text:
\begin{document}
\sc
\title[Multiple Hermite and Laguerre polynomials]
{Integral representations for multiple Hermite and multiple
Laguerre polynomials}

\author{Pavel M. Bleher}
\address{Department of Mathematical Sciences, Indiana
University-Purdue University Indianapolis,
402 N. Blackford St., Indianapolis, IN 46202, U.S.A.}
\email{bleher@math.iupui.edu}
\author{Arno B.J. Kuijlaars}
\address{Department of Mathematics,
Katholieke Universiteit Leuven,
Celestijnenlaan 200 B,
B-3001 Leuven
BELGIUM}
\email{arno@wis.kuleuven.ac.be}

\dedicatory{Dedicated to Pierre van Moerbeke on his sixtieth
birthday}
\date{\today}

\thanks{The first author was supported in part by
NSF Grant DMS-9970625. The second author was supported by projects
G.0176.02 and G.0455.04 of FWO-Flanders.}

\rm

\begin{abstract}
We give integral representations for multiple Hermite and
multiple Hermite polynomials of both type I and II. We also
show how these are connected with double integral representations
of certain kernels from random matrix theory.
\end{abstract}

\maketitle

\section{Multiple orthogonal polynomials}

Multiple orthogonal polynomials are an extension of  orthogonal
polynomials that play a role in the random matrix ensemble with an
external source
\begin{equation} \label{ensemble}
    \frac{1}{Z_n} e^{- \Tr (V(M) - AM)} dM
\end{equation}
defined on $n \times n$ Hermitian matrices, see \cite{BK1,BK2,DK}.
Here $A$ is a fixed $n \times n$ Hermitian  matrix and $V : \mathbb R \to \mathbb R$ is
a function with enough increase at $\pm \infty$ so that the
integral
\[ Z_n = \int e^{- \Tr (V(M) - AM)} dM \]
converges. Random matrices with external source were introduced and studied
by Br\'ezin and Hikami \cite{BH1,BH2,BH3,BH4,BH5}, and
P. Zinn-Justin \cite{ZJ1,ZJ2}.

In what follows, we assume that $A$ has $m$ distinct eigenvalues
$a_1, \ldots, a_m$ of multiplicities $n_1, \ldots, n_m$. We consider
$m$ fixed and use  multi-index notation $\vec{n} = (n_1,
\ldots, n_m)$ and $|\vec{n}| = n_1 + \cdots + n_m$.

The average characteristic polynomial $P_{\vec{n}}(x) = \mathbb E \left[\det(xI - M) \right]$
of the ensemble (\ref{ensemble}) is a monic polynomial of degree $|\vec{n}|$
which  satisfies for $k=1, \ldots, m$,
\begin{equation} \label{typeII}
    \int_{-\infty}^{\infty} P_{\vec{n}}(x) x^j w_k(x) dx  = 0,
        \qquad j=0, \ldots, n_k-1,
\end{equation}
where
\begin{equation} \label{weights}
    w_k(x) = e^{- (V(x) - a_k x)},
\end{equation}
see \cite{BK1}.
The relations (\ref{typeII}) characterize the polynomial $P_{\vec{n}}$ uniquely.
For $A = 0$, we have that $P_{\vec{n}}$ is the usual orthogonal polynomial with respect
to the weight $e^{-(V(x))}$, which is a well-known fact from random
matrix theory. For general  $m$, the relations (\ref{typeII}) are multiple
orthogonality relations with respect to the weights (\ref{weights})
and the polynomial $P_{\vec{n}}$ is called a multiple orthogonal polynomial of type II.

The multiple orthogonal polynomials of type I consist of a vector
\begin{equation} \label{degreeAk}
    (A_{\vec{n}}^{(1)}, A_{\vec{n}}^{(2)}, \ldots, A_{\vec{n}}^{(m)}),
    \qquad \deg A_{\vec{n}}^{(k)} \leq n_k - 1,
\end{equation}
of polynomials such that the function
\begin{equation} \label{Qfunction}
    Q_{\vec{n}}(x) = \sum_{k=1}^m A_{\vec{n}}^{(k)}(x) w_k(x)
\end{equation}
satisfies
\begin{equation} \label{typeI}
    \int_{-\infty}^{\infty} x^j Q_{\vec{n}}(x) dx = \left\{ \begin{array}{ll}
        0, & \quad j = 0, \ldots, |\vec{n}|-2, \\
        1, & \quad j = |\vec{n}|-1.
        \end{array} \right.
\end{equation}
The polynomials $A_{\vec{n}}^{(k)}$ are uniquely determined by the
degree requirements (\ref{degreeAk}) and the type I
orthogonal relations (\ref{typeI}).

By the Weyl integration formula, the random matrix ensemble (\ref{ensemble})
has the following joint eigenvalue distribution
\[ \frac{1}{\tilde{Z}_n}
    \prod_{j=1}^n e^{- V(\lambda_j)} \left( \int e^{A U \Lambda U^*} dU \right)
        \prod_{j <k} (\lambda_j - \lambda_k)^2 \ d\lambda_1 d\lambda_2 \cdots d\lambda_n \]
where $dU$ is the normalized Haar measure on the unitary group $U(n)$
and $\Lambda = \diag(\lambda_1, \ldots \lambda_n)$. Using
the confluent form of the Harish-Chandra / Itzykson-Zuber formula
\cite{HC,IZ} to evaluate
the integral $\int e^{A U \Lambda U^*} dU$, we find that the
joint eigenvalue distribution can be written as
\begin{equation} \label{biorthogonal1}
    \frac{1}{Z'_n} \det\left(f_j(\lambda_k)\right)_{1\leq j, k \leq n}
    \det\left(g_j(\lambda_k)\right)_{1\leq j,k \leq n} \ d\lambda_1 d\lambda_2 \cdots d\lambda_n,
\end{equation}
where $f_j(x) = x^{j-1}$ and $g_1, \ldots, g_n$ are the functions
\begin{equation} \label{basisfunctions}
    x^j e^{-(V(x)-a_kx)}, \qquad k = 1, \ldots, m, \quad j=0,1, \ldots, n_k-1
\end{equation}
taken in some arbitrary order. Then (\ref{biorthogonal1}) is a
biorthogonal ensemble in the sense of Borodin \cite{Bor}, which in
particular implies that the eigenvalue point process is determinantal,
that is, there is a kernel $K(x,y)$ such that
the $k$ point correlation functions have determinantal form
\begin{equation} \label{correlations}
    \det\left( K(\lambda_i, \lambda_j) \right)_{1\leq i,j\leq k}.
\end{equation}

The multiple orthogonal polynomials of types II and I can be used to
biorthogonalize the functions $f_j$ and $g_j$ and to give an explicit
formula for $K$. Choose a sequence of
multi-indices $\vec{n}_0, \vec{n}_1, \ldots, \vec{n}_n = \vec{n}$
such that $|\vec{n}_j| = j$ and $\vec{n}_{j} \leq \vec{n}_{j+1}$
and define
\[ p_j(x) = P_{\vec{n}_j}(x), \qquad q_j(y) = Q_{\vec{n}_{j+1}}. \]
Then $p_{j+1} \in \span\{f_1, \ldots, f_j\}$, and
$q_{j+1} \in \span\{g_1, \ldots, g_j\}$ for a suitable ordering
of the functions (\ref{basisfunctions}).
In addition we have the biorthogonality
\[ \int_{-\infty}^{\infty} p_i(x) q_j(x) dx = \delta_{i,j}. \]
As in \cite{Bor} it then follows that
\begin{equation} \label{kernelK}
    K(x,y) = \sum_{j=0}^{n-1} p_j(x) q_j(y).
\end{equation}
By the Christoffel-Darboux formula for multiple orthogonal
polynomials \cite{DK} the kernel $K$ satisfies
\begin{equation} \label{CDkernel}
    (x-y) K(x,y) =
    P_{\vec{n}}(x) Q_{\vec{n}}(y)
    - \sum_{k=1}^m \frac{h_{\vec{n}}^{(k)}}{h_{\vec{n}-\vec{e}_k}^{(k)}}
    P_{\vec{n}-\vec{e}_k}(x) Q_{\vec{n} + \vec{e}_k}(y)
\end{equation}
where
\begin{equation} \label{hnk0}
    h_{\vec{n}}^{(k)} = \int_{-\infty}^{\infty} P_{\vec{n}}(x) x^{n_k} w_k(x) dx
\end{equation}
and $\vec{e}_k$ is the $k$th standard basis vector in $\mathbb R^m$.

