« first day (3207 days earlier)      last day (100 days later) » 

12:01 AM
@Semiclassical Okay. The actual function was $\frac{x+2}{\sqrt{x^3+x^5}}$. And this was simplified to $\frac{2+0}{\sqrt{x^3}+0}$. When I draw this in Desmos, they are almost similar, and so I can see why the definite integral would be the same.
Right, cause this function behaves very much like $1/x$, so it's the small x that probably contribute most to the integral. Thanks a lot!
 
12:20 AM
@schn: To be a bit more precise. When $x\approx 0$, $x^3+x^5$ looks like $x^3$. So the function looks like $2/\sqrt{x^3} = 2x^{-3/2}$.
This sort of reasoning is very important for looking at infinite series and improper integrals.
 
@TedShifrin Hi uncle Ted
 
12:57 AM
Is anyone here familiar with LyX (the LaTeX editor)?
 
LOL, hi @Jacksoja.
 
@TedShifrin hehe
I was meaning to ask you about a smal problem
 
What's that?
 
in trying to prove that there cannot be a surjective map between S and S* ( the subsets of S)
we are not assuming finetness of S
 
Ah, this is a classic problem.
 
1:05 AM
am not sure how to tackle it
the idea is , one element subsets would exaust all of S
hence there are no room for 2 or more element sets
but this arguement does not work for infinity right?
 
Nope, it doesn't work.
 
i thought so
can you drop me a hint?
 
There are fancy ways with infinite arithmetic, but I think you can do a Bertrand's paradox type proof. You suppose you have a surjection, and you invent a subset that cannot possibly be in the image of it.
 
okay
i have to google this Bertnand
 
well, that gives it away.
 
1:12 AM
okay then later I shall
 
I'm desperate for a pun but that would give it away too
 
LOL, @ÍgjøgnumMeg. Bad boy.
 
@ÍgjøgnumMeg Hi , go ahead do that pun
:D
 
I shan't!
I cannot, nor will I
 
What Ted said holds true then
 
1:13 AM
@Jacksoja: Suppose you have a map $f\colon S\to S^*$. Can you think about whether $s$ and $f(s)$ are related?
That's a generous hint.
 
well i do want that s to be in f(s)
so f(s) is a proper name of it
the subset of S that contains our element s
but what else does it contain ?
 
Maybe. That's not quite the idea. You want to make a subset of $s$'s with some property.
 
if just a singelton, then i cannot do much with it
 
Make a subset $X = \{s: \text{something holds}\}$. Can $X=f(t)$ for some $t\in S$?
 
okay I see
so the idea I want to make a subset of S , that is not reached by my funtion
 
1:18 AM
Right.
 
no matter what function is
hmm that is quite evolved
 
@ÍgjøgnumMeg: Have I misled?
 
No that's what the standard method is I think?
 
I'm just trying not to give it away entirely. Yes, I know only the standard method.
 
It's just a nice choice of something holds I guess
 
1:21 AM
That's for @Jacksoja to contemplate.
BBIAB
 
would this be good, if i consider a subset of T of S such that, for each a in S, f(a) and T differ in at least one element
@TedShifrin what is BBIAB?
darn it ,I assume you had to go, thanks anyway all
 
@Jacksoja I assume "Be back in a bit"
 
@ÍgjøgnumMeg such a short sentence why not spell it out as it is
old generations are driving me crazy with these abbriviations
 
1:36 AM
Let $M$ be an $R$-module and $\varphi : R \to S$ a ring hom. Then $S$ is an $R$-module with $R$-action given by $rm := \varphi(r)m$. Thus we can form $M^S := M \otimes_R S$.. this is an $S$-module with $S$-action given by $(m \otimes s) \cdot s^\prime = m\otimes ss^\prime$
so I wanna check that this $S$-action is well defined
 
@Jacksoja LOL ... Mostly they come from your generation, guy.
 
and I guess $\varphi$ lets me do that?
 
Well, of course you need to use $\varphi$ :P
@Jacksoja: My hint was that your condition should involve relating $s$ and $f(s)$.
What you wrote is a good idea, but how would you define such a $T$?
Oh, @Rithaniel snuck in.
 
