Mathematics - 2017-12-01 (page 5 of 6)

It's much harder to fold from Balarka's configuration compared to TastyRomeo

45 mins and still no progress yet

Perturbative

12:20 PM

Does it seem geometrically obvious that $\overline{\mathbb{B}^n}/\mathbb{S}^{n-1}$ is homeomorphic to $\mathbb{S}^n$?

D^2/S^1 $\cong$ S^2 real time

Perturbative

Ahhh thanks Balarka!

Balarka: Just for a santiy check, in your fold, is the above twin cube config what you get before it becomes a cube?

Thank wikipedia, the free encyclopedia!

@Secret I fiddled with that configuration for a long time, yes. You need to revert a fold to get the cube

It's not as obvious as Steamy's config

RIP, my net is falling apart with a tear in one of the crease. Guess that mean I have to try again later and take some rest

FuzzyPixelz

12:37 PM

If $f'$ is continuous on the closed interval $[a,b]$

Is $f$ differentiable on $(a,b)$ or $[a,b]$

max

1:05 PM

guys plz help i got a question about finding curvature of a function in 3d

when i find magnitude of tangential vector, i get like \sqrt e^-2t + 0 + e^2t

and i dont know if i should plug in t values from the point or not

the book doesnt have that example it onley has example with sin t cos t which dissappear when squared

plz help

full problem is to find curvature of function r(t) = \sqrt 15 t, e^t, e^-t

at point (0,1,1)

1:32 PM

Steamy's 9 square net cube folding:

Q: Let $f$ be an analytic function defined on $D = \{ z\in\mathbb{C}\colon|z| <1\}$. Then $g\colon D\to \mathbb C$ is analytic if

2

Let $f$ be an analytic function defined on $D = \{ z\in \mathbb{C}\colon |z| <1\}$. Then $g \colon D\to \mathbb{C}$ is analytic if $g(z) = f( \bar{z}) $ for all $z\in D$ $g(z) = \overline{(f (z))}$, for all $z\in D$ $g(z) = \overline{(f (\bar{z}))} $ for all $z\in D$ $g(z) =\bar{i} f (z...

complex-analysis

Cauchy_riemann equation is $u_x=v_y$ and $u_y=-v_x$ right?

here in the answer it is not satisfied.

Am I correct?

please see (3)

1:54 PM

royalcaribbean.com/cruise-ships/allure-of-the-seas

The sea, the final frontier

@Secret what is this?

Mare Tranquillitatis

Mare Tranquillitatis (Latin for Sea of Tranquility or Sea of Tranquillity (see spelling differences)) is a lunar mare that sits within the Tranquillitatis basin on the Moon. The mare material within the basin consists of basalt formed in the intermediate to young age group of the Upper Imbrian epoch. The surrounding mountains are thought to be of the Lower Imbrian epoch, but the actual basin is probably Pre-Nectarian. The basin has irregular margins and lacks a defined multiple-ringed structure. The irregular topography in and near this basin results from the intersection of the Tranquillitatis...

google is super not helpful when googling "Cauchy Sea"

and btw, what you wrote is indeed the cauchy riemannian equations

@Secret yes

in 3, $u_y(x,-y)=v_x(x,-y)$

Right?

@Secret I don't understand this :(

2:13 PM

sign of y is flipped because of the conjugation on z

and when you differentiate that, you get a minus sign in front of the u_y

The answer is checking if g satisfy the cauchy riemann equations

ok. Now, i understood. Why didn't I think !!! :'(. Thank you very much

@Secret

3:01 PM

So the Spanish word for "paramedic" (paramédico) literally means "doctor stopper?

(Similar to how the word for "parachute" (paracaídas) means "fall stopper")

Leaky Nun

@BalarkaSen you’re not really quotienting S1 though

you’re just gluing a copy of S1 to a point

hmm. what kinds of mappings preserve convexity of sets...

@AkivaWeinberger (parachute means the exact same thing, it's just that English stole the French word)

Hey @AkivaWeinberger are you familiar with the construction of the (3-dimensional) lens spaces?

