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2:08 AM
@Ultradark ping me
 
 
3 hours later…
4:46 AM
Copy-pasted from elsewhere, about ½!
Remember the Wallis product?

π/2 = 2/1 · 2/3 · 4/3 · 4/5 · 6/5 · 6/7 · ...

We can rewrite that in a really neat way. Take the partial product:

2/1 · 2/3 · 4/3 · 4/5 · 6/5 · 6/7 · ... · (2N)/(2N−1) · (2N)/(2N+1)

Divide numerators and denominators by two:

1/0.5 · 1/1.5 · 2/1.5 · 2/3.5 · 3/3.5 · 3/4.5 · ... · N/(N−½) · N/(N+½)

Divide them into two groups, alternatingly:

[1/0.5 · 2/1.5 · 3/2.5 · ... N/(N−½)] ·
[1/1.5 · 2/2.5 · 3/3.5 · ... N/(N+½)]

The numerator is the product from 1 to N twice, so it's (N!)^(2).
 
did ted name this book? bookstore.ams.org/gsm-68
 
@AkivaWeinberger what's the "probability" that a "random" integer is square-free?
 
Something like $1/\zeta(2)=6/\pi^2$ if I remember right
 
how about don't remember
also great copypasta
 
Aight so let's say we're picking from $1$ to $N^2$
Hm. Inclusion-exclusion?
 
5:00 AM
want a hint?
 
$\frac1{n^2}$ have $n^2$ as a factor
but that'll overlap and be messy
Hm. Primes?
I'm guessing if zeta is my goal I'm gonna want the product form of it
$\frac1{p^2}$ have $p^2$ as a factor. So $\prod(1-\frac1{p^2})$… nah that doesn't sound right
or wait
Yeah it does sound right 'cause I bet I can say those are independent
I don't know how to justify that they're independent but they should be
 
yeah precisely
 
And then $\prod(1-\frac1{p^2})$ is $6/\pi^2$ as claimed
 
surely they're independent
 
Hm I don't need independent, I just need nice enough inequalities
which I'm sure is doable
so I'm not gonna
Does the same work for cubefree?
Should be
 
5:03 AM
yeah I guess
 
5:49 AM
Mornin'
 
@ÍgjøgnumMeg hi
 
Hiya @Leaky
 
@ÍgjøgnumMeg brought any math with you?
 
Uhhm, I have a question that might be interesting but hard to prove lol
So suppose I have a number field $K$ and a nonzero ideal $\mathfrak{a} \subseteq \mathcal{O}_K$ with the property that $\mathfrak{a}^m = (a)$ for some $a \in \mathcal{O}_K$. Can I find a finite extension $L/K$ such that $(m, h_L) = 1$?
(and such that $h_L \neq 1$)
The question stems from an exercise to prove that there exists an extension $L/K$ such that $\mathfrak{a}\mathcal{O}_L$ is already principal. I can take $K(\sqrt[m]{a})$ and that should be fine but it's kind of a boring answer, so I thought about taking the induced group hom. on the class groups of $K$ and $L$, then the image of $[\mathfrak{a}^m]$ under that hom is trivial, and if $(m, h_L) = 1$ then the image of $[\mathfrak{a}]$ itself is trivial :P
(I could also take the Hilbert class field of $K$ but for the purposes of the exercise I'd have to prove that exists rofl)
 
@LeakyNun what are the two definitions of separable degree again
it's that $f(x) = f_{sep}(x^{p^n})$ thing right
 
5:59 AM
@BalarkaSen 1. the number of embeddings to a big enough field (normal closure / algebraic closure)
2. the degree of the separable component in the decomposition into purely inseparable on top of separable
 
like K/F extension, sepdeg being degree of E/F where K/E is the unique purely inseparable guy s.t. E/F is separable
 
you reversed them
K/E is purely inseparable and E/F is separable
 
ah yes
 
6:20 AM
@LeakyNun do you mean in a fixed cardinality?
 
