TITLE: Calculating and estimating the size of a minimal normal form
QUESTION [2 upvotes]: Each Boolean function $f:\{0,1\}^n \rightarrow \{0,1\}$ of $n$ variables has a minimal normal form. The number of terms of the minimal normal form is the minimal number of hidden neurons in a feed forward neural network that calculates the function. So one has to apply something like the Karnaugh map to find the minimal size of the hidden layer for a Boolean function, which may be given by its truth table. The truth table has $2^n$ entries which is also the size of the hidden layer of a canonical neural network to compute the function.
The Karnaugh map immediately gives you not only the size of the minimal normal form, but the minimal normal form itself. But what if you are only interested in the size (because you want to train the network)? How do I calculate the size of the minimal normal form possibly more quickly than by applying the Karnaugh map?
Furthermore: Are there any heuristics for estimating the size of the minimal normal form of a truth table even more quickly (with a given error distribution)?

REPLY [2 votes]: I can offer some results / sources:

The problem you pose is hard. In fact, it has been shown in 2008 (see here) that the unrestricted minimization problem is $\Sigma _{2}^{P}$-complete.
Karnaugh maps and equivalent Quine-McCluskey algorithms need prime implicants and an upper bound for their number has been shown (see here) to be $3^n/n$, so even higher than the number of minterms.
Because of (2.), the standard heuristic nowadays is the Espresso algorithm and variants thereof, which for larger numbers of $n$ (as you  will most likely be interested in for neural networks) has completely replaced Karnaugh maps and Quine-McCluskey algorithm.
A design question's prominent article is  "How many hidden layers and nodes?" by D. Stathakis - International Journal of Remote Sensing, 2009, [pdf]. In there is a discussion based on Kolmogorow complexity, resulting in a typical hidden layer size of (2n +1) neurons, which is fine if the correct activation function is chosen.
An extremely well cited, yet older article is "What Size Net Gives Valid Generalization?" by Eric B. Baum and David Haussler, Neural computation, 1989. [pdf]. The authors argue in a probabilistic fashion, relying on Vapnik-Chervonenkis dimensions. The main finding is that in a network with $N$ nodes and $W$ weights, with some error $e$ you can expect the full Boolean mapping (all conjunctive normal forms CNF) to be correctly assigned  if $m = {\cal O} (\frac We \log \frac Ne)$  randomly picked samples (CNFs) are correctly mapped. This holds for any distribution of the samples.
A very recent overview is given in "Toward Deeper Understanding of Neural Networks: The Power of Initialization and a Dual View on Expressivity" by
A Daniely, R Frostig and Y Singer,  2016 [pdf]. This heavily draws on kernel techniques.


TITLE: Find extension into higher dimensions for a matrix equation representing finite difference formulation of the heat equation
QUESTION [0 upvotes]: On page 9 of this pdf describing the finite difference formulation for the heat equation, there is a convenient tridiagonal matrix equation to represent equation 17 (which is on page 8). This represents the finite difference solution for 1D space heat conduction over time. However, the extension into 2D space would make this equation not applicable because there would be 2 more terms added for the y-dimension. I visualized this as the $(1-2r)$ term being changed to $(1-4r)$ and then surrounded on all 4 sides by $r$, so not only left and right but also into and out of the page. And then the $\phi$ term is no longer a vector but represents a 2D sheet of temperature values.
From what I can find, it seems that using tensor math would be the way to properly extend this equation into 2D, but I don't have a full grasp of it yet.
My question is, given my limited understanding of tensors:
How would I (1) organize a tensor equation that would properly collect the 5 terms in the 2D heat equation and (2) create a Matlab implementation of the tensor equation ?

REPLY [1 votes]: To extend the discretisation to
$$
u_t = \alpha(u_{xx} + u_{yy}),
$$
let's first leave time continuous and concentrate on the spatial terms. If $D_{xx}$ and $D_{yy}$ are matrices approximating the second derivative operator in the $x$ and $y$ directions respectively, then an appropriate discretisation is
$$
\mathbf{u}_t = \alpha \left[(D_{xx} \otimes I_y) + (I_x \otimes D_{yy}) \right]\mathbf{u}.
$$
Here, $\otimes$ denotes the Kronecker product, defined for the $m \times n$ matrix $A$ and the $p \times q$ matrix $B$ as the $mp \times nq$ matrix
$$
A \otimes B =
\begin{pmatrix}
a_{11} B & \cdots & a_{1n} B \\
\vdots & \ddots & \vdots \\
a_{m1} B & \cdots & a_{mn} B
\end{pmatrix},
$$
and $I_{x,y}$ are identity matrices of appropriate sizes. If we let $i = 1, \dots, n_x$ and $j = 1, \dots, n_y$ be indices counting the grid points in the $x$ and $y$ directions respectively, then the elements in $\mathbf{u}$ are ordered in the way that would be obtained if the columns of the matrix $(u_{ij})$ were stacked on top of each other, i.e.
$$
\mathbf{u} = (u_{11}, \dots, u_{n_x1}, u_{12}, \dots, u_{n_x2}, \dots, u_{1n_y}, \dots, u_{n_xn_y})^\top.
$$
In Matlab, the Kronecker product is performed through the command kron(). Thus, the matrix on the right-hand side of the discretisation can be obtained by typing
alpha*(kron(Dxx, eye(ny)) + kron(eye(nx), Dyy)).
Rather than having me type out a big matrix here, I encourage you to play around with this formulation on your own and get a feeling for what the resulting matrix looks like for different values of $n_x, n_y, \Delta x, \Delta y, \alpha$ and so on (start with small matrices). Once you feel comfortable with the Kronecker product and the resulting discretisation, you can discretise in time in the same way as in the pdf you linked.


TITLE: Dimension of linear system of divisor of two points on curve of genus greater than 2
QUESTION [2 upvotes]: This should not be hard, but I am stuck on it nonetheless, so I would much appreciate a solution.
Suppose $C$ is a projective non-singular curve of genus $g\geq 2$ and $P,Q$ are distinct points on $C$.
The notes of my algebraic geometry course now claim: "since $g\geq 2$, the dimension of the linear system $|P+Q|$ must be equal to 0 or 1."
Why is this?

REPLY [2 votes]: Assertion:
If $D$ is an effective divisor of degree $d\geq 1$ on  a smooth projective curve $C$ of genus $g\geq 1$, then $l(D)\leq d$.
The proof is by induction on $d$:
Initial step d=1
If $d=1$, then $D=1. P$ for some $P\in C$ and $L(1.P)=k$ (= base field) so that the assertion is true. [Here we have used that $g\geq 1$, i.e. that $C$ is not isomorphic to $\mathbb P^1$]
Inductive step $d-1\to d$
Choose some $P\in \operatorname {supp} D$ and consider the exact sequence of $k$-vector spaces $$0\to L(D-P)\to L(D)\to k$$ It immediately implies $l(D)\leq l(D-P)+1\leq (d-1)+1=d$
The assertion, which is much stronger than what you ask about, is thus proved.
Note carefully that this proof is from first principles and does definitely not involve grandiose concepts like Riemann-Roch, sheaf cohomology or Serre duality...


TITLE: Finding a limit of a quotient with trig
QUESTION [1 upvotes]: I'm studying for my Cal 2 exam, but I can't remember how to do the following question:
Find the limit, as n approaches infinity, of the sequence or prove that the limit does not exist:
$$\lim_{n \to {\infty}}{{3 \cdot n^2 \cdot \cos(n \cdot \pi)} \over {\sqrt{1+4 \cdot n^4}}}$$
Please guide me in the right path. Thanks!

REPLY [1 votes]: $$(1) \quad \lim_{n \to {\infty}}{{3 \cdot n^2 \cdot \cos(n \cdot \pi)} \over {\sqrt{1+4 \cdot n^4}}}$$
This limit is undefined...
Proof:
Let's divide both the numerator and denominator by $n^2$ we'll get...
$$(2) \quad \lim_{n \to {\infty}}{{3 \cdot \cos(n \cdot \pi)} \over {\sqrt{1/n^4+4}}}$$
Apply the quotient rule...
$$(3) \quad {\lim_{n \to {\infty}}{3 \cdot \cos(n \cdot \pi)} \over {\lim_{n \to {\infty}}\sqrt{1/n^4+4}}}$$
Ok, the denominator clearly goes to $4$, but the numerator is indeterminate. Keep in mind that cosine is periodic, but since we approach infinity, we can't define its value. The most we can say is that it's between $-1$ and $1$.
Thus, the limit is undefined.


