Mathematics - 2018-10-11 (page 3 of 3)

Mike Miller

20:05

I don't know. Sounds like you're getting caught up in definitions.

The equations $x_2 = x_3 = 0$ define a 2-plane in $k^4$ and a projective line in $\Bbb P^3_k$.

Karl

Perfect

what if I have a point (0:0:s:t) and this projective line (a:b:0:0). How do I parametrize the plane containing both the point and this line?

mercio

o..o

is it on purpose that your notation for points and your notation for lines look like the same thing ?

Karl

No, maybe it is the source of confusion actually

these are homogeneous coordinates

so the point is written up to a scalar constant

in the line instead a and b can vary independently

Mike Miller

Yeah that's terrible notation. What you want is the set $\{[z_1, z_2, z_3, z_4]\}$ so that $z_3 t = z_4 s$.

Here $s,t \in k$ are fixed constants.

The problem is it was unclear in your formulation what is variable and what is constant.

Karl

absolutely

Will Hunting

20:13

Hello @MikeMiller I hope you are making progress in your work.

Karl

ok, so the set you wrote, satisfies the requirements because the projective line I wrote trivially satisfies it.

having z_3 and z_4 null

is my reasoning correct?

Ted Shifrin

20:47

@Karl, yes. I would comment that ordinarily people number indices starting with $0$ when doing homogeneous coordinates. You can also think of it by taking parametric equations of the various lines through the fixed point and an arbitrary point on the line.

BTW, please use early letters for constants and late letters for variables :P

Karl

oh hi @TedShifrin

Yes thanks.

Semiclassical

i'm at this week's physics and astro colloquium, and the speaker has the Gauss-Bonnet theorem showing up on a slide

Ted Shifrin

Good taste :P

Semiclassical

yep

Ted Shifrin

2-D or general?

Semiclassical

20:50

he's citing it in the context of area as motivation for more genreal stuff

Karl

I am trying to prove that the set of lines intersecting three lines in projective space P^3 is a quadric

Ted Shifrin

Ah, well, it's classic motivation for Chern classes and other characteristic classes, and those have physical impact.

Semiclassical

yep

Ted Shifrin

@Karl: I was about to give you that exercise :P

Semiclassical

i'm expecting Chern classes to at least get mentioned

Karl

20:51

:D

Ted Shifrin

One of my favorite classic questions (which I even did in my algebra text) is to find the number of lines meeting 4 lines in general position in $\Bbb P^3$.

Karl

well it's two

Ted Shifrin

(assuming $k$ is alg. closed, yes).

Karl

we've been doing it in class

the thing is I am trying to solve it differently

Ted Shifrin

Ah ... That motivates all sorts of mathematics ...

Yeah?

Karl

20:52

in class we assumed it was a quadric and just eliminated the terms, imposing that a generic quadric vanished on the lines

now instead I am trying to find a plane through a point in the first line and containing the second line

and then intersecting it with the third line

Ted Shifrin

Yes, the classical construction is beautiful. I've even done it in lectures with high school kids.

Karl

do you think it makes sense

?

Ted Shifrin

You can, of course, use $PGL$ to put a few of your lines in a nice convenient position.

Karl

already did

Ted Shifrin

OK, I figured.

Karl

20:54

x_0 = x _ 1 = 0

x_2 = x_3 = 0

Ted Shifrin

Wait, slow down.

Karl

tell me

something wrong?

Ted Shifrin

So we fix $\ell_1$. You take a point $P\in\ell_2$. $P$ and $\ell_1$ span a plane, and you intersect that plane with $\ell_3$. Now what?

Semiclassical

a probability exercise for myself. Suppose I've got three r.v.s Xk with k=1,2,3 such that E[Xk]=0, E[Xk^2]=1, and E[Xj Xk]=-1/2 for j=/=k.

Karl

yeah that's my problem

Ted Shifrin

20:55

@Semiclassic: Aren't you paying detention to colloquium?

