Mathematics - 2018-02-25 (page 2 of 4)

Daminark

5:00 PM

Orthogonal group? I don't think it's enough to have the determinant by $\pm 1$

Eric Silva

having det 1 is SL

Balarka Sen

That's a much larger group, known as $\text{SL}^{\pm}_2(\Bbb R)$

sniped

rip me

Skortya

yeah that's why I'm asking, first time working with them, no idea really how to write it as a set ?!

Daminark

Orthogonal matrices are those such that $AA^t = I$

So it's just $\{A\in M_2(\mathbb{R}) : AA^t = I\}$

Eric Silva

@BalarkaSen r e v i s i o n i s t h i s t o r y

Daminark

5:02 PM

But yeah as an example, take $\begin{pmatrix} \frac{1}{2} & 0 \\ 0 & 2\end{pmatrix}$

Skortya

OK, I guess I will try to work with that.

Daminark

This isn't orthogonal, but it has determinant 1

Skortya

Yes I understand

Balarka Sen

@Eric ur too deep into politisociology fam

OK gotta go now

Cya

Eric Silva

i actually am and it's driving me insane

Balarka Sen

5:03 PM

the solution is to MAGA harder

Eric Silva

uh oh

Mike Miller

i don’t know why daminark didn’t say “columns hav norm one and dot product zero”

Daminark

Seems it's a bit too late for me to edit the message and say that. Oh well, revisionist history has a 3 minute timer, it seems

Eric Silva

@Daminark if only real life had such rules

user193319

2

Q: Complicated Proof of Free Group Generators by $n$ elements implies $\operatorname{Rank}(F) \le n$

First I must point out the following corollary in my book: If $G$ is a finitely generated abelian group generated by $n$ elements, then every subgroup $H$ of $G$ may be generated by $m$ elements with $m \le n$. Here is the problem I am working on: If $F$ is a finitely generated free gr...

Akiva Weinberger

5:14 PM

So this is how a pump works

You can get from the left to the right but not from the right to the left

And you've got springs in there pushing the things out

Another type of vacuum pump

Gasses can get from the outside in, but not from the inside out

so you put like a hole in the middle for gasses to exit through

Alessandro Codenotti

@Mathei @Balarka are you here?

MatheinBoulomenos

yeah

Abcd

$\approx$

Akiva Weinberger

5:32 PM

I just watched a whole video on pumps 'cause why not

Abcd

@AkivaWeinberger It's cool!

Yuriy S

Scroll pumps are great. I have a few in my lab. They don't need any oil or other liquids to work

So good as a first stage of ultra high vacuum pumping system

Daminark

Hey @Mathein, @Alessandro, and @Akiva

MatheinBoulomenos

Hey @Daminark

Kasmir Khaan

Mathein :D

Can i invite you fast to our room ?

MatheinBoulomenos

5:38 PM

okay

DarkRunner

@LeakyNun are you here

Tanuj

Guys can someone help me over my physics problem ? It's about a disc undergoing pure rolling

DarkRunner

I want to find ${ 2 }^{ 100 }(mod\quad 1000)$, so I do:$\phi (1000)=400\\ { 2 }^{ k*400+1 }(mod\quad 1000)\Rightarrow k=2\\ { 2 }^{ 100 }(mod\quad 1000)\equiv { 2 }^{ 801 }*{ 2 }^{ 199 }(mod\quad 1000)\Rightarrow { 2 }^{ 200 }(mod\quad 1000)$. But now I'm stuck with $2^{200}$. If I repeat the process, I'll have a fractional $k$; Am I doing something wrong?

Daminark

@Mathein oh so one thing I've been wondering

In commutative algebra we put the Zariski topology on Spec(R)

MatheinBoulomenos

yeah

Daminark

5:50 PM

But what exactly does that topology do for you? I've heard it as a buzzword before but it feels like it's an iffy topology

e.g. not being Hausdorff

Alessandro Codenotti

So Mathei remember how you told me you can think about Noetherian normalization in terms of branched coverings? I'm trying to make sense of a short paragraph on the same topic in Reid's book

Hi @Dami

MatheinBoulomenos

Hmm, you can think of the topology as enconding information about the partial order of ideals (that's just a reformulation)

but you can translate algebraic properties of your ring into topological statements about Spec(R)

and then you can use topological reasoning

Mike Miller

The point is that algebraic functions are continuous, and (iirc) vice versa

DarkRunner

If someone can help that would be great

Mike Miller

Your topology shouldn’t have many more continuous functions than the ones that are relevant

MatheinBoulomenos

5:53 PM

for example, Spec(R) is connected iff it's impossible to write R as a product of two non-zero rings

@MikeMiller are you talking about the Zariski topology on the prime ideals or classical algebraic geometry?

