Mathematics - 2021-06-22 (page 1 of 2)

1:33 AM

I now work for a multinational with a large presence in the US. As a consequence I am obliged to learn how to respond in the event of an active shooter event

Yorch

how does one?

Ted Shifrin

Lovely.

Yorch

how do you prove a linear transformation of size $3$ has an eigenvector without using almost anything

including determinants

real*

Andrew Micallef

One runs, if one cant run, one hides, if there is nowhere to hide, one must accept that their life is in immenent danger and do what they can to fight back

The video was actually chilling in how matter of fact it was

Yorch

1:49 AM

They played me a video on lab safety when I started using the lab in high school

it was very chilling

Ted Shifrin

@Yorch without determinants, how do you define eigenvalues?

Rephrasing: Do you know or not what a characteristic polynomial is?

Yorch

A number which has an eigenvector

I don't know what a characteristic polynomial is

Ted Shifrin

If we don't have the char polynomial (which Axler defines without determinants), I claim it’s impossible.

Yorch

I know the usual $\det(A-\lambda I)$ definition but I don't know if there is a more universal characterization

Ted Shifrin

Huh?

Go read Axler’s book.

Yorch

1:53 AM

I don't remember Axler's definition

Ted Shifrin

Hence, go read.

Yorch

Oh ok

it's like similar to the frobenius form

Ted Shifrin

No clue what that means.

Yorch

Frobenius normal form

In linear algebra, the Frobenius normal form or rational canonical form of a square matrix A with entries in a field F is a canonical form for matrices obtained by conjugation by invertible matrices over F. The form reflects a minimal decomposition of the vector space into subspaces that are cyclic for A (i.e., spanned by some vector and its repeated images under A). Since only one normal form can be reached from a given matrix (whence the "canonical"), a matrix B is similar to A if and only if it has the same rational canonical form as A. Since this form can be found without any operations that...

Ted Shifrin

Everyone else says RCF.

NO, much more basic.

Yorch

1:57 AM

what is RCF?

Ted Shifrin

Read the first line.

Yorch

dope

Ted Shifrin

Look at the ideal in the polynomial ring of all polynomials annihilating the linear transformation.

That’s what Axler does. Perfectly reasonable for an advanced student.

Yorch

thats just the multiples of the minimal polynomial right?

Ryan Unger

algebra hard

@TedShifrin hm?

Yorch

2:13 AM

the determinant is op wtf

Ryan Unger

don't use determinants

Yorch

ok

Ryan Unger

you have to use determinants for change of variables etc

that's sad

Yorch

I'm going to watch a Wildberger video now

Ryan Unger

good stuff

Yorch

2:18 AM

and then some Terry Davis

Russian Bot Who Knows Your IP

2:46 AM

i got banned for two days

because sometimes the truth is hard to digest

Yorch

2:58 AM

that's a brave move

copper.hat

it did seem more space efficient for a while.

Yorch

you know, with you knowing their IP and such

copper.hat

well, it does not help when you suggest one liners that would flatten someone's machine. that is pure rude.

Yorch

like sudo rm -r /bin ?

oh u need an f

sudo rm -rf /bin

I haven't tried it though

copper.hat

@Yorch not that, but similar, some silly one on windoze.

Yorch

3:02 AM

alt f4 will cripple ur hardware

copper.hat

i am ok with a bit of fun, but that is not nice

Yorch

and alert the police

copper.hat

i'm mostly linux based.

Yorch

same, but with carbon

copper.hat

we used to do stuff to friends (after we had made a backup) like putting a file named * in their home dir.

Yorch

3:05 AM

one of my friends also did a tun of stuff to my system

thankfully it was almost new so I just reinstalled it

copper.hat

decades ago there was very little security.

