Mathematics - 2020-11-12

schn

00:57

Is it true that the arithmetic mean of uniformly distributed i.i.d. r.v.s approaches a normal distribution faster than if they were exponentially distributed?

demonstrations.wolfram.com/…

Leaky Nun

01:11

youtube is down!

Vasting

Anyone have statistics on what percentage of grad students from top 20 programs end up tenure-track at a research uni?

Balarka Sen

07:47

@user2103480 Learn the proof of the Blakers-Massey theorem

It's the homotopy version of the excision theorem; says if $X$ is a CW-complex, $X = A \cup B$ for some subcomplexes $A, B$ intersecting at $A \cap B = C$, (1) $(A, C)$ is $m$-connected (2) $(B, C)$ is $n$-connected, then the inclusion $(A, C) \to (X, B)$ (or you think about the arrow in the other direction, excising the "interior" of $B$ out) is an isomorphism in homotopy groups of degree $\leq m+n$

Well, on degree $< m + n$ and a surjection at $m + n$.

(The proof is a PL topology trick using graphs of functions)

skullpatrol

@LeakyNun it's up now, pal

epic_math

08:05

Hello!

RewCie

it was up whole time

skullpatrol

hi pal

RewCie

hi pal

epic_math

helli @skullpatrol

RewCie

reading CLRS

epic_math

08:15

@RewCie hella

RewCie

Insane book tho!

@epic_math hi

skullpatrol

@epic_math hello

epic_math

recently saw a mandelbulb. Amazed.

Mandelbulb

The Mandelbulb is a three-dimensional fractal, constructed by Daniel White and Paul Nylander using spherical coordinates in 2009.A canonical 3-dimensional Mandelbrot set does not exist, since there is no 3-dimensional analogue of the 2-dimensional space of complex numbers. It is possible to construct Mandelbrot sets in 4 dimensions using quaternions and bicomplex numbers. White and Nylander's formula for the "nth power" of the vector v = ⟨ x , y , z ⟩ {\displaystyle \mathbf...

skullpatrol

Mandelbulb 3d How To Start, And Why!

►

epic_math

no don't hypnotize me

why are fractals so conceited?

RewCie

08:21

I'm sick now... Tonsillitis... fever, headache, dizziness,

took PCM, let's see if fever goes away or not...

skullpatrol

:(

epic_math

because they are so full of themselves

lulz sculz I am laughing out my hulz

RewCie

has anyone played Imposter?

^Among Us

I mean

epic_math

it's laaame

RewCie

Oh okay

epic_math

08:25

Does someone want enjoyment?

skullpatrol

ok

epic_math

journals.elsevier.com/journal-of-number-theory

ask me if you want a journal of some other field.

enjoy!

skullpatrol

thnx

epic_math

my pleasure

okay so now I think skullpatrol if busy

@RewCie are you also reading it?

Eran

09:47

How should I define the random variables of maximum degree in a random graph?

epic_math

11:14

How can we prove that $$\frac{\zeta(s)}{\zeta(2s)}=\sum_{n=1}^{\infty}\frac{|\mu(n)|}{n^s}$$ where $\mu$ is the mobius function?

@LeakyNun hello! can you help me to solve this problem?

oh I did it. We can use euler product formula

Sophie

12:19

Why is the Bonferroni correction the value that it is, intuitively? I understand the proof but consider the following argument:

Imagine I have $n=10$ hypotheses and I a desired $p=0.05$. Then if the null hypothesis holds then the probability that at least one of them will appear true is $1-(1-p)^n=0.4$. We wish this was $p$ instead so we have to divide $p$ by $(1-(1-p)^n)p^{-1}$ so our new threshold is $p^2(1-(1-p)^n)^{-1}=0.006$. This is more than the true value $p/n=0.005$, so the argument fails. Why?

love_sodam

13:50

If G/H = K where G and K are groups and H is a normal subgroup of G then G = K\oplus H?

