Mathematics - 2018-09-26 (page 3 of 3)

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7:00 PM

I wanna show that the characteristic map from interior of n ball is homeomorphisim to its image , can I give Xn the subspace topology to show the characteristic map is an open map onto its image ?

I saw an interesting article that day about how all languages call tea in essentially one of two ways: something that sounds like tea, or something that sounds like cha, and both ways essentially come from Chinese @LeakyNun, because teh is tea in Hokkien and cha is tea in Mandarin.

interesting!

Of course, we are just saying that the action of $\Bbb Z/3$ on $\Bbb R^3$ given by permuting the coordinates splits as the sum of a 2-dimensional irreducible representation (the subset of sum-0) and the trivial representation (the possible sums)

It's a quadratic equaton right? So one can complete the square and use the 2a+b+c=0 thing to work out where the roots will lie with the help of the discriminant?

@JasperLoy I see

7:00 PM

@TedShifrin As a HW problem for someone reviewing a lot of different math subjects, I find this problem artificial but not terrible. Being able to plot stuff in Mathematica helps

interessant!

As a test question, though, it's stooopid

@JasperLoy and of course, the sound in "tea" before the Great Vowel Shift would have been exactly like Hokkien

@Secret could, yeah

@MikeMiller heh...

7:02 PM

but the IVT approach is simpler

@Semiclassical Is there any other approach too?

@LeakyNun i got confused for a half-second about why an abelian group had a 2d irrep

but only a half-second

@Secret My favourite text editor for TeX is TeXworks. =)

@Abcd i dunno, probably

Torskun diert ba man comm goa

7:03 PM

@MikeMiller because R isn't algebraically closed :P

we found a method that worked and went with it

@JasperLoy Processing 3 is not a text editor, it is a programing console. But sure, I use texwork and texlive for my LaTex report typing

@LeakyNun whence half-second :)

@JasperLoy @LeakyNun The one from Chinese that really surprises me is "Ketchup" dictionary.com/browse/ketchup?s=t

that's Cantonese, my native tongue :P

7:04 PM

now let's make sure abcd writes the fact down on his homework that Z/3 has a 2d real irrep

truly crucial

@JasperLoy I tend to use TeXworks offline as well

Tobias Kildetoft

@TedShifrin Apparently my TAs had a bit of trouble solving one of the exercises I had posed. It was showing that finite abelian groups satisfy converse Lagrange

Que whal sur jo?

@TobiasKildetoft I could do that

@Alessandro: I told you so!

7:05 PM

but that's mostly because the other tex editor I downloaded ends up crashing pretty much immediately

@LeakyNun Are cantonese speakers really aware "ketchup" originates from a chinese language?

sure

Oh, that's some sort of induction problem, @Tobias. I've done it before.

@rschwieb If you are interested in etymologies, etymonline.com is a great free resource.

highly skeptical

Tobias Kildetoft

7:05 PM

@TedShifrin Yeah, exactly.

@JasperLoy Yeah, I've seen it berore

I have solved the catalan conjecture

Really? Well, good for you.

Thanks

Wiktionary is actually very well-developed

7:06 PM

@JasperLoy @rschwieb vocaroo.com/i/s10V8449ltuA

Tobias Kildetoft

I had even given a hint to use induction and the correspondence of subgroups with the quotient

welcome to the chat preda mihăilescu

@JasperLoy I always find it funny when there are familiar words whose etymology isn't known

like strawberry

I'm a bit surprised your TAs didn't get it, @Tobias — not surprised the students didn't.

@TobiasKildetoft oh, we aren't allowed to use classification?

7:07 PM

@MikeMiller Americans would like merriam-webster.com while British would like en.oxforddictionaries.com.

Tobias Kildetoft

@LeakyNun No, that would make it trivial

:(

no one really knows where the word strawberry comes from

well what if I prove it?

Tobias Kildetoft

@TedShifrin Yeah. Possibly they just hadn't thought that much about it, since the exercises are usually much easier, so this harder one caught them off guard

7:08 PM

wasn't it just invented?

a berry from the straw fieldsd

that is red

strawberry.

nope

blog.oup.com/2015/04/strawberry-word-origin-etymology

user image

ok I am off, this is not going to finish in 2 hours and it is already 5:07

secret

sleep is important

However, I find that the dictionary with the best definitions is the American Heritage Dictionary of the English Language, at ahdictionary.com.