TITLE: Finding the determinant of a $4\times4$ matrix
QUESTION [1 upvotes]: How does one find the determinant of a $4\times 4$ matrix? I am using Cramer's rule to solve a system of linear equations but don't know how to find the determinant of a $4\times 4$ matrix. Our matrix is with variables and not actual values so the answer will be in terms of the variables. Is there a rule/formula that I can use to get the determinant without using co-factor expansion?
The matrix given is
\begin{bmatrix}
1 & 1 & -2 & 3 \\
3 & 1 &-6 & 8 \\
0 & 7 & 3 & 0 \\
-1 & 8 & 7 & -4
\end{bmatrix}
, but then to use cramers rule and then to find the subsequent determinants the matrix will change? so was hoping for a general formula?

REPLY [2 votes]: When using Cramer's rule, there will be no variables whatsoever, rather, you know that the coordinates of the solution to $Ax=b$ are given by $$x_i=\frac{\det{(A\mid i)}}{\det A}$$ where $(A|i)$ is the matrix obtained by replacing $A$'s $i-th$ column with the column vector $b$. I give you an example on how to compute $\det A$, and maybe you can compute the remaining determinants yourself. Recall how how the detereminant behaves:
$({\rm i})$ It remains unchanged if we sum a multiple of a row (column) to another row (column)
$(\rm ii)$ It changes sign if we permute two rows (columns)
$(\rm iii)$ Scalars hop off the determinant.
Thus you may partially triagulate your matrix and operate as follows
$$\begin{align}
  \det \left( {\begin{matrix}
   1 & 1 & { - 2} & 3  \\
   0 & { - 2} & 0 & { - 1}  \\
   0 & 7 & 3 & 0  \\
   0 & 9 & 5 & { - 1}  \\
 \end{matrix} } \right) &=  - \det \left( {\begin{matrix}
   1 & 1 & { - 2} & 3  \\
   0 & 2 & 0 & 1  \\
   0 & 7 & 3 & 0  \\
   0 & 9 & 5 & { - 1}  \\
\end{matrix} } \right) \cr
  \text{cofactor} &=  - \det \left( {\begin{matrix}
   2 & 0 & 1  \\
   7 & 3 & 0  \\
   9 & 5 & { - 1}  \\
 \end{matrix} } \right) \cr
    R_2-3R_1\to R_2'\;,R_3-9R'_2\to R_3'\;\;  &=  - \det \left( {\begin{matrix}
   0 & { - 6} & 7  \\
   1 & 3 & { - 3}  \\
   0 & { - 22} & {26}  \\
 \end{matrix} } \right) \cr
  \text{ permute rows } &= \det \left( {\begin{matrix}
   1 & 3 & { - 3}  \\
   0 & { - 6} & 7  \\
   0 & { - 22} & {26}  \\
 \end{matrix} } \right) \cr
  \text{ cofactor } &= \det \left( {\begin{matrix}
   { - 6} & 7  \\
   { - 22} & {26}  \\
 \end{matrix} } \right) =  - 2 \end{align} $$
You can also triangulate it, and just calculate $\prod a_{ii}$ which is almost what I did above. Here is W|A's computation.

ADD There is a formula that does not involve cofactors, but it involves $4!=24$ terms, namely $$\det A=\sum_{\sigma \in S_4}\operatorname{sgn}\sigma a_{1\sigma(1)}\cdots a_{n\sigma(n)}$$


TITLE: Help needed with converting sentence to predicate logic
QUESTION [1 upvotes]: Say x represents real numbers and E(x) represent Even numbers and I(x) represents Integers.
How would you convert this sentence to predicate logic?
No even integers are odd
Here's my attempt, but I have a feeling it could be wrong
¬ ∀x ( E(x) ^ I(x) => ¬E(x) )

REPLY [2 votes]: Your sentence $\neg \forall x ( (E(x) \land I(x)) \rightarrow \neg E(x) )$ is saying that 'not all even integers are not even' ... which is the same as saying that 'some even integers are even'.
However, 'no even integers are odd' can be understood as 'all even integers are not odd', and so it should be:
$\forall x ( (E(x) \land I(x)) \rightarrow \neg E(x) )$


TITLE: Multitape Turing machine with multiple non-blank tapes
QUESTION [0 upvotes]: A multitape Turing machine is defined to have input only appear on one tape, with the rest of the tapes blank.
Are there any formulations of a Turing machine that allow other tapes to be not blank? Why does the condition that the input only appears on one tape exist?

REPLY [1 votes]: It is just simpler to describe the situation with a single input tape -- in particular, if you want to compare the power or efficiency of machines with differing number of tapes, it is convenient that the format of the inputs to all the machines are the same.
But it is not a particularly deep choice -- if you have input on one tape and want it spread across several, or have input on multiple tapes and want it combined into one, it is generally trivial to specify a machine that starts by rearranging the inputs as you wish.


TITLE: Can Frobenius traces jump like crazy in non-geometric Galois representations?
QUESTION [5 upvotes]: If I have a continuous representation $\mathrm{Gal}(\overline{\mathbb{Q}}/\mathbb{Q})\to \mathrm{GL}_n(\mathbb{Q}_l)$ ramified at finitely many places how can the Frobenius traces behave?
Assuming they lie in an algebraic extension of $\mathbb{Q}$ can they grow exponentially with $p$? How much can $\lim\sup$ and $\lim\inf$ diverge?

REPLY [2 votes]: "Assuming they're in an algebraic extension of $\mathbb{Q}$ can they
grow exponentially with ?"?

I'm not sure this statement will have a truth value, because I suspect the assumption never occurs: non-geometric $\ell$-adic representations are fundamentally $\ell$-adic analytic objects and I don't know of any mechanism which would force their Frobenius traces to be in $\overline{\mathbb{Q}}$.
[Here I'm assuming that your representations are irreducible; one can easily construct examples of extensions $0 \to \chi_1 \to V \to \chi_2 \to 0$ with $\chi_i$ geometric which are unram almost everywhere but non-geometric at $\ell$. But then the traces are the same as $\chi_1 \oplus \chi_2$ so the question is not interesting.]
There are examples of irreducible 2-dim'l non-geometric representations arising from non-classical overconvergent modular forms. I tabulated a bunch of these in my first ever paper (paywall, arxiv version here). The result is a list of $\ell$-adic numbers (computed modulo some high power of $\ell$), which are the q-expansion coefficients, or equivalently Frob traces at the first few primes; and they don't like they come from $\mathbb{Q}$ in any recognisable way.


text:
\begin{document}
\begin{abstract}
We prove a continued fraction expansion for the reciprocal of a certain $q$-series.
All the specialists in the world are asked whether it is new or not.
\end{abstract}

\maketitle

\section{Introduction}

Let
\begin{equation*}
(x;q)_n=(1-x)(1-xq)\dots(1-xq^{n-1}).
\end{equation*}
Series of the type
\begin{equation*}
\sum \frac{(a_1;q)_n\dots(a_s;q)_n}{(b_1;q)_n\dots(b_t;q)_n}z^n
\end{equation*}
are called $q$-series (basic hypergeometric series), and the summation might be either over the nonnegative integers or over
the whole set of integers.

There are several results about such series and \emph{continued fraction expansions};
to give one example, set
\begin{equation*}
_2\psi_2(\alpha, \beta;\gamma,\delta\mid z)=\sum_{n\in\mathbb{Z}}
\frac{(\alpha;q)_n(\beta;q)_n}{(\gamma;q)_n(\delta;a)_n}z^n,
\end{equation*}
then
\begin{equation*}
\frac{_2\psi_2(\alpha, \beta;\gamma,\delta\mid z)}{_2\psi_2(\alpha, \beta q;\gamma,\delta q\mid z)}
\end{equation*}
is expanded into a continued fraction in \cite{Denis90}.

The examples that I could trace in the literature usually deal with \emph{quotients} of \emph{similar} functions.  The is a school of indian mathematicians who deal with such questions; \textsf{MathSciNet} will easily produce a list of such papers (which I found hard to get!).

The continued fraction that I will present here only deals with \emph{one} $q$-series, not a quotient. I have not been able to find it in the literature, although it might exist.

