Mathematics - 2021-01-31 (page 2 of 3)

BigSocks

17:03

@user3559014 cool cool good luck

Thorgott

Yeah, this map makes sense and the fibers are complex lines (they're parametrized by the last homogenous coordinate), but how to identify this with the tautological bundle in a well-defined manner is what I'm struggling with.
To identify a fiber of the tautological bundle with C, I have to pick a non-zero element and then the elements of this fiber scale inversely with this choice of generator, which is awkward (this is what I've been mitigating with picking a generator on a sphere and then complex conjugating in the second coordinate).

Mike Miller

@Thorgott Easier to go the other way, I think. I haven't thought through the details though; maybe you're right and it's hard to right down?

I think the map from the tautological bundle to this bundle should be explicit since you've made the choice by picking a point in the tautological bundle.

Could be wrong though.

Thorgott

A point in the tautological bundle looks like $([x],v)$ with $v\in\langle x\rangle$. I want a point in $\mathbb{CP}^{n+1}$ that looks like $[(x,\lambda)]$ for some $\lambda\in\mathbb{C}$, so the question is how to get some $\lambda$ from the $v$ in a well-defined manner, which runs into the "identifying the fiber with $\mathbb{C}$" issue I mention (at least I don't see a better way).

but, I mean, there has to be a better way

or wait

Mike Miller

I think there should be something not too painful, but you may have to get your hands dirty. After all, "projection" uses a dot product. So maybe you're right that you have to use a sphere somewhere.

It wasn't clear to me that you had this picture of what the natural map from CP^{n+1} - pt to CP^n was, and that's what I wanted stress

Ted would know the answer right away

Thorgott

here's the thing: my explicit form involves complex conjugation, so it doesn't work for RP instead of CP, but the picture still suggests the analogous thing is true for RP, doesn't it?

Mike Miller

17:15

Well, complex conjugation shows up in the dot product over C

So I'd just expect it not to show up over R

Anyway, let me say how to finish the job. E is homeomorphic to an open disc bundle in E. D(E)/S(E) is a one-point compactification of that open disc bundle, and hence homeomorphic to a one-point compactification of E. But CP^{n+1} is a one-point compactification of CP^{n+1} - pt.

Thorgott

right, that makes sense

BigSocks

So for $A$ a commutative domain, $\cap_{P \in Max(A)} (A_P)_Q \subset A_Q$ where $Q$ a particular max ideal.

but why

nvm it's bc $A_Q$ = one of the $(A_P)_Q$, when $P = Q$, and the intersection of all of them is gunna be in there

Charlie

17:38

Quick question, why is the following invalid?: $$-1=\sqrt{(-1)^2}=\sqrt{1}=1$$

oh wait sqrt(1) is multivalued

nvm

Thorgott

because $\sqrt{x^2}=|x|$, not $=x$

BigSocks

similar question to the above one. also start with $A$ a commutative domain, $S$ a multiplicative subset of $A$.

$S^{-1}\left( \bigcap\limits_{P \in Max(A)} A_P \right) \subset \bigcap\limits_{P \in Max(A))} S^{-1}(A_P)$

but I'm struggling to see why

Thorgott

17:58

take an element in the LHS, note it's contained in the RHS

by virtue of writing it down exlpicitly

BigSocks

$m \in LHS \Leftrightarrow m \in \{ (a,s) \vert a \in A_P \forall P \in Max(A), s \in S \}$ whereas $n \in RHS \Leftrightarrow n \in \{ (a,s) \vert (a,s) \in S^{-1}(A_P), \forall P \in Max(A) \}$

Charlie

@Thorgott ah, thank you

BigSocks

@Thorgott yeah I was trying to do that but the note part is where I got stuck

Thorgott

write things as fractions instead of this obfuscating notation

BigSocks

obfuscating is a matter of opinion

but I will give it a go

Thorgott

18:01

well, this notation is also plain wrong here, because you have to think of all of these $A_P$ as embedded in the fraction field of $A$, otherwise taking the intersection doesn't even make sense

(or well, you can technically do it, but it's rather meaningless)

BigSocks

I don't understand...