Tobias Kildetoft

@LeakyNun In this case, the students have also not seen group actions yet

7:10 PM

what is the etymology of conglomerate

Well at least I got 70% of the introduction done, so that will make me less likely to procrastinate on my thesis

Another big surprise is the surname Yuengling which looks and sounds chinese but is a weird anglicization

@Ultradark etymonline.com/word/conglomerate#etymonline_v_18204

what is the origin of species

@Ultradark It is a book by Darwin.

7:12 PM

can we stop feeding the troll?

3

why does everyone put periods at the end of the sentence.

CaptainAmerica16

@LeakyNun I have a question about a proof.

@Ultradark Did you go by another username in the past? I don't remember you.

@CaptainAmerica16 go ahead

@JasperLoy he is geocalc33

I am Ultradark

I am the ultra darkest wizzi

7:13 PM

@LeakyNun Oh, I don't know that one either.

wizzi = wizard

wow what a dark wizard

I went by Geocalc before

but now I am not just dark. I am ultra dark

I wanted to turn a new leaf

I like dark chocolate.

And I want to turn into a new leaf.

I like it too. What's percent of darkness do you prefer your chocolate to be In terms of the amount of cocoa?

CaptainAmerica16

7:16 PM

@LeakyNun I haven't started yet, but it's a question about starting. It's a set theory proof: Prove that, for a function f:X→Y, we always have A⊂B ⇒f (A)⊂f(B). I presented it to a teacher today and he said it may be too abstract, to begin with.

I like 60%

I don't know. I am not an expert.

@CaptainAmerica16 high school math teacher is the worst

mercio

CaptainAmerica16

@LeakyNun I'm familiar with set theory and figured it was worth a shot, although it is a bit a more abstract than I'm used to. What do you think? Does it seem out of range for a high school student? It's a logical statement at first glance.

7:17 PM

@CaptainAmerica16 Why did you present it to your teacher?

I have been thinking about the problem you gave to me

@CaptainAmerica16 just ignore your teacher. this problem is fine.

CaptainAmerica16

@JasperLoy I don't really have any guidance with my math studies, so I go to him sometimes with questions.

@LeakyNun Ok, I'm going to start on it today.

ok

@CaptainAmerica16 I see. This is a very simple question, can't be too abstract.

7:19 PM

@JasperLoy for a high school student studying "conventional" maths it might be too abstract

I think the teacher just doesn't know how to do it

@mercio I have been thinking about the problem of Navier Stokes. It seems like a very difficult problem. Do you have any suggestions. Where would you begin?

idk I'm not a specialist about ODEs

@LeakyNun Yes, but a high school teacher should not tell a student trying to write this proof that it is too abstract, if that is the situation here.

and if it were so easy maybe someone would have claimed the million dollar prize by now

4 mins ago, by Leaky Nun

@CaptainAmerica16 high school math teacher is the worst

7:22 PM

I'm looking at weak vector fields

that seems to be the best way forward

@Ultradark I suggest you begin by writing to Navier and Stokes.

2

okay

i believe they are not alive anymore

but I definitely would if they were still around

Seems I've stumbled upon some nerds

2

:0

Soham Chowdhury

@Daminark this is probably the most me_irl thing i've ever seen in chat

you are correct

Soham Chowdhury

7:25 PM

user image

3

Gold

Diamond

by asking to ask a stupid question he was asking a stupid question because no question is a stupid question except asking to ask a stupid question

that's one interpretation

I dunno if that was the point, more asking whether Borel and Langlands knew rep theory

I was just diverging

7:28 PM

Though I usually think of "Borel" as analysis

Like in Borel sets

Probably not the same guy, right? I feel the Borel of Borel sets was a bit too old to be in the presence of Langlands

@Daminark Emile is measure theory, Armand is representation theory

4

"He used to answer the question of whether he was related to Émile Borel alternately by saying he was a nephew, and no relation."

Heh, interesting

Emil and Michael Artin are father and son.

I am used to saying "Borel equivariant homology" on a daily basis so I am used to the idea of him as a G-action guy

How would you succinctly show that $\frac{z^3+i}{z^2-3z+2}$ is analytic except where z=1 and z=2?

