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\begin{document}
	\begin{CJK}{UTF8}{song}
\title{ The stabilizer for $n$-qubit symmetric states }
	\author{Xian Shi}
	\affiliation{Institute of Mathematics, Academy of Mathematics and Systems Science, Chinese Academy of Sciences, Beijing 100190, China
		\\
		University of Chinese Academy of Sciences, Beijing 100049, China
		\\
		UTS-AMSS Joint Research Laboratory for Quantum Computation and Quantum Information Processing,
		Academy of Mathematics and Systems Science, Chinese Academy of Sciences, Beijing 100190, China}
	\email{shixian01@gmail.com}
	\pacs{03.67.Mn, 03.65.Ud }
	\begin{abstract}
	\indent The stabilizer group for an $n$-qubit state $\ket{\phi}$ is the set of all invertible local operators (ILO) $g=g_1\otimes g_2\otimes \cdots\otimes g_n,$ $ g_i\in \mathcal{GL}(2,\mathbb{C})$ such that $\ket{\phi}=g\ket{\phi}.$ Recently, [Gour $et$ $al.$ 2017 J. Math. Phys. (N.Y.) \textbf{58} 092204] presented that almost all $n$-qubit state $\ket{\psi}$ own a trivial stabilizer group when $n\ge 5.$ In this article, we consider the case when the stabilizer group of an $n$-qubit symmetric pure state $\ket{\psi}$ is trivial. First we show that the stabilizer group for an n-qubit symmetric pure state $\ket{\phi}$ is nontrivial when $n\le 4$. Then we present a class of $n$-qubit symmetric states $\ket{\phi}$ with trivial stabilizer group when $n\ge 5$. At last, we prove that when $m\ge 5$, almost all $n$-qubit symmetric pure states own a trivial stabilizer group, due to the results in [Gour $et$ $al.$ 2017 J. Math. Phys. (N.Y.) \textbf{58} 092204], we have that almost all $n$-qubit symmetric pure states are isolated.
	\end{abstract}
\maketitle
\end{CJK}
\section{Introduction}
\indent Quantum entanglement \cite{HHH} is a valuable resource for a variety of tasks that cannot be finished by classical resource. Among the most popular tasks are quantum teleportation \cite{BB} and quantum superdense coding \cite{BS}. Due to the importance of quantum entanglement, the classification of quantum entanglement states is a big issue for the quantum information theory.\\
\indent Entanglement theory is a resource theory with its free transformation is local operations and classical communication (LOCC). As LOCC is hard to deal with mathematically, and with the number of the parties of the quantum systems grows, the classification of all entanglement states under the LOCC restriction becomes very hard. A conventional way is to consider other operations, such as stochastic LOCC (SLOCC), local unitary operations (LU), separable operations (SEP).\\
\indent Two n-partite states $\ket{\phi}$ and $\ket{\psi}$ are SLOCC equivalent \cite{WGJ} if and only if there exists n invertible local operations (ILO) $A_i,i=1,2,\cdots,n$ such that
\begin{align}
\ket{\phi}=A_1\otimes A_2\otimes \cdots\otimes A_n\ket{\psi}.
\end{align}
Then the classification for multi-qubit pure states under SLOCC attracts much attention \cite{MBDM, CC,L,ZZH, LL, GW}. However, there are uncountable number of SLOCC inequivalent classes in $n$-qubit systems when $n\ge 4,$ so it is a formidable task to classify multipartite states under SLOCC. \\
\indent SEP is simple to describe mathematically and contains LOCC strictly, as there exists pure state transformations belonging to SEP, but cannot be achieved by LOCC. The authors in \cite{GRW} presented that the existence of transformations under separable operations between two pure states  depends largely on the stabilizer of the state. Recently, Gour $et$ $al.$ showed that almost all of the stabilizer group for 5 or more qubits pure states contains only the identity \cite{GKW}. And the authors in \cite{SW} generalized this result to n-qudit systems when $n>3,d>2.$\\
\indent Symmetric states belong to the space that is spanned by the pure states invariant under particle exchange, and there are some results done on the classifications under SLOCC limited to symmetric states \cite{ TSPM,PSM,PR,PJM,BBM}. The authors in \cite{PSM} proved if $\ket{\psi}$ and $\ket{\phi}$ are $n$-qubit symmetric pure states, and there exists n invertible operations $A_i,i=1,2,\cdots,n$ such that $\ket{\psi}=A_1\otimes A_2\otimes \cdots \otimes A_n\ket{\phi},$ then there exists an invertible matrix A such that
\begin{align}
\ket{\psi}=A^{\otimes n}\ket{\phi}.
\end{align}
Moreover, P. Migdal $et$ $al.$ \cite{PJM} generalized the results from qubit systems to qudit systems.\\
\indent In the article, we consider the problem on the stabilizer groups for $n$-qubit symmetric states. This article is organized as follows. In section \uppercase\expandafter{\romannumeral2},
we present preliminary knowledge on $n$-qubit symmetric pure states. In section \uppercase\expandafter{\romannumeral3}, we present our main results. First, we present the stabilizer group for an $n$-qubit symmetric state is nontrivial when $n\le 4$, then we present a class of $n$-qubit symmetric states whose stabilizer group is trivial when $n\ge 5$, at last, when $m= 2,$ the pure state $\ket{\psi}$ owns a nontrivial stabilizer group, when $m=4,$ there exists only one case when $\ket{\psi}$ owns a nontrivial stabilizer group, when $m\ge 5,$ the stabilizer group of almost all $n$-qubit symmetric pure state is trivial. In section \uppercase\expandafter{\romannumeral4}, we will end with a summary.
\section{Preliminary Knowledge}
\indent In this section, we will first recall the definition of symmetric states, and then we present the Majorana representation for an $n$-qubit symmetric pure state.\\
\indent A pure state $\ket{\psi}\in \mathcal{H}_2$ can be represented by a point on
a Bloch sphere geometrically as $\ket{\psi} = \cos \frac{\theta}{2}
\ket{0} + e^{i\phi} \sin\frac{\theta}{
2}
\ket{1},$ here two parameters $\theta\in[0, \pi], \phi\in [0, 2\pi).$