they do that anyway, but it still makes sense yeah

they say $A \subset A_P \subset A_{(0)} = K$

Thorgott

yeah, that's the only reasonable way to do it

BigSocks

sure yeah, I guess I had that in mind

but uh, yeah, $m \in LHS \Leftrightarrow m \in \{ a/s \vert a \in A_P \forall P \in Max(A), s \in S \}$ and $n \in RHS \Leftrightarrow n \in \{ a/s \vert a/s \in S^{-1}(A_P), \forall P \in Max(A) \} \Leftrightarrow \{ a/s \vert a \in A_P \forall P \in Max(A) \wedge s \in S \}$

kinda weird. I wrote it and they look like the same set

Thorgott

cause you're being sloppy with the RHS

BigSocks

that's gotta be it

Astyx

18:14

hi chat

BigSocks

hiya

Thorgott

hi

Astyx

Any progress on the associate problem?

BigSocks

not yet

Thorgott

I gave up on it after a bit

Astyx

18:19

I think the C(R,R) example is the right way to think about it

In the sense that the answer lies in the fact that you have an "orientation" on the spectrum, and enough elements in the spectrum to construct similar examples

Fargle

That makes it make sense that it fails in general, for me, but not that it works in any PIR.

user2103480

@Thorgott this chat is the scorpion:

Astyx

lol

Sayan Chattopadhyay

@MikeMiller Yeah that's the point, like while I was explaining through examples they are like, there's an easier way, so I will do that. So I did examples but I kept on insisting the pivot variables and free variables picture so they have something to fall upon.

Thorgott

:deepfried_joy:

Sayan Chattopadhyay

18:27

But doing examples is the key thing. I had to show them the mess so that they were convinced that it is a mess and I wasn't taking good examples or using some trickery

Thorgott

actually, the subring generated by $(1,0,1),(1,0,-1),(1,1,1)$ is just the entire ring, isn't it

Astyx

Yeah mb

Thorgott

but wait, that is a PIR, no?

Astyx

Yes, but $(1,0,-1) = (1,0,1) \times (1,1,-1)$ so they're associates

Thorgott

oh, the example doesn't work

oops

user15072279

18:42

1

Q: How is this possible regarding theta?

Here, $\theta = d\theta$. Therefore, we can say $s$ is almost like a line. Then , that makes it a triangle. I have read that to find $\theta$, we say it is $\theta = S / H$. Now , let us say theta is a final velocity at of particle $P$ at $A$ and initial velocity of particle $P$ at $B$. Then , ...

Please help in this

monoidaltransform

hi

dc3rd

Tautological bundles, holomorphisms, subrings.................now things have gone way above my paygrade........whew.

monoidaltransform

I'm thinking, what would be an example of a geodesic metric space that is never uniquely geodesic?

so between any two points you can always find more than one geodesic?

amWhy

@robjohn, @TedShifrin, @anyoneWhoLikesWordGames!!:

in Cafe and Tavern on the math.se, 5 mins ago, by amWhy

Let's anticipate Valentine's Day! How many words with four or more letters can we create from the letters in VALENTINE'S DAY? Word Challenge!!!

Thorgott

@monoidaltransform R^2 with the taxicab matric

there you have uncountably many geodesics between any two distinct points, even

monoidaltransform

18:53

how do you formally prove this?

@Thorgott

Thorgott

you can just write enough of them down explicitly

if you understand this visually, it will be clear how to

I suggest drawing a picture

BigSocks

so for my question up there, it's not quite right to say $a \in A_p \forall P \in Max(P)$ in the $RHS$ I think

@Thorgott so this was the problem I think

Thorgott

more specifically, you can't just require that every fraction be in a form with the same denominator

eh, that's badly worded; I mean you can't require that an element of the RHS has a representation as element of $S {-1}A_P$ with the same denominator for every $P$

BigSocks

right, so intuitively the RHS is "bigger" since it could contain various representations of a given element in the LHS

monoidaltransform

19:26

I don't see why with the taxi cab metric (0,0) and (0,1) have more than one geodesic connecting them

@Thorgott

Hanno

Hi all

Thorgott

oh, you're right, they don't

BigSocks

seems like finitely many

monoidaltransform

so how would we resolve that issue?

