TITLE: Relation between height and volume of frustum
QUESTION [1 upvotes]: Suppose I have a physical coffee mug that is in the shape of a frustum. I can find the physical dimensions like smallest and largest radius and the height of the frustum. My question is till what height should I fill my cup to get a specific volume of water in it. For example I want 150 ml of water inside my cup so what height do I fill it till.The radius of the liquid at this level is also not know by me.
A,B,C are all known.
X is unknown.
Cup |Drawing

REPLY [0 votes]: If you know the formula of volume of frustum (as @skyking mentioned) you can easily solve your problem$$V=\pi\frac{h}{3}(R^2+r^2+Rr)$$If you know the volume, and both radius greater and smaller, then $$h=\frac{3V}{\pi(R^2+r^2+Rr)}$$Put the values and get your answer


TITLE: Finish the proof by induction: Prove $\exists k\in \mathbb Z \Big(\left(11^{n+2}+12^{2n+1}\right)=133k\Big)$.
QUESTION [0 upvotes]: I've got to prove, by induction, that  $$133\mid (11^{n+2}+12^{2n+1})$$
Basis for this is $n=0;$ So LHS = 133, and RHS = 133;
Assume there exists $k\in \mathbb Z$ $$(11^{n+2}+12^{2n+1})=133k\tag{inductive hypothesis}$$
We should prove that it then follows there exists a $k'\in \mathbb Z$ such $(11^{n+3}+13^{2n+3}) = 133k'$.
What to do with this next?
As I understand, I need to work with $(11^{n+3}+13^{2n+3})$, and try to get k by assumption. Can you help me?

REPLY [1 votes]: I think we need to prove that $11^{n+2}+12^{2n+1}$ divided by $133$.
I think a proof without induction is better.
Indeed, since $a^n-b^n=(a-b)\left(a^{n-1}+...+b^{n-1}\right)$, we have
$$11^{n+2}+12^{2n+1}=121\cdot11^n+12\cdot144^n=12\left(144^n-11^n\right)+133\cdot11^n$$
and we are done!


TITLE: Closed property of nonempty finite set
QUESTION [7 upvotes]: I came across this text in Rudin's book where it has been mentioned that a non-empty finite set is closed.
But a closed set is a set which contains all of it's limit points in the set itself but none of the elements of a non-empty finite set can possibly have a limit point because a neighborhood of a limit point has infinite points . So how come a non-empty finite set be a closed set ??

REPLY [4 votes]: Alternatively a set $A$ is closed iff $A=\bar A= A \cup \{\text{set of all limit points}\}.$
In your case, $\{\text{set of all limit points}\}=\emptyset.$ Thus $A=\bar A$.


TITLE: Find necessary and sufficient conditions on A so that $e^{At}$ is bounded for all $t>0$
QUESTION [0 upvotes]: Let A be an n × n real matrix. Find necessary and sufficient conditions on A so that $e^{At}$ is bounded for all $t>0$.
I am trying to figure this out. Can anyone give me a hint? Does it have to do with the eigenvalues of A?

REPLY [3 votes]: Yes. It has to do with the eigenvalues of $A$. Start by answering the question when $A$ is diagonalizable over $\Bbb C$. In the general case, Jordan canonical form will help.


TITLE: What is the characteristic function used for?
QUESTION [0 upvotes]: Im totally new to statistics , but what is the characteristic function for ?
I do not get that. I was reading about the bell curve and the Central Limit Theorem , but I did not get what the characteristic function is suppose to be , where it comes from or what it is used for.
It seems to appear in the proof of the Central Limit Theorem, but I do not know why.
It seems to have a Taylor series but again I do not know why.
I have been told all this relates to sum of 2 or 3 dices but again I do not understand how.
I know Bayes theorem (+proof) but I guess its unrelated ?
Sorry for the noob question. Thanks.

REPLY [0 votes]: Characteristic functions are great way to understand the sum of independent random variables.  If $X_1,X_2\,\dots,X_n$ are independent random variables, and $Y = \sum_{k=1}^n X_k$, then $\chi_Y(s) = \prod_{k=1}^n \chi_{X_k}(s)$.
Note also that $\chi_{\alpha X}(s) = \chi_X(\alpha s)$.
Also its Taylor's series is $\chi_X(s) = \sum_{r=0}^\infty \frac{(is)^r E(X^r)}{r!}$.
This can be used to prove the central limit theorem.  Here is the outline (with a lot of the rigour removed).  Suppose $X_k$ is a sequence of iidrv with mean zero and variance $\sigma^2$.  Let $Y_n = \frac1{\sqrt n}\sum_{k=1}^n X_k$.  Then
$$ \chi_Y(s) = [\chi_{X_1}(s/\sqrt n)]^n = \left[1-\frac{\sigma^2s^2}{2n} + O(n^{-3/2})\right]^n \to \exp(-\tfrac12 \sigma^2 s^2) $$
as $n \to \infty$.  And $ \exp(-\tfrac12 \sigma^2 s^2)$ is the characteristic function of a $N(0,\sigma)$ random variable.


TITLE: pre-calculus based word problem
QUESTION [0 upvotes]: A certain species of tree grows an average of 4.2 cm per week. Write an equation for the sequence that represents the weekly height of this tree in centimeters if the measurements begin when the tree is 300 centimeters tall.

REPLY [2 votes]: HINT: On average the tree is adding a constant amount, $4.2$ cm, per week; that’s an example of linear growth. Linear growth is described by a function of the form $f(n)=a+bn$, where $n$ is the number of the week. The amount of growth from week $n$ to week $n+1$ is therefore
$$f(n+1)-f(n)=\big(a+b(n+1)\big)-(a+bn)=\dots ?$$
If you answer that question correctly, you should be able to figure out what $b$ should be for your tree’s growth. And the height at the beginning is the height at week $0$, so $a=\dots$?


TITLE: Good book on Bolza problem and Hamilton systems
QUESTION [0 upvotes]: I am working with problems that characterize the solutions with the Bolza problem and Hamiltonian systems. I think this belongs to the area of Optimal Control and calculus of variations. Need a good resource to learn about these topics.  Any suggestions?
A reference for the Bolza Problem: https://www.encyclopediaofmath.org/index.php/Bolza_problem
Hamiltonian Systems: https://www.encyclopediaofmath.org/index.php/Hamiltonian_system

REPLY [0 votes]: I found a book that seems pretty good. Will confirm quality once I use it a little.
"The Calculus of Variations" - Bruce van Brunt
Also recommended "Stochastic Differential Equations: An Introduction with Applications" - Bernt Oksendal


TITLE: Convergence of $\sum_{n=1}^\infty\left(\frac{\sqrt{n+1}}{\zeta(n+1)}-\frac{\sqrt{n}}{\zeta(n)}\right)^\lambda$ for a suitable and small $\lambda>0$
QUESTION [0 upvotes]: I would like to get feedback about next exercise that I've considered, I don't know if it has a good mathematical content, but I would like to know how get a solution.
We consider series of the form $$\sum_{n=1}^\infty\left(\frac{\sqrt{n+1}}{\zeta(n+1)}-\frac{\sqrt{n}}{\zeta(n)}\right)^\lambda\tag{1}$$
where $\lambda$ is a real number and $\zeta(s)$ is the Riemann zeta function.
The exercise that I've created is to find and justify the existence of a $\lambda$ for which our series $(1)$ is convergent.

Question. Can you find a ray of positive real numbers $(\lambda_0,\infty)$ for which our series
  $$\sum_{n=1}^\infty\left(\frac{\sqrt{n+1}}{\zeta(n+1)}-\frac{\sqrt{n}}{\zeta(n)}\right)^\lambda$$
  is convergent when $\lambda_0<\lambda$? I am asking about what work can be done to get the smallest $\lambda_0>0$ as is possible. Many thanks.

REPLY [2 votes]: In my opinion the fact that $\zeta(n)\to 1$ is not sufficient for our purpose.
It is not true that if $(a_n)$ and $(b_n)$ are positive sequences which tend to $1$ then
$$\frac{\sqrt{n+1}}{b_{n}}-\frac{\sqrt{n}}{a_n}\sim  \frac{1}{2\sqrt n}.$$
Take for example $b_n=\sqrt{1+\frac{1}{n}}$ and $a_n=1$.
Here we need an estimate of how fast $\zeta(n)\to 1$,
$$1\leq \zeta(n)=1+\frac{1}{2^n}+\sum_{k=3}^{\infty}\frac{1}{k^n}\leq 1+\frac{1}{2^n}+\int_{2}^{\infty}\frac{dx}{x^n}=  1+\frac{1}{2^n}+\frac{1}{(n-1)2^{n-1}}.$$
Therefore $\zeta(n)=1+O(1/2^n)$ and we may conclude that
$$\frac{\sqrt{n+1}}{\zeta(n+1)}-\frac{\sqrt{n}}{\zeta(n)}\sim  \frac{1}{2\sqrt n}.$$


TITLE: Measure theory problem involving metric density function $\rho(E, x)$
QUESTION [3 upvotes]: Working on the real line $(\mathbb{R})$, let $\mu : \mathscr{M} \rightarrow [0, +\infty]$ represent the Lebesgue measure ($\mathscr{M}$ is the set of measurable subsets of $\mathbb{R}$).

For $E \in \mathscr{M}$ and $x \in \mathbb{R}$, we define $$\rho(E, x) = \lim_{\delta \to 0+} \frac{\mu(E \cap (x - \delta, x + \delta))}{2\delta},$$ if the limit exists. The above limit is called the metric density of $E$ at $x$.

$(1)$ Given $E=(1,2)\cup(2,5]\cup\{6\}$, find the metric density of $E$ for all $x \in E$.
$(2)$ Let $\alpha \in (0, 1)$. Construct a set $E\subset\mathbb{R}$ such that $\rho(E, 0) = \alpha$.

My work and thoughts:
$(1)$ Since $x$ can be any real number, the intersection $E \cap (x - \delta, x + \delta)$ can be empty.
Also, if $(x - \delta, x + \delta) \subset E$, as $\delta \rightarrow 0+$ the intersection $E \cap (x - \delta, x + \delta) = x$.
In either case the measure equals zero. Is this correct?
For $(2)$ I have no idea.

REPLY [1 votes]: For Q(1). For $x\in E, $ if $x\in (1,2)\cup (2,5)$ then for all sufficiently small $d>0$ we have $$E\cap (x-d,x+d)= (x-d,x+d)$$ and $$(2 d)^{-1}\mu (E\cap (x-d,x+d))=(2 d)^{-1}\mu ((x-d,x+d))=(2 d)^{-1}\cdot 2 d=1.$$ You should be able to to show that $\rho (E,5)=1/2$ and $\rho (E,6)=0.$


text:
\section{General preliminaries}
\label{sec:general_preliminaries}

In this preliminary section we gather some definitions and results required in the following considerations.

\subsection{Notation and conventions}

In the present paper, unless stated otherwise, all schemes are defined over a perfect field $k$ of characteristic $p>0$.  By $W_n(k)$ (resp. $W(k)$) we denote the ring of Witt vectors of length $n$ (resp. infinite length) and by $K$ the fraction field of the ring $W(k)$.  A \emph{variety} is a separated scheme of finite type defined over $k$.  We say that a morphism $f \colon X \to Y$ is a \emph{fibration}, if it is proper and satisfies $f_*\sO_X = \sO_S$.

