Mathematics - 2019-06-20 (page 2 of 3)

3:00 AM

And you wouldn't believe the trouble it took to unentomb that from a computer decades lost.

So now pick a d>c>0 and then you find a function $h\colon\mathbb R\to\mathbb R$ which is -1 on $[-c,c]$ and 1 outside of $[-d,d]$.

humn

You bad!

anakhro

Just a bump function kind of thing.

humn

Fist bump to you.

anakhro

You can draw it. Goes from 1, dips, then at -1, then goes back up to 1.

humn

3:04 AM

I gotta tell my homeless friends about you, @anakhro. These friends think i'm behind everything they plug into.

anakhro

Then take the plane defined by $z = \varepsilon\,h(x)y$ and we look at how the surface foliates under the effect of the plane field.

And voila, you get a hyperbolic point (x_0) and an elliptic point (x_1).

humn

Okay, @anakhro, the gloves are drawn.

anakhro

Your picture above had a hyperbolic point.

humn

Point to infinity.

anakhro

But the point is that we introduced singularities by deforming it.

The surface, si.

And moreover, we can do this in as small of a neighbourhood of the x-axis as we want.

humn

3:07 AM

Most mathematicians i know are no fun. You, @anakhro, are too much fun! Thank you for playing along.

anakhro

No problem.

Glad you enjoy the pictures.

@BalarkaSen Giroux's elimination lemma, in case you wanted a name.

humn

(And now the computer is acting up. By the time this sentence ends we'll be okay.)

Balarka Sen

The idea is you can cancel off an elliptic and a hyperbolic point, @anakhro?

anakhro

Yes.

They also have to be the same sign.

i.e. orientation coincides (positive) with regards to the surface/foliation or doesn't coincide (negative)

Balarka Sen

Got it. Cute!

Is there an intrinsic description of the isotopies of the characteristic foliation you are doing by isotoping the embedded surface in the contact 3-fold?

anakhro

3:20 AM

intrinsic in what sense? Like in terms of the contact structure?

For compact surfaces with Legendrian boundary, the characteristic foliation gives a germ of a contact structure in its neighbourhood.

So it gives some data of the contact structure.

humn

I'm starting to realisze, @anakhro, that i can help with (La)TeX more than with mathematix.

Balarka Sen

Apriori the characteristic foliation is just a structure on the surface, so I was wondering if you can say what the isotopies induced by the isotopies of the embedding are without referencing the embedding in the first place

@anakhro Oh that's cool

I love that

So maybe you're studying isotopy classes of germs of contact structures near S x 0 in S x [-1, 1]

humn

@BalarkaSen, nice choice of words, numbers and symbols.

Incidentally, my native language isn't what gets typed.

anakhro

That is indeed what I do, @BalarkaSen. Study contact manifolds by their so-called "movies" $S\times[0,1]$ induced on surfaces $S$.

(i am rlly bad at it tho)

Balarka Sen

Very cool stuff though

Send me your thesis after it's done so I can familiarize myself with it for future conversations :)

anakhro

3:31 AM

Heh

It's kind of an embarrassing thesis.

I failed to prove anything that I wished to prove.

Balarka Sen

I think most everyone starts out with a paper that's lower than their desired bar, unless you're John Pardon or something

Math is hard, it takes a lot of effort to prove something. But that's why you have to stick to it, I think: keep writing stuff which are far weaker than what you want to prove until one day you prove it

3

Ryan Unger

@BalarkaSen I haven't seen him around

I've actually seen very few professors even at tea

this might be a summer thing

the princeton math dept is very vertical so the only way to guarantee seeing famous people is by camping out front

Balarka Sen

hah

Ryan Unger

I saw Fefferman getting lunch

I have some messages to deliver from Thislethwaite

but I haven't seen the people around :(

Balarka Sen

I still struggle to spell his name

Ryan Unger

3:36 AM

I fucked it up haha

Thistlethwaite

Balarka Sen

Lmao what a disgrace

Ryan Unger

I misspelled Tennessee on one of my grad school apps

multiple times

I copy pasted something

anakhro

Cute.

@BalarkaSen with a similar method to the elimination lemma, you can turn elliptic points into hyperbolic and vice versa, too.

By introducing points around them

You can do it $C^0$-close in an arbitrary neighbourhood, specifying your separatrices.

Secret

5:28 AM

Rudyrucker: Infinity, self containment and the Absolute

it does seemed that M={M} will naturally generate an infinite regress such that there is no way reaching it from simpler things

Secret

5:43 AM

But this has the same structure as any fixed point of functions f(x)=x or more generally, any fixed point morphism Hom(X,X)=X

But I guess, many functions which do have fixed points, attract or repel nearby points under iteration, thus x is still reachable

But M is different, there is no way to get it without having it in the first place

Subhasis Biswas

hi hello

Let $S= \{f: \mathbb{R} \to \mathbb{R}: \exists \epsilon>0 \ $such that $ \forall \delta >0, |x-y|< \delta \implies |f(x)-f(y)|< \epsilon\}$, Then, $S= \{f: \mathbb{R} \to \mathbb{R}: f$ is continuous/uniformly continuous/bounded/constant $\}$?

