Mathematics - 2015-07-18 (page 1 of 3)

pjs36

00:11

Hey there Mr. @AlexClark, haven't seen you in a while! How's the Australian winter treating you?

Karim Mansour

Hi guys

pjs36

Hello Karim

Karim Mansour

00:33

how r you doing @pjs36

pjs36

Good, thank you @Karim! Finally got (loaned) Ted's Multivariable book, having fun looking it over, for now.

Yourself?

Karim Mansour

@pjs36 doing mechanics and will study analysis afterwards but last few days I have been busy about the schools that I want to apply for in graduate school

I considered applying to 3 universities for masters

pjs36

Ah, applications, that's fun. I need to think about something similar as well.

Abroad, or in your home country?

Karim Mansour

home country

yeah I have to deal with applications and all that mombo jumbo

pjs36

Yeah, it's never my favorite thing. Perhaps we'll learn to love applying for things, some day :)

Karim Mansour

00:46

yeah xD :D

Philip Hoskins

03:14

Anyone have an idea for this question? math.stackexchange.com/questions/1365126/integrability-of-xy-3

Stan Shunpike

03:25

@PhilipHoskins how did you find the first function?

Philip Hoskins

@StanShunpike It's just Tonelli. You integrate once with respect to y and can see the integral is a bounded continuous function for a>= 3

Soham Chowdhury

04:51

@Balarka, I'm rereading Aluffi's groups chapter. Also, SB postponed to Thursday.

Okay, exercise: find the orders of the symmetry groups of the five Platonic solids. Seems like simple combinatorics.

Jeff

Hi folks. How can I undo a bad edit that I made to an answer?

Remember me

What's platonic solid @Soham
I could do the question if I know what that means

Jeff

Oh yeah, and how do I integrate $\int \sec (t) dt$?

Soham Chowdhury

Platonic solid

In three-dimensional space, a Platonic solid is a regular, convex polyhedron. It is constructed by congruent regular polygonal faces with the same number of faces meeting at each vertex. Five solids meet those criteria, and each is named after its number of faces. Geometers have studied the mathematical beauty and symmetry of the Platonic solids for thousands of years. They are named for the ancient Greek philosopher Plato who theorized in his dialogue, the Timaeus, that the classical elements were made of these regular solids. == History == The Platonic solids have been known since antiquity...

Remember me

Convex polyhedra?

Soham Chowdhury

05:07

open the fucking wiki page

3

Cristopher

@Jeff Multiply and divide by $\sec(t)+\tan(t)$ then make the substitution $u=\sec(t)+\tan(t)$

Soham Chowdhury

@Cristopher ah, the old black magic trick

Remember me

No cussing :p@Soham

Jeff

@Cristopher yup. that works. got it

@Cristopher even before your edit

Cristopher

Lol.

Glad you got it.

Remember me

05:13

Will the order of the symmetric group of the tetrahedron be 16?

@Soham

@Soham you are online ?

Soham Chowdhury

06:07

yes, now.

@Rememberme "group of symmetries", not "symmetric group". and, no, not 16.

this is basic combinatorics. how the hell are you able to solve all the ISI admit problems? :P

@r9m, is that you?

mwahahaha

r9m

@SohamChowdhury yes ... chibi itachi :p

Soham Chowdhury

I don't know Japanese, if that's what that is.

r9m

Chibi means short ... ltachi is a character from nqruto anime .. (thats what my avatar is)

skill patrol

06:24

Which, imho, is one of the best avatars in here.

r9m

@Chris'ssistheartist @robjohn :) have you seen my most recent answer in main re a limit problem ? :) (sorry cant link right now ... chatting from tab)

@skillpatrol :D thanks!!!

Balarka Sen

06:55

@SohamChowdhury okie.

@Rememberme The full group of tetrahedral symmetry is of order $24$. A tetrahedra has $4$ vertices. Pick any vertex, and pick a codimension 1 plane through the vertex to the middle of the opposite face. You can rotate $3$ times through this axis. So there are a total og $3 \cdot 4= 12$ rotational symmetries. For each rotational symmetry, you can compose with a reflection through a codimension 2 plane through the middle of the tetrahedra.

