Mathematics - 2016-10-12 (page 2 of 4)

@Ramanujan Draw a right triangle triangle with one angle $x$, and hypotenuse of length 1. $\sin(x)$ is perp/hypotenuse = perp. As $x \to 0$, perp becomes $x$.

user228700

@DHMO Can u tell how distance has a sign?!

DHMO

11:49 AM

@Kaumudi if the point is at one side of the line, it is positive.

user228700

@DHMO What the heck? Which side?!

Ramanujan

@BalarkaSen do you mean when Θ is small it obeys l=rΘ ?

DHMO

@Kaumudi that depends on the coefficient

Balarka Sen

Yes. You can justify this by drawing an arc w/ radius being the hypotenuse.

Ramanujan

@Kaumudi I think distance is always positive, but it can be negative when we take some point of reference and conventions

Damion M. Terrin

11:51 AM

then its called something more generic like position, i thought.

Ramanujan

@BalarkaSen oh,, since sine Θ is nearly Θ it cancels and become 1?

Balarka Sen

More or less.

Ramanujan

Less

DHMO

@BalarkaSen you know, I must say that i prefer the area approach more

Ramanujan

When Θ is small

DHMO

11:53 AM

@Ramanujan "more or less" is a phrase

Ramanujan

@DHMO share with us

DHMO

it means "you are correct"

Ramanujan

@DHMO oh,sorry for my bad English

Damion M. Terrin

its a weird property of sine that as the angle gets really close to 0 the sine of the angle converges to the angle itself.

user228700

@DHMO Can u please explain this?

user228700

11:54 AM

@DamionM.Terrin Yep. @Ramanujan: This approximation is used to derive many important equations in physics too.

DHMO

@Kaumudi looks like physics is your forte

Balarka Sen

@DHMO Shrug. My definition of sine is usually a power series, in which case this is trivial anyway. One has to do some kind of graphical proof for high school as the definition of sine is itself graphical.

user228700

@DHMO You could say that...sort of :-)

DHMO

@Kaumudi sorry, I don't know how to explain it

@BalarkaSen yes, but I have to say I prefer the area approach more

user228700

@DHMO Sigh. Okay, thanks for trying. Does anybody else know..? :-(

Ramanujan

11:56 AM

@DHMO share with me that area approach

DHMO

@Ramanujan do you know what an arc is?

Damion M. Terrin

also a good way to test the accuracy of calculators i guess, see how close to 0 you can get before the calculator cant tell the difference between an angle and its sine.

Ramanujan

Yes

user228700

@Ramanujan: Do u understand that formula? (Angle bisector)

Ramanujan

Which? @Kaumudi

Damion M. Terrin

11:58 AM

you could probably till how many iterations of the series it does based how accurate it is,

DHMO

@Ramanujan and a sector?

Damion M. Terrin

im going to bed.

DHMO

@DamionM.Terrin a much simpler way is to see how many digits of 1/7 is calculated

user228700

@Ramanujan The one I was asking about before. That's why I asked if u're in 11th and if u go for coaching and all.

Ramanujan

@Kaumudi no(ーー;)

@DHMO yes I know sector

user228700

12:00 PM

@Ramanujan U're not preparing for JEE..?

Ramanujan

@Kaumudi can you tell formula name? I can understand that matjax

DHMO

Draw a sector OBC with centre O and with angle BOC being as small as you can. the area of the sector is (1/2)(r)(r)θ while the area of the triangle OBC is (1/2)(r)(r)(sinθ)

user228700

@Ramanujan Never mind >.< My brain is dying. I'll look into this tomorrow again. Time for some chemistry.

DHMO

as θ approaches 0, the two areas become the same

Ramanujan

@Kaumudi OK, but iam not shure I will come tomorrow here

@DHMO do you men 1/2 in area to be semicircle?

DHMO

12:03 PM

@Kaumudi should i be waiting you in the chemistry chatroom?

@Ramanujan not semicircle

Ramanujan

@DHMO area of sector is n/360π(r^2)?

DHMO

@Ramanujan oh... we are using radian here

Ramanujan

I don't know in case of radian

DHMO

alright

Ramanujan

OK so replace 360 by 2π will give what you mentioned!

