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12:28 AM
Then display. Stop whining! @Rithaniel
12:56 AM
who extended the Gaussian distribution to 3d?
I guess it makes sense to revolve it about the y axis
and not the x-axis
because of a probability axiom
1:26 AM
I think the answer is not quite right. Since there is that constant C you can't say g(x) is gonna grow as fast or faster than f(x)
I am really confused in Big Oh notation
He should be saying $Cg(x)$ will grow as fast or faster than $f(x)$
the solution is to not do in-line math
Also that second answer is good it mention upper bound but then it says $2^x$ will always be bigger than $f(x)$? May be he forgot to include or equal.
Am I right or wrong?
please tag me if you got explaination
2:23 AM
So I just read that arbitrary coproducts exist in the category of compact topological spaces. What do they look like?
1 hour later…
3:49 AM
@Thorgott compact sounds false, are you sure it wasn't compactly generated?
I'm pretty sure the coproduct of infinitely many singletons doesn't exist in compact spaces
4:48 AM
what term would you use to describe a situation where your conclusion requires an additional, previously unstated axiom?
2 hours later…
6:51 AM
2 hr no reply ;_; ok
2 hours later…
9:02 AM
@Stupidquestioninc you point to the question, but ask about an answer. It would be better to link to the answer you are asking about. In any case, the constant of proportionality is not specified, so if $f(x)\le4g(x)$, we might also say $f(x)\lt5g(x)$.
So, "less than" is not wrong
Is there a surjective parametrization of the invertible real matrices (GL(n, R)) a la parametrization of the real special orthogonal matrices via the exponential of the skew-symmetric part of an arbitrary matrix?
@Rithaniel You have your choice: take up more vertical space or use the inline form (which is not wrong; e.g. some people use $\int\limits_0^1$ and others $\int_0^1$)
The exponential of arbitrary real matrices yields the identity component of GL(n, R).
@BalarkaSen However, if it's the one white petal among infinitely many red petals, it will be noticed.
9:38 AM
@robjohn I didn't mean that "less" than is wrong. Anyway I understand that answer contains some ambiguity to me. I focus too much in minor detail.
@robjohn are you professor?
@Stupidquestioninc used to be
@Stupidquestioninc Nope
@robjohn if somebody wants to join university by using special way which involves writing an article and you are a professor there to review it to see whether he/she is worthy to join the university you teach then what kind of article will make you think he is special, likes math, extraordinary, prodigy and worthy to join the university ?
please no sarcasm
At the university where I taught, the admission process was not up to individual professors. I imagine that a professor could write a letter of recommendation to admissions that might help, especially if the professor has high standing.
9:51 AM
So I need to convince some topnotch professor to write letter of recommendation? Dank. So what about the article?
@Stupidquestioninc It couldn't hurt. Any good letter of recommendation helps.
whether it's enough to get admitted is up to admissions
@robjohn OK so first I need to make relationship with a top notch professor then convince him to write a letter of recommendation. And for article in your opinion what will tell a student is special, likes math, extraordinary, prodigy and worthy to join the university ?
You don't NEED to do that. I just meant that the letters of recommendation help (the higher standing of the letter writer, the more the letter helps).
@robjohn OK thanks.
@robjohn do you know some stragety to convince some professor to write that kind of letter?
@Stupidquestioninc I'm sure different people are impressed by different things. Just present yourself and your work as best you can.
10:05 AM
Sounds like I need to write bunch of original thesis in different fields in math or solve unsolved problems.
Ok I will try my best. Thanks for you tips.
10:17 AM
@robjohn where do I find these so called high standing prof
That's something you'll need to do for yourself. It depends on the school you're trying to get into and what your area is.
@robjohn I'm half-joking the comment, not to worry
${}_0^1 \int$
$\int[0 \to 1]$
10:37 AM
@robjohn Well my teacher said no need to have recommend letter but I need to write articles where I and some professor working in math problem to see my perspective in math and some of my original ideas to show how much I love math. I don't know how to show my original ideas since for now I am mostly learning math since I am just a student and if I need to show I have finished understanding a book called a real analysis and other are work in progress.
It sounds i need to pick few good long problems in real analysis so what do you think which are good problems to start with?
