TITLE: If matrix $A$ in $\mathbb{R}^3 $ such that, $A^3 = I$, $\det A = 1$. Is there a such matrix which is not orthogonal, rotation and identity?
QUESTION [4 upvotes]: I tried to use Cayley–Hamilton theorem to learn something about the matrix.
Using the theorem we have:
$p(A)=0 = - A^3 + \text{tr} A\cdot A^2 -\left(\begin{vmatrix}a_{11} && a_{12} \\ a_{21} && a_{22} \end{vmatrix} + \begin{vmatrix}a_{22} && a_{23} \\ a_{32} && a_{33} \end{vmatrix} + \begin{vmatrix}a_{11} && a_{13} \\ a_{31} && a_{33} \end{vmatrix}\right) A + \det A$.
Since $-A^3 + \det A \cdot I = 0$, then $A = \frac{\left(\begin{vmatrix}a_{11} && a_{12} \\ a_{21} && a_{22} \end{vmatrix} + \begin{vmatrix}a_{22} && a_{23} \\ a_{32} && a_{33} \end{vmatrix} + \begin{vmatrix}a_{11} && a_{13} \\ a_{31} && a_{33} \end{vmatrix}\right)}{\text{tr} A}I$. With this in mind I cannot obtain even rotation matrix to say nothing about something else. Any hints?

REPLY [6 votes]: Here's a quick example:
$$
A = \pmatrix{\cos(2\pi/3) & -10\sin(2 \pi /3) & 0\\
\frac 1{10}\sin(2 \pi /3) & \cos (2 \pi /3) & 0\\0 & 0 & 1}
$$
Is like the usual rotation, but with the off-diagonal entries slightly altered.  Show that $A^3 = I$, but $A$ is not orthogonal (or a rotation) since $A^TA \neq I$.

Another easy example:
$$
A = \pmatrix{\cos(2\pi/3) & \sin(2 \pi /3) & 2\\
\sin(2 \pi /3) & \cos (2 \pi /3) & 3\\0 & 0 & 1}
$$
the same argument applies.


TITLE: Nullspace of a matrix
QUESTION [2 upvotes]: I think I lack understanding in this area as I tend to get stumped on the idea of 'null space' when it comes to understanding proofs/answering questions.
Even though I have some understanding already, could someone explain:

what exactly is meant by the null space of a matrix $A$? Is it the kernel of the linear map $T$ associated with $A$ or the dimension of the kernel of $T$?
how does one find the null space of a matrix $A$ and a basis for it. Is the null space a vector space?
any other details you think are good to bear in mind when thinking about null spaces

REPLY [3 votes]: The null space is the kernel of the transformation.
One finds the null space by solving $A\vec{x}=\vec{0}$ for $\vec{x}$.  Typically, this is done through Gaussian elimination.  And yes, the null space is a vector space: note that if $\vec{x},\vec{y}$ are in the null space and $\alpha,\beta\in\mathbb{C}$, then
$$
A(\alpha\vec{x}+\beta\vec{y})=\alpha(A\vec{x})+\beta(A\vec{y})=\alpha\vec{0}+\beta\vec{0}=\vec{0},
$$
so that $\alpha\vec{x}+\beta\vec{y}$ is also in the null space.


TITLE: The ubiquitous Planewave Ansatz
QUESTION [6 upvotes]: In physics, the planewave ansatz (meaning: an educated solution guess) is very ubiquitously used, when solving differential equations, in different domains of physics. E.g. to solve the dispersion relation of magnons or phonons in solids, also very commonly in QM e.g. in semiclassical approximations, and so on. The examples above all correspond to systems of 2nd order linear differential equation.
My question:

Is the use of the planewave ansatz purely motivated by mathematical reasons? (i.e. form/type of differential equation).
Knowing that planewaves don't really exist in reality, as they would require infinite propagation, how is it that it still remains a theoretically suited ansatz most often (for describing physical phenomena)? (e.g. phonons to propagate to infinity surely) I guess this last question could be formulated differently: In which cases one can be sure that a planewave ansatz (type solution) would absolutely fail?

REPLY [5 votes]: John has answered this partially, however the fundamental mathematical idea is missing:
If we think of a function being member of some vectorspace, then basisvectors exist. This concept will surely be familiar to you from Quantum mechanics.
But also from there we remember, that problems were alot easier to handle, if we know the Eigenbasis of the Operators that we throw on functions. Then we just change the functions basis into the operators Eigenbasis and we're done.
Just how do we do this? How do we construct the Eigenspace for differential operators that make up differential equations? Let's think about it... First let's state the Eigenvalue Problem in abstract notation
$$ \hat{O} f=cf $$
and then put in that we use $\hat{O} = d/dx$:
$$\frac{df}{dx} = c f$$
And we immediately see, that this is a known equation! For which the solution is $f \sim exp(cx)$
Thus we know that if we deal with linear differential operators, an exponential basis will always faciliate our calculations. All we need to do is to use proper arguments $c$ to fit the order of the equation (i.e. add some $i$s or $\pm$).
The freedom in choosing $c$ for this method still to work has then given not only rise to the Fourier transform and its by John described properties, but also to the Laplace-transform.
By chosing $c$ appropriately we are ofc making also an 'educated guess' about the form of our exponential solution. For example a DEQ
$$\partial_t^2 f + \partial_x^2 f=0$$
will give oscillatory solutions for $c=i(kx \pm \omega t)$, but something that has a 'starting' point and then decays or grows for real c's. That's the idea of a Laplace-transform. You can also look up the examples on the wikipage (scroll down).

Concerning your other question:
The plane-wave-ansatz surely fails as soon as nonlinearities enter the differential equation, like this maximally simplified 1-D Navier-Stokes-Equation:
$$\partial_t v + v \partial_x v = 0$$
Already any attempt to use a Fourier-transform of $v$ will not result in an algebraicly solvable equation.
Numerically modellers adress this problem e.e. by pseudospectral methods, where all the equation is fourier-transformed except for nonlinearities. Those are calculated in real-space and then later numerically transformed.
The other problem you mentioned of infinity, can be also seen as problem of scale: With a box-size $L$ on which you want to solve your problem you will always get discretization effects of the waves that wobble around in this box. However if the wavenumbers $k << L$ then your density of waves is high enough to still be able to obtain a physically meaningful result.
The correctness then increases the better the above condition is fulfilled.


text:
\begin{document}

\begin{frontmatter}
\title{The characterization of perfect Roman  domination stable trees}

\author[LanZhou]{Zepeng Li}\ead{lizp@lzu.edu.cn},
\author[GZHU]{Zehui Shao}\ead{zshao@gzhu.edu.cn},
\author[GZHU]{Yongsheng Rao}\ead{rysheng@gzhu.edu.cn},
\author[CDU]{Pu Wu}\ead{puwu1997@126.com},
\author[Adelphi]{Shaohui Wang}\ead{shaohuiwang@yahoo.com}

\address[LanZhou]{
    School of Information Science and Engineering, Lanzhou University, Lanzhou 730000, China}
\address[GZHU]{
   Institute of Computing Science and Technology, Guangzhou University, Guangzhou 510006, China}
\address[CDU]{ School of Information Science and Engineering,
 Chengdu University,
  Chengdu 610106,  China}
\address[Adelphi]{Department of Mathematics,
Savannah State University, Savannah, GA 31419, USA}
\corauth[cor]{Corresponding author: Shaohui Wang.}

\begin{abstract}
A \emph{perfect Roman dominating function} (PRDF) on a graph $G = (V, E)$ is a function $f :
V \rightarrow \{0, 1, 2\}$ satisfying the condition that every vertex $u$ for which $f(u) = 0$
is adjacent to exactly one vertex $v$ for which $f(v) = 2$.
The weight of a PRDF is the value $w(f) = \sum_{u \in V}f(u)$. The minimum weight of
a PRDF on a graph $G$ is called the \emph{perfect Roman domination
number $\gamma_R^p(G)$} of $G$.
A graph $G$ is perfect Roman domination domination stable if the perfect Roman domination number of $G$ remains unchanged
under the removal of any vertex.
In this paper, we characterize  all trees that are perfect Roman domination stable.

\noindent{\bf Keywords:} Domination, Roman domination number, perfect Roman domination number, tree, stable.\\
\noindent{\bf MSC 2010:} 05C05, 05C31, 05C35, 05C69, 05C90
\end{abstract}

\end{frontmatter}

\newtheorem{thm1}{Theorem}
\newtheorem{clm1}{Claim}
\newtheorem{rem1}{Remark}
\newtheorem{lem1}{Lemma}
\newtheorem{cor1}{Corollary}
\newtheorem{def1}{Definition}
\newtheorem{observ1}{Observation}
\newtheorem{ex1}{Example}
\newtheorem{prop1}{Proposition}

\section{Introduction}
In this paper, we shall only consider graphs without multiple edges or loops.
Let $G$ be a graph, $S \subseteq V(G)$, $v \in V(G)$, the \emph{neighborhood} of $v$ in $S$ is denoted by $N_S(v)$.
That is to say $N_S(v) =\{u| uv \in E(G), u \in S\}$.
The closed neighborhood $N_S[v]$ of $v$ in $S$ is defined as $N_S[v]=\{v\} \cup N_S(v)$.
If $S=V(G)$, then $N_S(v)$ and $N_S[v]$ are denoted by $N(v)$ and $N[v]$, respectively.
Let $S \subseteq V(G)$, we write $N_G(S)= \cup_{x \in S}N_G(x)$.
The degree of $v$ is $d(v)=|N(v)|$.
We will omit the subscript $G$, that is to say, $N_G(T)$ is denoted by $N(T)$.
For a tree $T$ and a vertex $v\in V(G)$, we denote by $L(v)$ the set of all leaves of $v$.
 A tree $T$ is a double star if it contains exactly two vertices that are not leaves.
A double star with respectively $p$ and $q$ leaves attached at each support vertex is denoted by $DS_{p,q}$.

