groups-rings-fields

[carmine fossil]

That's kinda bad. You no longer have primes

[chilly ocean]

wtf are primes?

so?

[carmine fossil]

A prime is a prime number

[chilly ocean]

never heard of those

[golden pasture]

prime is a element such that quot is domain

[chilly ocean]

a prime is an element whose generated ideal is maximal

[subtle current]

[chilly ocean]

(all rings are pids)

[carmine fossil]

Make it Euclidean

[golden pasture]

eww

imagine having $h_K=1$

[cloud walrus]

a cute cat ٩(˃̶͈̀௰˂̶͈́)و

[golden pasture]

lame

[cloud walrus]

mirzathecutiepie

[carmine fossil]

Yes

[golden pasture]

a very funny way to write it tho

usually id see <a|a^n>

[scarlet estuary]

yes

this is basically the definition of a cyclic group btw

a group generated by a single element

(though this leaves off the case of the infinite cyclic group, i.e. the case where there is no such n; we call that group Z under addition)

[viscid pewter]

not quite

[carmine fossil]

Z

[scarlet estuary]

they arent by definition finite, although the only infinite one is Z

note that Z is generated by 1 element <1>

1 + 1 + 1 + .... makes any integer

[viscid pewter]

i mean it's like

[scarlet estuary]

or 1-1 - 1 - 1 -1...

[viscid pewter]

the term 'generated' doesn't mean 'you can make everything from one element', it means 'you can make everything from one element and its inverse', right

[hexed vortex]

I think it's a convention that depends on the country.

If I recall correctly, in undergrad in France I learned that a cyclic group is finite and we called an infinite cyclic group a "monogenic"
In english lanhuage, though, cyclic groups can be infinite

[carmine fossil]

Yes

Or a^n as n ranges over Z

[viscid pewter]

urgh france

is it true that you guys include 0 in N? hate it hate it hate it

[carmine fossil]

a is a generator

a^n is just a random term

[hexed vortex]

We do, and we are proud of it

[viscid pewter]

burn

[carmine fossil]

it depends on the group

[unique juniper]

unsure about this

since A is a subset of B

[scarlet estuary]

intuitively, commuting with everything in B is "harder" to do than commuting with everything in A

since there are more elements in B than A

can you formalize this idea?

[unique juniper]

no

if x and y are in the Centre of A in G, then x and y are also in the centre of B in G right?

if B has (possibly) more elements then it also has more elements in G that can commute ?

[viscid pewter]

i don't think that's correct terminology

centraliser, right

[scarlet estuary]

i think youre getting your definition of centralizer mixed up a bit

[unique juniper]

oof

ok

[scarlet estuary]

$C_G(A) = {g \in G \mid ag = ga\forall a \in A}$

[cloud walrus]

Namington

[scarlet estuary]

in particular, elements of C_G(A) come from G

not from A

[unique juniper]

yes

i have the same definition errrr

let me rephrase

[scarlet estuary]

so suppose $g \in C_G(B)$; then this means $bg = gb$ for all $b \in B$. but since $A \subset B$ this means that $ag = ga$ for all $a \in A \subset B$

[cloud walrus]

Namington

[scarlet estuary]

hence $g \in C_G(A)$

[cloud walrus]

Namington

[scarlet estuary]

all that's left to prove, if you havent already, is that the centralizer is indeed a subgroup

[unique juniper]

er

maybe i have more notation mixed up, A is a subset of B or superset?

[scarlet estuary]

A is a subset of B, so each element of A is also in B

bg = gb for each element of B

but each element of A is also an element of B

which means ag = ga for each element of A (since a is also an element of B)

this is what i meant by saying that it's "harder" to commute with evevrything in B than it is to commute with everything in A

B is larger than A, so in order to commute with everything in it, you need to commute with "more" stuff

(or at least as many)

which is a tighter condition

hence we expect C_G(B) to be smaller than C_G(A)

my proof formalizes that idea

(here "larger"/"smaller" and related terms are meant to be understood in a non-strict sense)

[hexed vortex]

A good intuition from linear algebra here is that C_G(A) behaves kind of like an "orthogonal" of A in G. And in linear algebra, if A is a subspace of B, the orthogonal of B is a subset of the orthogonal of A.

Here the role of your multilinear form is played by the commutator, [a,g]=aga^(-1)g^(-1)
I can't really recall right now if there's a way to make this intuition a bit more rigorous/meaningful

[unique juniper]

got it

[unique juniper]

interesting

[hexed vortex]

I guess my comparaison is a bit too ambitious though.
Maybe talking in terms of preimage of a function (the commutator) would be more sound than my idea of a bilinear map.
I think I was mixing it with memories of something elsd

[shut halo]

Hi, can anyone show me how to do the following question:

Let $$C: y^2 z + yz^2 + = x^3$$ over the field $\matbb{F}_2$ of integers modulo 2. Find $C(\matbb{F}_2)$.

[cloud walrus]

snypehype46
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[hexed vortex]

You have only 8 points to check out (or 7 if you see C as a projective curve) so the simplest way is probably to try them all out by hand

[shut halo]

Ok I see, but I don't quite what does it mean to compute C(F)?

[oblique river]

It means "find all the solutions with coefficients in F"

in this case there are 3 variables, so you're looking for triples (x, y, z) in F^3 which satisfy that equation

where here F is F_2

[shut halo]

Ok I see, do you know any references/textbooks that discuss this?

[oblique river]

yes, all of algebraic geometry

any book on algebraic geometry will do this

[shut halo]

Ok I'll check them out

[oblique river]

http://www.math.lsa.umich.edu/~wfulton/CurveBook.pdf

this book is on algebraic curves

your equation defines a projective curve

but also like, to answer the question you posted, you don't need anything fancy

just test all the points and see which ones are solutiosn

[chilly ocean]

short book

is this the calculus on manifolds of AG

it looks like it has many exercises

[oblique river]

I think it's a pretty standard intro AG course because curves are pretty concrete but also give you a chance to introduce some more high-powered AG tools

[next obsidian]

I want to learn curve theory

[oblique river]

I never officially learned any AG. everything I learned I picked up on the streets

[next obsidian]

Cringe

Imagine not cutting out 1 year of ur life to dedicate to hartshorne

cope

[shut halo]

@oblique river if in this F was F_20 then how would one check this?

