text:
\begin{document}

\title{Vector-valued covariant differential operators
for the M\"obius transformation}

\author{Toshiyuki KOBAYASHI,\footnote{Kavli IPMU (WPI)
and Graduate School of Mathematical Sciences,
The University of Tokyo, 3-8-1 Komaba, Meguro, Tokyo, 153-8914 Japan;
E-mail address: \texttt{toshi@ms.u-tokyo.ac.jp},} \;
Toshihisa KUBO,\thanks{Graduate School of Mathematical Sciences,
The University of Tokyo, 3-8-1 Komaba, Meguro, Tokyo, 153-8914 Japan;
E-mail address: \texttt{toskubo@ms.u-tokyo.ac.jp},}\;
Michael PEVZNER\thanks{
Laboratoire de Math\'ematiques de Reims,
Universit\'e de Reims-Champagne-Ardenne,
FR 3399 CNRS, F-51687, Reims, France;
E-mail address: \texttt{pevzner@univ-reims.fr}.}}

\date{}

\maketitle

\begin{abstract}
We obtain a family of functional identities satisfied by vector-valued
functions of two variables and their geometric inversions. For this we introduce particular differential operators of arbitrary order
attached to Gegenbauer polynomials. These differential operators are
symmetry breaking for the pair of Lie groups $(SL(2,\mathbb C), SL(2,\mathbb R))$ that arise from conformal geometry.

\end{abstract}

\medskip
\noindent
\textit{2010 MSC:}
Primary
          22E46;
Secondary
33C45, 53C35

\medskip
\noindent
\textit{
Key words and phrases\/}:
branching law, reductive Lie group,
symmetry breaking,
conformal geometry,
Verma module, F-method,
time reversal operator.

\setcounter{tocdepth}{2}

\tableofcontents

\section{A family of vector-valued functional identities}\label{sec:1}

Given a pair of functions $f$, $g$ on $\mathbb{R}^2\backslash \{(0, 0)\}$,
we consider
a $\mathbb{C}^2$-valued function
$\vec{F}:= \begin{pmatrix}f \\g \end{pmatrix}$.
Define its ``twisted inversion"
$ \vec{F}_\lambda$
with parameter $\lambda \in \mathbb{C}$ by
\begin{equation}\label{eqn:invF}
\vec{F}^\vee_\lambda(r\cos \theta, r\sin \theta)
:=r^{-2\lambda}
\begin{pmatrix}
\cos 2\theta & -\sin 2\theta\\
\sin 2\theta & \cos 2\theta
\end{pmatrix}
\vec{F}\bigg(\frac{-\cos\theta}{r}, \frac{\sin\theta}{r} \bigg).
\end{equation}
Clearly, $\vec{F} \mapsto \vec{F}^\vee_\lambda$
is involutive, namely,
$(\vec{F}^{\vee}_\lambda)^\vee_\lambda
= \vec{F}$.

\vskip 1pc
A pair of differential operators $\mathcal{D}_1$,
$\mathcal{D}_2$ on $\mathbb{R}^2$
yields a linear map
\begin{equation*}
\mathcal{D}: C^\infty(\mathbb{R}^2) \oplus C^\infty(\mathbb{R}^2)
\to C^\infty(\mathbb{R}),
\quad (f,g) \mapsto (\mathcal{D}_1f)(x,0) + (\mathcal{D}_2g)(x,0).
\end{equation*}
We write
\begin{equation*}
\mathcal{D}:=\Rest{y=0} \circ (\mathcal{D}_1, \mathcal{D}_2).
\end{equation*}

Our main concern in this article is the following:

\begin{question} \label{quest:A}
(1) For which parameters $\lambda, \nu \in \mathbb{C}$,
do there exist differential operators $\mathcal{D}_1$
and $\mathcal{D}_2$ on $\mathbb{R}^2$
with the following properties?
\begin{itemize}
\item $\mathcal{D}_1$ and $\mathcal{D}_2$ have constant coefficients.
\item For any $\vec{F} \in C^\infty(\mathbb{R}^2) \oplus C^\infty(\mathbb{R}^2)$,
the functional identity
\end{itemize}
\begin{equation}\label{eqn:A}
(\mathcal{D} \vec{F}^\vee_\lambda)(x)
= |x|^{-2\nu}(\mathcal D\vec{F})\left(-\frac{1}{x}\right),
\quad
\text{for $x \in \mathbb{R}^{\times}$} \tag{$\mathcal{M}_{\lambda,\nu}$}
\end{equation}
holds, where
$\mathcal{D} = \Rest{y=0} \circ
(\mathcal{D}_1, \mathcal{D}_2)$.
\vskip 0.05in

\noindent
(2) Find an explicit formula of such
$\mathcal{D} \equiv \mathcal{D}_{\lambda, \nu}$
if exists.

\end{question}

Our motivation will be explained in Section \ref{sec:2} by giving
three equivalent formulations of Question \ref{quest:A}.
Here are some examples of the
operators $\mathcal{D}_{\lambda, \nu}$ satisfying \eqref{eqn:A}.

\begin{example} \label{ex:QA}
(0) $\nu = \lambda$:
\begin{equation*}
\mathcal{D}_{\lambda, \nu}:= \Rest{y=0} \circ
\left(\text{id}, 0 \right),
\end{equation*}
namely,
\begin{equation*}
\mathcal{D}_{\lambda, \nu}\begin{pmatrix} f \\ g \end{pmatrix}(x)
=f(x,0)
\end{equation*}
satisfies \eqref{eqn:A} for $\nu = \lambda$.\\

\noindent
(1) $\nu = \lambda+1$:
\begin{equation*}
\mathcal{D}_{\lambda, \nu}:= \Rest{y=0} \circ
\left(\frac{\partial}{\partial x}, \lambda \frac{\partial}{\partial y} \right),
\end{equation*}
namely,
\begin{equation*}
\mathcal{D}_{\lambda, \nu}\begin{pmatrix} f \\ g \end{pmatrix}(x)
=\frac{\partial f}{\partial x} (x, 0) + \lambda \frac{\partial g}{\partial y} (x,0)
\end{equation*}
satisfies \eqref{eqn:A} for $\nu = \lambda+1$.\\

\noindent
(2) $\nu = \lambda+2$:
\begin{equation*}
\mathcal{D}_{\lambda, \nu}:= \Rest{y=0} \circ
\bigg(2 (2\lambda+1) \frac{\partial^2}{\partial x \partial y},
(\lambda-1)\frac{\partial^2}{\partial x^2} + (\lambda+1)(2\lambda + 1)
\frac{\partial^2}{\partial y^2} \bigg),
\end{equation*}
namely,
\begin{equation*}
\mathcal{D}_{\lambda, \nu}\begin{pmatrix} f \\ g \end{pmatrix} (x)=
2(2\lambda+1) \frac{\partial^2 f}{\partial x \partial y }(x, 0) +
(\lambda-1) \frac{\partial^2 g}{\partial x^2}(x, 0) +
(\lambda+1)(2\lambda+1)\frac{\partial^2 g}{\partial y^2}(x, 0)
\end{equation*}
satisfies \eqref{eqn:A} for $\nu=\lambda+2$.
\end{example}

Given $\mathcal{D} = \Rest{y=0} \circ
(\mathcal{D}_1, \mathcal{D}_2)$, define
\begin{equation}\label{eqn:Ddual}
\mathcal{D}^{\vee}:=\Rest{y=0}
\circ (-\mathcal{D}_2, \mathcal{D}_1).
\end{equation}
Clearly, $\mathcal{D}^\vee$ is determined only by
$\mathcal{D}$, and is independent of the choice of
$\mathcal{D}_1$ and $\mathcal{D}_2$.
Proposition \ref{prop:Ddual} below
shows that the map $\mathcal{D} \mapsto \mathcal{D}^\vee$
is an automorphism of the set of
the operators $\mathcal{D}$ such that
\eqref{eqn:A} is satisfied.

\begin{proposition}\label{prop:Ddual}
If $\mathcal{D}$ satisfies \eqref{eqn:A} for all $\vec{F}$,
so does $\mathcal{D}^{\vee}$.
\end{proposition}

\begin{proof}
For $\vec{F}=\begin{pmatrix} f \\ g \end{pmatrix}$,
we set
${}^\vee \! \vec{F}: =
\begin{pmatrix} g \\ -f\end{pmatrix}$.
Then we have
\begin{equation}\label{eqn:FF}
^{\vee\vee}\!\vec{F}
=-\vec{F},
\quad
\mathcal{D}\left({}^\vee\! \vec{F} \right) =
(\mathcal{D}^\vee)\vec{F},
\quad
({}^\vee\!\vec{F})^\vee_\lambda
={}^\vee\!(\vec{F}^\vee_\lambda).
\end{equation}

To see this we note that
$w :=\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$
commutes with
$\begin{pmatrix}
\cos 2\theta & -\sin 2\theta \\
\sin 2\theta & \cos 2\theta
\end{pmatrix}$
and that $\mathcal{D}^\vee$ and ${}^\vee\! \vec{F}$ are expressed as
$\mathcal{D}^\vee = \mathcal{D} w^{-1}$ and ${}^\vee \!\vec{F}
= w^{-1}\vec{F}$,
respectively. Therefore,
\begin{align*}
\left( \mathcal{D}^\vee \vec{F}^\vee_\lambda\right)(x)
&= \mathcal{D} \left({}^\vee \!\vec{F}\right)^\vee_\lambda(x)\\
&=|x|^{-2\nu}\left( \mathcal{D}\; {}^\vee \!\vec{F}\right)
\bigg( -\frac{1}{x} \bigg)\\
&=|x|^{-2\nu}\left( \mathcal{D}^\vee\vec{F}\right)
\bigg( -\frac{1}{x} \bigg),
\end{align*}
where the passage from the first line to the second one is justified by the fact that ${}^\vee\! \vec{F}$ satisfies \eqref{eqn:A}.
\end{proof}

In order to answer Question \ref{quest:A}
for general $(\lambda, \nu)$,
we recall that the Gegenbauer polynomial or
ultraspherical polynomial
$C^\alpha_\ell(t)$
is a polynomial in one variable $t$ of degree $\ell$ given by
\begin{equation*}
C^\alpha_\ell (t) =
\sum_{k=0}^{[\frac{\ell}{2}]} (-1)^k
\frac{\Gamma(\ell - k + \alpha)}{\Gamma(\alpha)\Gamma(\ell-2k + 1) k!}
(2t)^{\ell - 2k},
\end{equation*}
where $[s]$ denotes the greatest integer that does not exceed $s$.
Following \cite{KP}, we inflate $C^\alpha_\ell(t)$ to a polynomial of two variables by
\begin{equation}\label{eqn:Gtwo}
C^\alpha_\ell(s,t):= s^{\frac{\ell}{2}} C^\alpha_\ell
\bigg(\frac{t}{\sqrt{s}}\bigg).
\end{equation}
By formally substituting $-\frac{\partial^2}{\partial x^2}$
and $\frac{\partial}{\partial y}$ to $s$ and $t$
in $C^\alpha_\ell(s,t)$, respectively,
we obtain a homogeneous differential operator
$\mathcal{C^\alpha_\ell}:=
C^\alpha_\ell\left(-\frac{\partial^2}{\partial x^2}, \frac{\partial}{\partial y} \right)$
of order $\ell$ on
$\mathbb{R}^2$.
Here are the first four operators:
\begin{enumerate}
\item[] $\mathcal{C}^\alpha_0=\text{id}$,
\item[] $\mathcal{C}^\alpha_1 = 2 \alpha \frac{\partial}{\partial y}$,
\item[] $\mathcal{C}^\alpha_2 =
\alpha \left(-\frac{\partial^2}{\partial x^2}+ (\alpha + 1) \frac{\partial^2}{\partial y^2}\right)$,
\item[] $\mathcal{C}^\alpha_3 = \frac{2}{3} \alpha (\alpha +1)
\left( 3 \frac{\partial^3}{\partial x ^2 \partial y} + 2(\alpha + 2) \frac{\partial^3}{\partial y^3}\right)$.
\end{enumerate}

\begin{thmalph}\label{thm:A}
Suppose that $a:=\nu - \lambda$ is a non-negative integer.
For $a > 0$, we
define the following pair of homogeneous differential operators
of order $a$ on $\mathbb{R}^2$ by
\begin{align*}
\mathcal{D}_1 &: = a (2\lambda+a -1) \frac{\partial}{\partial x}\circ
\mathcal{C}^{\lambda+\frac{1}{2}}_{a-1}\\
\mathcal{D}_2 &:=\left(2\lambda^2 + 2(a-1)\lambda + a(a-1) \right)
\frac{\partial}{\partial y} \circ\mathcal{C}^{\lambda+\frac{1}{2}}_{a-1}\\
&\hskip 1.2in
+(\lambda-1)(2\lambda+1)
\left(\frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2}\right)
\circ\mathcal{C}^{\lambda+\frac{3}{2}}_{a-2}.
\end{align*}
For $a=0$, we set
\begin{equation*}
\mathcal{D}_1 := \emph{id}, \quad \mathcal{D}_2:=0.
\end{equation*}
Then $\mathcal{D}: = \emph{Rest}_{y = 0}
\circ(\mathcal{D}_1, \mathcal{D}_2)$
and $\mathcal{D}^\vee := \Rest{y=0} \circ
(\mathcal{D}_2, -\mathcal{D}_1)$ satisfy
the functional identity \eqref{eqn:A}.
Moreover, when $2\lambda \notin \{0,-1,-2, \cdots\}$,
there exists
a non-trivial solution to \eqref{eqn:A}
only if $\nu - \lambda$ is a non-negative integer.
and any differential operator satisfying
\eqref{eqn:A} is a linear combination of
$\mathcal{D}$ and $\mathcal{D}^\vee$.
\end{thmalph}

\textbf{Notation:} $\mathbb{N} := \{0, 1, 2, \ldots\}$\\
\phantom{a} \hskip 0.96in $\mathbb{N}_+ := \{1, 2, \ldots\}$

\section{Three equivalent formulations}\label{sec:2}

Question \ref{quest:A} arises from various disciplines
of mathematics. In this section we describe it in three equivalent ways.

\subsection{Covariance of $SL(2,\mathbb{R})$ for vector-valued functions}

For  $\lambda \in \mathbb{C}$, we define a group homomorphism
\begin{equation}\label{eqn:Fhl}
\psi_\lambda: \mathbb{C}^\times \to GL(2, \mathbb{R}),
\quad z=re^{i\theta} \mapsto r^\lambda
\begin{pmatrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{pmatrix}.
\end{equation}
For a $\mathbb{C}^2$-valued function $\vec{F}$ on
$\mathbb{C} \simeq \mathbb{R}^2$, we set
\begin{equation*}
\vec{F}^h_\lambda(z) : =
\psi_\lambda\left( (cz+d)^{-2} \right) \vec{F}
\bigg(\frac{az + b}{cz+d}\bigg)
\end{equation*}
for
$\lambda \in \mathbb{C}$,
$h^{-1} = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \in SL(2,\mathbb{R})$,
and $z \in \mathbb{C}$ such that $cz+d \neq 0$.

\begin{pquestion}\label{quest:A1}
(1)
Determine complex parameters $\lambda, \nu \in \mathbb{C}$
for which there exist differential operators
$\mathcal{D}_1$ and $\mathcal{D}_2$ on $\mathbb{R}^2$
with the following property:
$\mathcal{D}= \Rest{y=0} \circ
(\mathcal{D}_1, \mathcal{D}_2)$ satisfies
\begin{equation}\label{eqn:A2}
(\mathcal{D}\vec{F}^h_\lambda)(x)= |cx+d|^{-2\nu}
(\mathcal{D}\vec{F})\bigg(\frac{ax+b}{cx+d}\bigg)
\end{equation}
for all $\vec{F} \in
C^\infty(\mathbb{C}) \oplus C^\infty(\mathbb{C}),
h^{-1} = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \in SL(2, \mathbb{R})$,
and $x \in \mathbb{R} \setminus \{-\frac{d}{c}\}$.\\

\noindent
(2) Find an explicit formula of such $\mathcal{D} \equiv
\mathcal{D}_{\lambda, \nu}$.

