text:
\begin{document}
\maketitle
\begin{abstract}
We construct weak (i.e. non-graded) modules over the vertex operator algebra $M(1)^+$, which is the fixed-point subalgebra of the higher rank free bosonic (Heisenberg) vertex operator algebra with respect to the $-1$ automorphism. These weak modules are constructed from Whittaker modules for the higher rank Heisenberg algebra. We prove that the modules are simple as weak modules over $M(1)^+$ and calculate their Whittaker type when regarded as modules for the Virasoro Lie algebra. Lastly, we show that any Whittaker module for the Virasoro Lie algebra occurs in this way. These results are a higher rank generalization of some results by Tanabe \cite{T1}.
\end{abstract}

\section{Introduction}

Whittaker modules were first studied by Kostant \cite{K} in the context of finite-dimensional complex semisimple Lie algebras and play an important role in representation theory and geometry. They have since been defined and investigated  in many different settings such as the Virasoro Lie algebra \cite{OW,GLZ,MZ}, affine Lie algebras \cite{ALZ}, a general framework for Lie algebras \cite{BM}, and generalized Weyl algebras \cite{BO,FH}.

Vertex operator algebras (VOAs) \cite{FLM,LL} form a mathematical framework for studying conformal field theory. One of the fundamental methods for obtaining new VOAs from known ones is to take the fixed-point sub-VOA with respect to a finite order automorphism. Namely,  let $V$ be a rational VOA and $G$ be a finite automorphism group of $V$. Then $V^G$  is the resulting VOA is called an orbifold VOA because it corresponds to the conformal field theory on an orbifold (quotient) space.

Let $V$ be a rational VOA and $G$  be a finite automorphism group of $V$. The orbifold conjecture says that $V^G$ is rational and every irreducible $V^G$ module occurs in an irreducible $g$-twisted $V$-module for some $g\in G$. Miyamoto and Carnahan prove that if $G$ is solvable, then $V^G$ is rational \cite{CM}.  Later it is proved that every irreducible $V^G$-module occurs as an irreducible $g$-twisted $V$-module for some $g\in G$ \cite{DRX}.

One of the most important VOA is the Heisenberg VOA, also known as the free bosonic VOA and is denoted by $M(1)$. It is constructed from a complex vector space $\mathfrak{h}$  with a non-degenerate bilinear form. The dimension of the vector space is called the rank of the VOA. The VOA $M(1)$ has a natural automorphism of order two, induced by multiplication by $-1$ on the vector space $\mathfrak{h}$. The corresponding orbifold VOA is denoted by $M(1)^+$. Irreducible modules for $M(1)^+$ were classified by Dong and Nagatomo \cite{DN}.

 The orbifold conjecture still makes sense for non-$\mathbb{N}$-graded weak modules. The conjecture was confirmed by Tanabe for a class of simple non-$\mathbb{N}$-graded weak $M(1)^+$-modules defined by using Whittaker vectors for the Virasoro algebra \cite{T1}. Namely, let $M(1)$ be the VOA with the Virasoro element $\omega$ associated to the Hesenberg algebra of rank 1. Tanabe proved that any simple weak $M(1)^+$-module with at least one Whittaker vector is isomorphic to some simple weak $M(1)$-module or to some $ \theta$-twisted simple weak $M(1)$-module.

The structure of such Whittaker modules for $M(1)^+$ becomes more complicated when we consider higher rank case. In this paper, we construct a class of simple weak $M(1)^+$-modules when $M(1)$ is the VOA with the Virasoro element $\omega$ associated to the Heisenberg algebra of rank $\ell$ where $\ell\ge 1$. These modules are generalizations of Tanabe's modules and they are Whittaker modules with respect to the Virasoro algebra. We also determine explicitly the subfamily of all constructed simple weak $M(1)^+$-modules that give rise to the same Virasoro algebra module. In the rank one case there are only two, while in our case we get an affine variety which we explicitly describe.

\section{The vertex operator algebra $M(1)$}

\subsection{Definition of vertex operator algebra}

We recall the definition of a vertex operator algebra \cite{LL}.

\begin{Definition}
A \emph{vertex operator algebra} $(V,Y,\boldsymbol{1},\omega)$ consists of
\begin{enumerate}[{\rm (i)}]
\item a $\Z$-graded vector space $V=\bigoplus_{n\in\Z} V_{(n)}$ called the \emph{state space},
\item a linear map
\[
\begin{aligned}
Y(\cdot,z):V &\to \End(V)[[z,z^{-1}]] \\
 v &\mapsto Y(v,z)=\sum_{n\in\Z} v_n z^{-n-1}
\end{aligned}
\]
called the \emph{state-field correspondence};
\item a distinguished vector $\boldsymbol{1}\in V_{(0)}$ called the \emph{vacuum vector};
\item a distinguished vector $\omega\in V_{(2)}$ called the \emph{conformal vector};
\end{enumerate}
such that the following properties hold:
\begin{enumerate}
\item \emph{grading restrictions}: $\dim V_{(n)}<\infty$ for $n\in\Z$ and $V_{(n)}=0$ for $n\ll 0$;
\item \emph{truncation condition}: $Y(u,z)v\in V((z))$ for all $u,v\in V$;

\item \emph{vacuum property}: $Y(\boldsymbol{1},z)=1$;
\item \emph{creation property}: $Y(v,z)\boldsymbol{1}\in v+ V[[z]]z$ for all $v\in V$;
\item \emph{Jacobi identity}: For all $a,b\in V$,
\[
\begin{aligned}
 & z_{0}^{-1}\delta\left(\frac{z_{1}-z_{2}}{z_{0}}\right)Y\left(a,z_{1}\right)Y\left(b,z_{2}\right)-z_{0}^{-1}\delta\left(\frac{z_{2}-z_{1}}{-z_{0}}\right)Y\left(b,z_{2}\right)Y\left(a,z_{1}\right)\\
 & =z_{2}^{-1}\delta\left(\frac{z_{1}-z_{0}}{z_{2}}\right)Y\left(Y\left(a,z_{0}\right)b,z_{2}\right).
\end{aligned}
\]
\item \emph{Virasoro algebra relations}: There exists $c_V\in \C$ such that for all $m,n\in \Z$:
\begin{equation}
[L(m),L(n)]=(m-n)L(m+n)+\frac{1}{12}(m^3-m)\delta_{m+n,0}c_V
\end{equation}
where $L(n)=\omega_{n+1}$ for $n\in\Z$;
\item \emph{$L(0)$-grading:} $L(0)v=nv$ for all $v\in V_{(n)}$ and all $n\in\Z$;
\item \emph{$L(-1)$-derivation property}: $Y(L(-1)v,z)=\frac{d}{dz}Y(v,z)$ for all $v\in V$.
\end{enumerate}

\end{Definition}

\subsection{Definition of the vertex operator algebra $M(1)$}

We recall the well-known construction of the vertex operator algebra $M(1)$, called the \emph{Heisenberg} (or \emph{free bosonic}) vertex operator algebra.

Let $\mathfrak{h}$ be the complexification of an $\ell$-dimensional Euclidean space with orthonormal basis $\{h_1,h_2,\ldots,h_\ell\}$, viewed as an abelian Lie algebra. Its affinization is
\begin{equation}
\hat{\mathfrak{h}} = \mathfrak{h}\otimes\C[t,t^{-1}]\oplus\C K
\end{equation}
with
\begin{equation} \label{eq:heisenberg}
[K,\hat{\mathfrak{h}}]=0,\qquad
[a(m),b(n)]=m\delta_{m+n,0}(a,b)K
\end{equation}
for $a,b\in\mathfrak{h}$, $m,n\in\Z$ and we put $a(n)=a\otimes t^n$. We use the form $(\cdot,\cdot)$ to identify $\mathfrak{h}$ with its dual space $\mathfrak{h}^*$.
Let $\C e^0$ be the one-dimensional module over the Lie algebra $\mathfrak{h}\otimes\C[t]\oplus\C K$ with action given by
\[ h(n)e^0=0,\quad  \forall h\in \mathfrak{h},\, n\ge 0;\qquad  Ke^0 = e^0. \]
Define the vector space $M(1)$ by
\begin{equation}
M(1) = U(\hat{\mathfrak{h}})\otimes_{U(\mathfrak{h}\otimes\C[t]\oplus\C K)} \C e^0.
\end{equation}
On $M(1)$ define the state-field correspondence by
\begin{equation}
Y(a^{(1)}(n_1)\cdots a^{(r)}(n_r)e^0,z)=a^{(1)}(z)_{n_1}\cdots a^{(r)}(z)_{n_r}\Id_{M(1)}
\end{equation}
for $a^{(i)}\in\mathfrak{h}$ and $n_i\in\Z$, (see \cite[Thm.~6.2.11]{LL} for details).
The vacuum vector is $\boldsymbol{1}=e^0$ and the conformal vector is given by
\begin{equation} \label{eq:virasoro-vector}
\omega = \frac{1}{2}\sum_{i=1}^\ell h_i(-1)^2\boldsymbol{1}.
\end{equation}
In particular,
\begin{equation}
Y(\omega,z) = L(z) = \sum_{n\in\Z} L_n z^{-n-2},\qquad
L_n=\frac{1}{2}\sum_{m\in\Z}\sum_i \nord h_i(-m)h_i(m+n) \nord
\end{equation}

\section{The fixed-point vertex operator subalgebra $M(1)^+$}

Consider the order two automorphism $\theta$ of the vector space $M(1)$ given by
\begin{equation}\label{eq:theta}
\theta\big( h_{i_1}(-n_1)h_{i_2}(-n_2)\cdots h_{i_k}(-n_k) \boldsymbol{1}\big) = (-1)^k h_{i_1}(-n_1)h_{i_2}(-n_2)\cdots h_{i_k}(-n_k) \boldsymbol{1}
\end{equation}
for $i_j\in\{1,2,\ldots,\ell\}$ for all $j$ and $n_1\ge n_2\ge \cdots\ge n_k>0$.
 Let $M(1)^+$ be the corresponding subspace of fixed-points with respect to $\theta$:
\begin{equation}
M(1)^+=\big\{ v\in M(1)\mid \theta(v)=v\big\}.
\end{equation}
Note that the Virasoro vector $\omega$ belongs to $M(1)^+$, by \eqref{eq:virasoro-vector}. It is well-known that $M(1)^+$ is a vertex operator subalgebra of $M(1)$.
Let
\begin{equation}
J_a= h_a(-1)^4\boldsymbol{1}-2h_a(-3)h_a(-1)\boldsymbol{1}+\frac{3}{2}h_a(-2)^2\boldsymbol{1},\qquad a=1,2,\ldots,\ell.
\end{equation}
The following statement is known, see \cite{DN}.

\begin{Proposition}
\begin{enumerate}[{\rm (a)}]
\item The VOA $M(1)^+$ is generated by the the following subset:
\begin{equation}
\{J_a\mid a=1,2,\ldots,\ell\}\cup\{h_a(-1)h_b(-1)\boldsymbol{1}\mid a,b=1,2,\ldots,\ell\}
\end{equation}
\item The Zhu algebra $A(M(1)^+)$ is finitely generated.
\end{enumerate}
\end{Proposition}

\section{The weak $M(1)^+$-modules $M(1,\boldsymbol{\lambda})$ and  $M(1,\boldsymbol{\lambda})(\theta)$}

\subsection{Definition of weak $g$-twisted $V$-modules}

\begin{Definition} \cite{AbeDongLi}
Let $V$ be a vertex operator algebra and $g$ be an automorphism of $V$ of finite order $T$ with polarization $V=\bigoplus_{r=0}^{T-1} V^r$, $V^r=\{v\in V\mid g(v)=e^{-\frac{r}{T}2\pi i}v\}$. A \emph{weak $g$-twisted $V$-module} is a pair $(M,Y_M)$ where $M$ is a vector space and $Y_M:V\to\End(M)[[z,z^{-1}]], v\mapsto \sum_{n\in\Z} v_nz^{-n-1}$ is a linear map such that for $0\le r\le T-1$, $a\in V^r$, $b\in V$ and $u\in M$:
\begin{enumerate}[{\rm (i)}]
\item $Y_M(a,z)u\in z^{-\frac{r}{T}}M((z))$,
\item $Y_M(\boldsymbol{1},z)=\Id_M$,
\item the twisted Jacobi identity holds:
\begin{alignat*}{1}
 & z_{0}^{-1}\delta\left(\frac{z_{1}-z_{2}}{z_{0}}\right)Y_{M}\left(a,z_{1}\right)Y_{M}\left(b,z_{2}\right)-z_{0}^{-1}\delta\left(\frac{z_{2}-z_{1}}{-z_{0}}\right)Y_{M}\left(b,z_{2}\right)Y_{M}\left(a,z_{1}\right)\\
 & =z_{2}^{-1}\left(\frac{z_1-z_0}{z_2}\right)^{-\frac{r}{T}}\delta\left(\frac{z_{1}-z_{0}}{z_{2}}\right)Y_{M}\left(Y\left(a,z_{0}\right)b,z_{2}\right).
\end{alignat*}
\end{enumerate}
When $g=1$ a weak $g$-twisted $V$-module is called a \emph{weak $V$-module}. A \emph{$g$-twisted weak $V$-submodule} of a $g$-twisted weak $V$-module $M$ is a subspace $N$ of $M$ such that $a_n N\subseteq N$ for all $a\in V$ and $n\in\Q$. If $M\neq 0$ and $M$ has no weak $g$-twisted $V$-submodules except $0$ and $M$ then $M$ is said to be \emph{simple}.
\end{Definition}

\subsection{Construction of $M(1,\boldsymbol{\lambda})$}

Let $\boldsymbol{\lambda}=(\lambda_0,\lambda_1,\ldots,\lambda_r)$ be a sequence where $\la_i\in \mathfrak{h}$ such that $\lambda_n=0$ for $n\gg 0$.
Let $\mathbb{C}e^{\boldsymbol{\lambda}}$ be the one-dimensional module over the Lie algebra $\mathfrak{h}\otimes \C[t]\oplus \C K$ with action given by
\[ h(n)e^{\boldsymbol{\lambda}}=(h,\lambda_n)e^{\boldsymbol{\lambda}},\; h\in \mathfrak{h},\, n\ge 0;\quad  Ke^{\boldsymbol{\lambda}} = e^{\boldsymbol{\lambda}}. \]
Consider the corresponding induced $U(\widehat{\mathfrak{h}})$-module
\begin{equation}
M(1,\boldsymbol{\lambda})=U(\hat{\mathfrak{h}})\otimes_{U(\mathfrak{h}\otimes\C[t]\oplus\C K)} \C{e^{\boldsymbol{\lambda}}}.
\end{equation}
$M(1,\boldsymbol{\lambda})$ is an example of a \emph{Whittaker module} because it is generated as a left $U(\hat{\mathfrak{h}})$-module by a \emph{Whittaker vector} $e^\la$, which is by definition a common eigenvector for $\mathfrak{h}\otimes\C[t]\oplus\C K$ such that $(\mathfrak{h}\otimes t^n\C[t])e^\la=0$ for $n\gg 0$.
By the PBW theorem, $M(1,\boldsymbol{\lambda})$ has a basis consisting of monomials
\begin{equation}\label{eq:monomials}
h_{i_1}(-n_1)h_{i_2}(-n_2)\cdots h_{i_k}(-n_k) e^\la
\end{equation}
where $i_j\in\{1,2,\ldots,\ell\}$ for all $j$ and $n_1\ge n_2\ge \cdots\ge n_k>0$.