Karl

now what

Semiclassical

lol

yeah

Karl

because this does not necessarily lead me to a quadric

Semiclassical

oh hey, "first chern number" just showed up

Ted Shifrin

So that plane, $\Pi_P$, intersects $\ell_3$ in a point $Q(P)$. What are you doing with that point?

Karl

20:56

the plane I find is $\{[x_0, x_1, x_2, x_3]\}$ so that $x_2 t = x_3 s$

Ted Shifrin

I don't want equations, @Karl. I want ideas.

Karl

if the point p is parametrised with (0"0:s:t)

sure

yeah that point

what do I do with that point

Semiclassical

The weird thing with the above is that you can't produce that if $X_k\in\{-1,1\}$

Karl

it surely belongs to the quadric

Ted Shifrin

What do you know about the line determined by $P$ and $Q(P)$?

Karl

20:57

and my idea was

that in this way I would get a parametrisation of the quadric

Ted Shifrin

Not quite.

Karl

by varying s and t in the parametrisation of the point

hmmm

Ted Shifrin

Remember, there's really only one parameter there, not two.

You're working with $[0,0,s,t]$.

Karl

yes which is one generic point of a line though

so to change point I would change both parameters right

Ted Shifrin

But that's only one parameter.

No.

Karl

20:59

well multiplying by a scalar would give me the same point

Ted Shifrin

This is really $[0,0,1,t]$ together with $[0,0,0,1]$.

Karl

oh

Ted Shifrin

A line is one-dimensional ;)

Karl

ok well then surely that has no hope of parametrising a quadric

Ted Shifrin

Go back to my line of questioning. You're being stubborn (like everyone else in here).

Karl

21:00

sorry I don't want to be

Ted Shifrin

LOL

Sometimes it's really good to be stubborn.

Karl

so what do I know about that line

you asked

Ted Shifrin

Indeed I did.

Karl

ok well, that line belongs to my quadric

because it intersects all 3

Will Hunting

I thought only I used LOL

Ted Shifrin

21:00

Bingo.

how egocentric of you, Jasper.

Now, as $P$ varies, that line sweeps out your quadric, @Karl.

Karl

ok, I am thinking how to describe this line

Ted Shifrin

It's now best to work parametrically. One parameter for $P$, one parameter along that moving line $\overleftrightarrow{P\,Q(P)}$.

But work with inhomogeneous coordinates to start, so you get less confuzled.

Karl

I sort of want to get confused or I will never become less clumsy with homogeneous coordinates

ok so basically the intersection of the plane I found with the last line will give me the point Q(P)

Ted Shifrin

I suggest you do it inhomogeneously first and then rework it homogeneously, to minimize numbers of letters flying around first.

Karl

so parametrising p with (0,0,1,t) for example?

Ted Shifrin

21:04

Yup.

Karl

ok I'll think about it for a few minutes, don't forsake me if you can you are sort of my light at the end of the tunnel right now

:D

Ted Shifrin

You'll get it.

Hi, Demonark.

Karl

hmm ok

so I parametrised that last line as (1,s,1,s) as its equations where x_0=x_2, x_1=x_£

x_3 sry

but imposing

that (1,s,1,s) belongs to the plane $x_2 t = x_3 s$

I get s = t

Ted Shifrin

You're messing up letters.

Karl

sorry, the plane $x_2 t = x_3 $

still messing up?

Ted Shifrin

21:14

Well, let's check. Does $(1,t,1,t)$ belong to the plane $x_3=tx_2$? Yes. So this is fine.

That's just saying you use the same parameter $t$ to parametrize $Q(P)$. That's expected.

Karl

but then I would be using the same parameter for P and Q(P)

Ted Shifrin

Precisely, as you must.

Where does our extra parameter for our surface come from? Review.

Karl

with varying p

P

Ted Shifrin

No.

$t$ is taking care of that.

Karl

oh, the line PQ(P)

Ted Shifrin

21:18

right

Karl

which is of course different from the line (1,t,1,t)

lol

Ted Shifrin

of course

Daminark

Hey Ted!

Karl

I am still missing something, Would it make sense to say that PQ(P) obeys the equations
$x_0 t = x_1 $ $x_2 t = x_3 $

?

it should be the line joining (0,0,1,t) and (1,t,1,t) right?

