_g(v) \coloneqq f(g,v)$ is continuous then f is continuous. Now this is what I've got, if $U$ is open in $V$, we can write $$f^{-1}(U) = \bigcup_{v \in V} {g \in G : \rho_g(v) \in U } \times \bigcup_{g \in G} {v \in V : \rho_g (v) \in V } $$
cloud walrusBOT
gabo.brg
hollow fjord
I have problems showing that $\bigcup_{v \in V} {g \in G : \rho_g(v) \in U }$ is open in G
cloud walrusBOT
gabo.brg
hidden haven
That pre image looks wrong
It should be the set of all pairs (g, v) such that ρ_g(v) ∈ U
This set is smaller I think and need not decompose as a product
As for the solution, you'd have to explicitly make use of the topology on GL(V). Endowing V with a norm and using the operator norm might make your life easier.
white oxide
help i proved a-c and i can't put shit together through d
im assuming that there exists a in R such that sigma(a) = a' such that a \neq a' and trying to derive a contradiction
can't find any
i know that any automorphism carries positive numbers onto the positive numbers and negative numbers onto the negative numbers, but it may be a permutation of the positive numbers and a permutation of the negative numbers
i'm not sure how the ordering argument in c is supposed to help with d
hidden haven
Think about things in ℝ which must be fixed by the automorphism
Do you know any such elements
hidden haven
Nice now infer that more elements are fixed
white oxide
ah i suppose i can consider the cases for b = 0
hidden haven
And repeat till you get everything in ℝ
glossy crag
just out of spite
t. Shing-Tung Yau
delicate orchid
t. yer dad
dim scaffold
imagine a Hilbert space which has 2 coordinate axis. now we got a state vector in this 2 dimensional space (like spin). now it's obvious that state vector has images on every axis. consider a rotation which conforms the state vector on one of the coordinates. then there is not any image of state vector on the other coordinate axis.
is there any rotation defining this ? ( yes )
now can we assume any physical instruction to explore the rotation which means being certain about state of quantum system ?
white oxide
help i proved a-c and i can't put shit together through d
lmao
i would have never got this in a million years
also seems to assume that you know that Q is dense in R
hidden haven
lmao
Yeah so what I was hinting at is that from 0 and 1 being fixed, you find that all integers, and hence all rationals are fixed. From there, you don't necessarily have to take sequences like this. Instead, consider a real number r. If it is greater than some rational q, then σ(r) > σ(q) = q. If it is less than some rational s, then σ(r) < s. A real number is uniquely determined by the set of rationals less than it (this is tautologous if your definition of the reals is using dedekind cuts), so σ(r) must be r.
Is almost the same but I'd say conceptually simpler and there's a simpler picture of this with the number line than if you took a sequence
south patrol
<@!356844891722219520> what is your question?
Oh was just smth about shuffles like
Smth I read seemed to make no sense basically lol
But probably a typo
Not sure if I should email about it lol
hidden haven
now can we assume any physical instruction to explore the rotation which means b...
You might want to ask this on the physics server instead
teal vessel
I'm not dumb, right? for non-abelian groups the fact that there exists some ga=/=ag directly implies that this isn't a left regular action, yeah? this seems nearly trivial.
it could be considered a right regular action, if you want to define it that way (you'd probably prefer to have it be a.g instead of g.a for consistency), but it's not a LRA
hidden haven
I'm not dumb, right? for non-abelian groups the fact that there exists some ga=/...
Nope
teal vessel
nope I'm not dumb or nope it's not nearly trivial?
coral spindle
These are two different multiplications.
Oh it is nearly trivial, but for different reasons.
hidden haven
g•a ≠ ga is not an issue
hidden haven
delicate orchid
What
coral spindle
The word regular is nowhere in your screenshot lol
teal vessel
it was right after a question about left regular actions
hidden haven
What's a left regular action 
teal vessel
don't worry, I still can't read
a left regular action is a group action such that g.a=ga for all a,g in G
hidden haven
Oh lol
coral spindle
So the left regular action rather
delicate orchid
||a . (b . g) = gba which isn’t (ab).g!!!! AAAAAAAAAAAAA||
It’s the action that defines the regular rep isn’t that funny hahahahaha
delicate orchid
The first split the field in the 1930s
teal vessel
see, this is why I post in here, because I'm stupid
hidden haven
☝️
coral spindle
Big tips for doing math.
teal vessel
🤣
teal vessel
given what I thought the question was, I wasn't stuck, I was done, just way too quickly. My "you're missing something" alarm went off
white oxide
Yeah so what I was hinting at is that from 0 and 1 being fixed, you find that al...
oh damn, yea I’m def not smart enough to think of that lol
teal vessel
ok, also nearly trivial, but seems much more substantial in demonstration than just prima facie defying the definition like I thought it was.
coral spindle
Yup.
teal vessel
and now for one where instead it's g.a=ag^-1, which seems more substantial to say it is a group action. I've managed to show that x.(y.a)=(xy)^-1.a but that's not quite satisfying the definitions
at least not in the obvious sense
coral spindle
What you write is false.
teal vessel
hold on, lemme do it backwards
when in doubt, read it twice and try it backwards
blech. missed something or other somewhere in there
(xy).a=a(xy)^-1=ay^-1x^-1=x.(ay^-1)=x.(y.a)
much easier going backwards
teal vessel
dunno why, probably got hung up on an error I thought was right and did it wrong twice.
