groups-rings-fields

[obsidian sleet]

but

oh

hmm

well how could a be in <a^2> its like

all of the elements in <a^2> are a^2k

k in integers

2k can't be 1

???

[carmine fossil]

Yes

[obsidian sleet]

oh

[viscid pewter]

so basically it's not in it, right

[obsidian sleet]

i see

[carmine fossil]

So that's a proper subgroup

[obsidian sleet]

and then you keep going

[carmine fossil]

And it has to be smt different from the above mentioned 3 subgroups

[obsidian sleet]

oh this was enough

oh

i see

right we just generated a proper subgroup of G that was not one of the three which is a contradiction

G is finite

Good

and i guess to show that |a^2| is not 7, i can just use the fact that |a| is infinite

since (a^2)^7 = a^14 which cannot be the identity

?

[woven delta]

Yeah, that is correct

[obsidian sleet]

nice

[woven delta]

Any infinite cyclic group has infinitely many subgroups

[obsidian sleet]

right i could keep going

<a^3>

and on and on

[woven delta]

And being an infinite group without having an infinite cyclic subgroup is pretty hard

And involves having infinitely many subgroups in a different way

(eg the group (Z/2Z)^N if you've seen that before)

[obsidian sleet]

o

idk what that one is

i only seen Z_n type things

[woven delta]

Basically imagine a sequence of binary elements which is eventually 0

You can add 2 such sequences componentwise

[old lava]

can't the direct sum of like infinitely many of any finite non-cyclic group work

as an example of an infinite group with no infinite cyclic subgroup

[woven delta]

Well that's an interesting question

Yeah, it can I guess

Any sequence of elements with only finitely many nonzero entries is only going to generate a subgroup of a finite direct sum

eg if your generators is (a_1, a_2, ..., a_n, 0, 0,...) Then it is going to generate a cyclic subgroup of G_1xG_2x...xG_n

Anyway metal back to your proof

[obsidian sleet]

ya

[unique juniper]

can someone help me out on this please? im stuck

[carmine fossil]

Hint,how many elements are there in <2>?where by 2,I mean the element 2+48Z

[wispy scaffold]

did you tried this?

[unique juniper]

not exactly sure what that means tbh

an expression for the order an element in Z/48Z ?

$|\overline{x}^{a}| = \frac{n}{(n,a)}$ ?

[wispy scaffold]

in general, you can compute the order of an element in Z/nZ

[cloud walrus]

Yes

[carmine fossil]

What does the order of an element mean wrt the group generated by the element?

[unique juniper]

well since its cyclic

$|Z/48Z| = |\overline{1}| = 48$

[cloud walrus]

Yes

[carmine fossil]

I was thinking about the relation between |x| and the group G=<x>

[unique juniper]

oh

the order of the generating element equals the order of the group

[carmine fossil]

Yes

And an element with order less than order of group does not generate the group

[unique juniper]

ye

[carmine fossil]

And an element with order=order of group generates the group

So,Just look for such elements

[unique juniper]

so values of a that are relatively prime to n

i did that initially, but the next question it Z/202Z so i was hoping there was a quicker way

[viscid pewter]

totient?

[unique juniper]

whats that

[carmine fossil]

No of coprime numbers below a number

[viscid pewter]

yet another thing named after euler

phi(n)

[unique juniper]

the euler function?

i think thats what its called in this book

[viscid pewter]

as if there's only one

[unique juniper]

lol

idk

[viscid pewter]

https://en.wikipedia.org/wiki/List_of_things_named_after_Leonhard_Euler#Functions euler is sodding everywhere in maths

but yes probably it's that

[unique juniper]

i should make a function and name it after me

[viscid pewter]

but anyway the main point is totient m*n = totient m * totient n

[delicate bloom]

when gcd(m,n)=1

[fringe lagoon]

https://cdn.discordapp.com/attachments/767576686324875265/812046352581787688/485820733537386526.png
are any of statements below correct? if they'ren't can anyone point out what exactly isn't true so that i can recheck my work. Thanks

[oblique river]

the first two are correct but redundant. the whole point of the "if and only if" is so that you only have to write one of those two!

the third statement is not correct; take beta = 0

Z/2Z is a field but gcd(0,2) = 2

also just a general note on mathematical writing, you generally want to use your variable names to express things

using beta and L is really confusing to me

[fringe lagoon]

i totally guessed the iff tho, i can't relly see the 2 directions

[oblique river]

why is one a greek letter and the other is a capital letter?

[fringe lagoon]

lower case l looked weird, mb

[oblique river]

I guess my point is that regardless of whether or not they're correct, statements 1 and 2 are equivalent to each other. so if one is correct, so is the other, so writing both of them was redundant

use \ell

for lowercase l

but also why are you using \beta and \ell?

why not \alpha and \beta?

or k and \ell?

[fringe lagoon]

how exactly can you state z/pz being a field in terms of gcd? is it possible?

[oblique river]

yes, just fix the problem I pointed out

[fringe lagoon]

or the only option is to say p>2

[oblique river]

beta = 0 is a problem for that statement, so just remove beta = 0

"for all beta not equal to 0, gcd(beta, p) = 1"

or "for all beta in Z/pZ, beta \neq 0 implies gcd(beta,p) = 1"

[fringe lagoon]

gotcha, also, as said i couldn't really prove that the non existence of an inverse implied gcd!=1, any tips?

[oblique river]

just prove that if gcd = 1 there is an inverse and if gcd > 1 there isn't an inverse

use bezout's identity

[tawny shell]

hey so im trying to prove if h is an ideal of g_2, then a^{-1} (h) is an ideal of g_1 .... we know that g1,g2 are lie algebras and a sends g1 to g2 (homomorphism), can i just say take an x in g_1 and a y in a^-1 (h)

then a([x,y]) = [a(x), a(y)] is in [g_2 , h} is closed in h since h is an ideal of g_2

does that make sense?

ahh kk got it

ah but i have another question

how comes der(g) is a subalgebra of gl(g)? the part where i am at is that

[der(g), der(g)] ⊆ der(g)

how does this imply der(g) is a subalgebra of gl(g)?

ah because gl(g) is the lie algebra of all linear endomorphisms of g?

[chilly ocean]

smells like jacobi identity to me

the bracket on der(g) is just going to be the commutator coming from gl(g)

you can check that a linear combination of derivations is a derivation, so der(g) is a vector subspace of gl(g). then, to say that it's a lie subalgebra is to say [der(g), set(g)] \subseteq der(g), i.e. that der(g) is closed under the bracket

([der g, der g] \subseteq der g is the exact same thing as saying der g is closed under the lie bracket)

[molten silo]

Would z_10 work here

[sturdy marsh]

yes

[molten silo]

say 2*5=0mod10

[scarlet estuary]

for the first sentence ye

[molten silo]

This is not a ring because its not closed under addition right

And this is not a ring since it has no additive inverse right?

