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12:05 AM
feel like I'm about to have a stroke
ran 2 miles to get my blood flowing
 
wtf @ForeverMozart how old are you
 
I sympathize, @Forever. Drink some water.
 
mid 20s
i'm not overweight I just don't run that much
 
I am underweight and I don't run at all
 
I am average weight and do not exercise
 
12:08 AM
Yeah I need to also start excerise
 
if you run then stand on your head, you get blood flow to the brain
 
Lol xD
 
i don't know if that is recommended ;)
 
@MikeMiller Isn't the actual (the one not involving Skein relations) definition already very geometric.
 
I have more or less never ever thought about these things.
 
12:21 AM
I mean you build the infinite cyclic cover, and evaluate the action of the induced "translation" transformation on H_1 of this thing. To me that that is quite geometric.
 
Hola qué pasa
 
Hi, Akiva
 
@MikeMiller Actually I don't know what's being asked.
Now I think he asking if theres some relation between numerical hyberbolic invariants and the alexander polynomial. No idea.
 
@AkivaWeinberger What are you doing these days?
I hardly see you here.
 
12:30 AM
School stuff mostly
 
Me too. I got a largish holiday though
 
@PVAL Is the new SG paper interesting?
 
@BalarkaSen Suppose K and H are groups such that $\psi : K \rightarrow Aut(H)$ is a homomorphism. Suppose
$p \in Aut(K)$ and set $\psi_2 = \psi \circ p$. Prove that $H \rtimes_{\psi} K$ ~ $H \rtimes_{\psi_2}K$. Is this problem even correct ?
 
@Adeek Sorry, I think I am done with group theory for today.
 
ok
 
12:33 AM
It does sound correct at a glance
 
@MikeMiller Maybe. It's certainly not real progress towards the question in the title. I'm very inclined to not take it that seriously given what the title is and what they actually prove.
Potentially you'd like to know these aren't symplectically fillable, and they don't even talk about that.
 
Maybe I'll stay awake today and reboot my sleep cycle
Or maybe I won't. Shrug.
 
1:00 AM
got my new dvd in the mail
McCabe and Mrs. Miller
 
@Adeek yes. apply p^-1 to the K component.
 
@artic tern to K component ? what is K component ? the problem here is that p lives in Aut(K) right ?
essentially what I want here is to show that $\psi$ and $\psi_2$ are conjugates
 
@Adeek send (h,k) to (h,p^-1(k))
 
@Adeek why would you?
 
oh
I thought maybe to use the first part which I proved
 
1:13 AM
that's only for cyclic $K$.
 
maybe not yeah I guess $(h,k) \mapsto (h,p^{-1}(k))$ would maybe define a isomorphism
I am just in the car now I will check the details when I go home.
oh I see @BalarkaSen
I didn't notice that
 
Let's use $\phi=\psi\circ p$ instead. Then $H\rtimes_\psi K$ is $H\ast K$ mod the relations $khk^{-1}=\psi_k(h)$, and $H\rtimes_\phi K$ is $H*K$ mod the relations $khk^{-1}=\phi_k(h)$. since $\phi_k(h)=\psi_{p(k)}(h)$, replace $k$ with $p^{-1}(k)$ and these relations are $p^{-1}(k)hp^{-1}(k)^{-1}=\psi_k(h)$. to go from the relations $khk^{-1}=\psi_k(h)$ to $p^{-1}(k)hp^{-1}(k)^{-1}=\psi_k(h)$ inside $H\ast K$, apply $p^{-1}$ to $K$ and $\rm id$ to $H$
 
going to get some coffee. back shortly
 
(Since $H\rtimes_\psi K$ and $H\rtimes_\phi K$ are both quotients of $H\ast K$, to exhibit an isomorphism between them it suffices to exhibit an automorphism of $H\ast K$ which sends one kernel to the other.)
 
oh i see
@arctictern that is nice way to see it
i.e seing it as mod relations
seeing *
 
1:19 AM
mmhmm
 
brb few min just going to apartment
 
Really quickly fact-checking here; for the first order d-e $\frac{dy}{dx}=\frac{1}{\sec ^2y}$, the solution is $y=\tan ^-1x$, right?
 
does the banarch tarski paradox work on a discrete space?
 