[next obsidian]

F_20 isn’t a thing

But you could like

[hexed vortex]

F_20 is not a field, but if it was F_23 I'd use a computer

[next obsidian]

Theoretically write a program

[shut halo]

Oh ok so there is some kind of algorithm?

[oblique river]

yes, test every point

[shut halo]

Lol that's brutally simple

[prisma ibex]

this is unironically how a lot of implementations do this

• break after you hit the Weil bound and so on

[chilly ocean]

If E is an infinite algebraic extension over F and X is a basis for E as a vector space over F, then can every element of E be written uniquely as a linear combination (over F) of products of elements of X? If so, in any given product can an element x of X only appear d times where d is the degree of the minimal polynomial of x over F?

[next obsidian]

If X is a basis then any element of E is uniquely a linear combination of elements of X

You don’t look at products, and it doesn’t matter what E is over F, this is just the definition of a basis for a vector space

[void knot]

,tex Let $V$ be a vector space over $\mathbb{R}$ of dimension at least $3$, and let $T\colon V\to V$ be a linear mapping. Prove that there is a non-trivial vector subspace of $V$, say $W$, that is invariant-$T$.\
I've got some proofs but I don't remember the one which associates to $V$ a $\mathbb{R}[x]$-module as follows:
$$(a_0+a_1x+...+a_nx^n).v:=a_0v+a_1T(v)+...+a_nT^n(v)$$
It is known that if $W$ is $\mathbb{R}[x]$-submodule of $V$ then it is vector subspace that's invariant-$T$.\
Can I have some light, please? Thx.

[cloud walrus]

RaD0N

[latent anvil]

Do you know anything about like a decomposition of V as an R[x] module?

I'm thinking of something specific but I don't want to confuse you if you haven't seen it yet

[void knot]

don't bother. You can say it (:

[latent anvil]

lol alright

Use rational canonical form

Ie the structure theorem for modules over a pid

[void knot]

Ok. I undo my sentence lmfao

I didn't see it yet xD

[latent anvil]

alright lol

[hexed vortex]

If I may suggest a more linear algebra minded answer,
do you know the connection between factors of the characteristic polynomial of an endomorphism and invariant subspaces?

[latent anvil]

think about the complex version of this problem

Yes, this is secretly the same answer I gave

[hexed vortex]

Sorry if I gave too much

[latent anvil]

(but phrased in a more appropriate way for your level)

No it's okay, my thing would immediately give it away if they knew the material

[void knot]

yes. I solved this way.

But I wanted to review a more "adult" approach. @hexed vortex

[latent anvil]

I never took like a second course on linear algebra so I think about this stuff in terms of ring theory

the adult approach would be rational canonical form, but it boils down to the same argument

(and there's nothing wrong with the invariant factor argument)

[hexed vortex]

How does the fact that a polynomial of degree >= splits in R appear in that argument, btw?

[latent anvil]

hmm let me think about it

[hexed vortex]

Does it come naturally?

[latent anvil]

I think the idea I had in mind is slightly off, I didn't think about what happens if the minimal polynomial = the characteristic polynomial

[hexed vortex]

I would say that when you have an approach by linear algebra, unless you're trying to generalise a result, you really don't need to be ashamed to use it.

Sometimes it feels like a good half of advanced math is trying to reduce problems to linear algebra anyway 🙂

[latent anvil]

ah okay, I see

Not rational canonical form but Jordan canonical form

this gives you an isomorphism between V and a direct sum of things like k[t]/(p_i(t)^r_j)

as R[t] modules

If there are multiple factors in the sum you're done

Since any of them will be a proper submodule

Otherwise there's exactly one factor, so V = k[t]/(p(t)^r)

Note that p is degree 1 or 2 since it's prime and we're working over R

Then r > 1 as if r = 1 then dim V = dim k[t]/(p(t)) = deg p < 3

so (p(t))/(p(t)^r) is a proper submodule of V

I think I've overcomplicated things by introducing canonical forms here

Hmm possibly not, idk

[latent anvil]

You can see where I use it here

My argument fails for V = k[t]/(p(t)) for p an irreducible

oops k should be R

But we know this is impossible for dimension reasons

the key of my argument isn't really anything about canonical forms, it's just the statement that V ≈ k[t]/(p(t)) for some polynomial p

but I don't know how to prove this without invoking the structure theorem for modules over a pid

[hexed vortex]

It doesn't sound unreasonable that the structure theorem would be necessary

[latent anvil]

I agree

[unique juniper]

help pls

only some power of r can be in the center

and all r's commute so i need to do something with the sr^i's

[carmine fossil]

Try finding the centraliser of s

No power of r can be in that if n is odd

r^k s r^-k=s implies r^(2k)=1

[unique juniper]

centraliser of s in D2n ?

[carmine fossil]

Yes

[unique juniper]

emmmm

[carmine fossil]

Center is the intersection of centralisers

[unique juniper]

wouldnt you also have to find the centraliser of all the s's

[carmine fossil]

How did you show this?

[unique juniper]

er

like this

$sr^i = rsr^i \Rightarrow sr^{i+1} = sr^{i-1}$

and only can happen if n =2 right?

but n>3

[carmine fossil]

How did you get that eqn?

[unique juniper]

well supposing that some sr^i does commute

[carmine fossil]

Nvm

Rhs should be sr^{i+1}

[unique juniper]

lol yeh mb

i mistype

one sec

[carmine fossil]

*if n is even

[unique juniper]

$sr^ir = rsr^i \Rightarrow sr^{i+1} = sr^{i-1}$

[cloud walrus]

Yes

[unique juniper]

i mean

only when n is two?

[carmine fossil]

mb

Yes n=2

Now do (r^i)s=s(r^i)

[unique juniper]

yeah so no sr^i can be in center

ok

[carmine fossil]

r^i is some element in center

[unique juniper]

i dont get why only s

[carmine fossil]

If r^i were in the center,that condition should hold

[unique juniper]

ok

[carmine fossil]

It's a only if,not a iff(r^i is in centre only if r^i s=s r^i)

[unique juniper]

ok yes

dont get it

[carmine fossil]

r^i s=s r^i

Implies r^i s=r^-i s

Implies r^(2i)=1

Implies order of group is not odd

[unique juniper]

ahh yes

got it

ty

i think you mean n, group order is always even :D

unless i have my definition mixed up

[languid meteor]

Hey all, how would I (in general) construct a basis for a space using the bases for its subspaces

I know this is pretty out of context but in my quantum course we're trying to construct a basis for a state space. Each E(j,m) is spanned by eigenvectors with eigenvalues which are determined by j,m

so these subspaces should be distinct/disjoint

do I just take the set of all of the basis vectors for the subspaces and it makes up a basis for the whole space?

