Mathematics - 2017-08-05 (page 4 of 4)

Perturbative

18:00

My keyboard wants to snipe

Daminark

Okay I'm tired of that, on to my phone

Balarka Sen

you know you're going to get rekd for that d key right?

@TedShifrin He defined it as a stable homotopy theory of a spectrum

Ted Shifrin

pfeh

Semiclassical

What'd that make cohomology?

Daminark

At the end of the k-theory talk, Mark showed us that homology had this property, and was like "So if you want to be really funny, you write a colon here"

Araske

18:02

maps to eilenberg maclane spaces

Daminark

Yo @Araske

Balarka Sen

Araske got his g key busted

Araske

hopotopy classes thereof

lel

Balarka Sen

oh, so he's got his m and p keys switched

fucked up keyboard

Daminark

Welcome to the chat, I don't know if I've seen you around here before

Araske

18:02

heck ok

cannot type

Daminark

You need to do homotopy type theory to figure this problem out

Perturbative

I'm curious, what sort of "machinery" is used to prove the invariance of dimension for topological manifolds in the general case?

Balarka Sen

@Perturbative You need at least homology theory, I believe

Araske

can you do it with invariance of domain?

Balarka Sen

Unless you develop a notion of topological dimension?

Eric Silva

18:05

hi chat

Araske

hoi

Perturbative

@BalarkaSen Well I'll have to come back to the proof in a couple months then :p

Daminark

Hey @EricSilva

Perturbative

Yo @EricSilva

Araske

@Perturbative yeah you'd use invariance of domain, the only proof I've seen of that uses homology

Balarka Sen

18:06

@Araske Yeah. But that in turns need homology to prove it

Perturbative

@Araske, doesn't invariance of domain only apply to Euclidean spaces tho?

Balarka Sen

gah sniped

Araske

lol

Perturbative

Hehehehe

Balarka Sen

@Perturbative Manifolds are locally Euclidean

Araske

18:07

gets sniped in turn

yeah you can make a contradiction by poking charts

gian

Are there any video lectures for general topology?

Good ones.

Perturbative

Dammit @Balarka, you could've let me basked in five minutes of my sniping celebrations before mentioning that

Balarka Sen

You have to be quick

Daminark

@Araske they're delicate though, so don't poke them too hard

Eric Silva

I don't think I like the terminology of being locally Euclidean

Ted Shifrin

18:10

@EricSilva: I haven't worked it out for myself, but I'm pretty sure there's a typo in Robert's paper. In (3.14), the $h$ should be an $n$.

What's wrong with "locally Euclidean"? It's just like "locally constant" or "locally connected."

Daminark

@EricSilva for the confusion about Riemannian manifolds?

Eric Silva

cause in my brain i want euclidean to mean something about geometry

Perturbative

@gian I've never seen any on YouTube, the only good thing I've seen was a couple Differential Topology lectures by Milnor

gian

Okay thanks anyways.

Eric Silva

although I really don't have a better term for "looking locally like $\mathbb{R}^{n}$

but personally when you equip $\mathbb{R}^{n}$ with the standard Euclidean inner product I like to notate it $\mathbb{E}^{n}$

Araske

18:12

I think you can make the geometry carry locally?

idk tho

Ted Shifrin

I think you're protesting too much, Eric.

Eric Silva

i mean it's not a big deal at all

Ted Shifrin

Go read!

Araske

"local geometry" what am I even saying

Ted Shifrin

Oh, did you figure out Hopf?

Eric Silva

18:13

I'm working on it rn, I just woke up lol

Ted Shifrin

Good grief.

<-- going back to watch tennis

Perturbative

@TedShifrin I hope it's not the Citi Open :p

Alessandro Codenotti

Hi @Ted

Bye*

Daminark

Your sleep schedule might just be worse than mine @EricSilva

Eric Silva

@Daminark I slept at 1 am, I just slept for 12 hours

Alessandro Codenotti

18:14

Balarka is still the king of bad sleep schedules

Daminark

Damn, nice

Astyx

Why did Stackoverflow just send me a job proposition ?

