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17:26
it's confusing because why would a proof use choice if there's a proof that's just as simple without choice
1
Q: Check the differentiability and continuity of the function at $(0,0)$

Maneesh Narayanan Check the differentiability and continuity of the function at $(0,0)$ $$ f(x,y)= \begin{cases} x^2\sin\frac{1}{x}+y^2\sin\frac{1}{y},x\neq 0, y\neq 0\\\\ x^2\sin\frac{1}{x},x\neq 0, y= 0\\ y^2\sin\frac{1}{y},y\neq 0, x= 0\\ 0, x=0,y=0 \end{cases} $$ In all cases, $|f(x,y)|\le|x^2+y^2|\leq...

Am I correct?
please verify.
where'd you get the error term, remind me how it's calculated idr
In the answer key, it is given that $f$ is not continuous :(
@GFauxPas are you convinced of the proof now?
idk
I'm not convinced the reasoning isn't tacitly assuming $f$ isn't empty
@ManeeshNarayanan it looks continuous to me???????
also looks differentiable to me
what do I know
17:38
for me too. nobody is replying in the comment box ☹
maybe that means no one sees anything wron
i upboated
where's the exercise from
one of my friends sent me this question.
is your friend the kind of evil person who tricks you with wrong answer keys
because if so, you should reconsider your friends
Done with exams woot
@Daminark grats
17:44
@GFauxPas 😂
@GFauxPas she had put down from a coaching class :)
I think that teacher might be wrong.
Thanks!
Grats @Dami! Are you on summer break now?
So you have no excuses to avoid p-adics now :P
18:01
hi @TobiasKildetoft
@LeakyNun Hi
@TobiasKildetoft I've learnt a lot since I last talked to you a week ago
Awesome Dami
@ManeeshNarayanan I can't find anything wrong with your work
it's just triangle inequality and $|\sin|\le 1$
@TobiasKildetoft do you know about Weil restriction?
18:05
okay. Thank you very much. That tuition teacher might be wrong.
@GFauxPas
@LeakyNun No
Hi @LeakyNun @TobiasKildetoft
@MatheinBoulomenos Hi
hmm, I think Weil restriction should be irrelevant
Given K/F finite Galois, there should be an antiequivalence of categories between
1. Torus over F that splits over K
2. finitely generated free abelian groups with a right $\Gamma$-action
the latter should just be homomorphisms $\Gamma \to \operatorname{GL}_n(\Bbb Z)$
18:12
With $\Gamma$ being the Galois group?
@TobiasKildetoft we kind of did modular representation theory in ANT today, although we didn't call it that. We used the standard argument that if $G$ is a $p$-group, then every representation on a finite-dimensional $\Bbb F_p$ vector space has nontrivial fixed points to see that $\Bbb F_p[G]$ is a local ring. This allowed us to do some Nakayama arguments and in the end we obtained some cool results on cohomologically trivial $G$-modules
@MatheinBoulomenos Cool
18:28
Does $S^3$ contain any (embedded) nonorientable surface?
@TobiasKildetoft
For a $p$ group $G$, the following are equivalent and a $\Bbb Z[G]$-module $A$ such that $A$ is $p$-torsion, the following are equivalent:
- $A$ is a free $\Bbb F_p[G]$-module
- There exists a $i \in \Bbb N$ such that $\operatorname{Tor}^i_{\Bbb Z[G]}(\Bbb Z,A) = 0$
- There exists a $i \in \Bbb N$ such that $\operatorname{Ext}^i_{\Bbb Z[G]}(\Bbb Z,A) = 0$
- For every subgroup $H \subset G$ and all $i \in \Bbb N$, we have $\operatorname{Ext}^i_{\Bbb Z[H]}(\Bbb Z,A) = 0$ and $\operatorname{Tor}^i_{\Bbb Z[H]}(\Bbb Z,A) = 0$
Here $0 \not \in \Bbb N$
18:53
@AndersonFelipeViveiros no, you can see this by Alexander duality
 
