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12:01 PM
Actually I think I misremember it. Even though the square of $\tilde f\to k\tilde f$ is $-\Delta$, it's not its square root
 
Suppose $\lim x_n = a$, is it always true that if $x_{n+1} = f(x_n)$, then $\lim f(x_n) = a$?
 
I don't understand. After uncovering notation, you are just asking: if $\lim_{n \to \infty} x_n = a$, then is $\lim_{n \to \infty} x_{n+1} = a$?"
The $f$ serves no purpose except to obfuscate, so I removed it
 
yea pretty much
the first option
nvm its true im dumb
 
I don't agree you're dumb because of the question. Why don't you tell me the definition of $\lim_{n \to \infty} x_n = a$, and then explain to me why your desired result follows from the definition?
 
12:17 PM
Because for all epsilon there's an N so that for all n≥N $|x_n-a|< \epsilon$, so $|x_{n+1}-a| < \epsilon$ right
 
Let's say it carefully (it is always good to be very careful/precise when learning something new, and to be less careful once you have mastered it).
You've just told me that for all $\epsilon$, there exists an $N = N(\epsilon)$ so that for all $n \geq N(\epsilon)$, we have $|x_n - a| < \epsilon$.
From there I think you want to say: "if $n \geq N(\epsilon)$, then $n + 1 > N(\epsilon)$, and so this condition holds for $n+1$ as well; we have that $|x_{n+1} - a| < \epsilon|$ for all $n \geq N(\epsilon).$
And therefore that same $N(\epsilon)$ is also a suitable $N$ for the sequence $x_{n+1}$.
 
Ah right
 
You have to change things slightly if your second sequence was instead $y_n = x_{n-1}$. Do you see how to fix it in that case?
 
yea, I'd define N = $N_\epsilon + 1$ right
 
Yeah!
 
12:21 PM
thanks
 
@MikeMiller "I don't agree you're dumb because of the question." Then why? ;-p
 
for sure
I hope it's clear that there is something gained in being so precise, that it's not just pedantry. I think the point of being so precise is that you understand the why more, and it helps you generalize to other situations once you understand the reasoning.
@robjohn Good one... I've mostly trained myself to avoid ambiguity like this, after a student was deeply hurt after mishearing "And you're done".
 
@LHC2012 It is always good to practice new things, and being extra precise offers good practice.
 
yea ic
 
hello people.
any hints on solving
$$ \frac{dV}{dr} = \frac{-1}{4 \pi r^2 \epsilon}\left( q - k \int_R^r V(r') 4\pi r'^2 dr'\right) $$
multiplying by r^2, and then differentiation both sides wrt r gives:
$1/r^2 \frac{d}{dr} (r^2 \frac{dv(r)}{dr})= cv(r)$, where c is a constant
substituting r=1/t results in
$t^4*d^2v/dt^2=cv$
any hints on solving this?
 
12:41 PM
@copper.hat Oh, thank you. That confirms what I thought afterwards.
 
1:39 PM
@AlessandroCodenotti Off the top of my head, no. But it seems counterexamples to your specific situation are abundant: take the Denjoy counterexample T : S^1 -> S^1 which is topologically transitive but has an invariant Cantor set, and then put mass 1 million on the Cantor set and Lebesgue outside. This seems to be an invariant measure which is not ergodic to me
It seems to me that you can always "blowup" some generic topologically transitive map to construct examples of this sort.
 
2:26 PM
The Denjoy counterexample?
 
Hello, math people. Can I ask a question about calculus?
 
0
Q: Is this correct for pressure exerted by a liquid?

Physics is my life $P_0$ is atmospheric pressure and P1 is height * density * gravity. So , is the diagram correct for saying that Total pressure in the liquid = P0 +P1. Since the direction of P0 vector in downwards?