Then that's a total of $12 \cdot 2 = 24$ symmetries.

Soham Chowdhury

07:18

"codimension 1" lol, seriously?

also, a tetrahedron, two tetrahedra.

and why did you spoil the problem?

FreeMind

@SohamChowdhury Do you seriously think that he has spoiled the problem?! :O

Soham Chowdhury

07:41

well, he gave him the answer.

wow, I'm really warming to this idea of second and third readings. seems like it works as advertized.

hi, Mike.

The Substitute

off topic. An exercise from Dummit/Foote (#3 of 14.2) asks to find the Galois group of $(x^2-2)(x^2-3)(x^2-5)$ over the rationals. Isn't this not well-defined since the polynomial isn't irreducible?

Balarka Sen

08:40

@SohamChowdhury I waned to emphasise that you want a codimension 1 thing to rotate and codimension 2 thing to reflect.

I spolied it since Remember doesn't know how to do it.

@TheSubstitute Not at all. Why would the Galois group be ill-defined if the polynomial isn't irreducible?

The Substitute

@BalarkaSen I am not sure why I thought that.

Balarka Sen

The Galois group is defined whenever the extension is Galois. In this case, the splitting field is $\Bbb Q(\sqrt{2}, \sqrt{3}, \sqrt{5})$, which is of course Galois over $\Bbb Q$

dREaM

plato means plate in spanish

Balarka Sen

09:10

@SohamChowdhury You can even make the work easier by noting that octahedron is dual to the cube, and the dodecahedron is dual to the icosahedron.

dREaM

09:38

and what is the icosahedron dual to?

Remember me

09:53

Oh I see now.. @BalarkaSen

@Soham I just gave a rough idea though.. I didn't actually sit and do the question

That might have lead to the answer... Who knows :)

Balarka Sen

what might have lead to the answer? I have no idea how you found 16.

and you didn't give any rough idea either, @Remember. you just said "is the answer 16?"

@dREaM the dodecahedron.

Remember me

10:09

I miscalculated 4!

@BalarkaSen I thought of the four vertices and thought of S_4 .. It was just a guess... Sorry for inconvenience.. It should have been 24..4!

Balarka Sen

You calculated 4! = 16?

wut.

dREaM

4!=12

Balarka Sen

anyway, so you didn't have an idea "that might have lead to the answer". what you had is a guess. now prove that the group is really $S_4$, @Remember.

Remember me

Balarka you can have mistakes .... Sometimes

dREaM

@BalarkaSen S4 is the only group of order 12

@BalarkaSen wait, doesn't your exlanation of why the order is 24 already prove it is $S_4$?

Balarka Sen

10:18

no.

there are a lot of groups of order 24.

Remember me

@BalarkaSen showing that the group G is isomorphic to S_4 will do the job right?

dREaM

I meant your explanation

not the fact it has order 24

Remember me

By G I mean the symmetric group of the tetrahedron

dREaM

you already showed you can permute vertices anyway you like

Balarka Sen

@Rememberme that's precisely what I want you to prove.

@dREaM I didn't show it. You have to prove it if you want to go the hard way.

But there are easier ways to settle that isomorphism.

Remember me

10:21

Okay... I am serving the proof in minutes... :p

Balarka Sen

It'd take you more than just a few minutes... if you won't google it right away.

dREaM

I fail to see what remains to be done, we know the symmetries coincide with the vertex permutations.

Remember me

Obviously I wont ..

Balarka Sen

@Rememberme I hope so. It's not "obvious" that you won't, knowing that you have already done so a couple times before.

:P

Remember me

@BalarkaSen please will you stop saying that I don't do that anymore :)

dREaM

10:27

@Rememberme that's low dude

Balarka Sen

@Rememberme I can only believe that. Once you have proved that it's S_4, compute the order of the group of symmetries of the cube.

You can do that in a similar fashion, but it's harder.

Alyosha

Let $G$ be the additive group generated by 1 and $\zeta_{p^2}^i$ for all $i$ with $(i,p)=1$. Is the intersection $G \cap \mathbb{Z}[\zeta_p]$ equal to $\mathbb{Z}$?

