Mathematics - 2018-10-27 (page 1 of 3)

Mike Miller

00:00

Maybe I should have said facetious and avoided all this? :)

Ted Shifrin

ROFL

It would have saved my snide remark, yes. :P

Sorry, I'm in a grumpy mood, irregardless.

Mike Miller

No need to be sorry.

I'm always in a grumpy mood.

Ted Shifrin

I'm sorry :( ...

Nick

I have Cryptography and Digital Image Proc. today. Wish me luck.

Mike Miller

@TedShifrin What did I just say? :P

Ted Shifrin

00:07

keeps quiet

Leaky Nun

@MatheinBoulomenos why is Bayernhymne not in Bavarian

Mike Miller

@loch do you remember the theorem that if $G \circlearrowright \Bbb C^n$ is a finite group action on a (complex) vector space, $\Bbb C^n/G$ is smooth iff $\Bbb C^n/G \cong \Bbb A^n$ iff $G$ is generated by pseudo-reflections?

Leaky Nun

$\circlearrowright$ \circlearrowright

Mike Miller

00:22

i have that in the wrong direction

rip

imo it should be flipped along the line x=y

Ted Shifrin

I bet LaTeX only gives you flipping along $x=0$

Mike Miller

I meant x=-y, but yeah, you can only get the vertical flip

There is a package that has more that I went through great pains to add that specific symbol from into my usual preamble

Ted Shifrin

I actually coded my own transversality symbol because I wasn't going to settle for LaTeX's standard

I've done a few others.

Mike Miller

$\pitchfork$

I dunno, it's ok. I guess I'd like the cup smaller and lower.

Ted Shifrin

I wanted the bar on top, too.

Mike Miller

00:26

So you wanted upside down $\Psi$?

Ted Shifrin

I forget if I shrank that and added the top and whatever vertical was needed.

Nah, that's not gonna do it.

Mike Miller

Oh, is the joke that it's the letter T that's transverse to a curve?

Ted Shifrin

I know no jokes.

loch

@MikeMiller i remember you told me about this before

Mike Miller

ha

i forgot why it's true

and was going to ask you

loch

00:33

ah welp
seems like the latter isomorphism has a name - https://en.wikipedia.org/wiki/Chevalley%E2%80%93Shephard%E2%80%93Todd_theorem

Mike Miller

interesting

(A) = (B) = (B') is what I was asking at least

loch

@MikeMiller yeah

henceproved

01:00

minimal polynomial and charecteristics polynomial have same root: simple proof

c(x) = det( x - t)

putting x = t, we have c(t) = 0

so m(x) / c(x)

conversely

if y is a root of c(x)

then m(T)v = 0v = 0

= v(m(y)) = 0 implies m(y) = 0

Is my proof right?

Ted Shifrin

What you've written basically makes no sense for the first part. I haven't gotten to the second part.

So $X$ is the matrix? Then $c(X)$ is a polynomial in $t$.

You certainly can't set $X=t$. That makes no sense.

Mike Miller

That's a common (false) proof of Cayley-Hamilton.

Oh, it's Friday. I can't get excited for new papers. :(

Not that I read them anyway.

Ted Shifrin

People don't post papers on the weekend?

Mike Miller

They go up for Monday (on Sunday night).

Ted Shifrin

Oh, see ... I know none of this.

Mike Miller

01:11

No need to!

I'm gonna try to get some early sleep. Getting up at 4 to get to the airport.

Ted Shifrin

Damn, you're off again!

One of these days we need to get together.

I've realized that one of the liabilities of living on the west coast is 6:30 or 7:00 AM flights ... I never dealt with those in Georgia.

Mike Miller

No, heading home. I just finished up two talks in NJ.

But luckily that means I'll land early enough that the day won't be completely shot, assuming I get sleep on the plane. I'm usually ok at that, not great.

I get a month off (ha! no, I need to get my paper out by the end of the month.) and then one more in NYC in Dec.

Ted Shifrin

Ohhh ... well, it's great for you that you're getting around so much. Seriously.

Mike Miller

Yeah, I agree. Talks were well-received, which was good, especially given the locations.

Maybe you have something to say here. I have no idea what the commenter was thinking.

Ted Shifrin

I have no idea what the OP was thinking.

Mike Miller

01:23

How does one encode some version of smooth function, etc? I guess you have charts and elementary functions and whatnot.