7:32 PM

I have a question: Why do the pictures on the internet of lie groups look like graphs?

because they are not pictures of lie groups

what are they pictures of

I don't remember

they are lieing

Soham Chowdhury

In the mathematical field of Lie theory, a Dynkin diagram, named for Eugene Dynkin, is a type of graph with some edges doubled or tripled (drawn as a double or triple line). The multiple edges are, within certain constraints, directed. The main interest in Dynkin diagrams are as a means to classify semisimple Lie algebras over algebraically closed fields. This gives rise to Weyl groups, i.e. to many (although not all) finite reflection groups. Dynkin diagrams may also arise in other contexts. The term "Dynkin diagram" can be ambiguous. In some cases, Dynkin diagrams are assumed to be directed,...

7:32 PM

yes they are

You can't draw any of the interesting pictures of Lie groups. Instead you say "simple Lie groups are classified by these pictures, called root systems". Those are the graphs.

Soham Chowdhury

this?

yes

yeahh

pretty cool diagrams

@SohamChowdhury You seem to have aged a bit.

I remember your previous photo.

Soham Chowdhury

7:34 PM

Uh, thank you?

I don't remember what it was :/

You looked 10, now you look 20.

user image

that kind of looks like a Dynkin diagram

the picture in the thumbnail is not a dynkin diagram

@MikeMiller the one I posted is not

That picture is of a root system. A Dynkin diagram is a lot simpler

7:37 PM

This is a picture I drew for a professor

yeah it's a root system.

only took me about half an hour to draw it

@Rithaniel differentiability?

Yes, but it seems like a lot of work to prove that one is differentiable. Does not look like fun.

Separately show that polynomials are differentiable and quotients of differentiable functions are differentiable when the denominator is not 0

^

Thank you. That does sound much easier than what I was going to attempt.

7:45 PM

I call that develop the theory general enough until it's trivial

Also you can show that polynomials are differentiable via product rule

And addition

Since $f(z) = z$ is easy

(You could also do binomial stuff I guess, but I feel doing that directly for an arbitrary polynomial will be unpleasant)

Hmm, the torus is the only compact 2-dim Lie group, right?

It's the only compact 2-d parallelizable manifold, so yes

Ah makes sense

are all parallelizable manifolds also Lie groups?

not at all

all compact orientable 3-manifolds are parallelizable, including eg $SU(2)/G$ for any finite subgroup of $SU(2)$ - there are infinitely many such

however the only compact connected 3-dimensional lie groups are $SU(2), SO(3)$, and $T^3$

there are very few lie groups.

7:54 PM

right, I guess you do have a full set of globally-define vector fields which is closed under the Lie bracket, but to get a Lie group you need the structure coefficients to be constants

i would anticipate that's still true for $SU(2)/G$, no?

yes, so there's a local criterion which I mentioned, and a topological one which you did :)

I don't understand what you mean. I'm trying to say there's probably more or less no criterion you can apply to compact manifolds to see if they are Lie groups other than "is it on the list of compact Lie groups?"

although of course the question is "Can I find a basis whose Lie brackets have construct structure coefficients?", and I'm sure the answer is topological

*constant

8:15 PM

in theory salon, 2 hours ago, by vzn

Atiyah Riemann attack/ post mortem/ autopsy, Primes torch carried on / TMachine blog

Soham Chowdhury

8:33 PM

@vzn Riemann attack, now that's a band name

This might be a weird question to ask here, but here goes nothing; quantum computing is just linear algebra in disguise with some group theory (on the matrices) thrown in. Is there some nice write-up on this? Sort of like "quantum computing for mathematicians"

4

Most write-ups either assume that I'm a researcher in both quantum and mathematics, or assume that I am a physicist who has done some mathematics

9:10 PM

@Krijn I only know some books about quantum theory with a mathematical treatment

@MikeMiller indeed the issue with what I asked about the other day was that my foliation is defined by the ambient space, not by the surface alone.

So the vector field $(\phi_t)_*\frac{\partial}{\partial x}$ does not have the leaves as flow lines.

Which means back to square one with this continuity thing.

Hey guys, can someone help me understand why if I have k circles on the plane, that divide it into k^2-k+2 parts, and draw another circle, then the maximum numbers of divisions it will add is 2k? I can't seem to comprehend it.

