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12:00 AM
BTW, you should definitely see the main ideas in the proof of Gromov-Tischler. It's really very soft.
 
I wrote multivariable calculus notes for MIT students when I was an undergrad, but I never wrote things other than seminar lectures as a grad student. Somehow, though, I had decent exposition skills — or so I've believed.
Well, when you write that up, send it to me.
 
Well there is no doubt that you have decent exposition skills :)
@TedShifrin Absolutely.
In fact, "decent" is an understatement
 
Well, I wasn't asking for compliments, but I definitely don't have to work at it as hard as a lot of people. The creative math ... different.
 
Writing is the painting of the voice. -Voltaire
 
I think it's a shame that new mathematics keeps exponentially growing but the exposition is done so poorly (people want to write quickly and move on to newer projects, etc) that it is nearly impossible to understand for someone outside a narrow field of expertise.
I genuinely believe that it would be untenable to do math in a few years if people don't step back and try to communicate more than proving newer stuff
 
12:05 AM
Well, it's sorta like promotions value research over teaching at almost every institution of higher learning. Similarly, journals prefer short articles to ones that have exposition that less expert readers can learn from. I fought to keep a reasonable amount of exposition in my published thesis.
I'm surprised I got away with as much as I did.
 
Yeah
 
couldnt agree more. its a sad state. also it seems like those who dont have access to profs that are "in the know" have no hope of learning by themselves as there is too much info that is not explicitly stated but implicitly "understood" by the experts.
anyway thats my 2 cents. total trigger topic for me ha. im off! have a nice weekend everyone
 
You too
 
Take care, @Quin.
 
cya
 
when i need to write something, i often find it helps to create a few slides & pictures to get the general flow and then backfill.
 
I can now write at the computer, but certainly not in slide format. If there are involved calculations, I need them on paper before I type.
 
unfortunately, despite growing up with computers, i am still firmly in the pen & paper camp for initial outlines & pictures.
 
Yeah, I have to use pen and paper. I generally have to be wordy from the beginning to clear the story up in my own head first.
 
I actually typed a lot of my multivariable math book as I went. Of course, revising is a lot easier in the computer age than it was in the old typewriter days.
 
12:18 AM
I cannot understand symbols, sometimes even when I am writing them.
 
The stories are that Lang (and Kobayashi) wrote their graduate texts at the (old-fashioned) typewriter.
 
Insanity
 
i need to use specific, familiar symbols. i am not flexible in that regard.
 
My differential geometry text I wrote starting with my very cryptic lecture notes from teaching the course a few times.
 
i liked Kobayashi a lot.
 
12:21 AM
Yes, he was a sweet man. I now contribute to the fund in his name every year.
 
very gentle fellow.
 
His daughters took after his wife more ... sharp and more outgoing.
Gotta love the nerve of some people. Scroll down to the end.
 
lol
the other answers literally do not even contain the phrases "clearly" or "as you can see"
guy is tripping
 
so strange
 
trollers gotta troll
 
12:25 AM
entitlement riles me. the same way bullying does.
 
@Thorgott he said answers on this forum
by which he means all of MSE
not particularly that thread
 
ah, of course
mans calling out the entire site under a single question
 
^
 
massive legend yeah
 
i girl whose last name was Kobayashi took some matrix class with me (M221?), I assumed some relation.
 
12:26 AM
the only one brave enough to speak out against big clarity
 
big clarity lmfaooo
 
That's possible, @copper, although I think it's not an uncommon name. The girls were a few years younger than me, as I recall.
 
big pharma, big government, big clarity - the deep state trinity
 
i've has too much negative mse energy lately so i'm not actively entering frays for the moment.
 
I entered for you.
 
12:28 AM
@TedShifrin I am fairly sure she was, it's been a long time and I am not at all shy when it comes to asking such things.
 
I'm trying to remember their names. His wife was Grace. She was terrific.
 
I only met some of Kahan's family members.
I mean in the math detpt
 
Take a monic integer polynomial. Now take the polynomial whose roots (with multiplicity) are squares of the roots of the polynomial we started with (all over C). This polynomial has integer coefficients again. is there a direct way of seeing this?
 
Does this follow from symmetric functions?
Newton formulas or some such ...
 
The only argument I have is that the polynomial is the characteristic polynomial of some integer matrix and the polynomial with the roots squared is the characteristic polynomial of the square of that matrix.
 
12:33 AM
Oh, that's a fine argument.
 