Lukas Heger

@love_sodam no

consider G=Z/4Z and H=2Z/4Z, then K=Z/2Z, but G is not isomorphic to HxK

love_sodam

14:16

I see thanks

stackex33

14:32

Hi everyone, could someone verify my answer here. I posted this on the constructive feedback room but haven't (yet) recieved any feedback (constructive or otherwise). Thanks

0

A: Jump of an infinite step function

Instead of having a constant on $[0, 1/n]$, take a line from $(1/n, 1/n)$ to $(-1, 0)$ and then for $x<-1$, keep all $a_n=0$ (do the same thing for the other family of functions). This way the $a_n$ form a family of continuous functions. Now, fix some integer $n\geq 2$. Since $a_n$ is continuous ...

Edward Evans

14:52

this is a dumb dumb brainfart question, but is an isomorphism of fields also a linear isomorphism of 1-dim vector spaces?

user2103480

@BalarkaSen meh, maybe sometime later

Mike Miller

@EdwardEvans Vector spaces over what field?

There's a category of pairs (k, V) where V is a k-vector space, and morphisms are pairs (f: k -> K, T: V -> W) where f is a ring homomorphism, T is additive, and T(cv) = f(c) T(v)

user2103480

Is there a faster way to prove that the n-th homotopy group of the n-sphere is nontrivial, possibly not using brouwer, than this?
Assume it is trivial, and consider the homotopy group of R^n\{0}, which is the same by homotopy equivalence. Then the identity S^n -> R^n\{0} is homotopic to a constant map and hence we can extend it to a map D^{n+1} -> R^n\{0} that is the identity on the boundary. By dividing by the norm, we obtain a map to the n-sphere that is still the identity on the boundary, and if we concatenate this with some rotation, we get a map D^{n+1} -> D^{n+1} without fixed points

Edward Evans

I'm just brainfarting, because someone in my problem class claimed that if I have an isomorphism of fields $K \to L$ then that is also an isomorphism of $1$-dimensional $K$-vector spaces but I couldn't say why I find that suspicious

Mike Miller

Well why is L a K-vector space

The above framework packages in change-of-field but without that L is just some dipshit field

@user2103480 Brouwer is equivalent to the statement that the n-sphere is not contractible which is equivalent to saying that pi_n(S^n) is non-trivial

Edward Evans

15:02

right, I couldn't say why $L$ isn't a $K$-vector space lol

Mike Miller

Yeah he's wrong unless you use your iso to make L a K-vector space

user2103480

@MikeMiller okay fair enough, then I guess I can just use that

Mike Miller

@user2103480 In particular it's a little bit silly to say that you're "using" Brouwer. To prove Brouwer you're presumably setting up some machine, and that machine just as well proves the statement about pi_n(S^n)

Since they're straightfwdly equivalent

user2103480

@MikeMiller Gotta use something, right? Can't reprove everything

Mike Miller

I'm just saying that they're the same theorem

If A = B and you prove A then I wouldn't say you're using A to prove B

I'd just say you've already proved B

I think invoking Brouwer obfuscates the point because Brouwer is the less important claim

user2103480

15:05

I think this won't be a valid strategy in an exam

It always depends on the theorem that's proven first

Mike Miller

MF how did you prove Brouwer

Edward Evans

Maybe @Lukas can help: my kommilitone claims that if $\Bbb Q_p \cong \Bbb Q_\ell$ then the isomorphism $\varphi : \Bbb Q_p \to \Bbb Q_\ell$ is in particular a continuous linear map, and $|\varphi(\ell^{p^n})|_\ell \leq C|\ell^{p^n}|_p$ and that the RHS converges to 0 but the LHS diverges to $\infty$, which makes sense bis auf $\varphi$ is a linear map

Mike Miller

@EdwardEvans It's certainly Q-linear

Edward Evans

sure, but they claim that it is a linear map of $1$-dimensional $\Bbb Q_p$-vector spaces

shrugs

Mike Miller

The point above is that now that you are given the field iso you can make Q_l into a Q_p-vector space

