Mathematics - 2016-06-29 (page 3 of 4)

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6:27 PM

@PaulP Hi.

@BalarkaSen Hello

@BalarkaSen hii

Hi @Anubhav

what's upp

Not much.

6:28 PM

slow internet :(

What are you upto, @PaulPlummer?

Figuring out what I wan't to do today, I think I am going to start studying some analysis (I have a qual exam in about 6 weeks) @BalarkaSen

How about you

Ah, good luck on the qual.

I am not really doing anything productive.

Yah, I am not super worried about it, but I definitely need to study for this one. Although right now I wan't to figure out a "different way to study" for it, I don't just want to read a book and look at past quals. I wan't to find something that would get me interested and excited, about it

(maybe find an interesting problem or an avenue that captures my interests)

Sounds like a good idea.

6:37 PM

@BalarkaSen That has basically been the past two weeks for me

My knowledge of analysis (esp real analysis) is scattered and unsystematic. I started (re)learning some complex analysis a couple weeks ago but stopped.

expected

I have a hard time making sense of a lot of the inequalities that come up, so I think that will be one of the goals in my studying, trying to get a good intuition for them

@user349357 huh?

Yeah, making estimates are not something I am good at.

I meant aren't you a high school student? You cannot expect to know all subjects thoroughly before having studied at university

@PaulPlummer what inequalities are you referring to?

6:47 PM

I do not expect to be a know-it-all. I learn math because I like to learn.

Basically all of them (Holder etc)

all of them"

well in analysis there are very technical ones

but hölder is not one of them

many can be viewed as " interpolation inequalities"

it is also just a generalization of Cauchy Schwartz

@PaulPlummer I forget what Holder says. $\int |fg| \leq \sqrt[p]{\int |f|^p} \sqrt[q]{\int |g|^q}$, yeah?

Yah

i would just learns ome applications of it

in the end it all about the applications

Akiva Weinberger

6:58 PM

Not all immersions of the sphere in $\Bbb R^3$ have a regular homotopy to the usual embedding, right? (Seriously hope those are the right terms)

@PaulPlummer Kinda sorta sounds like AM-GM to me. Do something to the $|fg|$ term, I guess.

Akiva Weinberger

Kinda like how there's no regular homotopy to from a figure-eight to the usual circle in the plane

interesting application is that if the measure space is finite, then Lp lies in L1

No, every immersion of the sphere in $\Bbb R^3$ is regularly homotopic to any other.

Akiva Weinberger

Hm.

7:00 PM

if you want to learn about inequalities stuff, "The Cauchy-Schwarz Master Class" is really really good

quite readable

Akiva Weinberger

Consider the ⌘ shape, rotated around its horizontal line of symmetry

@Semiclassical Yah, I was thinking about looking into that book, it does look good.

I guess not every immersed sphere in R^4 are regularly homotopic to the usual one though. Take a self-transverse one with nontrivial self-intersection number, say. (is that right?)

Still regularly homotopic to the standard immersion. I couldn't give you an explicit homotopy.

Akiva Weinberger

Right, I just can't visualize it

I'll think about it

7:02 PM

Presumably that corrugatkon method will be helpful.

to understand the hölder equality

it is perhaps psychologically helpful to know that there are much worse inequalities

Gagliardo–Nirenberg interpolation inequality

In mathematics, the Gagliardo–Nirenberg interpolation inequality is a result in the theory of Sobolev spaces that estimates the weak derivatives of a function. The estimates are in terms of Lp norms of the function and its derivatives, and the inequality “interpolates” among various values of p and orders of differentiation, hence the name. The result is of particular importance in the theory of elliptic partial differential equations. It was proposed by Louis Nirenberg and Emilio Gagliardo. == Statement of the inequality == The inequality concerns functions u: Rn → R. Fix 1 ≤q, r ≤ ∞ and a natural...

The Smale-Hirsch theorem says that immersions satisfy an h-principle, meaning that the space of immersions is homotopy equicalent to a space of sections of a bundle (which can be calculated via algebraic topology). It is rarely possible to get good visualizations of results that come from an h-principle; we are indeed lucky that we can see it explicitly for sohere wversion.

