« first day (3634 days earlier)      last day (1370 days later) » 
00:00 - 15:0015:00 - 23:00

Balarka Sen
Balarka Sen
12:28 AM
@user10478 I wrote down a list of important distributions, their pdf/pmf, mean, variance and moment generating functions in my first course in probability
You should not just memorize this list you should be able to derive these things by yourself quickly. For example, compute variance of the hypergeometric distribution; it's honestly a good exercise and you'll end up noticing something (hypergeometric is a sum of weakly dependent bernoulli distributions)
its important to understand some continuous distributions are scaled limits of discrete ones, so you can derive stuff about the latter from stuff about the former (eg characteristic function)
it should be a healthy mix of examples and theory. too much of the former will make you an insane statistician, too much of the latter will make you an insane measure theorist. there's a middle ground, called being a probabilist
just play with lots of examples. sit down and see if you can prove the central limit theorem for the bernoulli distribution one day.
 
Konformist Liberal
Konformist Liberal
1:19 AM
Hi! Do you think $\langle f_i(A)\rangle^{n-1}_{i=0}$ is the correct way to write the group that's generated by the images of $f_i$?
I mean the group that's generated with all of them together
 
ronak jain
ronak jain
2:09 AM
Please tell me how to use math jax in mobile
I am having problems in understanding maths problems and answers just because of symbols used in chat
 
ronak jain
ronak jain
2:29 AM
@KonformistLiberal can you please tell me how to use math jax in mobile
 
Konformist Liberal
Konformist Liberal
math.ucla.edu/~robjohn/math/mathjax.html here at the end of the page there's some explanation
 
ronak jain
ronak jain
@KonformistLiberal can you suggest me a PDF containing question for the topic "Function" . So that I can strengthen this. I am having problems in this while preparing for jee
 
 
1 hour later…
Yuvraj
Yuvraj
3:34 AM
@robjohn hi sir
i need your help
 
robjohn
robjohn
@Yuvraj yes?
 
Yuvraj
Yuvraj
@robjohn this one sir
 
robjohn
robjohn
@Yuvraj is $\frac{\sin(x)}{x}$ an even function, an odd function, or neither?
 
Yuvraj
Yuvraj
3:51 AM
it is a even function
@robjohn
 
robjohn
robjohn
@Yuvraj so what would be the coefficient of $x$ in an even function?
 
Yuvraj
Yuvraj
it should be even
 
robjohn
robjohn
@Yuvraj is $x$ an even function, an odd function, or neither?
 
Yuvraj
Yuvraj
odd
@robjohn
 
robjohn
robjohn
@Yuvraj so what would be the coefficient of $x$ in an even function?
 
Yuvraj
Yuvraj
4:06 AM
got it sir @robjohn
 
robjohn
robjohn
so?
 
Yuvraj
Yuvraj
it should be odd
 
 
3 hours later…
Edward Evans
Edward Evans
7:18 AM
0
A: Equality of fields being deduced from isomorphism of Galois groups

Edward EvansTwo proofs (Following TokenToucan's comment): Conceptual proof: We have $E_\sigma \subset K^{\text{ur}}E_\sigma \subset K^{\text{ab}}.$ By Galois theory we have $$K^{\text{ur}}E_\sigma = (K^{\text{ab}})^{\operatorname{Gal}(K^{\text{ab}}/K^{\text{ur}}E_\sigma)}$$ and $$\operatorname{Gal}(K^{\text{...

rofl
 
Knight
Knight
I want to space two images within a line.
Can somebody suggest something
??
 
Calvin Khor
Calvin Khor
What is the context, @Knight? In a picture? In a word document? In an arts and crafts project? In a question on MSE?
 
Knight
Knight
Hello Calvin, How are you?
 
Calvin Khor
Calvin Khor
so-so, sleepy, busy
 
Knight
Knight
I'm writing an answer and I need two blue squares and then a proper space and then three blue squares
How can I do it?
 
Calvin Khor
Calvin Khor
7:25 AM
$$ {\color{blue}\blacksquare} {\color{blue}\blacksquare} \qquad {\color{blue}\blacksquare} {\color{blue}\blacksquare} {\color{blue}\blacksquare}?$$
 
Knight
Knight
Yes
Yes, can you please make them a little bigger?
 
Calvin Khor
Calvin Khor
$$ {\Huge \color{blue}{\blacksquare\blacksquare \qquad \blacksquare \blacksquare \blacksquare}} ?$$
 
Knight
Knight
Wow! Thats more than best
 
Calvin Khor
Calvin Khor
horizontal spacing can be added with \, \ \quad \qquad and if you need finer control \hspace{x em} where x can be negative also and have idk 3 decimal places. also \! removes one \ space
Also \phantom{} adds spacing equal to whats in the {}, so for instance if you want the empty space to fit one of the cubes exactly, you can do $${\Huge \color{blue}{\blacksquare\blacksquare \phantom{\blacksquare} \blacksquare \blacksquare \blacksquare}} $$
 
Knight
Knight
Oh
 
Taufi
Taufi
7:35 AM
@Ted
Shifrin: I see the connection now to the binomial theorem, but what to do about the missing binomial coefficients that come up in the expansion?
 