\begin{figure}
\centering

\includegraphics[width=60mm]{1.jpg}\\
\caption{Bloch Sphere}\label{Figure 1}
\end{figure}
We call an $n$-partite pure state $\ket{\psi}$ symmetric state if it is invariant under permuting the particles. That is, for any permutation
operator $P_{\pi}, P_{\pi}\ket{\psi} = \ket{\psi}.$
Generally, there are two main characterizations for an $n$-qubit symmetric pure state $\ket{\psi}$, Majorana representation \cite{M} and
Dicke representation . The Majorana representation for an n-qubit symmetric pure state is that there exists single particles $\ket{\phi_i},i=1,2,\cdots,n,$ such that
\begin{align}
\ket{\psi}=\frac{e^{i\theta}}{\sqrt{K}}\sum_{\sigma\in \textit{perm}} \ket{\phi_{\sigma(1)}} \ket{\phi_{\sigma(2)}}\cdots\ket{\phi_{\sigma(n)}},
\end{align}
where the sum runs over all distinct permutations $\sigma$, $K$ is a normalization prefactor and the $\ket{\phi_i}$ are single qubit states $\ket{\phi_i}=\cos \frac{\theta_i}{2}
\ket{0} + e^{i\phi_i} \sin\frac{\theta_i}{2}
\ket{1},i=1,2,\cdots n,$ we would denote $\ket{\phi_i}$ as $\ket{\phi_i}=a_i\ket{0}+b_i\ket{1}$ below.
 An n-qubit symmetric pure state $\ket{\psi}$ can also be characterized
as the sum of the Dicke states $\ket{D(n, k)}$, that is,
\begin{align}
\ket{\psi}=\sum x_k\ket{D(n,k)}.
\end{align} Here the Dicke states $\ket{D(n,k)}$ are defined as
\begin{align}
\ket{D(n,k)}=\frac{1}{\sqrt{C_n^k}}\sum \ket{\underbrace{0\cdot\cdot\cdot 0}_{n-k}\underbrace{1\cdot \cdot\cdot 1}_k},
\end{align}where the sum runs over all the permutations of the qubits. Up to a global phase factor, the parameters $a_i/b_i$ in a pure state $\ket{\phi_i}= a_i
\ket{0} + b_i
\ket{1}$
are
the roots of the polynomial $P(z) = \sum(-1)^k x_k\sqrt{C_n^k} z^k,$ here $C_n^k$ denotes the binomial coefficient of $n$ and $k.$  \\
\indent Next we introduce an isometric linear map
\begin{align}
\mathcal{M}: \mathcal{L}(\mathcal{H}_1,\mathcal{H}_2)\longrightarrow \mathcal{H}_1\otimes \mathcal{H}_2\\
\ket{b}\bra{a}\longrightarrow \ket{b}\ket{a}
\end{align}
here we denote that $\mathcal{L}(\mathcal{H}_1,\mathcal{H}_2)$ is the set of all linear maps from the Hilbert space $\mathcal{H}_1$ to the Hilbert space $\mathcal{H}_2.$ This map is useful for considering the stabilizer group for a $2$-qubit symmetric state. Now we introduce some properties of this map:\\
$(1).$ Assume $A\in \mathcal{L}(\mathcal{H}_1),B\in \mathcal{L}(\mathcal{H}_1),$ we have that
\begin{align}
A\otimes B\mathcal{M}(X)=\mathcal{M}(AXB^{T}).
\end{align}
$(2).$ Assume $\ket{\psi}\in\mathcal{H}_1,\ket{\phi}\in\mathcal{H}_2,$ then
\begin{align}
\mathcal{M}(\ket{\phi}\bra{\psi})=\ket{\phi}\ket{\bar{\psi}}.
\end{align}
\indent Then we recall the definition and some important properties of $M\ddot{o}bius$ transformation, which is useful for the last part of this article. $M\ddot{o}bius$ transformation is defined on the extended complex plane onto itself \cite{TN}, it can be represented as
\begin{align}
f(z)=\frac{az+b}{cz+d},
\end{align}
with $a,b,c ,d\in \mathbb{C},ad-bc\ne 0.$ From the above equality, we see that when $c\ne 0,$ this function $f:\mathbb{C}-\{-d/c\}\rightarrow \mathbb{C}-\{a/c\}, f(-d/c)=\infty,f(\infty)=a/c,$ when $c=0,$ this function $f:\mathbb{C}\rightarrow\mathbb{C},f(\infty)=\infty.$ The $M\ddot{o}bius$ transformation owns the following properties:\\
$(1).$ $M\ddot{o}bius$ transformation map circles to circles.\\
$(2).$ $M\ddot{o}bius$ transformation are conformal.\\
$(3).$ If two points are symmetric with respect to a circle, then their images under a $M\ddot{o}bius$ transformation are symmetric with respect to the image circle. This is called the "Symmetry Principle."\\
$(4).$ With the exception of the identity mapping, a $M\ddot{o}bius$ transformation has at most two fixed points.\\
$(5).$ There exists a unique $M\ddot{o}bius$ transformation sending any three points to any other three points.\\
$(6).$ The unique $M\ddot{o}bius$ transformation $z\rightarrow M(z)$ sending three points $q,r,s$ to any other three points $q^{'},r^{'},s^{'}$ is given by
\begin{align}
\frac{(M(z)-q^{'})(r^{'}-s^{'})}{(M(z)-s^{'})(r^{'}-q^{'})}=\frac{(z-q)(r-s)}{(z-s)(r-q)}.\nonumber
\end{align}
$(7).$ The $M\ddot{o}bius$ transformation forms a group, $M\ddot{o}bius$ transformation is isotropic to the projective linear group $PSL(2,\mathbb{C})\cong SL(2,\mathbb{C})/\{I,-I\}.$\\
\indent
As we know, the stereographic projection is a mapping that projects a sphere onto a plane. This projection is defined on the whole sphere except a point, and this map is smooth and bijective. It is conformal, $i.e.$ it preserves the angels at where curves meet. By transforming the majorana points of a pure state $\ket{\psi}$ to an extended complex plane, we may get the following proposition \cite{MW}.
\begin{Lemma}
	Assume $\ket{\psi_1},\ket{\psi_2}$ are two pure symmetric states, if $\ket{\psi_{1}}$ and $\ket{\psi_{2}}$ are SLOCC-equivalent iff there exists a $M\ddot{o}bius$ transformation $(10)$ between their Majorana points.
\end{Lemma}
\indent At last, we recall two parameters defined in \cite{TSPM}, diversity degree and degeneracy configuration of an $n$-qubit symmetric pure state. Both two parameters can be used to identify the SLOCC entanglement classes of all $n$-qubit symmetric pure state. Assume $\ket{\psi}$ is an $n$-qubit symmetric pure state, $\ket{\psi}=\frac{e^{i\theta}}{\sqrt{K}}\sum_{\sigma\in \textit{perm}}\ket{\phi_{\sigma(1)}}\ket{\phi_{\sigma(2)}}\cdots\ket{\phi_{\sigma(n)}},$ up to a global phase factor, two states $\ket{\phi_i}$ and $\ket{\phi_j}$ are identical if and only if $a_ib_j-a_jb_i=0,$ and we define their number the degeneracy number. Then we define the degeneracy configuration $\{n_i\}$ of a symmetric state $\ket{\psi}$ as the list of its degeneracy numbers $n_i$ ordered in decreasing order. We denote the number of the elements in the set $\{n_i\}$ as the diversity degree $m$ of the symmetric state, it stands for the number of distinct $\ket{\phi_i}$ in the Eq.$(3).$ For example, a $3$-qubit GHZ state $\ket{GHZ}=\frac{\ket{000}+\ket{111}}{\sqrt{2}},$ we have $\ket{\phi_1}=\frac{\ket{0}+w\ket{1}}{2},\ket{\phi_2}=\frac{\ket{0}+w^2\ket{1}}{2},\ket{\phi_3}=\frac{\ket{0}+\ket{1}}{2},$ $w$ is roots of $P(z)=1-z^3,$ the degeneracy number of $\ket{GHZ}$ is 3, the degeneracy configuration of $\ket{GHZ}$ is ${\{1,1,1\}}.$
 \section{Main Results}
\indent First we present the stabilizer group for a two-qubit symmetric pure state $\ket{\psi}\in\mathcal{H}_2\otimes \mathcal{H}_2$ is nontrivial.
\begin{Theorem}
	Assume that $\ket{\psi}$ is a $2$-qubit symmetric pure state, $\ket{\psi}=\sum_{i=0}^2c_i\ket{D(2,i)},$ then the stabilizer group for the state $\ket{\psi}$ is nontrivial.
\end{Theorem}
\begin{proof}
Assume $A\in \mathcal{L}(\mathcal{H}_1)$ and $B\in\mathcal{L}(\mathcal{H}_2)$ and $A\otimes B\ket{\psi}=\ket{\psi},$ then we have $\mathcal{M}^{-1}(\ket{\psi})=\mathcal{M}^{-1}(A\otimes B\ket{\psi}),$ according to the equality $(8),$  we have that
\begin{align}
AX=X(B^{T})^{-1}.
\end{align}
here we assume that
$
X=\left(
\begin{array}{cc}
	c_0 & c_1 \\
	c_1 & c_2 \\
\end{array}
\right),$ $A=\left(
\begin{array}{cc}
	 a&  b  \\
	 c& d \\
\end{array}
\right),B^{T}=\left(
\begin{array}{cc}
	h & -f \\
	-g &  e\\
\end{array}
\right)
$ and  $det(B)=1,$ then we can obtain the following equation set:
\begin{align}
\left\{
\begin{aligned}
c_0(c-a)=c_1(b-g)\\
c_0f-c_2b=c_1(a-h) \\
c_0c-c_2f=c_1(d-e)\\
c_1(f-c)=c_2(d-h)\\
\end{aligned}
\right  .
\end{align}
As there are four equations and eight variables, then we know that the rank of the solution vectors is more than 1, that is, the stabilizer group for the state $\ket{\psi}$ contains more than the identity.
\end{proof}
\indent Now we present a lemma to show an $n$-qubit symmetric pure state owns a nontrivial stabilizer group when $n=3,4$.
\begin{Lemma}
	Assume $\ket{\psi_{1}}$ and $\ket{\psi_2}$ are symmetric pure states, and $\ket{\psi_1}$ is SLOCC equivalent to $\ket{\psi_2},$ if there exists a nontrivial ILO $g$ such that $g^{\otimes n}\ket{\psi_{1}}=\ket{\psi_1},$ then the stabilizer group for $\ket{\psi_2}$ is nontrivial.
\end{Lemma}
\begin{proof}
	As $\ket{\psi_{1}}$ and $\ket{\psi_2}$ are symmetric states, then there exists an ILO $h$ such that $\ket{\psi_{2}}=h^{\otimes n}\ket{\psi_1}.$ $\ket{\psi_{2}}=h^{\otimes n}g^{\otimes n} (h^{-1})^{\otimes n}\ket{\psi_{2}},$ that is, the stabilizer group for $\ket{\psi_{2}}$ is nontrivial.
\end{proof}
\indent In \cite{TSPM}, the authors presents a three-qubit symmetric pure state is SLOCC equivalent to $\ket{W}$ or $\ket{GHZ}.$ As we know, when we choose $g=\left(\begin{array}{cc}
exp(i\pi/4) &0\\0
& exp(-i\pi/2)\end{array}\right), g^{\otimes 3}\ket{W}=\ket{W}, \sigma^{\otimes 3}_z\ket{GHZ}=\ket{GHZ}.$ From Lemma 2, we see that the stabilizer group for all three-qubit pure symmetric states is nontrivial. And a four-qubit symmetric state is SLOCC equivalent to one of the elements in $S=\{\ket{D(4,0)},\ket{D(4,1)},\ket{D(4,2)},\frac{\ket{D(4,2)}+\ket{D(4,0)}}{2},\frac{\ket{GHZ}+\mu\ket{D(4,2)}}{\sqrt{1+\mu^2}}\},$ then we present a nontrivial stabilizer for the elements in the set $S,$ for the state $\ket{D(4,0)},$ we have $\sigma_z^{\otimes 4}\ket{D(4,0)}=\ket{D(4,0)},$ for the state $\ket{D(4,1)},$ we choose an ILO $g=\left(\begin{array}{cc}