[sturdy marsh]

yes

to both

[molten silo]

Can someone check this

[tawny shell]

ahh right, k makes sense. Ty sir

[chilly ocean]

[tawny shell]

so if we want to show ⊆ der(g) for a lie algebra g do we just slap a jacobi identity into it

see if it works

and if it does

⊆ der(g)

[chilly ocean]

what you need to do is check that, with the commutator coming from gl(g), der(g) is closed

meaning that the bracket (commutator) of any two derivations is another derivation

i said jacobi identity because the proof probably uses the jacobi identity

[tawny shell]

that makes sense,

its just that now ive proved for another case

that the ad(g) ⊆ der(g)

and i just slapped jacobi in there

and it seemed all gucci

let g be a lie algebra, prove ad(g) is an ideal of der(g)

i guess ideal is a special case of subalgebra

[chilly ocean]

yeah

[tawny shell]

hence why we slap jacobi

[chilly ocean]

all lie ideals are lie subalgebras

so you wanna prove that ad(g) is a subspace (easy) and that if a is in ad(g) and b is in der(g) then [a, b] is in ad(g)

[tawny shell]

yeah i said basically

x,y,z is in g

and slapped the jacobi

and now i have

[ad(x)y,z] + [y, ad(x)z]

and its like gg

cuz ad(g) ⊆ der(g)

[chilly ocean]

[tawny shell]

I think

we should change the standard text of "Use the Jacobi Identity"

to

"Slap the Jacobi"

[chilly ocean]

sure

would make lie groups even more exciting than they already are

[tawny shell]

exactly

and then we get to have even more fun name with clifford algebras

because boi do i have some names for the gamma bois

[chilly ocean]

[vestal snow]

Can someone explain why the rational canonical form of a transformation stays the same even if we move to a larger field?

[sturdy marsh]

do you know how RCF is constructed?

like the proof of existence of RCF

[vestal snow]

invariant factors

[sturdy marsh]

yes

and invariant factors have nothing to do with the field

[vestal snow]

So I guess I'm asking why are the invariant factors unchanged

[sturdy marsh]

they do not take the field into account at all

[vestal snow]

Hold on

[sturdy marsh]

elementary divisors do

but invariant factors don't

[vestal snow]

If V is viewed as a k[t] module and wriiten in that form

we get that the f_i are elementary factors

Why is it true then that if F contains k, then V = F[t]/(f_1) + ... + F[t]/(f_m)?

As a F[t] module

multiplication by t = multiplication by the transformation

[sturdy marsh]

do you know where those f_i come from

[vestal snow]

PID?

Characterization of modules over a PID?

[sturdy marsh]

okay hold on gimme a sec

[sturdy marsh]

do you mean F tensor V

[vestal snow]

Uh I don't think so

[sturdy marsh]

if V is a k-vector space, the map you have make sense

[vestal snow]

My bad

V in the first equation is k^n

and in the second it is F^n

[sturdy marsh]

so it is F tensor V

F^n is F tensor k^n

[vestal snow]

We haven't done tensor products yet

[latent anvil]

Brofib is avoiding choosing a basis, basically

[vestal snow]

I think my question is basically this

[sturdy marsh]

okay, so ignoring the iso, if you work through the construction, RCF doesnt care about the field

the entries will be in the smallest field containing all the entries of the matrix

[vestal snow]

I don't think it would be very difficult to prove that the invariant factors remain the same after we move to a bigger field

[latent anvil]

Isn't there a uniqueness statement about invariant factors?

[sturdy marsh]

yes, from modules over PID

[latent anvil]

Right

I imagine you can use that

[vestal snow]

Yeah I though about doing it that way

But I think it might not be correct

I think it reduces to this

[latent anvil]

It is certainly cleaner if you can write it out with tensor products, oh well (sorry, this isn't helpful)

[vestal snow]

Lets say we write the matrix A as a matrix of the form (block matrices with 1's right under the diagonal, some constants r_i in the last column, and 0's everywhere else)

then is this necessarily the rational canonical form?

[sturdy marsh]

yes

[vestal snow]

and the polynomials on the columns are the invariant factors?

[sturdy marsh]

yes

[vestal snow]

Oh okay

[latent anvil]

This is the uniqueness theorem I was talking about

[vestal snow]

Then this problem seems straightforward

Thanks

[sturdy marsh]

i.g the important bit isnt just the uniqueness theorem, but there is no factorization going on when looking at invariant factors

so if given a matrix, over a field F, you could view this as a linear operator on a vector space over k

where k \subset F is the smallest subfield containing all the of the matrix

and then over k[t], you have an RCF by modules over PID

and then by the uniqueness thm you see that the RCF doesnt change for any other field

[tawny shell]

so im a bit stuck if anyone can explain to me why using the trace can prove that sl(n,k) is a subalgebra of gl(n,k)? namely using for x,y in gl(n,k) and tr(xy)=tr(yx) and hence tr([x,y])=0

I understand why these properties work, but why does this strictly prove sl(n,k) is a subalgebra of gl(n,k)?

oh

are we assuming near the identity hear?

so the determinant behaves like the trace?

oh wait i think

i might have it

so sl(n) consists of precisely the traceless matrices

heh

i was overthinking it

[tawny shell]

im confused on the lie algebra sp(n,k) ive expanded it out using the lie bracket but dont understand how they've gone from the step $y^T x^T J_n - x^T y^T J_n$ to $-y^T J_n x + x^T J_n y$

[cloud walrus]

𝓐eteer

[tawny shell]

J_n is just our block matrix (lie algebra version)

and x,y are in the algebra

[tawny shell]

@chilly ocean any ideas?

[coarse forge]

Hi,
Suppose that $\operatorname{ord}(g) = m$. Then $g^n = e$ iff $m \mid n$.
For the forward implication, is the following proof valid?

Suppose that $g^n = e$. Firstly, we must have that $m \leq n$. Let $k$ be a positive integer such that $g^{mk} = (g^m)^k = e^k = e$. By the uniqueness of the identity, we must have that $g^{mk} = g^n \implies mk = n$. Since $k$ is an integer, then we have that $m \mid n$ as required.

[cloud walrus]

opengangs

[coarse forge]

I understand the way my professor proved the forward implication (n = mq + r and show that r = 0) but was wondering if this method is also valid

[mild laurel]

No this doesn't make much sense

You say that "Let k be a positive integer such that..." but all positive integers k satisfy that condition so I'm not sure what you mean by that

Later, its not true that g^(mk) = g^n implies that mk = n

I mean, it's true that e^2 = e, but not true that 2 = 1 for example

@coarse forge

[coarse forge]

Okay I see my flaw, thanks

[unique juniper]

an automorphism is just an isomorphism from one group to itself?