Like long hard proofs?
The only one I have seen so far is construction of real number from rational and something related to blanc mange function
In math most students applying to programs (even PhDs) have never published an article
@AlessandroCodenotti what? don't you need to defend your thesis? and it should be original that no one has ever think of
Originality is not a big requirement for a masters thesis
i most certainly won't be producing anything new in my masters thesis rofl
almost certainly
in all but finitely many realities
@LeakyNun I'm gonna flag this for a moderator :P
10:44 AM
Math is already hard producing new results sounds kinda impossible
I wasted my weekend sleeping and now that it's 4 PM in Sunday I feel I need to get work done
just approach research like a pset and get it done in 1 week
so what does research actually mean if it is not to produce new original Mathematik?
I have heard master students conduct research so I don't know what does it means
research is when you get paid to do math
everyone's pipe dream
who gonna pay for a pure math research? may be you mean applied ones
10:56 AM
right, so there's no such thing as pure math research
this is the water, and this is the well. drink full and descend.
the horse is the white of the eyes and dark within
dear math students stop procrastinating and go get your job done ;-)
we will never get jobs so it wont ever get done
@BalarkaSen just listen to some Bach
11:02 AM
playing Bach, Bach-to-Bach?
@BalarkaSen harsh reality
I think math students are flexible enough to get good job
may be part time job
@Alessandro I'm sure I read "compact". Apparently the abstract nonsense reason is that reflective subcategories of cocomplete categories are cocomplete.
thorgott you will never get a job
11:05 AM
@BalarkaSen weekends aren't for work
someone can get me help
a wonderful morning/midday to you as well
do you happen to know an explicit description of the coproduct in the category of compact topological spaces
@ACuriousMind yeah, the weekdays are too tiring. but then i have to find some time to do anything outside assignments
@Thorgott Nope, sorry. I'm very surprised that coproduct of points exist
11:08 AM
so am I
guess I'll have to chase these abstract nonsense proofs and see how this works explicitly
apparently it has something to do with the Stone-Cech compactification (it's the left adjoin of the inclusion into Top)
does the 'compact' change anything? The coproduct of topological spaces is just the disjoint union.
Yeah, but Thorgott wants to say it for an infinite collection of topological spaces; the disjoint union of infinitely many points won't be compact
yeah, because the disjoint union of an infinite family of non-empty compact spaces isn't compact anymore
for finite collections, the coproduct in Comp is just good old disjoint union
ah, why can't people ever be content with finite stuff
what does it mean by isomorphic groupoids are essentially the same?
does it means carrier and binary operation are same?
11:11 AM
Ok I'll spend the next hour staring at the proof of Weierstrass preparation theorem
equality ?
didn't you already do that Malgrange stuff
I never went through the proof, it's too weird.
I want to give a talk on it so I decided it's a good idea to understand the proof
sounds like a good idea
11:25 AM
Understanding the proofs you explain is for rookies
@Stupidquestioninc applying for the program
Just speak fast so nobody has a chance to ask questions and then pack your things and leave before the end
@EdwardEvans Proof by intimidation works well
Yeah that too, that’s why I always pick talks near the end of the reading course
it is well-known that
11:27 AM
It is well known that the zeros of the zeta function have real part 1/2
a standard folklore result assures us that
This live grenade allows us to conclude that
11:44 AM
Oh fudge
I got confused because of a persistent typo in the book I am reading
That's so maddening
11:56 AM
Ok I have an argument for why the coproduct of points doesn't exist @Thorgott
I warn you that I slept 3 and a half hours tonight and I just moved between different countries, so it will be nonsense
Let $\{x_n\}_{n\in\Bbb N}$ be our family of singletons and suppose that $K$ is their coproduct in compact spaces. Passing to a subsequence we can assume that $x_n$ converges in $K$ (by compactness of $K$)
But now by universal property to give a function from $K$ to some compact $X$ we only need to fix a sequence in $X$
Find $y_n$ in $X$ not convergent, then the map sending $x_n\to y_n$ doesn't extend to a continuous map $K\to X$, which is a contradiction
@BalarkaSen I'm going to sign the contract tomorrow, I'll let you know how that feels
lmao yes feel free
@Thorgott To talk about Stone-Cech you must restrict to something like (completely?) regular spaces
@BalarkaSen evidently, he'll tell you tomorrow if he feels free...
