What is $(\mathbb{Z} \oplus \mathbb{Z})/ \langle (2,2) \rangle$?

Is it even a group?

Yes it is. You should understand why.

because it satisfies the group axioms lol

but it is really weird

That's garbage reason. Why does it satisfy the group axioms?

because it has an identity and every element has an inverse and group action is associative

you are repeating the group axioms, not proving anything

alright

$(0,0)$ is the identity

not all quotients are groups.

because $\langle (2,2) \rangle$ is a subgroup?

G/H need not be a group if H is merely a subgroup of G

normal subgroup?

Yes.

That is the real reason.

let H be an abnormal subgroup of G

(i just made that term up)

then exists g such that $gH \ne Hg$

In any case, it's not weird. Choose an appropriate basis of Z^2 with (1, 1) as one basis vector.

Then the quotient becomes Z/2 oplus Z

I'm visualizing the thing as a cross

where only elements near the x-axis and the y-axis are valid

and once you get out of the cross you get wrapped back

Visualizing which thing as a cross?

the group

Well it's a lattice in R^2

With your cross thing (1, 1) and stuff like that are nonexistent

what do you mean by (1,1) is nonexistent?

let us be clear

i mean i don't understand your cross picture

elements in the group are of the form (x,0), (x,1), (0,y), (1,y), right?

which group?

at least i can represent them that way

Z^2/(2, 2)?

$(\mathbb{Z} \oplus \mathbb{Z})/ \langle (2,2) \rangle$

Elements of the group are cosets, not ordered pairs.

I know

that's why I'm talking about their representation

Like

I can represent them using those elements

i.e. every element of the group is equivalent to one of the elements I list

Sure, that is true.

so those elements in the real plane form a cross

oh wait

x and y must be positive

so there's no cross

They're a bunch of dots, too

never mind, I can visualize why it is Z/2 x Z now

I can represent them with $\{(x,y) \ | \ x \in \Bbb Z, y \in \Bbb Z_2\}$

Correct.

Well, y must be positive, not x.

In the representation.

I mean, in my previous representation

not the new one

then I mapped the L-shaped representation to a straight line representation

now G/H is defined by an equivalence

a~b iff a=hb for h in H

hi @BalarkaSen

@BalarkaSen what does it mean for $gH \ne Hg$?

Hi @Jaynot.

Are you busy cos I see that you are in a middle of a convo with DHMO?

@BalarkaSen

@DHMO The two sets are not the same? I don't know what's the question.

@Jaynot Not busy per se. Did you have a question?

Yes. I do have a question.

Ask; don't ask to ask. If I don't answer someone will.

@BalarkaSen can they have different sizes?

do they have any (useful) properties at all?

No, they can't have different size.

@BalarkaSen I am trying to understand that the constant map $\sigma_2$ is null homologous. In the solution provided. I couldnt compute $\delta f = \sigma_2$

@BalarkaSen because $x \mapsto g^{-1}xg$ is a homomorphism?

@DHMO Multiplication by g, on either sides, is an injection H \to G, right? gh_1 = gh_2 means h_1 = h_2.

and same for the other side

I'm having trouble understanding why that is an injection

@BalarkaSen I used the boundary map formula but I cant see how that helps getting $\sigma_2$ as explained in the picture I uploaded.

oh right

you can just multiply by $g^{-1}$ on the left

smacks myself

@Jaynot OK. Suppose $\sigma : [0, 1] \to X$ is the constant loop at $x$. You want to prove $\sigma$ is nullhomologous?

Yes.

Well, its a path not loop.

Expanding...

$$\Large{{}^{{}^0_0 \overset{0}{\underset{0}{0}}{}_{0}^0}_{{}^0_0 \overset{0}{\underset{0}{0}}{}_{0}^0} \overset{{}^0_0 \overset{0}{\underset{0}{0}}{}_{0}^0}{\underset{{}^0_0 \overset{0}{\underset{0}{0}}{}_{0}^0}{0}}{}_{{}^0_0 \overset{0}{\underset{0}{0}}{}_{0}^0}^{{}^0_0 \overset{0}{\underset{0}{0}}{}_{0}^0}}$$

@Secret use \Huge

Sure, constant paths and loops are the same objects.