Like, since when is Stackoverflow used as such ?

SteamyRoot

Sure it's an actual proposition and not an ad for their "stack overflow jobs" thingy?

Balarka Sen

Jake Paul - It's Everyday Bro (Song) feat. Team 10 (Official Music Video)

►

Astyx

I got an email about this

Alessandro Codenotti

18:16

They have a jobs/career thing on SO

Balarka Sen

Need to go over 9000 to get the crown of a bad sleep schedule from me

SteamyRoot

Maybe your status is set to something else than "not looking" :P

Alessandro Codenotti

Over 9000 hours of missed sleep?

Astyx

It's quite well paid actually

Not that I'm interrested

Nor qualified

I prefer doing things with no concrete applications

Alessandro Codenotti

You should answer though, maybe it was meant to be sent to someone else

Daminark

18:19

@Astyx my brother

Astyx

I doubt so @AlessandroCodenotti

Balarka Sen

posting that video was a mistake

Astyx

Indeed

Balarka Sen

I am cringing at myself for doing that now

it's so bad that i want to throw myself into another galaxy

Daminark

@BalarkaSen that was truly a new level of trash, doesn't even sound nice

skullpatrol

18:21

^

Balarka Sen

With great trash comes great cringes

Daminark

Eh, I don't think the physics allows it, sorry

Balarka Sen

I can't handle it, @Daminark

It's too powerful for me

Daminark

Look up "Never Gonna hit those notes" for comparison

Balarka Sen

I mean, that was at least a thing

Astyx

18:22

@Daminark That's gold

Balarka Sen

@Astyx no

it's good but not gold

look up Never Gonna Follow That Train

Astyx

Dammit CJ

4

Eric Silva

@Daminark has anyone emailed you guys about what to lecture on for Ted's book yet

Balarka Sen

ok gotta get non-meme dinner

Perturbative

@Daminark I kinda liked the Rick roll song, you have permanently ruined it for me

I'm off to study for a Numerical Analysis test, cheers everyone!

Astyx

18:28

Seeya

Kirill

So, $f=x^2+y^2$ (red function in the picture). We define $g$ as $g: f(ty+(1-t)x)$, and that will be the gray curve. MVT says, there is a $t \in(0,1)$ such that $g(1)-g(0)=g'(t)$, so $f(y)-f(x) = g'(t)$. I want to show the MVT for scalar fields, means there is a $\xi$ with $f(y)-f(x) = \langle \mathrm{grad}f(\xi), y-x\rangle.$ How to transform $g'(t)$ $\textbf{formally}$ correct into the scalar product I need, like $g'(t) = \ldots = \langle \cdot , \cdot \rangle$?

Daminark

@Eric yeah, 2.1 on Monday, 2.2 on Wednesday, 2.3 on Friday

Are you not on the Google group?

Eric Silva

no

Daminark

Oh weird, that's where all the announcements are

Eric Silva

yeah id rather not be in it

Daminark

18:35

@BalarkaSen does that even exist?? Eh well, see you

Lolol

@Perturbative mwahaha, and see you

Mike Miller

mystical greetings, strange travelers

Jasper Loy

I got an email from a Mike yesterday, and I thought it was you.

Mike Miller

@BalarkaSen lol you have no idea what gta is but you still know CJ memes

ah, I don't email much

Daminark

Also everyone check out "Soviet army dancing to hard bass"

Balarka Sen

@MikeMiller I have played VC and SA back when I was 7 or something

I know, inappropriate games to be playing at 7

Mike Miller

18:41

Wow, you're way more of an edgy teen than I thought.