Mike Miller

i think he would have suggested a new example if that was immediately clear

I thought for a minute and don't see a good trick

Thorgott

19:38

yeah, this example isn't salvageable

and I don't have an alternative off the top of my head

Balarka Sen

what is the question

monoidaltransform

@BalarkaSen what would be an example of a geodesic metric space that is never uniquely geodesic?

copper.hat

what makes a metric space a geodesic metric space?

monoidaltransform

so between any two points, we can always find more than one geodesic between them

Thorgott

@copper.hat there's a geodesic between any two points

Mike Miller

19:41

Symbol-free definition
A geodesic metric space is a metric space if it satisfies the following equivalent conditions:

Given any two points, there is a path between them whose length equals the distance between the points

Thorgott

hmm, I actually have to think for a sec whether that's equivalent to the defn I have in mind

copper.hat

distance being path length in a $\mathbb{R}^n$ sort of way?

monoidaltransform

what's the definition you have in mind, @Thorgott ?

Mike Miller

copper.hat

i mean could i take a 2 point graph with 2 equal length edges between them?

Balarka Sen

19:43

i mean of course, locally geodesics are unique there

copper.hat

@MikeMiller Thanks!

Balarka Sen

I dunno. What if you paste two copies of R along Q?

monoidaltransform

@BalarkaSen I'm not sure I understand

Thorgott

I had an isometric embedding of an interval in mind, but I came to the conclusion that's equivalent

Mike Miller

Non Hausdorff junk. Hate it!

monoidaltransform

19:47

what that space looks like

Balarka Sen

Not a metric space, just a thought.

Thorgott

non-Hausdorff metric space, you love it

Mike Miller

I think you want to do a transfinite construction

Start with R

Add infinitely paths between any two points, all of which have the same length as the original distance between those two points

Balarka Sen

You have to do it at every point, which does not seem possible.

Mike Miller

Do the same on each of the new paths

monoidaltransform

19:48

what's a transfinite construction?

Mike Miller

Rinse lather repeat

Thorgott

@copper.hat take two points on the same edge, neither of them being one of the vertices you started with, there's only one geodesic between them: the part of the edge they lie on that's between them

Mike Miller

I described roughly what I mean, I'm not gonna be more precise than that

Thorgott

I doub't that's gonna be metric, Mike

Mike Miller

I'm content that this ought to work

Thorgott

19:49

sounds very non-first countable

Balarka Sen

Yeah this is not gonna be a metric space

monoidaltransform

@MikeMiller where can I find a more precise argument for that

Mike Miller

Do you think I would have written this if I had a reference man

monoidaltransform

you used the term transfinite construction

I didn't mean reference for the construction

for the term

Mike Miller

I'm not interested in googling for you

monoidaltransform

19:51

all I found was something on free algebras, not topological spaces

Balarka Sen

None of us know the answer, bottom line. I bet it's not possible, but I am also not willing to bet too much.

You should ask in the main site

monoidaltransform

thank you any ways @BalarkaSen

Thorgott

the concept of transfinite recursion should be covered in any text on Set theory

e.g. Halmos

monoidaltransform

@BalarkaSen could you perhaps please elaborate on why you think the construction Mike mentioned isn't a metric space... and first countable @Thorgott, please?

@Thorgott thanks!

at an intuitive level

Thorgott

honestly, I think the bigger issue that Mike's construction doesn't seem to converge to anything

monoidaltransform

19:59

I see

copper.hat

but is it Cauchy?

BigSocks

@Thorgott so let's say we just have 2, $a/s \in S^{-1} A_P \cap S^{-1} A_Q$. It could be the case that $A_P = 4 \Bbb Z$ and $A_Q = 6 \Bbb Z $ and $S^{-1} = \Bbb Z_{>0}$. So we could write $4/2 = 6/3$, but $6 \notin A_P$ and $4 \notin A_Q$. So $a \notin A_P \forall P \in Max(P)$ necessarily, but you could find $s$ such that $sa \in A_P$ I guess.