@DHMO Nah, that will flood the screen

@Jaynot Consider the map $f : \Delta^2 \to X$ from a 2-simplex which sends $f(p) = x$ for any $p \in \Delta^2$. So it's the "constant 2-simplex".

Do you understand why $\delta f = \sigma$?

I get the first part. Its the second part that I dont get. I dont get why $\delta f = \sigma$ @BalarkaSen

@BalarkaSen is it that quotient is group iff normal subgroup?

Yes, @DHMO

the converse is easy to prove i think

let me try the positive

@Jaynot What is $\partial f$, by the boundary formula?

is this correct? @BalarkaSen

I understand why we want to show that because it is the same thing as showing that \sigma is homologous to 0.

Yes.

Why don't you use a textbook instead of learning through internet?

It'd really be much easier than going through a bunch of definitions, asking us, then proving something etc etc

@Jaynot That is true. But you need the boundary formula to understand $\partial f$.

I like to prove things by myself

not reading the proof out of a textbook

@DHMO Textbooks have a bunch of exercises.

@BalarkaSen but they often prove the important theorems

You can try to prove the theorems without reading the proofs out of the book, and once you're done checking it out with the book instead of with us.

Like I like to do, eg.

By the boundary formula $\delta f = f^0 -f^1 +f^2$

@BalarkaSen

Where $f^i$ are?

if a=hb for h in H

i can't actually prove that a=bh for h in H

because H isn't normal

but let's ignore this

The notations are kind of confusing. Let me readjust it here. let $c : \triangle ^2 \rightarrow X$. Then $\delta(c) = c \circ f_0^2 - c \circ f_1^2 + c \circ f_2^2 $

@BalarkaSen

Well I still don't understand what $f_i^2$ are.

Shall I suggest a better notation?

and $f_i^n : \triangle^{n-1} \rightarrow \triangle ^n$

Ah, ok.

Yeah

a~a because a=ea and e in H

a~b implies a=hb, which implies h'a=b where h' is inverse, and h' in H, so b~a

if a~b and b~c, then a=hb and b=kc, where h and k in H, then a=hkc, and hk in H, so a~c

so the equivalence relation is valid

But let me simplify this. Call $\Delta^2 = [012]$ be the 2-simplex with vertices marked $0, 1, 2$. Then $\delta f = f|_{[01]} - f|_{[02]} + f|_{[12]}$.

Here 01, 02, 12 are the various edges of the triangle.

Yeah I get that

okay

$f|_{[\text{stuff}]}$ are all the same as $\sigma$, right? Because they are all constant.

So those are the constant simplices at $x$.

Okay, so we get that $\delta f =f $?

No, it's $\sigma$ (I typo'ed earlier). That's the constant 1-simplex at $x$, the path we started with.

$\delta f = \sigma - \sigma + \sigma = \sigma$.

Does that make sense?

Yeah thats right. Bingo. God bless you @BalarkaSen

existence of identity: if e~f then f is an identity

f=he=h

@BalarkaSen I appreciate your help. One more question. This one is actually from a HW.

Ok, go ahead

let gh=kg where h and k in H

How can I use excision and homology isomorphims associated to each of the pairs $\mathbb{S_+^{n+1}, \mathbb{S}^n$ and $(\mathbb{S}^{n+1}, \mathbb{S}_{-}^{n+1}$ to construct generators of $H_n \mathbb{S}^n$ inductively

I'm running out of letters very quickly @BalarkaSen

@DHMO lol rip

*Using excision and homology isomorphisms associated to each of the pairs $(\mathbb{S}^{n+1}_+, \mathbb{S}^n)$ and $(\mathbb{S}^{n+1}, \mathbb{S}^{n+1}_-)$

@BalarkaSen

then g=kgh'

let m be another element of H @BalarkaSen this letter no longer makes sense

@Secret Tier 8 is far beyond Gamma level. As I tried to tell you, $\psi_0(\alpha)\gg\varphi(...,.......)$

@Jaynot Consider the top and bottom hemispheres as singular 2-simplices $e_1, e_2$ on $S^n$. I claim $e_1 - e_2$ is a generator of $H_n S^n$.