 
1 hour later…
Thorgott
Thorgott
9:00 AM
Topology fact of the day: For each $k\in\mathbb{N}$, a $4k$-dimensional connected, orientable, compact smooth manifold with Betti numbers $b_0=b_{4k}=1$, $b_1=...=b_{2k-1}=b_{2k+1}=...=b_{4k-1}=0$ and $b_{2k}$ odd exists iff $k=1$ or $k=2^a+2^b$ for some $a,b\in\mathbb{N}$.
 
Balarka Sen
Balarka Sen
That follows from CGB?
 
Thorgott
Thorgott
no
I don't think that has anything to do with CGB
 
Balarka Sen
Balarka Sen
Why is it true
 
Thorgott
Thorgott
do I look like I know research-level mathematics
I can't tell you man
 
Balarka Sen
Balarka Sen
why is that research level math
seems like a cool exercise unless its very hard
 
Thorgott
Thorgott
9:04 AM
it's very hard
 
Balarka Sen
Balarka Sen
ref?
 
Thorgott
Thorgott
the reason this is true is because the existence of such a manifold with specified $b_{2k}$ is equivalent to the solubility of a certain equation involving Bernoulli numbers and that equation can be solved mod $2$
 
Balarka Sen
Balarka Sen
that doesnt tell me anything tho
what kind of tools are used?
 
Thorgott
Thorgott
for all multiples of $4$ smaller than $2^{25}$, $b_{2k}=5$ is realizable, but for dimension $2^{25}$, the smallest odd $b_{2k}$ is $34511$
 
Balarka Sen
Balarka Sen
Just give me a reference instead of random numbers man
lol
 
Thorgott
Thorgott
9:06 AM
idk, I'm just here for the fun facts
I don't know the specific reference, these are joint results from Kreck and Zagier
 
Balarka Sen
Balarka Sen
did Kreck tell you this stuff
 
Thorgott
Thorgott
yes
 
Balarka Sen
Balarka Sen
weird guy
 
Thorgott
Thorgott
he decided to make the last lecture about which Betti numbers are realizable
it has sth to do with the Hirzebruch signature theorem and Pontryagin classes, I think
 
Balarka Sen
Balarka Sen
Ah OK
That makes sense
The signature theorem is a ramped up CGB that's what I had in mind
 
Thorgott
Thorgott
9:11 AM
orientable and of dimension 4k is the hardest case for some reason
for non-orientable manifolds, the only obstruction is that the Euler characteristic has to vanish in odd dimensions
 
Balarka Sen
Balarka Sen
I think there's some general theory here. If you have a bunch of candidate Betti numbers for a potential orientable manifold it has to satisfy Poincare duality
Then you make a complex with those Betti numbers
That's a rational PD space; then you surger to modify it to be a manifold
So here's the question to be precise. If $X$ is a, say, an integral PD space, i.e., has cohomological dimension $n$ and $H_k(X; \Bbb Z) \cong H^{n-k}(X; \Bbb Z)$, when is $X$ a manifold?
Can you modify $X$ in some canonical way without changing homology so that it's a manifold?
I don't know how to do this but I think this has good answers
 
Thorgott
Thorgott
sounds like a hard question
 
Balarka Sen
Balarka Sen
yeah
Ask Kreck man
 
Threnody
Threnody
I'm sorry to invade - I have a somewhat subjective question so I don't think it's a fit for the main site
Why do we necessarily tie the concept of counting with $|\mathbb{N}|$ instead of ordinal numbers?
I'm a beginner in this idea of ordinals but they seem so much more expressive, so why limit counting to the naturals?
 
Thorgott
Thorgott
9:29 AM
Conceptually we don't, that's precisely why ordinals exist. Terminologically, it's probably for historical reasons, since counting with natural numbers preceded the conceptualization thereof.
 
Threnody
Threnody
The reason I ask is because we also do this limitation in computer science when it comes to what is computable and what isn't.
 
Balarka Sen
Balarka Sen
@Thorgott How do you formulate Poincare duality
 
Thorgott
Thorgott
I don't know any CompSci, but I assume when you actually try to get results from a computer, you want it to finish doing it's computation in finite time/a finite amount of steps, no?
 
Threnody
Threnody
Yes, we want it to "halt" in real life
But, I care more about the theory
 
Thorgott
Thorgott
@Balarka $H^k(M)\cong(H_c^{n-k}(M))^{\ast}$
 
Threnody
Threnody
9:33 AM
https://cs.stackexchange.com/questions/128356/whats-wrong-with-this-proof-that-mathbbr-is-enumerable

For example, I asked this ^. The proof, I think, works if we accept counting beyond $\omega_0$. So my issue is whether or not a theoretical computer can actually go through $\mathbb{R}$ - since by the axiom of choice we can well-order any set hence we can ask what's next - am I wrong?
 