In the following two sections we study two special cases related to
multiple Hermite polynomials and multiple Laguerre polynomials.
These cases correspond to the random matrix model (\ref{ensemble})
with $V(M) = \frac{1}{2}M^2$ and $V(M) = M$
respectively (in the latter case we restrict to positive
definite matrices). Br\'ezin and Hikami \cite{BH2} and Baik, Ben Arous,
and P\'ech\'e \cite{BBAP} gave double integral representations for
the correlation kernels for these cases. We will derive integral
representations for the multiple Hermite and multiple Laguerre
polynomials, and use that to show that the kernels agree with
the multiple orthogonal polynomial kernel (\ref{CDkernel}).

\section{Multiple Hermite polynomials}

The special case $V(M) = \frac{1}{2} M^2$ was considered in a series
of papers of Br\'ezin and Hikami, \cite{BH1,BH2,BH3,BH4,BH5}. This case corresponds to
$M = H + A$ where $H$ is a random matrix from the GUE ensemble
$ (1/Z_n) e^{-\frac{1}{2} \Tr H^2} dH$
and $A$ is  fixed as before. In \cite{BH2} the following expression
for the kernel was derived
\begin{equation} \label{BHkernel}
    K(x,y) = \frac{1}{(2\pi i)^2}
    \int_{-i\infty}^{i \infty} ds  \oint_{\Gamma} dt \
        e^{ \frac{1}{2} (s-x)^2 - \frac{1}{2} (t-y)^2 } \prod_{k=1}^m
        \left(\frac{s-a_k}{t-a_k}\right)^{n_k} \frac{1}{s-t}
\end{equation}
where $\Gamma$ is a closed contour encircling the points
$a_1, \ldots, a_m$ once in the positive direction, and the
path from $-i\infty$ to $i \infty$ does not intersect $\Gamma$,
see also Johansson \cite{Joh}.

When $V(x) = \frac{1}{2} x^2$, the multiple orthogonal polynomials
are called multiple Hermite polynomials, since they clearly
generalize the usual Hermite polynomials \cite{Apt,ABV,VAC}. We derive integral
representations for the multiple Hermite polynomials of
type I and type II, which resemble the integral representation (\ref{BHkernel})
of the kernel.

\subsection{Multiple Hermite polynomials of type II}

The multiple Hermite polynomial $P_{\vec{n}}$ is the monic
polynomial of degree $|\vec{n}|$ that satisfies (\ref{typeII})
with $w_k(x) = e^{-\frac{1}{2}x^2+a_kx}$, $k=1, \ldots, m$.

\begin{theorem} The multiple Hermite polynomials of type II has the
integral representation
\begin{equation} \label{intrepH2}
    P_{\vec{n}}(x) = \frac{1}{\sqrt{2\pi}i}
    \int_{-i\infty}^{i \infty} e^{\frac{1}{2}(s-x)^2}
        \prod_{k=1}^m (s-a_k)^{n_k} \, ds
\end{equation}
\end{theorem}
\begin{proof}
Let us denote the left-hand side of (\ref{intrepH2}) by $P(x)$.
After performing the change of variables $s = t+x$, we get
\begin{equation} \label{intrepH2b}
    P(x) = \frac{1}{\sqrt{2\pi} i} \int_{-i\infty}^{i \infty} e^{\frac{1}{2} t^2}
    \prod_{k=1}^m (t+x-a_k)^{n_k} \, dt,
\end{equation}
which shows that $P$ is a polynomial of degree $|\vec{n}|$
with leading coefficient
\[ \frac{1}{\sqrt{2\pi} i} \int_{-i\infty}^{i \infty} e^{\frac{1}{2} t^2} dt = 1. \]
So $P$ is a monic polynomial.

Now we use (\ref{intrepH2b}) to compute for $k=1, \ldots, m$,
and $j = 0, 1, \ldots$,
\begin{eqnarray} \nonumber
    \lefteqn{ \int_{-\infty}^{\infty} P(x) x^j e^{-\frac{1}{2} x^2 + a_k x} dx
     = } \\
     && \label{intrepH2c}
     \frac{e^{\frac{1}{2} a_k^2}}{\sqrt{2\pi} i} \int_{-i \infty}^{i\infty} \int_{-\infty}^{\infty}
     e^{\frac{1}{2} (t^2 - (x-a_k)^2)} x^j \prod_{l=1}^m (t + x-a_l)^{n_l} \, dx dt.
\end{eqnarray}
Switching to polar coordinates
$x-a_k = r \cos \theta$, $t = ir \sin \theta$, we find
that the right-hand side of (\ref{intrepH2c}) is equal to
\begin{equation} \label{intrepH2d}
     \frac{e^{\frac{1}{2} a_k^2}}{\sqrt{2\pi}}
     \int_0^{\infty} e^{-\frac{1}{2} r^2} r^{n_k+1}
     \left[\int_0^{2\pi} (a_k + r\cos \theta)^j  e^{i n_k \theta}
        \prod_{l \neq k} (r e^{i \theta} + a_k - a_l)^{n_l} d\theta \right] dr.
\end{equation}
The $\theta$-integral vanishes for $j =0, \ldots, n_k-1$, since the integrand
can be written as a linear combination of $e^{ip\theta}$ with integer $p \geq n_k-j$.
Hence $P$ is the multiple Hermite polynomial of type II and (\ref{intrepH2}) follows.
\end{proof}

\begin{remark}
Evaluating (\ref{intrepH2d}) for $j= n_k$ we find that the $\theta$-integral
is $2\pi \left(\frac{r}{2} \right) \prod_{l \neq k} (a_k-a_l)^{n_l}$ and
\begin{eqnarray} \nonumber
    h_{\vec{n}}^{(k)} & = &
        \int_{-\infty}^{\infty} P_{\vec{n}}(x) x^{n_k} e^{-\frac{1}{2}x^2+a_kx} dx \\
        & = & \sqrt{2\pi} (n_k)! e^{\frac{1}{2} a_k^2} \prod_{l \neq k} (a_k-a_l)^{n_l}.
    \label{hnk}
\end{eqnarray}
\end{remark}

\subsection{Multiple Hermite polynomials of type I}
The multiple Hermite polynomials are polynomials $A_{\vec{n}}^{(k)}$
as in (\ref{degreeAk}) such that the linear form
\begin{equation} \label{QfunctionH}
    Q_{\vec{n}}(x) = \sum_{k=1}^m A_{\vec{n}}^{(k)}(x) e^{-\frac{1}{2} x^2 + a_k x}
\end{equation}
satisfies (\ref{typeI}).

\begin{theorem} The multiple Hermite polynomials of type I have the
integral representation
\begin{equation} \label{intrepH1}
    A_{\vec{n}}^{(k)}(x) e^{-\frac{1}{2} x^2 + a_kx} = \frac{1}{\sqrt{2\pi} 2\pi i}
    \oint_{\Gamma_k} e^{-\frac{1}{2}(t-x)^2}
        \prod_{l=1}^m (t-a_l)^{-n_l} \, dt
\end{equation}
where $\Gamma_k$ is a closed contour encircling $a_k$ once in the positive direction,
but not enclosing any of the other points $a_l$, $l\neq k$.

