« first day (3053 days earlier)      last day (1958 days later) » 
00:00 - 17:0017:00 - 00:00

Secret
Secret
5:00 PM
yup
 
Semiclassical
Semiclassical
Are two 2-adic numbers considered to be close if their last few digits are the same? obv. with base-2 numbers it's the opposite
 
Jasper Loy
Jasper Loy
@Secret THat is the most interesting news I read this year actually.
Here comes @MikeMiller!
 
Semiclassical
Semiclassical
@JasperLoy a common variation on that question: "Where do you see yourself in 5 years?"
hate that question
 
Secret
Secret
> Where do you see yourself in 5 years?
This question make no sense as we can no longer predict deterministically where we are in even 2 days later
 
Rithaniel
Rithaniel
I think I'm seeing messages out of order. Either that or I've lost the flow of the conversation.
 
Secret
Secret
5:02 PM
the society have gone so complicated that prediction is no longer possible
 
Jasper Loy
Jasper Loy
@Rithaniel Is it because you are ignoring some users?
 
Rithaniel
Rithaniel
No, I don't ignore people, generally.
 
Semiclassical
Semiclassical
i think this conversation has just been a bit disjointed
 
Rithaniel
Rithaniel
Well, to date, I've not ignored anyone on this chat.
 
Jasper Loy
Jasper Loy
Then maybe the flow of the conversation is just free association or random. =)
 
Secret
Secret
5:04 PM
another possibility is that sometimes lag can cause messages to switch order at the cilent's end
 
Jasper Loy
Jasper Loy
I mean, there is no need to stick to one topic strctly at one time right?
 
Semiclassical
Semiclassical
@Secret I suppose the point of the question is more for you to indicate your life goals and comment on how much progress you hope to have made
 
Jasper Loy
Jasper Loy
@Secret Yes, some messages might not display on some computers till hours later...
 
Rithaniel
Rithaniel
Yeah, I think Semiclassical was saying that he hated the "5 years" question, but he said it before the question was posted on my screen.
 
Semiclassical
Semiclassical
no, that's the order in which it happened
 
Jasper Loy
Jasper Loy
5:05 PM
The 5 year question is just an interview question, standard fare...
 
Semiclassical
Semiclassical
I said I hated that question, then Secret repeated the question and indicated why it was nonsense
 
Jasper Loy
Jasper Loy
Some of these interviewers will ask you stupid questions, because they are stupid.
However, I think the 5 year question is an OK question.
 
Rithaniel
Rithaniel
Ah, gotcha
 
Secret
Secret
If there is a question that ask the interviewee to ask the interviewer a question, I will ask them the following:
 
Semiclassical
Semiclassical
I'm not sure it's a bad question, it's just one I really don't like because I don't have that mindset
 
Secret
Secret
5:07 PM
> What do you think of the typical interview questions and their textbook responses. Can you think of an interview question that is resistant to textbook solutions?
My experience told me people tend to go O_O whenever they heard meta statements
and then they enter some kind of trance state
 
Jasper Loy
Jasper Loy
I think the 5 year question is to see if your personal goals align with those of the company that is hiring you.
 
Semiclassical
Semiclassical
probably
 
Jasper Loy
Jasper Loy
Like you want to be CEO in 5 years, OK, we will make you CEO in 5 years.
 
Semiclassical
Semiclassical
(or "no, gtfo")
 
Secret
Secret
One of my wild desires is to ask someone a question and then said person enters nirvana as they tried to answer it
that will be a cool sight to behold
 
Jasper Loy
Jasper Loy
5:11 PM
@Secret That sounds incredible. I mean, I didn't expect you to say that.
 
Semiclassical
Semiclassical
hi @ted
oh, two fun things relating to actual previous research of mine
one is that there's another paper out which cites ours, which is fun
 
Jasper Loy
Jasper Loy
@Semiclassical Ours? You cowrote?
 