Yorch

I remember something we would do to our colleagues iphones was put it in chines

invert the colors

and then add double tap for zoom and triple tap for tts or something

copper.hat

one company i worked for had all hr dbs available without security

Yorch

was there juicy stuff on them?

copper.hat

salaries, etc

Yorch

3:07 AM

oh

no sexaholic records

copper.hat

in an earlier life i used to connect my terminal lines to one nearby and watch what people typed. early keystroke logger

Yorch

they call that ethical hacking now

copper.hat

probably had those sorts of things, we had a few domestic violence cases, etc

i wasn't really trying to hack anything, i was just curious if it worked

because i had a useful purpose to it

Yorch

academic hacking then

copper.hat

it never occurred to me that i would see the plaintext passwords. obvious of course, but that was not why i was doin git

i wanted to connect a single board computer to the mainframe so i could do some data logging for experiments i was running. this was before such things were done.

Yorch

3:11 AM

that sounds exciting

copper.hat

i worked spectacularly, i should have made a company out of the idea...

it was fun.

Yorch

I only used windows until high school

I thought being good at computers was turning the computer off with the windows shortcuts

copper.hat

this was around 1980

i liked hardware, in fact more so in some way. but software pays the bills.

my first language was basic :-)

Yorch

My first language was also basic

because we used it along with a parallel port for cheap robotics stuff

copper.hat

it was pretty cool. i used to to plot things like $(f(x+h)-f(x))/h$ to see how close it came to $f'(x)$, nerdy stuff like that

i love the hardware interface stuff.

Yorch

3:14 AM

I never learned it, I just sort of had someone over my shoulder who spoonfed me

copper.hat

its amazing, an arduino has more clout than our first computer.

Yorch

yeah I think everyone uses arduinos for that now

copper.hat

we experimented and i think got our hands on a byte magazine.

they had programs (mostly games) that you could type in

my brother loved that stuff, i was more nerdy and wanted to do little experiments

the idea that you had something that could react in a microsecond or so was fascinating to me

well, a little longer, that was the clock period

Yorch

how did you learn it?

with the byte magazine?

and experimenting?

copper.hat

i was studying elec eng, so had an idea of digital electronics, but the basic was self taughtt

when you are trying to figure stuff out even little hints are very helpful. so i would see stuff used in byte and try i out. things like peek & poke, that sort of stuff

when we found out that the screen was memory mapped that was a big deal

Yorch

3:18 AM

I hadn't even heard of that

copper.hat

we did not have an assembler unfortunately

but i figured out a little machine language

Yorch

sometimes I analyze the executable of my c++ code

copper.hat

silly things like measuring reflect response :-)

you mean you look at the assembler?

Yorch

yeah although I don't remember how I did it

copper.hat

you can do something like -s, i have forgotten, if you are using gcc/c++

but reading assembler for risc based machines is tough

Yorch

3:21 AM

oh yeah

copper.hat

its tough enough for 'normal' processors.

Yorch

risc is harder than cisc?

copper.hat

i like real time stuff. but have no use for it in my day job

harder because of pipelining

so with cisc you write an instruction and next cycle its done

with risc its spread out over a few cycles

so when you read the code you have to remember more stuff

a little ironic

i only look at assembler when i have some unusual issue or oi i need speed.

Yorch

I watched this interview the other day

Jim Keller: Moore's Law, Microprocessors, and First Principles | Lex Fridman Podcast #70

►

It was pretty good

this one was also pretty good

Computers Barely Work - Interview with Linux Legend Greg Kroah-Hartman

►

copper.hat

smart fellow

Yorch

3:25 AM

both?

copper.hat

keller

Yorch

I've got to cook some asparagus now

copper.hat

enjoy :-) speculative execution is interesting. i used to have a company that did formal verification.

Russian Bot Who Knows Your IP

3:40 AM

Hehe

We can make a religion out of this

rostader

5:30 AM

It got pretty late here last time.. about $:30 in the morning...had to catch up on some sleep

Last time I guess things were really unclear for mostly my fault. I will try to sort of clear things up.

Let T:\R^n \rightarrow \R^n be a function(non-linear). ONe thing for sure is I dont know whether lim _{\theta \tends \infty} T(\theta) is bounded or not.