Mike Miller

The latter, but with prime ideals in mind still

MatheinBoulomenos

You can also show that if you divide out by the nilradical, then the topology on Spec(R) doesn't change

so this shows that $R$ is a product of two non-zero rings iff $R/\operatorname{nil}(R)$ is a product of two non-zero rings

you can show this also algebraically

but it's a bit messy to lift idempotents

(of course, there's still algebra involved in proving the statements that relate algebra and topology)

Alessandro Codenotti

How does one get a ring homomorphism $k[W]\to k[V]$ from a polynomial map between varieties $V\to W$?

MatheinBoulomenos

you pull back regular functions

DarkRunner

Hi can anyone help me out with my problem please

Alessandro Codenotti

5:59 PM

makes sense

MatheinBoulomenos

@Daminark there's other stuff you can do as well. For example, irreducible components of Spec(R) correspond to minimal prime ideals, so if you can show that Spec(R) has only finitely many irreducible components, then you proved that R has only finitely many minimal prime ideals

this works e.g. for Noetherian rings

Also Spec(R) is Hausdorff iff R is 0-dimensional, i.e. every prime ideal is maximal

there are a couple more of these facts that relate algebraic properties of R and topological properties of Spec(R)

they can be really handy

but I think the main reason to care about the Zariski topology is that you can define sheaves on it

Daminark

@MatheinBoulomenos I can see this for finite rings for sure, since every point is closed in a finite space => discrete

MatheinBoulomenos

Spec(R) might be finite, but not discrete

Kasmir Khaan

can someone tell me what is a compact group ( in easy way )

like very short description

MatheinBoulomenos

it's a topological group whose underlying topology is compact?

Akiva Weinberger

6:04 PM

I know what a compact topological space is

MatheinBoulomenos

that probably doesn't help

Kasmir Khaan

it does not since I did not take topology

><

Akiva Weinberger

Right, yeah, topological groups are a thing

Kasmir Khaan

our teacher talked about that in rep theory

MatheinBoulomenos

maybe you know compactness from analysis?

Kasmir Khaan

6:04 PM

yes

MatheinBoulomenos

the thing with open coverings?

Akiva Weinberger

The group $\Bbb R/\Bbb Z$, or addition mod 1, can be thought of as a circle

Kasmir Khaan

Well i have a small idea about that

Akiva Weinberger

A circle is closed and bounded and thus it is compact

Daminark

I meant if R was finite, then Spec(R) is a finite $T_1$ space, that should imply discreteness, no?

Kasmir Khaan

6:05 PM

did not get it very well from analysis either

akiva

the way our teacher presented compactness

was more than bounded and closed

MatheinBoulomenos

@Daminark yeah that's true

Kasmir Khaan

who is starring the names i write?

is that you dami ?

Daminark

Nope

Kasmir Khaan

-.-

Ok i wont bother you guys

Daminark

I usually star witty or r/nocontext things

Kasmir Khaan

6:07 PM

I ll keep doing my algebra thing , see ya later :D

Daminark

@Mathein How do sheaves come into play?

(Though note that my idea of sheaves is very vague, just as a way to track local data that somehow has to do with contravariant functors from the poset of open sets of a space to sets)

MatheinBoulomenos

@Daminark going to eat dinner now, I'll explain that later

Daminark

Yeah sure, see you around!

watchme

I am a student and interested in Math - Can I ask some questions if something is unclear? Even if maybe I can google it easily? Because so often it comes that a tutorial doesn't explain it as detailed as I want it. And here I'd have the possibility to ask as much as I want.

... and as detailed as I want :-)

... hopefully...

Xander Henderson

don't ask to ask; just ask

watchme

6:22 PM

clear enough - thanks :D

Skortya

do isomorphism preserve open sets?

Balarka Sen

@AkivaWeinberger Eureka Forbes gang Eureka Forbes gang Eureka Forbes gang / spent 10 racks on a new pump / etc

lush

Skortya: isomorphisms of what? topological spaces?

