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12:03 AM
@copper.hat because its about differentiating the integral
are you talking about the ae. existence of $\lim_{r \downarrow 0} {1 \over m B_r } \int_{B_r} f dm$?
bounded & measurable means (locally) integrable, so yes.
What do you mean
i must be missing something. if $f$ is BV it is integrable (locally) so ae. point is a Lebesgue point.
12:13 AM
I don't want to differentiate the integral of f
I want to differentiate f
presumably you meant no rather than Yeah above
a monotonic function is differentiable ae., and since a BV function is the difference of monotonic functions then yes.
12:29 AM
I live in Hawaii now
@copper.hat I mean yeah because I understand it as you asking what I mean by Lebesgue differentiation theorem
still at parents though
@copper.hat A BV function is a sum of two monotone functions, so what I want to show is that a monotone function is differentiable almost surely
12:36 AM
@Shaun we moved finally, living in HI now
Hilo somehwere in the jungle
It's very πŸŒˆπŸŒ§β˜”πŸŒ¦πŸŒ¦ rainy so far
Also haven't bought a pack or a lighter since here
got some lozenges tho
This one uses Vitali covering theorem
Guess there's no escaping, its either that or rising sun lemma
I did learn rising sun lemma once I think its beautiful name for a beautiful theorem
Anyone have an idea how to show algebraically that $(0,0)$ is the only point of intersection of the two curves above?
12:56 AM
@ShaVuklia The eigenvalues of [1 -2 ; 2 1] are not real.
@copper.hat What's the connection between the eigenvalues of the Jacobian and the zeros?
Following copper, set $\lambda = x^2+3y^2$.
@TedShifrin I haven't done a interesting proof by induction.
but I should though :) .
Want an interesting one?
@TedShifrin The only thing I can think of is to write everything in terms of $x$ and $\lambda$. Is that what you're hinting at?
1:05 AM
Just substitute $\lambda$ and look,
@TedShifrin of course.
@EM4 OK. Prive that whenever I pick $n+1$ numbers from $1$ to $2n$ inclusive, then among them there are $a$ and $b$ with $a|b$.
@TedShifrin ah, gotcha. thanks
Thank copper :)
Hm, but I still don't see the connection with the Jacobian though...
(though I'm glad I've solved it with that substitution)
1:10 AM
Write it as a system of linear equations. Jacobian is tangential (pun intended).
thank you!
EM, it’s one of my favorites. There are other approaches to proving it, but I’m suggesting you do induction.
okay that's a hella cool approach
thank you copper :D
1:15 AM
geen dank
2 hours later…
2:54 AM
Let $f(x) = \left(\log{1\over|x|}\right)^\alpha\varphi(x)$ in $\Bbb R^2$ where $\varphi$ is a smooth bump function which $=1$ near the origin and $=0$ for $|x|\geq 1/2$. Let $0<|\alpha|<1/2$. How can I show $\partial f/\partial x_i\in L^2(\Bbb R^2)$ in the weak sense?
3:34 AM
@DanielDonnelly That's awesome! Congratulations!
Prove that if $d|n$ then $\varphi(d)|\varphi(n)$, where $\varphi$ is the Euler function. My attempt: since $d|n$, there exists $k \in \mathbb{N}$ such that $n=kd$. For the fundamental theorem of arithmetic, there exists primes $p_j,q_j$ such that $k=p_1^{\alpha_1}\dots p_m^{\alpha_m}$ and $n=q_1^{\beta_1} \dots q_s^{\beta_s}$.

Thus, for the multiplicativity of $\varphi$, we have $\varphi(n)=\varphi(p_1^{\alpha_1}\dots p_m^{\alpha_m} q_1^{\beta_1} \dots q_s^{\beta_s})=\varphi(p_1^{\alpha_1}\dots p_m^{\alpha_m})\varphi( q_1^{\beta_1} \dots q_s^{\beta_s})=\varphi(p_1^{\alpha_1}\dots p_m^{\alp
in Helpful Commentary, 20 secs ago, by Shaun
Can I get some feedback on this, please?
Ups, I made some mistakes with the letters in my latter message: here is the corrected one, sorry. Prove that if $d|n$ then $\varphi(d)|\varphi(n)$, where $\varphi$ is the Euler function. My attempt: since $d|n$, there exists $k \in \mathbb{N}$ such that $n=kd$. For the fundamental theorem of arithmetic, there exist primes $p_j,q_j$ such that $k=p_1^{\alpha_1}\dots p_m^{\alpha_m}$ and $d=q_1^{\beta_1} \dots q_s^{\beta_s}$.