\subsection{$F$-splitting and ordinarity}
\label{ss:F_splitting_ordinarity}

Let $f \colon X \to S$ be a morphism of schemes over a perfect field $k$.  We recall that the $e$-th relative Frobenius $F^e_{X/S}$ morphism is defined by the following commutative diagram:
\[
\xymatrix@C=50pt@R=13pt{
  X \ar[rd]|{F^e_{X/S}}\ar@/_1.2pc/[rdd]_{\pi} \ar@/^1.2pc/[rrd]^{F^e_X}& & \\
   & X^e_S \ar[r]\ar[d]_{\pi'}\ar@{}[dr]|-{\square} & X \ar[d]^{\pi} \\
   &  S \ar[r]_{F^e_S} & S.
}
\]

\begin{definition}
 We say that $f$ is \emph{globally $F$-split} if there exists a splitting of the natural map $\sO_{X^e_S} \to F^e_{X/S,*}\sO_X$, as a homomorphism of $\sO_{X^e_S}$-modules.  In particular, a scheme $X$ defined in characteristic $p$ is globally $F$-split if the natural map $X \to \Spec(\FF_p)$ is \emph{globally $F$-split} (equivalently, the natural homomorphism $\sO_X \to F_*\sO_X$ is globally split).
\end{definition}

\noindent Suppose $f \colon X \to S$ is a globally $F$-split morphism over $k$, and let $T \to S$ be a morphism.  Taking a base change of a splitting we see that $f_T \colon X_T \to T$ is also globally $F$-split.  In particular, fibres of a globally $F$-split morphism are globally $F$-split.  We now recall the following result.  In the proof, we adopt a few definitions included in \cite{Ejiri_When_is_the_Albanese_morphism_an_algebraic_fiber_space_in_positive___characteristic?}.  In particular, we say that a $\QQ$-divisor $\Delta$ is $\ZZ_{(p)}$-Weil (resp. $\ZZ_{(p)}$-Cartier) if there exists number $N$ coprime to $p$ such that $N \cdot \Delta$ is Weil (resp. Cartier).  We note that such conditions can always by exhibited using a number of the form $p^f - 1$, for $f \in \NN$.

\begin{proposition}
\label{prop:f_split_fibration_normal}
Let $X$ be a globally $F$-split normal projective variety over $k$ such that $K_X$ is $\bQ$-Cartier with index prime-to-$p$ and let $f \colon X \to S$ be a fibration.  Then there exists a non-empty open subset $U \subset S$ such that $f_{|U} \colon X_U \to U$ is a globally $F$-split morphism.  Moreover, there exists an open subset $V \subset X$ such that $V \to U$ is smooth and $\codim_X(X \setminus V) = 2$.
\end{proposition}

\begin{proof}
First, using the technique of \cite[Proposition 3.12]{Schwede_Smith_Globally_F_regular_an_log_Fano_varieties} we observe that there exists a divisor $\ZZ_{(p)}$-Cartier divisor $\Delta$ such that $(X,\Delta)$ is an $F$-split pair and $K_X + \Delta \sim_{\ZZ_{(p)}} 0$.  Localizing the splitting of $X$ at the generic point $\eta$ of $S$ and then using the argument of \cite[Proposition 2.6]{Gongyo_Li_Patakfalvi_Schwede_Tanaka_Zong_On_rational_connectedness_of_globally_F_regular_threefolds} for $D = 0$, we see that the geometric generic fibre $(X_{\overline{\eta}},\Delta_{\overline{\eta}})$ is $F$-split.  By \cite[Theorem B]{Patakfalvi_Schwede_Zhang_F_singularities_in_families} the same condition holds for all geometric fibres in the neighbourhood of $\eta$.  Using \cite[Proposition 5.8 (3)]{Ejiri_When_is_the_Albanese_morphism_an_algebraic_fiber_space_in_positive___characteristic?}, we may now assume that $X \to S$ is actually locally $F$-split (as defined in \cite{Ejiri_When_is_the_Albanese_morphism_an_algebraic_fiber_space_in_positive___characteristic?}) and therefore application of \cite[Proposition 5.7]{Ejiri_When_is_the_Albanese_morphism_an_algebraic_fiber_space_in_positive___characteristic?} yields the desired result.
\end{proof}

\begin{corollary}\label{cor:f_split_normal_fibres}
Let $X$ be a globally $F$-split normal projective variety over $k$ such that $K_X$ is $\bQ$-Cartier with index prime-to-$p$, and let $f \colon X \to S$ be a fibration.  Then there exists a non-empty open subset $U \subset S$ such that for every geometric point $u \in U$ the fibre $X_u$ is globally $F$-split and normal.
\end{corollary}

In the following statement, we need notions of a dualizing complex $\omega^\bullet_X$ and a dualizing module $\omega_X$ of a variety $X$ defined over $k$.  For a detailed description, we refer to \stacksproj{08XG} and \stacksproj{0A85}.  We note that by definition $\omega_X \isom \cH^{-\dim X} \left( \omega_X^{\bullet} \right)$, and that for normal varieties $\omega_X$ is in fact an $S_2$ divisorial sheaf $\cO_X(K_X)$ corresponding to the canonical divisor $K_X$.

    \begin{proposition}\label{prop:duality_for_S_k}
    Let $X$ be a projective variety over $k$.  Assume that $X$ satisfies Serre's condition $S_k$.  Then  there is a functorial isomorphism $H^{d-i}(X,\omega_X) \isom H^i(X,\cO_X)^\vee$ for $i<k-1$.  Moreover, there exists a natural surjective morphism $H^{d-k+1}(X,\omega_X) \to H^{k-1}(X,\cO_X)^\vee$.
    \end{proposition}
    \begin{proof}
        First, we observe that duality theory yields a functorial isomorphism $\Ext^{-i}(\cF,\omega^\bullet_X) \isom H^i(X,\cF)^\vee$, for every coherent sheaf $\cF$ on $X$.  In the particular, for $\cF = \cO_X$ one obtains an isomorphism $H^{-i}(\omega^\bullet_X) \isom H^i(X,\cO_X)^\vee$.
        We set $d = \dim X$ and consider the distinguished triangle:
        \[
            \omega_X[d] \ra \omega^\bullet_X \ra \cE \xrightarrow{+1}.
        \]
        Taking cohomology, we obtain an exact sequence
        \[
 H^{-i-1}(X,\cE) \ra H^{-i}(X,\omega_X[d]) \isom H^{d-i}(X,\omega_X) \ra H^{-i}(X,\omega^\bullet_X) \isom H^i(X,\cO_X)^\vee \ra H^{-i}(X,\cE)
        \]
        However, noting that $\cH^{-q}(\cE) = \cH^{-q}(\omega^\bullet_X)$ for by $q<d$ and using \stacksproj{0ECM} along with the spectral sequence:
        \[
          H^p(X,\cH^q(\cE)) \Rightarrow H^{p+q}(X,\cE),
        \]
        we see that $H^{-j}(X,\cE) = 0$ for $0 \leq j < k$. This finishes the proof.
    \end{proof}
    We also recall that Serre properties behave well with respect to the procedure of taking geometric generic fibres.
    \begin{proposition}\label{prop:property_S_k_inherited_by_fibres}
      Let $f \colon X \to Y$ be a flat morphism of varieties over $k$.  Assume that $X$ satisfies Serre's condition $S_k$.  Then the geometric generic fibre of $f$ satisfies Serre's condition $S_k$.
    \end{proposition}
    \begin{proof}
      The generic fibre is $S_k$ by the definition of the property $S_k$.  Taking an extension to the geometric generic fibre is valid by \cite[Corollaire 6.4.2]{Grothendieck_Elements_de_geometrie_algebrique_IV_II} (flat extension).
    \end{proof}

Concerning the relation between global $F$-splitting and ordinarity, we need the following result.  We provide the proof for the reader's convenience and because the classical reference does not treat the singular case.  A slightly different argument also already appeared in \cite[Section 2.4]{Joshi_Rajan}.

\begin{proposition}[{\cite[Proposition 9]{Mehta_Ramanathan_Frobenius_splitting_and_cohomology_vanishing_for_Schubert_varieties}}]\label{prop:ordinary_is_fsplit}
Let $X$ be a normal projective variety over $k$ with $K_X \sim 0$.  Then $X$ is weakly ordinary if and only if it is globally $F$-split.
\end{proposition}

\begin{proof}
Let $U$ be the regular locus of $X$, and set $d:= \dim X$. Then we have:

$\sO_X \to F_* \sO_X$ splits

$\explshift{-30pt}{\Longleftrightarrow}{$U$ is a big open set, and all sheaves are $S_2$}$

$\sO_U \to F_* \sO_U$ splits

$\expl{\Longleftrightarrow}{duality}$

$F_* \omega_U \to \omega_U$ splits

$\explshift{50pt}{\Longleftrightarrow}{$U$ is a big open set, and all sheaves are $S_2$}$

$F_* \omega_X \to \omega_X$ splits

\\[0pt]

\hspace{120pt} $\expl{\Longleftrightarrow}{$K_X \sim 0 \Rightarrow H^0(X, \omega_X) \cong k$}$

$H^0(X, \omega_X) \cong H^0(X,F_* \omega_X)  \to H^0(X, \omega_X)$ is surjective

\\[0pt]

\hspace{120pt} $\expl{\Longleftrightarrow}{$X$ is normal, hence \autoref{prop:duality_for_S_k} applies with $k=2$}$

$H^d(X, \sO_X) \to  H^d (X,F_*\sO_X) \expl{\cong}{$F$ is affine} H^d(X, \sO_X)$ is injective

\\[0pt]

\hspace{120pt} $\expl{\Longleftrightarrow}{a $p$-linear map between finite dimensional $k$-vector spaces, and $k$ is perfect}$  $H^d(X, \sO_X) \to  H^d (X,F_*\sO_X) \cong H^d(X, \sO_X)$ is bijective

\end{proof}

config: arxiv


TITLE: Law of total variance with i.i.d. positive continuous r.v.s.
QUESTION [1 upvotes]: So law of total variance states that:
$$Var(Y) = E[ Var(Y | X)] + Var(E[Y|X])$$
Now it's given that $X,Y$ are positive i.i.d. continuous r.v.s.
As $X,Y$ are independent, it means $E[Y|X] = E[Y]$, which is a constant. Let it be $\mu$.
Then, $$Var(E[Y|X]) = Var(E[Y]) = Var(\mu) = 0$$
Thus, $$Var(Y) = E[ Var(Y | X)] + 0 =  E[ Var(Y | X)]$$
Is my reasoning right?