Balarka Sen
Balarka Sen
Ah, sucks, so the isomorphism comes from the pairing $(\omega, \eta) \mapsto \int_{[M]} \omega \wedge \eta$
Hm
 
Thorgott
Thorgott
ye
I don't know what a "theoretical computer" can or cannot do.
 
Threnody
Threnody
A Turing machine, essentially. Although this might be more cs than math, in the sense that I don't think mathematicians cover Turing stuff?
 
Edward Evans
Edward Evans
$\text{CS} \subset \text{Math}$
 
Threnody
Threnody
Idk, I definitely would ask my professors about this if I could actually go to them :D
Yeah I agree that theoretical cs is a branch of math
 
Balarka Sen
Balarka Sen
9:39 AM
I don't think any of us know the theory well enough to respond to your question properly.
 
Thorgott
Thorgott
the Turing machines I know are finite objects
 
Edward Evans
Edward Evans
Truth
 
Threnody
Threnody
Yeah neither do I :D
 
Edward Evans
Edward Evans
I might take theoretical cs next semester
 
Threnody
Threnody
Turing machines, are in theory described as having infinite memory
The crux is that the literature takes that to mean "countably infinite"
Although I'm just an undergrad, I have no authority on what people think things mean...
It's just how it was always taught to me
ofc in reality computers have finite memory, but they're, in theory, as capable as finite memory turing machines
 
Edward Evans
Edward Evans
9:41 AM
@Balarka old Dimmu Borgir is cool af
 
Thorgott
Thorgott
yeah, what I mean is that they have finitely many states, use finite alphabets and halt in finitely many steps when they do so
 
Threnody
Threnody
@Thorgott yeah, when they do so :D
 
Balarka Sen
Balarka Sen
Turing-recognizability is by definition a countable phenomenon but what do I know.
You only call subsets of $\Bbb N$ to be Turing-recognizable
 
Drathora
Drathora
Do you understand what's wrong with the proof you wrote now Threnody? The one that you linked
 
Threnody
Threnody
yes, my question is what's up with that limitation
 
Balarka Sen
Balarka Sen
9:43 AM
Anyway we don't know this stuff I am sure. if anyone does it's one of @Alessandro, @LeakyNun or @AkivaWeinberger
 
Thorgott
Thorgott
I think the point is that a Turing machine's reader does things step by step and only moves left or right on the tape each step, so it can't go beyond anything countably infinite
 
Threnody
Threnody
@Drathora I really think I do but I'm not "convinced" in a sense
@Thorgott perhaps a better abstraction is a type-0 grammar
@Thorgott they're equivalent, but i think grammars are more ...digestible?
 
Thorgott
Thorgott
while I do have a limited understanding of what a Turing machine is, I don't know what that is anymore, I'm afraid
 
Drathora
Drathora
So your issue is that if you can't "enumerate" over the entirety of $S_\infty$ the way you've done
 
Thorgott
Thorgott
my point is that you can't go beyond anything countable by just repeatedly doing $+1$
 
Drathora
Drathora
9:45 AM
Because there are $2^{\omega_0}$ Dedekind cuts, not $\omega_0$
 
Threnody
Threnody
@Thorgott but isn't that what ordinal numbers do?
 
Drathora
Drathora
So you're trying to do a countable enumeration of something that isn't countable
 
Balarka Sen
Balarka Sen
No, the point is ordinal numbers need not have immediate predecessors
 
Threnody
Threnody
@Drathora Yes i understand that, but that's why I said "keep going"
 
Balarka Sen
Balarka Sen
The Turing machine only turns left and right, and that's all it does. You have both immediate successor and predecessor
 
Threnody
Threnody
9:46 AM
@Drathora as in, going from $\omega_0$ to $2 \omega_0$ doesn't change much when it comes to describing $\mathbb{R}$
 
Thorgott
Thorgott
no, ordinal numbers go further, but not by just doing $+1$
 
Drathora
Drathora
Going from $\omega_0$ to $2\omega_0$ doesn't change anything, that's correct
But going from $\omega_0$ to $2^{\omega_0}$ changes a whole bunch
 
Thorgott
Thorgott
but I'll leave this in the hands of the actual computer scientist here
 
Threnody
Threnody
@Drathora yes - that's what I'm trying to convey
@Drathora and assumingly, you could go beyond $\mathbb{R}$ if you were to accept that somehow you can "keep going" even further than $2^{\omega_0}$
 
Drathora
Drathora
You can keep going indefinitely with this process and never even leave the interval $[0,1]$
 
Threnody
Threnody
9:49 AM
@Drathora well yes, if you look at it as a step by step process
@Drathora hmm, I see your point
 
Balarka Sen
Balarka Sen
@Thorgott Anyway I wanted to say a PD space is slightly more general than how I described it because it specified what map $H^k(X) \to H_{n-k}(X)$ should be an isomorphism, but maybe this is a little tricky for me to explain with your formulation of PD
Because it's a cohomology-to-cohomology duality; you can do it but I'm too lazy to spell it out
Anyway if $X$ is a simply connected PD space of cohomological dimension at least $5$ and odd, it's homotopy equivalent to a manifold.
By a very explicit homotopy equivalence
 
Thorgott
Thorgott
idek what a PD space is
 
Balarka Sen
Balarka Sen
"Satisfies Poincare duality"
It makes sense to say a PD space is a space $X$ such that $H^k(X) \cong H_{n-k}(X)$. Just take that as a definition; it's a little more than that but this is ok for now
 
Drathora
Drathora
@Threnody one final remark about this. A way to think about this is "is there an algorithm I could theoretically write that will "eventually" cover everything I'm going to enumerate. When enumerating the rationals there are well-known ways of doing this via diagonalization. If you think about how you'd enumerate S_i though, no algorithm you can ever give me wouldn't get "stuck" in a state where there's something it's clearly never going to reach.
 