Thus, I post this draft now, and ask all the specialists in the world for feedback. If nobody will
identify it within a month, say, then I will dare to submit it to a journal. Otherwise, I will label it
\emph{written for my own benefit}.

\section{Main result}

\begin{theorem} Let
\begin{equation*}
G(z)=\sum_{n\ge0}\frac{(y;q)_nz^{n}}{(x;q)_n}.
\end{equation*}
Then we have the continued fraction expansion
\begin{equation*}
\frac{z}{G(z)}
=
\cfrac{z}{a_1+
	\cfrac{z}{a_2+
		\cfrac{z}{a_3+
			\cfrac{z}{a_4+\cfrac z{\ddots}
			}
		}
	}
}
\end{equation*}
with $a_1=1$, and for $k\ge1$
\begin{align*}
a_{2k}&=\frac{(x;q)_{k-1}(1-xq^{2k-2})(yq;q)_{k-1}}{(1-y)(yq)^{k-1}(\frac xy;q)_{k-1}(q;q)_{k-1}},\\
a_{2k+1}&=-\frac{(1-y)y^{k-1}(\frac xy;q)_{k-1}(1-xq^{2k-1})(q;q)_{k-1}}{(x;q)_{k}(yq;q)_{k}}.
\end{align*}
\end{theorem}
\emph{Proof.}
We write
\begin{align*}
\frac {z}{G(z)}&=\cfrac{z}{N_0}=\cfrac{z}{a_1+\cfrac{z}{N_1}}=\cfrac{z}{a_1+\cfrac{z}{a_2+\cfrac{z}{N_2}}}=\dots,
\end{align*}
and set
\begin{equation*}
N_i=\frac{r_i}{s_i}.
\end{equation*}
This means
\begin{equation*}
N_i=a_{i+1}+\frac{z}{N_{i+1}}
\end{equation*}
or
\begin{equation*}
\frac{z}{N_{i+1}}=\frac{zs_{i+1}}{r_{i+1}}=N_i-a_{i+1}=\frac{r_i}{s_i}-a_{i+1}
=\frac{r_i-a_{i+1}s_i}{s_i}.
\end{equation*}
We may identify numerators and denominators and deduce the recursion
\begin{equation*}
zs_{i+1}=s_{i-1}-a_{i+1}s_i.
\end{equation*}
Since $a_1$ is an exceptional value, we only start the recursion with
\begin{equation*}
s_0=1, \qquad s_1=\sum_{n\ge0}\frac{(y;q)_{n+1}z^{n}}{(x;q)_{n+1}}.
\end{equation*}
Note that the numbers $a_i$ are uniquely determined by annihilating the constant term in
$s_{i-1}-a_{i+1}s_i$, making $s_{i+1}$ a power series expansion. Our claim follows now
by the following \emph{explicit} formul\ae{} (for $k\ge0$)
\begin{align*}
s_{2k}&=\frac{(-1)^kq^{ \binom k2}}{(q;q)_{k-1}}\sum_{n\ge0}\frac{z^n(q^{n+1};q)_{k-1}(yq;q)_{n+k}}{(xq^k;q)_{n+k}},\\*
s_{2k+1}&=(1-y)y^k(\tfrac xy;q)_k(-1)^kq^{\binom{k+1}2}\sum_{n\ge0}\frac{z^n(q^{n+1};q)_{k}(yq^{k+1};q)_n}{(x;q)_{n+2k+1}},
\end{align*}
provided we are able to establish these formul\ae{} by induction via the recursion.
The initial values follow by inspection, and the induction step must be split into two
computations, according to the parity of the indices.
\small
\begin{align*}
s_{2k}&-a_{2k+2}s_{2k+1}=\frac{(-1)^kq^{ \binom k2}}{(q;q)_{k-1}}\sum_{n\ge0}\frac{z^n(q^{n+1};q)_{k-1}(yq;q)_{n+k}}{(xq^k;q)_{n+k}}\\
&-\frac{(x;q)_{k}(qy;q)_{k}(1-xq^{2k})}{(qy)^{k}(\frac xy;q)_{k}(1-y)(q;q)_{k}}(1-y)y^k(\tfrac xy;q)_k(-1)^kq^{\binom{k+1}2}\sum_{n\ge0}\frac{z^n(q^{n+1};q)_{k}(yq^{k+1};q)_n}{(x;q)_{n+2k+1}}\\
&=\frac{(-1)^kq^{ \binom k2}}{(q;q)_{k-1}}\sum_{n\ge0}\frac{z^n(q^{n+1};q)_{k-1}(yq;q)_{n+k}}{(xq^k;q)_{n+k}}-\frac{(1-xq^{2k})}{(q;q)_{k}}(-1)^kq^{\binom{k}2}\sum_{n\ge0}\frac{z^n(q^{n+1};q)_{k}(yq;q)_{n+k}}{(xq^k;q)_{n+k+1}}\\
&=\frac{(-1)^kq^{ \binom k2}}{(q;q)_{k}}\bigg[
(1-q^k)\sum_{n\ge0}\frac{z^n(q^{n+1};q)_{k-1}(yq;q)_{n+k}}{(xq^k;q)_{n+k}}-(1-xq^{2k})\sum_{n\ge0}\frac{z^n(q^{n+1};q)_{k}(yq;q)_{n+k}}{(xq^k;q)_{n+k+1}}\bigg]\\
&=\frac{(-1)^kq^{ \binom k2}}{(q;q)_{k}}
\sum_{n\ge0}\frac{z^n(q^{n+1};q)_{k-1}(yq;q)_{n+k}}{(xq^k;q)_{n+k+1}}\Big[(1-q^k)(1-xq^{n+2k})-(1-xq^{2k})(1-q^{n+k})\Big]\\
&=\frac{(-1)^kq^{ \binom k2}}{(q;q)_{k}}
\sum_{n\ge0}\frac{z^n(q^{n+1};q)_{k-1}(yq;q)_{n+k}}{(xq^k;q)_{n+k+1}}(-1)q^k(1-xq^k)(1-q^n)\\
&=\frac{(-1)^{k+1}q^{ \binom {k+1}2}}{(q;q)_{k}}
\sum_{n\ge0}\frac{z^n(q^{n};q)_{k}(yq;q)_{n+k}}{(xq^{k+1};q)_{n+k}}\\
&=zs_{2k+2};
\end{align*}
\begin{align*}
s_{2k-1}&-a_{2k+1}s_{2k}=
(1-y)y^{k-1}(\tfrac xy;q)_{k-1}(-1)^{k-1}q^{\binom{k}2}\sum_{n\ge0}\frac{z^n(q^{n+1};q)_{k-1}(yq^{k};q)_n}{(x;q)_{n+2k-1}}\\
&\quad+\frac{(1-y)y^{k-1}(\frac xy;q)_{k-1}(q;q)_{k-1}(1-xq^{2k-1})}{(qy;q)_{k}(x;q)_{k}}\frac{(-1)^kq^{ \binom k2}}{(q;q)_{k-1}}\sum_{n\ge0}\frac{z^n(q^{n+1};q)_{k-1}(yq;q)_{n+k}}{(xq^k;q)_{n+k}}\\
&=
(1-y)y^{k-1}(\tfrac xy;q)_{k-1}(-1)^{k-1}q^{\binom{k}2}\sum_{n\ge0}\frac{z^n(q^{n+1};q)_{k-1}(yq^{k};q)_n}{(x;q)_{n+2k-1}}\\
&\quad+(1-y)y^{k-1}(\tfrac xy;q)_{k-1}(1-xq^{2k-1})(-1)^kq^{ \binom k2}\sum_{n\ge0}\frac{z^n(q^{n+1};q)_{k-1}(yq^{k+1};q)_{n}}{(x;q)_{n+2k}}\\
&=(1-y)y^{k-1}(\tfrac xy;q)_{k-1}(-1)^{k-1}q^{\binom{k}2}\\&\quad\times\bigg[\sum_{n\ge0}\frac{z^n(q^{n+1};q)_{k-1}(yq^{k};q)_n}{(x;q)_{n+2k-1}}-(1-xq^{2k-1})\sum_{n\ge0}\frac{z^n(q^{n+1};q)_{k-1}(yq^{k+1};q)_{n}}{(x;q)_{n+2k}}\bigg]\\
&=(1-y)y^{k-1}(\tfrac xy;q)_{k-1}(-1)^{k-1}q^{\binom{k}2}\sum_{n\ge0}\frac{z^n(q^{n+1};q)_{k-1}(yq^{k+1};q)_{n-1}}{(x;q)_{n+2k}}\\
&\quad\times\Big[(1-xq^{n+2k-1})(1-yq^k)-(1-xq^{2k-1})(1-yq^{k+n})\Big]\\
&=(1-y)y^{k-1}(\tfrac xy;q)_{k-1}(-1)^{k-1}q^{\binom{k}2}\sum_{n\ge0}\frac{z^n(q^{n+1};q)_{k-1}(yq^{k+1};q)_{n-1}}{(x;q)_{n+2k}}\\&\quad\times(-1)q^{k-1}(1-q^n)yq(1-\tfrac xyq^{k-1})\\
&=(1-y)y^{k}(\tfrac xy;q)_{k}(-1)^{k}q^{\binom{k+1}2}\sum_{n\ge0}\frac{z^n(q^{n};q)_{k}(yq^{k+1};q)_{n-1}}{(x;q)_{n+2k}}\\
&=zs_{2k+1}.
\end{align*}
\normalsize