prolly just missed an inverse or something
I love that they're asking me to deduce something that I've already previously proven for all isomorphisms (that x and f(x) have the same order). I guess the last step isn't quite trivial but it seems to follow pretty much directly from the definition of isomorphism
because it's an isomorphism F, the image of A under F should have the same cardinality, especially for finite groups
delicate orchid
yeah these are all immediate from showing it's an isomorphism
teal vessel
I'm realizing that they use "deduce" to really say "this trivially follows but here's a reminder anyway"
reminder: isomorphisms are still isomorphisms
FINALLY getting to Lagrange's theorem lol
"using the previous two exercises, prove Lagrange's theorem"
(this is using 18 and 19, which I haven't shown)
dim scaffold
You might want to ask this on the physics server instead
alright
teal vessel
is the intersection of any two orbits under an operation empty?
south patrol
~~Intersect it with the same orbit ~~
teal vessel
oh hush. I meant distinct orbits. I've tried a couple things and I'm pretty sure the orbits under an action constitute a partition, but something about the definition of an orbit I'm given is weirding me out regarding the identity of the group
coral spindle
Again, try proving/disproving it.
teal vessel
oh, duh. equivalence relation, transitivity of equivalence dictates that if 1 is in every orbit, then the orbits would be non-distinct.
coral spindle
Yup.
teal vessel
man I really be asking the stupid questions today. Now I know what it feels like to have been that kid in class
night onyx
lol nothing wrong with asking questions, the annoying kids are the knowledge-shamers
teal vessel
it's a bit hard sometimes when the text gives you "show this equivalence relation, by the way, this is called an orbit." "now show that for any arbitrary subgroup H in a finite group G, H is in bijective correspondence with each orbit under H on G." "now deduce Lagrange's theorem from these two statements" after I've known what I'm dealing with for approximately 5 minutes
coral spindle
That's the point, you learn by doing.
If you spend time with them – and resist the urge to look at solutions – exercises are extremely rewarding.
teal vessel
it is. I just prefer to have definitions in the body of the text rather than buried in the exercises
night onyx
you also develop a sense of the right questions to ask, like showing that certain subsets are disjoint you'll start to think "oh, are these things just the classes of some equivalence relation"
steady apex
Sorry I realised I should've taken that to a help channel my bad
delicate orchid
no it was fine here
steady apex
oh ok
yeah I was getting stuck on where to start for part c on hw1 pinned here
I'm a newbie to abstract algebra so excuse me if I'm miss some obvious things 😅
obtuse bear
yeah I was getting stuck on where to start for part c on hw1 pinned here
think about noncommuting elements of S3
delicate orchid
we can have x, y with finite order and xy infinite order
obtuse bear
we can have x, y with finite order and xy infinite order
in a finite group?
delicate orchid
oh lol
delicate orchid
think about noncommuting elements of S3
yeah this works
steady apex
think about noncommuting elements of S3
S3? (unfamiliar with notation sorry)
is that the symmetric group
delicate orchid
symmetric group on 3 elements
alpine island
we can have x, y with finite order and xy infinite order
Wrong operation 
alpine island
Love the idea of calling people in help channels patients, I’m gonna steal that
steady apex
is a counter example good enough for that question or would I need a rigourous proof of some sort?
delicate orchid
a counter example is a rigorous proof
alpine island
is a counter example good enough for that question or would I need a rigourous p...
Not forall = exists not,
So a concrete counterexample is more than enough
white oxide
how would i go about showing associativity for this example? do i just draw a diagram of arrows and put paranthesis around it or some shit lmfao
i mean it just follows from the axioms of morphisms in C right
white oxide
kinda too lazy to draw everything out
and give names to all thes emorphisms
but oh well must be done
south patrol
Tbf composition is just induced by that of C right
white oxide
yeah true
white oxide
oh yea facts
south patrol
Also lol rare case where cat theory isn't in cat theory rather than other way round lol
Lol
white oxide
bro there are like 16 million morphisms in this example that i have to give names to holy shit
white oxide
Also lol rare case where cat theory isn't in cat theory rather than other way ro...
huh
south patrol
Oh lol I meant the q you gave was cat theory
but often I see people ask basically algebra questions in the cat theory channel
white oxide
oh lmfao
nah i feel like this shit way too elementary to ask in that channel
that channel intimidates me
white oxide
Hungerford
this is good enough for showing existence of identity right lol
idk how to write a curly D
wait
oops it shouldn't be g
wait
now i'm confused
ignore the clearly
south patrol
I'd write it differently
Well I don't think you've technically show that this is an identity
white oxide
yea i haven't
this is confusing lol
i mean it's just a matter of definitions but it's so weird
coral spindle
#category-theory proofs tend to be like this tbh
delicate orchid
not in this case. This one is literally just ||1_Af = g1_B => f = g|| lol
or
the other way
you know what I mean
anyway what we doing defining the arrow category
night onyx
this is good enough for showing existence of identity right lol
that's the identity, but you'd need to show it's the identity with a picture more like this to show it's actually the identity morphism
white oxide
that's the identity, but you'd need to show it's the identity with a picture mor...
hm okay thanks that makes more sense
night onyx
lol I was a bit sloppy in paint but that's the rough shape of it
but everything just works out because the underlying objects are proper morphisms
white oxide
thank you, yeah i'll definitely revisit this example later
glossy crag
thank you, yeah i'll definitely revisit this example later
That's Scriabin as your pfp, right
glossy crag
Any recs? I've listened to some of his stuff, but not too much
It was good ofc, but left no lasting impression
Poem of Ecstasy was good
white oxide
i really enjoy all of his piano sonatas
if you liked poem of ecstasy, i would highly recommend his piano concerto, fantasie in b minor, sonata number 2, or sonata number 5
his first symphony is also like poem of ecstasy
sorry this is off topic lol
coral spindle
You read my mind
glossy crag
myself i'm a RachChad
celest furnace
myself i'm a RachChad
Fellow Chad?
glossy crag
Fellow Chad?
can you feel my heart?
celest furnace
can you feel my heart?
