TITLE: Isomorphism of two simple normal subgroups of a group $G$, given that $G$ has composition series
QUESTION [1 upvotes]: Let $G$ be a group having composition series. Suppose that $H$ and $K$ are normal simple subgroups of $G$ and that $G/H\cong G/K$. Then $H\cong K$.
My attempt (edited with hints from @DerekHolt):
Since $G$ has composition series and $H$ and $K$ are normal in $G$, then $G/H$ and $G/K$ and $H$ and $K$ have composition series. Choose composition series for $G/H$ and $G/K$:
\begin{align*}
\{1_{G/H}\}\subseteq H_{1}/H\subseteq H_{2}/H\subseteq \dots\subseteq H_{n-1}/H\subseteq H_{n}/H&=G/H
\\
\{1_{G/K}\}\subseteq K_{1}/K\subseteq K_{2}/K\subseteq \dots\subseteq K_{m-1}/K\subseteq K_{m}/K&=G/K
\end{align*}
and composition series for $H$ and $K$
\begin{align*}
\{1_{G}\}\subseteq L_{1}\subseteq L_{2}\subseteq\dots\subseteq L_{s}=H
\\
\{1_{G}\}\subseteq M_{1}\subseteq M_{2}\subseteq\dots\subseteq M_{t}=K
\end{align*}
Combining the composition series for $H$, $G/H$ and $K$, $G/K$, we get
\begin{align*}
\{1_{G}\}\subseteq L_{1}\subseteq L_{2}\subseteq\dots\subseteq L_{s}=H\subseteq H_{1}\subseteq\dots\subseteq H_{n}&=G
\\
\{1_{G}\}\subseteq M_{1}\subseteq M_{2}\subseteq\dots\subseteq M_{t}=K\subseteq K_{1}\subseteq\dots\subseteq K_{m}&=G.
\end{align*}
By the Jordan-Holder Theorem, $m=n$. (This is where I am stuck)

I tried to consider a smaller case where the length is $4$ (things work out well for length $2$ and $3$.
If we have (noting that $H$ and $K$ are simple, so there are no nontrivial normal subgroups of $H$ and $K$)
\begin{align*}
\{1_{G}\}\subseteq H=L_{1}\subseteq L_{2}\subseteq L_{3}\subseteq L_{4}&=G
\\
\{1_{G}\}\subseteq K=K_{1}\subseteq K_{2}\subseteq K_{3}\subseteq K_{4}&=G
\end{align*}
then by the third isomorphism theorem
\begin{align*}
G/K_{\sigma(3)}\cong(G/K)/(K_{\sigma(3)}/K)\cong(G/H)/(L_{3}/H)\cong G/L_{3}
\end{align*}
for some $\sigma\in S_{n}$. (I don't think I'm supposed to use the isomorphism in this manner, but I cannot see how else to use it)

A similar question has been posted to How are composition series for two isomorphic groups related? but the answer states that the composition factors are isomorphic (which is precisely the Jordan-H\"older theorem), but nothing much is said about the individual subgroups $G_{i}$ $H_{j}$ (e.g. is $G_{i}\cong H_{j}$ for some $i,j$?) in the composition series of $G$ and $H$.
I would appreciate any comments, hints on this problem.

