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12:03 AM
Hey
 
hey @Thorgott! While on my walk today I realized why separability is necessary in terms of the degree of the extension and the number of embeddings (which is basically what you've said, but each step in my mind). Thank you so much!
 
12:45 AM
nice!
 
did you take Galois theory at uni?
 
I am still taking it
it's a major constituent of a lecture that's just called "Algebra"
 
1:01 AM
I would appreciate if all the folks who use mobile phones would vote for math.meta.stackexchange.com/questions/31167 and meta.stackexchange.com/questions/253036 so as to increase the chances that the bug is seen and fixed by the SE developers. Thank you in advance!
 
 
10 hours later…
10:53 AM
$V(X) = \sum (x-\mu)^2 .p(x)$

$SD = \sqrt{(V(X))}$
I feel that it makes more sense to define SD as $\sum\sqrt{(x-\mu)^2} .p(x)$ instead...
 
11:29 AM
Take a look at this
 
11:53 AM
Good day. I am reading an article and this piece of information is confusing me. It states that the sum of real numbers can be expressed using the following sum. However, how can the index be out of the neutrals. Does not that imply that there exist a bijection between the naturals and the reels. How do you even sum the reels since you do not know which element you will have. It is unlimitedly absurd.
 
the index is in the natural numbers?
 
It's just notation
The RHS is a shorthand for the sum on the left
 
But is it a legitimate notation. As you do not know which element you can have after the element 1, it could be 1.1 or 1.01.
Or is it common to abuse the notation in order to express this.
 
It's not abuse, it's the opposite
The RHS is much better notation than the LHS
The indices only go through the natural numbers
 
What is RHS and LHS.
 
11:57 AM
right-hand side and left-hand side
 
Ok. If the indices go through the natural numbers. Then it means you are not summing the reels. Since the reels are bigger than the naturals.
 
you are summing a countable number of real numbers
indexed by naturals
 
yes, you are just summing over the a_i where the index i ranges over the naturals from m to n
 
No. It is stated you are summing the whole Reels.
 
what
 
11:59 AM
In the article it states all the reels
 
share the article
 
Its g erman
 
Ich kann deutsch
 
Oh wait. You are right. i misread. he is summing a finite amount of reels. Ok that brings me to the question, can you express the sum of reels somehow.
 
well an integral is like the uncountable analogue of a sum but .. I'm not an analyst so anything analytic I say is to be taken with a pinch of salt
 
12:01 PM
Oh. Smart.
Is it also allowed to choose an index out of the reels.
 
Well yes, you can do that in the sense of integration once you choose a particular measure
However, neither the Lebesgue measure nor the counting measure support that
Yes, your indices can be any set as long as its countable (and in case it's infinite, you need to worry about the order of summation)
 
So you can not choose out of the reels
Since you cant count them.
 
Also, for non-negative summands, you can extend the notation to uncountable index sets by taking the supremum over the sums over all countable subsets, but this turns out to not be very interesting
The naturals are a subset of the reals
You can choose out of the reals, in fact you've already done that
 
but for $i \in \mathbb{R}$ how do you know whats the next index.
It could be 1 or 1.01 or 1.001 ……..
and what would that even mean for an index to be 1.01.
 
Well, there's no "next index" so to speak, it's important how you define these things
sensible summation happens over countable sets and it better happen in a sensible order
you should just think about finite summation for now
 
12:10 PM
Ohkey.
Thx for the tips
 
@sevdaicmis Hi, yeah I got it. Its differentiable. In fact max(0,f)^2 is differentiable (only points to worry about are when f=0 but taking one side of derivative and limits work out). Thanks.
 
