Mathematics - 2017-05-29 (page 2 of 7)

Ted Shifrin

03:00

Oh, it actually was lectures that never got published in French. I wonder how good Goldberg's English translation is.

Eric Silva

oh cool

Ted Shifrin

Ah, Chern wrote the forward for that ...

See y'all later.

Eric Silva

the array of topics definitely strikes my fancy, but every once in a while something is said and I have to think about it for days and do loads of computation

see ya @Ted

Daminark

See you @Ted!

Dair

cya @Ted

Daminark

03:03

Also @EricSilva backtracking quite a lot, do you think Marianna will cover harmonic functions next year?

Eric Silva

idk ask her

you still have class with her

Dair

@Eric Out of context it sounds like you're telling Daminark to ask a girl out lol.

Eric Silva

LOL

Balarka Sen

er

Eric Silva

that is very much not what's going on

Daminark

03:06

Oh God

Eric Silva

I'm sure she probably isn't really thinking about it right now though @Daminark

Also backtracking again, are there people in your cohort who you think would be very receptive to diff geo stuff? Are you the only one taking manifolds?

Daminark

There are 5 of us doing manifolds

One is not doing the bootcamp though

But yeah I mean, I think a lot of people will enjoy it enough

Eric Silva

Wow that's a lot more than in my cohort

Only me and Brian did

Daminark

For what it's worth, not everyone will be ecstatic

But on average people will see it as at least reasonably interesting

We also have 3 people taking complex, I think

Eric Silva

We had a bunch

Me and 5 or 6 others I think

Daminark

03:20

I think Webster teaching and the knowledge that it'd happen in bootcamp got many people to not do it yet

The people who did complex now are wishing they did manifolds now and complex next year

Last quarter would've been good because Calegari but most people who doubled did so with topology

Dair

there is a class dedicated to manifolds?

Eric Silva

it's a basic differential topology class

Dair

oh diff topo. i remember that class. I had the same teacher for diff topo as complex analysis. all in the same semester too.

Eric Silva

I took them in the same quarter as well

It was a fun term

Daminark

Lol, I imagine. Neves difftop was dank

Eric Silva

03:32

I would've preferred your version of the course, but I'm glad I took it my first year, that quarter would've driven me crazy without it

Daminark

Lol, I imagine

At some point I will need to go over the stuff, the stuff was a bit scattered and my understanding was vague

And lol you had the professor who did sheaves, right?

Eric Silva

Yeah

Daminark

Lmao

I mean I guess you gotta learn them eventually

Balarka Sen

meh. i mean it's an ok perspective

Mike Miller

03:52

Warner mentions them even vOv

Fargle

What should I know before trying to learn what a sheaf is?

inb4 "everything"

Daminark

Everything except what a sheaf is

Lol jk

I don't know what a sheaf is

Balarka Sen

@Fargle I say everything to answer the question "Why do I care?"

I mean I could suggest topology, and groups/rings but you'd only understand the definition and not why you'd want to know the definition

which is really more relevant imo

Fargle

That's a fair point.

Balarka Sen

start by knowing what a smooth manifold is

(easiest examples of sheaves)

Daminark

03:56

"A smooth manifold is a locally ringed space such that..."

Lol jk

Fargle

I think I know what a smooth manifold is.

Daminark

Sheaves are apparently certain types of categories so I imagine that by the time you care about them you at least know algebraic topology

Balarka Sen

nah man

sheaves may be a category but that's only how nlab would define them.

so yeah no

Daminark

I vaguely remember seeing that definition on wikipedia

Fargle

A smooth manifold is a set which is locally homeomorphic to (a fixed) $\Bbb R^n$ for which each transition map in the atlas is smooth.

Daminark

04:00

Yeah, preferably second countable and $T_2$ if I remember right

Because you want uniqueness of limits and partitions of unity

Wait

"We want it not only to be Hausdorff, but aLSo $T_2$!

Balarka Sen

@Fargle Ok. Fix a smooth manifold $M$. To each open subset $U$ of $M$ assign $C^\infty(U)$, the ring of smooth $\Bbb R$-valued functions on $U$ (why does this make sense?). This assignment is called a presheaf.

Or at least is an example of a presheaf.

Fargle

@BalarkaSen The ring with composition as multiplication?

No, that doesn't make any sense.

Balarka Sen

Indeed.

It's pointwise multiplication and addition.

In any case, why does "smooth function on $U$" make sense?