@BalarkaSen Maybe, idk. I think I need to think about them more, I can follow a proof of inequalites, but that hasn't really translated to understanding why it works or what it means so far.

Akiva Weinberger

Sphere eversion still breaks my brain

the book I mentioned does discuss Holder, btw

7:05 PM

@Akiva If what I said was right, then it should at least partially explain why the figure eight in plane is different than some weirdo immersion of S^2 in R^3 ('cause the same argument doesn't generalize).

You might look at the first chapter of Eliashberg-Misachev to see how strange h-principles can be. Avoid the proofs if you like, there are plenty of good examples of what h-principles can do for you.

Namely, "self-intersection number doesn't make sense for an immersion of a 2-manifold in R^3".

I'd quote some but I don't have my copy.

@user349357 If you think that is a good application to get a feeling, then I guess I can think about it. I don't really find "it is just a generalized C-S" to be that helpful.

I do not claim I know how to prove that every immersion of S^2 in R^3 is regularly homotopic to the standard one though.

7:09 PM

@PaulPlummer why not? are you comfortable with C-S?

Hölders inequality can also be viewed as saying that a certain multiplication operator is continuous

Akiva Weinberger

I wonder if there's a good way to build sphere eversions in Minecraft (where everything is just made up of cubes).

if you feel this is more intuitive

in the end i think you should try to do some proofs on your own to get a good feeling for inequalities

@user349357 This is analogous to how people didn't find your "sheaf cohomology helps prove de Rham's theorem" helpful.

Akiva Weinberger

If so, then sphere eversion would be a finitistic phenomenon, because only finitely many cubes would be involved.

hullo

Akiva Weinberger

7:10 PM

Hyello

@BalarkaSen well don't you think that it is often useful to view something as a generalization of something simpler?

at least psychologically?

well i agree that it might not be helpful for intuition

but it gives a motivation to learn these things

and that i find very important in learning math

I don't really want to talk about pedagogy right now. But I am just pointing out which of your things people aren't finding helpful, because you argued to the point of being rude that day about your point of view.

Quick Question: When self-studying math how strict should I be on reading through every proof?

@user1618033 I can't actually prove it, but numerical experimentation makes me think that the term in parens is $1/(n+1/2)$. if i assume that, then using mathematica to sum that up gives $4\log 2+2\pi+8C-16$ where $C$ is Catalan's constant

Hi @Danu.

7:14 PM

@Stephen In my opinion, very, unless you don't really mind forgetting what you read.

I've read some books without looking at all the details and I've forgotten most of everything in them.

Also, exercises (I know, it sucks to have to do them, but it's such a big difference!!!)

@Danu That's what I have been doing but sometimes it takes up an obscene amount of time

@Akiva That would be more in line with studying (locally flat) PL immersions, which is not the same story. Maybe it's a homotopy equivalent space in this case; that's probably true. I don't know off the top of my head.

not sure i'd call the above a 'derivation' of the sum, but i think it does suffice as a numerical computation of it :p

@user349357 I am somewhat comfortable with C-S, it isn't at the forefront of my mind though, and I have been away it is a generalization of C-S, doesn't mean that it makes the inequality click with me (basically what Balarka said). I completely agree I should go through proofs on my own and apply them to get a feeling, which is part of what I am going to do. The continuous function perspective sounds interesting, and maybe more up my ally.

I don't think my "problems" with inequalites is going to be cleared up in a one or two sentence statement, it is something I am going to work at.

That's another selling point of the book I mentioned: it has exercises at the end of each chapter

(and answers in the back, but one should of course resist the temptation to look at those)

7:22 PM

Is $TS^2$ diffeomorphic to a manifold I know?

What does the $T$ stand for?

Tangent bundle.

...I know what $ST^2$ would mean... :D

Oh, haha

@BalarkaSen No.

I think I "heard" about it's diffeomorphism type being computed before. Like, look at the unit tangent bundle sitting inside. I vaguely recall it being RP^3.

@MikeMiller Ah, ok.

I guess I misremembered.

7:24 PM

The unit tangent bundle is $\Bbb{RP}^3$, yes.

Hmm.

@Danu WHat about exercises after the chapter?

important

but all of them?