[rich ravine]

Yes if the whole space is decomposed as a direct sum of subspaces, you can concatenate bases of the subspaces to get a base of the whole space

It the subspaces are moreover orthogonal to each other, then you can concatenate orthonormal bases to get an orthonormal basis

[languid meteor]

perfect, thanks so much!

[tawny shell]

I'm having a little trouble proving the unitary group is a linear lie group, do we have to use the closed subgroup theorem?

[latent anvil]

sorry, what's a linear lie group? I might be able to help but I'm not familiar with the terminology

I would have assumed it means "admits a faithful lie group representation" but that's obvious here

[tawny shell]

A closed subgroup of GL(n,K)

[latent anvil]

K = C here?

[tawny shell]

mhm, it hasn't defined K to be C

but Im guessing R or C

[latent anvil]

yeah I mean unitary doesn't make sense over R and "lie group" only makes sense if the base field is R or C

so like, do you see why U(n) is closed in GL(n, C)?

and why it's a subgroup?

[tawny shell]

I see why it's a subgroup

what do we mean by closed?

[latent anvil]

sure

well, what's the topology on GL(n, C)?

since we're talking about lie groups it's probably the classical topology and not the zariski topology

so like, $GL(n,\C)$ can be identified with some subset of $\C^{n^2}$, right?

[cloud walrus]

Shamrock emoji ☘

[latent anvil]

just list out the entries of a matrix in a row

[tawny shell]

By topology

[latent anvil]

what?

[tawny shell]

sorry go back a few steps

what do we mean by toplogy on GL(n,C)?

[latent anvil]

do you know what a topological space is?

[tawny shell]

yeah

hausdorf

second countable

[latent anvil]

?

are you defining what a topological manifold is?

[tawny shell]

oh mbmb

[latent anvil]

what context are you encountering this problem in?

[tawny shell]

lie theory, but ive been working with manifolds/submanifolds all day

on these questions

so i just assumed

since a lie group is nonlinear

[latent anvil]

sure, so you know how GL(n, C) is a lie group then?

[tawny shell]

ye

[latent anvil]

well then it has a topology lol

lie groups are topological spaces with extra structure

[tawny shell]

right right

ooooo

i think i kind of get it

do we have to define an inner product here

[latent anvil]

no

[tawny shell]

then how do we show U is isomorphic to some A in GL(n,C) defined as A^T B \bar{A} = B?

[latent anvil]

it is not lol

what? groups are not isomorphic to individual matrices

[tawny shell]

ok im being lazy one minute

let me write this out

[latent anvil]

$U(n) = {A \in GL(n, \C) : A^\dagger A = I}$ by definition

dagger being the conjugate-transpose

[cloud walrus]

Shamrock emoji ☘

𝓐eteer

[tawny shell]

Let $B \in M(n,\mathbb{C})$ and $\beta$ be the sesquilinear form on $\mathbb{C^n}$ defined by $\beta (x,y) = x^T B \bar{y}$

[cloud walrus]

𝓐eteer

[tawny shell]

The Isometry group $U(C^n, \beta)$ is isomorphic to $A \in GL(n,\mathbb{C}) : A^T B \bar{A} = B$

[cloud walrus]

𝓐eteer

[tawny shell]

is this right so far?

[latent anvil]

why isomorphic? aren't those just equal?

also I think you're missing some curly braces

[tawny shell]

yeah i am

[latent anvil]

$U(\C^n, \beta) = {A \in GL(n, \C) : \forall B \in GL(n, \C).A^T B \overline{A} = B }$

[cloud walrus]

Shamrock emoji ☘

[tawny shell]

ty

[latent anvil]

I haven't actually seen this before, but it seems plausible

ah wait, taking $B = I$ we recover my definition

[tawny shell]

then i just define a continuous map

[latent anvil]

I believe this

[cloud walrus]

Shamrock emoji ☘

[tawny shell]

which one

[latent anvil]

$U(n) = {A \in GL(n, \C) : A^\dagger A = I}$

[cloud walrus]

Shamrock emoji ☘

[tawny shell]

ah yes

then we define some continous map from GL(n,C) to M(n,C)

okok take the inverse of this at 0 and hence closed subset

kk think i get it now, ty for ur help

[latent anvil]

um

the inverse at 0?

not at the identity?

what's your map?

[tawny shell]

f:GL(n,C) ---> M(n,C) via A---> A^T B \bar{A} - B

[latent anvil]

what is B lol

[tawny shell]

B=I_n

[latent anvil]

then yes, this works

I was thinking f(A) = A^T \bar{A}

and then f^{-1}(I)

[tawny shell]

mhm

[chilly ocean]

ALGEBRA

[obsidian sleet]

monoid with unique inverses for each element

[gritty adder]

groupoids with one object

[chilly ocean]

rings without a multiplication

[fathom delta]

does anyone have any references on algebras equipped with infty-ary maps? Is there a common name for them?

[viscid pewter]

loops with associativity

[latent anvil]

are you talking about loops in the sense of topology or loops in the sense of quasigroups?

I don't see how either fits here

[viscid pewter]

the meme before this

[latent anvil]

Oh sorry lol

[gritty adder]

symmetric groups

[uncut girder]

Any one know what is a reductive group?

and why we care about them?

[chilly ocean]

Do you smart guys keep notes

Or just work from books and do problems

[latent anvil]

I don't keep notes

Mostly just read books and talk to people

(and do problems)

[nova plank]

I handwrite notes for everything, but I want to stop because I hate it and it takes forever 😂

[chilly ocean]

@latent anvil

What a king

Loving the new profile pic

[latent anvil]

It's so good dude

Ty so much for making it

[ivory cosmos]

I use a spaced repetition system

so I have flashcards for various things

[uncut girder]

I do take notes

for classes

But i never look at them later

its mostly for processing the material as I am receiving it live

I think it also helps me remember the material because I am doing a physical activity (writing) while learning it

But yeah, I never reference my own handwritten notes later. I just look at the course textbook

[paper flint]

Pretty much the same. Notes for me are basically writing down definitions, theorems and their proofs, some important exercises with proofs. I throw in some insightful points sometimes, but I still haven't developed the habit of using notes as a quick reference, I continue to use textbook for that.