Balarka Sen
Balarka Sen
In even dimension 2k there's only signature restriction which vanishes when k is odd
So 4k is the only troubling dimension
 
Threnody
Threnody
9:58 AM
@Drathora Hmm, well of course such an algorithm would never halt even for just $\mathbb{N}$ but that's "ok" since I'm staying theoretical. So the problem then is the word enumerating.
 
Drathora
Drathora
It's not that it would never halt. It's that I can give you an actual example of a Dedekind cut that your algorithm would never reach
 
Thorgott
Thorgott
can I understand curvature without understanding Riemannian geometry?
 
Balarka Sen
Balarka Sen
Just learn some altopo so we can read this shit man
 
Drathora
Drathora
Whereas in the natural or rational case, there is no such example for some algorithms
 
Balarka Sen
Balarka Sen
Seems classical
 
Threnody
Threnody
10:00 AM
@Drathora I'm actually intrigued to see that...
@Drathora the single number is fine I can extrapolate the partitioning, somewhat
 
Balarka Sen
Balarka Sen
@Thorgott Yeah for sure. Curvature to you is just the curvature 2-form, right? Do you define a Levi-Civita connection or use connection 1-forms?
I can tell you the geometry
(I think)
 
Thorgott
Thorgott
yeah, the latter, and I would love to hear it
 
Drathora
Drathora
Well, the problem is that I can give you $2^{\omega_0}$ many Dedekind cuts that you'll never reach with any algorithm you give me :P
 
Thorgott
Thorgott
I wanna amp my understanding rate of CGB to like 30% at least
 
Threnody
Threnody
@Drathora I do know of the uncomputable numbers - but they're so called "uncomputable" because of the drawn line of enumerability being "no more than $|\mathbb{N}|$
 
Balarka Sen
Balarka Sen
10:02 AM
Ok let me think how to phrase this, I am not super versed with the moving frames language. I'll do this in exchange of you explaining CGB to me btw
 
Threnody
Threnody
@Drathora When I asked about this on the CS site I was led to "hypercomputation" and subsequently told that it is... a "crank-adjacent" subject :D
 
Thorgott
Thorgott
I can tell you about it once I fill my 5 or 6 large understanding gaps
 
Threnody
Threnody
@Drathora perhaps because some people think it's physically realisable... but again i'm fine with it not being so
 
Drathora
Drathora
Yeah I'm unfamiliar with the topic. Although the quote I found did give me a good laugh
"if non-computable inputs are permitted, then non-computable outputs are attainable"
 
Threnody
Threnody
@Drathora yeah precisely
 
Balarka Sen
Balarka Sen
10:06 AM
@Thorgott Can you tell me how you define the connection 1-form?
 
Threnody
Threnody
@Drathora so my dilemma is why do we draw the line to what is computable and what is not if it's just a matter of ... "just stop at the cardinality of N it's enough"
@Drathora if anything i'm trying to find where the rigor is, if it is a matter of rigor and not just a convention/definition
 
Thorgott
Thorgott
Take a frame $X_1,...,X_n$ and pass to the dual frame $\theta^1,...,\theta^n$. The connection forms are the unique forms $\omega_j^i$, $i,j=1,...,n$ satisfying $\omega_j^i=-\omega_i^j$ and $d\theta^i=\sum_{k=1}^n\theta^k\wedge\omega_k^i$ for all $i,j=1,...,n$.
 
Balarka Sen
Balarka Sen
Ah ok great.
 
Thorgott
Thorgott
apparently the "better" framework is to construct these on the frame bundle instead, but we didn't prove that and only got a reference in the 5th volume of Spivak for it that I haven't looked up yet
 
Leaky Nun
Leaky Nun
@Threnody would your issue be resolved if instead of saying "infinite memory" we say "arbitrarily large memory"?
 
Balarka Sen
Balarka Sen
10:09 AM
Nah locally is good enough
 
Threnody
Threnody
@LeakyNun eh, ... if we allow such memory... I'm not too well versed in ZFC to make statements like these to be honest :D
 
Thorgott
Thorgott
yeah, it's probably better to understand it in terms of frames first before passing to the frame bundle
 
Threnody
Threnody
@LeakyNun but presumably, why not?
 
