morning

Hello. I'm doing a math phd (just finished quals). If one specializes in pure math (say dynamical systems with no particular application in mind) how difficult can it be to transition into industry after the PhD? I am not particularly interested in applied math, but I'm not too optimistic about the future of the pure mathematician (unless he/she solves some millenial problem!!) and I would like to think that I have the option of switching from academia to industry. Any input is appreciated.

morning

That would be a good academia.SE question

I can only think of one regular in this room who would know the current answer to that question and he's out of town

thanks mike I will put it there

Let me know when you do, I want to save it.

will do

Thanks

you're welcome

Hey guys

Can I ask a question about sets and relationships between objects?

How do we talk about "is a" vs. "has a" in formal logic?

unsure what you mean, can you extrapolate

@MikeMiller I have hair but the hair is not me

Where as yellow hair is yellow and hair but is also both individually

But this is not simple disjunction because I am not hair OR I am me

No; I have hair, but I am also not it

Therefore, it cannot be a super set of me because the hair is not me; it is a quality of myself

So when I program, I can't always assume that a Human that has an eat() function can extend to all humans

All humans can eat --> except if they have their mouth disabled

Therefore, my construction of what a Human is and can do is flawed and any inheritance built from a flawed object will also carry that flaw

fair warning so you don't waste your time on me: I'm not really listening anymore; someone else might

@MikeMiller, do you have literally 10 seconds?

Don't ask to ask @KajHansen

Ah, well maybe someone here can help me out. Why does Hungerford add the "free group" hypothesis in this problem? This appears to be something true of all groups. My solution is extremely suspiciously short, but I honestly don't see anything wrong with it: i.imgur.com/FY7Rd3v.png

Hi Professor @TedShifrin

@KajHansen what you've written is correct, but is that sufficient for the group generated by?

Hi Rigor. Do I know you by a different name?

Oh wait, you're right @AlecTeal. Derp.

SkullPatrol Ted.

hi derp @Kaj

Oh, skull

@TedShifrin yes skull patrol :-)

Yeah you could have a bunch of different members of the generating set, at different orders there @KajHansen - unless (random order of powers of different elements)=$x^n$ for some $x$ it isn't true. Hope that helps.

That was a really bad mistake actually. Ugh, right when @Ted arrives too.

Kai, I'm on my phone on

Skull, you've never been rigorous.

I'm working on it.

Let me know how you get on @KajHansen I'd like to see the whole proof.

Sure thing @AlecTeal

marg nsl

oh, ugh ... On my phone on marginal wifi in Big Sur.

I have to take into account the time change when it appears that you're on later than usual at night @TedShifrin

haha

Yup, indeed, @Kaj. Plus, I don't have to be in my office by 8:30 ... Not that anyone misses me;)

@AlecTeal, I'm trying a different route altogether

Suppose $F$ is the free group on a set $X$, and let $I$ be an indexing set where $|I| = |X|$. Then the group $\displaystyle \prod_{i \in I} \mathbb{Z}_n$ can be realized as the homomorphic image of $F$, and hopefully the kernel matches the supposed normal subgroup. Just thinking through my reasoning and making sure it's airtight.

@KajHansen You have only shown for the generating set $S$ of $G$ that $gSg^{-1}\subset S$. To show the subgroup is normal you're going to need to do that for any $g$, $gGg^{-1}\subset G$. These two statements are not equivalent.

Yeah, I noticed my mistake pretty quick in that original image @PVAL, but what of the idea right above your post? I think I've filled out the details, and it seems to work.

I take my generating set on that product to just be $\{ \mathbf{e}_i \}$, which denotes a tuple zeros except a $1$ in the $i^\text{th}$ position. Then $\phi$ is induced by the inclusion $X \hookrightarrow G$.

Your kernel is the normal closure of the desired group. You need to show the normal closure is actually the desired subgroup. In general if you have a quotient of a free group relators $r_i$, the kernel of the map is just the normal closure of the subgroup generated by the $r_i$. You are asked to show that in this case the normal closure is the same as the group. Start with a generic element in $G$ (not a generator) and show that after conjugating, the element is still in $G$.

