Mathematics - 2021-02-19 (page 1 of 2)

robjohn

3:11 AM

Gads! I left 3 hours ago and no one has spoken since.

user 85795

3:29 AM

you've always been the life of the party in here :P

how was the rover landing?

Akiva Weinberger

3:46 AM

@BalarkaSen Heyy guess what

9

Q: What is the degree of an n-fold branched cover over a trefoil?

The order-2 cyclic branched cover over a trefoil has degree 6, meaning the preimage of any point off the trefoil has cardinality six. (You can find a wonderful video of this here, made by Moritz Sümmermann.) The order-3 cyclic branched cover over a trefoil has degree 24. What is the formula for t...

Ted Shifrin

@user85795 A bit of an exaggeration!

Rover

4:15 AM

@user85795 Rover landed safely!

User 873110

Hi, I ahave question Let $F$ be a free group of rank at least two and $a,b\in F$ be non-trivial elements such that $b\not\in \{a^n:n\in \Bbb Z\}$. Is it true that $a^kba^l\not =b$ for all $k,l\in \Bbb Z$ but finitely many?

Rover

4:34 AM

@Astyx I still didn't got the logic behind it can anyone help out...@robjohn

robjohn

@Rover what's the question?

Rover

For a group of 50 male workers the mean and standard deviation of their daily wages are 630 and 90 respectively and for a group of 40 female workers this are 540 and 60 respectively the standard deviation of this 90 workers is...?

Leaky Nun

you would benefit from knowing the E(X^2) - E(X)^2 formula @Rover

Rover

@LeakyNun ok, what's the formula?

Leaky Nun

the variance of {x1, ..., xn} is (x1^2 + ... + xn^2)/n - mean^2

Rover

4:48 AM

@LeakyNun Ah yes ,but we don't have x1, x2....xn

Leaky Nun

you don't need them

you only need to find out x1^2 + ... + xn^2 for the male workers and for the female workers

then you would be able to find it for the 90 workers

Rover

@LeakyNun Yes got it !

satan 29

5:06 AM

hello everyone

Find the volume bounded above by the sphere

x^2+y^2+z^2=32

and below by the

paraboloid

x^2+ y^2= 4z

Leaky Nun

find the intersection first

satan 29

the intersection is z=4, x^2+y^2=16

i.e a cirlce

Leaky Nun

then just do a double integration

dxdy

satan 29

I am getting a wrong answer.....

wont it be better to use cylindrical coordinates>

Leaky Nun

sure it would

satan 29

5:12 AM

$dv= sdsd(\phi)dz$

$\phi$ will be from $0$ to $2\pi$ no problem

Ted Shifrin

Put the $dz$ on the inside.

Leaky Nun

salve @TedShifrin

satan 29

@TedShifrin right

what my approach was

to divide the region into two parts: one of them being a hemisphere , lying in the region where x,y,z is negative. This volume is trivial

Ted Shifrin

No.

Too complicated. Because it probably isn't a hemisphere, is it?

satan 29

hang on, may I post a picture of what I think the region looks like?

Ted Shifrin

5:17 AM

No. Think about What I said .

satan 29

"because it probably isnt a hemisphere" this part?

Ted Shifrin

Yes.

satan 29

I maybe getting confused about what exactly the region is , because in my figure there is definitely a hemisphere...

and also, my answer differs from the answer by exactly the volume of the sphere

Ted Shifrin

Look at the math.

satan 29

@TedShifrin then what do you think the z limits are

Ted Shifrin

5:21 AM

Your notion of hemisphere is indeed sloppy.

Tell me the correct limits.

satan 29

@TedShifrin well that's where i am struggling then....I think I am not considering the correct region

Ted Shifrin

For fixed $s$, where does $z$ start and where does it end?

satan 29

isnt the volume simply region 2 + region 1?

Ted Shifrin

Start over.

Forget about two regions.

satan 29

ok

Ted Shifrin

5:25 AM

Don't be stubborn when you're clearly wrong.

