Mathematics - 2016-11-28 (page 4 of 6)

dsillman2000

7:00 AM

In the x-y plane, the region vaguely resembles that of the area under a quadratic. In the x-y-z space, the volume represents a bowled protrusion.

Null

maybe closure of scalarmultiplication?

dsillman2000

Am I talking about the wrong region?

Ted Shifrin

You're not understanding. The function you're integrating is not relevant.

dsillman2000

Ok. So by the region I am integrating you mean...? One integrates a function, right? You can't know what region you're integrating w/o the function

Ted Shifrin

I think you should wait until you study multivariable calculus carefully.

Null

7:02 AM

since $(-1)(a,a^2)=(-a,-a^2)\notin B_2$

dsillman2000

Ok... you've successfully discouraged me... Thank you...

I have to go to bed

g'night

Kaj Hansen

$\text{Heis}(\mathbb{Z}_3)$ is a weird group. The fact that it's of order $27$ and all elements are of order $3$, yet is not isomorphic to $(\mathbb{Z}_3)^3$ is more bizarre to me than the existence of nonabelian groups with every subgroup normal.

Ted Shifrin

Night.

Mike Miller

Really? There are always nonabelain groups of order $p^3$.

Kaj Hansen

Is it always the Heisenberg group @MikeMiller? Are there others?

Mike Miller

7:04 AM

math.uconn.edu/~kconrad/blurbs/grouptheory/groupsp3.pdf

Kaj Hansen

Ah, good ol' Keith Conrad

Let's check this out

At least the order $2^3$ examples have elements of order $4$

I wasn't aware of $G_p$. Interesting

Adeek

can we do calculus over finite fields @TedShifrin ?

Null

are rectangle brackets used for something specific? i use them right now for vectors

Kaj Hansen

[v] ? @Null

Null

well, like in the picture, [] and inside it the coordinates vertically

Kaj Hansen

7:16 AM

Oh. That's fine. I do that too. Vertical representation of vectors is superior to the seemingly more common horizontal representation for a number of reasons

IMHO

Null

yeah ,just asking if rectangle brackets are ok, i saw () too!

Kaj Hansen

I use rectangle brackets. Ted does too...I learned the notation from him, lol

Null

hehe

@KajHansen mmh, how do i determine the span of a set of vectors. I seem to only find examples with finite sets.

Adeek

..

Kaj Hansen

In $\mathbb{R}^n$, a collection of $m < n$ vectors will span a space homeomorphic to $\mathbb{R}^m$.

In $\mathbb{R}^3$, you can find the plane spanned by two vectors explicitly by finding a vector normal to the plane (via cross product of the two vectors)

Null

7:28 AM

Let's rephrase, what is the span of: $B_2:=\{(a,a^2)|a\in\mathbb{R}\}$

Kaj Hansen

You can use that information to find the equation of the plane

Null

ah well

Kaj Hansen

Not a subspace @Null

Null

2 vectors are enough to determine the span in a plane

Alessandro

It's important to notice that Kaj is assuming independent vectors, Null

Kaj Hansen

7:29 AM

(a, a^2) + (b, b^2) = (a + b, a^2 + b^2)$

And $a^2 + b^2$ won't be a perfect square in general

Indeed. What Alessandro said

I should be more careful

Null

@KajHansen why is a perfect square interesting? we are working in R

Kaj Hansen

The only subspaces of $\mathbb{R}^2$ are $\{0\}$ and a line through the origin

Because subspaces need to be closed under addition @Null

Adeek

Suppose our field $K$ is finite and $char(K) > 3$. Suppose E is the elliptic curve on the field K defined in same way as before. Following same reasoning as before we can define a group operation on the points of E, which will make E into a group. We will not proof why the operation defined is a group, since it follows from same ideas we developed last section. For the interested reader, I have included a reference for why this operation as defined below is a group

is this good english ?

@KajHansen @Null what do you guys think ?

Alessandro

Last sentence, the operations is not a group, it makes the elliptic curve into one, or the elliptic curve equipped with the operation is a group

Kaj Hansen

@Adeek, "Following the same reasoning"; "We will not prove",

You'll also want to say "Suppose E is an elliptic curve" unless you defined a specific individual one, as opposed to one with general coefficients.

Null

7:33 AM

@KajHansen . math.stackexchange.com/questions/2033161/… if you want to look at it. Not urgent tho ;)

Adeek

@KajHansen yeah I did it for rational field first.

