I apologize if this isn't the appropriate place to ask this. Anyone know what's going on with Project Euler? I try to access their forum and get a certificate error telling me I should close the site immediately. The only contact info they give is through the forum. They have an email for new problems, but they've stopped posting problems until the fall.

I'm half-considering going to a problem I've completed, making a post in the solution thread about what's going on and reporting it if this doesn't clear up.

@Mike maybe you went to the wrong site

the correct site is projecteuler.net

hey all, I'm somehow struggling with what seems should be basic algebra

I'm trying to make logarithmic scales more intuitive for myself, in the case where both x/y are logarithmic

I'd have thought in that context, a straight line would still mean a function in the form y=cx

but my paper calculations end up with a completely different function

@towc you want log(y) = c log(x), i.e. y=x^c

oh hold on, I think I got it

Py = log(y)
Px = log(x)

y = f(x)
x = e^Px
Py = log(f(x)) = log(f(e^Px))

if f(x) = x
Py = log(e^Px) = Px

so f(x) = x would give a straight line

yet I was reading this book that described a power-law distribution as something that would be plotted as a straight line in this kind of graph

but f(x) = x doesn't seem to have anything to do with powers

oh but way, maybe as you say, x² will also give a straight line

oh right, x² would give 2 Px

I see, thanks :)

another bit of confusion is that it sounds like when I plot log(e^log(x)) on a same-interval scale, I should still get a straight line, as if mimicking the logarithmic version of y=x

but that's now what I see

Morning all

if pB=Bp we say p normalises B or that B normalises p?

I always mix it up

p normalises B

if you mean groups

i mean algebras but the wording has to be the same :)

sure

@s.harp answering a *-algebras question, I see :P

I was keeping an eye on that question, I was sure it had to be false, but I didn't have a simple example

matrix algebras are horrible when it comes to subalgebras

so you can always build counterexamples from them :)

you must be working in... the wrong category then

is every section to the tangent bundle open?

@BalarkaSen

as a map? no, sections are almost never open

for example?

well for one thing if you have a section $U\to TU$ compare the dimensions of the right and the left object

oh what am I thinking about

how about the espace etale of the sheaf associated to the tangent bundle

or just espace etale of any sheaf

here's the claim: given a sheaf $\mathcal F$, any continuous section to the espace etale is open

do you remember how the topology of the etale space is defined?

(hint: the topology is trash, and has something to do with sections)

yeah

in this case the statement is true, to see that its true you need to unpack definitions

cool

why is the topology trash?

it depends on your perspective, but i like topological spaces where the topology has something to do with the setting (beyond being just compatible with the maps you need or some algebraic operations)

for example the fibres of the etale space are all discrete

Stalks are discrete

It's not how you recover the bundle from it's sheaf of sections

is Set(X) CCC?

Oh, by the way @s.harp do you see a simple proof that $K(H)$ has no characters? We were thinking about it the other day

it follows from $K(H)$ having no self-adjoint two-sided ideals, as the kernel of a character must be such an ideal

if you suppose you have such an ideal and take a self-adjoint element and conjugate by the appropriate projection you get a rank one map, the associated projection must lie in the ideal, by conjugating with appropriate rank 2 maps you can assume any rank 1 projection is in the ideal

the span of the rank 1 projections is dense in $K(H)$ however

(and two sided ideals are closed in $C^*$ algebras, although the kernel of a character is also automatically closed)

@s.harp why is any two sided ideal closed?

(just give me a reference, I'll look it up)

its in bratteli and robinson in the first non introduction chapter

Thanks

hold up, the statement is false

What a plot twist

x-files music plays

yeah, the finite dimensional range elements of $B(\ell^2)$ are obviously a two-sided non-closed ideal

the correct implication is twosided -> self-adjoint

not two-sided -> closed

but the kernel of a morphism must be a closed two-sided ideal, which is the statement we need

@s.harp makes sense

I see, thanks

@RyanUnger got some dumb functional analysis questions for you

Ie stuff i’d surely know if I actually knew functional analysis proper and not just the handwavy physics version

Namely, I’m trying to get a handle on the Hilbert-Schmidt class of operators on a Hilbert space and a C-star algebra of operators on a Hilbert space

Is it just a matter of what norm you use (H-S vs operator norm) and how nice the resulting structure is?

