Mathematics - 2017-10-17 (page 5 of 6)

MatheiBoulomenos

20:00

I guess that works. I was thinking of something less artifical. Can you give an example in the category of groups (where morphisms are group homomorphism?)

Leaky Nun

@MatheiBoulomenos associativity failed before the edit I presume?

Balarka Sen

In particular you're making yourself aware that you are making Mathei aware that you are making yourself aware about concrete categories

Leaky Nun

@BalarkaSen oh god

Evinda

Hello!!! Does $-a \leq b \leq a$ imply that $b^2 \leq a^2$ ?

Leaky Nun

@Evinda yes

@MatheiBoulomenos I think A and B are groups, lol

Balarka Sen

20:02

that hTop is nonconcretizable always intrigued me

Semiclassical

first equality implies $0\leq a+b$. second implies $0\leq a-b$.

MatheiBoulomenos

@Balarka thank you for making me aware that Leaky is making himself aware that he is making me aware that he is making himself aware of concrete categories

Semiclassical

RecursionOverflow

and the last inequality is equivalent to $0\leq (a-b)(a+b)$.

MatheiBoulomenos

@BalarkaSen yes, that's pretty cool

Leaky Nun

@Semiclassical nice

Balarka Sen

20:03

@MatheiBoulomenos good game. I'm too lazy to respond

MatheiBoulomenos

@LeakyNun I didn't notice it at first, but id(B->A) doesn't make sense, an identity always goes from an object to itself

Leaky Nun

@MatheiBoulomenos well I mean f:B->A given by, you know, f(b) is the copy of b in A

Sha Vuklia

Guys, let $I\subset R$ be an ideal (where $R$ is a commutative ring). Someone said that for $a\in R$, we have $Ia=I$. I don't see why $I\subset Ia$. Because we don't know if $a^{-1}\in I$?

Balarka Sen

@MatheiBoulomenos Somehow, it means homotopy theory needs to be modeled on more complicated categories than Sets

Unlike homology theory, I guess, which is modeled on Ch_*

Which are concretizable in Mod_R, like any abelian category

That's the Freyd-Mitchell embedding

Leaky Nun

@ShaVuklia $a^{-1}$ doesn't need to be a thing

MatheiBoulomenos

20:09

Freyd-Mitchell is pretty mind blowing to me

Balarka Sen

True man

Leaky Nun

@ShaVuklia Let $2\Bbb Z \subset \Bbb Z$ be an ideal. Let $a=2 \in \Bbb Z$. Then, $(2\Bbb Z)(2) = 4\Bbb Z \ne 2\Bbb Z$.

Sha Vuklia

@Leaky then say $x\in I$, how do we show $x\in Ia$? My initial idea was to have $x=xa^{-1}a$, but I can't do that

Leaky Nun

@ShaVuklia who said that?

Sha Vuklia

oh!

so it's not true:l

Leaky Nun

20:10

@ShaVuklia don't trust people randomly

"someone said that"

Sha Vuklia

maybe I misinterpreted

Leaky Nun

$Ib$ isn't defined as pointwise multiplication here

Balarka Sen

It is, isn't it? That's a typo, but the content of the post is not wrong, I do not think

Leaky Nun

I got confused by that also

@BalarkaSen it isn't

Sha Vuklia

wth is pointwise multiplication?

Kasmir Khaan

20:12

@MatheiBoulomenos still here? :)

Balarka Sen

@LeakyNun What do you think it is

Leaky Nun

@ShaVuklia I mean, $Ib \ne \{ib \mid i \in I\}$ here

Sha Vuklia

ohhh

Leaky Nun

Let $I$ be an ideal of $R$. Then, $R/I$ is the partition of $R$ by the equivalent relation $x \sim y \iff x-y \in I$ with the operations:
- $(I+a)+(I+b) = I+(a+b)$
- $(I+a)(I+b) = I+(ab)$

(credits to wiki)

Balarka Sen

I don't get it. Define $Ib$ for me.

Leaky Nun

20:14

wait, then what does $Ib$ mean there

lol

Sha Vuklia

it means nothing

Balarka Sen

it is pointwise multiplication

Sha Vuklia

they shouldn't even have included it

Balarka Sen

Leaky is confused and should stop talking

Sha Vuklia

if $a\in I$, then $ab\in I$, end of story:P (for this post)

Balarka Sen

20:14

@Sha It's a typo.

Sha Vuklia

yea

Leaky Nun

@BalarkaSen :c

did they mean $ab \in Ib \subset I$?