It... works. But it doesn't please me.
 
I was thinking of $\sigma_1 = \sum \lambda_i$ and finding the standard formula for $\sum\lambda_i^2$, etc.
I think your argument is far cleaner.
 
@TedShifrin Yeah that's what I would do
I mean, if $\alpha$ is a root of the original polynomial, $\Bbb Z[\alpha]$ is integral over $\Bbb Z$, so in particular $\alpha^2$ is also integral over $\Bbb Z$, and the Galois conjugates are $\beta^2$ where $\beta$ is a conjugate of $\alpha$.
Because Galois automorphism are field automorphisms, it commutes with multiplication
 
That's a nice abstract proof.
 
the engineer in me prefers concrete...
 
12:38 AM
ah right, that's the clean version of what I was saying
 
LOL so by clean version you mean topos
Got it. That's why you weren't pleased.
 
LOL
 
hey Balarka and Ted, how familiar are you with the K3 surface?
 
This is totally Ted's cup of tea, not mine.
 
I'm just speaking topologically
 
12:45 AM
Ted used to be more familiar. He's old and rusty.
Quartic in $\Bbb P^3$. So we can figure out stuff.
What topological are you wishing to say?
 
how well do you know its singular homology?
 
The betti numbers and intersection pairing on $H^2$ are well known.
 
do you know the intersection pairing explicitly?
 
Some stuff comes from Lefschetz hyperplane theorem.
 
usually, it's deduced by abstract classification
 
12:48 AM
I don't remember it, but it's all classical stuff. It's all in Griffiths/Harris, for example.
So what specifically are you asking ... before I disappear to go cook dinner?
 
there was no further punchline, just whether you know an explicit description of the intersection form (where by explicit I mean something like giving a concrete basis of H_2 by explicit generators which realizes an isomorphism to E8+E8+3H)
 
Ah. That is all well-known, but not by me at this stage of my life.
 
I will be looking at Griffiths/Harris. It's a great book from what little I've read, but it leaves me in the intellectual dust.
 
I have issues with the second part though.
 
12:53 AM
@Koro: First of all, for the first direction, can't you prove quite generally that if you take $f=(f_1,f_2)$, then $f$ is continuous if $f_1,f_2$ are both continuous?
 
@Ted: I thought that but then thought that's probably true in R^n only
 
You can do the open set proof or the metric space $\epsilon$ proof. But contradiction is excessive.
No, it's true in total generality.
 
Then it's okay. The first part reduces to two to 3 lines then. Done!
 
OK. Now let's move on.
 
What about the second part?
 
12:55 AM
So my hint was to consider $f^{-1}(C)$ for $C$ closed. How do we get this from the graph?
 
So I considered inverse image for inverse of C.
closed set.
 
$f\colon E\to Y$? What's the range space?
 
Ted, I'm new to topology and metric spaces so my question may sound trivial also to you.
 
I was told that it might not actually be well-known this explicitly, though I wouldn't know. The reference I've looked at so far deduced the isomorphism type only from an abstract classification of unimodular lattices, which reduces the problem to computing rank, signature and whether it's even or odd.
 
No, I'm not putting you down here. I want to get the set-up right.
 
12:57 AM
Ted: range space would $f(E)$
 
No. That's the image. What space is it a subspace of?
@Thor: To be honest, I thought it was more concrete than that, but I may be misremembering after decades.
 
I never used "subspace" term in continuity as far as Rudin's chapter on continuity is concerned.
 
I will let you know once I find out
 
I don't know answer to your question Ted.
 
If it hasn't been done before, I will give doing it myself a serious try
 
1:00 AM
I went back to Rudin. I think he's considering only real-valued functions, so the answer to my question is $Y=\Bbb R$. It's true in total generality here.
Maybe my question was irrelevant.
 
Ahh you were saying that. No he's not considering only real valued functions though
Let me state the question once again as stated in the book
 
I just looked it up. It's totally vague.
 
classic Rudin
 
I think he wants us to first prove it for any $E$ and then in particular for real valued. That's what I understood
 
I didn't remember that as an issue with Rudin, but it really is. OK, I want to think of the graph as a subspace of $E\times Y$. No, he's giving a specific example. You have that backwards.
 
1:03 AM
Ok Ted. I got your point. But is it really required?
I mean we can have consider $E\times f(E)$
It only complicates things to introduce Y. No?
 