\end{pquestion}

The equivalence between Questions \ref{quest:A} and \ref{quest:A1}
follows from the following three observations:
\begin{itemize}
\item The functional identity \eqref{eqn:A2}
for $h = \begin{pmatrix}1 & t \\ 0 & 1 \end{pmatrix} $
$(t \in \mathbb{R})$ implies that
$\mathcal{D}=\Rest{y=0} \circ
(\mathcal{D}_1, \mathcal{D}_2)$ is a translation invariant operator.
Therefore, we can take $\mathcal{D}_1$ and $\mathcal{D}_2$ to have constant coefficients.

\item $\vec{F}^\vee_\lambda = \vec{F}^w_\lambda$.

\item The group $SL(2,\mathbb{R})$ is generated by
$w$ and
$\left\{\begin{pmatrix} 1 & t \\ 0 & 1 \end{pmatrix}: t \in \mathbb{R}\right\}$.

\end{itemize}

\subsection{Conformally covariant differential operators}\label{subsec:conf}

Let $X$ be a smooth manifold equipped with a Riemannian
metric $g$. Suppose that a group $G$ acts on $X$ by the map
$G \times X \to X$, $(h, x) \mapsto h \cdot x$.
This action is called \emph{conformal} if there is a positive-valued
smooth function (\emph{conformal factor}) $\Omega$ on $G \times X$
such that
\begin{equation*}
h^*(g_{h\cdot x}) = \Omega(h,x)^2 g_x
\quad \text{for any $h \in G$ and $x \in X$.}
\end{equation*}

Given $\lambda \in \mathbb{C}$, we define a $G$-equivariant
line bundle $\mathcal{L}_\lambda\equiv \cL{\lambda}$
over $X$ by letting $G$
act on the direct product $X \times \mathbb{C}$ by
$(x, u) \mapsto (h\cdot x, \Omega(h, x)^{-\lambda}u)$
for $h \in G$. Then we have a natural action of $G$ on the
vector space $\mathcal{E}_\lambda(X) := C^\infty(X, \mathcal{L}_\lambda)$
consisting of smooth sections for $\mathcal{L}_\lambda$.
Since $\mathcal{L}_\lambda \to X$ is topologically a trivial bundle,
we may identify $\mathcal{E}_\lambda(X)$ with $C^\infty(X)$, and
corresponding $G$-action on $C^\infty(X)$ is given as the multiplier
representation $\varpi_\lambda \equiv \varpi_\lambda^X$:
\begin{equation*}
\left(\varpi_\lambda(h)f\right)(x)
=\Omega(h^{-1},x)^\lambda f(h^{-1}\cdot x)
\quad \text{for $h \in G$ and $f \in C^\infty(X)$.}
\end{equation*}
See \cite{xkors} for the basic properties of the representation
$(\varpi_\lambda, C^\infty(X))$.

\begin{example} \label{ex:SL2}
We endow $\mathbb{P}^1\mathbb{C} \simeq \mathbb{C}\cup \{\infty\}$
with a Riemannian metric $g$
via the stereographic projection of the unit sphere $S^2$:
\begin{equation*}
\mathbb{R}^3 \supset S^2 \stackrel{\sim}{\to} \mathbb{C}\cup \{\infty\},
\quad
(p,q,r) \mapsto \frac{p+\sqrt{-1}q}{1+r}.
\end{equation*}
Then $g(u,v) = \frac{4}{(1+|z|^2)^2}(u,v)_{\mathbb{R}^2}$
for $u, v \in T_z\mathbb{C} \simeq \mathbb{R}^2$,
and the M\"obius transformation, defined by
\begin{equation*}
\mathbb{P}^1\mathbb{C} \to \mathbb{P}^1\mathbb{C}, \quad
z \mapsto g\cdot z =\frac{az+b}{cz+d}
\quad
\text{for $g = \begin{pmatrix} a & b \\ c&d\end{pmatrix} \in
SL(2,\mathbb{C})$},
\end{equation*}
is conformal with conformal factor
\begin{equation}\label{eqn:czd}
\Omega(g,z) = |cz+d|^{-2}.
\end{equation}
Therefore,
\begin{equation*}
\left( \varpi_\lambda(h)f\right)(z)
= |cz+d|^{-2\lambda}
f\bigg(\frac{az+b}{cz+d}\bigg)
\quad
\text{for $h^{-1} = \begin{pmatrix}a & b \\ c & d \end{pmatrix}$.}
\end{equation*}
This is a (non-unitary) spherical principal series representation
$\Ind_{B_\mathbb{C}}^{G_\mathbb{C}}(1\otimes \lambda \alpha \otimes 1)$
of $G_\mathbb{C}= SL(2,\mathbb{C})$, where $\alpha$ is the unique
positive restricted root which defines a Borel subgroup $B_\mathbb{C}$.
\end{example}

Let $\wedge^i T^*X$ be
the $i$-th exterior power of the cotangent
bundle $T^*X$
for $0\leq i \leq n$, where $n$ is the dimension of $X$.
Then sections $\omega$ for $\wedge^i T^*X$ are
$i$-th differential forms on $X$, and $G$ acts on
$\mathcal{E}^i(X) = C^\infty(X, \wedge^i T^*X)$
as the pull-back of differential forms:
\begin{equation*}
\varpi(h)\omega = (h^{-1})^*\omega
\quad
\text{for $\omega \in \mathcal{E}^i(X)$.}
\end{equation*}
More generally, the tensor bundle
$\mathcal{L}_\lambda \otimes
\wedge^i T^*X$ is also a $G$-equivariant vector bundle
over $X$, and we denote by $\varpi_{\lambda, i}^X$
the regular representation
of $G$ on the space of sections
\begin{equation*}
\mathcal{E}^i_\lambda(X):=
C^\infty(X, \mathcal{L}_\lambda \otimes
\wedge^i T^*X).
\end{equation*}

By definition $\mathcal{E}^0_\lambda(X) = \mathcal{E}_\lambda(X)$.
In our normalization we have a natural $G$-isomorphism
\begin{equation*}
\mathcal{E}^0_n(X) \simeq \mathcal{E}^n_0(X),
\end{equation*}
if $X$ admits a $G$-invariant orientation.

Denote by $\text{Conf}(X)$ the full group of conformal
transformations of the Riemannian manifold
$(X, g)$. Given a submanifold $Y$ of $X$, we define
a subgroup by
\begin{equation*}
\text{Conf}(X; Y) :=
\{\varphi \in \text{Conf}(X) : \varphi(Y) = Y\}.
\end{equation*}
Then the induced action of
$\text{Conf}(X;Y)$ on
the Riemannian manifold $(Y, g|_Y)$
is again conformal.
We then consider the following problem.

\begin{problem}\label{prob:form}
(1) Given $0\leq i \leq \dim X$ and
$0 \leq j \leq \dim Y$,
classify $(\lambda, \nu) \in \mathbb{C}^2$ such that
there exists a non-zero local/non-local operator
\begin{equation*}
T: \mathcal{E}^i_\lambda(X) \to \mathcal{E}^j_\nu(Y)
\end{equation*}
satisfying
\begin{equation*}
\varpi_{\nu,j}^Y(h) \circ T
= T \circ \varpi_{\lambda, i}^X(h)
\quad
\text{for all $h \in \text{Conf}(X;Y)$.}\\
\end{equation*}

\noindent
(2) Find explicit \text{formul\ae}  of the operators
$T \equiv T_{\lambda, \nu}^{i,j}$.
\end{problem}

The case $i = j = 0$ is a question that was raised
in \cite[Problem 4.2]{Kob14a} as a geometric aspect of the
branching problem for representations with respect to
the pair of groups $\text{Conf}(X) \supset \text{Conf}(X;Y)$.

As a special case, one may ask:

\begin{ppquestion}\label{quest:A2}
Solve Problem \ref{prob:form} for covariant differential operators
in the setting that $(X,Y) = (S^2, S^1)$ and $(i,j) = (1,0)$.
\end{ppquestion}

We note that, for $(X,Y) = (S^2, S^1)$,
there are natural homomorphisms
\begin{alignat*}{3}
   G_\mathbb{C}&:=SL(2,\mathbb{C})&& \rightarrow  &&\operatorname{Conf}(X)
\\
    &\hphantom{mmmm}\cup &&     && \hphantom{mm}\cup
\\
   G_\mathbb{R}&:=SL(2,\mathbb{R})&&  \rightarrow &&
   \operatorname{Conf}(X;Y),
\end{alignat*}
\noindent
and the images of $SL(2,\mathbb{C})$ and $SL(2,\mathbb{R})$
coincide with the identity component groups of
$\operatorname{Conf}(X)\simeq O(3,1)$
and $\operatorname{Conf}(X;Y)$, respectively.
Question \ref{quest:A} is equivalent to Question \ref{quest:A2}
with $\operatorname{Conf}(X;Y)$ replaced by its identity component
$SO_0(2,1) \simeq SL(2,\mathbb{R})/\{\pm I\}$.
In fact, the differential operator
$\mathcal{D}=\Rest{y=0} \circ (\mathcal{D}_1, \mathcal{D}_2)$
in Question \ref{quest:A} gives a $G_\mathbb{R}$-equivariant
differential operator
\begin{equation*}
\mathcal{E}^1_{\lambda-1}(S^2)
\to \mathcal{E}^0_\nu(S^1) \equiv \mathcal{E}_\nu(S^1)
\end{equation*}
in our normalization, which takes the form
\begin{equation*}
\mathcal{E}^1(\mathbb{R}^2) \to C^\infty(\mathbb{R}),
\quad fdx + gdy \mapsto
(\mathcal{D}_1 f) (x, 0) + (\mathcal{D}_2g)(x,0)
\end{equation*}
in the flat coordinates via the stereographic projection.

\subsection{Branching laws of Verma modules}\label{subsec:2.3}

Let $\mathfrak{g} = \mathfrak{sl}(2, \mathbb{C})$, and $\mathfrak{b}$
a Borel subalgebra consisting of lower
triangular matrices in $\mathfrak{g}$.
For $\lambda \in \mathbb{C}$, we define a character of $\mathfrak{b}$,
to be denoted by $\mathbb{C}_\lambda$, as
\begin{equation*}
\mathfrak{b} \to \mathbb{C},
\quad
\begin{pmatrix}
-x & 0 \\ y & x
\end{pmatrix}
\mapsto \lambda x.
\end{equation*}
If $\lambda \in \mathbb{Z}$ then
$\mathbb{C}_\lambda$ is the differential of the holomorphic
character $\chi_{\lambda, \lambda}$ of
the Borel subgroup $B_\mathbb{C}$, which will be defined in
\eqref{eqn:chi} in Section \ref{subsec:3.1}.

We consider a $\mathfrak{g}$-module, referred to as a Verma module,
defined by
\begin{equation*}
M(\lambda) := U(\mathfrak{g}) \otimes_{U(\mathfrak{b})}\mathbb{C}_\lambda.
\end{equation*}
Then $\mathbf{1}_\lambda := 1 \otimes 1 \in M(\lambda)$
is a highest weight vector with weight $\lambda \in \mathbb{C}$, and it generates
$M(\lambda)$ as a $\mathfrak{g}$-module. The $\mathfrak{g}$-module
$M(\lambda)$ is irreducible if and only if $\lambda \notin \mathbb{N}$.

We consider the following algebraic question:

\begin{pppquestion}\label{quest:Verma}
(1) Classify $(\mu, \lambda_1, \lambda_2) \in \mathbb{C}^3$ such that
\begin{equation*}
\operatorname{Hom}_\mathfrak{g}
\left(M(\mu), M(\lambda_1) \otimes M(\lambda_2)\right)
\neq \{0\}.
\end{equation*}
(2) Find an explicit expression of $\varphi(\mathbf{1}_\mu)$
in $M(\lambda_1) \otimes M(\lambda_2)$ for any
$\varphi \in \operatorname{Hom}_\mathfrak{g}
\left( M(\mu), M(\lambda_1) \otimes M(\lambda_2)\right)$.
\end{pppquestion}

An answer to Question \ref{quest:Verma} is given as follows:

\begin{proposition}\label{prop:Verma}
If $\lambda_1 + \lambda_2 \notin \mathbb{N}$ then
the tensor product $M(\lambda_1) \otimes M(\lambda_2)$
decomposes into the direct sum of Verma modules as follows:
\begin{equation*}
M(\lambda_1) \otimes M(\lambda_2) \simeq
\bigoplus_{a=0}^\infty M(\lambda_1 + \lambda_2 -2a).
\end{equation*}
\end{proposition}

For the proof, consult \cite{K12} for instance.
In fact, in \cite{K12}, one finds the (abstract) branching laws of (parabolic)
Verma modules in the general setting of the restriction with
respect to symmetric pairs. By the duality theorem
(\cite{KOSS}, \cite[Theorem 2.7]{KP})
between
differential \emph{symmetry breaking operators}
(covariant differential operators to submanifolds)
and (discretely decomposable) branching laws of
Verma modules, we have the following one-to-one correspondence
\begin{align}
&\{\text{The differential operators
$\mathcal{D}$ yielding the functional identity \eqref{eqn:A}}\} \nonumber\\
&\leftrightarrow
\operatorname{Hom}_\mathfrak{g}
\left(M(-2\nu), M(-\lambda-1)\otimes M(-\lambda+1) \right) \label{eqn:KP}\\
&\hskip 1.6in \oplus
\operatorname{Hom}_\mathfrak{g}
\left(M(-2\nu), M(-\lambda+1)\otimes M(-\lambda-1) \right), \nonumber
\end{align}
because
$T_o (G_\mathbb{C}/ B_\mathbb{C}) \otimes \mathbb{C}
\simeq \mathbb{C}_{-2} \boxtimes \mathbb{C} + \mathbb{C} \boxtimes \mathbb{C}_{-2}$
as $\mathfrak{b} \otimes \mathbb{C} \simeq \mathfrak{b} \oplus \mathfrak{b}$-modules.
Combining this with Proposition \ref{prop:Verma}, we obtain

\begin{proposition}\label{prop:Verm}
If $2\lambda \notin - \mathbb{N}$ then a non-zero differential operator
$\mathcal{D}$ satisfying \eqref{eqn:A} exists if and only if
$\nu - \lambda \in \mathbb{N}$,
and the set of such differential operators
forms a two-dimensional vector space.
\end{proposition}

Owing to Proposition \ref{prop:Ddual}, we get the two-dimensional
solution space as the linear span of $\mathcal{D}$ and $\mathcal{D}^\vee$,
once we find a generic solution $\mathcal{D}$.

\section{Rankin--Cohen brackets}\label{sec:3}

As a preparation for the proof of Theorem \ref{quest:A},
we briefly review the Rankin--Cohen brackets, which
originated in number theory \cite{C75, EZ85, Ra56}.

\subsection{Homogeneous line bundles over $\mathbb{P}^1\mathbb{C}$}
\label{subsec:3.1}

First, we shall fix a normalization of three homogeneous line bundles over
$X=\mathbb{P}^1\mathbb{C}$, namely,
$\cL{\lambda}$ (Section \ref{sec:2}),
$\hL{\lambda}$, and $\mathcal{L}_{n,\lambda}$.

We define a Borel subgroup of $G_\mathbb{C} = SL(2, \mathbb{C})$ by
\begin{equation*}
B_\mathbb{C} := \bigg\{ \begin{pmatrix} a & 0 \\ c & \frac{1}{a} \end{pmatrix} :
a \in \mathbb{C}^\times, c \in \mathbb{C} \bigg\},
\end{equation*}
and identify $G_\mathbb{C} / B_\mathbb{C}$ with $X = \mathbb{P}^1 \mathbb{C}$
by $h B_\mathbb{C} \mapsto h \cdot 0$.