A \emph{dominating set} of $G$ is a subset $D$ of $V$ such that every vertex in $V- D$ is adjacent to at least one vertex in $D$. The \emph{domination number} $\gamma(G)$ is the minimum cardinality of a dominating set of $G$. A \emph{Roman dominating function} (RDF) of $G$ is a function $f: V(G)\rightarrow \{0, 1, 2\}$ such that every vertex $v\in V$ with $f(v)=0$ is adjacent to at least one vertex $u$ with $f(u)=2$. The weight of an RDF $f$ is the value $f(V)=\sum_{v\in V} f(v)$. The Roman
domination number $\gamma_R(G)$ is the minimum weight of an RDF of $G$. The problems on domination and Roman domination of graphs have been investigated widely, for example, see the list of references, \cite{Haynes1998} and \cite{Chambers2009,Cockayne2004,Favaron2009,WW6,WW7,WW8}, respectively.

The affections of vertex removal on domination number and Roman domination number in a graph
have been studied in \cite{Bauer,Rad1} and \cite{Hansberg,Rad2}, respectively. Jafari Rad and Volkmann \cite{Rad2} introduced the concept
of Roman domination stable graphs and these graphs had been further studied in \cite{Hajian,Samodivkin,WW1,WW2,WW3,WW4,WW5}.

Furthermore, Henning, Klostermeyer and MacGillivray\cite{Henning} introduce a perfect version of Roman
domination.
A \emph{perfect Roman dominating function} (PRDF) on a graph $G = (V, E)$ is a function $f :
V \rightarrow \{0, 1, 2\}$ satisfying the condition that every vertex $u$ for which $f(u) = 0$
is adjacent to exactly one vertex $v$ for which $f(v) = 2$.
The weight of a PRDF is the value $w(f) = \sum_{u \in V}f(u)$. The minimum weight of
a PRDF on a graph $G$ is called the \emph{perfect Roman domination
number $\gamma_R^p(G)$} of $G$.
A PRDF $f$ is call a $\gamma_R^p$-function of $G$ if $w(f)=\gamma_R^p(G)$.
A graph $G$ is \emph{perfect Roman domination stable} if the perfect Roman domination number of $G$ remains unchanged
under removal of any vertex.

Recently, many research are working in this topic. For instance,
Rad et al.\cite{Rad2} studied the changing and unchanging the Roman domination number of a graph. Henning et al. \cite{Henning} explored some trees about perfect Roman domination. Favaron et al. \cite{Favaron2009} found some Romain domination number of a graph. Chambers et al. \cite{Chambers2009} deduced some extremal results on Roman domination.
Motivated by the above results,  we continue to study Roman domination and characterize all perfect Roman domination stable trees.

\section{Perfect Roman domination stable trees}
In this section, we will give some lemmas and properties.
\begin{observ1}\label{PRO1}
Let $T$ be a perfect Roman domination stable tree and $f$ be a $\gamma_R^p$-function of $T$. Then \\
(i)  $f(v)\neq 1$ for any $v \in V(T)$.\\
(ii) $f(v) \neq 2$ for any leaf $v$ in $T$.\\
(iii) If there exists a vertex $x_4 \in V(T)$ is adjacent to a star with vertex set $\{x_1,x_2,x_3,y_1\}$ and center $x_2$ and $x_4x_3 \in E(T)$,
then $f(x_2)=2$ and  $f(x_1)=f(y_1)=f(x_3)=f(x_4)=0$.
\end{observ1}
\begin{proof}
(i) Suppose there exists a vertex $v\in V(T)$ for which $f(v)=1$.
 Let $T'=T-\{v\}$. Since $T$ is a perfect Roman domination stable tree, we have $\gamma_R^p(T)=\gamma_R^p(T')$.
Then we have $f|_{T'}$ is a PRDF on $T'$.
Thus
$\gamma_R^p(T')\leq w(f)-1= \gamma_R^p(T)-1$

a contradiction.\\
(ii)
Otherwise, let $u \in N(v)$ and $f(u)=0$.
Now we can obtain a PRDF with the same weight by assigning 1 to $u$ and $v$, a contradiction with (i).\\
(iii) By the results of (i) and (ii), we have $f(x_2)=2$, $f(x_1)=f(y_1)=0$ and $f(x_3) \in \{0,2\}$.
If $f(x_3)=2$, then we can obtain a new $\gamma_R^p$-function of $T$ by changing $f(x_2)$ to 0 and $f(s)$ to 1 for any $s\in \{x_1,y_1\}$,
contradicting with (i).
Thus, $f(x_3)=f(x_4)=0$.
\end{proof}

In this section we give a constructive characterization of all perfect Roman domination stable trees under vertex removal.
For a tree $T$, let
\begin{eqnarray*}
W(T) & = & \{u \in V(T)\mid \textrm{ $f(u)=0$ for any $\gamma_R^p$-funnction $f$ of $T$}\}.
\end{eqnarray*}

In order to presenting our constructive characterization, we define a family of trees as follows.
Let $\mathcal{T}$ be the family of trees $T$ that can be obtained from a sequence $T_1$,
$T_2$, $\ldots$, $T_k$ of trees for some $k \geq 1$, where $T_1$ is $P_3$ and $T=T_k$.
If $k \geq 2$, $T_{i+1}$ can be obtained from $T_i$ by    the following operation.

\begin{description}
  \item [Operation ${\mathcal O}_1$: ] If $u \in W(T_i)$, then ${\mathcal
O}_1$ adds a path $v_3v_2v_1$ and an edge $uv_3$  to obtain $T_{i+1}$.
\end{description}

\begin{lem1}\label{AddP3}
Let $G$ be a graph and $u \in V(G)$.
If $G'$ is a graph obtained by adding a  path  $v_3v_2v_1$ and an edge $uv_3$ from $G$,
then $\gamma_R^p(G')= \gamma_R^p(G) +2$.
\end{lem1}

\begin{proof}
Let $f$ be a $\gamma_R^p$-function on $G$.
If $f(u)\neq 2$, we define $f' :V(G') \rightarrow \{0, 1, 2\}$ by $f'(v_1)=0$, $f'(v_2)=2$, $f'(v_3)=0$ and $f'(v)=f(v)$ if $v\in V(G)$.
If $f(u)=2$, we
define $f' :V(G') \rightarrow \{0, 1, 2\}$ by $f'(v_1)=1$, $f'(v_2)=1$, $f'(v_3)=0$ and $f'(v)=f(v)$ if $v\in V(G)$.
Then in each case $f'$ is a PRDF function on $G'$ and $w(f')=w(f)+2$. Thus we have
$\gamma_R^p(G')\leq w(f')=w(f)+2 = \gamma_R^p(G)+2$.

Conversely, let $f$ be a $\gamma_R^p$-function on $G'$.  We consider the following cases.\\
\textbf{Case 1:} $f(u) \leq 1$. Clearly, we have $f(v_2)=2$ and $f(v_1)=f(v_3)=0$. Then $f|_G$ is a PRDF function on $G$.
Thus $\gamma_R^p(G')= w(f)= w(f|_G)+2 \geq \gamma_R^p(G)+2$, as desired.\\
\textbf{Case 2:}  $f(u)=2$. In this case we have $(f(v_3),f(v_2),f(v_1)) \in \{(0,1,1),(0,0,2)\}$.
 Now $f|_G$ is a PRDF of $G$ and so $\gamma_R^p(G) \leq w(f|_G)=w(f)-2=\gamma_R^p(G')-2$, as desired.
\end{proof}

\begin{lem1}\label{OneV}
Let $T$ be a perfect Roman domination stable tree and $u \in W(T)$.
If $T'$ is a tree obtained by adding a single vertex $v$ and an edge $uv$ from $T$,
then $\gamma_R^p(T')= \gamma_R^p(T) +1$.
\end{lem1}
\begin{proof}
Let $f$ be a $\gamma_R^p$-function on $T$. Define $f' :V(T') \rightarrow \{0, 1, 2\}$ by $f'(v)=1$ and $f'(x)=f(x)$ if $x\in V(T)$. Then $f'$ is a PRDF function on $T'$ and $w(f')=w(f)+1$. Thus we have
$\gamma_R^p(T')\leq w(f')=w(f)+1 \leq \gamma_R^p(T)+1$.

Conversely, let $f$ be a $\gamma_R^p$-function on $T'$, by Observation \ref{PRO1}  we have $f(v)\neq 1$ for any $v \in V(T)$.
Now we consider the following cases. \\

\textbf{Case 1:} $f(v)=2$. In this case we have $f(u) = 0$ and we consider a $\gamma_R^p$-function $f'$ on  $T'$ with $f'(v)=f'(u)=1$ and $f'(x)=f(x)$ otherwise. Then  $f'|_T$ is a PRDF of $T$ and we have $\gamma_R^p(T)\leq \gamma_R^p(T')-1$, as desired.\\
\textbf{Case 2:} $f(v)=0$. In this case we have $f(u) =2$.
Since $u \in W(T)$, and  $f|_T$ is a PRDF of $T$.
Then we have $\gamma_R^p(T)< w(f|_T)=\gamma_R^p(T')$.
Thus $\gamma_R^p(T)\leq \gamma_R^p(T')-1$, as desired.
\end{proof}

\begin{lem1}\label{AddP2}
Let $T$ be a perfect Roman domination stable tree and $u \in W(T)$.
If $T'$ is a tree obtained by adding a  path  $v_2v_1$ and an edge $uv_2$ from $T$,
then $\gamma_R^p(T')= \gamma_R^p(T) +2$.
\end{lem1}
\begin{proof}
Let $f$ be a $\gamma_R^p$-function on $T$. Define $f' :V(T') \rightarrow \{0, 1, 2\}$ by $f'(v_2)=0$, $f'(v_1)=2$ and $f'(v)=f(v)$ if $v\in V(T)$. Then $f'$ is a PRDF function on $T'$ and $w(f')=w(f)+2$. Thus we have
$\gamma_R^p(T')\leq w(f')=w(f)+2 = \gamma_R^p(T)+2$.

