Mathematics - 2016-12-10 (page 5 of 6)

Semiclassical

18:00

I have no idea what those mean, so I have nothing more to say than that

meow-mix

$W^{1,\Phi}(\Omega)=\overline{C_0^{\infty}(\Omega)}$

Vrouvrou

sorry i mean $W^{1,\Phi}_{0}(\Omega)=\overline{C_0^{\infty}(\Omega)}$

meow-mix

what is $C^{\infty}$?

infinitely differentiable functions?

Adeek

hey @arctictern here ?

arctic tern

yeah

Vrouvrou

18:09

Bump function

In mathematics, a bump function is a function f : Rn → R on a Euclidean space Rn which is both smooth (in the sense of having continuous derivatives of all orders) and compactly supported. The space of all bump functions on Rn is denoted C 0 ∞ ( R n ) {\displaystyle C_{0}^{\infty }(\mathbf {R} ^{n})} or C c ...

Adeek

I am just reviewing some stuff in group theory. If we we classifying groups of order pq. Then there is the abelian one and there is the one given by semi direct product but why are all semi direct product isomorphic ?

@arctictern ?

arctic tern

pretty sure we've covered this

or maybe not all the way

Adeek

I don't remember.

arctic tern

the semidirect product comes from homos Z/pZ->Z/(q-1)Z, and all elts of Z/(q-1)Z of order p (if they exist) are related by some automorphism of Z/(q-1)Z which can be extended to an isomorphism between the two semidirect products

Adeek

oh

MickLH

18:19

@meow-mix Trying to answer this question has led me to an interesting conjecture! much appreciated :) I believe that if we were to assume the existence of an imaginary element $i = 2^{-1}$ then the set and operators would form a full field, though I'm not provably sure about that

Though it's very consistent with the intuition that spawned the idea in the first place, so my gut is telling me to just conjecture it lol

meow-mix

@MickLH write out your full conjecture; i'd like to see it :)

arctic tern

@MickLH $\Bbb F_{2^n}$ is not the notation for integers mod $2^n$

SAWblade

If some $H \leq G$ is non-Hopfian, does that imply that $G$ is also non-Hopfian? :0

Anybody have an idea? xD

MickLH

@arctictern pls halp

arctic tern

what's your question

meow-mix

18:22

you would write $\Bbb Z / 2^n\Bbb Z$

MickLH

@arctictern "What do I call it?" is one of the biggest parts of the question :P

arctic tern

what do you call what?

MickLH

@meow-mix I've encountered this notation before, but I've always had to confirm the meaning from context in each situation. I'm not sure how I'm supposed to read it.

meow-mix

@MickLH it's the quotient group of the integers over the left coset $2^n$

arctic tern

nevermind

MickLH

18:25

@meow-mix Thank you! this seems like exactly what I'm looking for, though of course give me a second to review some terms

arctic tern

@MickLH https://en.wikipedia.org/wiki/Modular_arithmetic#Integers_modulo_n

meow-mix

@MickLH the left coset $gH$ of $g \in G$ with subgroup $H$ is the set of all $gh$ for $h \in H$

MickLH

@arctictern In simple, it's the set of odd natural numbers less than $2^n$, endowed with + and * operators

meow-mix

@MickLH hmm

arctic tern

@MickLH but adding two odd numbers does not give an odd number, and multiplying two numbers less than 2^n may not be less than 2^n.

meow-mix

18:26

^

MickLH

@arctictern Right, so the addition is defined like $f(x, y) = x + y - 1 \left(\text{mod} 2^n\right)$

meow-mix

modulo $2^n$, right

MickLH

yes, sorry to be ambiguous

arctic tern

then you've just got a set with two operations

SAWblade

better question - is hopficity heriditary?

meow-mix

18:27

@MickLH you want to start a private room investigating the properties of this structure

MickLH

Sure, if that's appropriate

arctic tern

doesn't seem necessary

meow-mix

@arctictern i don't want to spam chat

with proving associativity and what not

Vrouvrou

with this definition can i say that $u\in W^{1,\Phi}_0$ impies that $u=0$ on $\partial\Omega$ ?