Balarka Sen

Yep

Leaky Nun

I got double confused

Balarka Sen

which is true for all ideals, definitionally

@LeakyNun *(credits to wiki for confusing me)

Leaky Nun

20:15

a few days ago I got confused over the apparently wrong $II=I$

now I got that right and got this confused

MatheiBoulomenos

@Balarka Take the weirdest topological space X you can think of, the weirdest ring R you can think of the and the weirdest small category C you can think of. Then the category of functors from C to the category of double complexes over the category of sheafs of R-modules over X is abelian. I can't wrap my head around how this is modeled by something as innocent as abelian groups

Sha Vuklia

@BalarkaSen loll that is wiki's daily job

Leaky Nun

@MatheiBoulomenos so what was the example you had in mind?

MatheiBoulomenos

@Leaky consider Z -> Z/2Z

Leaky Nun

hmm

oh

MatheiBoulomenos

20:17

@Kasmir yes, I'm here as you can see

Kasmir Khaan

:D

Sorry was reading did not notice

Leaky Nun

24 mins ago, by MatheiBoulomenos

@LeakyNun my question was if you can think of an example of a category where the morphisms are given by maps (i.e. a "concrete category") such that there is a surjection which is not a split epimorphism

35 mins ago, by MatheiBoulomenos

@LeakyNun try to show that f:A -> B is surjective iff it is a "split epimorphism", that is there exists a g:B -> A such that fg is the identity on B. That's in fact equivalent to AoC iirc

Kasmir Khaan

Ehm my question is how imporant is the notion of maps of G sets

Balarka Sen

@MatheiBoulomenos Hahaha

Kasmir Khaan

In understanding this topic am doing

We have a short lecture for our class and the teacher mentioned something about it

Like we have a square

DaenerysDracarys

20:19

0

Q: Investment into $n$ mutual funds and $m$ asset categories.

Consider $n$ mutual funds all of which invest in $m$ asset categories. Let $a_{ij}$ be the fraction of the capital $j$ invested in category $i$. Suppose we have initial capital $K$ and we want to invest all or part of it in these funds, and we denote by $x_{j}$ the amount invested in the $j$th f...

Kasmir Khaan

X ---- Y

X ---- Y

MatheiBoulomenos

@LeakyNun why are you quoting that?

Semiclassical

@DaenerysDracarys hsssss finance hssss

8

DaenerysDracarys

Yeah. I don't like it any more than you do.

Leaky Nun

@MatheiBoulomenos don't get Balarka started

Balarka Sen

20:21

@Semiclassical Instant cracked up

DaenerysDracarys

But I need to pay for my armies so I can retake the Iron Throne.

Even if you don't consider it as a finance problem, what in the bleep is a "hyper-pyramid"?

And don't tell me it's a pyramid jacked up on sugar.

Semiclassical

Now I'm loling at the image of Daenerys Targareyon at a blackboard doing linear programming

MatheiBoulomenos

@Leaky do you see why the projection Z -> Z/2 is not a split epimorphism?

Semiclassical

I think what they mean is that it's an n-simplex?

DaenerysDracarys

Jon Snow tries to help and she's all like "Just stop. You know nothing, Jon Snow".

What's an n-simplex?

Like a polygonal shape in n dimensions?

Balarka Sen

20:23

pls no game of thrones

Leaky Nun

@MatheiBoulomenos because all hom Z/2 -> Z is trivial

DaenerysDracarys

Thrones, y'all

Balarka Sen

life is too short to hear people talking about normie TV shows

DaenerysDracarys

@BalarkaSen are you a hipster?

Balarka Sen

@DaenerysDracarys A rather massive Beatnik

Semiclassical

20:24

well, consider the case of $x_1+x_2\leq 1$ with nonnegative x1,x2

DaenerysDracarys

So, like a ball

Balarka Sen

lmao

Semiclassical

then that's just a triangle with vertices (0,0), (1,0), (0,1).

DaenerysDracarys

Okay...

Balarka Sen

You can try drawing it. $x + y \leq 1$ is bounded by the $x$-axis, the $y$-axis and the line $x + y = 1$

DaenerysDracarys

20:25

Then, in the case of $x_{1}+x_{2} + x_{3} \leq 1$ with nonnegative $x_{1}$, $x_{2}$, $x_{3}$?

Kasmir Khaan

@MatheiBoulomenos did you see my question ?