Now I need to do it for finite fields.

Kaj Hansen

Anything with the zero vector will not be linearly independent @Null

Null

@KajHansen yep, because you can strip the 0-vector and have the same span

Adeek

@Null your showing LI not span.

follow the definition of LI

Kaj Hansen

A set of vectors is linearly independent if $c_1 \mathbf{v}_1 + \cdots + c_n \mathbf{v}_n = \mathbf{0} \iff c_1 = c_2 = \cdots = c_n = 0$

Adeek

7:36 AM

hint anything times 0 vector is still 0 vector.

Kaj Hansen

Applying this reasoning to those sets will get you your answer easily enough

Adeek

yeah good call @Alessandro

Null

eh, then let#s say it this way: $0\vec{a}=0$, therefore 0 is exprassable by some other vectors

Kaj Hansen

Let's look at $\{(1,1,0)(0,1,1)\}$. Suppose we have $c_1 \langle 1, 1, 0 \rangle + c_2 \langle 0, 1, 1 \rangle = \langle 0, 0, 0 \rangle$.

Null

@KajHansen do you really want to use the span there?

Kaj Hansen

7:38 AM

This simplifies to $\langle c_1, c_1 + c_2, c_2 \rangle = \langle 0, 0, 0 \rangle$

We're forced to have $c_1 = c_2 = 0$, so we can conclude linear independence.

@Null, there are a couple equivalent ways of defining linear independence. The easiest one to work with, IMO, is that a set of vectors are linearly independent $\iff$ you cannot get the zero vector out of a nontrivial linear combination of the set.

Null

@KajHansen yep

altho nonzero imo, so it's unambigious

Kaj Hansen

Apply this reasoning to $\{(1,1,0)(0,1,1)\}$ (as I did above), and you quickly discover that the only linear combination of these vectors that yields the zero vector is the trivial one

Null

@KajHansen my reasoning is actually different: the first coordinate cant be expressed by a linear combination of the other vector. same goes for the 3rd cordinate.

user227867

If a mosquito is a vector, then a colony of mosquitoes is a vector space, since it is closed under breeding.

Kaj Hansen

Hello again Jasper

Null

7:44 AM

@WillHunting trolol^^

user227867

@KajHansen Hi. Have you found a new girlfriend since the last time I asked?

Kaj Hansen

I've been single for ~11 months

Null

@NaCl moin

NaCl

hi

user227867

I see. I have been single for 35 years, lol.

Kaj Hansen

7:45 AM

I can't remember when you last asked, but I think the answer was "yes" at some point between then and now

I was away from the chat on here for about a year

user227867

In fact, I have not even gone on a single date.

Kaj Hansen

That's unfortunate. How come?

(I don't mean to pry)

user227867

I think it is largely because I have been mentally ill for the last 17 years.

user227867

So, I don't really actively seek out anyone.

Kaj Hansen

That'll do it. I'm sorry to hear that

Null

7:48 AM

@WillHunting certain members of our society will make you a happy evening for money. So a relationship is not necessary.I hope tho that you have friends.

user227867

Anyway I have stopped my meds for almost 2 months. I don't intend to go back to meds anymore.

Kaj Hansen

Depression makes me isolate myself socially (not so much on here, but very much so IRL). I feel like I don't have anything worth saying when I am with people.

Null

@KajHansen wait 'til you are maniac^^

user227867

I don't think meds work for me, but others can take them if they think it helps them.

Kaj Hansen

@WillHunting, let me guess...due to side-effects?

user227867

7:50 AM

@KajHansen Firstly, I don't think they really work, other than the drowsiness effect that sedates your mind. Second, they make me feel lifeless and emotionless. Third, I don't wanna depend on meds the rest of my life. Fourth, though they are cheap, they are not free of charge, and I can better spend that money on things I like.

Kaj Hansen

Nearly everyone I've talked to hates the side-effect profile of psychotropic drugs. I've avoided them for that reason, opting instead for exercise and creative outlets to the best of my ability.

That's pretty much in-line with a lot of what I've heard. Best of luck to you man.

Null

I think that the definition of "healthy" is not helping either.

user227867

I am hoping to start studying math on the first day of next year on my own.

Null

Conjecture: there are 0 persons on planet earth that are mentally healthy.