Skortya

Between $\mathbb{R}^4$ and the 2x2 real matrix.

Balarka Sen

homeomorphism always sends open sets to open sets

Skortya

6:32 PM

cool

lush

@Daminark: Given a ring A, one defines a sheaf of rings on Spec A. One has a nice basis for Spec A given by D(f) := Spec A \ V(f), f \in A. Now one can show that it is somewhat sufficient to define a sheaf on a base of the topology and that uniquely extends to a sheaf on the whole space. On the base you now define the sheaf O to work like O( D(f)) := A_f, the localization of A on the set {1, f, f², ...}
Now (Spec A, O) is what geometers call an affine scheme and the central object to study in Algebraic Geometry are ringed spaces (Spaces with a sheaf of rings on the space) that locally look …

Maybe that helps... if not I'm sure Mathein can tell you more than that :)

@Daminark ah it may be interesting, that given the construction above one finds that O(Spec A) = A. So you don't lose any algebraic information when changing from A <-> (Spec A, O). That is, for example why it is important to have the structure sheaf together with Spec A: Spec A itself certainly encodes algebraic information topologically, but it doesn't quite catch all the information of A

MatheinBoulomenos

6:48 PM

yeah, that was pretty much what I was about to say. If you're not familiar with the notation O(Spec A)=A means that you can think of elements of A as "functions" on a the space Spec(A)

Daminark

Okay so I'll just try to absorb this, give me a second

Adeek

did someone say schemes ?

do you want me to tell you a story about schemes ?

@Daminark

Daminark

Well wait for a bit, I'm still a noob at this and just wrapping my head around what's being said

Adeek

schemes are just topological spaces that have algebraic local data

nothing more nothing less

Zal Tukhara

Sup guys?

Adeek

6:56 PM

you can do cool things with schemes

Daminark

Well the point is more, I should understand the formality behind it and why that translates down to what you're saying

Adeek

yo @MatheinBoulomenos

your familiar with fibrations right ?

yeah @Daminark

MatheinBoulomenos

yo @Adeek

Adeek

yo yo yo

@MatheinBoulomenos I was wanted to discuss something related to fibrations and chow groups

Balarka Sen

@Adeek I'd give that a 0/10 if that's your answer to why schemes are interesting.

Zal Tukhara

6:58 PM

Anyone here familiar with surreal numbers?

Adeek

@BalarkaSen it is your personal opinion. Actually one can get a lot of interesting things out of that intuition

MatheinBoulomenos

@Adeek I know basically nothing about that

Balarka Sen

The statement "you can do cool things with schemes" carries no intuition. Don't be delusional.

Adeek

I am not delusional

I don't know why your rude sometimes

attacking people who don't hold similiar ideas as you is a form of nazism

Zal Tukhara

Lol

Adeek

7:02 PM

As far as I can tell this chat is for people to share their ideas not to get attacked for thinking in certain way.

@BalarkaSen For the record, I didn't say you can do cool things with schemes as the intuition. I said topological space with local algebraic data.

You can use this intuition for example which tells you schemes over $\mathbb{C}$ admits mixed hodge structure.

Zal Tukhara

Has anyone read "Analysis on Surreal Numbers?" One of the definitions in there gives me an interesting connection with schemes

Balarka Sen

You are not getting attacked for holding ideas different from mine. You are getting criticized for passing off word salad as mathematical pedagogy/motivation, which is basically a form of intellectual dishonesty (an opinion not just possessed by myself, but many others)

Daminark

@lush okay so to my understanding, sheaves are supposed to be the sorts of things that you can glue together (I'm getting all this by thinking of smooth functions on a manifold), so I sorta buy that yo can do this. Now let's say you're talking a finite union just to be a bit easier

MatheinBoulomenos

Schemes allow us to use "geometric" reasoning to study algebraic objects, among other things. They also generalize and shed light upon the objects of study in classical algebraic geometry. (E.g. there are moduli spaces for algebraic varieties which are schemes, but not varieties)

Adeek

I am saying what I believe is nice intuition of thinking about schemes

lush

7:06 PM

@Adeek wanna help me understand why projective space as a scheme over an alg. closed field behaves like you'd imagine?

Zal Tukhara

Arguing over someone else's personal intuition on a concept is pretty rude imo

3

lush

@Daminark a finite union of what?