In addition the function $Q_{\vec{n}}$ of {\rm (\ref{QfunctionH})} has
the integral representation
\begin{equation} \label{intrepH1b}
    Q_{\vec{n}}(x) = \frac{1}{\sqrt{2\pi} 2\pi i}
    \oint_{\Gamma} e^{-\frac{1}{2}(t-x)^2}
        \prod_{l=1}^m (t-a_l)^{-n_l} \, dt
\end{equation}
where $\Gamma$ is a closed contour encircling $a_1, \ldots, a_m$ once in the
positive direction.
\end{theorem}
\begin{proof}
By the residue theorem, we have that the right-hand side of
(\ref{intrepH1}) is equal to
\begin{equation} \label{H1proof1}
    \frac{1}{\sqrt{2\pi}} \frac{1}{(n_k-1)!} \left. \left(\frac{d}{dt}\right)^{n_k-1}
    \left[e^{-\frac{1}{2}(t-x)^2} \prod_{l\neq k} (t-a_l)^{-n_l} \right] \right|_{t=a_k}.
\end{equation}
It is easy to see that (\ref{H1proof1}) has the
form $A_k(x) e^{-\frac{1}{2} x^2 + a_kx}$ where $A_k$ is a polynomial of degree $n_k-1$.
Define the linear form
\begin{equation} \label{H1proof2}
    Q(x) = \sum_{k=1}^m A_k(x) e^{-\frac{1}{2} x^2 + a_kx}
    = \frac{1}{\sqrt{2\pi} 2 \pi i} \oint_{\Gamma} e^{-\frac{1}{2} (t-x)^2}
        \prod_{l=1}^m (t-a_l)^{-n_l} \, dt
\end{equation}
where $\Gamma$ encloses all the points $a_j$, $j=1, \ldots, m$, once
in the positive direction.

Then
\[ \int_{-\infty}^{\infty} x^j Q(x) dx =
    \frac{1}{2\pi i} \oint_{\Gamma}
        \left( \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty}
            x^j e^{-\frac{1}{2}(t-x)^2} dx \right)
            \prod_{l=1}^m (t-a_l)^{-n_l} \, dt.
\]
Since
\[ \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} x^j e^{-\frac{1}{2} (t-x)^2}  dx =
    \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} (y+t)^j e^{-\frac{1}{2} y^2} dy
\]
    is a monic polynomial of degree $j$ in the variable $t$,
we find
\begin{equation} \label{H1proof3}
    \int_{-\infty}^{\infty} x^j Q(x) dx =
    \frac{1}{2\pi i} \oint_{\Gamma}
        \pi_j(t) \prod_{l=1}^m (t-a_l)^{-n_l} \, dt, \end{equation}
where $\pi_j$ is a monic polynomial of degree $j$.
Deforming the contour $\Gamma$ to infinity, and using the fact
that the integrand is $s^{j-|\vec{n}|} + O(s^{j-|\vec{n}| - 1})$
as $s \to \infty$, we find that
\[ \int_{-\infty}^{\infty} x^j Q(x) dx =
    0 \qquad \mbox{for } j=0, \ldots, |\vec{n}|-2 \]
and
\[ \int_{-\infty}^{\infty} x^j Q(x) dx =
    1 \qquad \mbox{for } j=|\vec{n}|-1. \]
This shows that $Q = Q_{\vec{n}}$ and $A_k = A_{\vec{n}}^{(k)}$
so that (\ref{intrepH1}) and (\ref{intrepH1b}) follow.
\end{proof}

\subsection{The multiple Hermite kernel}

Let us now show that the Br\'ezin Hikami kernel (\ref{BHkernel}) agrees
with the multiple Hermite kernel (\ref{CDkernel}).
To that end we compute $\frac{\partial K}{\partial x} + \frac{\partial K}{\partial y}$
for the kernel (\ref{BHkernel}) in two ways.

First we have
\begin{eqnarray*}
    \lefteqn{ \frac{\partial K}{\partial x} + \frac{\partial K}{\partial y}
    } \\
    && = \frac{1}{(2\pi i)^2}
    \int_{-i\infty}^{i \infty} ds  \oint_{\Gamma} dt \
        e^{ \frac{1}{2} (s-x)^2 - \frac{1}{2} (t-y)^2 } \prod_{k=1}^m
        \left(\frac{s-a_k}{t-a_k}\right)^{n_k} \frac{-s+x+t-y}{s-t} \\
    &&= (x-y) K(x,y) - \frac{1}{(2\pi i)^2}
    \int_{-i\infty}^{i \infty} ds  \oint_{\Gamma} dt \
        e^{ \frac{1}{2} (s-x)^2 - \frac{1}{2} (t-y)^2 } \prod_{k=1}^m
        \left(\frac{s-a_k}{t-a_k}\right)^{n_k}
\end{eqnarray*}
The last double integral factors into a product of two single integrals,
which by (\ref{intrepH2}) and (\ref{intrepH1b}) leads to
\begin{equation} \label{Kx+Ky1}
    \frac{\partial K}{\partial x} + \frac{\partial K}{\partial y}
    = (x-y) K(x,y) - P_{\vec{n}}(x)Q_{\vec{n}}(y).
\end{equation}

For the second way we evaluate $\frac{\partial K}{\partial x}$ by
noting that $\partial_x e^{\frac{1}{2}(s-x)^2} = - \partial_s e^{\frac{1}{2}(s-x)^2}$,
and integrating by parts the $s$-integral
\begin{eqnarray} \nonumber
    \frac{\partial K}{\partial x}
    & = &
    -\frac{1}{(2\pi i)^2}
    \int_{-i\infty}^{i \infty} ds  \oint_{\Gamma} dt \
        \frac{\partial}{\partial s} e^{ \frac{1}{2} (s-x)^2 -\frac{1}{2} (t-y)^2 }
        \prod_{k=1}^m
        \left(\frac{s-a_k}{t-a_k}\right)^{n_k} \frac{1}{s-t} \\
    &= & \nonumber
    \frac{1}{(2\pi i)^2}
    \int_{-i\infty}^{i \infty} ds  \oint_{\Gamma} dt \
         e^{ \frac{1}{2} (s-x)^2- \frac{1}{2} (t-y)^2 }  \\
         && \qquad \qquad \times \label{Kx}
        \prod_{k=1}^m
        \left(\frac{s-a_k}{t-a_k}\right)^{n_k} \frac{1}{s-t}
        \left\{ \sum_{k=1}^m \frac{n_k}{s-a_k} - \frac{1}{s-t} \right\}.
    \end{eqnarray}
Similarly, we use $\partial_y e^{-\frac{1}{2}(t-y)^2} = - \partial_t e^{-\frac{1}{2}(t-y)^2}$,
and apply integration by parts to the $t$-integral, to obtain
\begin{eqnarray} \nonumber
\frac{\partial K}{\partial y} & = &
    \frac{1}{(2\pi i)^2}
    \int_{-i\infty}^{i \infty} ds  \oint_{\Gamma} dt \
         e^{ \frac{1}{2} (s-x)^2 - \frac{1}{2} (t-y)^2 }  \\
         && \qquad \qquad \times \label{Ky}
        \prod_{k=1}^m
        \left(\frac{s-a_k}{t-a_k}\right)^{n_k} \frac{1}{s-t}
        \left\{ -\sum_{k=1}^m \frac{n_k}{t-a_k} + \frac{1}{s-t} \right\}.
\end{eqnarray}
We add (\ref{Kx}) and (\ref{Ky})
\begin{eqnarray} \nonumber
\frac{\partial K}{\partial x} + \frac{\partial K}{\partial y} & = &
   \frac{1}{(2\pi i)^2}
    \int_{-i\infty}^{i \infty} ds \oint_{\Gamma} dt \
         e^{ \frac{1}{2} (s-x)^2 - \frac{1}{2} (t-y)^2 }  \\
&& \qquad \qquad \times \nonumber
        \prod_{j=1}^m
        \left(\frac{s-a_j}{t-a_j}\right)^{n_j} \frac{1}{s-t}
        \sum_{k=1}^m \left(\frac{n_k}{s-a_k} - \frac{n_k}{t-a_k}\right) \\
& = & - \sum_{k=1}^m n_k \label{Kx+Ky}
    \frac{1}{(2\pi i)^2}
    \int_{-i\infty}^{i \infty} ds \oint_{\Gamma} dt \
         e^{ \frac{1}{2} (s-x)^2 - \frac{1}{2} (t-y)^2 }  \\
         && \qquad \qquad \times   \nonumber
        \prod_{j \neq k} \left(\frac{s-a_j}{t-a_j}\right)^{n_j}
        \frac{(s-a_k)^{n_k-1}}{(t-a_k)^{n_k+1}}.
\end{eqnarray}
For every $k$, the double integral in (\ref{Kx+Ky}) factors into
a product of two single integrals, which by (\ref{intrepH2})
and (\ref{intrepH1b}) are given in terms of multiple Hermite
polynomials. It leads to
\begin{equation} \label{Kx+Ky2}
\frac{\partial K}{\partial x} + \frac{\partial K}{\partial y}
    = - \sum_{k=1}^m n_k P_{\vec{n}-\vec{e}_k}(x) Q_{\vec{n}+\vec{e}_k}(y).
\end{equation}

