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12:01 AM
by that argument bijection also exists
 
yes
I'm not sure what you're trying to get at
 
to construct one :)
 
a bijection?
 
yes
 
you can do the usual trick
fix a copy of $\mathbb{N}$ in there and shift that by one, leave the rest the same
I'm not sure why you'd want to construct such a bijection though
 
very very
 
because apparently you can compactify $\Bbb R^2$ and $\Bbb R^n$
it's incredible
it's not homeomorphic to $\Bbb RP^n$ I believe
 
won´t do that, i am focused on open sets :P
 
Nice what are open sets
 
i am just considering those in $\mathbb R^n$, you know what they are :P
In mathematics, particularly in topology, an open set is an abstract concept generalizing the idea of an open interval in the real line. The simplest example is in metric spaces, where open sets can be defined as those sets which contain a ball around each of their points (or, equivalently, a set is open if it doesn't contain any of its boundary points); however, an open set, in general, can be very abstract: any collection of sets can be called open, as long as the union of an arbitrary number of open sets is open, the intersection of a finite number of open sets is open, and the space itself...
 
12:15 AM
nice lol
 
in fact, it appears that an open subset of $\mathbb{R}^n$ is homeomorphic to $\mathbb{R}^n$ if and only if it is connected and simply connected at infinity (source: MSE)
 
yes exactly. what I've done is transferred the structure of $\Bbb R^n$ to the $n-$cube
and then mapped the $n-$cube to the $n-$ torus
 
now map n-torus to n-koch snowflake :D
 
no @Masterphile lol :)
 
@Thorgott Does that hold for all $n>2$ or $n>4$? (For $n<3$ you just want simply connected, not simply connected at infinity.)
 
12:34 AM
$n\ge3$, according to this answer
should've specified that
 
I'm using a metric that restricts contractibility
 
12:51 AM
@Thorgott Yeah, I just didn't know the bottom two cases. I guessed 4 was Freedman but forgot how to do 3.
Thanks for the link, explains what I needed.
 
np
 
 
1 hour later…
2:05 AM
in Probability and Statistics, 24 hours ago, by Simple
Let $\{S_n\,:\,n\geq0\}$ be a simple random walk with $p$ be the probability of stepping up and $q=1-p$ be the probability of stepping down. Let $T_k=\min\{n\geq1\,;\,S_n=k\}$. If $p<1/2$, is $E(T_1\,|\,T_1<\infty)<\infty$?
Currently I have the probability generating function, $G(s)=\frac{1-\sqrt{1-4pqs^2}}{2qs}$ Not sure how to proceed,
 
 
3 hours later…
5:05 AM
I got it.
 
 
4 hours later…
8:50 AM
I know the irreduible representations of $SL_2$ (one for each natural integer $m$, by acting on polynomials of degree $m$ - let's denote thtat representation $V_m$) and I want to deduce they are irr rep of $SU_2$ as well. I am asked to prove that a if $W\subset V_m$ is stable under the action of $SU_2$, then $W = \oplus_{i\in I}e_i$ for some finite set of indices I, where $e_i = X^iY^{m-i}$
I have done this already but I fail to remember the proof of this argument.
 
9:10 AM
Nevermind, I got it
By looking at the eigenspaces of diagonal matrices
 
 
1 hour later…
10:31 AM
Hi Can I ask a question?
I try to simplify the following singular integral:

>$$\int_{R^2}\hat{p}(\xi)\hat{q}(\xi)|\xi|^{-2}d\xi= A\int_{R^2}\int_{R^2}p(x) q(y) (-\log|x-y|)dxdy$$
where $p, q\in C_c^{\infty}$ and $\int q=0$ and $\hat{p}(\xi):= (1/(2\pi))\int e^{-ix\xi}p(x)dx$ which is the Fourier transform, and $A$ is a constant.

1. My way is LHS=
$$\int_{R^2\setminus B(\epsilon)}\hat{p}(\xi)\hat{q}(\xi)|\xi|^{-2}d\xi+\int_{B(\epsilon)}\hat{p}(\xi)\hat{q}(\xi)|\xi|^{-2}d\xi$$
But why the RHS appears $(-\log|x-y|)$ which is the log potential of Poisson equation.
 