12:33 PM
Hi chat! I have a question about something in the proof of the formula of the determinant of a block lower triangular matrix in Huffman - Kunze.
Huffman and Kunze first show that if $B$ is obtained from $A$ by adding the rows $c\alpha_j$ to $\alpha_i$ where $i<j$ then $\det A=\det B$. Then they define $D(A,B,C)=\det\begin{bmatrix}
A & B \\ 0 & C
\end{bmatrix}$, if $A$ and $B$ are fixed then $D$ is alternating and $s$-linear as a function of the rows of $C$, so by a previous theorem $$D(A,B,C)=(\det C) D(A,B,I)$$ where $I$ is the identity function with $s\times s$ dimension. Now they claim t
 
12:51 PM
Isn't this counter intuitive? I feel that we should compare 3.5 with the expected value of the number we could obtain on rolling the die (which is also 3.5)
Please ping me if you reply to the above question because I am on a different tab. Thanks!
 
1:18 PM
@Archer In one case you take the average of the outcomes before taking 1/x. In the other case you take the averages after applying 1/x. The point is that the order in which you take an average and apply 1/x matters.
For instance, take two numbers a and b. Then 1/Avg(a,b) is 2/(a+b), whereas Avg(1/a,1/b) = 1/2a + 1/2b. These are just different.
 
@MikeMiller I know that E(1/X) isn't same as 1/(E(X)) ...But here doesn't it make sense to calculate E(X)
Like I expect 3.5 to be the outcome of the die.
So I should take the guaranteed amount directly.
1/X doesn't have any physical sense while rolling the die.
 
huh? If the dice shows (X = 2) dots then the numerical quantity 1/X is 1/2.
Who cares if 1/dot has physical sense, the gambler doesn't need that. He just needs to know how much to pay out if the dice shows X dots. And he pays out 1/X dollars.
 
2:20 PM
is every finite extension of a finite field separable?
 
no
 
I guess so: let $K/F$ be our extension with $p^n$ elements. Let $f \in F[x]:\ f(x) = x^{p^n} - x$
 
Yes, it is
 
It's clear that $\forall \alpha \in K:\ f(\alpha) = 0$
So the minpoly of $\alpha$ divides $f$. But $f' \neq 0$, so it has no repeated roots. Thus $\alpha$'s minpoly has no repeated roots
 
What does mean "A significant fraction of non-squares can be quickly identified by checking whether the input is a quadratic residue modulo small integers" from gmplib.org/manual/Perfect-Square-Algorithm.html
 
2:29 PM
@LeakyNun why not?
 
@LucasHenrique because you changed your question two times
 
in the current form the answer is yes
 
Your argument is incomplete. Non-zero derivative implies separability only for irreducible polynomials
 
I don't get it
 
2:35 PM
X^2 has non-zero derivative, but it's separable
 
Well, $\forall \alpha \in K:\ f'(\alpha) \neq 0$
$\alpha$'s minpoly $g\in F[x]$ is a product of distinct linear factors since $x^{p^n} - x$ is the product of $x-t$ for each $t \in K$
(actually the derivative is unnecessary, since the number of roots of a polynomial over a field does not exceed its degree, and this one has $p^n$ roots, so it splits exactly into linear factors)
@northerner before bruteforcing to check if the number is a perfect square, it checks if it's not a perfect square using modular arithmetic
 
Is there an easy way to find integer value of a real functions?
 
@LucasHenrique how can something be checked for squareness using modular arithmetic ?
 
for example, take mod 9 and square each of $0,1,\ldots,8$.
$$\begin{align}
0^2 \equiv 0
\\1^2 \equiv 1
\\2^2 \equiv 4
\\3^2 \equiv 0
\\4^2 \equiv 7
\\5^2 \equiv 7
\\6^2 \equiv 0
\\7^2 \equiv 4
\\8^2 \equiv 1
\end{align}$$
So if $n = k^2$, then $n \equiv 0,1,4,7 \pmod 9$
Thus $12$ is not a perfect square because it leaves remainder $3$
And you might be lucky to check if $139013384091384$ is a perfect square just by checking the sum of its digits
lol. in fact, it leaves remainder 3 too
@rapasite what do you mean?
 
2:56 PM
ok thanks
 
@LucasHenrique if you have $Y1=coh(t)$ and $y2=sinh(t)$ (t in real) is there an easy way to find a value for t who make (Y1,Y2) a couple of integers?
 
AFAIK, no.
 
@LucasHenrique Ok thanks, too bad I guess!
 
