Mathematics - 2017-12-07 (page 1 of 4)

A particle moves along the x-axis so that at time t its position is given by $x(t) = t^3-6t^2+9t+11$ during what time intervals is the particle moving to the left? so I know that we need the velocity for that and we can get that after taking the derivative but I don't know what to do after that the velocity would than be $v(t) = 3t^2-12t+9$ how could I find the intervals

Manolis Lyviakis

12:28 AM

X metric compact space. f continuous X to X prove there exists A subset of X such that f(A)=A

am i suppose to use any constant point theorems?

mick

12:47 AM

Is it easier to factor a non-ufd Then to factor a ufd ?????????

Antonios-Alexandros Robotis

@ManolisLyviakis Study the function d(f(x),x).

Show that it is continuous, and use compactness to show that it has a zero.

MATH ASKER

could anyone help me with my problem

Antonios-Alexandros Robotis

1:03 AM

@MATHASKER $v(t)$ is as you have described. Solve for the set of $t$ so that $v(t)<0$. Note that your polynomial factors as $3(t^2-4t+3)=(3(t-1)(t-3))$.

Simply Beautiful Art

It ought to become $(u+1)\sqrt u$, not $u+1\sqrt u$, and when you do substitutions, you need to readjust the bounds — Simply Beautiful Art 21 secs ago

@Ted because of you, I'm seeing this a lot more often :(

Ted Shifrin

Oh sure, @SBA, blame it all on me.

Antonios-Alexandros Robotis

lol

Ted Shifrin

hi Antonios

Antonios-Alexandros Robotis

Hi, @TedShifrin !

still studying for finals... woo.

Simply Beautiful Art

1:10 AM

Yes, it has always been your fault. You've planned it all from the beginning haven't you?

Ted Shifrin

congrats :)

Daminark

How did the visit go?

Ted Shifrin

They put me back together temporarily.

Julius

2

Q: Confirming asymptotic equivalence of two diverging series

Fix $c\in\{0,1,\dots\}$, let $K\geq c$ be an integer, and define $z_K=K^{-\alpha}$ for some $\alpha\in(0,2)$. I believe I have numerically discovered that $$ \sum_{n=0}^{K-c}\binom{K}{n}\binom{K}{n+c}z_K^{n+c/2} \sim \sum_{n=0}^K \binom{K}{n}^2 z_K^n \quad \text{ as } K\to\infty $$ but cannot ...

Antonios-Alexandros Robotis

Actually, @TedShifrin if you have a moment I have a linear algebra problem I'm somehow not seeing.

Ted Shifrin

1:14 AM

oh oh

Antonios-Alexandros Robotis

So, the whole discussion is about some polynomial $p(A)$, for $A$ an $n\times n$ matrix with entries in $\mathbf{C}$, and eigenvalues $\lambda_1,\ldots, \lambda_k$.

Anyways, part (a) is talking about proving that $p(\lambda_1),\ldots, p(\lambda_k)$ are eigenvalues of $p(A)$. That's basically routine computation. No problem there. The next bit is to compute the dimension of the eigenspaces $E(p(A), p(\lambda_i))$.

Seems like this bit follows from the same argument. An eigenvector for $A$ is an eigenvector for $p(A)$, so the rest seems to follow.

Finally, the last part is to find the characteristic polynomial of $p(A)$. I guess this means in terms of the characteristic polynomial of $A$.

Well, we do know what the eigenvalues are...

The so-called Spectral Mapping Theorem tells us that the eigenvalues of $p(A)$ are exactly the $p(\lambda_i)$.

Ted Shifrin

Hmm, so we want the kernel of $p(A)-p(\lambda)I$. Can we prove that it's the same as for $A-\lambda I$?

If $Av=\lambda v$, then $p(A)=p(\lambda)v$, clearly.

Antonios-Alexandros Robotis

Let me think about this for another minute.

Ted Shifrin

Hi DogAteMy

Akiva Weinberger

Hi

Antonios-Alexandros Robotis

1:23 AM

Okay, so @TedShifrin, doing what you suggest will show us that eigenvectors of $A$ are eigenvectors of $p(A)$.

(Unless I'm misreading, LaTeX doesn't render in chat for me).

Ted Shifrin

That's correct.