Dair

because composition of smooth functions is smooth again and so is the addition of smooth functions?

Daminark

That's not the point

This definition of a manifold is a topological space

Balarka Sen

04:07

Oh, no, I'm literally asking how to define a smooth function on (an open subset of)a smooth manifold. This is a technical point of course.

Dair

ohh.

Daminark

So you don't a priori know you have the structure of a smoothness, so the idea is to define it properly

Fargle

Right, I'm trying to consider that.

Daminark

Actually I should probably find a book that defines manifolds as such and go through it carefully

GP does everything embedded in $\mathbb{R}^n$

Balarka Sen

Yeah. It's not really that of a big deal but you should pick it up at one point.

Dair

04:11

a smooth function has derivatives of all orders defined everywhere?

Balarka Sen

Then learn the proof that every abstract smooth manifold embeds in R^n from Hirsch, so you know nothing is lost when you do GP's definition

Fargle

I feel like it's going to be something to do with restricting an atlas.

Daminark

Might try Hirsch

Balarka Sen

@Fargle Yep, good idea.

@Daminark It's a reference book to me, to be honest. I can't read through it.

Fargle

But each chart is mapping to a different "copy" of $\Bbb R^n$, and the transition map only tells you how those copies interact where the charts overlap...

Balarka Sen

04:14

Yes, it's a technical detail you need to pay attention to. But try it on a rainy day when you have nothing better to do, maybe :)

Alessandro Codenotti

Hi chat

Balarka Sen

hey

Dair

hi

Fargle

@BalarkaSen That happens to be today (er, tonight)

Balarka Sen

lol oops

Dair

04:17

is it really raining tho?

SBM

hmm hello people

Fargle

It has been

Dair

hmm hello @SBM

SBM

Hope you're having a good day

Daminark

Hey @Alessandro and @SBM

And @Balarka lol, well, what you do you think is better for working through more?

Balarka Sen

04:27

I don't know a great reference unfortunately

Daminark

I haven't noticed this because I was going with the flow, but apparently GP has a reputation for being wishy-washy with certain technical details that are important to see precisely

Balarka Sen

I learnt differential topology in fragments

Daminark

I see

Makes sense, the vibe I got is that most "manifolds" books start with all the basic stuff about smooth structures, various types of maps, bundles, Sard, all that

But then most of them seem to go down less of the transversality/intersection theory/degree and more toward things like forms and cohomology

So references are tricky to find, I only know of GP, Milnor, and Hirsch

Semiclassical

@BalarkaSen re: the formal thing from earlier

I think the issue for me is what it means to have a 'good' intuitive answer.

In some fields, that can be done pretty often. In others, not at all.

Balarka Sen

@Daminark Lee is also good

Daminark

04:51

Ah, alright, will check it out

Balarka Sen

The exercises are disappointing, but it's thorough with some of the stuff (vector fields and flows, Lie theory, stuff) you won't find on GP

Semiclassical

Lee for Lie!

user74961284659036576321890754

Rudin for?

Semiclassical

Rigor?

user74961284659036576321890754

Dummit and Foote for?

Daminark

04:55

Dammit my foot

Credits to Balarka

But yeah I mean, that's exciting for sure. I've wanted to learn a bit about cake theory for some time

Balarka Sen

@Daminark Definitely the reactions when the book slips out of your hand and falls right onto your toes.

user74961284659036576321890754

So dummit and foote is for sustaining foot injuries :P

BAYMAX

0

Q: Let $f$ be a polynomial with at least $k$ different roots, prove that $f'$ hast at least $k-1$ different roots

I don't know how to write the proof to this problem: Let $f$ be a polynomial with at least $k$ different roots, prove that $f'$ has at least $k-1$ different roots. It's featured among a few exercises made to apply Rolle's theorem, but this one is about a general result and asks for higher d...

I thought of proving it by contradiction!

Can we use that integration of a $k−1$ degree polynomial is a $k$th degree polynomial !

in the proof

user243301

05:56

@TheGreatDuck good morning, this is an response about one of your comments: my comment is about the edited post of the new user (with a question and member for 3 days). Thanks.

Abhishekstudent

06:29

0

Q: Why are we solving the limits in this way?