In this specific case, you could try $t = \cosh^{-1}(y_1)$ and plug this $t$ into the second expression to find restrictions for $y_1$ and $y_2$
 
3:13 PM
I see but the true form is $\left(\sqrt{N}\cosh\left(t\right),\ \sqrt{N}\sinh\left(t\right)\right)$ with (N an integer) and (t a real)
 
it's the same
 
so $t=cosh^{-1}(\frac{Y_{1}}{\sqrt(N)})$
and so since I will take t positive
$Y_{2}=\sqrt{N}\sqrt{(cosh^{-1}(\frac{Y_{1}}{\sqrt(N)}))^2-1}$
 
3:35 PM
0
Q: Relation between two matrices associated with a positive definite function

Rajesh DachirajuLet $f:\mathbb{R}^N \to \mathbb{R}$ be a positive definite function. Let $$g(h) = \int_{\mathbb{R}^N}f(x)f(x+h)\mathrm{d}x$$ Due to Bochner's and Parseval's $g$ is a lso a positive definite function. Define the set $S$ = $\{x_1,x_2,...x_n\}$ containing $n$ distinct points in $\mathbb{R}^N$ Defi...

 
@LucasHenrique and then what?
 
3:59 PM
@Lukas Hey, kann man das Seminar zu nicht kommutativer Algebra im LSF sehen oder steht das nur auf dieser Tafel im Mathematikon? Oder hast du bloß von irgendwem gehört dass es das gibt? :D
 
hab's gehört
 
Ah okey :)
 
how does it work, seeing that gehört is the past participle of both hören and gehören?
 
Context
lol
gehört is also the present third person form of gehören
and the present second person plural form lol
 
heh, is ihr even used
 
4:06 PM
yeah, informally when speaking to multiple people
"Seid ihr fertig?"
 
why do I have a feeling that it's the least used pronoun
 
I mean, it's used reasonably often in a group of friends
 
I see
 
"Wollt ihr ins Kino gehen oder so?"
 
I had to solve this integral $$ \int \frac{5x^8 + 7x^6}{\left(x^2 + 1 + 2x^7\right)^2} dx $$
I tried so many things but I couldn't do it, I tried normal substitutions , trig substitutions . Then I looked at the answer and solution
In the solution they divided the numerator and denominator by $x^{14}$ and then the numerator became the derivative of the denominator.
But how to think of such divisors. I mean it's totally unpredictable, how to do these things in future.
 
4:32 PM
well I've seen this trick before
so add this trick to your arsenal
 
@LeakyNun do you know German?
I'm considering picking up learning German.
 
@LeakyNun Something like this also happens when we have to limit to infinity of a rational polynomial, isn't it?
 
@adeshmishra I don't know
@zacts I know a little German, but there are a lot of people here who speak German
 
oh cool
 
4:51 PM
@zacts cool language, I recommend it
 
thanks
 
@LeakyNun Please see that image
 
oh
but the goals are different
 
Yes.
 
5:02 PM
So... I'm considering majoring in Mathematics at the university, as an undergrad student. I'm curious what this kind of path might look like once I reach the Master's or PhD level?
(I hope this is the appropriate venue for this kind of question. If I'm off-topic let me know.)
 
I don't understand the question lol
 
What would I actually be doing for the M.S. / PhD. Would I be doing research?
 
I can't speak for the US, but in the UK there is such a thing as MRes and MSc, where the former is a masters by research and the latter a taught master's degree
PhD is a research degree
I'd say it's less common to do a research master's degree for pure mathematics at least
 
5:25 PM
ok thanks
 
5:43 PM
I have a stupid problem which I'm not able to solve
We have a set of 6 numbers from 0 to 10. In how many ways can I order my six numbers are decreasing? Examples: 10,5,4,3,2,1 ; 6,5,4,3,2,1 ; 10, 8 ,6 4 , 2 ,0
So that they are decreasing*
 
6:03 PM
@Curio: How many ways can you pick 6 different numbers?
@LeakyNun Think about the number of homophones (some with the same spelling) in English. Crazy.
 