Leaky Nun
Leaky Nun
what exactly is the question?
I haven't followed the entirety of the conversation
 
Drathora
Drathora
To me, it's that computations flow in discrete steps. We can label the steps of a computation using the natural numbers. To compute something of cardinality higher than that of $\mathbb{N}$ would give us an injection from this "something" to the naturals
 
Threnody
Threnody
10:12 AM
@Drathora effectively the ordinals, from what I understand?
@LeakyNun I can phrase the question as:
@LeakyNun "Why do we limit the definition of computable to the idea of enumeration up to $|\mathbb{N}|$ when ordinals are much more expressive"
@LeakyNun strictly staying in theory, not reality
@LeakyNun Don't get me wrong I'm absolutely fine if it's a definition or just a convention we chose in the past that stuck - What I'm worried about is if I'm lacking rigor if I talk about this.
 
Drathora
Drathora
So is what you have in mind to describe a program that has to enumerate the rationals and then do one final step using the ordinal $\omega + 1$?
Whereas a program that does one step and then enumerates the rationals is described by $1 + \omega = \omega$
 
Balarka Sen
Balarka Sen
@Thorgott I'll call the moving frame $e_1, \dots, e_n$, and everything else will be notationally same. The connection equation is (matrix form) $d\theta = \theta \wedge \omega$.
What's $d\theta^i(e_p, e_q)$ from this? Because $\theta^i(e_j) = \delta_{ij}$, $d\theta^i(e_p, e_q) = (\theta^p \wedge \omega^i_p + \theta^q \wedge \omega^i_q)(e_p, e_q)$
Which is what? $\omega^i_p(e_q) - \omega^i_q(e_p)$, yeah?
 
Drathora
Drathora
The point here is that as soon as we go beyond $\omega$, even to the successor $\omega + 1$, nothing is computable as I can pinpoint a step that will never be reached
 
Threnody
Threnody
@Drathora Well yes, I understand that but the definition of what is computable is tied to enumerability which is tied to the natural numbers not the ordinals.
 
Balarka Sen
Balarka Sen
On the other hand, remember for a $1$-form $\omega$, $d\omega(X, Y) = X \omega(Y) - Y \omega(X) - \omega([X, Y])$, so $d\theta^i(e_p, e_q) = - \theta^i([e_p, e_q])$. So we get $\omega^i_q(e_p) - \omega^i_p(e_q) = \theta^i([e_p, e_q])$
 
Drathora
Drathora
10:21 AM
Computability is tied to enumberability for the reason I said before. Computation happens in steps that are isomorphic to the natural numbers.
 
Balarka Sen
Balarka Sen
Wow dude so much CS happening
 
Threnody
Threnody
@Drathora Yep. So it is after all just a matter of definition?
 
Balarka Sen
Balarka Sen
@Thorgott come to garbology
 
Drathora
Drathora
Sorry haha
 
Threnody
Threnody
I'm sorry honestly
 
Drathora
Drathora
10:22 AM
Well, sort of. Of course you could say it in terms of ordinals
But as soon as you exceed the ordinal $\omega$, everything is non-computable
So it's the exact same theory, just with a relabelling, as far as I'm concerned
Not accounting for any of this strange hypercomputation
 
Threnody
Threnody
Mhm. I think I understand
Thank you, sorry for the CS spam but I think you guys would've been more likely to know what ordinals are :D
 
Drathora
Drathora
This all kind of ties back to the explanation I used for ordinals using queues the other day
If I give you an infinite length queue, and you walk down it, if I give you a person $n$, you'll eventually reach them, no matter which $n$ it is
On the other hand, if I give you an $\omega + 1$ queue, then you'll never reach that $\omega + 1$th person
If you think of walking past position $n$ as being computation step $n$
 
Edward Evans
Edward Evans
woo, seminar on elliptic curves was just announced
 
Alessandro Codenotti
Alessandro Codenotti
Turing machines have a tape which is $\Bbb N$ because you want to be able to move both ways
Longer well orders have elements with no immediate predecessor
(I don't really know much about them, but this seems like a reasonable guess to me)
 
Thorgott
Thorgott
yeah, I think this is essentially the "computation happens in discrete steps" idea that Drathora brought up
 
Balarka Sen
Balarka Sen
10:35 AM
@AlessandroCodenotti Yeah that's what I suggested earlier
 
Alessandro Codenotti
Alessandro Codenotti
Ah I didn't see that, I only looked at the messages around the one where you pinged me :P
 
Balarka Sen
Balarka Sen
Surely some nutcases do ordinal length Turing machines
some weird computation happens at the limit ordinals
 
Alessandro Codenotti
Alessandro Codenotti
I guess there are all sorts of generalizations
 
Balarka Sen
Balarka Sen
said nutcase from your uni
 
Alessandro Codenotti
Alessandro Codenotti
Oh I know him
He's retired now officially, but he mostly works on automated proof verification with Hales nowadays
 
Balarka Sen
Balarka Sen
10:42 AM
Of course he works in automatic proof verifications
NERD ALERT
 
Alessandro Codenotti
Alessandro Codenotti
Hales does to, but he is an honest geometer pfff
 
Balarka Sen
Balarka Sen
i learnt yesterday that Mike Freedman does quantum computing now
topological quantum computing
in microsoft research
lmao
 