Attempt:

Let $f(0)=a$. So, $|x|< \delta \implies |f(x)-a|< \epsilon \implies a-\epsilon<f(x)<a+\epsilon$. So, $f(x)$ is bounded on $\mathbb{R}$ [since $\delta$ can be varied arbitrarily].

But answer provided is "uniformly continuous". Counterexample to the answer $f(0)=1$, $0$ elsewhere

is this correct?

@BalarkaSen i am not confident about this

user681391

6:52 AM

@Secret Shouldn't that be $Hom(X,X)\cong X$ like $Hom(\Bbb Z,\Bbb Z)\cong \Bbb Z$? Why are you calling that a fixed point morphism?

Subhasis Biswas

7:06 AM

Prove that the set of all algebraic numbers are countable:

Attempt: We consider the set $\Omega = \displaystyle\bigcup_{n \in \mathbb{N}} \{(a_0,a_1,a_2,...,a_n): a_i \in \mathbb{Z} \}$. It is a countable union of countable sets. so $\Omega$ itself is countable.

Now suppose that $z$ is algebraic. Then for some $(b_0, b_2, b_3,...,b_k)\in \Omega$, $b_0z^0+b_1z^1+...+b_kz^k=0$.

Now, with each possible $k$-tuple, there are only $k$ possible "associated" algebraic numbers (by "associated", it is meant that the set of complex numbers which are roots of the polynomial $b_0x^0+b_1x^1+..+b_kx^k=…

can anyone please verify?

@user681391 .....

MatheinBoulomenos

@SubhasisBiswas your attempt was successful

Subhasis Biswas

which attempt? countability?

MatheinBoulomenos

right, you wrote "attempt" at the beginning of your proof

I was referring to that

I just wanted to say that the proof is fine :)

Subhasis Biswas

@MatheinBoulomenos thank you for taking a look :)

it really helps

@SubhasisBiswas now, can you please verify this little one

MatheinBoulomenos

@SubhasisBiswas I agree with your counterexample and your reasoning. I'd say bounded, too, but you still need to prove that every bounded function is in $S$ (that's not hard, though)

Subhasis Biswas

7:13 AM

@MatheinBoulomenos will do it. Let me try

MatheinBoulomenos

Buongiorno @Alessandro

Subhasis Biswas

$f$ being bounded, for some $M >0$, $|f(x)|<M$ for every $x \in \mathbb{R}$.

$|f(x)-f(y)| \leq |f(x)|+|f(y)| <2M$ $\forall x, y \in \mathbb{R}$. We set $2M = \epsilon$

MatheinBoulomenos

@SubhasisBiswas actually, thinking about it, the problem is not well-defined. At which point do you quantify over $x$ and $y$? The answer if it is bounded or uniformly continuous will depend on it

@SubhasisBiswas correct

Subhasis Biswas

@MatheinBoulomenos I think , "uniform continuity" here implies it is uniformly continuous on $\mathbb{R}$ and "continuous" implies that it is continuous at every $c \in \mathbb{R}$

MatheinBoulomenos

@SubhasisBiswas no what I mean is that it's not clear if $\varepsilon$ and $\delta$ can depend on $x$ and $y$

Subhasis Biswas

7:20 AM

@MatheinBoulomenos it has been mentioned that $\epsilon$ is fixed (i.e. global) and $\delta$ can be varied at will

MatheinBoulomenos

okay, then I agree with "bounded" as a result

Secret

@user681391 An identity morphism is Hom(X,X)=id_X if I recall, thus Hom (X,X)=X seemed different as the object itself is also a morphism. Probably that means my category theory knowledge need some brushing up

MatheinBoulomenos

@Secret the identity morphism is an element of Hom(X,X), but not equal to it

Alessandro Codenotti

@MatheinBoulomenos buongiorno

MatheinBoulomenos

@Alessandro hai corsi oggi? Non so se c'è "Fronleichnam" in NRW

Subhasis Biswas

7:26 AM

Prove that the set of all $n$-tuples taken over a countable set is countable.

Attempt: We know that the number of primes is infinite and countable. We enumerate it as $p_1, p_2, p_3,..., p_n, ...$. Now, define $f: B_n \to \mathbb{N}$ such that $f(a_1, a_2, a_3, ..., a_n)={p_1}^{a_1} {p_2}^{a_2} ... {p_n}^{a_n}$, which is injective. [$B_n$ is the set of all $n$ tuples]. Now, $R(f) \subset \mathbb{N}$, so it is at most countable

can this work?