?

arctic tern

sure, your f(-,-) is commutative and associative, but it's not distributive so it's not clear there's any nice interaction between f(-,-) and multiplication, so ... it's a set with two binary operations

Semiclassical

18:30

f(cx,cy)=c(x+y)-1 doesn't seem particularly nice, no.

meow-mix

nope

MickLH

@arctictern I've also defined * in a non-standard way though to match, namely $g(x,y) = \frac{(x-1)(y-1)}{2}+1$

Semiclassical

:/

meow-mix

@MickLH you may also consider the usual multiplication of integers

arctic tern

@MickLH perhaps you should call it g(x,y)

Semiclassical

18:32

could also do $\oplus$ and $\otimes$.

Adeek

@arctictern can you use semi direct product to classify groups of order 8. I classified it normally didn't use semi-direct product.

arctic tern

iunno

Semiclassical

Doing $x\oplus y=x+y-1$ also makes sense because that's how binary OR works.

SAWblade

okay hopficity is not hereditary

meow-mix

@Semiclassical no it isnt?

Semiclassical

18:34

But I feel like this is just the usual integers with + and *, taken mod 2^{n-1}, under the mapping $n\mapsto 2n-1$.

$1\oplus 1 = 1,1\oplus 0=0,0\oplus 1=0,0\oplus 0=1$ (-1 = 1 in boolean)

yeah, not OR

meow-mix

exclusive or

Ted Shifrin

Hi @Semiclassic, @meow, tern.

meow-mix

hello @TedShifrin

Semiclassical

actually, it's the negation of XOR

meow-mix

XNOR

Semiclassical

18:36

NOT XOR = XNOR ?

Ted Shifrin

I feel like I stepped into a logic spaceship ... sort of the opposite of the real world around us.

meow-mix

@Semiclassical yes

Semiclassical

mmkay

Ted Shifrin

NOT XOR = either both or neither

meow-mix

similarly, negation of OR is NOR, and negation of AND is NAND

Semiclassical

18:37

NAND is fun to say

Ted Shifrin

Anything projective to discuss, @meow, before I disappear? :)

Semiclassical

@TedShifrin Well, if you want a touch of familiarity, I've got this stack of lab reports I should be grading right next to me :)

Ted Shifrin

That sounds reassuring, @Semiclassic.

Semiclassical

Hah.

I should at least count how many I have

meow-mix

@TedShifrin just pondering the harmonic-y questions

Semiclassical

18:39

31 reports.

That...actually seems small. :/

Which is worrisome.

Ted Shifrin

@meow: They're not super important.

Maybe some were stuck together, @Semiclassic.

Semiclassical

Maybe.

Ted Shifrin

Gluons, you know.

Semiclassical

riiiight

Ted Shifrin

shrug ... I tried

meow-mix

18:41

oh, @ted!

how do you take the cross ratio of 4 lines?

Semiclassical

Amusingly, the phrase 'cross ratio' came up in a physics talk this week, albeit in passing

It was on conformal field theory, albeit at a very sketchy level.

Okay, my first section has all of their reports in. So that's good.

And my second is missing two, neither of which are a huge surprise.

Danu

Hey @Ted

Hey @Semi

Semiclassical

Hey @Danu

Ted Shifrin

@meow: You only do that if they live as four points on a line in $\Bbb P^2{}^*$, right?

Hi @Danu

Semiclassical

@MickLH The thing I mostly notice is that your addition and multiplication satisfy $f(2n+1,2m+1)=2(n+m)+1$ and $g(2n+1,2m+1)=2(nm)+1$

Danu

18:46

@TedShifrin So while I'm waiting for my supervisor to come up with a precise problem to look at, I thought it'd be a good idea to read the references of his paper to get some better backgrond info. I'm debating what to start on...

meow-mix

@TedShifrin so how do we define cross ratio in $\Bbb P^n$?

Ted Shifrin

Ideally, you should read and ask questions rather than waiting for him to pose them. He can always say, "good," "bad," or "totally understood."

You don't, @meow. It's only defined for four points on a $\Bbb P^1$.

meow-mix

oh wait

Semiclassical

Hence why it's interesting for Mobius transforms.

Sophie

Palmyra fell

Ted Shifrin

18:47

But note that it's projectively invariant, so you get a well-defined thing.

meow-mix

so you consider

in $\Bbb P^2$

4 points that are collinear

Ted Shifrin

The world is a mess, @Sophie. I am sorry.