Balarka Sen

I had edginess in me but it took time to be unleashed

had to install edgy.dll

Mike Miller

meh i'll ask elsewhere

Balarka Sen

@Daminark I found some Soviet trash today

let me find it

here

Mike Miller

recipe i just learned for producing branched covers: take a section of a holomorphic line bundle that's transverse to 0 & equipped with an iso $K^{\otimes 2} \cong L$ for some $K$

Balarka Sen

does that come in opposite orders

how did that even happen

Mike Miller

18:49

the wifi is bad

Daminark

Ah I've seen that

Mike Miller

there's tons of soviet and non-soviet stuff on leftbook, any page w labourwave in the name

rip @labourwave, they got zucked

Daminark

What did they do?

anakhronizein

Oddly enough I was just reading about vector bundles in order to finish writing my notes on tangent spaces.

Balarka Sen

@MikeMiller So you get a double branched cover, right? Hm

gian

18:56

How would I show that if a subset of a topological space contains all of its limit points, then that implies that its closed? (I'm not concerned with the other direction). Also, I want to prove it without using metrics or closure.

Alessandro Codenotti

what's your definition of closed?

Daminark

Actually wait a second now that I think about it, what is the topology on the Grassmannian anyway?

gian

@AlessandroCodenotti, the only definitions of closed I have met so far are that a closed set's complement in space $X$ is open and that a set is closed if it is equal to its closure in $X$.

Alessandro Codenotti

ok, so you should show that if a set contains all of its limit points then its complement is open

Daminark

In the k-theory talk he mentioned vector bundles are formalized via the Grassmannian, but never said what the topology is

gian

19:00

But how would I show that? Would I need to show that its an arbitrary union of open sets?

Alessandro Codenotti

you could show that every point in the complement has an open neighbourhood completely contained in the complement

Balarka Sen

@Daminark $\text{Gr}(k, n)$ is the set of $k$-dimensional subspaces of $\Bbb R^n$. "Nearby" planes constitute an open set - that's all :)

More formally, it's a quotient of a subset of $(\Bbb R^n)^k$ (of $k$-tuples which are linearly independent - this is itself a manifold; give it the subspace topology. It's called the Stiefel manifold $V(k, n)$)

Daminark

Well what does "nearby" mean in symbols is my point

Balarka Sen

By identifying two $k$-tuples of vectors if they span the same subpace.

Daminark

Oh, so then it's the quotient topology under that.

I think I'm happy with that

Balarka Sen

19:06

Yeah

gian

Okay @AlessandroCodenotti, so I have an arbitrary $x \in X - A$ and an open set $U$ such that $x \in U$. Since $x$ cannot be a limit point of $A$, $U \cap A = \emptyset$.

Mike Miller

@Daminark leftbook pages just get the zucc sometimes

they had a nice thing you could stick on your profile pictures that said "please dont divide the left on my profile"

gian

Oh but then this implies that $U$ must be contained in $X - A$.

Daminark

shrug

anakhronizein

Having already defined the tangent space at a point, what's a good way of bringing up the tangent bundle?

Daminark

19:10

And somehow Facebook decided that was against the rules or smth? idek

anakhronizein

Spivak starts with TM then talks about T_pM afterwards.

Mike Miller

Nah thats unrelated I just liked it

Daminark

Oh lmao

anakhronizein

And Lee & Tu just introduce it without much motivation.

Mike Miller

you're trying to put all the fibers together at once

vOv

Semiclassical

19:11

@anakhronizein Huh. How?

Alessandro Codenotti

yep @gian

Mike Miller

The way it works is right-wing people report the pages and eventually the mods take them down and the corresponding thing is not as true the other way around

anakhronizein

How what, @Semiclassical?

Mike Miller

he wants to know how spivak defines TM

Semiclassical

How do you start with TM and then T_pM?

gian

19:13

But @AlessandroCodenotti, why does that prove that $X - A$ is open?

Daminark

@Balarka okay so, once we want $Gr(n,\mathbb{R}^{\infty})$, will this still hold nicely?

Balarka Sen

Yup.

It's a quotient of a subset of $\Bbb R^{k\infty}$

lol

Daminark

Kek

Mike Miller

@BalarkaSen n\infty, right?