From (\ref{Kx+Ky1}) and (\ref{Kx+Ky2}) we get
\[ (x-y) K(x,y) = P_{\vec{n}}(x) Q_{\vec{n}}(y) -
    \sum_{k=1}^m n_k P_{\vec{n}-\vec{e}_k}(x) Q_{\vec{n}+\vec{e}_k}(y),
\]
which agrees with (\ref{CDkernel}) since
\begin{equation} \label{nk}
    n_k = \frac{h_{\vec{n}}^{(k)}}{h_{\vec{n}-\vec{e}_k}^{(k)}}
\end{equation}
because of (\ref{hnk}).

\section{Multiple Laguerre polynomials}

Complex Gaussian sample covariance matrices have a distribution
\begin{equation} \label{Wishart}
    \frac{1}{Z_n} e^{- \Tr (\Sigma^{-1} M)} \left(\det M\right)^p dM
\end{equation}
defined on $n \times n$ positive definite Hermitian matrices $M$.
The matrix $M$ arises as $M = X X^H$ where $X$ is an $n \times (n+p)$,
matrix whose independent columns are Gaussian distributed with covariance
matrix $\Sigma$. Here $p$ is a non-negative integer.
The distribution (\ref{Wishart}) is also called a Wishart ensemble.
We assume that $\Sigma^{-1}$ has eigenvalues
$\beta_1, \ldots, \beta_m > 0$ with respective multiplicities $n_1, \ldots, n_m$.

Writing $\Sigma^{-1} = I - A$, we see that (\ref{Wishart}) takes
the form (\ref{ensemble}) with $V(M) = M$, but restricted to positive
definite Hermitian matrices.
It follows that the ensemble (\ref{Wishart}) can be described with
multiple orthogonal polynomials, which in this case are multiple
Laguerre polynomials \cite{Apt,ABV,NS,VAC}. (To be precise, they are
called multiple Laguerre II  in \cite{ABV,VAC} to distinguish them
from another generalization of Laguerre polynomials, called
multiple Laguerre I.)

The eigenvalues of $M$ follow a determinantal
point process on $(0,\infty)$ with kernel $K(x,y)$
given by (\ref{CDkernel})
\begin{equation} \label{CDkernel2}
    (x-y) K(x,y) = P_{\vec{n}}(x) Q_{\vec{n}}(y)
    - \sum_{k=1}^m \frac{h_{\vec{n}}^{(k)}}{h_{\vec{n}-\vec{e}_k}^{(k)}}
        P_{\vec{n}-\vec{e}_k}(x) Q_{\vec{n} + \vec{e}_k}(y)
\end{equation}
where now $P_{\vec{n}}$ is the type II multiple Laguerre polynomial
and $Q_{\vec{n}}(x) = \sum\limits_{k=1}^m A_{\vec{n}}^{(k)}(x) x^p e^{-\beta_k}(x)$
is the linear form involving the type I multiple Laguerre polynomials
$A_{\vec{n}}^{(k)}$.

Baik, Ben Arous and P\'ech\'e \cite{BBAP} gave a double integral representation
for the correlation kernel
\begin{equation} \label{BBAPkernel}
    K(x,y) = \frac{1}{(2\pi i)^2} \oint_{\Sigma} ds \oint_{\Gamma} dt
    e^{xs-yt} \left(\frac{t}{s}\right)^{|\vec{n}|+p}
        \prod_{k=1}^m \left( \frac{s-\beta_k}{t-\beta_k}
        \right)^{n_k} \frac{1}{s-t}
\end{equation}
where $\Sigma$ and $\Gamma$ are disjoint closed contours both oriented counterclockwise
such that $\Sigma$ encloses $0$ and lies in $\{s \in \mathbb C \mid \Re s < \min_k \beta_k \}$
and $\Gamma$ encloses the points $\beta_1, \ldots, \beta_m$ and lies in the right half-plane.

In view of (\ref{BBAPkernel}) and our experience with multiple Hermite polynomials
we expect integral representations for the multiple Laguerre polynomials as well.
We will see that this is indeed the case, and we use this to study the connection
between the kernels (\ref{CDkernel2}) and (\ref{BBAPkernel}). It will turn
out that the two kernels are equal up to a multiplicative factor $x^p y^{-p}$.
However, this difference does not affect the correlation functions (\ref{correlations}).

\subsection{Multiple Laguerre polynomials of type II}
The multiple Laguerre polynomial of type II is a monic polynomial $P_{\vec{n}}$
of degree $|\vec{n}|$ such that
\begin{equation} \label{LaguerreII}
    \int_0^{\infty} P_{\vec{n}}(x) x^{j+p} e^{-\beta_k x} dx = 0,
    \qquad k=1, \ldots, m, \quad j=0, \ldots, n_k-1.
\end{equation}

\begin{theorem}
    The multiple Laguerre polynomial of type II has the integral representation
    \begin{equation} \label{intrepL2}
        P_{\vec{n}}(x) =
        \frac{(|\vec{n}|+p)! x^{-p}}{2 \pi i \prod_{k=1}^m (-\beta_k)^{n_k}}
         \oint_{\Sigma} e^{xs} s^{-|\vec{n}|-p-1}
            \prod_{k=1}^m (s- \beta_k)^{n_k} ds
    \end{equation}
    where $\Sigma$ is a closed contour around $0$ oriented counterclockwise,
    which does not enclose any of the $\beta_k$'s.
\end{theorem}
\begin{proof}
    Denote the right-hand side of (\ref{intrepL2}) by $P(x)$
    and write $C = \frac{(|\vec{n}|+p)!}{2 \pi i \prod_{k=1}^m (-\beta_k)^{n_k}}$.
    Then
    \begin{equation} \label{intrepL2b}
    P(x) = C x^{-p} \sum_{j=0}^{\infty} \frac{x^j}{j!}
    \oint_{\Sigma} s^{j-|\vec{n}|-p-1} \prod_{k=1}^m (s-\beta_k)^{n_k} ds.
    \end{equation}
    All terms in (\ref{intrepL2b}) with $j \geq |\vec{n}| + 1$ vanish by Cauchy's theorem,
    as well as all terms with $j \leq p-1$, as we can see by deforming the contour
    $\Sigma$ to infinity. It follows that $P$ is a polynomial of degree $|\vec{n}|$,
    whose leading coefficient is
    \[ C \frac{1}{(|\vec{n}|+p)!}
    \oint_{\Sigma} s^{-1} \prod_{k=1}^m (s-\beta_k)^{n_k} ds
    = C \frac{2\pi i}{(|\vec{n}|+p)!} \prod_{k=1}^m (- \beta_k)^{n_k} = 1. \]
    Hence $P$ is a monic polynomial of degree $|\vec{n}|$.

