The h Bar - 2016-07-16 (page 2 of 3)

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4:00 PM

To me, it feels like a silk sliding across my tongue, and has a mild taste quite different from the refreshing minty cool taste of bottled mineral water

and no, you won't die if you drink it unless you drink litres ,causing your body osmotic pressure to increase so much that your red blood cells started to burst

Murder, cannibalism

This what Sofia was warning us about a year ago

Communist script

Sock puppetry

@Secret What kind of bottled water tastes "minty"?!

bonaqua for example

Your taste descriptions make no sense to me at all

@ACuriousMind Mojito brand.

4:02 PM

@0celo7 Interesting idea ;)

Gonna get a Doner today B)

Just have to decide when to not be too lazy to move

...why did you feel the need to post a picture of a Bonaqua bottle?

...why did you post that?

I think that's a European thing

I don't recall seeing it in America?

At least not with that design

because you ask what water taste minty and I am not sure if you have heard the brand bonaqua and hence not knowing about it

This brand is very common in Hong Kong

4:05 PM

Maybe it's a Hong Kong water thing

It also exists in Europe.

Or you have some terminal mouth disease, sorry

@Secret I still don't see how the picture conveys more than you just saying "Bonaqua".

@ACuriousMind I'm in trouble. I still don't understand step 1 in Milnor...

I can't read on until I do...

@0celo7 The step that isn't a step?

4:08 PM

What?

@ACuriousMind (Not knowing that you heard of this brand before the pics is post and 0celo7 started talking about it) based on past experience in h bar, you might ask me something along the lines of "what do you mean by Bonaqua"

@0celo7 You asked before about "step 1" and I told you that it is just one of two conditions that the $\epsilon$ that is being chosen must fulfill.

It's not a "step"

But that's already in the hypothesis, no?

You just choose $\epsilon$ such that (1) and (2) hold.

@0celo7 The hypothesis is that there exists $\epsilon$ such that (1) holds.

(and seriously, h bar should have an option that allow me to refer to a previous message without pinging someone, since I generally reserve pinging for askign important questions)

(I have an idle mode and a question asking mode. The idle mode is well known in giving incoherent random rambles but the question mode is serious. My pings *generally* serves to distinguish whether I am serious on something or not)

4:11 PM

In general, an $\epsilon$ that you choose such that (2) holds might be too large to fulfill (1)

Or not all $\epsilon$s fulfilling (1) might fulfill (2).

How do you prove (2) can be fulfilled?

So you are being told to choose $\epsilon$ small enough so that both (1) and (2) hold.

@0celo7 Uh, since the image of $U$ contains $0$ and is open, there is some closed ball inside it?

But it has to contain the closed ball with radius $2\epsilon$.

oh

That's why you are told to choose $\epsilon$ small enough!

@ACuriousMind I'm doof

Maybe this book is too hard

@ACuriousMind what is the German's reaction to this

4:16 PM

@0celo7 The German reaction is that the GEMA does not allow us to watch it :P

Oh for the love of god

It's about German automobiles

@ACuriousMind How are Germans supposed to listen to music?

Let $H=(\vec{n} \cdot \vec{\sigma}) \otimes I$, and $\rho$ be some arbitrary density matrix for a two qubit system

In indices
\begin{align}
& H=\sum_{i}n_i\sigma_{i,jk}\delta_{lm}\\
& \rho=\rho_{ij,kl}
\end{align}

Then in indices
\begin{align}
[H,\rho] & =\sum_{ikm}n_i\sigma_{i,jk}\delta_{lm}\rho_{kp,mq}-\rho_{jk,lm}n_i\sigma_{i,kp}\delta_{mq}\\
& =\sum_{ik}n_i\sigma_{i,jk}\delta_{ll}\rho_{kp,lq}-\rho_{jk,lq}n_i\sigma_{i,kp}\delta_{qq}\\
& =\sum_{ik}n_i\left(\sigma_{i,jk}\rho_{kp,lq}-\rho_{jk,lq}\sigma_{i,kp}\right)\\

...not this again

@0celo7 There's plenty of things not blocked on YouTube; but besides, I don't think we're supposed to listen to music :P

You're supposed to be working to pay of Greece's debt, I see.