Kaj Hansen

I find that intense daily exercise gives me a sufficiently good endo-opioid rush to ward off dysphoria for a decent period of time.

user227867

7:52 AM

I have been saying that for many years in this chat, always the first day of next year.

user227867

I have these yearly milestones you see.

Kaj Hansen

Yeah @Null. Things are not well-defined.

user227867

Giving myself at most 5 more years to enter grad school.

Null

Let's invade psychology SE :D

Kaj Hansen

The calendar is an artificial, arbitrary construct @WillHunting. Try not to hold yourself to it. It doesn't matter when you start; just start

user227867

7:54 AM

Well, this starting on the first day of the year is very much an OCD thing in some ways.

Kaj Hansen

Even if it's just a little every day. Getting absorbed in hobbies (if I can; I've been having bad motivation problems) helps me get my mind off of bad frames of mind.

First day of a month instead @WillHunting ? I don't have OCD, so I'm probably not giving the best advice or empathizing as well as I'd like to

user227867

By the way, my math collection now has 30 books, all different topics.

Null

@KajHansen my personal opinion is that some of the best advices come from outsiders.

because a change of point of view can mean worlds

user227867

@KajHansen Well, in this case, it is not about empathising. This thing I mention is tied to many OCD-related things that I need to write a long, long essay to describe so that someone else can even try to understand it.

user227867

@Null I agree with that. It is true sometimes.

user227867

7:59 AM

Anyway, it's almost the new year.

user227867

Christmas is like the best time of the year for me.

Kaj Hansen

I hate the winter

user227867

I like the decorations and lights everywhere on the streets and in the shopping malls.

Null

if we live in a finite field then 2016<beginning of time

Kaj Hansen

My mood worsens in winter every year, almost certainly due to less sunlight.

Null

8:00 AM

modulo universe anyone?

user227867

@KajHansen What you can do is stare at bright light every time you wake up for a minute.

Kaj Hansen

That might actually help. I feel like utter shit in the mornings waking up.

user227867

And maybe there is something behind the depression. Maybe something is wrong in your life that needs to be fixed @kaj

Kaj Hansen

Possibly true as well. I've tried focusing on exercise and nutrition. If that's the case, it's probably my social life.

My social life's never been great.

Null

@KajHansen winter depression is actually a thing, just saying. Or do you feel wierd the whole year?

Kaj Hansen

8:03 AM

I feel bad the whole year, but often worse in the winter @Null.

user227867

For me, my depression is actually a result of the other problems I have. It is not the cause but the effect.

user227867

@null I must correct your spelling. It is weird and not wierd.

Kaj Hansen

haha

"I before E except after C." [Insert long list of exceptions here]

user227867

I hope my next life is better than this one. But I will try to live on and make the best out of the shit that remains in this life.

Kaj Hansen

Are you spiritual / religious @WillHunting ?

user227867

8:05 AM

@KajHansen These rules are only heard in ignorant circles, lol.

Null

@KajHansen The span of $B_2$ is $f(x)=x^2$? (a parabola)

$B_2:\{(a,a^2)|a\in\mathbb{R}\}$

(dunno if that makes sense)

user227867

@KajHansen Well, spirituality and religion are not well defined, so that is hard to answer. But I am quite amazed by the truths in Buddhism.

Kaj Hansen

That locus of points is a parabola @Null. It's not closed under vector addition though

I'm a recent Christian convert after a lifetime of atheism.

user227867

I am an ex-Christian.

user227867

But I am going to study the New Oxford Annotated Bible just as part of my continued quest for the truth

Null

8:08 AM

@WillHunting weird and wired look so similar tho haha

Kaj Hansen

I feel that Christianity tends to be very misrepresented in culture.

It's sad, and alienates people from it.

user227867

I am pretty sure though I will never be a Christian again.

user227867

I got a copy of the Qur'an as well, for the same purpose.

Kaj Hansen

Never say never @WillHunting. Questing for the truth is commendable though. I wish you the best.

Electric Octopus - This Is Our Culture (Full Album 2016)

►

^Been into this album recently. It's psychedelic jazz :P

user227867

Oh, recently I got all of Charles Dickens's completed novels. That is 14 in total. Wordsworth Classics, very cheap. Like less than 5 dollars each.

Null

8:10 AM

@KajHansen its not closed under scalar multiplication either. As $a^2\not=-1$ with $a\in\mathbb{R}$.