REPLY [1 votes]: I'm answering my question based on @DerekHolt 's hints.
Since $G$ has composition series and $H$ and $K$ are normal and simple in $G$, then $G/H$ and $G/K$ and $H$ and $K$ have composition series. Choose composition series for $G/H$ and $G/K$:
\begin{align*}
\{1_{G/H}\}\subset H_{2}/H\subset H_{3}/H\subset \dots\subset H_{n-1}/H\subseteq H_{n}/H&=G/H
\\
\{1_{G/K}\}\subseteq K_{2}/K\subseteq K_{3}/K\subseteq \dots\subseteq K_{m-1}/K\subseteq K_{m}/K&=G/K
\end{align*}
and composition series for $H$ and $K$ (which are simple and normal in $G$)
\begin{align*}
\{1_{G}\}\subset H
\\
\{1_{G}\}\subset K
\end{align*}
Combining the composition series for $H$, $G/H$ and $K$, $G/K$, we get
\begin{equation*}
\{1_{G}\}\subset H\subset H_{2}\subset\dots\subset H_{n-1}\subset H_{n}=G
\\
\{1_{G}\}\subset K\subset K_{2}\subset\dots\subset K_{m-1}\subset K_{m}=G
\end{equation*}
By the Jordan-H\"older theorem, $m=n$ and we have
\begin{equation*}
\{1_{G}\}\subset H\subset H_{2}\subset\dots\subset H_{n-1}\subset H_{n}=G
\\
\{1_{G}\}\subset K\subset K_{2}\subset\dots\subset K_{n-1}\subset K_{n}=G
\end{equation*}
Now, the composition series for $G/H$ and $G/K$ are
\begin{equation*}
\{1_{G/H}\}\subset H_{2}/H\subset H_{3}/H\subset \dots\subset H_{n-1}/H\subset H_{n}/H=G/H
\\
\{1_{G/K}\}\subset K_{2}/K\subset K_{3}/K\subset \dots\subset K_{n-1}/K\subset K_{n}/K=G/K
\end{equation*}
Since $G/H\cong G/K$, then there exists an isomorphism
\begin{equation*}
\psi:G/H\tilde{\longrightarrow} G/K
\end{equation*}
and if $H_{i}/H\leq G/H$, by the fourth isomorphism theorem for groups, $\psi(H_{i}/H)\leq G/K$ and moreover, if $H_{i}/H\triangleleft H_{i+1}/H$ in $G/H$, then $\psi(H_{i}/H)\triangleleft \psi(H_{i+1}/H)$ as normal subgroups in $G/K$. This gives us
\begin{equation*}
\psi(H_{i}/H)=K_{i}/K
\end{equation*}
for each $i=2,3,\dots,n$ and so $H_{i}/H\cong K_{i}/K$ for each $i=2,3,\dots,n$. In particular, $H_{2}/H\cong K_{2}/K$ as simple quotient groups. Then we have composition series for $H_{2}$ and $K_{2}$ as
\begin{equation*}
1_{G}\subset H\subset H_{2}
\\
1_{G}\subset K\subset K_{2}
\end{equation*}
so that $H\cong K$.


TITLE: How do I calculate integral of complex function around the circle?
QUESTION [2 upvotes]: For example I have a complex function $$f(z) = \frac{z^2+1}{(z+1)(z-2)^2}$$ and a contour $$L: |z+i|=1$$ I want to find $$\oint_Lf(z)dz$$
I'm not sure about using Cauchy's integral formula as $$\oint_L \frac{g(z)}{z-z_0}\, dz = 2\pi i \cdot g(z_0)$$ with $$g(z) = \frac{z^2+1}{(z-2)^2}$$ and $$z_0=-1$$ Particularly, will not the answer depend on contour radius?

REPLY [2 votes]: Assuming you can use the residue theorem, this becomes quite an easy problem.
You care about the radius of the circle you are integrating around only insofar as it tells you whether the contour contains singularities.
In this case, you are integrating in a circle of radius 1 around $-i$. Your function is holomorphic except at 2 and -1. Neither of these poles are contained inside the disc. Thus your integral is zero.


text:
\begin{document}

\sloppy

\pagestyle{empty}
\hfill SPhT-T04/077
\addtolength{\baselineskip}{0.20\baselineskip}
\vspace{26pt}
\begin{center}
{\large \bf {A short note about Morozov's formula}}
\end{center}
\vspace{26pt}

\begin{center}
{\sl B.\ Eynard}\hspace*{0.05cm}\footnote{ E-mail: eynard@spht.saclay.cea.fr }\\
\vspace{6pt}
Service de Physique Th\'{e}orique de Saclay,\\
F-91191 Gif-sur-Yvette Cedex, France.\\
\end{center}

\vspace{20pt}
\begin{center}
{\bf Abstract}
\end{center}

The purpose of this short note, is to rewrite Morozov's formula for correlation functions over the unitary group,
in a much simpler form, involving the computation of a single determinant.