The very top of a sphere is hardly a hemisphere.

satan 29

@TedShifrin XD I am sorry....ALthough I felt there wouldnt be any point in discussing if I dont have a clear idea of what the region is.... I will start over

@TedShifrin WAIT

Ted Shifrin

The picture is fine. Your thinking is not.

satan 29

@TedShifrin ok

Ted Shifrin

That's the intersection, yes.

satan 29

so for a fixed s, z ranges from s^2/4 to sqrt(32-s^2) ?

Ted Shifrin

5:29 AM

Almost.

Right.

And $s$ goes from ...

satan 29

0 to 4..?

Ted Shifrin

Right. Now you'll get the right answer.

satan 29

hmm yes I do

Ted Shifrin

Shocking!

satan 29

@TedShifrin i was indeed the considering the wrong region..

Ted Shifrin

5:33 AM

Now understand how wrong your thinking was.

Far from a hemisphere!

satan 29

the required region was sphere - (what I was thinking)

@TedShifrin I think you completely missed my point, Ted

Ted Shifrin

Oh, you had the sphere in the bottom?

satan 29

Yes!

anyways, thank you very much, I got my mistake.

Ted Shifrin

Above by the sphere and beliw by paraboloid.

Reading!

Ted is sooooo mean.

satan 29

5:49 AM

XD

@TedShifrin yeah it was a bad mistake

i have another one where I am completely flummoxed

Balarka Sen

@AkivaWeinberger I tried that group but that was obviously infinite, so I thought there was something wrong. So I guess I don't really understand what the "n-fold branched cover over a trefoil" means, and the question is nonsense to me.

satan 29

Find the volume bounded by the surfaces (x^2 + y^2/4 = 4-z) and (3x^2 +y^2/4=z)

Akiva Weinberger

@BalarkaSen How could it be "obviously infinite" if it's finite for $n\le5$

Balarka Sen

It's the 2, 3, n triangle group

Which is obviously mostly infinite

I immediately thought of the 2, 3, n triangle group actually, but discarded it :)

Akiva Weinberger

?

Wait what's the definition of the triangle group

Balarka Sen

6:03 AM

<x, y, z|x^m = y^n = z^l = xyz = 1>

Akiva Weinberger

Oh I see

Balarka Sen

Only for n <= 5 does it correspond to a positive curvature tessellation, hence is finite

n >= 6 we get hyperbolic tessllations, so they're obviously infinite groups

Akiva Weinberger

That doesn't match Wikipedia's definition?

Balarka Sen

Yeah but it's the same

Akiva Weinberger

Though it is the "Von Dyck groups"

Balarka Sen

6:05 AM

Yeah that's what I mean, the index 2 subgroup of the actual triangle group

The actual triangle group is generated by reflections along the sides

This is the one with orientation preserving isometries

Akiva Weinberger

Huh.

So $Q_n=D(2,3,n)$

I still wonder what the precise relation between $Q_n$ and $G_n$ is

Balarka Sen

Yup.

Leaky Nun

@AkivaWeinberger pues uno tiene un ku y el otro un ge

gracias por venir a mi ted talk

Akiva Weinberger

un qu?

Leaky Nun

Q se pronuncia cu no

Balarka Sen

6:10 AM

You should try to understand what the cover you want is actually going to look like to be honest

Leaky Nun

ene o pe cu ere ese te

Balarka Sen

To do which you should cut along Siefert surface "branch cuts" and glue in clever ways :)

robjohn

7:53 AM

@Rover did you get your answer? Sorry, I had to leave.

user 85795

Yup, leaky answered.

ery cerious matherfunker

8:10 AM

@PaulWhite make me CEO of SE

Paul White

@eryceriousmatherfunker I'm sorry Dave, I'm afraid I can't do that

4

robjohn

8:30 AM

@user85795 I didn't see the answer, I just saw him mention that the variance was the mean of the squares minus the square of the mean

Did they work it through from there?

ery cerious matherfunker

8:54 AM

@PaulWhite this one you can probably help

I run query in sql but it only runs one statement even after typing ;

returns no syntax error

I use android platform to make sql database

how to fix it

I have problem with sleep texting so please don't kill me

user 85795

4 hours ago, by Rover

@LeakyNun Yes got it !