Given this I want to assume \lim_sup_{\theta}\inf_{x \in S}\|Ax - T(\theta)\|_2 where A is \R^{n \times d} and S is comapact subset of \R^d is finite

To support my assumption I can think of two choices

Spectre

@rostader , do you know any problem generator that can ask proof-writing questions??

I know Wolfram Alpha has a Problem Generator but that's not what I need.

rostader

5:48 AM

Either \|T(\theta)\| = \exp(-theta) so that lim\|T(\theta)\| remains bounded. Or as I was trying yesterday assume, \sup_\theta\inf{x \in \R^d}\|Ax - T(\theta)\|_2 is bounded. And then claim that \sup_\theta\inf{x \in S}\|Ax - T(\theta)\|_2 is bounded for any compact set S

@Spectre Sorry I dont know what you mean...

Spectre

@rostader I mean, I would like to get some random proof-writing question generator

I am looking forward to write the PRMO and hence I need to prep myself in number theory so as to write the proof questions/numerical answer questions that may come up in later stages.

If you have no idea regarding such a thing, no problems, sir :)

It was out of a longing to get a hand in number theory that I asked you so.

rostader

Sorry.. I dont have any idea regarding it

Spectre

@rostader Ok thank you

Russian Bot Who Knows Your IP

6:36 AM

The only reason for time is so that everything doesn’t happen at once
- a famous guy with uncombed hair

robjohn

8:01 AM

@rostader I bet you don't have ChatJax installed (from the half-latex in your comments). The link is in the info for the room (at the upper right of the page in the desktop version or in the page info for the mobile browser).

Koro

8:22 AM

@rostader: That orange like fruit with the size of a volleyball is called Jambura.

I just found out.

Pomelo is probably an another name for it. Though, I'm not sure about whether they are exactly same or not.

robjohn

8:44 AM

Pomelo is sometimes called jambola. Looking up Jambura leads to Pomelo.

So they are probably all the same.

Kiro

9:44 AM

Hello! It is well-known that the index of a nilpotent N-by-N matrix is at most N. If you would write that and add a citation, what would you cite?

Prithu biswas

I have seen some "Anti-Pi Rant" videos where they say that Pi being a boring number. There argument is that "Sure pi is an irrational number , but so is root 2 , root 3 , the golden ratio phi etc".My confusion is that , isn't pi also a transcendental number too ? I heard that to this day , very few numbers were proven to be transcendental.

Kiro

Isn't virtually every real number transcendental though?

Prithu biswas

Yes .... but proving that a particular one is a transcendental number is hard.

satan 29

hi

I was reading about orthogonal functions

in some cases, they use a weight function w(x) inside the integral..and in some cases they dont..

just what difference does using a weight function make?

Omni man

10:28 AM

This is very hard math question. How to be millionaire?

Kiro

Own at least a million of monies

That was a trivial question

But if you want more practical advice, exchange your money to Iranian rials. At the moment you can get 420100 Iranian rials with 10 $USD

Omni man

(every $60$ second in Africa a minute passes)

Anyway I am already millionare. Thanks for advice.

Kiro

My pleasure!

satan 29

12:44 PM

i was supposed to write the fourier series for f(x)=|x| in (-pi,pi) and f(x+2pi)=f(x)

i got the cos and sin coefficients right

however, i get the constant term as pi

but the given answer has pi/2

and if I look at the final fourier series I got, substituting x=pi/2 gets the constant term

and f(pi/2) is pi/2

so that suggests the constant term should be pi/2

but calculating it from $1/ \pi \int_{-\pi}^{\pi} |x| dx$ gets $\pi$...whats going wrong?

satan 29

1:11 PM

nevermind, i got my mistake

Lucas Henrique

1:35 PM

hey chat

proof verification: given $H: X \times I \to Y$ continuous, where $I = [0,1]$, every $\gamma_{t_0} := (x \mapsto H(x, t_0))$ is continuous

proof: given $U_Y \in \tau_Y: \gamma_{t_0}^{-1}(U_Y) = \{x \in X: H(x, t_0) \in U_Y\} = \pi_X ( H^{-1}(U_Y)\cap \pi_I^{-1}(t_0) ) = \pi_X(H^{-1}(U_Y)) \cap \underbrace{\pi_X(\pi_I^{-1}(t_0))}_{= X} = \pi_X(H^{-1}(U_Y))$ which is open since projections are open mappings.

