trying to decide how little i'd understand :)

The birational geometry thing is interesting, because birational CY 3-folds will have the same Hodge numbers. So I guess if there are birational CY 3-folds one shouldn't expect a unique mirror.

@Semiclassical I'll be working on the physicists' side

gotcha

You may as well be doing homological mirror synmetry, since I think there's a recet series of papers that show homological mirror symmetry implies classical mirror synmetry

Sure

If only anyone in my surroundings would be conversant in homological algebra

Studying things called 'mirror symmetry' must be nice. Sounds applicable, so infinite funding.

"Applicable."

applicable in string theory, yes

@AndrewThompson Hahaha

applicable in anything actually testable...

@Semiclassical String theory is the cash cow that keeps on giving

@Semiclassical Hey, there's always CFT

hurr durr Ising model

Well, keeps on giving until the NSF catches on.

depends on what one means by 'testable' i guess

Nonono. "So, we can see these symmetries appearing at the ground of the ocean, which makes extracting oil easier."

$$$ follows.

pretty sure that's a bullshit too far

Not really how funding works.

Eh, depends on where you are. Half of our applied people are funded by Norways biggest oilcompany.

maybe so, but i'd be shocked if any of them are working on Calabi-Yau stuff

Ohyes, I was just messing around.

ah

Right, applied people. The people who decide funding aren't completely idiotic. :)

keyword in there being 'completely'? :>

Well, we'll see if they fund me. If so, yes, key word completely.

there's also a guy giving two talks on integrable systems stuff this week

Are you applying for anything?

"Hodge integrals and integrable evolutionary PDEs" today, "Bihamiltonian cohomologies and integrable systems" tomorrow

(If they just hand you money randomly the keyword is definitely applicable.)

Yeah, I applied for the NSF graduate fellowship.

nice

both of the talks mention Gromow-Witten classes, which I've...heard of?

Find out in early April.

@Semiclassical Nice

Classes? I thought they were numbers.

I know them as invariants

which could be either ;D

here's the abstracts

Oh, I guess they live in quantum cohomology or something. vOv

Applying for stuff made me think about recletters. I hope profs dont base them on things you said at exams. "The candidate does not know the Leibniz rule and displays significant difficulty adding one-digit integers" won't look to good.

First sentences of Wikipedia

first one: "For an arbitrary semisimple Frobenius manifold we construct an integrable hierarchy of Hamiltonian partial differential equations, which we call the Hodge hierarchy. In the particular case of quantum cohomology the tau-function of a solution to the Hodge hierarchy generates the intersection numbers of the Gromov--Witten classes and their descendents along with the characteristic classes of Hodge bundles on the moduli spaces of stable maps.

Rational numbers, but they may be "packaged" into (co)homology classes

For the one-dimensional Frobenius manifold the Hodge hierarchy is a deformation of the Korteweg-de Vries hierarchy depending on an infinite number of parameters. We conjecture that this hierarchy is a universal object in the class of scalar Hamiltonian integrable hierarchies possessing tau-functions."

and the other: "We consider the problem of classification of hierarchies of bihamiltonian integrable PDEs that are closely related to the theory of Gromov-Witten invariants and 2D topological field theory, and show how to solve this problem by computing the bihamiltonian cohomologies associated to semisimple bihamiltonian structures of hydrodynamic type."

They're probably making it all up.

probably

I would love to have a seminar about a mathgen paper.

Related

Ah, classic!

"I have no idea what the author is doing in the next line. More details would be helpful."

"A miracle happens."

A friend of mine reviewed a paper once and wrote something along the lines of "the author uses the notation $g_{it}$, which I find needlessly offensive."

Yes, it's outrageous they still use indices.

@AndrewThompson There is a nice site with a collection of reviews from MathSciNet which make for good reads

Exceptional math reviews or something like that

Can someone help clarify this discussion for me? mathoverflow.net/a/230193/82691

@JulianRachman Well, are you familiar with the ordered set $\omega + 1$?

@Tobias No. Can you define it please?

@JulianRachman It is simply the naturals plus an additional element $\infty$ which is larger than all the naturals

Ok. I still don't understand how his answer and comments solves my problem.

@JulianRachman I didn't read the rest very carefully, it just seemed like you needed clarification on that order

@Tobias I definitely need clarification because the statement in my question is quite important for me to kmow about in terms of wqo

@JulianRachman I am not even really sure what a wqo is

hello people

can anyone help me in linear algebra

@Tobias good Defn: en.wikipedia.org/wiki/Well-quasi-ordering

if i am given to write v in linear combination of v1 v2 v3 in R^3

then i must first check that v1 v2 v3 are L.I.?

by making a matrix [v1 v2 v3]?

and then if i get I, then they are L.I?

@JulianRachman Ahh, the quasi part confused me, as I am used to calling those preorders

now using gauss-jordan I reduce [v1 v2 v3 | v]

@JulianRachman So it is just a preorder which is wellfounded as well?

I get I_4, so what does that mean?