Right, @meow, for example.

meow-mix

so that you have this "copy" of $\Bbb P^1$ intersecting those points to perform the cross ratio on

Kasmir Khaan

Hi @TedShifrin I have this question about line integral in complex plane , integral dz/ (z-a ) over Y, when Y = circle in C , centred at a with radius R

Danu

I'm tempted by Thom's 1958 big paper, but hesitant because it might be not-most-efficient. Other options include other papers (not directly related, but on characteristic numbers and stuff) by my supervisor, one paper by Gromov, one by Gromov & Lawson...

Ted Shifrin

18:48

containing those points, yes, @meow.

Semiclassical

50 years later, and this song seems as relevant as ever

Barry McGuire - Eve of Destruction

►

Ted Shifrin

@Kasmir: So?

Danu

My preference goes out to:

Bob Dylan- Masters of War

►

Sophie

@TedShifrin are you watching the conflict? ISIS took the SAA completely by surprise, Assad really scrapped the bottom of the barrel bomb to end Aleppo

Kasmir Khaan

@TedShifrin can you explain to me what am supposed to do ?

Ted Shifrin

18:49

No, @Sophie, I was listening to the news all morning, but I'll catch up later. The next years are going to be horrible everywhere.

Semiclassical

Yeah.

Ted Shifrin

Parametrize the circle and do the integral, @Kasmir. This is one that shows up thousands of times and you'll just know it after this.

Semiclassical

"this whole crazy world is just too frustrating"

Danu

Let's see how many European countries will be taken over by right-wing populists in the coming years...

Semiclassical

Ugh.

At least Austria held that off this year?

Danu

18:51

Yeah, that's a good start I guess?

Semiclassical

It's something.

Sophie

the previous austrian election was much closer, I think it was 49.5% for the right wing party

Semiclassical

But I dunno what counts as better or worse nowadays.

I mean, my feelings re: Trump were/are ... not positive, to put it lightly

Mike Miller

What are we doing?

Semiclassical

Sighing about geography while doing geometry, I guess.

Ted Shifrin

18:53

Other than ruminating over the state of the world, trying to decide what @Danu will read.

Danu

About specific questions, @Ted, I am assuming the most feasible thing to do for me is to somehow try to construct some spaces which (i) satisfy some interesting criterion (ii) satisfy/violate an interesting condition on characteristic numbers.

Ted Shifrin

Oh, and meow is pondering cross-ratio.

Semiclassical

random question, @Danu, but how much do you know about conformal field theory?

Ted Shifrin

@Danu: First you need to assemble a menagerie of example spaces for which you can compute things. Complete intersections are a great source of examples.

Danu

The way to do it will probably be analogous/related to my supervisor's recent papers.

@Semiclassical I'd say a lot for a master's student.

Mike Miller

18:54

complete intersections are good, as are their resolutions

Semiclassical

neat.

Danu

I took a course specifically on CFT. I only know anything about stringy CFT though.

So only d=2

Mike Miller

in general resolutions of singularities (incl quotient singularities) are good examples

Semiclassical

Virasoro algebra stuff?

Danu

@TedShifrin I'm supposed to be using bundles, I think.

Semiclassical

18:55

The guy who was talking here this week was working on higher dimensional stuff.

Ted Shifrin

When they talk about characteristic numbers, that's usually referring to the tangent bundle, @Danu.

Danu

@Semiclassical Yeah, and extensions ($\mathcal W$-algebras, currents)

Semiclassical

right.

Kasmir Khaan

@TedShifrin are line integrals on plane same as line integrals on complex?

Ted Shifrin

yes, @Kasmir. But use complex arithmetic/calculus to make life easier.

Semiclassical

18:56

The most I've had to deal with CFTs myself is that certain systems act like them when they go through a phase transition.

Danu

@TedShifrin Sure; I should clarify: Kotschick's paper computes characteristic numbers of (the tangent bundle of) some total spaces of vector bundles.

meow-mix

@TedShifrin projections preserve cross-ratio?

Ted Shifrin

hmmm ...

Kasmir Khaan

@TedShifrin what do you mean ? been 2 years since we last used complex numbers

Danu

$\Bbb P^{2k}$-bundles over $S^4$.

Semiclassical

18:57

which isn't terribly surprising, since at a phase transition you find that various quantities either diverge or become zero, and that includes the correlation length diverging

Ted Shifrin

@Kasmir: If you're doing complex functions and integrals, you're using complex numbers!

Semiclassical

^

Ted Shifrin

Ah, ok, @Danu.