10:56 AM
Is the book written in English and entitled "Set Theory" by Hausdorff actually translation of his work written originally in German, and entitled "Grundzüge der Mengenlehre"?
 
thanks, although it seems it is a translation of not the first German edition but of some later one
 
If I got a matrix $A = X^T D X$ with $D$ being a diagonal matrix, are the diagonal entries in $D$ the eigenvalues of $A$?
 
11:12 AM
@MathStudent Not unless $X$ is orthogonal
 
@TobiasKildetoft Thanks. That's unfortunate.
Is there any other easy way to find out whether such a matrix would be positive definite? Or only by calculating the leading minors because it is symmetric which is not necessarily easy either, right?
 
@LeakyNun yes, it is not the translation of the first (which has the largest number of pages!) edition:
https://en.wikipedia.org/wiki/Grundz%C3%BCge_der_Mengenlehre
why, on Earth, wasn´t the first edition translated, instead of third?
does anyone has some clues?
 
11:34 AM
isn't it standard practice to just translate the newest edition
 
yes, but the third has serious omissions, and relatively much lesser number of pages than the first, i am thinking whether the first survived the happenings of WW2?
 
learn german
 
they did
 
Lohnt sich
 
according to Wikipedia it appears the first edition had sections on measure theory and topology, which were then considered set theory, that got removed in the later ones
It seems that the measure theory and topology chapters were removed, but the other chapters actually increased in content
Considering that measure theory and topology became even larger independent topics during the times between the first and third edition, this makes sense
 
11:44 AM
@MathStudent What do you mean "because it is symmetric"? You did not mention any of these being symmetric
 
@Thorgott you can also read the first edition, written in German?
 
Yes
 
you haven´t read it all never before?
 
no, I haven't read it
 
11:56 AM
In the entrance of the main building of the math department in Bonn there's a few framed posters telling the story of Hausdorff and his set theory book
 
Guys, I need some help with something I didn't understand.
Can you please help me with it?
 
@AlessandroCodenotti you were there?
 
Not if you don't tell us what it is @TechnoKnight
 
Thanks.
 
I'm still a student there even though I haven't been in Bonn physically for almost two months because of the coronavirus
 
11:57 AM
The equation of wave is:
A = Amax*sin(wt + f)
My question is, how we figured out that Amax is Amax?
 
What does that mean ?
 
I mean, let's write this equation: A = Bsin(wt + f)
How did we know that B is Amax?
 
Because sin only takes values in $[-1;1]$
 
As we know, B is the max positive value of A...right?
Yes. But did we set wt + f to 90 or 0 or 180 or which one?
I'm really sorry for my bad English.
 
It's the maximum as t varies I guess
 
12:00 PM
Yes, but which part of that equation is responsible to control the angle?
 
ah, he was a professor in Bonn
apparently
 
@TechnoKnight I'm not sure what you mean by that
 
@TobiasKildetoft I meant the matrix A is symmetric. I'm pretty sure that a matrix of such form is always symmetric, I am double checking to make sure now though
 
@Astyx I will try to write in a different way.
sin(wt + f) = 1 means that wt + f = pi/2
 
12:02 PM
@MathStudent Ahh, right, it is. And then you just need the eigenvalues, which will be those on the diagonal, once you pick $X$ to be orthogonal
 
Right?
 
no, just that wt+f = pi/2 + 2 k pi
 
Then that means wt + f is the responsible for setting the angle for wave, right?
 
@TobiasKildetoft Unfortunately, I can't "pick" X to be orthogonal. It's arbitrary I think.
 
All what I mean, is that when wt + f = pi/2 + 2kpi, it is when A = Amax...right?
 
12:06 PM
@MathStudent Right, but you can pick a different one which is, since the matrix will be orthogonally diagonalizable
 
Yes
 
Another question
It's the final one
What does sin(wt + f + pi/2) mean?
Does it equal to 1 or something else?
 
it's a function
 
@TechnoKnight it can take "much" values, 1 being only one of the possible values, try to find some info about geometric definition of sine function, with triangles and circle
 
@TobiasKildetoft Ah. I think I see what you mean. Do you know if the diagonal matrix would not stay the same then though if X isn't already orthogonal though, right?
 