@Tobias Bingo

@JulianRachman Hmm, not quite it seems

or maybe all are LI,

?

@JulianRachman That is basically what that page says.

Can you explain because I do not understand the main answer or the discussion.

@JulianRachman It is stronger than what I said, in order to be better behaved with respect to certain operations

What

@JulianRachman Well, as mentioned on that page, the naturals with "divides" is not a wqo, but it is clearly a partially well-ordered set.

But then consider any infinite ordered set

Hi -- I have a rather pragmatic and dumb question... did the voting system stop registering upvotes ~2hours ago?

@Tobias

Which is what my question asks

@JulianRachman Consider what about that set?

@ADG How could you get the $4\times 4$ identity matrix when the vectors are in $\mathbb{R}^3$?

@Tobias my question days it all

@Tobia sorry they are in r^4.

@JulianRachman Well, the example given is indeed an infinite ordered set

@ADG Well, there is no reason to look for linear dependence of the vectors. Just write up the matrix you did and see what you get. If you do get the identity, then indeed the fourth vector cannot be written as a linear combination of the other three

yup. thanks

btw, do you know a way to find basis using rref for column/row space of a matrix

@ADG Not sure what you mean

I'm doin.. Find a basis for the column space of V spanned by vectors {v1; v2; v3; v4}.

@Tobias which example?

@JulianRachman The one given in the answer

btw im also reading math.tamu.edu/~fnarc/psfiles/find_bases.pdf which was available from google search

@ADG Ahh, set up a matrix of the vectors and do row reduction. Pick those vectors that correspond to leading $1$'s.

oh! it now seems trivial :P

@ADG Pick from the original vectors of course

for row space i 'spose transposing and doing the same suffices and transposing the soln?

@ADG Indeed

thanks.

@JulianRachman But actually, it is even easier to see this. Just take any two wqo which are not isomorphic but have the same cardinality. Clearly they cannot both be isomorphic to the free monoid with the subword order

btw, what does this mean : 'Extend S to the basis of R4.'?

S is a set of 4 vectors in r^4

anyways this solves thestudentroom.co.uk/showthread.php?t=1739470

@Tobias then how about taking one wqo to its free monoid which might be (or is) wqo?

Excuse me?

Is this room alive?

@JulianRachman Not sure what you mean

@ADG Usually I would use that phrase to mean "find additional vectors such that you get a basis". But clearly that requires the original vectors to be linearly independent

$A\to A^*$

@JulianRachman What order do you want on the free monoid? It is not even a wqo in the subword order.

Well the main reason I am asking these questions are to find out "if $A$ is wqo then $A^*$ is wqo

@Tobias

@JulianRachman Well, you need to specify an order on the monoid for that, and the subword order will never work

And read the references in my question. It tells you there.

@Tobias then how about the dominating order then?

@JulianRachman Not sure what order that is. I am only used to that term from partitions

Domination*

@Tobias look at the reference 2. and look at Collorary 1.6 and 1.7

Page 4

Um hello

Can I ask a question for critique?

@JulianRachman Right, but it is clear that for infinite $A$, the subword ordering is never a wqo

Then let's take $A$ to be finite then

But no, $A$ and $A^*$ are also not isomorphic under the domination ordering, as one can see by considering the example given in the answer to the MO question, where $A$ has a unique element which has infinitely many smaller elements, whereas this is not the case for $A^*$

i think you can askthis on main site??

Then let's consider $A$ to be finite

@JulianRachman Ok. What is the question then?

Same thing except $A$ is finite

@JulianRachman about a bijection? But $A^*$ is infinite

Isomorphic?

Wait. It has to be bijective to be isomorphic right?

@JulianRachman Yes, was the question not about a bijective map?

It was about a bijection

right

Then how about order preserving?

And can we at all make the order-perserving bijection?

@JulianRachman Could you write out the precise question you are at now?

The above two

@JulianRachman Please write the full question, I am getting lost in which assumptions there are

Why is it that when I study intro to algebra, many diverse concrete examples are given immediately, whereas the introductory axiomatic set theory that I have encountered so far gives so few?

Could someone please explain this equality? If $a,b,c$ are distinct primes then $\gcd(a^3b^2c,a^2c^2)=\min(a^3,a^2)\min(b^2,1)\min(c,c^2).$

@user276387 why would you include min when you know which element gives the min?

Hi there

@TobiasKildetoft I didn't write this. I copied it to here because I don't understand it.

hi @ValeryBaturin

@idonutunderstand because introductory axiomatic set theory is by definition "axiomatic."

@idonutunderstand why don't you make up your own examples?

^

@user276387 well, the "proper" way to have written it would have been as $a^{min(3,2)}\cdots$ instead

If anyone is here I'd like to ask what needed to be googled for solving limit n to inf and defined integral. It's needed to proof some equation for home home assignment. Thanks

hi @Hippalectryon what do you think are the best (and affordable) schools in which one registers for cpge? thx in advance

@ValeryBaturin www.wolframalpha.com

Wolfram prooves nothing just result to see. I know equailty already from a task

user, why do you need cpge?