I forget how they're defined

i only want convex sets -> convex sets, mind. i don't care about the pre-image of a convex set being convex

3:13 PM

or why we care about them

ok

you haf da lens

and you haf da group

you up and down and round and round

and you haf da lens space

They were historically interesting cause their classification up to homeo is NOT equal to the classification up to homotopy

no idea how thats proved tho

Also they're very natural and simple, really

So Balarka

you start with an action of Z_p on one circle and Z_q on another

and then sorta from the start of the construction you're already modding out one of them

and I don't understand how it plays any role

how does it make any difference

3:16 PM

mostly I want to argue that if $K\in [0,\pi/2]^3$ is convex then $f(K)\in [-1,1]^3$ is convex where $f(\alpha,\beta,\gamma)=(\cos 2\alpha,\cos 2\beta,\cos 2\gamma)$

@Danu I don't understand. I define the lens space as Z_p acting on a 3-ball, and then taking a quotient. What is this construction?

main thing that seems apparent is that $\alpha\to \cos 2\alpha$ is strictly decreasing

So yours have just a single parameter?

There is actually a 2-parameter family

Oh I see

MatheinBoulomenos

No, two parameters, but just one action

3:17 PM

and I'm just confused what role the one parameter (that you forget about) plays at all

Yeah I get it

You take two torii

solid

And glue them along the boundary by a p/q diffeomorphism of T^2

That's L(p; q) I think

@Semiclassical I'm willing to bet only affine maps

I mean, they have to send line segments to line segments

@Danu On the lens the other parameter corresponds to the number of polyhedral faces that you skip in the lens before modding out by Z/p

Or I guess it's possible that it does a weird space-filling thing but I think that's probably impossible

That's q

3:19 PM

it's affine if you demand the pre-image of a convex set be convex as well

but I don't care about that

so there's no need for it to send line segments to line segments

@BalarkaSen I don't quite understand that sentence

so let's start from this picture

Ok, let me clarify

Alright, you go

@Semiclassical I mean, they have to send line segments to something convex, right?

'Cause line segments are convex, and you want the images of convex stuff to be convex

no.

(wait a sec)

3:21 PM

both the tetrahedron and the sphere are convex sets

Right, so are line segments.

@Semiclassical Akiva is arguing that because line segments are convex, you have to send line segments to something that is convex, because you require image of convex set is convex

if I map the tetrahedron to the sphere, the image of line segments in the former will be spherical arcs

So the red lines show the Z_p-identification

3:22 PM

and arcs aren't convex sets.

What would the q do?

@Semiclassical …Right, so that means that that mapping doesn't satisfy your requirement

It doesn't send convex sets to convex sets

hrm

it only sends a particular convex set to some other particular convex set

To be clear, you want a map $f:U\to V$ such that for all convex $X\subseteq U$ you want $f(X)$ to be convex?

eh. I was thinking it'd be true for my mapping

3:24 PM

@Danu Right, so the action of Z/p on that lens is defined by rotating the whole thing by angle 2pi/p * q, and reflecting. So one upper face gets identified to a lower face that's $q$-cells separated from it

Or at least, I think this is the action

oh, now I see it. yeah, I was asking for something too strong

@BalarkaSen I'm confused by this sentence.

Hi @Mike

@BalarkaSen you can still define these as quotients by the action $1 \cdot (z,w) =(e^{2\pi i/p}z, e^{2\pi i q/p}w)$

hi

The generator of $\Bbb Z/p$ acts on the complex you drew by rotating $2\pi/p \cdot q$ around the axis, then reflecting along the equator. Is this a confusing statement, @Danu?

@MikeMiller Yeah but I never remember the formula.

@BalarkaSen I thought it was just by 2pi/p

3:28 PM

Nope, definitely not.

That would be L(p; 1)

@BalarkaSen Standard in the first coordinate, both coordinates need order p, you need a q somewhere

Hah

that's how i remember

So are they always S^1-bundles over S^2?

$(z,w)\to (\omega z,\omega^q w)$

3:31 PM

@Danu yeah I think so because this action factors through the fibers of the Hopf bundle

given what Mike wrote

So what's the boundary map from the 2-cell to the 1-cell?