That's sort of weird. So it's double covered by the Hopf bundle over S^2. I am trying to figure out if I can see it by looking at the circle fibers.

7:29 PM

Prove that it's $SO(3)$.

To be clear, prove what is $SO(3)$?

Ah. I sort of see it. I'll ponder a bit more.

@Semiclassical The unit tangent bundle on S^2.

ah

keeping mouth shut

Tobias Kildetoft

Hi @TedShifrin

7:34 PM

hi @Tobias

and continuing my ignorance, what "it" was double covered by the Hopf bundle over S^2?

Hi @Ted.

G'night @MikeM

hi @ted

rehi @Semiclassic

Heya @AlexW !

7:35 PM

Hey @Ted! :)

Hi @AlexW

Hi @AlexWertheim

Hey @Mike, @Balarka

Feeling joyful?

Hi @TedShifrin

7:37 PM

hi @Danu

@Semiclassical RP^3, which is seemingly the same as the unit tangent bundle on S^2.

Hehe. I've got joy, joy, joy down in my heart...

I've started attending the Riemann surfaces lectures by Forster now; the exam is next week and they finally got past the point where I'd read the book already.

Today, we actually came back to that motivation thing you were on about: We discussed Mittag-Leffler distributions

Ah, cool.

and how they have solutions (does that actually mean they're integrable?) if and only if their coboundary is trivial in $H^1(X,\Omega)$

7:38 PM

I would not refer to this as integrability, no.

Tobias Kildetoft

@Danu I did not know there was a thing called that. To me, Mittag-Leffler is just a nice place to hang around and do some math

I was wondering in what sense these distributions are distributions; I only know distributions as subbundles of the tangent bundle

@TobiasKildetoft Wut?

Distributions are also "generalized functions."

Ah, in that sense

i'll admit, i'm secretly trying to see if there's a physical way for me to argue that that should be $SO(3)$ :>

7:39 PM

I know those kinds of distributions all too well, from physics.

Tobias Kildetoft

@Danu The Mittag-Leffler institute near Stockholm

ah...

though when I say physical I really mean 'spin'

I'm not sure why Forster uses that language. Maybe just a naïve notion of distribution in terms of spreading around ...

So what does the "solution" to a generalized function mean? :P

7:40 PM

Looks like a nice place indeed @TobiasKildetoft

@TedShifrin Hmm... Okay.

@AlexW: Did you pass an exam or something?

Tobias Kildetoft

@AlexWertheim It is fantastic

Quals are in a couple months. We're just chatting.

Um, OK.

7:41 PM

@TedShifrin: not since last spring. ;) Mike and I are making references to a game =P

What was the book you recommended me earlier? I have a few minutes to check it out.

OK, @AlexW, sorry to have interloped :)

@MikeMiller Is that directed at me?

Yeah. I think I found them, though.

I mentioned a lot. Kobayashi, Narasimhan, Krantz, and Griffiths.

OK, I found the first and last.

7:44 PM

I want to say that given a point p on S^2, I can take a unit tangent vector v at p, and rotate S^2 by moving p along the great circle in that direction which gives me an isometry on S^2. Or something.

clears throat ... Since you mentioned isometry, @Balarka ...

Oh, this is all the Kobayashi metric stuff. A deep set of ideas I don't think I have any idea about.

@TedShifrin I am not sure if I get the reference.

Ah, right. @MikeM ... the others definitely talk about negative curvature explicitly.

Ah, this is Griffith's generalized Nevanlinna theory you told me about.

7:46 PM

Yes, @MikeM.

@Balarka: Is $S^2$ a homogeneous space?

@user1618033 oh wow I think that approach is much more intuitive than isosceles triangles. Then yes as leaky put it (I think) the arc length is given by $s = r\theta$ and the height is simply $r$. Though it goes down to isosceles triangles when it becomes an infinitely small arc length. This reminds me of the centripetal acceleration proof I did in physics earlier this year.

@TedShifrin: Are there complex structures on $\Bbb C^n$ that don't arise as open subsets of $\Bbb C^n$?

@TedShifrin What's a homogeneous space, again?