So I have $m \in M$ and $r_1, r_2 \in R$, $s_1, s_2 \in S$ I have $m \otimes (r_1s_1 + r_2s_2)s^\prime$.. and the expression $r_1s_1 + r_2s_2 = \varphi(r_1)s_1 + \varphi(r_2)s_2$ and you get $m \otimes (r_1s_1s^\prime + r_2s_2s^\prime)$
and then by $R$-bilinearity $r_1(m \otimes s_1s^\prime) + r_2(m\otimes s_2s^\prime)$
Spam end
 
My sleep schedule is fucked
 
1:49 AM
S a m e
 
Let $A$ be a $4\times 4$ matrix over $\mathbb C$ such that $\operatorname{rank} A=2$ and $A^3=A^2\neq 0$.Suppose $A$ is not diagonalizable.
$\exists$ a vector $v$ such that $Av\neq 0$ but $A^2v= 0$. Is the statement true or false?
 
It's holidays for me so I also have no "real" reason to fix it which makes it worse because I won't even though I want to
 
I'm off work atm so I'm slowly ruining my life
 
I know that rank (A)=2. so Geometric multiplicity of ($\lambda=0$)=2
 
LOL, a @Balarka. How unusual.
 
1:53 AM
@ÍgjøgnumMeg I wonder if this corresponds to pushforward of the coherent sheaf defined by $M$ on $\text{Spec}\, R$ to $\text{Spec}\, S$.
The jay lower star you know
(This is extension of scalars, of course)
 
lol I'm not there yet, this is just me justifying that $k[X]/(f) \otimes_k \bar{k} \cong \bar{k}^{\deg f}$
 
So Eigen space corresponding to $\lambda=0$ consists of two linearly independent vectors. $E_{0}=span\{x_1,x_2\}$. I can rule out the cases and show that characteristic polynomial is $x^3(x-1)$.
 
@N.Maneesh: I vote false. What do you vote?
No, that isn't the right characteristic polynomial.
 
Then obviously it must be false.
 
Hmm.
What are you suggesting the Jordan normal form will be?
 
2:02 AM
@ÍgjøgnumMeg Oh, that's because the spectrum of the left hand side is the fibered product given by limit of the pullback diagram $\text{Spec}\, k[X]/f \stackrel{\pi}{\to} \text{Spec}\, k \leftarrow \text{Spec}\, \overline{k}$, which is nothing but the fiber over the basepoint $\text{Spec} \, \overline{k} \to \text{Spec}\, k$, i.e., $\deg f$ many copies of $\text{Spec}\, \overline{k}$ which is $\text{Spec}\, \overline{k}^{\text{deg}(f)}$!
 
What's the algebraic multiplicity of $0$ and of $1$, @N.Maneesh?
 
Covering spaces ftw
 
$$\operatorname{Spec}^{-\deg f}! = \operatorname{Spec}^{-\deg f}(\operatorname{Spec}^{-\deg f} - 1)(\operatorname{Spec}^{-\deg f} - 2) \cdots$$
uh oh
missed a $k$
 
$$
J=
\left[\begin{array}{rrrrr}
0 & 1 & 0 & 0 \\
0 & 0 & 0 & 0 \\
0 & 0 & 1 & 0 \\
0 & 0 & 0 & 1\\

\end{array}\right]
$$
 
@ÍgjøgnumMeg ripriprip
 
2:04 AM
@Balarka sad
 
So that has rank $3$, @N.Maneesh.
 
I guess this book is supposed to be giving me an explanation for the obvious similarity between the galois correspondence for fields and for covering spaces, but i'm only on chapter 1
 
I don't know the explanation to be honest. Also I don't know any field theory; I can pretend to because I know covering spaces and I know the statement of the correspondence >:)
It's good for party tricks if you are a topology enthusiast want to surprise people with your quick Galois theory skills
 
$$
J=
\left[\begin{array}{rrrrr}
0 & 1 & 0 & 0 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 1\\ \end{array}\right]
$$ sorry
 
Use \begin{matrix}
 
2:08 AM
Alas, I don't really know any topology "properly", only things I've gotten from reading number theory books
lol
 
Yes, that is the only possibility.
 
