@Koro I don't see a more convenient way than calculating the homology

@TedShifrin No, this time its orange juice on my phone

good thing its water resistant

the funny part is that those two spaces have the same homotopy groups

@冥王Hades But not sugar “

Darn, I didn’t think of that

it seems to be working fine but sugar getting into the charging port or speaker is a problem

🤷‍♂️🤷‍♂️

I wish I could watercooled it. My computer is watercooled

@冥王Hades Serious gaming machine

@robjohn Yep, it does consume a lot of power, so naturally it produces a lot of heat as well that needs to be dissipated quickly

or else the CPU and the GPU will reduce their clockspeeds and slow down performance

As I said, serious gaming machine.

How’re the pooches, robjohn?

The pictures of the one at home look good. The one here with us is doing well.

Oh, you ditched the adoptee?

We could only bring one

I’m confused.

Wonder what would happen if I were asked to do a heat transfer analysis for my PC

We are in Mammoth Lakes.

My wife and I on vacation

Ohhh, I knew that, but I didn’t comprehend the limit.

Hope vacation is good!

photos.app.goo.gl/woFTeWAW2Tid4vBcA

Took a trip out to Canyon de Chelly yesterday.

"Chelly" is pronounced like "shay", for the record.

There is an error here right?

In particular, there should be another outer $\varphi$ on the RHS in order to map that product back to $G'$?

@TedShifrin Yes! Rosie and I just took a small pre-dinner walk.

She got her dinner, I still have to make mine.

@EE18 yes

Errors in textbooks are THE worst

:( like dang, it's hard enough already

Thanks very much Leslie!

Get used to errors. It always happens.

@copper.hat Hi, how are you doing? Hope things all's right with you.

Ig I haven't seen you in a while :D

@ThomasFinley Thanks for asking. Last week was a bit rough for me. My visiting family left, I got Covid (no real symptoms, but limited my activities), my fridge died and needed to be replaced, I sold a car to which I had formed a sentimental attachment and my son returned to college.

First world problems, but they are issues nonetheless.

@copper.hat Of course, they are issues. Don't worry, things will change soon, for the better. I wish you a very speedy recovery. Stay optimistic because as they say, tough times don't last, but tough people do! Stay strong, stay well.

@ThomasFinley Thanks, I am usually of an indomitable nature, but am suffering a bit from life lag :-).

@XanderHenderson Very nice!

@XanderHenderson Is that you in the picture?

@Thorgott yes. Do you remember that few days ago, I used a fact that "if pi_* of X and Y are isomorphic, then there is a weak equivalence between them"? The example is a counterexample to that as weak equiv. always induces isomorphism on homology groups.

@copper.hat almost all market sectors in red here. It is so annoying.

Hi @Koro :-)

@SoumikMukherjee No. That is one of our deans.

ooh

How to prove that $n^{-1/2}\chi_{[-n,n]}(x)\rightharpoonup0$ in $L^2(\Bbb R)$? I want to show that $n^{-1/2}\int_{-n}^n f(x)\,dx\to0$ for all $f\in L^2(\Bbb R)$.

Holder's inequality doesn't help.

Nevermind

No, sorry this isn't done. I thought I proved this but that was wrong.

@PNDas this limit seems to be non zero.

you probably meant $n^{-1/2}\chi_{[-m,m]}(x)$

This is written in a Lecture notes: Suppose $p>1$, $u\in C_c(-1,1)$ then $u_n=n^{\frac1p}u(nx)\rightharpoonup0$ in $L^p(\Bbb R)$ and $v_n=n^{-\frac1p}u(\frac{x}{n})\rightharpoonup0$ in $L^p(\Bbb R)$.

I took $p=2$ and $u=\chi_{[-1,1]}$.

I know it's not $C_c(-1,1)$ but I think $\chi_{[-1,1]}$ should work too.

Kinda jealous of how some people can just naturally tolerate spiciness effortlessly

I want to prove that $x^2 \equiv 1 (mod p) has solution for every prime p. How can I do that? I know that x=1 satisfies it but how to prove it rigorously.

Actually this problem originated from the problem which says "prove that 6x^2 + 5x +1 \equiv 0 (mod p) has a solution for every prime p.

@LuckyChouhan wdym? If $1^2\equiv 1 \pmod{p}$ then isn't this rigorous enough?

@LuckyChouhan not sure how the two are related

@冥王Hades you mean in context of unpleasant situations?

@Jakobian No I just mean I wish I could tolerate spice better

@PNDas $u_n = n^{1/2}\chi_{[-1/n, 1/n]}(x)$ in this case

oh you mean $v_n = n^{-1/2}u(x/n)$

oh okay so you mean weak convergence

this seems subtle, Cauchy-Schwarz isn't of help here for example

something to do with Fourier transforms?

@Koro indeed

@Jakobian This is just Holder.

Every topological manifold $M$ is completely metrizable. Furthermore if $M$ is smooth then the restriction of the metric on the open subset $\{(x,y) \in M \times M \vert x \neq y \}$ is smooth

So this may not be true for general topological manifolds, any example of that?

What are you talking about?

Is this true for topological manifolds as well?

That is for a complete metric on a topological manifold, is the restriction of $d$ on the open subset must be smooth?

@SoumikMukherjee if it's a topological manifold, then how can we talk about the smooth structure? We didn't assume it admits it

Precisely. You can't talk about smooth functions on a topological space with no extra structure.