@AlessandroCodenotti no.
 
oh no galois theory
 
I don't think so by Löwenheim-Skolem then? What am I missing?
 
@AlessandroCodenotti that requires the existence of an infinite model
 
If you have finite models of arbitrarily large size you have an infinite one by compactness and if the size of the models is bounded there's finitely many ways to interpret each symbol
If you have an infinite language most of the symbols will be interpreted in the same way so I don't think that's an issue
 
6:31 AM
so you can't have finite models of arbitrary large size
 
No, look at $phi_n$ saying "there are at least $n$ distinct elements" and use compactness
(or take an ultraproduct of models of increasing size if you prefer looking at it from this point of view)
 
as in, that requires the existence of finite models of arbitrary large size
i.e. that doesn't prove it wrong, that just gives it more restrictions
 
Ok but if the models are bounded in size can you even have infinitely many up to iso?
 
why not
 
Ah ok so infinitely many can be done by having $n$-ary relations for every $n$ in the language
I don't see how to get past countable though, for every fixed $k$ there's a finite number of interpretations of every symbol and at most countably many symbols that can have different interpretations
 
6:48 AM
@LeakyNun given K/F is a finite extension, Hom_F(K, Falg) is equal to |Spec(K otimes Falg)| since the former counts the number of geometric points in Spec K lying over a specific geometric point in Spec F.
if K is purely inseparable finite extension of F, you can decompose it as a tower of 1-generated inseparable extensions of degree p, so note that if K = F[x]/(x^p - a) then K otimes Falg is isom to Falg[x]/(x^p), which is has a unique prime. So if K is purely inseparable finite over F, |Spec(K otimes Falg)| = 1
fiber is multiplicative so the only part that contributes for a general finite extension K/F is the separable part E/F
in which case its the degree
 
what is this K otimes Falg nonsense
 
The Turing test claims to distinguish a consciousness from a lack of one
 
number of F-hom K -> Falg is equal to |Spec(K otimes Falg)|, what's wrong
 
Is there a test that can distinguish two consiousnesses from one consciousness?
 
“Any AI smart enough to pass a Turing test is smart enough to know to fail it.” ― Ian McDonald, River of Gods
also have you watched Ex Machina
 
6:52 AM
Yeah
 
@BalarkaSen it's probably right, by "nonsense" I just mean it's some high-powered feces that I don't know and that you would have to explain to me
 
@LeakyNun You've passed the Turing test, I can reasonably be sure you have a consciousness
 
@AkivaWeinberger beep boop
 
but I don't know if I have any way to be absolutely sure that you and Balarka aren't controlled by the same conciousness, I think
or you accounts, at least
 
@LeakyNun i tried to tick you off with that ;)
 
6:55 AM
so would you like to explain it to me
 
Also in general is kinda hard to get $\aleph_1$-many things, for example if the language is countable then a theory can have finitely many, $\aleph_0,\aleph_1$ or $\mathfrak c$ countable models, but whether $\aleph_1$ is possible is provably is open
 
so an F-algebra homomorphism $K \to \overline{F}$ is in 1-1 correspondence with a morphism $\text{Spec} \overline{F} \to \text{Spec} K$ of $\text{Spec} F$-schemes, where $\text{Spec} \overline{F}$ has been given a $\text{Spec} F$ scheme structure by choosing an embedding $F \to \overline{F}$
 
@AlessandroCodenotti really? my logic professor was discussing this with me and we concluded that $\aleph_1$ is not possible for a countable language
 
These therefore correspond to commutative triangles of maps $\text{Spec} \overline{F} \to \text{Spec} K$, $\text{Spec} \overline{F} \to \text{Spec} F$ and $\text{Spec} K \to \text{Spec} F$.
 
@LeakyNun in total or for countable models?
 