12:10 PM
@MathStudent I am trying to think of whether we are in fact guaranteed that $X$ must be orthogonal, but I am not seeing it
 
@Masterphile I know. I was just confused by someone writing the same equation and finding it equal to 1
 
@MathStudent No idea what that means :)
 
the equation can be set to be equal to 1, but the function takes many values, you can set, for example A(sin(wt+r))=1 and try to solve that @TechnoKnight
 
Though it might be that there is a hidden assumption that $X$ is in fact orthogonal
 
An operator is positive definite it $\langle Qv, v\rangle > 0$ for nonzero $v$. Let's show that this is true for $Q = X^T DX$ so long as $D$ has all positive eigenvalues (or more generally, is positive definite itself) and X is invertible.
 
12:14 PM
@TobiasKildetoft I sent the message too early by accident sorry. I meant to add that I found out what the fisher information matrix looks like from here web.stanford.edu/class/archive/stats/stats200/stats200.1172/… but i guess if you dont have any idea what it means then it probably doesnt help either
 
We have $$\langle Qv, v\rangle = \langle X^T DX v, v\rangle = \langle DXv, Xv\rangle.$$
Now $X$ is assumed invertible, so if $v$ is nonzero, then $Xv$ is nonzero. Now we get that $$\langle Qv, v\rangle = \langle D(Xv), (Xv) \rangle > 0,$$ as desired.
 
@TobiasKildetoft Anyway, I'm pretty sure it is not assumed that it is orthogonal. Do you have any idea whether it would be easier to try to find out what D would look like with an orthogonal matrix or to look at the leading minors?
 
@MathStudent See what Mike wrote above
 
I just proved for you that you only need to know that X is invertible and D is positive
 
Oh, I'm sorry, I didn't see those messages. Thanks a lot, @MikeMiller ! However, unfortunately, I don't think I can assume X is invertible either.
I do know that D is positive though
 
12:20 PM
If $X$ is not invertible, then $A$ is at most positive semidefinite
 
@Thorgott So would I be right in saying that if a function is not absolutely convergent, it is one of these three cases? (1) Integrals of both the positive and negative parts of the function are infinite. It is not HK integrable on the region either. The two iterated integrals are different, and the double integral does not exist. e.g. f(x,y) = (x^2-y^2)/(x^2+y^2)^2 integrated with x between 0 and 1 and y between 0 and 1.
(2) Integrals of both the positive and negative parts of the function are infinite. It IS HK integrable on the region, so the iterated integrals are equal and the double integral does exist, e.g. f(x,y) = sinx/x integrated with x between 0 and infinity and y between 0 and 1. (3) Integral of positive part of function is infinite and that of negative part is finite, or vice versa. Iterated integrals are equal to +/- infinity, and the double integral does exist and is also +/- infinity.
 
12:32 PM
https://math.stackexchange.com/questions/3617190/minkowski-theory-an-isomorphism-of-bbb-r-vector-spaces-induces-a-scalar-prod

longboi for a dumb question
 
If $H\triangleleft G$, what's the link between $L^2(G/H)$, $L^2(G)$ and $L^2(H)$ ?
If there is one
 
@TobiasKildetoft Because then there exists a $v$ such that $Xv = 0$ and looking at Mike's proof that gives positive semidefinite. Very interesting, thanks a lot, you two @MikeMiller @TobiasKildetoft !!!
 
please explain this solution
 
@EdwardEvans when you said long boi I thought you meant long exact sequence
 
$L^2(G/H)$ are the elements of $L^2(G)$ that are compatible with the quotient.
 
12:34 PM
@Leaky hahaha
 
i can't understand 2nd line of soln "if dt/dx = p"
 
Hmm, I think it may be possible to have the function non-integrable, but with equal iterated integrals
 
@Leaky I think I'm confusing a couple of definitions but I can't pinpoint where lol
 
what definitions?
 