Ted Shifrin

The OP seems interested in discretizing. I don't see what the issue is.

Mathematica has no problems drawing pictures of manifolds.

Mike Miller

It's not so much the manifolds (as long as they're defined by zero sets of elementary functions) as smooth maps, I guess. But I am not sure exactly what he means.

Ted Shifrin

I have no earthly idea what he means.

Mike Miller

Maybe I shouldn't have commented.

Ted Shifrin

Or etherial, e(i)ther.

Mike Miller

01:26

I deleted my comment.

Zzzzzz.

Ted Shifrin

LOL, fine.

Eric Silva

sup nerds

Daminark

@EricSilva ready for the final round?

Eric Silva

of wut

dalbouvet

of beer on a nice friday

Daminark

01:47

GRE

dalbouvet

close enough

Startec

What are the steps to take the derivative of ln(1/x)?

Daminark

@Startec do you know the chain rule?

Eric Silva

oh ok

whatever

henceproved

suppose T is a linear tranf. over a vector space V over F sulch that all eigen value of T lies in F, then T can be written as triangular matrix? I think I weaken the hypothesis of the above theorem by requiring atleast eigen value is in F. Am I right in saying so?

Scheme of proof: One show that V has one invariant subspace W (which will turn out to eigen space), and then we use induction on V/W

Daminark

01:56

@henceproved one eigenvalue isn't enough, that would imply that every linear operator on an odd-dim vector space is triangularizable

henceproved

that would be an intresting example to check for me

Daminark

As a counterexample, take an operator whose char poly is (x^2 + 1)(x-1)

henceproved

yes for every step of induction we need one eigen value

now I got it

Daminark

For reference, I think it's a theorem that you're triangularizable over a field iff your minimal polynomial completely splits in the field

Startec

@Daminark yes. I sett, i can treat it like -1

Daminark

02:05

@Startec I'm gonna go to sleep now, sorry for cutting short, but yeah you want to consider f(x) = log(x), g(x) = 1/x, and then note that you're taking the derivative of f(g(x)), so use chain rule

Seth Taddiken

Is it possible to have a matrix that converges on itself after infinite transformations by itself, but which also has only imaginary eigenvectors?

It seems to me like that would not be possible... but I'm dealing with a matrix right now that appears to have those properties

Seth Taddiken

03:05

Found my problem, didn't realize that MatLab had zeroed out some of the imaginary components... so the eigenvectors were real hahaha

Secret

04:54

[Random]

$\frac{n}{2} = \text{die}, \text{odd}$

$3n \mod 2 = ?$

$(2+1)n \mod 2 = 2n \mod 2 + n \mod 2 = 0 + n \mod 2 n \mod 2$

$\therefore 3n \mod 2 = n \mod 2$

$\therefore 3n + 1 \mod 2 = n \mod 2 + 1$

$\frac{n}{2} = \text{die}, \text{odd}$

$3n +1 = \text{even}, \text{odd}$

Let $n \mod 2 = 1$

$(3n + 1) \mod 2 = 0$

$1 \to 0 \to \cdots \to 1 \to 0 \to \cdots \to 1 \to \cdots$

FAIL ROUTE. RESTARTING KERNEL

user2236

05:09

why are you shouting?

Secret

It's not shouting, most computers output error messages and system messages like this

user2236

btw, we should have a "CAPS-LOCK" day in here :-)

Secret

$n = 2^m \cdot \text{odd primes}^{(n,o,p,q,r,s,\cdots)}$

$m > 0$, run $\frac{n}{2}$ until $n = \text{odd primes}^{(n,o,p,q,r,s,\cdots)}$

Next: $3n +1 = \text{odd primes}^{(n,o+1,p,q,r,s,\cdots)} +1 = ?$

Let $\text{odd primes}^{(n,o+1,p,q,r,s,\cdots)} +1 =2^k \cdot \text{odd primes}^{(u,v,w,x,y,z,\cdots)}$

$\ln (mess) = k \ln 2 + \sum \text{exponents} \cdot \ln (\text{odd primes})$

FAIL ROUTE. RESTARTING KERNEL

$3n+1 = 2m$

$n\mod 2 +1 = 0 \mod 2$

$n \mod 2 = 1, \therefore 1+1 = 0 \mod 2$

$3n +1 = 2^km, \text{n odd}$

$3(2n+1)+1 = 6n +3 +1 = 6n + 4 = 2 (3n +2) = 2^k m$

$n \text{ odd}$

Rithaniel

05:32

I'm having a tough time finding a general approach to proving or disproving that some sequence is periodic.