Balarka Sen

You clearly haven't understood my argument if you think I was arguing over personal intuition.

Daminark

So would you have something like, $O(D(f_1)\cup \ldots \cup D(f_n))$ is the localization of $A$ at the set of products of the f_i?

Zal Tukhara

I mean this entire conversation is practically informal, not some academic conference here

If it was then I could see your aim

lush

7:08 PM

Daminark: In general it is a little bit more difficult to calculate the sections ( <-> O(...)) over non-base sets

Daminark: Actually you have D(f_1) \cap ... \cap D(f_n) = D(f_1 * ... * f_n)

Think of it like this: the "smaller" your open subset, the "more" functions you have on it. If your ring behaves well (e.g. A is a domain), A_f embeds into A_fg, so by building the intersection of the D(f_i)'s you're making your subset smaller and A_{f1*...*fn} is a "bigger" ring

MatheinBoulomenos

The theory of schemes created a displine that was able to absorb ideas (sometimes directly, sometimes as analogs) from classical algebraic geometry, commutative algebra, homological algebra, category theory, number theory, algebraic topology and differential topology. In some of those cases, algebraic geometry also "gave back" to the other subjects and in the case of number theory, it even shaped the subject decisively.

(That's not something I came up with, it's stuff I read. Though I've been able to make sense at least of parts of that claim.)

lush

The elements of O(D(f) \cup D(g)) "correspond to" elements of (x,y) \in O(D(f)) x O(D(g)) such that they are "equal" when "restricting" them to D(fg)

Balarka Sen

@Zal My aim is to point out that Adeek was hijacking a pretty perfect, coherent, concrete answer by @lush explaining why a certain mathematical object was interesting by posting a loose intuition that doesn't really carry much meaning or motivation, following with a comment which basically boiled down to "these objects are cool" patching up the failure of the previous one. This has happened multiple times before, otherwise I would not have commented.

I hope that gives some context to what I was arguing against.

If it's anything that irks me, it's unhelpful pedagogy.

Zal Tukhara

I see, I can understand your stance then\

Eric Silva

@Adeek wAT, this is actually like, a crazy thing to say dude

7

Balarka Sen

7:18 PM

@Eric even more so, given the context that it wasn't an attack, but a constructive (perhaps a little harsh) criticism.

But shrug

MatheinBoulomenos

so much drama

Antonios-Alexandros Robotis

well well that was an interesting read.

Daminark

To say the least

Zal Tukhara

Do regular maps ever raise the genus?

Balarka Sen

Hmm

I know that if a map raises genus it can be homotoped to live inside the 1-skeleton of the codomain surface

But

Zal Tukhara

7:30 PM

Let's assume that the regular map is constant

Edit: non-constant

Balarka Sen

Do all regular map between surface have to be branched covers? I think so

Yeah

It's locally polynomial

Covering maps always decrease genus (Euler characteristic of domain multiplies by degree of the map). I'm thinking if I can generalize the argument to branched covers (i.e if I throw away a bunch of points)

Perturbative

0

Q: Are all cell decompositions useful?

Based on the definitions and conventions used in the book Introduction to Topological Manifolds by John Lee, I can do the following. Let $X$ be a Hausdorff space, then define $\Gamma = \{\{x\} \ | \ x \in X\}$, that is $\Gamma$ is the collection of all singletons of $X$, and is trivially a parti...

general-topology algebraic-topology cw-complexes

Balarka Sen

lush is right

That is not a cell decomposition

The 0-skeleton of a CW complex always has discrete topology

A concrete example of a space which does not admit a cell decomposition is the cantor set.

Perturbative

@BalarkaSen But aren't you using the definition of a CW Complex being inductively built up (like in Hatcher)?

MatheinBoulomenos

if the topology of the CW complex is not the topology on your original space, then what you have is not a cell decomposition

lush

7:42 PM

@Perturbative how is a cell decomposition defined in Lee?

Balarka Sen

Can you reiterate Lee's definition? I don't have the book with me. Any sensible definition should imply that the 0-skeleton of a CW complex has discrete topology.

Perturbative

Damn I'm on mobile now :( lemme see if I can make a plan

Basically what Lee does is define a cell decomposition of a topological space X as being a partition of X into open cells of various dimensions

And then the open cells of dimension n >= 1 have characteristic maps defines the usual way

*defined

lush

Why is {x} open in X?