It's an isomorphism given by transporting structure

Anyway I gotta go teach

Edward Evans

15:08

okay np, thanks

user2103480

@MikeMiller by showing there's no retraction, and then using the smart proof that constructs a retraction if we have a map without a fixed point

the first claim is proven using simplicial stuff

n-th homotopy groups weren't mentioned in that course

I may also add that every theorem is a tautology in ZFC if you include the premises, so saying that theorems are equivalent is only actually valid if you take out or add axioms :P

Rephrased: If one follows in three lines from the other but it was never mentioned, then I feel no shame in proving exactly that implication

Still good to know that I better not spend time searching for a faster proof, using just our course materials

feynhat

15:27

@user2103480 No retraction => the identity map on S^n is not null homotopic. QED.

Why would you invoke Brouwer.

user2103480

It's the same, I could say no retraction up there above

meaning no retraction & brouwer. I just invoked what the obvious theorem we have at our hands is

Gosh

Brouwers fixed point is the same as weak königs lemma over RCA0

why would anyone invoke anything

its just WKL smh

Lukas Heger

@EdwardEvans is the goal to show that $\Bbb Q_p$ is not isomorphic to $\Bbb Q_{\ell}$?

there's no reason for an isomorphism $\Bbb Q_p \cong \Bbb Q_{\ell}$ to be continuous

I think a good proof that $\Bbb Q_p$ is not isomorphic to $\Bbb Q_l$ is to consider the pro-$p$ completion of the multiplicative group (considered as an abstract group)

$\Bbb Q_p^\times \cong \Bbb Z \times \mu(\Bbb Q_p) \times \Bbb Z_p$

no if we consider the pro-$p$ completion of $\Bbb Q_p$, we get a pro-$p$ group of $\Bbb Z_p$-rank $2$

but for $\Bbb Q_{\ell}$ we get a pro-$p$-group of $\Bbb Z_p$ rank $1$

if $\ell \neq p$

the good thing about that proof is that it generalizes nicely to finite extensions

if $L$ is a finite extension of $\Bbb Q_p$ and $K$ is a finite extension of $\Bbb Q_{\ell}$ and $L \cong K$, then $p=\ell$

feynhat

15:50

what is wkl

Rithaniel

If you have a $0-$dimensional ring $R$ and $r\in R$, how do you show that $$(r)\supseteq(r^2)\supseteq(r^3)\supseteq\ldots$$ must stabilize in finitely many steps?

Like, I don't get this one. It's being very elusive

epic_math

Hello!

Rithaniel

$0-$dimensional doesn't mean Artinian, right?
Also, greetings

epic_math

was reading titchmarsh's book. Sometimes his reasoning is hard to understand, and he skips a lot of steps.

Edward Evans

@Lukas I gave the following proof: if $p < \ell$ then $\Bbb Q_\ell$ has more roots of unity than $\Bbb Q_p$ (there are exactly $p-1 < \ell - 1$ roots of unity in $\Bbb Q_p$ by Hensel's Lemma)

well that works for $p, \ell$ odd

Lukas Heger

15:57

that fails for $p=2, \ell=3$

Edward Evans

right

but I mean

Lukas Heger

and it doesn't generalize to finite extensions

Edward Evans

I could just find some elements of $\Bbb Q_2$ that aren't in $\Bbb Q_3$

Lukas Heger

you don't know what kinds of roots of unity are in a general local field just from the residue characteristic

Edward Evans

@LukasHeger nise

Lukas Heger

15:59

the proof is again to consider the $\Bbb Z_p$-rank of the pro-$p$ completion of the multiplicative group

this is $1$ for $\ell \neq p$ and $>1$ for $p$

epic_math

you people are talking about some advanced (but not that much advanced) mathematics but it's useless and a shit that even though we humans have made mathematics so much advanced but are techniques are still too lame to even prove the simple looking collatz conjecture. I mean what is the use of making so advanced techniques but they are still not enough to prove a simple conjecture. It's like learning to write before learning to read.

lulz schulz i'm laughing out my hulz

Edward Evans

okay man

thanks for the pointer

@LukasHeger ty :)

epic_math

Many people can understand the abc conjecture but still it's (unverified) proof uses interuniversal teichemular theory which is hard af

I would surely have spelt that wrong.

