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12:00 AM
elements can't appear twice in a cycle
 
Ok... I thought that we pick the last elemeny which is 2 and then we go to 1 then to n then again to 1 and then to 2. How else do we do this? @Semiclassical
 
there's three cases of interest: the permutation can act on 1, 2, or n.
 
What do you mean? @Semiclassical
 
if it acts on 1, then we have 1->2->2->1. if it acts on 2, then 2->1->n->n. If it acts on n, then n->n->1->2.
so the product of those cycles is (2n).
 
Shouldn't it be 2->1->n->1 ? @Semiclassical
Also how did you find this: n->n->1->2 ? @Semiclassical
 
12:05 AM
why would n->1? the second cycle has already acted at that point
 
A ok
 
let's try this a bit differently. what's (12)(123)?
(i'm taking my cycles to act on the right, btw)
 
We have 3->2->1
2->3->3
 
1->2->1
 
12:06 AM
3->1->2.
 
@Semiclassical oh yes, right
 
2->3->3 is correct, as is 1->2->1
 
And now which is the product? @Semiclassical
 
well, 1->1. so that's the singleton cycle (1).
and (2,3)->(3,2). so that's the transposition (23)
 
I mean at your example @Semiclassical
 
12:09 AM
so the product is (12)(123)=(1)(23)=(23). (i've dropped the singleton cycle by convention).
 
@Semiclassical don't we have 1->2->1?
 
yes. but that's reflected in the last two permutations.
in the first way of writing it, i explicitly have the singleton cycle (1) i.e. 1->1
in the second, the fact that the permutation is (23) means that nothing happens with 1.
 
@Semiclassical So we just look at the first and last number?
 
in the sense of input and output, yes
you put a number in, and you look what comes out
 
So if we have ( 1 2) (1 n) (1 2) do we have the following?

2->1->n->n
1->2->2->1
n->n->1->2 ? @Semiclassical
 
12:16 AM
right.
 
So the product is (1 )( n 2) right? @Semiclassical
 
yeah. usually we drop singleton cycles, though. so just (2n)
 
@Semiclassical So we have that $( 1 n) \neq ( 1 2) (1 n) ( 1 2)$ right?
 
right. you can also note that the LHS maps 2->2, but the RHS has 2->n.
 
@Semiclassical And does this mean that the centre of $S_n$ is the empty set? :/
 
12:19 AM
not quite. there's one thing which always commutes with everything in the group.
 
A singleton commutes with everything in the group, right? @Semiclassical
 
yeah. and a singleton is really just the identity permutation.
which is to say, the identity element of a group always commutes with the rest of the group. so at the very least, the center contains the identity.
the question is whether it contains anything else.
 
Why doesn't it contain all the singletons? @Semiclassical
 
well, what's the difference between the permutations (1) and (2)?
 
At the first one 1->1 , at the second one 2->2 @Semiclassical
 
12:23 AM
well, yes. but what, for example, does the first permutation do to 2?
 
Nothing @Semiclassical
 
right. 2->2.
 
So (1)=(2) ? @Semiclassical
 
@Semiclassical Nice... And how can we check if it contains anything else?
 
12:25 AM
ehhhh
not sure how one proves it
back later
 
Ok... We will talk about it tommorow... I have to go to sleep now... Good night!!! @Semiclassical
 
Vi.
12:38 AM
There is knot theory. Is there similar for origami? Does it resemble a "2D knot theory"?
 
Knot theory is part of topology, but it seems to me that origami is closer to geometry.
 
Vi.
Do ropes in knot theory extend and contract freely, as if made from rubber?
 
I guess it's equivalent both ways
If I have two arrangements of rope, A and B, of the same length…
…and I can change A into B by a continuous transformation that changes the length of the rope throughout…
…I can also do it without changing the rope's length.
 
12:53 AM
@AkivaWeinberger There is no reason to believe this.
 
No? If $f:[0,1]\times S^1\to\Bbb R^3$ is such that $f(0,x)$ traces out $A$ and $f(1,x)$ traces out $B$…
 
Vi.
Maybe if the rope is considered infinitely thin.
 