Thorgott

$A_P=4\mathbb{Z}$?

BigSocks

yeah?

oh at the end $P$ varies

so I guess I am abusing notation

Nariman Asgharian

Hi guys,

I want to calculate gradient ability for cars. Can you help me?

BigSocks

20:12

Just kind of looking at how it would work in one example and how I should write out what the set contains. still can't write it down sadly

Nariman Asgharian

I found some formulas but they didn't work for me.

Thorgott

but $4\mathbb{Z}$ isn't a ring

BigSocks

oh right, we like rings with unity here. I forget a lot

I guess if we just had them be $A$-modules

shouldn't be too bad

Thorgott

well, for $A$-modules, localization commutes with intersection

BigSocks

that's cool, but I guess I still wanna prove the one inclusion

although it probably doesn't work for infinite intersection?

idk

not 100% on that

Thorgott

20:21

sure, it works for infinite intersections

most rings have infinitely many maximal ideals, I think

BigSocks

I mean... take my example to the extreme. $S^{-1}(n \Bbb Z) = \Bbb Q$.
$\bigcap^\infty S^{-1} n \Bbb Z = \bigcap^\infty \Bbb Q = Q$, but $S^{-1} \left( \bigcap^\infty n \Bbb Z \right) = (0)$

Balarka Sen

This is tangential, but there was some issue about localization and intersection that I am not able to remember. Was it that $S^{-1} A = \bigcap_{\mathfrak{p} \subset A \setminus S} A_{\mathfrak{p}}$ is not true for arbitrary multiplicative sets $S$ in general?

$A$ is a domain, and intersection happens in quotient field.

BigSocks

idk what happened to my message

$S^{-1}(n \Bbb Z) = \Bbb Q$.
$\bigcap^\infty S^{-1} n \Bbb Z = \bigcap^\infty \Bbb Q = \Bbb Q$, but $S^{-1} \left( \bigcap^\infty n \Bbb Z \right) = (0)$

monoidaltransform

does the maximum metric on $\mathbb{R}^2$

work fo my question

BigSocks

better

$S = \Bbb Z_{>0}$

just in case @BalarkaSen comes up with some dirty multiplicative set

Thorgott

20:29

@BalarkaSen yeah, I think that's it

Balarka Sen

Cool, thanks

copper.hat

@monoidaltransform that is just the taxicab metric rotated.

BigSocks

@Thorgott so I guess it is not necessarily the case

Thorgott

I don't know what you're trying to do, but $n\mathbb{Z}$ is not a localization of $\mathbb{Z}$

BigSocks

$n \Bbb Z$ is just a submodule of $\Bbb Z$

you localize wrt $S = \Bbb Z_{>0}$

So I guess you could say you localize at $0$? unsure of the terminology

Hi @TedShifrin

Ted Shifrin

20:34

Hi, Big.

Thorgott

localizing at $\mathbb{Z}_{>0}$ is the same as localizing at $\mathbb{Z}\setminus\{0\}$

BigSocks

correct

Thorgott

my "works for infinite intersections" comment was for the containment that you initially wanted

BigSocks

oh I thought you meant about localization commuting with intersection

Thorgott

localizing modules commutes with finite intersections, but not with infinite ones as your example shows (but their still is the same one-sided containment as you're trying to prove for rings)

Ted Shifrin

20:36

Seems like a hifallutin' way of saying $\Bbb Q$.

BigSocks

yeah cool cool. still kinda choking on simply writing down the rhs though lol

@TedShifrin what does?

Ted Shifrin

Localizing at $0$.

monoidaltransform

@copper.hat what do you mean? what two points have only one unique geodesic between them under the maximum metric?

BigSocks

oh yeah, the flutes are high there for sure

Thorgott

hey Ted, I have a question

Ted Shifrin

20:38

Oh oh

copper.hat

@monoidaltransform no, i mean the taxicab metric is essentially the same as the $l_\infty$ metric.

in $\mathbb{R}^2$.