@robjohn he appears to be satisfied

ery cerious matherfunker

9:09 AM

🤪

love_sodam

9:20 AM

Please recommend a Matlab book for beginners

ery cerious matherfunker

9:34 AM

@love_sodam amazon.com/gp/product/0124058760/…

go for pirate version if you are broke

porridgemathematics

9:53 AM

hi, could someone help me understand why BV is used to establish the first equality in this proof in royden: imgur.com/a/GFLZMht , I can see why we need it to establish the second one

but I dont see where in the steps of his proof he uses BV to establsh the first one

since $T = 2P - (f(b) - f(a))$ holds regardless of whether $T$ is finite or not, and then $P = N + (f(b)-f(a))$ also holds regardless of whether $T$ is finite or not, so shouldn't we immediately get $T = P + (N + (f(b) - f(a)) ) + (f(b)-f(a)) = P + N$?

here $T$ is the total variation of $f$ over $[a,b]$, $P$ and $N$ are its positive/negative variations respectively

Mike Miller

2:05 PM

If $a, b \in F_n$ then the subgroup $H = \langle a, b\rangle$ is free, because it is a subgroup of a free group, and hence it is free on either one generator or two. In the first case, assuming $a$ is non-trivial, your condition holds: then $a,b$ commute and $a^{k+l} b = b \implies a^{k+l} = 1$ implies $a = 1$ as free groups have no torsion.

In the second case the map $F_2 \to H$ given by sending one generator to $a$ and the other to $b$ is a surjective homomorphism. Finitely generated free groups are "Hopfian", so any surjective homomorphism from a finitely generated free group to anothe…

So your condition holds for literally any choice of $a,b$ except $a = e$.

I feel you must have meant something different. But if you did analyze the argument above: I am sure that this case-analysis will answer any question you're looking for like this.

Leaky Nun

2:20 PM

@Thorgott du bist hier?

Rover

2:42 PM

@robjohn Yes !

robjohn

@Rover great. What did you get?

Rover

I got standard deviation as 90

robjohn

Yes. That surprised me since it was the same as one of the two originals.

Rover

Yeah

What do we mean by dual of a statement exactly?

Is it negation to given statement or just changing some symbols like of 'or 'by 'and'

Thorgott

2:59 PM

@LeakyNun ja

Leaky Nun

@Thorgott das ist stupid; warum sagt man "des Hundes" aber "roten Hundes"

warum ist es nicht "rotes Hundes"

Balarka Sen

das ist stupid lol

"warum ist es nicht" = "why is it not"?

Leaky Nun

yeah

Balarka Sen

das ist stupid

Astyx

ich lieb dich nicht du liebst mich nicht

da da da

Balarka Sen

3:03 PM

was ist lieb

love

wrong answer

baby hurt mich nicht

right answer

correct answer

3:10 PM

"rotes" ist Nominativ und "roten" ist Genitiv

außerdem wird "rotes" eh nur im Neutrum verwendet und Hund ist im Maskulinum

Leaky Nun

das ist es nicht; der Nominativ ist "roter Hund"

Thorgott

ja, "rotes" ist Nominativ Netrum, "roter" ist Nominativ Maskulinum

Leaky Nun

was ich sage ist das die Endung "roten" ist nicht gleich von "des"

des Hundes, roten Hundes?

Thorgott

ja, das mit den Endungen ist nicht wirklich konsistent, ich habe da leider keine gute Erklärung für

es wäre "des roten Hundes", den Artikel lässt man eigentlich nie weg

satan 29

3:28 PM

A solid body of constant density
$\rho$
is obtained by revolving the cardioid

$r = a(1+ cos\theta)$

about the initial line. Find its M.I. about a straight line through the pole

and perpendicular to the initial line.

how do we approach this?