Does that coproduct stuff make sense to you Balarka?
Can you actually assume that $x_n$ converges in $K$? Of course, you can find a convergent subsequence, but then the coproduct of that convergent subsequence may be different, so what guarantees that subsequence convergent in $K$ also converges in its own coproduct?
12:03 PM
Say $x_{n_k}$ is the convergent subsequence, define a map into $X$ sending $x_{n_k}$ into a divergent sequence and the rest of the $x_i$ to whatever
Oh, I just noticed that I did in fact misread something. It's claimed that arbitrary coproducts exist in the category of compact Hausdorff spaces, not just compact spaces (but of course, singletons are Hausdorff)
My example still works (assuming it works for arbitrary spaces to begin with lol)
hmm, your argument does seem convincing
wish me luck, I'll try figuring out the abstract nonsense argument and see if I can make sense of this
Good luck, I don't know what reflective subcategory means, so I can't help you I'm afraid
I'm familiar with the Stone-Cech compactification if you have questions about that though
reflective = full + inclusion has left adjoint
the left adjoint here is supposed to be Stone-Cech
12:20 PM
@Thorgott That I agree with
So how does one prove that a reflective subcategory of a cocomplete category is cocomplete?
Hmm no wait my argument can't be right, even though I don't see which part is wrong
Because otherwise the same argument would show that $\beta\Bbb N$ can't exist, which is nonsense
12:40 PM
I understand Weierstrass preparation now
That's cool but do you understand reflective subcategories?
Me neither
Ok, next thing on my schedule is point processes. Gotta do probability homework, write part 2 on blogpost on Poisson processes, and watch a lecture by Ol'shanskii
12:46 PM
Ah, the argument goes as follows. Let $\mathcal{A}$ be a reflective subcategory of $\mathcal{B}$. Pick a small category $\mathcal{D}$ and a functor $H\colon\mathcal{D}\rightarrow\mathcal{A}$. Let $i\colon\mathcal{A}\rightarrow\mathcal{B}$ be the inclusion, then $iH\colon\mathcal{D}\rightarrow\mathcal{B}$ has a colimit $(L,(s_D)_{D\in\mathcal{D}})$ by hypothesis.
Let $r\colon\mathcal{B}\rightarrow\mathcal{A}$ be the left adjoint to $i$, then $(rL,(rs_D)_{D\in\mathcal{D}})$ is the colimit of $riH\colon\mathcal{D}\rightarrow\mathcal{A}$ (left adjoints preserve colimits), but $ri$ is naturally
3 hours of work up next
Speaking of Stone-Cech compactifications I'm trying to understand Cech-complete spaces, also with little success
so the coproduct in compact Hausdorff spaces is the Stone-Cech compactification of the disjoint union
Yuck ^
Yes that makes sense to me
This is all a complicated way of rewriting the property of $\beta X$ that any continuous map $X\to K$ with $K$ compact Hausdorff extends to $\beta X$
12:51 PM
so now what's wrong with your argument
Ah derp
Compact does not imply sequentially compact
The $x_n$ will have a convergent subnet, but that is not an issue now
that hurt
minecraft damage noises
12:53 PM
I guess my argument now is a weird way of showing that $\beta\Bbb N$ is not metrizable then
Nor sequentially compact
1 hour later…
2:04 PM
can a Lorenz surface be a Lorentz surface?
Apparently there's been claims that Satoshi Nakamoto, inventor of bitcoins, is Shinichi Mochizuki
Intergalactic currency yo
mutually alien copies of bitcoins
2:36 PM
Is anyone here familiar with basic measure theory?
Q: Using the principle of good sets to prove $\sigma(\mathcal{A}) = \mathcal{B}$

ClarinetistI have read through https://math.stackexchange.com/a/2854237/81560, but I'm not sure if I'm applying the principle correctly and how its implications work. Let $X$ be a set. Say that I, for example, have a $\sigma$-algebra $\mathcal{B}$ of subsets of $X$, and I have a collection of subsets of $X$...

I am really struggling to understand this principle in general.