\exp(\frac{\pi i}{6})&0\\
0&\exp(\frac{3\pi i}{2})
\end{array}\right),$ for the state $\ket{D(4,2)},$ we can choose an ILO $g=\sigma_z,$ and for the last element in the set S, we can also choose an ILO $g=\sigma_z,$ then due to the lemma 2, we have the stabilizer group for all four-qubit symmetric pure state is nontrivial. Note that for four-qubit pure states, the authors in \cite{BBM} also proposed the similar results. \\
 \indent Here we denote
 \begin{align}
 G\equiv SL(2,\mathbb{C})\otimes SL(2,\mathbb{C})\otimes \cdots\otimes SL(2,\mathbb{C})\nonumber\\
 K\equiv SU(2)\otimes SU(2)\otimes \cdots\otimes SU(2)\nonumber\\
 \tilde{G}\equiv GL(2,\mathbb{C})\otimes GL(2,\mathbb{C})\otimes \cdots\otimes GL(2,\mathbb{C})\nonumber\\
 \tilde{K}\equiv U(2)\otimes U(2)\otimes \cdots\otimes U(2)
 \end{align}  Next we will use the method proposed in \cite{GKW} to give a class of symmetric states with its stabilizer group containing only the identity. First we introduce the definition of $SL$-invariant polynomials. A polynomial $f:\mathcal{H}\longrightarrow \mathbb{C}$ is $SL$-invariant if $f(g\ket{\psi})=f(\ket{\psi}),\forall g\in {SL}(2,\mathbb{C})\otimes {SL}(2,\mathbb{C})\cdots\otimes SL(2,\mathbb{C}),$ here we denote $\mathcal{H}=\mathcal{H}_2\otimes \mathcal{H}_2\otimes \cdots\otimes \mathcal{H}_2$ and $SL(2,\mathbb{C})=\{A=\left(\begin{array}{cc}
 a_{11}& a_{12}\\
 a_{21}& a_{22}
 \end{array}\right)| a_{ij}\in \mathcal{C},i,j=1,2,det(A)=1\}$. $f_2$ is a $SL$-invariant polynomial with degree 2, which is defined as $f_2(\ket{\psi})=(\ket{\psi},\ket{\psi})=\bra{\psi^{*}}\sigma_y^{\otimes n}\ket{\psi},$ here $\sigma_y$ is the Pauli operator with its matrix representation $\left(\begin{array}{cc}0&-i\\
 i&0\\
 \end{array}\right)$. Due to the property of $\sigma_y,$ we have that when $n$ is odd, $f_2(\cdot)=0.$ Another polynomial, $f_4(\ket{\psi}),$ is a polynomial with degree $4$, it is defined as
 $f_4(\ket{\psi})=\det\left(
 \begin{array}{cc}
 (\ket{\psi_0},\ket{\psi_0}) & (\ket{\psi_0},\ket{\psi_1}) \\
 (\ket{\psi_1},\ket{\psi_0}) &  (\ket{\psi_1},\ket{\psi_1})\\
 \end{array}
 \right),$ here we assume that $\ket{\psi}=\ket{0}\ket{\psi_0}+\ket{1}\ket{\psi_1}.$ Below we denote the stabilizer group for a pure state $\ket{\psi}$ as $\tilde{G}_{\psi}=\{g_1\otimes g_2\otimes \cdots\otimes g_n\in\tilde{G}|g_1\otimes g_2\otimes \cdots\otimes g_n\ket{\psi}=\ket{\psi}\}.$
 \begin{Example}
 	Assume $\ket{\psi}$ is an $n$-partite symmetric pure state $(n\ge 5)$, $\ket{\psi}=x_k\ket{D(n,k)}+x_l\ket{D(n,l)}+x_n\ket{D(n,n)},k+l=n+1,k,l\ne \lfloor\frac{n}{2}\rfloor, gcd(k,l)=1,$ here we denote that $gcd(k,l)$ is the greatest common divisor of $k$ and $l,$ Then $\tilde{G}_{\psi}=\{I\}$
 \end{Example}
 \begin{proof}
 	 First we denote $ g_2\otimes g_3\otimes \cdots\otimes g_n$ as $h$ and let $g_1=\left(\begin{array}{cc}
 	 a_1& b_1\\
 	 c_1& d_1\\\end{array}\right),$ then
 	\begin{align}
 	&\hspace{3mm} g_1\otimes h\ket{\psi}\nonumber\\
 	&=\ket{0}h[a_1(x_k\ket{D(n-1,k)}+x_l\ket{D(n-1,l)})+b_1(x_k\ket{D(n-1,k-1)}+x_l\ket{D(n-1,l-1)}\nonumber\\
 	&+x_n\ket{D(n-1,n-1)})]+\ket{1}h[c_1(x_k\ket{D(n-1,k)}+x_l\ket{D(n-1,l)})\nonumber\\
 	&+d_1(x_k\ket{D(n-1,k-1)}+x_l\ket{D(n-1,l-1)}+x_n\ket{D(n-1,n-1)})]\nonumber\\
 	&=\ket{0}(x_k\ket{D(n-1,k)}+x_l\ket{D(n-1,l)})\nonumber\\&+\ket{1}(x_k\ket{D(n-1,k-1)}+x_l\ket{D(n-1,l-1)}+x_n\ket{D(n-1,n-1)})
 	\end{align}
  \indent Apply $\bra{0}\otimes I^{\otimes (n-1)}$ to the left hand side (LHS) and the right hand side (RHS) of the above equality, and we denote that $\ket{\zeta_1}=x_k\ket{D(n-1,k)}+x_l\ket{D(n-1,l)},\ket{\zeta_2}=x_k\ket{D(n-1,k-1)}+x_l\ket{D(n-1,l-1)}+x_n\ket{D(n-1,n-1)},$ then
 	\begin{align}
&h[a_1\ket{\zeta_1}+b_1\ket{\zeta_2}]=x_k\ket{D(n-1,k)}+x_l\ket{D(n-1,l)}.
 	\end{align}
 \indent Assume $n$ is odd, then by using $f_2(\cdot)$ to the equality $(15),$ we have $2b_1^2x_kx_l=0,b_1=0.$ As $\ket{\psi}$ is a symmetric state, we may assume $g_i=\left(
 	\begin{array}{cc}
 	a_i & 0 \\
 	c_i &  d_i\\
 	\end{array}
 	\right),i=2,\cdots,n,$ due to $g_1\otimes g_2\otimes \cdots\otimes g_n\ket{\psi}=\ket{\psi} $ and through simple computation, we have that $c_i=0,$ Applying $g_i=\left(
 	\begin{array}{cc}
 	a_i & 0 \\
 	0&  d_i\\
 	\end{array}
 	\right)$  to the equality $(14),$ we have $d_i=\lambda a_i,\prod d_i=1,\lambda^k=\lambda^l=1,$ as $gcd(k,l)=1,$ we have $\lambda=1,$ that is $\tilde{G}_{\psi}=\{I\}.$\\
 	\indent When $n$ is even, we use $f_4(\cdot)$ to the equality $(15)$, we have $2b_1^4x_k^2x_l^2=0,b_1=0.$ From the same method as when $n$ is odd, we have that $\tilde{G}_{\psi}=\{I\}$
  \end{proof}
\indent Then we present a class of symmetric critical states $\ket{\psi}$ with $G_{\psi}=\{I\}$. First we present the definition of critical states and a meaningful characterization of critical states. The set of critical states is defined as:
\begin{align}
Crit(\mathcal{H}_n)=\{\psi|\bra{\psi}X\ket{\psi}=0, X\in Lie(G)\}.
\end{align}
Here $Lie(G)$ is the Lie algebra of $G.$ The critical set is valuable, as many important states in quantum information theory, such as the Bell states, GHZ states, cluster states  and graph states, belong to the set of critical states. Then we present a fundamental characterization of critical states.
\begin{Lemma} (The Kempf-Ness theorem)\\
	1. $\psi\in Crit(\mathcal{H}_n)$ if and only if $||g\psi||\ge ||\psi||$ for all $g\in G,$ $||\cdot||$ denotes the Euclidean norm of $\ket{\psi}\in \mathcal{H}_n.$\\
	2. If $\psi\in Crit(\mathcal{H}_n),$ then $||g\psi||\ge ||\psi||$ with equality if and only if $g\psi\in K\psi.$ Moreover, if g is positive semidefinite then the equality condition holds if and only if $g\psi=\psi.$\\
	3. If $\psi\in\mathcal{H}_n,$ then $G\psi$ is closed in $\mathcal{H}_n$ if and only if $G\psi\cap Crit(\mathcal{H}_n)\ne \emptyset.$
\end{Lemma}
\indent Due to this lemma, Gour and Wallach in \cite{GNW} showed that $\psi\in \mathcal{H}_n$ is critical if and only if all the local density matrices of $\ket{\psi}$ are proportional to the identity. And by the lemma in \cite{GKW} and the theorem 2.12 in \cite{W}, for a class of pure states $\ket{\psi}\in Crit(\mathcal{H}_n),$ if we could show the set $K_{\psi}=\{U_1\otimes U_2\otimes \cdots\otimes U_n\in K|U_1\otimes U_2\otimes \cdots\otimes U_n\ket{\psi}=\ket{\psi}\}$ contains only the identity, then $G_{\psi}=\{I\}.$
\begin{Example}
	Assume $\ket{\psi}= a_0\ket{D(n,0)}+a_{n-1}\ket{D(n,n-1)}$ or $\ket{\psi}=a_1\ket{D(n,1)}+a_n\ket{D(n,n)}$ is an $n$-qubit symmetric critical pure state with $n>4,$ then $G_{\psi}=\{I\}.$
\end{Example}
\begin{proof}
First we prove $K_{\psi}=\{I\}.$
 When $\ket{\psi}=a_1\ket{D(n,1)}+a_n\ket{D(n,n)},$ assume $U^{\otimes n}\in K$ satisfies $U^{\otimes n}\ket{\psi}=\ket{\psi},$ then
	\begin{align}
	(U\otimes U)\rho_{1,2}(U\otimes U)^{+}=\rho_{1,2},
	\end{align} the equality $(17)$ can be changed as
	\begin{align}
&\hspace{3mm}	(U\otimes U)[\frac{(n-2)|a_1|^2}{n}\ket{00}\bra{00}+\frac{|a_1|^2}{n}(\ket{01}+\ket{10})(\bra{01}+\bra{10})+|a_n|^2\ket{11}\bra{11}](U\otimes U)^{+}\nonumber\\=&\frac{|a_1|^2(n-2)}{n}\ket{00}\bra{00}+\frac{|a_1|^2}{{n}}(\ket{01}+\ket{10})(\bra{01}+\bra{10})+|a_n|^2\ket{11}\bra{11}
	\end{align}
	As the formula on the RHS of the equality above can be seen as a Schmidt decomposition, and $U\otimes U$ cannot increase the Schmidt rank, then $U\ket{1}=u_1\ket{1},$ as $U$ is a unitary operator, then $U\ket{0}=u_0\ket{0}.$ At last, due to $U^{\otimes n}\ket{\psi}=\ket{\psi},$ then we have $ u_0^{n-1}u_1=1,u_1^n=1,$ $i.e.$ $u_0=u_1.$ The other case is similar.
\end{proof}
\indent  Here we present another proof on $G_{L_n}=\{I\},$ $\ket{L_n}$ is defined in the equality $(8)$ of the article \cite{GKW}.  Note that the examples above tells us $G_{\psi}=\{I\},$ however, $\tilde{G}_{\psi}\supset
\{I\}.$ It seems that this result is simple, However, this method is very useful to present nontrivial examples of states in n-qudit systems with nontrivial stabilizer groups \cite{SW}. \\
\indent At last, I would like to apply $M\ddot{o}bius$ transformation to show when the diversity number $m$ of an $n$-qubit symmetric pure state $\ket{\psi}$ is 5 or 6, the stabilizer group of $\ket{\psi}$ is trivial, when $m\ge 7,$ under a conjecture we make, the stabilizer group of $\ket{\psi}$ is trivial.\\
\indent Assume a pure symmetric state $\ket{\psi}$ can be represented in terms of Majorana representation:
\begin{align}
\ket{\psi}=\frac{e^{i\alpha}}{\sqrt{K}}\sum_{\sigma\in perm}\ket{\phi_{\sigma(1)}}\ket{\phi_{\sigma(2)}}\cdots\ket{\phi_{\sigma(n)}},
\end{align}
where the sum takes over all the permutations and K is the normalization for the state $\ket{\psi}.$ Due to the main results proposed by Mathonet $et$ $al.$ \cite{PSM}, we see that if $\ket{\psi}=g_1\otimes g_2\otimes\cdots \otimes g_n\ket{\psi},$ then there exists an ILO $g$ such that $\ket{\psi}=g^{\otimes n}\ket{\psi},$ $i.e.$ if we could prove $\{g|g^{\otimes n}\ket{\psi}=\ket{\psi}\}=\{I\},$ then the stabilizer group of $\ket{\psi}$ is trivial. From the Eq.$(19),$