I know MOI= $\int dm*(distance^2)$

Leaky Nun

@Thorgott was ist die Ordnung der Fälle, die man in Deutschland lehrt?

Thorgott

3:44 PM

Nominativ, Genitiv, Dativ, Akkusativ

Leaky Nun

danke

satan 29

@robjohn may I ask for your help

Vio Ariton

4:02 PM

would this amazon.com/Vector-Calculus-Linear-Algebra-Differential/dp/… be a good textbook for self studying multivariable calculus

or does it require some extra knowledge

robjohn

4:50 PM

@satan29 with what?

satan 29

with that question I posted

Akiva Weinberger

5:02 PM

$$\begin{bmatrix}-1&0\\0&-1\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}\lambda\\x\end{bmatrix}$$

2

This is a typography joke

(though I think LaTeX's font is suboptimal here)

robjohn

@satan29 what is the "initial line"?

satan 29

I think it means the x axis

which is thy symmetry axis for the cardioid

robjohn

okay

Using your MOI formula, can you figure out the moment of inertia of a disk of radius $r$?

satan 29

yes, its Mr^2/2

robjohn

@AkivaWeinberger akin to $$\begin{bmatrix}0&-1\\1&0\end{bmatrix}\infty=8$$

2

satan 29

5:17 PM

@robjohn LMAO

robjohn

@satan29 are you sure? I get $\frac\pi2\sigma r^4$

for each ring it is $r^2 2\pi\sigma r\mathrm{d}r$

integrate from $0$ to $r$

Oh yeah... $m=\pi r^2\sigma$

satan 29

@robjohn sigma pir^2 is just M , no?

right.

robjohn

but since we are going to integrate over disks of various radii, we should use the formula with constant $\sigma$

satan 29

absolutely.

robjohn

The $r$ here is $a(1+ \cos(\theta))\sin(\theta)$

Akiva Weinberger

5:28 PM

Alternate definition of orthonormal, given a norm: two vectors are perpendicular of norm 1 iff $\|\sin\theta\vec u+\cos\theta\vec v\|=1$ for all $\theta$

(Thinking about how someone who doesn't know about dot products might invent it)

satan 29

@robjohn why>

robjohn

@satan29 we are rotating $r=a(1+\cos(\theta))$ about the line $\theta=0$, so the radius of each disk is $a(1+\cos(\theta))\sin(\theta)$

The thickness of each disk is $a(\sin(\theta)+\sin(2\theta))\,\mathrm{d}\theta$

satan 29

@robjohn ah okay

@robjohn how?

it should just be (radius)(d\theta)

Monty

5:44 PM

Hi !!!

robjohn

@satan29 picture the disk that is generated between the points for $\theta$ and $\theta+\mathrm{d}\theta$

Monty

anyone around for a quick chat about vector fields in $\mathbb{R}^d$

robjohn

if it's not $2$ $\text{H}\alpha\mathbb{R}^\text{d}$

Monty

is there a way to perturb any vector field so that it can be written as the gradient of some function?

satan 29

@robjohn sorry If i sound condascending to someone of your stature, but, did you mistakenly take sin(2x) to be = sinx cosx ?

Mike Miller

5:47 PM

What does perturb mean?