I don't think that's a functional strategy
so they're trivially $\sigma$-algebras, but that doesn't tell you anything
Thanks for the input @Thorgott. So it looks like it really depends on the situation
The actual problem I have is this (unfortunately the notation slightly differs): show that if $\mathcal{B}$ is a $\sigma$-algebra of subsets of $X$ and $E$ is a subset of $X$, then $\sigma(\mathcal{B} \cup \{E\}) = \{(A \cap E) \cup (B\setminus E): A, B \in \mathcal{B}\}$. I've already shown the RHS to be a $\sigma$-algebra, but I'm not sure what to do from here.
I'm not interested in a complete solution. I'm just not aware of the next step I should take.
Trying to prove that the polynomial ring of a UDF is a UFD. I did this once before with ACCP, but now I'm trying to do the "factorization in polynomials in the quotient field" in tandem with Gauss's lemma, and it's giving me a headache
2:45 PM
Also, I have no idea what strategies people use in general for solving such a problem.
I'd advise for the general strategy to show two sets are equal, i.e. to show that each is a subset of the other.
The RHS being a subset of the LHS is straightforward.
I tried that approach, but yeah, it's the other direction that's getting me.
The LHS being a subset of the RHS is implied by the RHS being a $\sigma$-algebra containing $\mathcal{B}\cup\{E\}$, which you see to have shown already
Oh, I haven't shown it contains $\mathcal{B} \cup \{E\}$ yet
Then it's fairly straightforward there, thanks
Ugh, these containments are so confusing
I actually thought your statement above implied containment in the other direction @Thorgott
No, I'm wrong
Actually, what I thought was that sets in the form of the RHS set would be contained in a sigma-algebra consisting of $\mathcal{B} \cup \{E\}$ implied RHS contains LHS
I'm doubting myself there
$\sigma(\mathcal{B}\cup\{E\})$ is, by definition, the smallest $\sigma$-algebra containing $\mathcal{B}\cup\{E\}$, so if the RHS is a $\sigma$-algebra containing these, it contains the LHS
2:53 PM
@Thorgott Right, I understand that, so... why exactly does containment in the other direction hold?
cause a the elements of $\mathcal{B}$ and $E$ are contained in the LHS and $\sigma$-algebras are closed under intersections, unions and complements
and thus we've shown that the RHS should be in the LHS because the RHS exhibits operations that would be valid in a $\sigma$-algebra containing sets in $\mathcal{B}$ and $E$, correct? @Thorgott (and then use minimality of $\sigma$)
I think that's it
3:12 PM
@Rithaniel Hmm, how did this go again. Let $R$ be a UFD and $K=\operatorname{Frac}(R)$. Let $P\in R[X]$, then $P\in K[X]$ and there it factors as $P=P_1\cdot...\cdot P_n$ with $P_1,...,P_n\in K[X]$ irreducible. Rewrite this as $c(P_1)\cdot...\cdot c(P_n)\cdot (P_1/c(P_1))\cdot...\cdot(P_n/c(P_n))$ ($c$ denotes content) up to a unit or whatever.
Gauß Lemma says that $c(P_1)\cdot...c(P_n)=c(P_1\cdot...\cdot P_n)=c(f)$ is in $R$ since the coefficients of $f$ are in $R$ and the polynomials $P_1/c(P_1),...,P_n/c(P_n)$ are primitive, hence in $R[X]$ and irreducible there as well. Now $c(P_1)\cdot
@Clarinetist Yes, but you don't need minimality for this inclusion
@Thorgott Thanks
Yeah, that's the general gist of it, but the whole thing of c(f) being in R isn't clear to me, so I'm not entirely comfortable using it
Let's see. $c(f)=\prod \pi^{\nu_{\pi}(f)}$, where $\pi$ runs through a representative system of all primes up to units and $\nu_{\pi}$ is the $\pi$-valuation, right
So $c(f)\in R$ is equivalent to demanding $\nu_{\pi}(f)\ge0$ for all $\pi$, which in turn is equivalent to demanding $\nu_{\pi}(a_i)\ge0$ where $a_i$ are the coefficients of $f$
But this is of course satisfied if $a_i\in R$, pretty much by definition
3:27 PM
Ah, wild. I didn't know that approach
I think I can work with that
(But first, chess break)
4:06 PM
Suppose $\mathbb{V}$ is a finite dimensional vector space and that $L: \mathbb{V} \rightarrow \mathbb{V}$ is a linear operator. We denote by $L^{2}$ the linear operator on $\mathbb{V}$ definfed by $L^{2}(\vec{v})=L(L(\vec{v}))$ for every $\vec{v} \in \mathbb{V} .$ We say that $L$ is $i d e m$ potent if $L=L^{2} .$ Suppose that $L$ is idempotent and let $\mathbb{W}=\operatorname{Range}(L) .$ Prove that $\left.L\right|_{\mathrm{w}}: \mathbb{W} \rightarrow \mathbb{W}$ is an isomorphism.