Alessandro Codenotti
Alessandro Codenotti
I have a feeling that pays better than geometry
 
Balarka Sen
Balarka Sen
he has millions of those multi-authored papers in arxiv in quantum CS
it pays better than bing topology at least
standard career tricks once you get a fields medal
leave math and troll applied people or live on your moms pension
 
Thorgott
Thorgott
life goals
 
Balarka Sen
Balarka Sen
10:47 AM
Grisha reacts only
 
Alessandro Codenotti
Alessandro Codenotti
@BalarkaSen or run for mayor in Paris
 
Balarka Sen
Balarka Sen
Oh yeah that
 
Alessandro Codenotti
Alessandro Codenotti
Or you can go the Grothendieck way and retire in a cave in the mountains or something like that
 
Balarka Sen
Balarka Sen
you can also almost get the Fields medal and chill as a Senator in the US only to learn your results are all wrong
@AlessandroCodenotti you mean becoming a pickup artist
 
Alessandro Codenotti
Alessandro Codenotti
Who's that?
 
Thorgott
Thorgott
10:51 AM
lol I was about to reference that one
 
Balarka Sen
Balarka Sen
Daniel Biss
 
Alessandro Codenotti
Alessandro Codenotti
@BalarkaSen lol I have a classmate who calls Grothendieck "the Rasputin of math"
 
Thorgott
Thorgott
forgot to check simply connected for Seifert-van-Kampen, I think
 
Balarka Sen
Balarka Sen
@Thorgott lmfao really
i know he made some stupid error i didnt know what it was
just 1 error man
scary stuff
 
Thorgott
Thorgott
yeah, I think that was it
 
Balarka Sen
Balarka Sen
10:52 AM
LMFAO
 
Thorgott
Thorgott
Kreck likes telling that story
 
Balarka Sen
Balarka Sen
I love Kreck
 
Drathora
Drathora
Can confirm that quantum CS does pay well
Our quantum department is raking in money, I'm pretty sure the US Navy or something is involved somehow
 
Thorgott
Thorgott
Kreck is awesome
 
Drathora
Drathora
Which is hilarious to me, because the main person who's bringing all this money is this man:
Most bizarre lecture series I've ever attended
 
Knight
Knight
10:55 AM
@Drathora Hello
 
Drathora
Drathora
hey ^^
 
Knight
Knight
How was your meeting?
 
Drathora
Drathora
Good, my undergraduate student is ready to begin implementing my results, while I sit back and never program again
 
Balarka Sen
Balarka Sen
@Drathora what the hell is that outfit
 
Drathora
Drathora
I know lmao. He's a peculiar guy, although quite brilliant I guess
 
Knight
Knight
10:56 AM
lol
 
Balarka Sen
Balarka Sen
i'd be running in the opposite direction if a man in that outfit started writing category theory on the board
 
Drathora
Drathora
The course he teaches is really strange. It's quantum mechanics via box/string diagram, all via some symmetric monoidal category or something
 
Knight
Knight
I have written a long answer to this question, but I'm still not sure what OP wanted. Can you please have a look at it?
 
Drathora
Drathora
I forget the details, it's one of those things I learned and then never thought about again
 
Thorgott
Thorgott
lmao that's wild
 
Balarka Sen
Balarka Sen
10:58 AM
fucking nut
 
Drathora
Drathora
But my experience is that it teaches you all of this quantum stuff, but in a language that nobody from any other institution knows, since they use traditional notation
 
Thorgott
Thorgott
@Balarka is there any man you wouldn't run from when he starts writing category theory on the board except Gromov
 
Drathora
Drathora
So you know a lot of the same things as them, but you can't actually communicate with them
 
Balarka Sen
Balarka Sen
no but this man
this man...
 
Drathora
Drathora
But it does make a lot of the results in quantum computing etc fall out very easily
 
Balarka Sen
Balarka Sen
10:59 AM
scares me
My god
$(\infty, 1)$-masculinity
No soy for this guy
 
Drathora
Drathora
Now imagine the startled 4th year Computer Science students at Oxford when he walks into the room, with a cap on that looks like it's been made out of 10 other recycled caps glued together, a string vest and about 5 chains hanging from his belt
Quite a far cry from what they imagined when they applied I bet haha
 
Balarka Sen
Balarka Sen
I understand completely
 
Thorgott
Thorgott
@Balarka I misremembered the SvK thing, it seems. Check this: arxiv.org/pdf/0709.1291.pdf
 
Balarka Sen
Balarka Sen
oh my
 
Drathora
Drathora
It's hard to assess your answer @Knight, since I don't really know exactly what the question-asker needs. I wonder if they wanted things like distributivity, associativity and commutativity to also be expressed
It's not really made clear, so I think the person asking the question needs to give a better picture of what's required
Not sure if I'd be mortified or proud if something like that got released about my work
On one hand, damn, all that time was for nothing
On the other... at least someone read it?
 
