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12:00 AM
I am actually often in MLC
 
it's on the door of MLC
from the inside
 
What are you doing in the colloquim?
 
I just did a talk today lol
so you can still see my name on the poster I suppose
 
I have different seminars which I have to give a talk
About?
 
model theory and ax-grothendieck theorem
 
12:02 AM
I am so out of number theory stuff:D
 
Well people. I'm headed off to my doom. This may be the end of my mathematical journey. Au reviour.
 
@orbit-stabilizer good luck
 
@orbit-stabilizer farewell
 
@LeakyNun I have rare contact to pure math. The only group I am working with is the group of Hairer.
you might know him, is from pure math
 
oh, what do you do?
I've seen a poster about Hairer on the 6th floor
 
12:06 AM
stochastic differential equations on manifold
 
how is "manifold" not pure math though
 
The main point is stochastic differential equation:P
We just exploit some property of certain manifold to make calculation easier
allowing us to estimate certain things
 
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
 
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?
 
12:47 AM
Is it easier to factor a non-ufd Then to factor a ufd ?????????
 
@ManolisLyviakis Study the function d(f(x),x).
Show that it is continuous, and use compactness to show that it has a zero.
 
could anyone help me with my problem
 
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 $3(t^2-4t+3)=(3(t-1)(t-3))$.
 
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 :(
 
Oh sure, @SBA, blame it all on me.
 
hi Antonios
 
Hi, @TedShifrin !
still studying for finals... woo.
 
1:10 AM
Yes, it has always been your fault. You've planned it all from the beginning haven't you?
 
congrats :)
 
How did the visit go?
 
They put me back together temporarily.
 
2
Q: Confirming asymptotic equivalence of two diverging series

JuliusFix $c\in\{0,1,\dots\}$, 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 ...

 
Actually, @TedShifrin if you have a moment I have a linear algebra problem I'm somehow not seeing.
 
1:14 AM
oh oh
 
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 $p(\lambda_i)$.
 
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.
 
Let me think about this for another minute.
 
Hi DogAteMy
 
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).
 
That's correct.
 
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.
 
The issue is whether geometric multiplicity could go up for $p(A)$. I highly doubt it.
 
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 $p(t)=t^2$.
 
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?
 
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.
 
No worries :-).
 
But you probably want to think about the situation I just brought up.
 
oh ok thanks @Antonios-AlexandrosRobotis
 
@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.
 
1:28 AM
I always recommend examples :)
 
(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.
 
Oh, BTW, I learned from Facebook today that Sylvain just got a big award.
 
Yeah, I saw that !!
 
@Ted How do I understand the map $S^2 \times S^2 \subset \Bbb R^3 \times i\Bbb R^3$ to $\Bbb CP^2$?
 
I added my congratulations to the FB post.
 
1:29 AM
Are you guys friends on FB?
 
Nope.
I should send him a personal message.
 
Just wondering... I got rid of my fb, but I recall Ken Ribet was friends with everyone...
 
Anything interesting going on
 
He was my undergrad advisor.
 
Yikes, @PVAL ... The $S^1$ action looks yucky.
 
1:30 AM
I don't think there's an S^1 action.
 
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?
 
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).
 
most of the elements of S^1 don't fix S^2 \times S^2
 
Well, you need one to get down to $\Bbb CP^2$, PVAL. That's what I meant.
 
I think only powers of i
 
1:31 AM
Yeah, right.
 
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.
 
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.
 
I was trying to show Sym(S^2\times S^2) is CP^2
I think @Balarka asked about that today.
 
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.
 
but I guess I was more interested in looking at this map then actually trying to proof the result tbh.
 
1:35 AM
But this is a holomorphic isomorphism, actually.
So I don't want to think of it your way :P
 
yeah my S^2's are totally real.
 
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.
 
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)
 
I think I call that totally real.
 
That's Sym^2, Demonark.
 
1:37 AM
if $iTS^2 \cap TS^2=0$
 
I would imagine you would, PVAL. ducks
Oh, I don't know what your first thing is, Demonark.
 
Oh wait... We are using different notation
 
I considered saying Lagrangian, but that would probably have confused you even more.\
 
It shouldn't. I've written papers on Lagrangian singularities.
But everything confuses me now.
 
Yeah but admittedly that word has at least a dozen different meanings.
 
1:39 AM
almost as many as normal
 
set
 
at least those words have connotation.
 
You're weakening our point.
 
@Antonios: I'm gonna head out, but let me know how that linear algebra story turns out.
 
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
:-)
 
2:06 AM
@Julius Deleted?
 
@Antonios-AlexandrosRobotis what happens if it has a zero?
 
2:19 AM
can complex numbers be described as "whole", iff both a & b are both integers in a+bi?
 
Like "whole numbers" {0,1,2,...}? @Riker
 
yes
like, is a "whole complex number" a thing?
if not, is there a name for when a & b are integers?
 
Sure, why not. I've never heard that description.
 
@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).
 
I've proven a
how do I go about proving b?
 
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.
 
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
 
@Antonios-AlexandrosRobotis cant find any zeroes
 
@ManolisLyviakis Is it urgent?
 
?
if i can prove it has zeroes then im done
if u mean if the excercise is urgent no its not
 
3:08 AM
What have you done so far?
 
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
 
0
Q: Factoring non-ufd methods ??

mickI 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...

Any ideas ?
 
@ManolisLyviakis I forget. Is your function a contraction?
 
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.
 
 
2 hours later…
5:38 AM
@Narcissusjewel Not yet. Do you know the answer?
 
 
1 hour later…
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
 
@user8663905 That does look weird
I suppose you could just ignore that part and find a basis without using it
 
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}"
 
@user8663905 No, the basis should not be the span of something, it should be some (finite) set of vectors
 
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
 
@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
 
8:16 AM
Hey @Tobias! It's been a while
 
@Daminark Yeah
 
How's it going?
 