@ValeryBaturin it also shows the proof if u have an ccount

@ValeryBaturin symbolab.com shows the proof

@TobiasKildetoft I understand that $\gcd(a, b) = \prod p_i ^{\min(s_i, t_i)}$. However, I'm a bit confused as to how that gives $a^{\min(3,2)}\cdots $.

@ValeryBaturin just asking as a possibility, do u know wat the cpge is?

@user276387 Well, that is the $a$-part of the gcd

IIIII HAVEEEEEEEE A LOOOOOOOTTTTTT OF FUN HEREEEEEEEEEEE!!!

Never thought to laugh so well while calculating some series!

@I'manartist $$\Huge\text{YOU HAVE A LOT OF FUN!}$$

@I'manartist i can sense your enthusiasm ^^

@user153330 :-)))))

symbolab.com/solver/step-by-step/…

Help me to type. lol I couldn't

@ValeryBaturin which limit is it?

user, I googled it before the first answer. It's just some courses with cool name, isn't it?

to inf

@ValeryBaturin $$\lim\limits_{n \to \infty} \int_0 ^1 (\frac{x^n}{1+x})dx$$ correct?

seems right

@TobiasKildetoft So $\gcd(a^3b^2c,a^2c^2) = a^{\min(3,2)}b^{\min(2,0)}c^{\min(1,2)}=a^2c$

@user276387 Yes, precisely

@ValeryBaturin Illegal Input: Indefinite integrals must have both lower bound and upper bound.

Yes. the same shit

@Huy test test, 1, 2, 3, can you hear me?

@TobiasKildetoft Thanks.

@I'manartist lol what is that? vOv

@user153330 I'm in the middle of a party here, I celebrate my results! :-))))

@Huy can you heaaaaaar me?

@I'manartist so you went back in time?

@ValeryBaturin so try wolframalpha

@user153330 I'm caught in the past in some way ... :-)

I have no account

@ValeryBaturin okay so ask about it in math.stackexchange.com

I wanted just a hint what to google not the solution from a person, but online result would be fine)

@user276387 Are $a,b,c$ relatively prime?

@ValeryBaturin what if it doesn't exist? (although i doubt that)

@robjohn distinct primes.

The result is zero, it's quite obvious in my head. But I have no proof

@user276387 then the answer is "yes"

@ValeryBaturin i don't think it will be zero

But it will be defently so

definetly

I suppose unique factorisation kinda of trivialises the problem.

@user153330 If there would be a little chance to exist the possibility to reverse the time, I'd probably invest my whole energy, power into that, but according to the most important scientists it is not possible. All good stuff is in the past.

I'm gonna try to do it the hard way.

@ValeryBaturin after a bit of thinking you are right it will be 0

Cool

@user153330 Probably I would also finish my first school (where I entered the first according to the points - I got a perfect score in mathematics).

In my task it's said that I need to use mean-value theorem for integrals

@I'manartist so you're a school dropout?

@user153330 I graduated later from financial accounting, but the first school I went to was a great one.

@I'manartist ah right

Is any done with it?

$$\int_0^1 \frac{\text{Li}_3\left(\frac{x}{2}\right) \log ^3(1-x)}{x} \, dx$$

$$\int_0^{1/2} \frac{\text{Li}_3\left(\frac{x}{2}\right) \log ^2(1-x)}{x} \, dx$$

$$\int_0^{1/2} \frac{\text{Li}_3\left(\frac{x}{2}\right) \log ^3(1-x)}{x} \, dx$$

5 beautiful integrals to take

@I'manartist i meant to ask: do you really mean 'without pen and paper?' the only way I can imagine that's possible is if it were to vanish trivially, and mathematica doesn't support that.

@Semiclassical No need to use without pen and paper. My research shows that it is possible without pen and paper.

purely by inspection?

@RandomVariable does that variable change you told me about works for that integral above? You said that it might work but I didn't ask you if you tried to finish it.

@Semiclassical Yeah, and that's great.

@I'manartist I don't think I had a particular change of variables in mind.

@RandomVariable OK. I thought you had in mind a particular change of variable.

I couldn't do $\displaystyle \int_0^1 \frac{x^2}{\log^2(1-x)}\;{dx}$ without pen and paper, let alone THAT one.

@user276387 Why don't you note a simple fact there?

What's that simple fact? I tried symmetry, didn't get anywhere.

@user276387 $$\int_0^1 \frac{x^2}{\log^2(1-x)}\;{dx}=\int_0^1 \frac{(1-x)^2}{\log^2(x)}\;{dx}$$

@user276387 also note that $$\int_0^1 x^s \, ds=\frac{x-1}{\log(x)}$$

An elementary triple integral to do without pen and paper.

@I'manartist Ah, I was missing the second bit. :D

@user276387 ;)

It's quite beautiful! :]

Hello @TedShifrin