Semiclassical

in that case it's a tetrahedron with vertices (0,0,0), (1,0,0), (0,1,0), (0,0,1)

Balarka Sen

Not a regular tetrahedron, a right angled one, just to note

DaenerysDracarys

@BalarkaSen really, the $y$-axis doesn't do much in terms of bounding that in 2d.

So, it always has $n$ vertices, plus the origin.

MatheiBoulomenos

@Kasmir ah sorry, I forgot. That's not really something I can answer. Personally, I haven't used G-maps a lot, I think I never used them in my algebra class, but I don't know what your teachter considers important

Semiclassical

20:27

you're taking the positive orthant of R^3 and bounding it by the plane x+y+z=1

i am going to stab my internet connection

DaenerysDracarys

No stabbing.

Balarka Sen

burn it

Semiclassical

lol

Kasmir Khaan

@MatheiBoulomenos okay :D but the idea of G acting on G/H

@MatheiBoulomenos I think it can help me understand more G/Stab (x)

MatheiBoulomenos

That's pretty important, I use that a lot

Kasmir Khaan

20:28

:D

So can you explain that to me +

I still struggle with the notion of quotient groups

Semiclassical

actually, i was wrong to cite the n-simplex.

Kasmir Khaan

I mean they dotn feel natural yet

Semiclassical

that's just the triangular surface of the tetrahedron i indicated

DaenerysDracarys

Burning, stabbing...keep it up and Imma bring up some normie tv shows.

MatheiBoulomenos

@Kasmir G/H is not necessarily a group if H is not normal, but G still acts on it even if its just the set of cosets

Semiclassical

20:29

though i guess it matters whether I say i'm talking about an n-simplex in R^n or an n-simplex in R^(n+1), ugh

Kasmir Khaan

@MatheiBoulomenos so far so good :D

DaenerysDracarys

Anyway, I think I understand now.

Kasmir Khaan

but the idea of G "acting" on G/H

Semiclassical

yeah.

DaenerysDracarys

But, why are they saying the apex is at the origin? Isn't the apex at the top?

Kasmir Khaan

20:30

is it like g aH = gaH ?

MatheiBoulomenos

@Kasmir exactly

Sha Vuklia

Guys, my book shows that: $M$ is a maximal ideal of $R$ iff $R/M$ is a field. Let $M\subset J\subset R$.

They consider $R/J\cong (R/M)(J/M)$. They write $\overline R=R/M$ and $\overline J=J/M$. I understand that if we have an $R$-ideal $J$ that is a proper subset of $R$ (and $M$ is a proper subset of $J$), that the corresponding $\overline R$-ideal lies ‘properly’ between $\{\overline 0\}$ and $\overline R$, and vice versa. Now they claim the following: $M$ is a maximal ideal of $R$ iff $\{\overline 0 \}$ is a maximal ideal of $\overline R=R/M$. I’m kind of missing the link here…

Semiclassical

well, how do you define the apex of a tetrahedron?

DaenerysDracarys

Dunno...

Semiclassical

it's inherently a bit of a slippery definition, since I could pick any vertex

Kasmir Khaan

20:31

@MatheiBoulomenos am gonna try to show that it is really an action =p

DaenerysDracarys

It's the vertex that does not lie on the base...

Semiclassical

what they're getting at, though, is that you could take the triangular surface where x+y+z=1 as the base of the tetrahedron, and the origin as its vertex

DaenerysDracarys

Which the origin does...

MatheiBoulomenos

Try to show that it's well-defined first (you're dealing with cosets after all)

Semiclassical

take the base as the face which is opposite to the origin

DaenerysDracarys

20:32

OOoh...

ÍgjøgnumMeg

@ShaVuklia If by $\{\}$ you mean the ideal $(0)$, how does that correspond to something "field" related?

Kasmir Khaan

(G x G/H ) ---> G/H

thats how we write it right?

G is a group and G/H is a set

Sha Vuklia

yea that was a typo, I was about to correct it

Kasmir Khaan

ill start with well defined part ><

Semiclassical

the same idea holds in higher dimensions as well---but don't ask me how to visualize it!

Balarka Sen

20:33

@ShaVuklia The essential fact is ideals of $R$ containing an ideal $I \subset R$ correspond 1-1 to ideals of $R/I$

Semiclassical

i guess one intuition for it is that, if you take horizontal cross sections of the tetrahedron in 3D, you'll get triangles in 2D

MatheiBoulomenos

And under this correspondence, maximal ideals correspond to maximal ideals

Semiclassical

similarly, you can take cross sections of the 4D version and get tetrahedra

Sha Vuklia

@Balarka yea my book mentions that also, but I don't see that

Semiclassical

that's about as far as my intuition goes, though

Sha Vuklia

20:35

like, I can see that $R$ quotiented by $J$, corresponds to $\overline R$ quotiented by $\overline J$

but I don't see the direct connection between $J$ and $\overline J$

Balarka Sen

@ShaVuklia You should spend some time on thinking about that, in my opinion. It took me some time to see it the first time.