[next obsidian]

I write shit down as I read, I usualyl copy down the theorems and then the proof

This helps me remember it, and in particular with certain books which leave out tons of details I have the details worked out

I don’t tend to look back at them though since it’s easier to look up in a textbook usually, but if it’s from a professor and it’s a more original proof I will look back at them

Since I can’t find it other places

Depends tbh

I take them if I’m feeling up to it, not too tired

And the material is stuff I don’t know

I did this for AG and over the summer they were very nice since the arguments presented were different than the ones in Hartshorne and were very slick

So understanding them took time, I should have looked over them while I was in class and figured out what was going on

[woven delta]

I take notes when I talk to my advisor and when I'm in a talk I'm very interested in

It's not a bad habit to have

[latent anvil]

I just can't focus on lecture while taking notes

I mean I usually can't focus anyway but trying to take notes makes it worse

[woven delta]

Same unless I'm super interested

[chilly ocean]

I just want to double check understand this correctly

Let a,b be two points in the affine plane

Let Y be the set contain those points

Now the ideal generated by Y, which is an ideal of k[x,y]

Is all f such that f is zero on a and b

[latent anvil]

yes

[chilly ocean]

I am getting super confused trying to write explicit generators

Is this correct? I think I’m getting confused because k[x,y] has polynomials with coefficients in k, but affine plane as points that are pairs of things in k

[latent anvil]

I think this is wrong. What you've written is the union of four lines

Let's just think about one point a = (a1, a2)

What's I({a})?

[chilly ocean]

I thought for one point it was this

[latent anvil]

oh sorry I missed that

That's not quite right

[chilly ocean]

Jesus

[latent anvil]

X is a closed set, right ? So we would expect V(I(X)) = X

Does that make sense?

[chilly ocean]

Yes

But the common zero

Oooh

[latent anvil]

That's what V is, the set of common zeroes

Well everything in your I(X) is a multiple of x1-a1

Right?

[chilly ocean]

so then it’s xi-ai

[latent anvil]

your ideal is all things of the form (x1-a1)(x2-a2)f for a polynomial f

[chilly ocean]

But as two seperate generators

[latent anvil]

Yes!

So I(X) = (x1-a1, x2-a2)

[latent anvil]

Yup

okay, now suppose we have subsets S, T of A^2

How can you write I(S union T) in terms of I(S) and I(T)?

[chilly ocean]

f is zero on S union T if it’s zero on both so intersect the ideals

[latent anvil]

Yup

And in this case (x1-a1, x2-a2) and (x1-b1, x2-b2) are coprime, right?

Assuming (a1, a2) ≠ (b1, b2)

[latent anvil]

sure

[chilly ocean]

Oh that’s more clever

[latent anvil]

Wait what was your argument for why the intersection is a product of ideals?

[chilly ocean]

Ehh

I just thought if it’s zero on A and B, then it has to have a part of its it’s product being zero in A and B

And then I just enumerated the non stupid combos

[latent anvil]

In general the intersection of two ideals might not be the product

like if a = b here

[chilly ocean]

Yeah

thanks for reminding me though that is something I’d ignore

[latent anvil]

It usually doesn't matter because r(IJ) = r(I cap J) = r(I) cap r(J)

Where r is the radical

And since I(V(J)) = r(J) and stuff it's really only the radical that matters when doing geometry

(for varieties at least)

[chilly ocean]

Do you have a hint for this let Y1 Y2 be two closed disjoint subsets of A^n. Show that the sum of the ideals generated by the Y’s is the entire polynomial ring

I tried writing any polynomial in terms of its linear factors, and use that the two sets are disjoint

But that’s didn’t work

[latent anvil]

what does disjointness tell you?

like

you have this correspondence between closed sets and radical ideals

try to turn disjointness into a statement about ideals using that correspondence

[chilly ocean]

We have an order reversing bijection

[latent anvil]

sure, but you also know how it behaves with intersections and unions

sorry, maybe you don't

we just talking about I(X cup Y), yeah?

what about I(X cap Y)?

[chilly ocean]

Oh

That should be the sum

And if they are disjoint

Very gucci

[latent anvil]

yup

[chilly ocean]

Wait is I(X cap Y) always I (X) + I(Y)/

[latent anvil]

yup

Err

you may need r(I(X) + I(Y))

[chilly ocean]

Hmm

Yeah in general we need the radical

I wonder why we don’t seem to need it here

Ah

[chilly ocean]

We use this and see that the radical is the whole ring, hence I(X)+I(Y) is the the whole ring

[latent anvil]

yup

[tawny shell]

Hey can anyone help me with proving there exists an open neighbourhood V of I_n (identity) in GL(n,K) such that {I_n} is the only subgroup of GL(n,K) contained in V.

So far I know that exp is a local diffeomorphism but don't know really where to go from this.

Why is the exp map from M(n,C) to GL(n,C) where M is in M(n,C) defined as Exp(M)= I + M + o(M^2)?

[chilly ocean]

I think I know this one

Actually nevermind I’d only handwave it

[latent anvil]

I thought I had a proof but it was wrong

tterra saves the day

[chilly ocean]

exp is unique map \mathfrak{g} -> G with exp(0) = I and d/dt exp(tA)|_{t = 0} = A. the matrix exponential satisfies these two properties, as you can check

[latent anvil]

Nevermind

[chilly ocean]

lol

[latent anvil]

I can show any subgroup of GL(n, C) either has elements of arbitrarily small norm or every element has norm >= 1

Although I don't see how this helps

[chilly ocean]

well, exp is a local diffeomorphism (because its differential at the origin in \mathfrak g is just the identity map, under the usual identifications) of some nbhd U of 0 in \mathfrak g onto some nbhd V of I in G, so if you had a subgroup contained entirely in V, maybe you look at its preimage?

just throwing possible first steps out there

[latent anvil]

Yeah, I didn't try to use the exponential yet

Probably should lol

So like, the subgroup has some lie subalgebra associated to it

but I don't see how this is relevant

Especially because exp isn't a group homomorphism

[chilly ocean]

it is if the matrices commute look at centers maybe? not sure

[latent anvil]

Yeah I was just thinking about that. Hmmm

ah wait maybe this is trivial

What is |e^(λA)| as λ goes to infinity?