DHMO

12:07 PM

yes

Ramanujan

Thanks @DHMO bye👍

DHMO

bye

user228700

12:25 PM

@Ramanujan That's OK :-)

user228700

@DHMO No, no! I'll ping u if I have a question...is that OK?

DHMO

@Kaumudi yes

user228700

@DHMO OK, thank you :-) See ya.

Danu

12:45 PM

@BalarkaSen I'm reading the char. classes stuff now

Balarka Sen

@Danu Nice.

Danu

So in section 4, do you see a nice way of deriving that formula for the inverse in lemma 4.1?

It's not very interesting, of course, but I'd like to have a slick way to show either (i) how to get that expression (ii) verifying that it is indeed the inverse at arbitrary order

Balarka Sen

Write down a candidate for the inverse, multiply. It's straightforward.

Danu

Did you see the definition?

It's inductive

Balarka Sen

Yes, I am aware.

Danu

12:50 PM

But I'd like to avoid a big induction proof

Balarka Sen

it's not really that tedious.

Danu

So $(w\bar w)_n=\sum_{k=0}^n \sum_{j=0}^{k-1} \bar w_j w_{k-j}w_{n-k}$

And you say it's quick to see this is zero?

Balarka Sen

How did you get three terms there?

Danu

The formula for $\bar w_k$

I'm trying to either derive that formula, or verify it

The above is a verification (attempt)

Balarka Sen

I mean, don't plug that in. Verifying would probably be more tedious.

If $\bar{w}$ is a candidate for an inverse, the $n$-th term in $\bar{w}w$ is $w_n + w_{n-1} \bar{w_1} + \cdots w_1 \bar{w_{n-1}} + \bar{w_n}$.

And that's $0$.

That immediately gives you your formula.

Danu

12:58 PM

I guess so, sigh. I thought it'd be cool to derive that formula from the intuitive idea using the power series argument

But that turned out to be kind of a bitch

Balarka Sen

Don't make a simple thing hard :)

Danu

:(

But as physicist I like power series :P

I guess I'll just mention the power series thing on the side as a way to get the thing in terms of $w$ alone

I guess the annoying/tedious thing is proving that power series trick gives the right result

Martin Sleziak

1:48 PM

This was kind of nice. After an author was notified about a mistake in his text (discovered thanks to math.SE post), he came here and posted an answer.

It's not a book, just some lecture notes, but still it's nice thing to see.

Danu

2:02 PM

Yeah, awesome @Martin

Balarka Sen

That has happened at least once before.

So, yep, MSE is great

Danu

Jack Lee is great on here

Balarka Sen

Mhm

Also, Hatcher's here too (though not as frequent)

0celo7

2:26 PM

In old papers they didn't write squares apparently

$mv^2$ was $mvv$

Balarka Sen

i have seen that too

0celo7

The old paper from Laplace on black holes is pretty terrible when it comes to notation

$$\frac{drddr}{dt^2}$$

That's in the paper

also $$d\frac{ds}{dt}=\frac{dds}{dt}$$

Not sure what that means

PearlSek

2:41 PM

Hello there :)

Danu

2:53 PM

So @BalarkaSen MS's alternative proof for triviality of SW class for $TS^n$ only works for $n>1$ right, since $\Bbb RP^1=S^1$

Balarka Sen

$TS^1$ is still trivial...

Danu

That's not my question :P

Balarka Sen

I mean, I don't see why you're claiming the proof won't work for $n = 1$.

Danu

He is using that the canonical projection induces a trivial map on top degree cohomology

Balarka Sen

It's a double cover. It always will induce trivial map on Z/2 cohomology

Danu

3:03 PM

He uses naturality of the cup product to say $f^*(\alpha^n)=(f^*\alpha)^n=0$ because $f^*\alpha\in H^1(S^n;\Bbb Z_2)=0$.

which is not true for $n=1$

So this doesn't work for $n=1$. But $TS^1$ is easily trivialized so it's no problem

Balarka Sen

Huh? It is.

Note the coefficient of the cohomology.

Danu

@BalarkaSen $H^1(S^n;\Bbb Z_2)=0$ isn't true for $n=1$

The induced map is still zero, I'm sure, but not for that reason

Balarka Sen

Ah, I misread. Yes. $f^*\alpha = 0$, but $H^1(S^1)$ is not.

Danu

Yeah

Balarka Sen

He probably meant to place the "$=0$" somewhere else.

