Mathematics - 2020-07-03

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12:16 AM

@BalarkaSen Hopf link, of course. :)

Nope, clearly wrong.

Hmm, now I'm confused. It's two fibers of the Hopf map, so, yes, it is the Hopf link. But I'm not seeing the linking. Maybe I need a drink.

Oh, duh. I see my stupidity.

12:40 AM

Hey Balarka and Ted!

Hi Demonark!

How's it going?

Still bumbling along, and you?

Pretty much same here, was a bit slow last few days so hoping to buckle down a bit the next few

You're never good at buckling :P

12:47 AM

Haha, hopefully I'll improve a little bit

Hopefully :)

6 hours later…

6:50 AM

whistles into the wind

Alessandro Codenotti

7:03 AM

Hi Edward

Abdullah O. Alfaqir

7:18 AM

Good morning chatters

@Peter hi

7:42 AM

Hi @Alessandro

et al

Alessandro Codenotti

7:54 AM

How is it going?

uhh okay, surprisingly my algebra course is defeating me (the doctoral students said even they are out of their depth)

so I'm just focusing on writing my number theory talk atm

how about you?

Alessandro Codenotti

Oh, what's the course about?

Quite well, I'll do my thesis seminar soon

8:10 AM

Well we first did some easy module theory, which was fine, and then we did some category theory, which was fine until the Yoneda lemma, and then suddenly we were building up to abelian categories and now we're doing homological algebra over general abelian categories (note: this is a 2nd semester bachelor's course rofl)

Nice though, good luck!

Alessandro Codenotti

@EdwardEvans lol now I understand why Mathei knows so much crazy algebra

the prof is Lukas' thesis supervisor, and he's a maniac

once the homological algebra is out of the way we'll move on to commutative algebra, which I think will be totally fine, but this entire chunk of the module has wrecked me lol

number theory is going well though, I'm giving a talk on Hasse-Arf and Kronecker-Weber in a couple of weeks

Alessandro Codenotti

Cool

:D

Lukas' profile pic changed on WhatsApp recently, so he's obviously just doing his own thing for a while

Alessandro Codenotti

Oh ok that's good

8:18 AM

aye

Alessandro Codenotti

Is Heidelberg going to do courses in attendance next term?

I don't know, we haven't had any communication about it

Alessandro Codenotti

I see

Slightly annoying but who knows

I'm hoping we stay online, I'm enjoying the online semester hahaha

Alessandro Codenotti

I strongly prefer in person classes

I'm happy I didn't need to take any courses this term

8:27 AM

The best situation would be: recorded in-person classes

but only one prof has done that so far in Heidelberg

I'm not sure what makes it so difficult to just set up a camera and record the lectures lol

Alessandro Codenotti

I guess they don't want to do that in normal circumstances because people won't go to the lectures then

Fair, unfortunate though

9:17 AM

@Alessandro oh and my scholarship funding just got reapproved for the next year :D

Alessandro Codenotti

Noice

I see you got all the seals in order this time around

1 hour later…

10:28 AM

@Alessandro yes indeed :D

11:06 AM

@TedShifrin Yup

It is a Hopf link in the $S^3$, that is.

11:26 AM

hey @Balarka

Hi @Edward, @Krijn

wadduppppp

Hi @Bala, still not sleeping I see?

@Alessandro Das Wintersemester wird am 02.11.2020 starten. Aufgrund der Corona-Situation ist zum gegenwärtigen Zeitpunkt davon auszugehen, dass auch im Winter wesentliche Teile des Studienbetriebs im Online-Modus stattfinden werden.

(assuming you can understand this)

Alessandro Codenotti

Ah I see

11:38 AM

same here most likely

Alessandro Codenotti

Random cool fact of the day: let $R\subseteq \Bbb R^2$ be a well ordering. Then $R$ isn't Borel, in fact it doesn't even have the Baire property

That means I won't have to cycle the 20km to uni in the winter

hahaha

Another random fact: Let $K^\prime/K$ be a finite totally ramified galois extension of local fields. Then $\operatorname{Gal}(K^\prime/K)$ is supersolvable.