Monty

or better said, given a vector field how can I perturb it so that the result is the gradient of a function

anything really - but im thinking a "small shift by epsilon"

Mike Miller

Then no

Monty

like how one can perturb a singular matrix

ok

Mike Miller

Gradient vector fields form a closed subset of all vector fields --- anything not a gradient is not approximated by gradient functions

I'll work in R^2 for convenience of notation

Monty

ahh

Mike Miller

5:49 PM

A vector field F = <P, Q> on all of R^2 is a gradient iff P_x - Q_y = 0

(In general, the correct statement is: A vector field on a domain in R^2 necessarily satisfies P_x - Q_y)

Therefore P_x - Q_y is a quantitative measurement stopping a vector field from being a gradient

If that's nonzero and you wiggle P, Q a little bit, the result remains nonzero

You need big wiggles

Monty

sorry what is <,>

Mike Miller

I'm just listing the components

Clarinetist

Suppose I have a multivariate normal vector $(Z_1, Z_2)$ with each component independent and having zero mean and unit variance, and I want the probability $\mathbb{P}(a_1Z_1 + a_2Z_2 \geq 0, b_1Z_1 + b_2Z_2 \geq 0)$
with $a_i, b_i$ not all zero. How in the world do you calculate this?

Mike Miller

How do you write vectors

Monty

ok right sorry

(,) :)

Mike Miller

5:51 PM

I'm not being mean I'm just asking what your notation is

OK

Monty

and P_x is P evaluated at x?

doesnt make sense to me?

or you mean differential

partial derivative

Thorgott

it's the partial derivative of P with respect to x

Monty

right cheers :)

Balarka Sen

Just noticed I wrote 104 = 2 * 3 * 17 in my assignment

robjohn

@satan29 no, there is a $2$ in there

Monty

5:54 PM

ill think about it for a bit thanks for getting me started Mike and Thorgott !

Thorgott

@BalarkaSen you absolute fool

satan 29

@robjohn shouldnt the thickness just be (radius)*(d\theta) ?

robjohn

@satan29 no: $\mathrm{d}(a(1+\cos(\theta))\cos(\theta))=-a(\sin(\theta)+2\sin(\theta)\cos(\theta))\,\mathrm{d}\theta$

satan 29

and what is $(a(1+\cos(\theta))\cos(\theta))$

Leaky Nun

@Thorgott Was ist der Unterschied zwischen "mancher gute Hinweis" und "manch guter Hinweis"?

satan 29

5:56 PM

?

robjohn

@satan29 that is the $x$ position of the disk

satan 29

ahh

okay yes, now i can visualis properly

robjohn

13 mins ago, by robjohn

@satan29 picture the disk that is generated between the points for $\theta$ and $\theta+\mathrm{d}\theta$

satan 29

x=rcos(theta), the thickness is just dx

the mass of this disc is $$\rho * \pi * y^2*dx $$ ?

(y=radius of the disc)

and then thiss mass * y^2 /2 for MOI

i.e

$$\rho * \pi * y^4*dx /2$$ ?

@robjohn which is just this, but with $\rho dx$ instead of $\sigma$ ?

robjohn

Yeah, I wrote $\sigma$, but it is the density $\rho$

satan 29

6:04 PM

so that final expression

should be it according to you?

robjohn

where $y=a(1+\cos(\theta))\sin(\theta)$ and $\mathrm{d}x=a(\sin(\theta)+2\sin(\theta)\cos(\theta))\,\mathrm{d}\theta$

satan 29

yeah

Thorgott

@LeakyNun hmmm, da bin ich überfragt

ich bin mir nicht sicher, ob es einen gibt

satan 29

@robjohn but the question has this line:

"line perpendicular to initial line and perp. to pole"

here it seems we are culculating MOI about the initial line..?

Leaky Nun

@Thorgott benutzt man immer den Akkusativ folgend "es gibt"?

Edward Evans

6:12 PM

ja

unless

yes

wollte voll was dummes sagen und dann hab ich mich davon abgehalten

Leaky Nun

ok danke

was = etwas?

Edward Evans

Ja sorry, ziemlich umgangssprachlich geschrieben

Thorgott

ja

geocalc33

6:26 PM

how many consecutive years of serious math study have y'all been doing?

I think I'm going to peak at age 45

Edward Evans

I peaked like 3 years ago

user2103480

@EdwardEvans me as well brother

not mathematically, but man life was good

geocalc33

Have I peaked?