\begin{Example}\label{ex:r0case}
Suppose $\boldsymbol{\lambda}=(\la_0,0,0,\ldots)$ for some $\la_0\in\mathfrak{h}$. Then $M(1,\boldsymbol{\lambda})=M(1,\la_0)$ is a highest weight $\hat{\mathfrak{h}}$-module, the level $1$ Verma module of highest weight $\la_0$.
\end{Example}

\begin{Example}
Suppose $\boldsymbol{\lambda}=(\la_0,\la_1,0,\ldots)$ for some $\la_0,\la_1\in\mathfrak{h}$ where $\la_1\neq 0$. Then for $|n|>1$ we have the usual behavior:
\[h(n) e^{\boldsymbol{\lambda}} =0, \forall n>1,\; h\in\mathfrak{h}\]
while the set of vectors
\[h_{i_1}(-n_1)h_{i_2}(-n_2)\cdots h_{i_k}(-n_k)e^{\boldsymbol{\lambda}}
\quad\text{where $n_1\ge n_2\ge \cdots \ge n_k>1$ and $1\le i_j\le \ell$}\] are linearly independent.
Also $h(0)e^{\boldsymbol{\lambda}}=(h,\la_0)e^{\boldsymbol{\lambda}}$ for $h\in\mathfrak{h}$.
However, for $n=\pm 1$ something different happens:
\begin{equation}\label{eq:whittaker-n1}
h(1) e^{\boldsymbol{\lambda}} = (h,\la_1) e^{\boldsymbol{\lambda}} ,\; \forall h\in  \mathfrak{h}.
\end{equation}
The Heisenberg relation \eqref{eq:heisenberg} implies that
\[h(1)h(-1) - h(-1)h(1) = (h,h)K,\;\forall h\in \mathfrak{h}.\]
Acting on $e^{\boldsymbol{\lambda}}$ in both sides and using \eqref{eq:whittaker-n1} we get
\[h(1)h(-1) e^{\boldsymbol{\lambda}} = \big( (h,\la_1) h(-1) + (h,h)\big) e^{\boldsymbol{\lambda}}. \]
Now choose $h\in\mathfrak{h}$ such that $(h,\la_1)\neq 0$ (possible since $(\cdot,\cdot)$ is non-degenerate and $\la_1\neq 0$). Then we see that $M(1,\boldsymbol{\lambda})$ is not an $\N$-graded module: When acting by $h(1)$ on the homogeneous vector $h(-1)e^{\boldsymbol{\lambda}}$ we obtain a sum of two homogeneous vectors of different degrees:
\[h(1)\cdot h(-1)e^{\boldsymbol{\lambda}} = (h,\la_1)h(-1)e^{\boldsymbol{\lambda}}  + (h,h)e^{\boldsymbol{\lambda}}. \]
Also note that this means that $h(1)$ does not even act locally nilpotently: For any $N>0$ we have
\[h(1)^N\cdot h(-1)e^{\boldsymbol{\lambda}} \in  (h,\la_1)^N h(-1)e^{\boldsymbol{\lambda}}  + \C e^{\boldsymbol{\lambda}}. \]
\end{Example}

\subsection{Construction of $M(1,\boldsymbol{\la})(\theta)$}

The following result is well known \cite{FLM}.

\begin{Proposition}
Let $V$ be a vertex operator algebra, $g$ be an automorphism of $V$, and $M$ be a simple weak $g$-twisted $V$-module. Then $M$ is a simple weak module over the orbifold vertex operator algebra $V^g$.
\end{Proposition}

Let
\begin{equation}
\hat{\mathfrak{h}}[-1] = \mathfrak{h}\otimes t^{1/2}\C[t,t^{-1}]\oplus \C K
\end{equation}
with Lie algebra bracket
\begin{equation}
[K,\hat{\mathfrak{h}}[-1]]=0, \qquad
[a(m),b(n)]=m(a,b)\delta_{m+n,0}K, \qquad
a,b\in\mathfrak{h},\, m,n\in\frac{1}{2}+\Z,
\end{equation}
where $a(n)=a\otimes t^n$ for $a\in\mathfrak{h}$, $n\in\frac{1}{2}+\Z$.

Let $\boldsymbol{\la}=(\la_{1/2},\la_{3/2},\ldots)$ be a sequence of elements of $\mathfrak{h}$ such that $\lambda_n=0$ for $n\gg 0$, and
let $\C e^{\boldsymbol{\la}}$ be the one-dimensional $U(\mathfrak{h}\otimes t^{1/2}\C[t]\oplus\C K)$-module given by
\begin{align}
a(n) e^{\boldsymbol{\la}} &= (\la_n,a) e^{\boldsymbol{\la}}, \qquad a\in\mathfrak{h},\, n=\frac{1}{2},\frac{3}{2},\ldots\\
K e^{\boldsymbol{\la}} &= e^{\boldsymbol{\la}}.
\end{align}
Define the vector space $M(1,\boldsymbol{\la})(\theta)$ by
\begin{equation}
M(1,\boldsymbol{\la})(\theta) = U(\hat{\mathfrak{h}}[-1])\otimes_{U(\mathfrak{h}\otimes t^{1/2}\C[t]\oplus\C K)} \C e^{\boldsymbol{\la}}
\end{equation}
By the PBW theorem, $M(1,\boldsymbol{\la})(\theta)$ has a basis
\begin{equation}
\left\{h_{i_1}(-n_1)h_{i_2}(-n_2)\cdots h_{i_r}(-n_r)e^{\boldsymbol{\la}}\mid r\in\N, n_1\ge n_2\ge\cdots\ge n_r, i_j=1,2\ldots,\ell \right\}.
\end{equation}

Then $M(1,\boldsymbol{\la})(\theta)$ has the structure of a weak $\theta$-twisted $M(1)$-module as follows \cite{FLM}.
For $a\in\mathfrak{h}$, put
\begin{equation}
a(z)=\sum_{n\in \frac{1}{2}+\Z} a(n)z^{-n-1}
\end{equation}
and for $u=a_1(-n_1)a_2(-n_2)\cdots a_r(-n_r)e^{0} \in M(1)$, define

\begin{equation}
Y_0(u,z)=\nord \frac{1}{(n_1-1)!}\left(\frac{d}{dz}\right)^{n_1-1}\left(a_1(z)\right) \cdots \frac{1}{(n_r-1)!}\left(\frac{d}{dz}\right)^{n_r-1}\left(a_r(z)\right)\nord
\end{equation}
Let $c_{mn}\in\Q$ for $m,n\in\N$ be given by
\begin{equation}
\sum_{m,n=0}^\infty c_{mn}z^mw^n = -\log \frac{(1+z)^{1/2}+(1+w)^{1/2}}{2}
\end{equation}
and set
\begin{equation}
\Delta_z = \sum_{i=1}^\ell\sum_{m,n=0}^\infty c_{mn} h_i(m)h_i(n)z^{-m-n},
\end{equation}

\begin{equation}
Y_{M(1,\boldsymbol{\la})(\theta)}(u,z) = Y_0(e^{\Delta_z} u,z).
\end{equation}

\section{Bosonic Fock space realization of $M(1,\boldsymbol{\la})$ and $M(1,\boldsymbol{\la})(\theta)$}

\subsection{Untwisted case}

It will be useful to study the $M(1)^+$-module $M(1,\boldsymbol{\lambda})$ using a Whittaker analog of the bosonic Fock representation.
First, since $h_i(-m)$ and $h_j(-n)$ commute for all $i,j\in\{1,2,\ldots,\ell\}$ and all positive integers $m$ and $n$, there exists a $U(\widehat{\mathfrak{h}}_-)$-module isomorphism
\begin{equation}
\begin{aligned}
\varphi:M(1,\boldsymbol{\lambda}) &\to B=\C[x_{in}\mid i=1,2,\ldots,\ell;\; n=1,2,\ldots] \\
h_{i_1}(-n_1)h_{i_2}(-n_2)\cdots h_{i_k}(-n_k)e^{\boldsymbol{\lambda}} &\mapsto x_{i_1n_1}x_{i_2n_2}\cdots x_{i_kn_k}
\end{aligned}
\end{equation}
where $U(\widehat{\mathfrak{h}}_-)$ acts on $B$ via $h_i(-n).f = x_{in}f$.
Then there is a unique action of $U(\widehat{\mathfrak{h}})$ on $B$ such that $\varphi$ is a $U(\widehat{\mathfrak{h}})$-module isomorphism. Explicitly, it is given by

\begin{equation}
\left\{
\begin{aligned}
h_i(-n) f &= x_{in}f\qquad i=1,2,\ldots,\ell;\; n=1,2,\ldots \\
h_i(n) f &= \big(\partial_{in} + (\la_n, h_i)1\big)(f)\qquad i=1,2,\ldots,\ell;\; n=0,1,2,\ldots
\end{aligned}\right.
\end{equation}
for all $f\in B$, where $\partial_{in}=n\frac{\partial}{\partial x_{in}}$.
We denote $B$ by $B_{\boldsymbol{\la}}$ when regarded as a $U(\widehat{\mathfrak{h}})$-module in this way. In particular we have
\begin{equation}\label{eq:M1plus-action-on-B}
h_i(m)h_j(n)f = \big(\partial_{im}+(\la_m,h_i)1\big)\big(\partial_{jn}+(\la_n,h_j)1\big)(f)
\end{equation}
or, equivalently,
\begin{equation} \label{eq:himhjn-identity}
\big(h_i(m)h_j(n)-(\la_m,h_i)(\la_n,h_j)\big)f = (\la_m,h_i)\partial_{jn}(f) + (\la_n,h_j)\partial_{im}(f) + \partial_{im}\partial_{jn}(f)
\end{equation}
for all $f\in B_{\boldsymbol{\lambda}}$, $i,j=1,2,\ldots,\ell$ and positive integers $m$ and $n$.

Let $U(\widehat{\mathfrak{h}})^+$ be the subalgebra of $U(\widehat{\mathfrak{h}})$ generated by all quadratic elements $h_i(m)h_j(n)$ where $i,j=1,2,\ldots,\ell$ and $m,n\in\Z$.

\begin{Lemma} \label{lem:even-subalg}
$B_{\boldsymbol{\lambda}}$ is a simple $U(\widehat{\mathfrak{h}})^+$-module if and only if the weak $M(1)^+$-module $M(1,\boldsymbol{\lambda})$ is simple.
\end{Lemma}

\begin{proof}
Let $u\in M(1,\boldsymbol{\la})$ and let $m\in\N$ such that $h(i)u=0$ for all $h\in\Fh$ and all $i>m$.
Then, as in \cite{T1}, we have the identity
\begin{equation} \label{eq:binom1}
\begin{aligned}
&\big(h_a(-p-1)h_b(-q-1)\mathbf{1}\big)_{n+1}\,u \\
&\quad = \sum_{\substack{i+j\ge 0\\ i+j=2m-n}} \binom{i-m-1}{p}\binom{j-m-1}{q} \nord h_a(-i+m)h_b(-j+m)\nord \,u
\end{aligned}
\end{equation}
which holds for all non-negative integers $p,q$ and all $a,b\in\{1,2,\ldots,\ell\}$.
This shows that if $B'$ is a nonzero proper $U(\widehat{\mathfrak{h}})^+$-submodule of $B_{\boldsymbol{\lambda}}$, then $\varphi^{-1}(B')$ is a nonzero proper $M(1)^+$-submodule of $M(1,\boldsymbol{\la})$.

Conversely, let $W$ be any nonzero proper $M(1)^+$-submodule of $M(1,\boldsymbol{\la})$.
By \cite[Lemma 2.3]{T1} the coefficient matrix in \eqref{eq:binom1} is invertible and therefore $U(\widehat{\mathfrak{h}})^+\varphi(W)\subseteq \varphi(W)$ which shows that $\varphi(W)$ is a nonzero proper $U(\widehat{\mathfrak{h}})^+$-submodule of $B_{\boldsymbol{\lambda}}$.
\end{proof}

\subsection{Twisted case}

Analogously to the untwisted case, we find a bosonic fock space realization of the $U(\hat{\mathfrak{h}}[-1])$-module $M(1,\boldsymbol{\la})(\theta)$.

For $\boldsymbol{\la}=(\la_{1/2},\la_{3/2},\ldots)$, define
\begin{equation}
B_{\boldsymbol{\la}}^{\theta}=\C[x_{in}\mid i=1,2,\ldots,\ell; n\in\frac{1}{2}+\Z]
\end{equation}
equipped with the $U(\hat{\mathfrak{h}}[-1])$-module action
\begin{align}
Kf&=f, \\
h_i(-n)f &= x_{in}f, \\
h_i(n)f &= \left(\partial_{in}+(\la_n,h_i)1\right)(f),
\end{align}
for $i=1,2,\ldots,\ell$ and $n=\frac{1}{2},\frac{3}{2},\ldots$.

Then the linear map
\begin{equation}
\begin{aligned}
\psi:M(1,\boldsymbol{\la})(\theta) &\to B_{\boldsymbol{\la}}^{\theta} \\
h_{j_1}(-m_1)\cdots h_{j_r}(-m_r)e^{\boldsymbol{\la}} &\mapsto x_{j_1m_1}\cdots x_{j_rm_r}
\end{aligned}
\end{equation}
is an isomorphism of $U(\hat{\mathfrak{h}}[-1])$-modules.

In particular we have the identity

\begin{equation}
\left(h_i(m)h_j(n)-(\la_m,j_i)(\la_n,h_j)1\right)(f)=
\left(\partial_{im}\partial_{jm}+(\la_m,h_i)\partial_{jn}+(\la_n,h_j)\partial_{im}\right)(f)
\end{equation}

for $i,j\in\{1,2,\ldots,\ell\}$ and $m,n\in\frac{1}{2}+\Z$ and $f\in M(1,\boldsymbol{\la})(\theta)\cong B_{\boldsymbol{\la}}^{\theta}$.