Thanks for the advice @PVAL. That makes sense.

@r9m ^^^

@robjohn did you manage to work on it more? Later when I finish the solution in latex I put it db.

@Chris'ssistheartist no, I've been away and now I am doing some work.

I should propose it in some magazine, it's simply too nice, but at any rate it will appear in my book together with other such questions.

@robjohn OK

I am TA'ing linear algebra at the moment, and I realized that I cannot come up with a satisfactory definition of a linear equation. I have some ideas, but it just never seems quite "right" (and it also gets rather complicated)

can you elaborate on what you mean by a linear equation, @Tobias?

@BalarkaSen Well, that is the problem of course :)

But essentially it should be of the form $L(x) = y$ where $L$ is a linear function

heh. do you mean equations of the form $Ax = B$?

The problem is that "obviously" also $L(x) - y = 0$ should be linear

where $A, B$ are matrices, $x$ is a vector.

So now the question is whether an equation is linear if it is equivalent to an equation of the form $L(x) = y$, but this seems too weak somehow

since for example, over the reals $x^2 = 1$ would then be linear (being equivalent to $0=1$) and also over $\mathbb{F}_2$ (being equivalent to $x=1$), but not over $\mathbb{C}$.

which then bring in "is linear over any field in which it makes sense", but that just seems wrong somehow

Is there a good point to this exercise?

which exercise?

Definining linear equation.

@MikeMiller Mainly because there was an exercise asking them to state whether $2x_1 = 4x_2 - x_3$ was linear. And the book said "yes" while the answers from the previous lecturer said "no" (due to an imprecise definition of the term in the book)

OK, I see. The book said $Ax = y$ when this is not what was desired here.

Which book?

@MikeMiller Well, the book says "can be written on the form $a_1x_1 + \cdots + a_nx_n = b$"

Well, the above can be written in that form :D

and the "can be written on the form" is the problem

I think what you want is precisely "Equations that can be derived from $Ax = y$ by elementary linear transformations", where by that I mean you can multiply either side by a scalar or add terms from one side to the other.

@MikeMiller Ahh, that does sound like a good approach, limiting the equivalence of equations to those coming from "linear" things

@robjohn I sent the solution in db.

@robjohn I'm going to propose it in some magazine, it's simply a jewel that cannot be put aside and let the dust in it.

Back in 30-60 min.

finally reassembled my laptop. what a pain

yikes

not a single interesting question to answer. ugh.

nothing to do.

@TobiasKildetoft how about a polynomial equation of degree one?

I mixed up the chapters in the file I sent (I mean the one related to the name of the file).

@BalarkaSen: You never told me when $M/G$ is orientable, $M$ orientable, $G$ a group acting freely on $M$.

oh, right

well, it is reasonable to conjecture that only if $G$ acts by homeomorphisms preserving orientation.

@MikeMiller Are you sure I have the necessary tools to do this? Top dimensional homology of $M/G$ doesn't seem easy to compute.

I never said $M$ was compact, and yes.

ok, hrm.

I guess this should be definitional, then.

i'll ponder on this while i eat

@MikeMiller OK, I'm back. The similar path trick works.

I hope you mean that $G$ acts on $M$ both freely and properly discontinuously, otherwise this'd be a bit more pain.

Singing while discovering tons of new amazing results ... (that is now I mean). I should have one or two assistants that record all my stuff I create every day, this would be very nice (in terms of time management).

@Chris'ssistheartist It seems easy .. multiply the numerator and denominator with $\sec^2 x$

@r9m yeah ... but how long did you think of it? :-)

as for the triple integral .. it seems difficult (haven't tried it on pen and paper) but the double integral version seems easy

@Chris'ssistheartist I just woke up and saw the messages you sent me :) (I have a high fever)

@r9m are you sick?