Dunno if that's cool though

I just like the word supersolvable

@Krijn Haha yeah

But I just woke up

@EdwardEvans Lmao

I forgot what supersolvable means

solvable and all the quotients are cyclic

possesses a normal series with cyclic factors

11:43 AM

Gotcha

normal meaning that each subgroup is normal in the group

not only normal in the next one

oh you're right

lol

solvable and cyclic factors is just solvable

So, very close to nilpotency

ye

11:45 AM

at least for finite groups

yeah, the implication goes nilpotency => supersolvability => solvability

in any case, it follows basically immediately from the definition of the higher ramification groups

in fact, you can force one more thing in between

nilpotent => supersolvable => Lagrangian => solvable

all not strict

I just skimmed a note from Conrad about this 20 minutes ago lol

whatever these are

lol

Lagrangian means having subgroups of all orders dividing the group order

11:47 AM

Ah yeah supersolvable definitely has that

@EdwardEvans what if I quotient by the trivial subgroup

fun fact: supersolvability is equivalent to every subgroup being Lagrangian

Easy Sylow theory

I think being a product of Sylow subgroups has something to do with supersolvability

fun fact: in a non-commutative integral domain (what people call domain), $fg=1 \implies gf=1$

@Leaky :P stated the wrong def

11:48 AM

everyone knows that man, classic power series trick

nilpotent iff direct product of its Sylow subgroups

this is easy Sylow theory

I don't think supersolvability implying Lagrangian is trivial, I'd like to hear your argument

@BalarkaSen did you remember the wrong trick

@LeakyNun its the same trick for matrices no

what's the trick?

i dont want to think about it man its something

@Thorgott something man i have done it once not doing algebra now

i think tobias explained it to me

11:52 AM

smh

supersolvability for finite groups is equivalent to all maximal subgroups having prime index

so, under Galois correspondence, all minimal proper extensions will have prime degree

@Thorgott yeah i think then you use Schur-Zassenhaus starting from the minimal index guy

that one is normal and the group splits as a semidirect product over that by S-Z

and induct

I should learn real group theory some day

i tried but i forgot because most of it wasn't useful to me

@LeakyNun Oh right I was thinking about $1 - fg$ is a unit iff $1 - gf$ is

that one's the power series trick to guess the inverse

12:49 PM

@Thorgott I prefer the real group theory

3 hours later…

3:25 PM

"G be a finite group of order p^a q^b, then G is solvable" | I prefer the real group theory
"G be a semisimple Lie group, then ad : g -> End(g) is an embedding" | I said real group theory
"Let G be an algebraic group scheme over k, then G is quasiprojective" | Perfection

Manolis Lyviakis

user image

trying to understant what he means the homomorphism extending j and i

i mean whats the actual homomorphism

why, Balarka, why

I cry

Manolis Lyviakis

3:50 PM

ok ok i got it j is the extension homomorphism whose restriction to any of the subgroups that generate the free product is just the induced homomorphism j and i

Manolis Lyviakis

4:21 PM

actually im wrong

that would be true if the 2 fundamental groups where subgroups of the free product

the product on the theorem ia an external free product right?

Given homomorphisms out of two groups there's a canonical homomorphism out of their free product

something about direct sums

I don't understand the external comment. G_1 and G_2 sit as subgroups of G_1 * G_2 in a canonical way

Manolis Lyviakis

canonical way you mean isomorphic copies

Sure

Manolis Lyviakis

4:25 PM

munkres makes a distinction between external free products and free products out of subgroups

and proves some extension theorems in each case

Meh

If G_1 = <S_1 | R_1> and G_2 = <S_2 | R_2> are two presentations, G_1 * G_2 = <S_1 U S_2 | R_1 U R_2>. That's my definition of a free product

Manolis Lyviakis

and so to understand the proof above i need to understand in which case i am

No you don't it's a bad distinction

They are equivalent objects

Manolis Lyviakis

yeah your definition is preferable and more understandable

Use my above definition to prove given homomorphisms f_1 : G_1 -> H and f_2 : G_2 -> H there's a homomorphism f : G_1 * G_2 -> H which extends f_1 and f_2

The external vs internal distinctions are only useful when the internal thing gives you a way to recognize objects. For example the internal definition of semidirect product is useful to recognize that a group is a semidirect product of two subgroups

Manolis Lyviakis

4:31 PM

ok

This is a moot point in this scenario. It's nontrivial to recognize if a group is a free product. $\text{PSL}_2(\Bbb Z)$ is $\Bbb Z_2 * \Bbb Z_3$. How do you recognize that? Beats me!