Edward Evans

what's your point

epic_math

Take the rabbits out before digging the rabbithole.

You would have understood what I wanna say.

Edward Evans

16:06

I think you're talking junk but shrugs

epic_math

man truth is not called junk

and truth sounds harsh always

feynhat

@epic_math We now have a very clear understanding of the interuniversal Teichmueller theory.

BWAHAHHAHA

Edward Evans

MWAHHAHAHAHA

that worked surprisingly well

epic_math

where the hell is the link going

Edward Evans

I'm deeply hurt

and embarrassed

feynhat

16:09

Didn't think you'd fall for it lol.

epic_math

you kicked me out

that is very bad.

Edward Evans

@feynhat I was genuinely excited hahaha

epic_math

okay be I have to go.

byeeeeee

feynhat

That's what I said, yes.

epic_math

I am always here.

Biden kicks Trump out of White house

Edward Evans

16:12

good one

user2103480

@feynhat weak könig's lemma

geocalc33

on this day in history Mount Vesuvius erupts

Tobias Kildetoft

@geocalc33 You mean Thursday?

geocalc33

yeah

also Alan Turing's paper "On computable numbers," in which he introduced the "Turing machine," was published on this day in 1936

Mike Miller

16:30

@user2103480 Lol you proved fucking Sperner?

Hack of a course that's not topology

user2103480

16:43

@MikeMiller I don't know what that is but this kind of cute picture was used as visualisation of whats happening

And then there comes something about the centers of mass of the simplices and a bit of geometry

Ah yeah the proof of BFT using sperner's lemma looks like it's the closest one in spirit

@MikeMiller It's probably the only proof that is accessible using the developed methods. All the other ones in the wikipedia entry use more advanced methods

geocalc33

given this time dependent vector field with time parameter $t$, $\vec X_t=\langle tx,-ty\rangle$, can this vector field be thought of as having symmetry in the sense that as $t$ runs from $(0,\infty)$ the vector field looks the same?

and can I think of it as a mapping from the vector field to itself?

Mike Miller

18:00

@user2103480 Yes but you should develop those more advanced methods

I do not like this

Lucas Henrique

hey chat

love_sodam

Lucas Henrique

currently reading about bilinear forms. our professor told us the following claim:

love_sodam

For given string, what is the optimal way to to get the maximum area of land surrounded by the given string. Here, two end points of the string must intersects with the road

Mike Miller

Uh this is not an easy problem, do you have tools you know are relevant here or results relating perimeter to area?

love_sodam

18:10

Well, I heard it uses PDE

Mike Miller

That's a very fancy approach, but one can do so

The answer should be the half-ellipse with given boundary and perimeter

This is related to the so-called isoperimetric inequality

Lucas Henrique

Let $\phi$ be a symmetric bilinear form, $\mathbb{F}$ an algebraically closed field of char $\ne 2$ and the base vector space $V$ anistropic w.r.t. $\phi$; let $T$ be an self-adjoint operator. Then there's an orthogonal base of $V$ w.r.t. $\phi$ s.t. every element of the base is an eigenvector of $T$

love_sodam

Hmm.. I never heard about that

Lucas Henrique

this is neither one of the real nor the complex spectral theorems. is there any specific name to this theorem?

love_sodam

I heard that problem from my friend so I actually don't need to solve but want to know how to solve this problem

Mike Miller

18:14

@love_sodam The very fancy way to do this is to study the space of functions from the interval to the upper-half-plane with specified boundary values, and $\int_0^1 |\gamma'(t)| dt$ fixed (the length of the string).