\emph{Remark.} Computer experiments indicate that we cannot add additional factors $(u;q)_n$
etc. in either numerator or denominator, as then the expressions for $a_i$ become very messy and
don't factor nicely.

\medskip

We note two special cases explicitly. Set $y=0$, then
\begin{align*}
a_{2k}&=\frac{(x;q)_{k-1}(1-xq^{2k-2})}{x^{k-1}q^{\binom k2}(q;q)_{k-1}},\\
a_{2k+1}&=-\frac{x^{k-1}q^{\binom {k-1}2}(1-xq^{2k-1})(q;q)_{k-1}}{(x;q)_{k}}.
\end{align*}
Set $x=0$, then
\begin{align*}
a_{2k}&=\frac{(yq;q)_{k-1}}{(1-y)(yq)^{k-1}(q;q)_{k-1}},\\
a_{2k+1}&=-\frac{(1-y)y^{k-1}(q;q)_{k-1}}{(yq;q)_{k}}.
\end{align*}

\bibliographystyle{amsplain}

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\providecommand{\MR}{\relax\ifhmode\unskip\space\fi MR }

\providecommand{\MRhref}[2]{
  \href{http://www.ams.org/mathscinet-getitem?mr=#1}{#2}
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\providecommand{\href}[2]{#2}

config: arxiv


TITLE: Is there a connection between the closed forms of these two infinite products?
QUESTION [6 upvotes]: Take the following two infinite products that have closed forms.
Assume: $\gamma_n > 0 \in \mathbb{R};s,a,x \in \mathbb{C}; x \ne 0,a \pm ix\gamma_n \ne 0$
The first product:
$$\displaystyle H_{int}(s,a,x) := \prod_{n=1}^\infty \left(1- \frac{s}{a + i x \gamma_n} \right) \left(1- \frac{s}{{a - i x \gamma_n}} \right) = \frac{\xi_{int}(0 -\frac{a}{x}+\frac{s}{x})}{\xi_{int}(0-\frac{a}{x})}$$
has $\gamma_n = n$, so runs through the integers with: $\xi_{int}(s) = \frac{\sinh(\pi s)}{s}$.
The second product, for which the closed form can be derived assuming RH is true,
$$\displaystyle H_{rie}(s,a,x) := \prod_{n=1}^\infty \left(1- \frac{s}{a + i x \gamma_n} \right) \left(1- \frac{s}{{a - i x \gamma_n}} \right) = \frac{\xi_{rie}(\frac12 - \frac{a}{x}+\frac{s}{x})}{\xi_{rie}(\frac12 - \frac{a}{x})}$$
has $\gamma_n = \Im(\rho_n)$, so runs through the non-trivial zeros $\rho_n$ with: $\xi_{rie}(s)= \frac12 s(s-1) \pi^{-\frac{s}{2}} \Gamma(\frac{s}{2}) \zeta(s)$.
The products clearly have a comparable structure and share the following characteristics:
1) They are both reflexive: $H(s,a,x) = H(2a-s,a,x)$ (since this is fully independent of $\gamma_n$).
2) Their $\xi(s)$-functions are reflexive as well:
$$\xi_{int}(s)=\xi_{int}(2(0)-s)$$
$$\xi_{rie}(s)=\xi_{rie}(2(\frac12)-s)$$
3) They both have a similar 'base' Hadamard product (for $a=0$ and $a=\frac12$ respectively):
$$\displaystyle \xi_{int}(s) = \xi_{int}(0) \prod_{n=1}^\infty \left(1- \frac{s}{0+i n} \right) \left(1- \frac{s}{{0- i n}} \right) \rightarrow \xi_{int}(0)=\pi$$
$$\displaystyle \xi_{rie}(s) = \xi_{rie}(0) \prod_{n=1}^\infty \left(1- \frac{s}{\frac12 + i \Im(\rho_n)} \right) \left(1- \frac{s}{{\frac12 - i \Im(\rho_n)}} \right) \rightarrow \xi_{rie}(0)=\frac12$$
4) They are both entire functions and the "undesired" zeros/poles from their meromorphic components $\sinh(s)$ and $\zeta(s)$ are annihilated via $s$ and $\frac12 s(s-1), \Gamma(\frac{s}{2})$ respectively. Both meromorphic elements can be expressed as an infinite product as well as an infinite sum (over a certain domain):
$$\sinh(s) = s \prod_{k=1}^\infty \left(1+ \frac{s^2}{k^2\pi^2)} \right)=\sum_{n=1}^\infty \frac{s^{2n-1}}{\Gamma(2n)}$$
$$\zeta(s) = \prod_{p} \left(\frac{1}{1-p^{-s}} \right)=\sum_{n=1}^\infty \frac{1}{n^s}$$
5) And both of the above can be analytically continued throughout the entire complex plane via:
$$\sinh(0-s) = -\sinh(s)$$
$$\zeta(1-s) = \chi(s)\zeta(s)$$
My questions:
1) Is there any other possible (or known) set of values for $\gamma_n$ that could yield yet another closed form? Or is this all there is, i.e. are the ("via parameter $x$ linearly scalable") integers $\gamma_n=n$ at $a=0$ and $\gamma_n =\Im(\rho_n)$ at $a=\frac12$, the only possible choices?
2) Since the $\rho_n$'s contain information about the (distribution of) primes and the primes in turn can induce the integers via unique multiplication, could there be a connection made between (the closed forms of) these two products?
Thanks.

REPLY [3 votes]: In my quest for similar closed forms as the above, I did find:
$$\displaystyle H_{int^2}(s,a,x) := \prod_{n=1}^\infty \left(1- \frac{s}{a + i x \gamma_n} \right) \left(1- \frac{s}{{a - i x \gamma_n}} \right) = \frac{\xi_{int^2}(0 -\frac{a}{x}+\frac{s}{x})}{\xi_{int^2}(0-\frac{a}{x})}$$
with $\gamma_n = n^2$, running through the squared integers and $\xi_{int^2} = {\frac {\sin \left(  \left(\frac12-\frac12i \right)  \sqrt{2s} \ \pi \right) \sin \left(  \left(\frac12+\frac12i \right)  \sqrt{2s} \ \pi \right)}{s}}$
I do believe more closed forms exist, however only for $\gamma_n = n^{2k}$ with ($k= 1,2,3...)$ and would like to conjecture that for the 'base' Hadamard product it is true that:
$$\displaystyle \xi_{int^{2k}}(s) = \xi_{int^{2k}}(0) \prod_{n=1}^\infty \left(1- \frac{s}{0+i n^{2k}} \right) \left(1- \frac{s}{{0- i n^{2k}}} \right) \rightarrow \xi_{int^{2k}}(0)=\pi^{2k}$$
However, my (maybe too big) dream was to find a closed form for $\gamma_n = p_n$ with $p_n$ being the n-th prime. Via brute force calculations, I did find a few (all explainable) 'leads' for:
$$\displaystyle H_{prime}(s,a,x) := \prod_{n=1}^\infty \left(1- \frac{s}{a + i x p_n} \right) \left(1- \frac{s}{{a - i x p_n}} \right)$$
$$H_{prime}(1,1,1) = \frac{\pi^2}{15} = \frac{\zeta(4)}{\zeta(2)}$$
$$H_{prime}(1,\frac12,1) = 1$$
$$H_{prime}(i,i,1) = \frac{\pi^2}{6} = \zeta(2)$$
and the general rules, that also apply to the above:
$$H_{prime}(s,a,x) = H_{prime}(2a-s,a,x)$$
$$H_{prime}(s,a,x) = \frac{1}{H_{prime}(s,1-a,x)}$$
I can also prove that a closed (entire) form must have the shape: $\dfrac{\xi_{prime}(\beta -\frac{a}{x}+\frac{s}{x})}{\xi_{prime}(\beta-\frac{a}{x})}$, however the key snag is that the meromorphic component of $\xi_{prime}(s)$ that generates the zeros (similar to $\zeta(s)$), must do so only at the primes, whilst all extra "undesired and maybe more trivial" zeros need to annihilated by other components of the entire function $\xi_{prime}(s)$.
This brought me immediately to Wilson's formula, where I used the $\Gamma$-function instead of the factorial and replaced the $mod$ by the $cos$ to produce a function that generates zeros only at integers that are prime:
$$\cos\left(\dfrac{\pi}{2}\dfrac{\Gamma(s)+1}{s}\right)$$
however it unfortunately also generates a tremendous amount of 'undesired, but hopefully more trivial' (non-integer) zeros that I need to annihilate. Is there any known function (other than using the floor function) that could accomplish this? Any other ideas\steers?
Thanks.