$(-1)(1,1^2)=(-1,-1)$

Kaj Hansen

Charles Dickens is pretty good. So far, I've read Tale of Two Cities, Oliver Twist, Great Expectations, and A Christmas Carol

user227867

Charles Dickens was a great author. He wrote 14 completed novels, 1 uncompleted novel, and 5 Christmas novellas, among other things.

Kaj Hansen

Indeed @Null

user227867

I recently watched a film adaptation of The Picture of Dorian Gray, starring Josh Duhamel. Very sexy movie, LOL.

Null

@KajHansen I don't know how to say it good enough. $\langle B_2\rangle$ is a parabola. Seems not enough.

Kaj Hansen

8:13 AM

Depends on how you define a parabola, lol

It's commonly accepted that $y = x^2$ is the "parent" parabola

Null

Haha^^

Kaj Hansen

"A parabola is the locus of points in the plane that are equidistant from a given point and a given line."

user227867

I went to a shopping mall and there was a room called Mother's Room for mums to change diapers. They should call it Parent's Room because dads change diapers too. This is sexism.

Kaj Hansen

LOL @Will

Null

"$\langle B_2\rangle$ is the "parent"-parabola of $\mathbb{R}^2$. $f(x)=x^2$ will yield the same points. Which in turn can be viewed as vectors."

user227867

8:16 AM

Even if the room is for breast-feeding, it should also be made available for diaper-changing. So I have covered all possibilities.

Null

@WillHunting it is sexism that only women have to give birth. j/k

user227867

@Null Why is your username Null?

Null

@WillHunting it was a viable name. found it funny^^

user227867

I have been up for three hours and I forgot about my coffee! I should make some now.

Null

@WillHunting tea with honey is great too

user227867

8:23 AM

@Null I know some people drink tea mixed with coffee. It is sold in some cafes here.

user227867

I think Hartshorne's Algebraic Geometry is the most advanced book I have on math right now.

Null

@WillHunting me after waking up: youtube.com/watch?v=B2vH9VABHRI especially the stretching haha

user227867

Stop Unnecessary Cardio and Try This For Six Pack Abs

►

user227867

@kaj Try this ^ LOL

user227867

@Null I seldom take honey. But it can be nice with toast.

Kaj Hansen

8:28 AM

"Stop unnecessary cardio and try this for six pack abs"....haha, if you're doing cardio for abs, you're doing it wrong.

hey @KajHansen

Hey there @Adeek

@KajHansen stop speaking english for learning japanese and use gtranslate instead!

Adeek

I want to discuss here why kernel E is isomorphic to the kernel of the map ?

Kaj Hansen

8:30 AM

You're learning Japanese @Null ?

Null

We interpret $\mathbb{R}$ as a $\mathbb{Q}$-vectorspace. shouldnt it be Q as a R-vectorspace @KajHansen also no, i just tried to make an analogy to the abs thingy

Kaj Hansen

What is $\phi - 1$ denoting @Adeek ?

Since it's an automorphism, $\ker(\phi)$ is the singleton $\{0\}$.

Adeek

$\phi$ minus the identity map I think

Kaj Hansen

ohh hahaha

I guess cardio would help you burn midsection fat, which is necessary for abs. But not sufficient.

Adeek

oh I see @KajHansen

Kaj Hansen

8:33 AM

I'm a stick though, so that's not relevant to me :P

$\mathbb{R}$ can be interpreted as a $\mathbb{Q}$-vector space with infinite transcendence degree.

Null

so any irrational will be displayed infitly close to the actual value you mean?

Kaj Hansen

$\mathbb{Q}$ can't be an $\mathbb{R}$ vector space though. Not closed under scalar multiplication. $\sqrt{2} \cdot \text{any rational} \notin \mathbb{Q}$.

Transcendence degree

In abstract algebra, the transcendence degree of a field extension L /K is a certain rather coarse measure of the "size" of the extension. Specifically, it is defined as the largest cardinality of an algebraically independent subset of L over K. A subset S of L is a transcendence basis of L /K if it is algebraically independent over K and if furthermore L is an algebraic extension of the field K(S) (the field obtained by adjoining the elements of S to K). One can show that every field extension has a transcendence basis, and that all transcendence bases have the same cardinality; this cardinality...

It's an abstract algebra concept. Any field extension can be viewed as a vector space over the base field. $\mathbb{R}$ can be realized as a field extension of $\mathbb{Q}$.

Null

So R is Q, but the "holes" in Q are filled with all sorts of stuff. (mainly irrationals)

Kaj Hansen

It's not pretty though. There are infinitely many (possibly uncountable) linearly independent vectors.