If you can answer , then pls check
 
This is somewhat surprising, because in the topological setting if $G$ acts in a topologically transitive way on a space $X$ (with some assumptions that I don't remember) and $A\subseteq X$ is $G$-invariant with the Baire property, then it is either meagre or comeagre
Ah we want $X$ to be Baire which I guess doesn't have a good translation in measure theory terms
 
2:51 PM
@MikeMiller formalize the classification of surfaces then
 
@user2103480 You will see above that my advice was to do this for things you don't understand well
 
how can one claim to understand something without having formalized it in lean huh??
I'm just salty that I have to do topology now instead of staying in the comfortable lane of stochastic analysis
 
but topology is beautiful
 
3:09 PM
yeh but analysts often have a different standard of rigor and I still find it hellishly complicated to formally compute anything with CW complexes. We calculated the cellular homology groups of RP^n via the local degree and when it came to actually computing the local degree we just skipped it cause its tedious calculation in local coordinates
and this "yeah it works but its tedious and obvious" happens over and over again and it was the same for the HoTT course and I really really just want some details written down carefully
 
@user2103480 This is a failure on the part of those teaching if they can't fill in the details
The best topologist knows how to explain every step in detail but only explains the revealing steps
Sometimes people need the details but at this point they should be getting trained to learn which details really are trivial to fill in and that requires doing them
 
I'm sure they can but there seems to be more of a tendency to assume that others can do that as well. And it's not like I'm generally unable to do that, but the more details are left out, the more topics are there to go through in the lecture's time so the amount of details to fill in, or learn to fill in in various areas, gets larger and larger
Of the four different lecturers in topology/geometry-related courses I attended, it was like that without exception
Interestingly, one of those lecturers also did ODE, complex & functional analysis classes that were much more stringent
 
3:25 PM
@user2103480 Yeh I get your point
I don't disagree
There is supposed to be a transition
 
embrace the pictures
 
Not a jump discontinuity
 
@Thorgott I tried so many times.....
And I will definitely give the correctors of my exam a taste of their own medicine
"It's a perfectly clear 7-dimensional picture"
 
What textbook are they having you use
 
There's no clear textbooks. There's of course significant overlaps with (and mentions of) Hatcher and Fuchs & Fomenko
 
3:31 PM
Oh those are good on the details though
Ah well anyway I'm not gonna give you a hard time for having a hard time
 
Also, there's lecture notes by CW (the initials of the professor leading the research group, maybe you remember) and by another professor, alas in german
@MikeMiller Not denying that!
 
I definitely understand the importance of the transition and for students to absorb which details are easy and which are not
 
The problem is more the accumulation of things we have to check out. Our lecturer said himself before christmas that if we actually did all this in detail, we'd fill the remaining 8 weeks
 
The only thing that frustrates me is when people accuse standard textbook authors of being nonrigorous
Which is simply not true
 
But instead we have to reach poincare duality
That's not his fault though I think. Maybe it's the professor's standard so that later lectures can start at a "reasonable" level
 
3:34 PM
Yeah
I get the frustration though
 
It just doesn't feel like I understand and I don't
In Bonn they spend a whole semester on homology, with a bit more about homotopy sprinkled inbetween I think. I'm envious
 
thanks for reminding me I have to understand exotic 7-spheres
*dies*
 
@Thorgott why do you have to understand that
seminar?
 
yes
 
hf gl
 
3:40 PM
@MikeMiller Hatcher definitely isn't rigorous in some select places. This is just an observation, not a judgement.
 
Incorrect sorry
Well, ok, there are a few errors and a few places where the argument doesn't hold up, but nothing beyond the standard of textbook writing in general IMO
 
yeah, that's fair
 
I remember some hiccups in the PD section specifically
One of his fracture arguments I think needs to be patched
But I remember patching these on the go in a class, not something that takes a lot of extra work at home (unlike, say, G-H.......)
 
one thing that bothered me is how he doesn't address identification maps (well, what's needed implicitly follows from the discussion of the compact-open topology in the appendix, but he doesn't mention that some of these facts are actually needed to verify that some homotopies he writes down in the early chapters are in fact continuous). perhaps it's expected the readers figures that out themselves, but he usually is more careful.
which fracture argument do you mean?
 