Conversely, let $f$ be a $\gamma_R^p$-function on $T'$, by Observation \ref{PRO1}  we have  $f(v)=0$ for any $v \in V(T)$.
Now we consider the following cases. \\
\textbf{Case 1:} $f(v_1)=2$. In this case we have $f(v_2) = 0$.   Then $f|_T$ is a PRDF of $T$ and we have $\gamma_R^p(T)\leq \gamma_R^p(T')-2$, as desired.\\
\textbf{Case 2:} $f(v_1)=0$. In this case we have $f(v_2) =2$ and $f(u)=0$.

Since $T$ is stable,
we have $\gamma_R^p(T-u)=\gamma_R^p(T)$.
Since $f(u)=0$, we have $f|_{T-u}$ is a PRDF of $T-u$, we have
$\gamma_R^p(T)=\gamma_R^p(T-u) \leq w(f|_{T-u})=w(f)-2=\gamma_R^p(T')-2$, as desired.
\end{proof}

\begin{lem1}\label{first}
{\em If $T_i$ is a perfect Roman domination stable tree and $T_{i+1}$ is a tree obtained from
$T_i$ by Operation ${\mathcal O}_1$, then
$T_{i+1}$ is a perfect Roman domination stable tree.}
\end{lem1}
\begin{proof}
By Lemma \ref{AddP3}, we have $\gamma_R^p(T_{i+1})=\gamma_R^p(T_{i})+2$.
Let $v \in V(T_{i+1})$ be an arbitrary vertex and  $T'=T_{i+1}-v$. \\
If $v\in V(T_{i})\setminus\{u\}$,
since $T_i$ is  stable, we have $\gamma_R^p(T_i-v)=\gamma_R^p(T_i)$.
Then by Lemma \ref{AddP3},  we have $$\gamma_R^p(T_{i+1}-v)=\gamma_R^p(T_i-v)+2=\gamma_R^p(T_i)+2=\gamma_R^p(T_{i+1}).$$\\
If $v=u$, since $T_i$ is stable,
obviously, $\gamma_R^p(T')=\gamma_R^p(T_i-v)+2=\gamma_R^p(T_i)+2=\gamma_R^p(T_{i+1})$.\\
If $v=v_1$, By Lemma \ref{AddP2}, we have $\gamma_R^p(T')=\gamma_R^p(T_i)+2=\gamma_R^p(T_{i+1})$.\\
If $v=v_2$, By Lemma \ref{OneV}, we have $\gamma_R^p(T')=\gamma_R^p(T_i+{uv_3})+\gamma_R^p(K_1)=\gamma_R^p(T_i)+1+1=\gamma_R^p(T_{i+1})$.\\
If $v=v_3$, then we have $\gamma_R^p(T')=\gamma_R^p(T_i)+\gamma_R^p(P_2)=\gamma_R^p(T_i)+2=\gamma_R^p(T_{i+1})$.
Thus $T_{i+1}$ is a perfect Roman  domination stable and the proof is complete.
\end{proof}

\begin{lem1}\label{Lemma5}
Let $T$ be a perfect Roman domination stable tree  of order $n\ge 3$ with $diam(T) \geq 4$ and $T$ contain no pendent $P_3$, and $P=x_1x_2\cdots x_k$ be a longest path of $T$.
Then $d(x_2)=2$.
\end{lem1}
\begin{proof}
Let $f$ be a $\gamma_R^p$-function of $T$.
By Observation \ref{PRO1}, we have $f(v) \in \{0,2\}$ for any $v \in V(T)$.

\begin{figure}[H]
\begin{center}
\includegraphics[scale=0.20]{Tk.eps}
\caption{The tree in the proof of Lemma \ref{Lemma5}}\label{LV2}
\end{center}
\end{figure}

First we have $|L(x_2)| \leq 2$. Otherwise, we assume $L(x_2)=\{x_1, y_1, \cdots, y_{k_1}\}$ where $k_1 \geq 2$.
It is clear that $f(x_2)=2$, $f(x_1)=0$ and $f(y_i)=0$ for any $i=1,2,\cdots,k_1$.
Let $g$ be a $\gamma_R^p$-function of $T-x_2$ and so $g(x_1)=g(y_i)=1$ for any $i=1,2,\cdots,k_1$.\\
If $g(x_3)=2$,  we define a function $g'$ of $T$ as follows.
$g'(x_2)=2$, $g'(x_1)=g'(y_i)=0$ for any $i=1,2,\cdots,k_1$ and $g'(v)=g(v)$ for other vertex $v \in T$ and we have
$\gamma_R^p(T)\leq w(g')=w(g)-(k_1+1)+2<w(g)=\gamma_R^p(T-x_2)$, a contradiction.\\
If $g(x_3)=0$,  we define a function $g'$ of $T$ as follows.
$g'(x_2)=2$, $g'(x_3)=1$, $g'(x_1)=g'(y_i)=0$ for any $i=1,2,\cdots,k_1$ and $g'(v)=g(v)$ for other vertex $v \in T$. If $k_1\geq 3$, then we have $\gamma_R^p(T)\leq w(g')=w(g)-(k_1+1)+3<w(g)=\gamma_R^p(T-x_2)$, a contradiction. If $k_1=2$, then we have $w(g')=w(g)$ but $x_3$ is assigned with 1, a contradiction with Observation \ref{PRO1}.

Then we have $|L(x_2)| = 1$.
Suppose to the contrary, $|L(x_2)| = 2$, then we have
\begin{clm1}
$d(x_3)=2$.
\end{clm1}
\begin{proof}
First, we have $x_3$ has no leaf neighbor.
Otherwise, let $w_1 \in L(x_3)$.
Since $f(x_3)=0$ from Observation \ref{PRO1} (iii), we have $f(w_1)=1$, contradicting Observation \ref{PRO1}.

Now we have $x_3$ is not adjacent to a pendent star.
Otherwise, assume $x_3$ is adjacent to a pendent star centered with $w_1$ with $x_3w_1 \in E(T)$ and $L(W_1)$.
Since $f(x_3)=0$ and $f(x_2)=2$ from Observation \ref{PRO1} (iii), we have $f(w_1)=0$, and $f(x)=1$ for any vertex $v\in L(w_1)$, contradicting with Observation \ref{PRO1}.
\end{proof}

Now we have $x_4$ has no leaf neighbor.
Otherwise, let $w_1 \in L(x_4)$.
Since $f(x_3)=0$ and $f(x_2)=2$ , we have $f(x_4)=0$ and thus $f(w_1)=1$, contradicting Observation \ref{PRO1}.
Now  it follows from Observation \ref{PRO1}  that $f(x_1)=f(x_3)=f(y_1)=f(x_4)=0$, $f(x_2)=2$ and we have
\begin{clm1}\label{Claim1}
There exists no vertex $w_1 \in N(x_4)$ for which $|L(w_1)|=d(w_1)-1 \geq 1$.
\end{clm1}
\begin{proof}
Otherwise, we assume $w_1 \in N(x_4)$ for which $|L(w_1)|=d(w_1)-1 \geq 1$.
Analogous to the proof of $|L(x_2)| \leq 2$, we have $|L(w_1)| \leq 2$.
Let $T'=T-\{x_1,y_1,x_2,x_3\}-N[w_1]$ with $T'=T_1 \cup T_2 \cup \cdots \cup T_k$ for some $k$ and
$v_i = N(x_4) \cap T_i$ for $i=1,2,\cdots,k$ (see Fig. \ref{LV2}).
By Observation \ref{PRO1}, we have $f(x_2)=2$, $f(x_3)=f(x_4)=0$, $f(w_1)=2$ and $f(v_i)=0$ for any $i$.\\

Then we have
\begin{equation}\label{EQAdd4}
 \gamma_R^p(T)=w(f)= \sum_{i=1}^k w(f|_{T_i}) + 4.
\end{equation}

Since $f|_{T_i}$ is a PRDF of $T_i$,   we have $\gamma^p_R(T_i) \leq w(f|_{T_i})$ for any $i$.

Now we will show that
\begin{equation}\label{EQ1}
\textrm{$\gamma^p_R(T_i) = w(f|_{T_i})$ for any $i$.}
\end{equation}
Otherwise, we assume that $g$ is a $\gamma_R^p$-function of $T'$.
Now we suppose $\gamma^p_R(T_i) < w(f|_{T_i})$, then we have $w(g|_{T_i}) < w(f|_{T_i})$ for some $i$.
If $g(v_i)\leq 1$, then under $f$ we assign $g(t)$ instead of $f(t)$ to $t$ for any $t \in V(T_i)$ and obtain a PRDF with fewer weight, a contradiction.
If $g(v_i)=2$,  then under $f$ we assign  $g(t)$ instead of $f(t)$ to $t$ with for any $t \in V(T_i)$, $g(z)=1$ for $z \in N[w_1]-x_4$ and obtain a PRDF with weight at most $w(f)$. But $w_1$ is assigned with 1,  contradicting Observation \ref{PRO1}.

Since $f|_{T_i-v_i}$ is a PRDF of $T_i-v_i$,   we have $\gamma^p_R(T_i-v_i) \leq w(f|_{T_i-v_i})=w(f|_{T_i})$ for any $i$.