\vspace{26pt}
\pagestyle{plain}
\setcounter{page}{1}

\section{Introduction}

The main result of this paper is given in \ref{mainresult}.

In \cite{morozov}, A. Morozov proposed a formula for correlation functions of unitary matrices with Itzykson-Zuber's type measure:
\beq
\langle|U_{ji}|^2\rangle_{U(N)} :=
{1\over I(X,Y)}\,\int_{U(N)}{\rm d}U  |U_{ji}|^2 {\rm
e}^{\tr(XU^\dagger YU)}
\eeq
where ${\rm d}U$ is the Haar measure over the unitary group $U(N)$ (appropriately normalized, so that \ref{IZ} below holds with prefactor $1$), and $X$ and $Y$ are two given diagonal complex matrices:
\beq
X={\rm diag}(x_1,\dots,x_N)
\virg
Y={\rm diag}(y_1,\dots,y_N)
\eeq
and the normalization factor $I(X,Y)$ is the so-called Harish-Chandra-Itzykson-Zuber integral:
\beq\label{defIZ}
I(X,Y) :=
\int_{U(N)}{\rm d}U   {\rm e}^{\tr(XU^\dagger YU)}
\eeq
It is well-known \cite{IZ, Mehta} that:
\beq\label{IZ}
I(X,Y) =
{\det E\over \Delta(X)\Delta(Y)}
\eeq
where $E$ is the matrix with entries:
\beq\label{defE}
E_{ij}:=\,{\rm e}^{x_i y_{j}}\,
\eeq
and where $\Delta(X)$ and $\Delta(Y)$ are the Vandermonde determinants:
\beq
\Delta(X):=\prod_{i<j}(x_i-x_j)
\virg
\Delta(Y):=\prod_{i<j}(y_i-y_j)\,\, .
\eeq
Here we shall assume that $\det E\neq 0$ and $\Delta(X)\neq 0$ and $\Delta(Y)\neq 0$.

\medskip

Morozov's formula was proven in \cite{mixedcorrelators}, from Shatashvili's formula \cite{shata}.
Morozov's formula was originaly written as follows, for any arbitrary sequences of complex numbers $a_i,b_j$ one has:
\bea\label{morozov}
&&\sum_{i,j=1}^N {a_i}{b_j}
\int_{U(N)}{\rm d}U  |U_{ji}|^2 {\rm
e}^{\tr(XU^\dagger YU)} =  \frac 1{\Delta(X)\Delta(Y)}
\sum_{\rho\in S_N} \epsilon(\rho) {\rm e}^{\sum x_\ell y_{\rho(\ell)}} \cr
&& \sum_{n=0}^{N-1} (-1)^{n}
\sum_{i_1<i_2<\dots<i_{n+1}}
\hspace{-20pt}
 \frac {
\det\pmatrix{
1&\cdots&1\cr
x_{i_1}& \cdots& x_{i_{n+1}}\cr
\vdots& &\vdots\cr
x_{i_1}^{n-1}& \cdots & x_{i_{n+1}}^{n-1}\cr
a_{i_1} & \cdots & a_{i_{n+1}}}}{
\det\pmatrix{
1&\cdots&1\cr
x_{i_1}& \cdots& x_{i_{n+1}}\cr
\vdots& &\vdots\cr
x_{i_1}^{n}& \cdots & x_{i_{n+1}}^{n}}}
\frac {
\det\pmatrix{
1&\cdots&1\cr
y_{\rho(i_1)}& \cdots& y_{\rho(i_{n+1})}\cr
\vdots& &\vdots\cr
y_{\rho(i_1)}^{n-1}& \cdots & y_{\rho(i_{n+1})}^{n-1}\cr
b_{\rho(i_1)} & \cdots & b_{\rho(i_{n+1})}}}{
\det\pmatrix{
1&\cdots&1\cr
y_{\rho(i_1)}& \cdots& y_{\rho(i_{n+1})}\cr
\vdots& &\vdots\cr
y_{\rho(i_1)}^{n}& \cdots & y_{\rho(i_{n+1}) }^{n}}}\ ,\label{moroformula} \cr
\eea
where $S_N$ is the smmetric group of rank $N$.

