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4:00 PM
which also fits the original equation, so z = -y mod 503 should be the solution set
uh y = 0
in that case
z^3 = 0 so z = 0
 
@Balarka wtf is an index -1 critical point?
if your morse function is generic you definitely shouldn't be getting something 4-dim
 
Grr damn it. index 0 and index 2
I mean the minimum and the maximum
 
Oh, that there are four flowlines
 
thanks
 
4:02 PM
not a four dimensional family
Sorry, is this the tilted torus or a straight-up torus?
 
Tilted.
I am using blatantly nonsensical terminologies today
 
So you should get a single 2-dimensional family of flowlines
 
Here's a question for me: What's the most efficient way to plot the cross-sections of a tilted torus in Mathematica?
 
You can compactify this with broken flowlines if you like
 
Ok, what I meant is, I see the 2 flowlines which go along the maximum to the minimum. what are the other 2?
 
4:03 PM
Something is wrong about what you're saying
There should be (after modding out by translation) an ind(a)-ind(b)-1 dim'l family of flowlines
your thing should be 1-dimensional
and there's only one component
there are four flowlines that decrease index by 1, two per "paired" critical points
so two from the top to the index 1 closest to the bottom, and similarly for the other two
 
So let's say, the maximum to the saddle right below it
 
the algebraic / signed counts come out to 0
Yes, good. That has 2
 
I agree.
 
Same thing from the other saddle to the base.
 
True.
 
4:06 PM
That's your four.
 
AHH
 
Whenever I try to google for tilted torus google wants me to search tilted uterus.
 
Meh
@MikeMiller lololol
I was such a dopey there.
 
What is a family? I read on wikipedia that it's a collection of subsets of a set, so is it correct to say that a family is a set of subsets of a set? Perhaps I make things too complicated...is something more behind families?
 
wee mathematica
 
Zee
4:26 PM
@Felix.Cdeoends on context but usually it means a collection
 
A universal $G$ bundle is a $G$ bundle so that any other $G$ bundle is isomorphic to a pullback bundle over this bundle. Can this be expressed in a way that makes this seem like a universal property?
 
it's a choice of representation of the functor that sends a homotopy type to its set of principal G-bundles
hTop -> Set
 
@Fargle because that requires the assertion that the world is real. Nobody can claim that statement with 100% certainty. You say real world must be what is observed. I claim the lesser requirement of studying what is around them or the world they are a part of. After all, if there were a fictitious world, why wouldn't the residents of that world be justified in studying science?
 
usually ive seen universal properties to be expressed via a diagram, for example the stone cech compactification is universal in that any map from a space into a compacta factorises uniquely via the stone cech compactification
isnt this usually the meaning of universal?
(also cofunctor)
 
@Typhon Let me clarify, then: by "what we observe" I mean what comes to us through the senses, and therefore through experiment, and by "the real world" I mean that which can be said to exist independent of human observation.
 
4:32 PM
@Fargle ah ok fair enough
 
Science makes the tacit assumption that these two realms line up with one another.
 
i was thinking that like if it was a computer real world would only refer to the outside world even if there is a world inside a computer
 
Not necessarily.
 
so like there is some diagram which only this guy can fufill
 
Also, to clarify my own position, I think this is a perfectly fine assumption to make, but it is still answering a question of philosophy.
 
4:33 PM
no
 
@Fargle i never said it wasn't. i just said claiming its "realness" can be speculative. To some, what is imaginary is real. It is all relativity.
 
i don't really see the point in trying to make this fit with some language you know when you know precisely what the space does
classifies G-bundles, dawg
 
:D
 
i'm really hungry
 
its just my association with the word universal is so entrenched with diagrams
 
4:36 PM
@MikeMiller then eat food :D
@s.harp universal means applicable everywhere
 
i think that's bad
 
@Typhon yes but ive rarely seen something called a universal construction that did not have some kind of a categorical definition via uniqueness of some such diagram
 
@MikeMiller then go hungry
@s.harp im just referring to the word universal...
 
but in any case form the category of spaces with free proper G-action. the projection to the quotient shows that this is the same thing as a principal bundle
maps are G-equicariant maps
in the language of G-bundles, the maps are maps f: X -> Y with isomorphism f*P' = P
then EG is the terminal object in the homotopy category of this
 
siiiiick
 
4:39 PM
"Every G-bundle has aj wqiicariant map to EG, unique up to homotopy"
But that's the same as saying that the pullback of the universal bundle is isomoephic tonyour bundle
and the uniqueness statement says that there's only one map to BG doing that, up to homotopy
 
What are we talking about?
 