6:58 AM
oh didn't see that part
 
Exterior angles GIF
(add to 360)
 
The morphism $\text{Spec} \overline{F} \to \text{Spec} K$ therefore factors through the geometric fiber $\text{Spec} \overline{F} \times_{\text{Spec} F} \text{Spec} K = \text{Spec}(\overline{F} \otimes_F K)$
 
@AkivaWeinberger nice
 
Fails in hyperbolic and spherical geometry
Doesn't fail in Euclidean geometry because similarity is a thing there
Zooming out on 0 curvature still gets you 0 curvature
 
@BalarkaSen is this just the "graph"?
 
7:00 AM
(The GIF is best thought of as "zooming out", not "sliding the pieces")
 
Why did you ask me if you already knew the answer then lol
 
@AlessandroCodenotti well it's a fun question...
 
Hm. This generalizes to 3D but in a weird way I think
 
@LeakyNun It's the fiber over the point, right? If $f : Y \to X$ is a map and $x : \{*\} \to X$ is a point in $X$, the fibered product $\{*\} \times_X Y$ is precisely $f^{-1}(x)$
 
oh really
 
7:05 AM
It's a neat way to think about things, I think, and I don't think I am missing a subtlety here
From the above, we're really counting morphisms $\text{Spec} \overline{F} \to \text{Spec}(\overline{F} \otimes_F K)$ of $\text{Spec}\overline{F}$-schemes. These just correspond to $\overline{F}$-algebra homomorphisms $\overline{F} \otimes_F K \to \overline{F}$, so prime=maximal ideals of $\overline{F} \otimes_F K$, really.
 
interesting
 
That's why $|\text{Hom}_F(K, \overline{F})| = |\text{Spec}(\overline{F} \otimes_F K)|$
Now naively it's easy to see why your result holds, because for every degree of inseparability there's a nil algebra factor in $\overline{F} \otimes_F K$
and for every degree of separability there's a $\overline{F}$
But I'll write down a serious proof if this is too much pretension :P
 
no i like your proof
lol
 
thanks
 
8:09 AM
Morning people
 
 
2 hours later…
9:57 AM
Good Morning
 
10:15 AM
For each normed vector space $(X,\|\cdot\|)$, let $\Lambda_X:X\to X^{**}$ be the canonical isometric linear map defined as $\Lambda_X(x)(f)=f(x)$, for all $x\in X$ and $f\in X^*$. Assume $X$ is complete with respect to $\|\cdot\|$, so a Banach space.

How can you show $\Lambda_X:X\to X^{**}$ is surjective iff $\Lambda_X:X^*\to X^{***}$ is surjective?
 
 
1 hour later…
11:16 AM
Yo @Soham
 
11:26 AM
hi, @BalarkaSen
 
How's it goin
 
@BalarkaSen hey
 
Hi @LeakyNun
 
@Adam I have been classifying your content each time I've seen it. But I haven't been storing permalinks to all of your posts...
 
well long story short i did no maths for a very long time
and now here i am back where i started, recovering
 
11:29 AM
its your last year right
 
Last year?
 
three-year degrees in india
 
Oh, I see.
What are you doing a degree in?
 
11:30 AM
where're'u planning to apply
 
@BalarkaSen how about you, where you at
sitting for JAM, and the ISI/CMI entrances
 
Cool!
 
I'm confused. How long is 'very long time'? :P
 
about two years of me just somehow passing exams
 
Come to ISI, although I guess I won't see you because your batch will be at ISICal
 
11:30 AM
@BalarkaSen @SohamChowdhury do you speak the same language?
 
yeah
 
we do
@TedE hint: brain problems, er, depression
 
@SohamChowdhury back home for a month
 
semesters back there are a drab man, its stressful lol
 
11:32 AM
Oh sure, I just didn't understand what it meant to have just finished your second year of a maths degree, but not having done maths in a very long time (which I'd normally take to mean like 2 years or something)
 
me i'm reviewing tons of stuff that i only half-remember
 
Good idea.
 