@LeakyNun please help if you're free.
 
12:36 PM
Well, I think I'm confusing what I'm trying to show; I think I'm trying to show that $\langle \psi(z_\tau), \psi(z_\tau^\prime)\rangle_c = \langle z_\tau, z_\tau^\prime\rangle_M$ in the notation given in the question
 
@Thorgott oh even non HK integrable?
 
@MikeMiller I got another question about this after all.. How do you know that it would be $>0$ at the end and not just that it is nonzero?
ah wait a second..
 
I'm not sure, I'll have to think about it later
 
Because $D$ is positive, we have that $\langle Dv, v\rangle > 0$ for all nonzero $v$ --- including $Xv$
More precisely, if $c$ is the smallest positive eigenvalue of $D$, you will see by hand that $\langle Dv, v \rangle \geq c |v|^2$
which is positive if $v$ is nonzero
 
I didn't think of the definition of the standard scalar product and thus didnt get to it sorry, silly mistake and thanks again!
 
12:53 PM
No worries
 
the definition of limit superior an limit inferior confuse me, they are given without any examples of the sequences
without any concrete examples
any! :D
i think they are equal to the usual limit if they are equal
 
@Masterphile I was looking at those recently again as well. It is simply the limit if the limit exists. One example that helped me understand it for when the limit doesn't exist is $sin(n)$. the limsup is 1 and the liminf is -1. Just reading the wiki page for it helped me: en.wikipedia.org/wiki/Limit_superior_and_limit_inferior
 
@MathStudent so they are unique?
they seem to be lowest upper and greatest lower bound of the set of all accumulation points of a sequence
@Thorgott are they that?
 
@Masterphile Yes I think
@Masterphile the wiki article also gives a definition of them being inf sup and sup inf so that would fit to what you are saying there I think and it at least fits in the example of the sinus function
 
1:08 PM
yes, it seems it fits
 
yes, that's one characterization
 
1:31 PM
@Thorgott Okay thanks please do let me know.
 
user131753
1:42 PM
1
Q: What conditions on the topological spaces are necessary and sufficient to ensure the existence of the following kind of functions?

user 170039Suppose that $(X,\tau_X)$ and $(Y,\tau_Y)$ are two topological spaces where neither is given the discreet (or indiscreet) topology. Does there always exists a nonconstant function $f:X\to Y$ is such that for all $Z\subseteq X$ if $\operatorname{Cl}_Y{\left(f(Z)\right)}\subseteq f(\operatorname...

 
2:00 PM
@Masterphile I just thought of an example where this is not the case I think. Say we have a sequence that is defined as $x_n = 1/n$ for all $n$ except for $n = 5$ with $x_5 = 100$. Then the lowest upper bound is 100. However the limsup is still 0 because the definition isnt just supremum but lim sup with n going to infinity or infimum over $n \geq 0$ and supremum taken over $m \geq n$.
 
that lowest upper bound is not from the set of accumulation points
58 mins ago, by Masterphile
they seem to be lowest upper and greatest lower bound of the set of all accumulation points of a sequence
 
Ah okay. That could be it then.
 
How do I find the decomposition of the representation of $SO_3$ on $L^2(SO_3/SO_2) = L^2(\Bbb S^2)$ if I know $L^2(SO_3) = \bigoplus_m \left(W_{2m+1}\right)^{2m+1}$ ?
Where $W_{2m+1}$ is the irreducible rep of dimension $2m+1$ of $SO_3$
 
@Astyx What does it mean for a function in that space to be compatible with the quotient?
 
I think it has to only depend on $z$
But it seems weird
 
2:12 PM
what is $z$?
Ahh, so functions that are constant on the cosets?
 