Antonios-Alexandros Robotis

That bit i've got.

I actually "know" all of the eigenvalues for $p(A)$ by that so-called Spectral Mapping Theorem (which I happen to remember the proof of).

My concern is findign the exponents in the characterstic polynomial.

Ted Shifrin

The issue is whether geometric multiplicity could go up for $p(A)$. I highly doubt it.

Antonios-Alexandros Robotis

Right.

Ted Shifrin

Certainly if $A$ is diagonalizable, we know that can't happen.

Oh, I see ... The eigenspace may become a sum of eigenspaces. Consider $A$ with eigenvalues $\pm 1$ and $p(t)=t^2$.

Antonios-Alexandros Robotis

1:26 AM

Well, geometric multiplicity is the same thing as # of linearly independent eigenvectors for $\lambda$. Since eigenvectors carry over, this isn't an issue, right?

Ted Shifrin

So it's going to depend on the relation between $p$ and the minimal polynomial of $A$.

I don't think I want to think about it too much more, especially since I'm going out for the evening in a few minutes.

Antonios-Alexandros Robotis

No worries :-).

Ted Shifrin

But you probably want to think about the situation I just brought up.

MATH ASKER

oh ok thanks @Antonios-AlexandrosRobotis

Antonios-Alexandros Robotis

@MATHASKER np.

@TedShifrin I'll think about it some more, taking that into account. I think my way forward now is to compute some examples.

Ted Shifrin

1:28 AM

I always recommend examples :)

Antonios-Alexandros Robotis

(This was on a practice exam – was the only bit I didn't get.)

Seems tricky, but I keep feel like I'm missing something silly.

Ted Shifrin

Oh, BTW, I learned from Facebook today that Sylvain just got a big award.

Antonios-Alexandros Robotis

Yeah, I saw that !!

PVAL-inactive

@Ted How do I understand the map $S^2 \times S^2 \subset \Bbb R^3 \times i\Bbb R^3$ to $\Bbb CP^2$?

Ted Shifrin

I added my congratulations to the FB post.

Antonios-Alexandros Robotis

1:29 AM

Are you guys friends on FB?

Ted Shifrin

Nope.

I should send him a personal message.

Antonios-Alexandros Robotis

Just wondering... I got rid of my fb, but I recall Ken Ribet was friends with everyone...

Akiva Weinberger

Anything interesting going on

Antonios-Alexandros Robotis

He was my undergrad advisor.

Ted Shifrin

Yikes, @PVAL ... The $S^1$ action looks yucky.

PVAL-inactive

1:30 AM

I don't think there's an S^1 action.

Daminark

If we're computing the nth homology of something and find that it can be written as the mth suspension of another space where m>n, that'll be zero, right?

Ted Shifrin

Yeah, Ken is friends with everyone ... but I knew him in grad school (and ran into him in France and almost ran into him in France several times).

PVAL-inactive

most of the elements of S^1 don't fix S^2 \times S^2

Ted Shifrin

Well, you need one to get down to $\Bbb CP^2$, PVAL. That's what I meant.

PVAL-inactive

I think only powers of i

Ted Shifrin

1:31 AM

Yeah, right.

PVAL-inactive

I think there's a Z/4 action with no fixed points and an S^2 worth of fixed points on the square of the generator.

Ted Shifrin

Interesting that $S^2\times S^2 = \tilde G(2,4)$ ... and this is so different.

I've never pondered this, @PVAL. Is there a good reason I should?

I guess it's coming from Mike's question yesterday.

PVAL-inactive

I was trying to show Sym(S^2\times S^2) is CP^2

I think @Balarka asked about that today.

Ted Shifrin

Ohhhh ... $\text{Sym}^k \Bbb CP^1 = \Bbb CP^k$. I told Balarka to prove that.

Totally non-topological proof.

I mentioned that and how it doesn't generalize nicely past complex 1-manifolds.

PVAL-inactive

but I guess I was more interested in looking at this map then actually trying to proof the result tbh.

Ted Shifrin

1:35 AM

But this is a holomorphic isomorphism, actually.

So I don't want to think of it your way :P

PVAL-inactive

yeah my S^2's are totally real.

Ted Shifrin

This reminds me a bit of the real grassmannian of oriented 2-planes embedding as a complex hyperquadric in $\Bbb P^n$. Danu and I discussed that ages ago.