Secret
Secret
@JasperLoy that I say is a desire means no I don't have that ability (yet?)
 
Semiclassical
Semiclassical
yeah: me, my advisor, and a collaborator (in that order on the paper)
the other is that, on a hunch, I went to the probability seminar last week
 
Jasper Loy
Jasper Loy
I see. Me, mine, and I, lol.
 
Semiclassical
Semiclassical
5:13 PM
and the talk turned out to be directly related to stuff I presented in my thesis defense talk
 
Secret
Secret
But to understand the ultimate limit of human knowledge it is to comprehend these:
Ineffability
Ineffability is concerned with ideas that cannot or should not be expressed in spoken words (or language in general), often being in the form of a taboo or incomprehensible term. This property is commonly associated with philosophy, aspects of existence, and similar concepts that are inherently "too great", complex or abstract to be communicated adequately. A typical example is the name of God in Judaism, written as YHWH but substituted with "the Lord" or "HaShem" (the name) when reading. In addition, illogical statements, principles, reasons and arguments may be considered intrinsically ineffable...
 
Semiclassical
Semiclassical
in fact, one of the papers the guy wrote (and which he mentioned on one of his slides) cites our paper :)
 
Jasper Loy
Jasper Loy
@Secret Are you a student of Buddhism?
 
Secret
Secret
nope, but I do have some buddhist friends
 
Semiclassical
Semiclassical
Now if only I'd realized that ahead of time...
 
Jasper Loy
Jasper Loy
5:14 PM
@Semiclassical Oh so did you tell him that?
 
Semiclassical
Semiclassical
yeah, after the talk
 
Secret
Secret
and just recently I was reading the wikipedia article on buddhism cosmology
 
Semiclassical
Semiclassical
had I realized beforehand that he'd be doing so, I'd have dragged my advisor along
 
Secret
Secret
because I want to get more information about the concept of nothing
 
Jasper Loy
Jasper Loy
@Secret If you are, I recommend you get a copy of the 4 main Nikayas published by Wisdom.
 
Semiclassical
Semiclassical
5:15 PM
(and had i realized it a few days in advance, I'm sure we could've set up a meeting. missed opportunities)
 
Jasper Loy
Jasper Loy
@Semiclassical Sometimes, when you miss one chance, a better one comes. =)
 
Semiclassical
Semiclassical
a different one comes, anyways. not sure 'better' is always a good metric
 
Jasper Loy
Jasper Loy
Anyway @Semiclassical I stopped taking meds for a few months now. I am not sure if what I am feeling now are withdrawal symptoms or not, but I didn't feel any different for a month or two.
 
Rithaniel
Rithaniel
I would argue that "better" doesn't even constitute a metric. A pseudo-metric, perhaps.
 
Semiclassical
Semiclassical
@Rithaniel a poset, maybe
 
Jasper Loy
Jasper Loy
5:17 PM
Yeah because often one doesn't know if something is good or bad.
 
Sharath Zotis
Sharath Zotis
Is $<x^2+5>$ a prime ideal in $\mathbb{Z}_7$
 
Semiclassical
Semiclassical
@JasperLoy i just picked up a refill on one of mine this morning
 
Rithaniel
Rithaniel
Because two things might have a "betterness distance" of 0, but not be the same element.
 
Semiclassical
Semiclassical
namely, the one which I take for sleep stuff
that's one I notice if I forget to take
 
Sharath Zotis
Sharath Zotis
It is reducible as 3^2 +5 = 14 which is 0 in $\mathbb{Z}_7$
But I am having trouble finding what it reduces to
 
Jasper Loy
Jasper Loy
5:20 PM
@Semiclassical I am going to do more reading on meds and therapy before deciding on anything again. There is a lot of information from different places saying different things. =)
 
Semiclassical
Semiclassical
sounds right
 
Sharath Zotis
Sharath Zotis
Anybody have an idea?
 