[cyan marten]

Yes.

[unique juniper]

oke ty

im not too sure about what i got for this one

if we let Zn be generated by x

the |x| = n

using a theorem |x^a| = \frac{n}{(a,n)}, x^a will also generate Zn if n,a are relatively prime. So wen (n,a) = 1, <x> = <x^a>. thus the map is an automorphism

does this work at all?

[cyan marten]

Yes! But what about the converse?

[unique juniper]

trivial

im sure about the converse

just that the solution was completely different to what i had thought so i had to confirm it was also correcgt

[inner acorn]

Is there a Z[i]-module with 2 (or 3) elements?

[rich ravine]

See the discussion in #advanced-number-theory ; the prime (2) ramifies as (1+i), so we see that the Z[i]-module Z[i]/(1+i)=F_2 has 2 elements

[stark sigil]

I can prove this: Z/pZ has the structure of a Z[i]-module iff p=2 or p is 1 mod 4. This is because a Z[i]-module structure on Z/pZ is the same thing as a Z[x]/(x^2+1)-module structure, which is the same thing as a choice of automorphism x in Z/pZ such that x^2 = -1, which exists only if p=2 or (Z/pZ)^\times has order divisible by 4.

[inner acorn]

Ah, thankies

[cyan marten]

Yeah

[brittle hemlock]

anyone have any idea how to solve this?

[viscid pewter]

f(a) is at max o(a)

there's a start...

i wrote this down somewhere

[next obsidian]

Ah, shoot

if H was normal there's a really slick way to do this

[brittle hemlock]

right that makes sense but I don't see why it divides it

also what does H <= G mean?

just that H is a subgroup of G?

[next obsidian]

It's a subgroup

[viscid pewter]

subgroup

[brittle hemlock]

ok

[next obsidian]

I think you could maybe do this via like an induction

I haven't fleshed this idea out but

show that if f(a) is not equal to o(a) then f(a) <= o(a)/2

I feel like you ought to be able to do this

then you can replace a with a^2

and each time you do this you half the order of a effectively

Or something like that

[viscid pewter]

actually wait

ok wait wait wait

[next obsidian]

this probably isn't the best way to go about this

[brittle hemlock]

wait so what if the order is odd

[next obsidian]

¯_(ツ)_/¯

[viscid pewter]

no i think i've got it lemme have this

so if f(a) >= o(a)/2

ok well call the element in H h

well h-1 has order <= o(a)/2

contradiction

more or less?

[brittle hemlock]

oh I see kinda, so then f(a) must be < o(a) / 2?

[viscid pewter]

yeah

wait this doesn't prove it

f

[brittle hemlock]

why does it have to divide into it tho

idk

[next obsidian]

yeah so my initial idea doesn't work

[brittle hemlock]

new to this group theory stuff I got no idea what I'm doing

[viscid pewter]

honestly same

ok just pinning this down into text bc i need more space free in my brain

intuitively it should be lagrange's theorem wrt. <h> somehow?

[rich ravine]

Write o(a)=f(a)q+r

[severe osprey]

changing the question I was a bit vague: If i have an automorphism, can I refer to it as an endomorphism aswell?

[rich ravine]

Then a^r is in H so r=0

[viscid pewter]

daaamn

that's baller

[brittle hemlock]

so only the identity is in H?

[viscid pewter]

the identity is necessarily in H

[rich ravine]

No, the point is that r<f(a)

and f(a) is the minimal non-zero integer n such that a^n is in H

[brittle hemlock]

OHHH

I see

[rich ravine]

This is similar to the proof that subgroups of Z are the nZ

[brittle hemlock]

good stuff

[rich ravine]

@severe osprey what's your definition of automorphism ? of endomorphism ?

[severe osprey]

Automorphisms are endomorphisms which are also isomorphisms

endomorphisms are maps from itself to itself

such as

$g: X \to X$

[cloud walrus]

F.B.I

[severe osprey]

is an endomorphism

and automorphisms are g's which are also isomorphisms

@rich ravine

It's kinda hard to explain what im asking but

Could I refer to something as an endomorphism

even if it was an automorphism

aswell

[rich ravine]

It depends on the context

[severe osprey]

adjoint representation

[rich ravine]

what do you lose saying it's an automorphism

[severe osprey]

Oh I just don't like writing automorphism

[rich ravine]

does your argument use the property that it's an endomorphism or that it's an automorphism

[severe osprey]

;-;

i just don't like the word automorphism it sounds wrong idk why

:>

[rich ravine]

I'd say just get over it lol

[severe osprey]

:C

cries

@rich ravine well another reason is if I was to say adjoint representation as endormorphisms and automorphisms, could I just say endomorphisms instead of both

petition to rename automorphism

to...

indomorphism

cuz it's like endo and iso :>

[jagged dune]

If left and right cosets are equal, does it imply being abelian?

[chilly ocean]

no, let G be a non-abelian group, and consider the left and right G-cosets

[celest brook]

what does it mean by 'invertibility of [a] in Zn. I'm just confused what invertible means in a modulo class.

[jagged dune]

Is there such a value that a*a^(-1) = e?

I think that's what it's saying.

[celest brook]

ohhh

how come it's multiplication

[jagged dune]

sorry, depends on the operation on the group

[celest brook]

what operation is it talking about. There's two operations in modulo, addition and multiplication, and they both have identity

[jagged dune]

well it should be addition

[chilly ocean]

it's likely talking about multiplicativity

[jagged dune]

integers are closed under addition

lol take Terra's word

I'm still in the class so I could definitely be wrong

[chilly ocean]

[a] always has an additive inverse in Zn so asking about inverses there is kind of boring

but whether elements of Zn have multiplicative inverses is a more interesting question

and boils down to whether the a representative a is coprime with n (as in the exercise)

this is also why Zn is a field iff n is prime

[celest brook]

i'm just confused what 'invertiblity' is it talking about

:(

[chilly ocean]

multiplicative

i.e. can you find a [b] in Zn with [a][b] = identity element

[celest brook]

so where [a][b]=[1]?

or smth?

[chilly ocean]

yes multiplicative identity

rings

[celest brook]

gonna learn about rings soon

ring properties were already mentioned but not rings themselves

[chilly ocean]

look at it like this

[celest brook]

would I use chinese remainder theorem partly?

[chilly ocean]

if as + nt = 1, what happens to this equation if you take the cosets in Zn?

you get [a][s] = [1]

because [n] = 0 in Zn

[celest brook]

ahhh

[chilly ocean]

and that means you have a multiplicative inverse!