I meant $\tan ^{-1}x$, sorry
 
i would imagine so because i can go from all even numbers to all integers by division by 2 which is kind of the same idea
 
1:30 AM
@dsillman2000 that's a solution
there are many solutions
the others are $\tan^{-1}(x+C)$
 
Ok cool thx
 
@shaihorowitz the fact that an uncountable set can be bijected with any countable number of copies of itself is not very surprising. Banach-Tarski applies specifically with rigid motions of space (rotations and translations), which is why it's so surprising. Although really BT is a fact about group actions, and indeed it can be generalized to other group actions. (I'm not really familiar with the details, even though I love group actions.)
 
back
 
thanks, that suffices to answer what i was curious about @arctictern
 
1:37 AM
Hello I have a small question, the weak open unit ball in a hilbert space $\mathfrak{H}$ looks like: $B_\varepsilon(0) = \{x\in\mathfrak{H}|\,|(x|y)|<\varepsilon \forall y \in \mathfrak{H}\},$ right?
 
I am out of ink. Does that mean I have to go to sleep?
 
it looks like your cat is typing
 
No, get another pen
 
Maybe it's an ill omen. What if I don't listen to it? Will I die?
 
You're gonna die regardless
 
1:41 AM
not according to wuantum immortality
quantum*
 
what is the best halloween movie
 
the first
the sequel halloween movies sucked
 
i mean in general
 
@ForeverMozart The Nightmare before Christmas
 
doesn't have to be a "haloween"
 
1:42 AM
the rocky horror picture show
 
@BalarkaSen I have a question for you
 
that is a cartoon
 
Hatcher or video games?
 
Hatcher
@ForeverMozart it is indeed an animation.
 
1:43 AM
its an animation not a cartoon
 
and quite a good one, too!
 
@BalarkaSen yes but Fallout New Vegas
 
never played fallout.
 
real still frame pictures 30 pictures for one second of film thats dedication to art
 
yup
 
1:45 AM
I haven't played a video game in months
 
I also like his animation "Frankenweenie".
 
I think I'll take a break from math/physics tonight
 
definitely
I thought that question was for me.
 
are you talking to me?
I didn't ask a question
 
whats a question?
 
1:46 AM
"Hatcher or video games?" I was referring to that
 
im bored sorry....
 
oh
yes that was @BalarkaSen
Hatcher is too long, it's intimidating
 
I would read Hatcher. But you need not, taking breaks is perfectly fine
 
i cannot prove :(
 
I would probably be tempted if you had one of those Empire Earth-type strategy games. I used to like those, would have played if I had them
 
1:48 AM
@BalarkaSen Ever heard of "strong differentiability"?
 
anyone who solves this I give 100 dollars:
 
Not by that name
 
@BalarkaSen Me neither.
The next lecture in analysis is on "strong differentiability"
 
shrug
 
My prof has an obsession with really old books so maybe it was standard back then
 
1:49 AM
Is there a metric continuum $X$ and a point $x\in X$ such that every two points of $X\setminus \{x\}$ are contained in a nowhere dense subcontinuum of $X$, and every non-degenerate subcontinuum containing $x$ has nonempty interior.
 
@0celo7 it is strong
 
The proof is trivial! Just view the problem as a
combinatorial
poset
whose elements are
greedy
bijections
 
solve that problem and be famous
 
@BalarkaSen I do have a question
 
go on
 
1:51 AM
$|X|>1$
 
Is the "invariance of domain" theorem on page 172 of Hatcher the same one as on wiki?
The hypotheses seem slightly different
 
Who can solve that???
 
I don't beremember Hatcher's version. I think he parses it as "R^m is not homeom to R^n if m \neq n"?
 
probably Poincare
@BalarkaSen No
I actually do not understand hatcher's formulation
 
$100 I promise
 
1:53 AM
he says "if a subspace $X$ of $\Bbb R^n$ is homeo to an open set in $\Bbb R^n$, $X$ is open in $\Bbb R^n$"
Maybe I'm being dumb, but isn't that trivial?
 
I am interested in very simple problems like that. Easy to understand but unsolved.
 
@0celo7 How would you prove it?
 
@BalarkaSen Homeomorphisms are open maps.
Oh wait
 
So what?
 
subspace
So with the subspace topology?
 
1:55 AM
of course, what else?
 
Well wiki says something completely different!
 
No it doesn't
 
ok that's what I'm asking about
it seems completely different
In Hatcher, he starts with a homeo, then concludes it is open
Wiki starts with an open set, concludes the map is a homeo
 
$h:X\hookrightarrow \mathbb R ^n$ homeomorphic embedding and $h[X]$ open in $\mathbb R ^n$.
 
Hatcher says if $U$ is open in $\Bbb R^n$ and $f : U \to \Bbb R^n$ is an embedding , then $f(U)$ is open. Wiki says the same but with "injective continuous". Replace that by embedding (because that bit is obvious once you can show for injective continuous)
 
1:59 AM
is your $U$ Hatcher's $X$?
 
now let $x\in X$.
 
