TITLE: Similarities between non-Euclidean geometries
QUESTION [5 upvotes]: So I've looked into Euclidean, spherical/ellpitic, and hyperbolic geometry, and found some possible similarities. I'm not much of an expert, so I can't really verify them for myself. I'd like to know which are actual properties of those geometries, and which are just coincidences, and for the coincidences, what is the true relation?
To keep things simple I'll only consider 2 coordinate axes $x,y$ and an extra axis for the model $z$. I'll call the curvature $K$ and define $k=\sqrt{K}$.

Some of these seem to only apply for $K\in \{-1,0,1\}$, so it may be necessary to split it into sign and magnitude, or sign of curvature and radius of curvature. It also means these may just coincidentally work for $K\in \{-1,0,1\}$ and not any other values.
The true shape or correct model is defined like so:
$$z^2=1-K(x^2+y^2)$$
The origin is always the same:
$$O=(0,0,1)$$
The distance from the origin to a point is proportional to the area it'd sweep out:

Projecting from $(0,0,0)$ to the plane $z=1$ will preserve straightness (Gnomonic/Beltrami-Klein). Projecting from $(0,0,-1)$ to the plane $z=1$ will preserve angles (Stereographic/Poincare).
Using the complex definitions for trigonometric functions...
$$\sin(z)=\frac{e^{iz}-e^{-iz}}{2i}$$
...we define a special curvature specific function:
$$\sin_K(x)=\frac{1}{k}\sin(kx)$$
We can verify this produces $\sin_1(x)=\sin(x)$, $\sin_{-1}(x)=\sinh(x)$, and $\sin_0(x)=x$. $K=0$ needs to be evaluated using a limit.
It also appears to be a kind of inverse of distance to origin:
$$\text{distance}[(\sin_K(v),0,\cdots)]=|v|$$
The circumference and area of a circle and the surface area and volume of a sphere:
$$S_1(r)=2\pi\sin_K(r)$$
$$V_2(r)=4\pi\sin_K(\frac{1}{2}r)^2$$
$$S_2(r)=4\pi\sin_K(r)^2$$
$$V_3(r)=\frac{1}{K}\pi(2r-\sin_K(2r))$$
For volume of sphere, $K=0$ needs to be evaluated using a limit.

REPLY [4 votes]: A very nice writeup! HyperRogue uses formulas very similar to these to deal with all geometries uniformly, although it only uses $K=-1,0,1$. (I have not checked other values of $K$ in detail, also your volume formulas, but these are simply integrals of circumference and surface area.)
There are also more formulas that could be added, for example: (also not checked for other values of $K$ -- please check if you want to use them for other values)

$\cos_K$ which is the integral of $K\sin_K$, and $\cos_K 0=1$: this is constant 1 in Euclidean, cos for $K=1$, and cosh for $K=-1$. I would rather define $\sin$ and $\cos$ as the only functions which satisfy the system of differential equations: $\sin_K(0)=0$, $\cos_K(0)=1$, $\sin'_K(x)=\cos_K(x)$ and $\cos'_K(x)=-K\sin_K(x)$. This definition avoids complex numbers (a good thing because you do not have to say which root of $k^2=K$ to take).
if you have a line and a segment of length $1$ on it, the matching segment of an equidistant of radius $r$ is $\cos_K r$
the matrix of an isometry which moves the origin $x$ units to the right, in the  "true shape" model: $\left(\begin{array}{ccc}\cos_K x&0&\sin_K x\\0&1&0\\K\sin_K x&0&\cos_K x \end{array}\right)$
the distance between two points $(x_1,y_1,z_1)$ and $(x_2,y_2,z_2)$ in the "true shape" model: first you compute the distance in the three-dimensional space $d^2=|x_1-x_2|^2+|y_1-y_2|^2+K|z_1-z_2|^2$, and then the geodesic distance is $2{\rm asin}_K(d/2)$
Pythagorean theorem: $\cos_K a \cos_K b = \cos_K c$ (you get the Euclidean version in the limit)
more models: orthographic projection of the sphere and the Gans model of the hyperbolic plane are obtained the same way, Mercator projection of the sphere and the band model of the hyperbolic plane are obtained the same way (though you need to solve an integral here which makes the end formulas less similar), etc.