Well, one was totally imaginary.

Daminark

Wait a second... I'm confused here, why would $\text{Sym}(S^2\times S^2) \cong \text{Sym}^2(S^2)$? (excuse me if this is a dumb question)

PVAL-inactive

I think I call that totally real.

Ted Shifrin

That's Sym^2, Demonark.

PVAL-inactive

1:37 AM

if $iTS^2 \cap TS^2=0$

Ted Shifrin

I would imagine you would, PVAL. ducks

Oh, I don't know what your first thing is, Demonark.

Daminark

Oh wait... We are using different notation

PVAL-inactive

I considered saying Lagrangian, but that would probably have confused you even more.\

Ted Shifrin

It shouldn't. I've written papers on Lagrangian singularities.

But everything confuses me now.

PVAL-inactive

Yeah but admittedly that word has at least a dozen different meanings.

Ted Shifrin

1:39 AM

almost as many as normal

Antonios-Alexandros Robotis

regular :)

skullpatrol

set

Antonios-Alexandros Robotis

nondegenerate

PVAL-inactive

at least those words have connotation.

Ted Shifrin

You're weakening our point.

Antonios-Alexandros Robotis

1:40 AM

loll

Ted Shifrin

@Antonios: I'm gonna head out, but let me know how that linear algebra story turns out.

Antonios-Alexandros Robotis

I'll have figured it out by the end of the evening, I'm sure.

Will let you know.

Great. Thanks :)

cya prof @TedShifrin

hi/bye skull

1:42 AM

:-)

Simply Beautiful Art

2:06 AM

@Julius Deleted?

Manolis Lyviakis

@Antonios-AlexandrosRobotis what happens if it has a zero?

Riker

2:19 AM

can complex numbers be described as "whole", iff both a & b are both integers in a+bi?

user685252

Like "whole numbers" {0,1,2,...}? @Riker

Riker

yes

like, is a "whole complex number" a thing?

if not, is there a name for when a & b are integers?

user685252

Sure, why not. I've never heard that description.

Antonios-Alexandros Robotis

@ManolisLyviakis If your equation $d(f(x),x)=0$ has a solution $x\in A$, then what can we say about the relationship between $x$ and $f(x)$?

or $x\in X$ (whatever your metric space was called).

Theo

I've proven a

how do I go about proving b?

user685252

2:28 AM

Usually, by the time you start talking about complex numbers you consider the real numbers as a subset of them, since a and b are real in a + bi. But you could define it that way and call it a "standard form" like ax + by = c for linear equations :-) @Riker

"a + bi where a and b are integers" Complex numbers a + bi where a and b are integers are called Gaussian integers.

@Riker en.wikipedia.org/wiki/Gaussian_integer?wprov=sfti1

Riker

2:52 AM

@user685252 sorry I had to eat something

@user685252 I do, it's just that I wanted to konw if the set of whole complex numbers was countable

it's pretty trivial that whole imaginary numbers are, and that plain complex/imaginary numbers are not

@user685252 ah, cool

Manolis Lyviakis

@Antonios-AlexandrosRobotis cant find any zeroes

Antonios-Alexandros Robotis

@ManolisLyviakis Is it urgent?

Manolis Lyviakis

?

if i can prove it has zeroes then im done

if u mean if the excercise is urgent no its not

Antonios-Alexandros Robotis

3:08 AM

What have you done so far?

Manolis Lyviakis

3:21 AM

np im trying to find if it has zeroes

i know it goes to R

it is continuous so it is compact

ill need to use sequences so the subsequences will converge to a poitn

and maybe i can find a stationary point or something

mick

0

Q: Factoring non-ufd methods ??

I was wondering If it is easier to factor in a non-ufd then it is to factor in a ufd. I can come up with arguments for that , but I also have arguments in the opposite direction. For instance : It should be easier to factor When there are more possibilities ( multiple factorizations in a non-ufd...

ring-theory norm computational-complexity factoring unique-factorization-domains

Any ideas ?

Antonios-Alexandros Robotis

@ManolisLyviakis I forget. Is your function a contraction?