Get to work!

Those spinors won't anticommute themselves.

4:25 PM

By running through the indices $jp=\{00,01,10,11\}$ (too lazy to type), we showed that $[H,\rho]=0$ as expected for no communication

Now to redo this with an arbitary unitary matrix $U$

Let $U\otimes I$
Then
\begin{align}
[U,\rho] & =\sum_{jl}U_{ij}\delta_{kl}\rho_{jp,lq}-\rho_{ij,kl}U_{jp}\delta_{lq}\\
& =\sum_{j}U_{ij}\rho_{jp,kq}-\rho_{ij,kq}U_{jp}
\end{align}
Taking the partial trace of Bob's subsystem
\begin{align}
Tr_B([U,\rho]) & =\sum_{k=q}\sum_{j}U_{ij}\rho_{jp,kq}-\rho_{ij,kq}U_{jp}\\
& =\sum_{j}U_{ij}\rho_{jp,00}-\rho_{ij,00}U_{jp}+U_{ij}\rho_{jp,11}-\rho_{ij,11}U_{jp}
\end{align}
Now since $U$ is unitary $UU^{\dagger}=I=U^{\dagger}U$
Now I am stuck on what I can do to simplify this further

> Almost immediately after Trump named Pence, Hillary Clinton’s campaign blasted out an email Friday calling him “the most extreme VP pick in a generation.”

More extreme than Palin!?

Whew, might have to czech this guy out.

4:40 PM

@0celo7 Indeed they won't :( Still haven't found an elegant way to talk about Majorana spinors

The "reality condition" $\psi^\ast = B\psi$ looks elegant until one realizes that $\psi^\ast$ doesn't actually mean anything on an abstract complex vector space.

it doesn't?

what does it mean

what's $B$ again?

@0celo7 some matrix

I have purged my mind of spinors

@0celo7 Well, it's supposed to be complex conjugation

$UU^{\dagger}=I=U^{\dagger}U$ means
\begin{align}
\sum_{j}U_{ij}U_{jk}^*=\delta_{ik}=\sum_{j}U_{ij}^*U_{jk}
\end{align}
@ACuriousMind I cannot seemed to get anything useful from the unitary property of $U$ to allow me to simplify the $Tr_B([U,\rho])$ and hence showing it vanishes, any advice?

4:42 PM

@ACuriousMind I don't think it's supposed to be that in general (something something charge conjugation)

@Danu The $\psi^\ast$ is supposed to be complex conjugation in this condition. One can relate both the conjugation and the $B$ to charge conjugation in a chosen basis, but that doesn't make the whole thing prettier

@ACuriousMind Oh, and you're supposed to pick a basis and then c.c. the components?

@0celo7 Yeah

Ah, I can see why you find this unsatisfying.

Well, the whole issue starts a bit earlier with the definition of charge conjugation and this $B$, where both complex conjugation and taking the transpose are happily used.

4:45 PM

What is $B$

what matrix is it

@0celo7 Up to varying signs, the matrix such that $\Gamma_a^\ast = \eta B\Gamma_a B^{-1}$.

Now, if that were $\Gamma_a^\dagger$, I would be happy because the adjoint does exist as an operator on the same space basis free, but no, it's conjugation.

Majorana fermions

I generally find them unappetizing

@Danu we...finally agree on something

One can abstractly talk about the existence of an "invariant real form" on the representation, but I cannot prove the existence of that damn form without various matrix shenanigans

quick, explain why so we can disagree again

4:48 PM

@0celo7 You've said this phrase many times

Proof?

@Danu Well, that may be, but they are crucial e.g. for "real supercharges".