Alessandro Codenotti

Hi @Dami

Ted Shifrin

21:31

So, let's see. It's of the form $(0,0,1,t) + u(1,t,0,0)$, as $u$ varies.

Daminark

How's it going Alessandro?

Ted Shifrin

Hi @Alessandro

Alessandro Codenotti

Pretty well, what about you?

Rehi @Ted

Karl

so I can express it as $x_1/x_3 = x_0/x_2$?

which would indeed be correct?

Ted Shifrin

Aha. And there's your quadric!

Mike Miller

21:40

Cross-multiply to get a polynomial

Karl

YAY

Jesus I suck

ok.

Ted Shifrin

Some of it comes from being rusty with just basic multivariable computations.

Karl

Any advice on how to unjust?

unrust

Ted Shifrin

Practice. But remember how to parametrize lines with points and direction vectors :)

Karl

ok :D 4 years no math

forgive me

Ted Shifrin

21:43

I'm not cursing you!

Karl

would you say the existence of this quadric

would be somehow related

or helpful in answering your first question about how many lines cross four skew lines?

Ted Shifrin

Oh, of course.

I thought you said they'd done that in class already.

Karl

but without the quadric approach

the argument made use of the segre variety

and it was rather indirect

Ted Shifrin

So did your professor work in the Grassmannian of lines in $\Bbb P^3$? I mean, the Segre embedding of $\Bbb P^1\times\Bbb P^1$ is the quadric.

Daminark

Oh apparently our AG pset actually includes a problem about Segre embedding

Also the Veronese embedding is a problem

Ted Shifrin

21:49

General case, Demonark?

OK, good classic examples.

I bet @Eric ends up running into those both, too. :P

Karl

Veronese?

what sorcery is that

Daminark

Ted Shifrin

Excellent, Demonark.

Anyhow, yes, @Karl. Sort through why a line meeting all four lines corresponds to an intersection point of your fourth line with the quadric.

Eric Silva

i have indeed seen those bois

Daminark

Nice

Karl

21:54

Ok, but just so I get this straight the quadric whose equation you helped me find is the Segre embedding of P^1 x P^1 in P^3?

Ted Shifrin

Yeah (they're all isomorphic).

$(x_0,x_1,y_0,y_1)\rightsquigarrow(x_0y_0,x_0y_1,x_1y_0,x_1y_1)=(u_0,u_1,u_2,u_3‌)$ and $u$ satisfies $u_0u_3=u_1u_2$. Look familiar?

Karl

Yes :

:D

Well if I pick 3 lines I know that the quadric we just talked about is basically a set of lines that cross all 3.

A random fourth line will generically meet the quadric in two points

at each of those points it will meet one of the lines that constitute the quadric

so there's just two of them

does this make any sense?

Karl

22:19

Actually I am doubtful of my own statement: why does a fourth line intersect the quadric in only two points? Because of Bezout?

Alucard

22:30

is precession a thing in a mathematical context?

s.harp

Like from a spinning top?

Will Hunting

I remember reading about precession and nutation in Feynman.

It's certainly not about procession and mutation.

Alucard

@s.harp yeah

Leaky Nun

so the global langlands philosophy is that representations of automorphic representations correspond to representations of the global langlands group

mercio

representations of automorphic representations

Leaky Nun

22:36

oops

automorphic representations, I guess

@TedShifrin I just thought of motivating a vector space as a field action on an abelian group

I don't know if this is the best way of teaching it, in terms of pedagogy

Ted Shifrin

@Karl: Sorry. I was working on figuring out how to vote this fall.

Bezout, yes. But you need to understand the quadric a bit more closely.

Note from your equation that it is doubly ruled. I.e., through each point there pass exactly two lines. There are two families of lines. The ones in each family are skew to one another.

@Leaky: Not to me.

Leaky Nun

@TedShifrin what would you say then

Ted Shifrin

Sometimes in advanced work you want to think of modules that way.