\section{Rewriting Morozov's formula}

The purpose of this short note is to rewrite this formula in a much simpler form, in a way very similar to what was done in \cite{mixedcorrelators}.

\medskip

For any two complex numbers $x$ and $y$ (such that $|x|>{\rm max}\, |x_i|$ and $|y|>{\rm max}\, |y_i|$), consider the choice:
\beq
a_i={1\over x-x_i}=\sum_{r=0}^\infty {x_i^r\over x^{r+1}}
\virg
b_i={1\over y-y_i}=\sum_{s=0}^\infty {y_i^s\over y^{s+1}}
\eeq

Introduce the Schur polynomials (corresponding to hook diagramms):
\bea
S_r(x_{i_1},\dots,x_{i_{n+1}})  &:=&
\frac {
\det\pmatrix{
1&\cdots&1\cr
x_{i_1}& \cdots& x_{i_{n+1}}\cr
\vdots& &\vdots\cr
x_{i_1}^{n-1}& \cdots & x_{i_{n+1}}^{n-1}\cr
x_{i_1}^r & \cdots & x_{i_{n+1}}^{r}}}{
\det\pmatrix{
1&\cdots&1\cr
x_{i_1}& \cdots& x_{i_{n+1}}\cr
\vdots& &\vdots\cr
x_{i_1}^{n}& \cdots & x_{i_{n+1}}^{n}}} \cr
&=&  \sum_{a_1\leq
a_2\leq\dots\leq a_{r-n}} \prod_{k=1}^{r-n} x_{i_{a_k}}
=
 \sum_{j_1+\cdots+j_{n+1}=r-n} x_{i_1}^{j_1}\cdots
 x_{i_{n+1}}^{j_{n+1}}\label{hook}\  .
\eea

The formal generating function of these Schur polynomials is:
\beq\label{Schurgenerating}
\sum_{r=0}^\infty {1\over x^{r+1}}S_r(x_{i_1},\dots,x_{i_{n+1}}) = \prod_{k=1}^{n+1} {1\over x-x_{i_k}}
\eeq

Inserting that into \ref{morozov}, one gets:
\bea
&&\hspace{-1.6cm}I(X,Y)\,\sum_{i,j}
{1\over x-x_i}\,{1\over y-y_j}\, \langle|U_{ji}|^2\rangle_{U(N)} \cr
&& \hspace{-1.5cm}	=\int_{U(N)}{\rm d}U  \,\,
\tr \left({1\over x-X}U{1\over y-Y}U^\dagger\right)\,
{\rm e}^{\tr(XU^\dagger YU)}  \cr
&& \hspace{-1.5cm}=\frac 1{\Delta(X)\Delta(Y)}
\sum_{r,s=0}^\infty \sum_{\rho\in S_N} \epsilon(\rho) {\rm e}^{\sum x_\ell y_{\rho(\ell)}} \cr
&& \sum_{n=0}^{N-1} (-1)^{n}\!\!\!\!\!\!\!\!\!\!
\sum_{i_1<i_2<\dots<i_{n+1}}

{S_r(x_{i_1},\dots,x_{i_{n+1}})\over x^{r+1}}
\,\,
{S_s(y_{\rho(i_1)},\dots,y_{\rho(i_{n+1})})\over y^{s+1}}
 \cr
&& \hspace{-1.5cm}=\frac 1{\Delta(X)\Delta(Y)}
 \sum_{\rho\in S_N} \epsilon(\rho) {\rm e}^{\sum x_\ell y_{\rho(\ell)}}
\sum_{n=0}^{N-1} (-1)^{n}\!\!\!\!\!\!\!\!\!\!
\sum_{i_1<i_2<\dots<i_{n+1}}