@MikeM: I actually have no idea if there are nonobvious complex structures on $\Bbb C^n$.

A space with a transitive Lie group action on it, @Balarka.

I googled a little and all I could find is people finding domains with strange properties.

7:48 PM

@Obliv Have you ever heard of Hadjicostas's formula? mathworld.wolfram.com/HadjicostassFormula.html

no

The desired argument is as follows. Suppose $M$ is a Kahler manifold with nonpositive curvature. Then the universal cover is $\Bbb R^{2n}$; and our charts then cover the universal cover with copies of $\Bbb C^n$. If it's, say, a polydisc, then we've reached a contradiction.

Hmm. No, I don't think so. Any isometry of S^2 fixes an axis, no?

Yes.

@Balarka: So what? Isn't there an obvious group action?

@MikeM: I know the famous result that balls and polydisks are not biholomorphic. We also know all of $\Bbb C^n$ can be biholomorphic to neither.

7:51 PM

@TedShifrin Oh, sorry. Yes, given any two points, I can just join the two by a great circle and rotate.

@Balarka: Much easier. It's amazing how you put geometry everywhere now. Think of a linear transformation.

@TedShifrin Right, that's what I'd like. I don't think $\Bbb C^n$ is biholomorphic to any open subset of itself, is it?

@user1618033 so you want to calculate the integral of a function in rectangular coordinates by using polar coordinates? That's not very hard. You just need to convert the coordinates into polar form, no?

I've not thought about this before, @MikeM, but I presume that we could make some argument with Liouville's Theorem to prove the restriction to any complex line would be constant.

@Obliv Well, yeah, but this simple idea was never used before there (at least from the information I have) and all becomes very nice. Also a cool generalization can be done to extend that formula using some variable.

7:54 PM

@MikeM: I dunno about any proper subset, but bounded subset ...

@Ted: It might also be a nice source of examples that Kobayashi hyperbolic spaces cannot satisfy my property, because they don't even have any maps from $\Bbb C$, much less $\Bbb C^n$. :P

Right, that's the sort of thing I had in mind yesterday, @MikeM.

I see.

I guess one can probably also argue that Gromov hyperbolic Kahler manifolds don't satisfy my property.

Because then there's a Kahler metric on $\Bbb C^n$ whose Kahler form has a bounded antiderivative, and I don't believe that.

@TedShifrin Sure, I can write down a rotation matrix corresponding to that. I don't see how that's easier than the picture of the isometry, though.

@TedShifrin I'll admit, my temptation when seeing the questions you're asking Balarka is to think of the Riemann sphere

7:57 PM

So $S^2 = SO(3)/ ?? $ @Balarka

though that's probably overthinking it :/

Not sure how that helps, @Semiclassic.

SO(2), @Ted.

well, representing certain transformations of the Riemann sphere as linear fractional

Of course, @Balarka. Now do you see how this tells you $SO(3)$ is the unit tangent bundle?

7:58 PM

...meh, i dunno where I'm going with that.

In $\mathbb{Z}/48\mathbb{Z}$ write out all elements of $\langle \bar{a} \rangle$ for every $\bar{a}$ Is this asking me to write out the generators of this group?

Tobias Kildetoft

@Obliv No, it is asking you to write out the elements of those subgroups

Hmm. So you want me to see why SO(3)/SO(2) --> S^2 is the same map as projection of the unit tangent bundle to S^2 (i.e., forgetting the circle fibers).

Give me a minute.

i'm not doing this then. too many subgroups

Tobias Kildetoft

@Obliv most of them coincide

8:02 PM

there should only be $\phi(48)$ of them, if I remember my stuff

Tobias Kildetoft

@SamuelYusim no, that is the number if generators, not the number of subgroups

@TobiasKildetoft It's still a silly exercise, I think

ah

@Ted I don't think what you've said is sufficient to conclude.

Tobias Kildetoft

@Krijn now that I do agree with

8:03 PM

Conclude which, @MikeM? I'm not sure.