H-S is nice because you get another Hilbert space I guess

Yeah

What are you asking exactly?

The relationship between Hilbert-Schmidt class and C-* algebra I guess? Eg is the former a special case of the latter

I don’t think that’s right tho

HS operators form an ideal in the C*-algebra $B(H)$ of all bounded operators

In a sense HS is like $L^2$, while $B(H)$ is like $L^\infty$

Hmm. Nice.

Oops

For a compact operator $T$ let $\sigma_k(T)$ be a descending enumeration of the eigenvalues of $T^\ast T$

The Schatten $p$-norm of $T$ is $$\|T\|_p=(\sum \sigma_k(T)^p)^{1/p}$

If we write $S^p$ for the set of operators $H\to H$ with finite Schatten $p$-norm then $S^2$ are the HS operators and $S^1$ the trace class operators

Turns out that Hölder's inequality holds for the Schatten norms as well, and it still holds if we use the sup norm for $p=\infty$

$$\|T\|_p=(\sum \sigma_k(T)^p)^{1/p}$$ woops

Hey @Mathein, weisst du wieviele Leistungspunkte Seminare betragen?

How much difference is there between those sets of operators when H is finite-dimensional? My knee-jerk reaction would be that there can’t be much difference

Alessandro is obsessed with these Schatten norms

LoL

They came up in noncommutative geometry to define summability of a spectral triple or something

I’ve run into the p=2 case of Schatten but that’s comparatively familiar

Or was it p=1? I forget.

But, getting back: How much different can those sets of operators be in finite dimensions? Seems like that’d make things trivial

all operators are bounded in finite dimensions

and all norms are equivalent

Right

Ok. So these distinctions only become interesting in the infinite-dimensional setting

yeah

why is $\sin x \mod \cos x = \sin x$?

@ÍgjøgnumMeg 6

desmos.com/calculator/mpnvpmnh10

the above link shows this

@Mathein ah cool :) Danke

@ÍgjøgnumMeg do you know why is $\sin x \mod \cos x = \sin x$?

@Mathphile what do you mean by mod

and what is $x$?

I mean where is $x$

@ÍgjøgnumMeg $y \mod x$ is the remainder when we divide $y$ by $x$

lol what

how do you do mod for sin and cos

@ÍgjøgnumMeg $x \in \Bbb{R}$

@RyanUnger i don't know but desmos is able to do it

@ÍgjøgnumMeg is the question clear now?

Here's one for you. For what values of $x\in\mathbb{R}$ does $\text{sin}x=n\text{cos}x$ for $n\in\mathbb{Z}$

oh shit

i made a mistake

a mod b is usually meaningless for real numbers a, b but some computer science people take it to mean the least element in {a + kb}_{k in Z}

it actually is $\sin x \mod \csc x = \sin x$

@BalarkaSen least non-negative element

thanks, yup

I mean you can say things like "$a \equiv b \bmod 2\pi$"

@BalarkaSen is simply-connected captured by some sheaf?

an open subset of a simply-connected open isn't simply connected...

I mean I can give you a sheafy interpretation, I guess

@Mathphile I mean.. congruences are equivalence relations so $a \equiv a \bmod \text{anything}$

please go ahead

Any locally constant $G$-valued sheaf on $X$ is constant iff $X$ is simply connected. Because the hypothesis is equivalent to demanding any $G$-representation of $\pi_1(X)$ is trivial.

Note I am not fixing $G$

$a\text{ mod cheeseburger}$

lol

@BalarkaSen cool

how about locally connected?

I mean it's not that cool, really, if you think about it. Any space $X$ has a locally constant $\pi_1(X)$-valued (set-valued) sheaf on it, given by the left-regular representation of $\pi_1(X)$. The espace etale is the universal cover if $X$ was a sufficiently nice space.

> espace etale

Does a non-trivial fundamental group of a smooth manifold always admit a non-trivial $\Bbb R$-linear representation?