1 hour later…
20:12
$v : K^\times \to \Bbb Z$ is surjective, and $\Bbb Z$ is projective, so $K^\times = \mathcal O^\times \oplus \Bbb Z$
@LeakyNun there's a more elementary way to see this: any element in $K^\times$ can be uniquely written as $\pi^n u$ for $n \in \Bbb Z$ and $u \in \mathcal O^\times$ (just by unique factorization)
I see
when you put the discrete topology on $\Bbb Z$ and the one induced by $v$ on $K^\times$ and $\mathcal{O}^\times$ this is even an isomorphism of topological groups
nice
btw what's with this locally compact totally disconnected business
like the Cantor set
like it's clearly very disconnected
but somehow it's compact
any Galois group and complete local ring with finite residue field also satisfies this
why do you feel that disconnected and compact are contradictory?
20:26
@MatheinBoulomenos they don't look nice together
visually
20:43
I suspect that that question has been artificially generated. But I'm not sure.
I am inclined to agree
interesting
21:06
I guess you won't like it if I say that a lot of spaces of ultrafilters are totally disconnected and compact
21:16
t r i g g e r e d
If you throw in Hausdorff too this situation is common enough there's a special name for compact, totally disconnected Hausdorff spaces: they're known as Stone spaces
I would call them profinite
What are confinite groups? :thinking:
21:34
I'd say torsion groups
hey
Can someone tell me what a scheme vs a stack is given some polynomial p(x) ?
anyone?
hi
@MatheinBoulomenos what's the deal between profinite groups and torsion groups?
profinite groups are inverse limits of finite groups. torsion groups are direct limits of finite groups
you mean of finite groups
21:38
interesting
does it have a topology?
discrete
exercise: show that the category of abelian profinite groups is equivalent to the opposite category of abelian torsion groups
i.e. finite abelian group is equivalent to its opposite category?
Hi guys
more precisely, you can write down a functor $\mathbf{Ab}^{op} \to \mathbf{Ab}$ that sends colimits to limits and restricts to an equivalence between finite groups and their own opposite category
(possibly useful hint: why is the category of finite-dimensional vector spaces over a field equivalent to its opposite category?)
22:00
@MatheinBoulomenos If $T$ is a split torus over $K$, is $T$ and $\operatorname{Spec}(K[\operatorname{Hom}(T,\operatorname{GL}_{1,K})])$ canonically isomorphic?
I don't understand the notation
oh you take the group algebra
I don't know how canonical this is, but they are certainly isomorphic
If $T=(K^\times)^g$, then $T= \operatorname{Spec}(K[x_1,x_1^{-1}]) \times \dots \times \operatorname{Spec}(K[x_g,x_g^{-1}]) = \operatorname{Spec}(K[x_1,x_1^{-1}] \otimes_K \dots \otimes_K K[x_g,x_g^{-1}]) = \operatorname{Spec}(K[x_1,x_1^{-1},\dots, x_g,x_g^{-1}]$ $= \operatorname{Spec}(K[\Bbb Z^g]) = \operatorname{Spec}(K[\mathbf{Hom}(T,\mathbf{GL}_1(K))])$
I don't really want to write these isomorphisms into a natural transformation and check naturality
But it should work out. Hom(-,GL(1)) is contravariant, taking the group algebra is covariant and taking Spec is contravariant
thanks
22:51
Hi @KasmirKhaan
Hello :D
Mathein my hero and leaky :D
When is summer break for you guys ?
30th july or something like that
wow that is a bit late
you have 1 month only =?
or when do you start again ?
october 15th
I need some help with a few trivial concepts
can someone tell me what a scheme is?
please?
I know what a polynomial is
22:55
lol, if that's a trivial concept then what is a nontrivial concept for you?
say I have p(x)
@MatheinBoulomenos wow nice >< why in all countries only germany has that period for break ><
for us June to august
2 months and a week or so :D
@MatheinBoulomenos lol, well it is highly non-trivial for me :P
do you know what a ring is?
yes,
i think it is a
set with two binary operations and some rules
22:58
the two operation has to work together also
yes indeed
to give us the distributive law
I am looking at wikipedia now and I think it is the third axiom in that definition
yes many books uses abelian group with +
and every element has an inverse with *
but some uses a ring with 1
ok, so how does one go from this to a scheme?
23:01
@Cows how much algebra do you know?
have you taken a course in abstract algebra/commutative algebra?
@MatheinBoulomenos I can get around, I have some trivial abstract algebra, self-read of course, but I feel like I can feel my way around
you're going to need some serious abstract algebra and commutative algebra to make sense of schemes
Is there no simple engineer friendly version of schemes?
why would an engineer need schemes?
I feel like it has something to do withe solutions of polynomials no?
I was just using "engineer friendly" as a soft way of describing a down to earth account of the matter
@MatheinBoulomenos let's give it a go , I will try to follow, and I will ask for clarification if i don't understand something
23:06
Okay I can talk a bit about the relatin with polynomials I guess
That's a bit disingenuous I think, if you follow the history for a hundred years you find that schemes are about solutions of polynomials
But that is not what a scheme looks like, except in some very informal sense
some schemes allow you to think about solutions of polynomials where you're allowed to look at it in various rings
yes so let us start with a polynomial p(x)
but that's all there's to it
even people who have taken a year or more worth of algebra struggle with schemes typically
@MatheinBoulomenos I understand that the subject is very complex, can you give me a taste of the subject
23:09
@anon Anon :D
@anon where have you been man, i missed you :D
@Cows I don't see the point
@MatheinBoulomenos ok, can you share a helpful link on the subject, that you feel might be accessible with me?
just unchatty
haha classic =p
what are you up too these days?
and how is it with your math olympiade class?
just started helping some fresh blood with p-adic numbers, planning a potential lie theory thing next semester that hopefully I can trick some physics students into joining
the putnam group is mostly disbanded for the summer
23:12
@anon nice :D i wish i could be on that class ._, is there something for poor kas online ?
like you send me some Q's or notes :D
@anon can you tell me how to study very basic scheme theory?
if I finish the notes I can send you a link later in the summer.
@Cows no, I can't
@anon thanks ! :DDD
@anon do you know about schemes?
nope
23:17
@Cows there's no shortcut. If you're serious, learn some serious abstract algebra, commutative algebra, maybe even start with linear algebra first. (And do some other maths related like algebraic topology, algebraic number theory and manifolds)
It's like if you're asking a physicist to explain string theory without knowing mechanics or calculus
Hey nerds
This conversation reminds me of a blogpost by Mumford titled "how to explain schemes to biologists"
That sounds interesting
I think it was really a rant about how little math other people learn
Lmaooooo
you should teach group cohomology so that people properly understand grade school addtition
I actually thought the blog post was a little silly if I remember
It felt like he was insulted by the suggestion that biologists neither know nor need to know abstract vector spaces, or something
You write for your audience first and foremost, always. You can be sad at the state of math education or something but the audience is what it is
23:33
8
Q: Lifting a Diffeomorphism to the Cotangent Bundle