Exercise to the reader.

@ACuriousMind Look at all the fucks I give :P

that meme is overused

I don't know that meme

It's a pretty obvious expression though, yeah.

4:49 PM

@Danu is now a mathematician, he does not care about physics @ACuriousMind

And I want to understand their existence in general so that I can understand whether or not Wick rotation is admissible in supersymmetric theories

@ACuriousMind If you can't Wick rotate it, it's not worth looking at :D

(strings 101)

What even is wick rotation

Formally

Not $t\mapsto -\mathrm it$.

Real life

Or should I say imaginary?

@Danu can you help me on a no communication proof thingy?

4:51 PM

@Secret He will not.

@Secret Nope

Sawry

Gotta study for exams

Are you taking any exams @ACuriousMind?

@Danu Well, there is the fact that if there are self-dual forms in Minkowski, there are none for the Euclidean theory of the same dimension. Yet SUGRA theories with self-dual forms seem to worry no one exactly, but there's the cop-out that the self-duality is imposed only on the level of the e.o.m. anyway

But there's no such cop-out for the Majorana condition.

@Danu Maybe I'll do "topology of singular spaces", but I don't need any more credit points

@JohnRennie maybe you can help me?
Acuriousmind is too busy on majoranas to answer

I would need my profs to get on with giving me a master's topic, though :P

Or not. I'm not actually in any hurry

@Secret I doubt my knowledge of QM is good enough to be of any help

4:56 PM

ok nvm...

@ACuriousMind Mmm

How many courses do you need for your degree?

Could this be helpful, @ACM? maths.ed.ac.uk/~jmf/Teaching/Lectures/Majorana.pdf

@Danu Lectures worth 60 credit points.

@ACuriousMind OK

@Danu I'm already reading that :)

Are they usually 9 a pop for you guys, too?

Or 6?

5:01 PM

@Danu "Small lectures" (once a week) are 4, "Big lectures" (twice a week, usually with additional exercises) are 8

Hmkay.

Here it's 9/3

And there are some courses in the holidays that are one week of eight hour lectures that give 2

The 3 are usually student-presented courses ("seminars")

@ACuriousMind Yeah, I had one of those too, this semester.

@Danu Ah, our seminars give 6, usually

Hmkay.

Next semester I'm doing one on characteristic classes :D

5:03 PM

@0celo7 can you help me on that problem?

The only further restriction on the 60 CP we need is that two of the courses need to be so-called "core courses", basically the intros to their fields (GR, QFT, condensed matter, astro, whatever...)

Similar-ish for me

We just have diffgeo/math. QM/math. stat. phys. as core courses

@Secret No

guess I have to wait until acuriousmind's discussion has finsihed, or the other quanutm guys came by, as now that I have the index issue solved, my only way to massage the question into one that can fit the PSE policy is now gone

(i.e. since the index problem is solved, I cannot use DanielSank's advice to massage it into a main site question)

@Secret What you are doing makes no sense. Why are you taking the commutator w.r.t. to an unitary matrix?

The commutator is the "infinitesimal transformation" under the generators, i.e. self-adjoint matrices. Unitary matrices/"finite transformations" act by conjugation, not commutation, and the Wikipedia page on the no communication theorem already has the proof in density matrix language

As so often, I have no idea what you are doing.

@Danu Every time someone says stat.phys. I wished I actually know anything about it...

5:14 PM

@ACuriousMind I want to take a course on it but no tiiiiiime

Maybe in grad school

Previously daniel and I have worked together to show the expression $\dot{\rho}=-i[H,\rho]$

...

Using what you said , you mean that only the hamitonian can be used to compute the time derivative of the density matrix (i.e. $\dot{\rho}$) and you cannot compute that using the evolution operator (defined as $exp(-\frac{i}{\hbar}Ht)$)?

My course on it was horrible. I passed the exam with a very good grade without having any idea about statistical physics. The only thing I di was compute integrals and sums correctly.