One of the canonical variables of $\Bbb R^3$
I think I'm confusing a lot of things
 
I don't think you can use the information about the big decomposition directly without knowing something about how it comes about
 
@MathStudent if it isn´t that then it should be
 
But I have not looked at anything like this for a long time (and never very much)
 
Could someone please have a look at this question I posted. I also mentioned what I tried but the question has received very little attention. I'd be very grateful if someone could have a look. Thanks.
0
Q: Geometry, prove that $E$ bisects $\overline{HI}$

user8718165Here in the figure, we have: $\bullet$ The radii of both the circles are equal $\bullet$ $E$ bisects both $\overline{AC}$ and $\overline{FG}$ $\bullet$ $\overline{HI}$ is a line passing through $E$ We have to prove that $E$ bisects $\overline{HI}$ as well. (I think it is really obvious fro...

 
2:23 PM
Is there an elegant method to see that alternating group $A_5$ can be written as the disjoint union of its Sylow subgroups and the trivial element? I mean, without having to explicitly write the subgroups in cycle notation.
 
@SanchayanDutta The two ingredients are that the Sylow subgroups intersect trivially (even for same prime), and all elements have prime power order.
 
@TobiasKildetoft How do you prove trivial intersection for the same prime $p$?
(That is, without explicit decomposition...)
 
For that part you do need to consider the prime 2 explicitly
 
@TobiasKildetoft And not the primes 3 and 5 too? Could you explain why?
 
because those only divide the order once
 
2:39 PM
@TobiasKildetoft (Please bear with me here a bit, I'm new to this.) Let's take the 3-Sylow subgroup for instance. It has 10 conjugate subgroups. So wouldn't we also have to show that all the conjugate subgroups also intersect trivially with each other?
 
@SanchayanDutta Yes, but that is clear just from the order
 
@TobiasKildetoft How so? I'm apparently missing that argument. Does that follow from the Sylow theorems?
 
no, from Lagrange
 
@TobiasKildetoft Alright, how exactly? I'm probably being silly here, but Lagrange's just says that the order of every subgroup $H_i$ of a certain group $G$ divides the order of the group. Well, it is also true that the cosets of the subgroup $H_i$ partitions the whole group.
 
@SanchayanDutta So what can the intersection be if both the subgroups have prime order?
 
2:45 PM
0
Q: Injective Morphism of Commutative $C^*$-algebras

user193319Let $X$ and $Y$ be compact Hausdorff spaces, and let $\varphi : C(X) \to C(Y)$ an injective morphism of $C^*$-algebras. The book I'm reading through claims that there exists a continuous surjection $\alpha : Y \to X$ such that $$\varphi (f) = f \circ \alpha$$ for every $f \in C(X)$. But I am havi...

 
@TobiasKildetoft If they have the same prime order? Let's label two conjugate Sylow 3-subgroups of $A_5$ as $H_1$ and $H_2$. Both $H_1$ and $H_2$ have prime order 3. But can we directly conclude from there that $H_1 \cap H_2 = \{1\}$?
 
@SanchayanDutta Same or different prime order. Same conclusion
Because what does Lagrange say about the order of the intersection?
 
@TobiasKildetoft I'm not sure what Lagrange's says about the order of intersection...
This is probably what I'm missing though
 
What does Lagrange say about a subgroup?
 
@TobiasKildetoft That its order divides the order of the whole group...and its cosets partition the whole group...
 
2:50 PM
Given a geodesic line $L$ in the Poincaré disc and a point $P \in L$, how can one construct a point on $L$ that has a given distance $D > 0$ from $P$?
 
Hey @ShaVuklia
 
ola @Astyx !
how ya doin'
 
Good and you ?
 
@SanchayanDutta Right. So use the first part here
 
ye, pretty good (given the circumstances x))
 
2:53 PM
think about $H_1\cap H_2$ as subgroup of $H_1$ (or $H_2$, doesn't really matter)
 
also question: say we have a one-parameter subgroup $\gamma\colon\mathbb R\to G$ of a Lie group G. How do we know the push-forward $\gamma_*(d/dt)$ exists? (treating $d/dt$ as a left-invariant vector field on $\mathbb R$)
 
What do you do nowadays ?
 
I'm finishing my bachelor
basically
and you?
 
Finishing my year
 
nice, what was it again you were doing?
 