Remember me

Okay Balarka injectivity can be shown by the kernel ( A method which i really like) . So I think that the points of the tetrahedron are independent of each other .. So since they are independent they form the whole of R^3 and due to this fact they can only be mapped by the identity map... Sorry if this is not right..

Which shows the kernel is just the identity subgroup hence injectivity

Surjectivity will require some thinking

dREaM

10:44

wut?

Remember me

What happened @dREaM

Sorry if this is nonsense ...

Your reaction tells me I have to think again...

dREaM

no, sorry. I just didn't understand, although that may be my fault

Remember me

Do you understand now?@dREaM

dREaM

what changed?

Remember me

Non I mean ... Do you get it ...?

dREaM

10:54

uhmm no

although that doesn't necessarily mean it's wrong.

Boni

What are you trying to prove?

That the symmetries of a tetrahedron form a group isomorphic to $S_4$?

dREaM

yeah

I never proved it rigorously when I did that excercise.

Boni

There are theorems to make this easier.

dREaM

about isometries?

Boni

Some basic group theory theorems or theorems on symmetry groups.

dREaM

11:02

but the hard part is the geometric part

I never justified a vertex must map to another vertex under any symmetry

Boni

What is your definition of a symmetry?

Balarka Sen

@Rememberme injectivity/surjectivity of what? you haven't even specified the homomorphism to $S_4$.

dREaM

I guess you could call a symmetry a bijective isometry on $\mathbb R^3$ that maps the length 1 tetrahedron centered in the origin back to itself.

Balarka Sen

The hom sends vertices to the letters in the symmetry group, right?

dREaM

I really don't know though, the book doesn't explain.

Balarka Sen

11:06

In that case, I don't understand your "proof" either.

dREaM

When I proved it I just thought everything was obvious, I just proved that for every permutation of symmmetries we have a symmetry that permutes the vertices that way, so we can associate each symmetry with its permutation of vertices, and composition of aplication of symmetry corresponds to composition of permutation of vertices, so right there you have the isomorphism.

Balarka Sen

Yes, that is possible.

You could show that the homomorphisms $T \to S_4$ (sends an isometry to the corresponding perm. representation of the vertices) and $S_4 \to T$ (sends a permutation to a sequence of isometries that perms the vertices like that) are inverse to each other.

Thus, by general nonsense, $T \cong S_4$

dREaM

by general nonsense?

Boni

"Left as an exercise to the readers"?

Balarka Sen

google "abstract nonsense".

dREaM

11:12

you mean cat theory?

Balarka Sen

yes.

dREaM

Oh, I am an expert at that

being a cat myself

Balarka Sen

maps which are categorical inverses to each other gives an isom between objects

these kinds of things are called "general nonsensical proofs"

dREaM

oh ok.

Remember me

@Balarka yes the homomorphism sends the vertices to elements in the symmetric group

Balarka Sen

11:14

because you don't have to explicitly check injectivity/surjectivity elementwise : you can just use the structure of the objects of whatever category your are on.

dREaM

no.

Balarka Sen

@Rememberme huh?

dREaM

the homomorphism sends SYMMETRIES to elements in the symmetric group.

Remember me

@Balarka I cannot get an idea for surjectivity ?

Hippalectryon

Am I the only one for who the chat's font recently changed to horrible monospace ?

Balarka Sen

11:15

elements of the symmetry group of tetrahedron are permutations of the vertices

NOT the vertices

@Remember You can't get an idea for anything because you don't know what you're trying to prove.

The codomain and domains of your maps are muddled up.

Remember me

Oh.. Okay..

dREaM

@Hippalectryon I have a custom font for this chat, I use Lemon Chicken font

Hippalectryon

q_q how do you change it ?

Remember me

I have to think again as I said

dREaM

The quick brown fox jumps over the lazy dog

Hippalectryon

11:17

:(

dREaM

@Hippalectryon Oh, that's horrible indeed.

is it the same i other parts of the internet?

In fire fox you can go to preferences -> content and then change font.

Hippalectryon

I'm in chrome

dREaM

perhaps someone played a trick on your computer?