Manolis Lyviakis

nop

Antonios-Alexandros Robotis

Oh... Hmmm. I think the proof I had in mind used that fact. I'm trying to think if this has to be true without that fact.

trynalearn

5:38 AM

@Narcissusjewel Not yet. Do you know the answer?

user8663905

7:01 AM

Hey, could someone help me understand what this linear algebra question is asking?

It gets three vectors, v1, v2, and v3. v1 is {1,4,-9,-5}. v2 is {4,-7,2,5}. v3 is {1,-5,3,4}. It then says "It can be shown that v1-3v2+5v5=0. Use this to find a basis for span{v1,v2,v3}."

But v5 is never given, and it's not a typo because v3 doesn't fit that criteria

Tobias Kildetoft

@user8663905 That does look weird

I suppose you could just ignore that part and find a basis without using it

user8663905

Yeah, I'm still not entirely sure what the answer would be. It's my understanding that the basis would be written in the form "span{some vectors here}"

Tobias Kildetoft

@user8663905 No, the basis should not be the span of something, it should be some (finite) set of vectors

user8663905

7:17 AM

Ah, okay. So I arrange these into a matrix, get it into reduced row echelon form, and then each column with a pivot column is one of the vectors that forms a basis?

I still don't know what exactly he wants with the v5 thing

Tobias Kildetoft

@user8663905 My guess would be that the 5 is a typo and that there is a further typo somewhere

or maybe all the 5's should be 3's. Hard to say

Daminark

8:16 AM

Hey @Tobias! It's been a while

Tobias Kildetoft

@Daminark Yeah

Daminark

How's it going?

Tobias Kildetoft

Good. Getting back to normal now. The kids both had the chicken pox last week

Daminark

Ah, well, happy that they're feeling better now!

Daminark

8:45 AM

Is it bad in a paper that won't likely be printed out if there's an overfull hbox thing going on?

(Due to a commutative diagram)

TastyRomeo

Generally speaking, overfull or underfull boxes appear all the time

So not really

But it's likely to be easily fixable, though

Daminark

Well...

Hmm I might have a really hacky fix

TastyRomeo

Also you are making your diagrams with tikz-cd right? :P

Daminark

Yeah

tikzcd.yichuanshen.de

With this since I'm lazy

TastyRomeo

hmmm, that's pretty neat actually

a bit too limited for me, but still :O

Daminark

8:56 AM

Yeah I like it, and so far I'm not doing anything fancier than rectangles large enough to do the 5 lemma on so far so like, I'm good

YES my fix worked

TastyRomeo

What'd you do?

Daminark

Basically, the idea is that I previously had the diagram on its own paragraph

So it indented (as you can see from the following paragraph), which pushed it off the edge a bit

What I did was that I rewrote the previous paragraph so that it exactly used up the whole line, and then put the diagram in that paragraph

I'm not sure if this is much in the grand scheme of things but it's by far the hackiest fix for an overfull hbox that I have ever done before

Anyway, how's it going for you Steamy?

Also yo @Akiva

Jacksoja

hi chat

Daminark

Hey there @Jacksoja!

Jacksoja

if f : S-->T is onto, g :T-->U , and h :T--> U are such that gf =hf , i need to prove that g=h

dont we need here f to be a bijection ?

@Daminark Hello

Daminark

9:10 AM

No need, it's only one-sided cancellation. Aluffi?

Leaky Nun

f is an epimorphism, and it follows

:P

Jacksoja

hierstein

Daminark

@Leaky kek

The task here would then be to prove that surjections are epimorphisms in SET

Leaky Nun

@Daminark trivial

Jacksoja

this should be done without that

because it is in the chapter 0

Leaky Nun

9:11 AM

@Jacksoja I'm just being facetious

Daminark

Leaky's just taking the piss rn, let's just work it out

Leaky Nun

now let t in T. since f is onto, there is s in S such that f(s)=t

g(f(s)) = h(f(s)), i.e. g(t)=h(t)

was it that hard?

Jacksoja

my idea was to try to see fg(s) and hf(s) and they should agree

Daminark

I guess I got sniped

Jacksoja

so what is it

who sniped you ?

Leaky Nun

9:14 AM

@Jacksoja I literally just gave you the entire solution in two lines

Daminark

Leaky sniped me, it's right up there

@Leaky calm down

Jacksoja

I did not see it just yet

and yes calm down :)

skullpatrol

Leaky sniper nun

Jacksoja

can we really use that g(f(s) = h (f(s) ?

because that is what we want to prove?