@Chris'ssistheartist ya .. I am running a fever cartel :-)

@r9m :-)))

@Chris'ssistheartist last night I was hearing sounds that seemed like Doppler effect :P .. it's amazing and hilarious at the same time what a delirious brain can make you experience

@r9m Congrats, no one of the people I know and that are good at math (professors) succeeded with that one so far. :-)

@r9m Yeah, I know. I experienced something like that when I didn't sleep for 2 days or so.

@Chris'ssistheartist You have gotta be kidding me :P

@r9m No, no kidding.

Mother of GOD!!

@r9m Which of the double integrals you were referring to?

@Chris'ssistheartist $\displaystyle \int_0^1\int_0^1 \frac{1}{(x+y)(1+xy)}\,dx\,dy$

@r9m Ah, that one is extremely easy. OK.

@Chris'ssistheartist Have you tried 11852 (AMM)? :-) It seems like a nice problem :D

@r9m No, I worked on other type of problems (the ones for my book).

@r9m did you finish 11849?

@Chris'ssistheartist 'kay :) btw I am dying for your book on a regular basis :D

@Chris'ssistheartist nope .. not yet :(( .. It's the most slippery problem I have ever encountered ;)

BBL ..

@r9m OK

@r9m Let me know when you finish the cubic version. :-)

What on hell !!

He's not the real guy.

I guessed so....

Hi, Balarka

It's unlikely that Alexander Grothendieck would ask for a list of homological methods in algebraic geometry.

there've been threads on meta (or at least 1) about users that take on the names of famous people

Is his name the reason he is banned ?

don't think so

if such practice were to become against the rules, they'd force a name change

Oh.

It'd have been fun if Alexander Grothendieck actually used MSE :P

Or even MO.

I mean, lots of famous people used to be on MO. Bill Thurston was there.

We have terrence tao using MO

And I think Bill Thurston is as famous for his works on geometric topology as Alexander Grothendieck is in algebraic geometry.

who's the most well known physicist?

I've talked to/with/around Lubin a little on MSE. my advisor would be jelly.

yeah, he's pretty famous too!

something something Lubin-Tate. who's your adviser, @anon?

guy who works in local fields and Galois module structure, you wouldn't recognize his name. also trying to stay anon here.

ok, sounds pretty cool!

@MikeMiller $f : M \to N$ be a map between closed compact orientable manifolds of nonzero degree. I want to prove that it's surjective. Here's what I have done : assume it's not surjective. then $f$ factors as $f : M \to N - \{x\} \hookrightarrow N$. $H_nf$ factors as $H_n M \to H_n(N - \{x\}) \to H_nN$. $N - \{x\}$ is not compact so the middle group is zero, forcing $\deg \, f = 0$, and the whole world blows up.

is that right? just asking for a sanity-check

@Clarinetist thanks thu, exam is taken today,

@Chris'ssistheartist i know maria isnt someone to defeat easily

Yes

ok, cool. I have used this in my answer here.

@BalarkaSen Sure, that's what I meant. Explain what you mean by path trick?

Please write out some details on what you're talking about for both directions of your proof

Right, I meant to, I was waiting until you were online.

hmm, actually, I don't think that works.

why can't we do the following? $G$ acts on $M$ by orientation preserving action, thus orbit of a single point $x \in M$ all have the same local homology. Pushforwarding by covering map gives a point $y \in M/G$. Now $U$ be a small nbhd of $y$ which is evenly covered, and lift this to $M$ to get a bunch of disjoint neighborhoods of the orbits.

Now, $M$ is orientable, so any orbit s.t. every element lies in each of the disjoint nbhds above get the local orientation from the total nbhd

pushing this down gives a point inside $U$.

I think something along this should work.

@Agawa001 Yeah, but this time she will fight Victoria Azarenka (that was the number 1 in tennis) :-))))

That will happen in 2, 3 hours or so.

Right, what you mean is that the orientation of $y \in M/G$ is the pushforward of the orientation of one of the elements in its preimage. The choice doesn't matter.