Hippalectryon

Anyway, it's the same in the physics chat

Nah only I use it

No one knows my password 0123456789

dREaM

I remember I used to add ponify to my friend's browsers whenever they left me alone with their computers

Remember me

11:22

Every element of the full tetrahedral group permutes the vertices of the regular tetrahedron among themselves. So I send this permutation of vertices to $S_4$.
This way I construct the homomorrphism

Balarka Sen

OK, good. Now you have to show injectivity and surjectivity.

dREaM

dig ding ding

and homomorphicity?

Remember me

I have shown injectivity

Balarka Sen

Of course, you have to show that it's a homomorphism too.

Remember me

okay..

Will this be just product of cycles ?

Balarka Sen

11:25

will have be a product of cycles?

dREaM

How many groups of order $2^n$ have an abelian group of automorphisms.

can the answer be more than $1$?

Balarka Sen

What d'you mean by group of symmetries of a group?

dREaM

yeah, my bad.

Balarka Sen

oh.

dREaM

what is AUT of hamiltonian group?

sorry, quaternion group

Balarka Sen

11:29

S_4.

dREaM

oh yeah

Balarka Sen

there is a geometric proof of this as well

dREaM

of what?

oh ok

Balarka Sen

that Aut(Q_8) \cong S_4

dREaM

mhhm

I think I've seen it at some point

Balarka Sen

11:31

take a cube. mark opposite pair of vertices by i/j/k and -i/-j/-k. The auts of Q_8 are then permutations of the diagonal.

Remember me

@Balarka $\phi(a)\phi(b)$ will be just cycles .. $\phi(ab)$ will be the same cycle

Balarka Sen

proof, proof.

give me a proof.

Remember me

Hmm..

I have gone brain dead for the homomorphism thing

@Balarka I shouldn't be asking this but since I am getting nothing ... Can you give me some hint , I don't know why I am not that interested in this question

dREaM

for the fact it is a homomorphism?

Remember me

Off topic :
http://math.stackexchange.com/questions/1365457/how-do-i-prove-that-rn-setminus-rk-is-homeomorphic-to-sn-k-1-times-rk

How should one visualize it?@Balarka

I am not asking for the proof since I feel it might happen to require fundamental groups

Balarka Sen

11:47

@Rememberme it should be pretty straightforward. where are you stuck in proving that composition of symmetries of the tetrahedron composes the permutation of the vertices?

@Rememberme The proof has nothing to do with fundamental groups. Visualize this with lower dimensions first. Say, $\Bbb R^2 - \Bbb R$. This is just the plane with the x-axis removed.

As $S^0$ is just disjoint union of two points, that's homeomorphic to $S^0 \times \Bbb R^2$

dREaM

@Rememberme Algebraic topology doesn't help to prove homeomorphism of spaces. it helps to prove non-homeomorphism of spaces.

Balarka Sen

For $\Bbb R^3 - \Bbb R$, this is just the $3$-space with the $x$-axis removed.

dREaM

well, at least the homotopy part.

Remember me

Ahh..

Balarka Sen

@Remember Can you visualize the homeomorphism $\Bbb R^3 - \Bbb R \cong S^1 \times \Bbb R^2$?

Remember me

11:51

Yes thinking of the xyz with lets say z axis removed

Balarka Sen

But why is it $S^1 \times \Bbb R^2$?

Hint : look at cross-sections.

@dREaM Not true in general, but ok.

Remember me

Hmm .. It is just the unit circle product the x-y plane ...

Balarka Sen

?

dREaM

well, the homotopy part.

Balarka Sen

Not true either.

dREaM

11:54

really?

how?

Balarka Sen

Whitehead's theorem.

dREaM

how does it help establish homeomorphism?

Balarka Sen

oh, whoops. I thought you said homotopy equivalence.

No, indeed, homotopy doesn't distinguish between homeomorphism and homotopy equivalences.

Remember me

$\Bbb{R^2}$... I know that it doesn't necessarily mean the x-y plane but .. $S^1$ does means the sphere right. So the product of them should be in $\Bbb{R^2}$.