Good. Getting back to normal now. The kids both had the chicken pox last week
 
Ah, well, happy that they're feeling better now!
 
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)
 
Generally speaking, overfull or underfull boxes appear all the time
So not really
But it's likely to be easily fixable, though
 
Well...
Hmm I might have a really hacky fix
 
Also you are making your diagrams with tikz-cd right? :P
 
Yeah
With this since I'm lazy
 
hmmm, that's pretty neat actually
a bit too limited for me, but still :O
 
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
 
What'd you do?
 
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
 
hi chat
 
Hey there @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
 
9:10 AM
No need, it's only one-sided cancellation. Aluffi?
 
f is an epimorphism, and it follows
:P
 
hierstein
 
@Leaky kek
The task here would then be to prove that surjections are epimorphisms in SET
 
@Daminark trivial
 
this should be done without that
because it is in the chapter 0
 
9:11 AM
@Jacksoja I'm just being facetious
 
Leaky's just taking the piss rn, let's just work it out
 
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?
 
my idea was to try to see fg(s) and hf(s) and they should agree
 
I guess I got sniped
 
so what is it
who sniped you ?
 
9:14 AM
@Jacksoja I literally just gave you the entire solution in two lines
 
Leaky sniped me, it's right up there
@Leaky calm down
 
I did not see it just yet
and yes calm down :)
 
Leaky sniper nun
 
can we really use that g(f(s) = h (f(s) ?
because that is what we want to prove?
@LeakyNun in your proof
 
No, you already know that
You're trying to prove that $g = h$ when you know $g\circ f = h\circ f$
 
9:18 AM
we dont
 
@Jacksoja you see, it's so trivial that you mistake it as what we're trying to prove
they're literally trivially equivalent
 
Nun can out snipe the leaky :-D
 
you don't know $g \circ f = h \circ f$?
gf = hf
(gf)(s) = (hf)(s)
g(f(s)) = h(f(s))
 
@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
 
oh right we do now that
@Daminark well said daminark
this guy leaky allways rubs it in my face
 
9:19 AM
sorry :P
 
@LeakyNun Bad man leaky ! -.-
kasmir mad at you now
 
for being mean to jack soja, cool name btw
 
haha ^_^
 
Oh no you've incurred the anger of Kasmir. I am truly horrified at what fate is about to befall you
 
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?
 
9:22 AM
@KasmirKhaan january exam
 
@skull it's fine, I think he gets the idea at this point
 
@LeakyNun good luck :D
 
@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)
 
@Daminark you have to get good sleep Before exam :D
 
lol people stop rubbing it in my face that I'm being condescending :P
 
9:23 AM
True that
 
sorry
 
@skullpatrol :P
 
:-D
 
so guys
 
I have come to kick bubblegum and ... no wait
 
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?
 
@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)
 
aha so f(s) = t was arbitrary here
 
yes
 
right right :)
thanks guys
so surjective ==> existence of right inverse, injective ==> existence of left inverse
 
yes
 
9:54 AM
Hi @Balarka
 
Hey hey hey
 
Do you have time to help me understand a remark from Reid's book?
 
For sure!
 
Yo @Alessandro!
 
Ok, do you have a copy available?
 
9:58 AM
Not in front of me, no
 
@Alessandro Reid commutative algebra or algebraic geometry?
 
Commutative algebra
 
Do you want me to rig a copy?
 
Okay I don't have a copy of that, I guess [DATA EXPUNGED]
 
I'm trying to work out how to upload a screenshot for my phone, if you can find a copy that'd be faster
 
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 :( )
 
I got one @Alessandro
which page/paragraph/line/word?
/letter/inkspot
 
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
 
Oh yes
I love that picture
 
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
 
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 $f^*:k[Y] \hookrightarrow k[X]$?
 
10:15 AM
Remind me what do you mean with $k[X]$ here, is it $k[X_1,\cdots,X_n]/I(X)$?
 
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
 
Right, that makes sense
 
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 $g \circ f \in k[X]$.
 
Ok, I don't see where the density of the image is needed (dense here is meant in the Zariski topology on $Y$, right?)
 
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
 
10:23 AM
Ah, ok, that makes sense now
 
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.)
 
Wait, he's thinking about $k[X]$ as the coordinate ring of the variety $Y=0$ here?
 
Yep
It's $k[X, Y]/(Y)$
 
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
 
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
 
10:33 AM
Doesn't it follow from going up/down? I think the professor mentioned this fact at some point
 
Remind me what that says? I'm weak on commutative algebra
 
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 $Q\cap A=P$
 
nonsense, nevermind
 
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 $Q'\cap A=P'$
 
Hmm
 
10:43 AM
(Which is just the first version applied to $A/P\subset B/Q$ which is still an integral extension)
 
I'd have to think about this
 
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
 
@AlessandroCodenotti like a true topologist
 
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.
 
@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
 
@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 $V$
 
@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
 
@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
 
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?
 
@Max Yes
 
Max
11:23 AM
@TobiasKildetoft Thank you so much! You saved me a lot of thinking :D
 
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 $S$?
 
@Max I mean that if you define the map in the given way, then what I said holds
 
Max
Ah, I understand!
 
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
 
@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
 
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 $O(n-1)$
 
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
 
[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
 
@BalarkaSen So the intuition why we need an orbit stabilizer is that, otherwise the element $g*x$ will leave the surface?
 
@BalarkaSen such an $f$ means "an $f$ such that $f^*$ is closed"?
 
@BalarkaSen I mean for example $O(n)$ can give me any $R^n$ vector without being on the sphere.
 
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!
 
@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.
 

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