copper.hat

2:42 PM

@LucasHenrique it might be easier to show that $x \mapsto (x,t_0)$ is continuous and then use continuity of compositions?

more general at least, if not more straightforward

Lucas Henrique

2:53 PM

@copper.hat oh god

...

Xin Yuan Li

Does anyone know how are [0,1] and the real line isomorphic as measure spaces?

Alessandro Codenotti

3:10 PM

With the Borel sigma algebra? All uncountable standard Borel spaces are

Xin Yuan Li

yeah, but doesn't saying that R is a standard Borel space presupposes that it is isomorphic to [0,1] as a measure space?

Alessandro Codenotti

Standard Borel to me just means that it is a space with a sigma-algebra that is the Borel algebra of a Polish topology

Xin Yuan Li

oh, I was unaware of this definition...

I guess that makes sense if a Polish topology is a complete metric space

thanks!

Alessandro Codenotti

Complete and separable, yes

but in this specific case it should be easier than invoking this general result: obviously [0,1] embeds into R, and R embeds into [0,1], and for standard Borel spaces this is enough to have an isomorphism

”Cantor-Schroeder-Bernstein holds for standard Borel spaces”

Ok this is pretty much the same as proving the big result I mentioned earlier though so I’m not sure what’s my point

anyway all of this can be found in Kechris Classical Descriptive Set Theory if you want to see the details

Xin Yuan Li

XP ah, but that makes sense, I really just needed a measurable function whose inverse is measurable which turns out to be the obvious one (I was just hung up on the fact that [0,1] was compact while R was not)

thanks for the reference, btw

I never actually learnt set theory properly ':)

Yorch

3:23 PM

I think the hairy ball theorem can be used for the "matrices of dim 3 have eigenvectors" problem

Yorch

3:34 PM

3

A: Link between the hairy ball theorem and the fundamental theorem of algebra

This is a (very) partial answer: Suppose the polynomial is real and of odd degree. Then this polynomial is the characteristic polynomial of some matrix acting in an odd number of dimensions. Restricting this matrix to the sphere and then projecting onto the tangent space of the sphere defines a v...

copper.hat

4:00 PM

@LucasHenrique ?

Ted Shifrin

4:16 PM

@Yorch That certainly uses the characteristic polynomial.

@LucasHenrique Restrictions of continuous are always continuous, because subspace topology.

Yorch

in what part?

Ted Shifrin

To turn a polynomial into a linear map.

Yorch

you mean in the proof of the hairy ball theorem?

or in the post?

Ted Shifrin

No, in the thing you linked.

Yorch

oh

But that part isn't important

you can start with the matrix

that's just an extra step

We don't want to start with a polynomial, we just start with the matrix A

Ted Shifrin

4:25 PM

Oh, OK. So we need Stokes’ Theorem and/or serious algebraic topology!

Yorch

yeah, haha it isn't what I had initially hoped for

although it does sort of provide some goemetric intuition for me at least

although I guess that intuition could have been obtained by just thinking of orthogonal matrices

Thorgott

I feel like that is trying to explain an easier theorem with a harder one

Yorch

no doubt

I still think it's interesting though.

satan 29

hi everyone

i had a couple of fourier series questios

is there a function whose Fourier series is sin(x) + sin(2x) + sin(3x)..... ?

it "seems" like no, bcs it appears that this series diverges but I dont know how to prove it

Thorgott

4:41 PM

seems very plausible that there is, but I don't know

Ted Shifrin

@Thorgott i was happy with the intermediate value theorem.

