00:00 - 15:0015:00 - 00:00

12:09 AM
Is there a standard term for $\gcd(\{P(n) \mid n\in \Bbb Z^+$ when $P$ is a polynomial with integer coefficients?

i do not know of one. sometimes things that can be expressed in short formulas with things that do have standard terms, do not get their own standard terms.
but the fact that i don't know it doesn't mean anything. this has been my useless answer of the day.

I don’t think about these things, but is there an example where it’s different from the gcd of the coefficients?
Oh, I see an easy example.
Sum of an even number of odd coefficient terms.

12:28 AM
For $x^2+x$ it is 2, not 1.

12:39 AM
Yes, read what I said next.

I did.

1:04 AM
barely survivable heat in albany today

not too bad here, although i'd rather be in albany.

apparently gavin declared a heat emergency
there's a cool room in the french laundry i hear

my daughter's yelling about not being allowed to bring handfuls of dirt into the house

As long as it’s by the truckful!

1:27 AM
"sorry olivia, you can't go outside. you're not old enough." explanation given to the cat for why she couldn't go with us to check the mail.
probably hears that explanation a lot at day care.

2:13 AM
@robjohn Are you free right now?

It's just about dinner time. Is it something quick?

Hi chat
I mean math gurus
^_^

it isn't something quick. everybody be quiet and maybe the whole thing will just blow over.

yo yo yo

2:33 AM
Hi @leslietownes @EM4
What maths are you studying lately?

I took day off rest so I read environmental anthropology haha, you?

i study what i see on the site and what pops into this chatterbox. i am not an active user of my own math.

@EM4 I'm trying to work a problem in Hartshorne's AG. It's a very advanced book, but hopefully I can swim and not sink. I don't care if this problem takes me a week to get. I'm going to work on it without looking at this solution manual on github
The solution manual was written by someone 10x smarter than me at least. There is no learning if I just solution manual every problem.

good luck. my school had a class where the syllabus for a 15 week course was something like 'we will do chapter 1 of hartshorne and a few topics from chapter 2, time permitting.' i exaggerate but only a little.

@leslietownes yeah, I am skipping some problems that just aren't aesthetically pleasing to me. I hope doing that doesn't prevent me from learning the subject

2:39 AM
my impression is that there's enough in there to keep somebody off the streets and out of trouble for a good long while.

It's as if, if you could do all the problems in this book in 2 years. Then you already knew the material to begin with...
@leslietownes I'm starting Open University (OU) distance learning for a Bachelor's in Mathematics this upcoming September
The first course I chose is statistics, which I'm completely new to except I know a little about Lebesgue measure
After I get a bachelor's I want to persue a PhD or Master's. I think in 4 years, I will be mathematically mature enough to start up doing that
After the PhD, I want to just keep going to school for life, lol
I will get a job where I can work from home, and do distance learning when I'm not working.
I'm good in school as long as I'm not studying grassmannians

I agree with you @OlympicComputerChairSitter I am the same way.
STEM is a drug.

what's that
Science Tech, Engineering, Math?

STEM is Science Technology Engineering Mathematics.

Nice
Yes, it's drug-like, and so we should take ample breaks to recover and do other things
such as exercise

2:50 AM

hey congrats on obamacare yall

What's that, what happened?

supreme court ruled in favour of upholding

hahhahha @leslietownes

@shintuku that's good
I heard that in Oregon, they've decriminalized posession of small amounts of any drug. Which is how it should be

2:52 AM
i now understand what you meant by ample breaks to recover and do other things.

I don't do a lot of drugs, mostly coffee and nicotine gum
*gum

Cyberbulling is a great way to recover from math infatuation

i think coffee becomes drugs if you take enough

It is a drug

yes but not drugs

2:54 AM
80% of the world is addicted to caffeine

i'm not addicted, i just like the way my tea tastes. eight or nine cups a day.

I am water guy LOL.

i'll die if i don't get my water.

i do a 2 day detox of coffee once a month

2:57 AM
Also, the aire we breath is addictive

it is painful but it seems like drinking coffee adds up after a while

Therefore, you should hold your breath for 1 minute every hour to detox
Also, sleep can be addictive. You should stay awake 1 / week to detox

i think it's pretty dehydrating if it's all you drink but otherwise mostly OK. i vary in a lot of noncaffeinated herbal 'teas' so i'm not buzzing all day

@leslietownes that's good. Sometimes we just want a cup of hot water with herbs

why is tea in single quotes haha
sounds like it is something much worse

2:59 AM
@OlympicComputerChairSitter ... i dont think you are using that word properly

right now i'm drinking burdock root we'll see what happens.