Ted Shifrin
Ted Shifrin
@Sharath: If you know $3$ and $-3$ are roots of $x^2+5\in\Bbb Z_7[x]$, then you know how to factor.
 
Sharath Zotis
Sharath Zotis
ah yes so (x+3)(x+4) and since both are in $\mathbb { Z } _ { 7 } [ x ]$ then it is prime ideal
 
Ted Shifrin
Ted Shifrin
Whoa, slow down.
In $\Bbb Z$, is $\langle 6\rangle$ a prime ideal?
Review the definition of a prime ideal.
 
Sharath Zotis
Sharath Zotis
5:24 PM
yes as 2*3 = 6 and 2 or 3 is in $\mathbb{Z}$
 
Ted Shifrin
Ted Shifrin
No, write down the definition of a prime ideal.
 
Sharath Zotis
Sharath Zotis
The definition is: Let $R$ be a ring and $I \subseteq R$ be an ideal. We say $I$ is prime if $ab \in I$ then either $a \in I $ or $b \in I$
 
Ted Shifrin
Ted Shifrin
OK, now read that and understand it. Then read what you typed to me above.
 
Sharath Zotis
Sharath Zotis
Oh yes neither 2 nor 3 is in $<6>$
What about 12 ? $12 \in <6>$ and 6*2 = 12 and $6 \in <6>$
 
Semiclassical
Semiclassical
Needs to be true for all.
 
Sharath Zotis
Sharath Zotis
5:28 PM
I seee
 
Ted Shifrin
Ted Shifrin
Wait, wait. Where did 12 come from?
Oh, I see ... Right. Whenever $ab\in I$ you must know that $a\in I$ or $b\in I$. That's a "for all" statement.
 
Semiclassical
Semiclassical
Just because there exists some $ab\in I$ such that $a\in I$ or $b\in I$ doesn't mean that it's true for every $ab$
 
Sharath Zotis
Sharath Zotis
ok makes sense
so to correct the thing I said about my problem. neither (x+3) nor (x+4) are in $<x^2+5>$
so it is not a prime ideal
 
Ted Shifrin
Ted Shifrin
Right.
 
Semiclassical
Semiclassical
Hmm. Seems like that should be pretty generic, i.e. if $p(x)$ is reducible in some polynomial ring then $\langle p(x)\rangle$ isn't a prime ideal
 
Sharath Zotis
Sharath Zotis
5:31 PM
Now with maximal ideals, $<x^2+5> \subset <x+3> \subset \mathbb{Z}_7$
 
Mike Miller
Mike Miller
@Semiclassical reducible element in a ring
 
Sharath Zotis
Sharath Zotis
this implies that $<x^2 + 5>$ is not maximal, corect?
 
Ted Shifrin
Ted Shifrin
@Semiclassic: I would rather say "isn't a prime ideal in that polynomial ring" since you brought that phrase into it.
 
Mike Miller
Mike Miller
And that <p> isn't prime is as you say definitional
 
Ted Shifrin
Ted Shifrin
Correct, @Sharath.
 
Semiclassical
Semiclassical
5:32 PM
@TedShifrin yeah, fair
@MikeMiller ah, so that's the terminology? not surprising that I'd flub that tbh
 
Ted Shifrin
Ted Shifrin
@Sharath: Can you see that any maximal ideal must be prime?
 
Mike Miller
Mike Miller
@Semiclassical no, I am just saying polynomials are not actually relevant
 
Semiclassical
Semiclassical
ahhh
yeah, fair
hmm, how big of a sparse matrix do I need to make
3-by-3-by-n-by-n-by-m...welp
 
Sharath Zotis
Sharath Zotis
hmm @TedShifrin. why is that @TedShifrin
 
Semiclassical
Semiclassical
(m versions of a 3-by-3 block matrix with n-by-n submatrices)
 
Ted Shifrin
Ted Shifrin
5:35 PM
Try proving the contrapositive, @Sharath.
 