Take me to the land ?
long nebula
his first symphony is also like poem of ecstasy
SCRIABIN
chilly ocean
if you liked poem of ecstasy, i would highly recommend his piano concerto, fanta...
"what fantasy?"
echo thunder
would herstein's topics in algebra be too difficult for beginner in abstract algebra?
would it be better to stick to gallian or fraleigh?
coral spindle
Just start reading and you'll know soon enough
rustic crown
shuwi in #groups-rings-fields 
lethal dune
coral spindle
@coral shale should people not read books to work out if they're at their level or not?
coral spindle
Hey det, hope you're well
coral spindle
I got a copy of Matsumura recently. I might read it a bit over the summer
coral shale
i wasnt judging, dont mind me
coral spindle
I will never figure out the tone of emojis in this server, apparently. "
"
rustic crown
det no understand that emoji either >.<
rustic crown
I got a copy of Matsumura recently. I might read it a bit over the summer
me was planning that too
i read the first 5 sections and i'm convinced it's not a cute book >.<
coral spindle
Also got Rudin's book on fourier in groups, so that's cool. But it's also written by Rudin
rustic crown
i read the first 5 sections and i'm convinced it's not a cute book >.<
wasn't a fan of most proofs so far, but it had some cool stuff
coral spindle
It looks intense
rustic crown
it issss 
coral spindle
It's a bit of a pity but I'm going to have to focus on another book over the summer mostly
Probably Geck's book on algebraic groups
But hey I should have some time for maths less important to my research
lethal dune
ryu <:eeveeKawaii:556858499930259471>
konnichiwa 
I had planned to read a lot of things lol, didn't do shit
rustic crown
same uwu i can only manage like a third of what i plan 
lethal dune
that's still a lot than what I have done till now lol
rustic crown
(oh should have said that i don't plan a lot >.<)
rustic crown
all reps are confoosing
summer path
rustic crown
commie ring thy
lethal dune
ah
rustic crown
that's what chmuwu recc'ed
rustic crown
CA just looked even harder
barren sierra
but it works too well (at a basic level) it feels like
coral spindle
lie algebra representation is so confusing
They never did click with me. I find it harder than with ordinary modules lol
lethal dune
yeah bcz there weren't any good CA books available back then
barren sierra
maybe I'll peek at Atiyah if I get stuck somewhere in CRT
coral spindle
For a second there I thought "what ofc they're going to use the chinese remainder theorem"
lethal dune
They never did click with me. I find it harder than with ordinary modules lol
yeah what's up with Ad, ad whatever, notational hazard
lethal dune
yeah what's up with Ad, ad whatever, notational hazard
not even associative lmao
lethal dune
det gets scared when there is differential structure
idk why I'm reading it tho
rustic crown
idk why i read math either
barren sierra
anything differentiable is scary (taking differentiable manifolds next sem...)
lethal dune
I probably should read some K-therory or something
rustic crown
barren sierra
what is K theory
rustic crown
you study the letter "K"
lethal dune
idk why i read math either
barren sierra
you study the letter "K"
ah thanks for the clarification
lethal dune
i never find the motivation to actually do what i plan to do
I still struggle
rustic crown
like in highschool it was like "oooohhh e^(i * pi) = -1"
rustic crown
that feeling has changed a bit :p
lethal dune
how so?
south patrol
K theory
rustic crown
to prove anything you have to develop like so much theory
south patrol
like in highschool it was like "oooohhh <:eeveeKawaii:556858499930259471> e^(i *...
It's only amazing when you don't know what the LHS even means
lethal dune
is it like "ah there's another thing I need to read up, god damn it"
rustic crown
e^(i pi) = -1 + ai
open sluice
$\tau$
cloud walrusBOT
bladewood
rustic crown
is it like "ah there's another thing I need to read up, god damn it"
not fully, but a lil yea
next obsidian
Chmonkey
rustic crown
chmuwu
next obsidian
I’m very chmusy
coral spindle
He's literally a pokemon. He's a pokemon.
lethal dune
kinda like that
I still get amazed by some topological results, not so much in algebra
for example, last thing that really amazed me was Rochlin's theorem
summer path
people say you get a nice feeling when you work hard on a problem and then eventually get it
but that's just a bunch of lies
lethal dune
well
coral spindle
I think the classifications that Coxeter–Dynkin diagrams provide is pretty amazing. I'm still in awe over that.
lethal dune
I am yet to reach Dynkin diagrams
coral spindle
**bladewood**
Arguing about tau vs pi is the most pseud thing you can do. Talk about actual maths, not the aesthetics of people who do maths.
lethal dune
for example, last thing that really amazed me was Rochlin's theorem
but the proof is
summer path
either you get and feel nothing, or you work on it for a while and never get it and then someone gives a hint and now you feel like an idiot
open sluice
either you get and feel nothing, or you work on it for a while and never get it ...
how much math have you done?
summer path
i finished undergrad math major
lethal dune
die a hero or stuck long enough to feel like an idiot
summer path
going to be a grad student next year
rustic crown
summer path
i hope i do too, knowing i won't
coral spindle
I don't know about that, being a grad student is pretty fun
coral spindle
sounded like a warning
It's sincere
summer path
coral spindle
I proved something really nice about Schur indices the other day
If a character induces irreducibly then there's a nice pincer between the schur indices
Not the most useful but oh well
wraith cargo
If R is a local Noetherian ring that is complete wrt some ideal m, is m necessarily maximal?
rustic crown
if it's complete wrt m i think it should also be complete wrt m^2
wraith cargo
hm
wraith cargo
if it's complete wrt m i think it should also be complete wrt m^2
this doesn't sound very true
rustic crown
why so?
wraith cargo
(Here I'm talking abt the m-adic completion)
rustic crown
the natural map R --> lim (R/m^n) an iso right?
and m^2n would be a coinitial thingy
so they have same inverse limits
wraith cargo
the natural map R --> lim (R/m^n) an iso right?
yeah but how is it an iso if you only count the even powers in the inverse limit?
rustic crown
basically the idea is that information mod m^n is redundant when you also know it mod m^n+1
rustic crown
so knowing for arbitrarily large powers will give the same thing. rest is just a diagram chase proof
lethal dune
if it's complete wrt m i think it should also be complete wrt m^2
easiest way to see this is that they induce same topology so same completion
pine crescent
Hi, I am new to abstract algebra. Can somebody help with my problem?