MatheinBoulomenos

@SubhasisBiswas this works, indeed. Cool idea to use prime factorization!

È "non so se c'è" oppure "non so se ci sia"?

Alessandro Codenotti

@MatheinBoulomenos no, è vacanza anche qua

Secret

Hmm, so Hom(X,X) denotes all epimorphisms on X, and the fixed point map f(x)=x for some x will be an element of Hom(X,X)?

Alessandro Codenotti

@MatheinBoulomenos se ci sia, but most people would use è nowadays in spoken language even though it's wrong

MatheinBoulomenos

@AlessandroCodenotti capito, grazie!

Secret

7:30 AM

I am not sure how to promote such f to categories other than noting it is not necessary injective and is an epimorphism

MatheinBoulomenos

@Secret the set of all _endo_morphisms, but yes, that's the identity morphism in a lot of categories where the morphisms are even maps

morphisms don't need to be maps, though

@Alessandro se hai letto il post del blog, godrei del tuo giudizio

ÍgjøgnumMeg

Morning :)

Secret

But how can f be an identity morphism. That only a few inputs are fixed means you can find $g \circ f \neq f \circ g \neq g$? e.g. For the case where morphisms are maps, consider $f(x) = x^3$ and $g(x) = x+1$ then $f\circ g = (x+1)^3 \neq g \circ f = x^3+1 \neq g$? yet $f(x)=x$ when $x=\pm 1$

but for something to be an identity morphism, it has to act as an identity element under composition

MatheinBoulomenos

@Secret I thought $f(x)=x$ for all $x$

Morning @ÍgjøgnumMeg

Secret

no, I was wondering about generic functions $f$ with a few fixed points. $\forall x : f(x)=x$ of course is an identity morphism since it left every point unchanged

MatheinBoulomenos

7:42 AM

@Secret say we're working over the category of sets, then the set of fix points of $f$ would be the largest subset(=subobject) on which $f$ agrees with the identity morphism. Having a few fix points means that this set is non-empty. This is a type of equalizer, i.e. a limit (in the categorical set), I'll talk about that later in my blog

Secret

ah cool

MatheinBoulomenos

So we have two parallel arrows $f:X \to X$ and $\mathrm{id}_X: X \to X$ and we're looking for the universal object $Y$ with an arrow $g:Y \to X$ such that $f \circ g = \mathrm{id}_X \circ g = g$, so the universal arrow such that $f$ behaves like a left inverse wrt that arrow $g$

Balarka Sen

Hi @Mathein

MatheinBoulomenos

$g$ is just the inclusion of the set of fix points in $\mathbf{Set}$

Hi @Balarka

Balarka Sen

@Subhasis Seems like Mathein answered your question so I'm not taking a look

MatheinBoulomenos

7:47 AM

the above construction makes sense in any category, i.e. we can define fix points even when the morphisms are not mappings! @secret

depending on the category, morphisms might be homotopy classes of mappings or relations or whatnot

Secret

How should I think of morphism as a concept itself. From what I gathered from your blog (and also some other readings such as mclaine) morphism seemed to be objects that takes two objects as inputs, and they behave like a monoid wrt composition?

Since relations is just one of the possible forms of morphisms, it seems morphisms are even more general than relations, but I am not sure what that generic thing is

MatheinBoulomenos

@Secret think of a category as a big graph with a special operation called composition. morphisms are just arrows. It's less about what morphisms are and more about how they relate to each other

Secret

I see

MatheinBoulomenos

you could say that category theory is "behavioristic" vs. the "ontological" set theory

i.e. it's about how things behave (= relate to each other) and not what they "are"

I think I'll add something along those lines to my blog entry. Thanks for asking!

Secret

cool. I really seemed to think naturally in categories

no wonder some users said I somehow rediscover concrete category theory when back then I don't even know what that is

Looking forward to your future blog posts

MatheinBoulomenos

8:02 AM

@Secret added the following paragraph:
One should think of a category as a big directed graph in which objects are represented by nodes and morphisms are represented by arrows and where we have a special operation called composition that takes to consecutive arrows and produces a third arrow representing going along one arrow first and then along the other one.
Compared with set theory, which can be called "ontological", in the sense that it is about what objects actually "are", category theory can be considered "behavioristic", in the sense that it's more about how objects behave, i.e. rel…

Secret

" that takes to consecutive arrows " -> that takes two consecutive arrows

otherwise the new paragraph makes it much clearer

MatheinBoulomenos

right, thanks

Balarka Sen

@MatheinBoulomenos What's the exact statement of Dold-Kan correspondence, out of curiosity? I guess if $(G_\bullet, d_{\bullet, \bullet}, s_{\bullet, \bullet})$ is a simplicial abelian group then you naturally get a chain complex out of it by letting the differentials be $\partial_k = \sum_{0 \leq i \leq k} (-1)^i d_{k, i}$, but I suppose you quotient $(A_\bullet, \partial_\bullet)$ by the subcomplex of degenerate simplices to get the "right complex"?