@LeakyNun in your proof

Daminark

No, you already know that

You're trying to prove that $g = h$ when you know $g\circ f = h\circ f$

Jacksoja

9:18 AM

we dont

Leaky Nun

@Jacksoja you see, it's so trivial that you mistake it as what we're trying to prove

they're literally trivially equivalent

skullpatrol

Nun can out snipe the leaky :-D

Leaky Nun

you don't know $g \circ f = h \circ f$?

gf = hf
(gf)(s) = (hf)(s)
g(f(s)) = h(f(s))

Daminark

@Leaky look you don't need to rub it in someone's face that it's trivial, if you'll be condescending about it just ignore

Jacksoja

oh right we do now that

@Daminark well said daminark

this guy leaky allways rubs it in my face

Leaky Nun

9:19 AM

sorry :P

Kasmir Khaan

@LeakyNun Bad man leaky ! -.-

kasmir mad at you now

Leaky Nun

k

Kasmir Khaan

for being mean to jack soja, cool name btw

Jacksoja

haha ^_^

Daminark

Oh no you've incurred the anger of Kasmir. I am truly horrified at what fate is about to befall you

Kasmir Khaan

9:20 AM

:D

Kasmir is a peacefull guy so he wont do much yet

but dont poke the bear

Okay phew

pokes

grrrr -.-

Noooo slow motion

anyway handsome folks

what is up with you ?

leaky have you done some exams yet?

dami are you done with algebra?

Leaky Nun

9:22 AM

@KasmirKhaan january exam

Daminark

@skull it's fine, I think he gets the idea at this point

Kasmir Khaan

@LeakyNun good luck :D

Daminark

@Kasmir I've got it on Friday (technically tomorrow for me but I intend to go to sleep tonight so morally it's the day after tomorrow)

Kasmir Khaan

@Daminark you have to get good sleep Before exam :D

Leaky Nun

lol people stop rubbing it in my face that I'm being condescending :P

Daminark

9:23 AM

True that

sorry

@skullpatrol :P

:-D

so guys

I have come to kick bubblegum and ... no wait

Jacksoja

9:24 AM

knowning that g(f(s) = h(f(s) for 1 point

why does that tell us that g = h

dont they have to agree for all points?

Leaky Nun

@Jacksoja for any t, g(t) = h(t)

therefore g=h

for any t, you can find s such that f(s)=t, from which it follows that g(t) = g(f(s)) = h(f(s)) = h(t)

Jacksoja

aha so f(s) = t was arbitrary here

Leaky Nun

yes

Jacksoja

right right :)

thanks guys

so surjective ==> existence of right inverse, injective ==> existence of left inverse

Leaky Nun

yes

Alessandro Codenotti

9:54 AM

Hi @Balarka

Balarka Sen

Hey hey hey

Alessandro Codenotti

Do you have time to help me understand a remark from Reid's book?

Balarka Sen

For sure!

Daminark

Yo @Alessandro!

Alessandro Codenotti

Ok, do you have a copy available?

Balarka Sen

9:58 AM

Not in front of me, no

Daminark

@Alessandro Reid commutative algebra or algebraic geometry?

Alessandro Codenotti

Commutative algebra

Balarka Sen

Do you want me to rig a copy?

Daminark

Okay I don't have a copy of that, I guess [DATA EXPUNGED]

Alessandro Codenotti

I'm trying to work out how to upload a screenshot for my phone, if you can find a copy that'd be faster

Daminark

10:01 AM

Oh yeah I've got no clue how to do that from phone

(For reference I can't likely help, just that I have a copy of AG so I could just send it over if that was it, but no :( )

Balarka Sen

I got one @Alessandro

which page/paragraph/line/word?

/letter/inkspot

Alessandro Codenotti

Ok, in chapter 4, right after the alternative proof of the weak Nullstellensatz by Nagata there is an explanation and a graph involving hyperbolas that I don't understand

Balarka Sen

Oh yes

I love that picture

Alessandro Codenotti

I see why $K[X,Y]/(XY-1)$ is algebraic but not integral over $K[X]$ and I see why modifying it makes it integral as well, but I don't understand how is that related to the projection missing a point

Balarka Sen

So remember that we discussed that if $f : X \to Y$ is a morphism of varieties with dense image, you could talk about the extension of rings $f^*:k[Y] \hookrightarrow k[X]$?