kinda sad i wasted like 4 or 5 years doing shit with no rigour while everyone here warned me not to
but better late than never
 
Meh its ok
 
I know that feeling
 
11:33 AM
i gots that mathematical maturity tho
that's what i tell myself
what are you studying atm B
 
@Adam Also I don't know who that is, and I don't smoke weed (and in particular, I don't get the reference) [Also, you said you wouldn't be offended, which is the only reason I answered...]
 
going through Galois theory rigorously
 
just ordinary gal theory right
 
yea
from Dummit-Foote
basic stuff
 
@BalarkaSen where're you at
 
11:34 AM
nearly done with the field theory chapter, finishing up a few exercises
i want to be done with a substantial amount of the galois theory chapter before this month
 
unusually slow for the Balarka i remember, which i take it is a good thing
 
lol
 
i've been using Artin for linear algebra over the last couple weeks but i'm tempted to look at Axler; i remember his proofs, while not being very hands-on, really made me feel like i was getting something
 
i dont really like his matrix-free approach but its ok i guess
Artin is my fav for linear algebra
 
also the problem with Artin is that everything looks familiar and all the exercises are ticked and i want to look out the window
obviously the solution is to buy another copy
 
11:38 AM
yeah sort of the same with me doing galois theory from Dummit-Foote, the thing is i dont think i "understood" when i did these problems a long time ago
i usually go on detours thinking about related questions when im doing these problems
 
likewise
 
makes me slower but i think i am learning more
 
great minds detour alike
2
 
oh also talking to people here helps (@LeakyNun taught me a bunch)
 
i worked through Artin in a horribly unthinking fashion back in the day
oh neat
 
11:40 AM
srsly i learn more talking here than in semesters
 
what sem are you now
 
2nd year 1st sem done
 
@BalarkaSen which exercise are you at
 
um let me check
i am in section 13.6. i didnt look at the exercises yet
 
11:43 AM
cyclotomic polynomials and extensions?
 
cool
 
@LeakyNun Let me write down a very precise story for the separability stuff without the scheme-theoretic stuff anyway. Suppose $F$ is a field of characteristic $p$ such that $\alpha \in F \setminus F^p$. Then $x^p - \alpha$ is irreducible in $F[x]$ because in the algebraic closure of $F$, $x^p - \alpha = (x - \alpha^{1/p})^p$, so by unique factorization in $\overline{F}[x]$, any factor must be of the form $(x - \alpha^{1/p})^n$ where $n < p$ but this cannot be in $F[x]$ because $\alpha^{n/p} \notin F$ since if not, we take $m$ such that $mn \equiv 1 \pmod{p}$, and take the $m$-th power to o
Finally, let $\varphi : K \to \overline{F}$ be an $F$-homomorphism. Then $\varphi(\beta_1)^p = \beta_1^p$, but the only root of $x^p - \beta_1^p = (x - \beta_1)^p$ in $\overline{F}$ is $\beta_1$, so $\varphi(\beta_1) = \beta_1$. By induction, $\varphi(\beta_i) = \beta_i$ as $\varphi(\beta_i)^p = \beta_i^p$. So $\varphi$ fixes all the $\beta_i$'s, which means $|\text{Hom}_F(K, \overline{F})| = 1$
 
therefore, if $i:F \to K$ is purely inseparable, then $i$ is a/an _
 
unsure. hint?
 
11:50 AM
it's a category theory term
anyway it isn't very important you can go on
 
im curious, i dunno what you have in mind
 
epimorphism
 
Ah
Because F -> K extends uniquely to F -> K -> E, I see your point.
That's pretty nice.
I should store that in my brain somewhere for counterexamples
 
lol
 
But yeah now for any extension $K/F$, take the decomposition $K/E/F$ where $E/F$ is separable and $K/E$ is purely inseparable, then $|\text{Hom}_F(E, \overline{F})|$ is the degree of $E/F$ and any $F$-homomorphism $E \to \overline{F}$ uniquely extends to an $F$-homomorphism $K \to \overline{F}$ by whatever said above
So $|\text{Hom}(K, \overline{F})|$ is indeed $|E : F|$
 
12:01 PM
@BalarkaSen that sounds kinda like you have a second brain to store counterexamples
 
I store counterexamples in my left hemisphere and do math with my right hemisphere
 
Einstein did it that way too :P
 
@BalarkaSen why $|\operatorname{Hom}_F(E,\overline{F})| = [E:F]$?
 