I know, hyponatremia, or something
I heard that in order to fully understand Grassmann's work you need to be high on a certain drug...
And that drug is caffeine

what work of Grassmann?

Hermann Günther Grassmann (German: Graßmann, pronounced [ˈhɛʁman ˈɡʏntɐ ˈɡʁasman]; 15 April 1809 – 26 September 1877) was a German polymath, known in his day as a linguist and now also as a mathematician. He was also a physicist, general scholar, and publisher. His mathematical work was little noted until he was in his sixties. == Biography == Grassmann was the third of 12 children of Justus Günter Grassmann, an ordained minister who taught mathematics and physics at the Stettin Gymnasium, where Hermann was educated. Grassmann was an undistinguished student until he obtained a high mark on the...

what did he do?

3:05 AM
In mathematics, the Grassmannian Gr(k, V) is a space that parameterizes all k-dimensional linear subspaces of the n-dimensional vector space V. For example, the Grassmannian Gr(1, V) is the space of lines through the origin in V, so it is the same as the projective space of one dimension lower than V. When V is a real or complex vector space, Grassmannians are compact smooth manifolds. In general they have the structure of a smooth algebraic variety, of dimension k ( n − k ) . {\displaystyle k(n-k).} The...
It's a coincidance that the grassmannian is like a bundle of long blades of grass

i know a grass man who lives in oregon. i wonder if they're related.

Yes, they are related, since they are both of the same species
Manjul Bhargava (born 8 August 1974) is a Canadian-American mathematician. He is the Brandon Fradd, Class of 1983, Professor of Mathematics at Princeton University, the Stieltjes Professor of Number Theory at Leiden University, and also holds Adjunct Professorships at the Tata Institute of Fundamental Research, the Indian Institute of Technology Bombay, and the University of Hyderabad. He is known primarily for his contributions to number theory. Bhargava was awarded the Fields Medal in 2014. According to the International Mathematical Union citation, he was awarded the prize "for developi...
He holds a professorship at many different colleges. Einstein of our time
That's a paper of his that seems accessible to an undergrad

interesting. i used to love reading AMM for that reason, you'd see really good exposition in there. and sometimes from the mathematical equivalent of celebrities.

3:30 AM
Hi Leslie
What is the difference between Herstein's Topics in Algebra and Herstein's Algebra
?

3:44 AM
i think one is more advanced or comprehensive and the other isn't, but i never got too familiar with either one.
my officemate did own something that said herstein algebra.

I already had Herstein's algebra book and I have been reading it. Today in one of the proofs it said : for more proofs refer our book "Topics in Algebra"

or maybe it's a pure cash grab, an attempt to confuse consumers into buying the same thing twice. like after 2pac died and his label kept putting out 'new' songs with old verses on them.

It seems that in Herstein's algebra many theorems have been shifted to exercises
For example: the famous relation: $|HK|= \frac{|H||K|}{|H\cap K|}$ etc.

I already know that (xdy-ydx)/(x^2+y^2) is closed but not exact on $\Bbb R^2-0$ But what if closed 1-form $w$ on $\Bbb R^2-0$ is continuous on some disk centered at the origin?

No,Topics in Algebra was the original ballbuster book. The next was watered down for merely mortal students.

3:51 AM
I want to show in that case, it's exact

@barista if it’s smooth and closed on a simply connected region, then it's exact.

Ted, I didn't know that before, I thought both were same. I've almost finished the group part in Herstein's algebra and I use Gallian's book also along with Herstein's.

That said, Herstein is one of my least favorite texts ever.

I still feel like I struggle with isomorphisms...

@TedShifrin Yes but mine is smooth only on $\Bbb R^2-0$ not $\Bbb R^2$.

3:52 AM
Gallian is rings first, isn’t it?

@TedShifrin You have expressed that few days earlier also while suggesting Artin. I remember :-)
@TedShifrin Rings first? Rings are introduced after Groups in Gallian

Then we don’t have Green’s Theorem necessarily.
Oh, ok, I misremembered.