REPLY [1 votes]: $$\operatorname{Var}\left(Y\right) = \operatorname{E}\left[\operatorname{Var}\left(Y\mid X\right)\right] +\operatorname{Var}\left(\operatorname{E}\left[Y\mid X\right]\right)$$
If $Y$ is independent of $X$, then
$$\operatorname{E}\left[\operatorname{Var}\left(Y\mid X\right)\right]=\operatorname{E}\left[\operatorname{Var}\left(Y\right)\right]$$
And
$$\operatorname{Var}\left(\operatorname{E}\left[Y\mid X\right]\right)=\operatorname{Var}\left(\operatorname{E}\left[Y\right]\right)=0$$
Therefore
$$\operatorname{Var}\left(Y\right) = \operatorname{E}\left[\operatorname{Var}\left(Y\right)\right]$$
Which makes sense because the expected value of a constant is equivalent to the constant. So yes, as others have already mentioned, your reasoning is indeed correct.


text:
\begin{document}

\title{Some more amplituhedra are contractible}

\author[Blagojevi\'c]{Pavle V. M. Blagojevi\'{c}}
\address{Institute of Mathematics, Freie Universität Berlin, Arnimallee 2, 14195 Berlin, Germany\hfill\break
\mbox{\hspace{4mm}}Mat.\ Institut SANU, Knez Mihailova 36, 11000 Beograd, Serbia}
\email{blagojevic@math.fu-berlin.de}

\author[Galashin]{Pavel Galashin}
\address{Department of Mathematics, Massachusetts Institute of Technology, 77 Massachusetts Avenue,\hfill\break
\mbox{\hspace{4mm}}Cambridge, MA 02139, USA}
\email{galashin@mit.edu}

\author[Pali\'c]{Nevena Pali\'c}
\address{Institute of Mathematics, Freie Universität Berlin, Arnimallee 2, 14195 Berlin, Germany}
\email{palic@math.fu-berlin.de}

\author[Ziegler]{G\"unter M. Ziegler}
\address{Institute of Mathematics, Freie Universität Berlin, Arnimallee 2, 14195 Berlin, Germany}
\email{ziegler@math.fu-berlin.de}

\thanks{The research by Pavle V. M. Blagojevi\'{c} has received funding from the grant ON 174024 of the Serbian Ministry of Education and Science.\newline\indent
The research by Nevena Pali\'c has received funding from DFG via the Berlin Mathematical School.\newline\indent
This material is based on work supported by the National Science Foundation under Grant No.\ DMS-1440140 during the Fall of 2017, while all authors were in residence at the Mathematical Sciences Research Institute in Berkeley CA}
\date{\today}

\begin{abstract}
The amplituhedra arise as images of the totally nonnegative Grassmannians
by projections that are induced by linear maps.
They were introduced in Physics by Arkani-Hamed \& Trnka (Journal of High Energy Physics, 2014) as model spaces that should provide a better understanding of the scattering amplitudes of quantum field theories. The topology of the amplituhedra has been known only in a few special cases, where they turned out to be homeomorphic to balls. The amplituhedra are special cases of Grassmann polytopes introduced by Lam (Current Developments in Mathematics 2014, Int.\ Press).
In this paper we show that that some further amplituhedra are homeomorphic to balls, and that some more Grassmann polytopes and amplituhedra are contractible.
\end{abstract}

\maketitle

\section{Introduction and statement of the main result}
\label{sec : Introduction and the statement of the main results }

\subsection{Introduction}
Let $n$ and $k$ be integers such that $n \geq k \geq 1$. If $\Mat_{k,n}$ denotes the space of all real $k\times n$ matrices of rank $k$,
then the real Grassmannian $\gr_k(\R^n)$ --- the space of all $k$-dimensional linear subspaces of $\R^n$ ---
can be defined as the orbit space $\gr_k(\R^n) = \GL_k \backslash \Mat_{k,n}$.
The totally nonnegative part of the Grassmannian is defined quite analogously.

\begin{definition}[Postnikov {\cite[Sec.\,3]{Postnikov2006}}]
Let $n \geq k \geq 1$ be integers, let $\Mat_{k,n}^{\ge0}$ be the space of all real $k \times n$ matrices of rank $k$ all whose  maximal minors are nonnegative, and let $\GL_k^+$ denote the group of all real $k\times k$ matrices with positive determinant, which acts freely on $\Mat_{k,n}^{\ge0}$ by matrix multiplication from the left.
The \emph{totally nonnegative Grassmannian} $\gr_k^{\ge0}(\R^n)$ is the orbit space $\gr_k^{\ge0}(\R^n) = \GL_k^+ \backslash \Mat_{k,n}^{\ge0}$.
\end{definition}

The totally nonnegative Grassmannian was introduced and studied by Postnikov in 2006 \cite[Sec.\,3]{Postnikov2006},
building on works by Lusztig \cite{Lusztig1994} and by Fomin \& Zelevinsky \cite{Fomin1999}.
Subsequently, the geometric and combinatorial properties of the totally nonnegative Grassmannian were studied intensively.
Rietsch \& Williams showed that the totally nonnegative Grassmannian is contractible \cite[Thm.\,1.1]{Rietsch2010}; an earlier argument by Lusztig \cite[Sec.\,4.4]{Lusztig1998} can also be adapted to prove the same. Galashin, Karp \& Lam \cite[Thm.\,1.1]{Galashin2017} proved that $\gr_k^{\ge0}(\R^n)$ is indeed homeomorphic to a closed $k(n-k)$-dimensional ball.

In 2014, the physicists Arkani-Hamed \& Trnka \cite[Sec.\,9]{Arkani-Hamed2014} introduced the amplituhedra as certain images of the totally nonnegative Grassmannians.
They conjectured that their geometry describes scattering amplitudes in some quantum field theories.
For a gentle introduction to amplituhedra in physics and mathematics consult \cite{Bourjaily2018}.
Shortly after, Lam introduced Grassmann polytopes \cite{Lam2016}, which generalize amplituhedra.

Postnikov \cite[Def.\,3.2, Thm.\,3.5]{Postnikov2006} defined a CW structure on the totally nonnegative Grassmannian $\gr_k^{\ge0}(\R^n)$ such that each cell, also called a \emph{positroid cell}, is indexed by the associated matroid -- a positroid -- of rank $k$ on $n$ elements, see also \cite{Postnikov2009}.
Furthermore, Rietsch \& Williams \cite{Rietsch2010} showed that the closures of positroid cells are contractible and that their boundaries are homotopy equivalent to spheres.

\begin{definition}
\label{def:amplituhedron}
Let $k\geq 1$, $m\ge0$ and $n \geq k+m$ be integers, and let $Z$ be a real $(k+m)\times n$ matrix such that the assignment
\begin{equation}
\label{eqn:definition map Z}
\widetilde{Z}(\span(V))=\span(VZ^\top)
\end{equation}
induces a map
\[
\widetilde{Z} \colon \gr_k^{\ge0}(\R^n) \longrightarrow \gr_k(\R^{k+m}).
\]
Here $V\in \Mat_{k,n}^{\geq 0}$, $\span$ denotes the row span of a matrix, and $Z^\top$ is the transpose of the matrix $Z$.
The image $\widetilde{Z}(\bar{e})$ of a closed positroid cell $\bar{e}$ in the CW decomposition of the nonnegative Grassmannian $\gr_k^{\ge0}(\R^n)$ is called a \emph{Grassmann polytope}, denoted by $\GP{e}$.
If $e$ is the maximal cell, which for this CW decomposition means  $\bar{e} = \gr_k^{\ge0}(\R^n)$, and all $(k+m)\times (k+m)$ minors of the matrix $Z$ are positive, then the Grassmann polytope $\GP{e}$ is called an \emph{amplituhedron} and is denoted by $\A_{n,k,m}(Z)$.
\end{definition}

The previous definition in particular means that if $v_1, \dots, v_k \in \R^n$ are linearly independent row vectors, then
\[
\widetilde{Z}(\span\{v_1, \dots, v_k\}) = \span\{v_1Z^\top, \dots, v_kZ^\top\}.
\]
The map $\widetilde{Z}$ is said to be \emph{well defined} if $\span(VZ^\top)$ is a $k$-dimensional subspace of $\R^{k+m}$ for every $V \in \Mat_{k,n}^{\ge0}$.
The fact that the map $\widetilde{Z}$ is well defined when $Z$ is a matrix with positive maximal minors was established by Arkani-Hamed \& Trnka in \cite{Arkani-Hamed2014} and by Karp in \cite[Thm.\,4.2]{Karp2017b}. Lam \cite[Prop.\,15.2]{Lam2016}, however, considers a larger class of matrices $Z$ for which the map $\widetilde{Z}$ is still well defined.

The structure of the amplituhedron is known only in a few cases.
In the case $m=0$ all amplituhedra $\A_{n,k,0}(Z)$ are the point $\gr_k(\R^{k})$, whereas when $m=1$ Karp \& Williams \cite[Cor.\,6.18]{Karp2017} have shown that the amplituhedron is homeomorphic to a ball.
For $k=1$ the amplituhedron is a cyclic polytope of dimension $m$ on $n$ vertices \cite{Sturmfels1988}, and for $n=k+m$ the map $Z$ is a linear isomorphism, and consequently the amplituhedron is homeomorphic to the totally nonnegative Grassmannian $\gr_k^{\ge0}(\R^n)$, which is a ball by \cite[Thm.\,1.1]{Galashin2017}. Finally, Galashin, Karp \& Lam \cite[Thm.\,1.2]{Galashin2017} proved that the cyclically symmetric amplituhedra, amplituhedra arising from particularly chosen matrices $Z$, are homeomorphic to balls whenever $m$ is even.
The topology of other Grassmann polytopes is unknown.

\medskip

\subsection{Main results}

Our first result gives a family of contractible Grassmann polytopes.

\begin{theorem}
\label{thm:contractible}
Let $k \geq 1$ and $m \ge0$ be integers, and let $Z$ be a real $(k+m)\times (k+m+1)$ matrix such that the map $\widetilde{Z}\colon \nng{k}{k+m+1} \longrightarrow \gr_k(\R^{k+m})$ is well defined.
Then the Grassmann polytope $\GP{e}$ is contractible for every positroid cell $e$ in the CW decomposition of $\gr_k^{\ge0}(\R^{k+m+1})$.
\end{theorem}

The proof of Theorem \ref{thm:contractible} relies on classical results of Smale \cite[Main Thm.]{Smale1957} and Whitehead \cite[Thm.\,1]{Whitehead1949} combined with the fact that every Grassmann polytope admits a triangulation (as a topological space), see Theorem \ref{thm:triangulation}.

\medskip

The following is a consequence of Smale's result \cite[Main Thm.]{Smale1957}.

\begin{theorem}[Smale]
\label{thm:Smale}
Let $X$ and $Y$ be path connected, locally compact, separable metric spaces, and in addition let $X$ be locally contractible.
	Let $f\colon X\longrightarrow Y$ be a continuous surjective proper map, that is, any inverse image of a compact set is compact.
	If for every $y\in Y$ the inverse image $f^{-1}(\{y\})$ is contractible, then the induced homomorphism
	\[
	f_{\#} \colon  \pi_i(X)\longrightarrow \pi_i(Y)
	\]
	is an isomorphism for all $i \geq 0$.
\end{theorem}

Recall that a continuous map $f\colon X\longrightarrow Y$ between topological spaces $X$ and $Y$ is a {\em weak homotopy equivalence} if the induced map on the path connected components $f_{\#}\colon\pi_0(X)\longrightarrow\pi_0(Y)$ is bijective, and for every point $x_0\in X$ and for every integer $n \geq 1$ the induced map $f_{\#}\colon\pi_n(X,x_0)\longrightarrow\pi_n(Y,f(x_0))$ is an isomorphism.

\begin{theorem}[{\cite[Thm.\,1]{Whitehead1949}}]
\label{thm:Whitehead}
Let $X$ and $Y$ be topological spaces that are homotopy equivalent to CW complexes. Then a continuous map $f\colon X \longrightarrow Y$ is a weak homotopy equivalence if and only if it is a homotopy equivalence.
\end{theorem}

Since Theorem \ref{thm:Whitehead} requires that spaces have the homotopy type of a CW complex, the following theorem is a necessary ingredient in the proof of Theorem \ref{thm:contractible}.

\begin{theorem}
\label{thm:triangulation}
Every Grassmann polytope is semi-algebraic as a subset of a Grassmannian.
In particular, every Grassmann polytope is homeomorphic to a semi-algebraic subset of some real affine space, and admits a triangulation.
\end{theorem}

Note that Theorem \ref{thm:triangulation} claims that every Grassmann polytope $\GP{e}$ can be triangulated in a classical sense, thus there exists a simplicial complex $T$ and a homeomorphism $T \longrightarrow \GP{e}$.
A very similar argument to ours was also given by Arkani-Hamed, Bai \& Lam in \cite[Appendix~J]{Arkani-Hamed2017}.