Yeah
@TedShifrin but they have to be decreasing
 
Stop and think. Given any set of six different numbers, there's a unique way to write them in decreasing order.
 
Let $\theta$ be a root of $x^3+x+1$ which is irreducible over $\mathbb{Q}$.

Then a basis for $\mathbb{Q}(\theta)$ over $\mathbb{Q}$ is $\{ 1,\theta,\theta^2\}$ right? As $\mathbb{Q}(\theta)$ is of degree 3 over $\mathbb{Q}$.
 
Yes, Puddle.
 
A connection on a manifold is a map from X(M)* X(M) to X(M) (X(M) as the space of all vector fields), which satisfies certain conditions about linear in the first variable and satisfies leibniz rule in the second variable. If I think of X(M) as a collection of smooth sections then it can be given the compact open topology. So now given this topology all the interesting connections that we deal with (like Levi-Cevita) are the continuous maps?
 
6:17 PM
Nice, I got something right for once.
 
You should learn to use MathJax and write things that are readable, @Sayan.
But, sure, a connection is locally given by smooth $1$-forms, so better than continuous in the sense you're asking.
 
Sorry about that @TedShifrin, I was just feeling too tired to use Mathjax. How is it given locally by smooth $1$-forms
 
@TedShifrin of course!!! Many thanks
 
Because $\nabla e_i = \sum \omega_i^j e_j$ if $e_i$ is a smooth (local) basis of sections.
@Curio: Sure thing.
 
Oh I see, like that.
 
7:17 PM
2
Q: Reference Request and Category Theoretic Interpretation of a Result on Banach Spaces

user193319Today I learned in class the following result, which my professor stated without proof: Given a Banach space $V$, there exists a compact Hausdorff space $X$ such that $V$ embeds into $C(X)$ as a closed subspace. Recall that $C(X)$ is the space of all continuous complex valued functions on ...

 
Hello,
 
@Lukas ich kann jetzt die Seminarvorträge für Themen der Nichtkommutativen Algebra auf mampf sehen, schaut cool aus!
 
is x^3 is uniformly continuous only on compact set ?
 
@EdwardEvans ah, sehr gut
machst du das dann noch?
 
or it can be uniformly continuous on open sets ?
 
7:20 PM
Ja vielleicht, ich müsste mich eventuell noch etwas darauf vorbereiten lol
Kenn mich in der Darstellungstheorie überhaupt nicht aus
 
naja, es geht ja darum, das da zu lernen
das sollte also kein Problem sein
 
Nice :)
die Isomorphie von $H^2(G, L^\times)$ und $\operatorname{Br}(L/K)$ hab ich schon irgendwo in Milne's notes gesehen
$G$ sollte dabei die Galoisgruppe von $L/K$ sein vermute ich
Dann hätte ich ein Seminar darin, ein Seminar in Klassenkörpertheorie, und dann 3 Vorlesungen lol
 
Ja
Gruppenkohomologie ist ja auch wichtig für den algebraischen Beweis von Klassenkörpertheorie
 
genau
und ich kann dein Blog benutzen für die ersten Paar Vorträge ;)
 
7:51 PM
@PolineSandra it's uniformly continuous on all bounded sets
 
You guys know any good textbooks on free probability?
 
8:10 PM
@Thorgott even on open bounded sets ?
hi @AkivaWeinberger
 
8:27 PM
@EdwardEvans Listen to "... in Stille" by Frigoris
 
@AkivaWeinberger is $x^3$ is uniform continuous on ]a,b[ ?
 