Given $n \in \mathbb{Z}$ and $\lambda \in \mathbb{C}$,
we define a one-dimensional
representation of $B_\mathbb{C}$ by
\begin{equation}\label{eqn:chi}
\chi_{n,\lambda} : B_\mathbb{C} \to \mathbb{C}^\times,
\quad
\begin{pmatrix}
\frac{1}{re^{i\theta}} & 0 \\
c & re^{i\theta}
\end{pmatrix}
\mapsto
e^{in\theta}r^\lambda,
\end{equation}
and a $G_\mathbb{C}$-equivariant line bundle
$\mathcal{L}_{n,\lambda} = G_\mathbb{C} \times_{B_\mathbb{C}} \chi_{n,\lambda}$
as the set of equivalence classes of $G_\mathbb{C} \times \mathbb{C}$ given by
\begin{equation*}
(g, u) \sim (gb^{-1}, \chi_{n,\lambda}(b) u)
\quad
\text{for some $b \in B_\mathbb{C}$.}
\end{equation*}
The conformal line bundle $\cL{\lambda}$ defined
in Section \ref{subsec:conf} amounts to
$\mathcal{L}_{0, 2\lambda}$ by the formula \eqref{eqn:czd}.

On the other hand, if $\lambda = n \in \mathbb{Z}$ then $\chi_{\lambda,\lambda}$
is a holomorphic character of $B_\mathbb{C}$, and
consequently, $\mathcal{L}_{\lambda, \lambda} \to X$ becomes
a holomorphic line bundle, which we denote by $\hL{\lambda}$.
The complexified cotangent bundle
$(T^*X)\otimes \mathbb{C}$ splits into a Whitney sum of the
holomorphic and anti-holomorphic cotangent bundle
$(T^*X)^{1,0} \oplus (T^* X)^{0,1}$,
which amounts to $\mathcal{L}_{2,2} \oplus \mathcal{L}_{-2,2}$.
In summary, we have:

\begin{lemma}\label{lem:Lpara}
We have the following isomorphisms of $G_\mathbb{C}$-equivariant
line bundles over $X \simeq \mathbb{P}^1\mathbb{C}$.
\begin{align*}
\hL{\lambda} &\simeq \mathcal{L}_{\lambda, \lambda}
\quad \hskip 0.07in\text{for $\lambda \in \mathbb{Z}$,}\\
\cL{\lambda} &\simeq \mathcal{L}_{0, 2\lambda}
\quad \text{for $\lambda \in \mathbb{C}$,}\\
(T^*X)^{1, 0} &\simeq \mathcal{L}_{2,2},\\
(T^*X)^{0,1}  &\simeq \mathcal{L}_{-2,2}.
\end{align*}
\end{lemma}

The line bundle $\mathcal{L}_{n,\lambda} \to X$
is $G_\mathbb{C}$-equivariant; thus, there is
the regular representation $\pi_{n,\lambda}$
of $G_\mathbb{C}$
on $C^\infty(X, \mathcal{L}_{n,\lambda})$.
This is called the (unnormalized, non-unitary) principal
series representation of $G_\mathbb{C}$.
The restriction to the open Bruhat cell
$\mathbb{C} \hookrightarrow X=\mathbb{C} \cup \{\infty\}$
yields an injection
$C^\infty(X, \mathcal{L}_{n,\lambda}) \hookrightarrow C^\infty(\mathbb{C})$,
on which $\pi_{n,\lambda}$ is given as a multiplier representation:
\begin{equation*}
\left( \pi_{n,\lambda}(h) F\right) (z) =
\bigg(\frac{cz+d}{|cz+d|} \bigg)^{-n}
|cz+d|^{-\lambda}
F\bigg(\frac{az+b}{cz+d}\bigg)
\quad
\text{for $h^{-1} = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$}.
\end{equation*}
Comparing this with the conformal construction
of the representation $\varpi_\lambda$ in
Example \ref{ex:SL2}, we have $\varpi_\lambda \simeq \pi_{0,2\lambda}$.

Similarly to the smooth line bundle $\mathcal{L}_{n,\lambda}$,
we consider holomorphic sections for the holomorphic line bundle
$\hL{\lambda}$. For this, let $D$ be a domain of $\mathbb{C}$ and
$G$ a subgroup of $G_\mathbb{C}$, which leaves $D$ invariant.
Then we can define a representation, to be denoted by
$\hpi{\lambda}$, of $G$ on the space
$\mathcal{O}(D) \equiv \mathcal{O}(D, \hL{\lambda})$
of holomorphic sections, which is identified with a multiplier representation
\begin{equation*}
\left(\hpi{\lambda}(h) F \right)(z)
=(cz+d)^{-\lambda}
F\bigg(\frac{az+b}{cz+d}\bigg)
\quad
\text{for $F \in \mathcal{O}(D)$}.
\end{equation*}

\begin{example}
(1) $D = \{ z \in \mathbb{C} : |z| < 1\}$, $G = SU(1,1)$.\\
\noindent
\phantom{}
\hskip 1.08in
(2) $D = \{ z \in \mathbb{C} : \text{Im} z > 0\}$, $G = SL(2, \mathbb{R})$.\\
\noindent
(For the application below we shall use the unit disc model.)
\end{example}

\subsection{Rankin--Cohen bidifferential operator}\label{subsec:RC}

Let $D$ be a domain in $\mathbb{C}$. For $a \in \mathbb{N}$
and $\lambda_1, \lambda_2 \in \mathbb{C}$,
the bidifferential operator
$\adRC{\lambda_1}{\lambda_2}:
\mathcal{O}(D) \otimes \mathcal{O}(D) \to \mathcal{O}(D)$,
referred to as the \emph{Rankin--Cohen bracket} \cite{C75, Ra56},
is defined by
\begin{equation*}
\adRC{\lambda_1}{\lambda_2}(f_1 \otimes f_2)(z)
:=
\sum_{\ell = 0}^a (-1)^\ell
\binom{\lambda_1 + a - 1}{\ell}
\binom{\lambda_2 + a - 1}{a-\ell}
\frac{\partial^{a-\ell}f_1}{\partial z^{a-\ell}}(z)
\frac{\partial^{\ell}f_2}{\partial z^{\ell}}(z).
\end{equation*}

In the theory of automorphic forms,
$\adRC{\lambda_1}{\lambda_2}$
yields a new holomorphic modular form
of weight $\lambda_1 + \lambda_2 + 2a$ out of two
holomorphic modular forms $f_1$ and $f_2$ of weights
$\lambda_1$ and $\lambda_2$, respectively.

From the viewpoint of representation theory,
$\adRC{\lambda_1}{\lambda_2}$
is an intertwining operator:
\begin{equation}\label{eqn:RCcov}
\hpi{\lambda_1 + \lambda_2 + 2a}(h)
\circ \adRC{\lambda_1}{\lambda_2}
=
\adRC{\lambda_1}{\lambda_2} \circ
\left( \hpi{\lambda_1}(h) \otimes \hpi{\lambda_2}(h)\right)
\end{equation}
for all $h \in G$.

The coefficients of the Rankin--Cohen brackets look
somewhat complicated.
Eicheler--Zagier \cite[Chapter 3]{EZ85}
found that they are related to those of a
classical orthogonal polynomial.
A short proof for this fact is given by the
\emph{F-method} in \cite{KP}.

To see the relation, we define a polynomial
$\aRC{\lambda_1}{\lambda_2}(x,y)$
of two variables
$x$ and $y$ by
\begin{equation}\label{eqn:RCxy}
\aRC{\lambda_1}{\lambda_2}(x,y):=
\sum_{\ell=0}^a (-1)^\ell
\binom{\lambda_1 + a -1}{\ell}
\binom{\lambda_2 + a -1}{a-\ell}
x^{a-\ell}y^\ell,
\end{equation}
so that the Rankin--Cohen bidifferential operator
$\mathcal{RC}^a_{\lambda_1, \lambda_2}$
is given by
\begin{equation*}
\mathcal{RC}^a_{\lambda_1, \lambda_2}
=\Rest{z_1 = z_2 = z}\circ
\aRC{\lambda_1}{\lambda_2}\left(
\frac{\partial}{\partial z_1}, \frac{\partial}{\partial z_2}
\right).
\end{equation*}
The polynomial $\aRC{\lambda_1}{\lambda_2}(x,y)$ is
of  homogeneous degree $a$. Clearly we have:

\begin{lemma}\label{lem:RCdual}
$\aRC{\lambda_1}{\lambda_2} (x,y) = (-1)^a\aRC{\lambda_2}{\lambda_1}(y,x)$.
\end{lemma}

Second we recall that
the Jacobi polynomial $P^{\alpha,\beta}_\ell(t)$
is a polynomial of one variable $t$ of degree $\ell$
given by
\begin{equation*}
P^{\alpha, \beta}_\ell(t) =
\frac{\Gamma(\alpha + \ell + 1)}{\Gamma(\alpha +\beta + \ell+ 1)}
\sum_{m=0}^\ell
\frac{\Gamma(\alpha + \beta + \ell + m +1)}
{(\ell-m)! m! \Gamma(\alpha + m + 1)}
\bigg(\frac{t-1}{2}\bigg)^m.
\end{equation*}

We inflate it to a homogeneous polynomial of two
variables $x$ and $y$ of degree $\ell$ by
\begin{equation*}
P^{\alpha, \beta}_{\ell}(x,y):=
y^\ell P^{\alpha, \beta}_\ell \bigg(2 \frac{x}{y}+1 \bigg).
\end{equation*}
For instance, $P^{\alpha, \beta}_0(x,y)=1$ and
$P^{\alpha,\beta}_1(x,y) = (2+\alpha + \beta)x+(\alpha+1)y$.
It turns out that
\begin{equation*}
\aRC{\lambda_1}{\lambda_2}(x,y)=(-1)^a
P^{\lambda_1-1, -\lambda_1 - \lambda_2 - 2a + 1}_a(x,y).
\end{equation*}
In particular, the following holds.

\begin{lemma}\label{lem:RCJacobi}
We have
\begin{equation*}
\adRC{\lambda_1}{\lambda_2} = (-1)^a
\Rest{z_1 = z_2=z} \circ
P^{\lambda_1-1, -\lambda_1 - \lambda_2 - 2a + 1}_a
\left(\frac{\partial}{\partial z_1}, \frac{\partial}{\partial z_2}\right).
\end{equation*}
\end{lemma}

\section{Holomorphic trick}\label{sec:htrick}

In this section we give a proof for Theorem \ref{quest:A}
by using the results of the previous sections

\subsection{Restriction to a totally real submanifold}\label{subsec:real}

Consider a totally real embedding of
$X = \mathbb{P}^1\mathbb{C} \simeq \mathbb{C} \cup \{\infty\}$
defined by
\begin{equation}\label{eqn:PCzz}
\iota: \mathbb{P}^1\mathbb{C}
\to \mathbb{P}^1\mathbb{C} \times \mathbb{P}^1\mathbb{C},
\quad
z\mapsto (z, \bar{z}).
\end{equation}
The map $\iota$ respects the action of $G_\mathbb{C}$ via the
following group homomorphism (we regard $G_\mathbb{C}$ as a real group),
denoted by the same letter,
\begin{equation*}
\iota: G_\mathbb{C} \to G_\mathbb{C} \times G_\mathbb{C},
\quad
g \mapsto (g, \bar{g}).
\end{equation*}
This is because
$G_\mathbb{C}/B_\mathbb{C} \simeq \mathbb{P}^1\mathbb{C}$ and
because the Borel subgroup $B_\mathbb{C}$ is stable by the complex
conjugation $g \mapsto \bar{g}$. Then the following lemma is immediate
from Lemma \ref{lem:Lpara}.

\begin{lemma} We have
an isomorphism of $G_\mathbb{C}$-equivariant line bundles:
\begin{equation*}
\iota^*\left(\hL{\lambda_1} \boxtimes \hL{\lambda_2}\right)
\simeq \mathcal{L}_{\lambda_1-\lambda_2, \lambda_1 + \lambda_2}.
\end{equation*}
In particular,
\begin{align}
&\iota^*(\hL{\lambda+1} \boxtimes
\hL{\lambda-1} )
\simeq \cL{\lambda-1} \otimes (T^*X)^{1,0}, \label{eqn:pb1}\\
&\iota^*(\hL{\lambda-1} \boxtimes
\hL{\lambda+1} )
\simeq \cL{\lambda-1} \otimes (T^*X)^{0,1} \label{eqn:pb2}.
\end{align}
\end{lemma}

\begin{proposition}\label{prop:htrick}
The isomorphisms \eqref{eqn:pb1} and \eqref{eqn:pb2} induce
injective $G_\mathbb{C}$-equivariant
homomorphisms between equivariant sheaves:
\begin{align*}
&(\iota^*)^{1,0}: \mathcal{O}(\hL{\lambda+1})
\otimes \mathcal{O}(\hL{\lambda-1})
\to \mathcal{E}^{1,0}_{\lambda-1},
\quad f_1(z_1) \otimes f_2(z_2) \mapsto
f_1(z) f_2(\bar{z}) dz,\\
&(\iota^*)^{0,1}: \mathcal{O}(\hL{\lambda-1})
\otimes \mathcal{O}(\hL{\lambda+1})
\to \mathcal{E}^{0,1}_{\lambda-1},
\quad f_1(z_1) \otimes f_2(z_2) \mapsto
f_1(z) f_2(\bar{z}) d\bar{z},
\end{align*}
that is, $(\iota^*)^{1,0}$ and $(\iota^*)^{0,1}$ are injective
on every open set $D$ in $\mathbb{P}^1\mathbb{C}$, and
\begin{align*}
(\iota^*)^{1,0} \circ
\left( \hpi{\lambda+1}(g) \otimes \hpi{\lambda-1} (\bar g) \right)
&= \varpi^1_{\lambda-1}(g) \circ (\iota^*)^{1,0} \\
(\iota^*)^{0,1} \circ
\left( \hpi{\lambda-1}(g) \otimes \hpi{\lambda+1} (\bar g) \right)
&= \varpi^1_{\lambda-1}(g) \circ (\iota^*)^{0,1}
\end{align*}
hold for any $g$ whenever they make sense.
\end{proposition}

\begin{proof}
The injectivity follows from
the identity theorem of holomorphic functions because
$\iota: \mathbb{P}^1\mathbb{C} \to
\mathbb{P}^1\mathbb{C} \times \mathbb{P}^1\mathbb{C}$
is a totally real embedding. The covariance property is derived
from \eqref{eqn:pb1} and \eqref{eqn:pb2}.
\end{proof}

Fix $\lambda \in \mathbb{Z}$ and $a \in \mathbb{N}$,
and set $\nu = \lambda + a$.
We want to relate the Rankin--Cohen brackets
$\adRC{\lambda \pm 1}{\lambda \mp 1}$ to our
differential operator $\mathcal{D}$ (see Question \ref{quest:A})
in the sense that
both of the following diagrams commute:

\begin{alignat*}{3}
\mathcal{O}(\hL{\lambda+1}) \otimes\mathcal{O}(\hL{\lambda-1})
&\stackrel{(\iota^*)^{1,0}}{\lhook\joinrel\relbar\joinrel\relbar\joinrel\relbar\joinrel\relbar}
&
\!\!\joinrel\rightarrow
\mathcal{E}^{1,0}_{\lambda-1}(\mathbb{C})&
\subset \mathcal{E}^{1}_{\lambda-1}(\mathbb{R}^2)  &
&\simeq C^\infty(\mathbb{R}^2) \oplus C^\infty(\mathbb{R}^2)\\
{\scriptstyle \adRC{\lambda+1}{\lambda-1}} {\big \downarrow} \hskip 48pt
& &&     &&\hskip 60pt \big \downarrow {\scriptstyle\mathcal{D}}\\
\mathcal{O}(\hL{2\lambda+2a}) \phantom{mm}
&
\stackrel{\iota^*}{\lhook\joinrel\relbar\joinrel\relbar\joinrel\relbar\joinrel\relbar}&
\joinrel\relbar\joinrel\relbar\joinrel\relbar
\joinrel\relbar\joinrel\relbar\joinrel\relbar\joinrel\relbar&
\!\!\joinrel\rightarrow \hskip 2pt
\mathcal{E}_\nu(\mathbb{R}) &&
\simeq  \hskip 35pt C^\infty(\mathbb{R}),
\end{alignat*}
and
\begin{alignat*}{3}
\mathcal{O}(\hL{\lambda-1}) \otimes\mathcal{O}(\hL{\lambda+1})
&\stackrel{(\iota^*)^{0,1}}{\lhook\joinrel\relbar\joinrel\relbar\joinrel\relbar\joinrel\relbar}
&
\!\!\joinrel\rightarrow
\mathcal{E}^{0,1}_{\lambda-1}(\mathbb{C})&
\subset \mathcal{E}^{1}_{\lambda-1}(\mathbb{R}^2)  &
&\simeq C^\infty(\mathbb{R}^2) \oplus C^\infty(\mathbb{R}^2)\\
{\scriptstyle (-1)^a\adRC{\lambda-1}{\lambda+1}} {\big \downarrow} \hskip 48pt
& &&     &&\hskip 60pt \big \downarrow {\scriptstyle\mathcal{D}}\\
\mathcal{O}(\hL{2\lambda+2a}) \phantom{mm}
&
\stackrel{\iota^*}{\lhook\joinrel\relbar\joinrel\relbar\joinrel\relbar\joinrel\relbar}&
\joinrel\relbar\joinrel\relbar\joinrel\relbar
\joinrel\relbar\joinrel\relbar\joinrel\relbar\joinrel\relbar&
\!\!\joinrel\rightarrow \hskip 2pt
\mathcal{E}_\nu(\mathbb{R}) &&
\simeq  \hskip 35pt C^\infty(\mathbb{R}).
\end{alignat*}
Here we have used the following identification:
\begin{equation*}
\mathcal{E}^1_{\lambda-1}(\mathbb{R}^2) \simeq
C^\infty(\mathbb{R}^2) \oplus C^\infty(\mathbb{R}^2),
\quad
fdx+gdy \mapsto (f, g).
\end{equation*}