Something around p247
I remember that picture
@Thorgott Identification maps, meaning quotient maps? You do need a real mastery of that to get into algebraic topology, and this is something a lot of students leaving a first course in point-set are missing, since it is treated so quickly
I make my students explain in detail how to go from his pictures that show pi_1 is a group and turn those into formulas and then a proof that the things defined in formula are continuous
 
3:55 PM
Yeah, for example in chapter 0 he proves that the quotient map for a pair satisfying the homotopy extension property with the subspace being contractible, the quotient map is a homotopy equivalence. He does so by taking a homotopy $f_t$ extending a homotopy equivalence from the subspace to a point and then factors each $f_t$ individually to a quotient map $\tilde{f}_t$.
These $\tilde{f}_t$ are supposed to be the final homotopy, but he doesn't justify why they are still jointly continuous in both variables (this comes down to the fact that the product of a quotient map with the identity map
 
exotic 7 spheres don't exist
nothing in nature is 7 dimensional
 
@MikeMiller Hmm, I thought that was sound when I read it. Time to go back and figure out what I missed.
 
I read Suppose $U$ and $V$ are sets and $T:V→W$ is a function. Then $T$ has a left inverse ⟺ $T$ is one-to-one, and $T$ has a right inverse ⟺ $T$ is onto.. Why is so? I think if $TV=W$, then $T^{-1}W=V$, but why does it indicate that $T$ is one-to-one? And if $VT=W$, then $WT^{-1}=V$, but why does it indicate that $T$ is onto?
 
what does $VT$ even mean
 
@Thorgott Aha, OK, I get your point. This is something he is indeed assuming from a point-set course (I stated that fact in mine before doing any algebraic topology, so that they could work with homotopies on T^2 in terms of homotopies on the square). I understand why he doesn't mention it even if readers aren't familiar with that fact: it points you in the wrong direction (being paranoid about the joint continuity)
 
4:01 PM
@Thorgott but if it has no definition, then where comes the right inverse?
 
To define the inverse you need the operator to be onto and one-to-one in the first place
 
@Bohemianrelativist having a unique left inverse $T' : W \to V$ means that for all $T(v) = w, T'(w) = v$, so you write $T(v_1) = T(v_2) \Rightarrow T'(T(v_1)) = T'(T(v_2)) \Rightarrow v_1 = v_2$
 
Guys . I have a confusion related to hydrostatics.
Could someone please help me
 
yeah, I get that, just my inner pedant is irked whenever a subtlety floats unaddressed in the air like that
 
Let me know if I can ask here.
No one is there is physics SE
 
4:07 PM
@BigSocks then what does having a unique right inverse mean?
 
You can ask whatever you want here or really anywhere
Whether or not someone will answer is a different question which you cant know until you ask
 
is it just me or is the $T$ in $VT$ lower than it should be?
 
Means that there's some $T' : W \to V$ s.t. if $T'(w) = v$, then $T(v) = w$, so you write $ \forall w \in W, T'(w) = v \Rightarrow T(T'(w) = T(v) \Rightarrow w = T(v)$.
 
@LeakyNun You mean in general or in a specific instance ?
 
in my latex renderer
 
4:13 PM
fwiw I see it that way too @LeakyNun
 
For me it's perfectly aligned with the other $T$
 
@Astyx which latex renderer you use?
 
@Bohemianrelativist holding back the urge to mention AC
 
No idea
I followed the procedure of the top right link
 
@user2103480 lmao, shouldn't matter
 
4:15 PM
But maybe that's not linked (no pun intended)
 
whatever
 
@BigSocks it's equivalent to "surjective functions have right inverses"
 
 
me too leaky
 
yeah, but they already gave it to them, so in the context of the question it shouldn't matter
 
4:17 PM
Yeah same here, the $V$ is taller than the $T$
 
I guess I'm just better than you
 
it looks shorter from afar, but the V is just as tall AS T when I zoom in sufficientlly
 
1 pixel difference in height of $V$
 
@Thorgott not on Leaky's SS clearly
 
4:28 PM
on my screen the $T$ is too short as well, same as Leaky's
 
do you need like a certain point score to upload images directly?
 