You study the functional $F(\gamma) = \text{Area } \gamma \text{ bounds}.$ You then calculate the derivative of this, aka, what happens when you wiggle $\gamma$ by a little bit?

A max is a critical point --- aka, wiggling $\gamma$ cannot ever increase area, so $\frac{d}{dt} F(\gamma + t\eta) = 0$ for all $\eta$

This gives you a new equation in terms of $\gamma$ which you then try to solve

This whole method is called "calculus of variations". Instead of doing calculus on R^3, you are doing calculus on a space of curves

@LucasHenrique I would still just call it a/the spectral theorem

love_sodam

Right right. I heard that's related to calculus of variations

Mike Miller

I have not thought through the argument here but that's a recipe to proving that the desired curve is an ellipse.

geocalc33

what if the string is corrugated

love_sodam

Well I heard the answer is half circle

geocalc33

18:31

A fully occupied hotel with infinitely many rooms may accommodate infinitely many new guests. Say at time $t=0$ the hotel is initially full. Then new guests are admitted. Thus at $t=1$ the hotel is full again. What is the name for this symmetry? At $t=0$ and at $t=1$ the sets look the same in the sense that they are both "full."

this is related to Hilbert's hotel

Mike Miller

@love_sodam Clearly false, take a string of length 100000 and the two specified points 0 and 1

If the length of string is pi times the distance between the two points on the road I agree

Lukas Heger

18:44

@MikeMiller what if the end points are not fixed? the problem statement by @love_sodam is not very clear on this

Mike Miller

Ok, in that case I agree it's a circle again

Ted Shifrin

There's probably a clever reflection argument that reduces it to the classic isoperimetric result. Perhaps you said that.

Oh, if the endpoints are fixed, then I'm in trouble.

Mike Miller

This follows easily from the isoperimetric inequality in the case that the endpoints are not fixed. If the length of our curve is L and the area enclosed is A, we may "double" the curve (take it along with its mirror image) to obtain a closed curve in the plane with length 2L and area enclosed 2A. Applying the isoperimetric inequality to these 2 pi A <= L^2.

Ah, same thought same time.

Ted Shifrin

Precisely.

Mike Miller

Now the circle realizes this bound, hence A = L^2/2pi is the maximal possible area enclosed.

I find the second problem more interesting

Ted Shifrin

18:59

I wonder if you get a sector of a circle then.

Mike Miller

I still think ellipse.

But maybe you're right.

Ted Shifrin

Does intuition say that we should have right angles at the boundary? I can't decide.

I think not.

I guess I can work this out.

Mike Miller

You think the variational problem will still give constant curvature?

I don't have time today to do the calculation unfortunately.

Ted Shifrin

Well, I have time. Do I have the interest? Maybe.

I think this is an old question, but I don't berember the answer.

I wonder if the line integral is the way to go, rather than the usual area integral.

Mike Miller

It certainly is. That's how I'd deal with the variational problem.

Mike Miller

19:32

@TedShifrin I wonder if there is a clever approach by adding the 'opposite' circular sector down below and applying isoperimetric.

Ted Shifrin

I thought about that briefly, but with the points being variable, I didn't see how to do it immediately. I'm checking the details of the computation, but (without endpoint conditions) of course E-L leads to the usual constant curvature result, so I'm betting I'm right.

Mike Miller

I thought we were imagining the points are fixed. When they are variable, you indeed get a half-circle by isoperim.

Ted Shifrin

Oh, shoot. You're right.

Then I'm sure it's constant curvature.

Mike Miller

Great!

Ted Shifrin

It's the usual E-L argument for sure. Yeah, so I think with endpoints fixed you put the circular arc on, as you suggest, and then it follows. Yup.

I thought about that before but got confuzled, I think.

Yippee for my naive intuition.

Mike Miller

19:38

I am less familiar with this stuff.

Ted Shifrin

It's OK. You're familiar with tons of stuff I have no idea about.

Mike Miller

Fair enough.