2:55 PM
Engineering school, but mostly pure maths
 
oh right ye
 
@Thorgott Aha! So the intersection of (say) two Sylow 3-subgroups $H_1$ and $H_2$, i.e., $H_1 \cap H_2$ necessarily divides both $|H_1|=3$ and $|H_2|=3$. So $|H_1 \cap H_2|$ is either $1$ or $3$?
 
yes
 
(But it doesn't make much sense for $|H_1 \cap H_2|$ to be 3 because then $H_1$ and $H_2$ would be the same. So it's necessarily 1.)
Interesting. So now I need to show that the Sylow 2-subgroups all intersect trivially. How do I go about this?
24 mins ago, by Tobias Kildetoft
For that part you do need to consider the prime 2 explicitly
 
3:09 PM
I have now posted my question on SE: math.stackexchange.com/questions/3617405/…
 
Hello! i plotted a function in mathematica and i want to make the gridlines look such, that it isbetween each point and point like a graph paper
Or also called millimetric paper
What ist the command ? i am suffering since an hour with this
 
i think i found three statements such that 1) and 2) imply 3) and 1) and 3) imply 2) and 3) and 2) imply 1), is this unusual?
 
Hey @abenthy, you wouldnt happen to know your way around Minkowski theory would you? :)
 
Hi @Edward
 
Hey @Alessandro
 
3:16 PM
yeah, for analysis it´s unusual
classical elementary analysis
 
nvm got it
 
i did once transform the expression of Riemann zeta to some other function H such that, i think, H(1/2+epsilon)=H(1/2-epsilon), and i think i only needed to show epsilon=0, but i do not know where are those papers, i think i threw them away when cleaning a room, i do not tend to collect all that i write, so i lose some of my work without being published, that was when i was better in transforming functions so they take another form, i forgot what i exactly did
that wasn´t so much close to resolving RH but the expression was nice to work with
i remember i introduced some new functions there
they also had some nice properties
although i could try to recall the procedure exactly
and now i think of open sets!
what a downfall
 
@EdwardEvans What do you mean? Minkowski space time?
 
@abenthy I mean Minkowski theory within algebraic number theory hehe
 
I've used some of this stuff, but am far from an expert. You have a question on it?
 
3:30 PM
No sorry i did not get it. If anyone knows how to create a grid that is millimitric in mathematica for plotting without making it look like utter eye Pain please mark me in your reply i was aiming for something like this
 
I have a question posted on the main site, it's about the isomorphism between the Minkowski space and $\Bbb R^{r + 2s}$ and the scalar product it induces
0
Q: Minkowski theory: an isomorphism of $\Bbb R$-vector spaces induces a scalar product on $\Bbb R^{r + 2s}$

Edward EvansI'm following Neukirch's algebraic number theory. The situation is as follows: Let $K$ be a number field of degree $n$. Then $n = r + 2s$, where $r$ is the number of real embeddings $\rho : K \to \Bbb C$ (i.e. those embeddings $\rho$ such that $\rho(K) \subseteq \Bbb R$) and $s$ is the number of...

I think I'm just being a doofus but if you have 5 mins at some point then I'd be grateful if you took a look :P No rush
 
@EdwardEvans what if I don't have 5 mins at any point
 
Then don't take a look
 
I'll only be grateful if you spend exactly 5 mins looking at it
 
So if I opened it, read "Neukirch" and closed it immediately I won't be receiving any gratitude?
 
3:43 PM
You won't, and you'll also have to pay royalties to Neukirch's grave for having seen his name
 
you mean to the grave of he-who-must-not-be-named?
 
nah that's Voldemort you're thinking of mate
 
gasps
 
oh fu- is immediately killed by death eaters
 
4:03 PM
@EdwardEvans What should be happening is that $z\bar w + \bar z w = 2\text{Re}(\bar z w) = 2(x_1x_2+y_1y_2)$ where $z=x_1+iy_1$ and $w=x_2+iy_2$.
 
4:17 PM
Hello.
 