You're on the other side. Learn the correspondence and come to topologists' parties to surprise people with your quick covering space theory skills
 
So then the statement must be correct, @N.Maneesh.
You have to show that that is the only possible Jordan form.
OK, I'm disappearing.
 
@TedShifrin How can be from the unique jordan form statement correct?
 
I like whole numbers too much
Despite my poor numeracy skills
 
2:12 AM
Hey guys.
Take $n$ vectors $\{v_i\}$ in an $F$-vector space $V$.
Does the existence of linear maps $\phi_i:V\to F$ with $\phi_i(v_j) = \delta_{ij}$ imply linear independence of the $v_i$?
I mean, for any $\alpha_i\in F$ we have
$$\sum \alpha_iv_i = 0 \implies \sum \alpha_i\phi_j(v_i) = 0 \implies \alpha_j = 0$$
Is that the proof? haha.
I it's so simple I feel I'm missing something important hahaha.
 
2:29 AM
$$
J^2=
\left[\begin{array}{rrrrr}
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 1\\ \end{array}\right]
$$. For $J^2$, I can prove that for $J^2$ there is $v=(a,b,c,0)$. For suitable $a,b,c$, I can prove that $Jv \neq 0$ and $J^2v=0$. I don't know about $A$
 
2:47 AM
@BAYMAX We know that there is a matrix $P$ such that $J=P^{-1}AP$. So, $J^2=P^{-1}A^2 P$. Similar matrices have rank same. So, Rank of $A^2$ must be 1. So, Nullity($A^2)=3$. Nullity(A)=2. So, there is a vector $v$ such that $Av\neq 0$ and $A^2v=0$.
Is my argument correct?
 
I am thinking about
> Nullity($A^2)=3$. Nullity(A)=2. So, there is a vector $v$ such that $Av\neq 0$ and $A^2v=0$.
@N.Maneesh
 
3:52 AM
Ah yeah. I just leave this tab open on my computer. So I imagine that I just randomly pop up whenever I connect my internet. :] @TedShifrin
 
4:15 AM
Hey.
I have vector spaces $U$ and $V$, and I'm trying to show that the usual construction $F(U\times V)/S$ (where $F$ is the free functor from Set to k-Vec) together with the map $\xi:U\times V\to F(U\times V)/S$ given by $(u,v)\mapsto [(u,v)]$ satisfies the universal property of the tensor product.
So I take an arbitrary bilinear map $f:U\times V\to W$, and I want to show it factors through $\xi$ via a unique $\tilde{f}$.
Haha is anybody following?
 
This sounds like Category Theory, and, unfortunately, I do not personally know category theory well enough to be of any assistance.
 
It's linear algebra. I just use categorical jargon because it's easier haha.
But thanks :)
 
5:10 AM
[Random]
On the problem of the Category of Mathematics
Consider the following diagram:
Top -> Ord
..........|
.........V
 
Hi all, where is the playground?
 
..........Alg
The morphisms that links Top to Ord and Ord to Alg may not be associative, as the morphisms Top -> Alg may be in general very different in structure
Thus mathematics is something much larger than a category
May 11 at 16:52, by Tobias Kildetoft
If you lecture long into the void, the void starts lecturing back.
@Rudi_Birnbaum what playground, there is no chat room call that
 
@Secret where I can test mathjax code (regularly).
I do it usually in the answer windows in MSE
that would be irregular I guess
 

Sandbox

Where you can play with chat features (except flagging) and ch...
 
@Secret Sandbox, that was the term! Thanks!!
 
5:45 AM
I'm about to go to bed, but I don't even know how to ask this question in a front-page appropriate way. Take a hyperbolic plane ($H^2$), mapping it to a disk, and compactifying it (including the circumference). Adding a Euclidean dimension $X$, to turn it into a cylinder, what type of geometry is along the surface of the $\text{Circumference}\times X$?
I'm probably even talking nonsense at this point, in regards to "what type of geometry", but I'm just having trouble grasping a geometric concept discussed in a physic discussion with someone
And this shape came up
I can't even begin to reason about the Circumference of a compactified Hyperbolic Disk, let alone see how extending it with a euclidean dimension affects it.
What should I ask to accomplish learning anything?
 