@Mike might be able to explain this better than me

Actually, it's pretty easy to see as long as you're not trying to obtain a basis. You can see that it's closed under addition and scalar ($\mathbb{Q}$) multiplication

For any $c \in \mathbb{Q}$ and $v \in \mathbb{R}$, we have $cv \in \mathbb{R}$

And the sum of any two reals is another real number.

So it's a vector space over $\mathbb{Q}$. That other stuff is concerned with finding a basis for the vector space.

Null

the basis of R is an infinite collection of numbers?

Kaj Hansen

8:45 AM

Yeah

Any finite collection of numbers won't span $\mathbb{R}$

Null

can't the same be said about any infinite field? @KajHansen

Kaj Hansen

No @Null. The extension $\{a + b \sqrt{2} \ | \ a, b \in \mathbb{Q} \}$ of $\mathbb{Q}$ is spanned by exactly two vectors, namely $1$ and $\sqrt{2}$.

Have you studied field extensions yet @Null ? Can't remember

Null

@KajHansen only a very small example

Kaj Hansen

This is relevant: en.wikipedia.org/wiki/…

Null

@KajHansen ah so, Q can only be extended to R with infinite "holefillers"

Kaj Hansen

8:54 AM

Yeah, rational multiples of one hole-filler won't hit them all

Mike Miller

uncountable even

Kaj Hansen

haha, yeah

Mike Miller

I need to sleep mostly

Null

@MikeMiller good night then =)

Kaj Hansen

Don't let us keep you @Mike. I hope you find some relief from the RLS

Arthur

8:58 AM

hi...any linear algebra people in? Is math.stackexchange.com/questions/2032918/… obvious?

no comments so far

Null

Let $a,b,c,d\in\mathbb{Q}$.

$a+b\sqrt{2}+c+d\sqrt{2}=(a+c)+(b+d)\sqrt{2}$ which is obviously in $\{a + b \sqrt{2} \ | \ a, b \in \mathbb{Q} \}$.
$\pi(a+b\sqrt{2})$ is not in $\{a + b \sqrt{2} \ | \ a, b \in \mathbb{Q} \}$.

Thus $\{a + b \sqrt{2} \ | \ a, b \in \mathbb{Q} \}$ is no subspace of $\mathbb{R}$.

@KajHansen

Kaj Hansen

$\mathbb{Q}$ is the field of scalars

Alessandro

The relationship you seek is written in the wiki page you linked @Arthur

Mike Miller

@Arthur They are the same. This is defijitioj pushin.

Arthur

@Alessandro I think that is only for square matrices isn't it?

Null

9:01 AM

@KajHansen mmh, how do i know what is the field of scalars?

Arthur

@MikeMiller even for non-square matrices?

Alessandro

It's not defined for non square matrices

Arthur

@Alessandro follow the link. It has the definition

@Alessandro en.wikipedia.org/wiki/…

Kaj Hansen

For field extensions, it's necessary that the scalars come from a subfield of the field that we're considering to be the whole vector space @Null

Arthur

@MikeMiller It doesn't seem obvious to me but I am no expert

Null

9:07 AM

@KajHansen so if i want to check $C:=\{a+1|a\in B\}$, then for scalarmultiplication i only consider elements of B?

Danu

Hi @MikeMiller

@MikeMiller Dafuq?

Null

@Danu he is sleeping

Danu

I don't mind

Kaj Hansen

If $B$ is a field, then $C = B$

Null

@KajHansen it was a pathologic example. i just want to know what i use for scalarmultiplication.

Kaj Hansen

9:17 AM

$B$ would be the only thing that makes sense

There he is

Null

$B_4:=\{a\in\mathbb{R}|a\geq 0\}$

$(-1)(1)=-1\notin B_4$

Thus it's no subspace of $\mathbb{R}$ ?

Arthur

@Alessandro any ideas gratefully received

Kaj Hansen

Correct

The only subspaces of $\mathbb{R}$ will be $\{0\}$ and $\mathbb{R}$ itself.

Alessandro

I'm pretty sure it works for rectangular matrices now but I'm in the middle of a lecture so I can't think about it at the moment, sorry

Kaj Hansen

What country do you hail from @Alessandro ?

Alessandro

9:26 AM

I'm not sure what hail means, but Italy

Null

@KajHansen and what is the span of $B_4$? $\mathbb{R}\setminus\mathbb{R}^{-}$?