    We now verify the orthogonality conditions (\ref{LaguerreII}).
    Take $k = 1, \ldots, m$ and let $j = 0, \ldots, n_k-1$. Then
    \[ \int_0^{\infty} P(x) x^{j+p} e^{-\beta_k x} dx =
        C \oint_{\Sigma}
        \left( \int_0^{\infty} x^j e^{(s-\beta_k)x} dx \right) s^{-|\vec{n}|-p-1}
            \prod_{l=1}^m (s- \beta_l)^{n_l} ds \]
    where we have assumed that $\Sigma$ is so that $\Re s <  \beta_k$
    for every $s \in \Sigma$.
    The inner integral is  $j! (\beta_k-s)^{-j-1}$, so that
    \[ \int_0^{\infty} P(x) x^{j+p} e^{-\beta_kx} dx =
        C j! \oint_{\Sigma} s^{-|\vec{n}|-p-1} (s-\beta_k)^{n_k-j-1} \prod_{l \neq k} (s-\beta_l)^{n_l} ds \]
        The integrand in the last integral has no pole at $\beta_k$, since
         $j \leq n_k-1$. At infinity the integrand behaves like $s^{-j-p-2}$. So by deforming the
         contour $\Sigma$ to infinity, we conclude that the integral is zero.
         The conclusion is that $P$ is the multiple Laguerre polynomial
        of type II and (\ref{intrepL2}) follows.
\end{proof}

\subsection{Multiple Laguerre polynomials of type I}
The multiple Laguerre polynomials of type I $A_{\vec{n}}^{(k)}$, $k=1, \ldots, m$, have
degrees
\begin{equation} \label{LaguerreIa}
    \deg A_{\vec{n}}^{(k)} \leq n_k-1, \qquad k =1,\ldots, m
\end{equation}
and are such that
\[ Q_{\vec{n}}(x) = \sum_{k=1}^m A_{\vec{n}}^{(k)} x^p e^{-\beta_kx} \]
satisfies
\begin{equation} \label{LaguerreIb}
    \int_0^{\infty} x^j Q_{\vec{n}}(x) dx =
    \left\{ \begin{array}{ll} 0, & \quad j=0, \ldots, |\vec{n}|-2, \\
    1, & \quad j=|\vec{n}|-1.
    \end{array} \right.
\end{equation}
There is also an integral representation for the  multiple Laguerre polynomials
of type I.

\begin{theorem} The multiple Laguerre polynomials of type I have the
integral representation
\begin{equation} \label{intrepL1}
    A_{\vec{n}}^{(k)}(x) x^p e^{-\beta_k x} = -
        \frac{\prod_{l=1}^m (-\beta_l)^{n_l} x^p}{2 \pi i (|\vec{n}|+p-1)! }
         \oint_{\Gamma_k} e^{-xt} t^{|\vec{n}|+p-1}
            \prod_{l=1}^m (t- \beta_l)^{-n_l} dt
\end{equation}
where $\Gamma_k$ is a closed contour around $\beta_k$, which does not enclose $0$
nor any of the other points $\beta_l$, $l\neq k$.

In addition the function $Q_{\vec{n}}$ has the integral representation
\begin{equation} \label{intrepL1b}
    Q_{\vec{n}}(x) = -
        \frac{\prod_{l=1}^m (-\beta_l)^{n_l} x^p}{2 \pi i (|\vec{n}|+p-1)! }
         \oint_{\Gamma} e^{-xt} t^{|\vec{n}|+p-1}
            \prod_{l=1}^m (t- \beta_l)^{-n_l} dt
\end{equation}
where $\Gamma$ is a closed contour around $\beta_1, \ldots, \beta_m$, but which
does not enclose $0$.
\end{theorem}
\begin{proof}
Only the pole at $\beta_k$ contributes to the integral in (\ref{intrepL1}).
By the residue theorem, we have that the right-hand side of (\ref{intrepL1})
\[ const  \ x^p \left(\frac{d}{dt}\right)^{n_k-1}
    \left. \left[ e^{-xt} t^{|\vec{n}|+p-1} \prod_{l\neq k} (t-\beta_l)^{-n_l} \right]
        \right|_{t=\beta_k}
\]
which is easily seen to of the form $A_k(x) x^p e^{-\beta_k x}$
where $A_k$ is a polynomial of degree $n_k-1$.

Let \[ Q(x) = \sum_{k=1}^m A_k(x) x^p e^{-\beta_k x} \]
which is then equal to the right-hand side of (\ref{intrepL1b}).
We may assume that $\Gamma$ is entirely in the right half-plane.
Then
\[ \int_0^{\infty} x^j Q(x) dx = -
        \frac{\prod_{l=1}^m (-\beta_l)^{n_l}}{2 \pi i (|\vec{n}|+p-1)! }
         \oint_{\Gamma} dt \int_0^{\infty} dx x^{j+p} e^{-xt} \ t^{|\vec{n}|+p-1}
            \prod_{l=1}^m (t- \beta_l)^{-n_l}. \]
The $x$-integral is $(j+p)! t^{-j-p-1}$, so that
\begin{equation} \label{intrepL1c}
    \int_0^{\infty} x^j Q(x) dx = -
        \frac{\prod_{l=1}^m (-\beta_l)^{n_l} (j+p)!}{2 \pi i (|\vec{n}|+p-1)! }
         \oint_{\Gamma} t^{|\vec{n}|-j-2}
            \prod_{l=1}^m (t- \beta_l)^{-n_l} dt.
\end{equation}
Assuming $j \leq |\vec{n}|-2$, we can deform $\Gamma$ to infinity
without picking up a residue contribution at $t=0$.
The integrand behaves like $t^{-j-2}$ at infinity, and so (\ref{intrepL1c})
vanishes for $j \leq |\vec{n}|-2$. For $j= |\vec{n}|-1$, we pick up
a residue contribution at $t=0$, and the result is that
\begin{equation} \label{intrepL1d}
\int_0^{\infty} x^{|\vec{n}|-1} Q(x) dx = 1.
\end{equation}
Thus $Q$ satisfies the type I multiple orthogonality conditions (\ref{LaguerreIb})
and the theorem follows.
\end{proof}

\subsection{The multiple Laguerre kernel}
We finally compare the representations (\ref{CDkernel2}) and (\ref{BBAPkernel})
of the kernel.

We start from the double integral (\ref{BBAPkernel}) and evaluate $xK(x,y)$
using an integration by parts on the $s$-integral. The result is
\begin{eqnarray} \nonumber
 xK(x,y) & = & \frac{1}{(2\pi i)^2} \oint_{\Sigma} ds \oint_{\Gamma} dt
    e^{xs-yt} \left(\frac{t}{s}\right)^{|\vec{n}|+p}
        \prod_{l=1}^m \left( \frac{s-\beta_l}{t-\beta_l}
        \right)^{n_l}  \\
        & & \qquad \times \label{xKxy} \ \frac{1}{s-t}
        \left\{ \frac{|\vec{n}|+p}{s} - \sum_{k=1}^m \frac{n_k}{s-\beta_k} + \frac{1}{s-t} \right\}.
\end{eqnarray}
Similarly, after integration by parts on the $t$-integral,
\begin{eqnarray} \nonumber
 yK(x,y) & = & \frac{1}{(2\pi i)^2} \oint_{\Sigma} ds \oint_{\Gamma} dt
    e^{xs-yt} \left(\frac{t}{s}\right)^{|\vec{n}|+p}
        \prod_{l=1}^m \left( \frac{s-\beta_l}{t-\beta_l}
        \right)^{n_l}  \\
        & & \qquad \times \label{yKxy} \ \frac{1}{s-t}
        \left\{ \frac{|\vec{n}|+p}{t} - \sum_{k=1}^m \frac{n_k}{t-\beta_k}  + \frac{1}{s-t} \right\}.
\end{eqnarray}
Hence
\begin{eqnarray*}
\lefteqn{(x-y) K(x,y) } \\
&& = \frac{1}{(2\pi i)^2} \oint_{\Sigma} ds \oint_{\Gamma} dt
    e^{xs-yt} \left(\frac{t}{s}\right)^{|\vec{n}|+p}
        \prod_{l=1}^m \left( \frac{s-\beta_l}{t-\beta_l}
        \right)^{n_l} \\
        & & \qquad \times \
        \left\{ - \frac{|\vec{n}|+p}{st} + \sum_{k=1}^m \frac{n_k}{(s-\beta_k)(t-\beta_k)} \right\} \\
        & & = -\frac{|\vec{n}|+p}{(2\pi i)^2} \oint_{\Sigma} ds \oint_{\Gamma} dt
    e^{xs-yt} \frac{t^{|\vec{n}|+p-1}}{s^{|\vec{n}|+p+1}}
    \prod_{l=1}^m \left( \frac{s-\beta_l}{t-\beta_l}  \right)^{n_l} \\
    & & \quad + \sum_{k=1}^m \frac{n_k}{(2\pi i)^2} \oint_{\Sigma} ds \oint_{\Gamma} dt
    e^{xs-yt} \left(\frac{t}{s}\right)^{|\vec{n}|+p}
        \prod_{l\neq k} \left( \frac{s-\beta_l}{t-\beta_l}
        \right)^{n_l} \frac{(s-\beta_k)^{n_k-1}}{(t-\beta_k)^{n_k+1}}
\end{eqnarray*}
Now we have $m+1$ double integrals and they all factor into products
of two single integrals of the forms (\ref{intrepL2}) and (\ref{intrepL1b}).
The result is that
\begin{equation} \label{compareK}
    (x-y) K(x,y) =
    x^p y^{-p} \left( P_{\vec{n}}(x) Q_{\vec{n}}(y) -
        \sum_{k=1}^m n_k \frac{|\vec{n}|+p}{\beta_k^2}  P_{\vec{n}-\vec{e}_k}(x) Q_{\vec{n}+\vec{e}_k}(y)
        \right).
\end{equation}
It can be shown that
\[ n_k \frac{|\vec{n}|+p}{\beta_k^2} = \frac{h_{\vec{n}}^{(k)}}{h_{\vec{n}-\vec{e}_k}^{(k)}} \]
so that (\ref{compareK}) agrees with (\ref{CDkernel2}) up to the factor $x^p y^{-p}$.
However, this factor is not essential since it does not change the correlation functions
(\ref{correlations}). Hence (\ref{CDkernel2}) and (\ref{BBAPkernel}) are essentially the same.