Sha Vuklia

could you at least give me a hint? @Balarka

Arrow

Hello,
I am looking for a simple example of two bundles $\begin{smallmatrix}X\\ \downarrow\\ U \end{smallmatrix},\begin{smallmatrix}Y\\ \downarrow\\ U \end{smallmatrix}$ satisfying the first condition below, and for a simple example satisfying the second condition.

1. There's no arrow from the first to the second.
2. There is an arrow between them but there are no isomorphisms.

Kasmir Khaan

@MatheiBoulomenos is the notion of an action just a "idea" to introduce orbits and stabilizer and their interaction ? so it does not stand on its own as a concept?

@MatheiBoulomenos this is very important to me because , i feel like i wasted time trying to make sense of what an action is on its own

ÍgjøgnumMeg

@KasmirKhaan An action is just a special kind of map

MatheiBoulomenos

20:39

@Kasmir certainly not

Balarka Sen

@ShaVuklia Think explicitly about the projection map $R \to R/I$

Kasmir Khaan

Hmm can you please explain more guys

am going nuts -.-

@ÍgjøgnumMeg special kind of map?

ÍgjøgnumMeg

@KasmirKhaan Sure, do you know the definition of a group action?

Kasmir Khaan

Yes

G acting on a set X

we have two properties

identity of G dont do anything

1x =x for all x in X

and g (hx) = (gh)x

where in the left we doing the action twice

on the right gh is taking in G , and that element act on x in X

Sha Vuklia

@Balarka so something like an ideal $J\supset I$ is sent to an ideal of $R/I$?

Kasmir Khaan

20:41

this much I know

Sha Vuklia

oh wait..

Kasmir Khaan

The whole interaction of this , group actions, orbit stab theorem , conjugation and class equation to sylow

that chain is not clear to me at all

Sha Vuklia

we have an isomorphism, so ideals $J\supset I$ must be sent to different ideals in $\mathbb R$

hm, well I think at least I got it intuitively

MatheiBoulomenos

Actions give us a reason to care about groups, honestly. (Well I'd care about them anyway, as I'm an algebra maniac, but that's a different story) A group is something that acts on something by symmetries is a perfectly fine informal defition of a group. A group acting on something means that this group is realizing some of the symmetries of that object. That's how I think about group actions

Leaky Nun

@MatheiBoulomenos amen

Kasmir Khaan

20:44

this is well said :D

so the idea behind introducing actions

is that they have a good structure

am gonna copy what you said and save it in my notes

Leaky Nun

Oct 14 at 16:10, by Leaky Nun

@KasmirKhaan Consider the group action by conjugation from the group $\Bbb S_3$ to itself, i.e. $g \circ h = ghg^{-1}$. Write down the image of the permutation representation as homomorphism $\Bbb S_3 \to \Bbb S_6$.

@KasmirKhaan have you thought about this?

offer of what?

Kasmir Khaan

@LeakyNun I saved that question for later, still dont get the concept

DaenerysDracarys

Group actions speak louder than group words.

Kasmir Khaan

hmm

Leaky Nun

@KasmirKhaan well you already have three days to think about it...

Kasmir Khaan

20:48

@LeakyNun without understanding the concepts , i wont be happy solving that

Leaky Nun

@KasmirKhaan I mean, three days to understand the concepts

Kasmir Khaan

Leaky your not helping me this way

only brining me down

Leaky Nun

sorry

Kasmir Khaan

np

ÍgjøgnumMeg

21:02

@Maxwell You have a hint already, why don't you try and work a bit more on it?

Mary Star

21:18

Let $D$ be the space that is defined by $x=0, y=0, z=0, x+y=4, z=x+y=1$.
I want to write it in the form of a set.
We have that $z=x+y+1=4+1=5$, so $0\leq z\leq 5$.
From $x+y=4$ we get $y=4-x$, so $0\leq y \leq 4-x$.
From $$z=x+y+1$ we get $x=z-y-1$, and so $0\leq x\leq z-y-1$.

So, we get the set $\{(x,y,z)\mid 0\leq x\leq z-y-1, 0\leq y \leq 4-x, 0\leq z\leq 5\}$.