Sharath Zotis
Sharath Zotis
does it stem from the fact that a maximal ideal is not a subset of any other ideal
other than the ring itself
 
Ted Shifrin
Ted Shifrin
Yes, of course.
 
Sharath Zotis
Sharath Zotis
hmm I see we can use the theorem that $R/M$ is a field given $M$ is maximal ideal in ring $R$
 
Semiclassical
Semiclassical
nope, sagemath doesn't like what I'm doing
 
Ted Shifrin
Ted Shifrin
Don't use that theorem. Just do it directly, like you saw with the examples we discussed.
 
Sharath Zotis
Sharath Zotis
5:38 PM
Ok
 
Ted Shifrin
Ted Shifrin
If $I$ is not prime, show it cannot be maximal.
 
Alessandro Codenotti
Alessandro Codenotti
6:06 PM
Hi @Ted
 
OneRaynyDay
OneRaynyDay
Hi guys, following the proof on why $l^p$ is first category in $l^q$ if $p < q$: math.stackexchange.com/questions/1097869/…
I'm a bit confused on why we need to show that the individual $A_n$'s used in the construction needs to be closed?
So what if it's open? $l^p = \cup_{n \in \mathbb{N}} K_n$ where $K_n$ is the open ball centered around zero with radius $n$
To show $K_n$ is nowhere dense in $l^q$, we just need to show that there exists some sequence $x \in l^q$ such that $||x||_q < n$ and $||x||_p = \infty$ right?
Then simply take some $x \in (l^q \cap (l^p)^c)$, then $||x||_q = C$, then divide the sequence elementwise by some large enough constant and you get it's within $n$ radius of the origin
Yeah... not really seeing where the closedness is important here
 
Isa
Isa
6:40 PM
@Palimondo Seems like you are using 🤰 to denote two different things. You said, 🤰 is a pregnant woman and then you write this '👤 ≡ 👤: {🤝,📜}' which is supposed to be equivalent to 'and equal in dignity and rights.' ?
 
OneRaynyDay
OneRaynyDay
Ah, for anyone also wondering - it's because for a set $A$ to be nowhere dense, $(\overline{A})^o = \emptyset$
In this case if $K_n$s were open we would need to first take their closures, so finding closed balls were just 1 step-expedited
 
Palimondo
Palimondo
7:03 PM
@Isa I don’t know what you mean. The 🤰 is there just once as part of the triple for “born” on first line. Then there’s kind of second statement on the second line…
What is tripping you up?
 
Ted Shifrin
Ted Shifrin
@AlessandroCodenotti Hi, demonic Alessandro ;)
 
Isa
Isa
aha in the second line you wrote again the square 🤰, why? Or is it that I am seeing another square but it's not? Sometimes emojis 'change' depending on the system you work
 
Alessandro Codenotti
Alessandro Codenotti
@PaulPlummer Interestingly the wikipedia proof uses HB instead of a nonprincipal ultrafilter, so amenability of $\Bbb Z$ should be a very weak statement. Do you know about the strength of "every Abelian group is amenable"?
Also a sanity check: The Cayley graph of a finite group with $n$ elements is quasi isometric to the complete graph on $n$ vertices right? Because I can take the whole group as generating set and get a complete graph as Cayley graph wrt this generating set
 
Fargle
Fargle
7:24 PM
Hello everyone.
 
Semiclassical
Semiclassical
just asked Sage to compute the intersection of the 62-dimensional simplex in 63-dimensional space with a 56-dimensional polytope
let's see how much it hates me
apparently, it hates me too much to actually finish that computation :)
 
Paul Plummer
Paul Plummer
7:46 PM
@AlessandroCodenotti I have no idea, seems like an interesting question. Maybe integers being amenable (where Folner is not the definition)could be independent of ZF?! I don't really know much about amenable groups.
Also, yes they are quasiisometric
 
Alessandro Codenotti
Alessandro Codenotti
I wouldn't be surprised if it were independent
 
Paul Plummer
Paul Plummer
Finite groups are quasiisometric to points, and this holds for any bounded space
 
Alessandro Codenotti
Alessandro Codenotti
I might ask about it on main
 
Paul Plummer
Paul Plummer
Yah, seems like a cool question, I would be interested in an answer
 
Alessandro Codenotti
Alessandro Codenotti
Amenability is the most interesting thing we've seen so far in my opinion!
 