PVAL-inactive

I usually just think about the infinite one as a direct limit.

Balarka Sen

19:15

Ah, sorry, I was using my original notation

Yes, in Daminark's notation

anakhronizein

It's just without the consideration of T_pM as a vector space itself.

Daminark

Oh whoops

anakhronizein

For example, in R^n, it is most natural to discuss the tangent bundle than the tangent space since people are familiar with the whole "displacement of vectors" thing.

Alessandro Codenotti

@gian for each point in $X-A$ you construct such a set $U$, then take their union

anakhronizein

So how Spivak does it, is introducing the tangent bundle on R^n, then discussing vector bundles and the differential.

gian

19:17

Oh I see. Thank you.

Daminark

@PVAL huh

That makes sense, I think

anakhronizein

Which makes for a good motivation, I think.

Balarka Sen

$\Bbb R^\infty$ is a direct limit of $\Bbb R^n$'s anyway

anakhronizein

However, I don't know what would make a good motivation for working T_pM towards TM.

The best I can come up with is an alternative way of looking at vector fields.

But I have not introduced vector fields yet, so I kind of have writer's block.

Abcd

0

Q: Finding the value of trigonometric expression using given data.

If $\sin\theta + \sin\phi = a$ and $\cos\theta+\cos\phi = b$, then find the value of: $\cos2\theta+\cos2\phi$. My attempt: Squaring both sides of the second given equation: $\cos^2\theta+ \cos^2\phi + 2\cos\theta\cos\phi= b^2$ Multiplying by 2 and subtracting 2 from both sides we o...

Daminark

19:22

@anakhronizein you don't really have to provide the motivation before you introduce things, it's totally fair to say "Okay it'll seem like I'm defining this for no reason at all but be patient, it'll make sense soon"

anakhronizein

Of course I don't have to.

But I want to.

It's one of the thinks I find math classes lack a lot of the time.

A pet peeve of mine.

Daminark

I mean the issue with some things is that the motivation is that an intuitive idea of conceptual importance now makes sense once you define this thing

So you're kinda stuck, either you have to delay motivation or introduce something before it makes sense

anakhronizein

Well in the case for the tangent bundle, what might you suggest?

Daminark

Tangent spaces allow you to talk about derivatives of maps between manifolds

anakhronizein

Well yes, I have tangent spaces already.

But to arbitrarily take the disjoint union of the tangent spaces.

Balarka Sen

19:32

Derivative of a map $f : M \to N$ is a map $Df : TM \to TN$.

anakhronizein

I was thinking using the global pushforward, but it doesn't seem too important for the scope of the notes so far.

Daminark

I mean, you can give the intuitive idea of a vector field maybe?

anakhronizein

I guess. I guess I could also just go back to the displacement of vectors and do it like spivak as well.

I am not sue, I will think on it.

Balarka Sen

The tangent bundle is like the whole point of smooth manifold theory. A smooth manifold is literally equivalent to the information of the underlying topological manifold with it's tangent bundle.

A topological manifold comes only with a manifold structure. A smooth manifold comes with a manifold structure plus it's tangent structure.

The tangent structure is encoded in the tangent bundle.

What else do you want

Mike Miller

@BalarkaSen this is not true unless you're saying it in a silly sense

Daminark

19:36

You could do that. Again I don't think you can get around either introducing ideas which are as of yet sketchily-defined or asking people to suspend disbelief about the importance of something until later

anakhronizein

I could approach it with the 'unification of information' idea, but that seems a bit less interesting of a motivation

Balarka Sen

@MikeMiller I was saying it in a silly sense: As in, you could define tangent spaces for smooth manifolds. But I also had microbundles vaguely in my mind when I said it

Mike Miller

Yeah what you're thinking of is a very hard theorem and the way you stated it isn't quite accurate. :P I don't like "equivalent" there

especially "literally equivalent"

Balarka Sen

whenever I say equivalent I almost always mean $\implies$. I don't know why.

please cross that off with implies

@anakhronizein Here is, in my opinion, the very first major theorem which is proved using tangent bundles: every smooth manifold $M$ of dimension $n$ embeds in $\Bbb R^{2n+1}$.