TITLE: Proving continuity of a monotone map from a Banach space to its dual with the weak* topology
QUESTION [2 upvotes]: I'm following Brezis' book on functional analysis, and there is this problem. Let $E$ be a Banach space and $A: E \to E^{*}$ a monotone map defined in $D(A)=E$. Assume that $ \forall x,y \in E$ the map:
$$t \in \mathbb{R} \mapsto \langle A(x+ty),y\rangle $$
is continuous at $t=0$. We need to show that $A$ is continuous from $E$ to $E^{*}$ with the $σ(E^{*},E)$ topology (this is the weak* topology).
I was given the following proof, but I can't understand some of the steps.
Proof:
It suffices to show this for sequences. Lets take $(x_n)$ in $E$  and assume assume that $x_n \to x$ (converges strongly in $E$), but that  $Ax_n \not\rightharpoonup Ax$ for  $σ(E^{*},E)$.
Meaning we have a $y \in E$ such that $\langle Ax_n,y\rangle \not\to \langle Ax,y\rangle$.
We know $Ax_n$ is bounded.
So we have a sub-sequence such that $\langle Ax_{nk},y\rangle \to l \not=\langle Ax,y\rangle$, given that $A$ is monotone we have that:
$$\langle Ax_n-A(x+ty),x_{nk}-x-ty\rangle \geq 0.$$
Taking limits we get:
$$-tl+t\langle A(x+ty),y\rangle \geq 0.$$
Which would mean that $l =\langle Ax,y\rangle $; a contradiction. Hence we are done.
I don't see very clearly why $Ax_n$ would be bounded, but more importantly how do we know that $l \not = \langle Ax,y\rangle $, is this because $Ax_n$ does converges but just not to $Ax$?
Taking the limits at the end would mean evaluating the limit of $\langle Ax_{nk},x_n\rangle $ and $\langle Ax_{nk},x\rangle$, how do we know these two are the same limits?
And finally how are we using the continuity at $t=0$ of the second map given?

REPLY [1 votes]: I don't see very clearly why $Ax_n$ would be bounded.
$A$ is locally bounded (and thus bounded in a neighborhood of $x$). See Problem 2.6, page 50.

How do we know that $l \not = \langle Ax,y\rangle$?
Using that $Ax_n \not\rightharpoonup Ax$, we can take $(x_{nk})$ such that
$$|\langle Ax_{nk},y_0\rangle-\langle Ax,y_0\rangle|\geq \varepsilon_0$$
for some $y_0\in E$ and some $\varepsilon_0>0$. Therefore,
$$\left\{\begin{align*}
&\langle Ax_{nk},y\rangle \not\to \langle Ax,y\rangle\\
&\langle Ax_{nk},y\rangle \to l
\end{align*}\right.$$
which implies that $l \not = \langle Ax,y\rangle$.

Taking the limits at the end would mean evaluating the limit of $\langle Ax_{nk},x_n\rangle $ and $\langle Ax_{nk},x\rangle$, how do we know these two are the same limits?
Since $Ax_{nk}$ is bounded and $x_{nk}\to x$ strongly, the difference of these terms goes to zero:
$$|\langle Ax_{nk},x_n\rangle-\langle Ax_{nk},x\rangle|\leq \|Ax_{nk}\|\|x_{nk}-x\|\to 0.$$

And finally how are we using the continuity at $t=0$ of the second map given?
In order to obtain $l =\langle Ax,y\rangle$ from the last inequality, we have to take the limit as $t\to 0$.


TITLE: An inequality with the characteristic function of a Distribution
QUESTION [2 upvotes]: Am solving some exercises in probability theory in order to prepare myself for the exams and i came across the following inequality which i cant prove:

Prove that for every $\alpha>0$ the following inequality holds
  $$\mathsf{P}(|X| \geq 2/\alpha) \leq \frac{1}{\alpha}\int_{-\alpha}^{\alpha}(1-\phi(t))dt$$
  where $\phi(t) = \mathbb{E}(e^{itX})$ , there characteristic function of a random variable $X$.

I tried to use Fubini in order to evaluate the right hand side and i derived the expression
$$\begin{align*} \int_{-\alpha}^{\alpha}(1-\phi(t))dt &= \int_{-\alpha}^{\alpha}(1-\mathbb{E}(\cos tX))\, dt \\ &=\int_{-\alpha}^{\alpha}\mathbb{E}(1-\cos tX)\, dt \\ &\overset{1-\cos tX\geq 0}{\geq}
\int_{-\alpha}^{\alpha}\mathbb{E}\biggl((1-\cos tX)I_{\{|X|\geq 2/\alpha\}}\biggr) \, dt \\ &=\int_{-\alpha}^{\alpha}\biggl(\int_{\mathbb{R}}(1-\cos tx)I_{\{|x|\geq 2/\alpha\}} \, dF_X(x)\biggr) \, dt \\ &=
\int_{\mathbb{R}}\biggl(\int_{-\alpha}^{-\alpha}(1-\cos tx)I_{\{|x|\geq 2/\alpha\}}\, dt\biggr) \, dF_X(x)\end{align*} $$
But now I cant figure out what to do with integrable function inside.
If you have any idea on how to continue or if there is other way , let me know!
Thanks in advance!

REPLY [1 votes]: Your calculation shows
$$\int_{-\alpha}^{\alpha} (1-\text{Re} \, \phi(t)) \, dt \geq \int_{\mathbb{R}} \int_{-\alpha}^{\alpha} (1-\cos(tx)) 1_{|x|>2/\alpha} dt \, dF_X(x) \tag{1}$$
(note that there are two typos in your computations; instead of $\alpha/2$ it should read $2/\alpha$). Clearly,
$$\int_{-\alpha}^{\alpha} (1-\cos(tx)) \, dt = 2\alpha - 2\alpha \frac{\sin(\alpha x)}{\alpha x}.$$
Since
$$\forall |r| \geq 2: \qquad \left| \frac{\sin r}{r} \right| \leq  \frac{1}{2}  $$
we find that
$$\int_{-\alpha}^{\alpha} (1-\cos(tx)) \, dt \geq 2\alpha - 2\alpha \frac{1}{2} = \alpha$$
for any $x$ such that $|\alpha x| \geq 2$. Plugging this estimate into $(1)$ proves the assertion.