Manolis Lyviakis

i was trying to ask how to recognise elements of the free product

about*

the canonical way of sending G1 to G1*G2

is just the identity?

yes

Manolis Lyviakis

i mean G1*G2 is not a cartesian product just concatanated words

right

Manolis Lyviakis

4:33 PM

its silly i cant say G1 is a subgroup of G1*G2

i get why G1 is not a subgroup of G1xG2

i say both very often, its abuse of notation yada yada

Manolis Lyviakis

well on the second case i can make arguments like the objects in the cartesian product have more dimensions

but i dont see the distinction on the first case

and munkres proofs are really long on these things

just think categorically

who cares about what these sets look like, what matters are the canonical inclusion homs satisfying the universal property

give me a categorical proof that PSL2(Z) = Z2 * Z3 thorgott

use the universal property chop chop

PSL2(Z) satisfies the universal property of Z2*Z3 QED

4:41 PM

im slain

Manolis Lyviakis

i can use the universal property to define the free product of groups

wobbly-dobbler owned by category theorist.

2

Manolis Lyviakis

so that the extension im talking about exists by definition?

@feynhat lmao

well, you can't define objects by universal properties

you do need an explicit construction to prove an object satisfying the universal property exists at all

but if an object satisfying a universal property exists, it is always unique up to unique isomorphism

4:43 PM

@BalarkaSen Super-noob question: How do I see EG? Say G=Z_2. It seems all 3(and up)-simplices would be degenerate, right?

Ah

no this is a good question. which model are you using?

the standard [g0|g0g1|...|g0g1...gn] are the n-simplices, right?

Yes.

the "right" model for $E\Bbb Z_2$ should be $S^\infty$, which is contractible and on which $\Bbb Z_2$ acts by antipodal action

but this picture gives some junk, I assume.

Somnath taught me a model he likes once, let me see if I remember it

But how do we conclude EZ_2 is S^\infty from the simplicial construction?

you dont, i think you get some contractible junk. of course, we only care about the homotopy type of EG

4:49 PM

S^\infty should have a non-degenerate cell in each dimension, no?

oh ok.

yeah it's obtained from attaching two n-cells for each n

S^0 in S^1 in S^2 in S^3 in ... take direct limit

but I think Somnath's model gives a decent picture 1 sec

Manolis Lyviakis

i like that the chat almost always talks about algebra/topology

If i define the free product of a family of groups through the universal property dont i get the homomorphism J that munkres posted above for free?

user image

J is the unique homomorphism of the universal property from teh definition

you can't define something by a universal property

@feynhat So I think his models is as follows, and works for any topological group G not only discrete. Consider the space of "step functions" [0, 1] -> G with at most n+1 discontinuity S_n(G), i.e., there are n subintervals of [0, 1] on which it's constant

This can be appropriately topologized, by compact-open topology.

S_n(G) naturally sits inside S_(n+1)(G) so take the direct limit S(G), space of all step functions [0, 1] -> G

This is EG I think?

yeah it's contractible, and G acts freely on this by multiplication on the values

Ah this is actually equivalent to the so called bar construction for a discrete group G

Why is it contractible

CausingUnderflowsEverywhere

4:58 PM

Waves

Manolis Lyviakis

@Thorgott Let Gi be a family of groups we define the free product of Gi to be any ${( P,j_i)} $ P being a group and j a family of homomorphisms that Gi--> P such that for evey G and a map f:Gi-->G there is a unique homomorphism H such that the diagram commutes?

Yes, that's the universal property of a free product. But you can't define the free product by its universal property.

@MikeMiller I guess you take a step function and "equalize the levels of the steps" so it becomes constant

Manolis Lyviakis

@Thorgott why cant this be the definition?

Because, a priori, you do not know that such an object exists.

This is generally the way it is for all universal properties: if an object satisfying a given universal property exists, it is unique up to unique isomorphism, but to know it exists at all, you usually need to exhibit some explicit construction.

5:05 PM

@BalarkaSen That sounds like it would give G, not EG, but I'm not sure the idea works since how would you equalize them

Is the space of 1-jump functions supposed to be like a model for G * G?

Yeah

I forget all this stuff

Manolis Lyviakis

@Thorgott so suppose i make the mistake and define the product as above and then construct an example is that ok?

no, examples don't prove anything

you have to provide a general construction

I think the usual argument that $G\ast G\ast G\ast G...$ is contractible is to show the n-fold join of G contracts to a point in the (n+1)-fold join of G by a straightline homotopy

Or well, join increases connectivity

@feynhat Yeah so I think this is a better model. Take the infinite join of $G$'s, which is contractible by whatever, and admits a free action of $G$. That's $EG$.