Suppose $$ordG = 2n+1$$ Prove that for any a in G $$a = m^2$$
I tried to use property that states that ord(a) divides ord(G) but I don't know what to do next
cloud walrusBOT
mathlover
coral spindle
what is m?
barren sierra
Hi, I am new to abstract algebra. Can somebody help with my problem?
Suppose $$o...
Here's how you use the fact that $\text{ord}(a) \mid \text{ord}(G)$. Consider $a^{\text{ord}(G)}$
cloud walrusBOT
spamakin
barren sierra
what is m?
just show that every element is a square
coral spindle
Oh right I see. Nicely spotted Spam
pine crescent
**spamakin**
And then we get that $a^{\text{ord}(G)}=e$ and $a^{2n+2} = a$, so $(a^{n+1})^2 = a$ ?
cloud walrusBOT
mathlover
barren sierra
perfection
pine crescent
Thank you
ember grove
can anyone help me understand that last sentence? I don't understand what the author means by saying that F(c) is unique up to isomorphism, because as I see it if there's another such "smallest" (being contained in any other field that contains both F and c) then it would just be equal to F(c) since they'd contain each other
wraith cargo
can anyone help me understand that last sentence? I don't understand what the a...
there can be more than one minimal field
take like a ring
It can have more than one biggest proper ideal
(this is the same reasoning more or less since the set (or category idk) of fields is partially ordered by inclusion)
vagrant zinc
Guys a question the subgroup of R* generated by 7, would be all the exponentials of 7 ?
ember grove
there can be more than one minimal field
but how? If F(c) and say K both contain F and c and are "smallest" then by the definition of smallest above shouldn't they be equal because they are subfields of one another?
wraith cargo
but how? If F(c) and say K both contain F and c and are "smallest" then by the d...
OK
what does smallest mean
It means that there isn't another field between the two you're looking at right?
but fields aren't stacked on top of each other exclusively
They can overlap a bit and whatever
rocky cloak
can anyone help me understand that last sentence? I don't understand what the a...
Yeah, this is a very strange thing to say. F(c) is unique. Not just up to isomorphism
barren sierra
Guys a question the subgroup of R* generated by 7, would be all the exponentials...
should be, yea
rocky cloak
OK
what does smallest mean
It means that there isn't another field between the ...
The intersection of two field extensions is again a field.
wraith cargo
The intersection of two field extensions is again a field.
here I was just trying to convey the idea that there need not be one minimal field over another
formal ermine
Yeah, this is a very strange thing to say. F(c) is unique. Not just up to isomor...
,, ==\iso
cloud walrusBOT
ember grove
Guess i'll ignore that sentence and go on then...
past temple
generally speaking, when is "A/B + B = A" okay?
is it at least fine for free R-modules
wraith cargo
is it at least fine for free R-modules
finitely generated free modules
when they're not finitely generated there's something called the Eilenberg swindle that sorta breaks this
past temple
wait in that case
why is
this okay?
H_0 is free, but definitely not finitely generated
coral spindle
generally speaking, when is "A/B + B = A" okay?
This is very rarely true.
For an example, take the Z-modules A = Z/4 and B = Z/2.
wraith cargo
I think there's some splitting of SES going on here
coral spindle
The only case in which it is always true is when your ring is semisimple. There are lots of nice semisimple rings (e.g. lots of group algebras, and all fields) but in general this simply does not hold.
This is also very much not true even when A and B are f.g. free
In fact taking the Z-modules A = Z and B = 2Z, it obviously does not hold that Z is isomorphic Z/2 + 2Z
coral spindle
Yes it is, it is isomorphic to Z.
rocky cloak
The key thing here is that A/B is projective. Then it works
coral spindle
In fact every Z-submodule of Z is free.
past temple
what about free f.g. modules over a unital ring
actually nvm lol, ur example alrdy does that
coral spindle
Yeah.
As jagr mentions, this holds in general when A/B is projective. As it happens, any free module is projective.
past temple
i see
coral spindle
Z-submodules of free Z-modules are free
This is a highly nontrivial fact, but you can take it for granted.
In any case the quotient A/B here is Z, which is free and hence projective
wraith cargo
(this is usually called the structure theorem for free modules over PIDs)
past temple
ah, so the criterion is that the quotient needs to be free
wraith cargo
ye
coral spindle
Projective*, but yes.
wraith cargo
tbh when I said f.g. free I was thinking like R^n/R^m + R^m 
past temple
wait so this requires the structure theorem for free modules over PIDs??
hatcher just states it so nonchalantly
coral spindle
No it doesn't here, as it happens
you just need the fact that Z is a free Z-module. Remember, we only needed that the quotient is projective.
If you really want to you can look at the proofs that this holds when the quotient is projective
The key terms are "short exact sequences ending with projectives split"
vagrant zinc
Guys, could someone give me an example of a proper inverse?
barren sierra
Guys, could someone give me an example of a proper inverse?
can you give more context about inverse vs proper inverse?
lusty marlin
This doesn't use the term 'proper inverse' though. What do you mean by it?
vagrant zinc
F
void cosmos
yo
stupid question
let R be a ring and M be a maximal ideal
consider the R-module R/M
why does this have no submodules?
vagrant zinc
This doesn't use the term 'proper inverse' though. What do you mean by it?