Any countable group is fundamental group of a smooth manifold, for one.

I don't know how wild your group needs to be to not admit R-linear representations at all

Panagiotis Polymerakis

sounds like a fake name

@Ryan there is a panagiotis working at my university lol

with an even faker sounding surname

hahaha check this

@Ryan Xochiquetzal Voyatzis Hernandez

is that a Basque name

I think it's central american

aztec style

@AlessandroCodenotti link.springer.com/article/10.1007/s12220-019-00196-1

a covering is amenable iff it preserves the bottom of the spectrum of any Schroedinger operator

@Balarka apparently, there are some countable groups with no finite-dimensional $\Bbb R$-linear representation. This means that bundles with flat connection on a smooth manifold can't detect if the manifold is simply connected

there are even finitely presented examples, so even assuming compactness doesn't help

Right, every finitely presented group is a closed 4-manifold group

but if we assume a compact 3-manifold, then flat connections do detect simply-connectedness: any compact 3-manifold group is residually finite, so it admits a non-trivial homomorphism to some finite group, we can compose that with a faithful representation of that finite group

I didn't know that fact; interesting

apparently the proof uses geometrization

Hah, of course.

Hi everyone

@Semiclassical did you hear back from the lecturer position?

@RyanUnger One day I will to actually learn some Riemannian geometry

@Ultradark not yet.

i dunno what conclusion to draw from that yet tho. the phone conversation was Friday

and I don't have any insight as to how quickly they'd put forth an offer (or choose not to do so)

Yeah I know where your coming from

@BalarkaSen Does Mostow-type rigidity hold for spherical space forms?

According to Wiki, it's true for locally symmetric spaces with finite volume

But is there an elementary answer?

@RyanUnger Oh boy this I do not know. I am only familiar with the statement for complete hyperbolic manifolds of finite volume

I don't think this is true. There are spherical 3-manifolds which are homotopy equivalent but not homeomorphic.

@BalarkaSen do you know hypercohomology?

Staggeringly little

can you teach me?

I was going to ask you to do that!

I really don't know enough to teach you

@BalarkaSen must K(G,n) be a CW-complex?

It admits CW complex structures, yes

> Such a space exists, is a CW-complex, and is unique up to a weak homotopy equivalence

I mean, there are certainly models which are not CW-complex

Choose a presentation of $G$, and take a wedge of $n$-spheres indexed by the generators of $G$, and then attach $n+1$-cells indexed by the relators. Now glue higher cells to kill higher homotopy groups.

yeah that's what everyone says, but I don't know how to "kill higher homotopy groups"

If $X$ is a CW complex, with a chosen class $\alpha \in \pi_n X$, then consider a representative $S^n \to X$ of this class. By cellular approximation theorem make it a cellular map. Then glue a $D^{n+1}$ by considering this to be your attaching map.

You say in the resulting space $X'$ you have "killed $\alpha$"

so attach as many $(n+1)$-cells as $\pi_n(X)$?

Just choose a generating set of $\pi_n X$, and then attach $n+1$-cells along those.

it's just a bunch of approximation anyway

ok let's say I want to construct $K(\Bbb Z/2\Bbb Z,1)$, like any reasonable person would

so I start with a circle ($e^0, e^1$)

and there is only one relator so $(e^0, e^1, e^2)$ with $\phi^2 : S^1 \to e^0 \cup e^1 : z \mapsto z^2$

An $\Bbb{RP}^2$ is the first approximation, right.

so I need to compute $\pi_2(e^0 \cup e^1 \cup e^2)$

can I do that without knowing that it is $\Bbb RP^2$ (i.e. without knowing that $S^2$ is a double cover)

i.e. just by using the cellular structure

No, you cannot compute homotopy groups "just from the cellular structure". If that was the case homotopy groups of spheres wouldn't be a thing.

oh no

can we compute it from the cellular structure given the homotopy groups of spheres then?