We define homogeneous polynomials
$D_1$, $D_2$ with real coefficients so that
\begin{equation*}
D_1(x,y) + \sqrt{-1} D_2(x,y) =
2^{-a}\aRC{\lambda+1}{\lambda-1}
(x-\sqrt{-1}y, x+\sqrt{-1}y),
\end{equation*}
where $\aRC{\lambda_1}{\lambda_2}(x,y)$ is a
polynomial defined in \eqref{eqn:RCxy}.
We set
\begin{align}\label{eqn:DRC}
\mathcal{D}_1:=D_1\left(\frac{\partial}{\partial x},\frac{\partial}{\partial y}  \right),
\quad
\mathcal{D}_2:=D_2\left(\frac{\partial}{\partial x}, \frac{\partial}{\partial y} \right),
\quad
\mathcal{D}:=\Rest{y=0} \circ \left(\mathcal{D}_1, \mathcal{D}_2 \right).
\end{align}

\begin{lemma}\label{lem:RCreal}
For any holomorphic functions $f_1$ and $f_2$,
\begin{align*}
&\mathcal{D}\left( (\iota^*)^{1,0}(f_1 \otimes f_2) \right)
= \iota^* \adRC{\lambda+1}{\lambda-1}(f_1 \otimes f_2),\\
&\mathcal{D}\left( (\iota^*)^{0,1}(f_1 \otimes f_2) \right)
= (-1)^a\iota^* \adRC{\lambda-1}{\lambda+1}(f_1 \otimes f_2).
\end{align*}
\end{lemma}

\begin{proof}
Let $\omega:=(\iota^*)^{1,0}(f_1 \otimes f_2) = f_1(z)f_2(\bar{z})dz$.
If we write $\omega = fdx + gdy$ then $f(z) = f_1 (z) f_2(\bar{z})$
and $g = \sqrt{-1}f$. Therefore,
\begin{align*}
\mathcal{D}\omega &= \Rest{y=0} \circ (\mathcal{D}_1, \mathcal{D}_2)
\begin{pmatrix} f \\ g \end{pmatrix},\\
(\mathcal{D}_1,\mathcal{D}_2) \begin{pmatrix} f \\ g \end{pmatrix}
&=\left(\mathcal{D}_1 + \sqrt{-1}\mathcal{D}_2) (f_1(z) f_2(\bar{z})\right).
\end{align*}
If we write
$\aRC{\lambda+1}{\lambda-1}(x,y) = \sum_{\ell=0}^a r_\ell x^{a-\ell}y^\ell$
then
\begin{align*}
(\mathcal{D}_1 + \sqrt{-1} \mathcal{D}_2)\left(f_1(z)f_2(\bar{z})\right)
&= \aRC{\lambda+1}{\lambda-1}
\bigg(\frac{\partial}{\partial z},
\frac{\partial}{\partial \bar{z}} \bigg)
(f_1(z)f_2(\bar{z}))\\
&= \sum_{\ell=0}^a r_\ell
\frac{\partial^{a-\ell}f_1}{\partial z^{a-\ell}} (z)
\frac{\partial^{\ell}f_2}{\partial \bar{z}^{\ell}} (\bar{z}),
\end{align*}
because $f_1$ and $f_2$ are holomorphic. Taking
the restriction to $y=0$, we get
\begin{equation*}
\mathcal{D}(\omega)
= \sum_{\ell=0}^a r_\ell
\frac{\partial^{a-\ell}f_1}{\partial x^{a-\ell}} (x)
\frac{\partial^{\ell}f_2}{\partial x^\ell} (x)
=\iota^* \adRC{\lambda+1}{\lambda-1}(f_1 \otimes f_2).
\end{equation*}
Hence we have proved the first identity. The second
identity follows from Lemma \ref{lem:RCdual}.
\end{proof}

\begin{remark}
If we multiply the bidifferential operator
$\adRC{\lambda+1}{\lambda-1}$ by
$\sqrt{-1}$ then
obviously \eqref{eqn:RCcov} holds, where the role of
$(\mathcal{D}_1, \mathcal{D}_2)$
is changed into $(-\mathcal{D}_2, \mathcal{D}_1)$ because
\begin{equation*}
\sqrt{-1}
(\mathcal{D}_1 + \sqrt{-1}\mathcal{D}_2)
= -\mathcal{D}_2 + \sqrt{-1}\mathcal{D}_1.
\end{equation*}
This explains Proposition \ref{prop:Ddual} from
the ``holomorphic trick."
\end{remark}

\subsection{Identities of Jacobi polynomials}\label{subsec:sp}

For $a \in \mathbb{N}_+$, we define the following three meromorphic
functions of $\lambda$ by

\begin{align*}
A_a(\lambda)&:= \frac{2 \lambda^2 + 2(a-1) \lambda + a(a-1)}
{a (2\lambda+a -1)},\\
B_a(\lambda)&:=\frac{(\lambda-1)(2\lambda+1)}
{a(2\lambda+a-1)}, \\
U_a(\lambda)&:=
\frac{2 \left(\lambda+ [\frac{a}{2}] \right)_{[\frac{a-1}{2}]}}
{\left(\lambda + \frac{1}{2} \right)_{[\frac{a-1}{2}]}},
\end{align*}
where
$(\mu)_k:= \mu(\mu+1)\cdots (\mu+k-1) =
\frac{\Gamma(\mu+k)}{\Gamma(\mu)}$
is the Pochhammer symbol.

\begin{proposition}\label{prop:Jacob}
For $a\in \mathbb{N}_+$, we have
\begin{align*}
&(1-z)^aP^{\lambda, -2\lambda-2a+1}_a
\bigg(\frac{3+z}{1-z}\bigg)\\
&=(-1)^{a-1}U_a(\lambda)
\bigg(
(1-A_a(\lambda)z)C^{\lambda+\frac{1}{2}}_{a-1}(z)+
B_a(\lambda)(1-z^2)C^{\lambda+\frac{3}{2}}_{a-2}(z) \bigg).
\end{align*}

Equivalently,\begin{align*}
&P^{\lambda, -2\lambda-2a+1}_a(x-\sqrt{-1}y, x+\sqrt{-1}y)\\
&=(\sqrt{-1})^{a-1}U_a(\lambda)\bigg(
xC^{\lambda+\frac{1}{2}}_{a-1}(-x^2, y) +
\sqrt{-1}\left( A_a(\lambda) y C^{\lambda+\frac{1}{2}}_{a-1}(-x^2, y)
+ B_a(\lambda) (x^2 + y^2) C^{\lambda+\frac{3}{2}}_{a-2}(-x^2, y)\right) \bigg).
\end{align*}
\end{proposition}

Proposition \ref{prop:Jacob} will be used
in the proof of Theorem \ref{thm:A} in the next subsection.
We want to note that we wondered if
the first equation of Proposition \ref{prop:Jacob}
was already known; however,
we could not find the identity in the literature.

One might give an alternative proof of Proposition \ref{prop:Jacob}
by applying the F-method to a vector bundle case.
We will discuss this approach in a subsequent paper.

\subsection{Proof of Theorem \ref{thm:A}}\label{subsec:proof}

The relations in Lemma \ref{lem:RCreal}
and the covariance property \eqref{eqn:RCcov}
of the Rankin--Cohen brackets
imply that the differential operator
$\mathcal{D}$ defined in \eqref{eqn:DRC}
satisfies the covariance relations \eqref{eqn:A2}
on the image
\begin{equation*}
(\iota^*)^{1,0}
\big( \mathcal{O}(\hL{\lambda+1})\otimes
\mathcal{O}(\hL{\lambda-1}) \big)
+ (\iota^*)^{0,1}
\big( \mathcal{O}(\hL{\lambda-1})\otimes
\mathcal{O}(\hL{\lambda+1}) \big).
\end{equation*}

In order to prove \eqref{eqn:A2}, we need to show that
the image
is dense in $C^\infty(\mathbb{R}^2) \oplus C^\infty(\mathbb{R}^2)$
topologized by uniform convergence on compact sets.
To see this we note that the image contains a linear span
of the following 	1-forms
\begin{equation*}
z^m\bar{z}^n dz, \quad
z^m\bar{z}^n d\bar{z}, \quad
(m, n \in \mathbb{N}).
\end{equation*}
Since a linear span of $(x+iy)^m(x-iy)^n$ $(m,n\in \mathbb{N})$
is dense in $C^\infty(\mathbb{R}^2)$ by the Stone--Weierstrass theorem,
we conclude that $\mathcal{D}$ satisfies \eqref{eqn:A2}.
An explicit formula for  the operators $(\mathcal{D}_1,\mathcal{D}_2)$
is derived from the Rankin--Cohen brackets by using
Lemma \ref{lem:RCJacobi} and
Proposition \ref{prop:Jacob}
for $\lambda \in \mathbb{Z}$.
Then the covariance relations \eqref{eqn:invF}
are satisfied for all $\lambda \in \mathbb{C}$ because $\mathbb{Z}$
is Zariski dense in $\mathbb{C}$.

If $2\lambda \notin -\mathbb{N}$ then the dimension of solutions is
two by Proposition \ref{prop:Verma} and the one-to-one
correspondence \eqref{eqn:KP}.
Since $\mathcal{D}$ and $\mathcal{D}^\vee$ are linearly independent
for our solution $\mathcal{D}$, the linear span of $\mathcal{D}$ and
$\mathcal{D}^\vee$ exhausts all the solutions by Proposition \ref{prop:Ddual}.
Hence Theorem \ref{thm:A} is proved.

\subsection{Scalar-valued case}\label{subsec:scalar}

So far we have discussed a family of vector-valued
differential operators that yield functional identities
satisfied by vector-valued functions. We close this
article with some comments on the scalar-valued case.

Let $\lambda\in \mathbb{C}$. Given
$f \in C^\infty(\mathbb{R}^2\setminus \{(0,0)\})
\simeq C^\infty(\mathbb{C} \setminus \{0\})$,
we define its \emph{twisted inversion} $f^\vee_\lambda$ by
\begin{equation*}
f^\vee_\lambda(r\cos \theta, r\sin \theta)
:=r^{-2\lambda}f\bigg(\frac{-\cos \theta}{r}, \frac{\sin \theta}{r}  \bigg)
\end{equation*}
as in \eqref{eqn:invF}, and more generally,
\begin{equation*}
f^h_\lambda(z):=
|cz+d|^{-2\lambda} f\left(\frac{az+b}{cz+d}\right)
\quad
\text{for $h^{-1} = \begin{pmatrix} a & b\\ c & d \end{pmatrix} \in SL(2,\mathbb{C})$}
\end{equation*}
as in \eqref{eqn:Fhl}.

For a differential operator $\mathcal{D}$ on $\mathbb{R}^2$,
we define a linear operator
$\tilde{\mathcal{D}} : C^\infty(\mathbb{R}^2) \to C^\infty(\mathbb{R})$
by
\begin{equation*}
\tilde{\mathcal{D}} := \Rest{y=0} \circ \mathcal{D}.
\end{equation*}

Fix $\lambda, \nu \in \mathbb{C}$.
As in Questions
\ref{quest:A},
\ref{quest:A1},
\ref{quest:A2}, and
\ref{quest:Verma},
we may consider the following
equivalent questions:

\begin{question}\label{quest:B}
Find $\tilde{\mathcal{D}}$ with constant coefficients such that
\begin{equation*}
\left(\tilde{\mathcal{D}} f^\vee_\lambda\right)(x)
=|x|^{-2\nu} (\tilde{\mathcal{D}}f)\left(-\frac{1}{x}\right)
\quad
\text{for all $f \in C^\infty(\mathbb{R}^2)$ and $x \in \mathbb{R}^\times$.}
\end{equation*}
\end{question}

\begin{pquestion}\label{quest:B'}
Find $\tilde{\mathcal{D}}$ such that
\begin{equation*}
\left(\tilde{\mathcal{D}} f^h_\lambda\right)(x) =
|cx+d|^{-2\nu}(\tilde{\mathcal{D}}f)
\bigg(\frac{ax+b}{cx+d}\bigg)
\quad
\text{for all $f \in C^\infty(\mathbb{C}),
h \in SL(2,\mathbb{R})$,
and $x \in \mathbb{R}^\times$.}
\end{equation*}
\end{pquestion}

\begin{ppquestion}\label{quest:B''}
Find an explicit formula of
conformally covariant differential operator
$\mathcal{E}_\lambda(S^2) \to \mathcal{E}_\nu(S^1)$.
\end{ppquestion}

\begin{pppquestion}\label{quest:B'''}
Find an explicit expression of the element $\varphi(\mathbf{1}_{-\nu})$
for  any $\varphi \in  \operatorname{Hom}_\mathfrak{g} \left(
M(-\nu), M(-\lambda) \otimes M(-\lambda) \right)$,
where $\mathfrak{g} = \mathfrak{sl}(2, \mathbb{C})$.
\end{pppquestion}

An answer to Question \ref{quest:B''}
(and also in the case $S^{n-1} \subset S^n$
for arbitrary $n \geq 2$)
was first given by
Juhl \cite{Juhl}.
In the flat model (Questions \ref{quest:B} and \ref{quest:B'}),
if $a : = \nu-\lambda \in \mathbb{N}$
then
\begin{equation*}
\widetilde{\mathcal C{}^{\lambda-\frac{1}{2}}_a}
\equiv\Rest{y=0} \circ C^{\lambda-\frac{1}{2}}_a\bigg(-\frac{\partial^2}{\partial x^2},
\frac{\partial}{\partial y}\bigg)
:\mathcal{E}_\lambda(\mathbb{R}^2) \to
\mathcal{E}_{\nu}(\mathbb{R}^1)
\end{equation*}
intertwines the $SL(2,\mathbb{R})$-action.
There have been several proofs for this
(and also for more general cases) based on:

\begin{itemize}
\item Recurrence relations among coefficients of $\mathcal{D}$ (\cite{Juhl}),
\item F-method (\cite{K13, KOSS, KP}), and
\item Residue \text{formul\ae} of a meromorphic family of non-local
symmetry breaking operators \cite{Kob14a, KS}.
\end{itemize}

The holomorphic trick in Section \ref{sec:htrick} applied to this
case gives yet another proof by using the Rankin--Cohen brackets
and the following proposition analogous to
(and much simpler than) Proposition \ref{prop:Jacob}.

\begin{proposition}
For $a \in \mathbb{N}$, we have
\begin{align*}
(1-z)^aP^{\lambda-1, -2\lambda-2a+1}_a
\bigg(\frac{3+z}{1-z}\bigg)
=(-1)^a
\frac{\left(\lambda+ [\frac{a}{2}] \right)_{[\frac{a+1}{2}]}}
{\left(\lambda - \frac{1}{2} \right)_{[\frac{a+1}{2}]}}
C^{\lambda-\frac{1}{2}}_{a}(z).
\end{align*}
Equivalently,
\begin{align*}
P^{\lambda-1, -2\lambda-2a+1}_a(x-\sqrt{-1}y, x+\sqrt{-1}y)
=(\sqrt{-1})^a
\frac{\left(\lambda+ [\frac{a}{2}] \right)_{[\frac{a+1}{2}]}}
{\left(\lambda - \frac{1}{2} \right)_{[\frac{a+1}{2}]}}
C^{\lambda-\frac{1}{2}}_a (-x^2, y).
\end{align*}
\end{proposition}

\subsection*{Acknowledgments}
The first author warmly thanks Professor Vladimir Dobrev for his
hospitality during the tenth International Workshop: Lie Theory and its Applications
in Physics in Varna, Bulgaria, 17-23 June 2013.
Authors were partially supported by CNRS,
the FMSP program at
the Graduate School of Mathematical Sciences
of the University of Tokyo, and
Japan Society for the Promotion of Sciences through
Grant-in-Aid for Scientific Research (A)
(25247006) and Short-Term Fellowship (S13024).