I think I'm claiming by Liouville that any holomorphic map to a bounded region must be constant.

oh because most share common G.C.D's @Tobi

Tobias Kildetoft

@Obliv because the group is cyclic, so it has a unique subgroup of any order dividing the order of the gorup

Some day I need to write something in which I define something called a gorup, just so my typos will obtain meaning

6

haha didnt even notice the mistake

It had better be a universal object in the category of typo-ed creations, @Tobias :P

Tobias Kildetoft

@TedShifrin Of course.

8:07 PM

I think it should be univesral instead

It's a nice categroy

depensd on hwere the typso aer...

Any typo factors through the categroy of gorups

There are typos that are not rearrangements :P

pfftf

8:08 PM

Your lying

Their is no way that works

christ

The universal its, it's, they're, there, their, ... problems notwithstanding.

@TobiasKildetoft Gorup sounds like a beast from a fairy tale.

Ali G - RESTECP! (from Indahouse)

Tobias Kildetoft

8:09 PM

So on those that are rearrangements, we have an actoin of the symmetirc gorup

lol

@TedShifrin Also, "principle" and "principal"

@BalarkaSen Indeed, the geometry books/articles that misspell principal drive me nuts.

"principle bundle"

not to mention principle curvatures

so you desire a principled use of the word principal :p

8:13 PM

In principle, yes.

Though I guess if the work is otherwise good, the principal it represents makes for the debt incurred with your principal-principle.

Everyone is clearly way too bored today.

god yes

how am I to use an isomorphism defined by mapping a generator of one group to the generator of another group to list all subgroups of one group?

though i do like the phrase 'principal-principle'

8:16 PM

specifically $Z_{48} = \langle x \rangle$ isomorphic to $\mathbb{Z}/48\mathbb{Z}$ by $\bar{1} \to x$

whereas i guess 'principle-principal' would be something a politician used to have :p

@Ted: I stopped by because I'm waiting for a package, and of course they gave a long range of delivery hours. Must be something in the air today =P

I JUST GOT ANOTHER AMAZINNNNNNNGGGGGGG RESULT!!!!!!!!!!!

@TedShifrin I'm not bored at all, I have just derived such marvellous stuff!! Am I so lucky or what??

Time for a party! I really didn't expect this new way I created will lead me so fast to the desired result!!

BBL

Huh, is that Chris's Sis? (or something like that, been a while, forgot names)

yeah

8:31 PM

@PaulPlummer I'm just user 1618033

BBL

@user1618033 Okay, fair enough

Hello.

@TedShifrin Ohh, I see. SO(3) acts on S^2 by isometries: clearly such an action induces an action of SO(3) on the unit tangent bundle on S^2 - multiplication by an element is moreover transitive on the unit tangent bundle because only place it can mess you up is at the axis/poles it fixes, but it clearly rotates the unit circle at those points. So multiplication by an element of SO(3) should definitely be a diffeomorphism w/ the unit tangent bundle on S^2.

No wait, that's nonsense and not what I meant.

SO(3) acts transitively on the unit tangent bundle on S^2. That much is ok. So fix an element of the unit tg bundle on S^2, say (p, v). Multiplication of elements of SO(3) with (p, v) gives a diffeo with the unit tg bundle on S^2, I believe.

what

whuh-oh

8:36 PM

@PaulPlummer here is a more appropriate answer

I fixed nonsensical writing. See if it's ok now.

Ok so you've said a true fact with a completely unconvincing proof. :)

I'm the one that knows to very nicely give the answers to questions like What are the closed forms of? $$\sum_{n=1}^{\infty} (-1)^{n-1} \left(\frac{H_n}{n}\right)^3$$ or $$\sum_{n=1}^{\infty} (-1)^{n-1} \left(\frac{H_n H_n^{(2)} H_n^{(3)}}{n}\right)$$ @PaulPlummer

Why does it act transitively? Why does it act freely?

This description is better than Chris's Sis or user 1618033 @PaulPlummer

8:39 PM

Suppose I have an isometry of S^2. There are only two points it can fix, the codimension 2 sphere it fixes inside. So tngnt vectors of only those points you have to worry about fixing if you worry about freeness. But clearly an isometry of S^2 would rotate the circle fibers at those points. That proves freeness, right?