@meow: You mean you have two lines in $\Bbb P^2$ and you project from a fixed point from one line to the other?

Kasmir Khaan

@TedShifrin hmm if i let Y :z=a+Re^it is that good parametrization ?

Ted Shifrin

If so, you should work it out, but isn't that a projective transformation, @meow?

Yes, @Kasmir, perfect.

Semiclassical

18:58

So if you sit at the transition point and let the system volume go to infinity, it'll behave like a CFT. You also see it showing up if you allow the system volume to go to infinity while also moving closer and closer to the transition i.e. an appropriate scaling

Danu

It's just because there is a theorem that tells you they admit a metric of non-negative sectional curvature @TedShifrin. I guess there are a lot of theorem about metrics on bundles?

Kasmir Khaan

@TedShifrin thanks again Ted! :) ill keep working on the problem

Semiclassical

blah blah blah holographic law

Ted Shifrin

Ah, @Danu, I don't know all the Gromov-Lawson stuff I should.

Mike Miller

Beautiful stuff.

Ted Shifrin

18:59

Maybe Danu should run a seminar in here on it :)

Danu

That's one of the papers that's referenced. Do you think I stand a chance of understnading any of it?

Ted Shifrin

I dunno. There's a fair amount of geometry and analysis, but jump in and take some stuff as black-boxed (even though you hate to do that, you need to).

Danu

@Semiclassical Yeah. In my course, this blahblah relevant to phase transitions blahblah was just introduced as a way to rake in funding ;)

Semiclassical

lol

meow-mix

@TedShifrin but it maps to a line, not the projective plane... and don't projective transformations have to be in GL(3,R)?

Semiclassical

19:00

we did some calculations for it, in the case of a topological phase transition of a particular 1D system

Which was neat, as mathematical physics

Mike Miller

also, Gromov-Lawson is a number of papers

Danu

@TedShifrin I'll probably retake Riemannian geometry next year... HOpefully no characteristic classes in disguise this year :P

Ted Shifrin

@meow: Choose coordinates on each of the lines and write down a formula for the projection as a map from the line to the other line.

Semiclassical

but I won't pretend any of the stuff I do actually matters :P

Danu

@MikeMiller Classification of simply connected [...]

Mike Miller

19:01

I hated the class on Riemannian geometry I took even though I find much of the geometry beautiful

Ted Shifrin

Of course you both would have loved my course :P Except for all those damn differential forms.

Danu

From the talks I've attended on Riemannian geometry, it sounds too analytic to me to really love it.

meow-mix

@TedShifrin soy confuzled

Ted Shifrin

@Danu: Parts of it have become very analytic. But it's not all.

meow-mix

@TedShifrin "lines" there are more than one line in a projection?

Ted Shifrin

19:01

LOL @meow

Danu

@TedShifrin Just today I took a look at the paper by Chern and Simons on the Chern-Simons form (which is heavily used in physics). Lots of forms :P

meow-mix

i should say "confuzlando"

Ted Shifrin

Read what I typed up there. I talked about fixing a point $P$ in the plane and two lines. You get a map $\ell_1\to\ell_2$ by sending $Q$ to the intersection of $\overleftrightarrow{PQ}$ and $\ell_2$.

meow-mix

and i feel like i just made a fool of myself.. :P

Danu

Are there any widely accepted "good texts" on Riemannian geometry?

Mike Miller

19:03

@TedShifrin Would I? Maybe.

Ted Shifrin

Yes, Chern is always forms, and Chern-Simons is one of the most basic examples of transgression.

Mike Miller

@Danu Find the French book. Ted will know the authors.

My favorite book is not suitable for a first course.

Danu

@TedShifrin Transgression being... a technical term?

Ted Shifrin

Oh, three authors. I never looked at the book.

Mike Miller

Actually, Danu will hate it, since half of the book is exercises.

Ted Shifrin

19:04

Yes, @Danu, it's like when a form on a bundle gives you a well-defined form downstairs of a lower degree.

Danu

@MikeMiller Yo screw you too :P

Balarka Sen

@Danu Great song.

Semiclassical

@TedShifrin I don't suppose there's an easy simple example of that?

meow-mix

@TedShifrin im going to re-read some of this... my brain is having a confuzled party

Vrouvrou

@TedShifrin please i have this definition , can we say that $u=0$ on $\partial\Omega$ ?