Do it for $G = \Bbb Z_2$, you do in fact get that $E\Bbb Z_2 = S^\infty$

5:12 PM

So by your construction, EZ_2 is actually homeomorphic to S^\infty?

I think so, yeah. Join of $n$-copies of $\Bbb Z_2 = S^0$ is $S^n$

And the inclusions are equatorial inclusions, so the direct limit is $S^\infty$

No. I was talking about S(Z_2).

not this join thing that you started talking about now.

They're the same model for discrete groups so no worries I think

same as in homeomorphic

Manolis Lyviakis

@Thorgott ive seen definitions start with the universal property but you think the
constructive definition and the property are not interchangeable

A step function [0, 1] -> X with 1 discontinuity is parametrized by three things; the point t in [0, 1] of discontinuity, where [0, t] gets mapped to and where [t, 1] gets mapped to, which both contribute to a factor of X

Manolis Lyviakis

5:16 PM

definition by construction*

So the space of all step functions is basically a quotient of X x X x I

If t = 0 and t = 1, the two copies degenerate to one copy, so it's exactly the join X * X

Similarly, you should be able to see S_n(X) = X * X * ... * X (n times)

@BalarkaSen *with 1-discontinuity?

My point is that if you say "the free product of G and H is a group G*H satisfying....", then you are implicitly assuming that such a group exists and this is not obvious (and, for other universal properties, not always true)

of course, if you know that such a group exists, then you can define the free product (up to isomorphy) by the property, but there is no way around proving this "if" at one point or the other

@feynhat yeah thanks

Manolis Lyviakis

@Thorgott ok gotcha

CausingUnderflowsEverywhere

5:23 PM

does anyone know the series of steps to isolating a variable

@BalarkaSen Yes. This seems believable.

CausingUnderflowsEverywhere

should I start with decompositions

and simplify everything into addition and subtraction?

What does BZ_2 look like in the simplicial construction?

user image

please give a hint?

First apply $\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$

@BalarkaSen In your S(Z_2), the anti-podal pair will be functions with their values flipped, right?

I mean when you identify S(Z_2) with S^\infty, the pair of functions corresponding to a pair of anti-podal points will be functions which disagree on every sub-interval.

5:46 PM

[Topology] Let $X$ be a countable set with a topology consisting of the void set together with all sets whose complements are finite. What sequences converge to what points?

And, that is exactly what orbits under Z_2 action look like: functions (with same no. of steps) which disagree everywhere.

ded?

@flowian I think all sequences converge. Eventually constant sequences converge to the obvious point. Non-eventually sequences converge to ALL points.

Suppose $(x_n)$ is non-eventually constant, and $x \in X$ is any point. Then any neighborhood of $x$ can leave out only finitely many points of $(x_n)$. So, the sequence will eventually lie in that neighborhood.

Assuming by 'countable' you meant 'countably infinite'.

yes, countably infinite.

CausingUnderflowsEverywhere

so I have this equation and Im wondering what I should be thinking of doing to rearrange so y is on one side.

$y^2 - 2by + y^2 m^2 - 2aym - 2bym^2 = c$

Well, for finite spaces all sequence are eventually constant. So my statement still holds.

non-eventually constant means that eventually it is not constant?

6:04 PM

Oh... my bad. I meant having infinitely many points instead of 'non-eventually constant'.

I see.

@flowian $(x_n)$ is eventually constant if there exists an $N \in \Bbb N$ such that $x_n = x_N$ for all $n \ge N$.

@feynhat Replace 'non-eventually constant' by 'has infinitely many points'.

To rephrase you:
Suppose $(x_n)$ is infinitely countable, and $x \in X$ is any point. Then any neighborhood of $x$ can leave out only finitely many points of $(x_n)$. So, the sequence will eventually lie in that neighborhood. Assuming space is countably infinite.

Exactly.

What I said about 'non-eventually constant sequence' is not true if it has only finitely many points. For example let $X=\Bbb R$, consider the sequence $0, 1, 0, 1, \dots$. This sequence is not eventually constant, but it does not converge to say 1/2 in your topology.

Why isn't it possible for $(x_n)$ to converge to $y$ in neighborhood $U_y$ s.t. all neighborhoods of $x$ and $U_y$ are disjoint?