Now we will show that
\begin{equation}\label{EQ2}
\textrm{$\gamma^p_R(T_i-v_i) = w(f|_{T_i-v_i})=w(f|_{T_i})$ for any $i$.}
\end{equation}
Otherwise, we
assume that $g$ is a $\gamma_R^p$-function of $T'$.
and suppose $\gamma^p_R(T_i-v_i) < w(f|_{T_i})$. Then we have $w(g|_{T_i-v_i}) < w(f|_{T_i-v_i})$ for some $i$.
Then under $f$ we assign 1 to $v_i$  and $g(t)$ instead of $f(t)$ to $t$ for any $t \in V(T_i-v_i)$ and obtain a PRDF of $T$ with at most weight $w(f)$.
But $v_i$ is assigned with 1, contradicting Observation \ref{PRO1}.

Now let $g$ be a $\gamma_R^p$-function of $T-x_2$.
Now we have $\sum\limits_{z \in N[w_1]} g(z) \geq 3$.
Since $T$ is stable, we have $w(f)=w(g)$.
Therefore, we have
\begin{equation}\label{4}
\gamma_R^p(T)=w(g)=\sum_{i=1}^k w(g|_{T_i}) + 2+ \sum_{z \in N[w_1]} g(z) \geq \sum_{i=1}^k w(g|_{T_i})+5,
\end{equation}
By Eqs.(\ref{4}) and (\ref{EQAdd4}), we have
\begin{equation}\label{EQAdd5}
 \sum_{i=1}^k w(g|_{T_i})\leq \gamma_R^p(T)-5< \sum_{i=1}^k w(f|_{T_i}).
\end{equation}
Hence by Eq.(\ref{EQ1}), there must exist some $j$ ($1\leq j \leq k$) satisfying that
\begin{equation}\label{EQAdd6}
w(g|_{T_j})<w(f|_{T_j})=\gamma_R^p(T_j).
\end{equation}
If $g(v_j)\geq 1$, then $g|_{T_j}$ is a PRDF on $T_j$, a contradiction with Eq.(\ref{EQAdd6}).
If $g(v_j)=0$, then $g|_{T_j-v_j}$ is a PRDF on $T_j-v_j$. By Eq.(\ref{EQ2}), $w(g|_{T_j-v_j})=w(g|_{T_j})<w(f|_{T_j})=\gamma_R^p(T_j-v_j)$, a contradiction.

This completes the proof.
\end{proof}

\begin{clm1}\label{Claim2}
$x_4$ is not adjacent to a vertex $w_1$ with $|L(w_1)|\geq 2$.
\end{clm1}
\begin{proof}
Suppose to the contrary, there exists  a vertex $w_1\in N(x_4)$  with $|L(w_1)|\geq 2$.
By Claim \ref{Claim1}, there exists a vertex $w_2\in N(w_1)\setminus \{x_4\}$ such that $|L(w_2)|\leq 2$.
We deduce from Observation \ref{PRO1} that $f(x_4)=f(y)=0$ for any $y\in L(w_1)\cup L(w_2)$, and $f(w_1)=f(w_2)=2$.

Then we consider a PRDF $f'$ of $T$ with $f'(w_2)=0$ and $f'(x)=1$ for any $x\in L(w_2)$.
It is obvious that $w(f')\leq w(f)$ and $f'$ is a $\gamma_R^p$-function of $T$.
But there exists a leaf vertex which is assigned with $1$ under $f'$, a contradiction with Observation \ref{PRO1}.

\end{proof}

\begin{clm1}
 $d(x_4)=2$.
\end{clm1}
\begin{proof}
Suppose to the contrary $d(x_4) \geq 3$, then by Claims \ref{Claim1} and \ref{Claim2}, it is sufficient to consider the following four cases.

\begin{figure}[H]
\begin{center}
\includegraphics[scale=0.7]{T4.eps}
\caption{Four cases if $d(x_4) \geq 3$ in the proof of  Lemma \ref{Lemma5}}
\label{FourCases}
\end{center}
\end{figure}

\textbf{Case A:} $x_4$ is adjacent to a pendent $P_3$ (see Fig. \ref{FourCases}a), contradicting with the condition of $T$.\\
\textbf{Case B:} $x_4$ is adjacent to a $P_4=w_3w_2w_1w_4$ with $x_4w_1 \in E(T)$ (see Fig. \ref{FourCases}b).\\
  Easily, we have $f(w_1)=f(w_3)=0$ and $f(w_2)=2$. Hence $f(w_4)=1$, a contradiction.

\textbf{Case C:} $x_4$ is adjacent to a $P_5=w_3w_2w_1w_4w_5$ with $x_4w_1 \in E(T)$ (see Fig. \ref{FourCases}c).\\
By Observation \ref{PRO1}, $f(w_3)=f(w_5)=0$ and $f(w_2)=f(w_4)=2$.
If $f(w_1)=0$, a contradiction (a vertex in $V_0$ is adjacent to exactly one vertex in $V_2$). If $f(w_1)=2$, then under $f$ we assign
   2,0,1,0,1 to $w_1,w_2,w_3,w_4$ and $w_5$, respectively,
and obtain a PRDF of $T$ with fewer weight $w(f)$, a contradiction.

\textbf{Case D:} $x_4$ is adjacent to a star with vertex set $\{w_1,w_2,w_3,w_4\}$ centered at $w_2$ with $x_4w_1 \in E(T)$ (see Fig. \ref{FourCases}d).\\
We can obtain  $f(w_1)=f(w_3)=f(w_4)=0$ and $f(w_2)=2$.

Let $T'=T-\{w_2,w_3,w_4\}$. Let $g$ be a $\gamma_R^p$-function on $T'$.
Since $T$ is stable,  We have $w(f)=\gamma_R^p(T-w_2)=w(g)+2$.

If $g(w_1)=2$, under $g$ we assign  0  to $w_2$,  1 to $w_3,w_4$ and obtain a PRDF of $T$ with same weight $w(f)$. But $w_3$ and $w_4$ are assigned with 1, contradicting Observation \ref{PRO1}.

If $g(w_1)=1$, under $g$ we assign  2  to $w_2$,  0 to $w_3,w_4$ and obtain a PRDF of $T$ with same weight $w(f)$. But $w_1$ is assigned with 1, contradicting Observation \ref{PRO1}.

If $g(w_1)=0$, then $g(x_4)=2$. Hence we have $g(x_1)=g(x_2)=g(y_1)=1$ and $g(x_3)=0$.
Thus $g(x_5)=0$. Otherwise, under $g$ we assign 2 to $x_2,w_2$, 0 to $x_1,y_1,x_3,w_1,w_3,w_4$, 1 to $x_4$ and obtain a PRDF of $T$ with fewer weight $w(f)$, a contradiction.

Note that $g(x_4)=2$ and $g(x_5)=0$. Under $g$ we assign the weight 0 to $x_2,w_2$, 0 to $x_1,y_1,x_3, w_1,w_3,w_4$, and 1 to $x_4,x_5$ and obtain a PRDF of $T$ with at most weight $w(f)$. But $x_4$ and $x_5$ are assigned with 1, contradicting Observation \ref{PRO1}.

This completes the proof.
\end{proof}

Let $T'=T-x_2$ and let $g$ be a $\gamma_R^p$-function on $T'$.
Since $T$ is stable, $w(g)=\gamma_R^p(T-\{x_1,x_2,y_1\})+2=w(f)$.\\
If $g(x_3)=2$, then under $g$ we assign 1 to $x_1,y_1$, 0 to $x_2$ and obtain a PRDF of $T$ with same weight $w(f)$. But $x_1$ and $y_1$ are assigned with 1, contradicting Observation \ref{PRO1}.\\
If $g(x_3)=1$, then under $g$ we assign 0 to $x_1,y_1$, 2 to $x_2$ and obtain a PRDF of $T$ with same weight $w(f)$. But $x_3$ is assigned with 1, contradicting Observation \ref{PRO1}.

Thus $g(x_3)=0$ and $g(x_4)=2$.
If $f(x_5)\geq 1$, then under $g$ we assign 0 to $x_1,y_1$, 2 to $x_2$, 1 to $x_3,x_4$ and obtain a PRDF of $T$ with same weight $w(f)$. But $x_5$ is assigned with 1, contradicting Observation \ref{PRO1}.
Therefore $f(x_5)=0$. Under $g$ we assign 0 to $x_1,y_1,x_3$, 2 to $x_2$, 1 to $x_4,x_5$ and obtain a PRDF of $T$ with same weight $w(f)$. But $x_4,x_5$ is assigned with 1, contradicting Observation \ref{PRO1}.
Then the proof is complete.
\end{proof}

\begin{thm1}\label{perfiff}
{\em Let $T$ be a tree  of order $n\ge 3$. Then $T$ is a perfect Roman domination stable tree if and only if $T\in \mathcal{T}$.}
\end{thm1}
\begin{proof}
According to Lemma \ref{first}, we need only to prove necessity. Let $T$ be a perfect Roman domination stable tree of order $n\ge 3$.
The proof is by induction on $n$. If $n=3$, then $T=P_3 \in \mathcal{T}$. Let $n\ge 4$ and let the
statement hold for all perfect Roman domination stable trees of order less than $n$.

 Clearly, ${\rm diam}(T)\ge 2$.
If ${\rm diam}(T)= 2$,then $T$ must be a star of order $n\ge 4$. Clearly $T$ is not a stable tree, a contradiction.
If ${\rm diam}(T)= 3$, then $T$ is a double star. Let $T=DS_{p,q}$ be a double star with respectively $p$ and $q$ leaves attached at two support vertex. It easily proved that a stable tree is not a stable tree, a contradiction.

Consequently, we have ${\rm diam}(T)\ge 4$.
Let $P=x_1x_2\cdots x_k$ be a longest path of $T$ and $f$ be a $\gamma_R^p$-function of $T$.
According to Lemma \ref{Lemma5}, we have $d(x_2)=2$.