Thus, for the multiplicativity of $\varphi$, we have $\varphi(n)=\varphi(kd)=\varphi(p_1^{\alpha_1}\dots p_m^{\alpha_m} q_1^{\beta_1} \dots q_s^{\beta_s})=\varphi(p_1^{\
3:54 AM
you can edit the post for some short time afterwards. avoids littering the chat
4:05 AM
Very short time.
Sorry again, I tried to edit but I didn't make it in time.
Consider the ODE $x'=F(x)$ with $F:U\to\mathbb R^n$ ($U\subset\mathbb R^n$, $F$ $C^1$), and let $x_0$ be a fixed point which is an unstable focus. Is it then true that the unstable manifold of $x_0$ is just $\{x_0\}$? Intuitively I would say yes, since near $x_0$ solutions should spiral outward as for the linearized system, but I'm not sure how to prove this. (If this is true, then this simplifies certain Poincare-Bendixson arguments)

I also asked the question here. Not sure if this counts as cross-posting (if so, I'm happy to delete it here)
4:24 AM
With $x'=x$ the unstable manifold would be $\mathbb{R}$.
I vote with copper.
4:43 AM
O shoot, I meant stable* manifold... my apologies
(I see that in the post I wrote it correctly)
also, I disagree with your answer Copper (for unstable manifold)
I'm not dealing with a linear system, so other behaviour is possible
I do agree that it is reasonable that the unstable manifold contains a neighbourhood around the fixed point
however, I'm curious about a formal reason for this (especially for my question regarding the stable manifold)
i am not sure why you disagree. as $t \to -\infty$, $x_{x_0} (t) \to 0$ for any initial $x_0$.
that would be the definition of unstable manifold in my book.
I am not necessarily dealing with a linear system. That is why I disagree
i am giving a specific example which contradicts your statement above. not sure how you can disagree with that.
5:00 AM
Oh, I misread. I agree then
if you take $x'=x, y'=-y$ then the stable manifold is the $y$ axis, not just the origin.
that is not a unstable focus though
(which is what my question is about)
it is unstable. the eigenvalues are $\pm 1$.
my question is about an unstable focus
not just an unstable fixed point
unstable focus = outward spiral
5:25 AM
are you asking a question about the global behaviour? Hartman Grobman might be relevant?
1 hour later…
6:39 AM
wouldn't you like to know
a lot of bad global behaviour at atm
@Jakobian you can deduce that from the lebesgue differentiation theorem
let $f$ be monotone and and set $g(x) = \lim_{z \downarrow x} f(z)$, and then consider the Lebesgue-Stieltjes measure induced by $g$ on the real line
now you can apply the Lebesgue differentiation theorem to $dg$
this is just a rough sketch of the proof, but you can get the result you want by following this rough sketch
you should be able to prove that $g' = f'$ almost everywhere, and $g' dm$ is the absolutely continuous part of $dg$
i'm just happy to see europeans using 'hella'
you would need $g \ll \lambda$.
no you do not
the Lebesgue differentiation theorem works for any Radon measure
it recovers the absolutely continuous part of the measure in the lebesgue decomposition
that absolutely continuous part is the whole measure iff $dg \ll \lambda$.
By the Stieltjes measure, I mean $dg[a,b) = g(b) - g(a)$
6:49 AM
i stay away from radon because it is radioactive
thats wise
anyway, the whole thing is fairly intuitive, when taking $\lim_{r \rightarrow 0} \frac{dg[a-r,a+r]}{dm[a-r,a+r]}$ it ought to be the case that the part of $dg$ orthogonal to $dm$ gets killed in the limit
so what you are left with is the absolutely continuous part, which of course will also turn out to be $g'$
do you have a reference for the Radon measure version?