Alessandro Codenotti

10:15 AM

Remind me what do you mean with $k[X]$ here, is it $k[X_1,\cdots,X_n]/I(X)$?

Balarka Sen

Indeed. Alternatively, if you imagine $X$ to be an affine variety sitting inside $\Bbb A^n$, think of $k[X]$ as the ring of polynomial functions/regular functions $X \to \Bbb A^1$.

That's of course the same as your definition, because those which vanish on the variety are precisely elements of $I(X)$

So you have to mod out $k[\Bbb A^n] = k[X_1, \cdots, X_n]$ by that ideal

Alessandro Codenotti

Right, that makes sense

Balarka Sen

There are small subtleties but let's not get into that.

@Alessandro $f^*$ is the map which pulls a function $g \in k[Y]$ on $Y$ back to $g \circ f \in k[X]$.

Alessandro Codenotti

Ok, I don't see where the density of the image is needed (dense here is meant in the Zariski topology on $Y$, right?)

Balarka Sen

Ah, that's to get the injectivity of $f^*$. Remember my half-assed argument that if you had a function on $f(X)$ you could extend that to a function on $Y$ by density?

Yes, dense in Zariski-topology

Alessandro Codenotti

10:23 AM

Ah, ok, that makes sense now

Balarka Sen

That's why my half-assed argument doesn't work, but you can fix it, because all the morphisms are polynomial

@Alessandro So in particular projection of $XY = 1$ to the x-axis gives rise to an extension $k[X] \hookrightarrow k[X, Y]/(XY = 1)$ is the upshot.

Because that map has dense image in the x-axis (it's $\Bbb A^1 - \{0\}$, which is a dense open in Zariski top.)

Alessandro Codenotti

Wait, he's thinking about $k[X]$ as the coordinate ring of the variety $Y=0$ here?

Balarka Sen

Yep

It's $k[X, Y]/(Y)$

Alessandro Codenotti

So the two varieties are the hyperbola and the x-axis in the plane, it wasn't clear to me at all from the book that he was thinking about the x-axis as a second variety here

Balarka Sen

So there's a fact out there (Lying over theorem iirc) that says if $f : X \to Y$ is such a map I mentioned, $f^* k[Y] \subset k[X]$ is integral implies $f$ is a closed map.

I am trying to recall the proof of this

Alessandro Codenotti

10:33 AM

Doesn't it follow from going up/down? I think the professor mentioned this fact at some point

Balarka Sen

Remind me what that says? I'm weak on commutative algebra

Alessandro Codenotti

Hm, it says that if $B$ is integral over $A$ and $P\subset A$ is prime then there is $Q\subset B$ prime such that $Q\cap A=P$

Balarka Sen

nonsense, nevermind

Alessandro Codenotti

And that you can extend chains, if $P\subset P'$ are prime ideals in $A$ and $Q$ is prime in $B$ with $Q\cap A=P$ then there is a prime $Q'\supset Q$ in $B$ with $Q'\cap A=P'$

Balarka Sen

Hmm

Alessandro Codenotti

10:43 AM

(Which is just the first version applied to $A/P\subset B/Q$ which is still an integral extension)

Balarka Sen

I'd have to think about this

Alessandro Codenotti

Hm, I forgot the details (we saw that in the last lecture but it was quite confused)

@BalarkaSen anyway, let's assume this holds, no need to worry about its proof now

Daminark

@AlessandroCodenotti like a true topologist

Alessandro Codenotti

:P

Max

11:05 AM

Does anyone know if $T: V \to R^n$ is an inner product space isomorphism if $T(v) = (v)_S$, where $S$ is a basis for $V$? My book isn't saying so explicitly, but there was a theorem saying that an inner product isomorphism exists, and another theorem kind of suggesting that it should work.

Tobias Kildetoft

@Max What do you require for something to be an isomorphism of inner product spaces?