…then define $F(t,x)$ to be $f(t,x)/{\rm Length}(f(t,S^1))$
Sorry for horrible notation. I hope that's clear.
Um, assume $A$ and $B$ have length $1$.
 
Yeah, that's clever. Good point.
 
Point is, just rescale the rope at each instant to have the right length.
Um, yes, infinitely thin rope.
It's still topology, though, because the exact distances between the different points of the rope don't matter.
Typos
Fixed
 
1:00 AM
Of course in retrospect this is obvious because you can model knot theory with pieces of string, whose lengths do not change.
 
Vi.
So 2D analogue of knot theory would consider something like bubbles?
 
Maybe (2D) bubbles in 4D spaces?
 
I don't know what that means.
What part are you trying to make 2D?
The knot?
 
Can you knot a sphere in 4D space?
 
Yes.
 
1:02 AM
Interesting.
What about 5D space? I know regular (1D) knots don't exist in 4D
 
Of course reducing dimension is not interesting: I can tell you precisely what the embeddings of $S^0$ into a manifold are, up to isotopy. :P
 
Vi.
What happens with knots in 4D?
 
They're all isotopic to the trivial knot.
 
In any case, that's what I would call the 2D version of knot theory
Spheres in 4D
 
@AkivaWeinberger You're about to ask an interesting question, one that depends on what you think a "knot" is. If knots, to you, are continuous/PL embeddings up to locally flat (or, if you prefer, ambient) isotopy, then knots only exist in codimension 2, and they always exist in codimension 2.
If they're smooth embeddings up to smooth isotopy, the story is more complicated; Thom constructed a smoothly knotted $S^3$ in $S^6$.
 
1:04 AM
What's PL?
 
Piecewise linear.
(It's still true that every smooth $S^2 \hookrightarrow S^5$ is smoothly isotopic to the unknot.)
 
Wait. So, I can unknot his knotted 3-sphere in a 6-sphere, but only if I crease it?
 
Vi.
Does knot theory helps predicting how good would real life knots be (e.g. would they capsize, how much friction, etc.)?
 
@AkivaWeinberger Yes.
@Vi. No.
 
I don't know if knot theory ever talks about friction
 
1:06 AM
The keyword for something like that is "Physical knot theory".
 
That exists?
 
I don't know anyone who does it, but I'm assured people do.
 
it's funny how "physical knot theory" can mean two very different things
one: how do physical knots behave in the real world
two: knot theory in the context of gauge theory in theoretical physics.
(i know that the phrase really means the first, but still)
 
Well, that's not really physics, is it? :P
 
heh
it does depend on who you talk to, i suppose
 
1:13 AM
"What's your favorite part of math" "Knot theory" "Me neither"
2
 
but i would consider Ed Witten to be as much a physicist as a mathematician, for instance
 
I'm ignorant and am not allowed to have opinions on the matter.
Other than that I know who's funding his grants.
 
@MikeMiller Here's hoping you keep that attitude with regards to political opinions as well
 
Uh, is that an intentionally pointed comment?
 
It's nothing about you, it's about people in general.
Too many people have opinions on things they know nothing about, especially when it's a political opinion.
 
1:18 AM
Ok.
 
So, DogAteMy, who is officially allowed to have an opinion?
goodnight, @MikeM
hi @Semiclassic
 
hi @ted
 
@TedShifrin Your ping failed there.
 
Morning
@Pedro: Wasn't a ping.
 
I intentionally didn't ping, mr @Pedro :)
Lovely to see you, mr @Pedro!
 
1:23 AM
@TedShifrin Those who know about the thing
 
That's definitely not well-defined.
 
All sorts of uneducated people think they know everything.
I certainly don't know everything, but that's not going to stop me from reading and having an opinion and voting.
But most of our citizenry shouldn't be allowed to vote, if one requires actual knowledge.
@Pedro: Have any good questions for me to put on my point-set topology midterm and final? (I'm doing one last course, long distance, with 2 students.)
 
"Is $\varnothing$ connected?"
(Just kidding)
 
Too easy.
These are talented students.
 