i mean you can ignore the singular part anyway
i guess Folland has it
folland will have it
the point is that you can ignore the singular part
then its just the usual lebesgue differentiation theorem
uhhh, folland wont have the radon measure version actually, but he has it for radon measure in euclidean spaces
but then you miss the whole Vitali fun
this theorem actually extends to a bunch of metric spaces
6:55 AM
maybe Durrett with have something
in some sense vitali fun is included here
since you need some sort of covering lemma to get an estimate on the hardy-littlewood maximal operator
and you need that estimate to get the differentiation theorem
so its there, its just hidden
but you dont need the full strength of the vitali covering lemma to get an estimate
you can make do with a much weaker covering lemma
oh nvm, you just use the 'finite version' as stated in the wikipedia page for the vitali covering lemma to obtain an estimate on the maximal operator
Durrett is surprisingly quite on this
the first chapter of geometric measure theory, by leon simon, is all about our discussion right now
he has a proof of the radon measure version, to probably the most general case I've seen
my memory for some detail is poor, so i stick to easily regurgitated theorems
his proof does use the full vitali covering lemma of course
the version for euclidean spaces just requires the finite version of that covering lemma
yeah I mean its not necessarily that useful to know these details, whats important is knowing how to apply the differentiation theorem
7:01 AM
who needs infinite dimensions anyway. the earth is flat
actually the finite version here is just referring to getting a disjoint sub-collection of balls from a finite one
the full vitali covering lemma gives you a disjoint sub-collection from an infinite one
with a worse constant (5 instead of 3 in the finite case)
the infinite one also requires a separable metric space to start whereas the finite one works in any
people seem preoccupied with the 3 vs 5 in the coverings
yeah , I've never gone deep enough to understand why its important
characteristic reply to similar stuff: Your rectagles are long and narrow, while the question is about cubes/balls.
i think the wine is overtaking my ability to remain coherent, time to hit the sack
good night
7:11 AM
nurse, he's out of bed and talking about balls again
i hope i never meet anyone who cares about the constants in those lemmas
i guess i can understand caring about where such a constant 'comes from' if you want some generalization specific to some nightmarish thing, but if the value itself actually matters somewhere, i sure as hell don't want to know about it
those people really put the anal in analysis
i have a covering lemma with the constant $-{1 \over 12}$.
at some point in a GUI i was working with we had labels load, select & run. i didn't like load so i changed it to analysis. when it got shortened in the GUI it came up as ANAL. i thought it was hilarious but my Italian boss didn't quite see it the same way
7:17 AM
well, in a much earlier existence, i added some code to some beta ware that printed something like FU Grant (Grant was an application engineer) randomly about 1% of the time. of course, i forgot about the code until the message popped up at a customer site
Grant was not amused
there may be a reason i am on the brink of financial ruin
you should write some software that is nothing but those jokes and see if mr ex-twitter is buying
he doesn't like me.
he was pushing mcgregor as Irish president (a mostly ceremonial role)
was a bit shocked at his FU to advertisers, i think he is taking his behavioural cues from an orange haired fellow
lots of very stable geniuses about
manifold unstable geniuses
7:32 AM
stable manifold?
throw in a tesla in orbit while we are conjugating
1 hour later…
8:56 AM
Is conjugation the only group action whose stabilizer is the centre of the group?
1 hour later…
10:10 AM
no, you can just post-compose with an automorphism
10:50 AM
So what’s so special about conjugation in the class equation you can just use any such function
You can write a class equation for any action $G\curvearrowright X$ of a finite group $G$ on a finite set $X$, you get $|G|=|\mathrm{Fix}(G,X)|+\sum |G|/|G_{x_i}|$ where $\mathrm{Fix}(G,X)$ is the set of fixed points of the action and $\{x_i\}$ is a transversal for the set of orbits of nonfixed elements
What if we require that the stabiliser of each element is the set of elements that commute with it. Then is the action conjugation?
1 hour later…
12:12 PM
@copper.hat Quite what?
12:24 PM
@leslietownes I'm going to make this my lifetime goal
Basic question, but suppose $A$ has an eigenvalue $\lambda$ of multiplicity $2$. Why is it that there exists vectors $u,v$ such that $(A - \lambda I)u = v$ and $(A - \lambda I)v = 0$?
Did you forget to write non-zero somewhere
yeah, indeed, non-zero, $v$ seems to just be the eigenvector associated with $\lambda$, but I'm confused about $u$...
In linear algebra, a generalized eigenvector of an n × n {\displaystyle n\times n} matrix A {\displaystyle A} is a vector which satisfies certain criteria which are more relaxed than those for an (ordinary) eigenvector.Let V {\displaystyle V} be an n {\displaystyle n} -dimensional vector space and let A {\displaystyle A} be the matrix representation of a linear map from...
ok, so it's a definition...