Let $\nu$ be the involution on the Lie algebra $\hat{\mathfrak{h}}[-1]$ given by negation, and let $U(\hat{\mathfrak{h}}[-1])^+$ denote the corresponding fixed-point subalgebra.

\begin{Lemma}\label{lem:even-subalg-theta}
For any $\boldsymbol{\la}=(\la_{1/2},\la_{3/2},\ldots)$, the weak $M(1)^+$-module $M(1,\boldsymbol{\la})(\theta)$ is simple if and only if $B_{\boldsymbol{\la}}^{\theta}$ is an simple $U(\hat{\mathfrak{h}}[-1])^+$-module.
\end{Lemma}

\begin{proof}
Similar to the proof of Lemma \ref{lem:even-subalg}, using the twisted version of Borcherds identity.
\end{proof}

\section{Main theorem: Simplicity of $M(1,\boldsymbol{\lambda})$ and $M(1,\boldsymbol{\lambda})(\theta)$ as weak $M(1)^+$-modules.}

We now prove that the weak $M(1)^+$-modules $M(1,\boldsymbol{\lambda})$
and $M(1,\boldsymbol{\la})(\theta)$
are simple for any nonzero
$\boldsymbol{\lambda}$.

\begin{Theorem} \label{thm:simplicity}
For any non-negative integer $r$ and any sequence $\boldsymbol{\la}=(\la_0,\la_1,\ldots,\la_r,0,0,\ldots)$ (respectively $\boldsymbol{\la}=(\la_{1/2},\la_{3/2},\ldots,\la_{r+1/2},0,0,\ldots)$) of elements of $\mathfrak{h}$ with $\la_r\neq 0$ (respectively $\la_{r+1/2}\neq 0$), the weak $M(1)^+$-module $M(1,\boldsymbol{\lambda})$ (respectively $M(1,\boldsymbol{\la})(\theta)$) is simple.
\end{Theorem}

\begin{proof}
We treat both cases simultaneously.
Let $B=B_{\boldsymbol{\la}}$ (respectively $B=B_{\boldsymbol{\la}}^{\theta}$) and $U^+=U(\hat{\mathfrak{h}})^+$ (respectively $U^+=U(\hat{\mathfrak{h}}[-1])^+$).
Put $I=\{1,2,\ldots,\ell\}$ where $\ell=\dim\mathfrak{h}$.
By Lemmas \ref{lem:even-subalg} and \ref{lem:even-subalg-theta}, it suffices to prove that $B$ is a simple $U^+$-module.
We equip $B$ with the usual weight gradation where $\deg x_{in}=n$.
When $\partial_{im}$ acts on any monomial in the variables $\{x_{jn}\}$, the result is either zero or the degree has decreased by $m$.
Thus for any $a\in B$,
\begin{equation}
\deg \partial_{im}(a)\le\deg(a)-m
\end{equation}
with equality if and only if $x_{im}$ occurs in some leading monomial of $a$ ($\deg 0=-\infty$).

Let $a\in B$ be a nonzero arbitrary polynomial. Let $d=\deg a$.
We prove that $1\in U^+ a$ by induction on $d$.
It suffices to show that there exists at least one leading term in $a$ that is mapped to something nonzero under the differential operator. So we may without loss of generality assume that $a$ is a sum of monomials of the same degree, $d$.
Any set of distinct monomials, each containing the variable $x_{in}$, remains linearly independent when acted upon by $\partial_{in}$.

If $d=0$ then this is trivial since then $a$ is a nonzero complex number.

If $d>0$ then we show that there exists $(i,j,m,n)\in I\times I\times \Z_+\times\Z_+$ (respectively $(i,j,m,n)\in I\times I \times (\frac{1}{2}+\N)\times (\frac{1}{2}+\N)$)
such that
\begin{equation} \label{eq:pf1}
b=\big(h_i(m)h_j(n)-(\la_m,h_i)(\la_n,h_j)\big)a
\end{equation}
is nonzero. If we can show this, then since $\deg(b) < \deg(a)=d$ it follows that $1\in U^+ b$ by the induction hypothesis. Since
$h_i(m)h_j(n)-(\la_m,h_i)(\la_n,h_j)\in U^+$ we get
\[1\in U^+b \subseteq U^+ a\]
as desired.

To prove that we can find such $(i,j,m,n)$, note that by \eqref{eq:himhjn-identity}, the expression \eqref{eq:pf1} is equal to
\begin{equation}
(\la_m,h_i)\partial_{jn}(f)+(\la_n,h_j)\partial_{im}(f)+\partial_{im}\partial_{jn}(a).
\end{equation}
Since $a$ is a nonconstant polynomial in the variables $x_{im}$, there exists $(i,m)$ such that $\partial_{im}(a)\neq 0$.

Fix $m$ to be the smallest positive (half-)integer such that there exists $i_0\in I$ such that $\partial_{i_0m}(a)\neq 0$.
Furthermore, let $n$ be any non-negative (half-)integer such that $\la_n\neq 0$.
The case when $\la_n=0$ for all $n>0$ corresponds to the highest weight case, which is already known \cite[Proposition 2.2.2]{DN}. So we can assume that $n>0$.

We now consider three cases.

(Case 1): If $n>m$ then pick any $j_0\in I$ such that $(\la_n, h_{j_0})\neq 0$. This is possible since $\la_n\neq 0$, $h_j$ form a basis for $\mathfrak{h}$, and $(\cdot,\cdot)$ is non-degenerate.
Then with $(i,j,m,n)=(i_0,j_0,m,n)$ we get
\[(\la_n, h_{j_0}) \partial_{im}(a) + \text{lower degree terms}\]
which is has a nonzero leading term of degree $(\deg a) - m$.

(Case 2): If $n<m$ then again pick any $j_0\in I$ such that $(\la_n, h_{j_0})\neq 0$. Then with
 $(i,j,m,n)=(i_0,j_0,m,n)$ we get
\[(\la_n, h_{j_0}) \partial_{i_0m}(a) + (\la_m,h_i)\partial_{j_0n}(a) + \partial_{im}\partial_{j_0n}(a) = (\la_n, h_{j_0}) \partial_{i_0m}(a) + \partial_{im}\partial_{j_0n}(a)  \]
since $\partial_{j_0n}(a)=0$ by minimality of $m$. The term $\partial_{im}\partial_{j_0n}(a) $ has degree $(\deg a)-(n+m) < (\deg a)-m  = \deg \partial_{i_0m}(a)$, and the first term is nonzero by the choice of $(i_0,m)$ and since $(\la_n,h_{j_0})\neq 0$.

(Case 3): If $n=m$ there are two subcases:

(Case 3a) Suppose $(\la_m,h_{i_0})\neq 0$. Then take $(i,j,m,n)=(i_0,i_0,m,m)$ we get

\[2(\la_m,h_{i_0})\partial_{i_0m}(a) + \text{lower degree terms}\]

and the coefficient is nonzero.

(Case 3b) Suppose $(\la_m,h_{i_0})=0$. Then find any other $j_0\in I$ such that $(\la_m,h_{j_0})\neq 0$. This is possible since we are in the case where $m=n$ and $\la_n\neq 0$ by choice of $n$.
Then with $(i,j,m,n)=(i_0,j_0,m,n)$ we get
\[ (\la_m,h_{j_0})\partial_{i_0m}(a) + (\la_m,h_{i_0})\partial_{j_0m}(a) + \text{lower degree terms}=
(\la_m,h_{j_0})\partial_{i_0m}(a)  + \text{lower degree terms}\]
since $(\la_m,h_{i_0}) = 0$.

This covers all cases and proves the claim.
\end{proof}

\section{$M(1,\boldsymbol{\lambda})$ and $M(1,\boldsymbol{\lambda})(\theta)$ are $\omega$-Whittaker $M(1)^+$-modules}

\subsection{Definition of $\omega$-Whittaker modules}

There is no general definition of a Whittaker module for a vertex operator algebra. However, such a notion exists for Virasoro algebras. Following \cite{T1} we therefore make the following definitions.

\begin{Definition}
Let $V$ be a vertex operator algebra with conformal vector $\omega$ and let $M$ be a weak $V$-module.
\begin{enumerate}
\item A nonzero vector $v\in M$ is a \emph{Whittaker vector for $\omega$} if there exist a positive integer $r$ and a sequence of complex numbers $\zeta=(\zeta_{r+1},\zeta_{r+2},\ldots,\zeta_{2r+\epsilon})$ where $\epsilon\in\{0,1\}$ and $\zeta_{2r+\epsilon}\neq 0$ such that
\begin{equation}
\omega_i v = \begin{cases}
\zeta_i v,& j=r+1,\, r+2,\,\ldots,\, 2r+\epsilon\\
0,& i>2r+\epsilon.
\end{cases}
\end{equation}
The sequence $\zeta$ is called the \emph{type} of $v$.
\item We say that $M$ is an \emph{$\omega$-Whittaker module} if $M$ is generated as a weak $V$-module by a Whittaker vector for $\omega$.
\end{enumerate}
\end{Definition}

\subsection{$M(1,\boldsymbol{\lambda})$ and $M(1,\boldsymbol{\lambda})(\theta)$ are $\omega$-Whittaker $M(1)^+$-modules}

\begin{Proposition}
For any positive integer $r$ and any sequence $\boldsymbol{\la}=(\la_0,\la_1,\ldots,\la_r,0,0,\ldots)$ (respectively $\boldsymbol{\la}=(\la_{1/2},\la_{3/2},\ldots,\la_{r-1/2},0,0,\ldots)$) of elements of $\mathfrak{h}$ with $\la_r\neq 0$ (respectively $\la_{r-1/2}\neq 0$), the weak $M(1)^+$-module $M(1,\boldsymbol{\lambda})$ (respectively $M(1,\boldsymbol{\la})(\theta)$) is an $\omega$-Whittaker module.
\end{Proposition}

\begin{proof}
By Theorem \ref{thm:simplicity} these $M(1)^+$-modules are simple, hence in particular generated by $e^{\boldsymbol{\la}}$. Therefore it suffices to show that, in both cases, $e^{\boldsymbol{\la}}$ is a Whittaker vector for $\omega$. We first consider the case of $M(1,\boldsymbol{\la})$.
Since $m+n\ge r$ implies $\min\{m,n\}\ge 0$ if $\max\{m,n\}\le r$ we have for all $j=r+1,r+2,\ldots,2r+1$,
\begin{align*}
\omega_j e^{\boldsymbol{\lambda}} &= \frac{1}{2}\sum_{\substack{m,n\in\Z\\ m+n=j-1}} \sum_i \nord h_i(m)h_i(n)\nord \, e^{\boldsymbol{\lambda}}  \nonumber \\
&=\frac{1}{2}\sum_{\substack{m,n\in\N\\ m+n=j-1}} (\la_m,\la_n) e^{\boldsymbol{\lambda}} \in \C e^{\boldsymbol{\lambda}}
\label{eq:omega_j-on-vacuum}
\end{align*}
where we used that $\sum_i(\la_m,h_i)(\la_n,h_i)=(\la_m,\la_n)$. Taking $j=2r+1$ there is only one nonzero term in the sum, $\omega_{2r+1}e^{\boldsymbol{\lambda}}=\frac{1}{2}(\la_r,\la_r)e^{\boldsymbol{\lambda}}\neq 0$.
This shows that $M(1,\boldsymbol{\lambda})$ is a $\omega$-Whittaker module for $M(1)^+$.

Now consider the weak $M(1)^+$-module
$M(1,\boldsymbol{\la})(\theta)$. The action of the conformal vector is given by \cite[Example 18.9]{KRR}:

\begin{equation}
L^{\textrm{tw}}(z) = \frac{1}{2}\sum_{i=1}^\ell \nord h\,_i(z)^2 \,\nord +\frac{1}{16}z^{-2}
\end{equation}

This shows that
\begin{equation}
Y_{M(1,\boldsymbol{\la})(\theta)} (\omega,z) =  \sum_{n\in\Z} \omega_n z^{-n-1}
\end{equation}
where
\begin{equation}
\omega_j = \frac{1}{2}\sum_{\substack{m,n\in\Z+\frac{1}{2}\\ m+n=j-1}} \left(\frac{1}{16}\delta_{j,1}+\sum_{i=1}^\ell \nord\, h_i(m)h_i(n)\,\nord\right)
\end{equation}
Suppose that $j\in\{r+1,r+2,\ldots,2r\}$.
Acting by $\omega_j$ on $e^{\boldsymbol{\lambda}}$ using that $\nord\, h_i(m)h_i(n)\,\nord e^{\boldsymbol{\lambda}}=0$ unless $\max\{m,n\}\le r-\frac{1}{2}$ in which case $\min\{m,n\}=(m+n)-\max\{m,n\}\ge r-(r-\frac{1}{2})=\frac{1}{2}$, we get
\begin{equation}
\omega_j e^{\boldsymbol{\lambda}} = \frac{1}{2}\sum_{\substack{m,n\in\N+\frac{1}{2}\\ m+n=j-1}} (\lambda_m,\lambda_n)e^{\boldsymbol{\lambda}}
\end{equation}
In particular for $j=2r$ we have $m+n=2r-1$ and $m,n\le r-\frac{1}{2}$ in nonzero terms hence $m=n=r-\frac{1}{2}$ which means that \[\omega_{2r}e^{\boldsymbol{\lambda}}=\frac{1}{2} (\la_{r-\frac{1}{2}},\la_{r-\frac{1}{2}})e^{\boldsymbol{\lambda}} \neq 0\]
since $\la_{r-1/2}\neq 0$.
This proves the claim.
\end{proof}

\begin{Remark}
Note that $e^{\boldsymbol{\lambda}}$ is an eigenvector for $\omega_{2r+1}$ with eigenvalue $\frac{1}{2} (\la_r,\la_r)$ which is nonzero. Therefore the weak $M(1,\boldsymbol{\lambda})$ is not $\N$-graded.

On the other hand,
as noted in Example \ref{ex:r0case}, if $\boldsymbol{\lambda}=(\la_0,0,0,\ldots)$ then  $M(1,\boldsymbol{\lambda})$ is the level $1$ Verma $\widehat{\mathfrak{h}}$-module of highest weight $\la_0$, which is $\N$-graded.
\end{Remark}

\section{On the Whittaker type of $M(1,\boldsymbol{\la})$ and $M(1,\boldsymbol{\la})(\theta)$}

In the last section we proved that $M(1,\boldsymbol{\lambda})$ and $M(1,\boldsymbol{\lambda})$ give rise to Whittaker modules over the Virasoro algebra. In this section we show the map is surjective and calculate the fibers. That is, given a Whittaker type $\zeta$ for $\omega$, we find all possible $\boldsymbol{\lambda}$ giving rise to the Whittaker module of type $\zeta$ for the Virasoro algebra. Our results generalize the rank one case from \cite{T2}.