Well, I want to take a closed set $C$ in $Y$ and look at its preimage. By definition, we're taking the preimage of $f(E)\cap C$, and that is closed in $f(E)$.
No, it's natural to have $Y$ or else I have no idea where $f(E)$ even lives.
Rudin is being truly sloppy (and very old-fashioned, but the book is old).
At any rate, how do I see $f^{-1}(C)$ from the graph?
 
Ahh. Right. Where will $f(E)$ live then? I didn't think about that :) It makes sense to introduce Y
 
Any function needs to be given as $f\colon X\to Y$.
Otherwise, the function is not even defined.
 
functions have to have codomains. ted and i agree yet again.
 
set theorists might disagree
 
1:05 AM
really getting sick of this.
 
@Leslie I paged you earlier.
Are we making you sick?
 
Very true. I thought function from $E$ to $f(E)$ would give us an extra advantage of $f$ being a bijection (if it is one -one in this case)
 
Oh, I missed the reason.
 
i'm still puzzling over that. i wonder if they mean to ask a slightly different question.
 
Not bijection. Don't even go there.
 
1:06 AM
my daughter just said the f-word.
 
Alright. Not going there
 
frack?
 
It's all your fault, @Leslie. Behave.
 
@leslietownes food
 
LOL. frak for fraktur?
 
1:07 AM
it's very funny but we can't acknowledge that it's funny.
 
NO.
@Koro. Answer my question. Maybe a picture would be good. How do I see $f^{-1}(C)$ from the graph?
 
she was cussing out the cat, too. we didn't laugh but it's very funny.
 
can a lattice meet a ball of finite radius in infinitely many points?
 
I cuss out my kitten, but not with four-letter words.
 
i uttered the word once after missing a bart train and my 2yo daughter immediately started saying f*,ff,...
 
1:08 AM
Hell no, @Thor.
 
Alright so pre-image of $C$ under $f$ is $\{x \in E: f(x) \in C\ }$
 
Do you see a picture for that, @Koro?
 
anyway, now she uses the word entirely appropriately
 
I still don't have the graph in there.
Well, @copper, she's no longer 2.
 
ted, we'll eventually need to see photos of the kitten.
 
1:09 AM
Yes Ted. $f$ is pulling closed sets back
 
if there's one thing the internet craves, it's photography of cats.
 
that's what I am thinking too, but apparently I'm messing something up
 
true. she manages to get a few syllables into the word christ
 
In the worst case, $f$ will not have any image in $C$
 
I want you to give me the interpretation in terms of the graph.
We only know things about the graph.
 
1:10 AM
In which case, $f$ inverse is empty set but empty set is also closed.
 
Forget that. You get side-tracked so easily.
 
assume $\alpha\in\mathbb{C}$ is an algebraic integer of absolute value. then $\mathbb{Z}[\alpha]$ is a lattice and it contains all powers $\alpha^n$, which all lie on the unit circle. so these should only be finitely many values, so $\alpha^n=\alpha^m$ for some $n\neq m$ and cancelling shows that $\alpha$ is a root of unity.
however, it's not true that every norm 1 algebraic integer is a root of unity, so where am I being stupid
 
This picture is a few days old, @Leslie, but it'll give you an idea.
 
that's cute.
 
Yes, he's way cuter than I ever was.
 
1:12 AM
 
Oooh, seriously black. I always had black-and-white or tabby-ish cats. The orange calico is new for me.
 
she cannot be seen after 8 pm or so. she uses this to her advantage.
 
I bet.
@Koro: I'll give you a battle-plan, so you can work it out for yourself. I want to turn $f^{-1}(C)$ into the projection of some compact subset of the graph (onto the first coordinate).
 
Ted: please tell me what was the intuition for inverse image for C under f
I didn't really get that.
interpretation in terms of graph
 
@Thorgott Why does the lattice contain all powers of $\alpha$?
 
1:18 AM
ok duh, "lattices" can meet finite radius balls in infinitely many points
the issue is that lattice here just means finite rank free abelian subgroup, not necessarily discrete
 
does lattice mean lattice?
ok, that's what i was wondering.
 
yesn't
 
Oh, different notions of lattice. Damn number theory.
@Koro: Have you drawn a picture? Draw $C$ on the $y$-axis, and show how to get $f^{-1}(C)$ from the graph using that picture.
 