12:42 PM
hmm... not really
Either way... such non-zero vectors can not exist, they exist when the algebraic multiplicity is greater than the geometric one
1:16 PM
@Thorgott also what automorphisms does a general group have other than conjugation?
Or is this the distinction between β€œinner” and β€œouter” automorphisms ? Still I think my second question is interesting I think @Thorgott
2:12 PM
can anyone explain in more detail how the $\chi$ above relates to quantization
2:25 PM
well this is a maths group and the people here are not so well versed in physics as in mathematics, i am a physics student and i might be able to comment on this, however you are better off asking in the physics
it says, the spatial dependecy of chi is discrete intervals
so you are considering discrete areas of where your things might happen to be, instead of example, a discrete flow of possible states where the things could be , i believe. in this sense, it is quantisized
@TedShifrin hi
@TedShifrin Mind helping a poor shmuck ? :(
@VivaanDaga Maybe you can refer your question to exercises 6, 7 of Jacobson's basic algebra chapter 1.9, where he gives some insight about inner and outer automorphism.
Somehow outer automorphism is more interesting than the inner automorphism
@TedShifrin https://math.stackexchange.com/questions/4160103/un-diffeomorphic-to-sun-times-s1
i am trying to show that this mapping is diffeomorph, thus, it is smooth, i understand the definition for smoothness between manifolds that the chaining of functions is smooht for all charts.

however, i am not sure how i can apply this, like, i dont know how these charts look like for the given groups.
So i cant build $ \phi \circ f \circ \psi$ since i have no idea how phi and psi look like!
2:39 PM
@Mad here the word quantization has nothing to do with quantum physics
although I see it is misleading that in the text I sent the title of the section has quantum in it
@AlessandroCodenotti Yeah i know that. but when proving sylow theorems conjugation is specifically used cause the stabiliser of each element under conjugation is the set of elements that commute with it. so i wonder if conjugation is unique in that sense.
$\chi$ is the characteristic function maybe? The same as in the measure theory
I think it makes sense to consider it a characteristic function since it says $\Delta x_m = \epsilon \chi(x_m)$
no its general
general probability distribution
3:02 PM
@AlessandroCodenotti Is it still open if $N_n^m$ is universal for all $n$-dimensional subspaces of $\mathbb{R}^m$?
Remind me what the two indices in the Nöbeling space are. $m$=number of coordinates, $n$=number of coordinates that must be irrational?
Ah no, $n$=how many coordinates can be rational
the recent thing I could find was this arxiv.org/pdf/1712.03181.pdf
Anyway iirc it was proved in the 80s for compact subspaces of $\Bbb R^m$ (but I'm not sure if there is an english translation of the russian proof)
Ah yes this is mentioned in the paper you linked, it is the result by Stan'ko
@XanderHenderson quite quiet
3:17 PM
I'm actually interested more if an $n$-dimensional subspace of $\mathbb{R}^m$ has $n$-dimensional compactification which can be embedded in $\mathbb{R}^m$
This is known for $2n+1 \leq m$
and for $m\leq 3$ from what I know
@AlessandroCodenotti any ideas for my question
@Jakobian By Stan'ko's result this would be enough to solve the problem about $N^m_n$ so good luck!
@VivaanDaga Not off the top of my head. But I'm not a group theory person
@AlessandroCodenotti That's a lot of luck to put into me...
Dimension theory is too difficult
3:40 PM
Its more that, I still see myself as a baby in terms of how much I know about topology
3:54 PM
There is infinite topology to learn, (un)fortunately
Do you people think general topology is a dead field?
Not at all
but a lot of articles written are in obscure journals
4:10 PM
I feel the need to mention one of the most influential general topologists
General topology is viewed as some bizarre field, while a lot of it can actually happen in $\mathbb{R}^n$, intertwine with other fields like algebraic topology, and I don't think a lot of people think of it that way
that's why they say its dead
Its not really "dead"... its just mostly complete
as in there is a good progress on which questions are solved
I'd say this is very good
I bet its the same with ring theory for example
why is no one saying that ring theory is dead?
I find this very rude
i think most people beyond point-set, view general topology as a bunch of reasons we need to impose regularity hypothesis to capture our intuition of how things ought to behave
alternatively, we can think of general topologists as necromancers
i thought they are essentially set-theorists in disguise
4:45 PM
It's finals week.
