@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 $f(x)\lt5g(x)$.
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 $\int_0^1$)
@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 ?
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.
@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 ?
@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
@Alessandro I'm sure I read "compact". Apparently the abstract nonsense reason is that reflective subcategories of cocomplete categories are cocomplete.
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
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?
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)
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
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 …
I 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 $X$...
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
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
$\sigma(\mathcal{B}\cup\{E\})$ 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
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 $\sigma$)
@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
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
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.
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 $z$?
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.
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?
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 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 $\mu\ll\nu$
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
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.
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.
I guess the first question is whether an almost-everywhere differentiable surjection exists from $\mathrm{M}(n, \mathbb{R})$ to $\mathrm{GL}_+(n, \mathbb{R})$
$\int_A\frac{d\nu}{d\lambda}d\lambda=\nu(A)$ 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.
I 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: ...
This 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.
I guess I can ask the following question: For some $k$, does there exist a differentiable surjection from $\mathbb{R}^k$ to $\mathrm{GL}_+(n, \mathbb{R})$?
I 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: ...