Now let us prove $d(x_3)=2$. Suppose to the contrary that $d(x_3)\geq 3$.
First, we prove $L(x_3)=\emptyset$. Suppose that $L(x_3)\neq\emptyset$, let $w_1 \in L(x_3)$. By Observation \ref{PRO1}, we have $f(x_1)=0$, $f(x_2)=2$, $f(w_1)\not\in \{1,2\}$.
Then we have  $f(w_1)=0$, then $f(x_3)=2$.  Now under $f$ assign  1 to  $x_1$, 0 to $x_2$ and obtain a PRDF of $T$ with fewer weight $w(f)$, a contradiction.
Therefore we have $L(x_3)=\emptyset$.

Second, we prove that $x_3$ is not adjacent to a path $P_2=w_1w_2$ with $x_3w_1 \in E(T)$ and $w_2 \in L(w_1)$. Easily we have $f(x_1)=f(w_2)=0$ and $f(x_2)=f(w_1)=2$.
It follows from Observation \ref{PRO1} and definition of $f$  that $f(x_3)=2$.
Then  under $f$ assign 1 to $x_1,w_2$, 0 to $x_2,w_1$ and obtain a PRDF of $T$ with fewer weight $w(f)$, a contradiction.

Consequently, we have $d(x_3)=2$. Hence $x_4$ must be adjacent to pendent a path $P_3=s_1s_2s_3$ for which $s_3 \in N(x_4)$.
Let $T'=T-\{s_1,s_2,s_3\}$. By Lemma \ref{AddP3}, we have $\gamma_R^p(T)=\gamma_R^p(T')+2$.
By Lemmas \ref{AddP3},\ref{AddP2} and \ref{OneV}, we have $T$ is perfect Roman domination stable iff
$T'$ is perfect Roman domination stable.

Now we show that $x_4\in W(T')$.
Let $f'$ be a $\gamma_R^p$-function of $T'$ such that $f'(x_4)\neq 0$, and we have $w(f')=\gamma_R^p(T)-2$.
If $f'(x_4)=2$, then under $f'$ we assign  1 to $s_2,s_1$ and obtain a PRDF of $T$ with same weight $w(f)$. But $s_2,s_1$ is assigned with 1, contradicting Observation \ref{PRO1}.
If $f'(x_4)=1$, then under $f'$ we assign  0 to $s_3,s_1$, 2 to $s_2$ and obtain a PRDF of $T$ with same weight $w(f)$. But $x_4$ is assigned with 1, contradicting Observation \ref{PRO1}.

Therefore, $T$ is obtained by Operation $\mathcal{O}_1$ by applying $T_i=T'$.
\end{proof}

By the construction of $\mathcal{T}$, we have
\begin{cor1}
Let $T$ be an $n$-vertex perfect Roman domination stable tree, then $n \equiv 0$ (mod 3) and $\gamma_R^p(T)=\frac{2n}{3}$.
\end{cor1}

config: arxiv


TITLE: Find the coefficient of $x^{70}$ in the product $(x-1)(x^2-2)(x^3-3)…(x^{12}-12)$
QUESTION [2 upvotes]: Find the coefficient of $x^{70}$ in the product $(x-1)(x^2-2)(x^3-3)…(x^{12}-12)$.
The above equation is a polynomial of degree 78 .
But I am not able to find the coefficient of $x^{70}$

REPLY [5 votes]: So total polynomial degree is 78, and you want $x^{70}$. It means replacing some $x^i$ with their corresponding integer ($-i$), where $\sum i=8$. Noticing that $1+2+3+4=10>8$, there are very few cases, and we can list them.
When I say replace, I mean replacing the term in the product $x^1x^2\cdots x^{12}$.

When we replace 1 term:

We must replace $x^8$ by -8, and we get the term $\fbox{$-8x^{70}$}$ from this.

When we replace 2 terms:

We can replace $x^ix^{8-i}$ by i(8-i), where $i$ can range from $1$ to $3$. We get the term $(1\cdot7+2\cdot6+3\cdot5)x^{70}=\fbox{$34x^{70}$}$ from this.

When we replace 3 terms:

Scary? No! The only (unordered) triples that sum up to $8$ are $(1, 2, 5)$ and $(1, 3, 4)$! These contribute to the term $((-1)(-2)(-5)+(-1)(-3)(-4))x^{70}=\fbox{$-22x^{70}$}$.
To conclude, we get that $4$ is the coefficient of $x^{70}$.
Verification: https://tio.run/##JcwxCoAwEAXR6xjhg7sxJh7A3j4oBBW0MIikyO1XWKs31dypnMedyrUlkfm9colTfVLeY1NBpm3qymDVwqo9etXBqQMG1cOrAUEdMarUgbq/CPQ/iUFslkXkAw


TITLE: When is there a 1-1 correspondence between relative discs and those in a cover?
QUESTION [0 upvotes]: Let $X$ be a topological space with covering $p : \tilde{X} \to X$ and $A \subset X$. Consider the set of maps $M := \{u: (D^n, \partial D^n) \to (X,A)\}$ taking $n$-dimensional discs to $X$ with boundaries lying in $A$. Let $\tilde{A} \subset \tilde{X}$ be the "induced cover" of $A$ such that $p(\tilde{A}) = A$. What conditions ensure the set of maps $\tilde{M} :=\{\tilde{u}: (D^n, \partial D^n) \to (\tilde{X},\tilde{A})\}$ is in bijection with $M$?  (I am particularly interested in the case $n=2$). More generally, how do we relate the two sets?
For example: take $X = \mathbb{R}\times \mathbb{R} \times S^1 \times S^1$ and $A = \mathbb{R} \times \{0\}\times S^1 \times \{0\}$ so that  $\tilde{X} = \mathbb{R} \times \mathbb{R} \times \mathbb{R}  \times \mathbb{R} $ and $\tilde{A}= \mathbb{R} \times \{0\}\times \mathbb{R} \times \{0\}$. How does $\{u : D^2 \to X | \partial u \in  A\}$ relate to $\{\tilde{u} : D^2 \to \tilde{X} | \partial \tilde{u} \in  \tilde{A} \}$?

REPLY [0 votes]: Let $p: \tilde X \to X$ be a covering map. Let $f: D^n \to X$. Note $D^n \cong D^{n-1}\times I$ so $f$ can be thought of as homotopy between two maps $f_0,f_1:D^{n-1} \to X$. Moreover one can arrange the homeomorphism $h$ between $D^{n-1}\times I$ and $D^n$ so that $h(D^{n-1}\times 0)$ is contained in any given open set $U\subset D^n$ containing a point of $\partial D^n$. In particular, one can arrange $h$ so that there exists an open $V\subset X$ with $f (h(D^{n-1}\times 0))\subset V$ with $p|_{p^{-1}(V)}:p^{-1}(V)\to V$ being the trivial cover with the same number of sheets as the original cover of $X$.
Now, the lifts of $f_0=f|_{h(D^{n-1}\times 0)}:D^{n-1}\to X$ correspond exactly to the sheets of the cover. By the homotopy lifting property (proposition 1.30 Hatcher), there exists a unique lift of the homotopy $f$ between $f_0$ and $f_1$ for each lift of $f_0$. We have arranged so the lifts of $f_0$ correspond exactly to the sheets of the cover. So the push-forward map $p_*:C(D^n,\tilde X) \to C(D^n, X)$ is $n$ to $1$, where $n$ is the number of sheets of the cover $p$. It is trivial to see that $p_*$ restricts to an $n$ to $1$ map from maps $(D^{n-1},\partial D^n) \to  (\tilde X, p^{-1}(A))$ to maps $(D^{n-1},\partial D^n) \to (X,A)$ (this is basically archipelago's second comment). In particular the only times when $p_*$ is a bijection are when the cover is $1$-sheeted or $M = \emptyset$.


TITLE: Finite intersection of nonempty sets is not empty (Poincaré-Miranda theorem)
QUESTION [2 upvotes]: I am trying to understand a proof of the Poincaré-Miranda theorem given in this article.
So, the theorem states that if we have a continuous map $f = (f_1, \ldots, f_n) : I^n \rightarrow {\rm I\!R}^n$ where $I^n$ is a $n$-dimensional cube $[-1, 1]^n$ that satisfies the following condition:
$$
f_i(x_1, \ldots, x_{i-1}, -1, x_{i+1}, \ldots, n) \leq 0 \\
f_i(x_1, \ldots, x_{i-1}, 1, x_{i+1}, \ldots, n) \geq 0
$$
for each $i = 1, \ldots, n$, there exists a point $c \in I^n$ such that $f(c) = 0$.
To prove the theorem authors defined sets
$$
H_i^- := f_i^{-1}(- \infty , 0] \\
H_i^+ := f_i^{-1}[0, \infty)
$$
Then they say that "using a compactness argument we infer that the intersection
$$
H := \bigcap \{ H_i^- \cap H_i^+ : i=1, \ldots, n\}
$$
is not empty."
I see that each $H_i^-$ and $H_i^+$ is not empty as it follows from the propositions, but why are $H_i^- \cap H_i^+$ nonempty and why is $H$ not empty?
Additionally, if $H$ is indeed not empty, it is obvious that for every $x \in H$ it follows $f(x) = 0$, so we found the zero. Why does not the proof just end here?
Thanks to anyone who could help me understand.