First we treat the untwisted case.

\begin{Proposition}
Let $r$ be a non-negative integer.
Define a function $\Phi=\Phi_r$ as follows:
\begin{equation}
\Phi: \mathfrak{h}^r\times(\mathfrak{h}\setminus\{0\})\to \C^r\times \C^\times
\end{equation}
given by
\begin{equation}\label{eq:lambda-system}
\Phi(\la_0,\la_1,\ldots,\la_r)=(\zeta_{r+1},\zeta_{r+2},\ldots,\zeta_{2r+1}),
\end{equation}
where
\begin{equation}
\zeta_i=\frac{1}{2}\sum_{\substack{0\le m,n\le r\\ m+n=i-1}} (\la_m,\la_n),\qquad i=r+1,r+2,\ldots,2r+1.
\end{equation}
Then $\Phi$ is surjective and for any $\zeta=(\zeta_{r+1},\zeta_{r+2},\ldots,\zeta_{2r+1})\in\C^r\times\C^\times$, the fiber $\Phi^{-1}(\{\zeta\})$ is is isomorphic to the variety  $S^{\ell-1}_{\C}\times \C^{(\ell-1)r}$ where
\[S^{\ell-1}_{\C}=\{(z_1,z_2,\ldots,z_\ell)\in \C^\ell\mid z_1^2+z_2^2+\cdots+z_\ell^2=1\}.\]
\end{Proposition}

\begin{proof}
Taking $i=2r+1, 2r, 2r-1, \ldots, r+1$, the system \eqref{eq:lambda-system} written out is
\[
\left\{\begin{aligned}
(\la_r,\la_r) &= 2\zeta_{2r+1} \\
(\la_{r-1},\la_r)+(\la_r,\la_{r-1}) &= 2\zeta_{2r} \\
(\la_{r-2},\la_r)+(\la_{r-1},\la_{r-1})+(\la_r,\la_{r-2}) &= 2\zeta_{2r-1} \\
&\;\;\vdots \\
(\la_0,\la_r)+(\la_1,\la_{r-1})+\cdots+(\la_r,\la_0) &= 2\zeta_{r+1}
\end{aligned}\right.\]
which is equivalent to
\[
\left\{\begin{aligned}
(\la_r,\la_r) &= 2\zeta_{2r+1}\\
(\la_{r-1},\la_r) &= \zeta_{2r}\\
(\la_{r-2},\la_r) &= \zeta_{2r-1} - \tfrac{1}{2}(\la_{r-1},\la_{r-1})\\
&\;\;\vdots \\
(\la_0,\la_r) &= \zeta_{r+1}-\tfrac{1}{2}\sum_{\substack{1\le m,n\le r-1\\ m+n=r}} (\la_m,\la_n)
\end{aligned}\right.\]
Letting $\la_{ij}=(\la_i,h_j)$ denote the coordinates of the vector $\la_i\in\mathfrak{h}$ in the basis $\{h_j\}_{j=1}^\ell$, the first equation is equivalent to the quadratic equation $\la_{r1}^2+\cdots+\la_{r\ell}^2 = 2\zeta_{2r+1}$. Putting $z_j=\frac{\la_{rj}}{(2\zeta_{2r+1})^{1/2}}$ (for some choice of square root of the nonzero complex number $2\zeta_{2r+1}$) shows that the $\la_r$ are in bijection with $S_\C^{\ell-1}$. The remaining equations are affine equations which have $(\ell-1)$-dimensional affine solution spaces at each of the $r$ steps, since $\la_r\neq 0$. This proves that $\Phi$ is surjective and that each fiber $\Phi^{-1}(\{\zeta\})$ is in bijection with the affine variety $S_\C^{\ell-1}\times\C^{(\ell-1)r}$.
\end{proof}

Similarly in the twisted case we have the following result.

\begin{Proposition}
Let $r$ be a positive integer.
Define a function $\Psi=\Psi_r$ as follows:
\begin{equation}
\Psi: \mathfrak{h}^{r-1}\times(\mathfrak{h}\setminus\{0\})\to \C^{r-1}\times \C^\times
\end{equation}
given by
\begin{equation}\label{eq:lambda-system}
\Phi(\la_{1/2},\la_{3/2},\ldots,\la_{r-1/2})=(\zeta_{r+1},\zeta_{r+2},\ldots,\zeta_{2r}),
\end{equation}
where
\begin{equation}
\zeta_i=\frac{1}{2}\sum_{\substack{m,n\in\{\frac{1}{2},\frac{3}{2},\ldots,r-\frac{1}{2}\}\\ m+n=i-1}} (\la_m,\la_n),\qquad i=r+1,r+2,\ldots,2r.
\end{equation}
Then $\Psi$ is surjective and for any $\zeta=(\zeta_{r+1},\zeta_{r+2},\ldots,\zeta_{2r})\in\C^{r-1}\times\C^\times$, the fiber $\Psi^{-1}(\{\zeta\})$ is isomorphic to the variety $S^{\ell-1}_{\C}\times \C^{(\ell-1)(r-1)}$ where
\[S^{l-1}_{\C}=\{(z_1,z_2,\ldots,z_l)\in \C^l\mid z_1^2+z_2^2+\cdots+z_l^2=1\}.\]

\end{Proposition}

\begin{proof}
Taking $i=2r, 2r-1, 2r-1, \ldots, r+1$, the system \eqref{eq:lambda-system} written out is
\[
\left\{\begin{aligned}
(\la_{r-1/2},\la_{r-1/2}) &= 2\zeta_{2r} \\
(\la_{r-3/2},\la_{r-1/2})+(\la_{r-1/2},\la_{r-3/2}) &= 2\zeta_{2r-1} \\
(\la_{r-5/2},\la_{r-1/2})+(\la_{r-3/2},\la_{r-3/2})+(\la_{r-1/2},\la_{r-5/2}) &= 2\zeta_{2r-2} \\
&\;\;\vdots \\
(\la_{1/2},\la_{r-1/2})+(\la_{3/2},\la_{r-3/2})+\cdots+(\la_{r-1/2},\la_{1/2}) &= 2\zeta_{r+1}
\end{aligned}\right.\]
which is equivalent to
\[
\left\{\begin{aligned}
(\la_{r-1/2},\la_{r-1/2}) &= 2\zeta_{2r}\\
(\la_{r-3/2},\la_{r-1/2}) &= \zeta_{2r-1}\\
(\la_{r-5/2},\la_{r-1/2}) &= \zeta_{2r-2} - \tfrac{1}{2}(\la_{r-3/2},\la_{r-3/2})\\
&\;\;\vdots \\
(\la_{1/2},\la_{r-1/2}) &= \zeta_{r+1}-\tfrac{1}{2}\sum_{\substack{m,n\in\frac{3}{2}+\N\\ m+n=r}} (\la_m,\la_n)
\end{aligned}\right.\]
The remaining argument is identical to the untwisted case.
\end{proof}

\bibliographystyle{siam}


TITLE: Understanding proof when equality holds in Minkowski's inequality
QUESTION [4 upvotes]: I have a simple question about a fact that is constantly mentioned when we have $\|f+g\|_p = \|f\|_p + \|g\|_p$ in $L^p(\mu)$ space for $1<p<\infty$ (I hope someone find this useful).
I read the arguments given here, and here and a proof almost related here. In all of these proofs it's mentioned that when the two Hölder's inequalities below that result on the steps of Minkowski's inequality:
$$\int |f| \cdot |f+g|^{p-1} \, d\mu \leq \|f\|_p \cdot \|f+g\|_p^{p-1}$$
$$\int |g| \cdot |f+g|^{p-1} \, d\mu \leq \|g\|_p \cdot \|f+g\|_p^{p-1}$$
These two holds with equality thanks to the given hypothesis, and that's clear to me. However, then comes a fact that is not as clear to me as it should be (since I found no justification for this) that is: these two equalities implies that there is $\alpha,\beta \geq 0$ such that $|f|^p = \alpha |f+g|^p$ and $|g|^p = \beta |f+g|^p$ almost everywhere. Any hint or reference is appreciated. Thanks!

REPLY [4 votes]: I am gonna give it a try, let me know if this is what your question is about!
Holder inequality is a consequence of the Young's inequality, the latter stating that
for $p,q\in (1,\infty)$ conjugate numbers
$$AB\leq \frac{A^p}{p}+\frac{B^q}{q}$$
$A,B\geq 0$ where equality occurs only if $B=A^{p-1}$.
Then in order to obtain Holder's inequality set $$A:=\frac{|f|}{\|f\|_p}$$
$$B:=\frac{|f+g|^{p-1}}{\|(f+g)^{p-1}\|_q}$$
Then
$$AB=\frac{|f|\cdot |f+g|^{p-1}}{\|f\|_p\|(f+g)^{p-1}\|_p}\leq \frac{|f|^p}{p\|f\|^p_p}+\frac{|f+g|^{(p-1)q}}{q\|(f+g)^{p-1}\|^q_q}$$
Integrating both sides we obtain the Holder's inequality.
Now in order to have equality in the final expression we need equality on the latter, which is true if $B=A^{p-1}$.
$$\frac{|f+g|^{p-1}}{\|(f+g)^{p-1}\|_q}=\bigg(\frac{|f|}{\|f\|_p}\bigg)^{p-1}$$
$$|f|^{(p-1)}=\bigg(\frac{\|f\|_p^{(p-1)}}{\|(f+g)^{p-1}\|_q}\bigg)\cdot|f+g|^{p-1}$$
rising both sides to $p/(p-1)$
yields to the desired result
$$|f|^{p}=\bigg(\frac{\|f\|_p^{(p-1)}}{\|(f+g)^{p-1}\|_q}\bigg)^{p/p-1}\cdot|f+g|^{p}$$
calling $\alpha=\bigg(\frac{\|f\|_p^{(p-1)}}{\|(f+g)^{p-1}\|_p}\bigg)^{p/p-1}\geq 0$.
$$\alpha=\frac{\|f\|_p^p}{\bigg(\int |f+g|^{q(p-1)}d\mu\bigg)^{\frac{1}{q}\frac{p}{p-1}}}$$
notice that $q(p-1)=p$ then we obtain
$$\alpha=\frac{\|f\|_p^p}{\int |f+q|^p d\mu}=\frac{\|f\|_p^p}{\|f+g\|_p^p}$$


TITLE: Topelitz and matrix operators in $\ell_2$
QUESTION [3 upvotes]: Let $a,b$ satisfy $|a|,|b|<1$. We then define a vector $y = (\dots,b^2 ,b ,1 ,a ,a^2 ,\dots) \in \ell_ 1 (\mathbb{Z})$ with the ordering $y_0 = 1$.
We define a matrix operator $Y$ by $$Yx = \sum _{k=0}^\infty  \left( \sum_
{i=0}^\infty y_{i-k} x_i \right) e_k$$
Where $e_k$ is the standard orthonormal basis of $\ell_2$.
I need to show that the matrix $Y$ defines a bounded operator in $\ell_2$
I have so far tried using triangle inequality together with Holder inequality and only came up with: $$ \|Yx\| \leq \|x\| \sum_k \sum_i |y_{i-k}|^2 $$ which does not converge (unless $a=b=0$).
Edit: I also thought of the similarity to laurent operators. the fourier coefficients of a multiplication operator define such a bounded operator, and so we can trace back a multiplication operator from a matrix, and with the "regular" isometry between $\ell_2$ and $L_2$ - such an operator must be well defined and bounded. However, i am not familiar with Topelitz operators and such analogies for them.
I would like a hint on this.
Another question regarding matrix operators: Sometimes - the sums arent written in a specific order: meaning you can invert the summation (if the matrix indeed represents a bounded operator). Why can this be done?

REPLY [2 votes]: Your current definition if $Y$ is
$$
                 Yx=\sum_{k=0}^{\infty}\left(\sum_{i=0}^{\infty}y_{i-k}x_{i}\right)e_{k}.
$$
Consider
\begin{align}
   \|Yx\|_2^2 & = \sum_{k=0}^{\infty}\left|\sum_{i=0}^{\infty}y_{i-k}x_{i}\right|^2 \\
  & \le \sum_{k=0}^{\infty}\left(\sum_{i=0}^{\infty}|y_{i-k}|||x_i|^2\sum_{i=0}^{\infty}|y_{k-i}|\right) \\
  & \le \|y\|_1\sum_{k=0}^{\infty}\sum_{i=0}^{\infty}|y_{k-i}||x_i|^2 \\
  & \le \|y\|_1\sum_{i=0}^{\infty}\sum_{k=0}^{\infty}|y_{k-i}||x_i|^2 \\
  & \le \|y\|_1^2\sum_{i=0}^{\infty}|x_i|^2 = \|y\|_1^2\|x\|_2^2
\end{align}
Therefore,
$$
                 \|Yx\|_2 \le \|y\|_1\|x\|_2 = \left(\frac{1}{1-a}+\frac{1}{1-b}-1\right)\|x\|_2.
$$


text:
\section{Geometric realisation} \label{sec:geometric}

We informally stated that constructible $n$\nbd molecules have $k$\nbd boundaries shaped as $k$\nbd balls for each $k < n$. In this section, we make this precise, by defining the geometric realisation of a constructible directed complex and of a constructible polygraph.

First, we need to recall some basic notions from algebraic and combinatorial topology; we refer to any textbook, for example \cite{may1999concise} and \cite{goerss2009simplicial}, for more details. We will work with the ``convenient'' category $\cghaus$ of compactly generated Hausdorff spaces. We assume that the definitions of simplicial set, their category $\sset$, and the geometric realisation $\realis{-}_\Delta: \sset \to \cghaus$ are known by the reader. Throughout this section, we will not always distinguish between an oriented graded poset and its underlying poset; the context should decide which one we mean.

\begin{dfn}
Let $P$ be a poset. The \emph{nerve} of $P$ is the simplicial set $NP$ whose
\begin{itemize}
	\item $n$\nbd simplices are chains $(x_0 \leq \ldots \leq x_n)$ of length $(n+1)$ in $P$,
	\item the $k$\nbd th face map $d_k: NP_n \to NP_{n-1}$ is defined by
	\begin{equation*}
		(x_0 \leq \ldots \leq x_n) \mapsto (x_0 \leq \ldots \leq x_{k-1} \leq x_{k+1} \leq \ldots \leq x_n),
	\end{equation*}
	\item the $k$\nbd th degeneracy map $s_k: NP_n \to NP_{n+1}$ is defined by
	\begin{equation*}
		(x_0 \leq \ldots \leq x_n) \mapsto (x_0 \leq \ldots \leq x_{k} \leq x_{k} \leq \ldots \leq x_n),
	\end{equation*}
\end{itemize}
for $0 \leq k \leq n$.
\end{dfn}

The nerve extends to a functor $N: \pos \to \sset$. Precomposing with the forgetful functor $\globpos \to \pos$, we obtain a functor $k: \globpos \to \sset$. Finally, we can post-compose with $\realis{-}_\Delta: \sset \to \cghaus$ to obtain a functor $\realis{-}: \globpos \to \cghaus$, which we call the \emph{geometric realisation} of a constructible directed complex.