There's an isomorphism $H_n(S^n) \to H_n(S^n, S^n_{-})$ by the homology long exact sequence. Can you see what $e_1 - e_2$ gets sent to by that map?

Or rather for notation's sake just write $e_1 = S^n_{+}$ and $e_2 = S^n_{-}$ because that's what it is.

@Secret also tier 6 and 7 are not quite right. As I said, do not even think about $\psi_x$ yet

It gets sent to zero?

@BalarkaSen

Why do you say that?

let hn=m

gm = ghn = kgn

Each hemisphere is contractible and then have a zero homology

@BalarkaSen I am not sure

So?

@Secret Construct $C_0(\omega_1+1)_n$ for $n=0,1,2$ before moving on. And don't try comparing to the Veblen function much further, as it will eventually be impossible to do.

Just trust me on it :-)

(recall e_2 = lower hemisphere)

@BalarkaSe it sends $e_2$ to$ e_2$

So I guess $e

SO I guess $e_1- e_2$ should be sent to $e_2$

Why does it send $e_2$ to $e_2$?

In $C_n(S^n, e_2)$, $e_2$ is a boundary, right...?

Yes

@BalarkaSen

So $e_2$ should get sent to $0$. Do you agree?

Yes. I agree.

@BalarkaSen does gH≠Hg imply ¬∃h,k∈H[gh=kg]?

@DHMO It means there is an h in H such that gh is never kg for any k in H.

I know what it means

I'm asking what it implies

I don't want to read that logical statement.

if it follows from what i said you're good

does it imply that there does not exist h,k in H such that gh=kg?

@Jaynot What does $e_1$ gets sent to?

That should be sent to itself.

@DHMO That's obviously garbage. gH \neq Hg can happen even if they have two elements in common. Those common elements constitute a gh = kg for some h, k in H.

thanks

f=he=h

gf~g

@BalarkaSen I see where it fails

@BalarkaSen Do I make any sense?

@Jaynot Yep. Now can you prove that $e_1$ is a generator of $H^n(S^n, e_2)$?

if you can do that, you are done because $e_1 - e_2$ gets sent to $e_1$ by an isomorphism; that forces $e_1 - e_2$ to be a generator too if so is $e_1$.

@BalarkaSen I am sorry I cant.

I havent grasp this homology part of the course quite well

Isn't this your homework? You should give it more time than 2 minutes :)

$H_n(S^n, e_2)$ is isomorphic to $H_n(e_1, \partial e_1)$ (why?). Proceed from there.

Its actually due tomorrow thats why I am still staying up late to figure it out. its 4.09 am my time lol

Okay. Let me give it a shot

I'd suggest writing down what you know so far and figuring the rest out tomorrow. 4 AM is not the best time to do math (speaking from experience)

Thanks @BalarkaSen GN.

The fast growing hierarchy and the hyperoperations does help in grasping Bower's array function

But the Vablen functions still remains really arcane to me

@Secret Veblen*

Guys please answer this question

Ask, don't ask to ask.

The sound pressure of a sound wave is 14 Pascal and the pressure of air is 10^5 Pascal find the difference between max and min pressure

@Secret I think you will find the ordinal collapsing function more understandable than the Venlen function, and you can actually write the Veblen function using the ordinal collapsing function.

Lol@TimTheEnchanter

@satyatech Might be more appropriate for the H-bar

It is a bit of physics (sound) question ,but I think that some guys here at math chat maybe able to answer it but hopefully I am also gonna post it on h-bar but if you can pls answer here.Thanks

@satyatech If 14Pa is the amplitude of the pressure oscillations, then the max and min pressure ought to be 10^5 + 14 and 10^5 - 14 Pa imo.