MatheinBoulomenos

@BalarkaSen precisely

that's the normalized Moore complex which gives an equivalence not only of categories, but even of Quillen model categories

and you can replace $\mathbf{Ab}$ by any abelian category

Subhasis Biswas

@BalarkaSen :D no problem

Balarka Sen

8:13 AM

What's the model structure on the category of chain complexes over abelian categories? The fibrations are chain maps with zero cokernel, and the cofibrations are chain maps with zero kernel?

And weak equivalences are chain homotopy equivalences, of course

MatheinBoulomenos

weak equivalences are quasiisomorphisms, but otherwise fibrations are epis and cofibrations are monos, yeah

the factorization condition from the model category axioms becomes epi-mono factorization in an abelian category

Balarka Sen

Oh, ok, quasi-isomorphisms would make more sense.

MatheinBoulomenos

(there's also a model category structure with chain homotopy equivalences, but that's not the "standard" one)

Balarka Sen

Mmk gotcha

To get an isomorphism of the underlying homotopy categories the un-normalized thing is enough, right? Because if $D \subset A$ is the subcomplex of degenerate simplices it's clearly acyclic, so $A \to A/D$ is a weak equivalence

(a quasi-isomorphism)

Subhasis Biswas

balarka

what is a reading course?

Balarka Sen

8:28 AM

u pick a book and u read it

with someone to advise you through it ideally

MatheinBoulomenos

@BalarkaSen right

F.White

9:42 AM

If $f$ is a smooth function and $\omega$ is a closed $(k-1)$-form, I can still use $d(f\omega)=d(f)\wedge \omega +f\wedge d\omega$ right? Since smooth functions are zero forms. Or does one then omit the wedge?

Like $df\wedge \omega + fd\omega$

Balarka Sen

Wedging with a 0-form is just multiplying by the function, yes.

F.White

Also if $f(x)=0$ on the boundary, stokes tells us that:
$$\int_M df\wedge \omega = \int_{\partial M} f\omega = 0$$
right?

Balarka Sen

$d\omega = 0$, you don't need the boundary condition!

F.White

Wait why not?

Balarka Sen

Oh, to get the last integral to be zero, sure.

F.White

9:45 AM

Yeah

Also, I wanted to show that a $1$-form $\omega$ on a connected smooth manifold is exact if and only if $\int_\gamma \omega =0$ for any $\gamma$ a closed curve

The hint says to let $f(x)=\int_{\gamma(0)}^{\gamma(1)} \omega$ for $\gamma$ a path from some base point $m$ to $x$

Balarka Sen

Yup.

F.White

I can't see why this is smooth

Balarka Sen

Do it on a chart containing $x$.

F.White

10:01 AM

So I can define a family of paths $\gamma_x:[0,1]\to M$ by $\gamma_x(0)=m$ and $\gamma_x(1)=x$, and fix a chart $U$ containing $x$ and $m$, with local coordinates $x^1,\dots,x^n$, in which case $$f(x)=\int_{\gamma_x} \omega = \int_{\gamma_x}\sum_{i=1}^n f_idx^i = \sum_{i=1}^n \int_{\gamma_x}f_i dx^i$$

am I meant to consider $x=m$ where it's a closed loop and note it vanishes by assumption

Balarka Sen

You don't need a family of paths from $m$ to every point in $M$ to define $f$, since ultimately $f$ is path-independent. You just need existence of a path, which is given to you by connectedness.

Namely, if $\gamma_1, \gamma_2$ are two paths from $m$ to $x$, then $\int_{\gamma_1} \omega = \int_{\gamma_2} \omega$. This follows from considering the loop which follows $\gamma_1$ from $0$ to $1/2$ and then $\gamma_2$ backwards from $1/2$ to $1$ then integrating $\omega$ over this.

F.White

That makes sense

Balarka Sen

But your setup now reduces the problem to a connected open subset $U \subset \Bbb R^n$. You have a fixed basepoint $m \in U$ and $\omega \in \Omega^1(U)$ which integrates to zero over closed loops, so define $f(x) = \int_m^x \omega$ (where the notation is unambiguous by the path-independence comment)

Prove that $f$ is continuous, moreover differentiable, and $df = \omega$, therefore smooth.