And it's locally the same because one can pick a neighborhood as you say.

The converse?

Hi, maybe someone can help if the following math.stackexchange.com/questions/1428123/… makes any sense

honestly I've never working with this kind of problems and I'm not sure of the steps. Thanks

for me it'd be enough to say that any $A(t)$ commuting with any riemann sum of $A(t)$s implies, after taking limits, commuting with any $X(t)$

So @anon, do you think makes sense what I did.

sorry, internet is pretty down in here.

@JoseAntonio I didn't check your details with the riemann sum stuff but yeah

@MikeMiller ok, cool. i'm glad that works.

by converse, you mean $M/G$ is orientable iff $G$ acts by orientation preserving homoemorphisms?

@anon Ah ok, honestly I've never working with this kind of problems. I think and immediately corollary, is that $\dot{x}=A(t)x$ has the same solution as the scalar case, right?

@anon Thank you

@JoseAntonio with the [A(t),A(s)]=0 condition, yes. it's easy to overlook the condition though when you're first solving that equation!

sorry, I mean "only if"

@anon Did I improve my question successfully or is it still unclear?

@NaCl Still a bit .. off. For instance, If $\sum t_k=m$ is $\le\sum n_k$ and $\sum n_k\le m$ then $m=\sum n_k$. You're aware of this right? Also, when you say "I want to generate tuples from that," do you mean you want the components/coordinates of the tuple to be taken from the multiset?

Also, what's the purpose of having repeated elements in the multiset?

small sidequestion: is it normal that your inline-math is not rendered and just displayed as the latex code?

@anon I see where there confusion is, thank you.

@MikeMiller if that's the converse you are referring to, then it's clear. act on $S^2$ by $\Bbb Z_2$ by antipodal map

@NaCl see "LaTeX in chat" on the starboard ----->

@anon thanks

then $S^2/\Bbb Z_2 \cong \Bbb RP^2$ is not orientable. done

You've proved that if $G$ acts by orientation-preserving homeos, then $M/G$ is orientable. Now assume $M/G$ is orientable...

yeah, I presumed that's what you meant. having two converses confused me.

There's not really two converses... I'm not sure what you're saying

well, cover of an orientable manifold is orientable.

@MikeMiller the other converse is that if $M/G$ is orientable only if $G$ acts by orientation preserving homeo on $M$

That's the same thing as what I just said.

ok, I am totally confused. the two converses I have in mind is 1) $M/G$ is orientable only if $G$ acts by orientation preserving homeos 2) $M/G$ is orientable only if $M$ is orientable.

Huh? 2) is trivial, not a converse, and we have been assuming $M$ is orientable all along anyway.

@anon Are you free?

yes

ok, you're right. I'm just being stupid.

@Rememberme though not for long

Forget it . I found my flaw

@Rememberme There is no reason to get bases and matrices involved here.

anyway, I must go mow

Yes I got that

Thanks!

and yes that is mow, not now

@anon alright, I tried my best

byebye. :)

@NaCl Would it be correct to say you have some list of positive whole numbers $n_1,n_2,\cdots$ and a chosen whole number $m$ and you want to enumerate the lists of nonnegative integers $t_1,t_2,\cdots$ such that $t_k\le n_k$ for each $k$ and $\sum t_k=m$?

Yes, that does seem to be my statement. However, I want to know if the size of the permutation-list stays equal when you change $n_1, n_2, \cdots$

do you mean T, the collection of tuples, stays the same size?

stop using the word permutation

Isn't it a permutation of it's summands?

of what's summands?