 
1 hour later…
7:02 AM
Let $f:[0,\infty)\to \mathbb R$ be a real valued function such that $f(1)=1$ and forall $x\in \mathbb R$ $f'(x)=\frac{1}{x^2+f(x)^2}$. Then $\lim_{x\to \infty}f(x)$
I know that $f'(x)\ge 0$. So, Monotonically increasing
$f''(x)\leq 0 \implies f$ is concave down
 
7:47 AM
Let $f:[0,1]\to \Bbb R$ be a continuous function, define $g:[0,1]\to\Bbb R$ and $g(x)=(f(x))^2$, then $\int_0^1g\,dx=0\implies\int_0^1f\,dx=0$. How to show this? I first thought that it is because $\int fg\,dx=\int f\,dx\int g\,dx$ kind of rule, but there is none such rule.
 
8:10 AM
can someone help me with this
1
Q: Proving that the set function $m$ is regular

Subhasis BiswasI want to finish this proof. OP has considered the interval differently. I choose here, in $\mathbb{R}^p$, the interval $a_i - (\epsilon/2)^{(1/p)} ≤ x_i ≤ b_i + (\epsilon/2)^{(1/p)}$ (the closed set $F$). I am trying to show that $m(A)≤m(F)+ \epsilon$. Clearly, $F \subset A$. We break up the...

 
@Silent that's because $g$ is always nonnegative
so if its integral is zero, it must be identically zero
 
@Sil
@Silent, suppose $f(x)=k$ somewhere. Then, in some neighbourhood of $x$, $f(x) \neq 0$. Therefore, $g(x)>0$ in there. Let the nbd be $(e,f)$. Now, split the integral of $g(x)$ into three parts: from $0$ to $e$. $e$ to $f$. $f$ to $1$. middle term is positive and the other parts are nonnegative, thereby yielding $\int_0^1g(x)>0$. A contradiction
this is basically what @LeakyNun said, I guess
 
Suppose we have a function $f:B\rightarrow\ddot{B}$ where $B$ and $\ddot{B}$ are algebraic structures with the same underlying set and $f$ satisfies the property that for $a,b\in B$ then $f(a)f(b)=\phi(f(ab))$ where $\phi :\ddot{B}\rightarrow\ddot{B}$ is a permutation on $\ddot{B}$. Is there a term for this sort of relationship?
 
8:34 AM
can anybody on earth say why this is wrong?
$p$$L$$E$$a$$s$E$
 
8:45 AM
I don't see anything necessarily wrong with your work, Subhasis, though I always doubt myself when I see expressions like this.
 
for one last a time, this is my request to go through this
maybe you can find some errors here
very sorry for the handwriting
 
Hello everyone. I have a small query. Is writing a matrix in its Smith normal form a basis change?
 
 
1 hour later…
10:05 AM
@ÉricoMeloSilva there is one such function for the upper half plane though: G(r;r0) = 1/2π [log|r-r0| + log|r-r0'|]
the "divergence theorem" proof doesn't work because the test function 1 isn't $C_0(V)$
 
 
3 hours later…
12:42 PM
@LeakyNun @SubhasisBiswas, thank you.
 
1:35 PM
Let $f:\Bbb R\to\Bbb R$ be a polynomial such that $f(0)>0$ and $f(f(x))=4x+1$ for all $x\in \Bbb R$, then what is $f(0)$? The options given are: 1/4, 1/3, 1/2, 1
I have no idea what to do
 
1:49 PM
6
Q: What eigenvector-like tools are there for analyzing tensors of rank three and higher?

Emilio PisantyIf I have a rank-two tensor that I want to analyze ─ say, an electric quadrupole moment, or a moment of inertia ─ it can often be very easy to analyze by moving to its principal-axes frame: one rotates to a reference frame where the tensor is diagonal, and this simplifies all sorts of understandi...