Edward Evans

I defo peaked mathematically

user2103480

btw harvard qualifiying exams are a treasure trove of algebraic topology questions

Edward Evans

6:39 PM

ew

user2103480

dont judge me

I'll never touch this stuff again if I manage to solve a sizeable portion of the exam

analysis, take me into your warm womb

are there answers?

yes

are there questions?

math.harvard.edu/media/qf16sol.pdf

@Thorgott yes as well

both questions and answers

@Thorgott man I wish I could downvote this smugness

>flag as inappropriate

Thorgott

6:45 PM

gottem

geocalc33

now that's what I call a Harvard qualifying exam

user2103480

thor's probably bursting while trying not to answer "I'd also say that's a Harvard qualifying exam"

in itself the questions aren't too bad I'd think, but man you gotta retain a lot of knowledge to solve these

geocalc33

yeah

Balarka Sen

yeah this is much easier than other qualifying exams i have seen, which is kinda sus

geocalc33

and this is to get into the phd program at Harvard?

user2103480

6:55 PM

@BalarkaSen when will you apply boi

Thorgott

@user2103480 well, it does look like one

user2103480

@Thorgott curious, and yet you participate in exams

geocalc33

I'm going to hire someone to write me a qualifying exam

just so I can "show what I know"

Thorgott

I don't participate in exams

my grades are the result of extortion and bribery

user2103480

@Thorgott a self-respecting citizen's way of life

writing exams like a commoner... pfft

geocalc33

7:12 PM

has anyone tried Coca Cola with coffee?

it's a new coke taste infused with Brazilian coffee

Thorgott

hmm, am I being stupid or is the cup pairing on $H^{2n}(M,\partial M)$ for $M$ a $4n$-dimensional oriented compact manifold with boundary not necessarily non-degenerate?

L.K.

I have a naive question. Let us assume we have a D dimensional manifold where the translations do not commute. Can I construct a D+N dimensional manifold where they will commute?

geocalc33

so many questions - so little answers

Thorgott

what are "the translations"

L.K.

@Thorgott The operations that translate a point in the manifold. For instance, in 1D Euclidean geometry, x-> x+a

Thorgott

7:19 PM

I don't follow. What does "translate" mean here? How are you translating the points? I know translations in R^n as x->x+a, but you don't have something completely analogous on an arbitrary manifold.

Thorgott

7:39 PM

@Thorgott Lefschetz duality gives non-degeneracy of pairing $H^{2n}(M)$ with $H^{2n}(M,\partial M)$, so this comes down to observing that the kernel of the restriction map $H^{2n}(M,\partial M)\rightarrow H^{2n}(M)$ is in general non-empty, so yeah

user2103480

@Thorgott yes thats what I would have said too

obv

ok google how do cvp prodvccs work

geocalc33

The mapping $(x,y,z) \mapsto \left(\dfrac x{\sqrt{1+x^2+y^2+z^2}},\dfrac y{\sqrt{1+x^2+y^2+z^2}},\dfrac z{\sqrt{1+x^2+y^2+z^2}}\right),$ compactifies $\Bbb R^3$ with the boundary being $S^2.$ I'd like to map this compactification of $\Bbb R^3$ to $[0,1]^3.$ Is this called "cubing the sphere?"

Ted Shifrin

@Thorgott Something's fishy. What if $M=D^4$?

Thorgott

then you're pairing two trivial groups

Ted Shifrin

This is nondegenerate?

Vacuously?

Thorgott

7:47 PM

yeah?

non-degeneracy to me means $\langle v,w\rangle=0$ for all $w$ implies $v=0$

vacuous condition if $v=0$ is the only possible $v$

non-degeneracy means this and the same thing with $v,w$ switched*

Ted Shifrin

OK

Snapdragon-X

Has there been any incident where a student was indulging in academic dishonesty on MATH SE and the teacher caught them lol?

copper.hat

8:01 PM

that would be hard to prove, i imagine.