Any continuous function is uniformly continuous on every compact interval [a, b]. If it's uniformly continuous on [a, b] it is also uniformly continuous on (a, b) (the same $\delta$ for a given $\varepsilon > 0$ that worked on [a, b] will work on (a, b)).
@Rithaniel I would like to learn free probability at some point, I'm also interested in a reference.
 
if a function is uniformly continuous on a space, it is also uniformly continuous when restricted to a subspace thereof
and any bounded subspace of the reals is contained in a compact subspace
 
9:01 PM
@Rithaniel there is some references on the Wikipedia page
 
Yeah, I try to go for ones that people recommend/have read. Pointers like "this one was easy to read" or "this one went into precise detail with the proofs" are what I hope for
 
9:22 PM
can't help you ;), terrytao.wordpress.com/2010/02/10/245a-notes-5-free-probability look promising. But I really don't know I am just a math enthusiast lol
 
9:48 PM
0
Q: $ \sum_{n = 1}^{\infty} ( \frac{p_{2n-1}}{p_{2n}} - \frac{p_{2n}}{p_{2n +1}} ) = ?? $

mickLet $ p_n $ be the $n$ th prime. I was confused about the following idea. $$A = \sum_{n = 1}^{\infty} ( \frac{p_{2n-1}}{p_{2n}} - \frac{p_{2n}}{p_{2n +1}} ) $$ Very confused actually. Does this even converge ? Do we need a summability method ? Does its converge depend strongly on conjectures...

3 close votes hmm :/
 
10:12 PM
How can find the order of the pole of $f(z)=\frac{1}{e^z - 1}$ at 0
 
Use the definition. We have $\lim_{z\rightarrow0}f(z)=\infty$. What about $\lim_{z\rightarrow0}zf(z)$?
 
It's known that every integer greater than 6 is a sum of distinct primes
in which case, we surely allow a prime $p$ to simply be $p$, a trivial sum in itself. otherwise e.g. $11$ has no representation
 
sure, a sum with one summand is still a sum
 
10:29 PM
@Balarka listenin'
 
yeah, it makes sense. the "distinct" part just makes me think of at least two summands, since distinction is a comparison thing
vacuously distinct with nothing else
 
Dude my set topology assignment sucks
I have to determine the subspace topology on a straightline in the Sorgenfrey plane
 
I haven't fallen in love with topology yet
but I feel like it might be coming
 
It's like three different things you have to do when your slopes are positive, negative or {0, \infty}
How do I write this
 
interestingly, if we let 1 be "prime" then every positive integer is a sum of distinct "primes"
 
10:31 PM
I am just going to draw a picture and say "done"
Fucking garbage
Oh
 
and also, every positive integer is the average of two "primes" (assuming goldbach)
 
I can just say if $x \times y \in \Bbb R^2_{\ell}$ then for any basic $[a, b) \times [c, d)$ containing $x \times y$, $x \times y \subset [x, b) \times [y, c) \subset [a, b) \times [c, d)$
And the middle thing sandwiched there is either a left-open right-closed segment if slope is positive, 0, infty or a point if slope is negative
Much easier
 
number theory is weird, and hard
 
@Balarka this is incredible black metal
 
I know!
 
10:38 PM
@Corellian literally me every day
 
I'm considering taking elementary NT next semester
 
do it, it's weird and hard
 
oh hey balarka
 
how's life
 
10:42 PM
Good
 
yes. NT is weird, hard, fun, cruel, pleasant, and (any adjective)
yes, at times, sexy even
big applications in cryptography if you're into that
 
Screw NT
It's for nerds
 
what is it the cool kids do these days?
algebraic geometry?
 
Higher topos theory
 
mmm
 
10:51 PM
Hi, is it possible to prove that $t^{1/t}>y^{1/y}$ knowing that $1<t<y$ and without looking at the function $x^{1/x}$?
I tried with exponentials, logarithms, properties of logarithms, but concluded nothing..
It seems to be a circle
 
Fight me
also @Balarka I have placed this album straight into my sacred BM playlist
 
Lmao
It's a really good one
 
Do you know Путь ?
Blyat metal
 
Oh I might have heard about them
Reco an album
 
The song В чертоге белом is my favourite
 
11:03 PM
How do I type this shit in my phone
Cant play on laptop, headphone jack doesnt work
Fucking Russian
 
Blyat Russian
Get a new phone?
 
With Russian keypad?
 
sure
 
the opposite is worse, e.g. "B 4eptore 6enom"
 
lol add the Russian keyboard to your international keyboards I guess
lool
 
11:12 PM
Algebraic geometry is what I see all the kids online move towards.
 
Where online?
 
Reddit, discord, etc.
 
hmm...
 