Paul Plummer
Paul Plummer
7:49 PM
It is definetly cool, but it is in some sense opposite of hyperbolic behavior
 
Alessandro Codenotti
Alessandro Codenotti
I've seen those "universe of groups" pictures, but I don't know why they're opposite
 
Paul Plummer
Paul Plummer
Well one thing is that(non elementary) hyperbolic groups contain free subgroups which is an obstruction
@AlessandroCodenotti As to the quasiisometry question, it is interesting that some infinite groups(not finitely generated) have no cayley graph where they are unbounded
 
Alessandro Codenotti
Alessandro Codenotti
Uhm that's weird
But in that case the properties of the Cayley graph also depend heavily on the generating set I think?
 
Paul Plummer
Paul Plummer
Cayley graph always depends on the generating set, but groups like that you end up where the quasisometry class does not depend on the generating set
in most of GGT you restrict to finitely generated groups because the quasiisometry class does not depend on finite generating sets (for example you could choose the whole group as a generating set to get a bounded diameter graph)
 
Alessandro Codenotti
Alessandro Codenotti
Yeah, sorry, I wasn't very precise, with infinitely generating sets you might even get cayley graphs which are not quasiisometric, right?
 
Paul Plummer
Paul Plummer
8:05 PM
Yah
But there are groups where you always get quasiisometric graphs, like the ones where every cayley graph is bounded diameter, even though it is infinitely generated
 
Alessandro Codenotti
Alessandro Codenotti
Ah, cool
 
Paul Plummer
Paul Plummer
8:31 PM
The comments to this question point out that it is independent of ZF @AlessandroCodenotti
 
Alessandro Codenotti
Alessandro Codenotti
I see, interesting!
Now I wonder about its strength though, it sounds like it should be in between ZF and ZF+HB
It surely is stronger than ZF and weaker but possibly not strictly than HB
But since it only involves countable objects I don't see how could it be equivalent to HB
 
Paul Plummer
Paul Plummer
8:52 PM
Another interesting question is ZF+shift invariant finitely additive probability measure on integers can prove the equivalence of the Folner sequence and shift invariant finitely additive probability measure for discrete countable groups
basically is it enough to for integers to get the common equivalent definitions?
 
Alessandro Codenotti
Alessandro Codenotti
Good question indeed, I have no idea
 
Balarka Sen
Balarka Sen
@AlessandroCodenotti Whoosh collapse it to a point
and the foosh explode it to any finite graph you want
qed
 
Alessandro Codenotti
Alessandro Codenotti
Paul sniped you
 
Balarka Sen
Balarka Sen
did he specifically say woosh and foosh
otherwise it's not a true snipe
 
Alessandro Codenotti
Alessandro Codenotti
He didn't use the whoosh which is clearly the correct technical terminology though
Oh come on you can't snipe me saying you weren't really sniped
 
Balarka Sen
Balarka Sen
9:06 PM
rip
 
Alessandro Codenotti
Alessandro Codenotti
That's not how it works
 
Semiclassical
Semiclassical
welp. my pattern for vertex counts so far is 4,40,576
sooo my next one probably has thousands of vertices...which probably explains why Sage is taking so long :P
 
Balarka Sen
Balarka Sen
@PaulPlummer Is any sufficiently reasonable metric space with two ends quasi-isometric to $\Bbb R$?
My mental picture is, I end-compactify it, then take an arc going between the two components in the end
 
Semiclassical
Semiclassical
(I think I may be making things a touch more difficult than I need to, tho.)
 