I got an email from a student asking "How do I graph a polynomial?" I responded and said "Have you reviewed the lecture from 30 October (link)? Have you looked at the notes (link)? Did you look at the appropriate section of the textbook? I can't really try to reteach this over email, since it is a kind of visual thing to do, but if you can ask a specific question, I'd be happy to help."
xander: how about a big review study session from 5pm-whenever the day before the exam and
oh you're getting there
The response was "When do you have time to meet this week? Because of my other classes, I am only available for six and a half minutes from now until Friday."
lets also pre-book a six hour session the day after grades are submitted for a new form of appeals process where students have a right to counsel
@leslietownes Not really practical here, as (1) the class is an asynchronous class, (2) I have students at six different campuses (and would need to coordinate classroom access across those campuses), and (3) there isn't a time which is good for everyone, even if students were willing to drive to a campus.
xander i was kidding
4:49 PM
@leslietownes Ha!
the joke is that at least one student always wants the entire class (except they're paying attention this time) on an abbreviated pre-exam schedule
@leslietownes No, I actually like holding big-ass review sessions right before the final. I tried doing that during my first year here. No one showed up. :(
and if you did book such a thing nobody who wanted something would show up
you're anticipating my every move
I'm playing four-dimensional chess, sir!
Ugh... I'm just so tired of this semester. Can it be over yet?
@XanderHenderson dream response
from a student perspective
4:52 PM
CLASS IS OVER! if you want it
happy christmas from john & yoko
@leslietownes Mazel.
Channukah starts Thursday evening. I wonder if I should bring a menorah to my evening class, so that they can stare at the pretty lights while taking their final?
5:15 PM
@copper.hat Hartman-Grobman was indeed what I was looking for!
@VivaanDaga well, I'm a bit unclear about your set-up, is one group acting on another group? is a group acting on itself? are we talking about general actions or actions by automorphisms?
5:32 PM
@porridgemathematics My point is that general topology doesn't have to be set-theoretic topology. Some things that classify as general topology, concern subsets of $\mathbb{R}^n$, such as continuum theory and dimension theory. But when people think about what general topology is, they only think about some weird set-theoretic constructions, which is part of it sure... but not all of it.
5:49 PM
@Thorgott a group is acting on itslef
Given any elment in a group, say its stabiliser under an action is the set of all elements the commute with it, must the action be conjugation?
6:12 PM
@Vivaan What automorphisms of $G$ fix every element?
@ted well the identity
So the composition of your action with the inverse of conjugation must be the identity.
I'll be back later.
I'll be back... now!
*spins around in his chair*
I was back, but now I am front, again.
@TedShifrin I dont get how you concluded that
6:27 PM
@XanderHenderson having fun with your swivel chair?
This is what I imagine professors do all day long
@copper.hat Heck, yes!
No one showed up for this morning's final. I am not surprised---I expect that most of them are going to delay as long as possible and show up on Thursday night, but I had thought that maybe someone would show up.
they have a choice when to take the final???
@copper.hat Yes. Because this college doesn't actually set aside time for students to take finals.
that makes it hard for instructors!
6:32 PM
But still want to offer a 3 hour final, so I tend to say "Either show up at your usual class time, or show up at one of these other times."
@copper.hat Yup.
that is very flexible of you!
@copper.hat A lot of my students signed up for an asynchronous class, but I am not comfortable giving them a final which they can complete at any time during the week. So I force them to show up; I kind of feel like it isn't fair if I don't give them some flexibility, as they didn't sign up to come to campus.
If not for the asynchronous students, I would say "I teach classes at the following times: [times]. Show up to one of those. I don't care which."
makes sense.
Okay, no one showed up. I'm out.
6:59 PM
I think I answered my question take any abelian group, G, and let the action associated to each element in G be the identity mapping, then the stab of each element is the entire group G. but the action is not conjugation
it was a stupid question
maybe i should require the action to be faithful
then it would be more interesting
7:14 PM
@copper.hat A function is an indefinite integral of Lebesgue integrable function iff its absolutely continuous iff its of bounded variation and an indefinite integral of a Henstock-Kurzweil integrable function
@Jakobian I am not familiar with Henstock-Kurzweil integration.