It's the "meme" thing to do.
 
Those scheming algebraic geometers
 
11:15 PM
I think it is an attraction to the "french style" of mathematical thinking.
i.e. the one Arnold hates with a passion.
 
Bourbaki?
 
Similar to Bourbaki, yes.
 
even they said their books were not meant to be used as textbooks in uni
 
I object that Bourbaki be the representative for French style of mathematics
Thom is a total French whose style of thinking has been intuitive and influenced decades of mathematics.
 
@skullpatrol I don't know anyone who uses bourbaki as textbooks.
 
11:21 PM
How would you characterize French, then?
 
There are Thurston-influenced geometers in France, eg Laudenbach, Poenaru, so on who have contributed to the visual style of mathematics
There are a French school of probabilistic geometers who are actively doing great work right now even
I have met them personally. Chill dudes
Bourbaki was just some mutant gang who nobody liked to interact with
Fuck em
 
I don't think the existence of non-"french" mathematicians disproves the empirical observation that there was a certain movement in mathematics in french during the 20th century.
 
they redefined rigour
 
I just object to associating mutant algebraic geometers with France :) Sure, Bourbaki was from France, but so are these other groups of people I mentioned
 
then Russell and Whitehead tried
 
11:25 PM
I think one would need a lot more historical background to say to what extent Bourbaki was characteristic of French math of that era
 
Yes, but they are not as widely regarded/known as the movement of french algebraic geometers.
 
Thom and the Thurston-influenced French geometers are pretty widely known
 
I've never heard of them, despite being a fan of Thurston.
 
Ah, you don't know them, but that's a singleton sample.
A generic non-mathematician won't know what Bourbaki is either
Doesn't mean much
 
What is a generic "non-mathematician"?
Someone who actually isn't a mathematician, or a math student who is not a mathematician?
 
11:29 PM
Pick a person at random, and ask them about Bourbaki, I mean.
They are generically not mathematician, so won't know any of these guys
That's a problem of the sample
 
To be anything close to reasonable, you should be discussing people who do pure mathematics, either as an undergrad or beyond.
And I think with that in mind, your sample would yield many more knowing Bourbaki than your beloved Thurston-influenced french geometers.
Bourbaki simply is just more prolific.
 
I think most graduate students have heard the name of Thom at least once in their lifetime.
@anakhro That's not true
 
Likewise for Bourbaki.
What's not true?
 
do the kids online know this?
 
What you said. The three distinct groups I mentioned did very influencial work in geometry
 
11:32 PM
To them "bourbaki" is a meme
 
Bourbaki is popular because of the cult fanbase, not because they were somehow more influencial
lol a google trend plot won't tell you who were more influencial
 
Grrr
How do you get away with spaces in urls here
 
What about Grothendieck
 
@BalarkaSen nice strawman.
 
11:35 PM
Also have you actually read any work by the French algebraists or the French geometers? I feel like it's a crime to claim one is influencial than the other than reading at least some of their works
 
Imagine not knowing about bourbaki
` I feel like it's a crime to claim one is influencial than the other than reading at least some of their works` what
 
@BalarkaSen did they come to your school?
 
'I feel like it's a crime to claim that one is more influential than the other, without having at least read some of each of their work'?
 
Ya that's what I meant
 
coolio
 
11:37 PM
I don't know if thats true though
 
I've read (parts of) several volumes by Bourbaki. I also have read parts of (or tried to, in the least) independent works by Henri Cartan, Shevalley, Dieudonne, Weil, Eilenberg, Koszul, JP Sere, Grothendieck, Lang, Bennequin, and Connes.
 
I mean it's very hard to explain how influential these people are if you don't see their names popping up everywhere in other people's works
 
A historian could work out which of two authors were more influential, without having read any of their work, right?
 
clearly we just need a page rank algorithm for this
 
I think 'influential' not being well defined is the problem in this discussion
 
11:39 PM
@skullpatrol I met them in a recent workshop on probability and hyperbolic geometry
 
@tigre I completely agree.
 