[celest brook]

which is s

ohh

[chilly ocean]

[s], yes

[celest brook]

ic

thank you

[chilly ocean]

so in particular the set of all elements of Z_n which have representatives coprime to n is a group

sorry i forgot how to type english there

what about the other way around? if you have [a] in Zn, and you can find an [s] in Zn with [a][s] = [1], must a be coprime to n?

[tawny shell]

proving (R^3, x) is a lie algebra do i just vector product R^3 x R^3, as this shows we make a mapping back to R^3 which what a lie group does, and then just slap the jacobi in there and show its = 0

[chilly ocean]

you need to prove that the cross product (which is presumably the operation here) is bilinear, antisymmetric, and satisfies the jacobi identity

[tawny shell]

yeah

isnt that what a lie group is equipped with

though

the skew symmetric bilinear map

commutator,

and jacobi

[chilly ocean]

lie algebra

[tawny shell]

oh

wait whats a lie group equipped with

bare with me i just woke up

[chilly ocean]

a group operation lol

[tawny shell]

oh yeah

[chilly ocean]

[tawny shell]

just the commutator

G x G --> G

equipped with the operation of lie bracket

[chilly ocean]

well sure lie groups have commutators, but it's not the same commutator as for a lie algebra

[tawny shell]

(G, [ ])

[chilly ocean]

you're conflating lie groups and lie algebras

[tawny shell]

mhm

whats the difference here

[chilly ocean]

a lie group is a group with a smooth manifold structure making the group operations smooth

a lie algebra is a vector space with an antisymmetric, bilinear operation [,] satisfying the jacobi identity

[tawny shell]

o

oh wait

so a lie group is just some endomorphism of spaces

with a differential structure

i feel like when u stop typing ive done this to u

XD

[chilly ocean]

lol

nah i'm just thinking about that

[chilly ocean]

insofar as a group is the set of symmetries of some object, sure

i'm not too comfortable with that statement but i guess you can make it work

[tawny shell]

in other words locally?

[chilly ocean]

[tawny shell]

well you know i was thinking the other day if we have a level set right, this proves a vector isomoprhism right?

between manifolds

does the converse hold?

[brittle hemlock]

anyone got any ideas on how to do this?

[sturdy marsh]

just write them down

there are only 6 computations to do

take each element \sigma of S3, and compute the set \sigma H

$\sigma H = {\sigma h : h \in H}$

[cloud walrus]

Brofibration

[sturdy marsh]

oh ig they want right cosets

so compute $H \sigma = {h\sigma: h \in H}$

[cloud walrus]

Brofibration

[brittle hemlock]

my bad didn't see this @sturdy marsh

I'm just confused on what a right coset really is

could you give me one example of a right coset

[sturdy marsh]

uh

<3>

is a subgroup of Z

[brittle hemlock]

right

[sturdy marsh]

and the right coset <3>2

is the set of multiples of 6

i.e. numbers of the form (n)(3)(2)

[brittle hemlock]

oh I see

thanks

[mild laurel]

Uh wait

[sturdy marsh]

oh wait i screwed up lmao

add

not multiply

3n + 2

[brittle hemlock]

so the coset is just an offset?

[sturdy marsh]

not multiples of 6

my bad

[brittle hemlock]

you're offsetting by 2?

[sturdy marsh]

yes

[brittle hemlock]

ok I see

[sturdy marsh]

it's a "shift" of the subgroup by an element

[latent anvil]

yes!!! (exclamation points because this is exactly the right idea)

it's also good to think about this for vector spaces

[sturdy marsh]

that was a pretty big brainfart

[brittle hemlock]

LOL

[latent anvil]

say you have some plane P through the origin like x + y + z = 0 in R^3

[brittle hemlock]

you're good

[sturdy marsh]

[latent anvil]

this is a subgroup of R^3 (under vector addition)

does that make sense?

[brittle hemlock]

yeah that makes sense

[latent anvil]

what do you think the cosets of P are?

g e o m e t r i c a l l y

[brittle hemlock]

uhh the entire 3d plane?

[latent anvil]

well the union of all of the cosets is always the whole group, sure

but they're still distinct from one another

[brittle hemlock]

bc you can shift the diagonal x + y + z = 0 by some offset in any direction

hmm

[latent anvil]

so let's try an example, whats the coset of P by v = (1,1,1)?

just as a random example

[brittle hemlock]

would it be like x + y + z = -3

or am I stupid

[latent anvil]

That's almost right, but you have a sign error

any element of v + P looks like v + w = (x + 1, y + 1, z + 1) for some w = (x,y,z) in P, right?

then (x+1) + (y+1) + (z+1) = (x+y+z) + 3 = 3 because w is in P

[brittle hemlock]

oh ok I see

[latent anvil]

and conversely if x+y+z = 3 then (x-1,y-1,z-1) is in P and (x-1,y-1,z-1) + (1,1,1) = (x,y,z) so (x,y,z) is in v + P

[brittle hemlock]

right makes sense

[latent anvil]

(we don't need to worry about left vs right cosets because addition is commutative)

[brittle hemlock]

left and right just refers to where you're applying the offset?

either to the left or right?

[latent anvil]

right, it's like gH = { gh : h in H } or Hg = { hg : h in H }

[brittle hemlock]

ok I appreciate it

[latent anvil]

first is the left coset and the second is the right

okay so

v + P = {(x,y,z) : x + y + z = 3}

can you make a conjecture about what all the cosets of P are?

like what they look like

[brittle hemlock]

x + y + z = n?

[latent anvil]

where n is what?

like what kind of thing

[brittle hemlock]

any value

any integer?

[latent anvil]

not just an integer

[brittle hemlock]

or can all real numbers be applied to cosets

[latent anvil]

think about (0.5,1,1) + P

yeah, so any real number

because we're working in R^3

[brittle hemlock]

yeah right

[latent anvil]

if we worked in Z^3 the answer would be different, but also we'd lose some of the geometry

[brittle hemlock]

that's why the union isn't a discrete set but instead all of R^3

yup

[latent anvil]

exactly!

so

the cosets looks like {(x,y,z) in R^3 : x + y + z = c}

for a parameter c

what does this look like geometrically?

[brittle hemlock]

just shifts of x + y + z = 0?

across the x axis

idek

[latent anvil]

https://www.geogebra.org/3d/jjqaghwk

it's all the parallel planes!

which is the same as what you said

I think this is a good picture to keep in mind

for cosets

[brittle hemlock]

oh yeah that's what I meant, makes more sense

bet I appreciate it

thanks

so quickly going back to the question I posted

I'm still a little confused on what cyclic subgroups are, what would an example of a right coset of this be?