To do that I first tried to prove the integral of $w$ over the boundary of that disk is zero
Maybe I can use stokes' theorem but well I wonder if it's a smooth mfd

Oh, apply Green on an annulus. It works.
Green = Stokes here

@TedShifrin You mean like a Key hole thing?

3:56 AM
Show that the inner line integral goes to $0$ As you shrink the radius.
No, just plain annulus .

4:09 AM
Ok I let $\gamma(t)=(rcost, rsint)$ where $r$ is radius of inner circle $C_r$. Then $\int_{C_r}w = \int_0^{2\pi}\gamma^*w$. Writing $w = w_1dx+w_2dy$, $=\int_0^{2\pi}-w_1(\gamma(t))sint dt+ cost w_2(\gamma(t))dt$

I find it very difficult to prove two groups isomorphic (To be more specific: in writing the explicit isomorphism). I feel like I am missing something in isomorphisms despite having understood the concept. I am not confident when I see "isomorphism". For example: $\frac {Z_4 \times Z_{12}}{<(2,2)>}$ is isomorphic to which of these: 1) $Z_8$ 2) $Z_2\times Z_2 \times Z_2$ 3) $Z_4 \times Z_2$.

@TedShifrin I think continuity of $\gamma$ should work here

I have rejected 1) and 2). For, there is no 8 order element in the aforementioned quotient group and there is one order 4 element in the quotient group so 2) is also not correct. 3) seems to be but I don't know how to write the explicit isomorphism :)

Oh I missed $r$

for small groups the orders of elements can decide the issue, particularly if they are abelian groups. i wouldn't discourage you from thinking in terms of maps but a lot of finite abelian group theory can be approached one level back from that. at the cost of some investigation and maybe some machinery

4:16 AM
Ok I missed $r$. I can write $\int_{C_r}w=r\int_0^{2\pi}(cost w_2(\gamma(t))-sint w_1(\gamma(t)))dt$. Since $w$ is continuous on $\Bbb R^2$, the integrand is integrable, so as $r\to 0$ the whole integral converges to $0$. What do you think @TedShifrin ?

disclaimer, i am not an algebraist. just a guy who took a few algebra classes once.
yoour book may have interesting isomorphism theorems involving direct sums/products that bear on quotienting.

Open source

@leslietownes There are 1st homomorphism theorem etc. but not yet introduced..

one thought without any theorems is to figure out or recall that Z_4 x Z_12 is Z_4 x Z_4 x Z_3, see what (2,2) looks like under that isomorphism, and then see if that helps the quotienting operation look simpler to you.
these thoughts could be worthless.

@barista And where is Stokes?

4:32 AM
@TedShifrin If I let $D_r$ the be the annulus obtained by removing radius $r$ disk from the given disk of radius $R$, then by stokes, $\int_{D_r} dw = \int_{\partial D_r} w = \int_{C} w+\int_{C_r} w$ where $C$ is circle of radius $R$. $dw =0$ by assumption, letting $r\to 0$, $0=\int_{C}w$

OK, except watch orientation.

@leslietownes I wrote all the elements of the quotient group (tedious task) and then using Cayley table got some intuition... I hope this will help write the isomorphism

@TedShifrin You mean orientation of inner circle runs clockwise right?

Right.

it's generally pretty annoying to define maps from finite groups to one another without a tiny toolkit of theorems. you have to think very much in terms of elements or basically end up proving aspects of those theorems.

4:38 AM
@TedShifrin From this result, I'd like to use the following fact to conclude $w$ is exact : If $w$ is a smooth covector field in an open set $U\subset Bbb R^n$ then $w$ is exact if and only if the line integral of $w$ is path-independent.
My $U =\Bbb R^2-0$ is obviouly open

It seems so. @Leslie

So you have that theorem or do you need to prove it?

I have that theorem

OK.

I first thought if the region enclosed by to paths contains the origin then we can find some small ball centered at 0 that does not intersects two paths. Then some argument but I wonder if that works
I mean I only know the integral over the boundary of the given disk is zero

4:45 AM
It is an annoying proof to get rigorous. You have to worry about how many times a closed path goes around the origin .

@TedShifrin Well then would be easier to show $w$ is conservative?

That doesn’t make sense, but it is no different.

Right..