in fact, you can prove that $L\vert_W=\operatorname{id}_W$
4:41 PM
Is it possible to have a group $G$ in which there doesn't exist a $y$ such that $x\circ y = z$ for $x,y,z\in G$? In words there doesn't exist an element that takes $x$ to $z$?
@Charlie No, any such equation will always have a solution in a grou
In fact, a unique solution which is easy to find
@Charlie, $x (x^{-1}z) = z$
Ok great that's what I was expecting, I couldn't think up a convincing proof
@JoeShmo ah I see, tyvm
5:11 PM
Let $f\colon A\rightarrow B$ be a homomorphism of commutative rings. Via restriction of scalars along $f$, we view $B$ as $A$-module. Assume $1\otimes_Ab=b\otimes_A1$ for all $b\in B$. Let $g,h\colon B\rightarrow C$ be two homomorphisms of commutative rings such that $g\circ f=h\circ f$. Via restriction of scalars along $g\circ f=h\circ f$, we view $C$ as $A$-module.
Then the map $B\times B\rightarrow C,\,(b,b^{\prime})\mapsto g(b)h(b^{\prime})$ is $A$-bilinear, hence factors through an $A$-linear map $\varphi\colon B\otimes_AB\rightarrow C$. We then have $g(b)=\varphi(b\otimes_A1)=\varphi(
Ok, I believe this works, but why is that tensor condition also necessary for $f$ to be an epi
Oh wait, you just look at the inclusions $B\rightarrow B\otimes_AB$ in each factor. They agree when composed with $f$ since we're tensoring over $A$, so they're equal by $f$ being epi, which is precisely that condition.
5:47 PM
Don't really know where to go for this question: Suppose that $\mu,\nu$ are finite positive measures on $(X,A)$ and let $\mu\ll\nu, \nu\ll\mu$. If $\lambda = \mu + \nu$, then show $\frac{d\nu}{d\lambda} \in (0,1)$ a.e..
I notice that $d\mu/d\nu$ and $d\nu/d\mu$ are inverses a.e.
But I am kind of at a loss for how to use finiteness or even start bounding $d\nu/d\lambda$
I was thinking maybe $\int \frac{d\nu}{d\lambda}\,d\lambda = \nu(X) < \infty$ might be useful?
Let $A=\{d\nu/d\lambda\ge1\}$. Integrate $d\nu/d\lambda$ over $A$.
With respect to $\lambda$?
To yield $\nu(A)$.
6:04 PM
I want to show $\nu(A) = 0$ in this case.
I will think about it a bit I guess. Thanks @Thorgott
6:28 PM
@Thorgott I think I asked you 12 times by now, but where in Germany do you study precisely?
lol I only remember once
in Frankfurt
6:43 PM
one quick algebra question: given some $(\mathbb{Z}/n)^\times$ can I quickly determine if it's cyclic?
could someone sanity check this statement is true for me ( i think it should be): R U {+infinity,-infinity} with the order topology (setting +infinity to be larger than every other element and -infinity smaller than every other element) is homeomorphic to R_+ U {+infinity} as the one point compactification of R_+, where R_+ is the non-negative real numbers with the euclidean subspace topology
@Stupid for prime n it is cyclic, and apparently Gauss proved that it is only cyclic when n=1,2,4, p^k or 2p^k when p is an odd prime - see en.wikipedia.org/wiki/…
so if you can quickly determine whether n is of one of those forms, you can quickly determine if that multiplicative group is cyclic
@Thorgott ah ok
@porridgemathematics aaaah forgot about that number theory fact
I was wondering because I just moved to Münster today, but that's even further away than Bonn lol
@porridgemathematics thanks just got an ahaaa moment :)))
6:52 PM
no worries
ah, yeah
I feel like Münster is kinda far away from everywhere else in Germany to be honest
@porridgemathematics yes, both are homeomorphic to $[0,1]$
7:08 PM
@AlessandroCodenotti nice
Thorgott thanks!