Thorgott
Thorgott
11:14 AM
"My very unfortunate duty is.."
spoken like a true assassin
 
Balarka Sen
Balarka Sen
lmfao yeah
 
 
1 hour later…
Lelouch
Lelouch
12:26 PM
Why the unit tangent bundle is a sphere bundle (when for example the underlying mainfold is not locally flat) ?
 
Nike Dattani
Nike Dattani
May be of interest to some people:
13
Q: What are the types of DFT?

Nike DattaniSimilar to: What are the different types of charge analysis?, What are the types of bond orders?, and What are some recent developments in density functional theory?, I would like to ask: What are the different variations/flavors of DFT (density functional theory)? I ask users to stick to one of ...

density-functional-theory one-topic-per-answer td-dft terminology density-functional-tight-binding
 
Lelouch
Lelouch
The only condition I've trouble verifying is the local trivialization property. Don't we need local flatness for that ?
 
Nike Dattani
Nike Dattani
Very easy answers, just pick one "type of DFT" in the list and explain it in an answer.
 
Edward Evans
Edward Evans
12:37 PM
Just proved $[E_\sigma : E_\sigma \cap K^{ur}] = [E_\sigma : E_\sigma \cap K^{ur}]$
the dream
 
Alessandro Codenotti
Alessandro Codenotti
I think I have a very short proof of that result
 
Edward Evans
Edward Evans
go on
rofl
 
Alessandro Codenotti
Alessandro Codenotti
Well they are equal, therefore they are equal. QED
 
Edward Evans
Edward Evans
jesus
Corollary: $n = n$ for all $n n n$
 
Balarka Sen
Balarka Sen
@Lelouch Um uh, what is local flatness?
 
Alessandro Codenotti
Alessandro Codenotti
12:40 PM
Like flatness, but locally
 
Edward Evans
Edward Evans
Alessandro throwing out the proofs like a farmer throwing out cow shit today
 
Alessandro Codenotti
Alessandro Codenotti
"moves hands in a plane, but only locally"
@EdwardEvans lmao
 
Balarka Sen
Balarka Sen
alessandro on drugs
what are you smoking man i want some of that
 
Alessandro Codenotti
Alessandro Codenotti
General topology
 
Balarka Sen
Balarka Sen
Nevermind I don't want it
 
Alessandro Codenotti
Alessandro Codenotti
12:42 PM
I've been thinking about the pseudoarc the whole afternoon
 
Balarka Sen
Balarka Sen
You can have it
 
Alessandro Codenotti
Alessandro Codenotti
It'll do that to you
 
Edward Evans
Edward Evans
"Love to mate, love to, but this is mine and I want it all.. so.." - Super Hans
 
Alessandro Codenotti
Alessandro Codenotti
I'm procrastinating because I should read the proof of Borel determinacy, but I skimmed it in the book and three pages into the proof he starts splitting cases into subcases
 
Balarka Sen
Balarka Sen
I have done 0 work today
I'm supposed to read probability
 
Alessandro Codenotti
Alessandro Codenotti
12:44 PM
Why are you doing so much probability? How crazy are the probability courses at your uni?
My undergrad probability course was like "this is a gaussian and this is a Poisson and don't ever dare mention measure theory"
 
Drathora
Drathora
Yeah measure theory wasn't introduced until the 2nd year of my undergrad, and wasn't mandatory
As a formal concept at least
 
Edward Evans
Edward Evans
@Alessandro same
 
Alessandro Codenotti
Alessandro Codenotti
We had both measure theory and the probability course I was talking about in the 2nd year as mandatory courses, but then we also had a non mandatory 3rd year course on measure theoretic probability
 
Edward Evans
Edward Evans
Lukas hadn't graduated from his bachelor's at Heidelberg because he couldn't bring himself to take intro probability hahahaha
 
Balarka Sen
Balarka Sen
@AlessandroCodenotti Mahan is thinking a lot about percolations so he's forcing me to
 
Drathora
Drathora
12:46 PM
I forget what the first year probability course entailed. I remember being overwhelmed with notation because I'd never really written formal mathematics before my undergraduate
 
Balarka Sen
Balarka Sen
His last paper was percolation on hyperbolic groups
 
Alessandro Codenotti
Alessandro Codenotti
Ok now that sounds pretty cool
 
Balarka Sen
Balarka Sen
Lmao
 
Charlie Shuffler
Charlie Shuffler
Hi all, I have a question about the error term coming up when approximating an integral with the trapezoidal rule
the rule states that we can make the approximation: $\int^b_a f(x) dx \approx (b-a)*(f(a)+f(b))/2$
is there a way to bound the error term that comes up here with some sort of big O term?
like $O(b^3)$ or whatever
 
Balarka Sen
Balarka Sen
We don't have too much probability either despite being Indian Statistical Institute
 
Charlie Shuffler
Charlie Shuffler
12:49 PM
any help is appreciated
 
Balarka Sen
Balarka Sen
Last year (so next for me) has a Markov chains course
or well, stochastic processes
I like prob its intuitive; I need to get a hang of these inequalities though
 
Alessandro Codenotti
Alessandro Codenotti
You're the first person ever to tell me probability is intuitive
 
Balarka Sen
Balarka Sen
It's alright man the analysis is actually bearable
 
Thorgott
Thorgott
@CharlieShuffler You can bound it explicitly in terms of the second derivative of $f$ (this should make sense intuitively if you draw a picture). I don't see why you want to bound this in terms of $b$. The way both the integral and approximation will change when $b$ changes will depend drastically on how $f$ changes therewith and you usually want to keep $a,b$ fixed either way.
 