If you want you can translate, "inverso propio" and you will see that it appears as follows heh
void cosmos
why does this have no submodules?
oh yeah cuz a submodule owuld have to be an ideal
right?
barren sierra
Uh
the proof I know uses isomorphism thrm
where you show that simple modules are isomorphic to R / M, M maximal
I think?
lusty marlin
oh yeah cuz a submodule owuld have to be an ideal
Yeah, and R/M is a field since M is maximal. Therefore it can only have 0 and itself as ideals
lusty marlin
If you want you can translate, "inverso propio" and you will see that it appears...
Ah
void cosmos
Yeah, and R/M is a field since M is maximal. Therefore it can only have 0 and it...
yea i just got stupid at the fact that a submodule would have to be an ideal
so now if R/M were to be a free module
it must be of rank 1 right
cuz if it has 2 basis elements for example then one must be a submodule
Rx_2..
is that correct
hexed mantle
...Oh.
clever beacon
It's reflexive (given) and transitive (by construction); you just have to prove symmetry
hidden haven
It's reflexive (given) and transitive (by construction); you just have to prove ...
The given condition is not transitivity lol read it again
clever beacon
oh wait it's slightly different
hidden haven
It has to be modified slightly because reflexive transitive does not imply symmetric
Eg partial orders
frail summit
In fields like (Z/3Z[x])/(x^2+1), is there any way to keep or assign degrees to these polynomials in the quotient?
lusty marlin
What do you mean by 'keep or assign degrees'?
frail summit
A lot of the nice properties for polynomials over integral domains follow because of the niceness of the degrees, I'm just wondering if one can make these carry over somehow
lusty marlin
Any polynomial in (ℤ/3ℤ)[x] is represented by a constant or degree one polynomial in the given quotient ring.
frail summit
Yeah, but e.g. is x+1 irreducible in the quotient too?
lusty marlin
I'm sorry, but I don't understand what your main question is here.
frail summit
In (Z/3Z)[x] we can immediately conclude it is irreducible because it's a linear factor
delicate orchid
it's a field, no elements are irreducible
frail summit
And degrees add up
frail summit
it's a field, no elements are irreducible
eh...
delicate orchid
part of the definition of being irreducible is being a non-unit
frail summit
Oh ye
Oke, but in general, if we take the quotient and say some element is not a unit
Prolly just me misunderstanding smth, nvm - thx tho
delicate orchid
it's just a ring it doesn't matter if it's a quotient
you can use all the regular methods to see if an element is irreducible or not
i.e. is the ideal generated by that element maximal etc. etc.
coral spindle
In fields like (Z/3Z[x])/(x^2+1), is there any way to keep or assign degrees to ...
I think what you're looking for is the theory of graded rings. In order for a quotient of a graded ring to be graded, you need a homogeneous ideal. Unfortunately (x^2 + 1) isn't one, if I'm not mistaken.
delicate orchid
ah yeah gradedness could be a good generalisation
coral spindle
I believe an example of when you can have the degrees descend to a quotient is k[x,y]/(xy-x^2). I'm rusty on anything to do with graded rings, so perhaps someone can correct me.
Noticed a huge mistake, do forgive me lol
delicate orchid
I think what you're looking for is the theory of graded rings. In order for a qu...
it isn't, homogeneous ideals in polynomial rings are generated by homogenous polynomials (if it's the same definition as in alg geo)
((big if))
coral spindle
Yeah I corrected it to a homogeneous polynomial
delicate orchid
yur
coral spindle
I guess that example is a bit boring though
maybe a nicer one is k[x,y,z,w]/(xy - zw), that's kinda interesting
delicate orchid
coordinate ring of some plane in k^4.,, sopoppy
frail summit
Found the section in Dummit and Foote covering graded rings - will mark it for when I get there. Thanks
coral spindle
coordinate ring of some plane in k^4.,, sopoppy
OK nerd............ Z[x, y, z]/(x^3 + y^3 - z^3) what ya think of them apples
delicate orchid
OK nerd............ Z[x, y, z]/(x^3 + y^3 - z^3) what ya think of them apples
that's like a sphere that's somebody denotated a grenade inside
or is that +z^3
delicate orchid
irrelevantgardless the shape is similar
u just intersect it with the lattice
the algebraic nonsense will be very different
but I do not care
coral spindle
But wew...
there are no solutions to x^3 + y^3 = z^3 in the integers where at least one is nonzero
:)
delicate orchid
good thing there's a minus sign
we're looking at x^3+y^3-z^3 = 0 not x^3+y^3 = z^3 you fool
slim kayak
we're looking at x^3+y^3-z^3 = 0 not x^3+y^3 = z^3 you fool
You forgot to say Uno, this doesn't count
tribal moss
there are no solutions to x^3 + y^3 = z^3 in the integers where at least one is ...
Do you mean "... where all are nonzero"?
coral spindle
Lmao yeah said the total opposite
glossy crag
I was wondering how $|a_0|=|\Pi|^n$ implies the index is $\geq n$, does this work?
Suppose $z$ has infinite order and $m=|G:\langle z\rangle|<\infty$, we want to show $x^n=z\implies m\geq n$: since $m$ is the index $x^m\in\langle z\rangle\implies x^m=z^k\implies z^m=x^{mn}=z^{nk}\implies m=nk\implies m\geq n$. Now apply this to $|E^\times|\supset|K^\times|$ and $|a_0|=|\Pi|^n$ ($|a_0|$ is a generator of the infinite cyclic $|K^\times|$).
cloud walrusBOT
leave_no_norm
chilly ocean
Hello I have a problem and I want a hint. We have a commutative group G of order n>=2 and n is not prime. Let m = the greatest proper divisor of n. Let p be a prime number. If we have at least m elements from G with order p, then all elements from G{e} have order p.
tender wharf
try classification of finite abelian groups
chilly ocean
I think I know how to do it
If I take H the set of elements x in G with x^p=e then |H|>m. H is closed (because of commutativity) and finite so H is subgroup of G and |H| | n, but because |H|>m then |H|=n so H=G.
sly rain
Hi, I calculated the minimal polynomial of $a = \frac{1+\sqrt(5)}{2}$ and $b = (z+z^{-1})$ over $\mathbb{Q}$.