4:47 PM
$$ \begin{tikzcd}
& \Omega^*(U) \arrow{dr} & \\
\Omega^*(U \cup V) \arrow{dr} \arrow{ur}& & \Omega^*(U \cap V) \\
& \Omega^*(V) \arrow{ur} & \\
\end{tikzcd}
$$
ugh... How do I commutative diagram?
 
you google "commutative diagram latex", then copypaste the first result and try to adjust it without breaking it
the second part sounds easier than it is
 
tikz is not supported in this chat or MSE
 
5:02 PM
@Thorgott How do I not break it? I mean what I posted earlier is a valid latex code, but mathjax doesn't render it for some reason.
@AlessandroCodenotti oh... For that reason.
 
@feynhat No, it is not valid basic LaTeX
It uses a package
 
I see.
 
(also, MathJax is only a subset of LaTeX, so there are occasionally other stuff that also doesn't quite work)
 
There is a weird commutative diagram package usable on MSE and here but I forgot what it's called or how it works. It's also rather limited
 
5:14 PM
Is this a pullback?
All arrows are induced by inclusion.
Suppose there is a module $M$, with maps $\alpha: M \to \Omega^k(U)$ and $\beta: M \to \Omega^k(V)$. I have to come up with a map $M \to \Omega^k(U \cup V)$.
$\alpha(m)$ is a k-form in $U$, $\beta(m)$ in $V$.
Do I have to invoke partitions of unity?
Of course, $U$, $V$ are open sets such that $U \cup V = N$ (a smooth manifold).
by $M$ being a module I meant, $M$ is an $\Bbb R$ vector space.
choose a partition of unity for $\{U, V\}$. Define $\gamma : M \to \Omega^k(N)$ as $\gamma(m) = \rho_U\alpha(m) + \rho_V\beta(m)$.
(Sorry for the poor choice of symbols, I should have chosen M for the manifold)
Does this $\gamma$ work?
 
5:41 PM
@AlessandroCodenotti how best to describe the weak convergence on $L^\infty(\mathbb R)$?
Weak-* convergence is pretty easy to describe as $\int x_ny\,dx\to \int xy\,dx$ for all $y\in L^1(\mathbb R)$.
 
6:26 PM
Hm I'm not sure
The dual of $L^\infty$ is some space of measures, right?
 
Why do we need p values when we have rejection regions? or vice versa?
 
6:45 PM
is some branch of topology specially suited for research of $\mathbb R^n$?
 
@feynhat Yes, that works. On the overlaps $\alpha(m) = \beta(m)$ and $\rho_U(m) + \rho_V(m) = 1$, so that $\gamma(m) = \alpha(m) = \beta(m)$. On $U \setminus V$ the desired equality is automatic.
Similarly with $V \setminus U$. I wouldn't have done this with partitions of unity.
 
for example, this theorem seems to be specially suited for $\mathbb R^n$:

https://en.wikipedia.org/wiki/Invariance_of_domain
 
I would have just said $\gamma(m) = \alpha(m)$ on $U$ and $\beta(m)$ on $V$, with no ambiguity on the overlap by assumption.
 
it seems "algebraic topology" could be that branch
 
Invariance of domain is proved by algebraic topology. Most theorems of algebraic topology about $\Bbb R^n$ also extend to the context of manifolds much more generally, for instance invariance of domain.
Algebraic topology is not about Euclidean space in particular.
 
6:51 PM
i think there is also "geometric topology", whatever that branch is about, is it specially for $\mathbb R^n$
 
does \begin{CD}\end{CD} work in chat?
 
Not for me
 
sort of, as an introduction to the subject, i would like to do topology in $\mathbb R^n$ mostly
eventually also in the case $n= + \infty$
 
salut, @Astyx
 
Salut
 
6:57 PM
ah, doesnĖ™t matter much, i think i can find a way to handle even more general cases
when some efforts to prove NT result which would be an improvement failed, i started to do topology
 
"do"?
 
yes, meaning "to research"
 
sounds like the first step should be to study ...
 
@TedShifrin what's the possessive of "Tom and I"?
 
yes, i am learning basics, but i am always research-oriented, much of results can be improved, some slightly but much of them can, although i do not know how much i am going to be skilled in some particular branch
 
7:03 PM
@Leaky: "Tom's and my ..." most likely.
 
hmm
English is weird
 
I can also work around ... "that belongs to Tom and me."
 