TITLE: No closed surface diffeomorphic to the Moebius trip without boundaries
QUESTION [1 upvotes]: Consider the Moebius strip $M$ as a quotient space of $\mathbb{R}\times [0,1]$ with opposite lines glued together with reverse orientation. I would like to prove that a closed submanifold $N$ in $\mathbb{R}^3$ cannot be diffeomorphic to $M$.
It seems to me that, even if $M$ is not the classical Moebius band with its boundaries, $M$ is still a non-orientable manifold.
My plan is to prove that any closed 2-submanifold $N$ of $\mathbb{R}^3$ is orientable and thus may not be diffeomorphic to $M$.
It seems to me that answering the two following questions is a good way to prove it:
1) Show that any closed submanifold of dimension 2 in $\mathbb{R}^3$ can be seen as $f^{-1}(0)$ where $f : \mathbb{R}^3 \rightarrow \mathbb{R}$ is a smooth map such that $0$ is a regular value of $f$.
2) Show that any closed submanifold of dimension $2$ of $\mathbb{R}^3$ that is obtained as :
$f^{-1}(0)$ where $f : \mathbb{R}^3 \rightarrow \mathbb{R}$ is a smooth map, such that $0$ is a regular value of $f$, is orientable.
Any idea on how to tackle 1) and 2)?

REPLY [1 votes]: I'm not sure I fully understand what your $M$ is. It does not
look compact to me. Anyway it is well-known that a closed 2-submanifold
of $\Bbb R^3$ must be orientable. This is a consequence of the
Alexander Duality Theorem. See Corollary 3.45 of Hatcher's Algebraic
Topology.


TITLE: Fourier Transform of Various Tempered Distributions
QUESTION [0 upvotes]: Let $T$ be a tempered distributions, and let $S(x) = e^{i\alpha x}T(x)$, where $\alpha \in \mathbb{R}$.

Find a formula that relates the Fourier transforms $\hat S$ and $\hat T$.
Find the Fourier transform $\hat f_\alpha$ of $f_\alpha (x) = \sin \alpha x$.
Find $\lim_{\alpha \to \infty} \alpha \hat f_\alpha$ in the sense of (tempered) distributions.

REPLY [1 votes]: Choose $\phi \in \mathcal{S}(\mathbb{R})$. Then
$$\hat S \phi (k) = \hat T (e^{i\alpha x}\phi (k)) = T(\widehat{e^{i\alpha x}\phi}(k)) = T\big(\frac{1}{\sqrt{2\pi}} \int_{\mathbb{R}} e^{i\alpha t}\phi(t)e^{-ikt} dt\big) = T(\frac{1}{\sqrt{2\pi}}\int_{\mathbb{R}} \phi(t)e^{-it(k-\alpha)}dt\big) = T(\hat \phi(k-\alpha) = T(\tau_\alpha \hat \phi (k)) = \tau_\alpha \hat T\pi(k)$$

So $\hat S=\tau_\alpha \hat T$.

Note that $f_\alpha(x) = \sin \alpha x = \frac{1}{2i}(e^{i\alpha x}-e^{-i\alpha x})$. Using part 1 ($\alpha \in \mathbb{R}$), we infer that
$$\hat f_\alpha = \frac{1}{2i} \tau_\alpha\hat 1 - \frac{1}{2i} \tau_{-\alpha}\hat 1 = \frac{1}{2i} \tau_\alpha\sqrt{2\pi} \delta - \frac{1}{2i}\tau_{-\alpha}\sqrt{2\pi}\delta = \frac{1}{2i}\sqrt{2\pi} \delta_\alpha - \frac{1}{2i}\sqrt{2\pi} \delta_{-\alpha}$$

We see then that $\hat f_\alpha = \frac{\sqrt{2\pi}}{2i} (\delta_\alpha- \delta_{-\alpha})$.

Again choose $\phi \in \mathcal{S}(\mathbb{R})$.
$$ \lim_{\alpha \to \infty} \hat f_\alpha \phi= \lim_{\alpha \to \infty} \frac{\sqrt{2\pi}}{2i}(\delta_\alpha - \delta_{-\alpha})\phi = \lim_{\alpha \to \infty} \frac{\sqrt{2\pi}}{2i} (\phi(\alpha) - \phi(-\alpha)) =0$$
where the last is because $\phi \in \mathcal{S}(\mathbb{R})$, and thus should decay to zero as $\alpha \to \pm \infty$.


TITLE: "Seeing yourself" in a 3-manifold
QUESTION [8 upvotes]: In his book "The Shape of Space" (2nd ed), Jeffrey Weeks talks about a technique called "Cosmic Cristallography" to find the global topology of the space we live in.  Starting with the assumption that we're living in a closed 3-manifold and that the observable universe is larger than the real universe, he says that if we see multiple images of the Milky Way, the 3-manifold is multiconnected (like e.g. a 3-torus) while otherwise it would be simply connected.
I don't get this.  Wouldn't we see multiple images even if we lived in a 3-sphere?
Imagine you're living on a 2-sphere and your world is 2-dimensional and there's a southbound light ray starting at the North Pole, "traveling" along a geodesic (meridian) and reaching, say, Hamburg.  If there's another light ray in the opposite direction it'll travel around the South Pole but eventually also reach Hamburg, won't it?
EDIT: In reply to a comment, here a some more details.
In a previous chapter, Weeks says that the focus is on "homogeneous, isotropic 3-manifolds".  He then includes a list of possible candidates with their geometries which includes the 3-sphere (elliptic) and the 3-torus (Euclidean).
The relevant sentences then are: "In a multiconnected space we see multiple images of ourselves.  So testing whether the real universe is multiconnected or not is easy, right?  We just point our telescopes out into the night sky. If we see images of our Milky Way galaxy out there, then the universe is multiconnected. If we don't see images of the Milky Way, then either space is simply connected, or it's multi connected but on too large a scale for us to observe it."

REPLY [4 votes]: I emailed Jeffrey Weeks and asked him the same question.  He just replied and said that this is actually an error in the book and that you'd also see multiple images if you were inside a 3-sphere.  (He also said, BTW, that a 3rd edition of the book is due for 2019.  I can really recommend the book which explains topology and geometry in an informal and intuitive way.)


TITLE: What are some examples of self-adjoint functors? Is this an example?
QUESTION [13 upvotes]: I've been trying to figure out what some examples of "self-adjoint" functors are, or when this even happens,  since I've never seen this before. What I mean is if $F: \mathcal{C} \to \mathcal{C}$ is a functor, then $F$ is self-adjoint if we have the natural bijection
$$
\text{Hom}_{\mathcal{C}}(F(A), B) \cong \text{Hom}_{\mathcal{C}}(A, F(B))
$$
for all objects $A, B \in \mathcal{C}$.
In terms of the unit/counit language, the universal diagrams would be as below.

(Forgive the abuse of $f$ and $g$.)
The second half of this question is the following: Is $(-)^{\text{op}}: \textbf{Cat} \to \textbf{Cat}$ an example? I think so. This is because $(\mathcal{A}^{\text{op}})^{\text{op}}= \mathcal{A}$ for any category $\mathcal{A}$. Hence we can take the $\eta_{\mathcal{A}} = \epsilon_{\mathcal{A}} = 1_{\mathcal{A}}$, so the unit and counits are trivial. The unique existence of a functor completing the commutative triangles is just given by $(-)^{\text{op}}$, e.g., if we have a functor $g: \mathcal{A} \to \mathcal{B}^\text{op}$, then take $f = g^{\text{op}}: \mathcal{B} \to \mathcal{A}^{\text{op}}$. The diagram on the above left commutes, and similar reasoning gives us the counit diagram on the right. However, this example seems a bit trivial that happens to work out because $(-)^{\text{op}}$ is a nice functor; hence my question regarding more interesting examples.

REPLY [14 votes]: This is a relatively uncommon scenario, but here are a few more examples.

As you suggest, $(-)^\mathrm{op} : \mathbf{Cat} \to \mathbf{Cat}$ is self-adjoint. More generally, this should be true for the underlying category of any 2-category with a duality involution.
If $\mathscr C$ has biproducts, so products coincide with coproducts, then the composite $\oplus \circ \Delta_n$ of the (discrete $n$-ary) diagonal functor $\Delta_n$ with the ($n$-ary) biproduct functor $\oplus$ is self-adjoint.
For completeness, the identity functor is self-adjoint.

There are also similar examples that involve a variance change, i.e. functors $F : \mathscr C^\mathrm{op} \to \mathscr{C}$, such that $F \dashv F^\mathrm{op}$. These are called self-adjoint on the left. The name comes from the characterisation in terms of hom-sets of $\mathscr C$, i.e. we have natural isomorphisms $\mathscr C(F(A), B) \cong \mathscr C(F(B), A)$. Conversely, if $F$ is self-adjoint on the right, then we have natural isomorphisms $\mathscr C(A, F(B)) \cong \mathscr C(B, F(A))$.

The contravariant powerset functor $\mathcal P: \mathbf{Set} \to \mathbf{Set}^\mathrm{op}$ is left-adjoint to $\mathcal{P}^\mathrm{op} : \mathbf{Set}^\mathrm{op} \to \mathbf{Set}$, i.e. self-adjoint on the right.
More generally, in a symmetric monoidal closed category $(\mathscr C, \otimes, I, \multimap)$, for a fixed object $A$, the functor $(-) \multimap A$ is self-adjoint on the right.
In a similar vein, functors self-adjoint on the right are used in Thielecke's Categorical Structure of Continuation Passing Style to describe the structure of CPS (see Example 4.3.2).


TITLE: Why it is more dangerous to touch a high voltage line wire where current is actually less than households?
QUESTION [13 upvotes]: I have seen that power is transmitted from power stations to households at high voltage and low current to minimize the power loss.
That means the current in the transmission line is less than the current in the household wiring as there using a transformer we decrease the voltage and current is increased to keep the same power.
As it's the current that is dangerous, as it means how much charge flows per time unit: why are transmission lines more dangerous than household lines, even though voltage is high, but current is less?

REPLY [0 votes]: Your language is imprecise. Voltage is not a property of a location, it's a property of two locations; it's the difference between the electrical potential between two points. It's not quite true that "it's the current that is dangerous", as the current is not the only factor, and to the extent that it is accurate, it needs the qualifier "the current through your body". The current flowing through a point that you are touching, by itself, is completely irrelevant to how much danger you're in (a wire with more current may be more dangerous, but the danger comes from the voltage causing the current, not the current itself). If you have less resistance than the power line (including the transformers), which you almost certainly do, then the current through you will be higher than the current through the wire, so the low level of current in the wire says little about how much danger you are in.

That means the current in the transmission line is less than the current in the household wiring as there using a transformer we decrease the voltage and current is increased to keep the same power.

That doesn't follow. If several households are connected in parallel to the power lines, then the current in the households could be less than the current through the power line.


TITLE: Analogues of the Riemann-Roch Theorem
QUESTION [15 upvotes]: In his thesis, Tate derives a Poisson formula on the adeles and a theorem which he calls the "Riemann-Roch Theorem". More specifically, given a continuous, $L^1$ function $f$ on the adeles such that certain sums converge uniformly, then for all ideles $a$, we have
$\frac{1}{|a|}\displaystyle\sum_{\xi\in k}\hat{f}(\xi/a)=\displaystyle\sum_{\xi\in k}f(a\xi)$.
Tate further refers to this theorem as the "number theoretic analogue of Riemann-Roch". My question is how this relates to the geometric Riemann-Roch theorems and why this deserves to be called an analogue of these theorems.

REPLY [6 votes]: Perhaps you might want to look in the book project of Frankenhuijsen http://research.uvu.edu/Math/machiel/RH.pdf and search for Riemann Roch theorem.

Quote from page 3: The function $ζ_C$ (zeta function of a curve) satisﬁes the functional equation $ζ_C (1 − s) = ζ_C (s)$. This functional equation can be proved using the Riemann–Roch theorem
  $$l(D) = deg D + 1 − g + l(K − D),$$
  which is the analogue of (1) above.


TITLE: Recursive function
QUESTION [1 upvotes]: For a sequence $a$, $a_1=2$, $$a_n=\frac{n-1}{a_{n-1}}+n-1$$Express $a_n$ in terms of $n$.

I tried keep expanding, got many level of fraction until $n=1$, but I still can't see the pattern. Please help. Thank you.

REPLY [2 votes]: I doubt very much that you’re going to get anything very nice. If you do the obvious thing and begin by calculating a few values, you get this:
$$\begin{array}{rcc}
n:&1&2&3&4&5&6\\\\
a_n=\frac{p_n}{q_n}:&2&\frac32&\frac{10}3&\frac{39}{10}&\frac{196}{39}&\frac{1175}{196}\\\\
p_n:&2&3&10&39&196&1175\\
q_n:&1&2&3&10&39&196\\
a_n:&2&1.5&3.\overline{3}&3.9&\sim5.025641&\sim5.994898\\\\
\end{array}$$
The last line is just to get an idea of the actual size of $a_n$, and the previous lines just give the numerator and denominator of the rational representation in the second line. It’s immediately apparent that $q_n=p_{n-1}$, and indeed if we set $a_n=\frac{p_n}{q_n}$ we have
$$a_{n+1}=\frac{n(p_n+q_n)}{p_n}\;,$$
$p_{n+1}=n(p_n+q_n)$ and $q_{n+1}=p_n$, and finally $p_{n+1}=n(p_n+p_{n-1})$.
In short, the sequence is entirely determined by the sequence $\langle p_n:n\in\Bbb N\rangle$ (where we may take $p_0=1$). This turns out to be OEIS A003048, and the entry gives the closed form
$$p_n=2n!-\left\lfloor\frac{n!+1}e\right\rfloor$$
and the exponential generating function
$$g(x)=\frac{2-e^{-x}}{1-x}\;.$$
(It is clear that $a_n\approx n$ for even moderatly large $n$.)


TITLE: If $p$ is a rational and $x$ is irrational, prove that $p+x$ and $px$ are irrational.
QUESTION [1 upvotes]: If $p$ is a rational and $x$ is irrational, prove that $p+x$ and $px$ are irrational.
Pf: Let $p$ be a rational such that $p= \dfrac{m}{n}$, $m, n \in \mathbb{Z}$ and $n \neq 0$ such that $m,n$ are not both even. An irrational number $x$, by definition, cannot be expressed in this form.
$$p+x = \dfrac{m}{n}+x = \dfrac{m+xn}{n}$$
$$px = \dfrac{mx}{n}$$
We defined $m, n \in \mathbb{Z}$. We wish to show that $m+xn$ is not an integer.
$+$ is an operation on $\mathbb{Z}$, which implies it must map pairs of elements of $\mathbb{Z}$ to an element of $\mathbb{Z}$. If  $nx = x+...+x = z$ ($n$ times) for some $z \in \mathbb{Z}$, $x$ must be an integer. In this case, $x$ is non integer so is $nx$. Therefore, $m+nx$ is also non-integer. This shows that $px$ and $p+x$ cannot be expresed in terms of division of two integers $\square$.

