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3:01 PM
@0celo7 No idea what you're talking about.
@Danu do you know why you can dominate a smooth manifold by a CW complex?
@Secret they (GR+QM) are currently independent mathematically and (my understanding) that nearly causes contradictions in analysis eg at black hole boundaries etc, not an expert on this. as the paper is in line with, there are new experimental tests being proposed/ even implemented but so far nothing measurable
@0celo7 The $K$ is a "true" direct limit, but it is "almost" a union, once you have understood why he can write $K_i\subset K_{i+n}$ for positive $n$.
What is there to understand?
You just attach cells as you increase $i$
@0celo7 Deformation retract?
3:04 PM
I see, hmm, I think johnrennie and slereah might know more about the black hole analysis, might ask them later
@0celo7 Doesn't Milnor say this: Embed it in Euclidean space, choose a tubular neighbourhood
@ACuriousMind I feel like I've said that a billion times
I think I have a sketch of a proof for smooth, compact manifolds.
I clearly do not understand why that works.
@0celo7 Ah, okay, that's one version.
Then $K$ is the initial $K_1$ "with all cells attached".
3:06 PM
I understand that
Let's move on to the dominated thing
Formally you need to take the disjoint union of all the $K_i$ and then identify all the points that are the same under the inclusions $K_i\subset K_{i+n}$.
@ACuriousMind yes, as in the direct limit of abelian groups I outlined above
...so why did you ask me what $M$ and $K$ were?
Oh, right
I don't know what the topology of $K$ is
@Danu I'm assuming so, $M$ should have the homotopy type of some CW complex
so I'm assuming $M$ has the homotopy type of its tubular neighborhood, but I'm not sure
this proof is insane
@0celo7 The disjoint union carries the obvious topology - just the disjoint union of topologies. The quotient carries the usual quotient topology.
3:12 PM
@0celo7 I just don't know what you mean by "dominated"
That's why I'm asking if it means "defo retr".
@vzn : Sorry, I think there's outright errors in the paper too. This is the arXiv version: arxiv.org/abs/1602.06377
@Danu If $f:X\to Y$ and $g:Y\to X$ and $f\circ g\sim\mathrm{id}_X$, then $X$ is dominated by $Y$.
@JohnDuffield so maybe write the authors about those :)
@0celo7 $f\circ g$ is a map from $Y$ to itself...
@ACuriousMind I clearly meant $g\circ f$
3:16 PM
Well, you might also have meant $\mathrm{id}_Y$!
So it's not a defo retr
no, it's "half"
And you're trying to prove this to show that (co)homology and homotopy is finitely generated?
Why are you trying to prove this?
@Danu Morse theory
3:19 PM
What do you need it for?
I'm assuming you want some induced map to be injective
No, I want to apply a theorem of Whitehead.
@Danu That is not so easy to answer - Milnor's proof of a theorem uses a theorem of Whitehead that has that domination as a hypothesis.
OK. This is not easy to prove.
I think
@ACuriousMind So you agree with me that this is a hard proof?
Wait, for compact
I think it might not be too hard
3:21 PM
@Danu No, the $M$ needs to be dominated, and that is not compact
@vzn : I'd rather spend my time writing something else. Which reminds me, I must get on. I only looked in for a break.
@ACuriousMind What is not compact?
The manifold.
@0celo7 Yes
3:21 PM
@JohnDuffield how's the physics in this room today
@Danu The $M$ - it might be an infinite union of $M^a$, so it doesn't need to be compact.
The proof is not going to be easy
@Danu story of my life
Just don't try to prove it, lol.
If Milnor uses the result without proof
That tells you he does not want to bother showing it
Which means it's a bother
I don't even know how to apply the thing he uses without proof, lol
I can't prove you can even use the thing he uses without proof
and he does give a "Milnor style" proof
@ACuriousMind that makes me feel somewhat better
@ACuriousMind @Danu According to El Math Chat, an open set of $\Bbb R^n$ is an infinite $n$-dim CW complex.
So the tubular neighborhood is a CW complex.
This retracts onto the manifold, which is a domination.