Yeah, that’ll be a distribution but not a function. Page @leslie.

copper.hat

it diverges when $x \notin \pi \mathbb{Z}$. math.stackexchange.com/a/2047327/27978

satan 29

@copper.hat right

proof?

or maybe just a small hint

copper.hat

i added a link

Ted Shifrin

But $\sum e^{inx} = \delta$.

satan 29

4:55 PM

@copper.hat your link didnt render properly

Thorgott

it being divergent does not stop it from being the Fourier series of a function

Buraian

why do I feel like...

oh yeah

I know this

in the geometric series, put x=e^{i theta}

and check the imaginary part of the corresponding series

satan 29

@Buraian thats what i did actually

Buraian

ye that's right

have you heard of book called visual complex analysis?

it is discussed in there like this exact example

satan 29

i trued to derive sum of sin(x)+sin(x+d)..... usig that and it worked

using that

Buraian

4:56 PM

yah

satan 29

but i was wondering if there was a better method

Buraian

oh btw

copper.hat

@satan29 are you familiar with distributions?

satan 29

not really , no

Buraian

once you had that expresison, you can actually evaluate some nasty integrals. Multiply both side of the expression with sin(nx) and integrate from -pi to pi. You will see all terms die off in the right side except one, and left side you have a really ugly integral

This answer

satan 29

5:17 PM

thanks

Balarka Sen

5:28 PM

hi @Alessandro

Alessandro Codenotti

Hi @Balarka

shintuku

If $(0, 1)$ is a solution for $y = \pm \sqrt{e^{-x^2}}$, the author infers it implies $y = \sqrt{e^{-x^2}}$, i.e., without $\pm$. Does anyone have a clue why? Of course, it is obviously true if $x=0$ given our initial supposition, but how do we know this holds for other values of $x$?

Thorgott

oh look who's still alive

shintuku

the specter of socialism haunting europe?

Balarka Sen

@shintuku lmaoooo

@Thorgott actually its a miracle i am

terrible stuff in this part of the world

Thorgott

5:32 PM

well I'm glad you are

Balarka Sen

i got vaxxed

chip-ed in, all matrix now

copper.hat

hmm, microsoft in health care

Balarka Sen

well bill doesnt want to distribute chips to us

its probably because we run pirated windows in this part of the world

Thorgott

@BalarkaSen how does it feel to live inside a cohesive infty-topos

copper.hat

i can't believe one has to pay for windoze pos

Balarka Sen

5:40 PM

@Thorgott i actually have been thinking about $A_\infty$-algebras a lot

Thorgott

that's, uh, cool?

Balarka Sen

@Thorgott Are you an expert in characteristic classes yet

CommonerG

I have a quick group theory question. If a group G acts on a set S, I understand that an orbit is a minimal set of elements in S that G sends into itself. Is there a name for a maximal set that G always sends outside of S?

Yorch

$S$ minus an orbit

complement of an orbit

CommonerG

I'm not sure. The complement of an orbit contains other orbits. Thus, that complement has some elements that G does not send outside of the complement.

Yorch

5:49 PM

oh

Alessandro Codenotti

@CommonerG a transversal of the orbit equivalence relation

CommonerG

Ok! Let google that a bit.

Alessandro Codenotti

hm not quite, that works if there are no fixed points

Yorch

you want a maximal set such that $GT\cap T = \varnothing$ ?

oh but $G$ has $e$

CommonerG

Oh. ok. Maybe I need to rethink what I am after. Since G contains e (identity) this set doesn't exist.

I should really turn on mathjax.

Ted Shifrin

5:59 PM

@BalarkaSen excellent!

Balarka Sen

Hi @Ted!

Thorgott

nah, I'm not doing anything related to characteristic classes this semester, I still know almost nothing

Balarka Sen

Well I should say I only got my first dose in. Second dose in 12 weeks.

Thorgott

maybe I will compute some when I get to attempt working on my thesis

Alessandro Codenotti

@BalarkaSen astrazeneca?