1 hour later…
6:17 AM
3

In Fourier optics it is sometimes convenient to think of lenses as "Fourier transformers". For an imaging system between two planes with a pupil in the center, the amplitude in the pupil is the FT of the image. A simple way to think about this is that a lens swaps angles and positions (for a plan...

asked in Physics SE, perhaps I should have posted in Mathematics SE instead

7:12 AM
$f:X\rightarrow Y$ and is not surjective so some points in Y are not "hit". Then what is the preimage of these points that aren't hit? Would it be the empty set? Or is it invalid question?

1 hour later…
8:22 AM
@Larry it would be the empty set.
$E=\{x\in X:f(x)\in\{y\in Y:(\lnot\exists x\in X:f(x)=y)\}\}$
$E=\{x\in X:f(x)\in\{y\in Y:(\forall x\in X,f(x)\ne y)\}\}$
suppose that $e\in E\subset X$, then $f(e)\in\{y\in Y:(\forall x\in X,f(x)\ne y)\}$
that is, $f(e)\in Y$ and $\forall x\in X,f(x)\ne f(e)$
but $e\in X$, so we have a contradiction, and thus, $E=\{\}$

1 hour later…
10:01 AM
Fix $x\in H$ where $H$ is a Hilbert space. Let $\phi$ be a convex functional on $H$, why is it obvious that the above should hold?

1 hour later…
11:19 AM
@TedShifrin My mistake. I suspect you already know that the proof I stated before shows $\int_{C_r}w =0$ where $C_r$ is a circle of radius $r$ centered at $0$. But the fact that $w$ is exact is still in mystery.

11:50 AM
The teacher writes Ga-lo-is. I hear Ga-lo-a. My brain.exe stops working .Can someone explain why it is pronounced like that?

It's French
Not all languages are written phonetically

@EdwardEvans Oh ok.

@Prithubiswas I've always heard it pronounced with two syllables. gal-wah

12:57 PM
Hi all, is there a special name for the sum of prime power exponents of a natural number? Anything known on that function? I suppose one could express it with the standard number theoretic functions easily.
Hi @EdwardEvans!
How are you doing! Hope fine!
@Prithubiswas for example English, which is phonetically pretty messed up.

French pronunciation is actually far more regular than English. But the sounds of the vowels and diphthongs (vowel combinations) in French are a little different from those in English. Also, in French the final consonant is generally silent unless it's C, F, L, or R. French kids tend to master spelling somewhat earlier than English speaking kids do.

@RaphaelJ.F.Berger Jo grüß Gott!

Griass di!

Ajo s'loft guat bi mir, außer dass as 33 grad hot und i nümma existiera will, alls roger bi dir? :D
scho lange nümma do gsi

French makes writing makes much sense if you are speaker of some dialects. One actually should say that most languages writings when they were invented were phonetically desinged. Just languages have a tedency to diverge and slowly transform so that the gap between writing and spelling gets ever larger.
Ja basst oiss, mia daugs eh wenn so hoass is.
For example "classical Finnish" from the say 60s or 70s were practically completely phonetic each letter has exactly one pronounciation more or less a bijection in between.

1:06 PM
@RaphaelJ.F.Berger Maybe. ;) FWIW, the structure of the sequence of exponents is pretty important in the study of highly composite numbers. So I guess their sum and partial sums are important too.

This is because Finnish writing was invented were lately, only very few hundred yeras ago

:D Mir hond zum Glück gestern im Wald a Kneippbecken gfunda wo ma üs abkühla künna hond hahaha

$\Omega$ and $\omega$ function is pretty close.
@EdwardEvans "cool"
bisch no z Heidelberg?
oda wars Ulm?
oder Tübingen?

Na Heidelberg ischas gsi, bin no do, i d'Wohnung igsperrt lol
Du warsch in Wien oder?

Salzburg

1:11 PM
Ajo beids as gliche

lol
Taugts dr no
?
Geht wos mid de Mädls??

Ja scho, hon interessante Vorlesunga hür, kann mi guat uf die konzentriera wil nix mit da Mädels goht hahaha
Lernsch etz Zahlatheorie oder wia

Win-Loose?

Jo sowas

Super!
Wos fiar a Semeschda bisch?

1:14 PM
kann i da ned säga

glob 4 aber i muass a kle länger macha weil i an semester verkackt hon

gibts do Bachelor/Master?

oder no "Diplom"?

1:16 PM
Na Diplom gits sit dem Mittelalter nümma
bzw. 19irgendwas glob i
kA
aber i mach Master :D

Ah! OK! Guad dass es sogst!