\begin{align}
g^{\otimes n}\ket{\psi}=&\ket{\psi},\\
 \sum_{\sigma\in perm}\bigotimes_i g\ket{\phi_{\sigma(i)}}=&\sum_{\sigma\in perm}\bigotimes_i\ket{\phi_{\sigma(i)}}.
\end{align}
 Due to the uniqueness of the Majorana representation for a symmetric state and according to the equality (21), we see that there is a permutation $\sigma$ such that
\begin{align}
g\ket{\phi_i}=\lambda_i\ket{\phi_{\sigma(i)}}
\end{align}
with $\Pi_i \lambda_i=1.$ That is,
\begin{Theorem}
	Assume $\ket{\psi}$ is an $n$-qubit symmetric pure state, then the stabilizer group of $\ket{\psi}$ is nontrivial if and only if there exists nontrivial $M\ddot{o}bius$ transformation that permutes the Majorana roots of $\ket{\psi}.$
\end{Theorem}
\indent Assume  the diversity number of a symmetric state $\ket{\psi}$ is $m$, the divergence configuration for the state $\ket{\psi}$ is $\{k_1,k_2,\cdots,k_m\}$ with $k_1\ge k_2\cdots\ge k_m$ and $\sum_i k_i=n.$ Then from the Lemma $1$ and the property $(4)$ of $M\ddot{o}bius$ transformation, we have:
\begin{Corollary}
Assume $\ket{\psi}$ is an $n$-qubit symmetric state, its degeneracy configuration is $\{k_1,k_2,\cdots,k_m\},m\ge 3,$ if there exists $k_i,k_j$ and $k_l$ such that these three values are unequal to the residual elements in the set, then the stabilizer group for the state $\ket{\psi}$ contains only the identity.
\end{Corollary}
\indent At last, we talk about the stabilizer group for an $n$-qubit symmetric pure state $\ket{\psi}$ with the increase of the diversity number $m$ of $\ket{\psi},$ when $m=2$ the stabilizer group for $\ket{\psi}$ is nontrivial, when $m\ge 5,$ under a conjecture we make, the stabilizer group for $\ket{\psi}$ is trivial. \\
 $\textit{m=1}$ (separable states): When a symmetric state $\ket{\psi}$ is separable, then it can be represented as
 \begin{align}
 \ket{\psi}=\ket{\epsilon\epsilon\cdots\epsilon},
 \end{align}
 it is easy to see when an ILO $g$ satisfies that $\ket{\epsilon}$ is an eignvector of $g,$ $g^{\otimes n}$ is the stabilizer for the state $\ket{\psi}.$\\
 $\textit{m=2}:$ In this case, the state $\ket{\psi}$ can be represented as
 \begin{align}
 \ket{\psi}=\frac{e^{i\theta}}{\sqrt{K}}\sum \ket{\underbrace{\epsilon_1\cdots \epsilon_1}_{k_1}\underbrace{\epsilon_2\cdots\epsilon_2}_{n-k_1}},
 \end{align}
 where the sum takes over all the permutation of $k_1$ $\ket{\epsilon_1}s$ and $n-k_1$ $\ket{\epsilon_2}s$ in this case, when $k_1\ne n-k_1,$ first we denote a local operator $h=[\ket{\epsilon_1},\ket{\epsilon_2}]^{-1},$ this means if $\ket{\epsilon_1}=a_1\ket{0}+b_1\ket{1}, \ket{\epsilon_2}=a_2\ket{0}+b_2\ket{1},$ $h=\left(\begin{array}{cc}
 a_1&a_2\\
 b_1& b_2\\
 \end{array}\right)^{-1},$ then $g$ can be written as $g=h^{-1}\left(\begin{array}{cc}\lambda_1&0\\
  0&\lambda_2\end{array}\right) h.$ And when $k_1=n-k_1,$ the ILO $g$ can also be written as $g=h^{-1}\left(\begin{array}{cc}0&\lambda_2\\
  \lambda_1&0\end{array}\right)h, h=[\ket{\epsilon_1},\ket{\epsilon_2}]^{-1}$.\\
  \indent When $m\ge 3,$ due to Lemma 3, by searching a nontrivial M$\ddot{o}$bius transformation between the Majorana points of the pure state $\ket{\psi},$ we can see whether a pure state $\ket{\psi}$ owns a nontrivial stabilizer group.\\ \indent When $m=3,$ assume the degeneracy configuration of a pure state $\ket{\psi}$ is $\{k_1,k_2,k_3\},$ due to the property (5), we see that except when $k_1\ne k_2\ne k_3$, $\ket{\psi}$ owns a nontrivial stabilizer group.\\
  \indent Here we assume that each $z_i$ is not $\infty,$ as if there exists $k$ such that $z_k=\infty,$ we can always make $z_k$ be not $\infty$ by M$\ddot{o}$bius transformation. When $m=4,$ assume the degeneracy configuration of the pure state $\ket{\psi}$ is $\{k_1,k_2,k_3,k_4\},$ when all the four number are different from each other, the stabilizer group for $\ket{\psi}$ is trivial. First we show when $k_1=k_2,k_3=k_4,$ the stabilizer group is nontrivial, $i.e.$ there exists a nontrivial M$\ddot{o}$bius transformation $f$ that can permute $z_i,i=1,2,3,4.$ Let $f(z_1)=z_2,f(z_2)=z_1,f(z_3)=z_4,f(z_4)=z_3,$ from the property (6) of the M$\ddot{o}$bius transformation, we see $f$ exists.\\
  \indent Here we note that when $\ket{\psi}$ is a four qubit symmetric pure state, $\ket{\psi}$ always owns a nontrivial stabilizer group. As the diversity configuration can only be $\{1,1,1,1\},\{2,1,1\},\{3,1\}$ or $\{4\},$ from the above analysis, we see that the stabilizer group is nontrivial.\\
  \indent Next we consider a general case. Assume the stabilizer group of the state $\ket{\psi}$ is nontrivial, then we have a nontrivial M$\ddot{o}$bius transformation which permutes the majorana points, that is, we could assume $f(z_1)=z_2,f(z_2)=z_3,\cdots,f(z_{i_1})=z_1,f(z_{i_1+1})=z_{i_1+2},f(z_{i_1+2})=z_{i_1+3},\cdots, f(z_{i_2})=z_{i_1+1}$, if $i_1,i_2-i_1\ge 3,(i_1,i_2-i_1)=1,$ here we denote that $(i_1,i_2-i_1)$ is the greatest common divisor of $i_1$ and $i_2-i_1$, then we could have $f^{i_1}(z_j)=z_j,j=1,2,\cdots ,i_1$, that is, $f^{i_1}$ is trivial, when $i_1,i_2-i_1\ge 3,(i_1,i_2-i_1)=r\ne \min(i_1,i_2-i_1)$, $f^i(z_{j})\ne z_j, j\ge i_1+1$, so it is invalid.  However, we cannot prove the case when $(i_1,i_2-i_1)=min(i_1,i_2-i_1)$ is invalid. When $i_1=2$, $i_2-i_1\ge 3,$ then we have $f(z_1)=z_2,f(z_2)=z_1,f(z_3)=z_4,f(z_4)=z_5,$
  \begin{align}
  \frac{(z_1-z_2)(z_3-z_4)}{(z_1-z_4)(z_3-z_2)}=\frac{(z_2-z_1)(z_4-z_5)}{(z_2-z_5)(z_4-z_1)},\nonumber,
  \end{align}
that is,
\begin{align}
(z_3-z_4)(z_2-z_5)=(z_3-z_2)(z_4-z_5),\nonumber\\
(z_3-z_5)(z_2-z_4)=0.\nonumber
\end{align}
This case is invalid.\\
\indent  Assume $\ket{\psi}$ is an $n$-qubit symmetric pure state with its degeneracy configuration $\{k_1,k_2,\cdots,k_m\}$, and we divide the degeneracy configuration into $d$ parts according to whether $k_i$ are equal. Assume $D$ is a set with its elements $n_j,j=1,2,\cdots, d$, here $n_j$, $j=1,2,\cdots,d$ is the number of each part. Due to the above analysis, only when the $M\ddot{o}bius$ transformation $f$ is represented as $f_{\{{r,1}\}}(r\ge 1,r\in\mathbb{Z}^{+}),f_{\{\underbrace{r,r,\cdots,r}_l\}}(r\ge 2,l\ge 1,r,l\in \mathbb{Z}^{+})$, $f_{\{\underbrace{r,r,\cdots,r}_l,1\}}(r\ge 2,l\ge 2,r,l\in \mathbb{Z}^{+})$ or $f_{\{\underbrace{r,r,\cdots,r}_l,1,1\}} (r\ge 2,l\ge 1,r,l\in \mathbb{Z}^{+})$ may present $\ket{\psi}$ own a nontrivial stabilizer group.  That is, if $n_j$, $j=1,2,\cdots,d$ cannot divide into the form of subscript of $f$, then it owns a trivial stabilizer group. For example, if the degeneracy configuration of $\ket{\psi}$ is $\{5,5,5,5,5,3,3,3,3,3,3,3,2,2,2\}$, $D=\{5,7,3\}$, this set cannot be divide into the form of the above, that is, $\ket{\psi}$ owns a trivial stabilizer group.\\
\indent As the transformations of three points can determine a unique $M\ddot{o}bius$ transformation, and when we have a Majorana point in each part, we could determine whether the above configuration is valid, if all the permutations are invalid, then the symmetric state owns a trivial stabilizer group. So when the degeneracy number $m\ge 5$, the symmetric pure state $\ket{\psi}$ with trivial stabilizer group is of full measure among the subspace of symmetric states.\\
\indent This fact is interesting, as it implies much in the entanglement theory. As when an n-qubit symmetric pure state is entangled, it is fully entangled. Due to the results in \cite{GKW}, we have the following theorem.
  \begin{Theorem}
   Assume $\ket{\psi}$ is an n-qubit symmetric pure state with $\tilde{G}_{\psi}=\{I\}$, $\ket{\phi}$ is an n-qubit symmetric entangled state, then $\ket{\psi}$ can be converted deterministically to $\ket{\phi}$ by LOCC or SEP operations if and only if there is $u\in \tilde{K}$ such that $\ket{\psi}=u\ket{\phi}.$
  \end{Theorem}
\indent From the above theorem, we have if two states satisfy the conditions in the above theorem, if they are not LU equivalent, they cannot be transformed into other by LOCC, even SEP. And from the statement below the theorem 2, we have that when $m\ge 5$, almost all symmetric pure states $\ket{\psi}$ are isolated. This fact may represent the complexity of the structure of the multipartite entanglement.\\
\indent From the above results, we see that when two symmetric pure states satisfy conditions in the above theorems, a state can be transformed into the other by local transformations only with probability. At last, we present the maximal probability with which $\ket{\psi}\rightarrow \ket{\phi}$ can be converted by LOCC by using the theroem 7 in \cite{GKW}.
\begin{Theorem}
	Assume $\ket{\psi}$ is a symmetric pure state with $\tilde{G}_{\ket{\psi}}=\{I\}$, $\ket{\phi}=g^{\otimes n}\ket{\psi}$ is a symmetric pure state, then the maximal probability with which $\ket{\psi}$ can be converted to $\ket{\phi}$ by LOCC or SEP is given by
	\begin{align}
		p_{max}(\ket{\psi}\rightarrow\ket{\phi})=\frac{1}{\lambda_{max}({g^{\dagger}}^{\otimes n}g^{\otimes n})},
	\end{align}
	here we denote that $\lambda_{max}(X)$ is the maximal eigenvalue of the matrix $X$.
\end{Theorem}