$m$'s summands

(2,2,0) is not a permutation of (1,3,0)

the set of summands is not even fixed

you can choose to write m as a sum of summands in multiple ways

such as 4=2+2+0 or 4=1+3+0

certainly the second is not a permutation of the first

Sure, it isn't.

and when you allow summands to be repeated, we are also no longer talking about permutations

at least not in the mathematical sense

Oh, ok. Thank you

@anon And yes, I mean T

So how about this: say "Given whole numbers $n_1,n_2,\cdots,n_f$ and $m$, let $T$ be the collection of all tuples $(t_1,\cdots,t_f)$ with $0\le t_k\le n_k$ for $k=1,\cdots,f$ and $\sum_{k=1}^f t_k=m$. Based on examples, it seems to me that $T$ depends only on $m$ and the sum $\sum_{k=1}^f n_k$. Is this correct?"

that's a concise statement of your problem

I'll edit again

A big thank you!

@BalarkaSen: I assume you know you haven't proved it yet, right?

yes, I do. I am thinking about too many things at once, let me concentrate on this.

I've got a good idea, let me see if it works

@anon I really enjoy your precise and yet brief way of asking. I should get used to it!

It's really cool that the endofunction species is isomorphic to the composition of the permutation species with the rooted trees species.

not sure who that's supposed to be directed at, but whatever

where's your x on the RHS?

Huh?

Explain.

you should have said x=pi/4 to begin with.

anyway, have you tried using a double-angle formula, or converting to complex exponentials, or anything?

@Rememberme You just changed your profile icon ?

hi

@Chris'ssistheartist what's up ?

@Ramanewbie salut

@Gato Playing with funny stuff. You? :-)

@Gato answer me without thinking for long :D Is it trivial to you?

Hi @Gato !

@Chris'ssistheartist just back from holidays :)

@Gato Okay, but asking for an opinion only, not a solution ... :-)

@gato Where did you go

@Chris'ssistheartist I was answering the you? part

@Chris'ssistheartist but I don't use trivial, but I would say it's a medium problem :p

@Ramanewbie Liban

@Gato it's a funny one. :-)

@Ramanewbie t'es en première S cette année ?

@gato oui

@Chris'ssistheartist I am looking at your questions

@Gato Oh, OK. :-)

@Chris'ssistheartist math.stackexchange.com/questions/1414996/… :DD

@Ramanewbie et ton frère ?

@Gato il te dira lui même si il veut bien... Je ne peux pas te dire

@Gato et toi tu fais quoi cette annee*

@Ramanewbie Tu ne peux pas me dire si il est bien rentrée en 5/2 ?

@Gato Comment sais-tu ??

@Ramanewbie M1 maths pures

@Ramanewbie j'avais parlé avec lui ici même

@Gato ah bon ds ce cas...

bon, je dois déjà y aller, bonne soirée

@Gato bonne soirée

@MikeMiller Hi how are you doing? If you have a chance please look at my post.

@Gato :D

@Ramanewbie salut

Hi @Agawa001 !

@Agawa001 Tu te souviens de l'exo du javelot ?

@Chris'ssistheartist noone would know !

oui je le souviens

yavait une erreur grammaticale je pense

@Agawa001 oui il manquait un 's' a 'mettre'...

@Agawa001 The game will begin in a few minutes I think. :-)

@Agawa001 Mais surtt en fait la vrair question etait : "Apres combien de tps le javelot va-t-il atterrir" ?

@Ramanewbie ce nest pas ça

Hi @Chris'ssistheartist Watching tennis again ?

@Agawa001 Alors quoi ?

@Ramanewbie Yes. :-)

@Chris'ssistheartist Women tennis ? How vs how ?

on peut pas prevenir quel etendue peut atterir le javelot

je vois

@Ramanewbie Yeap, Kvitova - Pennetta is now, and later our player Simona Halep.

@Agawa001 C'etait une erreur du prof...

@Ramanewbie google them a bit :-)

@Ramanewbie meme un 'metre' doit s'ecrire tel, excuse moi mais ton prof est dupe,

@Ramanewbie :D

@Agawa001 Lol j'adore le changement de "ton prof est dupe" a "excuse moi mais ton prof est dupe" !

oui, c un prof qd meme

tu penses ?