@Secret What?

Why would you think that the evolution operator computes the time derivative?

because it gives the state at some later time...?

@Secret It does. You have $\rho(t) = U(t)\rho(0)U(-t)$.

I thought $H$ gives the time derivative

5:17 PM

If you take the time derivative of this equation, what you get is $\dot{\rho} = -\mathrm{i}[H,\rho]$.

@ACuriousMind I thought that was true for any time-dependent operator

Heisenberg formalism or something

@0celo7 The sign is different

sign shmime

Ordinary observables evolve as $A(t) = U(-t)A(0)U(t)$.

oh wow I really should go get food

5:21 PM

What's an extraordinary observable

@Slereah You want me to make another joke about your mom? :P

Rude

Sorry :(

5:35 PM

@ACuriousMind ...or not :D

@Danu Bah, you really have become jaded, haven't you? :D

@ACuriousMind Hehehe

\begin{align}
& \rho(t)=U(t)\rho(0)U(-t)\\
& = \sum_{jlmn}U_{ij}(t)\delta{kl}\rho_{jm,ln}(0)U_{mp}(-t)\delta_{nq}\\
& = \sum_{jmn}U_{ij}(t)\rho_{jm,kn}(0)U_{mp}(-t)\delta_{nq}\\
& = \sum_{jm}U_{ij}(t)\rho_{jm,lq}(0)U_{mp}(-t)
\end{align}
Taking the partial trace wrt Bob's
\begin{align}
Tr_B(\rho (t)) & = \sum_{l=q}\sum_{jm}U_{ij}(t)\rho_{jm,lq}(0)U_{mp}(-t)\\
& = \sum_{jm}U_{ij}(t)\rho_{jm,00}(0)U_{mp}(-t)+U_{ij}(t)\rho_{jm,11}(0)U_{mp}(-t)\\
& = \sum_{jm}U_{ij}(t)(\rho_{jm,00}(0)+\rho_{jm,11}(0))U_{mp}(-t)\\

It is true that $U(t)U(-t)=I$ but the indices does not seemed to do that for some reason, need to check the workings again

@Slereah I wanted to ask you something

Are there any messages from me @ you

My new profile image looks like JD, when I saw myself in the thread I though "oh god he's here"

chat.stackexchange.com/…

There are 48 pages of that

5:48 PM

I like talking to him I guess

@ACuriousMind Diagram 5 in Milnor is seriously unhelpful.

Never mind it's explained on the next page

user54412

@JohnRennie You're not being creative enough ;)

Correction: Taking the partial trace wrt Alice to get Bob's
\begin{align}
Tr_A(\rho (t)) & = \sum_{j=m}U_{ij}(t)\rho_{jm,lq}(0)U_{mp}(-t)\\
& = U_{i0}(t)\rho_{00,lq}(0)U_{0p}(-t)+U_{i1}(t)\rho_{11,lq}(0)U_{1p}(-t)\\
& = U_{i0}(t)U_{0p}(-t)\rho_{00,lq}(0)+U_{i1}(t)U_{1p}(-t)\rho_{11,lq}(0)\\
& = \delta_{ip}\delta_{ip}(\rho_{00,lq}+\rho_{11,lq})\\
& = \delta_{ii}Tr_A(\rho(0))\\
& = Tr_A(\rho(0))
\end{align}

...somehow I am missing a summation on 0 and 1 to collapse the UU* into deltas

6:06 PM

@0celo7 wot

@Secret For the last time, the Wikipedia page already has the density matrix version of the no-communication theorem with its proof. I have no idea what you are doing, the only thing that's clear to me is that you are not particularly good at doing computations. There are issues with your indices from the first line, but instead of noticing that and stopping, you just chug on. And then indices vanish into thin air. You need to sit down and think about what you're doing.

The one sacred rule of index computations is: The number of free indices on both sides of an equality sign has to be equal. Always.

As soon as that is untrue, you must stop.