The functor $\realis{-}_\Delta$ has the property that $\realis{K \times L}_\Delta$ is homeomorphic to the product of spaces $\realis{K}_\Delta \times \realis{L}_\Delta$, and $\realis{K \join L}_\Delta$ to the join of spaces $\realis{K}_\Delta \join \realis{L}_\Delta$, for any pair of simplicial sets $K$, $L$. The nerve functor also preserves products and joins, and the underlying poset of the lax Gray product of two oriented graded posets is the product of their underlying posets. It follows that $\realis{-}$ becomes a monoidal functor from $(\globpos, \tensor, 1)$ to $(\cghaus, \times, 1)$ and from $(\globpos, \join, \emptyset)$ to $(\cghaus, \join, \emptyset)$.

\begin{remark}
In combinatorics, it is more common to consider the \emph{order complex} of a poset, an ordered simplicial complex, rather than the nerve, a simplicial set. The two notions of geometric realisation coincide up to homeomorphism.
\end{remark}

In what follows, let $D^n$ be a model of the closed $n$\nbd ball, and $\bord{}{}D^n$ its boundary, homeomorphic to the $(n-1)$\nbd sphere; for a map $x: D^n \to X$, let $\bord{}{}x$ be its restriction to $\bord{}{}D^n$.
\begin{dfn}
A \emph{CW complex} is a topological space $X$ together with a non-decreasing sequence $\{\skel{n}{X} \hookrightarrow X\}_{n \in \mathbb{N}}$ of subspaces, the \emph{$n$\nbd skeleta}, and families $\{\atom{n}{X}\}_{n \in \mathbb{N}}$ of maps $x: D^n \to \skel{n}{X}$, such that $X = \bigcup_{n\in \mathbb{N}} \skel{n}{X}$ and, for all $n$,
\begin{equation*}
\begin{tikzpicture}[baseline={([yshift=-.5ex]current bounding box.center)}]
	\node[scale=1.25] (0) at (0,2) {$\displaystyle \coprod_{x\in\atom{n}{X}} \bord{}{} D^n$};
	\node[scale=1.25] (1) at (3.5,2) {$\displaystyle\coprod_{x\in\atom{n}{X}} D^n$};
	\node[scale=1.25] (2) at (0,0) {$\skel{n-1}{X}$};
	\node[scale=1.25] (3) at (3.5,0) {$\skel{n}{X}$};
	\draw[1c] (0) to node[auto,swap] {$(\bord{}{}x)_{x\in\atom{n}{X}}$} (2);
	\draw[1c] (1) to node[auto] {$(x)_{x\in\atom{n}{X}}$} (3);
	\draw[1cinc] (0) to (1);
	\draw[1cinc] (2) to (3);
	\draw[edge] (2.5,0.2) to (2.5,0.8) to (3.3,0.8);
\end{tikzpicture}
\end{equation*}
is a pushout diagram in $\cghaus$.

A CW complex is \emph{regular} if, for all $n$ and $x \in \atom{n}{X}$, the map $x: D^n \to X$ is a homeomorphism onto its image.
\end{dfn}

By analogy with polygraphs, we will call $x \in \atom{n}{X}$ a \emph{generating} $n$\nbd cell of the CW complex $X$, even though the term is more commonly associated to an algebraic setting.

\begin{dfn}
Let $X$ be a CW complex. The \emph{face poset} $\face{X}$ of $X$ is the poset whose elements are the generating cells of $X$, and for any pair of generating cells $x: D^k \to X$ and $y: D^n \to X$ we have $x \leq y$ if and only if $x(D^k) \subseteq y(D^n)$.
\end{dfn}

The face poset is, arguably, the simplest non-trivial combinatorial structure that one can associate to a CW complex, yet for regular CW complexes, it specifies the type of the underlying topological space up to homeomorphism.
\begin{thm}\emph{\cite[Theorem 1.7]{lundell1969topology}}
Let $X$ be a regular CW complex. Then $|N(\face{X})|$ is homeomorphic to $X$.
\end{thm}

In \cite{bjorner1984posets}, Bj\"orner studied criteria for a poset to be the face poset of a regular CW complex. The following is [Definition 2.1, \emph{ibid.}].
\begin{dfn}
A poset with a least element $P_\bot$ is a \emph{CW poset} if $P$ has at least one element, and, for all $x \in P$, the geometric realisation $|N(\bord{}{}x)|$ is homeomorphic to a sphere.
\end{dfn}

\begin{prop}\emph{\cite[Proposition 3.1]{bjorner1984posets}} A poset $P$ is the face poset of a regular CW complex if and only if $P_\bot$ is a CW poset.
\end{prop}

We will prove that if $P$ is a constructible directed complex, then $P_\bot$ is a CW poset. For this purpose, we go through the intermediate notion of a recursively dividable poset, defined in \cite{hachimori2000constructible}. The following is a rephrasing of [Definition 3.1, \emph{ibid.}]; note that we use a pure $n$\nbd dimensional subset $U$ where Hachimori would add an $(n+1)$\nbd dimensional greatest element.

\begin{dfn}
Let $P$ be a graded poset. We define a class of pure subsets $U$ of $P$, that we call \emph{recursively dividable}.

The empty subset is recursively dividable. Suppose $U$ is inhabited, pure and $n$\nbd dimensional. Then $U$ is recursively dividable if and only if $\bord{}{}x$ is recursively dividable for all $x \in U^{(n)}$, and, inductively on the number of maximal elements of $U$, either
\begin{itemize}
	\item $U$ has a greatest element, or
	\item $U = U_1 \cup U_2$, where
	\begin{enumerate}
		\item $U_1$ and $U_2$ are pure, $n$\nbd dimensional, and recursively dividable, and
		\item $U_1 \cap U_2$ is pure, $(n-1)$\nbd dimensional, and recursively dividable.
	\end{enumerate}
\end{itemize}
\end{dfn}

\begin{prop} \label{prop:recdivid}
Let $U$ be a constructible molecule. Then $U$ and $\bord{}{}U$ are recursively dividable.
\end{prop}
\begin{proof}
If $U$ is a 0\nbd molecule, this is immediate from the definition, so suppose $U$ is a constructible $n$\nbd molecule with $n > 0$.

We have $\bord{}{}U = \bord{}{-}U \cup \bord{}{+}U$, where $\bord{}{-}U$ and $\bord{}{+}U$ are constructible $(n-1)$\nbd molecules, and $\bord{}{-}U \cap \bord{}{+}U = \bord{}{}(\bord{}{\alpha}U)$ is the boundary of a constructible $(n-1)$\nbd molecule: all are recursively dividable by the inductive hypothesis. It follows that $\bord{}{}U$ is recursively dividable.

If $U$ is atomic, we are done. Otherwise, $U$ splits as $U_1 \cup U_2$, where $U_1$, $U_2$ are constructible $n$\nbd molecules with fewer maximal elements, and their intersection $U_1 \cap U_2$ is a constructible $(n-1)$\nbd molecule. All are recursively dividable by the inductive hypothesis, so $U$ is recursively dividable.
\end{proof}

We do not know, at the moment, whether constructible molecules are also strongly dividable in the sense of \cite[Definition 4.2]{hachimori2000constructible}.

\begin{thm} \label{thm:regularball}
Let $U$ be a constructible $n$\nbd molecule. Then $\realis{U}$ is homeomorphic to a closed $n$\nbd ball, and $\realis{\bord{}{}U}$ is homeomorphic to an $(n-1)$\nbd sphere.
\end{thm}
\begin{proof}
By Proposition \ref{prop:recdivid} and \cite[Proposition 3.6]{hachimori2000constructible}, the order complexes of $U$ and $\bord{}{}U$ are constructible simplicial complexes [Definition 2.7, \emph{ibid.}]. Moreover, by thinness and Lemma \ref{lem:basic}, each $(n-1)$\nbd simplex is a face of at most two $n$\nbd simplices. The statement then follows from \cite[Proposition 2.16]{hachimori2000combinatorics}.
\end{proof}

\begin{cor} \label{cor:cwposet}
If $P$ is a constructible directed complex, then $P_\bot$ is a CW poset.
\end{cor}

Therefore, the geometric realisation of a constructible directed complex $P$ admits the structure of a CW complex whose generators are the geometric realisations of the atoms of $P$, and this CW complex is regular.

\begin{remark} \label{rmk:nonconstructible}
Not all regular CW decompositions of balls have recursively dividable face posets; similarly, not all molecules whose geometric realisation is a regular CW ball are constructible. The following example of a non-constructible 3-molecule, whose atoms are constructible and whose geometric realisation is a 3-ball, was suggested to us by S.\ Henry; the shaded area in each diagram is the input boundary of the following 3\nbd atom.
\begin{equation*}
\input{img/s5_nonconstructible.tex}
\end{equation*}
\begin{equation*}
\input{img/s5_nonconstructible2.tex}
\end{equation*}
\end{remark}

\begin{cons}
Restricting the geometric realisation functor to $\globe$, and taking its left Kan extension along the Yoneda embedding of $\globe$ into $\cpol$, we obtain a functor $|-|: \cpol \to \cghaus$, which sends any regular polygraph $X$ to the coend
\begin{equation*}
	\realis{X} := \int^{U \in \globe} \realis{U} \times X(U);
\end{equation*}
we call this the geometric realisation of constructible polygraphs.

This functor has a right adjoint $S: \cghaus \to \cpol$, defined on a space $X$ by
\begin{equation*}
	SX(-) := \mathrm{Hom}_{\cghaus}(\realis{-}, X);
\end{equation*}
we call $SX$ the \emph{singular constructible polygraph} of the space $X$.
\end{cons}

\begin{prop}
The functor $\realis{-}: \cpol \to \cghaus$ is monoidal from $(\cpol, \tensor, 1)$ to $(\cghaus, \times, 1)$ and from $(\cpol, \join, \emptyset)$ to $(\cghaus, \join, \emptyset)$.
\end{prop}
\begin{proof}
The functor $\realis{-}$ is compatible with lax Gray products and joins on $\globe$, and the argument of Lemma \ref{lem:restrictmonoidal} applies.
\end{proof}

\begin{thm} \label{thm:georegpoly}
Let $X$ be a constructible polygraph. Then the topological space $\realis{X}$ admits the structure of a CW complex, whose generating $n$\nbd cells are indexed by the $n$\nbd cells of $X$.
\end{thm}
\begin{proof}
Since $\realis{-}$ is a left adjoint, it takes the sequence of pushouts of Proposition \ref{prop:polygraph_ext} to a sequence of pushouts
\begin{equation*}
\begin{tikzpicture}[baseline={([yshift=-.5ex]current bounding box.center)}]
	\node[scale=1.25] (0) at (0,2) {$\displaystyle \coprod_{x\in\atom{n}{X}} \realis{\bord{}{} U(x)}$};
	\node[scale=1.25] (1) at (3.5,2) {$\displaystyle\coprod_{x\in\atom{n}{X}} \realis{U(x)}$};
	\node[scale=1.25] (2) at (0,0) {$\realis{\skel{n-1}{X}}$};
	\node[scale=1.25] (3) at (3.5,0) {$\realis{\skel{n}{X}}$,};
	\draw[1c] (0) to node[auto,swap] {$(\realis{\bord{}{}x})_{x\in\atom{n}{X}}$} (2);
	\draw[1c] (1) to node[auto] {$(\realis{x})_{x\in\atom{n}{X}}$} (3);
	\draw[1cinc] (0) to (1);
	\draw[1cinc] (2) to (3);
	\draw[edge] (2.5,0.2) to (2.5,0.8) to (3.3,0.8);
\end{tikzpicture}
\end{equation*}
and by Theorem \ref{thm:regularball} the $\realis{U(x)}$ are homeomorphic to $D^n$ and the $\realis{\bord{}{}U(x)}$ to $\bord{}{}D^n$. By induction on $n$\nbd skeleta, we see that this gives $\realis{X}$ the appropriate structure of a CW complex.
\end{proof}

config: arxiv


TITLE: Using Chernoff bound to analysis the Lazyselect algorithm
QUESTION [0 upvotes]: It's my homework of the course of randomized algorithm. In the textbook (Randomized Altorithm by Rajeev Motwani et.al.), the author analyzed this algorithm using Chebyshev bound, but are there any ideas on how to analyze this algorithm using Chernoff bound?
  The LazySelect algorithm is presented as follows:
Input: A set of n elementss $S$ from a totally ordered universe, and an integer $k$ in$[1,n]$
output: The $k$th smallest element of $S,S(k)$

Pick $n^{3/4}$ elements from S,chosen independently and uniformly at random with replacement; call this multiset of elements R.
Sort R in $O($$n^{3/4}logn$$)$ steps using any optimal sorting algorithm.
Let $x=kn^{-1/4}$. For $l=max\{x-\sqrt n,1\}$ and $h=min\{x+\sqrt n,$$n^{3/4}$$\}$, let $a=R_l$ and $b=R_h$. By comparing a and b to every element of S, determine $r_s(a)$ and $r_s(b)$.
if $ k<n^{1/4}$, then $P=\{y∈S|y≤b\}$
else if $ k>n-n^{1/4}$ ,let $P=\{y∈S|y≥a\}$;
else if $k∈[n^{1/4},n-n^{1/4}]$, let $P=\{y∈S|a≤y≤b\}$;
Check whether $S(k)∈P$ and $|P|≤4n^{3/4}+2$. If not, repeat Steps 1-3 until such a set P is found.
By sorting $P$ in $O(|P|\log|P|)$ steps, identify $P_{k-r_s(a)+1}$.

REPLY [1 votes]: Using Chernoff bound to analyze this question is easy, you can just replace the Chebyshev method with Chernoff bound method. In fact, we can also decrease the number of samples for example at $O(\sqrt{n})$ samples. Poor English, forgive me!