I mean zoom in on the browser, not on the picture. Perhaps that changes how the rendering is done?
@BigSocks I think so
 
gg me
and I took a screenshot myself of the browser idk. looks like that for me too
 
@Thorgott Smart, that does change it for me
It clearly re-renders as you zoom (as TeX ought to)
 
oh yeah
if I zoom in once it becomes the same size
 
4:31 PM
It doesn't change for me
 
if you zoom out?
 
@Thorgott it remains too short for me even at maximum zoom
 
If I zoom out or in
 
500% zoom is beautiful
 
5:08 PM
@Mike I just read through it again and I don't see the error in Hatchers subdivision argument
 
0
Q: Why do we write weight of body here in two forms?

Physics is my life If a body is half submerged in liquid , the weight of the body = m1 +m2? Why do we have two masses here? Why not just M as mass of whole body.? Total force exerted on liquid = m1g + m2g + P0 (atmospheric pressure) Height from mass m1 and m2 is different .

 
@Thorgott I got reminded again of how the ECTS calculations are very much unhealthy for students
my last semester had 16 weeks including exam preparation, and 30 CP corresponds to 900 hours of work which results in 56.25-hour-weeks
 
@MikeMiller All that work to show that a square commutes.
...and you guys were pissed at me when I was trying to do the same in de Rham setting.
 
And in the winter semester it's more weeks but also more material so basically the same pressure but longer
 
5:27 PM
@Physicsismylife you need to work on asking questions. just writing down the pieces of information that are in your mind at that moment are not sufficient. you need to think from the perspective of the person who is reading the question.
 
And to answer (my understanding of) your question, we divide the masses to make the computations easier: we consider the mass that is submerged and the mass that isn't. My take is the body's mass is homogeneous, so you can write the volume as a function of mass, then apply Archimedes principle to compute the forces on the object and find the state of equilibrium
But I agree with copper.hat that your questions are very poorly written, and it's hard to both make sense of them and help you because we have too little information about the context.
 
If a topological space $X$ can be written as a union of two open, connected, locally path-connected subsets such that the intersection of these two subsets has exactly two path-components then there is a surjective group homomorphism from the fundamental group of $X$ to $\Bbb Z$.
Can any one help me with this problem?
 
5:43 PM
Last tip: I strongly advise to not include images except diagrams (and in this case, make them as neat as possible, using specialized software instead of taking pictures of what you draw if that's possible). Type out every relevant equation in Latex, don't just take a photo of what you wrote on paper
@User873110 Do you see the intuition to this problem?
 
failure of using van -Kampen theorem in computation fundamental group of circle
yeah intuition is clear
how to write formally( more or less)
 
@Physicsismylife if you ask a question on a physics site and implicitly equate weight and mass, you are just asking for trouble.
 
I don't think local path-connectedness is a necessary hypothesis here, but maybe I'm missing something
also <insert joke about fundamental groupoids here>
no wait, obviously local path-connectedness is necessary
actually no again, it should be ok to drop local path-connectedness when one demands path-connectedness (it does break down when one only has connectedness in that scenario)
I should stop and actually think for a second
 
6:02 PM
My take is you want to quotient everything by $\pi_1(U_1)$ and $\pi_1(U_2)$
I think van Kampen doesn't require local path-connectedness
 
@Astyx and @copper.hat thank you for your feedback.I will work on it now
 
Then write $U_1\cap U_2 = A\cup B$ where the union is disjoint (I don't know the latex)
 
@Thorgott Not interesting enough for me to catch what the subtlety is from 10 years ago :)
Maybe it's even been corrected!
 
Take $a\in A$, $b\in B$, then an element of the quotient group will be of the form $(u_1u_2)^n$ where $u_1$ is a path connecting a to b in $U_1$ and $u_2$ is connecting b to a in $U_2$
That n is your surjection
 
fair
 
6:07 PM
There are people far more qualified than me to answer your question though
 
0
Q: Covariant derivative do Carmo help clarification concept

monoidaltransformLet $M$ be a Riemannian manifold and let $\nabla$ be a connection compatible with the metric. Then for any vector fields $V,W$ along the smooth curve $c:I\rightarrow M$ , we have $\frac{d}{dt}\langle V, W\rangle =\langle \frac{DV}{dt},W\rangle+ \langle V, \frac{DW}{dt}\rangle$ However my question...