Ted Shifrin

I was proud of myself earlier, though. Someone posted a question and I thought immediately — Milnor.

Mike Miller

We just did covering space actions today in topology. On their homework they'll have to calculate pi_1 of the Klein bottle by showing it's a quotient of R^2 by a covering space action (I told them the group, it's up to them to cook up the action).

Ted Shifrin

This.

Mike Miller

19:40

Next week free groups, then SvK, then surfaces.

Ted Shifrin

Oh, when I ran the topology qualifying exam reviews in the summer, I always made the students go through understanding the deck transformations and how they actually operate. Somehow, they usually came out of the course without knowing that, even slightly.

Mike Miller

I just proved that pi_1(S^1) = Z, identified the key points in the proof, and stated that pi_1(X/G) = G by a generalized winding-number map whenever X is s.c. and the action is a "covering space action".

Ted Shifrin

Seems like you moved super fast through point set. Covering a lot of stuff.

I suspect you have some extremely good students.

Mike Miller

7 weeks for the basics, compactness, connectedness, Hausdorffness, quotients, etc, seems to me pretty reasonable.

That's about the pace it was when I took the class.

Ted Shifrin

A few years ago I did the topology course long distance with two former students/advisees and they insisted on covering all the fundamental group/covering space stuff in Munkres. We actually did it. I graded their homework and sent email comments, too. They were great. Both well on their way to PhDs now (one at Penn in geometry, the other I forget where in number theory, I think).

I always did Urysohn and Tietze when I taught the course. But with my "usual" students at UGA, I spent more than a quarter on the point-set.

Mike Miller

19:44

I wrote up Urysohn as a guided exercise sheet for anyone who wanted to do it.

I don't teach any point-set in class that they are unlikely to actively use in their careers, IMO.

Ted Shifrin

Well, Tietze gets used in algebraic topology :) And in functional analysis. I guess, being a geometer, I'm addicted to bump functions, and so they're a smooth version of all this stuff.

Mike Miller

My goals were more or less:

(1) Get people comfortable with the basics of point-set topology, strong enough that they can prove anything they need to in the future. Even if I do not cover metrizations, paracompactness, whatever, they can work it out.

(2) Give people a taste of what modern topology is actually like (so a little algebra, a little geometry).

(3) Get people to master the picture <---> proof translation.

(3) is hard to do with point set, so I think it is important to have enough time on these later topics to actually develop that skill

I never remember what Tietze is good for. Jordan curve, certainly. Do you need it for junk about CW complexes being NDRs or whatever?

Ted Shifrin

Yeah, I never did paracompactness, I confess, even though it's used in manifolds. I did enjoy teaching them to look for examples/counterexamples for a lot of the weird stuff, even if it's useless :P ... Interesting question for you — historically, even at Berkeley, less than 20% of the math majors go on in math (I don't know the modern number). It was way less at UGA. What is the number at Columbia?

Mike Miller

Don't know.

Ted Shifrin

I predict it's a lot lower than one suspects. So most of your students are just learning math not for a graduate or ultimate career in math. I don't know how that affects your calculus. Maybe showing them more of the big picture is more worthwhile. I always liked the Guillemin & Pollack course for that.

Mike Miller

19:48

It's an elective, so I teach it assuming people would like to understand topology pretty seriously. If they want to use it in math careers, cool. If they want to "get" pictures, also cool. They'll be prepared to either start working with more serious texts or to have fun on their own.

Ted Shifrin

Yeah, I think Tietze comes up immediately with AR and ANR.

Mike Miller

I hate that ANR crap. Technical junk.

Ted Shifrin

LOL, well, a lot of math sadly is technical junk. Another reason I like G&P :P

Much better proofs than those in algebraic topology for thinks like Borsuk-Ulam.

Mike Miller

@TedShifrin It doesn't, really. If anything it further justifies my decision to compress point-set.

You can't give the big picture with point-set topology, because there hasn't been a big picture in that subject for a half-century.