TITLE: trying to use the triangle inequality, with vectors
QUESTION [1 upvotes]: how can one show that
$(\bf{u}\cdot\bf{v})^2 \leq |\bf{u}|^2  |\bf{v}|^2$ and hence $|\bf{u}\cdot\bf{v}| \leq |\bf{u}| |\bf{v}|$
by using the inequality $|\bf{u}|^2 + 2x(\bf{u}\cdot\bf{v}) +x^2|\bf{v}|^2\geq 0$.
To use this inequality i tried to show that if $a+bx+cx^2 \geq 0$ for any real number x then $b^2 - 4ac \leq 0$ and to show this
If $y=a+bx+cx^2 \geq 0$ then $y|_{\min}\ge 0$. $$x_{min}=\frac{-b}{2c}$$ therefore $ y(x_{min})\ge0$. If we calculate  $$y\left(\frac{-b}{2c}\right)$$ we will obtain ${b^2 - 4ca}\leq 0$.
but i dont know how to use it in this context.

REPLY [1 votes]: I think you are overthinking this.
Plug the values for $a,b,c$ back into $b^2-4ac\leq 0$ and see what you have. (It is right in front of you!)


TITLE: Showing that the complex series $\sum_{n=1}^\infty \frac{z^n}{1+z^n}$ converges
QUESTION [0 upvotes]: Show the series $$\sum_{n=1}^\infty \frac{z^n}{1+z^n}$$ converges absolutely and uniformly in $$\{z: |z|\le r\}$$ for $r<1$.
Can I use the M-test here and does uniform convergence imply absolute convergence?

REPLY [2 votes]: $$\left|\frac{z^n }{1+z^n}\right|\leq \frac{|z|^n }{1-|z|^n}\leq\frac{r^n}{1-r}$$
But $$\sum_{n=0}^{\infty}\frac{r^n}{1-r} =1$$
therefore by Weierstrass test the series converges uniformly and absolutely in the set $$\{z: |z|\leq r\}.$$


TITLE: Variational Calculus Optimization Problem with x-independent Lagrangian Solution via Beltrami Identity
QUESTION [0 upvotes]: Optimize $$\int_0^1 y^2 (y')^2 dx$$
subject to $y(0)=0, y(1)=1$

For x-independent Lagrangians it is easier to use Beltrami Identity: $$ F-y'{\frac{\partial F}{\partial y'}}=C $$

REPLY [1 votes]: According to Beltrami's formula
$$
y^2(y')^2-y'(y^2 2 y')= -y^2(y')^2= C
$$
then
$$
\left(\frac 12 \frac{d}{dx}y^2\right)^2= C_1
$$
or
$$
\frac 12 \frac{d}{dx}y^2 = C_2
$$
and
$$
\frac 13 y^3=C_3x+C_4
$$
and with the boundary conditions
$$
y = \sqrt[3]{x}
$$


TITLE: Fundamental Theorem of Affine Transformations in 3D
QUESTION [1 upvotes]: Fundamental Theorem of Affine Transformations states that "Given two ordered sets of three non-collinear points each, there exists a unique affine transformation $f$ mapping one set onto the other."
There are several proof of the same but all of them consider collinear points in 2D. By that I mean that both the ordered sets of non-collinear points lies in the same plane.
I am wondering if things will change if the the ordered sets of non-collinear points lies in different planes Or the theorem is true for all pair of three non-collinear points.

REPLY [1 votes]: Let $V$ be a vector space of dimension $n$ and assume that $a_0,a_1,\ldots,a_n$ are affine Independent, i.e., the vectors $a_1-a_0,\ldots,a_n-a_0$ are linearly independent. Then given arbitrary points $b_0,b_1,\ldots,b_n\in V$ there is a unique linear mapping $g\colon V\to V $ such that $g(a_i-a_0)=b_i-b_0$ for all $1\leq i\leq n$. Then $f$ defined by $f(x):=g(x-a_0)+b_0=g(x)+b_0-g(a_0)$ is a (in fact the unique) affine mapping satisfying $f(a_i)=b_i$ for all $0\leq i\leq n$.


text:
\subsection{Computing of coordinate systems.}
In this section we assume that the graph $\mathcal G=\graph{V}{E}$ is a combinatorial representation of $f$ resulting from Algorithm~\ref{alg:enclosure}. We present a heuristic method for computing of coordinate systems $C_V$ at the vertex $V$,  for $V\in \mathcal V$. We assume that at the beginning the coordinate systems are not set, which we will denote by $C_V$=NULL.

The method consists of the following steps.

\vskip\baselineskip
\noindent\textbf{Step 1. Finding cycles in the graph.} First we try to find (non-rigorously) as much as possible of periodic points of $f$ in $\supp{\mathcal V}$. This is important, because given a period $p$ point, say $x$, a very natural choice of a coordinate system in a vertex containing this point is the Jordan basis of $Df^p(x)$. Of course it may happen that several periodic points belong to the same vertex. In this case we are choosing one from these points of the lowest principal period.

More formally; let us fix a positive integer $N$. This is an upper bound for the highest period of orbits we will search for. Since $\mathcal G$ is a combinatorial representation of $f$, all periodic points of $f$ must belong to some cycles in the graph $\mathcal G$. Let $\mathcal P_i$, $i=1,\ldots N$ be a set of vertexes $V$ such that there exists an $i$-periodic cycle in the graph $\mathcal G$ through $V$ and no lower period cycle containing $V$ exists. The sets $\mathcal P_i$ can be computed by means of standard graph algorithms and we omit details here.

\vskip\baselineskip
\noindent\textbf{Step 2. Refining cycles to periodic points.}
In this step we compute the sets of points $P_i$, $i=1,\ldots,N$ that are good approximations of periodic points.
For $V\in\mathcal P_i$ we proceed as follows. Let $x$ be an approximate center of $V$. The point $x$ is a seed point for the standard Newton method for zero finding problem applied to the map $g(x)=f^i(x)-x$. If the Newton method converges to a point $y\in \supp{\mathcal V}$ which is of principal period $i$ then we insert the point $y$ to the set $P_i$.

We would like to comment here, that we do not try to find all periodic orbits of low periods for the map $f$. Using our approach one can detect most of them if the set $\mathcal V$ consists of small enough sets. Clearly, a larger number of detected periodic points is better for setting the coordinate systems but the task of finding all low period orbits is not a goal for us.

\vskip\baselineskip
\noindent\textbf{Step 3. Setting coordinate systems in a neighborhood of periodic points.}
The goal of this step is to assign a coordinate system to vertexes that contain computed approximate periodic point.
For $y\in P_i$ we proceed as follows. Let $V\in\mathcal V$ be a vertex such that $y\in V$. Such a vertex exists since we know that $y\in\supp{\mathcal V}$ -- as guaranteed by the previous step.
\begin{itemize}
\item If the vertex $V$ has already assigned a coordinate system then we proceed with the next point from $P_i$.
\item We compute an approximate derivative $Df^i(y)$ and an approximate matrix $M_y$ of normalized  eigenvectors. The columns of $M_y$ are sorted by decreasing absolute value of the associated eigenvalue.
\item We assign to the vertex $V$ a coordinate system $V_C = M_y$.
\end{itemize}
We would like to comment here that the sets $P_i$ are proceeded by increasing index $i$. This guarantees that a vertex $V$ will have assigned coordinate system from some periodic point of the lowest principal period. Note, the resulting coordinates depend on the order chosen in the set $P_i$.

\vskip\baselineskip
\noindent\textbf{Step 4. Spreading coordinates from periodic points.}
The main idea is to propagate the coordinate systems from periodic points by the action of the derivative. It is well known that the direct propagation of coordinates by $Df$ in floating point arithmetic usually results in collapsing of these coordinates to a singular matrix. Therefore we will involve some orthonormalization process.

The Algorithm~\ref{alg:spreadCoordinates} describes the procedure.
\begin{algorithm}[htbp]\linesnumbered\label{alg:spreadCoordinates}

 \caption{spread coordinates from periodic points}
  \KwIn{$\mathcal G=\graph{V}{E}$ : graph,\newline
         $\mathcal S$ : subset of $\mathcal V$,\newline
         $k>1$ : integer
         }
  \KwResult{$\{C_V\}_{V\in\mathcal V}$ - coordinate systems at each vertex in the graph $\mathcal G$}
  \KwData{
  $u$ : vector;\newline
  $M$ : stack of float matrices;\newline
  $C$ : matrix;\newline
  $\mathcal P$ : subset of $\mathcal V$;\newline
  $i$ : integer
  }
  \Begin{
  \Repeat{$S=\emptyset$}
  {
    $\mathcal P\leftarrow \emptyset$\;
   \ForEach{$V\in \mathcal S$}{
        $u\leftarrow \mathrm{centre}(V)$\; // we will save on the stack derivatives along the trajectory\;
        $C\leftarrow C_V$\;
        \For{$i=1$ to $k$}{
          push($Df(u)$,$M$)\;
          $C\leftarrow Df(u)\cdot C$\;
          $u\leftarrow f(u)$;\
        }
        // then we compute coordinate system at the image\;
        $C\leftarrow \mathrm{GrammSchmidtOrthonormalization}(C)$\;
        \For{$i=1$ to $k-1$}{
          $C\leftarrow \left(\mathrm{top}(M)\right)^{-1}\cdot C$\;
          pop($M$)\;
        }
        pop($M$); // to free the memory\;
        $C\leftarrow\mathrm{normalizeColumns}(C)$\;
        \ForEach{$W\in\mathrm{out}(V,\mathcal G)$}
        {
          \If{$C_W\neq$NULL}{
            $C_W\leftarrow C^{-1}$\;
             $\mathcal P\leftarrow\mathcal P\cup\{W\}$\;
          }
        }
    }
    $\mathcal S\leftarrow\mathcal P$\;
  }
  }
\end{algorithm}

\begin{lemma}\label{lem:spreadCoordinates}
 Assume the Algorithm~\ref{alg:spreadCoordinates} is called with its arguments $\mathcal G=\graph{V}{E}$, $\mathcal S\neq\emptyset$ and $k>1$. Assume $C_V\neq NULL$ for $V\in \mathcal S$. If $\mathcal G$ consists of only one strongly connected component then the algorithm stops and $C_V\neq NULL$ for all $V\in\mathcal V$.
\end{lemma}
\begin{proof}
  We will prove that the algorithm always stops. Assume it is not the case. Then in each iteration of the main loop the set $\mathcal P$ is non-empty. But this set contains (lines 3 and 19-22) exactly these vertexes for which the coordinate system is assigned in the current iteration of the loop. Hence, the set of vertexes at which we have assigned the coordinate system will increase in each iteration. This contradicts the fact, that $\mathcal V$ is a finite set.

  We will prove that the algorithm sets coordinates in each vertex. Assume it is not the case.
  Take $\mathcal G^T=(\mathcal V',\mathcal E')$. Let $V$ be a vertex such that $C_V$=NULL. Then for $U\in\mathrm{out}(V,\mathcal V')$ holds $C_U=NULL$. Otherwise, $U$ will belong to the set $\mathcal S$ in some iteration and then the coordinates will be set in $V$. Denote $\mathcal V_1=\mathrm{out}(V,\mathcal G^T)$.

   In a recurrent way we can define $$\mathcal V_k = \bigcup\left\{\mathrm{out}(U,\mathcal G^T) : U\in \mathcal V_{k-1}\right\}.$$ The same reasoning proves that $C_U$=NULL for $U\in\mathcal V_k$.

  Since $\mathcal G$ consists of only one strongly connected component, $\mathcal G^T$ does too. Hence, for some $K>0$ we have $V_K=\mathcal V$ which contradicts the assumption that $C_U\neq$NULL for $U\in\mathcal S$ and $\mathcal S\neq\emptyset$.
\end{proof}

The input argument $\mathcal S$ is the set of vertexes that contain periodic points and for which the coordinate systems were already computed in Step 3. As we have seen, lines 5-18 are not important for the correctness of the Algorithm~\ref{alg:spreadCoordinates}. In these lines a heuristic method for computing of coordinate systems is proposed. We propagate the actual coordinate system $C_V$ by the derivative along a short part ($k$ iterations) of the trajectory of the center of $V$. Then we perform orthonormalization of columns and propagate the obtained matrix backwards $k-1$ times. This heuristic can work because for the inverse map the stable directions are attracting for $Tf$ and after few steps we get some approximation of stable directions at $f(\mathrm{centre}(V))$.

In this step we also assumed that the graph $\mathcal G$ consists of only one strongly connected component. It is not a restriction, since the method is dedicated for attractors. In practice, Algorithm~\ref{alg:enclosure} always returns such a graph if the initial set $V$ is chosen carefully.  Formally, it is easy to verify that a graph consists of one strongly connected component using for example Tarjan's algorithm \cite{Ta}.

\begin{remark}
The result of the algorithms described in steps 2,3,4 depend on the order chosen to index the sets of vertexes and points ($\mathcal P_i$, $P_i$  and $\mathcal S$).
Moreover, these algorithms are not deterministic, when implemented in parallel mode. The coordinate system at each vertex is computed by the first thread that reaches this vertex.

Anyway, the conclusion of Lemma~\ref{lem:spreadCoordinates} holds; we have computed some coordinate system at each vertex and we can start verification of the cone conditions.
\end{remark}


TITLE: Convergence almost uniformly implies convergence in $L_\infty$
QUESTION [0 upvotes]: Show that if $f_n \rightarrow f$ almost uniformly then $f_n \rightarrow f $ in $L_\infty$.

I know that convergence in $L_\infty$ implies converge uniformly and hence almost surely but I have problems to show the converse, because I not see how construct sets of 0 measure such that in our complements the $|f_n(x)-f(x)|< \varepsilon$.
Any hint or help I will be very grateful.

REPLY [2 votes]: This is false. $\chi_{(0,\frac  1 n)}$ is a counter-example.


TITLE: Mordell curves with many integral points
QUESTION [5 upvotes]: For $k\in{\mathbb Z},k\neq 0$, denote by $f(k)$ the number of integral points on the Mordell curve $y^2-x^3=k$. According to the data at http://tnt.math.se.tmu.ac.jp/simath/MORDELL , the largest value of $f$ on the interval $[-10000,10000]$ is 32, attained for $k=1025$.
Is it known/conjectured whether $f$ can take arbitrarily high values ?

REPLY [6 votes]: The largest value for $f(k)$ that I know of is an example due to Noam Elkies with
$k = 509142596247656696242225$, where there are (at least) $125$ pairs of solutions (so $f(k)=250$ in your notation). If ranks of elliptic curves over the rationals are absolutely bounded, then so is $f(k)$ (as long as one restricts to $6$th power free values of $k$ to avoid trivial scaling). Nothing is provable at this point, however.


TITLE: If $X$ is a Banach space any capable conditions to make sure that $X$ is also Hilbert?
QUESTION [1 upvotes]: Let $X$ be a Banach space and suppose that $X$ is isometric with a Hilbert space.Is it true to say that $X$ is also a Hilbert space?If I assume that $X$ is isomorphic with a Hilbert space is it still true?