Danu

19:05

All those colors

Ted Shifrin

Chern-Simons is the most interesting basic example. I can try to give you something more explicit later, but I have to leave now.

LeGrandDODOM

@Ted salut

Balarka Sen

Rather violent at the end for a successful artistic protest against violence, but great.

Ted Shifrin

@Vrouvrou: I have to leave and this is too technical. You need to ask your professor.

@LeGrandDodo: Salut et à bientôt.

Bye all.

Semiclassical

bye @ted

Danu

19:06

bye

Simple Art

What in the world are you guys talking about?

LeGrandDODOM

@SimpleArt hi

Simple Art

Lol, ok

Danu

@BalarkaSen Let out the rage!

It's not really against violence, per se. More against war :)

Balarka Sen

Sure, sure.

Simple Art

19:08

I don't like violence unless it's meaningful

Semiclassical

Eh, for a more violent rage against war, I prefer Black Sabbath :)

Simple Art

What are we raging about? I'm so confused

Semiclassical

(I have no idea what the video itself in there is from)

Simple Art

So I recently learned about multivariate chain rule, and I find it a bit sad that most of the single variable derivatives rules are special cases of it.

Danu

It's Gallot, Hulin, Lafontaine, @MikeMiller @TedShifrin.

Balarka Sen

19:13

I have heard praises of G-H-L.

No new and interesting math today?

Simple Art

Idk

Mike Miller

No. I'm not doing math for another three days. Then I'll be doing uninteresting math.

Semiclassical

Not unless you count the Biot-Savart law as interesting :/

Balarka Sen

Taking a break is good, I am going to do that after my exam. And by uninteresting math did you mean algebra? :)

Semiclassical

Though you can use the Biot-Savart law in conjunction with Ampere's law to get a physical derivation of Gauss's formula for the linking number :)

meow-mix

19:21

is $\Bbb P^{2*}$ isomorphic to $\Bbb P^1$?

Adeek

@arctictern is $Aut(H \rtimes K) = Aut(H) \rtimes Aut(K)$ ?

Mike Miller

No. That's not even true for products.

$\text{Aut}(G \times G)$ always has an automorphism swapping the two factors, for instance.

Balarka Sen

Look at say Z/2 x Z/2

Mike Miller

You might stand a chance of it being true if $H$ and $K$ have coprime orders or something?

Adeek

Yeah I guess

arctic tern

19:22

If $H$ and $K$ have no common direct factor then $\mathrm{Aut}(H\times K)$ can be described as a matrix group with entries in $\mathrm{Aut}(H),\hom(H,Z(K)),\hom(K,Z(H)),\mathrm{Aut}(K)$

Arrow

Sorry if this is very trivial. In a group, can anything interesting be deduced from the equation $abc=b$?

Mike Miller

cool @arctictern

well, one learns that $abc = b$

Adeek

also for example $Z_n \rtimes Z_n^{\times}= Aut(D_2n) = Aut( Z_n \rtimes Z_2) \neq Z_n \rtimes Z_2$

Balarka Sen

@Arrow You get stuff like that in knot groups.

But I don't know what specifically you can learn.

Adeek

lol @MikeMiller

brb computer is crashing have to restart

I have a habbit of having like million tabs

brb

arctic tern

19:24

@Arrow yes, a^-1 = bcb^-1 tells us a^-1 and c are conjugate by b

Arrow

Suppose $f:A\to B$ is a monoid homomorphism and $s$ is a section of $f$. Suppose moreover the kernel of $s$ and the image of $f$ generate $A$. I'm wondering whether one can deduce that every element $a\in A$ is a product of kernel elements times the element in the image of $s$ in the same fiber as $a$.

arctic tern

@Arrow the kernel of s:B->A and the image of f:A->B are subsets of B, how can they generate A?

Arrow

sorry, typo

Balarka Sen

Unrelated question. Suppose you have a group $G$ finitely generated (say by $g_1, g_2, \cdots, g_n$) with all the relators of the form $hg_ih^{-1} = g_j$ where $h$ is a word on $g_1, \cdots, g_n$. Need that be a knot group? I see that you can build the knot locally around the crossings, but how the picture will patch up is not clear to me.