@BalarkaSen geometry IIIIN SPACE!
 
It's a beautiful result that [X, BG] = Vect_G(X). Please don't use the Vulgarization Functor on this
 
whats the vulgarisation functor?
 
One which takes beautiful geometric theorems and turns it into obscene algebraic symbols
Functorially
 
4:47 PM
so something like riemann roch gets turned into $\frac{\tilde\Xi}{\bar \Xi}$?
 
sure
 
its better if you write it on paper
 
@BalarkaSen english please
:p
 
@Typhon Nyet.
 
niemals!
 
4:48 PM
nada
 
@s.harp Would have made a great rap duo--they got bars
 
-1
Q: Is there a multivariate integer function f(x,y) that returns the number of factors of y in x with a closed form?

TyphonI'm looking for a simple function in terms of the elementary and exponential functions that when passed an integer x and integer y, it returns the number of times that y can be divided from x such that the result is an integer. I believe that it is something along the lines of $\lfloor\log_y(x)\r...

help?
:p
 
5:20 PM
Worst math oral exam ever, probably
 
@Astyx sorry
 
What are you sorry about ?
 
my question
"Worst math oral exam ever, probably"
 
Oh no, I wan't refering to your question
 
suuure
XD
 
5:23 PM
@Astyx :(
 
how did it go? @Astyx
 
I've had exams today and my math examination was just terrible
 
Zee
@Fargle hey buddy
 
hello friendo
 
It went alright, but the exercise were so dumb
 
5:23 PM
@Astyx details?
 
Like, it feels my two years training didn't help at all
 
Zee
@Fargle I enjoyed our talk last night
 
ah I know that feeling:P
but hey, if you passed, then it's alright I guess
Anyhow, if anyone has time: Let $n\in\mathbb Z_{>0}$ and $d$ a positive divisor of $n$. It is clear that the subgroup $\{\overline d,\dots,\overline n\}$ of $\mathbb Z/m\mathbb Z$ has order $n/d$ and index $d$. Now my book proceeds to write that a cyclic group of order $n$ for each divisor $d$ of $n$ has a subgroup with index $d$. I don’t understand why.
I know that each cyclic group of order $n$ is isomorphic with $\mathbb Z/n\mathbb Z$. However, does that mean that if $\mathbb Z/n\mathbb Z$ has a subgroup of index $d$ for each divisor $d$ of $n$, then automatically the same holds for any cyclic group? I guess I should show this using the isomorphism between $\mathbb Z/n\mathbb Z$ and this cyclic group, which we could call $\langle x\rangle$. Not sure how to proceed; maybe use contradiction?
 
@Zee To be frank, I didn't, but again, that's nothing against you. I'd rather not revive it if that's not too much to ask.
 
Zee
@fargle it's too early to discuss that kinda of stuff, day time is for math
 
5:27 PM
First one was about proving for $n\ge 6$ that $\sum_{k=3}^{n+2} k^n \lt (n+3)^n$, the exercise advised to first prove $\left(1-{k\over n+3}\right)^n\lt{1\over 2^k}$ for $k\in \{1,\dots, n\}$. Convexity solves this one for $n\ge7$ and you have to do $n=6$ by hand (read : with a calculator, bah). Then find all integers $n\in\Bbb N$ such that $\sum_{k=3}^{n+2} k^n = (n+3)^n$.
Second one was proving there was a unique solution of $y'-y=e^{-x^2}$ going to $0$ at both infinities.
I didn't pass yet, I won't have the results till end of july (although I'm pretty confident I'll get this group of schools)
This being said, I have to go and eat
See ya
 
@Astyx is there?
 
oh I think I see it
I should just think in terms of the isomorphism, so when I see $\overline{kd}$, I should think $x^k$.
 
5:49 PM
@Typhon There is
 
hello people of math
I come in physics
 
Is that a declaration of war? :P
 
6:04 PM
the fire rises !
 
Guys, how can I show that $H=\{\overline p,\overline{2p},\dots,\overline{np}\}$, where $p\nmid n$, equals $\mathbb Z/n\mathbb Z$? Say $\overline x\in\mathbb Z/n\mathbb Z$. We can write $x=qn+r$ for $0\leq r<n$, so $\overline x=\overline r$. I was thinking it should follow from here that $\overline x\in H$, but I'm not sure how.
 