Also known as "Whitney embedding theorem"

ctoi

I think the point is that otherwise for p,q distinct, the tangent spaces TpM and TqM on their own have no relation whatsoever. That shouldn't be the case though, there is some kind of smooth dependence (motivated by considering vector fields in R^n, for example). TM provides a way of giving this 'bundle' of spaces a smooth structure, which allows you to define a notion of smoothness for example.

zed111

19:42

Hey @Semiclassical, have you read Noether's theorem from Ch. 1 of Altland, Condensed Matter Field Theory?

Semiclassical

Probably. I remember the content of it (continuous symmetries imply conserved charges) and I do have that book.

zed111

Great. I've read about it from other sources but the explanation in Altland is pretty unclear to me. It calculated the action difference between transformed and original Lagrangians and says the difference must be linear in the derivatives ∂_μ(ω_a). How do we get that? Here ω_a are the parameters of the transformation and μ is the index of cartesian coordinates.

Semiclassical

That's more than I remember.

zed111

Ah alright. I probably need to stare at it more carefully.

Jasper Loy

Wow there are so many people in this room now, but most are quiet.

Daminark

19:49

shrug

Jasper Loy

Anyone knows about Mr Eyeglasses? I haven't heard from him for so long. I hope he's OK.

anakhronizein

@ctoi that's a very good idea. Because considering the tangent spaces we don't have any connection whatsoever. Always sticking locally.

Danu

@zed111 I think simply because you want to take the variation "up to first order"

The point is that, in the end, you want to take a derivative with respect to some parameter $\epsilon$ and then set $\epsilon=0$ afterwards, so you only care about linear terms

zed111

20:17

@Danu thanks. But why linear in the derivatives of the form ∂_μ(ω_a) only. Why not terms like ∂_(ω_a)(x_μ) ?

Danu

@zed111 It is not clear what your parameters mean, but I doubt that the coordinates should depend on them.

zed111

okay. Seems reasonable.

Finally after calculating the action difference, $\Delta S=\int j_\mu^a(x) ∂_μ(ω_a)$ and saying that it should be zero for symmetry transformations, the book says $\partial _\mu j_\mu =0$. I guess this is obtained through integration by parts of $\Delta S$ and setting it to zero?

Semiclassical

sounds right. they'll be assuming that $\omega_a$ vanishes at infinity, I think?

Anonymous

@Semiclassical May I ask you a question about Schrodinger's equation? I was told that "energy eigenvalues are experimentally measurable values in a lab". My question was: how does one measure eigenvalues in lab?

Semiclassical

Well, suppose you've got a particle in a box, occupying an excited state.

Anonymous

20:30

@Semiclassical Okay? Like an electron in excited state (within an atom)?

Semiclassical

Right.

By emitting an photon, that electron can drop from the excited state to the ground state.

Anonymous

Yes....

Semiclassical

The wavelength of the emitted photon then tells you the energy difference.

Anonymous

@Semiclassical Okay?

Semiclassical

Now, to be fair, this only tells you the gap in energy.

But, eh, it's differences in energy that are typically physically meaningful.

Anonymous

20:33

@Semiclassical Got it till here. Then?

Semiclassical

Then you're done? You've measured the difference in energy between a particular excited state and the ground state.

Anonymous

Sorry. I couldn't understand what you are saying. Basically I was told this:

Anonymous

in The h Bar, 2 hours ago, by Avantgarde

$\hat{H}$ is an operator, which in some basis (say position basis) is a matrix. $\Psi$ is a vector (ray, actually, but forget that for the time being) in Hilbert space. In general, when an operator acts on a vector, it results in a different vector, which is unrelated to the original vector. In some cases, however, the resultant vector is just the old vector times a constant multiple. Such special cases are what observables are about in QM. That constant multiple is called the eigenvalue

Anonymous

in The h Bar, 2 hours ago, by Avantgarde

And the constant multiples are the experimentally measurable values in a lab

Anonymous

How is that "constant multiple" obtained from the energy difference?