Genuinely asking

Lol

oh I guess you can do it with λ = n

Then A commutes with A and so on

So it's (e^A)^n

But I don't see why those would blow up in norm

[chilly ocean]

if a lie subgroup is contained in V then it is given by exp(tA) for A in the lie algebra of that subgroup

[latent anvil]

yup

I'm just thinking like

can exp of an entire line in the lie algebra be bounded

Oh yes it can

Take any compact subgroup

Like the unitary group

[chilly ocean]

ah but they didn't specify lie subgroup did they

||||

[latent anvil]

Right so we need to say that somehow sufficiently small nbhds exclude all subgroups

so we can wlog to cyclic subgroups I think

Err

Is a cyclic subgroup a lie subgroup?

I feel like it is...

Yeah it's a 1 param subgroup

Err I don't mean cyclic lol

A cyclic subgroup is what I mean yes

Those are countable

So 0 dim lie groups

Yup

Okay so we can wlog to cyclic subgroups, right ?

So say A is such that |A|^n < 1+ε for all n

for a fixed constant ε

I feel like A is ±I

Idk though this now seems sus

Wait @chilly ocean what's the lie subalgebra corresponding to a cyclic subgroup

Something is messed up here

Ah mb the correspondence is only for connected lie subgroups

[chilly ocean]

identity component is always a connected subgroup so u could restrict

[latent anvil]

Yeah but I'm looking at the cyclic group

[chilly ocean]

my sense of humor is fucking fried, every time i see the word "sus" i think of that stupid fucking game and i start laughing

[latent anvil]

The identity component is just the identity which is fine

When the lie group is simply connected

hnnng

I wonder if we can conjugate things

Like

can we wlog the generator of our subgroup to be upper triangular or something without changing the norm

We can diagonalize by a hermitian matrix

How does conjugation by a hermitian matrix change norm?

Because the result is obviously false for diagonal matrices

err...is it?

[tawny shell]

I think i managed to do something with open balls 😄

[latent anvil]

I think so

Can you share your proof?

[tawny shell]

It's a little lengthy

but

What i did is take some \epsilon in R_+ so that the open ball U_2\epsilon (0) in M(n,K) with radius 2\epsilon and centre 0 induces a diffeomorphism exp|_U(2\epsilon(0) : U_2\epsilon (0) --> GL(n,K)

from U_2\epsilon (0) onto exp(U_2\epsilon(0)) ⊂ GL(n,K)

[latent anvil]

Right

[tawny shell]

Then we define some open nbhd V=exp(U_\episolon(0)) of I_n in GL(n,K)

Let G be a subgroup of GL(n,K) with G ⊆ V

Let g be in G

Then there exists A in U_\epsilon (0) with exp(A)=g

[latent anvil]

Yup

[tawny shell]

can u see where im going with this?

[latent anvil]

I think you're going to say exp(n A) = g^n for large enough n so that nA isn't in the ball

[tawny shell]

going to show a contradiction of maximality

[latent anvil]

Wait maximality of what?

what was chosen to be maximal?

[tawny shell]

Assume that A =/=0 and choose k in the natural numnbers maximal with kA in U_\epsilon (0)

[latent anvil]

Sure

[tawny shell]

then exp((k+1)A)=exp(A)^K+1=g^K+1 in G ⊆ V

[latent anvil]

Makes sense to me

[tawny shell]

and hence there exists B in U_\epsilon(0) with exp((k+1)A)=exp(B)

just got to write down the rest of this

[latent anvil]

sure, so exp(B) = g^(K+1) = exp((k+1)A)

[tawny shell]

ye

[latent anvil]

And (k+1)A is in U_2ε?

[tawny shell]

yes

and since injectivity

we get (k+1)A=B

[latent anvil]

Ah, because (k+1)A = kA * (k+1)/k

and (k+1)/k <= 2

nice!

Oh wait no

I don't believe this proof

why is there a maximal such k?

[tawny shell]

because

then

[latent anvil]

Oh right nvm

[tawny shell]

lol

[latent anvil]

This is happening in the ball

Not the subgroup

Nice!

[tawny shell]

sometimes i just get confused on the easy parts and then i just have to stare at it for a while

[latent anvil]

I think there's a simple proof in terms of eigenvalues

But I'm having trouble writing out the details

If A is an invertible matrix contained in a bounded subgroup then the eigenvalues of A lie on the unit circle

Because if Av = λv then A^n v = λ^n v, so |A^n v| = |λ|^n |v| goes to infinity if |λ| ≠ 1 as you make n either a very large positive or negative number

and you can relate the norm of A in M(n, C) ≈ C^(n^2) with the norm of its maximum eigenvalue

(this is straightforward if you know about the operator norm)

so if we start by making V the unit ball around the identity we automatically get that any subgroup contained in V has all elements with eigenvalues on the unit circle

and then I think you should further be able to require that their eigenvalues lie completely on the left half plane, but it's not obvious to me that this set is open

[uncut girder]

hi

teach me lie groups

[latent anvil]

They're pretty sweet

Here's a lie group problem we just did

Let G = GL(n, C)

[uncut girder]

are lie groups only defined over R and C?

[latent anvil]

Then there's a neighborhood of the identity such that the only subgroup it contains is {±1}

[uncut girder]

like you cant have a lie group over Q?

[latent anvil]

Only over R really

A lie group is a smooth manifold

[uncut girder]

only over R

[latent anvil]

Like the definition is "a smooth manifold where the multiplication and inversion maps are smooth"

[sturdy marsh]

you can have group schemes over any ring

[latent anvil]

yeah I mean if you want to work over more general fields than R or C you're not really working with manifolds lol

[uncut girder]

then what are lie groups over C?