@Masterphile Just study. Research means you just keep asking questions while being curious. Don't worry about the forefront of research. It is very far.
4
 
Well put, @MikeM.
 
@MikeMiller yes, but i am more researching than proving actually, either known theorems, or some new ones (of not-so-great importance), i know that that is usual, but i am worried about that, and i shouldn´t be, since i am not student nor a professor, i do it as a hobby, so, in some senses, i won´t perish if i do not publish, most probably not even if i publish, if you understand me well
 
7:18 PM
You seem to be saying you're immortal.
 
no :D, i just have no complete understanding of the word "perish", i tried much to understand that word but i cannot fully understand its meaning, English is not my native so some words to me are very difficult to fully grasp
i know some synonyms, but they are not equivalent fully to that word, it is a very interesting word
 
To perish means to die. The phrase "publish or perish" does not use it literally, but as an analogy: to lose your job is like to die. The reality is that you publish or you won't keep your job.
I understood you meant "I won't lose my job, whether or not I publish something unrelated". But I was taking you literally (as in to die), as a joke.
 
nah, "to perish" is different than "to die", it is one of the most interesting words that i know of, but i do not know much about genesis of English language, nor of its evolution to be competent as much as i would like to
 
It does mean "to die".
 
7:33 PM
Who knew I walked into english.stackexchange.com? xD
 
It used to be french/german ... too :)
 
yeah, it's the Wurst
 
Knockwurst
 
Reminders of mortality everywhere now. Me don't like it.
Everybody say wurst keep it going
 
did some of you read Elements, or at least, for example approx. 50-100 pages of it?
 
7:39 PM
I actually wrote it, @Masterphile
 
lol
 
@MikeMiller so we now know who is immortal here
 
Immortality jokes xD
 
well, i have read relatively much of it, why not, it is a classic
 
I actually do have Euclid here (in English translation), but I only read pieces of it when it's relevant to do so. :P
 
7:41 PM
Isn't it all available online?
 
I took a history of math course where we went over it.
@Khallil half a billion different translations, too.
 
A former student gave me the paperback book years ago to thank me ...
 
@anakhro yes, i actually wanted to ask Ted which one he has
 
I thought they sold that type of geometry ages ago and exchanged it for some algebra.
 
The devil was very happy with that deal, @NoName lol
 
7:43 PM
Mine is translated by Sir Thomas L. Heath.
 
@TedShifrin the best english scholar of greek math
 
Glad to hear that, @Eric !
I am woefully ignorant in such matters.
 
his two volumes on the history of greek mathematics are very very good
 
who decides if a mathematician will get the title "Sir"?
 
I think it's the queen for England and the head of state elsewhere
 
7:46 PM
@Khallil i observe that as not a joke
could be, Newton was "Sir" also, so it is relatively an old title
 
I don't know how much it meant back then but now it hardly means anything.
Sir Harry Kane lol
 
does anyone know how to put $m \times n$ into a subscript with mathjax? I can't figure it out.
 
i think Heath-Brown was (or is) a "Sir"
 
\text{stuff}_{m \times n} gives $\text{stuff}_{m \times n}$ @CaptainAmerica16
 
@Khallil Thanks!
 
7:52 PM
You're most welcome :)
 
it seems i thought wrong, or there is no information on Wikipedia, i really thought Heath-Brown is a "Sir"
 
How can I make $$\int_{\Bbb R^d}{\overline{\nabla u} \cdot xu(x)\over|x|^2}dx$$ look nice ?
 
Start with \displaystyle
 
What does it mean to look nice, @Astyx?
 
7:58 PM
@Khallil Cool :D I was using another one, but it didn
have all of the necessary info
 
As in, how can I solve it :P
 
You also need $u(x)$ at the front of the numerator or parentheses.
Oh.
 
I always use \text{ d}x for my differential forms but I'm slightly strange
 
I never bother, Khallil.
 
Oh x2 combo
 
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