[latent anvil]

so the cyclic subgroup is H = {e, h, h^-1, h^2, h^-2,...}

so on, forever

if h has finite order n then H = {e, h, ..., h^(n-1)}

[brittle hemlock]

what does h^n denote (I'm very new to group theory, my bad)

[latent anvil]

so you just keep multiplying h with itself

oh, ^n is like a power

h^3 = h * h * h

[brittle hemlock]

yeah but how does that apply to a group

[latent anvil]

for examples

well we can multiply elements of a group, right?

[brittle hemlock]

yeah

[latent anvil]

we can't talk about like, h^(0.5)

but h^6 is just repeated multiplication

[brittle hemlock]

oh so this is just the operation of the S3 applied n times

[latent anvil]

exactly

and that's what the cyclic subgroup contains

all of the powers of h

[brittle hemlock]

oh ok

so what is the operation of S3? Sorry again, idk what S3 really is

[latent anvil]

Do you have notes/a textbook?

[brittle hemlock]

not really I'm just searching some problems up online LOL

F

[latent anvil]

oh

then uh

don't do that?

why are you searching random group theory problems online without reading a textbook?

[brittle hemlock]

idk I didn't think it would be as abstract with all the notation

thought it would be more intuitive like other math classes

[latent anvil]

I mean idk how you would learn like, calculus without reading some notes/textbook/watching lecture

[brittle hemlock]

fair enough

I was more like comparing it to number theory, where you can learn a lot from just thinking abt the problem

[latent anvil]

you can do that in group theory too

[brittle hemlock]

sorry for wasting your time 😦 should've consulted notes before

[latent anvil]

but it's the same in number theory, you need to see a definition of "equal mod n" eventually

[brittle hemlock]

true

[latent anvil]

np haha, you didn't waste my time

But I think you are wasting yours a little if you don't use outside notes

[latent anvil]

and just to reiterate, doing problems is really important for learning math in general, but you need to also know definitions

[brittle hemlock]

for sure, do you have any decent resources?

yeah I get that you're right

[latent anvil]

I like Dummit and Foote's book

[brittle hemlock]

alright thank you

just for the satisfaction of finishing this problem though, what would some right cosets be?

I saw this on wikipedia

just the permutations of 123, is this what S3 is about?

[chilly ocean]

S_3 is the set of permutations of the numbers 1, 2, and 3

[summer geyser]

Any ideas for this? I played around with \alpha and noticed it was the negative golden ratio, so it's a root of X^2+X-1 in Q[X]. However, I only managed to get this b/c I plugged it into a calculator. Is there any efficient way of getting the minimal polynomial of \alpha?

[chilly ocean]

Maybe not a good strategy in general, but if you already know it's quadratic, you can square alpha and try to write it as a linear combination of alpha and 1

[summer geyser]

I tried this and got (writing beta for zeta) $$\alpha^2=\beta^6+2\beta^5+\beta^4=\beta^4+\beta+2.$$

[cloud walrus]

Athurus

[mild laurel]

It might also be helpful to note that this number is 2cos(2pi/5)

[summer geyser]

not sure how to express this in terms of alpha and 1

yeah I got that

so I thought of using chebyshev polynomials but that seems kinda overkill (although it worked)

[mild laurel]

yeah, so you need to use the fact that $\zeta^4 + \zeta^3 + \zeta^2 + \zeta + 1 = 0$

[cloud walrus]

Zopherus

[summer geyser]

oh yes of course, then I assume that reduces it to a degree 2?

[mild laurel]

something like that yeah

[summer geyser]

alright ill give it a shot

ah yes it works, thanks so much that was a lot easier than I had originally thought

[celest brook]

wait, it's all just matricies

[inner acorn]

Hmmm, how would I approach finding a basis for part (5)?

I know from experience that this ring R is a representation of the quaternions

that was pretty standard to show

and I know that M_2(C) is a free module over R because, every module over a division ring has a basis

I tried some basic guesses, but nothing seems promising, any tips?

[eager schooner]

would anyone help with in thinking with this problem (not give answer)

I know that the two subgroups mentioned are indeed cyclic

but I'm unsure how to get to G is cyclic from this fact though

it seems like it might be related to multiplying groups but we haven't covered that yet - perhaps someone could comment on if it does?

[old lava]

if there's only one subgroup of a given order, that subgroup must be normal

so you have normal subgroups of orders 3 and 11

[eager schooner]

I'm not sure what a normal sub group is

[old lava]

ah

[eager schooner]

we haven't covered that either

[thorn delta]

basic idea: let x!=0. There are 3 possibilities for the order of x: 3, 11, or 33. Show that the order of x is not 3 nor 11.

[eager schooner]

Okay sure that was an approach I was considering

The problem is, I don't know of anyway to show that an element can't be of order 3

Espec. here because it could be of order 3. In fact exactly one x is of order 3

[thorn delta]

oops, i made a mistake.

let x!=0 have maximal order in G. Then x has order 11 or 33.

I think this is probably the right place to start

[viscid pewter]

well, two x surely

at least x and x-1

[inner acorn]

#groups-rings-fields message <@&286206848099549185>

👀

[kindred rivet]

Yeah

Oh no

[inner acorn]

problem 4 is basically free marks, but I have no clue how to find a basis for M(C, 2x2) over the quaternion matrices qwq

[chilly ocean]

I don't know a lot about modules, but maybe: do free modules have a well defined dimension? The at least if it is free it must be dimension no more than 4. and you could probably show it's not 1

[latent anvil]

Not in general for noncommutative rings, but it should all work out here since we're working over a base field C

I think dim_C R = 2 and dim_C M_2(C) = 4 but I'm not sure (about the dimension of R)

[inner acorn]

yeah that sounds right

[latent anvil]

right, so this should force it to be 2d if it's free

2d over R I mean

[inner acorn]

so I just need to find 2 linearly independent matrices?

[latent anvil]

No, that's not sufficient

In Z×Z the elements (2,0) and (0, 2) are linearly independent but not a basis

It might be sufficient since R is a division ring?

[inner acorn]

yeah you're right, it's not sufficient because I could take [[1, 0], [0, 0]] and [[0, 1], [0, 0]]

even with a division ring it doesn't work

oh wait

it does! :U

[eager schooner]

Okay so for my question here's what I came up with

You have an element that generates a group with order 11

and an element that generates a group with order 3

at most there are a total of 14 elements between these groups

then take an element not in those 14 elements

This element must have an order of 33 since there is only 1 group of 11 and 1 group of 3

(and don't take the identity element)

But I'm no sure perhaps someone could help debunk this

[viscid pewter]

could it not have some other order

it doesn't say that those are the only subgroups

[eager schooner]

Right but since the order of an element has to divide the group the only options are 11, 3 and 33

11 and 3 are already taken

(and it says there's only one of them)

[viscid pewter]

oh right yes

derp, good one

[latent anvil]

nice! sorry if I wasn't much help

to my memory linear algebra works pretty much the same over a division ring

in some ways it's better than modules over a commutative ring

(in other ways it's worse)

[inner acorn]

[golden pasture]

wait srs?

i rmb there being very messed up things

like AB=1 doesnt imply BA=1 i think?

honestly idk noncom kinda pain

[tawny shell]

hey can someone explain to me the intuition behind lower central series of a lie algebra

I get its used to show nilpotency

but I dont actually get what's happening

do we keep quotienting the group till we get the centre?