I have had many courses in the uni now, and only a few of the courses will give a review at the end of the last few classes
@copper.hat Its like Riemann integration but instead of demanding that each interval in the partition has length less than some $\delta > 0$, we replace $\delta$ with a function $\delta:I\to (0, \infty)$ and demand that the interval $I_i$ in a given partition with a tag $t_i$ is such that $I_i\subseteq [t_i-\delta(t_i), t_i+\delta(t_i)]$
This small modification enables us to integrate more than just the Lebesgue integrable functions, and includes all improper integrable functions too
So the intervals don't have to be small uniformly
We can phrase Lebesgue integration on $I = [a, b]$ in a similar way
@Jakobian What I still don't understand is why being able to integrate "more" functions is a desirable thing.
No one said that?
7:27 PM
@Jakobian You did, several days ago, when I first asked why anyone should care about the HK-integral.
I did? I'm not so sure. Can you provide the comment
@Jakobian I am relying on my memory, and if you claim that you didn't say it, I will chalk it up to my own misunderstanding. Still, it seems implicit in the statement "This small modification enables us to integrate more than just the Lebesgue integrable functions, and includes all improper integrable functions too".
@Jakobian You can phrase Lebesgue integration using partitions. The difference here is that instead of tagged partitions we use free tagged partitions, where the tags $t_i$ don't have to belong to the corresponding interval $I_i$, we only demand that $I_i\subseteq [t_i-\delta(t_i), t_i+\delta(t_i)]$.
So the difference between Henstock-Kurzweil and Lebesgue integrals is that for Henstock-Kurzweil integrals we demand that $t_i\in I_i$
@XanderHenderson That's just to point out that even though its a small modification, ... blah blah blah
It'd be pretty disappointing if after modification we'd end up at Riemann intergation again
There is some benefits to Henstock-Kurzweil integrals, the main ones being that the functions admitting a $c$-primitive obey the fundamental theorem of calculus, and the other one is that there's no theory of improper integration
Sorry to cut the conversation, but I have a quick question: $f(z)=\frac{\cos z-1}{z^2}$ has an essential singularity at infinity since $f(1/z)=z^2(\cos(1/z)-1)$ has infinite terms in the Laurent series. Is this right?
@XanderHenderson the advantage of hk is thast it makes every derivative integrable which is quite useful in some cases
7:46 PM
$F$ is a $c$-primitive of $f$ if $F$ is continuous, $F'$ exists except for a countable set $C$, and $F'(x) = f(x)$ for $x\in I\setminus C$
@VivaanDaga Yes, I am aware. That is not what I was asking about.
so this is contained in what I said, and its a bit more general since $\int_a^b F' = F(b)-F(a)$ in this case too
@SineoftheTime sounds right
@SineoftheTime yes, if there were finitely many powers of (1/z) there, then the function f(1/z) would have a pole of some order at 0.
I forgot a proof during my exam today so I left that proof halfway.
I had studied that proof long time back but forgot a step in it.
@SineoftheTime in addition to this, you should also show that z=0 is not a removable singularity of f(1/z).
only then you can say essential singularity.
7:52 PM
it is not removable because it has no limit (?)
i.e., around 0 f(1/z) is not bounded.
@SineoftheTime yes that's fine too.
@XanderHenderson I'll be happy to talk about it more if you have any questions
@Jakobian No, that's okay. I went though an HK-integral obsession as an undergrad, too.
I'm not an expert but I feel like I'm more advanced into the topic than a lot of people
It's not a phase, dad!
7:56 PM
Parle a ta tete is a good song :).
@VivaanDaga But it is conjugation. It's just that in an abelian group the conjugation automorphism is the identity. This does not contradict my argument.
If $\phi\colon G\to \text{Aut}(G)$ is your action and $\sigma$ is the conjugation automorphism, then $\phi\circ\sigma^{-1}$ fixes everything and hence is the identity. From $\phi\circ\sigma^{-1} = \text{id}$ we get $\phi = \text{id}\circ\sigma = \sigma$.
@TedShifrin how is an action a map from G to Aut(G)?
That's the definition of a group action. If $G$ acts on $S$, then this means you have a homomorphism from $G$ to the permutation group of $S$.
Honestly I think I just don't like measure theory arguments but I like analysis. That's probably the appeal here
Analysis is far too broad a term. Measure theory is part of analysis. Complex analysis is part of analysis. You are far more narrow in your interests.
8:09 PM
That's true, I forgot complex analysis exists
Functional analysis...
but permutation group is not same as autmorphism group @ted for example left multiplication gives an action but not an automorphism
Yes, you're right, @Vivaan. Replace my $\text{Aut}(G)$ with $\text{Perm}(G)$. The same proof applies.