3:27 PM
...was it not clear to you that the tubular neighbourhood was supposed to be the dominating complex?
No, not at all.
I asked about 5 times but I got no answer.
Sorry, I thought your issue was proving the tub. neighbourhood is a CW-complex, or that it deformation retracts onto the manifold
So much talk, so little communication :D
@ACuriousMind Apparently you just take the open balls as your cells.
@Danu Yes, that is a recurring theme...
3:29 PM
I thought the whole point was to get a finite CW complex, haha.
@Danu That's where Morse theory comes in!
Yeah, it's supposed to be finite.
Well, whatever.
It's not finite
it depends on the number of critical points of the Morse function
@ACuriousMind So.
I think you can get finite in general.
Homeomorphisms to finite CW complexes.
But I will not bother you with it.
Maybe Milnor will prove that @Danu
@ACuriousMind So, now about this $g:M\to K$.
Why does it induce isomorphisms of homotopy groups in all dimensions
3:32 PM
@0celo7 I don't think so
36 mins ago, by ACuriousMind
@0celo7 You need to show that $g$ induces isomorphisms on homotopy. It should follow from all the $g_i$ doing that, so I think you need to show that two maps $S^n\to K$ that are homotopic in $K$ already are homotopic in some $K_i$.
please be more specific
I can't because I have not done that proof.
4:12 PM
@0celo7 ACM is talking about the cellular approximation theorem, like I said in the math room.
If $X$ is an arbitrary CW complex, $\pi_n(X)$ is isomorphic to $\pi_n(X^{n+1})$.
$X^{n+1}$ being the $n+1$-skeleton.
I am?
The $K_i$ are not the skeletons here, @BalarkaSen
Inclusions of $K_i$ to $K_{i+1}$ are cellular, right?
That suffices.
Each $K_i$ have dimension bounds.
@BalarkaSen yes
4:18 PM
What is the relation between the dimension of $K_{i+1}$ and dimension of $K_i$? Dim of the former greater than dim of the latter?
@BalarkaSen $K_{i+1}$ is $K_i$ with some $\lambda$-cell attached, $\lambda\leq n$.
That $\lambda$ can be anything? Not a cell of dimension higher than the cells in $K_i$?
@BalarkaSen Yes, it can be anything between $0$ and $n$.
Yikes. Then, I guess, there's some work involved.
@ACuriousMind perhaps more than one
depends on the number of critical points in the preimage
4:22 PM
Ah, yes, that too
@ACuriousMind I am not familiar with the context, but this then becomes an ambitious claim. I can take S^n, pinch an equator, get two spheres and send those guys to two different places. Stuff like those maps may mess me up.
But I should shut up because I don't know what's going on anyway.
@ACuriousMind I take it you didn't even read this proof when going through Milnor
@0celo7 Well, I read it, saw he referenced things, and then moved on. We already established that I did not read this on the level of understanding the proofs, why do you keep coming back to whether I "read" a particular part or not?
If you're going to keep doing this, just assume I didn't read Milnor at all.
5:23 PM
@ACuriousMind I came back to this answer I awarded a bounty to some time ago. One question pops op in my head: Why should we allow all projective representations?
@Danu Because the true space of states is the projective space, and all a physical transformation therefore needs to be is a ray transformation
@ACuriousMind It's quite natural, yes, I agree.
But it's not forced or anything like that
Once you are on the projective space, the better question is: Why do we get away by looking at unitary representations?
@Danu I'm not sure what you mean by "forced"
It's not inconsistent or anything like that not to allow some
But it's very natural so it's kind of a silly question
@Danu In many cases, once you have the fundamental, you have effectively allowed all since all other representations arise as subrepresentations of tensor products of the fundamental
5:27 PM
All of them do? Interesting
I think that is true at least for SU(N), and possibly for a larger class of groups
Hmm, could be.
Somethingsomething Young tableaux
5:55 PM
A: My star will explode as a supernova. What can I do in order to ensure that my planet survives that?