Ted Shifrin

I caught some of my own students posting homework. Long before COVID days, so exams were in class.

I caught them before answers were posted.

BigSocks

@Thorgott lol

Thorgott

this makes me want to repost

3

A: Universal properties of $\mathbb Z$

I'd be happy to discuss this in person either during or after your oral exam. Come find me in my office.

Ted Shifrin

LOL

I see a number of the geometry exam posts I bitched about have vanished.

geocalc33

lol

Thorgott

8:09 PM

recently saw someone who edited his question text into something unintelligible after accepting an answer and then even tried contesting a rollback

Ted Shifrin

Yeah, things going badly downhill here.

geocalc33

is it plausible to compactify R^3 into the interior of S^2 and then map to [0,1]^3 and then quotient into T^3?

BigSocks

so uh, what's the general name for a cardinal $\kappa$ that satisfies $\kappa ^ \lambda = \kappa$ ?

please don't call me names/say my question is lame, thank you

Alessandro Codenotti

what's $\lambda$

BigSocks

another cardinal mb

probably at least $\aleph_0$ or it's lame right

copper.hat

8:20 PM

@Thorgott excellent post!

never occurred to me that this site must be the bane of teachers everywhere.

Astyx

Especially nowadays

Alessandro Codenotti

@BigSocks yes if $\lambda$ is finite and $\kappa$ is infinite it always holds. For infinite $\lambda$ it's doesn't hold very often

BigSocks

right that was about as far as I had thought too

Alessandro Codenotti

Even whether $\aleph_1^{\aleph_0}=\aleph_1$ is already independent of ZFC

BigSocks

I guess if $\kappa = 2^{\aleph_0}$ and $\lambda = \aleph_0$ that's something

Alessandro Codenotti

8:24 PM

sure, if $\kappa=2^\mu$ and $\lambda=\mu$ then it works but that's very specific cases

BigSocks

yeah :/ but there's no word you remember for this either? like singular or one of those

fancy words

copper.hat

I used to think I was helping by answering questions, now I am not so sure...

user2103480

@BigSocks Idk why you're doing this to yourself but here you have a nice condition

"If $\kappa>\lambda, \forall \xi<\kappa: \, \xi^{\lambda}<\kappa,$ and $\operatorname{cof}(\kappa)>\lambda$ then $\kappa^{\lambda}=\kappa$"

Alessandro Codenotti

(In the $\mathrm{cof}(\kappa)\leq\lambda$ case you get $\kappa^{\mathrm{cof}(\kappa)}$ instead)

user2103480

Koepke brain

Alessandro Codenotti

8:31 PM

Nah that's a standard result

See 5.20 in Jech for a proper reference

Ted Shifrin

@copper.hat A good deed rarely goes unpunished?

user2103480

@AlessandroCodenotti Well of course its a standard result, I assume any little strengthening is undecidable or false

copper.hat

@TedShifrin Now I am paranoid :-).

user2103480

or shelah

BigSocks

@user2103480 very ominous

must observe for a while. my set theory past usual things is v rusty

@user2103480 "having a little shelah as a treat" is why i am doing this to myself

user2103480

8:34 PM

@BigSocks same question applies

Thorgott

sTaNdArD rESuLt

BigSocks

@Thorgott thank you for this. needed to hear it

@user2103480 the least element in the set of these pursuits is usually masochism anyway

so it's not that far off just to believe it is ill-founded

user2103480

8:55 PM

So how much do I have to understand for the cup product

This example is not very enlightening

Thorgott

amazing example

user2103480

So the homology groups are zero and there is the generator of the 0-th and n-th homology and that generator cupped with itself is 0. that much I understand

Thorgott

yeah, it's boring as there's only one non-trivial group in positive degree

here's something instructive one can do: prove explicitly that the cross product of a generator of $H^1(S^1)$ with itself gives a generator of $H^2(S^1\times S^1)$

user2103480

cross product?

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$Mathematics$ Mathematics