Secret

$2(3n+2) \to 3n +2 \to 3(3n+2)+1=...$
ERROR, COMPUTATION TOOK TOO LONG. TERMINATING OPERATION

Wild number

Originally, wild numbers are the numbers supposed to belong to a fictional sequence of numbers imagined to exist in the mathematical world of the mathematical fiction The Wild Numbers authored by Philibert Schogt, a Dutch philosopher and mathematician. Even though Schogt has given a definition of the wild number sequence in his novel, it is couched in a deliberately imprecise language that the definition turns out to be no definition at all. However, the author claims that the first few members of the sequence are 11, 67, 2, 4769, 67. Later, inspired by this wild and erratic behaviour of t...

$1$
$2$
$3$
$2^2$
$5$
$2*3$
$7$
$2^3$
$3^2$
$2*5$

$1$
$01$
$001$
$020$
$0001$
$0110$
$00001$
$03$
$002$
$0101$

there is absolutely no pattern at all

It is impossible to solve the Collatz conjecture unless one has the full knowledge of the distribution of the prime numbers because it is these prime numbers that dictate how often you will need to do $3n+1$

Anything that has something to do with prime numbers is frustrating because we don't have a closed form formula that can enumerate all prime numbers

Isana Yashiro

05:53

2

Q: 4 females and 6 males will be seated on 19 chairs

There are 4 females and 6 males students. They will be seated on 19 chairs. How many ways can we do this, if no two females are seated in adjacent? What makes me confuse is there are only 10 students but the chairs are 19. If the chairs are 10 only (equal to the number of persons) I could do ...

help me check my answer

And I can believe that everyone did it wrong...

And the most annoying thing is that this kind of question OP don't give answer first, lol.

@Secret: good morning, Secret~

Secret

If there is no mass pinging going on, then there is no help vamp

user1414

yeah, they like to "cry wolf" loudly

LoveWithMaths

07:30

Is anybody there ? I have a small doubt

user1414

askaway

LoveWithMaths

Would be difficult to explain since i cannot upload an image here to elaborate

I thought upload was disabled

anyways

so the question is ASA wont necessary prove that triangles are congruent in my opinion

for example

In above diagram Angle BCA= Angle CAD

and CA are common side for both triangle cba and cda

and angle CAB= ange DCA

According to ASA triangle CBA and CDA should be congruent

but they arent.

Ignore got my answer. My base was itself incorrect, i dont know why but i considered BC parallel to AD and CD parallel to AB

user1414

08:03

thinking out-loud helps :-)

Isana Yashiro

@LoveWithMaths why not?

anyway, I need a help about why RREF is unique?

is there any simple way of proving/thinking it?

LoveWithMaths

08:32

@IsanaYashiro because they arent congruent i knew that before start. Lets assume they arent so i was trying to prove that ASA theorem wont always stand true, but later realised that my base understanding was incorrect about parallel lines, where i assumed BC || to AD and BA || CD. (Sometimes we assume stupid & weird things)

Isana Yashiro

@LoveWithMaths I meant, I think it's congruent so why you said they aren't, and if i remember correct the definition of parallel fits if your "Angle BCA= Angle CAD" and "angle CAB= angle DCA" are given(known) by question. Then it's the graph wrong. And why call yourself assuming stupid thing...

LoveWithMaths

@IsanaYashiro So as i was learning ASA theorem and I was applying it on a kite shaped parallelogram, also as seem in above diagram, the kite has a longer tail body and its not a rhombus. And we all know by common sense that kite with longer tail body will never be a rhombus but then ASA was allowing triangle CBA = triangle CDA. The stupid thing was i considered AB || CD and BC || AD

Isana Yashiro

LoveWithMaths

08:48

@IsanaYashiro Ok so what are we saying here ?

Ohh i get it.. wait dont say anything

Isana Yashiro

@LoveWithMaths Apparently an approximation of circle(?)