Is it multiplication by p

or by pq?

actually, Balarka, I think you meant that the rotation is by 2pi q/p

You mean 3-cell to the 2-cell?

and then the boundary map is q/p

no, 2 to 1

@Danu Oh that's what I wrote but in a fucked up way

(2pi/p)*q

sorry

@BalarkaSen oh, I see, haha

3:40 PM

@MikeMiller I had a question about something in terms of geometry. There is this thing called a characteristic foliation which is also sometimes defined by a vector field $X$ such that---when given a volume form $\theta$ and a 1-form $\beta$ whose kernel is locally given as $\xi$, your contact hyperplane field---we have $\iota(X)\theta = \beta\wedge(d\beta)^{n-1}$.

It is usually defined however in terms of the tangent plane of a surface $S$. You have that $T_pS\cap\xi_p$ is either 1 or 2-dimensional. And you 'draw' the foliation in this manner, filling in where it is 2-dimensional with a point.

I don't know how one would approach showing these two definitions are equivalent. I don't intuitively understand how the first relates to the second.

woo, these animations of cubic surfaces are really nice: madore.org/cubic-dvd

see in particular "The Cayley cubic surface (having four A1 singularities) (simple rotation)." @BalarkaSen

Geiges shows that $X$ is in the kernel of $\beta$ and then that $d\beta(X,v) = 0$ for any $v\in T_pS\cap\xi_p$, but I don't see how this is sufficient either.

Cubic surface: the Cayley cubic surface (having four A1 singularities) (simple rotation)

actually, easier is probably this Youtube link

►

@anakhronizein Sounds nice!

Sounds nice but I still can't figure it out. ;)

3:48 PM

nice @Semi

yeah

4:05 PM

@BalarkaSen So when I have a basis of left-invariant vector fields on a Lie group, and I consider some Lie brackets, is it OK if I get a function times an element of the basis, instead of just a number times it?

Hm

I think so

why? :P

it corresponds to a number at the identity

but... idk

I feel like it shouldn't be OK

it's not going to be left-invariant, is it?

I am not at all in the right state of mind right now to think about it. Haven't slept all day :P

You're probably right

Yeah actually $dL [X, Y] = [X, Y]$ I think.

yeah because of naturality of the Lie bracket

so I must've made an error somewhere...?

This is disturbing

cause I'm pretty sure I didn't make an error.

oh lel

of course I did

:D

4:35 PM

@BalarkaSen Weird-ass identity of the day

$$4 \sin (\alpha +\beta +\gamma ) \sin (\alpha +\beta -\gamma ) \sin (\alpha -\beta +\gamma ) \sin (-\alpha +\beta +\gamma )=\begin{vmatrix}
1 & \cos 2 \alpha & \cos 2 \beta \\
\cos 2 \alpha & 1 & \cos 2 \gamma \\
\cos 2 \beta & \cos 2 \gamma & 1 \\
\end{vmatrix}$$

dangit mathematica

Hah, pretty cool

that's true for all alpha,beta,gamma

@Semiclassical I actually stumbled upon something like that today

and it guarantees that both sides vanish whenever $\pm \alpha\pm \beta\pm \gamma=\pi n$

for any combination of signs

that it works for any choice of signs and any n is obvious when you note that the RHS is invariant under 1) permutation of angles, 2) sign flips of angles, 3) shifts of pi in angles

so the LHS had better have those same symmetries

I still feel like I don't really understand why this identity is true, though, beyond the purely algebraic facts.

I mean, sure: Using Euler's formula, you can expand out the stuff on the left and simplify down to expressions in terms of cos(2alpha) etc alone

in that respect it's not especially shocking

but it still weirds me out

@Daminark I wasn't

4:42 PM

your gravatar woke up

I think I'd be happy with just a simple explanation for why that determinant ought to vanish when $\gamma=\alpha+\beta$

the rest of the cases follow from the facts above.

i guess one observation one can verify is that in that case the matrix has the zero eigenvector $(-\sin 2\gamma,\sin 2\beta,\sin 2\alpha)^T$

but eh

I can't get no satisfaction from an identity I only sorta understand :/

5:07 PM

Hi

HELLO.