   \section{Conclusion}
\indent In this article, we consider the stabilizer group for a symmetric state $\ket{\psi}.$ First we present that the stabilizer group for an $n$-qubit symmetric state $\ket{\psi}$ contains more than the identity when $n=2,3,4$, then similar to the method presented in \cite{GKW}, we give a class of states whose stabilizer group contains only the identity, we also propose a class of states $\ket{\psi}$ with $G_{\psi}=\{I\},\tilde{G}_{\psi}\ne \{I\},$ at last, we present that when $m\ge 5$, almost all $n$-qubit symmetric pure states owns a trivial stabilizer group, then due to the results in \cite{GKW}, we have that almost all $n$-qubit symmetric pure states cannot be transformed among symmetric pure entangled states by nontrivial LOCC transformations deterministically, and we also present the optimal probability of the local transformations among two SLOCC-equivalent symmetric states.
\section{ACKNOWLEDGMENTS}
	\indent This project was partially supported by the National Key Research and Development Program of China under grant 2016YFB1000902, the Natural Science Foundation of China grants 61232015 and 61621003, the Knowledge Innovation Program of the Chinese Academy of Sciences (CAS), and Institute of Computing Technology of CAS.


TITLE: Suppose $A \subset X$ where $X$ is possibly a metric space, must $A$ also inherit the subspace metric?
QUESTION [0 upvotes]: Okay so I don't know what i am overlooking at this simple issue. I was reading the Metrization theorem by Urysohn and the entire idea is focused on constructing a bi-continous one-to-one function that maps onto a subspace (hence this is an embedding mapping onto its image making it a homeomorphism) of metrizable space.
So the stated setup is like this, we have $F: X \to F(Y)$ where $F(Y) \subset W$ and $W$ is a metrizable space like $\mathbb{R}$ (actually it's just the Hilbert Cube).
Now usually when we talk about metrizable subspaces, we begin with the universal space $W$ first and talk about its subspaces, but aren't we doing it backwards and we are exhibiting a subspace that is in a Metric Space? So how do we know that given a subspace of a metric space, it must also contain a metric? It only inherits the topology and the open sets right? And metric induce topologies not the other way around right? I must be forgetting something important.

REPLY [0 votes]: The steps are as follows: we consider  the Tychonoff cube $[0,1]^{\mathbb{N}}$ in the product topology.
We can show in general that a countable product of metric spaces in the product topology is metrisable, using a product metric that induces the product topology.
Another general fact: if $(X,d)$ is a metric space (with induced topology $\mathcal{T_d}$) and $Y \subseteq X$ is a subspace of $X$ in the subspace topology $\mathcal{T}_Y$ w.r.t. $\mathcal{T}_d$, then letting $d_Y$ be the restriction of $d$ to $Y \times Y$ defines a metric on $Y$ (quite clear) which exactly induces $\mathcal{T}_Y$. This is probably what you ask about in the title of the post.
So the above 2 facts tell us that when we are in the class of metrisable spaces (so spaces whose topology can be induced by a metric in the space) are closed under the usual (general topological) operations of countable products and subspaces.
So if we show for some space $X$ that there is a map $F:X \to [0,1]^{\mathbb{N}}$ such that $F$ is a homomorphism between $X$ and $F[X]$ then $F[X]$ is metrisable, as a subspace of a countable product of metric spaces (and so $X$ is too).


text:
\begin{document}

\maketitle

\renewcommand{\thefootnote}{\fnsymbol{footnote}}

\footnotetext[2]{Authors address: Department of Applied Mathematics, Naval Postgraduate School, Monterey, CA 93943.  Email: borges@nps.edu}

\begin{abstract}
We develop an algorithm solving the $3\times3$ real symmetric eigenproblem. This is a common problem and in certain applications it must be solved many thousands of times, see for example \cite{tripref} where each element in a finite element grid generates one. Because of this it is useful to have a tailored method that is easily coded and compact. Furthermore, the method described is fully compatible with development as a GPU based code that would allow the simultaneous solution of a large number of these small eigenproblems.
\end{abstract}

\section{Reduction to arrow form}

The traditional first step in solving any real symmetric eigenproblem is to use unitary similarity transformations to reduce the matrix to tridiagonal (Hessenberg) form. For a $3\times3$ this would typically involve a Givens rotation that eliminates the $(1,3)$ and $(3,1)$ elements. We will stray from this approach and instead rely on a unitary similarity that eliminates the $(1,2)$ and $(2,1)$ elements and transforms our original matrix into an {\em ordered} $3\times3$ real symmetric arrow matrix.\footnote{I use an arrow structure rather than a tridiagonal for pedagogical reasons as I believe the derivations are easier to follow in this form. The two are equivalent under a simple permutation similarity.} Specifically, a matrix of the form
$$
A =
\bmat{ccc}
\alpha_1 & 0 & \beta_1 \\
0 & \alpha_2 & \beta_2 \\
\beta_1 & \beta_2 & \gamma \\
\endbmat
$$
with $\alpha_1 \geq \alpha_2$.

Although such a similarity can be constructed in several ways, we will use a Jacobi rotation (in effect, the eigenvectors of the principal $2\times2$) operating in the $(1,2)$-plane. The rotation can always be constructed so that the resulting arrow will have the elements of the shaft properly ordered. It is important to construct the Jacobi rotation carefully and we do so using the algorithm presented in \cite{borgesjacobi} which is demonstrably superior to the standard approach.

\subsection{Deflating the Arrow}

We note that if any $\beta_j = 0$ then it is possible to set $\l_j = \alpha_j$ and deflate the matrix since $\be_j$ is clearly an eigenvector \cite{wilkinson}.  This is known as $\beta$-deflation. A second type of deflation occurs if $\alpha_1 = \alpha_2$, in that case we can apply a $2\times 2$ rotation similarity transformation in the $(1,2)$-plane that takes $\beta_1$ to zero and creates a $\beta$-deflation. This is known as a combo-deflation.

Exact deflations are theoretically easy to handle, however, before proceeding we must address the issue of {\em numerical deflation} which happens whenever {\em tiny} changes in the matrix can lead to deflation. Because the matrix is only $3\times3$ we can deal with the problem in a very direct manner. We begin by constructing a Givens rotation in the $(1,2)$-plane that takes $\beta_1$ to zero. The first step is to compute $h = \sqrt{\beta_1^2+\beta_2^2}$. if $h = 0$ then the matrix is diagonal and we are finished. If not, then we let
$$
G = \bmat{c c c} \beta_2/h & -\beta_1/h & 0 \\ \beta_1/h & \beta_2/h & 0 \\ 0 & 0 & 1 \endbmat.
$$
Then
$$
G A G^T =
\bmat{ccc}
\frac{\alpha_1 \beta_2^2+\alpha_2 \beta_1^2}{h^2} & \alpha\frac{\beta_1\beta_2}{h^2} & 0 \\
\alpha\frac{\beta_1\beta_2}{h^2} & \frac{\alpha_1 \beta_1^2+\alpha_2 \beta_2^2}{h^2} & h \\
0 & h & \gamma \\
\endbmat
$$
where $\alpha = \alpha_1-\alpha_2$.

As the matrix is now tridiagonal we can adapt the deflation condition that is used in EISPACK (see \cite{gvl} pp.352-353) to our cause which would have us deflate if
$$
|\alpha\beta_1\beta_2| \leq C \epsilon \left( |\alpha_1 \beta_2^2+\alpha_2 \beta_1^2| + |\alpha_1 \beta_1^2+\alpha_2 \beta_2^2 | \right)
$$
for some constant $C$. We can simplify this condition by noting that
$$
|\alpha_1+\alpha_2| h^2\leq |\alpha_1 \beta_2^2+\alpha_2 \beta_1^2| + |\alpha_1 \beta_1^2+\alpha_2 \beta_2^2 |
$$
by the triangle inequality. This leads to a simpler but slightly more restrictive test where if
$$
|\alpha\beta_1\beta_2| \leq C \epsilon  |\alpha_1+\alpha_2| h^2
$$
we invoke the Wielandt-Hoffman theorem and ignore it yielding a deflation where we accept
$$
\frac{\alpha_1 \beta_2^2+\alpha_2 \beta_1^2}{h^2}
$$
as an eigenvalue. The rest of the spectrum can be recovered by solving the $2\times2$ eigenproblem for
$$
\bmat{cc}
\frac{\alpha_1 \beta_1^2+\alpha_2 \beta_2^2}{h^2} & h \\
 h & \gamma \\
\endbmat
$$

If the matrix does not deflate then we can assume that it is both ordered and {\em reduced}, that is the strict inequality $\alpha_1 > \alpha_2$ holds, and further $\beta_j \neq 0$ for $i=1,2$.

\subsection{Solving the Eigenproblem for a $3\times3$ Ordered and Reduced Arrow}

The interlacing property for real symmetric matrices combined with the fact that $\alpha_1 > \alpha_2$ implies that there is a rightmost eigenvalue $\l_1 > \alpha_1$ of multiplicity one with associated eigenvector $\bu_1$. If we shift $A$ by $\alpha_1$ we find that  $\mu = \l_1-\alpha_1$ is the only positive eigenvalue of
$$
A-\alpha_1 I = \bmat{ccc}
0 & 0 & \beta_1 \\
0 & -\ab & \beta_2 \\
\beta_1 & \beta_2 & \gamma-\alpha_1 \\
\endbmat
$$
where $\ab = \alpha_1-\alpha_2$. We shall call a symmetric matrix of this form (all zeros above the main counterdiagonal and a strictly negative middle element) a {fully reduced arrow}. The eigenvector associated with $\mu$ is identical with $\bu_1$ and it is easily verified that it is given by
$$
\bu_1 = \bmat{c} \beta_1 (\mu+\ab) \\  \beta_2 \mu \\ \mu(\mu+\ab) \endbmat
$$

Observe next that
$$
P(\alpha_2 I - A)P = \bmat{ccc}
0 & 0 & -\beta_2 \\
0 & -\ab & -\beta_1 \\
-\beta_2 & -\beta_1 & \alpha_2-\gamma \\
\endbmat
$$
where $P$ is the simple permutation  that swaps the first and second rows, is also a fully reduced arrow and that its only positive eigenvalue, $\nu$, satisfies $\nu  = \alpha_2 - \l_3$. One can verify that its associated eigenvector is
$$
\bmat{c}  \beta_2 (\nu+\ab) \\  \beta_1 \nu \\ -\nu(\nu+\ab) \endbmat
$$
and hence we can recover that corresponding leftmost eigenvector of $A$ by applying the permutation $P$ to get
$$
\bu_3 = \bmat{c} \beta_1 \nu  \\  \beta_2 (\nu+\ab) \\ -\nu(\nu+\ab) \endbmat
$$

Because the eigenvectors must be orthogonal we can compute the middle eigenvector directly by taking the cross product $\bu_1\times\bu_3$. This is better accomplished after a bit of algebra by using the form:
$$
\bu_2 = \bmat{c}
-\beta_2 \mu (\nu+\ab)\\ \quad\beta_1\nu(\mu+\ab)\\ \beta_1\beta_2\ab
\endbmat
$$
It is worth noting that, discounting errors accrued in finding $\mu$ and $\nu$, there is no cancellation in the construction of any of the eigenvectors beyond the benign cancellation in computing $\ab$.

Finally, we can use the trace identity
$$\mbox{tr}(A) = \l_1 + \l_2 + \l_3$$
to compute the corresponding eigenvalue. A bit of algebra yields
$$
\l_2 = \nu-\mu+\gamma
$$

In section \ref{algsec} we develop an algorithm that computes the dominant eigenvalue of a fully reduced arrow.

\subsection{Orthogonality of the Computed Eigenvectors}

In this section we show that the computed eigenvectors will be numerically orthogonal provided that
\begin{eqnarray}
\hat{\mu} & = & \mu (1+\delta_1) \nonumber \\
\hat{\nu} & = & \nu (1+\delta_2)
\label{munu}
\end{eqnarray}
where $|\delta_i| < \epsilon$ for some small positive number $\epsilon$.