Balarka Sen

6:02 PM

yeah

astrazeneca you meant

Alessandro Codenotti

I see, I got the first dose of pfizer and I only need to wait 4-6 weeks

Balarka Sen

aha

CommonerG

Here's what I'm thinking. Suppose we have C2 group acting on a set of 4 points $\{1,2,3,4\}$ with the orbits $\{1,3\},\{2,4\}$ what I want is a name for sets like $\{1,2\},\{3,4\}$. or $\{1,4\},\{2,3\}$.

Ted Shifrin

6:30 PM

@AlessandroCodenotti Odd. Here Pfizer was a 3- week wait.

Alessandro Codenotti

@CommonerG those are transversal of the orbit equivalence relation

you're picking one representative from each orbit

schn

If $k(u)$ is non-negative and $\int k(u)du=1$, does it hold that $\int |k^2(u)u|du \le \int k(u)|u|du$?

Lucas Henrique

@copper.hat your comment made me realize it's trivial, and I was strugling a while with something I shouldn't

copper.hat

@LucasHenrique the nature of mathematics is that many things go from from "don't know what to do" to "obvious" (clearly a bad characterisation) with little transition.

CommonerG

Perfect. Thank you so much @AlessandroCodenotti

Ted Shifrin

6:36 PM

@schn Why do you think it should?

schn

@TedShifrin Not sure really, but its related to an exercise about kernel density estimation. $k(u)$ is a kernel density estimator. There are more conditions on $k(u)$ that I left out.

Ted Shifrin

Well, as stated it’s definitely false.

schn

Can you show that it's false? :)

Ted Shifrin

Take $k=0$ except on $[3/2,2], where it’s $2$.

schn

6:57 PM

@TedShifrin Thank you! If I add the condition that $u\in[-1,1]$, does that change anything?

Ted Shifrin

Yes, of course. Think.

Oh, I was thinking of $k$, not $u$. Still false. Doesn’t my example still work if you shift it over?

schn

Yes, your example still works for say $[1/2,1]$, thanks for the help.

Ted Shifrin

Sure.

robjohn

8:43 PM

@schn Let $k_\alpha(x)=\frac{[0\le x\le \alpha]}\alpha$, then $k_\alpha^2(x)=\frac1\alpha k_\alpha(x)$, so you can adjust the ratio of the two integrals to be anything you wish.

robjohn

9:18 PM

Approach0 seems broken. Everything I've given it recently has returned an error.

Ted Shifrin

What's dat?

robjohn

@copper.hat That is true. When I see people ask what someone has done on something when they say they have no idea what to do, you know that even a small nudge in the right direction will help. However, if you give a hint, people complain that hints are not good.

@TedShifrin You've never used Approach0? it is a good, TeX-friendly search engine.

copper.hat

I suspect the takeaway is that some people will always complain :-).

Ted Shifrin

I do nothing but complain.

Interesting, @robjohn. Never hoyd of it.

robjohn

It's one of the things that people suggest to look for duplicates

Ted Shifrin

9:26 PM

Oh, I confess I only look for duplicates when it's a question I've answered.

And I'm still going to give hints and put leading questions. The hell with everyone else.

Oh, it's working now, @robjohn. Very cool.

robjohn

9:42 PM

I am trying to find $\sum_{k=1}^n(-1)^{k-1}\binom{n}{k}\frac1k$ and it returns an error.

I know that I have proved that this is $H_n$ somewhere, and I am trying to find it.

I just provided a new proof for a recent question, but I wanted to refer to the previous one

Ted Shifrin

Well, perhaps you reordered

robjohn

@TedShifrin I have tried all different ways to put it together. Approach0 usually does try various combos

Ted Shifrin

Ah

robjohn

it's very good at that stuff

Ted Shifrin

I can usually find things in my own answers pretty easily. You have thousands more than I, however.

Balarka Sen

9:53 PM

I need an example of a terrible topological space for a talk.

The best topological spaces are manifolds. The antithesis is a Cantor set but I would not call it a terrible space. What's a good space to dunk on?

Ted Shifrin

Double cone on the hawaiian earring.

Clearly it should not be semilocally 1-connected.