REPLY [2 votes]: Their proof is laid out backwards and is indeed confusing. Their general proof is that they can show their are an odd (and so at least 1) simplices of each given diameter that intersect both $H_i^{-}$ and $H_i^{+}$ for all $i$. They then effectively wave their hands and say that's sufficient to prove that $H$ is nonempty by an unstated compactness argument before going back and proving the existence of such simplices.
Here's one way of showing the compactness part. Let $S_k$ be a simplex with diameter $1/k$ which intersect $H_i^-$ and $H_i^+$ for all $i$. Then for all $k$ there are exist $a_k$  in $S_k \cap H_i^-$ and $b_k$ in $S_k \cap H_i^+$ for all i. Then, by compactness, there is some convergent subsequence of the $a_k$. Let that subsequence be $a_{k_i}$ and let its limit be $a$. Note that $a\in H_i^-$ for all $i$ since they are closed. Since the diameter of $S_{k_i}$ is $1/k_i$, the distance $d(a_{k_i},b_{k_i})<1/k_i$ goes to 0, so $a=\lim a_{k_i}=\lim b_{k_i}$. Since $b_{k_i}$ are in the closed set $H_i^+$ for all $i$, the limit $a$ is in $H_i^+$ for all $i$, so $a\in H$.


TITLE: Characterisation of Galois Group with the action of $\sigma \in S_n$ on the roots
QUESTION [2 upvotes]: Let $f \in K[X]$ be irreducible and separable with roots $x_1,...,x_n$ in a splitting field $L$ of $f$ over $K$. We identify $\text{Gal}(L|K)$ with $\text{Gal}(L|K)\cong G\subset S_n$.
How can I see the equivalence of the following two statements? (which means a characterisation of the galois group with the action of a $\sigma \in G$ on the roots $x_1,...,x_n$)
$(1)$ $\sigma \in G$.
$(2)$ If $P \in K[X_1,...,X_n]$ with $P(x_1,...,x_n)=0$, then for $P(X_{\sigma(1)},...,X_{\sigma(n)})$ it follows that $P(X_{\sigma(1)},...,X_{\sigma(n)})(x_1,...,x_n)=0$.

REPLY [2 votes]: I prefer using three different notations, say $\sigma$ for the permutation
of indices ,$\tau$ for the permutation of roots and $t$
for the field homomorphism extending $\tau$ (when it exists) : thus $\tau(x_i)=x_{\sigma(i)}$.
$(1) \Rightarrow (2) $ If $t$ exists, and $P(x_1,x_2,\ldots,x_n)=0$, we have
$$
P(X_{\sigma(1)},...,X_{\sigma(n)})(x_1,...,x_n)=
P(x_{\sigma(1)},...,x_{\sigma(n)})=
P(\tau(x_1),\tau(x_2),\ldots,\tau(x_n))=
t(P(x_1,x_2,\ldots,x_n))=t(0)=0.
$$
$(2) \Rightarrow (1) $ The permutation $\tau$ is defined on $\lbrace x_1,x_2,
\ldots ,x_n\rbrace$ and we sould like to extend it to the the whole of
$L=K[x_1,\ldots,x_n]$. The obvious definition which comes to our mind is
$$
t(A(x_1,x_2,\ldots,x_n))=A(x_{\sigma(1)},\ldots,x_{\sigma(n)}) \tag{1}
$$
for any $A\in K[X_1,\ldots,X_n]$. The problem with (1) is that it might be an
incorrect definition, with two different values set for the same argument.
However, if $A(x_1,x_2,\ldots,x_n)=B(x_1,x_2,\ldots,x_n)$ for two polynomials
$A,B$, then the polynomial $C=A-B$ satisfies $C(x_1,x_2,\ldots,x_n)=0$, so
$C(x_{\sigma(1)},...,x_{\sigma(n)})=0$ by (2), and (1) will therefore yield
the same value in both cases.
So $t$ is correctly defined, and it follows immediately from its definition
that it is a homomorphism.
Alternatively, you can define $t$ as a "quotient map".


TITLE: Decide if set of value can be image of continuous function
QUESTION [1 upvotes]: $f:[0,10]\cup[12,20)\to\mathbb{R}$ is continuous.  Decide if following sets of values can be image of $f$
a. $[0,1]$
b. $[0,1]\cup (3,5)\cup \{7\}$
c. $(0,1)$
a. is easy, for example $f(x) = \frac{1}{10}x,\ \ x\in [0,10]$ and $f(x)=\frac12\ \ \ \ x\in [12, 20)$
b. It is not possible.  Image of $f$ can have at most two intervals on its image.
c. No idea about this question. Can you help me, plesae ?

REPLY [1 votes]: For c. we can define $g$ to be any constant you like on $[0,10],$ and $g(x) = (1/(20-x))(\sin(1/(20-x))$ for $x\in [12,20).$ Then $g$ is continuous on $D = [0,10] \cup [12,20).$ Note that on $[12,20),$ $g$ takes on arbitrarily large positive and negative values. By the intermediate value theorem it follows that $g(D) = \mathbb R.$ The function
$$f(x) = \frac{1}{\pi}(\arctan g(x) + \pi/2)$$
then maps $D$ continuously onto $(0,1).$


TITLE: Standard form and basic solutions
QUESTION [0 upvotes]: Is it possible that for a general linear program none of the optimal solutions have to be basic? Meanwhile, for a program in standard form an optimal solution has be basic. If so, why ?

REPLY [0 votes]: For $$\min_{x \geq 0} \{ x : x \leq 1\}$$ the only basic feasible solution is $x=1$, which is not optimal.


TITLE: Are there integers $a, b$ such that $\pi^a = e^b$?
QUESTION [8 upvotes]: Is $\log \pi $ a rational number? That is, are there non-zero integers $a, b$ s.th. $\pi^a = e^b$ ?

REPLY [2 votes]: This is still an open problem. The algebraic independence of $\pi$ and $e$ is unknown, i.e., it is unknown whether there exist algebraic numbers $n_0, n_1, \ldots, n_k$ such that
$$
n_{0} \pi^{0} + \ldots + n_{k-1} \pi^{k-1} + n_k \pi^{k} = e.
$$


TITLE: One ball is drawn from 2 boxes of white/black balls. What is the probability of getting one ball of each color?
QUESTION [4 upvotes]: In box A there are $7$ white balls and $5$ black balls. In box B there are $2$ white balls and $4$ black balls.
  One ball is drawn from each box. What is the probability of getting one ball of each color?

In order to find the possible cases I multiplied the 12 balls in the box A by the $6$ balls in box B. There are $72$ possible cases.
Then I set $2$ different events:
A-"draw a ball from box A".
W-"draw a white ball".
Then I thought, there are $2$ situations where the drawn balls can have different colors.I can remove a black ball from box A and a white ball from box B or the inverse.
$P(\bar{W}|A) \cdot P(W|\bar{A})=\frac{5}{12}\cdot \frac{2}{6}=\frac{10}{72}$
$P(W|A) \cdot P(\bar{W}|\bar{A})=\frac{7}{12} \cdot \frac{4}{6}=\frac{28}{72}$
Then I added the two probabilities:$\frac{38}{72}$
But the result don't mix with the book solutions. Who is wrong? Thanks.
The book solution is $\frac{19}{72}$

REPLY [3 votes]: Say you draw from box A first. There are two opportunities: WB and BW, as The Chaz said. BW's chance is (5/12)*(2/6), or 10/72. WB's chance is (7/12)*(4/6), or 28/72. This is exactly what you got, and when you add then you do get 38/72, or 19/36, which is your answer. Brian is likely right, after a few people trying this and getting the same answer, the only likely possibility is an error in your book.


TITLE: Sequence of exponential random variables
QUESTION [2 upvotes]: Let $\mathcal{E}=\{E_n : n\geq 1\}$ be a sequence of independent exponential random variables with mean 1. How can I see the set $E=\{E_n(\omega) : n\geq 1\}$ dense in $[0,\infty)$? I imagine one can make an argument about the cdf being discontinuous if there's an open set that doesn't contain any $E_n(\omega)$, but I don't know if this is even the right direction.

REPLY [1 votes]: Let $a <b$. Then $$P(E_n \notin (a,b) \,  \forall n \in \{1,2,..,N\})=\prod_{k=1}^{N} P(E_n \notin (a,b))$$ $$ =\prod_{k=1}^{N} (e^{-a}-e^{-b})=(e^{-a}-e^{-b})^{N} \to 0$$ as $N \to \infty$. Hence, $P(E_n \notin (a,b) \,  \forall n )=0$. Taking union over all rational values of $a$ and $b$ we see that there is one null set $E$ such that $\omega \notin E$ implies $(E_n(\omega))$ intersects every open interval in $\mathbb R$.


text:
\begin{document}

\begin{abstract}
In this paper, we show that if Rabinowitz Floer homology has infinite dimension, there exist infinitely many critical points of a Rabinowitz action functional even though it could be non-Morse. This result is proved by examining filtered Rabinowitz Floer homology.
\end{abstract}

\maketitle

\section{Introduction}
Many recent studies have focused on the coisotropic intersection problem. In particular, there has been growing interest in the leafwise intersection property described below.\\[-1ex]

\noindent\textbf{Leafwise intersection.}
Let $(M,\om)$ be a $2n$ dimensional symplectic manifold and $\Sigma$ be a coisotropic submanifold of codimension $k$. Then the symplectic structure $\om$ determines a symplectic orthogonal bundle $T\Sigma^\om$, which is a subbundle of $T\Sigma$ by the definition of coisotropic:
\beqn
T\Sigma^\om:=\{(x,\xi)\in T\Sigma\,|\,\om_x(\xi,\zeta)=0 \textrm{ for all } \zeta\in T_x\Sigma\}
\eeq
Since $\om$ is closed, $T\Sigma^\om$ is integrable, thus $\Sigma$ is foliated by the leaves; we denote by $L_x$ the leaf through $x\in\Sigma$. We call $x\in\Sigma$ a {\em leafwise intersection} of $\phi\in\Ham_c(M,\om)$ if $x\in L_x\cap\phi(L_x)$. $\Ham_c(M,\om)$ is defined below in Conventions and Notations.