Thorgott

remove the point $[0\colon...\colon0\colon1]$ from $\mathbb{CP}^{n+1}$, the remaining lines project to lines on $\mathbb{CP}^n$, giving us a bundle $\mathbb{CP}^{n+1}\setminus\{[0\colon...\colon0\colon1]\}\rightarrow\mathbb{CP}^n$. what's the nicest way of seeing this is actually the same as the tautological bundle?

I was talking about this with Mike earlier, but we weren't sure how clear this is/should be

Ted Shifrin

I say it's the dual of tautological, I think.

BigSocks

the contradictory bundle

Balarka Sen

Nice one. We should call it that.

C.F.G

20:42

Hi

Is it true that trace commutes with Laplacian?

In my point of view it is true because $\Delta T_{kl}=g^{ij}\nabla_i\nabla_j T_{kl}$ and we know that any trace commutes with covariant derivative.

Mike Miller

I think you're right but the indices can only obscure your point. If this is true then it should be obviously true in more invariant notation by appeal to axioms (eg, how covariant derivative relates to the metric).

Balarka Sen

$O(-1)$ is a neighborhood of $\Bbb{CP}^{n-1}$ in $\overline{\Bbb{CP}^n}$. This follows from the description of blowup, for example. By changing orientation, you obtain $O(1)$ is a neighborhood of $\Bbb{CP}^{n-1}$ in $\Bbb{CP}^n$.

Mike Miller

@Thorgott This is why they were giving us such a hard time about O(-1) vs O(1). The first one is the one that's naturally equivalent! We were dualizing the whole time.

Ted Shifrin

This is clearly the normal bundle of $\Bbb P^n$ in $\Bbb P^{n-1}$1, via tubular neighborhood theorem, a @Balarka.

Mike Miller

Surely the line above a given line in CP^{n-1} is canonically identified via dot product as the line of functions on the given line.

Ted Shifrin

20:46

The normal bundle is $\mathscr O(+1)$ for "obvious" reasons. Now I would like to find a global holomorphic section.

Mike Miller

Line line line line.

C.F.G

@MikeMiller Hi Mike. thanks for response. so if a tensor were trace free then its Laplacian is also trace-free?

Ted Shifrin

I don't want to leave the holomorphic category, you damn topologists.

Mike Miller

I dunno. I don't like to think about this stuff. Was just raising my opinion that indices obscure.

Whose lines are on display here? ...

Balarka Sen

My lines. Line's line. Line line line line.

Thorgott

20:50

ok, I'm not familiar with the abstract viewpoint, but I assume this dualization is why there was a complex conjugation appearing in my explicit homeo?

Balarka Sen

That should be correct.

Mike Miller

@Thorgott Yeah, this is great. Notice that over RP^n there is O(n) for n in Z/2!

That's why O(1) = O(-1) there, no complex conjugation

Ted Shifrin

I have assigned this problem many times, but I haven't thought about it so I have to rethink. But I'm 99% sure it's $\mathscr O(1)$.

Yes, $\Bbb R$ is very boring. $\Bbb Z/2$ instead of $\Bbb Z$.

Balarka Sen

You are correct, of course.

Ted Shifrin

Mike's proof should be that a Möbius strip with two half-twists is a cylinder.

Balarka Sen

20:53

These guys want to see O(1) and O(-1) are isomorphic as real bundles, where their definitions for O(1) and O(-1) are such that it is not a tautology.

Hence the struggle.

Ted Shifrin

Tautological is, in fact, tautological, and $\mathscr O(1)$ is the dual.

Balarka Sen

:)

Thorgott

so that I can follow this conversation, what's this O(-) stuff

Ted Shifrin

sheaf of germs of holomorphic functions (polynomials) homogeneous of degree $k$.

Mike Miller

So the identification of ([z_0 : ... : z_n]) is with ([z] = [z_0 : ... : z_{n-1}], f), where f(z) = z_n.

Canonical as hell.

Balarka Sen

20:55

Haha, Ted rekt Thorgott.

Now he has to untangle the algebra to understand.

Perfect exercise.

Ted Shifrin

I can just feel a @Balarka smirking.

Mike Miller

@Thorgott Ted is being too canonical for my taste. O(-1) is the tautological bundle and O(1) is its dual. We were not being canonical enough to see the right answer.