Leaky Nun

so I told him to try to come up with axioms, one for each operation/constant in both sides, and then in the end he said he enjoys it and this might make him remember the axioms better

I don't know

Ted Shifrin

I teach vector spaces by adding and stretching vectors. For me, the motivation is largely physical, although I can give others.

Who's "he," Leaky?

Leaky Nun

22:41

the guy I motivated this to

@TedShifrin ah, I guess that's a good way to motivate it

and then generalize to arbitrary fields

Ted Shifrin

Sure.

CaptainAmerica16

So I started the prologue of Spivak - didn't know there was so much to the associative property.

Ted Shifrin

Ah, glad to hear it, @CaptainAmerica.

Well, before you read stuff, also try to figure out why $0\cdot a = 0$ for any real number $a$. That is not an axiom.

And why $(-1)\cdot a = -a$ (the additive inverse of $a$).

You'll see that I'm a bigger fan of the distributive property than of the associative property. Associativity is super important in linear algebra, though.

CaptainAmerica16

@TedShifrin Like, try to write a proof for it?

Mike Miller

Yes

Ted Shifrin

22:51

Yuppers.

CaptainAmerica16

@TedShifrin I hadn't thought of what my favorite property is. I do like distributing things though. I like when it makes the problem look neat, although I assume there's more to it than that :p

Ted Shifrin

Maybe so.

Leaky Nun

@TedShifrin I thought someone doesn't like foundations

Alucard

mmh, can somebody say how you say "though"? taff or sow?

Ted Shifrin

tho

Secret

22:53

@Ultradark I knew nothing about moduli space, almost nothing about algebraic varieties and curves and nothing about spec. The only thing I can say is Wikipedia said the generalisation of algebraic varieties are sheaves which I also know nothing about other than it needs a topology

Ted Shifrin

the th like in the or that

Alucard

ah, thanks @TedShifrin

Leaky Nun

@Alucard vocaroo.com/i/s19XWM0WrUjj

Ted Shifrin

@Leaky: I don't like belaboring formal logic. The basics of algebra are important for grounding in both algebra and analysis.

And you can ridicule me on this, too, of course. Of course I teach negations and converses, contrapositives, etc.

Secret

and, that is a lot of nothing in my paragraph. Looks like you asked me about something which I have almost no knowledge about other than names

Leaky Nun

22:56

@TedShifrin so, prove it from what?

Ted Shifrin

the list of properties of the real numbers that is sitting in front of @CaptainAmerica.

I would say 98% of my students over the years thought $0\cdot a = 0$ was an axiom.

Oskar Tegby

I get the feeling that the more math I learn, the less the rest of knowledge seems worthwhile.

CaptainAmerica16

@TedShifrin What kind of axioms can I use to prove such basic statements?

Semiclassical

Rant: an online bus schedule that doesn’t match reality is infuriating

CaptainAmerica16

OMG I was just asking that

Ted Shifrin

22:57

The list of properties you have in Spivak, @CaptainAmerica.

CaptainAmerica16

Ok, now I know where to start lol

Secret

(Unrelated) Last night dream took place in a white void where there is a set comstructed as follows:

Ted Shifrin

That's the spirit of his first chapter (and then you'll get to inequalities).

Leaky Nun

@TedShifrin I mean, I don't know what Spivak has, but I'm not sure all of them are as much qualified to be axioms as is $0 \cdot a = 0$

Ted Shifrin

you shouldn't list as an axiom things that follow from the axioms you already have.

Leaky Nun

22:58

but maybe you want the usual proof that group homs preserve identity (obscuring this to avoid being a hint)

CaptainAmerica16

@TedShifrin That should be interesting. I never really got deep into inequalities in school.

Leaky Nun

ah, so you want axioms to be minimal. I don't think that's really a principle in logic.

Secret

1. There is a set with a minimum and some cardinality A. Inside this set, there are A number of subsets with strictly smaller cardinality and each has a minimum

Ted Shifrin

Ah, but it's more than that, @Leaky. I already stressed the importance of the distributive property.

Oskar Tegby

Does the construction of numbers fail if we remove the axiom of the emptyset?

What else could we use to construct them?