Lelouch
Lelouch
@BalarkaSen I meant I thought I need to assume that it's a locally flat manifold
 
Balarka Sen
Balarka Sen
1:00 PM
What is a locally flat manifold
 
Alessandro Codenotti
Alessandro Codenotti
Sanity check, for an $A$-module $B$, to say $B\otimes_A A\cong B$ it's enough to note that $b\otimes a=ba\otimes 1$, right?
 
Lelouch
Lelouch
shit, I meant it's flat metric
@BalarkaSen
somehow in my head "locally isometric to Euclidean metric" became "locally flat"
 
Balarka Sen
Balarka Sen
Ah I see. No, the unit tangent bundle is a bundle regardless, although I see your point that locally isometric to R^n makes the proof immediate.
 
Lelouch
Lelouch
Hmm @BalarkaSen so how do you show the trivialization condition of the bundle ?
 
Thorgott
Thorgott
@Alessandro yeah, the iso is $b\otimes a\mapsto ab$ in one direction and $b\mapsto b\otimes1$ in the other direction; inverseness one way is obvious and inverseness the other way is what you're noting
 
Charlie Shuffler
Charlie Shuffler
1:02 PM
@Thorgott I am indeed familiar with such a bounding using the second derivative of $f$, but I don't see how that holds up when using only 2 points in my trapezoidal rule application
 
Alessandro Codenotti
Alessandro Codenotti
@Thorgott perfect, thanks
 
Balarka Sen
Balarka Sen
@Lelouch You just need to show the fiberwise unit bundle of $U \times \Bbb R^n$ with a pointwise smoothly varying metric $g_p$ on the fibers $\{p\} \times \Bbb R^n$ is diffeomorphic to $U \times S^{n-1}$, yeah?
 
Alessandro Codenotti
Alessandro Codenotti
And why is $B\otimes_A A^n\cong (B\otimes_A A)^n$ true? I don't understand anything about modules or their tensor products...
 
Thorgott
Thorgott
tensor commutes with direct sum (naturally)
 
Balarka Sen
Balarka Sen
@AlessandroCodenotti u can write down the same argument u were writing to see it's $B^n$
just componentwise
 
Alessandro Codenotti
Alessandro Codenotti
1:07 PM
So in general $B\otimes_A (C\oplus D)\cong (B\otimes_A C)\oplus (B\otimes_A D)$?
 
Thorgott
Thorgott
yes
even for infinite sums
 
Alessandro Codenotti
Alessandro Codenotti
With the iso being $b\otimes (c,d)\mapsto (b\otimes c,b\otimes d)$ then I suppose
 
Thorgott
Thorgott
@CharlieShuffler there's a corresponding error bound that works for any amount of nodes; it's on Wikipedia for example
@Alessandro yeah, and the inverse comes from the inclusions componentswise
 
Charlie Shuffler
Charlie Shuffler
Yeah I've seen it, but I actually just realized I'm trying to apply the trapezoidal rule to an improper integral
So i've got new problems to worry about
 
Alessandro Codenotti
Alessandro Codenotti
@Thorgott So if I write it out explicitely it looks like $(b_1\otimes c,b_2\otimes d)\mapsto b_1\otimes(c,0)+b_2\otimes(0,d)$
 
Thorgott
Thorgott
1:17 PM
yeah
 
Alessandro Codenotti
Alessandro Codenotti
Makes sense, thanks
 
Lelouch
Lelouch
@BalarkaSen Yes, exactly. The main difficulty is that if $(u, v)$ goes to $(u, \Psi_u(v))$, then defining $\Psi_u$ in a smoothly varying way. This is probably outright trivial (being the first exercise in Lee), but I don't see how ?
Doesn't the fact $(U, g)$ being isometric to $(O, \cdot_{\mathbb{R}^n})$ imply that the manifold must be flat ?
Shit, nevermind. It's $T^1(U)$ should be diffeomorphic to $U \times S^{n-1}$, not isometric.
So we can probably do it like this: Consider the map $h(p, T) = TG_PT^{-1}$, where $g_p(v, w) = v^T G_P w$. Now we can do some bullshit on the preimage on identity
 
beta_me me_beta
beta_me me_beta
2:14 PM
is any one online
need help understanding a solution
 
Alessandro Codenotti
Alessandro Codenotti
2:41 PM
@BalarkaSen karlin.mff.cuni.cz/~pyrih/e/e2001v1/c/ect/node93.html look at the beautiful pictures at the end of this page
 
Edward Evans
Edward Evans
tf man
 
Alessandro Codenotti
Alessandro Codenotti
They're just chains, but very wiggly
 
00:00 - 15:0015:00 - 23:00

« first day (3634 days earlier)      last day (1370 days later) »