Semiclassical

20:35

Well, let's take a special case---the ground state energy of the Hamiltonian.

Anonymous

@Semiclassical I'm a newbie to this. Can you say what you mean by ground state "of the Hamiltonian" ?

Semiclassical

The lowest stationary energy level.

For instance, if you take the Hamiltonian to be that of a harmonic oscillator $\hat{H}=-\frac{\hbar^2}{2m}\frac{d^2}{dx^2}+\frac{1}{2}m\omega^2 x^2$

Anonymous

@Semiclassical What do you mean by ground state of the "Hamiltonian"? I know that ground state means the lowest stationary energy level when talking about an electron in an atom

Anonymous

@Semiclassical Okay, what's the ground state of that Hamiltonian?

Semiclassical

I'm forgetting the precise paramters, but it's like $\psi(x)=Ae^{-x^2/2a^2}$.

With $a$ chosen properly, that'll satisfy $\hat{H}\psi = (\frac12 \hbar\omega) \psi$.

Anonymous

20:42

@Semiclassical Wait. How did you find $(\frac12 \hbar\omega)$ ?

Semiclassical

With an atom, you'd have $\frac{\partial^2}{\partial x}^2+\frac{\partial^2}{\partial y}^2+\frac{\partial^2}{\partial z}^2$ in place $\frac{d^2}{dx^2}$ and the potential would be a different function.

memory, if I'm honest.

blah, should've been $\partial x^2$ on the bottom

Anonymous

I didn't understand how you found the $(\frac12 \hbar\omega)$ though...

Semiclassical

I didn't.

Anonymous

@Semiclassical What?

Semiclassical

I'm saying that I remember that that's how the calculation works out.

Here's the computation. If you take that $\psi(x)=Ae^{-x^2/2a^2}$, and act on it with $\hat{H}$, you'll get...

Anonymous

20:47

Okay. Just by chance I'll get some constant times the $\psi$ ?

Semiclassical

$$\frac{d^2}{dx}^2 Ae^{-x^2/2a^2}=\frac{d}{dx}(-x/a^2)Ae^{-x^2/2a^2}=(-1/a^2)Ae^{-x^2/2a^2}+(-x/a^2‌)Ae^{-x^2/2a^2}$$

It's not exactly by chance. The condition $\hat{H}\psi(x)=E\psi(x)$ is an ODE.

As such, it can be solved to obtain $\psi(x)$.

What makes things tricky is that, if you just pick $E$ arbitrarily, that solution won't be physically reasonable.

The idea of $\psi(x)$, as a wavefunction, is that $|\psi(x)|^2$ should correspond to a probability density.

As such, you need $\int_{-\infty}^\infty |\psi(x)|^2\,dx=1$ (the probability of finding the particle somewhere is 1).

At a bare minimum, that only works if $\psi(x)\to 0$ as $x\to\infty$.

Astyx

(a square is misplaced in the big equation)

Semiclassical

erk, yes.

a few of them, in fact. sigh.

It turns out that, for most values of $E$ that you pick, the corresponding $\psi(x)$ will not go to zero at infinity.

In fact, the only values of $E$ for which this doesn't happen are: $E=\frac12 \hbar \omega,\frac32 \hbar \omega,\frac52 \hbar \omega,$ and so forth.

Anonymous

Why are you using the word "you pick"? Do we just run through all real values to check which one can be a suitable E ?

Semiclassical

Right.

Anonymous

20:52

Using computers?

Semiclassical

Nah, using math. One can solve that ODE as a series expansion, and show that series doesn't converge everywhere unless $E$ has those specific values.

A textbook will have that derivation.