 
@Astyx I have a doubt that when a body is partially submerged in a liquid. Then the whole weight of body reduces because of upthrust on its surface in water .
Then why did we divide the masses in two categories m1 and m2
How does it make computations easier
it is making me more confused
 
I don't know because I don't know the rest of the problem
 
@copper.hat
 
Give context
 
6:17 PM
In this image , question is to find pressure exerted on liquid
 
@Physicsismylife you need to start with clear labeling and diagrams.
 
I've already seen this photo on your question on physics SE
 
as we can see , that is what is in the answer
ok.Then I noticed ma*g was considered as atmospheric pressure
 
i am happy to help with solving a problem. i am not happy to try and unravel your convoluted approach to asking a question.
 
Ok.What is it now that you are not getting
I will clear it
 
6:20 PM
no. you need to step back and think about what it is that people are saying and asking and figure out what it is that you are not doing.
solving the problem is easy. but you need to learn how to ask a question.
 
Ok.I am writing . 1min give me pls
 
ok, local path-connectedness is in fact superfluous
 
I hope it’s better
@Astyx @copper.hat
Do help me pls
 
it seems like I'm the only one working on our SPDE sheets lol
found three mistakes in one exercise, which have now been corrected after the sheet has been uploaded for a month
 
6:37 PM
nerd
 
everybody there is a nerd it's probability IV
 
lmao truth
Intravenous probability
 
loooool
and I haven't been particularly ambitious either, I mean it took 1 month for me to even understand the exercise enough to find mistakes
 
I didn't do shit for probability during undergrad so
feel that
 
I did suspect there were mistakes making my life harder but the only thing making my life harder was my lack of understanding
the mistakes were only a bonus
 
6:41 PM
Is that not what makes all life hard
 
no I think if you really understand some things you go crazy
although that might be due to my lack of understanding
 
I'm crazy and I don't understand anything
drew the short straw
 
IT's nOT yOu tHat'S cReeeZy It'S tHe WoRLd
sike no its you
 
You're a beautiful princess Edward
 
hahaha the correct perspective
thank you bby x
 
6:45 PM
@EdwardEvans read more in my new book it costs only 19,99 and I explain that my thinking is just too unconventional to understand for most people :-)
unconventionality level: IQ 60
 
imagine marketing a book as something that patronises you
 
no no you're the elite cause you read the book and understand it
that is the trick
 
become one of the world elite for just 19,99
 
it's crackpot/alt-right 101
 
Now with free doctoral degree insert!
 
6:47 PM
by trump university
 
christ
imagine going to a university whose name is the sound a fart makes
 
@EdwardEvans that's called attending eton
 
@user2103480 oof
 
okay, it's not free but it's included in the fees
 
Prince Philip has an honorary doctorate from my alma mater
His robes are in a show cabinet in the marine building
 
6:49 PM
@EdwardEvans has he earned for anything else than mere existence?
Although the longevity of his mere existence is astonishing
 
I expect not
Yeah that's because he's an immortal vampire, as evidenced by literally any picture of him
 
lmao
prince philip: vampire
prince andrew: nonce
wonderful
 
Queen Elizabeth II: Galactic Overlord
 
Can't wait for when she molts and becomes Queen Elizabeth III
 
@Physicsismylife i answered what i think is your question.
 
6:59 PM
Hey @Ted could you ping me when you are around? I have a diffgeo question for you
 
are you suddenly doing humane math
 
70% chance it's some weird group action
 
@Thorgott no it's question someone asked me help with but it's way over my head
 
ah ok
@user2103480 that's what i was expecting
 
chance was calculated conditioned to it being something related to his interests
 
7:08 PM
thanks for explaining your methodology, mr. probabilist
 
transpacy is important
next time I'll train a neural net on alessandros questions and then generate my guess
 
pfff it's a completely honest question
 
it's an inception
it was planted into your head, that's not honest
 
@user2103480 Can you be sure Alessandro is not already a neural net to ask questions?
 