I just cannot see where I am going when everything is in indices. When I do the proof all in matrices, I basically end up what the wikipedia page shown

But a proficient person has to be able to derive the same thing using different languages, which is why despite knowing wikpedia has a no communication proof, I am trying to derive that in the index language to train myself

Except that I keep making silly mistakes here and there and I cannot really see where I am going

(You are also not the first person who pointed out the "You need to sit down and think about what you're doing". For the …

so my issue is: I know what the problem want me to solve, I know what is the goal I am trying to achieve, I know the method I need to use to reach the goal, but somehow, I keep losing track in the computation

6:27 PM

@ACuriousMind Correct.

@ACuriousMind Ignoring step 1 in this proof, I'm liking the argument a lot

It's very intuitive

Taking some work to go through the details

@ACuriousMind Deformation retractions are never smooth, right?

Not in general, but there's plenty of defo rets which are smooth.

Or at least they don't have to be...?

E.g., def. ret. of $\Bbb R^2 - 0$ to the unit circle.

That's pretty smooth.

Never mind, I was confused by something in the book

@0celo7 The identity looks pretty smooth to me :P

6:32 PM

proof

That was a dumb, imprecise question

Ignore it.

I guess once you have a topological deformation retract so that the retract is itself smooth (an obstruction here is that the subspace you retract to is a smooth submanifold - this is in, e.g., Hirsch and is a cool exercise), you can approximate that arbitrarily well by a smooth deformation retract.

In the uniform norm, I mean.

But this is off-topic.

@ACuriousMind I guess my real question is: is the retraction Milnor defined in this proof smooth?

It doesn't seem so. He argues that it's continuous at least (so it is actually a retraction)

Because it coincides on the boundaries of the cases

Curious, what is the thing we're talking about?

But that's not enough to guarantee smoothness, right?

@BalarkaSen Morse theory

One of the first results in Milnor

@0celo7 Why does that matter?

6:39 PM

on mobile, why does what matter

Whether it is smooth or not

I am curious.

@0celo7 Ah. I should flee then.

@BalarkaSen it's just differential topology at this point, no Riemannian geometry yet

@0celo7 What pg no?

6:46 PM

@ACuriousMind probably stupid question, but is the ratio x/y defined as x->0,y->0

I think it depends on the direction you approach the origin, no?

@0celo7 Not so much the direction as the speed - consider $x_n=1/n,y_n=1/n^2$ (and swapped).

(This is not a question)
Let me think... so for this expression:
\begin{align} Tr_A(\rho (t)) & = \sum_{jm}U_{ij}(t)\rho_{jm,lq}(0)U_{mp}(-t)\end{align}
I have free indices i and p and l and q, which is consistent with the fact that the resulting object is an operator with a matrix representation doubly indiced (hence order 2)
Therefore if I let W be the result of the sum it will look like $\sum_{jm}U_{ij}(t)\rho_{jm,lq}(0)U_{mp}(-t)=W_{ip,lq}$

So if I want to take the partial trace wrt Alice (i.e. space 1), I will need to sum up $ip=00$ and $ip=11$

Finally I can RIP and move on to the next page

@ACuriousMind So...Milnor asks the reader to verify that those functions are continuous as xi to epsilon and eta to 0, but it seems that limit is undefined

Page 19

Oh...maybe ui goes to zero as well

Maybe that limit is indeed defined.

I don't know how to prove it.

He says within the region $\xi \in [\epsilon, \eta + \epsilon]$. Saying $\xi \to \epsilon$ and $\eta \to 0$ are then not two different things.

Why is that limit defined then

6:53 PM

But ignore me, probably that's garbage. I haven't really read it.

Sorry, carry on.

It's because the ui that are multiplied by st also get sent to zero

Mu=0 corresponds to those latter ui=0

Not mu, I mean eta.

So ui and mu are coupled limits

@ACuriousMind my analysis isn't good enough to compute this limit

Never mind, got it.