[golden pasture]

C=R^2

[latent anvil]

I'm sure you can talk about a complex manifold whose multiplication and inversion maps are holomorphic

But GL(n, C) is just a smooth manifold over R

It's an open subset of R^((2n)^2)

[golden pasture]

group objects in category of manifolds with complex structure

[latent anvil]

So to me this is what a lie group is

[sturdy marsh]

the usual construction of an elliptic curve as a quotient of C is a complex lie group

[latent anvil]

I think lie groups automatically have a real analytic structure?

[uncut girder]

thats not a lie group?

[latent anvil]

so maybe real complex lie groups and complex complex lie groups are the same

[sturdy marsh]

C mod lattice is a lie group

[uncut girder]

or is it?

[latent anvil]

It is

[uncut girder]

its not an algebraic group

[sturdy marsh]

it is also an algebraic group

[uncut girder]

🤔

[chilly ocean]

LIE GROUPS

[latent anvil]

im just gonna note

> teach me lie groups
> only over R :chmonkey:

[sturdy marsh]

CATWIGGLE

[chilly ocean]

[uncut girder]

yeah

[latent anvil]

wtf why no emotes

[uncut girder]

i need lie groups over Q

[latent anvil]

Okay it works

[sturdy marsh]

group schemes

[uncut girder]

no group schemes

[latent anvil]

Good

Only lie groups

Stick to R kiddo

[uncut girder]

GL(n,Q) is good

[latent anvil]

I am considering making a giant baked pasta dish

[chilly ocean]

anyways the only lie groups you should care about are compact lie groups because then you can do things like average over them with integrating wrt a unique invariant probability measure and that gives you a ton of nice shit like bi-invariant riemannian metrics and they have nice group action properties and etc etc etc it's all true

[latent anvil]

but I'm also lazy

SL(n, R) btfo

[chilly ocean]

[latent anvil]

R^n btfo

[uncut girder]

is there a classification of lie groups?

[chilly ocean]

"shit" (noncompact) and "good" (compact)

[latent anvil]

I think there's a classification of simply connected ones

[sturdy marsh]

yes

[latent anvil]

There's an equivalence of categories between simply connected lie groups and lie algebras

and lie algebras encode a lot of info about lie groups

[uncut girder]

from the simply connected ones you can get all of the other ones by quotienting right?

[latent anvil]

I think so

via the universal cover

[uncut girder]

oh this is cool

[latent anvil]

yup

Lie groups good

[uncut girder]

but really we need lie groups over Q

[latent anvil]

It's actually stronger than that

if G is a simply connected lie group, lie group maps G -> H are the same as maps between their lie algebras

[chilly ocean]

literally the reverse of this

[sturdy marsh]

did you hear from UGA yet?

[uncut girder]

yeah waitlisted 😢

[latent anvil]

Rip

Better than nothing?

[sturdy marsh]

rip

[uncut girder]

true

[latent anvil]

shouldn't have written about lie groups over Q on your app

[uncut girder]

the lie algebra would be a lattice

instead of a vector space

[sturdy marsh]

early 2022 is going to suck

[latent anvil]

Actually, do group schemes have something like a lie algebra?

you do sort of have vector fields, right?

[sturdy marsh]

yes

[latent anvil]

like sections of the dual to the cotangent bundle or something

Presumably you still get pushfoward maps so you can talk about left invariance

but the lie bracket seems hard to generalize

[latent anvil]

Bc grad apps?

[sturdy marsh]

yes

[latent anvil]

yeah I'm uhh not looking forward to it

[sturdy marsh]

ill probably leave the server towards the end of 2021 lmao

[latent anvil]

Oh that's significantly more healthy than my idea

[sturdy marsh]

there seem to be a lot of people in our year

[latent anvil]

Which was to start a community to collaborate on grad apps

[sturdy marsh]

and all of them will be applying

[latent anvil]

but how do I do this without it becoming hopf/gradcafe

[sturdy marsh]

probably a bad idea

it's probably best to apply

and then forget about it

completely

[latent anvil]

I will be lucky if I can think of literally anything else

[sturdy marsh]

true

[latent anvil]

Might as well have people to talk to and like, remind eachother to work on apps/provide crit on essays/whatever

I'm very bad at applying for things

[sturdy marsh]

I am also bad at applying for things

so I usually don't apply

[latent anvil]

I went out to see a play with a friend the night my uw college app was due

she found out on the bus ride back that I hadn't submitted it yet and made me work on it on my phone

Might not have finished in time otherwise

Ugh I don't think I have enough cheese

[uncut girder]

this is classic

tfw you have anxiety about the nearing deadline so you avoid working on application because you keep worrying about it and worrying you wont finish in time instead of just working on it while you have time

[ivory dust]

if x = a mod (49) then does x = a mod (7)

[latent anvil]

yes

what's the definition of "x = a mod (49)"?

[delicate bloom]

don't cross post @ivory dust

[ivory dust]

sry mb and thanks guys

[obsidian sleet]

aiya this one problem is giving me wonky thoughts

im gonna post it and tell you guys what i think so far

no this is not a test due to the point count

anyways

[latent anvil]

that point count looks like group theory

[obsidian sleet]

h m m m?

[latent anvil]

like the class formula

(ddon't read into this)

[obsidian sleet]

oh ok

uh ok

where to begin

i wrote down what i got so far gimme a sec

for part a i said that |z| could be 1,5,11,55 (positive divisors of 55)

now for b i am a bit confused

[latent anvil]

this is true

[obsidian sleet]

if G has an element of order 55, and the order of G is 55...

then do we have a cyclic group or

[latent anvil]

for a you might want to exhibit a group with elements of these orders

[delicate bloom]

do you get distinct elements if you raise that element to a power relatively prime to 55

[obsidian sleet]

oh my i gotta produce a group of order 55.. monka

hmm

[latent anvil]

give me a group of order n metal

[obsidian sleet]

uhh

[latent anvil]

hint

[obsidian sleet]

huh

pbtbt im gonna need to back up

wait so for a my answer is correct but i might want to show an example of this

[latent anvil]

right

it depends on the meaning of "possible"

[obsidian sleet]

ok let me start thinking about that

im no good at producing examples for this class idk why

[latent anvil]

i get it

[obsidian sleet]

i know the following kinds of groups

i know dihedral groups and i know this group called U(n)

which is like

positive integers less than n coprime to n, and we use multiplication mod n

[latent anvil]

right

you also know another group

I promise you

[obsidian sleet]

uhh

Z_n

[latent anvil]

also yeah I think this is a big mood for anyone taking group theory for the 1st time

[latent anvil]

yes!