@TedShifrin That makes sense to me.
 
the simplest example is that $\mathbb{Z}+\sqrt{2}\mathbb{Z}$ is dense in $\mathbb{R}$
 
1:19 AM
@Thorgott I know that one :)
 
I wanted a rank $n$-lattice in $\Bbb R^n$.
 
i still have my sign ready to go, for any protest. FREE ABELIAN GROUPS.
 
Never mind.
 
@TedShifrin Alright. Let me try
I understand now what to draw in picture :)
 
yeah, sloppiness got the better of me this time
 
1:24 AM
Sloppiness is against your religion.
OK, I have to disappear to the kitchen to cook dinner. Back later.
 
Okay :)
 
Previet
 
Nah
Bonjour or namaskar would also work
Hola is also nice
 
Bonjour, ca va?
 
1:34 AM
ya
 
I have never seen anybody say namaskar in any chat. "Namaste "name here" ji" is more common. @RussianBotWhoKnowsYourIP
 
namaste name here ji to you
2
:)
 
what about ni hao
 
1:36 AM
Haha. No that's not what I meant @leslietownes
You have to put your name in place of "name here" :)
 
Does anyone here knows conlangs
 
namaste your name ji
am i doing it right
 
Lol
Namaste leslie ji
 
a friend of mine in grad school was very into esperanto.
 
Namaste Leslie ji
 
1:44 AM
i didn't know the ji part. i had only seen namaste. mostly from people running yoga studios.
 
ji adds politeness
 
i like that. i'm going to do that from now on.
 
It's like difference btw Mr. Leslie and Leslie
But much more than that.
 
that's a new thing i've learned today. thank you.
 
Adding ji after a name is very polite
 
1:45 AM
Oye leslie namaste
Arey maine kaha namaste
Lmfao
I love it when people take the format literally
Namaste name here ji to you
 
@TedShifrin I understand it now. So we can take projection on $x-$ axis.
Thanks a lot for your help :) @Ted.
 
2:20 AM
Hey, Ted. How was dinner?
 
I just finished preparing. Letting it marinate/rest for a few minutes.
@Koro: Glad you got it. Cool argument, isn't it?
 
Ah.
 
If I can just train people in here to draw the right pictures, they won't need us anymore.
 
I may not be able to draw the right picture, but I can draw the wrong conclusion.
 
2:36 AM
@robjohn Well, your talents are unparalleled.
 
@TedShifrin Ted: that was fantastic argument. That alongwith (f_1,f_2) is continuous iff f_1,f_2 are continuous solves my problem :)
Thanks a lot :)
 
3:11 AM
Tetris was made by a russian
 
Yup, you're welcome. Rudin doesn't necessarily teach you things you really need.
 
tetris is a great game.
rudin teaches you ways of looking at stuff, but it's not that citable as a reference. i ran into that several times, not quite finding the thing i wanted to use as a textbook theorem.
 
I hadn't paid that much attention, never having actually taught out of it (as a student, it seemed fine to me), but talking with Koro, I realize he's totally vague about defining a function with domain and range, etc.
 
i think sometimes rudin falls into the trap of not wanting to explain too much, or give people citable things they can just plug into other things. i don't know why. there's sometimes an impression that this sort of treatment dumbs math down. i should ask my friend who took classes from him whether he was like that in real life.
 
I'm just saying a function given in an exercise should have domain and range. But a book written in the 50's wasn't so carefully written.
Generally, my view was that Rudin is fine for our top students, but not for the rest.
 
3:22 AM
ahlfors is similarly weak on the kinds of fundamental things you would expect a modern book to do better.
complex analysis is tough. sometimes you don't want to define a series as only a function on an open disc, because it converges for lot of the boundary of the disc too and maybe continues beyond that in some sense. so maybe you leave out exactly where you're talking about it being a function.
 
Ahlfors didn't bother me, though, even though it's dated, because his mathematical taste was good and those were mature students studying from it. At least one year more mature.
 
too geometrical for my taste.
(:
 
Big shock there.
It's really not particularly geometric.
But more topological than Conway, say.
 
i like conway's focv i. focv ii is not that good.
one of the nevanlinnas had some notes on entire functions that were really good.
 
3:37 AM
I had R. Nevanlinna's book on value distribution theory. Don't know other.
 
that might be it.
 
@TedShifrin I know domain and range and codomain. I don’t know what made you say that. :’(
 
i think he meant that rudin was vague, not that you were.
 
:)
 
'codomain' is a very late 20th century concept. before that everybody was winging it.
i think some of the bourbaki books even fudge this.
 