\medskip
In order to apply Theorem \ref{thm:Smale} to the map $\widetilde{Z}$,
we need to understand its fibers. Thus we prove the following theorem.

\begin{theorem}
\label{thm:fibers}
Let $k \geq 1$ and $m \ge0$ be integers, and let $Z$ be a real $(k+m)\times (k+m+1)$ matrix such that the map $\widetilde{Z}$ is well defined.
Then for every positroid cell $e$ and for every point $y \in \GP{e}$, the inverse image $(\widetilde{Z}|_{\bar{e}})^{-1}(\{y\}) = \widetilde{Z}^{-1}(\{y\}) \cap \bar{e}$ under the restriction map $\widetilde{Z} |_{\bar{e}} \colon \bar{e} \longrightarrow \GP{e}$ is contractible.
\end{theorem}

The proof of Theorem \ref{thm:fibers} is postponed to the next section.
Here we show that Theorem \ref{thm:fibers} in combination with Theorems \ref{thm:Smale}--\ref{thm:triangulation} implies our main result.

\begin{proof}[Proof of Theorem \ref{thm:contractible}]
Let $e$ be a positroid cell in the CW decomposition of $\nng{k}{k+m+1}$. We apply Theorem \ref{thm:Smale} to the map $\widetilde{Z}: \bar{e} \longrightarrow \GP{e}$.
The spaces $\bar{e}$ and $\GP{e}$, as well as the map $\widetilde{Z}$, satisfy the assumptions of Theorem \ref{thm:Smale}.
Furthermore, Theorem \ref{thm:fibers} implies that for every $y \in \bar{e}$, the fiber $\widetilde{Z}^{-1}(\{y\})$ is contractible.
Thus, from Theorem \ref{thm:Smale} we have that the map $\widetilde{Z}$ is a weak homotopy equivalence.

The closed positroid cell $\bar{e}$ is a CW complex.
Furthermore, the Grassmann polytope $\GP{e}$ is a CW complex, by Theorem \ref{thm:triangulation}.
Thus, from Theorem \ref{thm:Whitehead}, we conclude that the map $\widetilde{Z}$ is a homotopy equivalence.
Hence, the Grassmann polytope $\GP{e}$ is homotopy equivalent to the closed positroid cell $\bar{e}$, which is contractible, see \cite[Thm.\,1.1]{Rietsch2010}.
\end{proof}

Theorem \ref{thm:contractible} in particular implies that all amplituhedra $\A_{k+m+1,k,m}(Z)$ are contractible.
Our next result shows that if in addition $m$ is even, they are homeomorphic to balls.

\begin{theorem}
\label{thm:homeomorphic}
Let $k \geq 1$ be an integer, let $m \ge0$ be an even integer, and let $Z \in \Mat_{k+m,k+m+1}$ be a matrix with all $(k+m)\times (k+m)$ minors positive.
Then the amplituhedron $\A_{k+m+1,k,m}(Z)$ induced by the matrix $Z$ is homeomorphic to a $km$-dimensional ball.
\end{theorem}

The proof of Theorem \ref{thm:homeomorphic} is presented in Section \ref{sec:homeomorphic}.  We remark that the combinatorics of the amplituhedron in the case $n=k+m+1$ with $m$ even is identical to that of a cyclic polytope, see \cite{Galashin2018}.

\subsection*{Acknowledgement}
The authors thank Rainer Sinn for sharing the knowledge about semi-algebraic sets, to Thomas Lam, whose great observations increased the generality of the results in this paper, and to Steven Karp for helpful comments.
We are grateful to the referee for careful reading of our manuscript and for useful suggestions that improved the quality of our paper.

\section{Proof of Theorem \ref{thm:fibers}}
\label{sec:contractible}

Let $k \geq 1, m \ge0$ and $n \geq k+m$ be integers and let $Z$ be a real $(k+m)\times n$ matrix such that the map $\widetilde{Z}$ is well defined.
Since the action of the group $\GL_k^+$ on $\Mat_{k,n}^{\geq 0}$ is free, there is a fibration
\begin{equation}
\label{eqn:fibration nonnegative Grassmannian}
\GL_k^+ \longrightarrow \Mat_{k,n}^{\ge0} \longrightarrow \gr_k^{\ge0}(\R^n).
\end{equation}

The matrix $Z$, as in Definition \ref{def:amplituhedron}, induces a map
\begin{eqnarray*}
\widehat{Z} : \Mat_{k,n}^{\ge0} & \longrightarrow & \Mat_{k,k+m},\\
V & \longmapsto & VZ^\top,
\end{eqnarray*}
which is again well defined, see for example \cite[Prop.\,15.2]{Lam2016}.

Let $e$ be a positroid cell in the CW decomposition of $\nng{k}{n}$, and let $I_e \subseteq \binom{[n]}{k}$ be the family of nonbases (dependent sets) of cardinality $k$ of the matroid that defines the cell $e$.
The maximal minors of a $k \times n$ matrix are indexed by the set $\binom{[n]}{k}$. Denote by $\Mat_{k,n}^{\ge0}(e)$ the set of all matrices $V \in \Mat_{k,n}^{\ge0}$ whose minors indexed by elements of $I_e$ are equal to zero.
Then every point in $\bar{e} \subseteq \nng{k}{n}$ is represented by a matrix in $\Mat_{k,n}^{\ge0}(e)$, and the row span of every such matrix lies in $\bar{e}$.
In other words, $\bar{e}= \GL_k^{+} \backslash \Mat_{k,n}^{\geq 0}(e)$.
Thus the restriction of the fibration \eqref{eqn:fibration nonnegative Grassmannian} is a fibration
\begin{equation}
\label{eqn:fibration positroid cell}
\GL_k^+ \longrightarrow \Mat_{k,n}^{\ge0}(e) \longrightarrow \bar{e}.
\end{equation}
Note that if $e$ is the maximal positroid cell, the set $\Mat_{k,n}^{\ge0}(e)$ is the whole set $\Mat_{k,n}^{\ge0}$.

Denote by $\widehat{\GP{e}}$ the image of the set $\Mat_{k,n}^{\ge0}(e)$ under the map $\widehat{Z}$. With a usual abuse of notation, we consider maps $\widehat{Z}: \Mat_{k,n}^{\ge0}(e) \longrightarrow \widehat{\GP{e}}$ and $\widetilde{Z}:\bar{e} \longrightarrow \GP{e}$. Then there exists a commutative diagram of spaces and continuous maps
\[
\begin{tikzcd}
 \Mat_{k,n}^{\ge0}(e)   \arrow{r}{\widehat{Z}} \arrow{d}  & \widehat{\GP{e}} \arrow{d}      \\
\bar{e}	  \arrow{r}{\widetilde{Z}}  & \GP{e},
\end{tikzcd}
\]
where vertical maps send any matrix to its row span.

\medskip

The proof of Theorem \ref{thm:fibers} splits into the following two lemmas.

\begin{lemma}
\label{lemma:fiber}
Let $k \geq 1$ and $m\ge0$ be integers, $n=k+m+1$, and let $Z$ be a real $(k+m)\times n$ matrix such that the map $\widetilde{Z}$ is well defined.
Then for every positroid cell $e$ in the CW decomposition of $\nng{k}{n}$ and for every $W \in \widehat{\GP{e}}$, the inverse image $\widehat{Z}^{-1}(\{W\}) \subseteq \Mat_{k,n}^{\ge0}(e)$ is nonempty and convex.
\end{lemma}

\begin{proof}
The matrix $Z$ induces a linear map
\begin{eqnarray}
\label{eqn:map corank 1}
\R^n &\longrightarrow& \R^{k+m}\\
v &\longmapsto& vZ^\top \nonumber,
\end{eqnarray}
where $v \in \R^n$ is a row vector. Since $n=k+m+1$, the kernel of the map \eqref{eqn:map corank 1} is $1$-dimensional. Fix a generator $a \in \R^n$ of that kernel.

Choose an arbitrary point $W \in \widehat{\GP{e}}$, and let $U$ and $V$ be any two points in $\widehat{Z}^{-1}(\{W\})$.
Our goal is to show that for every $\lambda \in [0,1]$ the convex combination $(1-\lambda)U+\lambda V$ also belongs to $\widehat{Z}^{-1}(\{W\})$.

Since $UZ^\top=VZ^\top=W$, the rows of the matrix $V-U$ belong to $\ker(Z)$.
Consequently, there exists a row vector $x \in \R^k$ such that $V-U = x^\top a$, where $a$ is also considered as a row vector.
Thus we have to show that for every $\lambda \in [0,1]$ the convex combination
\begin{equation}
\label{eqn:convex combination}
(1-\lambda)U + \lambda V = U + \lambda x^\top a
\end{equation}
belongs to the space $\Mat_{k,n}^{\ge0}(e)$, this means that every $k \times k$ minor of the matrix \eqref{eqn:convex combination} is nonnegative, and in addition that all the minors of the matrix \eqref{eqn:convex combination} indexed by the nonbases $I_e \subseteq \binom{[n]}{k}$ of the matroid corresponding to $e$ are equal to zero.

A $k \times k$ submatrix of the matrix \eqref{eqn:convex combination} is of the form
\begin{equation}
\label{eqn:matrix minor}
\left(
\begin{matrix}
u_{1i_1}+\lambda x_1a_{i_1} & \dots & u_{1i_k}+\lambda x_1a_{i_k} \\
\vdots & & \vdots \\
u_{ki_1}+\lambda x_ka_{i_1} & \dots & u_{ki_k}+\lambda x_ka_{i_k}
\end{matrix}
\right),
\end{equation}
where
\[
U=\left(
\begin{matrix}
u_{11} & \dots & u_{1n} \\
\vdots & &\vdots \\
u_{k1} & \dots & u_{kn}
\end{matrix}
\right), \
x=(x_1 \dots x_k), \  a=(a_1 \dots a_n),
\]
and $1 \leq i_1 < \dots <i_k \leq n$. The matrix \eqref{eqn:matrix minor} can be transformed using row operations into a matrix that contains the variable $\lambda$ only in one row. Therefore, every $k\times k$ minor of the matrix \eqref{eqn:convex combination} is a polynomial of degree at most $1$ in the variable $\lambda$. Since it takes nonnegative values for $\lambda=0$ and $\lambda=1$, it is also nonnegative for all $\lambda\in [0,1]$.
Thus for every $\lambda\in [0,1]$, the point $(1-\lambda)U+\lambda V$ belongs to $\Mat_{k,n}^{\ge0}$.
Similarly, if $\{i_1, \dots, i_k\}$ is a nonbasis of the matroid corresponding to $e$, then the determinant of the matrix \eqref{eqn:matrix minor} is zero for $\lambda=0$ and $\lambda=1$, so it is a constant zero-polynomial, meaning that the matrix \eqref{eqn:convex combination} belongs to $\Mat_{k,n}^{\ge0}(e)$ for every $\lambda \in [0,1]$. Consequently the set $\widehat{Z}^{-1}(\{W\})$ is convex.
\end{proof}

\begin{lemma}
\label{lemma:homeomorphism}
Let $k\geq 1, m\ge0$ and $n\geq k+m$ be integers.
For every positroid cell $e$ and for every $W \in \widehat{\GP{e}}$, the inverse images
\[
\widehat{Z}^{-1}(\{W\}) \subseteq \Mat_{k,n}^{\ge0}(e) \subseteq \Mat_{k,n}^{\ge0}
\qquad\text{and}\qquad
\widetilde{Z}^{-1}(\{\span(W)\}) \subseteq \bar{e} \subseteq \gr_k^{\ge0}(\R^n)
\]
are homeomorphic.
\end{lemma}

\begin{proof}
Let $\varphi \colon \widehat{Z}^{-1}(\{W\}) \longrightarrow \widetilde{Z}^{-1}(\{\span(W)\})$ be defined by $\varphi(U)=\span(U)$, where $U\in \widehat{Z}^{-1}(\{W\})$, and $\span$ denotes the row span.
We prove that $\varphi$ is a homeomorphism.