Ted Shifrin

Yeah, I'm ending up agreeing with you on showing them the flavors of math.

Mike Miller

19:49

If I teach this again one thing I will do is take your model of letting people choose their own difficulty level.

Ted Shifrin

I think you have the competence to do some knot theory stuff, which is very popular and current. I just don't know any of it.

Mike Miller

I prefer surfaces. You can do them very rigorously and carefully without knowing anything smooth (though I will have to punt a lot of that into bonus exercise sheets, eg the fact that all discs in surfaces are isotopic up-to-reflection, and that an orientation means a choice of disc-up-to-isotopy.)

We could definitely do knots and Wirtinger presentations and colorings as homomorphisms to S_3.

Ted Shifrin

Yeah, that technique worked pretty well in diff geo. I did have one supreme student — perhaps one of the most talented students I've ever taught — who opted too low all the time, and I was frustrated by that. She became a high school teacher, didn't go on in math, although she had more than enough talent to get a PhD.

Mike Miller

Good for her.

Ted Shifrin

I've never been excited about classification of surfaces, I confess.

Mike Miller

19:52

Unfortunately I've decided the "right" proof is too hard to present (cut down instead of build up).

Ted Shifrin

But I remember her as the only strong student who didn't push herself with the challenging problem options. But boy did she nail everything. I think she got better than 100% on the final. (She caught an error, on top of everything else.)

Mike Miller

Haha. Nice.

This is just a temporary game for me, anyway, since I'm not attempting to end up teaching at an R1 or something like that. Maybe CC. Who knows.

Ted Shifrin

I should look for her on FB or something. Have no idea what her name changed to, though.

Maybe R2?

Mike Miller

I'm gonna delete that, since it's on record.

Ted Shifrin

There's a lot of room between that and CC. Yes.

I think some of the fabulous-teacher grad students I worked with have ended up at Georgia colleges, teaching a large variety of things and doing some publications in pedagogy along with math. Less high-pressured.

Mike Miller

19:55

We'll see. I care more about being able to choose where I live. So I need to expand my options.

Ted Shifrin

Yeah, I get it 200%.

Mike Miller

Trying to get a very varied set of classes while I am here for the resume, even though that obviously means more work for me every year.

Give the good ones a few passes, like calc 1 or linalg or precalc.

Ted Shifrin

I worked my butt off teaching when I was a postdoc. Probably stoopid, but I loved it. And I loved doing the 350-person multivariable lectures twice.

Do you know about that question I linked above? (Flat bundle with nonzero Euler class.)

Mike Miller

That's what I'm doing now. Some weeks have been 60hr weeks. But it's worth it.

I've got 2 indep study students this sem, one last summer, and probably 3 next sem.

Ted Shifrin

Oh, how's the geometry kidlet doing?

Mike Miller

19:58

I missed him this week :( so I dunno.

Ted Shifrin

Just curious :)

Mike Miller

I dunno this stuff about non-O(n)-connections.

Ted Shifrin

Oh, OK. I think it's in Milnor/Stasheff, but I no longer possess that book. But I did remember that it was a famous issue that Milnor settled.

Mike Miller

20:15

I never remember that stuff, to be honest.

Milnor-Wood inequality is relevant?

Ted Shifrin

I never knew that. But this paper of Bill Goldman addresses it.

But, yes is the answer to your query.

love_sodam

What is <a_0,a_1,...,a_n>/<a_0,2a_1,...,2^na_n> where <> is a free abelian group

Mike Miller

Missed the word "abelian". It'z Z/2 x Z/4 x ... x Z/2^n. You can prove this.

love_sodam

Yes, but I saw that

Mike Miller

@TedShifrin Second cohomology of PSL(2,R) is relevant, as is Homeo(S^1). I knew this a half-decade ag.

love_sodam

20:28

the it's Z/2^nZ

Mike Miller

Yeah but that's not true.