[obsidian sleet]

hmm Z_n is nice and friendly

[latent anvil]

right

so what's a group of order 55?

[obsidian sleet]

Z_55 = {0,1,2,...,54}

[latent anvil]

right

[obsidian sleet]

closed under addition mod 55

[latent anvil]

so you claimed this

[obsidian sleet]

now i need to find them

[latent anvil]

do all of these orders occur?

also just to check

do you understand why I'm asking this?

[obsidian sleet]

no not really

[latent anvil]

sure, so like

it depends on how you interpret the question

> Q: what are the possible orders of the elements of G?
> A: any positive integer

does this sound right?

[obsidian sleet]

hmmm

i mean

it's certainly possible

we would just need to find one

[latent anvil]

exactly

[obsidian sleet]

that works

[latent anvil]

but I would say no, because the answer is too general

[obsidian sleet]

i see

[latent anvil]

because e.g. 3 can't be the order of an element of a group of order 55

[obsidian sleet]

right this is worrying about "possible"

[latent anvil]

so I'm asking you to make sure each of 1,5,11,55 are actually the order of some element of some G with |G| = 55

yup

[obsidian sleet]

and here we need to show that we can get a group that has elements of each one simultaneously in the group

instead of just saying it might be one of 1,5,11,55

this is to just be extra sure right

[latent anvil]

nah, that's not quite what I'm saying

[obsidian sleet]

oh

[latent anvil]

suppose that somehow no group of order 55 had an element of order 5

would your answer "1,5,11,55" still be correct?

[obsidian sleet]

sure we just have an extra 5 answer

[latent anvil]

it's a superset of the possible values, but it's not specific enough

well then why not say "all positive integers"?

[obsidian sleet]

right

so we want to be specific

[latent anvil]

so I'm asking you to show that
${n : \text{there exists a group } G \text{ of order } 55 \text{ and } g \in G \text{ of order } n} = {1,5,11,55}$

[cloud walrus]

Shamrock emoji ☘

[latent anvil]

you've shown $\subseteq$

[chilly ocean]

find a group such that for each nonnegative integer n, there is an element of order n, and such that there is an element of infinite order idk i just saw the word order so i wanted to post a problem

[cloud walrus]

Shamrock emoji ☘

[latent anvil]

sorry metal I sort of derailed you

[obsidian sleet]

no its ok

i need the practice

i am not good at this

[latent anvil]

but I think this is worth thinking about

[obsidian sleet]

this class is very hard for me

[latent anvil]

I was so lost in my first group theory class

I also couldn't ever think of examples

and it sucked

[obsidian sleet]

oh

[latent anvil]

eventually for me it just started to click

and in my second course I had more examples

and I felt a lot better about it

point being: don't worry about it

[obsidian sleet]

i am more inspired :3

hmm

[chilly ocean]

:3

[obsidian sleet]

so now i have to show the other way

subset in the other direction

er

[latent anvil]

take a big ol direct product of all cyclic groups

[chilly ocean]

ya

[latent anvil]

or direct sum, w/e

more interesting: the circle

if you want there to be no infinite order element, Q/Z

[chilly ocean]

no difference unless you're a stinky UGCT who cares about the distinction between product and coproduct

[latent anvil]

this is false

we're taking an infinite direct product

so it is different

[chilly ocean]

you might be right

[obsidian sleet]

55 has order 1, 5 has order 11, 11 has order 5, 1 has order 55 (in Z_55)

[chilly ocean]

but im also pulling words i don't know much about out of my ass

[latent anvil]

lol

yup!

so your answer is the best possible answer

[obsidian sleet]

nice

this makes more sense

[latent anvil]

@chilly ocean just to be clear, $(1,1,\ldots,)$ is an element of $\prod_{n=0}^\infty \Z/n\Z$ but not the subgroup $\bigoplus_{n=0}^\infty \Z/n\Z$

[cloud walrus]

Shamrock emoji ☘

[chilly ocean]

kindly explain the difference

[latent anvil]

the subgroup only contains tuples with finitely many nonzero entries

[chilly ocean]

ah

i see

[latent anvil]

it's like a polynomial vs a power series

[chilly ocean]

same vibes as product topology category time?

[latent anvil]

actually it's exactly that, $k[x] = \bigoplus_{n=0}^\infty k$ and $k[[x]] = \prod_{n=0}^\infty k$ as groups/modules/vector spaces/whatever

[cloud walrus]

Shamrock emoji ☘

[chilly ocean]

yeah actually ive seen this kind of construction a lot

[latent anvil]

ikr

[chilly ocean]

now that you mention it

[latent anvil]

it pops up a lot

like with the infinite grassmannian

in bundle theory

[chilly ocean]

now i know the distinction

[sturdy marsh]

petthecat

[obsidian sleet]

ok for part b i ask this because if i can show that G is cyclic i can reduce my answer from a positive multiple of \phi(55) to just being \phi(55)

[chilly ocean]

grassmannian...

[obsidian sleet]

o.o

chat moved fast lol

[latent anvil]

doxxed

[latent anvil]

yup

this is exactly the right line of thinking

[obsidian sleet]

okay

the reason i suspected i might be able to try this is because if the order of the group is 55 and i have an element with order 55, i feel like that element when applied to itself should go through all 55 things in the group

but

[latent anvil]

but

[obsidian sleet]

idk

it feels shaky

maybe there's a counterexample

[latent anvil]

it is shaky as stated

but it's the right intuition

🐦

tfw no pigeon emote

[obsidian sleet]

pigeon... pigeonhole?

🤔

[latent anvil]

[obsidian sleet]

h m m m

my class never went over this lol i have never formally learned it

[latent anvil]

ah yeah so actually I was in the same spot lol

[obsidian sleet]

huh

ok

[latent anvil]

so

[obsidian sleet]

lol

[latent anvil]

here is my version of pigeonhole

suppose $X,Y$ are finite and $f$ is a function from $X$ to $Y$

[cloud walrus]

Shamrock emoji ☘

[obsidian sleet]

ok

[latent anvil]

if $|X| < |Y|$ then $f$ cannot be surjective, if $|X| > |Y|$ then $f$ cannot be injective (this is the pigeony holey one), and if $|X| = |Y|$ then $f$ is injective iff surjective iff bijective

[cloud walrus]

Shamrock emoji ☘

[latent anvil]

does this intuitively make sense?