The question is- Find $ \lim_{x\to 5} f(x) $ if it exists $f(x)=\frac{x^2-9x+20}{x-[x]}$ where [.] is G.I.F. Now, my teacher solved it like this- $ \lim_{x\to 5^+} \frac{(x-5)(x-4)}{(x-5)}$ Now, (x-5) gets canceled and gives us $x-4$ $=$ $5-4=1$ Similarly, for $5^-$, we get the answer as 0...

limits functions limits-without-lhopital

SBM

easy

Balarka Sen

08:22

O_o

and that's my reaction to the entirety of the album

eg more O_o

Mike Miller

good album

Balarka Sen

um. lol.

I kinda like it.

Danu

Hey @MikeMiller

Leaky Nun

08:45

what axioms among the group axioms are needed to prove "if [xa=ya for all a] then [x=y]"?

Mike Miller

Hi @Danu

SteamyRoot

@LeakyNun Existence of an inverse.

You don't need $xa = ya$ for all $a$ anyway. A single $a$ suffices - just multiply $xa$ and $ya$ on the right by $a^{-1}$.

Danu

Hi @arctictern

arctic tern

or, without existence of inverses, it would suffice to take a=id (so applies to any binary operation with a right identity)

hi

Danu

So if I puncture $\Bbb C\mathrm{P}^1$ $n$ times, $n\geq 3$, and I look for automorphisms of the resulting Riemann surfaces, that's the same as looking for Mobius transformation with $n$ prescribed points

"Generically" it should not be possible to find any Mobius transformation that does that

(for $n> 3$)

arctic tern

08:55

yeah, modular group acts sharply 3-transitively

Danu

It would correspond to some kind of embedding of a subgroup of $S_n$ into the Mobius group

but it is possible "accidentally". Can I make this more precise somehow?

I am interested in $n=4$ first because I feel like you can maybe bootstrap those results (especially if they're mostly "negative")

So how many of the 24 possible permutations of the four points are actually possible as Mobius transformations?

How does it depend on the points I chose?

For instance, I know that if I take 3 of the four, permute them amongst themselves, then the fouth is a fixed point and that's possible by fine-tuning it to be one of at most 2 points

Is it easy to improve on this?

I feel like all of this is already known

arctic tern

wifi bad. all 24 are possible (whenever any are possible at all)

Danu

@arctictern So to get from 1 to all 24, what do you do?

arctic tern

Suppose the modular group stabilizes the set of points {a,b,c,d}. Then given any permutation, there is a unique g that matches that permutation when restricted to {a,b,c}, which must also send d to the same place as well.

err

I shouldn't be saying "modular group." not sure I'm making any sense.

Danu

Mobius group?

I mean PGL(2,C)

BUt what you're saying makes sense

so take {a,b,c} and take the g that sends them to prescribed points {x,y,z}

arctic tern

09:02

PGL(2,C) will never stabilize a 4-element subset

Danu

Then also d maps to w

@arctictern What do you mean by stabilize? I don't mean stabilize every individual element

It definitely can stabilize in the sense of permuting {a,b,c,d}

arctic tern

G stabilizes S if gS=S for all g in G

Danu

right

I mean not the entire group

I mean just a single element

I just mean there exists a g such that gS=S

(so an automorphism of the Riemann surface)

I wanna understand how many automorphisms there are as you vary n

For n=3 it's clearly exactly 6, but for n=4 what is it?

More, or less?

The symmetric group is larger but the system is now over-determined so typically one should have no solutions at all. What weirds me out is that it seems to depend on the choice of the points

arctic tern

what part of the world do you live in?

it's 4am here lol

try it with {0,1,inf,z} wlog

Danu

Yeah, I tried that

arctic tern

09:06

see if/how it depends on z

Danu

I'm in Germany

arctic tern

ah

Danu

you get solutions for some special case of z

For instance, I don't know, take $0\mapsto 1,1\mapsto \infty,\infty\mapsto 0,z\mapsto z$

This will be possible iff z is a fixed point of the unique transformation determined by the three other points, which is 1 or 2 points (too lazy to check)

arctic tern

1/(1-w) I think

Danu

yeah that's right

Right, so the fixed points of that

solutions of w^2-w+1=0

arctic tern

09:10

okay, let's distinguish the variable from the parameter z

Danu

right haha

so $z=\frac{1}{2}\pm i\frac{\sqrt 3}{2}$

are the only ones that will work, none of the others do

So say we take one of those two and do a different permutation of the 4 elements. Does it still work?

First order check: Does every permutation of $\{0,1,\infty\}$ fix the same points?