Where $ z = exp(\frac{2\pi i}{5}) $
In this context i want to show that $\mathbb{Q}(a) = \mathbb{Q}(b)$.
Does one calculate $z+z^{-1} = \frac{\sqrt(5)-1}{2}$ explicitly here and show that both fields are equal or how does one usually approach this (when working with roots of unity)?
Also if it wasnt for wolfram alpha, I wouldnt have known that $z+z^{-1} = \frac{\sqrt(5)-1}{2}$. "Should" one be comfortable calculating these in the context of abstract algebra (and later commuative algebra and whatever else might follow)? As in does this type of knowledge come in handy every now and then?
cloud walrusBOT
aabb
formal ermine
you should only be comfortable calculating these things if your exam tests you on them
otherwise just use a calculator for these "trivial" calculations
but that z + z^-1 can be done using the cos + isin representation of your e^ and some trig identities
but yeah since you've shown that 2b = 1 + sqrt 5 you can reduce Q(b) = Q(1 + sqrt5) = Q(sqrt 5) and similarly with Q(a)
which shows that they are equal
rustic crown
(you can cheat a lil, the extension Q(zeta_p)/Q is galois with group (Z/pZ)* which is cyclic, so there is a unique quadratic extension which can be shown to be Q(g) where g^2 = legendre(-1, p) * p, and noticing that for odd primes, L=Q(zeta + 1/zeta) is the unique extension with [Q(zeta_p) : L] = 2)
formal ermine
(you can cheat a lil, the extension Q(zeta_p)/Q is galois with group (Z/pZ)* whi...
I think we will do this in my next alg nt lecture
we did cyclotomic fields and legendre symbol last time
primes in extensions are still confooosing 
@rustic crown did you know that graded means graduiert in german
rustic crown
nope
formal ermine
weird word
rustic crown
det no know any german
formal ermine
you know zwischenkörper
sly rain
(you can cheat a lil, the extension Q(zeta_p)/Q is galois with group (Z/pZ)* whi...
Yeah im not sure If my tutor allows this
rustic crown
yee but if you wanan do it by hand, notice that z satisfies
1+z+z^2+z^3+z^4 = 0
divide by z^2 and write it as a poly in z+1/z, which gives you a quadratic in that and solve it uwu
formal ermine
what does "semigroup in ..." mean? also what exactly is a semigroup? I thought it was an associative magma (wikipedia), but my prof is saying ".... is a semigroup in ... with neutral element 0" so I'm confoosed, shouldn't it be a monoid then?
coral spindle
all monoids are semigroups
coral spindle
Why not
formal ermine
idk
coral spindle
You don't have to specify the most restrictive thing something is.
formal ermine
also what about the "in" part
coral spindle
You've omitted context so I have no idea.
I presume that your prof meant it was a subsemigroup.
rustic crown
(oh i was thinkign semi-group object in a category )
coral spindle
That might also explain why they didn't call it a submonoid, as typically submonoids are defined as sharing the same unit element as the thing they sit in.
rustic crown
(but that makes more sense)
formal ermine
let $\nu : A \setminus \set{0} \to \bQ^r$ be a valuation. then $S(A, \nu) = \set{v(a) | a \in A\setminus\set{0}}$ is a semigroup in $\bQ^r$ with neutral element $0$, the so called valuation semigroup
formal ermine
I presume that your prof meant it was a subsemigroup.
I think this makes the most sense
oh lemme fix my texit
low surge
comic sans
cloud walrusBOT
formal ermine
fixed
formal ermine
comic sans
COMIC NEUE
delicate orchid
comic sans
formal ermine
I think he calls it semi group because that sounds cooler than monoid in german
bewertungsmonoidgruppe
bewertungshalbgruppe
ok thanks boytijejie
delicate orchid
ermmmmmmmmm evaluates ur ring
coral spindle
bewertungshalbgruppe
oh no illum fell asleep on the keyboard 🙀
sly rain
Thanks, also reasonable answer in general. I'll ask the assistant. Also since our courses probably looked similar (as in standard german into to algebra class), did you just grind out past papers or are there other nice resources with exam esque questions? (everyone can answer)
sly rain
Thanks, also reasonable answer in general. I'll ask the assistant. Also since ou...
@formal ermine, meant as reply to your answer
formal ermine
Thanks, also reasonable answer in general. I'll ask the assistant. Also since ou...
I did the assigned homework
but for practice questions you can just take any algebra textbook and do the exercises in it
like d&f or artin
white oxide
that's the identity, but you'd need to show it's the identity with a picture mor...
ah so it would look like this:
silly question but is there any way to diagram a product for a family of three objects A1 A2 A3 in a category C?
rocky cloak
silly question but is there any way to diagram a product for a family of three o...