TITLE: number of derangements
QUESTION [3 upvotes]: In the normal derangement problem we have to count the number of derangement when each counter has just one correct house,what if some counters have shared houses.
A derangement of n numbers is a permutation of those numbers in which none of the numbers appears in its original place. For example, the numbers {1,2,3} can be deranged into {2,3,1} and {3,1,2}
But in our modified version no number in the derangement can be in the place that a number of the same value was in in the original ordering. So the numbers {1,1,2,2,3} could be deranged into {2,2,1,3,1}, {2,2,3,1,1}, {2,3,1,1,2}, and {3,2,1,1,2}.
How to go about it! any hints?

REPLY [2 votes]: This is a permutation with forbidden positions.  Your specific problem corresponds to the board below, where $X$ in square $(i,j)$ denotes that element $i$ is not allowed to go to position $j$.
XXOOO
XXOOO
OOXXO
OOXXO
OOOOX
Let $r_i$ denote the number of ways to place $i$ rooks on the $X$ squares, such that none attack each other.
$r_1=9$,  because there are $9$ squares marked $X$.
$r_2=4^2+4+4+2+2$, because there are $4^2$ ways to pick one square from each of the $2\times 2$ grids, there are $4+4$ ways to pick one square from one of the $2\times 2$ grids plus the lower-right square, and there are $2+2$ ways to pick two squares from each of the $2\times 2$ grids.
You still need to find $r_3, r_4, r_5$.  It's fairly tedious but possible.  Then, the answer is $$5!-r_14!+r_23!-r_32!+r_41!-r_50!$$


TITLE: Understanding EM Fields
QUESTION [1 upvotes]: I am an electronics engineering major and some questions arise when studying communications technology that utilizes wireless technology. In particular, I am more of a complete picture kind of person, as such I like to see how various things are connected. I want to understand, at least on the surface level, how the following things are connected: Phonons, photons, electrons, charge fields.
As far as I understand things you have phonons. These are quasi-particles representing the quantized energy packets of vibrations through a lattice structure. This seems intuitive enough, and after some further reading I see that photons can be spontaneously emitted from photon interactions resulting from absorption and re-emission of the quantized energy of the phonons. I assume the photon emission frequency/energy is directly related to the energy of the phonon. I believe that photons can also cause phonons when a photon is incident to a lattice structure, though I am not 100% on the details of this.
Photons are actual particles of quantized packets of energy propagating through the EM field medium. As to what exactly constitutes the EM field I am not sure. I believe magnetic fields are due to electrons? I also believe that electrons have an electric field due to the inherent charge electrons pose? However, I believe the charges reach/influence is based on the inverse square law. As such, is the electromagnetic field simply composed on the electron charges as the medium? If this is the case, then how can the EM field be so far reaching?
I apologize for the large amount of questions, and I am sure there are similar questions around, but none that seem to tie all of these things together.
EDIT: To add, am I correct to assume the EM Field is modeled as a continuous distribution of charge, as such the inverse square charge distance is a non-factor? And to answer one of the comments in the question, I have taken classes on partial differential equations, but my physics did not require them, thus I have only indirectly utilized Maxwell's Equations. I assume those will help connect a few of my questions mathematically, so I'll be sure to take a look at those.

REPLY [3 votes]: Actually the  framework where one can describe  electromagnetic fields is a classical framework. When one is talking of photons phonons etc one is in the quantum mechanical regime where the concept field, is different.
A classical field in physics:

A field can be classified as a scalar field, a vector field, a spinor field or a tensor field according to whether the value of the field at each point is a scalar, a vector, a spinor or a tensor, respectively. For example, the Newtonian gravitational field is a vector field: specifying its value at a point in spacetime requires three numbers, the components of the gravitational field vector at that point.

The electric and magnetic field are also vector fields in classical physics.
When one goes to the quantum mechanical frame the point in space time is characterized by  quantum operators, and to get a vector field one must operate with this operators on the state function of the system.
It aint simple.
Classically the electric field is due to a space distribution of charge, and can be described by a potential. The magnetic field due to the fact that we have never observed magnetic monopoles is generated by magnetic dipoles, or by moving charges . The electromagnetic field, light, is the result of changing magnetic fields and electric fields described experimentally well by the solutions of Maxwell's equations.
Going into the quantum mechanical framework of photons, elementary particles , we know that the classical fields emerge from a huge number of photons  whose phases are such as to build up what we see as  light/radiation. A single photon is an  excitation of the quantum mechanical field which is mathematically defined over all space and is a completely different story than the classical electric or magnetic field
Photons/radiation are emitted in various ways as I explained in answer to a different question. One of them is the energy transfer from the quantized collective kinetic energy in a lattice, phonons,  to a photon via interaction with a bound electron. A transition has to happen between energy levels of electrons for a photon to  be emitted. Phonons are not the main way photons are emitted (it might be so in black body radiation but I am not sure).
You should read on the links and if really interested attend a course on classical electricity and magnetism first and then quantum mechanics.


TITLE: Why is the independence of paths for complex integrals not sufficient enough to proof Cauchy's integral theorem?
QUESTION [0 upvotes]: I have seen multiple proofs of Cauchy's integrals theorem, mostly using the Deformation Invariance theorem or a version of Green's Theorem. But why exacly is this necessary?
From my understanding of the indepence of paths, isn't it obvious that if you take 2 random point z1 and z2 on a closed contour, then the integral from z1 to z2 and back from z2 to z1 cancel out and thus results in zero?
For closed contours consisting of multiple smooth curves, simply taking the sum of the integrals from each start and end point should result in a telescoping sum that again cancels out to zero.
All of this assumes the function is analytic in the domain D. Is there some other condition I am missing? Or is this a circular reasoning? (no pun intended). For reference, my math education is at the level of an undergraduate, with mostly self taught complex analysis.

REPLY [2 votes]: This reasoning is indeed circular, at least most of the time. Generally, Cauchy's integral theorem is used to prove the independence of paths for complex integration, so one can't use the independence of paths to prove Cauchy's integral theorem.


TITLE: If $f$ is a Morse function, then so is $f \circ \phi^{-1}$, where $\phi: U \rightarrow \mathbb{R}^k$ is the coordinate chart.
QUESTION [0 upvotes]: I am trying to show: if when $f^\prime = 0$, then $f^{\prime\prime} \neq 0 \Leftrightarrow (f \circ \phi^{-1})^\prime = 0$, $(f \circ \phi^{-1})^{\prime\prime} \neq 0$.
But the problem is, because $(f \circ \phi^{-1})^\prime = f^\prime(\phi(x))\phi^\prime(x)$, when $(f \circ \phi^{-1})^\prime = 0$, it could be $f^\prime = 0$ or $\phi^\prime = 0$. Can I simply regard $\phi$ as a change of coordinate function hence its Jacobian is nonsingular, so $\phi^\prime \neq 0$?
$(f \circ \phi^{-1})^{\prime\prime}
= (f^\prime(\phi(x))\phi^\prime(x))^\prime
= f^{\prime\prime}(\phi(x))\phi^\prime(x)\phi^\prime(x) + f^\prime(\phi(x))\phi^{\prime\prime}(x)$
Assuming $f^\prime(\phi(x))=0, \phi^\prime \neq 0$:
$(f \circ \phi^{-1})^{\prime\prime} = f^{\prime\prime}(\phi(x))\phi^\prime(x)\phi^\prime(x) \neq 0$
Assuming $\phi^\prime = 0:
(f \circ \phi^{-1})^{\prime\prime}  = 0$ Failed.
Thank you very much!