TITLE: Yoneda's lemma as a generalization of Cayley's theorem for groups
QUESTION [0 upvotes]: $ <G, *> $ a group and $ G $ is seen as a category denoted by $ C_G $ that has a single object, say "@".
$H: C_G\to [C_G, Sets]$ is defined as:
In objects: $ @\in {C_G} $ we define an object $h _@$ of $[C_G, Sets]$ in
the following way: $H_@: C_G\longrightarrow {Sets} $
$ @\mapsto H _ @@ = Mor_ {C_G} (@, @) $
$ @ \xrightarrow [s {}] \, {@}\mapsto h _ @@ \xrightarrow [H_ @ s] \, {h _ @@} $ such that $ H_ @ s (t) = t * s $ with $ t: @ \longrightarrow {@} $, defined

so $ H _ @ $ is a contravariant functor.
In arrows: For every $ f: @ \longrightarrow {@} $ we define a natural transformation $ H_f: H _ @ \longrightarrow {H _ @} $.
Thus defined $ H $ is covariant functor.
The embedding of Yoneda tells us that $ H $ is fully faithful and injective in objects.
Denote for $ F $ to the image of the functor $ F $ then $ F = Im (H) $.
Affirmation 1:
$ F: C_G \to Sets $ is a faithful contravariant functor.
$ F $ is faithful yes and only if $ T $ is injective.
$ T $ is defined as follows:
the application $ T: Mor_{C_G} (@, @) \to Mor_{Sets} (F _@, F _@) $ defined as $ T (g): F _ @ \to F_@ $ such that $ T (g)_@: F_@@ \to F _ @@ $ with $ T (g)_@ (h) = h * g $. (This definition of $ T $ is because of how $ H $ is defined)
MY QUESTION IS THIS:
Why when $ F $ is the image of $ H $ does it have to $F _ @@ = Mor_{C_G} (@, @)$?
Let's see that $ T $ is injective:
Let $ f, g \in Mor_ {C_G} (@, @) $ such that $ T (g) = T (f) $
(I'm matching natural transformations, does this make sense?)
I think that equality means $ T (g) _@ = T (f) _@ $, then
$ T (g) _@ (h) = T (f)_@ (h) $ and this is $ h * g = h * f $ and since
these are elements
of the group then we take inverse of $ h $, then $ g = f $. Therefore $ T $ is injective.
Now let's look at the following:
$ G = Mor_{C_G} (@, @) $ and $ Mor_{Sets}(F _ @, F _ @) = \left\{{G \to G} \right \}$ is a group with the composition operation, then $ T: G \to \left \{{G \to G} \right \} $.
 Missing to see that $ T $ is a homomorphism (this I got) and then I
already have a monomorphism osea $ G $ would be a subgroup of $ \left \{{G \to G} \right \} $.
It is necessary to see that the collection of applications $ {G \to G} $
are bijectives, for this last one what is the idea?
Thank you

REPLY [1 votes]: In a general category, we have $H_a:=\hom(\_,a)$ is a contravariant functor to $\mathcal{Set}$, which sends an object $x$ to $\hom(x,a)$ and a morphism $f:x\to y\ $ to the function $\ \underset{y\to a}g\,\mapsto \underset{x\to a}{gf}$.
If $\alpha:a\to b$ is any morphism, it induces a natural transformation $H_a\to H_b$ by left composition.
So, now we have a single object and its contravariant hom functor $H_@=\hom(\_,@)$ which maps $@\mapsto\hom(@,@)=G\ $ and $\ s\mapsto \,\underset{G\to G}{(g\mapsto gs)}$.
Moreover, we consider here $H:C_G\to [{C_G}^{op},\,\mathcal{Set}]$ by the above natural transformations.
Note also, that if a category $C$ has only one object $@$, and $F,G:C\to D$ are functors, then a natural transformation $F\to G$ is (given by) a single morphism $F(@)\to G(@)$ of $D$, which has a special property (has to satisfy the commutative diagrams).
To your questions:

As I understand your description, you take $F$ to be the image of $H$, so it's simply $H_@$ as $C_G$ has only one object? That would also answer your question.
What you call $T$ is the morphism part of $H$, so $T$ preserves the group operation exactly because $H$ is a functor, i.e. preserves composition, which is defined as to be the group operation. And the Yoneda embedding is a functor. Verify it in the general case, then you should immediately see it for the one object case.
If a morphism $g$ is invertible (=isomorphism) in a category $\mathcal C$, and $F:\mathcal C\to\mathcal D$ is any functor, then $F(g)$ is invertible as well. Since all morphisms of $C_G$ are invertible, it follows that their images are invertible in $[{C_G}^{op},\mathcal{Set}]$, i.e. those are natural isomorphisms, so their single component is an invertible $\mathcal{Set}$-morphism.


TITLE: Distance in the Poincare Disk model of hyperbolic geometry
QUESTION [12 upvotes]: I am trying to understand the Poincare Disk model of a hyperbolic geometry and how to measure distances. I found the equation for the distance between two points on the disk as:
$d^2 = (dx^2 + dy^2) / (1-x^2-y^2)^2$
Given two points on the disk, I am assuming that $dx$ is the difference in the euclidean $x$ coordinate of the points, and similar for $dy$. So, what are $x$ and $y$ in the formula? Also, I see some formulas online as:
$d^2 = 4(dx^2 + dy^2) / (1-x^2-y^2)^2$
I am not sure which one is correct.
Assuming that $X$ and $Y$ are coordinates of one of the points, I have tried a few examples, but can't get the math to work out. For instance, if $A = (0,0)$, $B = (0,.5)$, $C = (0,.75)$, then what are $d(A,B)$, $d(B,C)$, and $d(A,C)$? The value $d(A,B) + d(B,C)$ should equal $d(A,C)$, since they are on the same line, but I can't get this to work out. I get distances of $.666$, $.571$, and $1.714$ respectively.
MathWorld - Hyperbolic Metric

REPLY [11 votes]: As others have indicated one has to distinguish a metric $(x,y)\mapsto d(x,y)$ which measures distances between points $x$, $y$ in a space $X$ and a Riemannian metric which tells us how lengths of curves $\gamma$ in a manifold $X$ should be computed.
In the complex plane we have the usual euclidean metric $d(z_1,z_2):=|z_2-z_1|$ and at the infinitesimal scale the Riemannian metric
$$ds^2:=|dz|^2=dx^2+dy^2\ .$$
The latter formula says that the length of an arbitrary curve
$$\gamma: \quad t\mapsto\bigl(x(t),y(t)\bigr)\qquad(a\leq t\leq b)$$
should be computed as
$$L(\gamma)=\int_\gamma ds=\int_a^b\sqrt{x'^2(t)+y'^2(t)}\ dt = \int_a^b|z'(t)|\ dt\ .$$
This formula implies that the length of a segment $\sigma$ connecting two points $z_1$ and $z_2$ is just $|z_2-z_1|$.
On the "Poincare disc" $P$ we are given a priori only a Riemannian metric
$$ds:= {|dz|\over 1-|z|^2}$$
(resp., $ds^2=\ldots$). This metric allows us to compute the lengths of arbitrary curves in $P$:
$$L(\gamma)=\int_a^b{|z'(t)|\over1-|z(t)|^2}\ dt\ .$$ The particular definition of $ds$ is chosen such that this hyperbolic length is invariant under arbitrary conformal movements of $P$ and the curves therein.
A posteriori one can define a metric $d(\cdot,\cdot)$ on $P$ by letting the distance $d(z_1,z_2)$ between two points $z_1$, $z_2\in P$ be the hyperbolic length of the shortest curve $\gamma$ connecting $z_1$ and $z_2$. The actual carrying out of this idea shows that $d(z_1,z_2)$ can be written as an elementary function (using ${\rm artanh}$, etc.) in terms of  $z_1$ and $z_2$.


TITLE: How to find out the dimension of the following system of linear equation
QUESTION [0 upvotes]: I have to find out the dimension of the solution of this set of the equations with the system being an affine space of $\Bbb R^4$ and $t∈R$, $u∈\Bbb R^4$.
So I tried to row reduce the left side of the system and found out that the rank is $4$ if $t≠0$ and $3$ if $t=0$. With the formula $\dim(L)=n-\operatorname*{rank}(A)$, I think that the dimension is $1$ for $t=0$ or $0$ for $t≠0$. But I am completely unsure if any of this is correct.