Is this correct? But shouldn't y depend on z instead of x?

I want to calculate the volume over D. For that the definition of D that I wrote is not correct, is it?

Kasmir Khaan

21:32

@LeakyNun here? :)

Leaky Nun

@KasmirKhaan ?

@MaryStar rip latex

Kasmir Khaan

I found few questions about ring theory

on old exams

can you send me how you work them out , so when i take a look on the exams + solutions

I use those to understand?

If you have time ofc

Leaky Nun

$$\{(x,y,z)\mid 0\leq x\leq z-y-1, 0\leq y \leq 4-x, 0\leq z\leq 5\}$$

Kasmir Khaan

I am gonna try to go for the exam and if failed ill do the re exam

Leaky Nun

ok

Kasmir Khaan

21:35

but if would be good if i have solutions to them

thanks alot man :)

Mithrandir

*raises hand in salute* Goodbye and good luck to @arjafi. Thank you for your service to the Math.SE community and SE as a whole. Farewell!

Kasmir Khaan

please in details so i see how to work other questions out

Leaky Nun

@Mithrandir what happened?

Kasmir Khaan

Ill send 1 of each idea total of 3 questions

Mithrandir

@LeakyNun he stepped down

Leaky Nun

21:37

@MaryStar I would instead have $\{(x,y,z) \mid 0 \le x \le 4, 0 \le y \le 4-x, 0 \le z \le x+y+1\}$

mercio

what are $r,V,\phi,\psi,p,t$, and $m$ ?

Mary Star

From $z=0$ and $z=x+y+1$ we get the inequality $0 \le z \le x+y+1$.
From $y=0$ and $x+y=4\Rightarrow y=4-x$ we get the inequality $0 \le y \le 4-x$.
But how do we get the inequality $x \le 4$ ? Is is because $4-x$ (the upper bound of $y$) must be positive?

Leaky Nun

@MaryStar yes

ShrtTth

Hi guys anyone knows which is the most efficient algorithm which can compute the stoppage times in the hailstone sequence ?

Kasmir Khaan

@LeakyNun send them here?

I collected them into 1 pdf

3 questions about ring

Leaky Nun

21:44

send to my email

Kasmir Khaan

ok

Mary Star

@LeakyNun Ah ok! Thank you! If we want to draw this space, how can we start, what do we have to draw first?

Leaky Nun

@MaryStar the triangle on the x-y plane

then extend it upward with variable height

Kasmir Khaan

@LeakyNun Did not work via email

@LeakyNun was loading nonstop and did not go thru

slow connection or something

Annelise t.

A polytope is the convex hull of a finite number of points, let S be a polytope, I know that S is bounded but I don't see it intuitively, can someone explain this?

Kasmir Khaan

21:52

@LeakyNun It worked now

Leaky Nun

:o how did you delete a message that was posted 5 minutes ago

Kasmir Khaan

@LeakyNun haha dont know how it worked =p just tested

Mary Star

So at the xy-plane (for z=0) we have the triangle bounden by the lines x=0, y=4-x, y=0. So it the triangle with vertices (0,0), (4,0), (0,4). Using the inequality $0\leq z\leq x+y+1$, do we extend the triangle parallel to the plane $z=x+y+1$ ? @LeakyNun

Leaky Nun

@MaryStar I don't understand

you extend the triangle upwards

parallel to the xy-plane

@KasmirKhaan so what should I do?

Kasmir Khaan

@LeakyNun Just how you would answer on exam, i want bascilly to learn how to anwer on exam

meanwhile i understand from the answer

Mary Star

21:58

Yes, but how do we know how far we have to extend the triangle? Do we use for that the inequality $0\leq z\leq x+y+1$ ? If yes, how? @LeakyNun

Kasmir Khaan

we have some solutions that are clear and others that are not posted on our site @LeakyNun

Leaky Nun

@MaryStar extend each point $(x,y)$ to a height given by $x+y+1$

@KasmirKhaan which questions?

Kasmir Khaan

@LeakyNun I sent them via email

@LeakyNun you did not get?

Leaky Nun

I got them, I'm asking which questions do I need to do

Kasmir Khaan

@LeakyNun Any >< i dont undersand them all so :D

Leaky Nun

22:00

ok later

Kasmir Khaan

okay :D

thanks man :)

Simply Beautiful Art

@Mithrandir wrong room? :P

Mithrandir

@SimplyBeautifulArt nope

Simply Beautiful Art

Okay lol

For those who are unaware:

19

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