[median tree]

In the definition for a transitive group action, I am unsure of why we need to say for all x, rather than for any x in S. Is there an example of a situation where the orbit of one element in a set is equal to the set, but the orbit of another element in that set isn't equal to the set?

It seems a bit unnecessary, because it seems to me that if the orbit of any element is equal to the set then wouldn't the orbit of every element be equal to the set?

[scarlet estuary]

uh could you give the full definition

[scarlet estuary]

okay

[median tree]

yeah, on it, lol

I'm just thinking of this in terms of the basic example of say, the cyclic group C_3

[scarlet estuary]

why we need to say for all x, rather than for any x in S.
what's the difference?

oh by "any" you mean pick a particular x?

[median tree]

Sorry, I should have said, that there exists an x

[scarlet estuary]

sure

[median tree]

sure as in, yes that would also be correct or just with my clarification?

[scarlet estuary]

no, i mean your clarification

the canonical example AFAIK is an orthogonal group acting on, say, R^2 \ {0}

[mild laurel]

Yeah I agree that the definition is stronger than strictly necessary, but as you note, they're equivalent and this form is a lot earlier to use in practice

And its also more intuitive too

[scarlet estuary]

no this isnt stronger?

wait

its just as strong as you need for equivalence, right? unless im being a dumbass

because the action on the unit circle is transitive

but it isnt on R^2 \ {0}

[old lava]

if two definitions are equivalent, how can one be stronger than necessary

that makes no sense

[mild laurel]

I just mean the statement seems stronger, even though its not actually

[scarlet estuary]

oh lmao

[mild laurel]

Like usually we write things as "weak" as possible in some sense, because usually its easier to prove weaker statements

but we don't do so here because of what I said

[median tree]

this is the canonical example of what, exactly?

[scarlet estuary]

of a case where "there exists x" wouldnt be enough

there might be a simpler example but this is the first that comes to mind

[median tree]

Okay, so in the case of R^2 \ {0} there is an element where the orbit of that element under action with the orthogonal group is equal to R^2 \ {0}?

[mild laurel]

wait yeah uh, I'm pretty sure these are equivalent nami

[scarlet estuary]

...

oh my god im a dumbass

sorry

got wires crossed

[mild laurel]

A lot of transitivity proofs do go through the alternative statement you use though. Like choosing a special element x in S and showing that you can get from x to everywhere

showing that the action of a group on its Cayley graph is transitive for example often does this where you choose the identity element of your group/Cayley graph

[median tree]

Getting my wires crossed is a regular occurence, no worries

Okay! That's actually great to hear, because I was about to do a transitivity proof and it seemed like proving it for all elements was a bit unecessary

[uncut girder]

anyone know about the weight lattice for algebraic torus?

https://en.wikipedia.org/wiki/Algebraic_torus#Weights_and_coweights

So the weight lattice of an algebraic torus is the group of algebraic homomorphisms from your torus to the multiplicative group in your field

basically, the group of characters of your algebraic torus

where the group operation is pointwise multiplication

the weight lattice will be a free abelian group whose rank is the same as that of the torus

that's actually pretty cool imo, is there an obvious reason as to why that's true?

I guess each of the generators of this free abelian group will be the character with is just projecting onto the i'th coordinate, and sending the other coordinates to 1.

yep. makes sense

wikipedia says taking weights gives an anti equivalence between the category of algebraic tori and the category of free abelian groups

thats kinda cool 🤔

why would it be an anti-equivalence? i guess it has to do with taking maps out of tori? its like taking the dual space of a vector space

which are maps out of the original vector space

thats exactly why.

Say you have a morphism of algebriac tori T -> S.
Then form the weight lattices X* (T) and X*(S).

Take any character in X*(S) and precompose with the map T -> S to get a character in X *(T).

This gives a map X*(S) -> X *(T) between free abelian groups in the opposite direction.

i like this

How do you recover a map T -> S between algebraic tori from a map between their weight lattices X*(S) -> X *(T) in the opposite direction?

well lets just work with the rank 1 case for now

lets say T and S are both rank 1

then a map between them will just be taking an N'th power

this induces a map between weight lattices X*(S) -> X *(T), namely the one that's multiplication by N

(just think about it, ask me if you have questions lol)

I just realized that an map between tori T-> S will always be surjective

kinda like a covering map

wait is that true?

no

say you have a weight lattice Z^2 and another one Z, and the map between then is projection onto the first coordinate

this is definitely a valid map between abelian groups

so there should be a corresponding map between algebraic tori

and I claim it the injection of a rank one torus into a rank two torus, where its identity on the first coordinate and a constant 1 in the second

ok this actually makes a lot of sense

https://en.wikipedia.org/wiki/Weight_(representation_theory)

[chilly ocean]

you know after reading that definition I'm slightly disappointed that's not called an eigenweight

"In this context, a weight of a representation is a generalization of the notion of an eigenvalue, and the corresponding eigenspace is called a weight space. "

[stark sigil]

Another better term would be eigencharacter

[chilly ocean]

just something eigen

nothing says math phys like eigen every 20 or so syllables

[carmine fossil]

eigeneigen

[median tree]

I'm working a problem and I think I just need some clarification here... So I'm trying to prove Burnside's lemma

But it seems to me that one gets that |F|/|G| = the sum for all x of 1 / |orb(x)| which is not the same thing as the number of orbits in the action of G on X

[carmine fossil]

What is F exactly?

[median tree]

I think this may be an issue of parsing the question

but to me |F|/|G| being equal to the number of orbits in the action of G on X means that |F|/|G| = the sum for x in X of |orb(x)|

but I am pretty clearly getting the sum for x in X of 1/|orb(x)|

you may also want to note that

where

I just have a suspicion that this question is phrased in a way that is either counterintuitive, or is perhaps just wrong

[mild laurel]

I think it's just how many orbits there are

[median tree]

ah, I was missing the exact definniition for the set of orbits

I see now

[cyan marten]

How is A* a graded ring? For example, if x is an element of a, then x* = (0, x, 0, 0, ...) is an element of A*, so if A* is graded we would havw x*^2 in a^2, but it is not!