Balarka Sen

I was thinking of the Hawaiian earring actually.

Good call

Ted Shifrin

Glad to cooperate for once.

Balarka Sen

I'm going to give a series of talks on stratified Morse theory :)

I'm preparing lecture 1, and I start by explaining why stratified spaces

Thorgott

10:10 PM

the antithesis of manifolds are indiscrete spaces

Balarka Sen

That's not a topological space

Ted Shifrin

Finite CW structure?

Versus not. I guess the topology on the hawaiian earring makes it non-CW.

Balarka Sen

There are too many bad spaces to put qualifiers and explain in how many ways manifolds or spaces with controlled singularities are better

Hawaiian is quite pleasing though. Never liked that space.

Shouldn't offend too many people

Ted Shifrin

You’d better talk about my favorite example.

Balarka Sen

Whitney umbrella?

Ted Shifrin

10:19 PM

Bingo

And its different tangent bundles :)

Balarka Sen

Absolutely. Crucial, because I will sketch a proof that semianalytic varieties are Whintey stratified.

Whitney umbrella is the first nontrivial example which shows the singularity stratification is not Whitney regular

Ted Shifrin

If you do Macpherson Chern classes, let me know.

Balarka Sen

I actually want to pick up some of that as I give these talks. Long way from now though.

Ted Shifrin

:)

Lucas Henrique

hey chat

I was talking to a professor. I need to find a research area I like... someday, sometime...

topics in algebraic topology are really interesting and I liked algebra. I thought of something related to homological algebra or something but I'm really not sure

Thorgott

10:29 PM

@BalarkaSen it's a great space, you're a nerd

Ted Shifrin

The Whitney umbrella wins. Even motivates the notion of embedded components in schemes. (Not that I remember details.)

Lucas Henrique

the path of research is distressing, I'm sure I'm not mathematically qualified to choose what I'll be doing - probably - until I die

everything is too hard to know with deepness and the time is too short. there are technical matters like good subject but poor advisers in the area

really distressing...

Thorgott

Lucas, you don't have to settle so early

Lucas Henrique

I'd really appreciate any advice from how you guys chose your own areas while undergrads

Balarka Sen

@Ted @Thorgott Speaking of Chern classes, I realized a good way to think about them is as intersection of homotopies (homotopes between homotopies, etc) of sections of a given bundle with the zero section.

But then I found out Danny Calegari already knows this

Thorgott

10:39 PM

I haven't chosen anything yet

what you do during undergrad probably has less impact on what comes after than you're imagining

@BalarkaSen hmm, intersection of generic section and zero section is dual to the Euler class, which is the top Chern class, and you're saying something inductively based on the inductive construction of the Chern class or something?

Balarka Sen

That sounds like the right idea to me. Here is the penultimate Chern class for a complex vector bundle $E \to B$; take any section $s : B \to E$, multiply it with the fiberwise $S^1$-action in virtue of the complex structure to get a family of sections $B \times S^1 \to E$ which can be filled to a disk of sections $B \times D^2 \to E$ because the space of sections is contractible.

Intersect this with the zero section $E_0 \subset E$ and this is the representative for $c_{n-1}$, $n = \mathrm{rk}(E)$.

Thorgott

I mean, if you just scalar multiply fiberwise, isn't $B\times D^2\rightarrow E$ just given by scalar multiplication fiberwise as well?

Balarka Sen

No reason that will be transverse to $E_0$

You can choose a filling which is transverse to $E_0$, thanks to Thom.

Thorgott

10:56 PM

oh yeah, right, you want something transverse

this is cool, got a reference?

Balarka Sen

Nope!

Thorgott

f

Balarka Sen

It's relatively easy to check by naturality; I believe the inductive Gysin construction is related but have not checked. For Stiefel-Whitney classes you can do the same but with $S^0$-action. Which screams Wu's formula, because you're saying SW classes = homotopy-coherent self-intersection number of 0-section. Remember cup product does not commute, it commutes upto homotopy, and the failure is recorded by Steenrod squares.

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