Thanks

> (From wikipedia) If the density operator P ( σ ) is allowed to evolve under the influence of non-local interactions between A and B, then in general the calculation in the proof no longer holds, unless suitable commutation relations are assumed.

@Acuriousmind For a nonlocal hamitonian, does it mean it is $H$ such that it cannot be written as a tensor product of operators in the two hilbert spaces

i.e. Is all possible forms of nonlocal H cannot be decomposed into a sum of tensor products. I.e. has the form $H\neq \sum_i \sigma_i \otimes \tau_i$ where $\tau$ and $\sigma$ are components of the hamitonian that act only on Alice's and Bob's Hilbert spaces respectively,yet H will act on both hilbert spaces?

I proved its bounded above by a function that goes to zero

Taking baby analysis really helped me

Typo: I mean nonlocal interaction represented by U, not H

7:13 PM

@0celo7 I had faith you would :)

The reason I am guessing that is that if we have a $U=\sum_i \sigma_i \otimes \tau_i$ then when computing $\rho(t)$ we get
\begin{align}
& U(t)\rho(0)U(-t) \\
& =(\sum_i \sigma_i (t) \otimes \tau_i (t))\rho(0)(\sum_i \sigma_i (-t)\otimes \tau_i (-t))\\
& =(\sum_i (\sigma_i (t) \otimes I)(I \otimes \tau_i (t)))\rho(0)(\sum_i (\sigma_i (-t) \otimes I)(I \otimes \tau_i (-t)))\\
& =\sum_{ij} (\sigma_i (t) \otimes I)(I \otimes \tau_i (t)))\rho(0)(\sigma_j (-t) \otimes I)(I \otimes \tau_j (-t)))\\
\end{align}

@Secret "non-local" here means just that it is not the tensor product of two operators on the individual spaces. All operators on finite-dimensional spaces can be written as sums of tensor products of operators.

@ACuriousMind OTOH the Whitehead lemma is proving to be a difficulty

I don't know what the map l is supposed to be

Give me a moment to figure it out

I don't want to be @Obliv ;)

7:26 PM

@ACuriousMind
So I am guessing the hamitonian that describes how the two qubits causing the energy level splitting of each other
$$H=\frac{\omega}{2}\vec{\sigma}\cdot \vec{\tau}=\frac{\omega}{2}\left( \sigma_x \otimes \tau_x +\sigma_y \otimes \tau_y +\sigma_z \otimes \tau_z\right)$$

found in susskind will be an example of a nonlocal operator?
Is there a stronger theorem than no communication theorem (because wikipedia said the theorem breaks down for these operators) that take account of these operators to ensure no superluminal commuciation possible (i.e. even if interactions described by…

@Secret Stop thinking about "superluminal". Ordinary quantum mechanics has no notion of the speed of light.

The two qubits with that Hamiltonian are close so they actually can interact in that fashion.

The no-communication theorem is not about the absence of non-local interactions. It is about one measurement on one part being unable to influence the other part of the entangled stated.

@ACuriousMind I cannot come up with l that works. do you know what it should be?

He says "similar formulas"

I don't know if that means the exact same, or flipped somehow

I'm trying to engineer the homotopy inverse by hand now and it's not working.

@0celo7 My intuition would be to replace the $2-2t$ by $2t-2$, and perhaps switch the behaviour in [0,1/2] and [1/2,1]

That's mine, too.

ahhhhh

I might have figured out the trick

You need that k(u)=u because of the way the attachment is defined

in English Language & Usage, 4 hours ago, by TIPS

Fun fact: we're hitting memory ceilings on the Stack Exchange network due to the insane interest in Pokemon. https://t.co/cVr0NXDoKL

Tweeted by Nick_Craver on July 15, 2016 at 12:48 PM

7:40 PM

lol

It's not even that fun

Have I mentioned that I hate topology

@ACuriousMind
The no communication theorem showed how measurement on one part of an entangled system cannot influence another

But in light of the operators mentioned by wikipedia that mentioned the proof will break down (thus does not cover all possible cases of influence between two far away systems that interact nonlocally yet are not entanglement), what theorem can explain why we never observed experimentally a case where a nonlocal interaction can be used to transfer information without needing a classical channel?