@Astyx trained on "pacific journal of mathematics"
 
7:20 PM
@Astyx we met in person many times so I hope he's sure
 
lmao
 
replying to demonic @Alessandro ?
 
7:37 PM
Hi @Ted
 
Ah, there's the demon.
 
To fix the notation $M$ is a smooth manifold and $B$ is some bundle over it. I'm trying to understand the correspondence between connections on $B$ and sections of $J^1B$ and there are a couple of details that I'm not convinced by
 
What kind of bundle?
 
Just a fiber bundle, everything is smooth, but I'm not assuming my bundles to be vector bundles or principal bundles here
 
I don't know what a connection on a random bundle is.
 
7:42 PM
To me a connection is a choice, for every $b\in B$ of a complement $H_b$ of the vertical vectors in $T_bB$
 
Ehresmann connection?
 
OK, and no further conditions?
 
they should constitute a subbundle, of course
 
OK. I don't usually think about this generality, but what's the question?
 
@Thorgott right
 
7:47 PM
@TedShifrin seeing the answer you put up just now (re: the gram matrix) I'm reminded of a pet peeve of mine in physics
 
Have I pet peeved you?
 
Ok so to get this correspondence I want, given a connection on $B$, to build a section of $J^1B$. The way this happens is as follow: for $b\in B$ with $x=\pi(b)$ find $\sigma_b\colon U\subseteq M\to B$ a local section such that $\sigma_b(x)=b$ and such that $\mathrm{Im}T_x\sigma_b=H_b$. Then map $b\mapsto j^1_x\sigma_b$ and this is claimed to be well defined and the map I'm looking for
 
your answer illustrates the reverse, actually
 
I'm confused, demonic. What happens if I take a vertical jet? Then I can't do this.
 
7:49 PM
in solid state books, one common idea is that of constructing a reciprocal lattice. so you start with a 3D basis $\{v_1,v_2,v_3\}$ and choose $\{u_1,u_2,u_3\}$ such that $v_j \cdot u_k =C \delta_{jk}$ for some constant $C$. (typically $C=2\pi$)
 
The point is that $H_b$ doesn't contain vertical vectors
 
I know.
But there are vertical jets. Oh, you're not claiming you get all jets.
 
the way the books typically do that is to compute normalized cross products of the initial basis vectors
 
Oh no, I just get some section $B\to J^1B$ from $H_b$
 
which i sorta hate
 
7:50 PM
Right, OK, @Alessandro.
 
much rather compute the inverse of $V=\begin{pmatrix} v_1& v_2 & v_3\end{pmatrix}$
 
However I'm failing to see both why $\sigma_b$ with the prescribed properties always exists, and why the result of this map is independent of which local section I pick
 
Is your pet peeve even really honestly related to the Gram determinant, Semiclassic?
 
not the gram matrix, but constructing a matrix $A$ with specified columns and reasoning based on that
it's a good method and it bugs me that that's not the standard approach in our books
 
So can't you construct $\sigma_b$ easily using a trivialization, Alessandro?
 
7:54 PM
especially because it trivializes the fact that $[u_1,u_2,u_3]\cdot [v_1,v_2,v_3]=C^3$
 
You're not trying to prove that you have locally flat sections, Alessandro. You're just trying to do this at a single point, right?
 
So you just need to think of $T_xH_b$ as the graph of some linear map $T_xB\to T_b F$.
So the physics literature is hiding duality stuff by abusing the cross product, I guess, Semiclassic. I'm not thinking about it too seriously right now.
 
wait what is $F$ now
 
7:57 PM
$F$ is the fiber?
 
i get it insofar as "we really really want to have formulas"
but formulas should make things -easier-, not -harder-
 
It's related tangentially to the question about how to get $A(u\times v)$ when $A$ is orthogonal. Is it $Au\times Av$? Nope.
This comes up in elementary differential geometry.
 
7:59 PM
@TedShifrin Do you mean $H_b$?
 
Oh, sorry, yes.
 

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