What do you mean by "to diagram"?
night onyx
silly question but is there any way to diagram a product for a family of three o...
that's my naive thought lol
white oxide
What do you mean by "to diagram"?
like what kryo just posted
white oxide
that's my naive thought lol
ye that's kinda what i ended up w/ too
shit seems like a topological probelm
idk
rocky cloak
Sure, you can draw a diagram for the product of any number of factors
night onyx
you'd probably end up with different unique maps for (X x Y) x Z and X x (Y x Z) but they're probably naturally isomorphic in some sense (lol it's been a long time since I've thought about cat theory so take it with a grain of salt)
coral spindle
Yes, they differ by the unique natural isomorphism making the product associative
rocky cloak
And both of those are naturally isomorphic to the usual way of defining the product of three things
coral spindle
omg it's boytjie
no i'm not
agile burrow
Hm I'm feeling a little silly right now. I have a Lie group G whose center is S^1. The claim is that in any finite dimensional irrep of G, the center of G must be diagonalizable since it is compact and abelian. I'm not sure if this means S^1 is simultaneously diagonalizable, or just that the image of every element is diagonalizable. The latter is clear since every element has finite order, but the former seems more relevant and I'm not sure how to extend simultaneous diagonalization to all of S^1
white oxide
i'm struggling to see exactly how F and F' being free on X and X' and the cardinalities of X and X' being equivalent lead to the existence of a bijection from X to X', could i just take i: X --> X' to be f and fbar: X' --> X (since C is a concrete category and we can consider X' and X to be functions on the underlying sets; don't really know if this works) (to show injectivity)
in this definition
or am i just hella overcomplicating things again lmfao
coral spindle
The cardinalities being equal is equivalent to there being a bijection between the sets.
This isn't an algebraic fact, this is just a set theory fact.
round hull
btw as an aside, why are these types of questions not asked in the category channel
tribal moss
I'd say they're on topic in either channel.
rocky cloak
Hm I'm feeling a little silly right now. I have a Lie group G whose center is S^...
So since S^1 is in the center, multiplication by an element in S^1 defines a morphism of representations of G. So by Schurs lemma they are all just scalar multiples of the identity.
Also not every element in S^1 has finite order.
agile burrow
Ok yes I feel extraordinarily silly today. Thanks for your help
hidden haven
waltering
rocky cloak
Funnily enough, it didn't matter that it was compact
white oxide
This isn't an algebraic fact, this is just a set theory fact.
oh yea oops
forgot about that thank you
holy shit right that’s literally the definition fuck
summer path
It's okay to be silly occasionally
white oxide
honk
hidden haven
Too late for you 
summer path
Duck duck duck goose!!
slim kayak
or am i just hella overcomplicating things again lmfao
Pretty sure bijections are the definition of equal cardinality
white oxide
Pretty sure bijections are the definition of equal cardinality
yea i remembered tha tlater
how is tau inverse onto F bar? i really don't see it
nvm i'm stupid
wait huhhh
wait how is it onto Fbar never mind
what
the image of F under tau is F since it leaves it fixed
so how can the image of F be all of Fbar
nvm
delicate orchid
hello SARGE!!!
ebon gyro
i am in attendance
delicate orchid
wtf am I readddinggg
ebon gyro
oh finally some GOOD math it looks like
ebon gyro
not to you i meant to what ive been going through lately
cloud walrusBOT
Sloth King Daminark
delicate orchid
are cosets of open groups open dami
bleak abyss
Yeah
delicate orchid
oh yeah duh
bleak abyss
Multiplication by g is a homeomorphism
delicate orchid
multiplication by g is continou
ebon gyro
yes
translation is continuous Why do i even bother sending messages if imbgonna get sniped
delicate orchid
I'm reading these real fast so I might ask stupid things
cloud walrusBOT
Sloth King Daminark
ebon gyro
its pretty quick iirc
formal ermine
that clopenussy 
bleak abyss
Since if you contain any neighborhood U of the identity, you push U around by g in H and get that H is union of opens
bleak abyss
And then G splits into cosets of H, each is open since L_g is homeomorphism, so G\H is a union of all the nontrivial cosets and is thus open, so H is closed
(It is not true in general that closed subgroups are open ofc)
ebon gyro
something like H^2 = H and H bar is stuck between H and H^2 should do it
bleak abyss
Or what I said
ebon gyro
or what you said
delicate orchid
And then G splits into cosets of H, each is open since L_g is homeomorphism, so ...
this is so different to how I'm used to thinking 
but I follow completely yeah
bleak abyss
So yeah if v_i is stabilized by open subgroup H_i
Then v_1 + v_2 is stabilized by the intersection, which contains open neighborhood of 1, so it's open
delicate orchid
I have absolutely no idea why that last implication holds
bleak abyss
Since if you contain any neighborhood U of the identity, you push U around by g ...
What I said here
If H contains U contains 1, then for any g in H, H contains gU contains 1
and gU is open since L_g is homeo
delicate orchid
ah ok
yeah so I get why these vectors are called smooth I think
it's like their "preimage under the group action" is open kinda deal
bleak abyss
Functions on p-adic groups are considered "smooth" if they are locally constant of compact support
ebon gyro
schwartz bruhat time
delicate orchid
right makes sense
compact support so you don't get any nonsense "towards infinity" I presume
cloud walrusBOT
Sloth King Daminark
delicate orchid
I have the notes up u don't need to latex lol
this bit I understand completely because it's just rep theory lol
ebon gyro
ya lmao
bleak abyss
I'm also doing it to help me think, but I guess I can just type badly 😛
ebon gyro
this i can follow but i may not be p-adic group pilled enough specifically to get some of the later stuff
bleak abyss
So dual of smooth isn't smooth, I'll let some grad student find an example
ebon gyro
LOL
delicate orchid
hmmmm
bleak abyss
I mean it's prob like
Remember that the algebraic dual space of some infinite-dim vector space can be fucked
STrictly larger cardinality blah blah
delicate orchid
yeah I know
it would have to be infinite dim
but there is no such thing as an infinite dimensional vector space so we're all out of luck
bleak abyss
Tru
ebon gyro
wow. talk over
bleak abyss
p-adic groups are now d o n e
Actually the only irreps of GL(2,Qp) are characters in finite dimension
So as far as Wew is concerned automorphic forms are just ez
delicate orchid
the trick is I am simply not concerned
anyway lemme think about why the smooth part is fixed
if we pick a g outside of the isotropy group and consider vg
hmmm
bleak abyss
It's prob the general shtick that stabilizer of gv is gStab(v)g^{-1}
delicate orchid
yeah
bleak abyss
Conjugate of open is open gg
ebon gyro
is preservation under surjective morphisms some standard rep theory thing im just forgetting rn
bleak abyss
Well not just continuous but a homeomorphism, thus an open map
delicate orchid
wowie kazowie!