TITLE: Partial square root of an integer
QUESTION [3 upvotes]: I am looking for an efficient algorithm to compute what I'd call the "partial square root" of an integer. In more formal terms:
Given a positive integer $k$, find the pair of positive integers $(r,n)$ with the smallest $n$ such that $k=r^2n$.
Naively, one could start with $(r,n)=(1,k)$, iterate over all $p \in \{2,3,5,7,...,2i+1\}$ (as a stand-in for all the primes), and while $n/p^2$ is integer, compute $(r',n')=(rp,n/p^2)$, but something tells me that there should be a more more efficient way to accomplish this.

REPLY [2 votes]: One way to accomplish the task is to factor $k=\prod p_i^{a_i}$ where $p_i$ are primes that divide $k$.
Then, for each instance in which an exponent $a_i$ is odd, call them $a_j$, use the corresponding prime base to form the product $n=\prod p_j$.
The quotient $\frac{k}{n}$ will then have only prime factors with even exponents, and hence will be a square, $r^2$.
You cannot make $n$ smaller by removing one or more of the $p_j$ from its definition, as that will leave one or more of the remaining $p_i$ in $\frac{k}{n}$ with an odd exponent, and that cannot be a square.


TITLE: Quantum extremal surface and black hole evaporation inquiry
QUESTION [3 upvotes]: Recently there has been progress made in the black hole information paradox by using the tools of AdS-CFT correspondance. Specifically, the Page curve for an evaporating black hole has been successfully reproduced with the use of the so-called quantum extremal surfaces. The entaglement entropy of the black hole is given by:
$$S_{gen}(X)=\frac{A(X)}{4}+S_{quantum fields}(\Sigma_{X}) $$
where $X$ is the quantum extremal surface and $\Sigma_{X}$ is the volume bounded by $\Sigma$ and a cutoff surface far from the black hole. Initially, the QES is at the singularity so there is no area contribution. Thus the entaglement entropy grows as radiation is emitted and crosses the cutoff surface. At the page time, the QES lies just behind the horizon. Then, it is claimed that the surface term dominates the entropy and since that is decreasing, the Page curve is reproduced.
My question is: why does the surface term dominate at the Page time? At https://arxiv.org/abs/2006.06872 the authors say that the entropy of the quantum fields is small because it does not capture many Hawking quanta. Is this because most of them have escaped the cutoff surface?
Thank you.

REPLY [1 votes]: First of all you are looking at the calculation of the generalised entropy of the black hole radiation inside the cutoff region. The paper then goes to do the calculation of the region beyond the cutoff, employing the island trick, and showing that the two results match.
So let's focus on the region inside the cutoff.
Here the generalised entropy gets mainly two kinds of contributions: vanishing surface and non-vanishing surface contributions. We are interested in taking the minimum of these two and it will turn out that the vanishing surface is the minimum before the Page time, while the non-vanishing surface is the minimum after the Page time.
The vanishing surface term should be clear: it starts zero since the quantum extremal surface X can be shrunk to radius zero and the quantum state is pure, so the fine grained entropy is zero. Then it increases monotonically, due to the entanglement of the interior Hawking quanta on the region $\Sigma_X$ bounded by X and the cutoff surface, which is crossing the horizon, and those that escaped.
The non-vanishing surface term is where your doubt lies. Here the area term is non zero, since the surface X is close to the horizon, which is close to maximal, since the fine grained entropy is always smaller than the coarse grained entropy given by the usual Bekenstein area entropy.
Now the region $\Sigma_X$ bounded by X and the cutoff surface is much smaller though, since the horizon is not too far from the cutoff. Therefore the entanglement term of the region $\Sigma_X$ is negligible. The area term (relative to the non-vanishing surface contribution) is therefore dominant at all times, not just at Page time, but is monotonically decreasing, since the area of the black hole is decreasing due to the evaporation.
To prove the stability of the non-vanishing surface the authors further show how if we were to move X in the ingoing null direction the entanglement term would first decrease and then increase when the entanglement with outgoing quanta outside the black hole region kicks in.