That was my first thought @TimTheEnchanter but the answer is given to be 10 Pascal.

How can I construct a vector field whose curl is equal to another simple-ish vector field?

I think it can be done by comparing the coefficients of i, j and k, then perhaps integrating?

Never mind, I figured it out by letting one of the components be zero (hopefully WLOG).

Sum of perpendicular distances to the lines is equal to the perpendicular distance between the lines should work.

How do you guys go about taking notes and studying for math courses? Do you take notes during lectures, read ahead, use loose A4 paper or a notebook etc...?

Depends on the class

If lecture notes are provided, I wrote extra things in between or on the back side of the previous page

If no lectures notes were provided or a book was used, I just took notes separately

I write a lot of notes because I remember things much better if I write them

Even when I self study stuff from a book I need to write down some notes

@AlessandroCodenotti I think I'm much the same, I basically transcribe all my course notes

The first one is indeed injective but not surjective

The second one I would call "not well-defined"

The second one must not be called a function, I think

Ah ok!! Maybe it is meant $f:\mathbb{R}_+\rightarrow \mathbb{R}_-$, $f(x)=-|x|$ ? Then it would be injective and surjective, i.e., bijective right?

Yeah

Ok!! Thank you!! :-)

What are strict global extrema? Which is the difference betwenn them and global extrema?

@MaryStar they're the only global extrema?

The inequality has to be strict

$x$ is a global extremum if $f(x) \geq f(y)$ for all other $y$.

$x$ is a strict global extremum if $f(x) > f(y)$ for all other $y$.

You can do the same for local extrema

@DHMO Very close, actually. $x$ is a strict global maximum/minimum if and only if it is the unique global maximum/minimum

But it's possible to both have a strict global maximum and a strict global minimum

I see!! Thank you very much!! :-)

(I am currently afking, expect no response until I am back)

Does $C_3$ mean the cyclic group of order 3?

Sometimes it's denoted that way, yes

How many points do you require to determine a circle? 3?

Points on the circumference that is

yes, 3

perimeter I guess it's called

Hmm okay

Someone at the chemical engineering campus apparently has the same last name as me. This are now submitting all orders to chemical companies using my full name and email address. I've gotten a lot of mail about it, including one email from the chem campus insisting that I am not me.

@BalarkaSen how does that work?

How does what work? 3 points determining the circle?

Yeah

Draw it out, convince yourself

Or

Do some algebra and get bored and stuck

@Secret okay

Algebraically that makes perfect sense but what about with compass straightedge?

There's a slick proof I am not remembering. Given three points, join pairs of points by segments, and draw their perpendicular bisector.

Those should pass through a unique point, which is supposed to be the center.

I was messing around with Euclidea and did that accidentally but I don't quite fully understand why their intersection is the center

the proof that circumcenter exists should be quite enough to prove that 3 points determine a circle

i.e. there is a unique point which is equidistant from the 3 points

Oh neat okay yeah

Alright so I have an integral $\oint_L \textbf{A} \cdot d \textbf{r}$ for a given vector field $\textbf{A}$. $L$ is the curve given by the intersecting line between the cylinder $\cases{(x-a)^2+y^2=a^2\\ z\geq0}$ and the sphere $x^2+y^2+z^2 =R^2$ where $R^2>4a^2$.

So I want to calculate it using Stokes' theorem

But I'm having trouble finding the surface area given by the boundary

Or rather, how to parametrize it

@Krijn There should be an algebraic geometry proof. A circle is like $x^2 + y^2 + 2fx + 2gy + c = 0$; so the space of circles is, what, $\Bbb{R}^3$, determined by pairs $(f, g, c)$. Three points on the circle determine three hyperplanes in $\Bbb{R}^3$. Those intersect in a unique point generically; that's your circle :)

algebraic geometry in the sense that it's a "baby moduli space" proof.