F.White

Thanks Balarka, I'll try that

Balarka Sen

Forms which integrate to zero over closed loops are colloquially called "conservative" because of physics :P So you're proving exact = conservative

F.White

10:10 AM

@BalarkaSen Right :)

Balarka Sen

@Alessandro I think the keyword is "infinite feather"

Alessandro Codenotti

I found a construction, but I'm not fully convinced by one step

(actually I read in this paper that there is such a space and asked here about it, just to notice that it's actually described in the paper itself...)

Balarka Sen

If I remember right the feather looks like the "branching point" in $\Bbb R \times \{0, 1\}/\sim$ where $(x, 0) \sim (x, 1)$ for $x > 0$ at a dense subset or something

Alessandro Codenotti

@BalarkaSen Google only has tattoos suggestions for infinite feather :P

Balarka Sen

It's a really niche name

Alessandro Codenotti

10:19 AM

Start with a line
segment in the plane. Attach another line segment to the midpoint. Now there
are three line segments, the new one and two halves of the original. Attach two
more line segments to one midpoint, three more to another, and four more to the
last, ensuring that no new line segment meets the old set at any other point. Order
the midpoints of all the line segments now present, attach five line segments to the
first, six to the next, etc. Repeat infinitely often, and take the closure. If each

Copy/pasting from said paper

Now the idea is that the set of midpoints is dense and removing any one of them breaks the space into a different number of connected components, so every automorphism must fix all of them

Balarka Sen

Ya sounds right. No nontrivial self-homeomorphisms exists because of the valence obstruction

Cut points or whatever they are called

Alessandro Codenotti

I'm not convinced that taking the closure can't cause issues

Balarka Sen

Polish maniacs

@Alessandro Take the tree with a root node, which has one offspring, which has two offsprings, each of which has three offsprings, so on. This is a simplicial complex (not compact though). Collapse each half-open edge to a point. This has nontrivial automorphism group, yeah?

I'd try to end-compactify this but I'm not thinking too hard

Alessandro Codenotti

Sure, "swapping" branches below a node

Balarka Sen

Ah yeah

So maybe you need to modify the branch structures as well

Alessandro Codenotti

10:27 AM

This has a huge automorphism group :P

Balarka Sen

Root node has one offspring, which has two offsprings, the first of which has three offsprings, the second of which has four offsprings, ... :P

Alessandro Codenotti

If you take the infinite rooted binary tree $T$ its automorphism group $G$ is very weird, $(G\times G)\rtimes\Bbb Z_2\simeq G$

@BalarkaSen Right, this is very similar to the construction I posted above now!

Balarka Sen

So the point is it's a countable space with an enumeration such that the $n$-th node is an "$n$-cutpoint"

Which is why no point can map to a different point by a self-homeo for connectivity reasons

Alessandro Codenotti

Right

Balarka Sen

What if you end-compactify this thing now?

MatheinBoulomenos

10:32 AM

@AlessandroCodenotti it's the inverse limit of taking a wreath product with $C_2$ at each step

Balarka Sen

If $T_n$ is the root binary tree of height $n$ then $\text{Aut}(T_{n+1})$ is $\text{Aut}(T_n)$ wreath $\Bbb Z_2$ certainly

Alessandro Codenotti

Yes

It's an important group in GGT

Balarka Sen

You have two branches of $T_n$'s, so it's a subgroup of $\text{Aut}(T_n) \times \text{Aut}(T_n)$ and there's an action of the $\Bbb Z_2$ subgroup by flipping the two branches, i.e., flipping the two factors.

@Alessandro Oh interesting

Alessandro Codenotti

Grigorchuck's group can be constructed as a subgroup of it

Balarka Sen

Oh that's the thing of intermediate growth?

Alessandro Codenotti

10:37 AM

(this is not the original construction, but it's the easiest way to construct a group of intermediate growth)

Balarka Sen

Nice!

Alessandro Codenotti

arxiv.org/abs/math/0607384

That's a very well written and easy to follow exposition

Balarka Sen

Oh shit Igor Pak lmao

One of the first answers I wrote in MSE was a question by him

Now deleted because I didn't want my bullshit hypergeometric function stuff up for public consumption

@Alessandro So the end-compactification of my thing will essentially look the following: every "branch" in my "tree" looks like the Sierpinskified $\Bbb N$: $\Bbb R_{\geq 0}$ with $[n, n+1)$ collapsed to a point for every natural number $n$. So the end-compactified thing will have branches which are Sierpisnkified $\{0\} \cup \{1/n : n \in \Bbb N\}$: $[0, 1]$ with $(1/n, 1/{n+1}]$ collapsed to a point for every natural number $n$.

Alessandro Codenotti

What do you mean Sierpinskified?

Balarka Sen

$[0, 1]$ with $[0, 1)$ collapsed is the Sierpinski 2-point space, for example.