bleak abyss
Hmm let's see
delicate orchid
it's almost like I already knew it was invertible DAMI
anyway so we restrict to that doodad and get the contragredient reppo
cloud walrusBOT
Sloth King Daminark
delicate orchid
I believe the remark cause subquotients and direct sums are obvious I think
ebon gyro
OH
bleak abyss
So yeah surjective morphisms make stabilizers bigger
ebon gyro
ok this is very believable
bleak abyss
So if they were already open...
delicate orchid
yur can't make em smaller
bleak abyss
So yeah subrep because... it's got the same damn stabilizer
And that handles subquotients
delicate orchid
this holds just for a general group action which is why I believed it immediately
well, T has to be a G-set morphism
but that's irrelevantgardless
bleak abyss
And if (v,w) is in the direct sum, its stabilizer should just be Stab(v) cap Stab(w)
delicate orchid
yeah exactly
ebon gyro
just intersection of stabilizers and win
bleak abyss
Which is still open
delicate orchid
uhhhh but what if infintie direct sum ⁉️⁉️⁉️⁉️⁉️⁉️⁉️⁉️
ebon gyro
those dont exist actually
delicate orchid
yeah I know it still holds cause in an infinite direct sum only finitely many of the dudes are non-zero or WHATEVER
ebon gyro
all vector spaces are fd all direct sums are finite its true
delicate orchid
so it corresponds to a finite intersection I GUESS
bleak abyss
Well if you take Hilbert space direct sum now you might have to actually think 😛
delicate orchid
direct integral
ebon gyro
no stop
delicate orchid
anyway I'm getting distracted
bleak abyss
Tiem 4 parabolic induction
delicate orchid
let us peep back at the notes
ebon gyro
time to INDUCT
bleak abyss
So yeah condition 2 is just saying K_f becomes the stabilizer of f in this rep which we want to be open so that we get something smooth
ebon gyro
sensible
bleak abyss
And we should prob get that one theorem
delicate orchid
you've been hit by, you've been struck by
a smooth inducer
bleak abyss
Well idk about "agrees with tau" since H is acting on something different now?
delicate orchid
mans remains unbothered
bleak abyss
But that one thing about restriction and induction being adjoints
Whose name I'm forgetting for some ungodly reason
ebon gyro
frobenius reciprocity
delicate orchid
frobby recopo
delicate orchid
guessing it still works here
as it should
if ur induction functor isn't left adjoint to ur restrction functor u got ur definitions wrong
cloud walrusBOT
Sloth King Daminark
delicate orchid
I'm soyjaking rn
cloud walrusBOT
Sloth King Daminark
delicate orchid
yeah I thought this held automatically dude to the gStab(woiegfsd;if)g^{-1} dealio
I suppose not cause we don't automatically have that all of V_tau is smooth
cloud walrusBOT
Sloth King Daminark
delicate orchid
ne bovva
anyway I shouldn't be too surprised there's parabolic groups here
but it's something I already know about which is pog
bleak abyss
Compactly induced geez
delicate orchid
does the existence of the unipotent radical follow from the quotient being complete?
I'm like thinking you quotient everything again by the (???) of all the subgroups which plays nice because they're closed
and then consider some sequence and use the correspodence theorem? dunno
just spitballing
bleak abyss
Probably some algebraic group shit?
Complete might mean complete variety here
Rather than complete metric space
delicate orchid
what a bunch of BALONEY
bleak abyss
This is different from what I'm used to thinking lol
delicate orchid
oh lord there's an integral
bleak abyss
Borel subgroup is maximal Zariski closed and connected solvable subgroup
And parabolic means contains Borel
delicate orchid
I've literally never cared about it beyond it being the upper triangulars in GL_n
bleak abyss
Pretty much
delicate orchid
and parabolic to me is a subgroup of a reflection group that leaves (some set of or) a hyper-plane completely fixed
bleak abyss
I just think of parabolic as shit like
delicate orchid
but that's obviously equivalent to containg borel
cloud walrusBOT
Sloth King Daminark
delicate orchid
yus
this would be the parabolic group fixing the uhhh
not sure
xy-plane? possible
bleak abyss
Setwise stabilizes the subspace/incomplete flag <e_1,e_2>
delicate orchid
yussss
that's the one I think
calling a subspace an incomplete flag...
that's gonna be a 24h mute
bleak abyss
I'll just black box Levi decomposition
delicate orchid
I'm gonna blackbox parabolic induction a bit
actually I think I'm just gonna jump off here cause I got NO CLUE now
bleak abyss
Get back here @delicate orchid
And don't think I didn't notice your disappearance @ebon gyro
delicate orchid
you bring the haar measure in I :gottago:
bleak abyss
So the point of slide 7 is just that, the stuff that happens on slide 6 for algebraic groups are still gucci when we pass to F-rational points, and thankfully G/P is compact so smooth = compact
chilly ocean
I've never done abstract algebra but I have a question about the graph of this style
Or
I feel like abstract algebra is like black magic where you point out with arrows and weird symbols everywhere
delicate orchid
the top is a definition lol
chilly ocean
Like why do you do this kind of graphs?