BBL

As for transitiveness, I have (a, v) and (b, w) on the unit tg bundle on S^2. First take a great circle from a to b, rotate along that to get from (a, v) to (b, v). Then rotate around the axis passing through b to get from (b, v) to (b, w).

transitivity*

Thanks.

I hope what I said is not nonsense.

i get this feeling that most of the time, as much I worry about not being able to come up with proofs, I worry about coming up with nonsense proofs. :(

I forgot what the number denoted, I read it the other day but cannot find it. Can someone tell me what the 2 in: $\lvert \lvert r_{i} \rvert \rvert _{2}$ means?

8:49 PM

2-norm?

Yes. Thanks.

Hey, anyone here good with number theory?

What does "The multiplication of nonzero residues mod p is the same as addition of all residues mod p−1. " mean

9:04 PM

Do you know the isomorphisms between the multiplicative group mod p and the additive group mod p-1?

haha, I'm still in high school. No, I don't

Right now, I'm reviewing the transcript for an intermediate number theory class I had. The topic is primitive roots

Well you should probably provide some context, and maybe read the transcript more carefully, to so if it is expected for you to know that and if it is just pointing something out. Also I probably won't be able to help out, in the middle my own stuff

@widow did you mean $\mod (p-1)$ or $\mod p ... -1$?

Hmm, it seems I gotta read Solzhenitsyn someday.

@Obliv I just copy pasted it...

More context:Let me make that much clearer: The multiplication of nonzero residues mod p is the same as addition of all residues mod p−1. This correspondence is exactly what primitive roots give to us. The significance of this observation is unlikely to have fully sunken in at this point, but hopefully the next few bits of discussion will get you closer.

9:13 PM

the 'all' part is throwing me off.

So given a set $\mathbb{Z}/5\mathbb{Z} = \{\bar{1},\bar{2},\bar{3},\bar{4}\}$ from what I gather it's saying $1*2*3*4 \mod p = 1+2+3+4 \mod (p-1)$

i don't see that being true lol

wait a second let me read the rest of the transcript

@MikeMiller I am not sure if you saw my proofs of freeness and transitivity above.

That sounds like the following statement that occurs in the Wiki page on units in ring theory: "In the ring $\mathbb{Z}/n\mathbb{Z}$ of integers modulo $n$, the units are the congruence classes (mod $n$) represented by integers coprime to $n$. They constitute the multiplicative group of integers modulo $n$."

oops. Ring theory is in abstract algebra. Sorry, beyond the scope of my mind

that doesn't give you the n-1 anyways, so i'm not helping

is $p$ supposed to be prime, btw?

9:23 PM

yeah I think

mmkay

lol sorry if that wasn't clear

ituba 2016-05-05 20:31:05
In terms of our primitive root 3, how do we express 5 and 6 in mod 7?
ituba 2016-05-05 20:31:25
We have 5≡35(mod7) and 6≡33(mod7).
ituba 2016-05-05 20:31:36
Then 5⋅6≡35+3≡38(mod7). And what's that?

hmm. so for $p=5$ that'd be $(1)(2)(3)(4)\mod{5} = 4$ and $0+1+2+3\mod{4}=2$

yeah I know it doesn't make sense urgh

5=35 mod 7???

actually $3^5=243=7*30+33=7*34+5$

so presumably he means $5=3^5$ and $6=3^3$ mod 7, which are both true

and then the last statement is just $5\cdot 6=3^5\cdot 3^3 = 3^8$ mod 7

which is a heck of a lot more sensible

9:30 PM

hi

user image

and which would require $3^8=30=2$ mod 7, which is consistent with Euler's theorem since $3^8=3^6\cdot 3^2$ and $3^2=2$ mod 7

Can someone explain how in that it is easy to see there exist corresponding $a,n$?

@WidowMaven So that does make sense if one properly interprets the (terrible) notation

yeah, sorry

np

9:34 PM

it actually looks better with latex on my screen. I should've reformatted it

But my question is how does this tie in to the first statement?