I have seen proofs that just say:"It's obvious that this circle will intersect with each one of the others in at most 2 points, so we get 2k new parts from k circles". The fact that there are 2 intersections for each circles don't make it obvious to me, that 2k new divisions will be added.

@anakhro Yeah, by the end of that I was sure I wasn't saying what you wanted, but wasn't sure what the right thing to talk about was.

By the way, do you want to send me an email and kill your anonymity? :)

Sure, what about? :P

I dunno, foliations? Contact structures? World peace? World conquest?

I'm just curious. ;)

9:24 PM

@MikeMiller can I send you an email about 2k plane divisions? :D

You will be sorely disappointed when you find out I am literally a nobody, @MikeMiller :P

@anakhro Sure, but one day we may meet, no? I guess I would know closer to that time.

@Coder-Man My email is publicly available, but I don't know what that means, and I probably won't respond. <3

Maybe!

9:36 PM

@MikeMiller here's to hoping I sent it to the right Mike Miller.

Also, is a diffeotopy the same thing as a smooth isotopy.

I keep on forgetting to ask my supervisor.

Because I always see isotopy used interchangeably for continuous or smooth isotopies, but diffeotopy would really help out the resulting ambiguity.

@anakhro are you a graduate student?

No, I am a talented grade 8 student.

Balarka 2: Electric Boogaloo

@anakhro really? or are you trolling?

@Daminark where is balarka

I miss him

9:44 PM

in The h Bar, Jul 19 at 11:26, by Balarka Sen

It's all bullshit anyway. Not even sure why I am going to a university.

@anakhro I think it is but that you would serve to confuse your readers by using it.

Better to say smooth isotopy and then "all isotopies are smooth". If they are.

terminology is kind of the worst

Undergrad has taken him from us

Hi chat

Wassup ?

Yo @Astyx! How's it going?

9:50 PM

Good and you ?

Doing alright, thanks!

@Krijn old but this might work/ be close to what you want, ping me for more refs, have a bunch, need to write them up sometime... An Introduction to Quantum Computing for Non-Physicists Eleanor G. Rieffel, Wolfgang Polak arxiv.org/abs/quant-ph/9809016

10:03 PM

@Daminark I would have thought undergrad would have been the easiest thing in the world for him

Clearly he's in some dean's basement in chains.

Doing math on command.

Fourier transforms or something vicious like that.

That's what the GRE felt like

Except with ordinary integrals

So maybe they're training him

10:20 PM

@Daminark can you help me define some parametric curves

@vzn Thanks!

If $f$ is Riemann integrable, $g$ is monotonic, does it follow that $f \circ g$ is Riemann integrable?

Apologies if this is a naive question as I don't have a strong background in higher mathematics but does anyone know of a standardized formula for calculating permutations when using all possible values for k?

LOL, @user193319 doesn't believe me. :P

oh, I missed @Astyx.

@Phaeze: What do you mean, precisely?

Haha...I knew you were going to enter the room any moment!

10:31 PM

@user193319: Do you know the $\epsilon$ criterion for integrability in terms of upper and lower sums?

Yes, I have that criterion.

BTW, I meant g $continuous* and monotonic.

We were discussing continuous functions.

OK, then apply the criterion.

Oh, that changes things a lot. Let me give the problem another go.

Remember the context of our discussion — I backed down from general continuous $g$. (Do you have a counterexample to that, by the way?)

@TedShifrin I am generating test data for an encoding function that operates on a list of values. Given the count of values in the list I'm trying to find a simple way to find to find the total number of test cases generated.

10:33 PM

No, I don't actually. This isn't a homework problem, so I haven't had a whole lot of time to think about it; but I'll add that to my list.

OK, I think I have one, when you're ready to discuss it.

Heya chat.

Don't ever talk to me if you don't want a longer list :P

heya @Fargle

Haha, okay!

The function generates every possible subset and ordering of those values so for 5 items it would be 5! + (5!/4!) + (5!/3!) + (5!/2!) + (5!/1!)

10:36 PM

@TedShifrin given an algebraic variety, can you give me a fairly easy problem to work on, or do you know of anyone else who can give me a problem to work on?

The algebraic variety in question is a set of solutions to a system of polynomial equations and I was just wondering if you knew of any particular problem with regards to that

@ultradark: These are very hard problems. It's a whole field of research mathematics that uses a lot of material. You could start by reading about algebraic curves, but you need either some complex analysis or a good bit of algebra.