$f(0) > 0$
o great, f is not necessarily order preserving
trying again...
$ff(0) = 1$
 
2:01 PM
@Silent $f(x)=\frac13 + 2x$
 
Wow! how did you figure that out?
 
Just take an arbitrary polynomial $f(x)=a_0+a_1x+\dots +a_nx^n$ and you get $$f(f(x))=a_0+a_1(a_0+\dots+a_nx^n)^1+\dots + (a_0+a_1x^1+\dots +a_nx^n)^n=4x+1,$$
and over an integral domain you can deduce that $a_2=a_3=\dots=a_n=0$
 
hmm
 
So $f(x) = a_0+a_1x$ and $f(f(x))=a_0+a_1(a_0+a_1x)=a_0+a_0a_1 + a_1^2x=1+4x$
so $a_0+a_0a_1 =1 $ and $a_1^2=4$ which tells you that $a_1=\pm 2$
And thus you have two cases, either $a_1=2$ in which case $a_0=\frac13$ and you have $f(x)=\frac13 + 2x$ which satisfies $f(0)>0$
Or else you have $a_1=-2$ and thus $a_0=-1$ which gives you $f(x) = -1-2x$ which does not give you $f(0)>0$
Hence you can conclude that $f(x)=\frac13 + 2x$ yielding $f(0)=\frac13$
 
Thank you so much, @EarthCracks!
 
2:04 PM
Not a problem
 
2:18 PM
Afternoon
 
How does one estimate the behaviour of a function? Say I have the function $\frac{x+2}{\sqrt{x^3+x^5}}$, how come one can simplify this to $\frac{1}{\sqrt{x^3}}$. I'm asking cause I'm determining the convergence of an integral from 1 to infinity with this function, and this approximation is made. But according to my drawing, the approximated function is less than the original, isn't that a problem?
I would simplify the function to $\frac{2+x}{\sqrt{x^3}}$ cause for large x, this is GREATER than the original function, and so, if the integral with this simplified function convergence, the original has to as well, no?
But I can't see how one can go even further and approximate $\frac{2+x}{\sqrt{x^3}}$ to $\frac{1}{\sqrt{x^3}}$. This is making compromises on both numerator and denominator, which gets itchy.
 
2:36 PM
Can someone help me figure out how to turn this into a front-page appropriate question? I was thinking about this last night and I honestly, don't know what I'd ask if I posted it on the front page
9 hours ago, by Axoren
I'm about to go to bed, but I don't even know how to ask this question in a front-page appropriate way. Take a hyperbolic plane ($H^2$), mapping it to a disk, and compactifying it (including the circumference). Adding a Euclidean dimension $X$, to turn it into a cylinder, what type of geometry is along the surface of the $\text{Circumference}\times X$?
 
@schn Note that $\frac{x + 2}{\sqrt{x^3 + x^5}} \sim \frac{x}{\sqrt{x^3 + x^5}}$. Now $x^3 + x^5 = x^2(x + x^3)$ so you have $\frac{x}{\sqrt{x^3 + x^5}} = \frac{1}{\sqrt{x + x^3}}$
 
is 2, 3 the only prime of the form $(p-1)!+1$ where $p$ is prime?
if yes then why so?
 
probably smth to do with Wilson's Theorem
 
@ÍgjøgnumMeg Okay. But when approximating from $\frac{x + 2}{\sqrt{x^3 + x^5}}$ to $\frac{x}{\sqrt{x^3 + x^5}}$, aren't we approximating to a lesser function, and so a lesser integral over 1 to infinity? And how would one arrive at $\frac{1}{\sqrt{x^3}}$ from your last expression $\frac{1}{\sqrt{x + x^3}}$. It seems like here one would have to make the function greater again, no?
My intention is to determine the convergence of the definite integral over the function $\frac{x+2}{\sqrt{x^3+x^5}}$ from 1 to infinity, and the approximation $\frac{1}{\sqrt{x^3}}$ is made.
 