TITLE: Is this true: $\sum_{n=1}^\infty a_n = \sum_{n=2}^\infty a_{n-1}$ ??
QUESTION [2 upvotes]: Does this equality hold?
$$\sum_{n=1}^\infty a_n = \sum_{n=2}^\infty a_{n-1}$$

REPLY [0 votes]: Recall that, by definition
$$
\sum_{n=1}^{+\infty}a_n=\lim_{N\to+\infty}S_N
$$
where
$$
S_N:=\sum_{n=1}^{N}a_n\;.
$$
It's clear that
$$
\sum_{n=2}^{N}a_{n-1}=S_N-a_N=S_{N-1}\;.
$$
Since $\lim_{N\to+\infty}S_N=\lim_{N\to+\infty}S_{N-1}$ you can conclude they are equal.


TITLE: Evaluate the Jacobi Symbol $\frac{1111}{8193}$
QUESTION [2 upvotes]: Evaluate the Jacobi Symbol $\dfrac{1111}{8193}$
The way I approached this was:
$$\frac{1111}{8193} = \frac{416}{1111}=\frac{279}{416}=\frac{137}{249}= \frac{5}{137}=\frac{2}{5}=-1$$
And following the Jacobi properties $$\left(\frac{2}{n}\right)=(-1)^{\frac{n^2-1}{8}}= -1$$ if $n=3,5.$
Is my computation correct?

REPLY [1 votes]: It looks correct. As an alternative,
$$\left(\frac{1111}{8193}\right)=\left(\frac{1111}{3}\right)\left(\frac{11}{2731}\right)\left(\frac{101}{2731}\right)=\left(\frac{3}{11}\right)\left(\frac{4}{101}\right)=\left(\frac{11}{3}\right)=-1. $$


TITLE: Introducing probability vectors
QUESTION [2 upvotes]: When I studied probability I did it in a classical manner. My course used Loève's book on probability theory and I used Allan Gut's book for a modern version of it. Now I'm following a course in information theory and the notation, terms and concepts change so that it is difficult for me to connect these concepts with what I had.
I was given the following problem:

What is the minimum value of $H(p)$ as $p$ ranges over the set of
  n-dimensional probability vector.

I looked on Wikipedia and there it says that

A probability vector or stochastic vector is a vector with
  non-negative entries that add up to one.

My question is how does this concept relates to probability theory. What is a probability vector used for? Is it an extended manner of denoting the probability mass function of a random variable which takes a finite number of values?

REPLY [2 votes]: Let $X$ be a random variable with finite support, i.e. the possible values of $X$ are $x_1,\dots,x_n$. Let $p_i=P(X=x_i)$. Strictly speaking, the distribution of $X$ is the measure $\sum_{i=1}^np_i \delta_{x_i}$.
Once you know the support of $X$, in order to describe its distribution, you can just give the vector $(p_1,\dots,p_n)$, which has non-negative entries and adds up to $1$.
On the other hand, if you choose such a vector, it determines a probability distribution on $\{x_1,\dots,x_n\}$.


TITLE: Finding minimum of trigonometric function
QUESTION [5 upvotes]: Let $$v=\sqrt{a^2\cos^2(x)+b^2\sin^2(x)}+\sqrt{b^2\cos^2(x)+a^2\sin^2(x)}$$
Then find difference between maximum and minimum of $v^2$.
I understand both of them are distance of a point on ellipse from origin, but how do we find maximum and minimum?
I tried guessing, and got maximum $v$ when $x=45^{o}$ and minimum when $x=0$, but how do we justify this?