3:41 AM
One question Leslie: Did they use sets in 1930?
 
Remember that we never said where the values of $f$ were. All Rudin said was $f(E)$.
 
I have a book that uses the word “class” for sets
 
Codomain is NOT a late 20th century concept. What are you talking about?
I was taught domain and range in the 60's.
No, class is different.
There is a hierarchy of sets, collections, class.
 
would the 1960s be the late 20th century? i don't know of earlier references that really pin it down.
'class' did mean something bigger than set, even in the 30s.
 
Right. Like for dedekind cuts, class of rationals satisfying $r^2\lt 2$
Etc.
 
3:44 AM
i was taught domain and range in the 1990s in high school, codomain came later. it has not been absorbed into the hs curriculum.
 
But I think set word was not around back in 1930s
 
for a lot of people functions are just rules and where they end up is someone else's department.
it was. cantor was already very involved in it in the later 19th century.
 
Codomain is the same as range. I don't care a flip about the terminology.
 
you began to need it to discuss convergence of fourier series.
 
Codomain is not same as range.
 
3:45 AM
that's interesting, ted. for many people range is {f(x): x in domain} which might be a proper subset of the codomain.
it's all kind of a matter of opinion. is a function a set of ordered pairs, or is it more than that.
 
The how would you distinguish between surjectives and non surjectives.
 
it's not fully worked out even now.
 
I see function as a special type of relation.
 
i had an 'intro to proof' book that spent what i thought to be far too much time on a function being a tuple of domain, codomain, and then a subset of ordered pairs of domain x codomain. precisely so you could say f(x) = x^2 was surjective R to [0, infty) but not surjective R to R and that those were different functions.
it was hyper formalistic.
they're not different as sets of ordered pairs. i didn't understand the need for the attention. the sharp students could appreciate the distinction and the rest were completely confused by it.
 
ahh, the corange and antidomain
 
3:49 AM
bourbaki may have changed how they did this from one edition to another.
 
What??
I never heard corange before
Anti-domain?
 
you can co-anything, koro. and anti-anything.
reverse the arrows.
 
@Koro sorry, i am just joking, playing with words.
 
Ahh I see
 
as i am wont to do
 
3:50 AM
copper hat is engaging in something intrinsic to the irish national character, which is talking s---.
 
my son left for his prom, so i am all lonely now
i talk excellent $\text{sh}_1\text{t}$.
the chemical symbol
 
i generally love irish people. my dad's cousins are the funniest people i've ever met.
 
i am obviously very comfortable with irish culture, but broadly view people as much the same everywhere
 
there's something about 'the gift of gab' which is true. the culture encourages non substantive storytelling. nothing revelatory of emotions or intellect, just shaggy dog stories that take forever and go nowhere.
 
I refuse to think of myself as the "same" as Gaetz and Greene, for example. Deplorable excrement of human beings.
 
3:53 AM
maybe i'm wrong about that. it could just be my dad's cousins.
 
My friend recently went to Dublin for his higher studies
@copper
 
i have found that when irish people meet each other abroad (not while traveling, or temporary situations) they often seem to avoid each other :-(
 
ted and i agree for the ten billionth time.
 
irish people broadly love company
 
Oh dear. I'm surprised by that statement, @copper.
 
3:54 AM
i agree with ted, but much admit i am close to many people who hold questionable views
 
This is the far limits of "questionable views."
 
i presume that was in jest @TedShifrin :-)
i mean about the surprise
 
That Irish people want to avoid one another? No, that sincerely surprises me.
 
avoid might have been stronger than i intended, but not the kinship of movies
 
i want to avoid americans when i travel abroad. they seem louder and more garish than i am. or than i imagine myself to be.
 
3:55 AM
@Leslie, on that I agree, sadly. Most American tourists act like boors.
 
i think americans get the short shrift because of their broad wealth
 
I try to assimilate as much as I can, especially in countries where I can speak the language to some extent.
 
they're very loud. and they dress like s---.
 
@copper I think some Asians and Russians get it more.
 
my dad was never so flattered than he was when a native parisian came up and asked if he knew the way to something.
i told him later, it is because you have a gallic nose, not because your french is very good. he reminded me that i have that nose too.
 
3:57 AM
@copper.hat I do not have a lot of broads!
 
@robjohn :-) my response is curtailed as i do not want to cause unintended offence to theose who are not present :-)
america is a huge place with gazzillions of cultures, it is unfair to expect some degree of uniformity
 

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