My error.
@XanderHenderson There are some parts where functional analysis and measure theory intersect, like considering duals of some spaces, I suppose. Thankfully I never got to consider dual of such space
Integration theory and measure theory are quite different. See Segal and Kunze Integrals and Operators for a measure-theory absent course in real integration.
I actually took that course from one of the authors. Probably the worst course I had as an undergraduate, but that was because the author was too arrogant to prepare for his lectures and gave us a take-home final where he couldn't solve two of the questions.
8:15 PM
OK. @TedShifrin but i still dont understand why the map you give must fix everything
I got an A anyway. shrug
@Vivaan Because you told me that $\phi(g)(h) = h \iff \sigma(g)(h)=h$ for all $h$.
@Jakobian Yes... tools like the functional calculus, the Fourier transform, the structure of $L^p$ spaces, and so on are built on the Lebesgue integral.
What is functional calculus?
In mathematics, a functional calculus is a theory allowing one to apply mathematical functions to mathematical operators. It is now a branch (more accurately, several related areas) of the field of functional analysis, connected with spectral theory. (Historically, the term was also used synonymously with calculus of variations; this usage is obsolete, except for functional derivative. Sometimes it is used in relation to types of functional equations, or in logic for systems of predicate calculus.) If f {\displaystyle f} is a function, say a numerical...
I know, I'm reading it right now
8:18 PM
You asked...
So its considering things like $\exp(T)$ or $\sin(T)$ where $T$ is an operator
in a metric space X, is it true that for a closed set A, d(x,A)=0 iff x is in A?
Yes. Prove it
Hint: The proof should be less than one line.
Actually prove that d(x,A)=0 iff x is in the closure of A without assuming anything about A
8:22 PM
@Koro More generally, if $A$ is closed and $K$ is compact, both disjoint, then $d(A, K) > 0$
Koro's life would be simpler if he just thought more before opening the door for the lions here.
I see it is true. d(x,A)=0. Suppose on the contrary x is not in A. x is in A^c. There is a d>0 such that B(x,d) is contained in A^c. Since d(x,A)=inf_{a\in A} d(x,a), there is an a in A such that d(x,a)<d/2 which implies a is in B(x,d) whence a is in A^c. contradiction.
Unnecessary contradiction.
@TedShifrin Doesn't seem measure theory absent to me
Well, I no longer possess the book. But the whole thing is based on the Daniell definition of the integral, not on measure theory.
We never did inner and outer measures, etc.
8:35 PM
Are you sure this is the same book?
That question makes no sense.
They use measure and there's even a chapter about Haar measure
Haar measure is not measure theory.
Saying the word measure does not violate my statement.
I took this course in 1972 or 1973, so I am not exactly in the best position to debate with you.
So drop it.
It seems to me like they do use measure up to like chapter 6 or 7, where they introduce algebraic integration
Is the image of a continuous function $f:[0,1]^2\to\mathbb R^2$ always Jordan measurable? Lebesgue measurable?
8:38 PM
I took the course from the original edition of the book. I have no idea what was revised in the second edition.
oh I see. Maybe they changed a lot
@mr_e_man its always compact
that means its measurable
@TedShifrin I still dont get it. Can you please reframe in terms of individual maps for a fixed group element g
I mean i got your equation but i dont get the conclusion
@mr_e_man iirc it doesn't need to be Jordan measurable
Restate your original question. Did you not say that the stabilizer of every element was the commutator subgroup?
I agree with your equation but how did you finish with conclusion?
8:44 PM
I deduced that $\phi = \sigma$.
That says that $\phi(g) = \sigma(g)$ for every $g$, but $\sigma(g)$ is conjugation by $g$.
but how? lets fix an element g then we are saying that $\phi_{g} \sigma_{g}(x)=x$ how?
I mean $\sigma$ inverse
By $\sigma^{-1}$ I mean the inverse permutation, not the inverse function.
So I'm composing the permutation $\phi(g)$ with the permutation $\sigma^{-1}(g)$. The value on $h$ is $\phi(g)(g^{-1}hg) = \phi(g)(h) = h$.
so sigma inverse of $x$ is the element c whose conjugate with respect to g is $x$?
$\sigma^{-1}(g) = \sigma(g)^{-1} = \sigma(g^{-1})$.
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