@ShaVuklia $\nmid$= "doesn't divide" ?
 
yea
I want to show that $\langle\overline p\rangle$ generates the whole group @Hippa
 
It's clear that your group is cyclic, so you just want to prove that it contains exactly $n$ elements, no?
 
hm yes that is true @Steamy
I mean, it can't contain more than $n$ elements
so we could try $<n$ elements
and get a contradiction
ohh
right
why am I not awake :P
Oh, of course $np$ is the least common multiple of $n$ and $p$, because $\gcd(n,p)=1$
 
6:21 PM
Hai everyone!
@Steamy we must charge into battle! For Lurie!
:P
 
Ummm
I don't even know who Lurie is, but okay :P
 
The magnitude for the power of a lamp is $\sim 10 W$ right ?
 
Jacob Lurie, someone who I know wrote a book on higher topos theory
(Seems like one of the more abstract subjects out there, more contrast to physics I guess)
 
@Astyx Old light bulbs used to be 100W - I think LED's are about 10W nowadays, yes.
 
But that's the consumption right ? The actual power from the light of the lamp is far far less isn't it ?
 
6:25 PM
@Steamy 10 what?
@Astyx as well
 
Since LASER's only emit up to $1mW$
@Daminark 10 watts pretends he didn't get the joke
 
:P
 
@Astyx Yes, that's the consumption.
The emitted light is measured in "lumen"
and hence the efficiency is given in lm/W
 
Cause I had a physics oral exam where I figured you needed a 10W power transmitted to a small ball of metal in order for the light to be able to lift it
 
I see
 
6:28 PM
Hey everybody
 
Yo!
 
And the examiner asked me wether I thought that was much, and wether it was plausible to actually achieve that in a lab
Intuitively, obviously we do not know how to lift objects with light
 
Lift something with light - through radiation pressure?
 
Yup
 
0
Q: Geometrical Intution behind $\nabla(|f|^{p}) = p(p-1)|f|^{p-2}|\nabla f|^{2}$

ZophikelIn the text "Theory of Functions of One Complex Variable", I'm having trouble verifying the way I expressed the Laplacian Operator, and generalizing $(1.)$ to functions of more then one complex variable following Proposition in $(1.)$ Also I'm having trouble establishing the geometrical intuition...

 
6:29 PM
It'd help if that object is light
 
So 10W needs to be a lot (and I've been reading a laser is 1mW, which supports this assumption)
Badum tss @Dami
 
^ I'm having trouble trying to establish the geometrical intuition behind the following Theorem:
 
You can leave us now
 
How's thing going? @Dami
Heya @AlessandroCodenotti
 
Hey @Alessandro
I'm alright @Perturbative, how about you?
 
6:30 PM
$f$ is harmonic and real-valued on $\text{U} \subset \mathbb{C}$ and if $f$ is nonvanishing, then:
$$\nabla(|f|^{p}) = p(p-1)|f|^{p-2}|\nabla f|^{2}$$

$$\nabla(|u(x,y)+ iv(x,y)|)^{p} = p(p-1)|)|u(x,y)+iv(x,y)|^{p-2}| \nabla f|^{2}$$
^ anyone help me establish the geometric intuition behind this
 
@Daminark I'm good, mainly procrastinating at the moment :p
 
That question smells like Greene & Krantz? :P
 
But the fact that on lamps there is written power of order of magnitude 10W, I hesitated
 
Huh? @Steamy
 
Is all the remaining power converted into heat ?
 
6:31 PM
The complex analysis one.
@Astyx Pretty much, yes
 
Oh like a book?
 
So heat is a lot more expensive in energy than light ? That seems counterintuitive
Or does it ?
 
Hi @Daminark
 
Hey @Balarka!
 
What did you learn?
 
6:39 PM
So far today was Peter May's lecture
He reviewed categories/functors/natural transformations, and introduced adjoint functors
Talked about the smash product, which very much helped clarify what was up with loops and suspensions
Talked a bit about how free and forget functors are adjoint, and the $\eta$ and $\epsilon$ stuff creating a triangle which characterizes adjointness
Then he used the fact that $\Omega K(\pi,n+1) = K(\pi,n)$ to get that $\widetilde{H}^n(X:\pi) = \widetilde{H}^{n+1}(\Sigma X:\pi)$
Which he will soon show is one of the properties asked for by the Eilenberg-Steenrod axioms
 
Cool.
 