@feynhat Ah I see.

Got it

If Topological space has countably infinite points, then every sequence must also have countably infinite points in the range. Is that correct?

6:14 PM

Hi chat

Hi Astyx

Heya

How's it going ?

Topologically

I guess that's a good thing

6:16 PM

yes

Not much man... Balarka was explaining classification spaces to me but he ded now. I think.

classification spaces are from analysis?

Such is the fate of those who try to explain classification spaces

Topology afterall
https://en.wikipedia.org/wiki/Classifying_space

feyn: * needs help with classification spaces * | Balarka: * I sleep * (literally)
thor: * needs help with formula jiggling * | Balarka: * real shit? *

6:30 PM

Anyway I don't know how is BZ_2 a K(Z_2, 1). It should be some RP^n (n > 1), no? But BZ_2 has no non-degenerate 2(and up)-simplex... what.

I think I don't understand the simplicial construction of EZ_2.

6:42 PM

@feynhat I don't understand why you think it's some RP^n. The construction of EZ_2 above gives the infinite join of Z/2s.

It's the colimit of $*^n (Z/2) = S^n$, so $S^\infty$.

Anyway, EZ_2 isn't a space, it's only defined up to equivariant weak equivalence, more or less. It's a contractible space with a free Z/2 action, and that is the property it's defined to have.

Then you have the fibration Z/2 -> EZ/2 -> BZ/2, and the homotopy long exact sequence is what tells you that BZ/2 = K(Z/2, 1).

7:38 PM

@feynhat The simplicial BZ/2 has a horrifying number of simplices. It's homotopy equivalent to $\Bbb{RP}^\infty$

@feynhat Ya, a step function [0, 1] -> Z/2 is basically a binary sequence, antipodality is negating the binary string

I didn't sleep I bailed out because I had to read some probability lol. You asked a fine question

I don't think it's feasible to try to geometrically visualize the simplicial model though

What does it mean to say that a Haar measure on a locally compact group is nonzero? Does it mean that there exists a subset $A$ such that $\mu (A) > 0$, or must $A$ be measure? If $\mu$ is nonzero, can we actually show that there must exist an open set $U$ such that $\mu (U) > 0$?

for a triplet of numbers, does them all being pairwise coprime mean that all 3 terms have no overlapping factors

@user193319 That should follow from the Haar measure being outer regular

8:14 PM

Hi, a @Balarka.

Hi @Ted!

Your Hopf link almost tripped me up. But, as usual, I was being stooopid when I tried to see the linking number incorrectly by filling in a disk. Ha.

It's actually pretty surprising to me. But you were correct of course.

I like how changing $zw = 0$ to $zw = \varepsilon$ modifies the neighborhood of the singularity from a cone over the Hopf link to the Seifert surface of the Hopf link

Well, the reason it was "obvious" to me immediately is that you're taking the preimages of the points $[1,0]$ and $[0,1]\in\Bbb CP^1$ under the Hopf fibration.

Your $\epsilon$ perturbation is the standard Milnor number game for isolated singularities.

@TedShifrin Yeah

Also yeah I never got around to reading Milnor's short paper on complex singularities

I probably very much should

8:29 PM

There's a book.

I learned about this stuff with Griffiths and his Duke J. complex integral geometry paper that showed me how to do my thesis.

Too much math to read

So he was interested in curvature integrals when you deform from the singular to the smooth like this. The curvature integral converges on the singular thing the "defect" is interesting. This is a bit like the little paper I sent you on curvature integrals on the Whitney umbrella.

Yup, always too much to read. That's why I quit :P

@TedShifrin, I have a random question, do u know who MS Dhoni is?

@TedShifrin Ahh I see, so there's some actual concentration of curvature at singular points.

@Arjun. Nope.

8:34 PM

@TedShifrin , ok thanks:-)

@Balarka: Not surprising, if you think about the real case and corners.

Right.

So this is also where different blowings-up come in. To see finiteness of the curvature integral one way is to take the closure of the image of the Gauss map on the smooth part and recall that curvature comes by pulling back curvature of the tautological bundle on the Grassmannian.

Hmm, I remember something.

Ah, so this is why you do Nash blowup and the like.

That makes perfect sense

So then you're merging into Chern-Mather classes and MacPherson Chern classes ... :P

8:37 PM

Hahah

I had some of that stuff in that little horribly typewritten paper.

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