1:27 AM
Nah, the problem is that different people define it differently. It satisfies the normal definition, but some people think it shouldn't count.
Kinda like how $1$ isn't prime.
 
Well, $1$ isn't.
But sort of hard to write the empty set as a union of disjoint non-empty open subsets :P
 
Might be wrong about that, though, but that's what I've heard.
 
@AkivaWeinberger: I guess this ties in with your previous discussion: if someone is so foolish as to think the empty set is disconnected, they shouldn't be talking about the empty set. :)
 
Did you ever work out the cells for the half-bagels? :)
I hate to agree with @MikeM, but in this case I must.
 
I thought it was what I had said, but with the longitudes and meridians switched.
Two of the flat ones, two halves of a vertical one
 
1:29 AM
Hmm, definitely not the right answer for slicing an actual bagel horizontally. So how many cells of which dimensions?
 
@TedShifrin Hm. I guess I shouldn't disclose them here. At any rate, I cannot think of one now.
 
@Akiva: Are you an undergrad?
 
No
I'm 16
 
I have some nice problems from my midterms, though.
 
OK, @Pedro. Feel free to disclose any way you wish.
 
1:30 AM
Man, 16 year olds abound.
 
BTW, @Akiva, you never replied to my verse/averse comment yesterday :D
 
Did it need a reply?
 
"Need"? Why do you keep asking ill-defined questions? :D
 
Why not?
:P
 
I suppose I deserved that.
 
1:32 AM
Oh, you know what, I think I was answering the wrong question with regards to the torus thing.
Um, wait
Two $0$-cells (on the inner and outer boundary of the annular surface)
 
Why are you sighing?
 
Four $1$-cells (inner and outer boundary of annular surface, one connecting the $0$-cells, and one connecting the $0$-cells the other way — that's is, through the curved surface)
And two $2$-cells (annulus, curved surface)
That work?
 
Excellent, @Akiva, except for your description of the $2$-cells. Their boundaries are effectively the annulus.
The boundary of the annulus, that is.
 
@Ted: "Smooth manifolds" are not geometry, and the statement "Topology is more general than geometry" is a poor understanding of what the phrase 'more general' means.
 
1:36 AM
Well, more structure typically in geometry.
 
But, you're now on my side. I bitch every time someone calls a general manifolds course differential geometry.
I understood, @Akiva.
 
@Ted: Aye. Whenever someone asks me what a course on smooth manifolds is, topology or geometry, I tell them it's smooth manifolds.
 
Wait, what do you mean about the $2$-cells
 
The 2-cell is a disk, it's not an annulus.
 
1:38 AM
(I took that picture before I saw your replies)
 
Topology is more fun.
 
Some of us would disagree, @PVAL.
 
I'm still not sure what I do.
 
Well, for all we know, you walk the streets mumbling ...
 
@TedShifrin You don't?
 
1:39 AM
@TedShifrin Sure, I do.
 
Well, maybe if you take away the "mumbling."
 
Question is, do you mumble about the length of your path or its homotopy type?
 
Geometry is more general than Riemannian geometry (which has lengths), btw.
There's no notion of length in projective geometry, or in all sorts of differential geometry.
 
True
Right, Klein described what "a geometry" is
 
Well, that's a basic starting point, but I'm not sure it is quite general enough for the last 70 years.
 
1:42 AM
It's weird how we can have "a geometry," "a topology," "an algebra"… we turn so many fields into nouns.
Well, they're already nouns.
But you know what I mean.
 
I was about to say that.
You're back to being ill-defined. You're not qualified to speak ... or to have an opinion.
 
what i wonder, come to think of it, is what examples of geometry show up in physics which wouldn't contain a notion of length
 
I'm mostly convinced the separation of fields is somewhat semantics and personal preference. I have seen people with work which to me clearly labels them "topologists" to me call themselves "geometers" and vice versa.
 
Symplectic.
 
as an example? hmm.
 
1:44 AM
@PVAL, when I finished my Ph.D. and applied for jobs, U Wisconsin had no "geometry" checkbox, and I had to say I was a topologist (was I going to say logician?). But complex integral geometry, despite being filled with Chern classes, was not topology.
 

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