@Agawa001 Que quoi ? Pour le record ou la duree ?

c'est un examen special de demander a letudiant de corriger la question

ou choisir une version (plus convenante)

@Ramanewbie pour le record

@Agawa001 C'etait juste une erreur involontaire de la part de mon prof...

tu dis que ca netait pas un hasard

@Agawa001 je n'ai pas vérifié, c' est ce que nous a expliqué mon prof mais je pense que c'est vrai.

@Agawa001 Quand ai-je dit ca ?

@Agawa001 Oui, le chiffre etait choisi volontairement, mais le fait de demander de trouver la distance au lieu du temps est bien une erreur involontaire.

@Chris'ssistheartist wish i rather watch a live-stream in interet

@Agawa001 It's on Eurosport (in a few minutes)

Hello

@Chris'ssistheartist what about jsc sport ?

@Agawa001 Never heard of.

I'm feeling ashamed I'm not better at this today.

hello @Owatch

Given I have to use X and Y amounts of A and B to reach a sum of 10,000, and given I must use a combination of exactly 100 X and Y to get this, I'm left with:

xa + yb = 10000

And x+y = 100 of course.

So I solved it to y = -200, x = 300

But this makes no sense practically. Because I can't have -200 cash.

So I'm trying to think of how I'd use another combination.

Hold on.

I may have thought of something.

ax+b(100-x)=10000?

Yes yes, I've done that.

I'm saying the results work, but aren't logical in real life. You don't have negative money.

(a-b)x+100b=10000

of course because you chose a wrong couple (a,b)

the couple which verifies 100=>x=>0 is true

x=100(100-b)/(a-b)

a-b>100-b , a>100 gives you likely wanted result

lets try a = 101

@Agawa001 the game has begun ... :-)

x=100(100-b)/(101-b)

@Chris'ssistheartist i havent this channel in my tv

@Agawa001 maybe you can see it on internet

@Chris'ssistheartist i must run a proxy , which i havent, gonna search another providers

@DanielFischer That exercise doesn't even mention anything about the continuity of $f(x)$. It simply says, "Show that if $F(s) = \int_{0}^{\infty} e^{-sx} f(x) \, dx $ converges for $s=s_{0}$, then it converges uniformly on $[s_{0}, \infty)$."

This doesn't make sense.

You can't just put ask someone to give in three values for the price of three items, and output combinations of them that give you a precise amount for a precise price.

If you must spend $100 to get 100 items, and you input prices 0.1, 0.1, 0.1. There is no way you can spend exactly $100 to get 100 of them, you must get more.

@Agawa001 do anything to watch it, it's a great game you might not like to miss. ;)

@RandomVariable Hm. I'm not even sure that is true. Could be, but since $f$ could be ugly, I expect that it won't be quite easy to show.

@Chris'ssistheartist :'(

eurosportplayer.fr/denied.shtml

@Agawa001 Uh ... (bad)

racist website

x=100(100-b)/(101-b)=100(100-1)/(101-1), for (a,b)=(101,1) , x=99, y=1 and ax+by=101x+y=10000 which verifies

@Owatch ....

Thanks, I will look at that momentarily. I've trying to figure something out here briefly, and I'm very frustrated.

this solution is just a drop from a sea

too many other (x,y) (non negative )couples

It looks like given I need (In my example case), 100 of a and b, where price(a) * a + price(b) * b = 10,000

If I choose a = 105, and b = 25, there is no combination.

@DanielFischer If it is indeed true, would it directly imply $\lim_{s \to 0^{+}} \int_{0}^{\infty} e^{-sx}f(x) \, dx = \int_{0}^{\infty} f(x) \, dx$ as long as $\int_{0}^{\infty} f(x) \, dx$ converges as an improper Riemann integral? Or is there something I'm missing here.

@RandomVariable It would.

@Owatch You said just x+y = 100 , not about a and b

@Owatch (price(a),price(b))=(101,1) so (a,b)=(99,1) there is combination

@DanielFischer It would explain why I've never been able to find a counterexample.