Let $y=(0,\cdots, 0, 1,\cdots, 1, 0,\cdots)$. What is the norm of the operator $A\:\ell_p(\mathbb N)\to \mathbb C$ given by $A(x) = \langle x, y \rangle$ ?

Where $y$ has, say, $n$ nonzero elements

Hi! This is a geo. series: ${b_(n) | b_(n) \in \Bbb R}$. i wanna add 1 detail that it is a growing series. How can i represent the term "growing" symbolically inside the above given representation in symbols?

@CRYPTONEWBIE what have you tried?

Well it's the composition of the "multiplication by $y$" operator and a sort of sumation operator

The former has norm $\|y\|_{\infty}$ if I'm not mistaken

But I don't know how useful that is

5:19 PM

How do you define the norm of A?

It ought to be $\|A\| = \sup_{\|x\|=1} |Ax|$ right?

Yes that is fine.

So what does |Ax| typically look like for ||x||=1?

If $x$ has, say, one $1$ in one of the slots where $y$ has a $1$ too, then $|Ax|=1$

That doesn't really answer your question though, I suppose

Well it is a start.

How is $\ell_p(\mathbb N$ defined?

@Tug'Tegin if you wanted that to have subscript (n) you'd want to do b_{(n)}

5:28 PM

Infinite sequences $(x_n)\subset \mathbb C$ such that $\sum_n |x_n|^p <\infty$ right?

@Semiclassical yeah, got it, i meant that.

as for how to denote it: $b_n=a r^n$ for $n\in \mathbb{N}$?

So $|Ax|= |x_n+\cdots + x_{2n}|$

no, i meant how to represent the symbol "growing" mathematically @Semiclassical

If the 1's in y start at y_n, yes

5:32 PM

That's what I meant, sorry

n+k not 2n also

So I think you have most of what you need.

Just need to find the sup.

Hmmm

then I really don't know what you mean @Tug'Tegin

Keeping in mind that ||x|| = 1, maximize |Ax|

Right, I'm trying

5:36 PM

@Semiclassical o-o-h sorry, you were right, i didn't pay close attention at your answer! Gracias! i starred that.

So certainly it is at least $1$

Yes.

If $p=1$ it is also at most $1$

Because $|Ax| = |x_n +\cdots x_{n+k}| \leq |x_n| + \cdots + |x_{n+k}| \leq \|x\|_1 = 1$

I don't know what to do for other values of $p$

Hi @anakhronizein

If $\|x\|_p = 1$ and $x\neq (0,\dotsc,0,1,0,\dotsc)$ , then what can you say, @CRYPTONEWBIE?

Hi Mike!

5:47 PM

I'm a little confused. Let's set notation. $(M,\xi)$ is contact, $\alpha$ a contact form. What is its dimension and what are we taking the characteristic foliation of?

$2n+1$ is the dimension, and let $S$ be a surface in $M$.

The first thing seems to give me a global vector field and the second a foliation of a surface (presumably you set $n=1$ for the second one)

Oh I see.

So we are discussing the characteristic foliation of $S$ in $M$.

Those were two definitions I gave. The first one is in terms of a vector field which I do not understand, the second in terms of the tangent bundle of $S$. I understand the second, but the first I don't understand, nor do I see how it is equivalent. Even with Geiges's proof.

If it helps, I can reiterate the definitions more clearly?

@anakhronizein I really can't figure that out, can you give me another hint?

Well @CRYPTONEWBIE can you make |Ax| > 1 in that case?

5:59 PM

Ahh

Yes we cna

E.g. when p=2 and $x$ has two nonzero entries $1/ \sqrt{2}$

The norm is $2/\sqrt{2}$

So my guess for the general norm is $p/\sqrt{p}$ for all $p\leq k$ where $k$ is the number of nonzero entries in $y$

Note that p/sqrt(p) = sqrt(p).

Yes. Is the result correct? I still need to prove that the norm can't get any bigger

I have not worked out the problem myself, but I would say you are on a good path.