@Secret It's a formal assumption in quantum field theory that operators at spacelike separation commute, which may be shown for certain theories, and must so far be taken on faith for others.

I see

Why the hell doesn't he just wrote out what l is

@ACuriousMind No clue how to do it :(

7:47 PM

@0celo7 He gave a reference "for further details".

Have you seen this lemma before?

Perhaps? Can't remember where if I have, though

[Quantum self study]
Issues related to my understanding on no communication theorem and nonlocal interaction discussions are now resolved.

Susskind's book chapter 7-8 (which basically went through von neumman measurement) caused me to have an interpretation related to measurement. However, my interpretation might be nonsense unless I can make sure I understood this paper fully (since the paper explain why and how entangled states are fragile, which is an important ingridient in accessing whether my interpretation and possible experimental detection make sense)

I also read the article on quantum decoherence, and my interpretation relies heavily on the concepts in the decoherence theory

(All of the above to be checked after fisnih reading the noncommuntative probability PSE post)

8:11 PM

@ACuriousMind I'm at the store, if you're bored could you look up that paper please :P

@0celo7 I'm kinda bored but I am already combatting that boredom by other means :P

1

Q: Should a question be voted down for poor formatting or editing?

My question is, whether one should downvote a question just because he dosen't understand it or he finds the O.P's formatting or editing as poor. I will like to reason something with my example.The place I live is 80% natural environment.Only recently there has been a craze to use smartphones,and...

discussion editing down-votes

8:40 PM

@ACuriousMind This paper is crazy advanced :/

@ACuriousMind I think he proves it using homology or something

you chat in here from the store?

@skillpatrol No

I went to the store and was thinking about it

I wondered if ACM would be kind enough to figure it out while I was at the store.

sure, sure

apparently it's nontrivial?

nice avatar, btw

8:46 PM

from an album that came out yesterday

or sometime in the recent past

I just got it yesterday

@ACuriousMind Nope, that paper is way too long and seems unrelated

@0celo7 Well that's not my fault. (You did look specifically at the lemma Milnor points to, right?)

@ACuriousMind I cannot find that lemma in that paper

Just a bunch of cohomology of cell complexes stuff

How can you fail to find "lemma 5"?

lol

I did not see that

Aha :D

8:51 PM

oh, that explains it too

the notation is totally different

@ACuriousMind I was looking through the whole paper...

user image

good lord

what is a centroid?

@ACuriousMind help I don't understand old-timey algebraic topology

I don't either!

no idea what a "centroid" is

:D

Seems like you found another rabbit hole :P

how did you do this proof when you read it

Have we not established that I did not read this that closely?

8:54 PM

Well how closely did you read it?

oh, I see what's going on

I think I looked at that map, thought "seems reasonable" and moved on

Milnor defines a $k$-cell in $\Bbb R^k$

so the centroid is $0\in\Bbb R^k$

Whitehead might be using some more abstract definition

centroid of a cell?

@ACuriousMind Thanks for pointing me to lemma 5

user image

centroid of simplices are what I know exist

8:56 PM

Milnor, was that so hard to write?

@BalarkaSen Is that another word for barycenter?

yes

Also, do you have some script that alerts you when we start talking about Milnor here? :P

The same whiteheaad who worked with Russell on principia? @0celo7

lol, no

8:57 PM

I know what a barycenter is

ok whitehead, give me the magic

wonder if whitehead was an albino

@skill no, different guy

or maybe had albino ancestors

a k-cell is just a k-dimensional disk

Thanks @BalarkaSen

I know what a k-cell is

I don't think that's how Whitehead defines it though

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