We begin by noting that since $\bu_2$ is a cross product it will be necessarily be numerically orthogonal to $\bu_1$ and $\bu_3$ provided that these two are themselves numerically orthogonal to each other (see \cite{kahan}) and so we only need demonstrate that. To that end we note that
\begin{eqnarray*}
\hat{\bu_1} & = & \bmat{c} \beta_1 (\hat{\mu}+\ab) \\  \beta_2 \hat{\mu} \\ \hat{\mu}(\hat{\mu}+\ab) \endbmat \\
& = & (1+\delta_1) \bu_1 + \bmat{c} 0 \\  0 \\ \mu^2(\delta_1+\delta_1^2) \endbmat
\end{eqnarray*}
and similarly
$$
\hat{\bu_3}  = (1+\delta_2)\bu_3 +  \bmat{c} 0 \\  0 \\ \nu^2(\delta_2+\delta_2^2) \endbmat
$$
This leads us to the useful fact that
\begin{eqnarray*}
\hat{\bu_1}^T \hat{\bu_3} & = & \nu(\nu+\ab)\mu^2(1+\delta_2)(\delta_1+\delta_1^2)
+ \mu(\mu+\ab)\nu^2(1+\delta_1)(\delta_2+\delta_2^2)
+ \mu^2\nu^2(\delta_1+\delta_1^2)(\delta_2+\delta_2^2)\\
& = & \nu(\nu+\ab)\mu^2\delta_1 + \mu(\mu+\ab)\nu^2\delta_2 + O(\epsilon^2)
\end{eqnarray*}
Finally we note that $\Vert \bu_1\Vert \geq \mu(\mu+\ab)$ and $\Vert \bu_2\Vert \geq \nu(\nu+\ab)$ whence
$$
\frac{\hat{\bu_1}^T \hat{\bu_3}}{\Vert \bu_1\Vert \Vert \bu_2\Vert} \leq
\frac{\mu^2}{\mu(\mu+\ab)}\delta_1 + \frac{\nu^2}{ \nu(\nu+\ab)}\delta_2 + O(\epsilon^2) \leq 2\epsilon + O(\epsilon^2)
$$
and the computed eigenvectors are numerically orthogonal provided that condition \ref{munu} is satisfied.

\subsection{Finding the Rightmost Eigenvalue of a Fully Reduced Arrow}
\label{algsec}

In this section we develop two stable and efficient methods for finding the rightmost eigenvalue of a fully reduced arrow matrix
$$
\Ab =
\bmat{ccc}
0 & 0 & \beta_1 \\
0 & -\ab & \beta_2 \\
\beta_1 & \beta_2 & \gb \\
\endbmat
$$

Note that the block Gauss factorization of $\Ab -\l I$ is
$$
\Ab- \l I =
\bmat{ccc}
1 & 0 & 0 \\
0 & 1 & 0 \\
\frac{-\beta_1}{\l} & \frac{-\beta_2}{\l+\ab} & 1 \\
\endbmat
\bmat{ccc}
 - \lambda & 0 & \beta_1 \\
0 & -(\l + \ab) & \beta_2 \\
0 & 0 & -f(\l) \\
\endbmat
$$
where $f$, the {\em spectral function} of $\Ab$, is given by
$$
f(\lambda) =
\l - \gb  - \frac{\beta_1^2}{\l} - \frac{\beta_2^2}{\l+\ab} .
$$
This is a rational Pick function with a pole at infinity. Inspection of the graph of the spectral function reveals that the elements of the shaft interlace the eigenvalues
\bne
\l_1 > 0 > \l_2 >  -\ab > \l_3 .
\label{interlace}
\ene
Moreover,  the derivative of the spectral function is
\bne
f^{\prime}(\l) = 1 + \frac{\beta_1^2}{\l^2} + \frac{\beta_2^2}{(\l + \ab)^2} .
\label{derspec}
\ene
and is clearly bounded below by one so that its zeros are, in a certain sense, well determined. Furthermore, we note that the second derivative of the spectral function
$$
f^{\prime\prime}(\l) = -2\left(\frac{\beta_1^2}{\l^3} + \frac{\beta_2^2}{(\l + \ab)^3}\right)
$$
is strictly negative over the interval $(0,\infty)$ and therefore $f^{\prime}(\l)$ is strictly decreasing over the same interval.

\subsubsection{Using the Borges-Gragg zero finder}

One approach to finding the unique zero of $f$ in the interval $(0 , +\infty)$ is to use the zero-finder developed in \cite{borgesgragg}.  Let $x_0$ be an initial approximation to the eigenvalue. If $x_j$ is known let our approximating function be
$$
\phi_j (x) = \omega_0 x - \sigma - \frac{\omega_1}{x}.
$$
If we select the constants $\sigma$, $\omega_0$, and $\omega_1$ so that
\bne
\phi_j^{(i)}(x_j) = f^{(i)}(x_j) , \quad i = 0,1,2.
\label{rat_conds}
\ene
then we will be able to guarantee cubic convergence. Therefore, we solve
$$
\bmat{r c r}
-1 & x_j & -1/x_j \\
0 & 1 & 1/x_j^2\\
0 & 0 & -2/x_j^3
\endbmat
\bmat{c}
\sigma \\
\omega_0 \\
\omega_1
\endbmat
=
\bmat{c}
f(x_j) \\
f^{\prime}(x_j) \\
f^{\prime\prime}(x_j)
\endbmat
$$
and find
\begin{eqnarray*}
\omega_1 & = & \beta_1^2 + \beta_2^2 \left( \frac{x_j}{x_j + \ab} \right)^3, \\
\omega_0 & = & 1 +  \beta_2^2 \frac{ \ab}{(x_j + \ab)^3} , \\
\sigma & = & \omega_0 x_j - \frac{\omega_1}{x_j} - f(x_j).
\end{eqnarray*}
Note that $\omega_1 > \beta_1^2 > 0$ and also that  $\omega_0 > 1$. The inequalities are strict and since both $\omega_0 > 0$ and $\omega_1 > 0$ it follows that $\phi_j$ is a Pick function and has a unique zero $x_{j+1} \in (0 , +\infty)$.

Casual inspection of the error function
$$
f(x) - \phi_j(x) = (1-\omega_0) x - (\gb - \sigma) -  \frac{\beta_1^2 - \omega_1}{x} - \frac{\beta_2^2}{x + \ab}
$$
over the interval $(0,+\infty)$ reveals that it converges to $+\infty$ at the left boundary (as $x\rightarrow0^+$) and to $-\infty$ at the right boundary (as $x\rightarrow  +\infty$) and therefore crosses zero in the interval. Moreover, it is not hard to verify by differentiation and a bit of calculus that its derivative never changes sign in the interval. These facts imply that it crosses zero exactly once in the interval and further that we obtain monotonic convergence from any starting guess whatsoever $x_0 \in (0,+\infty)$. The cubic rate of convergence follows from (\ref{rat_conds}).

Successive iterates can be found by solving quadratic equations.  Rather than solve $\phi_j (x) = 0$ for $x_{j+1}$ it is better to solve
$$
\phi_j (x_j - \Delta) = 0
$$
for the {\em increment} $\Delta = x_j - x_{j+1}$.  Some rearrangement using (\ref{rat_conds}) reduces this to
\bne
a \Delta^2 + b \Delta - f = 0 ,
\label{del_quad}
\ene
with
\begin{eqnarray*}
a & = &  -\frac{\omega_0}{x_j} , \\
b & = &  f^{\prime} (x_j) + \frac{f(x_j)}{x_j} .
\end{eqnarray*}

 If $x_0 > \l_1$ then $f(\l) > 0 $ so $b > f^{\prime} > 1$ and $\Delta$ may be computed stably using
\bne
\Delta = \frac{2f/ b}{1 + \sqrt{1 + \frac{2a}{b}
\frac{2f}{b}}} ,
\label{increm_eq}
\ene
and we are therefore inclined to start to the right of $\l_1$. We can guarantee this by using the fact that the Borges-Gragg zero finder can start from $+\infty$ and choosing the first iterate from $+\infty$ to be our starting point. To do so, note that as $x \rightarrow +\infty$ the approximate Pick function tends to
\bne
\phi (x) = x - \gb - \frac{\beta_1^2+\beta_2^2}{x}.
\label{ext_start}
\ene
We propose to take $x_0$ to be the zero of (\ref{ext_start}) in $(0, +\infty )$  which is
$$
x_0 = \frac{\gb}{2} + \sqrt{ \left( \frac{\gb}{2} \right)^2 + \beta_1^2+\beta_2^2}.
$$
If $\gb<0$ then we may wish to multiply top and bottom by the conjugate to avoid cancellation.

Termination in this case is straightforward. The mean value theorem gives
$$
 \fp(c) = \frac{f(x_j) - f(\l_1)}{x_j - \l_1} = \frac{f(x_j)}{x_j - \l_1}
$$
for some point $c\in(\l_1,x_j)$. However, since the iteration is monotonic our iterates are on the right so that $x_j > l_1$. Since $\fp(\l)$ is strictly decreasing over the same interval we can use $ \fp(x_j)$ as a lower bound on $ \fp(c)$
$$
 \fp(x_j)  < \frac{f(x_j) }{x_j - \l_1}
$$
and we conclude that the absolute error\footnote{The absolute value is unecessary here but we leave it in for pedagogical reasons.} is bounded by
$$
|x_j - \l_1| < \frac{f(x_j)}{ \fp(x_j) }.
$$
It is therefore reasonable to terminate when
$$
\frac{f(x_j)}{ \fp(x_j) } < C \epsilon x_j
$$
for some constant $C$ since that will imply that $x_j$ has been found to high relative precision.

\subsubsection{Using Newton's Method}

It is remarkably easy to find the unique zero of $f$ in the interval $(0 , +\infty)$ using Newton's method. It is worth observing that because the derivative of the spectral function \ref{derspec} is strictly decreasing and bounded below by $1$ on the interval $(0 , +\infty)$ Newton's method will converge {\em monotonically} to the unique zero in the interval provided our initial guess, $x_0$, lies to the left of the zero. We can find an appropriate starting guess by finding the point where the spectral function intersects
$$
 -\frac{\beta_2^2}{x+\ab}
$$
since the graph of this function lies strictly below zero over the interval. After a bit of algebra and invoking the quadratic formula we find that this leads to a starting guess of
$$
x_0 = \frac{\gb}{2} + \sqrt{\left(\frac{\gb}{2}\right)^2 + \beta_1^2}.
$$

Termination in this case is also straigthforward. Since this iteration is monotonic on the left we use the fact that $f^{\prime}(\l) > 1$ over the interval $(x_j,\l_1)$ and we conclude that the absolute error is bounded by $ |x_j - \l_1| < | f(x_j)|.$ It is therefore reasonable to terminate when $ |f(x_j)| < C \epsilon x_j$ for some constant $C$ since that will imply that $x_j$ has been found to high relative precision.