That's what makes these problems a bit of a pain, tbh. When doing the wave equation, one can always solve the eigenvalue problem in a formal sense. But it won't be the right kind of function (square-integrable, as one says) unless $E$ is chosen particularly.

Anonymous

@Semiclassical Do you know any website which discusses that method?

Semiclassical

Wikipedia almost certainly does. Look up their page on the 'quantum harmonic oscillator'

(I should also note that, if one is doing a particle in an infinite square well, the issue is a bit different: In that case, one demands that $\psi(x)=0$ at the edges of the well. So you're trying to solve an ODE subject to certain boundary conditions, and that won''t work unless $E$ is an energy level.)

(I don't remember the energy levels of the infinite square well, so don't ask. :P)

But I'll have to be going for now.

bye

Anonymous

@Semiclassical Thanks for the help. Bye!

Kirill

21:42

Where does the tangent vector on the level set start and end?

For some function $\mathbb{R}^2 \to \mathbb{R}$ and for some point $(x,y)$ the tangential space should be a plane. Is that plane a level set?

anon

what do you mean by the tangential space? the tangent plane as a graph in R^3? if so, then of course the plane is not a level set, it's not even located in the domain.

Kirill

hi @anon, $T(x)=\{v \mid v \text{ is a tangent vector on }N_c(f) \text{ in }x\}, f$ function, $x \in \mathbb{R}^2, c=f(x), \mathrm{grad}f(x)\ne 0$

anon

$N_c(f)$?

Kirill

oh, $N_c(f)=\{x \mid x \in \mathbb{R}^2 \text { (here) }, f(x)=c \}$

again, we have $f(x,y) = x^2 + y^2$ (x,y coordinates here), the point $F=(1,2)$ and the point $G=(1,2,5)$ that is on the surface.

tangent vectors are $\in \langle \begin{pmatrix} 2\\1\end{pmatrix}\rangle$

I just want to draw some of them, but don't know where they should start and end.

here: $N_5(f) = \{ x \mid x \in \mathbb{R}^2, f(x)=5\}$

Kirill

22:16

oh, $\begin{pmatrix} 2\\-1\end{pmatrix}$ two lines before

Wheat Wizard

23:10

Hey could I get some tagging help? I have this question but I don't know the tag system on Math.se well enough to know what tags should be added. I have added recreational-mathematics but I'm not sure if that tag is even supposed to be there.

It is not professional mathematics, and it seems to loosely fit the tag definition, but I am still wary because I haven't used the tag before.

Akiva Weinberger

23:40

\begin{align} 2^7&=5^3\\ 2^4\times5&=3^4\\ 2^2\times3^2&=5\times7\end{align}

Daminark

@AkivaWeinberger wao reax only

anon

$$ 2^4+2\cdot 4=24=4\cdot\binom{4}{2}$$

(counted from the vertices of the 24-cell and its dual)

Akiva Weinberger

That's legit true, get out

Ali Caglayan

Suppose, for a fibration $F\to E\to B$ I know the integral and mod-n cohomology rings of some fiber $F$ and base $B$, is this enough to determine the cohomology rings of the total space $E$?

This is really a question about group cohomology. Equivalently, given some split extension of $G$ by $H$ is it possible to determine cohomology rings of $G\rtimes H$ from the cohomology rings of $G$ and $H$.

Daminark

Hey @Ali! How've you been?

Ali Caglayan

23:54

But I propose it in two forms incase one sticks

I found this question mathoverflow.net/questions/250991/…

but it has no answers :(

and Hi @Daminark

I have been good how are you?

Daminark

I'm alright, thanks!

Ali Caglayan

@Daminark How has your summer been so far?

Daminark

It's been a whole lot of fun, very busy :P

Yours?

Ali Caglayan

Too much to do, too little time

What topics are you seeing next year?

Daminark

I'm still in the middle of figuring it all out

Algebra for sure

$>$ 90% complex analysis

Likely gonna audit rep theory

Combo for sure

Algorithms for sure

Transcript for

$Mathematics$ Mathematics