Ted Shifrin

23:00

I don't care about Peano axioms. :P

Secret

2. The construction is repeated for each subset indefinitely

Ted Shifrin

And I have never taught Dedekind cuts. The first 3 weeks of my Rudin analysis course were sheer horror.

Oskar Tegby

I agree that analysis and geometry is more compelling.

Leaky Nun

@OskarTegby the empty set follows from the infinite set and specification

Secret

The dream claims the whole set is dedekind finite

Oskar Tegby

23:00

Specification?

Leaky Nun

@OskarTegby aka the axiom schema of subsets

Secret

Thus what we end up with is a set such that there are elements forming a tree of infinite height and A, A^2, A^3 etc. levels each

Oskar Tegby

Is that a part of ZFC?

I find it amusing that for recreation I simply do other parts of mathematics. Things that I don't have to do feel more fun to do.

Secret

Reality check: The set is actually dedekind infinite. Let the parent set be X and it's minimum be X(0) and let Xi(0) be the minimum of the corresponding subsets

Alucard

is "we" interchangable with "one" in scientific papers or does the "we" have a implication of some sort?

Oskar Tegby

23:05

What would the difference possibly be?

CaptainAmerica16

Oh man, my mom has the best friends. One of them literally just brought us burgers and fries because she's pregnant and just happened to be at a fast food place.

Ted Shifrin

Fast food be badddddd for you.

Oskar Tegby

I think that the answer is yes, @Alucard. They are interchangeable.

Secret

Then what we have here is X (0) > X1 > X2 > X3 > ....This is a well ordered collection of subsets in X thus it is at least countable. Thus we have a countable subset within X and hence it is dedekind infinite

Ted Shifrin

"We" is supposed to be more casual and less formal. It's sometimes artificial when there is just one author, but it is still used, nevertheless.

CaptainAmerica16

23:07

@TedShifrin sigh I'll change - eventually. I can't waste it now...

Alucard

@TedShifrin maybe to draw the reader in, but this isn't a novel so I don't know..

Oskar Tegby

Ah! I found it. It is a part of ZFC.

What other axiom systems are common to work with?

Secret

By the same reasoning, all subsets of the form Xi are also dedekind infinite

CaptainAmerica16

@TedShifrin Wait...we is informal? I put that in my proofs because I thought it was supposed to be more formal. My stuff must look so amateur.

Ted Shifrin

Compared to "one."

You're fine. Stop overreacting and eat your poison :P

CaptainAmerica16

23:11

@TedShifrin Lol, ok.

If I get a stomachache later, I'll remember what you said.

micsthepick

I want to show that $$\prod_{n=1}^{\infty}{\Bigl(1+\frac{1}{k^2}\Bigr)}<4$$

Leaky Nun

@OskarTegby one of the axioms in one of the axiomatizations, yes

@OskarTegby ZFC+universes is the second most common among mathematicians, I think

micsthepick

I thought of tightening that inequality to $$4-\frac{a}{k}$$

Oskar Tegby

@LeakyNun: Okay. Can we remove axioms?

micsthepick

but there is no a that satisfies the first case, and allows for the inductive step at the same time

Leaky Nun

23:15

@OskarTegby in order to what?

your question is incomplete

you can do whatever you want

micsthepick

I found the following inequality when applying the inductive step: $$0<(a-4)n+2a$$

Secret

What really is axiomless mathematics

micsthepick

so, a must be greater than or equal to 4, yet a needs to be less than two to satisfy the first case

Alucard

@Secret True=True maybe

Secret

That is a boring foundation lol

Oskar Tegby

23:18

Different things, @LeakyNun. First, what's the least we can have? Second, what do we keep with different combinations of axioms? Last, what's the most powerful axiom system there is?

Leaky Nun

@OskarTegby still incomplete question. what's the least we can have so that what?

Oskar Tegby

Anything.

Leaky Nun

also "powerful" is subjective

Oskar Tegby

If we have the empty set axiom, then we can't really do anything.

Leaky Nun

error. your questions are unanswerable.

Oskar Tegby

23:19

Only the empty set axiom, I mean.