[obsidian sleet]

yes

two things mapping to one element

is the pigeon holey thing

two birds in a slot or w/e

[latent anvil]

right

failure of injectivity

so your logic was

if it has order 55

then it should go fill out the whole group

when you take powers

can you phrase that conclusion in terms of surjectivity of a function?

[obsidian sleet]

so if |z| = 55 for some z\in G, then the cyclic group <z> has order 55, that is |<z>| = 55. Then we also know that |G| = 55. so from your last line we can form a function that maps elements from <z> to G that is surjective and hence also injective ... ?

h m mm

[latent anvil]

oh wait uh

if you know |<z>| = 55 you're sort of just done lol

<z> is a subset of G and both have the same (finite) size

yeah?

[obsidian sleet]

yea we proved that |z| = |<z>|

yes

oh yes eys

ahaha

yes

ok this is much easier now

[latent anvil]

lol

I was going to prove that |<z>| = 55 basically

[obsidian sleet]

ah

[latent anvil]

define a map {0,...,54} -> G sending n to z^n

argue it's injective

thus it's surjective

[obsidian sleet]

ah

[latent anvil]

so everything in G is a power

[obsidian sleet]

i assume that last thing we got to would be the reasoning prof expects, since we didn't do pigeonholing ever haha

this is very good

[latent anvil]

nice!

okay so to take stock

what have we shown?

[obsidian sleet]

my answer to part a was correct, but i needed to show that there was a group of order 55 that contained elements of each order 1,5,11,55 , so that i could confirm that this set of positive integers was the set that contained nothing extraneous (as in there is not some positive integer in that list i gave that would never appear as an order of some element in a group of order 55)

i hope that is worded okay

haha

[latent anvil]

yeah!

[obsidian sleet]

and then for part b i started with the right idea that we know that the number of elements of order 55 is going to be some multiple of \phi(55), but i had a suspicion that the given information (that G had an element of order 55) could allow me to refine the answer

because if G contains an element z of order 55 then G can be concluded as cyclic since <z> is a subset of G

[latent anvil]

Right

so G is the cyclic group of order 55

yeah?

[obsidian sleet]

yes

[latent anvil]

and then you know the answer

from facts about cyclic groups

[obsidian sleet]

<z> contains every element that G would

yes

then that allows me to go straight to \phi(55) as opposed to some multiple since we know G is cyclic under this condition

[latent anvil]

100%

[obsidian sleet]

now i think i did c and d correctly, i said those would be k\phi(5) and j\phi(11) for positive integers k and j

but then e is what i am working on now

but

[latent anvil]

can you repost the image?

[obsidian sleet]

yeah

so the thing is I proved a similar thing in the preceding problem

right

there is a line saying the group G has an element a of order n

but i do not see a similar line in part e

so i want to apply this but i cannot

unless somehow the thing from part B carried through to part E

[latent anvil]

hmm

I mean I assume not...

what's the corollary?

like, does the assumption carry from (b) to (c) and (d)?

[obsidian sleet]

i have no idea

from here i might just email the professor to be sure

but i might try this problem as it is written just to see how it would work

[latent anvil]

I would email, seems ambiguous

[obsidian sleet]

okay cool

im good to go then

thank you so much

i learned a lot

[vestal snow]

Can someone verify a proof?

Let k be a field contained in the bigger field K. Then A and B are similar over k if they are similar over K

[sturdy marsh]

a matrix may not have a jordan form over k

you need k to be alg closed for that

you can work with RCF tho

[vestal snow]

Thanks

[unique juniper]

"By the minimality of d it follows that r = 0" i dont get this part

[sharp sonnet]

d is the smallest positive integers such that x^d is in K

if r were not 0, it would be a positive integers smaller than d and x^r in K

[unique juniper]

i see

so if r was not zero, r would be in the set P

and nt the smallest

i get it i think

[chilly ocean]

Anyone good with operator algebras? I’m trying to show that if a is invertible, and both a and it’s inverse have norm less than 1, then their spectrum is the unit circle

Working over a Unital banach algebra

[rich ravine]

Is there not some story about the norm of elements in the spectrum being less than the operator norm ?

[chilly ocean]

Sorry yes I should have says norm less than or equall

[cloud walrus]

slimvesus

slimvesus

[unique juniper]

is there a theorem that lets me do this quicker?

cuz rn the only way i can do this is by counting

[rustic crown]

can you find an explicit expression for the order of an element?

[unique juniper]

yea

but

actually

is Z/nZ cyclic?

[rustic crown]

yep!

[unique juniper]

lol

ok actually i dont know lol

[unique juniper]

what did you mean here?

[rustic crown]

in the group Z/nZ you have a very nice expression for order of a+nZ. Try to find it!

it seems you are starting group theory... so I guess I shouldn't use the notation a+nZ... that just means the element a mod n.

[unique juniper]

:3

[rustic crown]

:3

[unique juniper]

lol

[rustic crown]

[obsidian sleet]

i have returned

so i know that i have to take two cases: 1 where |G| is finite and another where |G| is not finite

when |G| is finite i can apply the fundamental theorem of cyclic groups

and part of it says that for every divisor of |G| there is exactly one subgroup of that order, and we have that there are subgroups of order 1,7,and say n

so the only value n that works is 49 since there would be no other divisors of n (n = |G|)

therefore |G| = 49

does that sound right?

[carmine fossil]

Yes

[obsidian sleet]

o kgood

then for |G| infinite i was told to show that this cannot be

by inspecting the following subgroups (a in G where <a> = g), <a>, <a^2>, <a^3>, ...

somehow i gotta figure out how this Cannot be

i dont really get it tho

cus u have infinite group

[chilly ocean]

||it might help to notice that every infinite cyclic group is isomorphic to the integers with addition||

[obsidian sleet]

but there are only 3 subgroups

wiggity wack

[carmine fossil]

How many elements does <a^2> have?

[obsidian sleet]

hmm

not .. sure

[carmine fossil]

And is a in <a^2>?

[obsidian sleet]

i know <a> will have infinite elements