celtschkCreate a pocket dimension and move your planet in there. During the nova, make the entry to the pocket dimension (that is, the size of it as seen on the outside) small enough that the energy that enters is no larger than the planet can handle. After the nova, you can widen the entry again and get...

Is this answer (the 2nd answer) physically plausible?
@Mostafa Uh..."pocket dimension" is not a physical notion.
@ACuriousMind Yet 15 upvotes! :)
I think his idea is to reduce the *interaction cross section* of the planet with what is coming to hit. Basically the same idea of the cloaking metamaterials in the first answer.
Sorry if it sounds very naive (I'm not familiar with GR), but I was thinking, Isn't it possible to somehow bend the space around the planet in order to direct the energy and mass *around* that?
@Mostafa Well, it's worldbuiling.SE, not what-if-physics.SE
A pocket dimension is a totally viable solution for a (soft) Sci-Fi story
Although I'm not exactly sure how that is "worldbuilding", but I never really understood the scope of that site anyway, I guess
Is $\langle U\rangle=\langle \psi \lvert U\rvert \psi \rangle$ physically meaningful, given that we cannot really directly see the time evolution of a state?

Are there other examples of operators where the notion of expectation value is not a physically sensible notion?
6:08 PM
@Secret Time evolution is not an observable (it is not self-adjoint), so taking its expectation value makes no sense.
@Mostafa I...don't think so, but maybe other people more proficient in weird GR stuff can say for sure
@Slereah ^see above
6:23 PM
@ACuriousMind Suppose we have two $n$-cells and we identify them along their boundaries. This is $S^n$. Why?
@0celo7 Because the map sending each cell to a hemisphere and the boundary to the equator is a homeomorphism?
@ACuriousMind Clearly. Why?
I don't understand the question.
(I have no idea how to prove the general case rigorously as my current level of topology is not high enough) There's an intuition behind that, think of the case of two discs, if you deform and glue them along the circular boundary, you get a sphere
similarly if you glue two $D^n$ discs along their $S^{n-1}$ boundary, you get a n sphere, which is $S^n$
@ACuriousMind Why is it a homeomorphism?
6:27 PM
@0celo7 How exactly you show that depends on what exactly your definitions of $D^n$ and $S^n$ are. And I'm not going to do it for you.
I don't even know how to go about proving it.
What's the idea?
I'm convinced the boundary map to the equator is a homeomorphism.
It's just the identity.
Now, as for the others...perhaps projection?
Yes, that works.
Now how to prove continuity...
What is le question?
Glue two n disks along the boundaries
Prove that this is a sphere
Along the boundaries by what homeomorphism?
6:31 PM
It is true for arbitrary homeomorphism, but less obvious than identity.
Q: About merging of questions

lucasThis question of mine merged with this one while the questions were pretty different with each other. And this answer of Gennaro Tedesco (before is edited by me, even after that!) is completely meaningless. Why must those questions be merged? I could explain clearly why those questions were m...

@0celo7 Write $\Bbb R^n$ as union of the unit ball and closure of it's complement. Then compactify.
That seems really elaborate and I don't know what it means.
Too bad.
@Mostafa Google wheeler bag of gold spacetime or something similar. If you had access to exotic matter geometries like this could be produced.
6:50 PM
@0celo7 I see; you were trying to prove gluing two arbitrary $n$-cells along boundary by identity is $S^n$, not that $S^n$ admits such a cell structure.
In which case, construct the obvious continuous bijection as ACM suggested and note that continuous bijection from compact to Hausdorff is a homeomorphism.
But I guess you already figured it out.
@BalarkaSen It's far from obvious, which is the issue.
I'm not sure how to show continuity.
I've been trying to get out of the habit of writing pseudo-answers in the comments. But I'm not happy keeping my mouth shut, so all of a sudden I'm writing a lot more real answers.
Don't know if that is a good or bad thing.
Probably good
@dmckee It's good if before you were just too lazy to leave a proper answer. It's kinda bad if you were leaving comments because you thought the question didn't deserve more effort on your part.
7:30 PM
do whatever best fits your schedule :)
7:43 PM
:chuckles:: @CuriousOne I see what you did there.
fighting fire with fire?
8:12 PM
eg. "Thank you for your time"? No... thank you for wasting ours.
8:23 PM
@0celo7 : I am reminded of what my nephew Owen said. He and his fiancée did a physics degree. That's kind of how they met. They came out of it feeling kind of jaded, saying there wasn't much physics, it was all maths. This chatroom seems somewhat similar. As you doubtless know. You not being a physicist and all. ;)
Why does Qmechanic not mark questions as duplicate while he/she completely sure about duplication of them? Does he/she respect to trusted users votes more than his/her own opinion? Does he/she respect to the main policy (democracy) of the SE? I don't know his/her reason, but I respect to him/her very much! physics.stackexchange.com/questions/234084/…
One thing I do noticed is that of the maths, topology tend to be the most frequent topic.

However because I am currently self studying basic quantum and not QFT or GR or other related things, I have not revise the relevant material thus most of those conversations are incomprehensible to me
@JohnDuffield what do you want to talk about?
A: Why is the application of probability in QM fundamentally different from application of probability in other areas?