Clearly, $\varphi$ is continuous, so it suffices to find a continuous map $\psi:\widetilde{Z}^{-1}(\{\span(W)\}) \longrightarrow \widehat{Z}^{-1}(\{W\})$ such that $\varphi \circ \psi$ is the identity map on $\widetilde{Z}^{-1}(\{\span(W)\})$ and $\psi \circ \varphi$ is the identity map on $\widehat{Z}^{-1}(\{W\})$. Let $L \in \widetilde{Z}^{-1}(\{\span(W)\})$. Then there exists a matrix $K \in \Mat_{k,n}^{\geq0}(e)$ whose rows span the subspace $L$. Since
\[
\span(KZ^\top) = \span(W),
\]
there exists a unique $C \in \GL_k$ such that $KZ^\top=CW$. Now define $\psi$ as $\psi(L)=C^{-1}K$. It can be seen using Cauchy--Binet formula that $\det(C)>0$, thus, $C^{-1}K \in \Mat_{k,n}^{\ge0}(e)$.
Even though we have defined the map $\psi$ using an arbitrarily chosen matrix $K$ such that $\span(K)=L$, it can be checked directly that the definition of $\psi$ does not depend on a choice of $K$.

In order to prove that the map $\psi$ is continuous, we need to show that the choice of a matrix $K$ can be made continuously on $\widetilde{Z}^{-1}(\{\span(W)\})$.
The choice of a matrix $K$ is equivalent to the choice of a positively oriented basis for the subspace $L \subseteq \R^n$.
Therefore, we need a continuous section of the fiber bundle \eqref{eqn:fibration positroid cell} restricted to the set $\widetilde{Z}^{-1}(\{\span(W)\})$.
Since the base space $\bar{e}$ is contractible, the fiber bundle \eqref{eqn:fibration positroid cell} is trivial.
In particular, its restriction on $\widetilde{Z}^{-1}(\{\span(W)\})$ is also trivial, so it admits a continuous section.
Therefore, the bases for elements of $\widetilde{Z}^{-1}(\{\span(W)\})$  can be chosen continuously.
On the other hand, the matrix $C$ is a solution of the linear system $KZ^\top=CW$, which depends continuously on $K$, thus it also depends continuously on $L$.

Lastly,
\[
\varphi(\psi(L))=\varphi(C^{-1}K)=\span(C^{-1}K)=\span(K)=L,
\]
holds for every $L \in \widetilde{Z}^{-1}(\{\span(W)\})$, and
\[
\psi(\varphi(U))=\psi(\span(U))=C^{-1}U,
\]
for every $U \in \widehat{Z}^{-1}(\{W\})$, where $C$ is the unique $k \times k$ matrix such that $W=\widehat{Z}(U)=UZ^\top=CW$, hence $C$ is the identity matrix.
\end{proof}

Finally, Lemma \ref{lemma:fiber} and Lemma \ref{lemma:homeomorphism} complete the proof of Theorem \ref{thm:fibers}.

\section{Proof of Theorem \ref{thm:triangulation}}
\label{sec:triangulation}

Let us fix an arbitrary positroid cell $e$ in the CW decomposition of the totally nonnegative Grassmannian $\nng{k}{n}$.

Furthermore, let $d=\binom{k+m}{k}$, and consider the Veronese embedding $\nu \colon \RP^{d-1} \longrightarrow \R^{d\times d}$ given by
\[
x=(x_1: \ldots :x_d) \longmapsto \left( \frac{x_i x_j}{x_1^2+\dots +x_d^2} \right)_{1\leq i,j\leq d},
\]
where $x=(x_1: \ldots :x_d)\in \RP^{d-1}$.
The embedding $\nu$ maps every line $x \in \RP^{d-1}$ to the matrix of the projection $\R^d \longrightarrow x$.
For more details on the Veronese embedding see for example \cite[Sec.\,3.4.2]{BochnakEtAl1998}.

Consider next, with obvious abuse of notation,  the continuous map $\nu\colon \R^d{\setminus}\{0\} \longrightarrow \R^{d\times d}$ given by
\[
(x_1, \dots, x_d) \longmapsto \left( \frac{x_i x_j}{x_1^2+\dots +x_d^2} \right)_{1\leq i,j\leq d} \in \R^{d \times d}.
\]

In this way, we obtain the commutative diagram of spaces and maps
\[
\begin{tikzcd}
 \Mat_{k,n}^{\ge0}   \arrow{r}{\widehat{Z}} \arrow{d} & \Mat_{k,k+m} \arrow{r}{\gamma} \arrow{d} & \R^d \setminus \{0\} \arrow{d}{\pi}  \arrow{r}{\nu} & \R^{d\times d} \arrow{d}{\id}     \\
\nng{k}{n} \arrow{r}{\widetilde{Z}} & \gr_k(\R^{k+m}) \arrow{r}{\gamma} & \RP^{d-1} \arrow{r}{\nu} & \R^{d\times d} ,
\end{tikzcd}
\]
where $\gamma\colon \gr_k(\R^{k+m}) \longrightarrow \RP^{d-1}$ is the Pl\"ucker embedding, $\gamma\colon  \Mat_{k,k+m} \longrightarrow \R^{d}\setminus \{0\}$ maps every matrix to the tuple of its $k \times k$ minors, and $\pi\colon \R^{d}\setminus \{0\} \longrightarrow \RP^{d-1}$ is the quotient map.

The Grassmann polytope $\GP{e}=\widetilde{Z}(\bar{e})$ is embedded into $\RP^{d-1}$ via $\gamma$, the projective space $\RP^{d-1}$ is embedded into the Euclidean space $\R^{d \times d}$ via $\nu$, and thus the image $\nu(\gamma(\GP{e}))$ is homeomorphic to $\GP{e}$.

First, we prove that the homeomorphic image of the Grassmann polytope $\nu(\gamma(\GP{e}))$ is semi-algebraic.
The commutativity of the diagram above implies that
\[
\nu(\gamma(\GP{e}))  = \nu(\pi(\gamma(\widehat{\GP{e}}))) = \nu(\gamma(\widehat{\GP{e}}))= \nu(\gamma(\widehat{Z}(\Mat_{k,n}^{\geq 0}(e)))).
\]
The set $\Mat_{k,n}^{\ge0}(e) \subseteq \R^{k \times n}$ is semi-algebraic, even algebraic.
Since the map $\widehat{Z}$ is multiplication by a matrix, the set $\widehat{\GP{e}}$ is also semi-algebraic \cite[Cor.\,2.4(2)]{Coste2002}.
Furthermore, the map $\gamma\colon \Mat_{k,k+m} \longrightarrow \R^{d}\setminus \{0\}$ is a restriction of a polynomial map $\R^{k\times (k+m)}\longrightarrow\R^d$, and thus $\gamma(\widehat{\GP{e}}) \subseteq \R^{d} \setminus \{0\}$ is semi-algebraic by \cite[Cor.\,2.4(2)]{Coste2002} as well.
Finally, the map $\nu\colon\R^d \longrightarrow \R^{d \times d}$ is a regular rational map, and consequently it maps semi-algebraic sets to the semi-algebraic sets, see \cite[Prop.\,2.2.7]{BochnakEtAl1998} \cite[Cor.\,2.9(1)]{Coste2002}.
Hence, we have proved that $\nu(\gamma(\GP{e}))$ is semi-algebraic in $\R^{d\times d}$, and consequently the Grassmann polytope $\GP{e}$ is homeomorphic to a semi-algebraic set.
In particular, since $\GP{e}$ is compact and homeomorphic to a semi-algebraic set it admits a triangulation according to \cite[Thm.\,9.2.1]{BochnakEtAl1998} \cite[Thm.\,3.11]{Coste2002}.

\medskip
Second, notice that we obtained a bit more.
The $\pi$ inverse image of the embedded Grassmann polytope $\GP{e}$ via $\gamma$ can be presented as follows
\[
\pi^{-1}(\gamma(\GP{e}))=\gamma(\widehat{Z}(\Mat_{k,n}^{\ge0}(e)))\cup \big(-\gamma(\widehat{Z}(\Mat_{k,n}^{\geq 0}(e)))\big).
\]
Since we proved that $\gamma(\widehat{Z}(\Mat_{k,n}^{\geq 0}(e)))$ is semi-algebraic we can conclude that $\pi^{-1}(\gamma(\GP{e}))$ is also a semi-algebraic subset of $\R^d$.

Having in mind that every real projective variety is affine,  we can define that a subset $X$ of the real projective space $\RP^{d-1}$ is semi-algebraic if, for example, its  preimage $\pi^{-1}(X)\subseteq\R^d$, via the defining quotient map $\R^d\longrightarrow \RP^{d-1}$, is semi-algebraic.

Thus, we proved that the Grassmann polytope $\GP{e}$, when embedded in $\RP^{d-1}$ via the Pl\"ucker embedding, is a semi-algebraic subset of the real projective space.

\qed

\section{Proof of Theorem \ref{thm:homeomorphic}}
\label{sec:homeomorphic}

Let $ k \geq 1$, $m \ge0$ and $n \geq k+m$ be integers, and suppose in addition that $m$ is even. Let $S \in \GL_n$ be given by
\[
S(x_1, \dots, x_n) = (x_2, \dots, x_n, (-1)^{k-1}x_1).
\]
Denote by $Z_0 \in \Mat_{k+m,n}$ the matrix whose rows are the eigenvectors of the matrix $S+S^\top$ that correspond to the largest $k+m$ eigenvalues.
It was shown in \cite[Lemma~3.1]{Galashin2017} that all $(k+m)\times (k+m)$ minors of the matrix $Z_0$ are positive, thus it defines an amplituhedron $\A_{n,k,m}(Z_0)$, called \emph{cyclically symmetric amplituhedron}.
Galashin, Karp \& Lam \cite[Thm.\,1.2]{Galashin2017} showed that $\A_{n,k,m}(Z_0)$ is homeomorphic to a closed $km$-dimensional ball whenever the parameter $m$ is even.

We conclude the proof of Theorem \ref{thm:homeomorphic} by showing that the amplituhedra $\A_{n,k,m}(Z)$ and $\A_{n,k,m}(Z_0)$ are homeomorphic.

From \cite[Cor.\,1.12(ii)]{Karp2017b} we know that entries of every nonzero vector of $\ker(Z_0)$ and of $\ker(Z)$ are nonzero, and they alternate in sign.
Since $n=k+m+1$, the kernels of matrices $Z$ and $Z_0$ are $1$-dimensional.
Let $a=(a_1, \dots, a_n)\in \R^n$ be a generator of the kernel of $Z$ and let $b=(b_1, \dots, b_n)\in \R^n$ be  a generator of the kernel of $Z_0$ (it follows from the cyclic symmetry of $Z_0$ that $b_i=(-1)^{i-1}$ for $1\leq i\leq n$, see~\cite{Galashin2017}).
Choose them in such a way that $a_1$ and $b_1$ have the same sign.
Consequently, for every $1 \leq i \leq n$, the entries $a_i$ and $b_i$ have the same sign.
Let $D$ be an $n \times n$ diagonal matrix $D=\diag(\frac{a_1}{b_1}, \dots, \frac{a_n}{b_n})$.
The matrix $ZD$ has the same kernel as the matrix $Z_0$, and since the diagonal entries of the matrix $D$ are positive, all maximal minors of the matrix $ZD$ are positive.
The fact that the matrices $ZD$ and $Z_0$ have the same kernel implies that they have the same row spans, as well.
In particular, there exists a matrix $C \in \GL_{k+m}^+$ such that $Z_0  = CZD$.