\subsection{Testing}

In this section we test the algorithm we have developed herein, which we call {\tt threig}, versus the \matlab function {\tt eig} which uses routines from \textsc{LAPACK}. Our test is designed to compare the performance of our algorithm to that of the industry standard and therefore the protocol is simple.
\begin{enumerate}
\item
Create a set of 100,000 random real symmetric test matrices where the elements of each test matrix $T$ are {\em randomly} drawn from a known distribution (the distributions we test are uniform, normal, and chi-square).
\item
Run both {\tt threig} and {\tt eig} on each test matrix $T$.
\item
Compute two measures of accuracy for each case:
\begin{enumerate}
\item
$\Vert I - V^TV \Vert_F$ which measures the orthogonality of the computed eigenvectors.
\item
$\Vert TV - V\Lambda \Vert_F$ which measures the accuracy of the computed spectral factorization (i.e. how well it reconstructs the test matrix.)
\end{enumerate}
\item
Finally we compute difference between the results from {\tt eig} and those from {\tt threig}, so that the ouptut will be negative when the measured error in {\tt threig} exceeds that of {\tt eig} and positive otherwise. We then sort these values and generate a plot of the results. Since the results are sorted before being displayed the plot reveals how often and by how much one algorithm is {\em superior} to the other.
\end{enumerate}

\begin{figure}[ ht ]
\begin{center}
\includegraphics[width=6in]{unitest}
\caption{Test matrix elements distributed $U(0,1)$.}
\end{center}
\end{figure}

\begin{figure}[ ht ]
\begin{center}
\includegraphics[width=6in]{normtest}
\caption{Test matrix elements distributed ${\cal N}(0,1)$.}
\end{center}
\end{figure}

\begin{figure}[ ht ]
\begin{center}
\includegraphics[width=6in]{chitest}
\caption{Test matrix elements distributed $\chi^2_1$.}
\end{center}
\end{figure}

On viewing the figures we see that the proposed algorithm is consistently more accurate in both measures.

config: arxiv


TITLE: Moon and the moving train
QUESTION [1 upvotes]: While travelling by train (travelling West to East), it seems the moon is moving in opposite direction when seen from the window, but then it reverses its direction, after a certain amount of time and moves along the train, but then flips back again. And it kept oscillating, as the train kept moving. Can anyone point out the reason?

REPLY [2 votes]: The Moon was moving in relation to what?
I had to think about this when you described your experience. Because of the distance of the moon compared to objects on the horizon, it should appear to stand still in comparison to everything else closer whizzing past you.
But then I thought of your description of "oscillations" and thought "AHA! The window."
If you were travelling exactly west to east, and you were observing the moon through a window exactly facing south (or north, depending which side of the train you are on and the time of year and where on Earth you are,) it would appear to be standing still in your window. (Neglecting for the moment the comparatively slow apparent progression of the moon through the sky.)
But I reckon your train track was not exactly straight for the duration of your trip and the moon was never exactly perpendicular to your train's direction. But it was probably quite close to perpendicular. So any slight deviation would cause you to observe the moon travelling towards one edge of the window, and then a slight deviation in the other direction would cause the moon to apparently move towards the other edge of the window.
It's simply the winding s-shaped route your train took, heading generally east but never exactly, that caused these oscillations.


TITLE: Euler Characteristic of Coverings via Sheaf Theory
QUESTION [4 upvotes]: Let $X$ be a nice space (compact $CW$-complex or triangulated space, compact manifold, whatever works),
$f:Y\to X$ be a finite covering of degree $n$, and $\chi(X)$ be the euler characteristic.
By the Lefschetz fixed point theorem (or by counting cells of $X$ and $Y$ with the induced cell structure) one obtains
$$\chi(Y)=n \cdot\chi(X)$$
I wonder if there is  a sheaf theoretic proof of this statement, or even better a sheaf theoretic generalization of this statement  by replacing the constant sheaf by any other local system.

REPLY [5 votes]: Generalization: Euler characteristic with coefficients in a rank $n$ local system equals $n$ times ordinary Euler characteristic. Then using Leray (as in Gunnar Magnusson's comment) gives the case of a finite covering (and more generally that of a fibre bundle).
The following `sheafy' proof of the generalization should work. Pick an acyclic cover that trivializes the local system and use the associated Cech resolution. An essentially equivalent way would be to do an induction argument using the Mayer-Vietoris distinguished triangle.


TITLE: In Hilbert space: $x_n → x$ if and only if $x_n \to x$ weakly and $\Vert x_n \Vert → \Vert x \Vert$.
QUESTION [6 upvotes]: Assume that $H$ is a $\mathbb K$-Hilbert space, $(x_n)_{n \ge 1}$ a sequence in $H$ and $x ∈ H$.
  Show that $x_n → x$ if and only if $x_n \to x$ weakly and $\Vert x_n \Vert → \Vert x \Vert$.

I'm trying to prove this statement.
The $\Rightarrow$ is basically clear, since a  strongly convergent sequence is also weakly convergent. But how can I show the other direction $\Leftarrow$ ?

REPLY [9 votes]: To show that $x_n\to x$, you must show that $||x_n-x||\to 0$ as $n\to\infty$. Since $x_n\to x$ weakly, we know that
$$\langle x_n,y\rangle \to \langle x,y\rangle$$
for all $y\in H$, so in particular $\langle x_n,x\rangle\to\langle x,x\rangle$. Hence
$$\langle x,x_n\rangle=\overline{\langle x_n,x\rangle}\to\overline{\langle x,x\rangle}=\langle x,x\rangle$$
as well.
Therefore
$$ ||x_n-x||^2=\langle x_n-x,x_n-x\rangle=||x_n||^2-\langle x_n,x\rangle-\langle x,x_n\rangle +||x||^2\to 2||x||^2-2\langle x,x\rangle=0$$
so $x_n\to x$.


TITLE: Differential equation of 3rd order
QUESTION [0 upvotes]: I need some help solving the differential equation $$y'''=x+y\\   y(1)=3\\  y'(1)=2\\  y''(1)=1 $$ and $h=0.5$  with Euler's method
I don't know how to rewrite the equation to a system of equations of the first order..
Thanks!

REPLY [0 votes]: Here is how you linearize the system, we are given:
$$y'''(x) = x + y , y(1)=3,  y'(1)=2,  y''(1)=1$$
We can proceed as follows:

$x_1 = y \implies x_1' = y' = x_2$
$x_2 = y' \implies x_2' = y'' = x_3$
$x_3 = y'' \implies x_3' = y''' = x + y = x + x_1$

So, our new system is:
$$\begin{align}
x_1' & = x_2 \\
x_2' & = x_3 \\
x_3' & = x + x_1
\end{align}$$
Note, the given initial conditions, similarly follow as:

$y(1) = 3 \implies x_1(1) = 3$
$y'(1) = 2 \implies x_2(1) = 2$
$y''(1) =1 \implies x_3(1) = 1$

Now, apply Euler's method to this system using $h = 0.5$.


TITLE: If $\Bbb C^\infty$ is the only infinite-dimensional Hilbert space, why do other space names need to be used?
QUESTION [0 upvotes]: In Dym and McKean book, the map ^ originally start out as an isomorphism $L^2(Q)$ ($Q$ is any finite measure interval) to $L^2(\mathbb Z^+)$ (which is just another name of $\mathbb{C}^\infty$), therefore the coefficients are summed from $1$ to $\infty$. As I understand here it comes from the definition of Lebesgue integral. Later, when $Q=S^1$, it maps to $L^2(\mathbb Z^1)$, making the summation ranges from $-\infty$ to $\infty$. The problem is

Up to isomorphism, $\Bbb C^\infty$ is the only infinite-dimentional Hilbert space there is.

If so, why do we need to separate $L^2(\mathbb Z^+)$ and $L^2(\mathbb Z^1)$?

REPLY [1 votes]: If you start with a concrete mathematical problem, such as

for given function $p,q$ on an interval $I$, find functions $f$ that satisfiy $\lambda f = -(p(f')' + qf$

then you start with a function space and you can ask this question for $f \in L^2(I)$ and end up with a differential operator (in this case a Sturm-Liouville operator) in $L^2(I)$. You can write down what the operator does in that setting. However, if you use the isomorphism which maps $L^2(I)$ to $\mathbb C ^\infty$, you need to transform the operator and the pre-images as well and you lose easy access to some information:

What's the action of the differential operator?
We can't integrate by parts to manipulate algebraic statements depending on the operator
How does a boundary value of a function translate to $\mathbb C ^\infty$
Given two functions, how is the scalar product translated?

To wrap up, different Hilbert spaces are driven by applications they are used in and some properties of elements of these are possily easier worked with than an element the image of an isomorphism.


TITLE: Why can't we differentiate $x^x$ like we differentiate $x^3$ (e.g.)?
QUESTION [0 upvotes]: I know the derivative of $x^x$ is $x^x(1+log(x))$.
 So that must mean we cannot differentiate it like we do, say $x^5$.
The reason I read explaining why this is so is that the proof of the formula that the derivative of $x^n$(where n is a constant) $=$ $n.x^{n-1}$ involves binomial theorem.
But when I read the proof, I couldn't identify why there'd be any difference in applying that theorem for $x^x$.
So, I want to ask why can't we use the binomial theorem for justifying the same formula for, $x^x$ or $a^x$?
EDIT: I am explicitly asking why the 'binomial theorem proof' doesn't act as a proof for $x^x$, $a^x$ and so on.
I appreciate the points given in the already existing answers, but I want a direct answer to my actual question? Thank you.

REPLY [0 votes]: Differentiation is a simple procedure and it starts with a very simple definition and leads to laws of differentiation : sum/differencr, product, quotient, chain rule, implicit differentiation. Another part is establishing the derivatives of $x^{n}$, circular functions (and their inverses), exponential and logarithmic functions. These are building blocks of what we call elementary functions which are built using these blocks via a finite number of algebraic operations and a finite number of compositions.
Thus it is clear that using the derivatives of these building blocks and the rules of differentiation one can evaluate the derivative of any elementary function in an almost mechanical fashion. The function under consideration $f(x) =x^{x} $ involves base and exponent and we see that there are two building blocks which look similar to it namely $$(x^{n}) '=nx^{n-1},\,(a^{x})'=a^{x}\log a$$ Note that in both these formulas either the base or the exponent is a constant and hence these formulas can't be used directly to deal with $f$ where both the base and exponent are variable. Why don't we include functions like $f$ (where both base and exponent are variable) in the list of building blocks for elementary functions? Because it is unnecessary. This function $f$ can be expressed using the existing building blocks and composition. Thus $x^{x} =e^{x\log x} $ and we can apply the rules now to get its derivative.
Also note that the building blocks are defined via very simple formulas so that their derivatives can be easily evaluated using the limit definition. It is simply not possible to apply the limit definition to evaluate derivative of $f(x) =x^{x} $ in a manner which does not mirror the application of chain rule.


TITLE: Subspace test for space of $2\times2$ matrices, determine basis for $W:=\{M \in M_{2,2}(\mathbb R) | M^T = -M\}$
QUESTION [1 upvotes]: I'm sorry I don't know where to start on this question
Let $M_{2,2}(\mathbb R)$ denote the vector space of 2 by 2 matrices.
Define: $W:=\{M \in M_{2,2}(\mathbb R) | M^T = -M\}$ and $U := \{M \in M_{2,2}(\mathbb R) | M^T = M\}$
where $M^T$ denotes the transpose of $M$. Use the subspace test to show that $W$ is a subspace of $M_{2,2} (\mathbb R)$. Find a basis for $W$. Determine U∩W. Find a basis $B$, for $U$. Extend $B$ to a basis $M_{2,2}(\mathbb R)$.
Thank you in advance

REPLY [1 votes]: First note that for $w\in W$, $w$ is in the form
$\begin{bmatrix}0&a\\-a&0\end{bmatrix}$. Do you see why the diagonal has to be $0$?
Next, look at $u \in U$, where $u$ is in the form $\begin{bmatrix}a&b\\b&d\end{bmatrix}$. Can you prove to yourself that both of these forms represent elements of $W,U$ respectively?
Now, what do you know about the subspace test? Take any $w_1,w_2\in W$ in the form above and prove to yourself that $W$ is a subspace of $M_{2,2}(\mathbb{R})$.
Given the representation of elements in $W$, how many elements do you need to form a basis? Can you figure out what this basis looks like? Use similar ideas for $U$.
Lastly, what do elements in $W$ and $U$ have in common? Use that to solve for their intersection.