\prod_{k=1}^{n+1} {1\over x-x_{i_k}}

\,\,
\prod_{l=1}^{n+1} {1\over y-y_{\rho(i_l)}}

 \cr
&& \hspace{-1.5cm}=\frac 1{\Delta(X)\Delta(Y)}
 \sum_{\rho\in S_N} \epsilon(\rho) {\rm e}^{\sum x_\ell y_{\rho(\ell)}}
\left[
1-\prod_{i=1}^{N} \left(1-{1\over x-x_{i}}{1\over y-y_{\rho(i)}}\right)
\right]
 \cr
&& \hspace{-1.5cm}=\frac 1{\Delta(X)\Delta(Y)}
 \sum_{\rho\in S_N} \epsilon(\rho)
\left[\prod_{i=1}^{N}\,{\rm e}^{x_i y_{\rho(i)}}\,
-\prod_{i=1}^{N} \left(\,{\rm e}^{x_i y_{\rho(i)}}\,
-{1\over x-x_{i}}
\,{\rm e}^{x_i y_{\rho(i)}}\,
{1\over y-y_{\rho(i)}}\right)\right]
 \cr
&& \hspace{-1.5cm}=\frac 1{\Delta(X)\Delta(Y)}
\left[
\det\left(\,{\rm e}^{x_i y_{j}}\,\right)
-\det\left(\,{\rm e}^{x_i y_{j}}\,-
{1\over x-x_{i}}
\,{\rm e}^{x_i y_{j}}\,
{1\over y-y_{j}}\right)
\right]
 \cr
\eea

Thus, Morozov's formula can be rewritten:
\beq
\int_{U(N)}{\rm d}U  \,\,
\tr \left({1\over x-X}U{1\over y-Y}U^\dagger\right)\,
{\rm e}^{\tr(XU^\dagger YU)}
=
{\det E
-\det \left(E-{1\over x-X}E{1\over y-Y}\right)
\over \Delta(X)\Delta(Y)}
\eeq
or:
\beq\label{mainresult}
\encadremath{
\begin{array}{rcl}
{\displaystyle  \left<\tr \left({1\over x-X}U{1\over y-Y}U^\dagger\right)\, \right>}
& = &
{\displaystyle 1-{\det \left(E-{1\over x-X}E{1\over y-Y}\right)\over \det E} }\cr
& = &
{\displaystyle 1- \det \left(1-{1\over x-X}E{1\over y-Y}E^{-1}\right)}
\end{array}}
\eeq

\section{Concluding remarks}

From that expression of Morozov's formula, it is rather easy to recover any individual correlator by taking residues:
\beq
 \left< U_{ij}\, U^\dagger_{ji} \right>
 =
 \mathop{\rm Res}_{x\to x_i}
 \mathop{\rm Res}_{y\to x_j}
 \left<\tr \left({1\over x-X}U{1\over y-Y}U^\dagger\right)\, \right>
\eeq

\medskip

Notice also that \ref{mainresult} is very similar to what was found in \cite{mixedcorrelators} after integration over $X$ and $Y$.

\medskip

Notice that by expanding the determinant in \ref{mainresult} along its last column, one can find a recursion relation relating
$U(N)$ to $U(N-1)$ integrals, which is equivalent to Shatashvili's approach \cite{shata}.

\bigskip

{\em Aknowledgements:} the author wants to thank M. Bertola and J. Harnad for discussions about that topic.
The author thinks that this short note should be seen as an addendum to the article \cite{mixedcorrelators}.

config: arxiv


TITLE: Expectation value of $1/r^3$ in the $2p$ state of a Coulomb potential
QUESTION [1 upvotes]: I've been having difficulty calculating this quantity.
We have an electron in the 2p state of a hydrogen atom (Coulomb potential). As a preliminary step to finding the energy difference between the $2p_{1/2}$ and $2p_{3/2}$ states, I am attempting to find expectation values for $\vec S \cdot \vec L$ and $\frac 1{r^3}$.
However, when I attempt to calculate $<\frac 1{r^3}>$, I run into a non-convergent integral.
If I understand the rules with expectation values correctly, the operator for $<\frac 1{r^3}>$ would be $\frac 1{r^3}$; thus,
$$<\frac 1{r^3}>=\int \Psi^* \frac 1{r^3} \Psi dV $$
over all space. However, in the 2p state, the radial wave function is of a form
$\Psi = C r e^{-r/2a}$ where $a$ is the Bohr radius and C is a mess of constants. Inserting this into the above equation gives me
$$<\frac 1{r^3}> = 4 \pi^2 C^2\int_0^\infty \frac 1r e^{-r/2a} dr $$
which does not converge at $0$.
Any help would be appreciated.