Good catch. You're a careful reader, @Danu :)

Danu

3:06 PM

Anyways, so this thing you said

double cover kills $\Bbb Z_2$ cohomology

Mike said it too recently

I see that if you have a generator of a double cover's cohomology then it should project down to two times the generator in the base

Is that all you're saying?

Balarka Sen

Yes.

Danu

How do I formally prove it?

Balarka Sen

Think homology because that'd be easier. Suppose $p : \tilde{M} \to M$ be your double cover. You want to prove $p_* : H_n(\tilde{M}; \Bbb Z/2) \to H_n(M; \Bbb Z/2)$ is zero. Triangulate $M$ so that $\tilde{M}$ also gets a triangulation with two triangles over each triangle in $M$.

Then on chain level a representative of the fundamental class $[\tilde{M}]$ is given by $\tilde{M}$ (as a simplicial complex). That gets mapped to a chain in $M$ such that each singular simplex appears in pairs (by construction). Mod 2, that's $0$.

Danu

Okay, but then I just need to konw that I can do such a triangulation :P

Balarka Sen

Once you know $M$ is triangulable, that's not hard. Barycentrically subdivide till your simplices fit inside a specific open cover of $M$ over which $p$ trivializes.

That's Lebesgue number lemma.

(also, I am doing this on top homology, note)

Danu

3:20 PM

I'm thinking about what the best way is to draw a picture

I want to draw that the inclusion $\Bbb RP^1\hookrightarrow \Bbb RP^n$ is covered by a bundle map between taut. line bundles

What I ended up drawing intuitively is just including a circle in a 2-sphere with a red line through the center in the circle mapping to a red line through the sphere's center

Balarka Sen

That's exactly my mental picture.

Danu

Yeah?

Good

I'm just thinking if there is anything to make it look less like a normal sphere bundle :P

DHMO

Sorry for a stupid question, but what is the genus of a solid cube? @BalarkaSen @Danu

Balarka Sen

Only surfaces have a well defined genus.

DHMO

surfaces?

I mean the surface of a cube then

Danu

3:24 PM

Do you know a bit of topology?

Balarka Sen

Surface of a cube is topologically just the 2-sphere.

DHMO

@Danu not really, sorry

Danu

^shaddap Balarka :P

Mary Star

Hello!!

Let $f = x^4−2x^2−1 \in \mathbb{Q}[x]$.
We have that the polynomial $f(x+1)$ is Eisenstein. This means that $f(x+1)$ is irreducible, or not?
Knowing that, can we conclude that $f(x)$ is irreducible.

Let $\rho\in \mathbb{C}$ be a root of $f$. I want to find a basis and the degree of the extension $\mathbb{Q}[\sqrt{2},\rho]$.

Do we have that $\text{Irr}(\rho, \mathbb{Q})=f$ and $\text{Irr}(\sqrt{2},\mathbb{Q}[\rho])=x^2-2$ ?
If this is true we have that the degree of the extension $\mathbb{Q}[\sqrt{2},\rho]$ is $\deg f \cdot \deg (x^2-2)=4\cdot 2=8$.

Balarka Sen

@Danu I don't think there's any harm fiddling with surfaces and their genus and Euler characteristic completely non-rigorously before learning any topology at all.

DHMO

3:25 PM

@BalarkaSen because I'm a little bit confused about what "manifold" means

> In mathematics, a manifold is a topological space that locally resembles Euclidean space near each point.

Danu

@BalarkaSen I just meant you shouldn't spoilerino

DHMO

How does a vertex resembles Euclidean space?

Balarka Sen

@Danu oh, I see.

Danu

It doesn't, but you can stretch it out to be smooth

Balarka Sen

^

DHMO

3:26 PM

I see

Danu

So in a certain strict sense (if you're not allowed to stretch) it's not a manifold

But if you're allowed to stretch/smoothen then it's fine

DHMO

> The genus of a connected, orientable surface is an integer representing the maximum number of cuttings along non-intersecting closed simple curves without rendering the resultant manifold disconnected

so this definition is wrong?

Danu

No, that's okay

And it will tell you the genus of a cube's surface

DHMO

but it isn't a manifold as you said

Balarka Sen

"In a strict sense"

Danu

3:27 PM

It's a topological manifold

It's not a smooth manifold

DHMO

I see

as you see, I have confusion manifold

Danu

Best to just learn stuff rigorously

then it won't be so confusing

DHMO

@Danu trying to

how do you prove that the genus of a sphere is 0?

sorry if i'm asking too much

Balarka Sen

Any closed simple curve disconnects the sphere.