Multiplication by $D$ on the right gives a homeomorphism $\widehat{D}: \Mat_{k,n}^{\ge0} \longrightarrow\Mat_{k,n}^{\ge0}$, which induces a homeomorphism $\widetilde{D}:\nng{k}{n} \longrightarrow \nng{k}{n}$. Furthermore, multiplication by $C^\top$ on the right gives a homeomorphism $\widehat{C}:\Mat_{k,k+m} \longrightarrow \Mat_{k,k+m}$, thus the induced map $\widetilde{C}: \gr_k(\R^{k+m}) \longrightarrow \gr_k(\R^{k+m})$ is also a homeomorphism. Hence, we obtain the commutative diagram of spaces and maps
\[
\begin{tikzcd}
 \Mat_{k,n}^{\ge0}   \arrow{r}{\widehat{D}} \arrow{d} & \Mat_{k,n}^{\ge0}   \arrow{r}{\widehat{Z}} \arrow{d}  &\Mat_{k,k+m} \arrow{r}{\widehat{C}} \arrow{d} & \Mat_{k,k+m} \arrow{d}      \\
\nng{k}{n} \arrow{r}{\widetilde{D}} & \nng{k}{n} \arrow{r}{\widetilde{Z}}  & \gr_k(\R^{k+m}) \arrow{r}{\widetilde{C}} & \gr_k(\R^{k+m}).
\end{tikzcd}
\]
The image of the composition $\widetilde{C}\circ\widetilde{Z}\circ\widetilde{D}$ of the maps in the lower row of the diagram is the cyclically symmetric amplituhedron $\A_{n,k,m}(Z_0)$ and the image of the map $\widetilde{Z}$ is the amplituhedron $\A_{n,k,m}(Z)$. Since the maps $\widetilde{C}$ and $\widetilde{D}$ are homeomorphisms, these two amplituhedra are homeomorphic. Finally, the fact that the cyclically symmetric amplituhedron $\A_{n,k,m}(Z_0)$ is homeomorphic to a $km$-dimensional ball \cite[Thm.\,1.2]{Galashin2017}, when $m$ is even, concludes the argument that every amplituhedron $\A_{n,k,m}(Z)$ is homeomorphic to a $km$-dimensional ball whenever $n=k+m+1$ and $m$ is even.\qed

\medskip
The proof of Theorem \ref{thm:homeomorphic} gives even more.
Let us say that two Grassmann polytopes $\GPZ{e}{Z},\GPZ{e'}{Z'}\subseteq \gr_k(\R^{k+m})$ are \emph{projectively equivalent} if there exists a matrix $M\in \GL_{k+m}$ such that
\[
\GPZ{e'}{Z'}=\{\widetilde{M}(x)\mid x\in \GPZ{e}{Z}\}.
\]
Here $\widetilde{M}$ denotes a map $\gr_k(\R^{k+m})\longrightarrow \gr_k(\R^{k+m})$ induced by the natural action of $M$ on $\R^{k+m}$.
For $k=1$, this coincides with the standard notion of projective equivalence for polytopes in the projective space $\RP^{k+m-1}$.
The proof of Theorem \ref{thm:homeomorphic} actually shows that for $n=k+m+1$, any two amplituhedra are projectively equivalent.


TITLE: In the Group $(S_{3},\circ)$, what are the elements of the group $\Big(\big((123)\big), \circ\Big)$?
QUESTION [0 upvotes]: Given the Group $(S_{3},\circ)$
What are the elements of the group $\Big(\big((123)\big), \circ\Big)$?
Also, why does $\big((123)\big)$ have two brackets around it?

REPLY [2 votes]: I suggest avoiding the round-bracket notation for generated substructures at all costs, since they are already used for grouping expressions and delineating function inputs, and using it for anything else almost surely will result in unnecessary ambiguity.
That said, in a group $G$ and given $g \in G$, "$\langle g \rangle$" denotes "$\{ g^k : k \in \mathbb{Z} \}$". In an infinite group, it is important to have both positive powers and negative powers (powers of the inverse), otherwise the set 'generated' may not be a subgroup. In a finite group the inverse of every element is some positive power of that element itself because of Lagrange's theorem, or equivalently because the positive powers must repeat cyclically.
In general, "$\langle x,y,...,z \rangle$" denotes the smallest subgroup of $G$ that contains $x,y,...,z$. We can prove (and you should try) that the smallest such subgroup exists because $G$ is such a subgroup and the intersection of all such subgroups is also such a subgroup. Alternatively, it can be defined to be $\{ \prod_{k=1}^n {a_k}^{c_k} : n \in \mathbb{N} \land \forall k \in [1..n]\ ( a_k \in \{x,y,...,z\} \land c_k \in \mathbb{Z} ) \}$. It is not hard to prove that this is indeed the smallest subgroup we are looking for.


TITLE: Creating a Hermitian function
QUESTION [1 upvotes]: Say I have an operator $A$ such that $A^\dagger = B$.
I want to construct a Hermitian function, $f$, of these operators, $f(A,B)^\dagger = f(A,B)$. Is it possible to construct a function $f$ such that $f$ is not a function of $A+B$?:  $f(A,B)\neq f(A+B)$ but $f(A,B)^\dagger = f(A,B)$?

REPLY [4 votes]: Observe that
$$(\mathrm{i}(A-B))^\dagger = -\mathrm{i}(A^\dagger - B^\dagger) = \mathrm{i}(A-B)$$
so $f(A) = \mathrm{i}(A-A^\dagger)$ is a Hermitian function of $A$ that is not a function of $A+A^\dagger$.

REPLY [2 votes]: A trivial answer might be any product of $A$ and $A^{\dagger}=B$ is an hermitian function take $f_1(A,B)=AB=AA^{\dagger}$ and than since also every power of $AB$ is an hermitian function: ${(AB)^n}^{\dagger}=B^{\dagger}A^{\dagger}B^{\dagger}A^{\dagger}....B^{\dagger}A^{\dagger}=ABAB....AB = (AB)^n$ and since you can define $$f(A,B)=\sum_{n}\alpha_n(AB)^n$$ with $\alpha_n \in \mathbb{R}$ there you have a large class of functions of $A$ and $B$ which are $f(A,B)=f(AB)$. Another possibility that comes to my mind is a slight generalization of what you wrote(which takes into account just linear combinations of $A$ and $A^{\dagger}$): take any function $g(A,B)$ and define $f(A,B)=g(A,B)+g(A,B)^{\dagger}$ or $f(A,B)=i(g(A,B)-g(A,B)^{\dagger})$.
To my knowledge these are the most common cases without any further specification on the properties of $A$ ,edit: beyond the trivial cases of symmetrization and antisymmetrization $A+A^{\dagger}$ and $i(A-A^{\dagger})$


TITLE: If $\lim_{x\to n}f(x) = 0$ and $f(n)=0$, does $\lim_{x\to n}\frac{\sin(f(x))}{f(x)} = 1$ always?
QUESTION [1 upvotes]: If $\lim_{x\to n}f(x) = 0$ and $f(n)=0$, does $\lim_{x\to n}\frac{\sin(f(x))}{f(x)} = 1$ always?
I have been playing around with some graphs on desmos and there's always the indication that the limit equals $1$. I know that $\sin x$ becomes very linear, is there any function with the potential to "un-linearize" it?

REPLY [1 votes]: Yes, assuming that:

$\exists \delta' > 0$ such that if $|x - n| < \delta'$, then $|f(x)| > 0$.

Since $f(n) = \lim_{x\to n}f(x) = 0$ and $\lim_{y \to 0} \frac{\sin y}{y} = 1$, we know that:

$\forall \varepsilon_1 > 0$, $\exists \delta_1 > 0$ such that if $|x - n| < \delta_1$, then $|f(x)| < \varepsilon_1$
$\forall \varepsilon_2 > 0$, $\exists \delta_2 > 0$ such that if $0 < |y| < \delta_2$, then $|\frac{\sin y}{y} - 1| < \varepsilon_2$

Now given any $\varepsilon > 0$, let $\delta > 0$ be $\min({\delta', \delta_1})$, where:

$\delta_1$ corresponds to taking $\varepsilon_1 = \delta_2$
$\delta_2$ corresponds to taking $\varepsilon_2 = \varepsilon$

Then observe that if $0 < |x - n| < \delta$, then:

We know that $|x - n| < \delta \leq \delta'$, so $|f(x)| > 0$.
We know that $|x - n| < \delta \leq \delta_1$, so $|f(x)| < \varepsilon_1 = \delta_2$.
We know that $0 < |f(x)| < \delta_2$, so $|\frac{\sin f(x)}{f(x)} - 1| < \varepsilon_2 = \varepsilon$, as desired. $~~\blacksquare$


TITLE: PDF of $f(x)=1/\sin(x)$?
QUESTION [1 upvotes]: What is the probability density function (PDF) of $f(x)=1/\sin(x)$ when $x$ is uniformly distributed in $(0,90)$?
$f(x)=\sin(x)$ has a known PDF, which has the form $2(\pi\sqrt{1-\sin(x)^2})^{-1}$, but I cannot find the PDF for $1/\sin(x)$. The latter would have  interesting applications in astronomy, specially for so-called "luminosity functions".
Thank you very much
Sebastian

REPLY [3 votes]: Let $z=1/\sin(x)$, then $x=\sin^{-1}(1/z)$ on $(0,\frac{\pi}{2})$ interval. The pdf you seek, is the differential of uniform c.d.f., which is $\frac{2}{\pi} dx$
$$ \frac{2}{\pi} d( \sin^{-1}( z^{-1} ) ) = \frac{2}{\pi z} \frac{1}{\sqrt{z^2-1}},$$ where variable $z>1$.
By the way, you could have used Mathematica to directly find this:
In[100]:= FullSimplify[PDF[TransformedDistribution[1/Sin[x],
       x \[Distributed] UniformDistribution[{0, Pi/2}]], z]]

Out[100]= Piecewise[{{2/(Pi*z*Sqrt[-1 + z^2]), z > 1}}, 0]


TITLE: Show that any plane whose normal lies on cone $(b+c)x^2+(c+a)y^2+(a+b)z^2=0$ cuts the surface $ax^2+by^2+cz^2=1$ is rectangular hyperbola
QUESTION [0 upvotes]: Show that any plane whose normal lies on cone $(b+c)x^2+(c+a)y^2+(a+b)z^2=0$ cuts the surface $ax^2+by^2+cz^2=1$ is rectangular hyperbola
My attempt:
let $\frac {x}{l} = \frac {y}{m} = \frac {z}{n}$ be normal of plane $lx+my+nz = 0$. Now the normal is generator of given cone. So we get condition $(b+c)l^2+(c+a)m^2+(a+b)n^2=0$ --->(1)
THe given plane cuts the conicoid $ax^2+by^2+cz^2=1$.
The intersecting conic :
Eliminating z between plane equation and conicoid equation
$z = \frac{-lx-my}{n} $ Subsittuting this in conicoid equation we get,
$ax^2+by^2+c\frac{(-lx-my)^2}{n^2} =1 $ --->(2)
This represents rectangular hyperbola if
coefficient of $x^2$ + coefficient of $y^2$ = 0
$(2)$ implies
$a+b+c\frac{l^2}{n^2}+c\frac{m^2}{n^2} =0 \\ \fbox {$n^2(a+b) + c(l^2+m^2) = 0$}$ for conic to be rectangular hyperbola
If i eliminate x, i get $\fbox {$l^2(b+c)+a(m^2+n^2)=0$}$  for conic to be rectangular hyperbola
If i eliminate y, i get $\fbox {$m^2(c+a)+b(n^2+l^2)=0$}$  for conic to be rectangular hyperbola
But what i have is  equation (1) with me for conic to be rectangular hyperbola
How to proceed further?