@schn the denominator can be further factored to give $\sqrt{x^3}\sqrt{\frac{1}{x^2} + 1}$
Now in the limit $\frac{1}{x^2} \to 0$
Also, in the limit the $2$ in the numerator becomes irrelevant (think $x \to \infty$, so $x + 2 \to \infty$)
Could've done this all in one go by writing $x^3 + x^5 = x^5\left(\frac{1}{x^2} + 1\right)$
 
3:02 PM
@ÍgjøgnumMeg True. Would you say then that $\frac{x + 2}{\sqrt{x^3 + x^5}}\geq \frac{1}{\sqrt{x^3}}$ or the other way around? Would this matter in determining the definite integral of $\frac{x + 2}{\sqrt{x^3 + x^5}}$ from 1 to infinity?
I'm thinking, wouldn't one want to find a function that behaves very similar to the one in question but that is greater than it, so if the integral with the approximated or simplified function, $\frac{1}{\sqrt{x^3}}$ in this case, is converging, then the integral with the original function has to as well
 
Well you can think of $\int_{1}^\infty$ as $\lim_{c \to \infty} \int_1^c$
so that probably helps idk
I can't do analysis
 
4:02 PM
@TedShifrin on a final exam I just took, there was another typo :\
 
 
3 hours later…
6:46 PM
is infinitation of x equal to the infinite tetration of x?
Lets introduce a new notation for this question:
let $x^{/1/}=x+x$ (addition)
$x^{/2/}=x.x$ (multiplication)
$x^{/3/}=x^x$ (exponentiation)
$x^{/4/}=^xx$ (tetration)
so infinitation according to my notation is $x^{/\infty/}$
infinite tetration of x is $x^{x^{x^{x.....}}}$
so my question is whether $x^{/\infty/}=x^{x^{x^{x.....}}}$
 
When people talk about the tails on box plots, are they talking about just the whiskers or are they talking about anything after or before the median?
 
7:00 PM
Why does anything I read involving a topological group specify a basis of open neighbourhoods of the identity? What's useful about having that information?
 
Context? Having a local basis at identity gives a local basis at every point which might come in handy
 
I see, context is varied, I've just seen it a lot of times and wasn't sure what the use of it was
The kernels of the connecting maps in a profinite group give a basis of open neighbourhoods of 1
No not the connecting maps
The projections down from the inverse limit
 
So in particular if it's a countable inverse system, the inverse limit is a first countable profinite group. That's always a plus.
 
ergh how far can I get without actively learning any topology and just picking up the relevant definitions as and when I need them
lol
 
Very far
I do that all the time with point set topology
 
7:06 PM
@ÍgjøgnumMeg because you can just translate it around and describe a basis around every point, hence the whole topology
 
So having a local basis at every point lets you piece together the whole topology?
@alessandro ah sniped
 
That's nice
 
The idea is that you only need to know what happens near a point to talk about convergence
(random remark: if you're convinced of this fact it's obvious that the metrics $d(x,y)$ and $d'(x,y)=\min\{1,d(x,y)\}$ induce the same topology)
 
Throw away all the big balls and you're still fine
 
7:12 PM
lol
 
It is sometimes useful to know that every metric is equivalent to a bounded one
Proving that a countable product of Polish spaces is Polish is the first example I could think of but there's surely more
 
sounds way further from the surface than I care to venture
 
Even more fundamentally you need that to metrize the product topology, if you have a countable family of spaces
 
Oh right, of course, that's how it's used in my example too
 
Right
 
7:15 PM
might go ahead and learn p-adic analysis before I learn real analysis properly
 
@ÍgjøgnumMeg why not both
write a program to calculate exp in 3-adic :P really cool
 
and then just claim that the p-adic metric is more natural and that real numbers don't make sense to me
 
if i want to consider the antipodal map on the n-sphere, i can consider it as being restricted from $-Id_{n\times n}$ on $R^{n+1}$. can I make rigorous through this embedding taking the map on tangent spaces on the sphere as just coming from the Jacobian $-Id_{n+1\times n+1}$
or should I instead take the stereographic projection charts on the n-sphere, and determine the Jacobians in local coordinates
 
whenever I get stuck on something I go back in time and remember when I didn't know the difference between $\cap$ and $\cup$
 
What's the subgroup of PGL(2,C) consisting of transformations leaving the unit circle invariant?
 