Arrow

meant the kernel of $f$ and the image of $s$

Mike Miller

19:29

@BalarkaSen no, you further need $H_2(G) = 0$

actually hm

i think that's only known for sufficiently high-dimensional knots, and is open for dim 1

Balarka Sen

interesting; nonetheless good point bringing up that criterion

Mike Miller

(of course one has $H_2(G) = 0$ but there should be further restrictions - for instance, the binary icosahedral group times Z is not the fundamental group of a knot complement)

Balarka Sen

ah. funny.

I should also have said there's a relator like that for all $i, j$. You'd want to have the right abelianization.

Mike Miller

you can probably check that $\Bbb Z \times 2I$ does that

(it does have the right $H_1$, which is why I chose it)

Balarka Sen

yep, agreed.

Mike Miller

19:44

so it does appear as a knot complement in dimensions 5 and up, but not 3 (and idk about 4)

Balarka Sen

Why does it appear as a knot complement in dimension => 5?

Mike Miller

kervaire proved it

maths.ed.ac.uk/~aar/papers/kervhknt.pdf

oh there's one more necessary fact

you can find an element that normally generates the group

which i guess automatically disqualifies $\Bbb Z \times 2I$ :(

oh no nvm, it's satisfies for that, since $2I$'s only normal subgroup is order 2, so it seems like picking $x$ to be any element with order larger than 2 $(1,x)$ normally generates the group? maybe? worried about this

but i'll let you think about it

Balarka Sen

if $\Bbb Z \times 2I$ has the property that for all $i, j$ you have a relator $g_i = hg_jh^{-1}$ like you claimed, it should be normal closure of a single element just fine, I think.

Mike Miller

you can try to prove it if you like

i think it doesn't tho

Adeek

hey @arctictern is it true that when we have a group G such that $|G| = n\phi(n)$ and we have a subgroup of index $\phi(n)$ then it is normal

I am classifying $Aut(D_{2n})$ where this is true but I was wondering if it is true in general.

Mike Miller

19:56

i would guess no. try groups of order 20.

Adeek

alright

Tobias Kildetoft

The form $n\varphi(n)$ does not seem like it is "special" enough to say much about things like this

arctic tern

@MikeMiller the sylow thms say n|4 and n=1mod5 for n the number of conjugates the the cyclic group of order 5 in any group of order 20

Mike Miller

dang

Tobias Kildetoft

Yeah, as long as all prime divisors of $n$ are greater than $\varphi(n)$ then the result works out

(such as when $n$ is prime)

Balarka Sen

20:00

@MikeMiller I pass; that's too much algebra for me on antibiotics.

Neat results on that paper however.

Mike Miller

ok, take $n=8$ and you're probably fucked there

Tobias Kildetoft

Agreed

Semiclassical

For reference: groupprops.subwiki.org/wiki/Groups_of_order_32

meow-mix

im feelin stupid

Alessandro

Why?

meow-mix

20:07

i can't do this exercise :(

Alessandro

What exercise?

Adeek

hey @arctictern so I am little bit confused about this so I proved that $Aut(D_{2n}) = Z_n \rtimes (Z_n)^{\times}$. But I was wondering I am trying to compute the map here this gives the semi direct product. So, I proved that the maps given by $r \mapsto r^k$(gcd(k,n) = 1) and $s \mapsto s$ and $s \mapsto sr^l$ (fixes r) where $0 \leq l \leq n - 1$ gives this classification where the second one is normal in $Aut(D_{2n})$.

meow-mix

@Alessandro i have to prove the medians of a triangle in projective $2$-space are concurrent using desargues' dual

Adeek

to get the explicit automorphism map we conjugate the normal group by the non-normal one. But the problem here we don't have explicit generator for the $Z_n^{\times}$

I guess I can transfer everything to $Z_n$ and $Z_n^\times$ and work with them ?

yeah I think it is easier that way coz I think the maps picture are confusing me.

Semiclassical

@meow-mix How're you setting it up?

meow-mix

20:16

i don't knowwww

i have to prove these 3 points collinear to use desargues dual

Semiclassical

Hm

I don't know projective geometry, so I can only say stuff that's likely uninformed

Adeek

maybe I should compose them and see relations among the powers @arctictern ?

Semiclassical

But, you start with three vertices $[\alpha],[\beta],[\gamma]$ in P^2.