TITLE: Probability measure on naturals
QUESTION [0 upvotes]: Does there exist a probability measure $μ$ on a sigma algebra of subsets of $\mathbb N = \{ 1,2,3,...\}$ such that, $$μ(\{ n\in\mathbb N :n\equiv j \bmod{N}\})=\frac{1}{N}$$ for every $j,N\in \mathbb N$?
By intuition i think that it can't be found such  measure but i don't see a contradiction..Any hint will be very helpful.

REPLY [1 votes]: $$\mu(\{n\})\le \lim_{k\to \infty} \mu( \{ an, \gcd(a,k!)=1\})= \lim_{k\to \infty} \frac1n\frac{\phi(k!)}{k!}=\lim_{k\to \infty} \frac1n\prod_{p\le k}(1-p^{-1})=0$$
The countably infinite additivity implies that $$\mu(\Bbb{Z}_{\ge 1})= \sum_{n\ge 1}\mu(\{n\})=0$$


TITLE: Why does physical quantity misinterpretation result in near-identical torque results?
QUESTION [2 upvotes]: I am going to use an equation
$$\text{torque} = \frac{\text{power}\times 5252}{\text{RPM}}$$
derived on Wikipedia.
Suppose that $\text{power} = 100\ \mathrm{hp}$ and $\text{RPM} = 5252\ \mathrm{rpm}$
$$\text{torque} = \frac{100\ \mathrm{hp} \times 5252}{5252\ \mathrm{rpm}} = 100\ \mathrm{ft\,lb}\tag{a}$$
So far, so good.
Suppose, however we first convert power from $\mathrm{hp}$ to $\mathrm{kW}$
$$1.0\ \mathrm{hp} = 0.746\ \mathrm{kW}$$
And use it as-is in our original equation with no regard to other units:
$$\frac{5252 \times 74.6\ \mathrm{kW}}{5252\ \mathrm{rpm}} = 74.6\ \mathrm{\frac{kW}{rpm}}$$
So far so good (aside from out-of-place hp to kw conversion)
Now, the bad:
However, say we interpret that last number as $\mathrm{ft\,lb}$ and convert it to $\mathrm{N\,m}$
$$74.6\ \text{(interpreted as ft lbs)} = 101.1\ \mathrm{N\,m}\tag{b}$$
Using $1\ \mathrm{ft\,lb} = 1.36\ \mathrm{N\,m}$ for conversion.
Question:
Why is the result gotten in (b), $101.10$, so close to the one gotten in (a), $100$?  (they are off by 1.1%)
Is there a deeper meaning here or is it just a physical coincidence?
I got this example from a real world scenario where I did the above misinterpretations  and conversions at the wrong time, but then got a similar looking results to where at first I thought I had rounding errors introduced by a computer.  After some research I saw that this is not an a computer error.
I was curious to see if I just hit on this by accident or if this closeness in results has some deeper meaning.

REPLY [6 votes]: Just an accident.  You've discovered that the ratio of $1\text{hp}$ to $1\text{kW}$ (0.7457) is pretty close to the ratio of a $\text{N m}$ to a $\text{ft lb}$ (0.7376).  So if you apply one of them and the reciprocal of the other, the answer doesn't change much.
As the value for horsepower isn't derived from other units (it's a measured quantity), there's no deeper meaning to be found.
There are many, many different ratios you can form from different conversions.  To find that one particular pair falls within 1% of each other is to be expected.

REPLY [1 votes]: However, say we interpret that last number as ft * lb

A kilowatt hour per rpm is not a foot-pound - you cannot "interpret" it as that.
$$W=\tau\cdot\theta$$
$$P=\frac{W}{t}$$
$$P=\frac{\tau\cdot\theta}{t}=\tau\cdot\omega,$$
where $\omega$ is your angular frequency in rad/s (about 500 rad/s, in the case of 5000 RPM).
Plug in 100 horsepower for your power to find the torque is
$$\tau=\frac{100 HP}{523 rad/s}\approx0.191\textrm{ foot-pounds,}$$
which is very different than your interpretation.


TITLE: Proof term for predicate calculus.
QUESTION [1 upvotes]: Proofs of propositional calculus may be written in simple typed lambda calculus(λ→).
(as shown on p.52-53 in https://www.cs.ru.nl/~freek/talks/lc-2012/lambda5.pdf )
I think that this calculus may be enriched with 0,1,*, + and constant c:A\/~A for proofs of classical reasoning, which represents(roughly) contradiction, truth, concatenation of proofs, combining of proofs and LEM.
1) Could you please recommend me a paper which contains detailed presentation of such enriched calculus?
2) I am also  interested in similar calculus but for predicate calculus. If it exists, how it is called, which paper about it should I read?

REPLY [1 votes]: This question hints at a lot of major mathematics, as the intersection of lambda calculus, type theory, and logic is a major talking point right now.
One point you are sure to run into when reading about this is the LEM - most people working in type theoretic logics prefer to forgo the LEM. Intuitively this is because we want a very strong form of disjunction, which says that we must have proven either $A$ or $B$ in order to conclude $A \lor B$. This is in direct opposition with the idea that we can know $A \lor ~A$ without having proven either!
In this vein, here is a lovely set of lecture notes from a logic class at CMU which explains what you are after (without the LEM). It is from 15-317, Constructive Logic, and the website is here in case you want to read more.
If you insist upon having LEM, there are a few avenues available. Here is another set of lecture notes from a prior iteration of the same course, which discusses classical logic. For completeness, here is the website from this iteration of the course.
As for quantification, here is another set of lecture notes from the same course which talks about one approach.

If you are interested in lambda terms acting as witnesses for the truth of propositions, Type Theory is a worthwhile keyword to start searching! There are two sides to the type theoretic coin, with proof theory on the math side, and programming language theory on the other. You can find information about both here. It is a github repo including PDFs relating to type theory, subcategorized by topic!
An excellent reference for the proof theoretic side of Type Theory is Proofs and Types by Girard. I've also been told Type Theory and Formal Proof, by Nederpelt and Geuvers is good, though I have not read it.
As for programming language theory, I cannot recommend Practical Foundations for Programming Languages by Harper highly enough. It is excellently written (if extremely opinionated), and does a good job surveying the entire field.
Finally I feel personally obligated to mention Martin-Lof type theory, which serves as a potential replacement for set theory as the foundation of mathematics. It is a dependent theory, which requires some care in setting up. The results, however, are beautiful. If you have a background in Algebraic Topology, The HoTT Book is an excellent example of Martin-Lof type theory in the wild.

I hope this helps ^_^


TITLE: Why the excess pressures are equated in hydraulic press?
QUESTION [0 upvotes]: Let's say we have a closed chamber filled with incompressible liquid and there are two area's(smaller(a) and larger(A)) where we can apply force or keep objects.We keep an object of weight W on larger area.Due to this,pressure at every point in a liquid will increase by W/A ,according to pascal's law.The situation is depicted in picture below.

Now,we apply force f on smaller area .Due to this the pressure at every point will even further increase by f/a.Now the situation looks something like the picture below.
Now from what I have read in books
W/A=f/a
Why the pressure exerted by the box W has to be equal to the pressure exerted by force f?

REPLY [2 votes]: This equality is a equilibrium condition, i.e. a condition for the fluid to be in rest. In general, inertia force terms must be added if the fluid is not in equilibrium.


TITLE: Notation in Fundamental Groups in Topological Spaces
QUESTION [1 upvotes]: I was reading about Fundamental Groups in Topological Spaces at WikiPedia. In the middle of the text, they define the following set $\pi_1(X,x_0):$

I guess I don't understand the meaning of "$\text{/homotopy}$". What does this mean?

REPLY [2 votes]: If you have a set $S$ and an equivalence relation $\sim$ on $S$, then
$$S / \sim$$
denotes the set of all equivalence classes with respect to $\sim $. In your question we have $S =$ set of all loops $\gamma : [0,1] \to X$ based at $x_0$. You have the relation $\gamma$ is homotopic to $\gamma'$ which is an equivalence relation on $S$. If you write $\gamma \simeq \gamma'$ for this equivalence relation, then
$$\pi_1(X,x_0) = S/\simeq \phantom{.}= S/\text{homotopy} .$$


TITLE: Is coin tossing a Markov process?
QUESTION [1 upvotes]: When we toss an unbiased coin, the probability of observing both heads and tails is 1/2. I take that to mean that over a really large number of coin tosses the number of times the coin will turn heads will be almost equal* to the number of times the coin will turn tails.
My question is, if we witness a series of coin tosses that happens to have many more number of, say heads, than tails then will we not expect the upcoming coin tosses to be 'mean-reverting' i.e. more inclined to produce tails than heads - only in order to maintain the definition of probability being 1/2 as per the previous paragraph?
I think what I am really asking is whether an unbiased coin-tossing is a Markov process. I would add that if it is, then my understanding of why the probability of heads/tails is 1/2 is wrong.
[*] - If not, then we need more coin tosses such that the ratio of number of heads (or tails) to the total number of coin tosses approaches 1/2 as the number of coin tosses approaches infinity.

REPLY [1 votes]: Intuition can be gained from the fact that these averages are completely determined by the tail values, and not the first few terms. That is, for any fixed $k∈\Bbb N$ and any finite number of terms $b_1,…,b_k$ (even if $b_i ≠ a_j$ for every $i,j$)
$$ \lim_{n→∞} \frac{1}{n}(a_1+…+a_n) =  \lim_{n→∞} \frac{1}{n}(b_1+…+b_k+a_{k+1} + … + a_n) $$
Hence, there is no need for the $a_j$s to 'correct the bad behaviour of $b_i$s'; the $b_i$s simply don't matter in the long run.


TITLE: What can we infer about the connectivity and edge connectivity of a graph given the number of edge disjoint paths?
QUESTION [1 upvotes]: So the number of edge disjoint u,v paths in a graph is x. Doing this problem, I thoguht back to Menger's theorem, and thought that the graph is x-edge-connceted and so the number of cut sets between any two vertices is also x.
However I'm not sure how to find find out anything about the vertex connectivity since all I was given was the number of edge disjoint paths.

REPLY [1 votes]: I think this is because there is nothing you can find out in general i.e. you can have many edge disjoint paths for every pair of vertices but it is still possible for the graph to have vertex connectivity $1$.
For example suppose you have two complete graphs $G, G'$ on $n$ vertices each. Now let's merge these two graphs in the following way: Construct a new vertex $v$ and connect $v$ with every vertex in $G$. Also connect $v$ with every vertex in $G'$. Call this new graph $G''$. If you choose any two vertices $u, w$ from $G''$ you will find that there are always at least $n - 1$ (actually it is even more, $n$ would work too for example) edge disjoint paths between $u$ and $w$. But obviously the graph $G''$ has vertex-connectivity $1$.