And I don't know that the metric on $M$ that comes from point set topology necessarily has any smoothness. The natural construction is to use a Riemannian metric on $M$ to get a metric space structure.

@Jakobian We are talking about the smoothness of the metric function, why do we need a smooth structure on the manifold to discuss it?

What does it mean for a function to be smooth?

If it has continuous derivative of all order

What is a derivative?

Why do you need topology on a set to discuss continuity?

It's pretty much the same

@TedShifrin $\lim_{t \to 0} \frac{d(f(x+t) ,f(x))}{t}$

Oh, wait... I didn't see your distance function.

@Jakobian I get this but why smoothness is required, we can talk about continuity without smoothness

@Soumik Have you even studied rigorous multivariable calculus?

@TedShifrin Sorry but I am getting confused, am I making some silly mistake?

@SoumikMukherjee continuity was just an example

you can't talk about smoothness without smoothness

I understand that

my point is a topological manifold is completely metrizable, so it admits a complete metric

Can't we ask about the smoothness of this metric function?

sure, but for example, norm $x\mapsto \|x\|$ of a vector space is usually not differentiable at $0$

or well, never I think

oh okay

you cannot ask about the smoothness of a function unless its domain and codomain are smooth manifolds

(or a more general class of spaces for which a notion of smoothness exists, but let's not go there)

@Thorgott I get it now, thank you

This is the definition of group action my text gives me, whereas I am more familiar with the definition as a homormorphism from $G$ into the set of bijections on $X$. Is one preferable, or more sophisticated than the other?

Ah, answered here (math.stackexchange.com/questions/127736/…). Nice to see Prof. Shifrin's book come up in the question :)

either definition is fine, there are some contexts in which one of them generalizes better and other contexts in which the other generalizes better

Got it, thanks very much thorgott

what does it mean if $K$ is a field and they wrote $K^x$?

is $x$ a natural number or a set?

@Jakobian @TedShifrin I can now understand the source of my confusion. I just forgot the fact that on a Manifold $M$, a function $f$ is smooth if each $f \circ \phi_i^{-1}$ is smooth. So the underlying manifold must have a smooth structure to begin with.

@EE18 no it's like $\times$ so like $K^\times$

I was thinking about $\Bbb{R}^n$ which has an obvious smooth structure so there is no need to consider about those $\phi_i$'s, so one only considers the smoothness of $f$

@user123234 The group of nonzero ekements.

But for an arbitrary topological manifold, the smooth structure of the manifold is needed

@Soumik Look back even at what you wrote when I asked what the derivative is. Even in $\Bbb R^n$ it makes no sense.

Ah OK, yes $\times$ is different than $x$ :) yes, I agree w Prof Shifrin

@TedShifrin Yes I get it now, sorry I was being lazy and just wrote the $\Bbb{R}$ case

@TedShifrin so $K^\times:=K\setminus \{0\}$?

if $K$ is a field

@user123234 yes

if $R$ is any unital ring, then $R^\times$ denotes the group of all invertible elements of $R$

when $R = F$ is a field, those are all non-zero elements

@user123234 Yes, but typically the notation suggests that you're thinking of this as a group under multiplication.

okey thanks

Gonna need a new whiteboard

and a new orange-juice-proof phone

salt on wounds

But it has degree $m$.

so $q<m$ needs to hold. No?

Almost.

is $q\leq m$?

Why $q$?

q is the number of roots of $x^m-1$

No.

hmm why no?

By the way, have you already proved the order of every element divides $m$?

no I have not but I am still wondering why $x^m-1$ does not have $q$ roots

why $q$?

Reread what we've said.

Because I have shown that in $K^\times$ every element has order $m$ so it satisfies $x^m=1$ but this means that $q-1$ elements satisfy the equality in $K^\times$ now since the only element in $K$ not contained in $K^\times$ is $0$ and $0$ satisfies the equality qe know $q$ elements satisfy the equation

$0$ satisfies the equation?

$0^m=1$

!

Calling Munchkin for help!

@user123234 This is a pretty serious error. In any ring, what is $0\cdot a$ for any $a$ in the ring?

So apparently spinning a donut on your finger isn't a good idea. It can go flying and end up somewhere nasty. Speaking from (recent) experience.

$0\cdot a=0$ no?

But I mean does $x^m-1$ then has $q-1$ roots in $F$?

@冥王Hades You're apparently learning all sorts of lessons today that a 3-year old might already have learned.

Yes, @user123234.

@TedShifrin What can I say, having a bad week

ah and q-1 needs to divide m

Why is that?

I asked a much more important question. If $m$ is the maximal order, why does every other order divide $m$?

if I have an element of order $n<m$, which means $x^n=1$ then clearly $x^{lcm(n,m)}=1$ but since $lcm(n,m)\geq m$ so in particular $lcm(n,m)=m$ since $m$ is the maximal order but this means that $n|m$ @TedShifrin

@TedShifrin but I don't see how this can help me futher

Hi. So this is the math chat

I guess so

And the chemistry chat

@冥王Hades eh?

wouldn't that be The Periodic Table?

There is not much chat going on there, so there might be more chemistry talked about in this room simply because there is so much more chat going on here.

@robjohn I was trolling him. He's someone from ComicVine

Hades will end up with some more orange juice on his computer, next.

Oh come on

Not my computer please

Unlike my phone, even a small drop of orange juice on the PCB will destroy it

@user123234 Your “clearly” seems wrong. $m$ is for a different element, $y$.