Bullshit name

Alessandro Codenotti

10:49 AM

Oh sure

Balarka Sen

Any self-homeomorphism must send an end to an end. But that would entail sending some point in one branch to some point in another branch, which is not possible by the cutpoint argument again.

So I think that does it

Alessandro Codenotti

This works, but I wanted a metrizable example

Balarka Sen

Ah fuck lmao

Alessandro Codenotti

While the construction from the paper is since it is a subset of $\Bbb R^2$

But it's a cool example nonetheless

Balarka Sen

11:16 AM

I mean the Sierpinsky space is already a non-metrizable example, so @Alessandro

Metricity really is the point

Alessandro Codenotti

@BalarkaSen oh right, of course

ÍgjøgnumMeg

12:28 PM

Hey @Mathein :) Hast du Zugriff auf Uebungsblaetter vom ANT1 Kurs? Ich hab versucht sie herunterzuladen aber man muss einloggen... lol

Ryan Unger

@ÍgjøgnumMeg where are you going to school?

ÍgjøgnumMeg

@Ryan Heidelberg

Ryan Unger

damn there should be a stack exchange Heidelberg meetup

ÍgjøgnumMeg

lol actually @Alessandro is in Bonn and we mentioned going for a drink at some point

Ryan Unger

3

Q: Proof $|u|\leq 1$ for pde $\Delta u=u^3-u$

Let $\Omega \subseteq \mathbb{R}$ be open and bounded and let $u\in C^2(\Omega)\cap C(\overline{\Omega})$ be solution of \begin{align} \Delta u=u^3-u\quad &\text{in } \Omega\\ u=0 \quad &\text{at } \partial \Omega \end{align} Show that $|u|\leq 1$ in $\Omega$. Now to proof that $u\geq 1$ ist ...

lmao

this person must be at my summer school

actually looking his second question, maybe not

but it's suspicious

Joshua Ronis

12:54 PM

Hey, I'm not sure if it's okay to request answers to questions here, but if someone has a great geometric understanding of Linear Algebra, I would REALLY appreciate an answer to this question: https://math.stackexchange.com/questions/3266970/a-geometric-question-on-the-commutativity-of-inner-products-with-symmetric-matri
If it's not okay to request answers (which makes sense, since this could get flooded), I won't post requests again, just a moderator let me know. I just really want to understand this, because I feel like it should be so intuitive, but its not (at least not to me)!!!!

Michael.P

1:06 PM

Hello, guys

Anyone familiar with Dedekind cuts?

Secret

1:23 PM

depends on what context

Alessandro Codenotti

1:34 PM

Just ask

Thorgott

I wish I had enough geometric intuition to answer that.

Michael.P

Where the hell did that come from?

(the screen is from original paper)

Thorgott

Obligatory: What is $D$?

Secret

how is that related to dedekind cuts, looks more like ODE stuff to me...

Michael.P

Number which square is to be 2 (for instance)

It's bound by $$\lambda ^{2}< D<(\lambda+1) ^{2}$$

Thorgott

1:50 PM

I don't think that's a Dedekind cut

Michael.P

It's in section IV (creation of irrational numbers) of essays
on the theory of numbers by Richard Dedekind

Thorgott

2:06 PM

Ah, I see. So what exactly is the question?

Michael.P

Variables $x$ and $y$ don't appear before that in the paper. And they are introduced in these equations (marked by arrows) that are not foreshadowed themselves. I suspect they have to do with the narrowing of the bounds around $D$ but can't connect the two together. So, my question is: "Where the hell did that come from?"

Thorgott

Dedekind elaborates in the next paragraph. $x$ is some rational number and $y$ is constructed depending on $x$. Then, if $x$ is taken to be a member of $A_1$, $y$ is a greater member of $A_1$ and if $x$ is taken to be a member of $A_2$, $y$ is a smaller member of $A_2$. Since this works for any $x$, this proves that $A_1$ has no largest and $A_2$ has no smallest element.

Michael.P

Yes, however, I hope you understand, his elaboration on the variables doesn't justify the costruction of $y=\frac{x(x^2+3D)}{ 3x^{2}+D }$

Thorgott

2:23 PM

What is there to justify? This is the definition of $y$.

Michael.P

Okay

Thorgott

And the definition is good (read: useful) precisely because it allows these conclusions.

Michael.P

Equations don't just come into existance because they will support your further hypothesis

He constructed $y$ in that way based on some reasoning. I ask if anyone has any clue on what that reasoning is

MatheinBoulomenos

2:38 PM

@ÍgjøgnumMeg ich schick dir eine Mail

Thorgott

Probably through some numerical square root approximation method. Not one I'm familiar with though.

MatheinBoulomenos

@ÍgjøgnumMeg I sent you an email to your student email address, is that still your current email?