Alessandro

Wait isn't Desargues theorem its own dual?

meow-mix

@Semiclassical ok so let me give an outline

so we have points $P,Q,R$ forming the triangle

$P'$ is the midpoint of $QR$, $Q'$ is the midpoint of $PR$, $R'$ is the midpoint of $PQ$

i have to prove that $\vec{PQ} \cap \vec{P'Q'}$, $\vec{QR} \cap \vec{Q'R'}$, and $\vec{PR} \cap \vec{P'R'}$ are collinear

those are lines by the way

Semiclassical

20:19

Collinear, or concurrent?

meow-mix

collinear

to apply desargues' dual

Semiclassical

Am confuzed. Do PQ and P'Q' intersect in that case?

meow-mix

all line in projective $2$-space intersect

Semiclassical

Yes, you're right

Makes more sense now.

I guess my mentality is rather brute-force. If you've got points a,b,c in P^2, then their midpoints are the 3 pairwise averages of those points.

Adeek

yes so I think I figured it out. So here is the explicit way to construct $Aut(D_{d2n})$

Semiclassical

20:23

And from that you get three lines.

But this probably isn't the Desargues' route.

Adeek

Suppose that $Z_n = <a>$ and $Z_n^{\times} = <b>$ then $Aut(D_{2n}) = <a,b : a^n = e, b^{|Z_n|^{\times}} = e, bab^{-1} = a^{|Z_n^{\times}|}>$

haha that is some crazy group @arctictern

Mahmoud

Hi.

meow-mix

20:39

@Semiclassical i have another problem i dont know how to do

Balarka Sen

21:17

"Between the idea / And the reality // Between the motion / And the act / Falls the Shadow: for Thine is the Kingdom
Between the conception / And the creation // Between the emotion / And the response / Falls the Shadow : life is very long"

Rüdiger

no spam plz

Semiclassical

for thine is/ life is/ for thine is the

(I know a lot of TS Eliot by memory) @BalarkaSen

Balarka Sen

@Rüdiger You consider that spam? Pshaw. I can do better.

Semiclassical

Though I should really have done

between the desire / and the spasm / between the potency/ and the creation/ between the essence/ and the descent

falls the shadow / (for thine is the kingdom)

Balarka Sen

@Semiclassical Cool. I only started reading Eilot very recently.

Rüdiger

21:29

Thou

shalt

Semiclassical

I've been reading him since high school

Rüdiger

not spam*

Semiclassical

If this is what you consider spam, you need better taste in literature.

So...about 12 years?

@BalarkaSen Which of his stuff have you read so far? Early or later?

"The Hollow Men" is right at the end of early TS Eliot, I suppose

with "Ash Wednesday" at the beginning of late TS Eliot

Balarka Sen

I read two (The Hollow Men and Love Song of J Alfred Prufrock) about 2 days ago :)

Semiclassical

Ahh.

Balarka Sen

21:32

Our high school syllabus isn't nearly as good to include Eliot. Oh well.

Semiclassical

Pfff

Fun fact, while I was in St Louis (for reasons) I went to see the house he grew up in. It's not marked as historical---someone owns it nowadays---but it was still neat.

Balarka Sen

I don't know much about his personal life either except that he was post-WWI.

MickLH

I'm afraid that one company I tend to buy pre-processed food from lacks any real engineer on their product development team :/
(I mean the recipes are great! but it's the small things, like this frozen meal that is marketed on its short prep time, yet it's frozen into a brick when clearly it could have much higher surface area)

Balarka Sen

I wouldn't have read him if I didn't hear J Alfred Prufrock being recited by someone a few days ago; it immediately sounded like something I could connect to (what came to mind was Dostoyevsky in poetry :P).

Alessandro

@Balarka I can draw the trefoil on a torus so I can cut its (or a?) Seifert surface from the surface of the torus, it looks something like a Möbius strip with more twists I'd guess, I need to get better at drawing :P the Hopf link is painful because its surface can't be embedded in R^3

Balarka Sen

21:43

@Alessandro Good approach on trefoil. You're wrong about the Hopf link!

Alessandro

Ah, that's a surprise

Balarka Sen

Actually, more rigor is needed on your trefoil thing. If you remove a trefoil from the torus you still got a 2-manifold, noncompact, but without boundary.

I am not sure anymore if that approach works.

Alessandro

If my drawing is right removing a treefoil from the torus divides it into 2 connected components

Balarka Sen

That sounds right. So you take one of those connected components, together with the boundary knot at the end, yeah?

Alessandro

Yes

Perturbative

21:58

Hey everyone, has anyone here used Lang's 'Algebra'?

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$Mathematics$ Mathematics