In this article, we focus on the case that $\Sigma$ is a restricted contact type hypersurface with contact form $\lambda$, i.e. $d\lambda=\om$ and $\lambda\wedge\om^{n-1}|_\Sigma\neq0$, which bounds a compact region in $M$. The argument developed in this article would continue to hold in general contact case with minor modification (see \cite{Ka1} for the generalized Rabinowitz Floer theory). In that case, $T\Sigma^\om$ is nothing but the characteristic line bundle spanned by the Reeb vector field $R$. A leafwise intersection $x\in\Sigma$ is called {\em a periodic leafwise intersection} if the leaf $L_x$ through $x$ is a closed Reeb orbit.\\[-1ex]

The problem of finding leafwise intersection points was initiated by Moser \cite{Mo} and pursued further in \cite{Ba,Dr,EH}. Recently there are three different techniques to this problem; \cite{Zi} approached the leafwise intersection problem with Lagrangian Floer homology and \cite{Gi,Gu} took advantage of Symplectic Floer homology. Another effective way to investigate leafwise intersections is Rabinowitz Floer homology developed by Cieliebak-Frauenfelder \cite{CF}. Albers-Frauenfelder \cite{AF1} observed that critical points of a perturbed Rabinowitz action functional give rise to leafwise intersections; \cite{AF1,AF2,AF3,AF4,AMo,AMc,Ka,Ka1,Ka2,Me} obtained many results on the leafwise intersection problem using Rabinowitz Floer homology. Rabinowitz Floer homology is the Floer homology of the Rabinowitz action functional $\AA^H_F$. Thus, the dimension of Rabinowitz Floer homology gives a lower bound on the number of leafwise intersections as in the Morse inequalities in case the Rabinowitz action functional is Morse. Interestingly enough, Rabinowitz Floer homology has infinite dimension in some cases \cite{CFO,Me,AMc,Ka2}. In such cases, we know that for $F\in\Ham_c(M,\om)$ which makes $\AA^H_F$ Morse, $\AA^H_F$ has infinitely many critical points. It turned out that the Rabinowitz action functional is Morse for a generic $F\in\Ham_c(M,\om)$ \cite{AF1}; therefore a generic perturbation has either infinitely many leafwise intersections or a periodic leafwise intersection (see Proposition \ref{prop:critical point answers question}); furthermore, \cite{AF2} showed that generically there is no periodic leafwise intersection points.\\[-1ex]

\noindent\textbf{Main Theorem.} {\em If Rabinowitz Floer homology has infinite dimension, then for $\phi_F\in\Ham_c(M,\om)$ there exists infinitely many leafwise intersections or a periodic leafwise intersection.}\\[-1ex]

It is noteworthy that Main Theorem does not assume any kind of non-degeneracy for $\phi_F$; that is, the perturbed Rabinowitz action functional $\AA^H_F$ is not necessarily Morse. When it is Morse, the theorem follows immediately from the Morse inequalities.
\begin{Rmk}
An analogous result continues to hold without doubt in any other Morse or Floer homology whenever the action functional is a Morse (resp. Morse-Bott) and it has only finitely many critical points (resp. finitely many compact critical components) in a compact action interval.
\end{Rmk}
\begin{Rmk}
The Main theorem subsumes a result in \cite{AF3} which showed the above result for a (unit) cotangent bundle by means of the spectral invariants in Rabinowitz Floer homology.
\end{Rmk}

\noindent\textbf{Convention and Notations.}
\begin{itemize}
\item The {\em Reeb vector field} $R$ is characterized by $\lambda(R)=1$ and $i_R d\lambda=0$.
\item The {\em Hamiltonian vector field} $X_F$ associated to a Hamiltonian function $F\in C^\infty(S^1\x M)$ is defined implicitly by $i_{X_F}\om=dF$.
\item $\phi_F$ is the time one map of the flow of $X_F$ and called a {\em Hamiltonian diffeomorphism}.
\item We denote by $\Ham_c(M,\om)$ the group of Hamiltonian diffeomorphisms generated by compactly supported Hamiltonian functions.
\end{itemize}

\subsection{Filtered Rabinowitz Floer homology}
Since $\Sigma$ is a contact hypersurface, there exists a Liouville vector field $Y$ such that $L_Y\om=\om$ and $Y\pitchfork\Sigma$; we denote by $\phi_Y^t$ the flow of $Y$ and fix $\delta>0$ such that $\phi_Y^t|_\Sigma$ is defined for $|t|<\delta$. Since $\Sigma$ bounds a compact region in $M$, we are able to define a Hamiltonian function $G\in C^\infty(M)$ so that
\begin{enumerate}
\item $G^{-1}(0)=\Sigma$ is a regular level set;
\item $G(\phi_Y^t(x))=t$ for all $x\in\Sigma$ and $|t|<\delta$;
\item $dG$ has compact support.
\end{enumerate}

\begin{Def}
Given time-dependent Hamiltonian functions $H\in C^\infty(S^1\x M)$ and $F\in C_c^\infty(S^1\x M)$, a pair $(H,F)$ is called a {\em Moser pair} if it satisfies
\begin{enumerate}
\item $H$ is a weakly time-dependent Hamiltonian function. That is, $H$ is of the form $H(t,x)=\chi(t)G(x)$ for $G\in C^\infty(M)$ defined above and $\chi:S^1\to S^1$ with $\int_0^1\chi dt=1$ and $\Supp\chi\subset(\frac{1}{2},1)$;
\item their time supports are disjoint, i.e.
\beqn
H(t,\cdot)=0 \quad\textrm{for}\,\,\, \forall t\in\big[0,\frac{1}{2}\big] \quad\textrm{and}\quad F(t,\cdot)=0 \quad\textrm{for}\,\,\, \forall t\in\big[\frac{1}{2},1\big].
\eeq
\end{enumerate}
\end{Def}
\begin{Rmk}
We can easily check that every element in $\Ham_c(M,\om)$ is generated by a compactly supported Hamiltonian function with time support on $\big[0,\frac{1}{2}\big]$.
\end{Rmk}

We define a perturbed Rabinowitz Floer functional $\AA^H_F:C^\infty(S^1,M)\x\R\pf\R$ for a Moser pair $(H,F)$ as follows:
\beqn
\AA^H_F(v,\eta)=-\int_0^1v^*\lambda-\eta\int_0^1H(t,v)dt-\int_0^1F(t,v)dt.
\eeq
Albers-Frauenfelder observed that a critical point of $\AA^H_F$ gives rise to a leafwise intersection.
\begin{Prop}\label{prop:critical point answers question}
\cite{AF1} Let $(v,\eta)\in\Crit\AA_{F}^{H}$. Then $x:=v(0)\in\Sigma$ and satisfies $\phi_F(x)\in L_x$. Thus, $x$ is a leafwise intersection point. Moreover, the map
\beqn
\Crit\AA^{H}_F\pf\big\{\textrm{leafwise intersections}\big\}
\eeq
is injective unless there exists a periodic leafwise intersection.
\end{Prop}

\begin{Def}
We define the {\em action spectrum} for a Moser pair $(H,F)$ by
\beqn
\mathrm{Spec} (H,F):=\big\{\AA^H_F(v,\eta)\in\R\,\big|\,(v,\eta)\in\Crit(\AA^H_F)\big\}.
\eeq
\end{Def}

Now, we roughly describe a filtered Rabinowitz Fleor homology for a Morse action functional $\AA^H_F$ and $a<b<c\notin\mathrm{Spec}(H,F)$ (see \cite{Ka1} for a rigorous definition); it is proved in \cite{AF1} that $\AA^H_F$ is Morse for a generic $F\in C_c^\infty(S^1\x M)$ with time support in $[0,1/2]$. Let us set the $\Z/2$-module by
\beqn
\CF^{(a,b)}(\AA^H_F):=\Crit^{(a,b)}(\AA^H_F)\otimes\Z/2
\eeq
where
\beqn
\Crit^{(a,b)}(\AA^H_F):=\big\{(v,\eta)\in\Crit\AA^H_F\,\big|\,\AA^H_F(v,\eta)\in(a,b)\big\}.
\eeq
Then it becomes a complex with the boundary operator $\p$ defined by counting solutions of a nonlinear elliptic PDE. Then the filtered Rabinowitz Floer homology is defined by
\beqn
\HF^{(a,b)}(\AA^H_F)=\H\big(\CF^{(a,b)}(\AA^H_F),\p\big).
\eeq
In the case $a=-\infty$ and $b=\infty$, we get a full Rabinowitz Floer homology $\HF(\AA^H_F)$ and it is invariant under the choice of a perturbation $F\in C_c^\infty(S^1\x M)$ by the standard argument in the Floer theory. Thus we denote by
\beqn
\RFH(\Sigma,M):=\HF(\AA^H)\cong\HF(\AA^H_F)
\eeq
where $\AA^H=\AA^H_0$. The Rabinowitz action functional $\AA^H$ is Morse-Bott for a generic contact hypersurface; in that case Rabinowitz Floer homology $\RFH(\Sigma,M)$ is well-defined. Otherwise, we define $\RFH(\Sigma,M):=\HF(\AA^H_f)$ for $\AA^H_f$ being Morse. Moreover we have canonical homomorphisms
\beqn
i_a^{b,c}:\CF^{(a,b)}(\AA^H_F)\pf\CF^{(a,c)}(\AA^H_F),\quad \pi_{a,b}^c:\CF^{(a,c)}(\AA^H_F)\pf\CF^{(b,c)}(\AA^H_F)
\eeq
inclusions and projections respectively. They fit into a short exact sequence
\beqn
0\pf\CF^{(a,b)}(\AA^H_F)\stackrel{i_a^{b,c}}{\pf}\CF^{(a,c)}(\AA^H_F)\stackrel{\pi^c_{a,b}}{\pf}\CF^{(b,c)}(\AA^H_F)\pf0,
\eeq
and we obtain a long exact sequence
\beqn
\cdots\stackrel{\pi_{a,b*}^c}{\pf}\HF^{(b,c)}(\AA^H_F)\stackrel{\delta_*}{\pf}\HF^{(a,b)}(\AA^H_F)\stackrel{i_{a*}^{b,c}}{\pf}\HF^{(a,c)}(\AA^H_F)\stackrel{\pi_{a,b*}^c}
{\pf}\HF^{(b,c)}(\AA^H_F)\stackrel{\delta_*}{\pf}\cdots.
\eeq