ÍgjøgnumMeg

2:54 PM

@Mathein ja danke sehr!

MatheinBoulomenos

it has 1MB worth of pdfs as attachments

ÍgjøgnumMeg

Yeah I'm looking at it now lol

MatheinBoulomenos

for the course you'll be doing, there will be different exercise sheets of course

ÍgjøgnumMeg

Yeah I figured :P I just wanted something to do

Ryan Unger

3:15 PM

sup nerds

wagwan

$\sup$

$\mathfrak{henlo}$

$\inf$

h a

3:29 PM

esssup

Alessandro Codenotti

Found the analyst

Balarka Sen

lol

Ryan Unger

you can tell that from the fact that the most algebra in what I'm doing now is "real vector space of symmetric nxn matrices"

Alessandro Codenotti

$C^\ast$-algebra, can't be analysis, right?

Ryan Unger

3:45 PM

it's not really

there's hard analysis and soft analysis

Alessandro Codenotti

I think you mean Russian analysis and soft analysis

Balarka Sen

gRoMov

@RyanUnger Gromov proves h-principles in his book by using sheaves of topological spaces

Akash. B

I have a question

Ryan Unger

Russian analysis is completely incomprehensible to me

Akiva Weinberger

I have an answer @Akash.B

Probably not to your question though

Akash. B

3:52 PM

@AkivaWeinberger How does that help me?

Balarka Sen

Akiva is being mean

Akiva Weinberger

It was an attempt at humor

Just write your question

Rithaniel

You should probably ask the question to see if anyone can indeed help.

Akash. B

If we take the square of 25

We get 625

But if i take square of 0.1

We get 0.01

Akiva Weinberger

0.01

Akash. B

3:58 PM

Oops sorry

How strange?

Akiva Weinberger

Yeah x²>x iff x>1

(or if x is negative)

Akash. B

If we take the square of huge numbers we get numbers more than that

Akiva Weinberger

x²<x iff x is between 0 and 1

Akash. B

But if we take small numbers we get numbers smaller than that number

Alessandro Codenotti

@Mathei why are irreducible representations called irreducible?

Akiva Weinberger

4:01 PM

And if we take x=1, we end up with the square being neither bigger nor smaller

Balarka Sen

@Alessandro It means there are no proper invariant subspace, right?

Alessandro Codenotti

proper closed subspaces in the case I'm interested in (I'm looking at representations of a $C^\ast$-algebra on a Hilbert space)

Akiva Weinberger

I just got an ad for Masterclass

"Chris Hadfield teaches Space Exploration"

Balarka Sen

LOL

More like he teaches you how to be a table

Akiva Weinberger

It's like Masterclass just up and went, "We know you're not gonna do any of this stuff anyway"

4:05 PM

"You saw the Itzhak Perlman one on the violin, you think that's any more realistic?"

@BalarkaSen Wat?

Also I would hope that if you were about to become an astronaut

you'd be given better training than you would get from an online video series

Balarka Sen

4:34 PM

@AkivaWeinberger It's from "Lulu" by Lou Reed & Metallica

also known as Art

MatheinBoulomenos

5:02 PM

@AlessandroCodenotti if it is not irreducible, then you can "reduce" it by an exact sequence $0 \to U \to V \to V/U \to 0$ to smaller representations. In case the representation is unital $V/U$ can be identified with $U^\perp$ and you even have $V=U \oplus U^\perp$

Alessandro Codenotti

Wait what are $U$ and $V$?

MatheinBoulomenos

$V$ is the non-irreducible space and $U$ is a proper nontrivial closed invariant subspace

Balarka Sen

Your representation is $\rho : G \to \text{GL}(V)$ and $U \subset V$ is the invariant subspace

Alessandro Codenotti

Oh, I see

Balarka Sen

If you have a $G$-invariant Hermitian metric on $V$ then for any $x \in U$ and $y \in U^\perp$, $\langle x, gy \rangle = \langle g^{-1}x, y \rangle = 0$ since $g^{-1} x \in U$ (by $G$-invariance of $U$), so $gy \in y$, aka $U^\perp$ is also $G$-invariant.

Which is the $G$-invariant decomposition $V = U \oplus U^\perp$ Mathein mentioned

Ryan Unger

5:23 PM

lol

I prove an "algebraic estimte" using compactness

Balarka Sen

5:34 PM

@MatheinBoulomenos In particular if $G$ is a finite group, I can always cook up a $G$-invariant metric on $V$ by averaging that makes my representation unitary, so such a decomposition into irreducible representations always exists, yeah?