If you assume that, you just say that X is discrete

Perturbative

And then he says a Hausdorff space X together with a cell decomposition of it is a cell complex

Balarka Sen

Eric Wofsey just posted an answer saying Lee's definition is not useful.

Strange.

I have never heard anyone distinguish cell complexes from CW complexes

Ohhhh

I see what Lee does

He defines a cell complex as a pure partition into cells, and defines the weak topology + closure-compactness later

Or at least I bet that's what's happening. Let me copyleft the book

Perturbative

7:48 PM

Yeah

Balarka Sen

Nailed it

Perturbative

Sniped me as I was typing out a similar message

Balarka Sen

Hahah

So yeah that definition is worthless

Perturbative

Aight

Balarka Sen

Well, only of conceptual use anyway

Good question (+1)

Perturbative

7:49 PM

Thanks

I'm off to bed now, thanks for the help everyone!

Night

Zal Tukhara

gn

Antonios-Alexandros Robotis

@BalarkaSen for characteristic classes should I do milnor?

Balarka Sen

heheheh

Hatcher dawg

Antonios-Alexandros Robotis

I'm not wild about hatcher's Alg Top book

Balarka Sen

I think Milnor-Stasheff has more material. I don't know much about char classes beyond the basics

Antonios-Alexandros Robotis

7:54 PM

(i don't hate it either)

Balarka Sen

I think you algebraists would like M-S's style better

But get VBKT as a side-reference while you're at it :)

Antonios-Alexandros Robotis

im probly not an algebraist lul

thanks for the link :)

Balarka Sen

>im probly not an algebraist lul
>is reading Szamuely

Antonios-Alexandros Robotis

I just read the stuff that seems interesting at the time

I was pretty into analysis at some point, I arguably still am. I just haven't really done much in a while.

Balarka Sen

reading stuff that seems interesting is totally what algebraists do

we topologists only read the bible

Antonios-Alexandros Robotis

7:56 PM

loll

Balarka Sen

chastity dawg

lush

lol

Balarka Sen

@Daminark can you confirm this one?

lush

I think most topologists read the book "how to not really prove a result but just claim its intuitively clear"

:-P

Daminark

I dunno if the manga of topologists is analogous to a holy book but...

Antonios-Alexandros Robotis

7:58 PM

I have no problem with intuitive arguments for examples

But when the proofs get too handwavey, I get a bit spooked

Balarka Sen

Minecraft - Damage Sound

►

MatheinBoulomenos

just draw a picture of a 2-sphere and prove it in that context. The case for general topological spaces that satisfy the right adjectives (which are not stated explicitly) follows in the same way

2

lush

^^

our algebraic topology class 2 in a nutshell

Antonios-Alexandros Robotis

hahaha @MatheinBoulomenos shots fired

Balarka Sen

y'all have seen Serre's diss on topology papers, right?

lush

8:01 PM

nope

Zal Tukhara

Hatcher's font style is funky

lush

really wanna see that now :D

Balarka Sen

"How to write mathematics badly" by Jean Pierre Serre (noise removed)

►

(Got it timed)

MatheinBoulomenos

that's from a person who got a field medals for topology

that lecture is so great

11/10

Balarka Sen

@MatheinBoulomenos (mumbles) really tho, homotopy theory...

MatheinBoulomenos

8:02 PM

okay, "modern topology" :P

Balarka Sen

Well super influential to topology nonetheless

Daminark

Lol I'm taking AT next quarter, taught by my commalg prof, so that'll be fun

Balarka Sen

@Daminark you should have gotten Farb

sadface

lush

ahaha :D

Balarka Sen

@ZalTukhara I think that font is 10/10

Zal Tukhara

8:05 PM

Funky in a good way of course

Balarka Sen

ah then you are spared

Skortya

is $\left(\begin{matrix}0&a\\ b&c\end{matrix}\right)$ $a,b,c\in\mathbb{R}$ open in $\mathbb{R}^{2\times 2}$ it's not right?

Balarka Sen

iirc it's Lucida Bright

Tobias Kildetoft

@Skortya Right, because it is closed and the space is connected

Skortya

Thanks

Eric Silva

8:09 PM

@Daminark who?

@BalarkaSen one time benson thought he discovered the standard construction of the universal cover and it was p funny

Tobias Kildetoft

@MatheinBoulomenos Had a chance to read about positive bases yet?