I don't see an $X$ in page 172. I have the online version open in front of me.
 
wtf
that's completely different than my version
in my version, Hatcher says exactly what I quoted above.
 
Sounds like the same thing to me, with $X = f(U)$
$f$ is the homeo with the open set $U \subset \Bbb R^n$.
 
@BalarkaSen I still don't get it. Hatcher starts with a homeo as a hypothesis
is Wiki saying the converse?
 
2:06 AM
This can't be right
 
@0celo7 No, just replace wiki's hypothesis "injective continuous" with "homeomorphism".
Then compare.
 
Unless I'm missing something, those are not the same hypotheses...
 
Sure, it's a bit more general in that sense. But part of the statement of the theorem (the version in wiki) says that it doesn't matter, @0celo7.
@enthdegree What are you referring to?
 
@BalarkaSen Well, yeah
that's all I was wondering about
 
Can anyone explain how on earth you get your hands on the weak closure of a subset of a space?
I have a normed vector space X and its dual X*, and X* induces a 'weak topology' on X (the weakest topology where all the functionals in the dual are continuous, i.e. a sequence (xn) --> x weakly in X iff (psi(xn))--> psi(x) for any psi in X*)
 
2:31 AM
hey @arctictern I am classifying all groups of order pq where p and q are distinct prime. THe way I am solving this is as follows we can assume that q < p. Let H be a sylow p subgroup and K be a sylow q subgroup. We can prove H is normal in G and HK = G.
then this means that $G = H \rtimes K$
If K is also normal in G then we get $G = H\times K$
why is it true that we will have a non-trivial homomorphism iff q divides p - 1 ?
 
is an image of hitler stuffed with straw
 
@Adeek One direction is clear from Sylow's 3rd theorem.
 
what about <--
ok let us say I know that the theorem above is true
hm yeah I guess I would be done
 
You need to explicitly work out a homomorphisms K --> Aut(H). H is a cyclic group so that shouldn't be hard.
 
I see
 
2:39 AM
I was messing up a matrix multiplication for the last few minutes. Maybe staying awake wasn't a great idea.
 
yeah ok I proved it @BalarkaSen
what are you doing anyway @BalarkaSen
 
Differential geometry.
As in, surfaces in $\Bbb R^3$.
 
a non-trivial homomorphism gives a non-abelian group right ?
we can prove that
I guess I am trying to show that only two groups exist with order pq where they are distinict and show that they are not isomorphic.
@BalarkaSen can we actually show what kinda group is the semi-direct product ?
or it is too general to answer.
in the case where it is non-trivial homomorphism
 
@Adeek That's right. If it was abelian, H and K both would have been normal.
 
yeah
 
2:47 AM
I don't know what you mean by "what kinda group". You can write down a presentation for it: is that sufficient?
 
yeah maybe
yeah
 
Then the presentation comes from the very definition of semidirect products.
 
i see
@BalarkaSen I attended this seminar today I thought it might interest you
Title : Calculus of functors and the De Rham complex

Abstract : The calculus of functors, pioneered by Goodwillie in the 1980’s, refers to a way of examining functors which topologists and algebraists often use to distinguish and classify algebraic or topological objects like algebras. The De Rham complex of a commutative algebra over a field, the algebraic analogue of De Rham cohomology of a manifold, is precisely one of these functors. The De Rham complex is closely related to cohomology theories for non-commutative algebras, such as cyclic (co)homology and crystalline cohomology.
was very interesting
 
I know nothing of Goodwillie's calculus, but I have heard of it.
 
Does anyone have any idea how to complete this? imgur.com/Zpxdwjo.png
 
2:54 AM
yeah it was kinda of interesting but too advanced
for me
 
I can't figure out how to show that the epsilon ball intersects \mathfrak{I}, even given the hint!
I guess i should add, I^{-w} denotes the closure of I in the weak topology
 
3:08 AM
Hey guys
Is $\int 3x^2y\ dx = x^3(yx) = x^4y$ because y would be treated as a constant, as in partial differentiation?
 
Or by that I meant yx^3
Why not?
 
So it would be yx^3?
Because I think someone made a fatal error in their answer to someone's question but I don't have enough rep to comment xD
 
you can edit your answer
 
3:13 AM
alright, it looks like i have some fundamental misunderstanding of what the weak* topology is
does anyone online right now know functional analysis?
 
3:24 AM
@BalarkaSen this question is really cool
classify all groups of order 8
very fun
 
I don't see why that's interesting. There's just one.
 
order 8
I dunno why I said order 5
 
Hi @TedShifrin
 
fixed it
hi @TedShifrin
I was actually wanted to talk to you
 
Sylow hard.
 