REPLY [0 votes]: If $nx=x+...+x=z$ ($n$ times) for some $z∈\mathbb Z$, x must be an integer.

This is incorrect. Consider $2\cdot\frac12=\frac12+\frac12=1$. This would actually tell us that $x=\frac{z}{n}$ and is rational. Personally, I think your proof is fine if you remedy this. But you must also take into account that $p\ne0$.

We defined $m, n \in \mathbb{Z}$. We wish to show that $m+xn$ is not an integer.
  $+$ is an operation on $\mathbb{Z}$, which implies it must map pairs of elements of $\mathbb{Z}$ to an element of $\mathbb{Z}$. If  $nx$ $\color{red}{\text{is an integer and $n$ is nonzero}}$, $x$ must be $\color{red}{\text{rational}}$. In this case, $x$ is $\color{red}{\text{irrational}}$ so $\color{red}{\text{$nx$ cannot be an integer}}$. Therefore, $m+nx$ is also non-integer. This shows that ${\color{red}{\text{if $p\neq0$}\text{, then}}}$ $px$ and $p+x$ cannot be expressed in terms of division of two integers $\square$.


TITLE: I am having a doubt in graphs of $4πr^2|\psi|^2$ vs $r$ and $4πr^2|R(r)|^2$ vs $r$
QUESTION [0 upvotes]: To show radial probability, in some sources they used the graph $4πr^2|\psi|^2$ vs $r$
In some other sources, they used the graph $4πr^2|R(r)|^2$ vs $r$
However, $\psi(r,\theta,\phi)=R(r)Y(\theta,\phi)$, where $R(r)$ is the radial function, and $Y(\theta, \phi)$ is the angular function
But the same graph is drawn for both of these equations. So does $Y(\theta,\phi)$ have no significance in graph? If not, what is the explanation?

REPLY [0 votes]: For Hydrogen atom, the wave function look like
$$\psi(r,\theta ,\phi)=R_{El}(r)Y_l^m(\theta,\phi)$$

When one is interested in only radial distribution, that is How the probability changes with radial distance.
$$|\psi(r)|^2=\int |\psi(r,\theta,\phi)|^2d\Omega=\int|R_{El}(r)|^2|Y_l^m(r,\theta)|^2d\Omega=|R_{El}(r)|^2\int Y_l^m(r,\theta)|^2d\Omega \propto|R_{El}(r)|^2$$
So It doesn't matter.


TITLE: To what does energy belong in an energy diagram?
QUESTION [0 upvotes]: I was reading this answer and I saw an energy diagram:

Does the energy belong to the system or to the outside?

REPLY [1 votes]: The big arrow labelled energy is just an axis. It's supposed to indicate the upwards is positive, i.e. energy is being added to the system, and downwards is negative, i.e. energy is being emitted from the system.
Re the extra question:

where does the energy released in a nuclear reaction go, is it kinetic energy or heat?

Heat is a collective phenomenon. It doesn't make sense to ask about how much heat a single object has. At the individual particle level all reactions produce either photons or kinetic energy. In the latter case the energy is released as the kinetic energy of the reaction products. There is a discussion of this in my answer to Can we make usable energy from subnuclear particles?.


TITLE: Determine the ideal $\langle 2i \rangle$ of the Gaussian integers $\Bbb Z[i]$. Describe the quotient $\Bbb Z[i]/\langle 2i\rangle$.
QUESTION [1 upvotes]: Determine the ideal $\langle 2i \rangle$ of the Gaussian integers $\Bbb Z[i]$. Describe the quotient $\Bbb Z[i]/\langle 2i\rangle$.

I am having confusion with the theorem for $\langle 2i \rangle$. It's proved in my course book that for a ring $R$ and $a \in R$ the ideal $\langle a \rangle = \{ra \mid r  \in R\}$. But I've also seen that $\langle a \rangle = \{na \mid n  \in \Bbb Z\}$.
So either $$\langle 2i \rangle = \{2ri \mid r \in \Bbb Z[i]\}=\{2(a+bi)i \mid a+bi \in \Bbb Z[i]\}=\{2ai -2b \mid a+bi \in \Bbb Z[i]\} = \{2(ai-b) \mid a+bi \in \Bbb Z [i]\}$$ or $$\langle 2i \rangle = \{2ni \mid n \in \Bbb Z\}$$
which one is it here?

REPLY [0 votes]: You should check whatever course or book you are using for the definition, but the standard one (for a commutative ring $R$) is the former: $\left\langle a\right\rangle=\left\{ra\mid r\in R\right\}$. If $R=\mathbb{Z}$, this is of course equivalent to the latter.


TITLE: Strategy In auction bidding
QUESTION [1 upvotes]: I was at an auction recently where three autograph signatures of Marilyn Monroe were up for auction, not as a single lot, but in three separate lots. The three lots were virtually identical, and all had the same opening bid amount ($1000). Assuming the number of interested bidders is unknown, but greater than three, what is the best bidding strategy for acquiring one of them at the lowest price?

REPLY [0 votes]: As Robert Israel points out, you'll need to specify the conditions.  For example, let's say that all the other bidders have a maximum amount that they can bid per lot, and you know that the maximum among these is $B_{max}$.  If you have unlimited money, and your goal is to acquire at least one of the lots at the "lowest price," then simply bid $B_{max}+2$ on the first two lots, then bid $B_{max}+1$ on the third lot.  If this is not what you had in mind, then you should specify the conditions that you are thinking of.


TITLE: Integral inequality given the bounds for derivative
QUESTION [3 upvotes]: Let $f:[0,1]\rightarrow \mathbb{R}$ be a continuously differentiable function such that $$\int_0^1 f(x) \, dx=0$$ and $m \leq f'(x) \leq M$ on $(0,1)$. Prove that $$\frac{m}{12} \leq \int_0^1 xf(x) \, dx \leq \frac{M}{12}.$$

This is just the last bonus question in our test yesterday. I wasn't able to answer of course. Though I did verify it by letting $\displaystyle f(x)= \left( x−\frac{1}{2} \right)^3$ for which I found $0\le f′(x)\le 3/4$ on $(0,1)$ and $$0 \le \int^1_0 xf(x) \, dx=\frac{1}{80} \le \frac{3}{48}.$$
I do not know how to prove this.

REPLY [3 votes]: One proof goes via integration by parts. I show you how to start:
\begin{align}
\int_0^1 1 \cdot x f(x)\,dx &= [(x-1)xf(x)]_0^1-\int_0^1(x-1)(f(x)+xf'(x))\,dx\\
&=-\int_0^1 xf(x)\, dx+\int_0^1(1-x)xf'(x)\, dx.
\end{align}
Here, we used that $\int_0^1f(x)\,dx=0$. Thus, we get
$$
2\int_0^1 xf(x)\,dx = \int_0^1 x(1-x)f'(x)\,dx.
$$
Can you proceed from here?
Update:
Next, since $x(1-x)>0$ on $(0,1)$ and $m\leq f'(x)\leq M$,
$$
m\int_0^1 x(1-x)\,dx \leq \int_0^1 x(1-x)f'(x)\,dx\leq M\int_0^1 x(1-x)\,dx.
$$
The argument is complete by noting that $\int_0^1 x(1-x)\,dx = \frac{1}{6}$.
(By the way, $f(x)=x-1/2$ gives the sharp estimate in both ends. Indeed, then $f'(x)=1$ so $m=M=1$. The integral $\int_0^1 xf(x)\,dx=1/12$.)


TITLE: Distinct equivalence classes of the relation $R$ defined on $\mathbb Z$ defined by $mRn$ if $5 \mid m^3 - n^3$
QUESTION [0 upvotes]: I am wondering how to describe the distinct classes of $[2]$ and $[3]$.
For $[2]$ I got $[2] = \{5k + 3: k\in \mathbb{Z}\}$.
For $[3]$ I got $[3] = \{5k + 2: k\in \mathbb{Z}\}$.
But apparently these are switched around. How come? This is from my exam that I recently wrote and I really thought I had it. Any help?
PS: Forgive me for the poor formatting. I am still trying to learn how to do it as I am relatively new to the site.

REPLY [1 votes]: $$0^3 \equiv 0 \pmod{5}$$
$$1^3 \equiv 1 \pmod{5}$$
$$2^3 \equiv 3 \pmod{5}$$
$$3^3 \equiv 2 \pmod{5}$$
$$4^3 \equiv 4 \pmod{5}$$
\begin{align}[2]&=\{n:2^3\equiv n^3 \pmod{5}\}\\&=\{n:3 \equiv n^3 \pmod{5}\}\\
&=\{n: n \equiv 2 \pmod{5}\} \end{align}


text:
\begin{document}

\title[The Cayley-Menger determinant is irreducible for~$n\geq3$]
{The Cayley-Menger determinant is irreducible
  for~$\boldsymbol{{\mathnormal n} \geq 3}$}

\author[Carlos~D'Andrea]{Carlos~D'Andrea}

\address{Department of Mathematics,
University of California at Berkeley,
Berkeley CA~94720, USA.}

\email{cdandrea@math.berkeley.edu}

\urladdr{http://math.berkeley.edu/\~{}cdandrea/}

\author[Mart{\'\i}n~Sombra]{Mart{\'\i}n~Sombra}

\address{Universit{\'e} de Lyon 1,
Laboratoire de Math{\'e}matiques
Appliqu{\'e}es de Lyon,
21 avenue Claude~Bernard,
69622 Villeurbanne Cedex,
France.}

\email{sombra@maply.univ-lyon1.fr}

\urladdr{http://maply.univ-lyon1.fr/\~{}sombra/}

\date{\today}

\subjclass[2000]{Primary 12E05; Secondary 52A38.}

\keywords{Volume of a simplex, Cayley-Menger determinant, irreducible
  polynomial.}

\thanks{C.~D'Andrea was supported by a Miller Research Fellowship.}

\begin{abstract}
We prove that the Cayley-Menger determinant of an $n$-dimensional simplex is an absolutely irreducible polynomial for $n\geq3.$
We also study the irreducibility of polynomials associated to
related geometric constructions.
\end{abstract}
\maketitle

\overfullrule=1mm

\vspace{-8mm}

\setcounter{section}{1}

\section*{}

Let $\{d_{ij} \, : \ 0\le i < j \le n\}$ be a set of
$\displaystyle \frac{n\, (n+1)}{2}$ variables and consider the square
$(n+2) \times (n+2)$  matrix
\begin{equation} \label{pn}
\CM_n:=
 \left[ \begin{array}{cccccc}
0 & 1         & 1        & 1        & \cdots & 1        \\[2mm]
1 & 0         & d_{0 1}^2 & d_{0 2}^2 & \cdots & d_{0 n}^2 \\[2mm]
1 & d_{0 1}^2  & 0        & d_{1 2}^2 & \cdots & d_{1 n}^2 \\[2mm]
1 & d_{0 2}^2  & d_{1 2}^2 & 0        & \cdots & d_{2 n}^2 \\[2mm]
\vdots &      &          &          & \ddots &          \\[2mm]
1 & d_{0 n}^2  & d_{1 n}^2 & d_{2 n}^2 & \cdots & 0
\end{array} \right]
\enspace.
\end{equation}
The multivariate polynomial $\Gamma_n:=\det(\CM_n)\in\Z[d_{01},d_{02},\ldots,d_{(n-1)\,n}]$
is the \textit{Cayley-Menger determinant}.

\smallskip

\par\noindent
Let $v_0, \dots, v_n \in \R^n$ be  a set of $n+1$ points and
denote by $S$ its convex hull in $\R^n$. This determinant gives a formula for the $n$-dimensional volume of $S$ in terms of
the Euclidean distances
$\{{\delta_{ij}}:=\dist(v_i, v_j) \, : \ 0\leq i<j\leq n\}$
among these points.
We have~\cite[~Sec. IV.40]{Blu53}, \cite[~Sec. 9.7]{Ber87}
$$
(-1)^{n+1}\, 2^n \, (n!)^2 \, \Vol_n(S)^2 =  \Gamma_n({\delta_{01}, \delta_{02}, \dots,
\delta_{(n-1)\, n}}) \enspace.
$$
This formula shows that $\Gamma_n$ is a  homogeneous
polynomial of degree $2\, n$.

The second polynomial $\Gamma_2$ can be completely
factorized, giving rise to
the well-known {\it Heron's formula}
for the area $A$ of a triangle with edge lengths $a$, $b$, and $c$:
\begin{equation} \label{chavo}
16 \, A^2 =
- \Gamma_2(a,b,c) =  (a+b+c)(-a+b+c)(a-b+c)(a+b-c)
\enspace.
\end{equation}

\vspace{20mm}

\begin{figure}[htbp]

\begin{picture}(0,0)

\put(-60,0){\epsfig{file=triangulo.eps, height= 23 mm}}

\end{picture}

\end{figure}

\vspace{-5mm}

Note also that the equation $\Gamma_n(\delta_{01},\delta_{02},\ldots,
\delta_{(n-1)\,n})=0$ gives a necessary and
sufficient condition for the points
$v_0,\ldots,v_n$ to lie in a proper affine subspace of $\R^n.$
\par The Cayley-Menger determinant can be also used for deciding
whether a set of positive real numbers $\{\delta_{ij}:\,0\leq i<j\leq n\}$
can be realized as the set of edge lengths of an $n$-dimensional simplex in $\R^n$:
in \cite[~Sec. 9.7.3]{Ber87} it is shown that this condition is
equivalent to $(-1)^{n+1}\,\Gamma_n(\delta_{01},\delta_{02},\ldots,
\delta_{(n-1)\,n})>0$.

\smallskip
\par The matrix $\CM_n$ also gives a criterion to determine if $n+2$ points in $\R^n$ lie in an $(n-1)$-dimensional sphere, and         to solve the related problem
of computing the radius of the sphere circumscribed around a simplex.
To do this, consider the $(1,1)$-minor
$$
\Delta_n:=\det\left[ \begin{array}{ccccc}
0         & d_{01}^2 & d_{02}^2 & \cdots & d_{0 n}^2 \\[2mm]
d_{01}^2  & 0        & d_{12}^2 & \cdots & d_{1 n}^2 \\[2mm]
d_{02}^2  & d_{12}^2 & 0        & \cdots & d_{2 n}^2 \\[2mm]
\vdots &            &          & \ddots &          \\[2mm]
d_{0 n}^2  & d_{1 n}^2 & d_{2 n}^2 & \cdots & 0
\end{array} \right] \in\Z[d_{01},d_{02},\ldots,d_{(n-1)\,n}]\enspace.
$$
From this expression we see that this is a homogeneous polynomial
of degree $2\,n+2$.

Assume now that $v_0,\ldots,v_n$ do not lie in a proper affine subspace,
so that $S$ is an $n$-dimensional simplex. The
radius $\rho(S)$ of the sphere circumscribed around $S$ is given by
\begin{equation}\label{kiko}
\rho(S)^2=-\frac12\frac{\Delta_n(\delta_{01},\delta_{02},\ldots,
\delta_{(n-1)\,n})}{\Gamma_n(\delta_{01},\delta_{02},\ldots,
\delta_{(n-1)\,n})}\enspace.
\end{equation}
Also, the condition for $n+2$ points $v_0,\ldots,v_{n+1}$ in $\R^n$ to lie in
the same sphere or hyperplane is given by the annulation of the
$(n+1)$-th polynomial
$\Delta_{n+1}(\delta_{01},\delta_{02},\ldots,
\delta_{n\,(n+1)})$ $=0$, see \cite[~Sec. 9.7.3.7]{Ber87}.
\par\smallskip
The third polynomial $\Delta_3$ factorizes  as
\begin{eqnarray} \label{donramon}
\Delta_3&=&-(d_{01}\, d_{23} + d_{02} \, d_{13} + d_{03} \, d_{12})\,
(d_{01}\, d_{23} + d_{02} \, d_{13} - d_{03} \, d_{12})\\[0mm]
&&(d_{01}\, d_{23} - d_{02} \, d_{13} + d_{03} \, d_{12})\,
(-d_{01}\, d_{23} + d_{02} \, d_{13} + d_{03} \, d_{12}) \nonumber
\enspace.
\end{eqnarray}
This is equivalent to Ptolemy's theorem,
which states that a convex quadrilateral with edge lengths $a,b,c,d$
and diagonals $e,f$ as in the picture,
is circumscribed in a circle if and only if $a\,c+b\,d=e\,f$.