7:18 PM
We had a lecture on Lean yesterday, but I slept in... @Leaky
 
@AlessandroCodenotti quel dommage
 
@LeakyNun {1}?
 
definitely not
 
you mean taking the unit circle into itself?
 
they arise in the context of classifying Aut(D)
 
7:19 PM
or fixing the unit circles elements
 
maps like [z-a]/[1-conjugate(a)z]
$\dfrac{z-a}{1-\overline{a}z}$
@ÉricoMeloSilva I'm wondering whether we can use these maps to generate a Green's function for the Laplacian in the unit disc
 
Does anybody have a nice sounding soundbite as to why a geometric structure on a manifold should always have a finite dimensional symmetry group?
 
Remind me what the Green's function is again? Solution to Laplacian f = delta-mass at the boundary?
 
ie if the symmetry group were infinite dimensional "its topology and not geometry"
 
@s.harp what's a geometric structure?
 
7:23 PM
@BalarkaSen $G(r;r_0)$, $\nabla^2 G = \delta(r-r_0)$, $G|_{\partial M} = 0$
 
@F.White here just a structure that is about geometry, so the question is vague and I'm trying to motivate why it should be so.
there is a notion by Gromov called geometric structure, but thats not what I mean
(or it is what I mean, but I don't want to say it)
 
0
A: References for actions of infinite-dimensional Banach-Lie groups on infinite-dimensional Banach manifolds

Alexander SchmedingHere are some partial answers: It is known that Banach Lie group actions on finite-dimensional manifolds are quite restricted. What I mean by this is: Due to a theorem by Omori, see Hideki Omori, On Banach-Lie groups acting on finite dimensional manifolds, Tohoku Math. J. (2) 30 (2), 223-25...

 
Myers-Steenrod theorem states for example that Isom(M) for any Riemannian manifold M is actually a finite-dimensional Lie group. I don't know a good argument for it to be finite-dimensional; you might want to argue that the tangent space to Isom(M) at the identity is finite dimensional - which I suppose is equivalent to claiming that the space of Killing fields is finite dimensional.
That should be doable but I haven't tried it
 
@s.harp

"if a Banach-Lie group acts smoothly, effectively and transitively on a finite-dimensional manifold, then it automatically is finite-dimensional. Hence for many manifolds which come up, one can only consider actions which do not satisfy these requirements or one is forced outside of the class of Banach Lie groups (e.g. if one considers diffeomorphism groups of finite-dimensional manifolds).

Quotient theorems, or smooth structures on orbits for infinite-dimensional group actions are in general much harder to establish than in the finite-dimensional case. Something along the lines y
 
@F.White that is very interesting, thank you for that link. Just a remark: Diffeo(M) is infinite dimensional modelled on a Frechet space and acts on M, for me to motivate with this remark would require me to distinguish between Frechet spaces and Banach spaces, which is too subtle I believe
 
7:31 PM
@BalarkaSen rip me I got topology exam tomorrow
year 2
 
What kind of topology?
 
Viel Glück :)
 
Good luck!
 
@AlessandroCodenotti "metric space and topology"
 
Oh, way easier.
 
7:33 PM
@BalarkaSen I believe that that theorem follows from you being able to see that an isometry is uniquely determined by its derivative at a point (if manifold is connected). So the idea that geometric structures have to be rigid (that is symmetries are uniquely determined by a fixed amount of derivatives at a point) would be enough to motivate finite dimensional
 
@LeakyNun I remember when I took the exam one question asked us to state and prove Baire's theorem and I was afraid I remembered the statement wrong :p
 
yeah I'm just afraid of missing some things that are considered basic :P
oh no
 
@s.harp Yeah, that's it, I figured it out right before you mentioned.
 
@loch that's hardcore
lemme dig out your year's paper...
 
2015 I think
 
7:35 PM
yeah found it
this is not good
 
Fixing a point $x \in M$, you can write down a map $\text{Isom}(M) \to M \times GL_n(T_x M)$
This is going to be injective.
 