3:26 AM
no @BalarkaSen
by structure theorem
 
Hi, all. Order 8 is still pretty easy ... For the non-abelian, anyhow.
 
you can classify 3 abelian group of order 8
 
@Adeek Structure theorem? Huh?
 
then Suppose now that G is not abelian
 
That's for abelian.
 
3:26 AM
then there is an element of order 4
 
i have killed so many trees
 
because otherwise it have an element of order 8 which would make it cyclic
or all elements will have order 2
 
Sure.
 
i just write and write until I finally discover something
 
which mean it would be isomorphic to $Z_2 \times Z_2 \times Z_2$
 
3:27 AM
and there is a pile of paper
 
denote this element by a so <a> is subgroup of G of index 2 so it is normal
 
No, Karim, that's not clear
 
?
 
Why must it be a direct product?
 
what is not clear ?
 
3:28 AM
@TedShifrin If it has no element of order 4, there are only two choices for the orders.
 
yeah
 
Balarka, hush.
 
Oops.
I'll scurry, as usual.
 
Why if every non-1 element has order 2 must it be .... .
Go do interesting diff geo, Balarka.
 
okay so by lagrange we can only have elements of order 2 or 4 or 8. If we have all elements of order 2 then it would be isomorphic to $Z_2 \times Z_2 \times Z_2$
 
3:30 AM
I need interesting stuff. I am going to keep myself awake today and reboot my sleep cycle.
 
You keep saying that. That doesn't suffice.
Do you need a list of exercises, B?
 
@TedShifrin I don't understand what is the issue ?
 
@Adeek He's asking you why all elts having order 2 imply the group is Z/2 x Z/2 x Z/2
 
What justifies your claim, Karim?
 
oh
 
3:32 AM
@TedShifrin Yeah, definitely.
 
oh I see
yeah I can't answer that hmm
 
-30
 
pokes on "I don't suppose anyone knows of any decent textbooks on Banach Lattices? Having trouble getting into the one my professor gave me."
 
No clue, Alan.
 
alright i cant believe i was being this dense before, i was basically not interpreting my definition of the weak* topology
 
3:34 AM
Are we doing anything fun?
 
oh I see
I got it @TedShifrin
 
Pretty sure the classify all groups of "x" order when "x" has a simiple factoroization is easiest done via semidirect products, but it's been a few years since I did algebra qual prep
 
Karim's classifying groups. I'm trying not to fall asleep.
none of these are fun though.
 
I'm trying to find a text on Banach Lattices that I can understand :(.
 
with the weak topology we assert that sets of the form {x: |phi(x-z)|<\varepsilon, z in X} (for phi in the dual, varepsilon positive)
are open
 
3:35 AM
I have no idea what a banach lattice is
 
what is a banach lattice? is that just a banach algebra with more structure
 
if we have every element has order 2 then take the group generated by the elements a,b,c then <a> <b> <c> all intersect trivially and they all generate the whole group.
 
@Balarka: Make sure you know how to do #2 (linear alg, of course), 5 (note that problems with * have hints/answers at the back), 10, 12, 13, 14 (important), 15, 17, 18, 19 (sorta cool), 21 (might interest you because there's a blowup in there).
 
so this means that $G = <a>\times<b>\times<c>$
 
Total handwave, Karim.
 
3:36 AM
I worked out #2 a couple hours ago.
 
it's a banach space (Not necessariolly an algebra) that has a lattice order that is compatible with the vector operations. A lattice is a partial order in which every two elements has an inf and a sup (though not necessarily every subset, so think of it as an lcm and gcd)
 
Why is it abelian? You keep missing the main point.
OK, cool, @Balarka.
 
Thanks for the list, by the way!
 
23 is actually interesting, Balarka, but it might take you down a rabbit hole.
 
the compatibility requirements being if $x<y$ then $x+z<y+z$ and if $a\in mathbb R , a>0$, $x<y$ means $ax<ay$
 
3:39 AM
@Alan Why is this interesting?
 
@MikeMiller I've yet to get to that part. I wanted to do something in Functional Analysis, and the main professor at my school who does that turns out to work in this area.
 
I bookmarked that too for kicks.
 
:P I think step one in studying anything is to find out why you care.
 
@BalarkaSen What?
 
If you can't find that, I suggest trying something else.
 
3:40 AM
@TedShifrin Problem 23. I bookmarked it along the others.
 
Students always complain when professors don't show them why things are interesting, @MikeM, and professors shrug.
 