Scheiß drauf

Ein Meisterwerk der Poesie

Hahahahaha!!!

Raphael j.f. burger

1:18 PM
Was is des?

Many European and Indian languages are in the Indo-European family. (Finnish and Hungarian, and a few others like Estonian are in the Finno-Ugric family. And there other ones, like Basque).
The oldest known Indo-European language is Sanskrit, which is quite regular in its phonetics and grammar, although of course we don't know exactly how it was pronounced in ancient times. But the modern Indian languages aren't quite as regular as Sanskrit. The further you get from the root of the ancestral tree, the greater the variation and irregularity.

Basque is an isolate right?

linjusitisc

oida hoid di goschn

@EdwardEvans Very. :)

1:20 PM
I never understood that, it's not exactly in an isolated place on the continent
I don't really know its history though

Modern linguists have a little isolate bias.

Korean is also considered an isolate I think

in the 19th centrury they null hypotheses was that all languagages are related

Again, I know too little about the Korean language to know why

nowadays it seems to have bceome the opposite

1:21 PM
Well the Basque region is in the Pyrenees mountains between France and Spain, which made it fairly inaccessible before modern roads.

The discuss it also for Japanese
iirc

@PM2Ring I see, so it just did its own thing

and some of the native american languages.
Some huy woke up in the morning: Hey I have invented a new language! Lets use that!!

And why ever not

And since the basque are funny people they immediately agreed

1:23 PM
And lo, the Basque separatist movement was spawned

I once read a bit about Basque grammar it seems quite complicated

In the old days, the Basques had a reputation for being highly moral and well-behaved. They had a saying that their language is so hard, even the Devil couldn't learn it, so he could tempt them to be evil. :)

Nice? I remember "ez" means no, is this correct? Edward, what are currently learning/reading for number theory?
Nice! (I mean)

Sorry, I don't know any Basque vocabulary or grammar. I just know that it's hard. :) But I suspect that part of the reason that it has the reputation for being hard is that it's different to the Romance languages of its neighbours.

Yes that makes sense. I was just thinking about this primefactor-power sum and how it might look like. Its 1 for the primes and reaches "somehow maximal" values for n=2^m
and then I was wondering if it ever anywhere becomes "somehow smoothish".

1:31 PM
@RaphaelJ.F.Berger I'm currently writing some talks on Galois cohomology and global class field theory, and also a talk on modular curves
I don't really know anything about "analytic" number theory though

I never really the concept of cohomolgy.
got

eh I barely know what I'm doing either tbf
for me it's currently a bunch of formal garbage that streamlines the stuff I care about
but I wouldn't say I understand why it streamlines stuff I care about

OK! :-)
what is the stuff you care about then?

But speaking of languages used in the Iberian region, I've long been fascinated by Silbo Gomero, which is a whistled version of Spanish used in the Canary Islands. It almost died out, but happily it's now undergoing a revival.

Oh yeah I heared and saw that! Thats really cool!

1:34 PM
is Galois cohomology sheaf cohomology on some appropriate Grothendieck topos

most likely

There is something very fascinating in that direction: There is a native "indian" american language in brasil which can be withou loss of infurmation: whistled or even murmurmed (was heisst murmeln?) with full mouth and closed lips.
The language/and people are "Pirahan"

Mumble
or murmur I guess?

Its an totally fascinating language in many respect. It seems they have (or better to say had) no way of counting. You could ask someone how many childer he has but he could not communicate thats its three. They would say something like some...

Also I "care" about Brauer groups of fields atm, these are equivalence classes of central simple algebras over the field with tensor product as the group operation, it turns out that the Brauer group is isomorphic to a second Galois cohomology group and this totally streamlines stuff (we spent like 5-6 weeks proving lots of facts about the Brauer group by hand and interpreting it through Galois cohomology these facts all become 2-3 line proofs)

1:40 PM
@RaphaelJ.F.Berger You should definitely check out the highly composite numbers, and the superior highly composite numbers, "a natural number which has more divisors than any other number scaled relative to some positive power of the number itself". Ramanujan did some nice work on them, but there's been some progress since then, particularly by Erdős, and Jean-Louis Nicolas & Guy Robin.