REPLY [2 votes]: YES for the first question, and NO for the second. If $T$ is an isometry from $X$ and $Y$ then $ \langle Tx, Ty \rangle=\langle x, y \rangle$ gives an inner product on $Y$ which makes it a Hilbert space.  $\mathbb R^{2}$ with any norm  is isomorphic to $\mathbb R^{2}$ with the Euclidean norm . Take the norm $\|(x,y)\|=|x|+|y|$ for a counter-example to  the second statement. I leave it to you to verify that this norm does not satisfy Parallelogram Law which means that the space is not an inner product space. For Parallelogram Law see https://en.wikipedia.org/wiki/Parallelogram_law


TITLE: $\int_{0}^{x} f(t) dt \ge {f(x)}$ holds or not on $[0,1]$
QUESTION [12 upvotes]: Let $$f\colon [0,1]\rightarrow [0, \infty )$$  be  continuous. Suppose that
$$\int_{0}^{x} f(t)\,\mathrm dt \ge {f(x)} \quad\text{for  all }x\in[0,1].$$
Then
$A.$ No  such  function  exists.
$B.$ There  are  infinitely  many  such  functions.
$C.$ There is  only  one  such  function
$D.$ There  are  exactly  two  functions.
Now  I  was  thinking,  since $\int_{0}^{x} f(t) dt$  is  the  area  under  the  graph  of  $f(t)$  from $0$  to  $x$ so it  can  be  easily  made  greater  than  the  value  of  $f(t)$  at  one  point, namely  $x$.  So  there   will  be  infinitely  many  such  functions  satisfying  this  condition.
Is  my  reasoning  correct or  not?
Thanks.

REPLY [9 votes]: Let us modify the problem slightly.
Assume that the function
$$
f:[0,a]\to[0,\infty)
$$
is continuous and that
$$
\int_{0}^{x}f(t)\, dt \ge f(x) \quad \text{for all } x\in[0,a].
$$
Then $f(x) = 0$ for all $x\in[0,a]$.
Proof. We can rewrite the inequality as
$$
\dfrac{d}{dx}\left(e^{-x}\cdot\int_{0}^{x}f(t)\, dt\right) \le 0 \quad \text{for all } x\in[0,a].
$$
Consequently the function
$$
x\mapsto e^{-x}\cdot\int_{0}^{x}f(t)\, dt
$$
is decreasing and it takes its maximum at $x=0$.
Thus
$$
e^{-x}\cdot\int_{0}^{x}f(t)\, dt \le 0 \Leftrightarrow \int_{0}^{x}f(t)\, dt \le 0.
$$
However $f$ is non-negative and continuous. We conclude that
 $f(x) = 0$ for all $x\in[0,a]$.


TITLE: Given $\forall X\, p(X)$, use the Fitch System to prove $\lnot \exists X\, \lnot p(X)$
QUESTION [1 upvotes]: I've tried to solve this exercise based on a similar question that was asked some years ago, but I'm stuck in step 5.
Any help? Thanks in advance.
By the way I'm using Stanford's system.
1.  AX:p(X)            Premise
2.  ~p(X)              Assumption
3.  EX:p(X)            Assumption
4.  ~p(X)              Reiteration: 2
5.  EX:p(X) => ~p(X)   Implication Introduction: 3, 4

REPLY [3 votes]: The idea is that you assume $\exists x \,\lnot P(x)$ and show that it leads to a contradiction with the premise $\forall x \, P(x)$, by means of  the rules for the elimination of existential and universal quantifiers. This means that the negation of the assumption  $\exists x \,\lnot P(x)$ holds.
This intuition is a little watered-down in Stanford's platform because its rules for negation introduction and existential elimination are slightly clumsy and unusual for natural deduction.
Anyway, the correct formalization of a proof of $\lnot \exists x \, \lnot P(x)$ from $\forall x \, P(x)$ in the Fitch-style natural deduction system implemented by Stanford's platform is the following:


TITLE: Closed surfaces of prescribed mean curvature
QUESTION [1 upvotes]: Let $D\subset\mathbb R^n$ be a smoothly bounded open domain and $0\in D$. For any $x\in\partial D$ there holds
\begin{eqnarray*}
2 \,a'(\vert x\vert)\,(x\cdot\nu(x))+(n-1)\,a(\vert x\vert) \, H(x) = \text{const.}
\end{eqnarray*}
Here $a:(0,\infty)\to (0,\infty)$ is a positive smooth function, $a'$ denotes its derivative, $\nu(x)$ is the outer unit normal vector in $x\in\partial D$ and $(n-1)H(x)$ denotes the mean curvature in $x\in\partial D$. Does this imply that $D$ is a ball?

REPLY [3 votes]: If you let $\partial D$ be a hypersurface of revolution whose generatrix $\gamma: I \to [0,\infty) \times \mathbb{R}$ is such that $|\gamma|$ is monotonic, you then have $H(x)$ and $x\cdot \nu(x)$ can both be written as functions of $|x|$ only. One can further arrange that $x\cdot \nu(x) > \epsilon > 0$ for all positions.
Let $G(|x|) = H(x) / x \cdot \nu(x)$.
If you choose $a(s)$ to be any solution of
$$ (\ln a(s))' = \frac{1-n}{2} G(s) $$
your desired equation is satisfied with constant 0.


TITLE: The difference between consecutive primes in arithmetic progressions
QUESTION [9 upvotes]: Let $\pi(x)=\sum_{p\leq x}$ denote the prime counting function. A well known result of Baker, Harman, and Pintz on prime gaps states that for $x\geq y\geq x^{0.525}$ we have that
$$\pi(x+y)-\pi(x)\gg \frac{y}{\log x}.$$
Let $\pi(x;q,a)$ denote the number of primes $p\leq x$ such that $p\equiv a\pmod{q}$. Is there an analogous version of this theorem for $\pi(x;q,a)$?

Question: What is the smallest growing function $f(x,q)$ such that for any $x\geq y\geq f(x,q)$ we have that $$\pi(x+y;q,a)-\pi(x;q,a)\gg \frac{y}{\log x}?$$

Corollary 13.8 of Montgomery and Vaughn's Multiplicative Number Theory states that under Generalized Riemann Hypothesis $$\psi(x;q,a)=\sum_{\begin{array}{c}
n\leq x\\
n\equiv a\ (q)
\end{array}}\Lambda(n)=\frac{x}{\phi(q)}+O\left(x^{1/2}\log^{2}x\right).$$
From this Corollary, it follows that under GRH the desired lower bound holds for $$f(x,q) =  \phi(q)x^\frac{1}{2}\log^{2}(x).$$
I'm aware of a 1933 result of Heilbronn (MR1545353), which I believe (my German isn't great) proves that for fixed $q,a$, there exists $\delta>0$ such that for $x\geq y\geq x^{1-\delta}$ $$\pi(x+y;q,a)-\pi(x;q,a)\sim \frac{y}{\phi(q)\log x}.$$
Are there any more recent results on this topic?

REPLY [5 votes]: Unconditionally, one can take $f(x,q)=x^{0.525}$, provided that $q$ is in the Siegel-Walfisz range $q\leq (\log x)^A$.  There exists a (suitably small) constant $\delta>0$ such that one can take $f(x,q)=x^{0.525}$ for all except a density zero subset of $q\leq x^{\delta}$.
This follows from a certain Bombieri-Vinogradov-type result for a carefully constructed lower bound for the indicator function of the primes (due to  Kumchev, but with $0.525$ replaced by $0.53$) combined with the second moment bound for $L(s,\chi)^2$ times a Dirichlet polynomial (due to Harman, Watt, and Wong).  Full details on how to combine these two results can be found in the work of Alweiss and Luo (see Theorem 2.3 therein).  One then takes $Q = (\log x)^A$.


TITLE: Do we suppose that $y$ is the variable at which $f$ is differentiable in $\mathbb{R}$ ?
QUESTION [1 upvotes]: In my notes there is the following:
The solution of the problem
$$u_{tt}-c^2u_{xx}=0, x \in \mathbb{R}, t>0 \\ u(x, 0)=f(x), x \in \mathbb{R} \\ u_t(x,0)=g(x)$$
is $$u(x,t)=\frac{1}{2}[f(x+ct)+f(x-ct)]+\frac{1}{2c}\int_{x-ct}^{x+ct}g(s)ds$$ where $f \in D^2(\mathbb{R}), g\in D^1(\mathbb{R})$
When we want to differentiate $f(x+ct)$ in respect to $t$ do we set $y=x+ct$ and suppose that $y$ is the variable at which $f$ is differentiable in $\mathbb{R}$ ??

REPLY [0 votes]: Here we have that $\frac{\partial}{\partial t}(f(x+ct))=\frac{\partial}{\partial t}(x+ct)f'(x+ct)=cf'(x+ct)$, this obtained by the composite function differentiation rule $(f\circ g)'=(f'\circ g)g'$. You do not need a substitution, the fact that $f$ is differentiable on all of $\Bbb R$ is enough.


TITLE: There are infinitely many vectors such that $\|\mathbf u-(1,1,1)\|\le 3$ and $\|\mathbf u+(1,1,1)\|\le 3$
QUESTION [1 upvotes]: I want to prove that there are infinitely many solutions in $3$-space for $\|\mathbf u-(1,1,1)\|\le 3$ and $\|\mathbf u+(1,1,1)\|\le 3$ (where bold refers to vectors). My proof:

What is wrong with my current proof? If you can prove this question please  post your proof.

REPLY [0 votes]: You proved the following statement:

If $\|u-(1,1,1)\| \le 3$ and $\|u+(1,1,1)\| \le 3$, then $\|u\| \le \sqrt{6}$.

The converse of that statement:

If $\|u\| \le \sqrt{6}$ then  $\|u-(1,1,1)\| \le 3$ and $\|u+(1,1,1)\| \le 3$

is not true.
If you try to reverse your steps, you will see that $\|u\| \le \sqrt{6}$ implies $\|u-(1,1,1)\|^2+\|u+(1,1,1)\|^2 \le 18$. But this does not imply that $\|u-(1,1,1)\|^2 \le 9$ and $\|u+(1,1,1)\|^2 \le 9$.
As you pointed out by testing $u = (1,1,1)$, it is possible to have $u$ satisfy "$\|u-(1,1,1)\|^2+\|u+(1,1,1)\|^2 \le 18$" and not satisfy "$\|u-(1,1,1)\|^2 \le 9$ and $\|u+(1,1,1)\|^2 \le 9$".

To prove that infinitely many $u$ satisfy $\|u-(1,1,1)\| \le 3$ and $\|u+(1,1,1)\| \le 3$, first, show that there are infinitely many vectors $u$ such that $\|u\| \le 3-\sqrt{3}$. Then, by the triangle inequality, all of those vectors satisfy $\|u-(1,1,1)\| \le \|u\|+\|-(1,1,1)\| \le (3-\sqrt{3})+\sqrt{3} = 3$, and $\|u+(1,1,1)\| \le \|u\|+\|(1,1,1)\| \le (3-\sqrt{3})+\sqrt{3} = 3$. Therefore, there are infinitely many vectors $u$ which satisfy $\|u-(1,1,1)\| \le 3$ and $\|u+(1,1,1)\| \le 3$.


TITLE: Break RSA given a correct and faulty implementation
QUESTION [0 upvotes]: Suppose I have two machines, $A$ and $B$. $A$ encrypts a message $m$ and outputs the ciphertext $m^e \pmod n$. $B$ outputs $c$ such that $c = m^e \pmod p$ and $c = m^e + 1 \pmod q$. How can I use $A$ and $B$ to find $p$ and $q$? I am allowed to choose $m$ and $n$.

REPLY [3 votes]: HINT $\ $ Consider $\rm\ gcd(A(m)-B(m),\:pq)$


TITLE: unresolved partial fraction decomposition
QUESTION [4 upvotes]: I'm having some trouble doing this  partial fraction decomposition:
$$\frac{1}{t^3-2t+1}$$
using Ruffini rule i get:
$$\frac{1}{t^3-2t+1}= \frac{1}{(t-1)(t^2+t-1)}$$
i would like to decompose the previous result into partial fraction.
I did in this way:
$$\frac{1}{(t-1)(t^2+t-1)}=\frac{A}{t-1}+\frac{B}{t^2+t-1} \leftrightarrow$$
$$\leftrightarrow t^2A+t(A+B)+(-A-B)=1$$
comparing the coefficients i get the following system of equations:\begin{cases} A=0 \\ A+B=0 \\ -A-B=1 \end{cases}
that are not true..
what am i doing wrong?

REPLY [2 votes]: Notice, In general $$\frac{1}{(ax+b)(px^2+qx+r)}=\frac{A}{ax+b}+\frac{Bx+C}{px^2+qx+r}$$ Now, factorizing the expression, we have $$\frac{1}{(t-1)(t^2+t-1)}=\frac{A}{t-1}+\frac{Bt+C}{t^2+t-1}$$ $$\implies \frac{1}{(t-1)(t^2+t-1)}=\frac{A(t^2+t-1)+(Bt+C)(t-1)}{(t-1)(t^2+t-1)}$$ $$\implies (A+B)t^2+(A-B+C)t-(A+C)=1$$ Now, comparing the corresponding coefficients of both the sides, we get $$\begin{cases}
A+B=0\\
A-B+C=0\\
A+C=-1
\end{cases}$$ On solving the above three equations, we get $A=1, B=-1$ & $C=-2$
Hence, the required partial fractions are as follows $$\frac{1}{(t-1)(t^2+t-1)}=\frac{(1)}{t-1}+\frac{(-1)t+(-2)}{t^2+t-1}$$$$=\color{blue}{\frac{1}{t-1}-\frac{t+2}{t^2+t-1}}$$


TITLE: Using reduced mass to solve problems
QUESTION [0 upvotes]: A small block of mass $m$ rests on the bottom of a big box also of mass $m$. If the small block is then given a velocity $V$ to the right, how far has the box moved once the block has come to rest within the box. The coefficient of friction between the block and box is $\mu$ and all collisions are elastic.

My Attempt:
I tried to solve the question by setting the velocities of both objects equal then finding the acceleration of the box and using kinematics. When I tried to find the acceleration of the box, I thought it would be $\mu mg/m$. The answer says it should be $$\frac{\mu mg}{\frac{m^2}{2m}}$$ aka reduced mass of the system. I assume my method was wrong because of a non-inertial reference frame, but I'm not sure exactly why. Why does using the reduced mass give the correct answer? What does the reduced mass represent in this problem?