JonHermanBoth Abraham-Marsden and Da Silva seem to imply that given a symplectomorphism $g:T^\ast X\to T^\ast X$ which preserves the tautological $1$-form $\alpha$, it must be that $g$ is fibre preserving. In fact, this is addressed in the question Cotangent bundle lift theorem. The answer given to thi...

@MikeMiller in the answer to this question, he says that the $\alpha$-limit set $S_{0_x} = T^*_xM$.
I don't see why he concludes this.
What is the alpha limit set?
I think he does bring up the point on math being disconnected with others (even within STEM) which I think is a valid point and something should be attempted to be addressed, although I don't really know how this could be improved
@anakhronizein Are these the set of points which flow to z along the flow of the dual of alpha, either as t goes to plus or minus infinity?
Or rather, the set of points whose flowlines' closure includes z.
alpha is $t\to-\infty$ I think
And then omega is $t\to+\infty$
I don't know this language. But fair.
I think if it's "the set of points whose flowlines' closure includes z", this set gets preserved by g because g commutes with flowing
(or rather, S(z) gets sent to S(gz) )
(I worry about flowing to +-inty specifically because it seems clear you can only flow to a zero?)
23:46
On this note.. @Cows if you want to get an 'idea' on why schemes are developed you should learn a little bit about varieties first - then you'll see how there are certain things which are rather awkward to talk about (for example - multiplicities) and become more natural if you know schemes

At least from the geometric perspective, people in algebraic geometry really care about things like curves, surfaces and higher dimensional varieties - so I think it's a good idea to get a feel of them first.
ok
I am back in the room
let me get a few things done first
@MatheinBoulomenos thanks for displaying arrogance. 1) I have taken most of the classes you listed in you mockery of my mathematical capabilities. 2) the reason I am not in school right now is a bit personal and I suggest you reconsider and your words and show decorum next time.
@MatheinBoulomenos I attempted showing humility by asking in a meek manner
How did you go from "I have some trivial abstract algebra, self-read of course" to that?
but if you want to show how tough you are get on youtube and let's talk math so the world can see if you are as bright as you claim
23:49
lmao
@loch indeed I have been playing with some varieties in my recent work
math throwdown
If you know varieties, then why did you ask by saying "I know what a polynomial is"?
@MatheinBoulomenos get on skype and let's go live on youtube
talk is cheap bro
23:51
I am going to immediately step out of this conversation, but while I respect a desire for respect Mathei did not say anything arrogant. At most, he made suggestions that did not know your background.
@MatheinBoulomenos pick your math battles bro
You responded extremely aggressively for no reason. I am certainly that you have gained no 'allies' in this approach.
wtf is this
I was nice until I got disrespected lol
@loch What are you up to lately?
23:53
I just want to learn some about schemes and you just start running up about my deficiencies and other things, I mean dude, you don't even know my background
I think the problem is that some math people grow up with everyone telling them they are the brightest people on the planet and lavishing praise. I'm not hatting but don't . . . talk to people like they are dogs
@MatheinBoulomenos learn some related math like manifolds? wtf
jesus christ man
Well anyway time for some Szamuely commentary
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