There was a big dispute on that language between Chomsky and the guy who first decribed it (Dan Everett) because Chomsky had made this dogmatic claim or "Axiom" that any language has some "recursive" structure. where recursion means something different as it does in maths.

and it seems Pirahan doesn't have this

Any ideas on how to show this? Looks like it has something to do with probability and the standard Gaussian

Thank you fir the hint, I will have a look at it!
@epsilon-emperor you could try it by taking the epsilon derivative

1:46 PM
@RaphaelJ.F.Berger Interesting
Page 3, there's an upper deviation inequality.
I guess that solves my question exactly

@EdwardEvans I get lost at "central simple algebras"

Plenty of languages used by nomadic peoples don't have formal ways of talking about large integers. There was no practical use for numbers >1000. But when people settled down, and got organised into larger groups, bigger numbers were necessary to track livestock & grain, etc, for accounting and taxation purposes.

they're pretty simple
weyyy nah that's cringe

@PM2Ring One, two, three, many, more, lots.

miss a few, 99, 100

1:51 PM
With the Pirahan its just that they do not have the concept of numers at all, if everett is right. There is no way to say even 1. They cannot propery and reliably destinguish 1 from 2 it seems
at least before the were "missionized" and "cultivated" and so on.
They understand that all their children are in the hut and if there are some children missing but they cannot represent how many it is that are missing. they will rather tell the names.
But they already seem to have difficulties in speaking about anything which is not relly present. If you like to read about it: Dan Everett, "Don't sleep there are snakes""

So kids in such nomadic societies learn to count, but they don't get hung up on concepts like "a million". However, they do learn about other mathematical structures. Years ago, I was told that (in the old days, before modern standardised education) that a 5 year old kid in Central Australia may not be able to count beyond 20. But he can easily verbalise the tree of his kin and the moiety of all of the people in that tree, to 5 or 6 generations.

I guess it's hard to understand how one would live without numbers without having lived without numbers

@epsilon-emperor It's false for $\epsilon\lt0$, so it should be stated that $\epsilon\ge0$ before setting people to work on it.

@robjohn I'm sorry I forgot to mention that

1:59 PM
I see that is connected with dual organisation which seems very important on the whole pre-columbian americas
The separate their family or tribe in quite a few dual categories. With marriage-restrictions and so on (I mean exceeding the normal close blood relation ones).

Right. The basic moiety calculations are essentially multiplication in a group of order 4. They aren't hard, but I still think it's cool that little kids can do it as easily as other kids can do simple addition.

And this concept seems to be even more wide spread than only the Americas I think it also exists over large parts of Asia,
Yes, fascinating.

"This bound is sharp up to polynomial factor corrections"
What does that mean?

@PM2Ring I think "nomadic" is the wrong characterization. It is about the size and structure of a society.
Larger groups require greater craft specialization, which engenders a need to trade for those things one cannot produce oneself. This requires accounting, which requires numbers.
Though there is a correlation between larger, more hierarchical societies and sendentism, it is not a perfect correlation.
E.g. the peoples of the Pacific Northwest lived a hunter-gatherer lifestyle at the time of European contact, but happened to be quite sedentary (thanks to abundant resources there). And early Jews were relatively nomadic (they were pasoralists, who herded sheep), but they had a writing system and were relatively numerate (though they were surrounded by larger agrarian societies, so that likely had an influence).

@XanderHenderson "Nomadic" is a bit of a simplification. But there's a strong correlation, because nomadic & semi-nomadic groups tend to be smaller: you generally need settled agriculture to support high population densities & numbers. And nomadic people simply don't accumulate large numbers of things. They may have a large amount of intellectual property, though: songs and stories.
And I guess if you're following a herd of reindeer (or whatever), it is useful to be able to count them. But that doesn't really need large magnitude numbers >1000.

2:11 PM
@PM2Ring I am very aware. I am merely pointing out that social stratification and craft specialization (things which drive economic specialization) are far more important than how much a group moves around.

@XanderHenderson True. I was actually thinking of the Pacific Northwest (with its seasonal abundance of salmon) as one of the exceptions when I said "generally need settled agriculture". :)
(I'm on my phone, so it's tricky to read posts while I'm writing).
@XanderHenderson I think we probably agree more than we disagree. :)

@PM2Ring Indeed. I am being pedantic. I started my academic career in anthropology (archaeology of the American Great Basin, in particular), my father was a cultural anthropologist, and my mother an archaeologist.

Cool.