No, it doesn't I think

Because $\{0,1,\infty\}\mapsto \{1,0,\infty\}$ is given by $f(w)=1-w$

Manolis Lyviakis

Guyz

can u help me with complex analysis

i need to proof check my sollution

Danu

@MikeMiller Any comments on the automorphism group of $n$-times punctured $\Bbb C\mathrm{P}^1$ ($n>3$)? It depends on where you puncture... Take $n=4$ if you like :P

Manolis Lyviakis

math.stackexchange.com/questions/2301120/… can you plz?

Danu

So you don't get all 24 @arctictern... Maybe even just 1?

I conjecture either 1 (for the special choices of $z$) or 0 automorphisms

arctic tern

09:17

@Danu did you check permutations where $z$ is not a fixed point?

Danu

@arctictern No

But I can try

arctic tern

Suppose $g$ is an automorphism of $\Bbb CP^1-X$, where $X$ is a finite set of punctures. So I guess it lifts to a auto of $\Bbb CP^1$ with $gX=X$. This means $X$ is a finite union of finite cycles of $g$. It might be useful to look at what finite cycles of mobius tranformations are possible.

$x$ is in a finite cycle of $g$ of length dividing $k$ iff $x$ is a fixed point of $g^k$ (not sure if helpful)

Danu

So $(0,1\infty,z)\mapsto (z,\infty,1,0)$ is given by $\frac{aw-az}{aw-a}=\frac{w-z}{w-1}$

@arctictern Exactly, that was my second line of thought when I started yesterday

What's it called when $g^k=id$ for some power?

arctic tern

torsion

Danu

I was thinking of the subset of $S_n$ that has that property

arctic tern

09:21

but an individual $x$ fixed by $g^k$ is not the same as $g^k=\rm id$

Danu

No, of course not

Leaky Nun

@SteamyRoot thanks

Danu

But if you can permute let's say 3 of them in a torsion way

You would get it

arctic tern

oh, you mean solutions to $\{\sigma\in S_n:\sigma^k=\rm id\}$

Danu

yeah

So maybe say we know PGL(2,C) has no torsion of order k then we can rule out stuff like that

arctic tern

09:22

I don't see how. Having torsion as a map on CP^1 doesn't follow from restricting to finite order permutation of X.

Danu

Also that other permutation I gave above also doesn't preserve the fixed points

Does any permutation ever preserve fixed points?

arctic tern

${\rm PSU}(2)\subset {\rm PSL}(2,\Bbb C)$ should be able to act on $\mathbb{CP}^1\simeq S^2\subset\Bbb R^3$ by any possible 3D rotation. Suppose we take a set of points in $S^2$ with lots of symmetry (like a regular tetrahedron for the $n=4$ case)...

Danu

Actually it's probably important to note that that permutation $(0,1,\infty,z)\mapsto (z,\infty,1,0)$ does work.

So it's at least 2 non-trivial automorphisms

and it doesn't even depend on z (!) that this second one works

Manolis Lyviakis

if i take the limit $\lim_{z \to\ z_0} (z-z_0)^mf(z)$ and it is zero is it a pole or essential singularity?

Danu

Urgh, I'm doing something st00pid

Manolis Lyviakis

09:32

does any1 know?

Danu

@ManolisLyviakis If it is zero for every $m$ then it is essential

Manolis Lyviakis

lets take $\lim_{z\to\ 0} \frac{z^m}{e^z-e^{-z}}$

Danu

So @arctictern for any $z\neq 1$ the map $f(w)=\frac{w-z}{w-1}$ definitely works, and sends $(0,1,\infty,z)\mapsto (z,\infty,1,0)$. I double-double checked :P

Manolis Lyviakis

since it is 0/0 if i use D'hospital rule it is 0 for every m i think so that mean 0 is essential for the function $f(z)=\frac{1}{e^z-e^{-z}}$ @Danu or i cant use d'hospital

btw @danu

i think ive seen u here before

arctic tern

@Danu If I did my math right, for $\{\infty,\frac{1}{\sqrt{2}},\frac{\omega}{\sqrt{2}},\frac{\omega^2}{\sqrt{2}}\}‌$ we should get $A_4$.