REPLY [0 votes]: Disclaimer: I'm a high school student
We use that formula because the box is co-dependant on the block inside it, so we can't look at it as a single-body problem. Everything that affects the block affects the box too, so we can look at it as a one-body problem, plus it's easier than doing extensive math that's required. I hope this help, and if I'm wrong please correct me


TITLE: Equivalence of categories of vector bundles and locally free sheaves
QUESTION [24 upvotes]: Let $X$ be a scheme. It is known that the category $\mbox{Vec}_r(X)$ of vector bundles of rank $r$ on $X$ and the category $\mbox{Loc}_r(X)$ of locally free sheaves of rank $r$ on $X$ are equivalent (The equivalence is given by the functor $F\colon \mbox{Loc}_r(X)\rightarrow \mbox{Vec}_r(X)$ such that $F(\mathcal{E})=\textbf{Spec}(\mbox{Sym}(\check{\mathcal{E}}))$, where $\mbox{Sym}(\mathcal{F})$ is the symmetric algebra of $\mathcal{F}$ and $\textbf{Spec}(.)$ is as defined in Hartshorne p.128.
My confusion is the following. In many books it is written that a locally free subsheaf $\mathcal{E}'$ of a locally free sheaf $\mathcal{E}$ does not always correspond to a sub-vector bundle of the vector bundle $F(\mathcal{E})$. However, since the categories are equivalent, the monomorphism $\iota\colon\mathcal{E}'\rightarrow\mathcal{E}$ maps to a monomorphism $F(\iota)\colon F(\mathcal{E}')\rightarrow F(\mathcal{E})$ in $\mbox{Vec}_r(X)$. So, I would expect to get a sub-vector bundle of the vector bundle $F(\mathcal{E})$.
A monomorphism in a category is not always an injective map in the usual sense. Is it maybe the point that $F(\iota)$ might not be an injective map, although it is a monomorphism?
PS: While I was trying to sort out this confusion, I looked in Hartshorne and was even more confused, because he does not define what a morphism of vector bundles is, although he defines what an isomorphism is (again on page 128). So my second question is what is the definition of a morphism of vector bundles in algebraic geometry?

REPLY [34 votes]: Let me tell the whole story of vector bundles because in many books this subject is not developed properly or only in an ad-hoc manner.
Definition. Let $X$ be a scheme. A pre-vector bundle on $X$ is an $X$-scheme $V$ such that for every $X$-scheme $T$ the set of $T$-valued points $V(T)=\hom_X(T,V)$ carries the structure of a module over $\mathcal{O}_T(T)$. This structure belongs to the data. For $X$-morphisms $T \to T'$ we require that $V(T') \to V(T)$ is linear over $\mathcal{O}_{T'}(T') \to \mathcal{O}_T(T)$. There is an obvious notion of a morphism of pre-vector bundles. Thus, we obtain a category. Every morphism $X' \to X$ induces a pullback functor $V \mapsto V|_{X'}$ (the underlying scheme is $V \times_X X'$) from pre-vector bundles on $X$ to pre-vector bundles on $X'$.
An example is $\mathbb{A}^n_X$ with the usual module structure on $\mathbb{A}^n_X(T) = \mathcal{O}_T(T)^n$. Any pre-vector bundle isomorphic to $\mathbb{A}^n_X$ for some $n$ is called trivial. A vector bundle over $X$ is a pre-vector $V$ bundle over $X$ which is locally trivial, i.e. there is an open covering $\{X_i \to X\}_i$ such that $V|_{X_i}$ is trivial for each $i$. We obtain a category of vector bundles.
If $E$ is a quasi-coherent module on $X$, then $\mathbb{V}(E):=\mathrm{Spec}(\mathrm{Sym}(\check{E}))$ is an affine $X$-scheme with $\mathbb{V}(E)(T) = \hom_{\mathcal{O}_T}(p^* \check{E},\mathcal{O}_T)$ for $p : T \to X$. This set has a canonical $\mathcal{O}_T(T)$-module structure, so that $\mathbb{V}(E)$ becomes a pre-vector bundle. Observe that $\mathbb{V}(\mathcal{O}_X^n)=\mathbb{A}^n_X$ is the trivial vector bundle. Hence, if $E$ is locally free, then $\mathbb{V}(E)$ is a vector bundle. Clearly this defines a functor $\mathbb{V}(-)$ from the category of locally free sheaves to the category of vector bundles over $X$.
Theorem. The functor $\mathbb{V}(-)$ is an equivalence of categories.
Proof. First, $\mathbb{V}(-)$ is fully faithful: Let $E,F$ be locally free sheaves on $X$. We claim that $\hom(E,F) \to \hom(\mathbb{V}(E),\mathbb{V}(F))$ is a bijection. Actually we have a morphism of sheaves $\underline{\hom}(E,F) \to \underline{\hom}(\mathbb{V}(E),\mathbb{V}(F))$ on $X$ and we claim that this is an isomorphism. This way we can work locally and thereby assume that $E,F$ are trivial, say $E=\mathcal{O}_X^n$ and $F=\mathcal{O}_X^m$.
Then $\underline{\hom}(E,F)$ becomes a sheaf of matrices $\mathcal{O}_X^{m \times n}$. The sheaf of $X$-morphisms $\mathbb{A}^n_X \to \mathbb{A}^m_X$ is $\mathcal{O}_X[t_1,\dotsc,t_n]^m$. One checks (!) that a morphism $\mathbb{A}^n_X \to \mathbb{A}^1_X$ corresponding to a polynomial in $\mathcal{O}_X[t_1,\dotsc,t_n]$ is a morphism of vector bundles iff this polynomial is linear. This shows $\underline{\hom}(\mathbb{V}(E),\mathbb{V}(F)) \cong (\mathcal{O}_X^m)^n$, which finishes the proof.
Secondly, $\mathbb{V}(-)$ is essentially surjective: Let $V$ be a vector bundle on $X$. Choose a cover $\{X_i \to X\}$ and a  trivialization $\phi_i : V|_{X_i} \cong \mathbb{V}(\mathcal{O}_{X_i}^n)$ (here $n$ may depend on $i$). By fully faithfulness, the resulting isomorphisms $\phi_{ij} := \phi_i |_{X_i \cap X_j} \phi_j^{-1}  |_{X_i \cap X_j}$ are induced by isomorphisms $\psi_{ij} : \mathcal{O}_{X_i \cap X_j}^n \cong \mathcal{O}_{X_i \cap X_j}^n$. Since $\phi_{ij}$ satisfy the cocycle condition, the same is true for the $\psi_{ij}$. Hence, we have a gluing data $(\mathcal{O}_{X_i}^n,\psi_{ij})$ and obtain a locally free sheaf $E$ such that $\mathbb{V}(E) \cong V$ by construction. $\square$
As already mentioned, every equivalence of categories preserves (and reflects) monomorphisms. In particular, a homomorphism of free sheaves of modules $\mathcal{O}_X^n \to \mathcal{O}_X^m$ is a monomorphism (in the category of locally free sheaves) if and only if the induced morphism of trivial vector bundles $\mathbb{A}^n_X \to \mathbb{A}^m_X$ is a monomorphism. However, monomorphisms of locally free sheaves should not be confused with monomorphisms in the larger category of all sheaves of modules, and monomorphisms of vector bundles should not be confused with monomorphisms of schemes, let alone morphisms of schemes whose underlying map of sets is injective, or injective on the fibers - which seems to be the usual ad-hoc definition of a subbundle.
For example, consider a regular global section $s$ of $\mathcal{O}_X$. Then $s : \mathcal{O}_X \to \mathcal{O}_X$ is a monomorphism of sheaves, hence also of locally free sheaves. The induced morphism of vector bundles $\mathbb{A}^1_X \to \mathbb{A}^1_X$ corresponds to the polynomial $s * t$. If $s$ is invertible, it is an isomorphism. If not, choose some $x \in X$ such that $s_x \in \mathfrak{m}_x$, i.e. $s(x)=0$ in $\kappa(x)$, so that on the fiber of $x$ we get the the zero morphism $\mathbb{A}^1_{\kappa(x)} \to \mathbb{A}^1_{\kappa(x)}$, whose underlying map of sets is not injective.


TITLE: Can an electric car be usefully powered (or at least supported) by solar energy?
QUESTION [0 upvotes]: There are cars on the road that run on electricity instead of gasoline.  And that's great, it really is.  But you have to charge them, which is a problem if there are no charging stations in your area.
But what if your car could charge itself with the sun?  It would be refuelling while you parked, or even while you're driving.  My question: would it be practical?  More specifically, let's assume the following:

The car's feature set, weight, and charging time are comparable to that of a Tesla Model 3
The car can still be charged with conventional charging stations
The car can be designed around solar panels (as opposed to haphazardly grafting them on)

Given this, can an electric car be powered with solar energy in a manner that is practical, either as a primary source or as a supplement?

REPLY [3 votes]: As far as I know, this is your current state of solar-powered cars:

(See Wikipedia: Solar Car)
They're designed completely around the solar panel, and they're not really sedans.

REPLY [1 votes]: what if your car could charge itself with the sun?

It would take a long time to charge.  My plug-in hybrid (Cheverolet Volt) requires a modest 3.8 kilowatts for charging, and it can go about ten miles for every hour that it spends on the charger.
I'm not totally up-to-date with the efficiency of modern solar cells, but even if they were 100% efficient (they're not), and even if I covered every non-glass square inch of the car with solar cells, and even if I lived in the tropics; the best I could hope for would be maybe two or three hundred Watts of power at mid-day.
A Tesla, "standard" charger requires almost ten thousand Watts, and the "fast" charger needs even more.
You can charge a car from solar power, you just can't carry the charging station around with the car itself.


TITLE: continuous random variable example
QUESTION [0 upvotes]: Let $x$ be a random variable with the following distribution function $f$:
$$f(x) = \begin{cases} \dfrac{1}{2x}, & \text{ if } 0 \leq x \leq 2 \\
                  0, & \text{ elsewhere }
         \end{cases}$$
P(1 <= X <= 1.5) = limit 1.5 and 1, 1/2x dx = 1/4 dx^2 ] 1.5 and 1 = 5/16
I do not get how they got 5/16?

REPLY [1 votes]: The shaded area is a trapezoid. The two bases of the trapezoid have lengths of $f(1) = \dfrac{1}{2}$ and $f(1.5) = \dfrac{3}{4}$. The height of the trapezoid is $1.5-1 = \dfrac{1}{2}$.
So the area is $A = \dfrac{1}{2}(b_1+b_2)h = \dfrac{1}{2}\cdot\left(\dfrac{1}{2}+\dfrac{3}{4}\right)\cdot\dfrac{1}{2} = \dfrac{5}{16}$.


TITLE: Does this Euler product have a name?
QUESTION [2 upvotes]: Let
$$\displaystyle f(s) = \prod_p \left(1 - \frac{1}{2p^s}\right), \Re(s) > 1.$$
This Euler product converges. Does it have a name?

REPLY [0 votes]: It could help to visualize this product by rewriting it as $$f(s)=\prod_{p} (1-q^{-s})$$ where $$q=2p$$
Recall that $$\frac{1}{\zeta(s)}=\prod_{p} (1-p^{-s})=\sum_{n=1}^\infty \frac{\mu(n)}{n^s}$$
So if you expand the product into a sum you should get an answer very similar to that, which is $$\sum_{n=1}^\infty \frac{\mu(n)2^{-\omega(n)}}{n^s}$$
Where $\omega(n)$=amount of prime factors of n.
This looks to be similar to the identity $$\frac{\zeta^2(s)}{\zeta(2s)}=\sum_{n=1}^\infty \frac{2^{\omega(n)}}{n^s}$$
but I could not find any way to relate your function to the above sum.


TITLE: Given polynomial $P(x) = x^2 + ax + b$ and that there only exists one $c$ such that $P^2(c) = c$. Calculate the minimum value of $a + b + c$.
QUESTION [2 upvotes]: Given polynomial $P(x) = x^2 + ax + b$. Knowing that there only exists one value of real $c$ such that $P^2(c) = c$, calculate the minimum value of $a + b + c$.

Notation: $P^2(x) = (P \circ P)(x)$.

We have that $P^2(c) = c \iff (c^2 + ac + b)^2 + a(c^2 + ac + b) + b = c$
$$\iff [c^2 + (a - 1)c + b][c^2 + (a + 1)c + a + b + 1] = 0$$
For the above equation to have only one solution, $$(a - 1)^2 - 4b, (a + 1)^2 - 4(a + b + 1) \le 0 \implies b \ge \frac{(1 - a)^2}{4}$$
$$\implies \left[c^2 + (a - 1)c + \frac{(1 - a)^2}{4}\right] \cdot \left[c^2 + (a + 1)c + a + \frac{(1 - a)^2}{4} + 1\right] \ge 0$$
$$\implies (a + 2c - 1)^2[(a + 2c + 1)^2 + 4] \ge 0$$
It could be implied that 'the only solution' is $c = \dfrac{1 - a}{2}$, where $b = \dfrac{(1 - a)^2}{4}$.
$$a + b + c = a + \frac{1 - a}{2} + \frac{(1 - a)^2}{4} = \frac{a^2 + 3}{4} \ge \frac{3}{4}$$
Is the above solution correct? If not, could you provide with one? Thanks for your attention.

REPLY [4 votes]: Just a less computational argument for the first part of the solution: if $P(x) = x$ had no real solutions, then $P(x)>x$ for all real $x$ and hence $P(P(x)) > P(x) > x$, contradicting the existence of $c$. Any solution of $P(x) = x$ is also a solution of $P(P(x)) =x$, hence $P(x) = x$ must have exactly one solution, $c$. That implies $P(x) = x+(x-c)^2$.
Therefore $a=1-2c$, $b= c^2$ and $a+b+c = c^2-c+1 = (c-\frac{1}{2})^2 + \frac{3}{4} \geq \frac{3}{4}$. Equality holds for $c=\frac{1}{2}$, $a=0$, and $b = \frac{1}{4}$.


TITLE: Does a conservativity conjecture imply the standard conjectures?
QUESTION [9 upvotes]: Does a conservativity conjecture (e.g. Conjecture 2.1 of http://user.math.uzh.ch/ayoub/PDF-Files/Article-for-Steven.pdf) imply the standard conjectures? Specifically I am confused with Beilinson's article  arXiV:1006.1116 "Remarks on Grothendieck’s standard conjectures", where it seems to show that the conservativity of a realization functor implies the standard conjectures on algebraic cycles.

REPLY [9 votes]: Here is a partial answer: In characteristic 0 it is known that conservativity + algebraicity (edit: modulo rational equivalence) of the Künneth projectors implies the remaining standard conjectures.
Indeed, if the Künneth projectors are algebraic, then one may use conservativity to show that the inverse of the Lefschetz operator is algebraic (i.e. Lefschetz standard conjecture). Once that is known, the hom=num standard conjecture also follows. For example, André's category of “motivated” motives would be equivalent to the categories of homological/numerical motives.