Ted Shifrin

ROFL

Mike Miller

So my taste was not canonical enough for this problem, and you had to get a complex geometer to fix my mess.

Thorgott

all this canonicity is overwhelming me

Balarka Sen

21:00

It seems I have to learn what an Ising model is now.

Ted Shifrin

I need to do some stuff in the kitchen, but will try to find a holomorphic section whilst there.

Mike Miller

You're a categorist, I would think that canonicity is ... natural ... for you!

No applause or boos? Ah well.

Balarka Sen

Your joke sucks

Mike Miller

Give me that T-shirt

Balarka Sen

Lol

Good reference

Jack Zimmerman

21:11

What did Jordan say to Chad?

Balarka Sen

Such a clean room.

Sebastiano

21:23

Good evening to all users connected to the chat.

Please, is there someone that is also an user of Physics.SE?

Ted Shifrin

@Balarka @MikeM @Thor: I'm stooopid. Of course there are obvious holomorphic sections which are zero precisely on a hyperplane (divisor) of $\Bbb P^n$. Join the point you're at in $\Bbb P^n$ to the point of projection $[0,\dots,0,1]$ and intersect that line with the plane $z_j = 0$ ($j\ne n+1$).

Sebastiano

0

Q: Different frequencies and the same $n-$revolutions

The frequency $\nu$ it is defined as: $$\nu=\frac1T \tag 1$$ where with $1$ indicate one cycle and $T$ the period, i.e. when a material point completes a complete circle. If I want generalize the $(1)$, I write: $$\nu^*=\frac{n}{\Delta t}$$ where $n\in \Bbb N\smallsetminus\{0\}$ is the number of ...

frequency rotational-kinematics

Ted Shifrin

One can check easily that (a) this is a section of the given line bundle; (b) it is holomorphic.

Sebastiano

This is my question....

monoidaltransform

@Sebastiano this is DiffGeometry.SE

Sebastiano

21:27

@monoidaltransform Oh excuse me very much. Then I have wrong the room.

Ted Shifrin

@monoidaltransform, no that was AlgGeo.SE

monoidaltransform

my bad @Sebastiano. I meant this is AlgGeo.SE

Sebastiano

@monoidaltransform Don't worry :-)

Ted Shifrin

@monoidaltransform With your name, there should never be any doubt.

@Balarka @Thor: So this line bundle (which we were earlier calling by its sheafical name) is usually called the hyperplane section bundle because its sections vanish literally on hyperplanes.

monoidaltransform

@TedShifrin everything is diff geo up to canonical iso

Ted Shifrin

21:30

No, that's definitely not so.

Even in my world, it's not so.

Astyx

Trend is that everything is AlgGeo

Balarka Sen

Hi @Astyx

Astyx

hi

Ted Shifrin

Just a few days ago @polite was doing baby analysis. I wonder what happened.

Salut, @Astyx.

Astyx

Salut

Balarka Sen

21:32

I'm thinking a lot about mass-transport lately.

Ted Shifrin

OK, @Balarka, mr topologist, you buy what I just typed (in the two things)?

Balarka Sen

Yeah I agree of course

Astyx

I haven't at all, but I'd love to hear it

Ted Shifrin

LOL, OK. You and your silly orientation reversals.

PDE or probability, Balarka?

Balarka Sen

Closer to probability, but I don't know PDE manifestations really

Thorgott

21:35

I see

Ted Shifrin

@Thor: Was that not seeing directed at moi?

Balarka Sen

@Astyx You remember the grandparent graph, right?

Astyx

Yes

Ted Shifrin

@Thor: In what context are you learning about these bundles? What course?

Balarka Sen

Consider the following automorphism-invariant random subgraph of the grandparent graph: Do not pick any grandparent-grandchild edge. At any vertex $v$, there are three parent-child edges, and one of them points towards the end $\xi$ that has been marked in the grandparent graph. Do not pick this edge. So you have to decide, out of 2 edges, which one to pick. Toss a coin to decide this.