For eg let $(x,y)\in \Bbb{R^2}$ and $(x_0,y_0)\in S^1$ then the product is $\{(x,x_0),(y,x_0),(x,y_0),(y,y_0)\}$ and these all are still in $\Bbb{R^2}$ @Balarka

Balarka Sen

$S^1 \times \Bbb R^2$ is not $\Bbb R^2$

$S^1$ is not the sphere. It's a circle.

Remember me

12:00

I really think today is my bad day.. I cannot think anything :(

Balarka Sen

I recommend you to focus at a single problem instead doing two things at once. You're muddling up trying to do both.

Remember me

I think so...

dREaM

It's vacation time.

Remember me

@Balarka What is the problem with the example I mentioned? Is $(x_0,y_0)\notin S^1$?

Balarka Sen

I don't even understand your "example". I just see written down that "product of them should be $\Bbb R^2$" which is just false.

Remember me

12:05

@Balarka is a circle homeomorphic to $\Bbb{R^2}$?

Balarka Sen

no.

Boni

I think @BalarkaSen is just looking for a intuitive graphical description?

Balarka Sen

I am not looking for anything :P

Boni

As an answer from @Rememberme.

Balarka Sen

?

Remember me

12:07

A circle times the x-y plane... what does it give me....

Balarka Sen

@Remember asked for a way to visualize it, I gave it as an exercise to visualize it for $\Bbb R^3 - \Bbb R$.

Boni

@BalarkaSen Sorry, nevermind; just ignore me. I shouldn't mess up your hints.

Remember me

@Balarka Will it give me something in $\Bbb{R^3}$?

Wait....

Balarka Sen

@Rememberme Consider the 3-space with the x-axis removed. Fill up this space with a bunch of punctured planes perp to the removed axis. So your space is homeomorphic to $\Bbb R \times (\Bbb R^2 - \{0\})$

Can you prove that $\Bbb R^2 - \{0\}$ is homeomorphic to $S^1 \times \Bbb R$? If so, you're done.

Remember me

12:21

@Balarka Something just did strike my brain.. I think I have got a picture and why it has to be homeomorphic...

Balarka Sen

Ok?

Remember me

We have the point 0 in $\Bbb{R}$ , but when we take the product with $S^1$ it will always give me a point which will not be the origin . For eg : (0,2) here 0 is in $\Bbb{R}$ but the cartesian product is not the orgin. This will fill up the entire space but 0 will still remain@Balarka

I think I have hit the nail...

Balarka Sen

I don't even understand what you just wrote.

Remember me

Ahh... Okay once more

Balarka Sen

You have to prove that $\Bbb R \times S^1$ is homeomorphic to $\Bbb R^2 - \{(0, 0)\}$. To do that, you have to write down a map between them. What's your map?

Sorry, but I don't think you understand product spaces well enough.

You should think of product $X \times Y$ as sticking a copy of $X$ at each point of $Y$.

Remember me

12:26

@Balarka I am just thinking of using a picture why it has to be homeomorphic .. I havent written the map yet

If you want I can just cook up a map for you

Balarka Sen

OK, then what is your picture? Start with R^2 - {0, 0}. Where are S^1s?

D'you know how $S^1 \times \Bbb R$ looks like?

Remember me

I am starting with R \times S^1

Balarka Sen

Ok, then tell me how it looks like.

Remember me

So since it is the product I will think of every $x\in S^1$ with $R$ . So let us say some $y\in S^1$ and we know that $0\in R$ . But when we take the Cartesian product of these two it will be (y,0) . Similarly if I take any point in $S^1$ and think of its cartesian product with 0 it will never give me the origin.
But if I take any other point in $S^1$ and any other point (except 0) in R it will give me the whole of $R^2$ though except 0 whose reasoning I gave above

Balarka Sen

Just tell me how $\Bbb R \times S^1$ looks like, @Remember

If you can't tell me that, there's not much point doing all this.

Remember me

12:32

Because of my reasoning above it will look like $\Bbb{R^2}-\{0,0\}$ because it will never give me the origin@Balarka

Balarka Sen

no, there's a simple, very easy description

Remember me

I have given you my idea.. Is my idea wrong?

Balarka Sen

if you understand product spaces well, you should be able to answer that

Your idea is not really much sensible. You're thinking of $\Bbb R \times S^1$ as product of their underlying sets, forgetting completely about the topology.