LoveWithMaths

@IsanaYashiro Circle ? Dont know what you mean by that in above context. But The above diagram cannot be a rhombus since you cannot prove the diagonals are perpendicular bisectors.. isnt it ?

BAYMAX

09:05

0

Q: Checking whether $w = f(z)$ is conformal and its mapping?

I was thinking about the conformal map $w =f(z)$ in which $f$ satisfies $w = f(z) = \frac{df}{dz} - e^z$. Is ths conformal, how can i find the conformal mapping of $z$ plane? Similarly what will be the case like how it changes if we take $w = f(z) = \int f(z) dz - e^z$? Seeing the question it...

complex-analysis conformal-geometry

any help?

user1414

saying "please" may improve your chances of getting help

Secret

die

die.

die.die

diediedie

QED

user1414

E

D

LoveWithMaths

@Secret what does above mean

Secret

just rambling because of seeing the chat is not flowing

LoveWithMaths

09:18

@Secret And how does that help :D

In fact that will make Baymax question get lost in above unnecessary chat.

@BAYMAX Someone please look into this if anyone knows ?

Secret

Well Baymax is not a RHV though some professors do get annoyed. My ramble is short because of this

If there is an actual RHV on, I do not mind flooding their question into oblivion. One of my friends will be able to walk you through what RHV means

To be very clear: My ramble is triggered by the chat discontinuity caused by the next message after Baymax's, not because of Baymax's question

user1414

Real Help Vampire

BAYMAX

Inever got so much baymax since start of my MSE

Secret

Meanwhile, I don't know enough complex analysis to figure out that question. In fact this is probably the first time I saw a conformal mapping question where a differential operator is involved

Leaky and other complex guys were asleep however

BAYMAX

haha

Isana Yashiro

09:33

@LoveWithMaths sorry, i should not make any joke since my english is not very good.

LoveWithMaths

@Secret I really dont understand the logic, may be some one can explain so i can get onboard on how things work on stack exchange.

Isana Yashiro

But if your known is that it's a rhombus then your statements "Angle BCA= Angle CAD" and "angle CAB= angle DCA" are not true.

@Secret lol

LoveWithMaths

@IsanaYashiro In fact they would have been true if it was a rhombus since opp side would have been parallel.

@IsanaYashiro ??

Secret

@LoveWithMaths that's just my personal logic, not the logic of SE

Isana Yashiro

@LoveWithMaths right, I wanted to say kite actually

That's why I said I should not answer anything before double check my noob english...

LoveWithMaths

09:38

@IsanaYashiro A kite can also be a rhombus too, i guess you meant a kite with long lower tail body.

Isana Yashiro

yes, you shouldn't continue the discussion with me lol

sorry for that

LoveWithMaths

No dont be since discussion helps in finding out new discoveries ;)

btw how do we write triangle symbol while we ask questions on SE

Secret

$\triangle$

Isana Yashiro

@LoveWithMaths but misused words lead to war sometimes.

sorry it's my bad english

LoveWithMaths

@IsanaYashiro true.

@Secret thanks what about angle symbol ?

Secret

09:47

$\angle$

LoveWithMaths

Cool

LoveWithMaths

10:00

0

Q: Is my approach correct & optimal way of proving that a parallelogram is a rhombus?

Consider midpoint as O And I will try to prove that the diagonals are perpendicular bisectors of each other. Given that $DC || AB$ and $AD || BC$ we can say that $\triangle CDB = \triangle ABD$ by Angle Side Angle theorem. Thus $CD=AB$ and $CB=AD$. $\angle COD = \angle AOB$ & $CD = AB$ (prov...

geometry angle quadrilateral

Abcd

12:17

0

Q: Combinatorics and number of functions

Let functions are defined from set A to set B where B = $\{\alpha, \beta\}$ and $\alpha , \beta$ are the roots of the equation $t^2 - \sqrt 2 t - \pi = 0$ , then the number of functions which are (A) discontinuous only at each even integers if $A = [0, 11]$ is $682$ (B) discontinuo...

calculus combinatorics

Can someone please give a look to the above question?

Mary Star

13:40

Given the euclidean transformation $\phi (x,y)=\frac{1}{5}(3x-4y+10, 4x+3y+5)$ ans the line $\epsilon$ with equation $2x+y+5=0$, then we the equation of the line $\phi (\epsilon)$ by substituting $x'=\frac{3x-4y+10}{5}$ and $y'\frac{ 4x+3y+5}{5}$ in the equation of $\epsilon$ : $2x'+y'+5=0$, or not?