Valter MorettiThe theory of probability used in QM is intrinsically different from the one commonly used for the following reason: The space of events is non-commutative (more properly non-Boolean) and this fact deeply affects the conditional probability theory. The probability that A happens if B happened is ...

How to understand a huge wall of text?
Read the whole wall of text, then copy just the formulae in order, do some computation, think about it and then read the key parts again
The point is: I am much better understanding someting if it is presented vertically
This is why I like matrices, because they are all vertical presentation of formulae
8:44 PM
@skullpatrol : I'm afraid I've got to go. Sorry.
1 hour later…
9:46 PM
@lucas That sounds about right, but you could always ping qmechanic on chat and ask.
@ACuriousMind For the love of god, I'm stuck again
Since when does Poincare duality work on noncompact manifolds and what does it have to do with Morse functions -.-'
Well, maybe it works with compactly supported cohomology.
@ACuriousMind Apparently if you have a Morse function on an $n$-manifold with 3 critical points, then the indices must be $0$, $n/2$, and $n$.
And apparently this is due to Poincare duality.
@ACuriousMind Also, is there any link between homotopy type and homology?
@0celo7 Sure, (co)homology is homotopy invariant and thus only depends on the homotopy type
Oh, right.
As for the rest of my questions?
@0celo7 Yes.
@lucas Many mods are wary of singlehandedly closing a question, and prefer to let the community decide. It does happen on straightforward cases, but it's often best to let the reviewers choose. (Just speculating on the matter.)
10:01 PM
I think Milnor means for compact, even-dimensional.
Otherwise $n/2$ is nonsense.
@ACuriousMind @BalarkaSen What does Poincare duality have to do with the indices of Morse functions?
Hell if I know. I think the critical points are related to betti numbers.
Wtf Milnor
relevant keyword is probably Morse homology
He didn't say anything about that
@0celo7 So any Morse function on a odd dimensional manifold has to have # of critical points unequal to 3? Nice fact.
10:06 PM
@BalarkaSen I'm not sure.
Milnor does not prove the statement, I do not know if the converse is true.
@0celo7 If the manifold has a Morse function with three critical points on it, then you must have that it is homotopic to a CW-complex with 3 cells. But taking cellular homology and Poincaré duality, you see that for each $k$-cell, you need a $n-k$-cell, and you know that you need to have one 0-cell to generate the 0th homology, so the only consistent option is to have cells of dimensions 0,n/2,n, and thus indices of that dimension.
Oh, good point, ACM.
I was thinking about something like that but couldn't make it pan out.
I don't know cellular homology.
What does Poincare duality say for cellular homology?
The same as for all homology theories.
The $k$-th cohomology is isomorphic to the $n-k$-th homology.
I don't see how that applies...
Well, isn't the $k$-th cohomology isomorphic to the $k$-th homology?
10:10 PM
@0celo7 The k-th cohomology is the dual of the k-th homology
But they're all $\mathbb{Z}$ or zero here, so the distinction doesn't matter
The point is that the three cells make exactly three (co)homology groups non-zero. Two of those need to be the top and bottom ones, and duality forces the third to sit in the middle.
@ACuriousMind Not sure what you mean by that. Dual as in dual modules? Only for field coefficients though.
@BalarkaSen Hm, I thought of its construction by taking the dual of the chain complex. Not very helpful or precise in this case though, you're right.