Balarka Sen
Balarka Sen
I should be able to collapse the space to the arc but how do you do this, formally?
 
Leyla Alkan
Leyla Alkan
9:14 PM
Hi all! Is this proof correct? How does the underlined sentence follow by the previous one?
 
Semiclassical
Semiclassical
Actually, quick sanity check
 
Balarka Sen
Balarka Sen
@Semiclassical You are insane
 
Semiclassical
Semiclassical
damn
Suppose I've got two convex sets and I intersect them. Is this preserved by linear transformations?
i.e. if $C=A\cap B$ for convex $A,B$ and $F$ is some linear transformation, should $F(C)=F(A)\cap F(B)$?
 
Paul Plummer
Paul Plummer
@BalarkaSen Depends on what you mean by sufficiently nice. For example n^3 for integers n is not quasiisometric to the integers (with induced metric).
Maybe if there are enough quasiisomeries, or it is geodesic
 
Balarka Sen
Balarka Sen
Oh, because the distances grow far
Cannot linearly scale it down
 
Paul Plummer
Paul Plummer
9:17 PM
Yah
 
Balarka Sen
Balarka Sen
rip. Good example
 
Paul Plummer
Paul Plummer
Actually geodesic is not enough
 
Semiclassical
Semiclassical
according to a random internet paper: "For any linear transformation T and convex closed sets A and B, T(A ∩ B) = T A ∩ T B if and only if A ∪ B is path-connected with respect to ker(T)."
now to figure what that actually means...
 
MatheinBoulomenos
MatheinBoulomenos
@Semiclassical take $A=\langle (1,1) \rangle$ $B=\langle (1,0) \rangle$ and let $F$ be the projection onto the first coordinate
 
Semiclassical
Semiclassical
Hmm. So $A\cap B=\{(0,0)\}$
then $F(A)\cap F(B)=\langle (1)\rangle \cap \langle (1)\rangle=\langle (1)\rangle$ whereas $F(A\cap B)=\{(0)\}$
checks out (as a counterexample, I mean)
in the context of the statement I just gave, that makes sense because $A\cup B$ is certainly not convex
So my approach is probably not going to be suitable
 
Paul Plummer
Paul Plummer
9:28 PM
Something like being uniformly quasigeodesically "connected", and every triangle is in a uniform neighborhood of one of the uniform quasigeodesics should get it. whoosh foosh.(but that is basically saying it is quasiisometric to a line so that isn't really interesting) @BalarkaSen
But I think that is isolating the main features
 
MatheinBoulomenos
MatheinBoulomenos
quasiisometry seems like a very coarse notion
 
Paul Plummer
Paul Plummer
Oh and have two ends
 
Balarka Sen
Balarka Sen
@PaulPlummer Aha, I see
 
Paul Plummer
Paul Plummer
@MatheinBoulomenos Haha
 
 
1 hour later…
Liad
Liad
10:47 PM
Hi, $X$ is a normed space. i want to show that $x_1,\dots,x_n$ are independent iff for all $\alpha_1,\dots,\alpha_n \in \Bbb F$ there is $\phi \in X^*$ s.t $\phi(x_i)=\alpha_i$. im stuck on this question, any ideas?
$\Bbb F\in \{\Bbb R , \Bbb C\}$
 
Richard
Richard
can someone help me with morphological gray-scale gradient ?
 
Ted Shifrin
Ted Shifrin
11:11 PM
@Liad: Surely one direction is easy. Where are you stuck?
 
Ultradark
Ultradark
Are most 3-manifolds hyperbolic
 
Ted Shifrin
Ted Shifrin
Certainly not.
I have no idea what "most" means, anyhow.
 
Balarka Sen
Balarka Sen
11:26 PM
@TedShifrin A random mapping torus is hyperbolic, if I recall correctly
 
Ted Shifrin
Ted Shifrin
I know very little about 3-manifolds, actually, since I live in the complex world by choice :P
Or lived.
Hi, a @Balarka.
 