Isana Yashiro

13:57

In row reduced echelon form, (except for last column) it's impossible to have a column entirely zero right?

or it will mean that there is no such variable?

Abcd

Anyone? Please?

I am very worried about it.

user76284

14:20

@MikeMiller So the exponential of a $C^r$ vector field is always a $C^r$ diffeomorphism?

@TedShifrin Meant to say $C^r$ diffeomorphism.

Mike Miller

C^{r+1}. The vector field is the derivative of the flow.

These things are somewhat subtle and I don't know a reference.

user76284

math.berkeley.edu/~kpmann/algdiff.pdf on page 4 says $\mathrm{Diff}^r(M)$ is a Lie group with the $C^r$ vector fields as its Lie algebra

Alessandro Codenotti

14:37

Hi @Mathei

Topologicalife

14:54

Hi.

Must an undefined function diverge?

Alessandro Codenotti

As stated I can't understand your question, can you elaborate?

Topologicalife

I have $f,g:\mathbb{R}\longrightarrow{\mathbb{R}}$ both continuous and $g$ bounded. I also have the ODE's system $\begin{cases}x'=f(t)g(x)\\x(t_0)=x_0\end{cases}$ Let $\phi$ be the solution of the sytem defined in $(-\infty, b)$.

Now I have $\displaystyle\lim_{t \to{+}b}{\phi'(t)=f(t)g(\phi(t))=f(b)K}$ with $k \geq g(\phi(t))$ since g is bounded.

And now, my notes claim that $\phi(t)$ goes to $\infty$ in $b$ since is not defined there.

That doesn't make sense to me. Is like saying if a funtion is not defined on a point then must diverge at that point.

I don't see why $\phi \to \infty$ in $b$.

MatheinBoulomenos

Hi @Alessandro

Alessandro Codenotti

I can see this by unraveling the definition, but is there an intuitive explanation of why sheafification doesn't change the stalks?

Jake S

15:34

Can it be that the dot product < (1-i, 2, 4i), (1, i, 0) > gives (1+i)?

Unknown MathMan

Can anyone explain what does this means: "a group homomorphism from $\mathbb{Z}$ to any other group is determined by the image of 1." ?

Jake S

To expand upon my question, I feel like it should be ((1-i) -2i)) as I would be taking the conjugate of (1,i,0) and multiplying it by 2, yet the answer seems to indicate that I should be adding 2i.

Isana Yashiro

I thought the proof of row reduced echelon form's uniqueness is easy, but not, it holy cowly has a tricky step.

Jake S

Multiplying the second component by 2 rather.

Isana Yashiro

16:10

Is it possible to tag a person not in this room?

Will him/her get a ping?

Alessandro Codenotti

Only if that person has been in the room in the last couple of weeks or so

Mike Miller

16:43

@user76284 Katie Mann is better at this than I am, so trust her over me.

Isana Yashiro

@AlessandroCodenotti thanks

Mike Miller

Observe though the warning that the smooth charts are not the same as the usual Lie theoretic exponential map. You cannot realize every small diffeomorphism as the flow of a vector field.

I am uncomfortable with the $C^r$ case. Sorry for asserting my statement so confidently.

MatheinBoulomenos

@AlessandroCodenotti not sure how intuitive this is to you, but sheafification is left adjoint to the forgetful functor from sheaves to presheaves and the stalk functor is left adjoint to the scyscraper sheaf functor, so composing sheafification and stalk functors is another left adjoint to skyscraper sheaf functor (technically composition of skyscraper sheaf functor and inclusion from sheaves to presheaves), so the two left adjoints are isomorphic since left adjoints are unique up to natural isomorphism

Alessandro Codenotti

Uhhh let's just say that unraveling the definitions was more intuitive for me

But I'll reread your message after actually learning some category theory

Isana Yashiro

17:00

man, wtf is this: ams.org/journals/proc/2001-129-01/S0002-9939-00-05538-6/…

Alessandro Codenotti

Some hardcore set theory

Isana Yashiro

it's looks like he's talking to God...

holy

holy moly

Akiva Weinberger

Got myself a copy of Dummit & Foote's Abstract Algebra by ~~stealing it from a friend~~ asking a friend if I could borrow it

Isana Yashiro

lol

but the power of internet gives us freedom to asking any ebook... why borrow

ÍgjøgnumMeg

@Akiva nise

Isana Yashiro

17:05

or if you like paper book, I like paper book.