REPLY [0 votes]: This is just a calculus question. The given condition "coordinate chart" is strong. I found this in my calculus book: Advanced Calculus of Several Variables, Edwards
Definition: Let $\phi: U \rightarrow \mathbb{R}^n$ be a $\mathscr{C}^1$ mapping defined on an open subset $U$ of $\mathbb{R}^k (k \leq n)$. Then we call $\phi$ regular if the devrivatire matrix $\phi^\prime(u)$ has maximal rank $k$, for each $u \in U$.
Definition: Given $p \in M$, there exists an open set $U \subset \mathbb{R}^k (k<n)$ and a regular $\mathscr{C}^1$ mapping $\phi: U \rightarrow \mathbb{R}^n$ such that $p \in \phi(U)$, with $\phi(U^\prime)$ being an open subset of $M$ for each open set $U^\prime \subset U$. Then the mapping $\phi$ is called a coordinate patch for $M$ provided it is injective.
Certainly, a full rank matrix can't be an empty matrix.


TITLE: What electric field will act in the vacuum between two closely placed metal plates with saw-tooth voltage applied to them?
QUESTION [0 upvotes]: If two parallel metal are placed very close to each other and a saw-tooth voltage is applied in the plates, how will the electric field behave in the vacuum in between the plates?
As the separation between the plates is very small, I think we can assume the field will be homogeneous.
As we know, $\vec E = - \frac{\partial V}{\partial x}$ (assuming the parallel plates are placed along the x axis), then as the voltage $V$ is a linear function of time i.e. $V(x,t)= a \cdot t \cdot f(x)$ where $a$ is some positive constant); then it should be true that $\vec E(x,t) = t \cdot g(x)$, as the defining equation for $\vec E$ doesn't contain a derivative wrt time. Now, as we are assuming the field $\vec E$ being homogeneous as the separation between plates is very small; $\vec E(x,t) = k \cdot t$ (i.e. $g(x) = k$).
Is my analysis right?
I have one point of confusion on whether the electric field will change instantaneously inside the vacuum between the plates as the voltage changes linearly in time. If it does, I think my analysis is right.

REPLY [1 votes]: As long as the quasi-static approximation is valid (the plate dimensions are much smaller than the wavelength $\lambda = c/f$, where $c$ is the speed of light and $f$ is the bandwidth of the waveform), you can safely assume that $\vec{E}$ has the same time variation as $V$, the potential difference between the plates. So if $V$ has a sawtooth time variation, so should $\vec{E}$.
Realistically, a bigger concern should likely be whether the potential difference between the plates (which act as a capacitor) is equal to the voltage waveform generated by whatever circuit/signal generator you are using. At high frequencies, the impedance of the capacitor is no longer infinite as it is at DC, so you may have significant potential drops along the wires connecting your signal generator to the plates, due their resistance and inductance. The output impedance of the signal generator might be an issue at high frequencies for the same reason.


text:
{\bf Problem.} The product of two positive consecutive integers is 506. What is their sum?
{\bf Level.} Level 2
{\bf Type.} Algebra
{\bf Solution.} We are given that $x(x+1) = 506$, so $x^2 + x = 506$, which means $x^2 + x - 506 =0$.  The prime factorization of $506$ is $2\cdot 11 \cdot 23$, so we see that the quadratic factors as $(x + 23)(x-22)=0$.  The positive solution is $x=22$, so the two numbers are 22 and 23.  Their sum is $22+23 = \boxed{45}$.

set_name: MATH


TITLE: Explanation on examples of non-divisible module
QUESTION [0 upvotes]: So the multiplicative group of complex numbers, $\mathbb{C}^*$, is divisible as an abelian group.
Why is the multiplicative group of real numbers, $\mathbb{R}^*$, not divisible as an abelian group?
What is the difference?
My lecturer remarked this as an explanation : "$\mathbb{R}^*$ contains $-1$ which is not divisible by $2$".
I don't really understand this, because you can clearly write $-1 = 2(-\frac{1}{2})$

REPLY [1 votes]: "Divisible" means with respect to the $\Bbb Z$-module structure of the abelian group $(G,*)$, id est the action $$n\odot z=\begin{cases}\underbrace{z*z*\cdots*z}_{n\text{ times}}&\text{if }n\ge 0\\ \underbrace{z^{-1}*z^{-1}*\cdots*z^{-1}}_{(-n)\text{ times}}&\text{if }n< 0\end{cases}$$ Since the operation, in this cases, is multiplication, the action of $\Bbb Z$ is $n\odot z:=z^n$.
Concretely, here being divisible by $2$ means having a square root.


TITLE: Prove that $9 \vert (1-10^{i})$ without Mathematical Induction
QUESTION [0 upvotes]: I’m an undergraduate student taking a Discrete Mathematics module. I have a proof question from a past year paper to prove that the digit sum, $s$  of a non-negative integer,$n$ satisfies the following result, $$s \equiv n\ (mod\ 9) $$.
I have approached it initially using a direct proof. A sketch of my proof would be deducing that $s-n$ reduces to $$
\sum_{i=0}^k d_{i}.(1-10^{i})$$, where $k$ is some non-negative integer and $d_{0},d_{1},...,d_{k}$ are the decimal digits of the number $n$. Afterwards, we need to prove that $9 \vert (1-10^{i})$ and by the linear combination property of integers, $9 \vert (s-n)$ which is then equivalent to saying $s \equiv n\ (mod\ 9) $.
 (By the way, $ \vert$, means "is a divisor of", in case this is not a universal notation)
My initial plan to prove this lemma: $9 \vert (1-10^{i})$ is by Mathematical induction, but it looks like this question is supposed to be simple due to the marks awarded, so Mathematical Induction might be an overkill.
Alternatively, I think a result I learnt from tutorial, which is that $9 \vert n \iff 9 \vert s$, might be useful here. However, I don’t know how I could link this result to modular arithmetic. I am also restricted to quoting results from the discrete mathematics textbook:Discrete Mathematics with applications by Susanna S. Epp, without proof, as well as tutorial results. Can someone share how this result might make my proof of the Lemma: $9 \vert (1-10^{i})$ simpler, or is there any other way of making this part simpler besides Mathematical Induction. Thanks.

REPLY [1 votes]: You need to calculate $1-10^i \pmod 9$. Notice that you can reduce the number $10 \pmod 9$, since $a^i \equiv b^i$, if $a\equiv b$.
But
$$10^i \equiv 1^i = 1 \pmod 9$$
so what you get is
$$1 - 10^i \equiv 1 - 1 = 0 \pmod 9$$
which is exactly what you want.


text:
\begin{document}

\newcommand{\K}{\ensuremath{\mathbb{K}}}

\bibliographystyle{alpha}
\title{Connected  subgroups of $SO(2,n)$ acting irreducibly on  $\R^{2,n}$}
\author{Antonio J. Di Scala\thanks{{\sc  Dipartimento di Matematica,
Politecnico di Torino,
Corso Duca degli Abruzzi 24, 10129 Torino, Italy}, {\tt
 antonio.discala@polito.it }}\ \  and  Thomas Leistner\thanks{{\sc  Department Mathematik, Universit\"at Hamburg, Bundesstra{\ss}e 55, D-20146 Hamburg, Germany}, {\tt leistner@math.uni-hamburg.de} \newline {\em  Date: June 16, 2008}}}
 \date{ }
\maketitle
\begin{abstract}We classify all connected subgroups of $SO(2,n)$ that act irreducibly on $\rr^{2,n}$. Apart from $SO_0(2,n)$ itself these are $U(1,n/2)$, $SU(1,n/2)$,  if $n$ even, $S^1\cdot SO(1,n/2)$ if $n$ even and $n\ge 2$, and $SO_0(1,2)$  for $n=3$. Our proof is based on the Karpelevich Theorem and uses the classification of totally geodesic submanifolds of complex hyperbolic space and of the
Lie ball. As an application we obtain a list of possible irreducible holonomy groups of Lorentzian conformal structures, namely $SO_0(2,n)$, $SU(1,n)$, and $SO_0(1,2)$. \\[1mm]
{\em MSC:}  22E46;

 53C35;

53C40;

53C29;

53A30.