Balarka Sen
Balarka Sen
In the sense that there is a dense subset of the mapping class group of a surface such that the corresponding mapping torus has a hyperbolic metric
 
Ted Shifrin
Ted Shifrin
With the Thurston classification of 3-manifolds, I realize hyperbolic structures play a huge role. But a random negatively curved manifold isn't hyperbolic, at any rate.
Presumably you mean a surface of genus $g\ge 2$?
 
Balarka Sen
Balarka Sen
Yeah I believe so
 
Ultradark
Ultradark
Is an orbit another name for an integral curve
 
Ted Shifrin
Ted Shifrin
11:28 PM
No.
You need to learn some mathematics systematically and not just come in here asking random questions.
 
Ultradark
Ultradark
In dynamical systems the integral curves for a differential equation that governs a system are referred to as trajectories or orbits
 
Balarka Sen
Balarka Sen
An integral curve is an example of an orbit
 
Ted Shifrin
Ted Shifrin
The word "orbit" is far more general.
 
Ultradark
Ultradark
Just out of curiosity, what is geodesic flwo
 
Ted Shifrin
Ted Shifrin
Ultra: I'm not going to play this game.
Go read books.
 
Balarka Sen
Balarka Sen
11:39 PM
@TedShifrin I'm planning to give a student talk on group cohomology the coming semester, and learn something about it in the process
 
Ted Shifrin
Ted Shifrin
Cool, @Balarka. You probably already know more than I do.
Is this a grad student seminar?
 
Balarka Sen
Balarka Sen
Nah, just an undergrad math club
 
Ted Shifrin
Ted Shifrin
That's pretty fancy for undergrads.
Unless all the undergrads are like you, which I know they aren't.
 
Balarka Sen
Balarka Sen
Haha yeah I don't plan to go into the derived functors formalism
 
Ted Shifrin
Ted Shifrin
But do they even have any idea what cohomology is?
 
Balarka Sen
Balarka Sen
11:42 PM
No. I plan to explain what $H^2(H; N)$ is and why group extensions of $H$ by $N$ are classified by it
Pretty concrete stuff
I'll draw some pictures for the semidirect product
 
Ted Shifrin
Ted Shifrin
Ah, all formal algebra. You see?! You have reverted! :D
 
Balarka Sen
Balarka Sen
I said I'll draw pictures, Ted!
 
Ted Shifrin
Ted Shifrin
I typed what I typed before you typed that.
:P
 
Balarka Sen
Balarka Sen
Lol
 
Ted Shifrin
Ted Shifrin
So semidirect products are like nontrivial fiber bundles.
 
Balarka Sen
Balarka Sen
11:43 PM
Ya exactly
 
Ted Shifrin
Ted Shifrin
I'm not sure I know how to "draw" a simple example.
 
Balarka Sen
Balarka Sen
The general picture I have in mind is $N$ semidirect $H$ looks like an $N$-bundle over $H$ (picture a lots of sticks, all the cosets of $N$, over $H$), and I'll explain why the multiplication $(n_1, h_1)(n_2, h_2) = (n_1 \varphi_{h_1}(n_2), h_1, h_2)$ in the group the way it is by basically explaining that the action $\varphi : H \to \text{Aut}(N)$ that's conjugation inside the semidirect product is giving monodromy on the fibers of my bundle picture
 
Ted Shifrin
Ted Shifrin
So is there a concrete topological picture we have for some examples? I guess the Euclidean group is the easiest non-discrete example.
 
Balarka Sen
Balarka Sen
Right, that's $O(2)$ semidirect $\Bbb R^2$
 
Ted Shifrin
Ted Shifrin
I always tried to draw (at least schematic) pictures for $G\to G/H$ as a bundle ... even in my algebra book.
 
Balarka Sen
Balarka Sen
11:48 PM
Well, the isometries of the plane, I mean
 
Ted Shifrin
Ted Shifrin
Right, sure.
 