Yeah :). I really enjoyed Kreyzig's Functional Analysis. I'm year 4 of grad school, I may switch thesis advisors at the end of this semester if I can't seem to get a hold of tghis
 
Oh, ok.
 
Maybe I should see if anyone's working in topological data analysis....I found that fascinating when we had a talker come in about it, and I enjoyed both general and algebraic toplogy
err, speaker. Yeah. I can think right now :)
 
@Alan Your professor would be a better person to ask than us. This sounds like pretty specialized stuff.
 
3:42 AM
It's almost late, @Alan.
 
Top. data analysis is fantastic.
 
@Ted: When a student asks me why something is interesting, I certainly don't shrug. I think that's one of the best questions you can ask.
 
@MikeM: I don't disagree. I say that in general teachers do a terrible job of making math motivated/interesting.
 
Yeah. I...think I'm coming to that point where I should switch tracks before I sink too much time/energy into this.
 
One of my colleagues when I first went to UGA did lattice ordered groups. It left me cold.
But the math I spent my life on doesn't interest plenty of people ... so it's a matter of individual taste and strengths.
 
3:44 AM
I know I always have real world examples of interesting stuff for when I teach. Sometimems little things, like the hash code for Florida's driver's license.
Yeah, I think this was a case of me thinking I'd like something then trying it and finding out I don't.
 
Well, yeah, talk to other people, @Alan.
 
@ted I don't understand how to make it more rigorous the way I see it is that if we have all orders of order 2, then take a particular element of G and denote that by a. Then $<a> = Z_2$. suppose b not in <a> and go like that and we can then construct 3 groups of order 2, which I call <a>,<b>,<c>
 
Time to talk to my advisor/the graduate studies head. Thanks for the feedback.
 
There's lots of other things in functional analysis, but if no one there can advise you on other things, talk to other people.
 
Since they are all p groups so they intersect trivially
 
3:45 AM
OK, Karim. But you're far from a direct product.
 
@Ted: Part of why I don't judge people's interests. I couldn't spent my time writing down estimates for gradient flows of analytic functionals on Banach manifolds, but I'm glad someone can, since I like to use them.
 
We know the following theorem
 
There's no denying taste, @MikeM.
 
@Alan Good luck.
 
Yes, @Alan, we're rooting for you.
 
3:46 AM
Is there a metric continuum XX and a point x∈Xx∈X such that every two points of X∖{x}X∖{x} are contained in a nowhere dense subcontinuum of XX, and every non-degenerate subcontinuum containing xx has nonempty interior.
 
De de gustibus non est disputandum
 
if (1)$H \cap K = \{1\}$ (2) H and K are both normal (3)HK = G then $G = H\times K$
 
i might have a way to solve this problem
 
You're far from having hypotheses for that, Karim.
 
yeah
yeah I see what the trouble I am facing now
 
3:48 AM
Have any of you worked with extremely disconnected spaces? I ran into them trying to understand tghis stuff.
 
@ForeverMozart, perhaps?
 
is there a more general version of this theorem
I guess by induction
 
definition is the closure of every open set is open.
 
You haven't justified normality, Karim.
 
yes it is deep
usually induction doesnt work in topology
 
3:49 AM
normality comes from having index 2
 
in continuum theory anyway
 
@TedShifrin
 
You have order 2, not index 2, Karim.
 
Do you have semidirect products to work with?
 
The induction comment wasn't aimed at you, @Forever.
 
3:49 AM
yeah
 
(whoever's working on this classification problem, I lost track :))
 
Am I sleepy or is this conversation becoming confuzzling
 
yeah @Alan
 
Yes @Balarka
 
@adeek I'm about 95% sure you can recognize any $p^3$ group as a semidirect product, and work from there.
 
3:51 AM
Stop trying to kill a fly with a cannon, guys.
 
grin But.....cannons are fun! :)
 
I won't even go there ...
 
yeah I thought about it @Alan but then the problem should have been more simpler
hm
 
There's an exercise that appears early in any undergraduate group theory course.
 
why is subgroups of order 2 in group of order 8 are normal.
hm maybe by sylow theory ? let us see
 
3:52 AM
That's false.
$D_4$ has plenty of subgroups of order 2 that are not normal.
 
yes
 
Don't $p$ groups have subgroups of every power of $p$ up to the group?
 
Stop it with all this generality. We're talking $p=2$.
 
Normal subgroups even
 
How the hell do I get dragged into undergraduate algebra? I'm leaving ...
 
3:54 AM
And now, the whole room is talking about group theory.
 
laugh Sorry, reminds me of one of my professors who always started with the most general case.
snorts Just saw the bordism thing on the sidebar, had to email a link to my topology professor
 

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