The obvious choice, I guess, is using spherical coordinates to cut out a small piece of the sphere since I know the radius and all I need is a loop for radius to travel in

If that makes any sense...

Oh yeah, nice, but that's really the same as "draw it out, convince yourself"

@Bala what are you doing nowadays

Stuff and things

@BalarkaSen Yikes that's one way to do it, I literally just sanity checked the circle equation and noticed there were 3 free variables...

@Excalibur42 Actually, similar proof shows that 5 points determine a conic, or 9 points determine a cubic, or...

Typically you'd want to do this all in projective spaces to remove non-generic stuff.

@BalarkaSen Any ideas?

@BalarkaSen pick an algebraically closed field as well, just in case :P

Wait, I'll make a picture

(@Krijn To elaborate, mostly Riemannian geometry and foliations. Still no algebra though :P)

@BalarkaSen :(

So there's the pesky surface

it's for the best

@Lozansky I am not really helpful on these realms. I'd use spherical coordinates to parametrize the sphere and restrict that to those curves, maybe.

Parametrizing is what I'm bad at

@BalarkaSen Well finding those curves is the tricky part :P

Wait I think I got it

Let $0\leq \phi \leq cos(2a/R)$ and $0 \leq \theta \leq 2 \pi$ and cut out the piece

Hey everybody!

morning

How's it going?

not bad. doing some high school stuff

Not getting done what I should be getting done :/

Isn't this what we all (don't) do?

Same, I should have made more progress on my difftop pset than I did... And on analysis...

Morning

Hey @Akiva!

Is Z^2/<(3,2)> isomorphic to Z+Z3 or Z+Z2?

this is troubling me very much

My instinct is to say the latter but I'm not sure, let's see

Maybe it's isomorphic to $\Bbb Z$

@AkivaWeinberger nah

With generator $(2,1)$

consider (2,3)

I think it has order 3

but I haven't been able to find an element with order 2

so I guess it is Z+Z3

but I can't prove anything

How? $(2,3)\cdot3=(6,9)=(0,5)\ne(0,0)$

Hold on

heh, never mind

$$(2,1)\cdot(-5)=(-10,-5)=\\ (-10,-5)+(12,8)=(2,3),$$right? @DHMO

that is witchcraft

So I maintain that this is probably homeomorphic to $\Bbb Z$, with generator $(2,1)$

I think (1,0) is also a generator. Prove me wrong.

Since $\Bbb Z$ has two generators, the other one would have to be $(-2,-1)=(1,1)$

Yeah, that one's simpler :P

2n+3k=x
 n+2k=y

the matrix is invertible with determinant 1

@DHMO The $y$-coordinate of any multiple of that is even

no matter what representation you choose

@AkivaWeinberger nice point

So $(2,1)$ and $(1,1)$ are the generators, and the isomorphism $f:\Bbb Z/\langle(3,2)\rangle\to\Bbb Z$ is $f(a,b)=2a-3b$.

missing some squares there :O

@SteamyRoot squares?

That has $f(2,1)=1$ and $f(1,1)=-1$.

$\mathbb{Z}^2/\langle (3,2) \rangle$ :P

I think Z^2/(m, n) is isomorphic to Z $\oplus$ Z/d where d = gcd(m, n).

Many group theorists would interpret $\mathbb{Z} / \langle (3,2) \rangle$ as modding out $\mathbb{Z}$ by the gcd of $2$ and $3$

@BalarkaSen interesting

@SteamyRoot Yeah, sorry, typo

You can calculate the quotient of Z^n (R^n, R a PID) by any subgroup (submodule) by finding a basis for the submodule and putting that into something something normal form, a diagonal matrix where each term divides the next

It's easy to show the quotient contains at least $\mathbb{Z}$ and $\mathbb{Z}_{\gcd(m,n)}$ as subgroups

@DHMO Re: the thing you just commented on, don't just give him the answer like that…

@AkivaWeinberger ok

Should I delete it?