Next, we briefly recall the definition of Hofer norm.
\begin{Def}\label{def:Ham}
Let $F\in C^\infty_c(S^1\x M,\R)$ be a compactly supported time-dependent Hamiltonian function. We set
$$
||F||_+:=\int_0^1\max_{x\in M} F(t,x) dt\qquad||F||_-:=-\int_0^1\min_{x\in M} F(t,x) dt=||-F||_+
$$
and
$$
||F||:=||F||_++||F||_-.\;
$$
For $\phi\in \Ham_c(M,\om)$ the Hofer norm is
$$
||\phi||:=\inf\{||F||\mid \phi=\phi_F,\,\, F\in C^\infty_c(S^1\x M,\mathbb{R})\}.\;
$$

\end{Def}

The following proposition is well-known in the standard Floer theory and easily shown by energy estimates (see e.g. \cite{Ka1}).
\begin{Prop}\label{prop:continuation}
There exist continuation homomorphisms for the filtered case:
\bean
\Phi&:\HF^{(a,b)}(\AA^H_f)\pf\HF^{(a+||F-f||_-,b+||F-f||_-)}(\AA^H_F),\\
\Psi&:\HF^{(a,b)}(\AA^H_F)\pf\HF^{(a+||F-f||_+,b+||F-f||_+)}(\AA^H_f).
\eea
\end{Prop}

\section{Proof of Main Theorem}
\begin{proof}
Assume on the contrary that $\RFH(\Sigma,M)=\HF(\AA^H_f)$ is infinite dimensional and $\AA^H_F$ for $F\in C^\infty_c(S^1\x M)$ has finitely many critical points. Then there exist $A,B\in\R$ such that
$$
\mathrm{Spec}(H,F)\subset[A,B].
$$
Since $F$ and $f$ are compactly supported, $||F-f||_-$ has finite value, so we can pick $b\in\R$ so that $b>B-||F-f||_-$. Then $\CF^{(b+||F-f||_-,\infty)}(\AA^H_F)$ is empty and thus its homology $\HF^{(b+||F-f||_-,\infty)}(\AA^H_F)$ is well-defined although $\AA^H_F$ could have degenerate critical points with action outside $(b+||F-f||_-,\infty)$. Due to Proposition \ref{prop:continuation} together with the ``homotopy of homotopies'' argument, we have the following commutative diagram \eqref{eq:diagram}.
\bea\label{eq:diagram}
\xymatrix{ \HF^{(b,\infty)}(\AA^H_f) \ar[dd]^{\big(\pi^\infty_{b,b+||F-f||}\big)*} \ar[dr]^\Phi & \hspace{8cm}  \\
 &  \HF^{(b+||F-f||_-,\infty)}(\AA^H_F) \ar[dl]^\Psi \\
\HF^{(b+||F-f||,\infty)}(\AA^H_f) & \hspace{8cm} }
\eea
Since $\HF^{(b+||F||_-,\infty)}(\AA^H_F)$ vanishes, accordingly $\big(\pi^\infty_{b,b+||F-f||}\big)*$ also vanishes and we have the following short exact sequence:
\beqn
0\stackrel{\pi_*}{\pf}\HF^{(b+||F-f||,\infty)}(\AA^H_f)\stackrel{\delta_*}{\pf}\HF^{(b,b+||F-f||)}(\AA^H_f)\stackrel{i_*}
{\pf}\HF^{(b,\infty)}(\AA^H_f)\stackrel{\pi_*}{\pf}0
\eeq
where $\pi_*=\big(\pi^\infty_{b,b+||F-f||}\big)_*$, $i_*=\big(i_{b}^{b+||F-f||,\infty}\big)_*$, and $\delta_*$ is a connecting homomorphism. Since $||F||<\infty$, the dimension of $\HF^{(b,b+||F-f||)}(\AA^H_f)$ is finite. Since $i_*$ is surjective, $\HF^{(b,\infty)}(\AA^H_f)$ is finite dimensional.

In the same way, we choose $a\in\R$ so that $a<A-||F-f||_-$ and then the following diagram commutes.
\bean
\xymatrix{ \HF^{(-\infty,a)}(\AA^H_f) \ar[dd]^{\big(i_{-\infty}^{a,a+||F-f||}\big)_*} \ar[dr]^\Phi & \hspace{8cm}  \\
 &  \HF^{(-\infty,a+||F-f||-)}(\AA^H_F) \ar[dl]^\Psi \\
\HF^{(-\infty,a+||F-f||)}(\AA^H_f) & \hspace{8cm} }
\eea
Analogously, $\big(i_{-\infty}^{a,a+||F-f||}\big)_*$ vanishes and thus we have
\beqn
0\stackrel{i_*'}{\pf}\HF^{(-\infty,a+||F-f||)}(\AA^H_f)\stackrel{\pi_*'}{\pf}\HF^{(a,a+||F-f||)}(\AA^H_f)\stackrel{\delta_*}{\pf}\HF^{(-\infty,a)}(\AA^H_f)
\stackrel{i_*'}{\pf}0.
\eeq
where $i_*'=\big(i_{-\infty}^{a,a+||F-f||}\big)_*$, $\pi_*'=\big(\pi^{a+||F-f||}_{-\infty,a}\big)_*$, and $\delta_*$ is a connecting homomorphism.
Therefore $\HF^{(-\infty,a)}(\AA^H_f)$ has finite dimension. Next, we compare the dimensions of homologies and this is contradictory to the our assumption that $\AA^H_F$ has finitely many critical points. Consider a part of the long exact sequence:
\beqn
\cdots\stackrel{\delta_*}{\pf}\HF^{(b,\infty)}(\AA^H_f)\stackrel{i_{b,a*}^\infty}{\pf}\HF^{(a,\infty)}(\AA^H_f)\stackrel{\pi_{a*}^{\infty,b}}{\pf}\HF^{(a,b)}
(\AA^H_f)\stackrel{\delta_*}{\pf}\cdots.
\eeq
Then we know that
\bea\label{eq:dim cal}
\dim\HF^{(a,\infty)}(\AA^H_f)&=\dim\im\,\pi_{a*}^{\infty,b}+\dim\ker\pi_{a*}^{\infty,b}\\
&=\dim\im\,\pi_{a*}^{\infty,b}+\dim\im \,i_{b,a*}^\infty\\
&\leq\dim\HF^{(a,b)}(\AA^H_f)+\dim\HF^{(b,\infty)}(\AA^H_f)<\infty.
\eea
In addition, we have another long exact sequence of the form
\beq\label{eq:les}
\cdots\stackrel{\delta_*}{\pf}\HF^{(-\infty,a)}(\AA^H_f)\stackrel{i_{-\infty*}^{a,\infty}}{\pf}\RFH(\Sigma,M)\stackrel{\pi_{-\infty,a*}^\infty}{\pf}
\HF^{(a,\infty)}(\AA^H_f)\stackrel{\delta_*}{\pf}\cdots.
\eeq
Using the computation \eqref{eq:dim cal} together with \eqref{eq:les}, we derive the following contradiction and it finishes the proof of theorem.
\bean
\infty=\dim\RFH(\Sigma,M)&=\dim\im\,\pi_{-\infty,a*}^\infty+\dim\ker\pi_{-\infty,a*}^\infty\\
&=\dim\im\,\pi_{-\infty,a*}^\infty+\dim\im \,i_{-\infty*}^{a,\infty}\\
&\leq\dim\HF^{(a,\infty)}(\AA^H_f)+\dim\HF^{(-\infty,a)}(\AA^H_f)<\infty.
\eea
\end{proof}

\subsubsection*{Acknowledgments}
I owe a debt of gratitude to Urs Frauenfelder for valuable discussions. This work is partially supported by the Basic Research grant Nr. 2010-0007669 from the Korea government.


text:
\section{Conclusion}
\label{conclusion}

There are a great many other graph polynomials equally interesting to
those surveyed here, including for example the F-polynomials
\index{F-polynomial}\index{polynomial!F} of Farrell, the Hosaya or
Wiener polynomial, the clique/ independence and adjoint polynomials,
etc.  In particular, Farrell~\cite{Far79b} has a circuit cover
polynomial (different from the circuit partition polynomial of
Subsection~\ref{sec:cpp}) with noteworthy interrelations with the
characteristic polynomial. Also, Chung and Graham developed a
`Tutte-like' polynomial for directed graphs in \cite{CG95}.  The
resultant cover polynomial \index{cover polynomial}
\index{polynomial!cover}is extended to a symmetric
function generalization, like those in section \ref{subsec:symmetric},
by Chow \cite{Cho96}. Similarly, Courcelle~\cite{Cou08} and
Traldi~\cite{Tra} have also very recently developed multivariable
extensions of the interlace polynomial.  Some surveys of graph
polynomials with
complementary coverage to this one include: Pathasarthy ~\cite{Par89},
Jaeger ~\cite{Jae90}, Farrell ~\cite{Far93}, Fiol ~\cite{Fio97},
Godsil ~\cite{God84}, Aigner ~\cite{Aig01}, Noy ~\cite{Noy03} and Levit
and Mandrescu ~\cite{LM05}.