REPLY [1 votes]: If your wave function is $\Psi = Cre^{-r/2a}$, then you mustn't forget the additional $r^2$ you get from the Jacobi determinant (actually $r^2\sin\theta$, but the $\sin\theta$ is accounted for with the $4\pi$) when changing coordinates:
$$
\left<\frac{1}{r^3}\right> = 4\pi C^2\int_0^\infty\mathrm{d}r \;r^2\cdot\frac{1}{r^3}\cdot(re^{-r/2a})^2 = 4\pi C^2\int_0^\infty\mathrm{d}r \;r\cdot e^{-r/a} = 4\pi C^2a^2
$$


TITLE: The idea on how to show that the cone is not a topological surface
QUESTION [1 upvotes]: I just started taking Topology classes and we've been introduced to the concept of a topological surface, that is by Definition:

Def: A topological surface $S$ is a Hausdorff set such that $\forall x \in S, \exists  U \ni x $ open neighborhood such that $U $ is homeomorphic to a subset of $\mathbb{R}^2$

My Professor then gave us a few examples and one particular counter example that I still can't wrap my head around:

Example: The set $C= \lbrace (x,y,z) \in \mathbb{R}^3 \mid z^2 = x^2 + y^2 \rbrace$ is not a topological surface

When he asked the class I answered that the origin, i.e. the point $(0,0,0) \in C$ would cause trouble but I would not know on how to prove it. He affirmed that idea to me and said that I should think about removing the point and think about

Lemma: $f: X \to Y$ homeomorphism, $X$ connected $\implies Y$ connected

So that is what I did for the last couple of hours and I have confused myself utterly.

My problems:
Let me first start with the things that are clear to me:

If $U \subset \mathbb{R}^2$ is connected, then $U\setminus \lbrace c \rbrace$ for $c \in \mathbb{R}^2$ is still connected
If I take $U \ni (0,0,0)$ open neighborhood and consider $C \cap U \setminus \lbrace (0,0,0) \rbrace$ then this set is clearly disconnected, I did split up the cone locally around its origin in the upper half and lower half

Now to my problems:

I cannot prove that $C \cap U\setminus \lbrace (0,0,0) \rbrace$ is not homeomorphic to a subset of $\mathbb{R}^2$ although it is intuitively very clear
I don't see how to apply the Lemma
Far worse: I don't see why I want to do anything of the above.

If I negate the statement of a topological surface I read it as: $$ \exists x \in S, \forall U \ni x \text{ open neighborhood}: U \text{ is not homeomorphic to } \mathbb{R}^2 $$
So my point of interest (i.e. $(0,0,0)$) should still be in the neighborhood $U$ I consider right? I am sure that I need to raise a contradiction with all the listed ideas above, but I don't see the connection to $C$ because I have removed informations from $C$.

REPLY [3 votes]: Assume that $C$ is a topological surface.
It is our aim is now to find a contradiction.
Open sets $U\subseteq C$ and $V\subseteq\mathbb{R}^{2}$ exists together with a
homeomorphism $\phi:U\rightarrow V$ and $\langle 0,0,0\rangle\in U$.
For $\langle a,b\rangle:=\phi(\langle 0,0,0\rangle)\in V$ some open ball $B\subset\mathbb{R}^{2}$
centered at $\langle a,b\rangle$ exist with $\langle a,b\rangle\in B\subseteq V$.
Let $W:=\phi^{-1}\left(B\right)$ and prescribe $\psi:W\rightarrow B$ by $w\mapsto\phi\left(w\right)$.
Then $\psi$ is a homeomorphism.
However, it sends the not connected set $W-\left\{ \langle 0,0,0\rangle\right\} $
to the connected set $B-\left\{ \langle a,b\rangle\right\} $.
A contradiction has been found. Actually I used the converse of the lemma (a homeomorphism sends not connected sets to not connected sets) wich is also true. This because homeomorphisms have inverses, and these inverses are also homeomorphisms.