Balarka Sen

the space of paths thing?

starting at a basepoint

Eric Silva

yeah

MatheinBoulomenos

@TobiasKildetoft ah sorry, I've been a bit buy

Balarka Sen

that's a 10/10 construction.

Eric Silva

8:10 PM

he got really excited thinking and then someone told him it was the standard def and he was just like "o"

Tobias Kildetoft

@MatheinBoulomenos That's fine

Balarka Sen

ayy lmao

actually i understand him

i never read the definition beyond the first two lines until i figured out why it was natural

Eric Silva

yeah it sort of goes to show you that it's a good construction

Balarka Sen

I guess the way to think about it is

if you have $X$ with an open cover $U_\alpha$

look at $\bigsqcup U_\alpha \times \pi_1(X)$

Mike Miller

I mean, you know what it should be as a set

Daminark

8:15 PM

@EricSilva Rohit

@BalarkaSen I don't think the guy often does AT, usually rep theory or AG

Balarka Sen

and then identify along the overlaps by requiring if $\gamma$ is a loop at $x_0$ of nontrivial homotopy class then it should lift to a path above

Eric Silva

oh sweet

Mike Miller

The fiber above x being pi_1(b, x)

Balarka Sen

right

Mike Miller

The open cover just helps you know what it is as a space

Balarka Sen

8:18 PM

Actually it's exactly like the construction of analytic continuation of a holomorphic function. If $\gamma$ is a path on $X$ starting at $x_0$, make up the space above little-by-little by gluing a Cech-path of $\{U_\alpha\}$ that hits $\gamma$

If you have a loop below it just opens up above because... there's only backward identification not forward

(U_k is glued to U_{k-1})

So that should justify why we should appropriately topologize a space of paths

Jyrki Lahtonen

You guys are great. Enjoying math! Wish I were young(er). Sanity seems to prevail here (a nice reprieve from some of the stuff in main). I'm glad I checked the chatroom out tonight. Thanks for the link to Serre's lecture @BalarkaSen. G'night!

Mike Miller

Nobody but me is gonna read that message Balarka

user193319

Let $G=\sum_{i \in I} \Bbb{Z}$ be the direct sum of an arbitrary number of copies of $\Bbb{Z}$. Let $k \in I$ and define $H_k = n \Bbb{Z}$, and define $H_i = \Bbb{Z}$ for all $i \in I \setminus \{k\}$. Is $H = \sum_{i \in I} H_i$ a subgroup of $G$?

Balarka Sen

Haha @JyrkiLahtonen glad to see you around! Please come here more often

@MikeMiller Oops lol I should point to a more coherent discussion here for those interested.

please excusefully my hiccupy incoherence topology

Tobias Kildetoft

@JyrkiLahtonen Hi. Don't usually see you here

@user193319 Yes, that is a subgroup

user193319

8:24 PM

@TobiasKildetoft Thanks!

@TobiasKildetoft Actually...the only thing that bothers me is that $H$ contains elements which have infinitely many nonidentity components. Isn't that a problem?

Tobias Kildetoft

@user193319 why would it contain those?

user193319

Well, if $I = \Bbb{N}$, and $k =1$, then $H$ contains the element $(2,1,1,....)$ which apparently isn't in $\sum_{i \in I} \Bbb{Z}$.

Also, I'm taking $n=2$.

Tobias Kildetoft

@user193319 No, it does not contain that element, since you have defined it to be a direct sum, rather than a direct product

user193319

So $H$ isn't even in $\sum_{i \in I} \Bbb{Z}$...Shoot...

Tobias Kildetoft

yes it is

Unless you mean to use the symbol $\Sigma$ in two different ways here

user193319

8:32 PM

It is? But surely it's not a subgroup...I'm using $\sum$ however Hungerford uses it. He uses it in the definition of the (external) weak direct product.

Tobias Kildetoft

right, and you use that both for the group and for the subgroup, so why would you get things in the subgroup not in the group?

user193319

Oh...Yeah. I see. So $H$ is a subgroup of $G$. Then is it true that $G/H \simeq \Bbb{Z}_n$, given how $H$ is defined?

Tobias Kildetoft

yes

user193319

Very good! Thanks for the help.

Adeek

8:54 PM

Hey @BalarkaSen just wanted to apologize for saying what I did about with holding my ideas

I thought you were criticizing my intuition of things I am sorry for that