Okay, so is the instantaneous velocity of a Brownian motion process undefined?
@Mikhail What exactly do you mean by that?
Are you thinking about the typical path of Brownian motion being non-differentiable?
10:21 PM
I'm stumbling upon something like this
I'm tying to figure out how fast my camera needs to photograph to capture diffusive motion, but I can't quite get a handle on how fast the particles are moving
Ahhhh, yes, the Brownian motion will not help you to get an answer to that, the typical path of Brownian motion indeed has no notion of velocity.
(Yes, Brownian motion/the Wiener process is weird)
But it seems like the problem is that the autocorrelation function doesn't converge to a single mean
@ACuriousMind Uhhhh, crap, what?
I still don't see how the cohomology relates :/
But wait - isn't Brownian motion suppoesd to be thermal motion? The velocity should just obey the M-B distribution
Cells give cohomology groups. This is the content of the theory of cellular cohomology/homology.
10:25 PM
We know that the homotopy type is that of a CW complex with 3 cells.
We pick one of them.
@0celo7 That is, in my opinion, inevitable if you study something which is going to use algebraic topology without learning algebraic topology.
@ACuriousMind So, its also the motion of small particles, so you can show that given the specific information about the particles, they will become Brownian as time goes up. But their exact velocity then depends on their physical characteristics, which is no good for me.
Not that my opinion matters. Feel free to ignore that last thing.
@Mikhail Googling gives me lots of results on the average velocity of Brownian motion, and some rather recent experiments that did measure an "instantaneous" velocity for Brownian motion.
@BalarkaSen No, I know.
I'm upset my professor wants me to read this
I clearly cannot
10:32 PM
I'd talk to him if I were you.
@0celo7 Call him out to a fist fight
He literally said it's "trivial"
What he did he say is trivial, precisely?
@ACuriousMind Yeah, but notice that the instantaneous (predicted) measurements are object specific (bead at a temperature). On the other hand the averaged rms velocity is supposedly not related to the instantaneous one.
@BalarkaSen The first 40 pages of Milnor
10:34 PM
@Mikhail ...so you expect there to be a non-object-specific not-temperature dependent prediction for the instantaneous velocity?
And yes, the averaged velocity is a property of the statistical process, the instantaneous velocity probably not so much
@ACuriousMind I think the divergence in velocity doesn't preclude some kind of velocity estimation, although perhaps not the mean velocity
Well, clearly, to him (?). I have talked to it about many people (who are genuine topologists) and while they did say with appropriate prerequisites it's easy to read (same applies to any of Milnor's expositions, or so I heard), no one ever said it's trivial. Instead, I have heard in the first 40 pages Milnor lays foundations on one of the deepest and most useful tools of topology.
He's not even a topologist.
Perhaps he thinks I'll give it a cursory reading like @ACuriousMind did.
@0celo7 Probably it's been a decade since he's last read it and he doesn't actually remember what's in it :P
Talk to him.
10:39 PM
@Mikhail maybe? I don't know anything that could help you, sorry. That could make a question for the main site, though!
What does he work on, out of curiosity?
@BalarkaSen Ricci flow, geometric analysis.
Mean curvature flow.
I see.
Riemannian geometry.
A geometer, in essence.
10:42 PM
General relativity. Some complex analysis.
@BalarkaSen Yup.
Got it.
He's pretty big on PDEs and functional analysis, too. (Geometric analysis.)
@0celo7 @ACuriousMind