That's a good question, but I still doubt it. $\pi_n(S^2 \vee S^2)$ are very complicated groups.

oh no

ok so $2 \to S^2 \to \Bbb RP^2$

$\pi_2(2) \to \pi_2(S^2) \to \pi_2(\Bbb RP^2) \to \pi_1(2) \to \pi_1(S^2)$

wait the fibre isn't path-connected

how does this work?

oh, galois theory

no that's only for $\pi_1$

This is just map lifting lemma. Since $S^2$ is simply connected, any map $S^2 \to \Bbb{RP}^2$ lifts to a map $S^2 \to S^2$ in the universal cover.

In general $\pi_n X \cong \pi_n \widetilde{X}$ if $n > 1$ and $\widetilde{X}$ is the universal cover of $X$.

cool

and $\pi_2(S^2)$...

This partially explains your problem as well. Consider instead $S^2 \vee S^1$. It's universal cover is homotopy equivalent to an infinite-wedge of $S^2$'s, so their higher homotopy groups are really really unpleasant compared to that of $S^2$ or $S^1$.

we did that before :P

$\pi_2(S^2) = [S^2,S^2] = [\Sigma S^1, S^2] = [S^1, \Omega S^2] = \pi_1(\Omega S^2)$.... ok this isn't leading anywhere

oh and using the pathspace fibration gives you $\pi_1(\Omega S^2) = \pi_2(S^2)$ so this leads to nowhere

@BalarkaSen help how $\pi_2(S^2)$

do I know this?

$\pi_2(S^2)=H_2(S^2)$ by Hurewicz

you what

eh...

Well, you can instead compute $H_1(\Omega S^2)$ (which is the same thing, because abelian) using the spectral sequence :P

ok that was a let-down

ok let's do the spectral sequence

so $\Omega S^2 \to PS^2 \to S^2$

you can also run long exact sequence on S^1->S^3->S^2, if you know that \pi_2(S^3) = 0

why would I know that

cellular approximation

oh no

:P

see three people have proved my stupidity now

ok let's carry on

$S^2 = e_0 \cup e_2$

@LeakyNun It appears the problem has been corrected. The site itself was fine. The problem was with the forums. Both old bookmarks and the link to projecteuler.chat from the About page on projecteuler.net had the issue. It now loads fine with no issues.

$\cdots \to 0 \to 0 \to 0 \to \Bbb Z \to 0 \to \Bbb Z$ is the cellular chain complex

@Mike cool

so the spectral sequence becomes $\begin{array}{c} H_0(\Omega S^2) & 0 & H_0(\Omega S^2) & 0 & 0 & \cdots \\ H_1(\Omega S^2) & 0 & H_1(\Omega S^2) & 0 & 0 & \cdots \\ \vdots \end{array}$

aha, so $H_1(\Omega S^2) = H_0(\Omega S^2) = \Bbb Z$

yay

Mhm.

ok that took long enough

should I try $\pi_3(S^3)$?

wait that means $H_n(\Omega S^2) = \Bbb Z$ for any $n$!

I ain't seen such a space before

Sounds like you need Hurewicz to relate $\pi_2 \Omega S^3$ to $H_2 \Omega S^3$ anyway.

@LeakyNun No!

This was explicitly about $H_1$.

but the spectral sequence continues

Can't you also compute $\pi_n(S^n)$ with differential topology? Smooth approximation + Hopf theorem should do it

Oh, I misread. Yeah, that is correct.

and $PS^2$ is contractible

@MatheinBoulomenos $\pi_n(S_n) = 1$

$\Omega S^n$ has $\Bbb Z$ homology in degree multiples of $n-1$, and zero everywhere else.

@LeakyNun lol, good point

@MatheinBoulomenos Yes, that's the standard argument.

@MatheinBoulomenos also, differential topology

I can't English

(that took longer than I expected)

You can also compute homology of $\Omega S^n$ using Morse theory

ich kann kein Deutsch

@BalarkaSen that sounds like $K(\Bbb Z,n)$

@MatheinBoulomenos You can also argue, like, $\pi_n S^n = [S^n, K(\Bbb Z, n)] = H^n(S^n)$. Basically topologized Hopf degree theorem.

what

oh cellular approximation?

Right.

yay

@LeakyNun That doesn't have simple homology. Lots of torsion in lots of places.