\vspace{30mm}

\begin{figure}[htbp]

\begin{picture}(0,0)

\put(-60,0){\epsfig{file=ptolomeo.eps, height= 33 mm}}

\end{picture}

\end{figure}

\vspace{-3mm}

The key sources for the Cayley-Menger determinant are the classical
books by L. Blumenthal \cite{Blu53} and by M. Berger \cite{Ber87}.

\smallskip

This polynomial plays an important role in some problems of metric geometry.
It was first applied by K. Menger in 1928,
to characterize Euclidean spaces in metric terms alone~\cite[~Ch.~IV]{Blu53}.
It also appears in the metric characterization of
Riemannian manifolds of constant sectional curvature obtained by M. Berger \cite{Ber81}.

Another important result based on the Cayley-Menger determinant is the proof of the invariance of the volume
for flexible $3$-dimensional polyhedra
(the ``bellows'' conjecture), see
\cite{Sab96,CSW97,Sab98}.
There is also a huge literature about
applications to the study of
spatial shape of molecules (stereochemistry), see e.g. \cite{KD80,DM00}.

\bigskip

It is natural to ask whether Heron's formula~(\ref{chavo})
generalizes to higher dimensions,
that is whether $\Gamma_n$ splits as a product of linear forms.
Note also that $\Gamma_1= 2 \, d_{01}^2$.
The purpose of this paper is to prove that this is not possible for $n
\ge 3$.
Moreover, we show that for $n \ge 3$
the only factors
of $\Gamma_n$ in $\C[d_{0 1}, d_{0 2}, \dots, d_{(n-1)\, n}]$
are the trivial ones, that is either a constant or a constant multiple of
$\Gamma_n$.
In other words $\Gamma_n$ is {\it absolutely irreducible}.

\begin{thm} \label{irreduc}
The polynomial \ $\Gamma_n$
is irreducible over $\C[d_{0 1}, d_{0 2}, \dots,  d_{(n-1)\, n}]$
for $n\ge~3$.
\end{thm}

In a similar way, one may wonder whether $\Delta_n$ splits as a product of
simpler expressions, as in (\ref{donramon}).
Note that $\Delta_1=-d_{01}^4$ and $\Delta_2= 2\, d_{01}^2\,
d_{02}^2\, d_{12}^2$.
Again we can show that this is not possible for $n\ge 4$.

\begin{thm} \label{chilindrina}
The polynomial $\Delta_n$
is irreducible over $\C[d_{0 1}, d_{0 2}, \dots,  d_{(n-1)\, n}]$
for $n\ge~4$.
\end{thm}

As a straightforward consequence of this,
we find that the determinant of the general symmetric $n \times n$
matrix  with
zeros in the diagonal is an absolutely irreducible polynomial for
$n\ge 4$, see Remark \ref{nionio}.

\medskip

We can verify that $\Gamma_3$ is {\it twice} an integral polynomial
and the same holds for $\Delta_4$.
This does not affect  their irreducibility over $\C[d_{ij}]$:
2 is trivial factor as it is a unit of $\C[d_{ij}]$.
Nevertheless it is interesting to
determine how they split over $\Z[d_{ij}]$.
Recall that the {\it content} of an integral polynomial is defined as
the gcd of its coefficients.

\begin{thm} \label{contenido}
Let $n \in \N$, then both $\Gamma_n$ and $\Delta_{n+1}$
have content 1 for even $n$ and 2 for odd $n$.
\end{thm}

Let us denote  $\ov{\Z} $ the ring of algebraic integers, that is
the ring formed by  elements in the algebraic closure $\ov{\Q}$
satisfying a {\it monic} integral
equation.
It is well-known that an integral
polynomial is irreducible over
$\ov{\Z}[d_{ij}]$
if and only if it
is  irreducible over
$\C[d_{ij}]$ and has content 1.
Set
$$
\begin{array}{ll}
I_n := \left\{ \begin{array}{ll}
\Gamma_n&  \hspace{10mm} \mbox{ for {\it n} even} \enspace \\[2mm]
\Gamma_n/2  & \hspace{10mm}  \mbox{ for {\it  n} odd}  \enspace
\end{array} \right.
\quad, & \quad \quad
J_n := \left\{ \begin{array}{ll}
\Delta_n/2&  \hspace{10mm} \mbox{ for  {\it n} even} \enspace \\[2mm]
\Delta_n  & \hspace{10mm}  \mbox{ for {\it n} odd} \enspace
\end{array} \right.
\end{array}.
$$
Hence
Theorems~\ref{irreduc}, \ref{chilindrina} and \ref{contenido}
can  be equivalently rephrased as the fact that
$I_n$ and $J_{n}$ are irreducible over
$\ov{\Z}[d_{ij}]$
(and in particular over ${\Z}[d_{ij}]$)
for $n\ge 3$ and for $n \ge 4$, respectively.

\bigskip

Let $t_n$ be a new variable and set
$$
\Lambda_{n,n-1}:= \Gamma_n\big( d_{i n} \mapsto t_n \, : \, 0 \le i \le  n-1
\big)
\in \Z[d_{0 1}, d_{0 2}, \dots, d_{(n-2)\, (n-1)}][t_n] \enspace.
$$
Up to a scalar factor, $\sqrt{\Lambda_{n, n-1}}$ is the
formula for the volume of an
isosceles simplex $S(\tau) \subset \R^n$ with base
$B:= \Conv(v_0, \dots, v_{n-1})$ and
vertex $v_n$ equidistant at distance $\tau$ to the other
vertices.

\vspace{21mm}

\begin{figure}[htbp]

\begin{picture}(0,0)

\put(-38,-15){\epsfig{file=tetra.eps, height= 29 mm}}

\end{picture}

\end{figure}

\vspace{1mm}

In~\cite[Sec. 9.7.3.7]{Ber87} it is shown that
\begin{equation} \label{recurrencia}
\Lambda_{n, n-1}= -2 \, \Gamma_{n-1} \, t_n^2 - \Delta_{n-1} .
\end{equation}
The dominant term in this expression  corresponds with the geometric
in\-tui\-tion
$$
\Vol_n(S(\tau)) \sim \frac{\tau}{n} \, \Vol_{n-1} (B)
\quad \quad \mbox{for }  \tau \gg
0 \enspace.
$$
Assuming $\dim(B)=n-1$, note that when
$\tau=\rho(B)$ is the radius of the circle circumscribing $B$
we have $\Lambda_{n, n-1}=0$
and thus we recover (\ref{kiko}).

More generally, let $ 1 \le p \le n$
and set
$$
\Lambda_{n,p}  :=
\Gamma_n\big( d_{i \ell} \mapsto t_\ell \, : \
p+1 \le \ell \le n, \ 0 \le
i \le \ell-1 \big)
$$
where $\{t_2 ,\dots, t_n\}$ denotes a further group of variables;
in particular $\Lambda_{n, n} = \Gamma_n.$
It turns out that $\Lambda_{n,p}$ is a homogeneous evaluation of $\Gamma_n$, and
so $\Lambda_{n,p} $ is a homogeneous polynomial
in $\Z[d_{ij} \, : \ 0 \le i < j \le p ] [t_{p+1}, \dots, t_n]$
of degree $2\,n$.

\smallskip

Set $B_p:= \Conv(v_0, \dots, v_p) $ a $p$-dimensional
simplex with edge
lengths $\{ \delta_{i j} \, : 0 \le i < j \le p \}$, and
for $0 \ll \tau_{p+1} \ll \cdots \ll \tau_n$
set $S(\tau_{p+1}, \dots, \tau_n) \subset \R^n$ the
$n$-dimensional simplex
built from $B_p$ by
successively adjoining a vertex $v_\ell$ equidistant to the previous
vertices $v_0, \dots, v_{\ell-1}$
for $\ell = p+1, \dots, n$.
Up to a scalar factor, $ \sqrt{\Lambda_{n,p}}$ is
the formula for the volume of $S(\tau_{p+1}, \dots, \tau_n)$.
We have the recursive relation:

\begin{lem} \label{seniorbarriga}
$\displaystyle
\Lambda_{n, p } = -2 \, \Lambda_{ n-1,p}\, t_n^2 - \Lambda_{n-2,p}\,
t_{n-1}^4 $
for $ n \ge p+2$.
\end{lem}

\begin{proof}
From the determinantal expression of $\Delta_n$ we get
\begin{equation} \label{brujadel71}
\Delta_{n-1}(d_{i\, (n-1)}\mapsto t_{n-1} \, : \ 0 \le i \le n-2) = t_{n-1}^4 \,
  \Gamma_{n-2} \enspace,
\end{equation}
and so
by (\ref{recurrencia}) we have   $\displaystyle
\Lambda_{n, n-2 } = -2 \, \Lambda_{ n-1,n-2}\, t_n^2 - \Lambda_{n-2,n-2}\,
t_{n-1}^4 $
for $n\ge 2$.
The general case follows by evaluating $
d_{i \ell} \mapsto t_\ell$ for
$ p+1 \le \ell \le n-2$ and  $  0 \le
i \le \ell-1$ in both sides of this identity.
\end{proof}

\medskip

Theorem \ref{irreduc} is a particular case of the following:

\begin{prop} \label{doniaflorinda}
The polynomial \ $\Lambda_{n,p}$
is irreducible over $\C[d_{0 1}, d_{0 2}, \dots,  d_{(n-1)\, n}]$
if and only if $n\ge~3$ and $ 2 \le p \le n$.
\end{prop}

The following is a graphical visualization of this proposition.
We encircle the integral points $(n,p)$ such that
$\Lambda_{n,p}$ is
absolutely irreducible, and we mark with a cross the points where it is not.
The behavior of $\Gamma_n$ is read from the diagonal.

\vspace{30mm}

\begin{figure}[htbp]

\begin{picture}(0,0)

\put(-38,-15){\epsfig{file=lambdas.eps, height= 38 mm}}

\end{picture}

\end{figure}

\vspace{2mm}

\begin{proof}
First we will prove by induction that $\Lambda_{n,2}$ is absolutely
irreducible for $n\ge 3$.
Let $n=3$, and to simplify the notation set
$d_{01} \mapsto  a, d_{12} \mapsto  b, d_{02} \mapsto c, t_3\mapsto \tau$.
Identity (\ref{recurrencia}) and Heron's formula imply
\begin{equation} \label{Q3}
\hspace*{10mm} \frac{\Lambda_{3,2}}{2}= - \Gamma_2 \, \tau^2 - \frac{\Delta_2}{2}=
 (a+b+c)(-a+b+c)(a-b+c)(a+b-c) \, \tau^2 - a^2 \, b^2 \, c^2 \enspace.
\end{equation}
The polynomials
$
f:= (a+b+c)(-a+b+c)(a-b+c)(a+b-c)$ and $g:= - a^2 \, b^2 \, c^2 $
have no common factors over $\C[a,b,c]$
and so a (non trivial) factorization of $\Lambda_{3,2}$ should be of the
form
$$
\Lambda_{3,2} =  (\alpha \, \tau + \beta) ( \gamma \, \tau + \delta)
$$
with $\alpha \,  \gamma = 2\, f$, $\beta \,  \delta = 2\, g$ and $\alpha \, \delta +
\beta \, \gamma = 0$.
But this is impossible since  $\alpha,\gamma, \beta\,\delta$
have no common factor; we conclude that $\Lambda_{3,2}$ is irreducible.

\smallskip

Now let $n \ge 4$ and assume that  $\Lambda_{n-1,2} $
is irreducible.
By Lemma \ref{seniorbarriga}
$$
\Lambda_{n, 2 } = -2 \, \Lambda_{ n-1,2}\, t_n^2 - \Lambda_{n-2,2}\,
t_{n-1}^4 \enspace.
$$
The polynomials $\Lambda_{ n-1,2}$ and $\Lambda_{n-2,2}\,
t_{n-1}^4 $ are coprime,
since $\Lambda_{n-1,2}$
is irreducible
of degree $2\, n -2$
and  $\deg(\Lambda_{n-2,2}) = 2\, n -4$.
As before, this implies that any non trivial factorization
of $\Lambda_{n,2}$ should be of the form
$$
\Lambda_{n,2} = (-2 \, \Lambda_{n-1,2} \, t_n + \beta) ( t_n + {\delta})
$$
with $\beta, \delta \in
\C[ d_{i   j} \, : 0 \le i < j \le 2 ] [t_{3}, \dots, t_{n-1}] $
such that
$\beta \, \delta = -\Lambda_{n-2,2}\,
t_{n-1}^4 $ and
$-2\, \Lambda_{ n-1,2} \, \delta + \beta=0$.
But this is impossible
because $\Lambda_{ n-1,2}$ and  $\Lambda_{n-2,2}\,
t_{n-1}^4$ are coprime.
We conclude that $\Lambda_{n,2}$
is irreducible.

\smallskip

Now let $n\ge 3$ and $3 \le  p \le n$.
Suppose that we can write
$\Lambda_{n,p}= F \cdot G $ with
$F,G \in  \C[d_{ij} \, : \ 0 \le i <j \le p][t_{p+1} , \dots, t_n]$
homogeneous  of degree $\ge 1$.

The evaluation map $d_{k  \ell} \mapsto t$
($0 \le k \le \ell, p+1 \le \ell \le n$)
is  homogeneous and so
\begin{eqnarray*}
F'&:=&
F\big(d_{i\ell} \mapsto t_\ell \, : \ p+1 \le \ell \le n, \
0 \le i \le \ell-1\big)
\enspace , \\[2mm]
G'&=& G\big(d_{i\ell} \mapsto t_\ell \, : \ p+1 \le \ell \le n, \
0 \le i \le \ell-1\big)
\end{eqnarray*}
are also homogeneous polynomials of degree $\ge 1$, which would
give a non
trivial factorization
of $\Lambda_{n,2}$.
This shows that $\Lambda_{n,p}$ is also irreducible.

\smallskip

To conclude, we have to verify
that $d_{01} | \Lambda_{n,1}$ for all $n$, which follows by
checking that  $\Lambda_{n,1}(d_{01} \mapsto 0) =0$, due to the fact that the
second and third rows in the matrix defining $\Lambda_{n,1}(d_{01} \mapsto 0)$ coincide.
The remaining case $n=p=2$ corresponds to Heron's formula.
\end{proof}

\medskip

\begin{Proof}{Proof of Theorem \ref{chilindrina}.}
Set
$$
\Delta_n':= \Delta_n( d_{in}\mapsto 1 \, : \ 1 \le i \le n-1 ) \in
\Z[d_{01}, d_{02}, \dots, d_{(n-2)\,(n-1)}, d_{0n}]\enspace.
$$
From the determinantal expression of $\Delta_n$ we get
\begin{equation} \label{lapopis}
\Delta_n' =
d_{0n}^4 \, \Gamma_{n-1}\Big(\frac{d_{01}}{d_{0n}}, \dots,
\frac{d_{0\,(n-1)}}{d_{0n}}, \, d_{12}, d_{13}, \dots, d_{(n-2)\,(n-1)}\Big)
\enspace.
\end{equation}
Note that the partial degree of $\Gamma_{n-1}$ in the
group of variables
\begin{equation} \label{profesorgirafales}
\{ d_{0i} \, : \ 1\le i \le n-1\}
\end{equation}
is four.
Hence $\Delta_n'$ is the homogenization
of $\Gamma_{n-1}$ with respect to these variables,
with $d_{0n}$ as the homogenization variable.
This follows again from the same determinantal expression.