Mike Miller

@AlessandroCodenotti Those answers never help my intuition and I know plenty of category theory vOv

It's a matter of taste

MatheinBoulomenos

@MikeMiller I don't disagree with that

Mike Miller

Yup, I know for some people it's very helpful

MatheinBoulomenos

Hey @Ted

Ted Shifrin

hi @Mathein, @MikeM

oh, and demonic @Alessandro

and @ÍgjøgnumMeg

Alessandro Codenotti

17:07

Hi @Ted

Ted Shifrin

oh, and DogAteMy

(too many peoples)

Mike Miller

I don't understand why he's demonic

Ted Shifrin

Cuz he kept trying to run me over with his parents' car.

Mike Miller

I was here long enough to know DogAteMy though

Ted Shifrin

Or during driving lessons. I forget.

Alessandro Codenotti

17:08

Why not both?

Ted Shifrin

Precisely.

Mike Miller

Ah. I thought maybe his avatar.

Ted Shifrin

Nah.

Mike Miller

Anyway, I am sure many people want to run you down. Mostly students.

Ted Shifrin

Good thing I retired across the country ...

Someone has me stumped with a question on one of my answers (where I cited some result from Dieudonné).

Leaky Nun

17:09

you cited some god-given result?

Ted Shifrin

Indeed, Leaky. A necessary and sufficient condition for multivariable differentiability.

I guess today is the day Eric and Demonark retake the GRE.

Mike Miller

Yup.

MatheinBoulomenos

@MikeMiller can you explain to me why homology with coefficients in local systems corresponds to taking homology of a suitable covering space?
I can (via "topological/categorical hogwash") see how a local coefficients system gives rise to a covering space: local coefficients systems correspond to functors $\Pi_1(X) \to \mathbf{Ab}$, so if we compose with the forgetful functor we get a functor $\Pi_1(X) \to \mathbf{Set}$ which corresponds to a covering space.

Alessandro Codenotti

Fargle too I think

Mike Miller

I think Fargle too.

Ted Shifrin

17:11

Leaky: Here if you wanna think about it.

Oh yeah, @Fargle, too.

Alessandro Codenotti

sniped

Mike Miller

Lol

From my end you sniped me

Ted Shifrin

So it looks to me, too.

Alessandro Codenotti

@MikeMiller Yeah that's what I meant, sorry

MatheinBoulomenos

(sorry for using category language)
But because all this stuff is just nonsense, I have no idea what happens if I take the (co)homology of that associated covering space

Alessandro Codenotti

17:12

We actually wrote the exact same sentence modulo some action of $S_4$

Ted Shifrin

A Klein sub-action?

MatheinBoulomenos

@AlessandroCodenotti I like how you think

Ted Shifrin

Hey, @Mathein. You should give me credit for my comment :P

MatheinBoulomenos

gives credit

Mike Miller

@MatheinBoulomenos Okay, so I have to be very careful - I once did the computation for coefficients in real and complex line bundles, but not for arbitrary systems

Allow me to think

Or rather find the computation and translate

Ted Shifrin

17:20

If you look at a local system in terms of sheaf cohomology, I think it's pretty clear you're getting a $\pi_1$ representation. I understood this in grad school.

Mike Miller

@TedShifrin the question is more subtle than that - I'm writing out the computation

Let $\rho: \pi_1 Y \to \text{Aut}(A)$ be the monodromy of a local system on a connected space.

Let $C_*(Y; \Bbb Z)$ be the cellular chain complex of some finite CW-decomposition of $Y$, and $C_*(\tilde Y; \Bbb Z)$ the cellular chain complex with $\pi_1Y$-action. The cohomology groups $H^*(Y; \rho)$ are (essentially by definition) isomorphic to the homology groups of the cochain complex $\text{Hom}_{\pi_1 Y}\left(C_*(\tilde Y; \Bbb Z), A\right)$, where $\pi_1 Y$ acts on $A$ via $\rho$.