Kaj Hansen

You can't really talk about it because it's not a subspace

Usually you talk about the span of a collection of vectors

Null

$B_4$ is a collection of vectors

Kaj Hansen

Like, $Span( \langle 1, 0 \rangle, \langle 0, 1 \rangle) = \mathbb{R}^2$.

Oh yeah, I see what you're saying. It'll be all of $\mathbb{R}$ actually @Null

Span of a collection of vectors is always a subspace

Null

so in esscence it is possible that the span of a set of vectors is the parentspace, but the collection of vectors are not a subspace.

Kaj Hansen

9:29 AM

@Alessandro, dictionary.cambridge.org/us/dictionary/english/…

Correct

Not necessarily a subspace

Null

so $\{a|a\in\mathbb{R}\}$ would be $\mathbb{R}$?

Kaj Hansen

The span of that, yeah

And that set itself

Adeek

I am almost done with my project

hurray

Balarka Sen

Hi

Adeek

hi @BalarkaSen

Alessandro

9:34 AM

@Kaj Italy was the correct answer then!

Hi @Balarka

Adeek

@KajHansen would you like to read my project when I am done ?

Kaj Hansen

I'll give it a look-see

Won't have time tonight though

Null

@KajHansen $\langle B_4\rangle _{\mathbb{Q}}$ is R? i dont think so yet :/

Adeek

it is no problem tomorrow or something @KajHansen

Null

@KajHansen let me rephrase where my problem is: what's the difference between $\langle B\rangle _{\mathbb{Q}}$ and $\langle B\rangle $ ?

Kaj Hansen

9:37 AM

You can get all of $\mathbb{R}$ by taking a positive number and multiplying by $-1$

You can consider $B$ as a subset of a vector space over either $\mathbb{Q}$ or $\mathbb{R}$

Adeek

cool I didn't know RSA uses elliptic curve cryptography

Kaj Hansen

It's standard implementation doesn't, but there might be a version that does

Have you seen factoring integers with elliptic curves?

Null

@Adeek i really only know of elleptic curve decryption because of the game Uplink lol. and my understanding is neglible

Adeek

@Null it is really cool

I think also NASA uses elliptic curve cryptography

Balarka Sen

I don't remember the ECPP/ECM algorithm anymore but I remember being pretty surprised when I read the algorithm

Adeek

9:42 AM

I could give you my project once I am done @BalarkaSen

Null

@KajHansen so $\langle B\rangle _{\mathbb{N}}$ would be quite different from $\langle B\rangle _{\mathbb{R}}$ ?

(at least if B is not R)

Kaj Hansen

Scalars have to come from a field

Balarka Sen

@Adeek Sure

I think you already gave me a thing on elliptic curves a few months ago tho

Adeek

yes but I edited that

Balarka Sen

ah, ok

Null

9:45 AM

@KajHansen $\langle B\rangle _{\mathbb{Z}_3}$ would be sensical then?

Kaj Hansen

Multiplication of elements from Z_3 and positive real numbers isn't defined though

Null

@KajHansen so Q is really the smallest field in that regard?

Kaj Hansen

Potentially

$\mathbb{Q}$ is the smallest subfield of $\mathbb{R}$. There might be some way of re-defining scalar multiplication to make finite fields work idk

DHMO

Isn't Z_2 smaller?

Null

Z_2 doesnt work with real number multiplication. Unless defined beforehand.

DHMO

9:51 AM

I see

Does anyone have a polynomial bijection between QxQ and Q?

Balarka Sen

I think that's an open problem on MO or something

Alessandro

That's open I think, there was a question about it

Balarka Sen

392

Q: Polynomial bijection from $\mathbb Q\times\mathbb Q$ to $\mathbb Q$?

Is there any polynomial $f(x,y)\in{\mathbb Q}[x,y]{}\ $ such that $f:\mathbb{Q}\times\mathbb{Q} \rightarrow\mathbb{Q}$ is a bijection?

DHMO

let's have it between NxN and N instead?

Balarka Sen

10

Q: What polynomials biject from $\mathbb{N}^{2}$ to $\mathbb{N}$?

Perhaps there are none with integral coefficients; so let us admit rational coefficients. The map $(x, y) \mapsto x + \frac{1}{2}(x + y)(x + y + 1)$ is well known, and swapping $x$ and $y$ in the formula yields another, so we have two for starters.

nt.number-theory open-problem

Null

10:56 AM

Let $K$ be a field, $V$ a $K$-vectorspace and $v\in V$.