\\[1mm]
{\em Keywords:} Irreducible, orthogonal representations; Lie ball; complex hyperbolic space; totally geodesic submanifolds of symmetric spaces; conformal holonomy; Berger list.
 \end{abstract}

\section{Background, result, and applications}
One of the results  at the origins of modern differential geometry is Marcel Berger's classification of irreducible connected  holonomy groups of complete semi-Riemannian manifolds. The most striking feature of this {\em Berger list} is that it is rather short. This is more surprising in some signatures and more natural in others. For example,  that the only possible connected irreducible holonomy group of Lorentzian manifolds is $SO_0(1,n)$ is due to the fact that there are no proper connected subgroups of $SO_0(1,n)$ that act irreducibly on $\rr^{1,n}$ \cite{olmos-discala01}. In Riemannian signature, on the other hand, it is more surprising that only so few groups occur as holonomy groups, taking into account that any representation of a compact group, in particular irreducible ones, is orthogonal with respect to a positive definite scalar product. For a recent proof of Berger's theorem for Riemannian manifolds see \cite{Olmos05}.

In this paper we consider the case of signature $(2,n)$ and study connected subgroups of $SO(2,n)$ that act irreducibly on $\rr^{2,n}$. We give a classification  of these groups:
\btheo\label{list}
Let $G\subset SO(2,n)$ be a connected Lie group that acts irreducibly on $\rr^{2,n}$. Then $G$ is conjugated to one of the following,
\bnum
\item for arbitrary $n\ge 1$:  $SO_0(2,n)$,
\item
for $n=2p$ even:  $U(1,p)$, $SU(1,p)$, or $S^1\cdot SO_0(1,p)$  if $ p>1$,
\item for $n=3$: $SO_0(1,2) \subset SO(2,3)$, .
\enum
\etheo
Our interest in signature $(2,n)$ is twofold. One aspect is the more general interest in the Berger list. Our result shows that there is only one group, namely $S^1\cdot SO(1,n)$,  that does not appear in the Berger list, i.e. that is not a holonomy group for a metric of signature $(2,n)$.

 More important is the relation to conformal  Lorentzian structures.   To a Lorentzian conformal structure in dimension $n$, which is defined as an equivalence class Lorentzian metrics differing by a scaling function, one may assume a conformally invariant Cartan connection, the holonomy group of which is contained in $SO(2,n)$. For the so-called {\em conformal holonomy} the algebraic restrictions are much more difficult to handle than the Berger criterion in case of metric holonomy algebras. Hence, it was natural to ask first: What are possible connected subgroups of $SO(2,n)$ that act irreducibly on $\rr^{2,n}$? Our answer to this question gives a list of possible candidates for {\em special conformal Lorentzian structures}, a name which refers to --- in analogy to special Riemannian structures ---  Lorentzian conformal structures with {\em irreducibly} acting conformal holonomy group (for indecomposable, non-irreducible Lorentzian conformal structures we refer the reader to \cite{leistner05a}). Now,  two of the groups Theorem \ref{list} are known to be Lorentzian conformal holonomy groups,  $SO_0(2,n)$ itself and $SU(1,n/2)$, the first being the generic conformal holonomy, the second being that of a Fefferman space (see for example \cite{baum07}).
In \cite{leitner08} it is proven that if a {\em connected} conformal holonomy group is contained $U(1,n/2)$ then it is already contained in $SU(1,n/2)$. Hence,  $S^1\cdot SO_0(1,n/2)$ cannot occur as connected conformal holonomy group of a Lorentzian conformal structure, because it is not contained in $SU(1,n/2)$. We get the following consequence.
\bfolg
Let $G\subset SO(2,n)$ be the connected conformal holonomy group of a Lorentzian conformal structure. If $G$ acts irreducibly on $\rr^{2,n}$, then
\[ G=SO_0(2,n)\text{, or }G=SU(1,n/2)\text{ if $n$ is even, or }G=SO_0(1,2)\text{ if   $n=3$.}\]
\efolg
Unfortunately, we cannot yet exclude the exceptional case of $SO_0(1,2)\subset SO(2,3)$ as a possible conformal holonomy of a $3$-dimensional Lorentzian manifold. We only know that
$SO(1,2) $ does not define a {\em conformal Cartan reduction} in the sense of \cite[Section 3.3]{alt08}.
Such a conformal Cartan reduction of $SO(p+1,q+1)$ to a group $G\subset SO(p+1,q+1)$ exists if and only if $G$ acts transitively on the pseudo-sphere $S^{p,q}= SO(p+1,q+1)/P$, where $P$ is the parabolic subgroup defined as the stabiliser of a light-like line in $\rr^{p+1,q+1}$. Examples of conformal Cartan reductions are given by $SU(p+1,q+1)\subset SO(2p+2,2q+2)$, see \cite{baum07} or \cite{cap-gover06},  the non-compact $G_{2(2)}\subset SO(3,4)$ in \cite{nurowski04,nurowski07}, and $Spin(3,4)\subset SO(4,4)$ in \cite{bryant06}, and they are linked to so-called {\em Fefferman constructions}.
Now, the action of $SO_0(1,2)$ on $S^{1,2}=SO(2,3)/P$  is not  transitive (see Appendix \ref{so12}) and hence does not define a conformal Cartan reduction, but we do not know
if this already excludes $SO_0(1,2)\subset SO(2,3)$ as an irreducible conformal holonomy. To clarify this question lies beyond the scope of this paper and  will be subject to further studies.

\bigskip

Our proof of Theorem \ref{list} is based on the Theorem of Karpelevich and Mostow.
\begin{theo}\label{Karp}(Karpelevich \cite{karpelevich53}, Mostow \cite{mostow55}, also \cite{discala-olmos07})
Let $M =  {Iso}(M)/K$ be a Riemannian symmetric space of
non-compact type. Then any connected and semisimple subgroup  $G$
of the full isometry group ${Iso}(M)$ has a totally geodesic orbit
$G\cdot p \subset M$.
\end{theo}
We will apply this theorem to a connected subgroup $G$ of $SO(2,n)$ that acts irreducibly on $\rr^{2,n}$ and to  the Riemannian symmetric spaces that are related to $SO(2,n)$: the complex hyperbolic space $\C H^{n}=SU(1,n)/U(n)$ and the Grassmannian of   negative definite planes in $\rr^{2,n}$ given as $SO_0(2,n)SO(2)\cdot SO(n)$ and as  $SO(2,n)/SO(2)\cdot SO(n)$ if one considers {\em oriented} negative planes. The latter has two connected components and can be realised in $\C P^{n+1}$ as the  submanifold of  negative definite lines in $\C^{2,n}$. Its connected component is called {\em Lie ball}. In applying Karpelevich's Theorem we have to deal with two difficulties that are related to each other: First, we cannot assume that $G$ is semisimple, and secondly, if $T$ is a totally geodesic orbit with isometry group $H=Iso (T)$ our group $G$ in question can be the product of $H$ with the group $I(T)$ that is defined as
\[I(T):=\{A \in G\mid A|_T=\mathrm{Id}_T\}.\]
$I(T)$ is a normal subgroup in $Iso(T)$.
We know that $G$ is reductive but it may be that its semisimple part does not act irreducibly. On the other hand, it might happen that $T$ is the orbit of a group $H$ that does not act irreducibly but that $I(T)\cdot H$ acts irreducibly.
Overcoming these difficulties,
our proof will consist of  three main steps:
\bnum
\item Show that if $G\subset SO_0(2,n)$ acts irreducibly, then it is {\em simple} or contained in $U(1,n)$.
\item Classify connected subgroups of $U(1,n)$ acting irreducibly on $\rr^{2,n}$ using:
\bnum
\item $G$ is reductive with possible centre $S^1$,
\item By Karpelevich's Theorem
 applied to $\C H^n$, the orbits $T$ of the semisimple part are isometric to either $\C H^k$  or to real hyperbolic spaces $\rr H^k$ for $k\le n$.
\item $I(T)$ can be calculated.
\enum
\item If $G$ is not in $U(1,n)$, $G$  is simple and we apply Karpelevich's Theorem to the Lie ball
$SO_0(2,n)/SO(2)\cdot SO(n)$. Then we use
the classification of totally geodesic orbits in the complex quadric $SO(n+2)/SO(2)\cdot SO(n)$ by \cite{klein08}, transfer it by duality to the Lie ball  and obtain $G$ as isometry group of these orbits. As $G$ is simple, $I(T)$ can be ignored.
 \enum

\section{Algebraic preliminaries}
\subsection{Irreducible representations of real Lie algebras}
Most of the groups appearing in the theorem are well known. However, regarding $S^1\cdot  SO_0(1,p)$  we will make some remarks about irreducible representations of real Lie algebras and about symmetric space which will be useful in what follows. The transition to Lie algebras is justified by the restriction to connected subgroups of $SO(2,n)$.