@EdwardEvans I'm still waiting for a nice set theory conference in Heidelberg to use it as an excuse and come there
@Thorgott I have a paralysis with that problem. I tried to play around with the $\frac{d\nu}{d\mu}\frac{d\mu}{d\nu} = 1$, but nothing novel seemingly popped out.
It makes sense though that if $\nu(A) = 0$, then $\lambda(A) = 0$.
Both via that integral, and intuitively, knowing $\mu\ll\nu$
7:24 PM
How can I sample from $\mathrm{GL}_+(n, \mathbb{R})$ uniformly at random? I'm parameterizing through the matrix exponential.
@user76284 Bertrand's paradox
Apart from that, I think you can't, because it isn't compact
Oops I messed up my statement.
"Parametrizing through the matrix exponential"? exp : M_n(R) -> GL+(n, R) is not surjective
You cannot just pushforward the Lebesgue measure on M_n(R)
Forget the uniformly at random part.
@LeakyNun Is there any content to that paradox
7:28 PM
@BalarkaSen I believe it is surjective?
Do you have a counterexample?
@BalarkaSen uniform has no meaning
@LeakyNun uniform on [0, 1] to me means the lebesgue measure on [0, 1]
What I originally meant was sample from $SO(n)$ uniformly at random. And @LeakyNun I mean with respect to the Haar measure.
oh I guess if you have a compact topological group...
yeah that was what I was missing
@user76284 yes, it's not
7:30 PM
you do have a notion of uniform coming from the topological group structure?
translation invariance
@user76284 I guess you could choose each column?
ah but how do you choose the first column
how do you choose from the sphere uniformly
Lebesgue measure dude
@BalarkaSen Ah nevermind. You're right.
its the uniform random variable on the sphere
7:32 PM
I bet the $\mathbb{C}$ version is.
I believe they mean actually writing a program to choose an element of SO(n) uniformly?
Yup, @user76284
@LeakyNun you discretize the measure
yes that's what I was asking about
how do you do that
On the other hand, $\exp : \mathsf{skewsymmetric} \rightarrow \mathsf{special orthogonal}$ is surjective, I think?
exp is always surjective for compact Lie groups
"orthogonal" lol... i need sleep
7:35 PM
"QR decomposition of independent normally distributed random entries"
with positive diagonal
as Leaky pointed out, the Haar measure on SO(n+1) is the same as choosing a random point on S^n, then choosing a random point on the orthogonal S^(n-1), then on the orthogonal S^(n-2) inside that, etc
I wonder what's a distribution on skew-symmetric matrices that yields the uniform distribution on SO(n)?
I don't know a good description
Formally you can pullback the Haar measure by exp
Yeah I wonder what it is.
Under the principal logarithm.
So $A \mapsto \exp(A - A^\top) : M_n(\mathbb{R}) \to SO(n)$ is a surjective parametrization of the latter from $M_n(\mathbb{R})$, right?
And differentiable.
Is there anything similar for $\mathrm{GL}_+(n, \mathbb{R})$ rather than $\mathrm{SO}(n)$? For $\mathrm{GL}(n, \mathbb{R})$ I think it's impossible because it's disconnected.
What do you mean by "similar"
7:46 PM
Also, does the image of $M_n(\mathbb{R})$ under $\exp$, which is a subgroup of $\mathrm{GL}_+(n, \mathbb{R})$, have a name?
@BalarkaSen I mean a differentiable surjection or something like that.
I want to use unconstrained optimization on something like $M_n(\mathbb{R})$ to perform optimization on $\mathrm{GL}_+(n, \mathbb{R})$, if that makes sense.
Sorry, I don't know.