6:15 PM

@anakhronizein I was just distracted for a bit; I haven't thought yet.

It's fine!

user8663905

Hi all

Anyone have the time to help me with a differential equation?

Thank you Ana

@anakhronizein So it seems to me that Geiges wants you to consider $\text{ker}(i_X(d\beta)) \cap \xi$, and call that the characteristic foliation. Because $\beta(X,-) = 0$, Cartan's magic formula identifies this as $\text{ker}(\mathcal L_X \beta) \cap \xi$.

Geiges shows that this contains what you would like to call the characteristic foliation. The question, then, is whether it is that.

The only way that could fail is if this kernel was larger than $T_p S \cap \xi$.

Ugh I guess I'm just really confused as to how the surface even shows up in your first setup.

6:41 PM

Oh sorry, I guess I neglected to say that in the first set up $\theta$ is a volume form on $S$. Not on $M$.

@anakhronizein Oh, I see.

A volume form on S that extends to a global form?

Or just everything on S I guss.

Eef I have to go. That Cartan magic formula was the only thing that was useful I saw.

Ask PVAL. He's better.

2

Sure I will take a look at what you said and ask PVAL if I see him

sup ppl

hey

Silent

Hey! Let $F$ be an archimedean ordered field. Suppose $\bar F$ be Dedekind completion of $F$. Does $\bar F$ has to be archimedean?

7:05 PM

Let $G = C_{25} \times C_{25 }\times C_5$. How many elements of order 5 does $G$ have? Is it 100?

hrair

hrair?

yep.

Sorry, I don't know what that means.

Oh wait.

Yeah, 100.

tbh i was making a joke that i was pretty sure would only make sense to me

7:10 PM

everything2.com/title/hrair cool.

yeah

so basically just "i dunno, a lot?"

Haha, okay

7:35 PM

Hey

hello

@Daminark i might end up having a pseudo-reading course pop out of low dim top

p exciting

smbc-comics.com/comic/a-math-lesson

reddit.com/r/math/comments/7gqhlc/…

2

Ah nice, with Danny?

yeah

it was his suggestion when i spoke w him after class

7:43 PM

That's gonna be p sick

8:17 PM

$p < 0.05$

@Semiclassical person 1 "Many." person 2 "Some!"

@orbit-stabilizer but $np > 0.05$ so $p\ne np$ qed

wooooaaahhh. stop the presses. someone just won hrair dollars!

That famous open problem, $“\exists k:k(P-NP)=1?”$

@Daminark turns out, we're not working in a field with the Archimedean property

8:23 PM

:O

Alternatively, $“(1-N)P=0?”$

And it's always kinda confusing that the N in NP doesn't stand for "not"

I heard a talk yesterday by the Nobel laureate Barry Barish who was involved with LIGO. They discovered gravitational waves. It was pretty cool.

Nice

scratched that off of my bucket list: Hear a Nobel prize winner speak in person

Leaky Nun

@orbit-stabilizer every element there is either of order 1, 5, or 25.

8:27 PM

Next item on the list: Win the Nobel prize

@LeakyNun yeah. And I think there are 100 elements of order 5. Because 100 = 5 x 5 x 4.

Hey!

Anybody up?

no im sitting down

Haha... Can we discuss about a problem? @orbit-stabilizer

@ffahim depends. what's the question?

Finding the rank of an elliptic curve? Sorry no can do

Gradient @orbit-stabilizer

8:32 PM

Just ask

Suppose that the steepest slope on a hill is 40%. If a road goes directly up the hill, then the steepest slope on the road will also be 40%. If, instead, the road goes around the hill at an angle, then it will have a shallower slope. For example, if the angle between the road and the uphill direction, projected onto the horizontal plane, is 60°, then the steepest slope along the road will be 20%, which is 40% times the cosine of 60°. ... Why 40% is multiplied by cos60° @orbit-stabilizer

I'm assuming you're in university, yet you're using degrees in your math course - that's surprising.

I'm assuming steepest slope being 40% means 40% of 90 degrees.

Actually I am in 11th grade @orbit-stabilizer

You're learning multivariable calc in grade 11? Wow, that's cool. Good for you