@BalarkaSen nice
 
'infamy' is easy
 
Like influential to who, and at what time, and how
I can imagine a situation where someone writes a false paper, it generates lots of work, and is considered super influential, and then it's realised that it's wrong, killing all of the work built on it
 
I didn't claim influence is comparable, anakhro said Bourbaki is somehow more prolific and linked a google trends page
 
11:40 PM
I am simply referring to a movement in the french school of mathematics towards algebraic geometry that occurred during the 20th century. I am not suggesting that this movement was any more influential than any other movement.
 
Local influence
 
Bourbaki is more widely known than Thom.
I think that much can be guessed.
Hence the google trends page, which was meant to take a shot at addressing your "ask a generic non-mathematician" idea.
 
@anakhro Yes, but then you shouldn't identify that movement with "French mathematics" in general, no?
 
It was predominantly in France, so yes you can definitely identify it with French mathematicians.
 
I think my algebra prof likes Serre a lot
 
11:42 PM
Knowledge isn't immune to geography, Balarka.
People start to do research in areas because people around them are doing (or not doing!) research in them.
 
No, you weren't claiming Bourbaki was a French movement. You claimed French style of mathematical thinking is represented by Bourbaki
 
I made no such claim.
 
France is a big country.
 
I said "Similar to Bourbaki" and you took to a strawman, it seems.
 
@anakhro This is what I refer to
 
11:44 PM
29 mins ago, by anakhro
I think it is an attraction to the "french style" of mathematical thinking.
 
You will notice I never mentioned Bourbaki there.
Indeed, I wasn't the one who brought up Bourbaki.
 
I did.
 
@Thorgott tbf your algebra prof is Stix and Serre was a very influencial number theorist
or rather, very influencial in number theory
 
"i.e. the one Arnold hates"?
Arnold only ever commented on Bourbakists, not other French mathematicians. In fact, he was a collaborator and a good friend of Thom
They were the two major people involved in singularity theory
 
No, Arnold was hyper critical of french mathematics at the time in general.
Whether in jest or not, he often mocked "french" people.
 
11:47 PM
[citation-needed]
 
-_-
 
"For example, these students have never seen a paraboloid and a question on the form of the surface given by the equation xy = z2 puts the mathematicians studying at ENS into a stupor."
- Arnold
"Drawing a curve given by parametric equations (like x = t3 - 3t, y = t4 - 2t2) on a plane is a totally impossible problem for students (and, probably, even for most French professors of mathematics)."
 
ENS was a very Bourbaki inspired institute. I have read the article you refer to.
 
"How could this happen in France, which gave the world Lagrange and Laplace, Cauchy and Poincaré, Leray and Thom? It seems to me that a reasonable explanation was given by I.G. Petrovskii, who taught me in 1966: genuine mathematicians do not gang up, but the weak need gangs in order to survive."
" They can unite on various grounds (it could be super-abstractness, anti-Semitism or "applied and industrial" problems), but the essence is always a solution of the social problem - survival in conditions of more literate surroundings."
Notice how Arnold himself even mentions how this is happening "in France".
 
The article is a transcript of a speech by Arnold directed towards the Bourbakists who were present there
 
11:49 PM
Heh: "From my French friends I heard that the tendency towards super-abstract generalizations is their traditional national trait. I do not entirely disagree that this might be a question of a hereditary disease, but I would like to underline the fact that I borrowed the cake-and-apple example from Poincaré."
 
It was meant to insult them.
 
It's fine, Balarka, deny it all you want.
This movement existed, was associated with the french, and this association hasn't stopped today.
It's definitely not as strong as it was, but it's still someone living mathematicians refer to.
 
That's because people are ignorant enough to not read other French mathematical works
Like you are being right now
 
What am I being ignorant of? You still haven't actually addressed what my original point was.
I know France has many mathematicians that do not work in AG.
I have already affirmed this.
 
I won't continue this argument but only suggest you to read this: arxiv.org/abs/1912.03115
 
11:55 PM
Sure, have a good night, then.
 
@Edward yeah, he's also hosting a seminar with Serre as the main reference material next semester
 
A local fields seminar?
 

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