Its square is (0, x^2, 0,...), so still not an element of the summand a^2.

I am missing something, but I don't know what it is..

[sturdy marsh]

ye that's not the multiplication on A^*

x lives in degree 1

x^2 lives in degree 2

I guess the example to keep in mind is just a polynomial ring over a field

say k[t]

we have a module iso $k[t] \rightarrow \oplus_{0}^\infty k$

[cloud walrus]

Brofibration

[sturdy marsh]

but the multiplication on the ring is not componentwise multiplication on the thing on the right

it's multiplication of polynomials

idk if that made sense @cyan marten

im super sleepy

but lemme know if youre still confused

[chilly ocean]

go to sleep

[sturdy marsh]

I am typing up homework

[chilly ocean]

😔

i know the struggle

good luck

[celest brook]

word problems come back with a vengeance

[chilly ocean]

combinatorics is just the study of word problems

[carmine fossil]

just spam gen functions

[celest brook]

i have no idea what a generating function is

[sturdy marsh]

people seem to like the book generatinfunctionology

[celest brook]

i wish I knew more computer science and coding

[chilly ocean]

generatingfunctionological generatingfunctionology

[latent anvil]

Functional generatingfunctionology

It's weird how we have both geometric algebra and algebraic generatingfunctionology

[sturdy marsh]

functionological generating theory

[chilly ocean]

function

[latent anvil]

fonction

[chilly ocean]

🥖

[carmine fossil]

fun

[latent anvil]

I have some very exciting and tense news

[chilly ocean]

⠀

[latent anvil]

I got a really great sealed deck on magic arena and am currently 1 game from winning the league

no losses so far

[sturdy marsh]

[latent anvil]

It is the biggest/best thing going on in my life rn

[cyan marten]

It did. I am still annoyed they didn't how multiplication works..

[sturdy marsh]

graded stuff is confusing

[latent anvil]

I have a tweet about this

It's so dumb

It's literally like, nothing. Graded rings have so little info

And yet

So confusing

[chilly ocean]

graded rings

[latent anvil]

https://twitter.com/grassmannian/status/1257768033449209857?s=19

Free Brendan On One Generator (@grassmannian)

Why are graded rings/things so confusing for me. It's literally just indexes and extra sign terms. There's nothing to be confused by!!!!

Twitter · May 5, 2020, 8:24 PM

I was doing singular cohomology stuff I think

[chilly ocean]

i will reply

"because you are brain smooth"

[latent anvil]

yes do it

[chilly ocean]

so you can find my account?

[latent anvil]

[sturdy marsh]

graded stuff confused the crap out of me while doing proj for the first time

[chilly ocean]

i can reply but make it so you can't see the reply

[latent anvil]

That too brofib

I think like, twisting sheaf/shifting the degree on O_X modules was when I totally checked out of my AG class

[sturdy marsh]

yup

[latent anvil]

I mean I was very behind and lost anyways but something about the graded stuff was too much

Alright wish me luck gang

[cyan marten]

Good to know!

[latent anvil]

I'm gonna beat this league without any losses

[sturdy marsh]

[cyan marten]

Is my intuition for a-filtrations of bounded difference (a an ideal) correct?

M_i and M'_i are of bounded difference iff they define the same topology

[latent anvil]

You did not wish me enough luck

One more chance

[sturdy marsh]

[carmine fossil]

[latent anvil]

Currently at 1 loss 0 wins in my second go at the final round...

okay I won this round

it all comes down to this

welp I lost lol

Opponent had a crazy deck

gahhh

[vestal snow]

Can someone tell me about a non-trivial use of Yoneda's Lemma?

[latent anvil]

i had something interesting come up with yoneda the other day. there's these things called characteristic classes in topology, and they're pretty much natural transformations from a specific contravariant functor F to the cohomology ring functor Top^op -> Set. It's a nontrivial theorem that F is representable, but once you know it you know characteristic classes are really just elements of the cohomology ring of the representing object

F(X) is the set of isomorphism classes of vector bundles over X, if that is meaningful to you

(you actually need to add some topological niceness constraints but if you eg restrict to cw complexes everything I've said is valid)

Oh and this is interesting because you can actually just compute the cohomology ring of that representing object explicitly

[vestal snow]

Hmm okay

Thanks

[latent anvil]

Sorry, I'm trying to think of examples without blackboxing things

[vestal snow]

Do you know of any to algebraic geometry?

It's all good

[latent anvil]

Yoneda is very important in AG but I'm going to butcher it if I try to explain

@next obsidian if you're up

[vestal snow]

I'd still like to hear it

[latent anvil]

So one thing yoneda tells you is that you can always work with the functor that an object represents instead of that object

Right?

you lose no information

[vestal snow]

Yup

[latent anvil]

Terminology aside: if X is a scheme, Hom(-, X) is the functor of points of X

so it turns out that a lot of schemes are awkward to work with but have very nice functors of points

Have you seen the fibered product of schemes?

[vestal snow]

Yes

[latent anvil]

Right

It's miserable

Awful construction

Annoying to work with

but!

It represents a very very nice functor

what's the functor of points of X ×_S Y?

[vestal snow]

Hom (-,X x_S Y)

and then we can use the universal property of fibered products

[latent anvil]

right exactly

So it's the fibered product of the two functors of points of X and Y

So yoneda can reduce messy statements about fibered products to nice statements about functors

Why is the functor nicer? Because it's valued in sets

And we know how to prove things in sets

and actually you can use this to construct the fibered product

There are certain criteria that tell you when a functor Sch^op -> Set (or even Ring -> Set) is representable by a scheme

so one way to avoid the tedious construction of the fibered product of schemes is by showing the fiber product functor is representable

There are other useful representable functors, like A^1_Z represents the functor X |-> O_X(X)

Or even better A^1_k if you're in the category of k schemes

[vestal snow]

Okay that's pretty cool

[latent anvil]

Yeah!

Do you see how to prove it?

[vestal snow]

I think so

[latent anvil]

I guess for me like, yoneda can be a powerful tool but it's most important philosophically

it tells you you go look at representable functors

Okay so here's where it gets really out there

Do you know anything about manifolds?

[vestal snow]

Uh not really

[latent anvil]

hmm okay

[vestal snow]

locally R^n

that's about it

[latent anvil]

sure

Well we have very nice theorems about when a quotient of a manifold by a lie group/finite group action is still a manifold

Very handy for eg constructing the grassmannians as smooth manifolds

it would be nice if we had theorem like this for schemes

so assuming G is a finite group acting nicely on X, is there a scheme Y such that maps Y -> S are the same as maps X -> S which are invariant wrt the action

Does that make sense?