This gives a random subgraph, by tossing an unbiased coin independently and identically at every vertex $v$ to pick out one of these two parent-children edges pointing away from the end.

Thorgott

21:40

topology

Astyx

So if I get this right, you get a bunch of semi lines and one line right?

Balarka Sen

Yes, but no lines, right? There will be no bi-infinite line, at best you will get a ray as an infinite connected component.

Thorgott

this was part of a homework sheet, the point of which was to derive that the powers of a generator of $H^2(\mathbb{CP}^n)$ generate $H^{2k}(\mathbb{CP}^k)$ for $k=1,...,n$ from the Thom isomorphism with the help of noting that the Thom space of the tautological bundle is $\mathbb{CP}^{n+1}$

Balarka Sen

At worst you will have a bunch of segments

So some rays union some segments

Ted Shifrin

Ah, so very much a topology course.

Astyx

21:43

Well, there is a probability event that you get a line, but its negligible (probability zero)

Balarka Sen

Yeah

Astyx

Ok, we agree

LSS

Hey guys, someone is physicist here?

Astyx

What do you call a physicist?

LSS

Well, someone with knowledge of physics in at least undergraduate nivel, to answer a question i have XD

Astyx

21:45

Ask your question and you'll see if someone answers

LSS

Ok thx

It is about EPR paradox

Balarka Sen

So it is possible to argue with mass-transport that in a unimodular transitive graph, you will never have such scenarios. You can never have an automorphism-invariant model of a random subgraph which has infinite connected components consisting of a single leaf and all other vertices degree 2.

Astyx

unimodular = ?

Balarka Sen

Oh sorry, that's a name for "mass-transport holds", $\sum_x f(o, x) = \sum_x f(x, o)$ for any vertex $o$. This was equivalent to some condition about stabilizers.

$|\text{Stab}(x)y| = |\text{Stab}(y)x|$ I think

For any pair $x, y$.

Astyx

Ok ok

Balarka Sen

21:48

Proof: Given any vertex x, send mass 1 from x to the leaves of the components of x in the random subgraph. If there are no leaves, don't send any mass.

This is a automorphism-invariant transportation scheme, because the random subgraph is automorphism-invariant.

Astyx

By the way, by automorphism invariant random subgraph, you mean that $P(\phi(G) =x) = P(G=x)$ where X is the random graph?

Balarka Sen

I think what you mean to say by that is the same as saying, given a collection of edges e_1, ..., e_n, P(e_1, ..., e_n is in the random subgraph) = P(ge_1, ..., ge_n is in the random subgraph) for any automorphism g, yes?

LSS

I am not sure yet what exactly is the paradox they talk about,
It is the paradox of know the position and momentum of one particle at the same type?
Or it is the fact that information will travel faster than light using this point of view?
Or, for example, two particles go out a singular particular, conserving angular and linear momentum, if an experimenter A measure Sx, we know what will be Sx of ther other particle, but the other experimenter B can measure Sz directly from the other particle, so we tecncially know Sz and Sx at the same time

Astyx

Yes that's it

Balarka Sen

Cool yeah so that's exactly it

LSS

21:51

singular particle*

Balarka Sen

No I didn't set it up correctly.

Given any vertex x, send mass 1 from x to the unique leaf of the component of x in the random subgraph, if there is in fact a unique leaf. Otherwise, send no mass.

Astyx

Is that probabilistic? in the sense that you don't consider a realization of the random subgraph but you take the expected value?

Balarka Sen

Exactly!

The expected mass transferred is the correct transportation scheme to look at

Astyx

Ok I follow

Balarka Sen

So the expected mass sent from any vertex has to be at most 1, right? I mean hell the random mass sent from any vertex is at most 1.

Astyx

21:58

Yeah

Balarka Sen

But if an infinite component with a unique leaf appears with positive probability, then the leaf will receive infinite mass some time, from all the other vertices in the component

I guess one thing isn't clear; why is a certain vertex such a unique leaf with positive probability?

if that's argued we're done, because then with positive probability a certain vertex will receive infinite mass hence infinite expected mass as well

Astyx

I mean, the probability of that happening is 1 no?

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$Mathematics$ Mathematics