Danu

09:37

Yeah, you have. But sorry I'm not going to work on your problem now @Manolis

Manolis Lyviakis

ok

Danu

@arctictern $\omega$ anything, or unit norm?

arctic tern

$\omega$ is cube root of $1$

Danu

ah

arctic tern

that's the regular tetrahedron in the Riemann sphere under stereographic projection

the whole SU(2)->SO(3) thing

Danu

09:38

Ah

@arctictern So, why is it $A_4$ and not $S_4$ or something else? Any insight?

arctic tern

orientation-reversing isometries of the sphere would require conjugation and be anti-holomorphic

Danu

And the odd permutations reverse orientation?

arctic tern

(the rotational symmetries of a regular tetrahedron are even permutations of the vertices; odd permutations come from reflections, which on the sphere are orientation-reversing, which back in the Riemann world are anti-holomorphic and would be mobius transformations of $\bar{z}$ instead of $z$)

right

Danu

Aight

cool

so for special values you get a lot

can one ever get all of $S_4$?

Can we prove that if one works, and you just switch around two elements, then it doesn't work?

First check: Does $\{0,1,\infty,z\}\mapsto \{1,0,\infty,z\}$ ever work?

Manolis Lyviakis

$\displaystyle \lim_{z \rightarrow z_o} f(z) = \infty$ does that mean it must be a pole for sure?

Danu

09:45

No, it doesn't for any $z\neq 0,1,\infty$ because $f(w)=1-w$ is the Mobius transformation

arctic tern

$\{\pm1,0,\infty\}$ should net you the dihedral group $D_4$ (at least)

Danu

How does that relate to $S_4$?

So those are the "largest cycle" ones, plus the ones given by reflections right?

arctic tern

multiplication by $-1$ is sufficient to transpose $\pm1$

otherwise yeah 90 degree rotation matrices cycle them

Danu

And $1/z$ transposes $0,\infty$ while fixing $\pm 1$

arctic tern

presumably because 1/z is conjugate to -z by a rotation matrix

but yeah

Danu

09:55

There should be a general result :P

@arctictern thank you so much for all this :D

I'm thinking what of this I can put on my exercise sheets for the course on Riemann surfaces...

arctic tern

Suppose g is in PGL(2,C) transposes two points and fixes some others (collect these things in a set X). Wlog (by conjugating g) we can ensure g transposes 0 & inf and fixes 1, so in other words g(z)=1/z. the only other fixed point is -1, which means for |X|>4 no automorphism can restrict to a transposition.

@Danu fun stuff

Danu

@arctictern That sound good!

It's funny because for |X|=4 I have this double transposition {a,b,c,d} -> {d,c,b,a} which actually works (that's the one I found earlier)---it doesn't fix anything but transposes

So if you add a 5th (and fix that one) this can't work is what you say

So it doesn't factor through S_5

arctic tern

you can get other odd permutations though. for instance 1/z on {0, inf, r, 1/r, -r, -1/r}. where r is not +/-1.

Danu

@arctictern Ah, but this is actually trivial because if that were the case then you have 3 fixed points $(|X\setminus\{0,\infty\}|\geq 3$) but not id, which is not possible trivially

That is, if you'r ejust transposing 2 elements and otherwise do nothing

arctic tern

consider the Iwasawa decomposition ${\rm SL}(2,\Bbb C)=KAN$, where $K=$ special unitary matrices, $A=$ positive real diagonal matrices, and $N=$ unitriangular matrices. everything in ${\rm SL}(2,\Bbb C)$ is conjugate to an element in one of these three subgroups. this should allow you to determine the conjugacy classes of ${\rm SL}(2,\Bbb C)$, and then determine the cycle types of a representative of each class

Danu

10:10

oompf

(btw we're working with PGL, not PSL, right?)

Wait... am I wrong?

arctic tern

doesn't mattress over C

Danu

right

because you just scale

arctic tern

inclusion PSL(n,C)->PGL(n,C) is an iso (not over R tho)

Danu

and it doesn't change the automorphism---the constant drops out

arctic tern

yeah

I think the iwasawa idea gives a complete solution

the finite cycles of things in A and N are obvious

which leaves torsion elements of K=SU(2)

which basically means you only get nonobvious permutations when you use polyhedral symmetry from special sets of points on the Riemann sphere stereographically projected into 3D space

err, actually have to see if nontorsion elements of SU(2) could have finite cycles first I guess

Danu

10:20

This is a little over my head

But thanks a lot

I don't really know much about SL(2,C)

arctic tern

KCd has a nice note which on SL(2,R)=KAN which generalizes pretty easily.

the wikipedia page on iwasawa decomposition is way too advanced

Danu

10:33

I read that KAN stuff at some point

BAYMAX

10:53

hi

can $\dot{x^2} = Ax + B\dot{x} + C$ can be written as two different set of or coupled or uncoupled differential equations ?

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