@LeakyNun $E/F$ is finite separable, so choose a primitive element $E = F(\alpha)$. $F$-automorphisms of $\overline{F}$ act transitively on the roots of the minimal polynomial of $\alpha$ by before, so
The minimal polynomial has $[E : F]$ many roots by separability
 
aha
nice
maths is nice
 
12:08 PM
Yeah, very much. Especially Galois theory
 
@BalarkaSen I suppose one should now prove the tower law for separable degree
$[M:L]_s [L:K]_s = [M:K]_s$
 
@BalarkaSen Fair enough
I just carry around a copy of Steen Seebach
 
Lol
 
that's cheating :P
 
I have to read some topology soon...
I have been doing 0 topology
 
12:11 PM
That's a book I'd actually love to have a paper copy of
Oh, it's 28€ on amazon, maybe I'll make myself a Christmas present
 
yeah its quite cheap here as well
 
ebooks are cheap
 
I'll go to libgen if I want a pdf tbh
 
@LeakyNun This is observing that once you choose an $F$-embedding $\varphi : L \to \overline{K}$ you can extend that in $\text{Hom}_L(M, \overline{K})$-many ways to an $F$-embedding $M \to \overline{K}$.
 
i really wish i had a paper copy of Amann-Escher volume 1
 
12:18 PM
@AlessandroCodenotti generation: lib
 
very nice analysis book
 
what kind of analysis
 
Rudin is the best analysis book from my experience honestly speaking
 
rudin doesn't have enough "fiddly" exercises
i need "stupid" exercises
 
12:20 PM
fiddly as in
ah ok
 
"calculate this limit"
 
yeah but the thing is that stuff isnt too important
whatever he has is enough i feel
 
i trick myself into thinking i understand the material because i'm better at doing handwavy conceptual exercises but then i can't actually do computations
 
Emmanuele Dibenetto's "Real Analysis" is an awesome book
Maybe not to study from, but as a reference it contains a lot of stuff that's hard to find elsewhere
 
i dont think Rudin's conceptual exercises are easy to handwave
 
12:21 PM
ah i misspoke
 
@SohamChowdhury this sums up my experience with differential geometry surprisingly well
 
cut the handwavy from that sentence
 
some of them are quite hardcore gritty analysis
no but i know what you mean
i mean to say
there are conceptual exercises in analysis which are gritwork
 
yeah they're not exactly like those abstract-nonsensey arguments that make you think you understand algebra but then you can't actually work with elements
that was the great mistake of my youth
sips tea
 
@BalarkaSen are you using baby rudin first?
 
12:24 PM
you can't not
 
i honestly have not read any other Rudin
 
big rudin doesn't contain everything from baby rudin
it assumes one-variable real analysis
 
@SohamChowdhury you might enjoy Stein-Shakarchi as well but I really have no experience with their real analysis book
their complex analysis exercises were a lot of gritwork
which i learnt quite a bit from
 
isn't that book also a bit handwavy in the text itself?
that's what i've heard
 
in the text yeah
 
12:26 PM
i have not heard anything about the exercises, i'll keep it in mind when i have to review complex (which should be like in a few weeks)
thank
 
@BalarkaSen where from?
 
but i think the handwavy text makes it easier to read and compensates with the hard exercises
@Alessandro i dunno man i havent done any topology in a long time
i will read Freedman's lecture notes
soon
 