Hmm, allow me to take $A = \Bbb C^n$ :)

MatheinBoulomenos

sure

Mike Miller

I was trying my best to just copy-paste my $U(1)$-system computation and it doesn't work like that lol

@AlessandroCodenotti $C_4$ even

Oh Ted sniped me

Topologicalife

Hello. Does anyone know what is going on here? math.stackexchange.com/questions/2973575/…

I think the derivative $\phi'(t)$ is not even defined in $b$.

Ted Shifrin

@MikeMiller Even better. Mathein gave me a nod of credit :P

Mike Miller

17:28

@MatheinBoulomenos OK, I think I was being somewhat unsubtle in my phrasing. The point is that the local system homology/cohomology is given by taking the universal covering space and taking the tensor product of chains (or hom) with a $\Bbb Z[\pi_1]$-rep. It is clear that if your action factors through a subgroup, you can compute on the corresponding covering space.

MatheinBoulomenos

@MikeMiller ah, that makes sense

Mike Miller

Now in general what I'm saying is "Take group (co)homology of this action on a chain complex." If things are nice, then group (co)homology is just the same as the (co)coinvariants. I originally came to the statement I had there via a spectral sequence argument.

Let me try to recover that.

Akiva Weinberger

A video popped up in my YouTube feed titled "Can We Block the Sun to Stop Climate Change" and I can't think of anything more ill-advised

Mike Miller

Let $C$ be a $G$-module in chain complexes. You always have a spectral sequence $H^*(G;H^*C) \implies H^*_G(C)$. This degenerates if the group cohomology in positive degrees of $H^*C$ vanishes for some reason.

Ted Shifrin

DogAteMy: That's on the level of Repugnican science.

Mike Miller

17:31

@AkivaWeinberger spending 50-60 years on PR campaigns to make climate scientists look out of touch so that any new laws don't impede your profits?

@MatheinBoulomenos Ah ok, so now here is the point. Let $C$ have coefficients in $\Bbb C$, and $G$ be finite. Then recall that the cohomology of a finite group vanishes on any divisible group.

So that spectral sequence only lives on the line $H^0(G;H^*(C))$, and so the group cohomology we are calculating is the invariants of the action on $H^*(C)$.

This works fine even for complicated local systems $\rho: \pi_1(M) \to GL_n \Bbb C$ so long as they factor through a finite group.

MatheinBoulomenos

very cool, thanks!

Mike Miller

I am sure there are many phrasings of this argument. The one I eventually wrote down for my paper is probably the most elementary, but I think only works for $\rho: H_1 M \to U(1)$.

I have been told, eg, that this should really just be Schur's lemma.

(for the divisible $G$-module thing up there I had in mind a transfer argument)

Akiva Weinberger

(Link to the video: https://youtu.be/tJFtdvrTZs4)

user 2646

did you rewrite the GRE? @MikeMiller

Akiva Weinberger

17:46

I dunno if climate change is an engineering problem or a politics problem

Ted Shifrin

Both.

Mike Miller

What does "an engineering problem" even mean?

"Well guess it's now time to geoengineer the shit out of everything and hope it works bc we have no other choice anymore lmao" I guess

Akiva Weinberger

@MikeMiller Throw metal at it until stuff happens

(Pretty sure that's what engineering is)

user 2646

aka applied metal throwing

Akiva Weinberger

There's also materials science, aka let's break stuff so we learn how much it can take

but that's what we're doing to the atmosphere already

user 2646

17:53

algebra was the science of reduction and comparison

Isana Yashiro

18:06

why a function $T$ is one-to-one then it has left-inverse? What about those element outside of range?

randomly assign them some values?

Mike Miller

@IsanaYashiro Yeah.

user 2646

Lunch break? @Daminark

Daminark

@user2646 just finished the GRE and came back home. About to have some breakfast

Krijn

Hi

user 2646

how did it go? @Daminark

Hi @Krijn

Krijn

18:21

Hi @user2646

Daminark

This probably would've gone better than the September one did if I took it in September, this time I was a bit sick, tired, didn't have time to eat, etc, so it probably didn't go as well

But eh first time was probably good enough for any school that I actually expect to get into

So even if I get a really shit score today I don't care that much

user 2646

hmmm

Krijn

@Danimark, apparently, you test better when you're hungry

Daminark

18:54

Turns out I had the right idea all along

user 2646

says who? @Krijn

Transcript for

$Mathematics$ Mathematics