Show that: $(-1)v=-v$

Where (-1) is the additive inverse of the 1 in K and -v is the additive inverse of v in V.

Would the following be sufficient?

$(-1)\cdot v=(-1)(x_1,x_2,...)=(-x_1,-x_2,...)$

Also note that $x_1,...$ are all elements of $K$, so $-x_1$ has to be the additive inserve of $x_1$.

Then: (x_1,x_2,...)+(-x_1,-x_2,...)=0

(standard vectoraddition and scalarmultiplication, as i might add)

Tobias Kildetoft

@Null You would need to show then that your equality holds for any choice of basis

(there is no "standard" multioplication and addition)

You need to look at the axioms for a vector space and derive it from them

Null

$(x_1,x_2,...)+(-x_1,-x_2,...)=(x_1-x_1,x_2-x_2,...)=0$ i mean

this is true for any choice for $x_1,...$ i don't know what there is to show

Tobias Kildetoft

@Null How do you know you can write the elements in that form?

Null

. en.wikipedia.org/wiki/Scalar_multiplication#Properties

Tobias Kildetoft

@Null Don't link a wiki page, but use your own words

Null

11:05 AM

so i have to argue/show those properties?

Tobias Kildetoft

No, you have to not try to write the elements in that form because once you define what it even means, it does not actually help

You need to look at the axioms for vector spaces and use only those to show this

Null

okay

Danu

@Semiclassical Did you see this? Seems super cool!

Null

@TobiasKildetoft you mean that vectoraddition and scalarmultiplication doesn't have to be THIS way. Therefore i have to show the statement not relying on this way?

Tobias Kildetoft

@Null yes

Danu

11:21 AM

@BalarkaSen Sorry for bailing on this yesterday

I suddenly had some stuff to do.

So I'm not entirely sure how to work with this---the fact that the odd-dimensional subbundle is not orientable doesn't imply that the space is not, does it?

Balarka Sen

@Danu I mean, you know that the tangent bundle cannot have an odd-dimensional subbundle for oriented manifolds, don't you?

So for nonorientable cases you look at the orientation double cover, is all. If you had an odd-dim subbundle below you'd get an odd-dim subbundle above.

And since the base has $e(M) \neq 0$, so would the cover because $e(\tilde{M})$ is just twice of $e(M)$.

I am not sure why you want to think about orientation of the odd-dimensional subbundle.

Actually, can you repeat the question?

Danu

@BalarkaSen No, actually.

That's all I'm missing.

Balarka Sen

11:36 AM

So you want to prove that oriented manifolds with $e(M) \neq 0$, $TM$ does not admit odd-dimensional subbundles? And you know this when the subbundle is orientable?

Is that what the question is?

Danu

@BalarkaSen Yes.

Balarka Sen

Ok, I see. So you gotta take the orientation double cover of the subbundle if it's not bundle-orientable.

Danu

I don't need the total space to be orientable (that's not the same thing as being an orientable vector bundle---actually, how do the two relate?!)

So for all I nkow the total space could already be oriented so taking the double cover wouldn't help anything

Or am I missing something?

Balarka Sen

No, I mean, you can take a double cover of a bundle. Look at each fiber, oriented positively and negatively, and take union over all those. Like you do for constructing the orientation double cover for a manifold.

That'd also be a vector bundle.

Danu

@BalarkaSen You mean, look at each fiber, take the positive and negative generator only, and take the union over those (that's how we constructed the orientation double cover in my course)

Actually, that's not what you mean

You are talking about a construction that differs from the one I saw.

So what would the fiber over one point be? THe disjoint union of positive and negatively oriented fiber of the point?

Balarka Sen

11:54 AM

@Danu Yeah, for each chart $U$ you take two copies of $p^{-1}(U)$ and then construct a vector bundle by gluing appropriately according to whether the transition functions of the original vect. bundle preserved or reversed orientation.

What I can't figure out is what the zero section would be. I.e., what would it be a vector bundle over?

Danu

I'm weirded out by this construction.

Balarka Sen

it's like the moebius band being double covered by the standard strip

It's what Milnor means by "double cover" in his hint, doesn't he? Otherwise I can't make sense of the hint. Does he not mention this construction before?

Transcript for

$Mathematics$ Mathematics