REPLY [0 votes]: As an alternative solution,
Let $/$=$/$=$/$ be normal of plane ++=0. Now the normal is generator of given cone. So we get condition $(+)^2$+$(+)^2$+$(+)^2$=0 -----------(eq. 1)
Thus, our required plane is lx+my+nx=p whose normal is $/$=$/$=$/$
We now show that it's intersection with conicoid: $^2$+$^2$+$^2$=1 generates rectangular hyperbola, i.e. the asymptotes are mutually perpendicular.
Consider the asymptotes in the plane (lx+my+nx=p) as
$(-x_0)/p$=$(-y_0)/q$=$(-z_0)/r$, having DCs <p,q,r> thus we should show that $p_1p_2$+$q_1q_2$+$r_1r_2$=0 then it will be a rectangular hyperbola.
Now, since the line (asymptote) with DCs <p,q,r> lie in plane lx+my+nz=p,
thus lp+mq+nr=0 --------(eq.2)
For the line $(-x_0)/p$=$(-y_0)/q$=$(-z_0)/r$ = k(say), we have x=kp+$x_0$ ; y=qk+$y_0$ ; z=rk+$z_0$ and we put this in conicoid equation $^2$+$^2$+$^2$=1
we will get a quadratic in $k^2$, here we can put the coefficient of $k^2$ equal to zero, since the solution will be at infinity (for asymptote).
Hence, we get another equation:
$ap^2$+$bq^2$+$cr^2$=0 ---------(eq.3)
from equation 2 and equation 3, putting $p_1p_2$+$q_1q_2$+$r_1r_2$=0 (here, we just need to show sum of product of roots is zero)
we will get our equation 1  i.e. $(+)^2$+$(+)^2$+$(+)^2$=0
Q.E.D.


TITLE: How to deal with this boundary condition when using separation of variables?
QUESTION [2 upvotes]: Consider Laplace's equation $\nabla^2u = 0$ on the region $A = [0,a]\times [0,b]\times [0,c]\subset \mathbb{R}^3$ and suppose we impose the boundary conditions:
$$u(0,y,z) = \sin \frac{\pi y}{b} \sin \frac{\pi z}{c}, \quad u(a,y,z)=0 \\ u(x,0,z)=u(x,b,z)=u(x,y,0)=u(x,y,c)=0.$$
In that case if we use separation of variables we set $u(x,y,z)=X(x)Y(y)Z(z)$ and the equation reduces to
$$\frac{X''}{X}+\frac{Y''}{Y}+\frac{Z''}{Z}=0.$$
In that case reorganizing gives
$$\frac{X''}{X}+\frac{Y''}{Y}=-\frac{Z''}{Z}=\lambda$$
for some constant $\lambda$. Thus we have two new equations, one of which is
$$\frac{X''}{X} = \lambda-\frac{Y''}{Y} = \mu,$$
for some constant $\mu$. Finally we end up having three equations
$$\begin{cases}Z''+ \lambda Z &= 0, \\ Y'' + (\mu+\lambda)Y&=0, \\ X'' - \mu X &=0\end{cases}.$$
The boundary conditions which are zero on the boundaries transfer nicely to the functions $X,Y,Z$. That is, except the first condition, all the other becomes
$$X(a)=0, Y(0)=Y(b)=0, Z(0)=Z(c)=0.$$
With this the equations for $Y$ and $Z$ with those boundary conditions becomes Sturm Liouville problems. On the other hand the first condition does not separate nicely. We have
$$X(0)Y(y)Z(z) = \sin \frac{\pi y}{b} \sin \frac{\pi z}{c}.$$
This boundary condition doesn't seem to transfer nicely to $X$. So how do we deal with this boundary condition? How do we use it to solve the problem in question?

REPLY [1 votes]: $$\frac{X''}{X}+\frac{Y''}{Y}+\frac{Z''}{Z}=0.$$
$$\frac{X''}{X}+\frac{Y''}{Y}=-\frac{Z''}{Z}=\lambda$$
$$\lambda-\frac{X''}{X} = \frac{Y''}{Y} = \mu,$$
so you solve 3 auxiliary problem 2 of them are  Sturm Liouville:
$$(1)\;\begin{cases}Z''+ \lambda Z &= 0 \\ Z(0)=0=Z(c)\end{cases}
\qquad (2)\;\begin{cases} Y'' + \mu Y&=0, \\ Y(0)=0=Y(b)\end{cases}$$
The solutions are (verify!) $Z_n(z)=\sin\lambda_n z$, $Y_m(z)=\sin \mu_m z$.
The last one is a "boundary value problem" with missing condition, you need two but here you still have only one. Note that at this stage $\mu$ and $\lambda$ are  known.
$$\begin{cases}X'' - (\mu_m+\lambda_n) X &=0\\X(a)=0\end{cases}$$
The solution  is (verify!)
$$X_{mn}(x)=A_{mn}  e^{a b_{mn}} \sinh (b_{mn} (a-t)),$$ where $b_{mn}=\mu_m+\lambda_n$.
Note that you  still have a constant ($A_{mn}$) to determine, for which you will use the last condition in the following way. The overall solution is now given by $$u(x,y,z)=\sum_{mn} A_{mn}  X_{mn}(x)Y_m(y) Z_n(z)$$
now for $x=0$ you get
$$u(0,y,z)=\sum_{mn} A_{mn}  X_{mn}(0)Y_m(y) Z_n(z) = \sin \frac{\pi y}{b} \sin \frac{\pi z}{c}$$
which you solve for $A_{mn}$  in general using orthogonality etc - you know, the formula with the integral. However, for this example matching of the coefficients should work, because you have the RHS given in the form of the series on the left side.


TITLE: If $f:S^1 \to \mathbb{R}$, defined by $f(x,y)=x-y$ and $X_p$ be a tangent vector at $p=(\frac{1}{\sqrt 2}, \frac{1}{\sqrt 2})$
QUESTION [0 upvotes]: If $f:S^1 \to \mathbb{R}$, defined by $f(x,y)=x-y$ and $X_p$ be a tangent vector at $p=(\frac{1}{\sqrt 2}, \frac{1}{\sqrt 2})$ given by $X_p=4 (\frac{\partial}{\partial t})_p$ where $t$ is the local coordinate on the chart $(U, \phi)$, where  $U=S^1-\{(0,0)\}$, and $\phi(x,y)=\frac{x}{1-y}$, then find $X_p(f)$.
Answer:
$X_p(f)=4(\frac{\partial}{\partial t})_pf=4(\frac{\partial}{\partial x})_p (x-y)+4(\frac{\partial}{\partial y})_p(x-y)=4-4=0$
But where is the need of $ \phi(x,y)=\frac{x}{1-y}$ ?

REPLY [2 votes]: The easiest way to proceed here is to solve for $x(t)$ and $y(t)$ [Hint: Use $x^2+y^2=1$.] and then use
$$\frac{\partial}{\partial t} = \frac{\partial x}{\partial t}\frac{\partial}{\partial x} + \frac{\partial y}{\partial t}\frac{\partial}{\partial y}.$$
In this case, at the point $(1/\sqrt2,1/\sqrt2)$, you should find that
$$\frac{\partial}{\partial t} = \frac{1-\sqrt2}2\,\frac{\partial}{\partial x} + \frac{\sqrt2-1}2\,\frac{\partial}{\partial y}.$$
Note that this vector is a multiple of $-\frac{\partial}{\partial x}+\frac{\partial}{\partial y}$, as is appropriate at a point on the circle with $x=y$.
Even easier, once you know $x(t)$ and $y(t)$, then rewrite $f$ as $f(x(t),y(t))$ and take $\partial/\partial t$ !!


TITLE: Which integers can be expressed as a sum of three cubes in infinitely many ways?
QUESTION [42 upvotes]: For fixed $n \in \mathbb{N}$ consider integer solutions to
$$x^3+y^3+z^3=n \qquad (1) $$
If $n$ is a cube or twice a cube, identities exist.
Elkies suggests no other polynomial identities are known.

For which $n$ (1) has infinitely many integer solutions?

Added

Is there $n$, not a cube or twice a cube, which allows infinitely
  many solutions?

Added 2019-09-23:
The number of solutions can be unbounded.
For integers $n_0,A,B$ set $z=Ax+By$ and consider
$x^3+y^3+(Ax+By)^3=n_0$. This is elliptic curve
and it may have infinitely many rational points
coming from the group law. Take $k$ rational points
$(X_i/Z_i,Y_i/Z_i)$. Set $Z=\rm{lcm}\{Z_i\}$.
Then $n_0 Z^3$ has the $k$ integer solutions $(Z X_i/Z_i,Z Y_i/Z_i)$.

REPLY [27 votes]: As stated by Dietrich Burde, it is known that there is no solution for $n\equiv \pm 4 \pmod{9}$, and conjectured that there are infinitely many solutions otherwise.
A cryptic aspect is that it is not even known that there exists one solution for all $n \not\equiv \pm 4 \pmod{9}$.
Today the smallest number for which the problem is open is $n=114$.
Here is a (non-exhaustive) history of the latest solutions found for $n \le 100$ (see here and there):
(1960s)

$87 = 4271^3 – 4126^3 – 1972^3$
$96 = 13139^3 -15250^3 + 10853^3$
$91 = 83538^3 – 67134^3 – 65453^3$
$80 = 103532^3 -112969^3 + 69241^3$

(1990s)

$39 = 134476^3 - 159380^3 + 117367^3$
$75 = 435203083^3 – 435203231^3 + 4381159^3$
$84 = 41639611^3 – 41531726^3 – 8241191^3$

(2000s)

$30 = 2220422932^3 – 2218888517^3 – 283059965^3$
$52 = 23961292454^3 - 61922712865^3 + 60702901317^3$
$74 = 66229832190556^3 − 284650292555885^3 + 283450105697727^3$

(2019)

$33 = 8866128975287528^3 - 8778405442862239^3 -2736111468807040^3$
$42 = 80435758145817515^3 - 80538738812075974^3 +  12602123297335631^3$
$3 = 569936821221962380720^3 - 569936821113563493509^3 - 472715493453327032^3$
$906 = 72054089679353378^3 -74924259395610397^3 + 35961979615356503^3$
$165 = 383344975542639445^3 -385495523231271884^3 + 98422560467622814^3$

Remark: for $n \le 1000$, the problem is still open only for $114$, $390$, $579$, $627$, $633$, $732$, $921$, and $975$ (see this paper and this paper, and also this).
Numberphile's videos:

The Uncracked Problem with 33
42 is the new 33
NEWS: The Mystery of 42 is Solved


TITLE: Gauss Law in conducting slab
QUESTION [1 upvotes]: I think the charge density should be 2E(o)ε0 via a rectangular Gaussian box oriented through slab but I am confused about the fields.Is it zero on first region and 2E(0) on second one?

REPLY [0 votes]: Think of it as the addition of three electric fields.
One is the external electric field and the other two are due to  the electric fields produced by the induced charges.

Equal numbers of positive and negative charges are induced on the surface of the conductor so the electric fields due to these charges have the same magnitude.
$|\vec E_+| =|\vec E_-|  $
Outside the conductor these electric fields due to the induced charges are in opposite direction and so cancel each other out leaving the external electric field $\vec E_o$.
Inside the conductor the electric fields due to induced charges are in the same direction but in the opposite direction to the external electric field and add up with a magnitude equal to the magnitude of the electric field.
Thus the net electric field in the conductor is zero.
$\vec E_+ + \vec E_-  + \vec E_o =0$
Now you can use a Gaussian cylinder across one of the surfaces to find the surface charge density of the induced charges.