REPLY [0 votes]: How does your set of solution looks like?
I think you are right, but I am unsure because of your statement for t = 0, because you can reduce further than just row echelon form until you reach the 4 canonical basis vectors of $\mathbb{R}^4$


TITLE: Where is the "tube" of a wormhole relative to normal space?
QUESTION [0 upvotes]: If a wormhole could be described as two mouths connected by a tube of some length shorter than the mouths are separated in normal space, how is the "tube" able to be shorter than the normal space distance between the mouths?  Is it travelling in a "straighter line" while normal space is curved, like cutting through a hill, or tunneling through a sphere, or are there extra spacial dimensions that connect them but disappear on the macro-scale?

REPLY [1 votes]: or are there extra spacial dimensions that connect them but disappear on the macro-scale?

No. Extra dimensions are not a feature of general relativity. Sometimes people draw embedding diagrams or conceptual diagrams that include extra dimensions, but this is just a visual device.

Is it travelling in a "straighter line" while normal space is curved, like cutting through a hill,

The surrounding spacetime can be flat (or, technically, asymptotically flat). The fact that a surface is represented as curved on an embedding diagram does not necessarily relate at all to whether it's actually curved. There is a distinction between instrinsic curvature and extrinsic curvature. Only intrinsic curvature is of interest in GR.
Distances in GR do not behave the way they do in flat spacetime. Embedding diagrams of wormholes really just show the topology, which is separate from the question of distance measurements. You can take the wormhole topology and outfit it with a variety of different metrics. The metric is the mathematical machinery that defines distances. If you want to, you can define the metric in such a way that the distance through the wormhole is longer than the distance through the ambient space. (And by default, a wormhole would probably transport you through time as well.)


TITLE: Ideal gas and diatomic gas with same temperature
QUESTION [5 upvotes]: If a box of ideal gas and another box of diatomic gas are in thermal equilibrium,

does it mean that the average translational energy of ideal gas particle (A) is the same as that of diatomic gas particle (B)?
or does it mean that A is equal to the sum of the average translational energy (B) + average vibrational energy (C) + average rotational energy (D) of diatomic gas particle?
Equilibrium is achieved when A = B = C = D?
or is it A = B + C + D?

REPLY [5 votes]: The Equipartition theorem states that each degree of freedom has an average energy of 1/2KT.
This is valid at large enough temperatures where quantum mechanics does not play a role.
A = 3/2 KT(3 degrees of freesom)
B = 3/2 KT
C = 1/2 KT (1 vibrational degree of freedom in a di-atomic molecule)
D = KT (2 axis of rotation, the third has very low Moment of Inertia, and will not be excited)


TITLE: More comprehensive General Topology's book than Engelking
QUESTION [3 upvotes]: Other than Engelking General Topology, I also come across other graduate general topology text such as Dugundji and Kelley, which I also find them interesting.
However, I find Engelking's book more comprehensive compare to the two graduate texts.

Question: Does there exist a general topology book which is more
  comprehensive than Engelking? If yes, what is the title?

REPLY [5 votes]: I'm a fan of "Handbook of Set-theoretic Topology", which I often use, and Engelking as well. It contains many things that are more recent than Engelking and is more research oriented.
Nagata's "Modern General Topology" is also nice. Also in the Japanese school "Topics in General Topology", by Morita and Nagata. The latter book is more like the Handbook. The first more of a text book, like Kelley or Dugundji (which are good as well).


TITLE: Solve in prime numbers the equation $p^q+q^r=r^p$
QUESTION [6 upvotes]: Find all triples of prime numbers $(p,q,r)$ such that $$p^q+q^r=r^p.$$

I proved that when $r=2$, the equation becomes $$p^q+q^2=2^p.$$ Then I tried to use reciprocity laws and Fermat's little theorem. I could prove that $p\equiv 7\pmod 8$ and that $p>q$.
The equation appeared in some olympiad . They asked to prove that $r=2$. So I am trying to find at least one triple.

REPLY [2 votes]: So $$p^q+q^2=2^p \qquad(*)$$  It is easy to show that $\boxed{p\gt q}$ or just look at the graph of the function $x ^ y + y ^ 2 = 2 ^ x$ for $x,y\gt1$.
If $p=4n+1$ then from $(*)$,
$$q=4m\pm1\Rightarrow4N+1+16m^2\pm8m+1=2N+8m^2\pm4m+1=2^{p-1}\space\text{ absurde}.$$
Then $p$ must be of the form  $\boxed{p=4n-1}$.
Put $p=an-1$ where $n$ is odd and $a=2^r,\space r\ge 2$.
$$(an-1)^q=(an)^q-1+\sum_{k=1}^{q-1}(\pm1)^k\binom qk(an)^{q-k}$$ $$(an-1)^q=(an)^q-1+an\left(\sum_{k=1}^{q-2}(\pm1)^k\binom qk(an)^{q-k-1}\right)+anq$$
For $\boxed{q\equiv\pm1\pmod4}$ put $q=bm\pm1$ where $m$ is odd and $b=2^s,\space s\ge 2$.
The equation $(*)$ becomes
$$(an)^q+an\left(\sum_{k=1}^{q-2}(\pm1)^k\binom qk(an)^{q-k-1}\right)+anq+(bm)^2\pm2bm=2^p\qquad(**)$$
This last condition, , implies
Note that the sum $\left(\sum\right)$ is even.
Now
►if $a\le b$ (i.e. $2^r\le2^s$) then dividing by $a$ we are done because $nq$ is odd.
►if $a\gt b$ (i.e. $\dfrac {a}{2b}=2^{r-s-1}\ge1$) then dividing by $2b$ we are done for all $p,q$ such that $r\gt {s+1}$ because $m$ is odd.
It remains the cases for which $r=s+1$, in other words and taking into account the above
$$p=2^{s+1}n-1,\space q=2^sm\pm1\space n,m \space \text{ odds},\space \text {with }p\gt q$$
This last condition, $p\gt q$ , implies
$$2^{s+1}n-1\gt2^sm\pm1\Rightarrow\begin{cases}2^sn\gt2^{s-1}m+1\\2n\gt m\end{cases}$$ Thus it remains to study the equation where $p=2^{s+1}n-1,\space q=2^sm\pm1\space n,m \space \text{ odds},\space \text {with }p\gt q$ both odd primes.
$$\boxed{\displaystyle(2^{s+1}n-1)^{2^sm\pm1}+(2^sm\pm1)^2=2^{2^{s+1}n-1}}$$
I will return to the possible proof of this last part. If someone wants to end this problem by proving the apparently probable impossibility of solution, let it go forward. In particular if @Servaes wants to finish his very interesting partial answer.


TITLE: Proving that the vector field X is tangent to hypersurfaces.
QUESTION [1 upvotes]: Consider the map $f: \mathbb R^6 \to \mathbb R$ ,where
$f=x_1 x_2 x_4 x_6+x_1 x_3 x_5 x_6-x_1 x_2 x_3 x_4 x_5 x_6$
The normal vector field is
$$N=(\frac{\partial f}{\partial x_1},\frac{\partial f}{\partial x_2},\frac{\partial f}{\partial x_3},\frac{\partial f}{\partial x_4},\frac{\partial f}{\partial x_5},\frac{\partial f}{\partial x_6})$$
Consequently, the vector field
$$X=(-\frac{\partial f}{\partial x_2},\frac{\partial f}{\partial x_1},0,0,0,0)$$
How would I go about proving that the vector field X is  tangent to hypersurfaces.
Thanks for help .

REPLY [2 votes]: When you say "the normal vector field," you mean the vector field normal to the level hypersurfaces $\{f=\text{constant}\}$ (wherever the normal is nonzero). What does "normal" mean? It means that this vector at point $p$ is orthogonal to the $\Bbb R^5$ that is the tangent space to the level hypersurface at $p$. Your vector field $X$ is orthogonal to $N$, which means it is tangent to the appropriate level hypersurface.