Now let $ p,q \in \C[d_{ij}]$ such that
$
\Delta'_n = F \cdot G$.
Since $\Delta_n'$ is homogeneous with respect to the variables
(\ref{profesorgirafales}), we have that $F$ and $G$ are also homogeneous with
respect to this group.
Now we dehomogenize this identity by setting $d_{0n}\mapsto 1$ and we find
$$
\Gamma_{n-1} = F( d_{0n}\mapsto 1)\cdot G( d_{0n}\mapsto 1).
$$
By Theorem~\ref{irreduc}, $\Gamma_{n-1}$ is irreducible for $n \ge 4$,
which implies that either $F( d_{0n}\mapsto 1) \in \C$ or
$ G( d_{0n}\mapsto 1) \in \C$.
This can only hold if $F$ or $G$ is a monomial in $d_{0n}$, but this
is impossible since  $d_{0n}$ is the homogenization variable.
We conclude that $\Delta'_n$ is irreducible.

\smallskip

Now suppose that $\Delta_n$ {\it can} be factorized, and let
$P, Q \in \C[d_{ij}]$ be homogeneous polynomials of degree $\ge 1$
such that $\Delta_n= P\cdot Q$.
This implies that
$\Delta_n'= P'\cdot Q'$ with
$$
P':=P( d_{in}\mapsto 1 \, : \ 1 \le i \le n-1 )
\quad , \quad \quad  Q':= Q( d_{in}\mapsto 1 \, : \ 1 \le i \le n-1 ) \enspace.
$$
Note that $\deg(\Delta_n') = \deg(\Gamma_{n-1}) +4= 2\, n+2$ and so
$\deg(\Delta_n') = \deg(\Delta_n)$.
This implies that both $\deg(P')= \deg(P)\ge 1 $ and $\deg(Q')=\deg(Q) \ge 1$, which
contradicts the irreducibility of $\Delta'_n$.
Hence $\Delta_n$ is irreducible.
\end{Proof}

\medskip

For the proof of Theorem~\ref{contenido} we need an auxiliary result.
Let $n \in \N$ and $\{ x_{ij} \, : \  1\leq~i<j\leq n\}$ a set of
${(n-1) \, n}/{2}$ variables.
Then set
$$
X_n:=\left[
\begin{array}{ccccc}
0&x_{12}&x_{13}&\ldots&x_{1n} \\
x_{12}&0&x_{23}&\ldots&x_{2n} \\
x_{13}&x_{23}&0&\ldots&x_{3n} \\
\vdots&&&\ddots& \\
x_{1n}&x_{2n}&x_{3n}&\ldots&0
\end{array}
\right]
$$
for the general symmetric matrix of order $n$ with zeros in the diagonal.

\begin{lem}\label{mod2}
For odd values of $n$,
the content of $\det(X_n)$ is divisible by 2.
\end{lem}

\begin{proof}
Set
$$
A_n:=\left[
\begin{array}{ccccc}
0&x_{12}&x_{13}&\ldots&x_{1n} \\
-x_{12}&0&x_{23}&\ldots&x_{2n} \\
-x_{13}&-x_{23}&0&\ldots&x_{3n} \\
\vdots&&&\ddots& \\
-x_{1n}&-x_{2n}&-x_{3n}&\ldots&0
\end{array}
\right]
$$
for the general {\it antisymmetric} matrix of order $n$.
Then
$$
\det(A_n) = \det(A_n^t) = (-1)^n \, \det(A_n) \in \Z[x_{ij}]
\enspace,
$$
which implies  $\det(A_n) =0$ because $n$ is odd; here $A_n^t$ denotes
the transpose of $A_n$.
On the other hand
$ X_n \equiv A_n \pmod{2}$  and so we conclude
$$
\det(X_n) \equiv \det(A_n) = 0 \pmod{2} \enspace.
$$
\end{proof}

\begin{Proof}{Proof of Theorem~\ref{contenido}.}
Let $c(n) \in \N$ be the content of $\Gamma_n$.
Lemma~\ref{mod2} shows that $2|c(n) $ for odd $n$, as the
Cayley-Menger matrix $\CM_n$
is symmetric of order $n+2$ with zeros in the
diagonal.
By Lemma \ref{seniorbarriga}
$$
\Lambda_{n, n-2} (t_n\mapsto 0) = -   \Gamma_{n-2} \,  t_{n-1}^4 \enspace.
$$
By definition
$ \Lambda_{n, n-2}(t_n\mapsto 0) $ is an integral
evaluation of $\Gamma_n$ and so $c(n)$ divides its content, that is
$c(n) | c(n-2)$.
We conclude by induction, checking  the statement directly for $n=1$ and
$n=2$.

\smallskip

Now let $c'(n) \in \N$ be the content of $\Delta_n$.
Lemma~\ref{mod2} shows that $2|c'(n) $ for even $n$, as the
matrix in the definition of $\Delta_n$
is symmetric of order $n+1$ with zeros in the
diagonal.
Identity (\ref{lapopis}) implies that $c'(n) |c(n-1)$, that is
$c'(n)=1$ for $n$ odd and $c'(n)|2$ for $n$ even; we conclude that
$c'(n)=2$ in this case.
\end{Proof}

\begin{rem} \label{nionio}
Set
$$
K_n := \left\{ \begin{array}{ll}
\det(X_n) &  \hspace{10mm} \mbox{ for {\it n} even} \enspace, \\[2mm]
\det(X_n)/2  & \hspace{10mm}  \mbox{ for {\it  n} odd}  \enspace.
\end{array} \right.
$$
As a byproduct of Theorems~\ref{chilindrina} and~\ref{contenido}, we find that
$K_n$ is an irreducible polynomial over $\overline{\Z}[x_{ij}\, : \ 1\le i <j\le n]$
for $n\ge 5$;
a direct verification shows that this is also true for $n=4$.

\end{rem}

\subsection*{Acknowledgments}
It is a pleasure to thank Victor Alexandrov from the Sobolev Institute of Mathematics
at Novosibirsk, who posed us this problem and constantly encouraged us to solve it.
We also thank him for pointers to the litterature on the Cayley-Menger
determinant.

We are grateful to Mat{\'\i} as Gra{\~n}a for suggesting us a
simplified proof of Lemma~\ref{mod2}.

\smallskip

Part of this paper was written while Mart{\'\i} n Sombra was visiting the University of California at Berkeley, supported by the Miller Institute for Basic
Research in Science.

\typeout{Referencias}

config: arxiv


TITLE: GCD in a subring is GCD in a bigger ring?
QUESTION [5 upvotes]: Let $R$ be a UFD which is a subring of an integral domain $S$. If $r_1$ and $r_2$ are two nonzero elements of $R$ with GCD $d$, is it true that $d$ is also a GCD of $r_1$ and $r_2$ in $S$?
I know this is true if $R$ is a PID.

REPLY [4 votes]: No, e.g. $\rm\:\gcd(2,x) = 1\:$ in $\rm\:\mathbb Z[x]\:$ but the gcd is the nonunit  $\:2\:$ in $\rm\:\mathbb Z[x/2]\subset \mathbb Q[x]$.
But gcds in a PID D persist in extension rings because the gcd may be specified by the solvability of (linear) equations over D   and such solutions always persist in extension rings, see here.


TITLE: Car Crash Scenario
QUESTION [1 upvotes]: Two vehicles travelling at 80mph in the same direction.
Vehicles are directly behind each other.
12 meter distance between them.
Front door of car in front rips of and hits the front window of the car behind.
At what speed did the door make contact with the car?
How would one go about calculating this?
====
My current guess is as follows: The car behind would take approximately 0.34 seconds to reach the door in front, the door in front would have a forward velocity starting at 80mph but would quickly be decelerating due to no longer being attached to the car in front. Considering the car behind reach the door in approximately 0.34 seconds, i would guess that the speed of the door would be still very close to 80mph.
But this is a pure guess, is there a formula i can use here to accurately calculate this?

REPLY [0 votes]: There's a good way to get an estimate out of this.
First, perform dimensional analysis to get an estimate.
Start with, finding ${{\Delta a} \over {\Delta v}}$, which is the change in the drag acceleration with respect to velocity. The argument that follows is physical rather than technical, so beware. The above expression, should be proportional to the density of air $\rho$ and the surface area exposed $A$. In addition, the more mass the object has, the greater its momentum, and the harder it is to accelerate. Thus, ${{\Delta a} \over {\Delta v}}$ is inversely proportional to mass. Finally, the faster an object changes its velocity the greater the change in force. So the above is proportional to $v$.
Putting the above together yields,
$${{d a} \over {d v}}=C_d \cdot {{\rho \cdot A} \over {m}} \cdot v$$
Where we have changed $\Delta$ to $d$ since we'll examine this at infinitesimal changes in the variables. $C_d$ is referred to as the "drag coefficient". Summing the above expression over all changes in v, will yield the drag acceleration.
$$da=C_d \cdot {{\rho \cdot A} \over {m}} \cdot v \ dv$$
$$\rightarrow a_d=C_d \cdot {{\rho \cdot A} \over {2 \cdot m}} \cdot v^2$$
We have to make a couple more assumptions now, and then we can get the total acceleration on the door. First of all, when the door detaches from the car, we assume it'll hit the window on the wide side of the door. Thus, we will simplify the door to a rectangular block. Secondly, we'll assume the door does not spin while in the air. Third, we assume that the speed is fast enough to ignore Reynolds number phenomena. Last, we will assume that door movement is purely horizontal. Using these assumptions, we get,
$$a=C_d \cdot {{\rho \cdot A} \over {2 \cdot m}} \cdot v^2=\lambda \cdot v^2$$
Where $\lambda=C_d \cdot {{\rho \cdot A} \over {2 \cdot m}}$.
To solve this equation, we notice that the rate of change of velocity is acceleration.
$${{dv} \over {dt}}=\lambda \cdot v^2$$
$$\rightarrow {{dv} \over {\lambda \cdot v^2}}=dt$$
We once again sum over the $dv$ and $dt$ and get, taking the integral.
$${{-1} \over {\lambda \cdot v}}=t$$
$$\rightarrow v={{-1} \over {\lambda \cdot t}}$$
Of course, now we note that we have an initial velocity term and proceed to add it in the denominator so that at $t=0$ the velocity is 80mph or $35.75$ meters per second.
$$\rightarrow v={{1} \over {\lambda \cdot t+1/v_0}}$$
Where we've eliminated a sign, and have created $v_0$ for the original velocity.
Note that velocity is the rate of change of position, and integrate again.
We get,
$$x={1 \over {\lambda}} \cdot \ln(\lambda \cdot v_0 \cdot t+1)$$
Sadly, after all this work, you can't solve the equation to get the time to impact the car.
The drag coefficient for a plate is $2$. The density of air is $1.2754$ kilograms per $m^3$. The surface area shall be estimated as 1 $m^2$. The weight will be $30$ kilograms. The initial velocity is $35.75$ meters per second.
Substituting this with the condition for collision gives.
$$v_0 \cdot t-12={1 \over {\lambda}} \cdot \ln(\lambda \cdot v_0 \cdot t+1)$$
Solving yields the time of collision at $t=0.836$. Hope you like overkill ;)
Here's a reference to the concept of free fall, and the forces that act on objects in mid fall.
Conclusion
In other words, this envelope calculation basically shows that the problem is ill-posed. Since the door is to hit the road before the car, there can be no scenario where it hits a window. At this point, the reader should accept this is just a home-work problem and move on...
(Posted from mobile)


TITLE: Algebraic way to see why only $n=3$ is a valid coefficient
QUESTION [1 upvotes]: I'm a bit of a sucker for brute force calculations. Say I want to calculate a coefficient with Fourier theory, in my case
\begin{align*}
a_n = \int_0^1 \sin (3\pi x) \cos (n\pi x) dx.
\end{align*}
Clearly, only $a_3$ is nonzero. But if I calculate this integral explicitly, I get
\begin{align*}
a_n = \frac{3(-1)^{n+1} - 3}{\pi(n^2-9)}.
\end{align*}
I'm not immediately seeing that only $n=3$ will yield a nonzero result here... So how can I evaluate the integral so that only $a_3$ remains?

EDIT: Some extra notes on my calculation. We have
\begin{align*}
I = \int \sin(3\pi x)\cos(n\pi x) dx = \int u \frac{dv}{dx} dx,
\end{align*}
where
\begin{align*}
u = \sin(3\pi x) \Rightarrow \frac{du}{dx} = 3\pi \cos(3\pi x),\\
\frac{dv}{dx} = \cos(n\pi x) \Rightarrow v = \frac1{n\pi}\sin(n\pi x),
\end{align*}
such that
\begin{align*}
I &= \sin(3\pi x) \cdot \frac1{n\pi}\sin(n\pi x) - \int 3\pi \cos(3\pi x) \cdot \frac1{n\pi}\sin(n\pi x) dx \\
&= \frac1{n\pi}\sin(3\pi x)\sin(n\pi x) - \frac 3n\int \cos(3\pi x)\sin(n\pi x) dx \\
&= \frac1{n\pi}\sin(3\pi x)\sin(n\pi x) - \frac 3n J.
\end{align*}
Now repeat the calculation with $J$,
\begin{align*}
J = \int \cos(3\pi x)\sin(n\pi x) dx
\end{align*}
where
\begin{align*}
u = \cos(3\pi x) \Rightarrow \frac{du}{dx} = -3\pi \sin(3\pi x),\\
\frac{dv}{dx} = \sin(n\pi x) \Rightarrow v = -\frac1{n\pi}\cos(n\pi x),
\end{align*}
such that
\begin{align*}
J &= \cos(3\pi x) \cdot \left(-\frac1{n\pi}\cos(n\pi x)\right) - \int 3\pi \sin(3\pi x) \cdot \frac1{n\pi}\cos(n\pi x) dx \\
& = -\frac1{n\pi}\cos(3\pi x) \cos(n\pi x) - \frac3n \int \sin(3\pi x) \cos(n\pi x) dx \\
& = -\frac1{n\pi}\cos(3\pi x) \cos(n\pi x) - \frac3n I
\end{align*}
Plugging in $J$ in the expression for $I$, we get
\begin{align*}
I &= \frac1{n\pi}\sin(3\pi x)\sin(n\pi x) - \frac 3n J \\
&= \frac1{n\pi}\sin(3\pi x)\sin(n\pi x) - \frac 3n \left[-\frac1{n\pi}\cos(3\pi x) \cos(n\pi x) - \frac3n I\right] \\
&= \frac1{n\pi}\sin(3\pi x)\sin(n\pi x) + \frac3{n^2\pi}\cos(3\pi x) \cos(n\pi x) + \frac9{n^2} I \\
I &= \frac{\frac1{n\pi}\sin(3\pi x)\sin(n\pi x) + \frac3{n^2\pi}\cos(3\pi x) \cos(n\pi x)}{1-\frac9{n^2}} \\
&= \frac1\pi \frac{n\sin(3\pi x)\sin(n\pi x) + 3\cos(3\pi x) \cos(n\pi x)}{n^2-9}.
\end{align*}
On the interval $0 < x < 1$, we then get
\begin{align*}
I \Big|_{0}^1 &= \frac1\pi \frac{n\sin(3\pi)\sin(n\pi) + 3\cos(3\pi) \cos(n\pi)}{n^2-9} - \frac1\pi \frac{n\sin(0)\sin(0) + 3\cos(0) \cos(0)}{n^2-9}\\
&= \frac1\pi \frac{3\cos(3\pi) \cos(n\pi)}{n^2-9} - \frac1\pi \frac{3}{n^2-9} \\
&= \frac1\pi \frac{-3(-1)^n - 3}{n^2-9},
\end{align*}
where in the last step I used $\cos(n\pi) = (-1)^n$.

REPLY [1 votes]: Considering $$a_n = \int_0^1 \sin (3\pi x) \cos (n\pi x)\, dx$$ we can make the calculations faster using $$\sin(a) \cos(b)=\frac 12\big(\sin(a+b)+\sin(a-b)\big)$$ So $$2a_n=\int_0^1 \sin\big((3+n)\pi x\big)\, dx+\int_0^1 \sin\big((3-n)\pi x\big)\, dx$$ which already shows that, for $n=3$, the second integral disappears.
So $$a_3=\frac{1+\cos (3\pi  )}{2 \pi  (n+3)}=0$$ For the other cases $(n\neq 3)$, after simplifications, $$a_n=-\frac{3 (1+\cos (\pi  n))}{\pi  \left(n^2-9\right)}$$ which implies $$a_{2n+1}=0 \quad \quad  \quad a_{2n}=-\frac{6}{\pi  \left(4 n^2-9\right)}$$