I don't know those
 
<<10 years later>>
oh i meant the bing topology stuff
 
i have some kind of deep phobia of handwave-ism now to the point where i bash myself over the head trying to justify everything
and honestly i think it's doing me more good than not
 
12:27 PM
Oh I see, I never read those
 
i've learned more analysis in the last month than in the five years before that
 
have you done spivak?
 
i still dont know any analysis
 
i rushed through spivak as a teenager, which is a long-winded way of saying "no"
 
the analysis courses in isi are the worst
 
12:29 PM
bad how
 
analysis-1 was taught by an algebraic number theorist who works with the indian math olympiad camp so it was just a lot of tricky stuff; its still fine i guess. no integration in first course, which is weird
analysis-2 was super weird, it was mostly metric spaces and then multivariable with a little bit of riemann integration done badly squished between.
 
@BalarkaSen yeah
 
it was like a review of topology but without mentioning topology and working with metrics instead
 
analysis-3 was horrendous; the instructor followed Apostol and the course was engineering calculus
 
12:32 PM
how big were the classes, on average?
 
we have like a little less than 40 people in our batch
@Soham yeah, contrary to popular belief, algebra is great here
there's an active community of algebraic geometers
 
what about the tutorials or recitations
 
none of that
 
how is the student community
 
12:34 PM
bad lol
 
feel like that's most of the experience ngl
oh no
 
nobody gives a shit about math
 
-_-
 
same here but, obviously, worse
 
theres like 4 people from every batch who are truly interested
i think its the same everywhere and this is our future
 
12:35 PM
lol
tbh i expected to pop up here and see you studying derived AG or something equally esoteric
with some prof
 
not sure if it's a pleasant surprise or not
 
i do basic things too
or rather, who says Galois theory isn't esoteric!
 
i want to review topology; i remember most or all of the definitions, i just need to work through a bunch of exercises which i never did
 
its about studying etale site over $\text{Spec} K$, after all
 
12:39 PM
is Munkres still the best place to look for this?
 
@BalarkaSen It's not the same everywhere lol.
 
ofc i was talking about my country
 
Oh okay.
I took everywhere to mean everywhere :P
 
clearly
 
12:46 PM
Would you like to learn some derived algebraic geometry btw?
 
rip
 
This complete algebraization of homotopy theory is so unappealing, how is it a trend
People are mutants
 
In what sense is it algebraization of homotopy theory?
It's more homotopification of algebraic theories if you wish
 
I wasn't referring to DAG (i dunno much about it) but rather the whole stable homotopy theory trend
 
12:50 PM
Oh right
I mean that should be as appealing as taking homology is
 
Homology is not fundamentally an algebraic idea, it has it's geometric roots in thinking in terms of cycles and cobordisms, a la Poincare.
 
In some sense, though, you can think of it as a machine that takes something geometric/topological, that is unwieldy, and converting it to a convenient algebraic form that one can actually do things with (and is invariant under whatever you want). I'd still say it's algebraization of homotopy theory in some sense.
How do you feel about the localisation of spaces?
 
That's beautiful
 
Is the adams conjecture (Or sullivans theorem, qulliens theorem, whatever you want to call it now) beautiful to you?
 
I don't know Adam's conjecture
What's the statement
 
12:56 PM
If $k$ is an integer, and $X$ is a finite CW-complex, $y\in K_R(X)$, then there exists a non-negative integer $e$ such that $k^e(\Psi^k-1)y$ maps to zero in $J(X)$.
 
$\Psi$ is some cohomology operation? Can you explain the symbols
 
$\Psi^k$ is the $k^{th}$ adams operation
$k^e\xi$ means a whitney sum of $\xi$, $k^e$ times
 
Stable, since it's taking values in the K-group, I assume?
 
Yep
 
I know what adding two vector bundles mean :P
 
12:58 PM
I should really say whitney join
 
That much you can assume
 
And $J(X)=im(\widetilde{KO}(X)\to \widetilde{KG}(X))$
 
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