Like, yeah I know that 3 is a primitive root of 7

*mod 7

do not know

hmm, maybe

actually, i think i see it

and then basically you can write the multiplication of any nonzero mod 7 residues in terms of powers of 3

but so what??

the point is that $3^1 3^2 3^3 3^4 = 3^{1+2+3+4}=3^{10}$

but by Euler's theorem, we know that $3^4=1$ mod 5

huh

hmm ok

wait why is mod 5 here

because that's the mod we used earlier

9:36 PM

? I thought we were talking about it in mod 7

as in 3 is a primitive root of mod 7

i'm referring back to when Obliv stated it

but, let's do the case of mod 7

ok

in that case a primitive root, as you say, is 3, and the nonzero residues mod 7 are 1,2,3,4,5,6

now, suppose we consider the elements 3^1, 3^2,...3^6 and multiply them together

ok and their product is just $3^{1+2+3+4+5+6}$

right

9:38 PM

yeah

but we know 3^6 = 1 mod 7 (Euler's theorem)

right

...is that Euler's theorem or Fermat? ugh, I can't remember

both actually

Euler's is generalization of Fermat's but in this case because 7 is prime it's fermat also

anyways, therefore we can write 1+2+3+4+5+6 = 6+6+6+3=3 mod 6

right, Euler is the one with totient

9:39 PM

...

yeah ok

oh I see and then yeah

hmm wait

is it saying that 3^3 mod 7 is the same as 1+2+3+4+5+6 mod 6?

nah,it's saying that 1+2+3+4+5+6 mod 6 = 3 and therefore 3^(1+2+3+4+5+6)=3^3 mod 7

more generally, suppose x+y=z mod 6

wait so

then (3^x)(3^y)=3^(x+y)=3^(z+6n) for some positive integer n (taking z to be a residue)

but 3^(6n)=(3^6)^n=1^n =1 by Fermat

basically take 1+2+3+4+5+6 mod 6 and that's your exponent of 3 when multiplying them mod 7?

so therefore (3^x)(3^y)=(3^z) mod 7

right. so instead of having to add them together, then take a huge power of 3 and mod by 7

you can instead add them together, mod by 6, then take that much smaller power of 3 and mod by 7

9:44 PM

:( @semiC can you help me determine when a homomorphism exists between $\mathbb{Z}/48\mathbb{Z}$ and $Z_{36}$ given by $\bar{1} \to x^a$? this solution crazyproject.wordpress.com/2010/01/26/… is so hard to follow and I'm sure there's a more attainable explanation.

wow ok that doesn't really sound like what the statement said but ok

yeah, it's not terribly well-stated

Also how do I render Latex in this chat??

see the link in the room description

$\varphi(a\star b) = \varphi(a) * \varphi(b)$ is my starting point for determining if a homomorphism exists between two groups but apparently that's not the right route

9:46 PM

I read it but I don't have a bookmark bar and I can't right click to add it on chrome browser?

hmm

get a bookmark bar

in settings in the top right

yeah haha whoops

yeah, that's really the easiest way :p

damn I'm out of time. I'll try this again later

9:47 PM

i can't really help with the homomorphism, sorry. best I can think of is that, when you want to disprove the existence of a homomorphism, you look for a property that'd be preserved by that mapping and show that it holds for one but not both of them

anyways, @WidowMaven, here's how I read their statement

suppose I have p=7 and I know that 3 is a primitive root

then any of the nonzero residues can be represented by a unique power of 3. so for instance one has 3^0=1, 3^1=3, 3^2=2, 3^4=6, 3^5=4, 3^6=5, and after that it loops back

yup

woops, should've been 3^3=6, 3^4=4, 3^5=5, and 3^6=1.

that makes more sense, heh

also I would appreciate it if you could proofread this proof:

latex.artofproblemsolving.com/miscpdf/…

anyways, i can therefore represent (5)(6) as (3^5)(3^3)=3^8

oh wait forgot to include the problem

9:56 PM

and then use FLT to say that 8 mod 6=2 and so 3^8 = 3^2=9=2 mod 7

Show that the cube roots of three distinct prime numbers cannot be three terms (not necessarily consecutive) of an arithmetic progression.

and indeed (5)(6)=30=2 mod 7

how can prime numbers have cube roots?

I mean, why not?

They don't have to be integers

fair enough

arithmetic progression differences also don't have to be integers

9:59 PM

i think i get it, actually

say the three primes are p<q<r

oh

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