Given that a separable topological space is one where there is a sequence of element such that at least one element is contained in any open subset of the space

and that a second countable space is one where there is a countable collection of open subsets such that any open subset can be written as a union of them

A space which is separable but not second countable will mean it is a space that any open subset of it contains at l
He told me he recently reread Bott & Tu.
He's been impressed with my progress in GP, and I wanted to learn Morse theory. But I'm clearly not prepared for it.
10:45 PM
Yes, knowing the background better will help you appreciate the content better, I think.
@0celo7 Understandable why he likes it.
@BalarkaSen Maybe I should read it for the remainder of the summer.
Then read Hatcher next summer.
Forms have plenty geometry. More geometry than topology.
Bott-Tu has a forms aspect on stuff.
I don't have anything against algebraic topology, and I have the background to read Hatcher.
OK, I should sleep now.
But I need theorems 200-300 pages in.
10:51 PM
@Secret Sorry, what exactly is your question? Second-countability is stronger than separability, and you can easily find examples by searching for "separable but not second countable".
@BalarkaSen Have you read Bott & Tu?
tfw you accidentally watch episode 10 instead of 1 and the whole season is spoiled in "previously on" bit
They claim you need to know geometry to read it, but algebraic topology is unnecessary.
@0celo7 lol
now I'm playing episode 1 lol
10:54 PM
what are you watching?
Milnor has sent me into a deep depression that can only be cured by TV
(no, I have not been trying to read Milnor and watch TV at the same time)
@ACuriousMind It's really good imo
Let's peruse Bott & Tu
Although I can no longer trust my advisor's recommendations
@ACuriousMind In general, when is $\pi_1$ abelian?
@0celo7 I don't think there's a good general condition for that
K, just wondering
BT begins by defining $\pi_n$, which is already better than Milnor.
11:10 PM
Hmm, now I'm really confused, so if you want to capture the motion of a diffusing particle (1um bead) you need to sample at the nano second scale, on the other hand if you want to calculate the diffusion coefficient using methods like FCS you can sample at the milisecond scale and get the correlation. Or even manual particle tracking and then calculate the MSDs. Now I'm really confused.
11:23 PM
@ACuriousMind I give up, I can't even understand this book
I'm trying to compute $H^1_c(\Bbb R)$
They explain that with compact supports, not every 1-form is a gradient, and I think I understand that
But why does that immediately imply it's $\cong \Bbb R$
Yes, I understand why not every 1-form is a gradient.
The condition on the 1-form is that $\int_\Bbb R\theta=0$.
@ACuriousMind Ah, so we can classify the elements (equiv classes) of $H^1_c(\Bbb R)$ simply by their integrals over $\Bbb R$?
If two forms integrate to the same thing, they can only differ by a form for which $\int_\Bbb{R}\theta=0$
which is equivalent to $\theta=df$.
Suppose $\theta=df$. Then $\int \theta=\int df=0$ by Stokes. We can define $f(x)=\int_{-\infty}^x \theta$, which has compact support by virtue of $\int\theta=0$.
@ACuriousMind They mention de Rham currents
The $\delta$ is the Poincare dual of a point
@0celo7 I know
The problematic object is $\delta\wedge\delta$.
@ACuriousMind I like to think you were letting me figure the above out on my own, not too lazy to respond :P
because you clearly saw the chat ;)
hopefully you'd tell me if I was doing something wrong
@0celo7 That was in fact exactly what I was doing. You tend to figure out quite a bit of your questions if you just stew in them for a bit ;)
@0celo7 If I was a nitpicker I would ask you why exactly $\int \theta = 0$ implies $f(x)$ has compact support.
@ACuriousMind Well, because $\theta$ itself has compact support. Thus the $x$s that cause $\int^x\theta$ to be nonzero form a compact set.
@0celo7 Carry on, then ;)
11:37 PM
@ACuriousMind Well, nonzero if it doesn't go to zero randomly somewhere in the support, I guess.
does that make any sense
support like $A\sqcup B$, where $\int_A\theta=0=\int_B\theta$
How do I make that "proof" precise, is my question?
@Mikhail I'd really advise you to phrase your confusion into a question on the main site instead of asking here in chat. You'll get more people who can answer that way
@ACuriousMind How about this: Since the support of $\theta$ is compact, we may start the integral where the support starts: $f(x)=\int_p^x\theta$.
Also, since the support is compact, we can end the integral where the support ends and still get zero, that is, $\int_p^q\theta=0$.
Thus, the only $x$s for which $f(x)$ could be nonzero form a compact set, namely, the support of $\theta$.
@ACuriousMind Does it take more work to show that rigorously?
I fear that's still just heuristic.
@0celo7 If my space is an H-space, $\pi_1$ is abelian.
H-space means the space has a continuous multiplication map, $X \times X \to X$ with an identity element.
Besides Lie groups :P
Lie groups.
Well, $S^7$.
Octonions don't associate, so that's no group.
11:52 PM
Must be lonely to be an octonion
Octonions are the mathematician's spirit animal
Silly question: is there a chartwise, generic, description of transverse manifolds?
@BalarkaSen did ACM ever tell you he's got a serious man crush on Urs Schreiber
@BalarkaSen what's wrong with the transversal equation?
That is to say, suppose $X$ and $Z$ subfolds of $Y$ intersect transversely. Take a point $z \in Z$: can I describe what happens near an appropriately choosed chart around $z$?
@0celo7 Oh, are we reviving old running jokes now?
11:56 PM
Analogous to submersion and immersion theorems, that is to say.
My bet is on no.
@ACuriousMind I was reminded by my comment
but if you want to, sure
@0celo7 Nope.
How so?
@BalarkaSen His goal in life is understand Urs' 900 page paper on categorical QFT or something
and Urs is his "spirit animal"
@0celo7 No, you're an adult now, we must behave seriously
@ACuriousMind ...I don't know what you're referencing
if you wish to resurrect my claim that I'm a 10yo girl
11:58 PM
If I were not a guy, I'd have said at this point that I like Lurie over that guy
But I am, hence I won't.
@BalarkaSen On the internet, no one knows you're a dog
Or an AI, for that matter
I have a man crush on a guy in Turn, nothing to be ashamed about
Lurie...let's see
Tao is such a nerd
he's literally an Asian math nerd
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