Oh also he mentioned that forgetful functor is adjoint to unioning a disjoint basepoint
So yeah that happened
It was very helpful, clarified a lot of going on yesterday
 
@Daminark Oh, so he did it backwards like you said he would.
shakes head
 
Interresting exercise (not too hard if you like algebra) I did not have today : If $\{A_1, \dots, A_p\}\subset GL_n(\Bbb C)$ is closed under product, show $Tr(\sum_{k=1}^p A_k) \equiv 0\pmod p$ ($p\in \Bbb N$).
 
hi chat
 
6:46 PM
Any product of them is also one of them
 
Thanks.
 
(removed)
 
(unremoved)
 
@Fargle That seems to be basically the Kantian distinction between things (phenomena) and things-in-themselves (noumena).
On a lighter note: The notes I'm looking at just kinda perplex me, and I can't figure if that's a problem with me or a problem with the notes.
 
@Dami You're so ahead of us
 
6:50 PM
(moved)
 
(I'm reading a bit on the pushforward of a map.)
 
God we are immature :p
 
Pah, this is precisely what mature people do!
 
lol, the last 4 messages in parenthesis.
 
I feel like the odd man out in this :P
 
6:51 PM
Oh god for a sec I thought you said "old man" and I'm like, dayum
 
I mean, what is normal anyway ?
 
pretty sure I'm that in this conversation too :/
 
You're cool to me @Semi
 
@Astyx Kernel of a homomorphism ofc
 
ofc, my bad
Actually that's a good one
 
6:52 PM
@Semi yeah if you're the grandpa, you're the fun skateboarder grandpa (is that a thing?)
 
@Daminark, you mean a perpendicular vector field to the tangent bundle.
 
The notes take $\alpha(t)$ to be a smooth map from a subset of $\mathbb{R}$ to $\mathbb{R}^3$, assigning to each real $t$ a point $(x^1(t),x^2(t),x^3(t))$ in $\mathbb{R}^3.$
 
(perhaps procrastinating all year wasn't such a good idea after all)
 
(it's working in R^3 because hey physics)
 
@Balarka nah I'm a finite group theorist, remember?
 
6:53 PM
I feel sorry for you.
 
Are you ?
 
Though I guess I acknowledge normal in the sense of closed sets being separated by disjoint open ones
 
Now, if I'm being naive, I'd understand $\alpha'(t)$ to just be the standard velocity vector
 
@Astyx did I never tell you that this was my line of work?
 
so just a vector of time derivatives.
 
6:54 PM
You might have
 
Jk I'm still a second year, though finite group theory is rather fun
Like, it's soon enough that I won't make too many bold claims with undue confidence
 
You're a second year ? How old does that make you ?
 
But they seem to want to understand $\alpha'(t)$ as being able to act on functions.
 
@Astyx Um, in particular you mean Tr(sum A_k) = sum Tr(A_k) is an integer?
 
But I'd probably go down an algebra route, either more number theory-y or algebraic topology. Those subjects seem to be most up my alley, as opposed to like, analysis
(We'll see about geometry)
Today I turn 20, second year college student, not grad school
 
6:56 PM
specifically, $\alpha'(t)(f)_{\alpha(t)}=\frac{d}{dt}(f\circ \alpha)_t$.
 
Yes @Balarka
 
I feel like I'm either missing something or they're abusing notation.
 
$Tr(\sum A_k) = lp$ for some integer $l$
And my internet connection is dying
 
Particularly since they go on to write the expression $\alpha_* (\frac{d}{dt})=\alpha'(t)$
where $\alpha_*$ is the push-forward.
Should I be thinking of $\alpha'(t)$ and $\alpha(t)$ as basically different kinds of objects? That's what would seem to make sense but it doesn't feel right...
 
{Tr(A_1), ..., Tr(A_p)} is then a subset of C closed under multiplication. Pretty sure that means it's, upto scale, a bunch of roots of unity (a Z/p subgroup of the circle group).
Maybe I'll think about it for a moment.
 
6:59 PM
Damn I already forgot all about Ted's lectures on differential forms
 
This isn't even forms yet :/
 
@Semiclassical $\alpha$ is just a curve, no?
 
All the math you've learned, you've forgotten? I'm shocked @Astyx
 
@BalarkaSen Are you sure about that ?
 
@SteamyRoot Yeah. a mapping from R to R^3 in this case.
 

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