Balarka Sen
Balarka Sen
@TedShifrin Yeah I just don't think normal subgroups and quotients are explained well enough in first year algebra
 
Ted Shifrin
Ted Shifrin
That's why I started with rings and quotients, where you don't have to stress over normality at first.
 
Balarka Sen
Balarka Sen
That's a good cause
 
Fargle
Fargle
I know I definitely had trouble with normality and quotients in my first pass through the subject
 
Ted Shifrin
Ted Shifrin
11:50 PM
well, hello @Fargle
 
Paul Plummer
Paul Plummer
@BalarkaSen In the random walk models this has been proven, although interesting in "counting" models it is still open
 
Ted Shifrin
Ted Shifrin
Anyhow, @Balarka, you'll have to send me your LaTeXed notes when you give the talk :P
 
Fargle
Fargle
Hey @Ted
 
Balarka Sen
Balarka Sen
@TedShifrin Uh oh
 
Ted Shifrin
Ted Shifrin
LOL
Or not.
 
Balarka Sen
Balarka Sen
11:51 PM
I can't draw pictures in LaTeX, I'm not Danu
 
Ted Shifrin
Ted Shifrin
I can't, either. I draw them otherwise and input them as .eps into my LaTeX document.
Well, scanned handwritten notes will suffice, @Balarka. Or not ...
 
Fargle
Fargle
I just use Paint. Professionality be darned.
 
Balarka Sen
Balarka Sen
@PaulPlummer What are those models
 
Ted Shifrin
Ted Shifrin
I'm just curious to see what you do.
 
Balarka Sen
Balarka Sen
@TedShifrin Cool, I can do that
I just didn't want to LaTeX that's all :P
 
Ted Shifrin
Ted Shifrin
11:53 PM
Understood.
checks to see if @CaptainAmerica is afire
 
CaptainAmerica16
CaptainAmerica16
No, just in pain. I just got back from the dentist.
He was stabbing my gums.
 
Ted Shifrin
Ted Shifrin
At the dentist at 7 PM?
 
CaptainAmerica16
CaptainAmerica16
No 4:30
 
Paul Plummer
Paul Plummer
@BalarkaSen I am not super familiar with random walks, but basically you take random walks in the Cayley graph, as the walks get longer the probability you land on a pseudo Anosov goes to 1. counting model would be taking balls in the cayley graph and counting the proportion of pseudo Anosov's. It is open that this goes to 1 as the radius goes to infinity(it is known that at least a positive proportion is pA though)
 
Ted Shifrin
Ted Shifrin
Oh, "just."
 
CaptainAmerica16
CaptainAmerica16
11:55 PM
I "just" got in the door.
 
Ted Shifrin
Ted Shifrin
still confuzled
 
CaptainAmerica16
CaptainAmerica16
Words mean nothing >:C
 
Fargle
Fargle
@CaptainAmerica16 What? I don't know what you mean.
 
CaptainAmerica16
CaptainAmerica16
I don't mean anything.
 
Fargle
Fargle
Huh?
 
CaptainAmerica16
CaptainAmerica16
11:57 PM
Exactly >:|
 
Balarka Sen
Balarka Sen
@PaulPlummer Oh cool shit
 
CaptainAmerica16
CaptainAmerica16
So, I found out my brother can count to 10. He's only 2 years old.
 
Ted Shifrin
Ted Shifrin
Soon you'll be trying to teach him limits.
 
CaptainAmerica16
CaptainAmerica16
Hehe. I already told my mom that I plan on teaching him addition and subtraction before he starts kindergarten.
 
Ted Shifrin
Ted Shifrin
Don't ruin his life.
 
CaptainAmerica16
CaptainAmerica16
11:59 PM
You think he won
*won't like it
?
 
00:00 - 17:0017:00 - 00:00

« first day (3053 days earlier)      last day (1958 days later) »