TITLE: Is there an orthogonal time dimension in Minkowski spacetime?
QUESTION [3 upvotes]: Space dimensions are orthogonal one to each other. But what about time in the Minkowski diagram?
At first sight, time seems to be orthogonal to space. But we have to consider that each Minkowski diagram is an observer's diagram. Thus, an object that is not moving with regard to the observer will describe a vertical worldline in the Minkowski diagram of the observer. However, from the point of view of another observer (inertial frame) the worldline described by the object might be inclined. Even the worldline of the observer himself which is the y-axis might be sloped from the point of view of other observers.
The conclusion seems to be that time is not orthogonal to space. Or is there an error?

REPLY [0 votes]: Any timelike vector has a 3-dimensional subspace orthogonal to it, orthogonal under the Minkowski inner product.  Such a subspace is spanned by vectors that are spacelike under the inner product, and all linear combinations of those vectors are spacelike.
Any timelike vector might be the direction a massive object follows at some point on its worldline.  The 3-dimensional subspace orthogonal to that four-velocity is what an observer following that trajectory would instantaneously consider "space".
An observer's "time" is always orthogonal to his "space", and further, all observers would agree that that is so.


TITLE: Utility of the coordinate free definition of the derivative on manifolds.
QUESTION [1 upvotes]: Preface: I am not an expert on the topic of smooth manifolds, nor do I have the perspective gained from knowing many theorems proven on smooth manifolds. Please try to look at the problem from the perspective of someone learning this topic for the first time.
For future reference let $f:M \rightarrow N$ be a smooth function where $M$ and $N$ are smooth manifolds. Let the derivative of $f$ at $p\in M$ be denoted as the usual $df_p:T_pM \rightarrow T_{f(p)}N$.
Question:
In what ways is it advantageous to define $df_p$ in a coordinate free way? I cannot see the motivation behind such a definition. To me there are three defining characteristics of the derivative at a point of a function between $\mathbb{R}^n$ and $\mathbb{R}^m$.

The derivative of a function offers an approximate relationship between perturbations of the input and corresponding perturbations of the output.

This relationship is linear.

This approximation is better than any other linear approximation.

Let's check how many of these properties actually survive the abstraction process needed for a coordinate free generalization of the derivative on manifolds.
Property 1: To interpret a tangent vector as a "perturbation" of the point p, we would need some sort of canonical mapping from $T_pM$ to $M$. I do not know of any such mapping which doesn't rely on a local coordinates. Since we can't interpret the inputs and outputs of $df_p$ as perturbations, I assume that the notion of local approximation is abandoned as well. How could $df_p$ "approximate" $f$ when we don't even have a way to relate their respective domains and ranges. Thus I will assume that this property of the derivative has been stripped away in the abstraction process.
Property 2: This is satisfied.
Property 3: Since we have determined that without coordinates $df_p$ does not act as an approximation of $f$, it cannot be the "best" approximation.
Thus only one property survives. $df_p$ seems to be (to my unexperienced eyes) a glorified linear mapping with no substantial importance until local coordinates are introduced. Thus why don't we simply define the derivative in a coordinate dependent manner? This would also make the definition of the tangent space much simpler. The "geometric tangent space", $\mathbb{R}^n_p$, as defined in Lee's "Smooth Manifolds" would suffice as a definition since we are assuming some coordinate representation. Is anything lost by doing this?
In response to nicrot000’s answer:
To summarize, nicrot000 states in his answer that it is the general trend of differential geometry to express concepts in coordinate free ways. This allows said concepts to be independent of arbitrary decisions (i.e. choice of local coordinates). /end of summary\
I guess my question is rooted in the fact that I’m not convinced arbitrary decisions are a bad thing in terms of defining the derivative. From my perspective, $df_p$ can only be interpreted in some meaningful way after local coordinates have been assumed.
An analogy:
Suppose we have a computer which we will refer to as the “idea machine”. The sole purpose of this computer is to run a program which, on command, generates a random idea. After you have clicked “generate idea”, the program will prompt you to choose from a set of languages (all of which can be “smoothly” translated between each other) in which the idea will be displayed on the screen (the choice of a language corresponds to choosing a local coordinate chart). There is however the option to choose no language (corresponding to coordinate free). If no language is chosen, nothing will be displayed on the screen and the idea will forever stay encoded by the ones and zeros in the hardware. We do however know that the idea exists independent of our arbitrary choice of language (hooray!). At the end of the day, if we want to know what the idea was then in some form or another we would have to use a “language” to interpret the ones and zeros in the hardware.
Interpretation of the analogy:
The idea produced by the idea machine represents the information encoded by the coordinate free function $df_p$. The set of languages which one can choose from represents the smooth structure. Choosing a language to display the idea corresponds to choosing a coordinate chart to represent $df_p$. We can choose to use a coordinate free representation however I do not see the purpose of this just as choosing no language is useless. From a developer standpoint, why would we go through the work of adding the no language option? i.e. why would we go through the work of developing a coordinate free definition of $df_p$.

REPLY [1 votes]: A coordinate free approach is important when dealing with infinite dimensional manifolds. In infinite dimensions, there is typically no natural coordinate system to work with.
You might think this is a niche topic because we live in a finite dimensional world. However, generally a space of functions that maps between two finite dimensional manifolds is an infinite dimensional manifold.
So, even when you start with finite dimensional manifolds, infinite dimensional manifolds arise naturally. This is in keeping with the modern trend in all of mathematics towards studying objects by instead studying spaces of functions that map between those objects.


TITLE: Using Mean Value Theorem, Prove that ${\tan x\over x}>{x\over\sin x}$
QUESTION [5 upvotes]: Prove using Mean Value Theorem,

$${\tan x\over x}>{x\over\sin x} \space\forall \space x \space\in(0, \pi/2) $$

Attempt::
$f(x) = x-\sin x$
$f^\prime(x) = 1-\cos x > 0 $
Hence $x-\sin x>0, {x\over\sin x}>1$
Similarly, I got ${\tan x\over x}>1$
But how do I compare them and get the required inequality?

REPLY [0 votes]: Suppose we have a function $f(t)=\sqrt{\sin{t}\tan{t}}$. Now, applying the mean value theorem in $(0, \frac{\pi}{2})$, we have $$\frac{f(x)-f(0)}{x-0}=f^\prime(c)$$ Or $$\frac{f(x)}{x}=f^\prime(c)\quad (where\quad 0<c<x<\frac{\pi}{2})$$ Or $$\frac{\sqrt{\sin{x}\tan{x}}}{x}=f^\prime(c)\tag{1}$$ Now we proceed to derivation of $f(t)$: $$f^\prime(t) = \frac{\sin{t}\sec^2{t}+\tan{t}\cos{t}}{2\sqrt{\sin{t}\tan{t}}}$$ $$=\frac{\tan{t}(\sec{t}+\cos{t})}{2\sqrt{\sin{t}\tan{t}}}$$ $$=\frac{\tan{t}(\sec{t}+\cos{t})\sqrt{\cos{t}}}{2{\sin{t}}}$$ $$=\frac{\sec{t}+\cos{t}}{2\sqrt{\cos{t}}}$$ $$=\frac{\sec{t}\sqrt{\sec{t}}+\sqrt{\cos{t}}}{2}$$ Now, $$f^{\prime\prime}(t)=\frac{3\sec{t}\tan{t}-\sin{t}}{2{\sqrt{\cos{t}}}}$$ Since $f^{\prime\prime}(t) = 0$ at $ t = 0$ we get that $f^\prime(t)$ is minimum at $t = 0$. As $f^\prime(0) = 1$, we conclude that for $c\in(0, \frac{\pi}{2})$, $f^\prime(c)>1$. Therefore, from (1) we have, $$\frac{\sqrt{\sin{x}\tan{x}}}{x}>1\quad x\in(0, \frac{\pi}{2})$$ or $$\frac{\sin{x}\tan{x}}{x^2}>1$$ or $$\frac{\tan{x}}{x}>\frac{x}{\sin{x}}\quad
 if \quad 0<x<\frac{\pi}{2}$$
$$***$$


TITLE: How to find the number of outcomes in an events with colored balls
QUESTION [0 upvotes]: An urn holds 13 identical balls except that 1 is white, 9 are black, and 3 are red. An experiment consists of selecting two balls in succession without replacement and observing the color of each of the balls.
What math would I do to find the number of outcomes?
What math would I do to find the out the event if a certain color is not selected?
I couldn't find anything relating to this. I have found basic examples of how to find the number of out comes, however nothing like this. If you found something similar, please tell me.

REPLY [1 votes]: The number of outcomes can be calculated by the number of balls.
The first ball is one of 13, the second is one of 12, so the number of outcomes is $13\times 12$.
To calculate the probability that a colour is not selected, work out $1 -
\Pr(A)$, where A = outcome includes that colour.


TITLE: What are all the concordant forms $n$ such that $a^2+b^2 = c^2,\,a^2+nb^2=d^2$ for $n<1000$?
QUESTION [4 upvotes]: Part I. The list of congruent numbers $n<10^4$ such that the system,
$$a^2-nb^2 = c^2$$
$$a^2+nb^2 = d^2$$
has a solution in the positive integers is known (A003273)
$$n = 5, 6, 7, 13, 14, 15, 20, 21, 22, 23, 24, 28, 29, 30, 31, 34,\dots$$

Part II. Strangely, for the similar concordant forms/numbers $n$ such that,
$$a^2+b^2 = c^2$$
$$a^2+nb^2 = d^2$$
is not even in the OEIS,
$$n = 1,7,10,11,17,20,22,23,24,27,30,31,34,\dots$$
The list of $104$ prime $n<10^3$ is known (by Kevin Brown, David Einstein (hm?), and Allan MacLeod) though several troublesome primes were excluded assuming the Birch/Swinnerton-Dyer conjecture.

Question: Anybody knows how to generate the list of all concordant forms $n<1000$? (Elkies describes a method here.)

P.S. Incidentally, the special case $n=52$ appears in equal sums of like powers. Let,
$$a^2+b^2 = c^2$$
$$a^2+52b^2 = d^2$$
Then for $k=1,2,4,6,8,10,$
$$(8b)^k + (5a-4b)^k + (-a-2d)^k +  (a-2d)^k + (-5a-4b)^k  + (-12b+4c)^k + (12b+4c)^k =\\
(4a+8b)^k + (3a-2d)^k + (-3a-2d)^k + (-4a+8b)^k + (-16b)^k + (a+4c)^k + (-a+4c)^k$$
found by J. Wroblewski and yours truly. An initial primitive solution is $a,b,c,d = 3,4,5,29$ and an infinite more.

REPLY [0 votes]: While individ in his answer has shown there is an infinite number of concordant forms $n$, it can also be shown there is an infinite number of the special case $n = \pm N^2$. Given the system,
$$a^2+b^2 = c^2\tag1$$
$$a^2+nb^2 = d^2\tag2$$
First, let $a,\,b,\,c = p^2-q^2,\,2pq,\,p^2+q^2$ to satisfy $(1)$. The second becomes,

$$(p^2-q^2)^2+n(2pq)^2 =d^2\tag3$$

Two parametric solutions to $(3)$ are,
$$n = -(2t^2-2)^2,\quad p =t(2t^2-1),\quad q=1$$
found by Will Jagy and,
$$n = (4t^2+3)^2,\quad p =t(4t^2+1),\quad q=1$$
by this OP.