DHMO

how to prove that?

Balarka Sen

3:30 PM

Jordan curve theorem.

Hard to prove rigorously - the simplest proof in the greatest generality that I know of is by homology theory.

DHMO

I forgot the quantity represented by the reciprocal of the radius

curvature?

@BalarkaSen thanks

Balarka Sen

Curvature of a curve at a point is the reciprocal of the radius of the circle which approximates the curve the best at that point. I am not sure how that's relevant here though.

DHMO

Is the surface inside a hollow sphere having negative curvature?

Sorry for stupid question

Danu

So this is what I drew @BalarkaSen

Balarka Sen

@DHMO No.

DHMO

3:33 PM

@BalarkaSen ok, thanks

Danu

Balarka Sen

You and your tikz :P But looks good. I'd draw a few more lines if I drew it on paper.

I stumbled upon a cool-looking definition of classifying spaces that I faintly recalled of hearing from a topologist once upon a long time ago, and a few minutes later realized the lecture notes are from his course :P

iwriteonbananas

Whazzzzzzzapppp

@BalarkaSen What is it

Balarka Sen

Hi bananas

@iwriteonbananas If you have a topological group G which admits a G-invariant metric, he's defining EG to be the space of step functions [0, 1] --> G (i.e., it is piecewise constant), modulo the equivalence relation identifying two of them if they agree at finitely many points.

Then G acts freely on EG by multiplication, and the quotient is your friendly neighborhood BG.

On, and the topology on EG is actually given by a metric, given by integrating the pointwise distance (makes sense, as G has a metric on it) between the step functions.

This is apparently Segal's model of BG.

Danu

@Balarka ^

Balarka Sen

3:48 PM

OK

Danu

All good? ;D I gave the arrow a bit more space (on the right), too

Balarka Sen

Sure

Danu

If I fail in math I'd love to be a math book editor

Balarka Sen

Nah you'll be fine

Puzzled417

is this true: A number $x \in (0,1)$ is rational if and only if its decimal expansion is preperiodic?

DHMO

4:05 PM

@Puzzled417 or terminates

(in which case the repetend is 0)

what is "preperiodic"?

Balarka Sen

I have to get schoolwork done. Ugh.

Mike Miller

4:20 PM

morning

Danu

eyyo

Balarka Sen

Good morning

Danu

BACK TO YOUR WORK KID

Mike Miller

ok

DHMO

Topology of a Twisted Torus - Numberphile

►

Secret

4:30 PM

A Hole in a Hole in a Hole - Numberphile

►

When one of the tube opening become 2, is it breaking rules. I thought you cannot change the number of holes when deforming a topological object...?

DHMO

@Secret which second?

no, you haven't changed the number of holes

the original thing doesn't have 3 holes to start with

the "hole" in the title "a hole in a hole in a hole" is not actually a real hole

Secret

4:37

DHMO

@Secret think of pulling the inner glass out

Secret

ah I see, makes sense

Danu

So @BalarkaSen I'm getting a bit confused here... In example 3 from section 4 of MS, they have $w(\gamma^\perp)=1+a+a^2+\dots + a^n$. Kinda fine---but then... $w(\Bbb R^{n+1})=w(\gamma_n^1)w(\gamma^\perp)$

But if I just write the right hand side out I get $1+a^{n+1}$...

Mike Miller

4:40 PM

And what is $a^{n+1}$?

Danu

Well, that's what I find weird

obviously the cohomology groups of $\Bbb R^n$ all vanish so there has to be just 0

But I can't use info about the cohomology of $\Bbb RP^n$ to say something about the cohomology classes which are supposed to live on $\Bbb R^n$, can I?

(of course I realize that in $\Bbb RP^n$ we have $a^{n+1}=0$)

Balarka Sen

@Danu $\Bbb R^{n+1}$ is the trivial vector bundle on $\Bbb{RP}^n$, not the Euclidean space.

Danu

@BalarkaSen Sorry, I think not

But you're right

:D

Never mind

That's great

I'm retarded :D :D

Also thanks @MikeMiller

Mike Miller