Max

That the geometric structure is preserved, which I assume is that the inner product can be used

Tobias Kildetoft

@Max That is not precise enough to answer this

Max

i.e. you can transform $v,w \in V$ into $R^n$ with $T$, take the inner product between them there, then transform back, making the result the same as taking the inner product in $V$

Tobias Kildetoft

@Max Then clearly not all isomorphisms of vector spaces will be isomorphisms of inner product spaces

Actually, I am not sure what you mean by "transform back" since you at that point just have a number

Max

11:16 AM

@TobiasKildetoft Sorry, I meant that they should be equal (accidently sent this before writing my answer. Writing it now)

Isn't there this theorem saying that if $v,w \in V$ ($V$ being an inner product space), then $||v|| = ||(v)_S||$? (where the left norm is defined as the norm in $V$ and the right norm is the euclidean norm) I thought that this would somehow result from isomorphism

Tobias Kildetoft

@Max No, that is definitely not true

simply because one side clearly depends on the choice of basis which the other does not

Did you mean for the basis to be orthonormal?

Max

@TobiasKildetoft Yes! That's what I'm missing.... Yes, for orhonormal basis, thank you

Tobias Kildetoft

right, then it all works out

Max

Even the isomorphism-part? The thing I proposed from the beginning? But with the addition that S is orhonormal?

Tobias Kildetoft

@Max Yes

Max

11:23 AM

@TobiasKildetoft Thank you so much! You saved me a lot of thinking :D

Tobias Kildetoft

in fact, the map will be an isomorphism of inner product spaces iff the basis is orthonormal

Max

@TobiasKildetoft Isn't there any other transformation that is an inner product isomorphism, aside from $v \to (v)_S$ with orthonormal $S$?

Tobias Kildetoft

@Max I mean that if you define the map in the given way, then what I said holds

Max

Ah, I understand!

Tobias Kildetoft

also, if you have an isomorphisms of inner product spaces, then there is a basis where the map will be given like this, and that basis will be orthonormal

Max

11:30 AM

Oh, interesting. There seems to be a stronger connection than I thought at first

Balarka Sen

@AlessandroCodenotti Actually, such a $f$ in fact needs to be surjective. Take any $y \in Y$; the maximal ideal of $k[Y]$ corresponding to that is $(Y_1 - y_1, \cdots, Y_n - y_n)$. The ideal corresponding to the subvariety $f^{-1}(y) \subset X$ in $k[X]$ is then nothing but $(f^* Y_1 - y_1, \cdots, f^* Y_n - y_n)$. If this is empty, weak Nullstellensatz kicks in to say that there are $g_1, \cdots, g_n \in k[X]$ such that $\sum_i (f^* Y_i - y_i)g_i = 1$.

Well, better to say that $(f^* Y_1 - y_1, \cdots, f^* Y_n - y_n)$ is the trivial ideal I guess. Hmm, I'm stuck again

quallenjäger

11:48 AM

How can I see that a $n-1$ sphere is isomorphic to $O(n)/O(n-1)$, I don't understand why do we need a stabilizer $O(n-1)$

Balarka Sen

orbit-stabilizer dude

O(n) acts transitively on S^(n-1) with stabilizer at a point O(n-1)

For any transitive G action on a set X with stabilizer H, G/H $\cong$ X set theoretically. In this case, as the action is a smooth action by a Lie group, you can prove this set-theoretic bijection gives a diffeomorphism

Secret

[To be deal with shortly] Leaky's metric question. My first step will be to draw the spheres specified by that metric space and then investigate

quallenjäger

@BalarkaSen So the intuition why we need an orbit stabilizer is that, otherwise the element $g*x$ will leave the surface?

Alessandro Codenotti

@BalarkaSen such an $f$ means "an $f$ such that $f^*$ is closed"?

quallenjäger

@BalarkaSen I mean for example $O(n)$ can give me any $R^n$ vector without being on the sphere.

Alessandro Codenotti

11:56 AM

I have lectures now, so I can't really think about this, but I'll do later, your considerations about $f^*$ where already very helpful though, thanks!

Balarka Sen

@Alessandro An $f$ such that $f^*$ has dense image and the inclusion on ring-level is an integral extension

Have fun with the lectures!

I'll think a little more about it later

@quallenjäger $O(n)$ does not act transitively on $\Bbb R^n$. It preserves length.

Transcript for

$Mathematics$ Mathematics