@loch what if they ask us to prove sequential compactness => compactness lol
 
That's actually surprisingly complicated if you think about it ^
 
yeah I remember the whole proof I think
and I think I also remember the skeleton of Baire
and also the filter proof for Tychonoff
but I'm kinda feeling that they're pointless
like who cares about proofs of technical result
 
Oh remembering them are, yes. Cramming to keep them in your head for an exam the next day is very pointless.
I hate doing it.
 
7:37 PM
@LeakyNun review the Hausdorff iff closed diagonal thing, a lot of people like to put it in exams apparently
 
@AlessandroCodenotti that's trivial
seriously
in Lean you just "follow your nose"
 
So is the filters proof for Tychonoff, so what?
 
it's true I don't remember most of these things since I don't use them
 
you're probably right
ok FIP characterisation for compact is also follow-your-nose
eh, what more technical results do you have in mind
@BalarkaSen et al
 
Baire is pretty simple it's just that some applications are weird.
 
7:39 PM
prove that isometry of a compact space is a compact group
 
@LeakyNun well I sure hope you have a metric space if they ask that
 
@AlessandroCodenotti yeah, metric space
 
I suppose that Baire theorem means "complete metric spaces are Baire" in this context then?
 
yeah
 
What's your proof that a complete metric space with no point isolated is uncountable?
@Leaky
 
7:41 PM
@BalarkaSen "BCT1" :P
 
Ah, good.
I didn't think of applying Baire when this came up in my exam. Just did the Rudin-style proof. Essentially proving Baire but still.
 
but isn't Z a complete metric space with every point isolated
 
@BalarkaSen without isolated points*
 
Thanks
 
With a bit more work you can prove it actually has cardinality $\geq |\Bbb R|$
 
7:44 PM
always assume continuum hypothesis is true to save yourself as much effort as possible
 
Con(ZFC) implies Con(ZFC+GCH) so I guess you can even assume GCH if you really want to :P
 
I dont know the contents of GCH so I'll assume that too why not
 
Also, $\Bbb R^n$ cannot be written as a countable union of disjoint closed balls. That's actually notoriously complicated.
 
$2^\kappa=\kappa^+$ for every infinite cardinal $\kappa$. $\kappa=\aleph_0$ is CH
 
It's "obvious" but you need Baire to prove this.
 
7:47 PM
can't $\Bbb R^1$ be done like that?
 
Thanks again :P
 
@BalarkaSen 0celo is back.
(as Ryan Unger ;)
 
I'm having trouble visualizing this,
if $\dot{\mathbf x} = \mathbf f(t,\mathbf x(t))$, what's the graphical interpretation of $\langle \mathbf x, \mathbf f \rangle < 0$?
 
Any metric space $X$ embeds into $C(X,\Bbb R)$
$X$ is complete iff the image is closed
 
could i get some help for this question:
1 hour ago, by Mathphile
is infinitation of x equal to the infinite tetration of x?
Lets introduce a new notation for this question:
let $x^{/1/}=x+x$ (addition)
$x^{/2/}=x.x$ (multiplication)
$x^{/3/}=x^x$ (exponentiation)
$x^{/4/}=^xx$ (tetration)
so infinitation according to my notation is $x^{/\infty/}$
infinite tetration of x is $x^{x^{x^{x.....}}}$
so my question is whether $x^{/\infty/}=x^{x^{x^{x.....}}}$
 
7:56 PM
@LeakyNun Busemann embedding, very important
I wouldn't say technical though
 
I'm just wondering if passing to the space $C(X,\Bbb R)$ would help me prove that any complete perfect metric space is at least continuum
 
@s.harp So actually $\dim \text{Isom}(M^n) \leq n(n+1)/2$, because that's the dimension of $M \times SO(T_x M)$. Which is interesting: this bound is tight, it's attained for eg $S^n$.
Is it true that it's attained iff $M$ is isotropic?
 
@Balarka its is also attained by euclidean space (with symmetry group $O(n)\ltimes \Bbb R^n$)
 

« first day (3207 days earlier)      last day (100 days later) »