@Remember Revise product spaces.

Get through some concrete examples.

Remember me

The plane minus the origin can be written as

$R^2\setminus\{0\}=\{(r\cost,r\sint):0<r<\infty,0≤t<2\pi\}$

Balarka Sen

I don't want an algebraic manipulation. I want a picture of $\Bbb R \times S^1$

sigh I really don't have the time to do this. Since you don't know, $\Bbb R \times S^1$ is a cylinder.

Remember me

12:36

What ?? Shock!!

Balarka Sen

If you're shocked by that, then you definitely do need to relearn product spaces.

Remember me

How?

Yes I have to do it again..

Balarka Sen

$X \times Y$ is, as I said above, the topological space obtained from sticking a copy of $X$ at each point of $Y$. $\Bbb R \times S^1$ is then a copy of $\Bbb R$ stuck at each point of $S^1$.

Remember me

Where $R$ is given the discrete topology ?

Balarka Sen

NO

Remember me

12:38

Or subspace

Balarka Sen

huh?

Remember me

I am really going crazy today

Balarka Sen

This is a result of jumping through several things at once and not really understanding any of them.

$\Bbb R$ has a topology of itself.

Remember me

Standard topology

Balarka Sen

And what is that topology?

i.e., what are the open sets?

Remember me

12:39

open intervals

Balarka Sen

right.

Remember me

i should have seen that

Balarka Sen

so anyway, $\Bbb R \times S^1$ is a real line stuck at each point of $S^1$.

Remember me

Which makes a cylinder

Balarka Sen

Or equivalently, a copy of $S^1$ stuck at each point of $\Bbb R$. You can visualize that as stacking circles over one another.

Yes, then that's a cylinder.

It's actually an open cylinder, i.e., cylinder with two ends heading off to infinity.

Mary Star

12:41

Hello!! Is someone of you familiar with existential theory?

0

Q: Existential theory

I am reading the following lemma about (positive) existential theory: Could you explain to me the last proposition? Why does this hold?

Remember me

Then as an exercise can I take $S^2\times R$?@Balarka

dREaM

@Rememberme how old are you?

Balarka Sen

You can show that it's homeomorphic to $\Bbb R^2 - (0, 0)$ by inflating the top and deflating the bottom so that the bottom hole becomes a removed point and the top hole is just very big.

Now project downwards to $\Bbb R^2$.

Done.

Remember me

But what is the problem in just thinking it as the punctured plane

Balarka Sen

@Rememberme $S^2 \times \Bbb R^2$ is too hard to visualize.

Remember me

12:43

$S^1 \times R$ is also the punctured plane

Balarka Sen

You have to prove that it's a punctured plane, so assuming that is the punctured plane is not of much use.

Anyway, I am abandoning this conversation. I have wasted the whole afternoon.

Remember me

Eesh .. disgust on myself

Sorry @Balarka ...

dREaM

:)

Remember me

@dREaM 15

dREaM

@Rememberme Can you prove a continuous function $\mathbb R \rightarrow \mathbb R$ is integrable?

without google?

Remember me

12:46

Yes why not... But I don't want to ... okay let me give you the skeleton of the proof

dREaM

ok

what happened?

Remember me

Okay got momentarily disconnected... On WIFI

Okay since f is continuous on [a,b] implies it is bounded

dREaM

ok

Remember me

this implies f has an upper integral and a lower integral

dREaM

ok

Remember me

12:54

if I can show that the upper integral and the lower integral are equal then the function is integrable on [a,b]

dREaM

ok

Remember me

Choose an integer $k=\dfrac{1}{m}$ , by small span theorem there has to be there a partition for it

of [a,b] obviously

dREaM

what is $m$?

Remember me

Okay ya, $m\geq 1$

Going well? @dREaM

dREaM

So you choose $k=\frac{b-a}{m}$? Or am I not understanding?

to make partition $\{a,a+k,a+2k,\dots ,a+mk=b\}$?

you lost me in the last sentence

Choose an integer k=1m , by small span theorem there has to be there a partition for it

of [a,b] obviously

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