Secret

I think nuking infinity, pairing and power set will get you quite stuck

Oskar Tegby

Everything is subjective. Haven't you read Wittgenstein?

Secret

because you can no longer define supersets and hence cannot even construct the finite sets other than emptyset

Oskar Tegby

That's what I mean: Worthless.

Ted Shifrin

@micsthepick That makes no sense. $k$ is a dummy variable on the left hand side.

Oskar Tegby

23:22

If we add the axiom of regularity we still don't have anything.

Secret

of the 8 axioms of ZF, only those 3 can make supersets

micsthepick

whoops

pretend that n=k

Oskar Tegby

I don't get the regularity axiom.

micsthepick

(as in they are the same symbol)

Secret

Regularity prevents the existence of infinite decreasing chains by having a set to contain itself

I don't fully understand ZF-R otherwise

Ted Shifrin

23:29

What tools do you have for a proof, @mics?

micsthepick

I am able to apply induction and algebra

Leaky Nun

@Secret right

apply induction is an easy thing

apply algebra :o

Ted Shifrin

OK, so no calculus techniques, @mics?

I don't see anything easy with induction for that question.

micsthepick

I think calculus would be fine

Ted Shifrin

Where did the question come from?

Alucard

23:31

man, translating is hard...

micsthepick

it is based off another question that I was given to practice for a maths competition

the other question involved $n^3$ instead of $n^2$

and it was less than $3$

Ted Shifrin

I don't see anything obvious.

My approach would be to take logs and use Taylor series arguments.

micsthepick

so if you take the log, you get a series

Ted Shifrin

Yup.

micsthepick

how do you apply the Taylor series arguments to that?

Ted Shifrin

23:36

Because essentially you're going to have $\log(1+1/n^2) \approx 1/n^2$ (and you can estimate the error quite well).

And we know $\sum 1/n^2$.

micsthepick

would it maybe help to change $\frac{a}{n}$ into $\frac{a}{n+k}$ and try my method again, where $k$ is perhaps equal to $1$?

Secret

Alien foundation to be analysed later: en.wikipedia.org/wiki/Mereology

Ted Shifrin

Huh?

micsthepick

i.e. use induction to show that it is less than $4-\frac{4}{n+1}$

Leaky Nun

@TedShifrin our modular forms lecturer proved that $\pi z \cot \pi z = \displaystyle \sum_{n=-\infty}^\infty \frac1{z+n}$ or something like that

Ted Shifrin

23:41

You mean the product up to $n$?

micsthepick

yes

Ted Shifrin

Yeah, that's standard stuff, Leaky.

Leaky Nun

just telling you

Ted Shifrin

Why?

Secret

34

A: Set theories without "junk" theorems?

Structural set theory, as described on the nlab page you linked to, is probably the best answer to your question. To avoid junk theorems, one must deviate somewhat from ordinary ZF-style set theory where everything is a set. That's because, once you decide, in the context of such a "material" s...

Oskar: this might also be rekevant

Leaky Nun

23:42

@TedShifrin because I feel like sharing to you what I learnt?

micsthepick

well, the original question I was asked was whether it worked up to arbitrary n, and the proof was a simple induction

Ted Shifrin

So you should rig your inequality so that the induction step is clear, then, @mics.

micsthepick

go on?

how could I do that, wouldn't I need to solve a recurrence relationship?

Ted Shifrin

I dunno. You want something like $\frac m{m+1}$ so that $\frac m{m+1}\cdot \frac{n^2}{n^2+1} < \frac {m+1}{m+2}$. Personally, I don't see it. But if you make your denominator quadratic, can you get it?

I don't see it.

Secret

@LeakyNun It';s amazing how sums like these have something to do with the circle of infinite radius (c.f. 3B1B)

micsthepick

23:50

is there a nice one for the original problem?

Ted Shifrin

I have no idea.

Will Hunting

Sometimes, all this math sounds like bullshit.

Leaky Nun

Other times, all this bullshit sounds like math.

Alucard

I know: if anyone knows about something, then this chat :D

SE is great =)

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$Mathematics$ Mathematics