REPLY [1 votes]: Let $v=\sqrt{a^2\cos^2(x)+b^2\sin^2(x)}+\sqrt{b^2\cos^2(x)+a^2\sin^2(x)}$
$v^2=a^2+b^2+2\sqrt{\left(a^2\cos^2(x)+b^2\sin^2(x)\right)\left(b^2\cos^2(x)+a^2\sin^2(x)\right)}$
$v^2=a^2+b^2+2\sqrt{a^2b^2\left(\cos^4(x)+\sin^4(x)\right)+\left(a^4+b^4\right)\cos^2(x)\sin^2(x)}$
$v^2=a^2+b^2+\sqrt{4a^2b^2\left(1-2\sin^2(x)\cos^2(x)\right)+\left(a^4+b^4\right)\sin^2(2x)}$
$v^2=a^2+b^2+\sqrt{4a^2b^2-2a^2b^2\sin^2(2x)+\left(a^4+b^4\right)\sin^2(2x)}$
$v^2=|a|^2+|b|^2+\sqrt{4a^2b^2+\left(a^2-b^2\right)^2\sin^2(2x)}$
To find minimumvalue of $v^2$
put $\sin^2(2x)=0$
to obtain $v^2=\left(|a|+|b|\right)^2$. So
$v_{min}=|a|+|b|$
And to obtain maximum value of $v^2$ put $\sin^2(2x)=1$
$v_{max}=\sqrt{2(a^2+b^2)}$


TITLE: Alternate proof for Arzela-Ascoli
QUESTION [0 upvotes]: Im trying to finish a beautiful excercise, which consist of giving an alternate proof for the following corollary of Arzela-Ascoli´s Theorem.
Given $X,Y$ metric spaces, $X$ compact, $Y$ complete, and let $\{f_n\}_{n\in \Bbb{N}} \subset C(X,Y)$, be an equicontinuous family of functions such that for every $x\in X$, $\{f_n(x)\}_{n \in \Bbb{N}}$ is relatively compact. Then by Arzela-Ascoli, there exist a subsequence of $(f_n)$ that converges uniformly.
So the alternate proof, is intended in the following way, take $\{x_m\}_{m\in \Bbb{N}}$ a dense subset of $X$. Show that there exist a subsequence of $(f_n)$, $(f_{n_{k1}})$ such that $(f_{n_{k1}}(x_1))$ converges. Show that there exist a subsequence of $(f_{n_{k1}})$, $(f_{n_{k1_{k2}}})$ such that $(f_{n_{k1_{k2}}}(x_2))$ converges.
Show that the diagonal sequence $f_{n_1},f_{n_{k1_{2}}},f_{n_{k1_{k2_3}}},...$ converges uniformly.
Ok, so I've done the following, since for every $x\in X$, $\{f_n(x)\}_n$ is relatively compact then, $\{f_n(x_1)\}_n$ is relatively compact and therefore there exist a subsequence $(f_{n_{k1}}(x_1))$ that converges in $Y$. Now, since $\{f_{n_{k1}}(x_2)\}_{k1}\subset \{f_{n}(x_2)\}_{n}$ then $\overline{\{f_{n_{k1}}(x_2)\}_{k1}}\subset \overline{\{f_{n}(x_2)\}_{n}}$ and therefore, $\overline{\{f_{n_{k1}}(x_2)\}_{k1}}$ is compact which implies that $\{f_{n_{k1}}(x_2)\}_{k1}$ is relatively compact and therefore, there exist a subsequence $(f_{n_{k1_{k2}}})$ such that $(f_{n_{k1_{k2}}}(x_2))$ converges.
Now, since $Y$ is complete, and $\{x_m\}_{m \in \Bbb{N}}$ dense, then if $f_n \to f$ uniformly in $\{x_m\}_{m \in \Bbb{N}}$ then it converges uniformly in $\overline{\{x_m\}_{m \in \Bbb{N}}}=X$.
Ok so I have to hyphotesis left to use, $X$ is compact and the family $\{f_n\}$ is equicontinuous, to show that the diagonal sequence, $f_{n_1},f_{n_{k1_{2}}},f_{n_{k1_{k2_3}}},...$ converges uniformly.
My intuition says that the $f$ to which the diagonal sequence converges, should be defined as $f(x_1)=\lim f_{n_{k1}}(x_1)$, $f(x_2)=\lim f_{n_{k1_{k2}}}(x_2),...$
However im not being able to use the compactness and equicontinuity (or uniform equicontinuity which is implied by the previous) to conclude the proof. How can I relate equicontinuity with convergence?

REPLY [1 votes]: Read the proof of Lemma 9.4.12 in H. H. Sohrab Basic Real Analysis 2nd edition (2014), pp. 448-449.
Reading the proofs of Lemmas 9.4.10 and 9.4.11 can be useful too.
For a discussion, I prefer to wait for your comments.