@Phaeze: So it's ordered lists, not just lists?

With unordered lists, you'd just get $2^5$.

Hmm ...

Yeah order matters as it changes the output of the encoding

I don't know if I've ever seen this.

Ahh okay, it did seem like a pretty rare scenario in all the googling I did. Thanks for giving it a thought.

I made a list of the first ten values on Mathematica. I don't see anything familiar.

For $6$ it's just a bit more than my birth year :P

For large $n$, I can tell you an approximation for what you have.

It's going to be roughly like $n^n/e^{n-1}$, which is huge.

10:47 PM

Yeah for n=7 it was ~14000 and n=8 jumps to ~109000

Yeah, my approximation isn't very good until we get to large $n$.

Ah, so I need better Stirling to nail it. It looks like $$\frac{n^n}{e^{n-1}}\sqrt{2\pi n}.$$

Very, very close.

For $n\ge 70$ it's right to two decimal places in exponential notation $x.yz... \times 10^N$.

heya @Krijn

@vzn This is exactly what I wanted

Heya @Ted

What are you up to today?

is $\ln(x)$ an algebraic curve

No

what is it

10:54 PM

awesome, thanks @TedShifrin

What is ln(X) or what is an algebraic curve?

both

is ln(x) a transcendental curve or something

you mean $y=\ln(x)$, which is $x=e^y$? Does that look like polynomial equations to you?

no

so what is that equation

is it just called non algebraic?

It's a transcendental equation. The graph is a perfectly lovely $1$-dimensional manifold, but NOT an algebraic anything.

10:56 PM

are there varieties with transcendental equations

Not algebraic varieties, no.

what if transcendental equations intersect and the solutiion space consists of a system of polynomial equations over the reals

what is that called?

You can talk about real analytic varieties or complex analytic varieties.

real please.

Real is worse than complex, it turns out.

10:59 PM

what do you mean by worse?

Harder to understand.

The complex ones are easier to grasp, for me at least

I don't know anyone who has studied the question you just asked. Other than trivial circumstances, I don't know that it ever happens.

I mean it's true, it's possible to construct a solution space in which real analytic functions intersect and the intersection points are found by solving a system of quartics or whatever. Not sure if that would be enough to define an analytic variety

Isn't then the system of quartics itself enough to define an affine variety?

11:11 PM

or maybe the algebraic varieties would just be defined from the intersection points themselves

yeah

so basically, each point would represent an affine variety, okay that makes much more sense

No, you can't define varieties just knowing the intersection points. Just think about the point $(0,0)$ in the plane. You can give it in countably many ways as the intersection of a pair of lines.

And, working over the real numbers, you can always make things of arbitrarily high degree because polynomials like $x^2+y^2+7$ have no real roots.

That's one reason that complex is so much better than real.

Ah yes, I remember now that alg. geom. was always trickier than I thought

LOL

But hey, I finished that sooooo...

and those beastly non-reduced points ...

11:22 PM

can you define a real analytic variety with just one variable $x$ or do you need it to be two variable

Whatever came out of that article on reduction modulo p?

Also, quantum entanglement is so mindblowingly weird in physics, but the linear algebra behind is just "oh ok yeah clear"

@Ultradark: Lots of variables.

11:37 PM

@TedShifrin okay. I 'm just kinda confused as to how to categorize the space of some special transcendental equations, where their intersections require you to solve nasty polynomial equations. I was discussing this with @Semiclassical

I haven't even considered the complex case

Whether you can solve the equations or not, polynomials still have zeroes ... in general, you cannot solve in any sense.

okay, thanks for helping me. I really learned a lot

you're welcome.

also I found a cool thing where one of the solutions equals $\phi -1$

so if you solve: $e^{2/\ln(x)}=e^{4/\ln(1-x)}$ you get $1/\phi$

11:54 PM

I'm claiming this is an artificial transcendental equation. It really comes from a polynomial that you then transform into something transcendental. But I consider that artificial.

Hi Ted Sir

user image

How should I get this extra condition?

You need the half of the circle corresponding to the argument condition. In the other half, the argument will be $-\pi/2$.

so I'm confused, it's neither algebraic nor transcendental?

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