Even if it is a compact Lie group, that's true

MatheinBoulomenos

@BalarkaSen right, that's one proof of Maschke over $\Bbb R$ or $\Bbb C$

Ryan Unger

havent heard that name in a while

Balarka Sen

whats maschke

MatheinBoulomenos

if you use the Haar measur for averaging, it works on any compact Hausdorff group works, it doesn't have to be a Lie group

Balarka Sen

Ah yeah fair point

I just automatically use the bi-invariant volume form because I'm a bum

Ryan Unger

5:38 PM

what about compact non-Hausdorff groups

MatheinBoulomenos

Maschke is the theorem that for finite groups (modulo the modular case when the characteristic divides the group order) all representations decompose as a direct sum of irreducible ones

Balarka Sen

Huh pretty simple fact to be called blank's theorem

MatheinBoulomenos

@RyanUnger if you look at continuous representations into a Hausdorff topology on $GL(V)$ (say Euclidean or discrete), then the representation factors over the quotient modulo the closure of the identity, so things work out.

@BalarkaSen it's simple, but important

Balarka Sen

I suppose in the end I'm really saying if $G$ is a compact Hausdorff group then there's a decomposition of any finitely generated module over $\Bbb C[G]$ into simple modules, which looks slightly nontrivial and deep now that I think about it

MatheinBoulomenos

@BalarkaSen you can get rid of the f.g. hypothesis by Zorn's lemma

Balarka Sen

5:42 PM

Ah

MatheinBoulomenos

also if $\Bbb C[G]$ is a semisimple ring, then any module is semisimple (that's just hiding the application of Zorn though)

the other standard proof just takes any projection onto an invariant subspace, averages that projection such that the kernel is an invariant complement

the advantage is that this needs less assumptions on the base field

Balarka Sen

Gotcha

MatheinBoulomenos

In fact, the averaging operator $f \mapsto \frac{1}{|G|}\sum_{g \in G}$ is a projection $\mathrm{Hom}_k(V,W) \to \mathrm{Hom}_{k[G]}(V,W)$ that behaves naturally wrt composition

Thus if you take the trace of the averaging operator, you get the dimension of the space $\mathrm{Hom}_{k[G]}(V,W)$, but using the $G$-equivariant isomorphism $\mathrm{Hom}_k(V,W) \cong V^* \otimes_k W$ you can also compute that trace as $\frac{1}{|G|} \sum_{g \in G} \chi_V(g^{-1}) \chi_W(g)$

that's the inner product of characters $\langle \chi_V, \chi_W \rangle$ which is thus equal to the dimension of the space of morphisms of reps

if $V$ is irreducible, then the dimension of $\mathrm{Hom}_{k[G]}(V,W)$ tells us how often $V$ appears as a summand in the decomposition of $W$ (by Schur's lemma)

but since we can compute that from the character of $W$, it follows that we can compute the multiplicity of all simple representations appearing in the decomposition of $W$ from its character

thus we have proved that the character determines the representation

(sorry for the rambling)

Balarka Sen

Damn nice

No I loved that

I should read representation theory someday

MatheinBoulomenos

it's really pretty

Balarka Sen

5:55 PM

Can you tell me what irreducible representations of $S_n$ look like?

Semiclassical

there's a wiki page:

Representation theory of the symmetric group

In mathematics, the representation theory of the symmetric group is a particular case of the representation theory of finite groups, for which a concrete and detailed theory can be obtained. This has a large area of potential applications, from symmetric function theory to problems of quantum mechanics for a number of identical particles. The symmetric group Sn has order n!. Its conjugacy classes are labeled by partitions of n. Therefore according to the representation theory of a finite group, the number of inequivalent irreducible representations, over the complex numbers, is equal to the number...

MatheinBoulomenos

these are classified by Young tableaus, combinatorial objects

Balarka Sen

@Semiclassical Ya I was hoping Mathein could explain it to me :)

Semiclassical

lol

MatheinBoulomenos

I have not studied irreps of $S_n$ in detail, just knowing that there's some relation to combinatorics

I'd have to read up on it myself

Balarka Sen

5:57 PM

@MatheinBoulomenos I know what those are but from different context

Aha OK gotcha

Semiclassical

looks like it's not so bad when you're doing irreps over the complex numbers

"The determination of the irreducible modules for the symmetric group over an arbitrary field is widely regarded as one of the most important open problems in representation theory."

MatheinBoulomenos

you might be interested in Schur-Weyl, though

Schur–Weyl duality

Schur–Weyl duality is a mathematical theorem in representation theory that relates irreducible finite-dimensional representations of the general linear and symmetric groups. It is named after two pioneers of representation theory of Lie groups, Issai Schur, who discovered the phenomenon, and Hermann Weyl, who popularized it in his books on quantum mechanics and classical groups as a way of classifying representations of unitary and general linear groups. Schur–Weyl duality can be proven using the double centralizer theorem. == Description == Schur–Weyl duality forms an archetypical situation in...

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