So in manifold world we can take a quotient of the sphere by a Z/2Z action to get projective space

[vestal snow]

I think I get what you're saying

Not the technical bit

But I can see that Yoneda is pretty useful

[latent anvil]

Sure, I won't go any further in this example

the point was going to be that these quotients don't exist as schemes, but do exist as functors

and yoneda might lead you to think, hey, schemes are just a certain type of functor

Maybe more general classes of functors are still geometric?

and that's how you end up getting into stacks

Which I know nothing about but chmonkey is currently taking a course on

[vestal snow]

Is it a class in algebraic geometry?

Or algebraic stacks specifically?

[latent anvil]

Stacks and moduli of curves I think?

Definitely stacks

[vestal snow]

Dang

I wish my school had classes as specialized as those

I think the highest we have is algebraic geometry

So I had a question

Nevermind it was a stupid question

[latent anvil]

hell yeah

[vestal snow]

so fibered product commutes with taking functor of points?

Is this correct?

[latent anvil]

Yup

Limits in a functor category are computed objectwise

(if the target category has all limits)

[vestal snow]

Is Yoneda an example of Grothedick's relative viewpoint?

They sound similar, but I just wanted to make sure I'm not completely misinterpreting them

[latent anvil]

Not as far as I see

But also I'm not AG brained

[vestal snow]

Let F be a function field in one variable over k. Is the ring of integral elements over F of any particular importance?

I have an idea for a project and I'm trying to see if anyone knows this is worthwhile

[wraith obsidian]

What's the „relative viewpoint“? I've heard of grothendieck's notion of „generalized elements“ by seeing elements of X as morphisms Y→X, is that it?

[vestal snow]

Yes that's exactly it

[wraith obsidian]

Then I'd say yes, at least that's how Yoneda has been introduced to me.
My view on that is that in set, we can recover what a morphism f: X→Y does by „probing“ it with singleton maps *→X.
Taking a more non-trivial example, like Grp, we see that this works when we replace the singleton by ℤ.
Yoneda now gives us that Hom(Hom(-,X),Hom(-,Y)) \simeq Hom(X,Y) (modulo some covariant-contravariant-swap I always do), which essentally means that morphisms X→Y can be recovered once we get a natural translation of the generalized elements, i.e. „probes“ of X to generalized elements of Y. So Yoneda is essentially saying that this „trick“ works no matter how absurd your category is, once you include all the objects of your category as possible probes.

[vestal snow]

Got it. Thanks for the detailed answer

[next obsidian]

Yoneda is cool because when you want to generalize a scheme you generalize it by working with certain nice functors which in a sense let you glue (have descent) the way a scheme does

In order to do this you’d need to know that a scheme is a type of functor, because how can you generalize a scheme to a special functor if it isn’t a special case of tha?

So to even define these things you end up implicitly using that morphism of the functors of points are exactly the same as normal morphisms between the schemes.

There’s also a lot of constructions which are easier to verify from the functor of points. If you want to define a group scheme it’s actually easiest (IMO) to consider it as a factorization of the functor of points through the category of groups. Then, to define an action of a group scheme on another scheme it suffices to actually define a on the nose group action on the functor of points

This for example makes it easy to see that the pullback of a principal G-bundle is again a principal G-bundle, and the G-equivariance of a map is again, easy to see via the functor of points

[uncut girder]

https://youtu.be/2TrFG032mGo

[tawny shell]

hey guys, if i and j are solvable ideals of g

why is 1 - j a solvable ideal of i + j

[sturdy marsh]

characteristic classes, cohomology operations

[latent anvil]

Lol char classes were my example

[tawny shell]

on a side note, how comes [i+j, i+j] = [i,i] + [j,j] + [i,j]? shouldn't there be a 2 on [i,j]?

[sturdy marsh]

https://ncatlab.org/nlab/show/motivation+for+sheaves%2C+cohomology+and+higher+stacks

relevant-ish

[latent anvil]

I'd never thought about this but it's a very interesting perspective on yoneda's lemma

wait no I thought about this a couple weeks ago in terms of simplicial sets I just forgot

Still very interesting

[sturdy marsh]

it's a really good article

gets me in the mood for category meming

and then 30min later the category meming gets too dry and I want to do something more concrete

[latent anvil]

lol

[chilly ocean]

What do we call division on R?

A word for it?

[next obsidian]

This notation is used by eg Alper

It’s a good notation

[latent anvil]

Well it's not really the notation I care about, it's the fact that maps h_U -> F are the same as elements of F(U)

Which is what we'd expect if F were maps into some fixed space

this is like how we can think of the elements of a simplicial set as like singular simplices, since Δ[n] -> X is the same as X_n

[uncut girder]

Division

[chilly ocean]

?

[sullen bloom]

multiplication by inverses?

[next obsidian]

Right, but I mean the notation of conflating X with h_X

It’s good

[hazy leaf]

So, $\mathbb{R}/n\mathbb{Z}={n\mathbb{Z}+r;|;r\in[0,1)}$, how do you in a similar way like $\mathbb{R}/\mathbb{Z}$

[cloud walrus]

亜城木 夢叶

[mild laurel]

Uh what? That's just the n = 1 case

[hazy leaf]

when n = 1, we have [0,1)

[mild laurel]

No, I mean that R/Z is just R/nZ if you let n = 1

[hazy leaf]

i am looking for a general form

[mild laurel]

I'm not really sure what you mean

$\bR/ \bZ = { \bZ + r \mid r \in [0,1)}$

[cloud walrus]

Zopherus

[mild laurel]

which is the same thing that you have if you set n = 1

[hazy leaf]

ok

i just make thing complicated

[snow flint]

its okay, dw ab it :)

[crystal vault]

Hello c: I had a simple group theory question that I asked a while ago but am still not clear on. The exercise given is this:

And the author defines powers like this:

When I did part (a) myself, I just argued that $x^{a+b}$ is $xxx...x$ (a+ b times), so you can split it into two groupings, one with "a" of them, and one with "b" of them. The product is independent of bracketing (which was proved earlier in the section), so the new product is the same.

But when I looked online, the solution given was to perform induction on "b" ONLY. My question is why you wouldn't also need to perform a double induction, using both "a" and "b".

[cloud walrus]

kirafa

[old lava]

is this from d&f lol

I remember this exercise

[crystal vault]

yes it is (:

Abstract Algebra 3rd Edition

[chilly ocean]

I'm not sure why induction is necessary at all if you already have associativity of multiplication (the general version you are allowed to multiply any number of terms in any order). But I guess if you don't have general associativity then you need to be more careful, and induction is necessary

[old lava]

well, all you have to do is induction on b, since x^(a+0) = x^a x^0 is obvious

so you just induct on b