Oh, and my grandfather was a psychologist, but worked briefly with Alfred Kroeber in the 40s. And I have a cousin who is a cultural anthropologist (his area of study is muslim peoples Indonesia and surrounding islands).

@EdwardEvans whisper?

2:17 PM
That's flüstern in German

I got corrected a lot about these kinds of things while growing up (my father, in addition to being a cultural anthropologist, was a lawyer---if you think I'm pedantic...).

I think mumble is the best translation for murmeln

murmeln Rapper

Yeah
Murmelrap
rofl

2:29 PM
Here's some nice French singing by Tatiana Eva-Marie + Sarah Lancman. Tatiana (on our left) is from Switzerland, but lived in Paris while she went to the Sorbonne, and now lives in New York. She often sings Gypsy style jazz, in English as well as French. Sarah is French, and is based in Paris, AFAIK.

Pirahan can be hummed as well. It has a pretty minimal phonetic repertoire.

hey chat
is there any general rule to determine whether an irreducible representation $\rho: G \to \mathrm{GL}(V)$ is an isomorphism?
or at least a monomorphism

you mean faithful?

(I highly doubt you'll be interested in studing representations of... a group of matrices)
@RaphaelJ.F.Berger yeah

does happen often with irreps?

2:39 PM
sorry, what do you mean?

Well irreps are the small bits and pieces from which you can build a faithful representation.

compute the kernel?

@RaphaelJ.F.Berger oh well. I thought that would happen :p

I think in that case the group has exactly two irreps

I'm not really acquainted with actual examples where you compute properties/characters etc of representations

2:41 PM
the trivial and the faithful.

@Thorgott haha.
You know, the regular representation is always faithful (?)
But it looks too big in general

but is it irreducible
?

nope, that's why I'm asking about irreducibles. I'm trying to find a more appropriate/computationally smaller representation of the same group

the tensor sum of all irreps?

for example, $S_n$ has $n!$ elements but you don't need a $n!$-dimensional space to get a faithful representation; in fact $\mathbb{C}^n$ does it

2:43 PM
sure

@RaphaelJ.F.Berger hmm..

that would be a bit too large I suppose
but that should be an upper bound
in size
dimension

apparently that's a hard problem
18

How does one compute the lowest dimensional faithful representation of a finite group? This question originated in the context of given a finite group $G$: trying to find the lowest dimensional shape whose rotational/reflection symmetries form $G$. (Formally stated as finding the lowest dimensio...

were talking over $\mathbb{C}$ I hope
this question is over a general field

yes, I'm taking an introductory course
nothing crazy with finite fields

2:46 PM
So over the complex numbers

yeah
here is the case over $\mathbb{C}$

What is your actual question now? This: "is there any general rule to determine whether an irreducible representation $\rho: G \to \mathrm{GL}(V)$ is an isomorphism?"?

when a representation (over \mathbb{C}) of a finite group is faithful
obviously it's never an isomorphism, I was thinking about the correct concept but being dumb

just a representation not an irreducible representation?

yeah

2:52 PM
Well take the group with two elements.
There you have the trivial rep. and the second irrep which is of course faithful.
So this is an example where a faithful representation is among the irreps.

Hi Folks! I'm trying to calculate the characteristic function for the normal distribution. I got stuck at this integral, I was trying to extract X but it doesn't really work... Would you have any hints?
$$\int_\mathbb{R}{e^{i t x - \frac{x^2}{2\sigma^2}}}dx, \ \sigma > 0$$

and its also the smallet faithful.

I think, however, that irreps "usually" aren't faithful

of course not

it's a rare phenomenon afaict

2:55 PM
Any rep can be built as a tenso sum of irreps right
(given a base field)
?

e.g. a finite group having a faithful irrep (over C) forces the center to be cyclic

thus abelian

what's a tensor sum?
do you mean a direct sum?

yeah
this is Maschke's theorem
well, it only works if the characteristic of the field is coprime to the group order

2:56 PM
So then aso your smallest faithful rep can be built from that
field is C
so its 0

@Thorgott uhh why tho

@epsilon-emperor If you take the derivative, it is easily positive on one side of $\frac1{\sqrt{2\pi}}$ and negative on the other.

the center necessarily acts by homotheties by Schur's lemma, so the rep induces a faithful one-dimensional irrep of the center, but finite subgroups of $\mathbb{C}^{\times}$ are cyclic

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