Thorgott

you're missing a factor of Z/1 :P

love_sodam

Actually I'm solving hatcher 2.1.6

The above is H_1(X)

Ted Shifrin

@MikeMiller Makes sense, but I actually never knew this stuff. I just knew about Milnor's result. :)

Howdy, Thor.

Mike Miller

I have my own topology stuff to write up, someone else should take over on this.

Ted Shifrin

20:29

Lunchtime for this Bonzo.

love_sodam

Thorgott

hey Ted, bon appétit

love_sodam

It turned out that H_1(X) = <a_0,a_1,...,a_n>/<a_0,2a_1-a_0,...,2a_n-a_{n-1}>

And other's conclusion is Z/2^nZ

schn

What is a criterion function?

Eduardo C.

Hello everyone. When should I start attending conferences and talks?

I'm just starting to get my hand into more interesting stuff, but still I'm very very far from understanding what is going on in a research talk, seminar, etc.

A professor from a well-respected university is organizing weekly seminars (online) in a topic that I would like very much to work with (and also with him). Should I ask to join, just to watch/listen as curious, even though I won't understand much? If I continue and decide to study this topic, will this help me to get in contact with him?

Mike Miller

20:37

@love_sodam But this is different than what you wrote before

Because in the new formulation there are relationships between a_{i+1} and a_i

Take n = 2 to be explicit

love_sodam

Oh yes my mistake

Mike Miller

<a,b>/<2a, 2b - a> is actually generated by b, and 4b = 0

Your group is generated by a_n

love_sodam

Why is that?

Mike Miller

you can write a in terms of b...

Thorgott

shouldn't it be <a,b>/<a,2b-a>

Mike Miller

20:40

No, that's the case n = 1

I just killed off the stupid generator

Thorgott

ah

then I agree

love_sodam

Oh I see

user10478

21:00

Is there an online tool that can graph complex functions as surfaces?

Leaky Nun

21:48

@user10478 for example?

user10478

@LeakyNun Like this: youtu.be/T647CGsuOVU?t=106

Leaky Nun

that picture is a bit misleading though

e.g. if you just look at the picture you would think that the zeros of x^2+1 form a semicircle

but algebraically there's only i and -i

it looks like they're using colour to encode the imaginary part or something like that

and showing the real part with the distance to the plane

if you look at a plot of the real part of z^2+1 in wolfram alpha you pretty much get the same thing

so I guess this is the answer to your question

user10478

Hmm, yeah you're right about it being misleading.

Leaky Nun

every diagram is wrong

one needs to understand the limitations of each diagram

user10478

I'm not sure if it's what I should be aiming for then. What I was trying to do is see is a visual representation of what a negative constant to a variable power looks like with complex numbers.

Leaky Nun

21:59

aha

user10478

If you plot, i.e., $y = (-2)^x$ on desmos, it's a discontinuous curve of dots.

But I suspect there's more continuity with complex numbers.

Leaky Nun

Visualizing the Riemann hypothesis and analytic continuation

►

3b1b is the answer

user10478

Wait what

Leaky Nun

(start from 8:00 and you'll see an animation of f(s) = s^2)

user10478

I've watched that particular video a few times. I didn't know it was the answer to this question XD

Leaky Nun

22:02

the function f(s) = s^2 is mostly a 2-to-1 function

it's like folding two copies of something to itself

more precisely, every circle of radius r is mapped to the circle of radius r^2

but not just enlarging: it's folded on itself 2-to-1

i.e. you travel around the circle at double speed

user10478

Hmm, I'm not sure if that's exactly the same problem. That video shows a variable base to a constant power. I'm considering the case of a negative constant base to a variable power. This is bizarre with real numbers because, for example, $(-2)^{.5}$ is undefined, so you get a graph that looks like this: i.imgur.com/oJzwCOO.png I wanted to see if there's an extra hidden dimension or two to that graph using complex numbers.

user2103480

22:38

Leaky Nun

23:08

@user10478 aha but the problem is that things like i^i is multi-valued

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$Mathematics$ Mathematics