Maybe someone can help though
@anakhro Estimate $\int_A\frac{d\nu}{d\lambda}d\lambda$ by the definition of $A$
I guess the first question is whether an almost-everywhere differentiable surjection exists from $\mathrm{M}(n, \mathbb{R})$ to $\mathrm{GL}_+(n, \mathbb{R})$
@Thorgott oh my
That was easy.
Thank you, @Thorgott !!!
Is anyone here familiar with Smith-Normal form of a matrix (lets say over a PID)?
7:54 PM
np :P
@Thorgott may I ask what your thought process was roughly when you approached the problem?
I imagine you just thought of finding the measure of that set to be 0 because it said a.e. is < 1?
And then just fiddled with it until you found that?
what's the conjugate of a complex valued distribution?
a $\overline{\text{complex valued distribution}}$
@user76284 why is it a subgroup?
but is there a concise way to write $\langle T_{\bar{f}}, \phi \rangle = \langle T_f, \cdot \rangle$?
$\int_A\frac{d\nu}{d\lambda}d\lambda=\nu(A)$ defines the Radon-Nikodym derivative, so I felt like that identity should be the key. Then I thought along the lines of contradiction, what happens when $\frac{d\nu}{d\lambda}\ge1$ and saw that then I could estimate that integral. Then I considered the specific set $A$, because the statement is equivalent to showing it has measure $0$ and the rest came pretty much together by itself.
8:06 PM
@LeakyNun Hmm maybe not.
@Thorgott Ah, I see, thanks!
Q: Non surjectivity of the exponential map to GL(2,R)

Andrea OrtaI was asked to show that the exponential map $\exp: \mathfrak{g} \mapsto G$ is not surjective by proving that the matrix $\left(\matrix{-1 & 0 \\ 0 & -2}\right)\in \text{GL}(2,\mathbb{R})$ can't be the exponential of any matrix $A \in \mathfrak{gl}(2,\mathbb{R})$. My proof (edited) Lemma: ...

A: Is the image of a Lie-subalgebra under the exponential map a subgroup?

Jack LeeThis is not true in general. The image of $\mathfrak h$ generates a subgroup of $G$, but the image might not be the whole subgroup. For example, if $\mathfrak g$ is the set of all $2\times 2$ real matrices, and $\mathfrak h$ is the subalgebra of trace-free matrices, then the smallest subgroup of ...

@LeakyNun Yeah that I already knew, since the image under exp can only have positive determinant.
@user76284 that matrix has positive determinant
Sorry, I was referring to the question title. Yes, the question I linked to also has an example.
8:13 PM
@LeakyNun Why did "a" escape conjugation?
hmm, good point
I suppose $\langle T_{\bar{f}}, \phi \rangle = \overline{\langle T_f, \bar{\phi}\rangle}$ will do..
its because im real
I guess I can ask the following question: For some $k$, does there exist a differentiable surjection from $\mathbb{R}^k$ to $\mathrm{GL}_+(n, \mathbb{R})$?
8:15 PM
hi Ted!
Happy New Year
Hi, a @Balarka, @JoeShmo. Yes, to you too. It can't be a worse year than we have been having :D
Apparently any matrix in $\mathrm{GL}_n(\mathbb{R})^+$ is a product of $n$ matrix exponentials: arxiv.org/pdf/2007.09651.pdf#page=3.
That's a contentless statement. Image of the exponential map always generates the group.
Ah, $n$. Yes, that's something.
So there exists a surjection $\mathbb{R}^{n \times n \times n} \to \mathrm{GL}(n, \mathbb{R})^+$.
8:31 PM
is it gonna be differentiable
Well, it's going to be an open subset, not the whole thing.
Oh, I guess you can make it map the whole thing. Duh.
of course, just any surjection exists for cardinality reasons
Right, I should say a "nice" surjection, where the level of "niceness" is yet to be determined :)
Continuous? Almost everywhere differentiable? Differentiable? Continuously differentiable?
Sounds smooth to me, if it's a product of exponentials.
That would be my guess.
8:53 PM
Why does a circle, as a manifold, require two charts?
Don't jinx it, but I am looking forward to a better year, too.. :)
I see why coordinates on the sphere fail at the poles, but why can't we cover the circle with one chart?
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