Let $\lag$ be a real Lie algebra and $E$ an irreducible {\em real} representation. We say that $E$ is of {\em real type} if $E^\C:=E\otimes\C$ is irreducible as well.  Otherwise we say that $E$ is {\em not of real type}.  In the latter case there is a splitting of $E^\C$ as
$E^\C=V\+ \ol{V}$ where $V$ is an irreducible complex representation and $\ol{V}$ the conjugate representation w.r.t. the real form $E\subset E^\C$.
By the conjugate representation we mean the representation of $\lag$ on $\ol{V}$ defined by
\[ A\cdot \ol{v}:= \ol{A\cdot v}.\]
In deed, if $V$ is a complex invariant subspace  of $E^\C$, the complex
subspaces  $V +\ol{V}$ and $V\cap\ol{V}$ are invariant as well. On the other hand, they are equal to their complex conjugate, and thus, complexifications of real invariant subspaces. As $E$ is irreducible, we obtain that $V\+ \ol{V}=E^\C$.

On the other hand, it is $\left(V_\rr\right)^\C=V\+\ol{V}$ and multiplication with $\mathrm{i}$ defines an invariant isomorphism $J$ on $V_\rr$ and by complexification on $(V_\rr)^\C=E^\C$ that squares to $-$Id. Hence, $E^\C$ splits into the invariant eigen spaces w.r.t. $\mathrm{i} $ can and  $-\i$, which are given by $V$ and $\ol{V}$, respectively. Then
\[x\ni E\ \mapsto\  x-i Jx \ \mapsto \ x+iJ \in \ol{V}\]
gives an isomorphism of real representations
$E\simeq V_\rr\simeq \ol{V}_\rr$.
 Complex multiplication with the imagninary unit on $V$ induces an invariant complex structure $J$ on $E$.

In the other case, where  $E$ is of real type,
$W:=E^\C$ considered as a real vector space,  denoted by $W_\rr$, is reducible,  with invariant real form $E$. This is equivalent to $W$ being {\em self-conjugate} with a conjugation that squares to the identity. Recall that a complex representation $V$ of a real Lie algebra is self conjugate if $W\simeq \ol{W}$ as a $\lag$-representation, i.e. there exists an invariant isomorphism between $W$ and $\ol{W}$. In case of a representation of real type, the invariant real form $E$ then is given as the $+1$ eigen-space of this conjugation.

After this change of  the viewpoint,  it is natural to say  that a {\em complex} irreducible representation is of {\em real type} if it is self-conjugate with a conjugation squaring to one. Otherwise it is called of {\em non-real type}.
Examples of representations of real type are complexifications of the standard representations of $\laso (p,q )$ on $\rr^{p,q}$. Examples of representations of non-real type are representations of $\mf{u}(p,q)$ and $\mf{su}(p,q)$ on $\C^{p,q}$ and $\rr^{2p,2q}$ respectively.

For a complex irreducible representation $V$ of $\lag$ there is a further distinction beyond being of real type or not. If $V$ is not self-conjugate, then $V$ is called {\em of complex type}. If $V$ is self-conjugate with respect to a conjugation $C$, then $C^2$ is a $\C$-linear invariant automorphism of $V$. By the Schur lemma, it is a multiple of the identity, say $C=\lambda \cdot$Id, with $\lambda\in \rr$ because
\[\lambda C (v)= C(\lambda v)=\ol{\lambda }C(v).\]
By scaling $C$ we can assume that $\lambda^2=\pm 1$. In one case, $V$ was of real type, in the case where $C^2=-$Id one says that $V$ is {\em of quaternionic type}, because $C$ defines another complex structure which anti-commutes with the multiplication with i.  To summarise these standard facts, a complex irreducible representation is either of real, complex or quaternionic type. If it not of real type, then $V_\rr$ is irreducible, if it is of real type, it is the complexification of an irreducible real representation.

The following  lemma is a standard result. We cite without proof.
 \blem
$\lag\subset \laso(p,q) $ is of  real type if and only if $p$ and $q$ are even and $\lag\subset \mf{u}(p/2,q/2)$.
\elem
We suppose that the next lemma is also a standard fact from representation theory. Nevertheless, for the sake of being self-contained we prove it here.

\blem\label{oslemma}
Let $\lag $ be a real Lie algebra and $V$ a complex irreducible representation of quaternionic type and of complex dimension $2m$.
\bnum
\item If $V$ is symplectic, then $\lag\subset \mf{sp} (p,q)\subset \mf{u}(2p,2q) $ with $p+q=m$.
\item If $V$ is orthogonal, then $\lag\subset \mf{so}^* (2m)\subset \mf{u}(m,m)$.
\enum
\elem
\bprf
Let $J$ be the anti-linear invariant automorphism of $V$ with $J^2=-1$, and let $V$ be of complex dimension $2m$
Assume that $\w$ is an invariant symplectic form on $V$.
First we show that we can assume the following relation between $\w$ and $J$:
\begin{equation}\w(Jx,Jy)=\ol{\w(x,y)}\label{compatw}\end{equation}
In fact, $\hat{\w}:=\ol{\w(J.,J.)}$ gives another invariant symplectic form on $V$. By the Schur lemma, they are a complex multiple of each other, i.e. $\ol{\w(J.,J.)}=\lambda \w$ for a $\lambda\in \C^*$.  This implies that
\[\w(J.,J.)=\ol{\lambda}\ \ol{\w(J^2.,J^2.)}=\ol{\lambda}\lambda \w(J.,J.)\]
i.e. that $\lambda=\e^{i\theta}\in S^1$. Rescaling $\w$ by $\e^{-i\frac{\theta}{2}}$ enables us to assume equation (\ref{compatw}).
Now note that equation (\ref{compatw}) implies that $\w(J.,.)=-\ol{\w (.,J.)}$. This enables us to define an invariant hermitian form $\la.,.\ra$ on $V$ via
$\la x,y\ra:= \w(x,Jy)$. This is in deed hermitian,
\[\la y,x\ra= \w(y,Jx)=-\ol{\w(Jy,x)}=\ol{\w(x,Jy)}=\ol{\la x,y\ra}\]
and compatible with $J$,
\[\la Jx,Jy\ra = -\w(Jx,y)=\w(y,Jx)=-\ol{\w(Jy,x)}=\ol{\w(x,Jy)}=\ol{\la x,y\ra }.\]
This shows that
 $\lag\subset \mf{u}(2p,2q)\cap \mf{sp}(m,\C)=\mf{sp}(p,q)$, with $p+q=m$.

Now assume that $\s$ is an invariant symmetric bilinear form on $V$. By the same argument as in the symplectic case we get that
\[ \s(Jx,Jy)=\ol{\s(x,y)}\  \text{ and }\  \s(Jx,y)=-\ol{\s(x,Jy)}.\]
A hermitian form is now defined by $\la x,y\ra := \mathrm{i} \s( x,Jy)$.  The compatibility with $J$ is given by
\[\la Jx,Jy\ra = -\mathrm{i}\s(Jx,y) = \mathrm{i}\ol{\s(x,Jy)} = -\ol{\la x,y\ra},\]
which shows that $\la .,.\ra $ has neutral signature $(m,m)$ and that an orthonormal basis of $\s$ is a light-like basis of $\la .,.\ra$.
A calculation in a basis then shows that
 $\lag\subset \mf{u}(V,\la.,.\ra )\cap  \laso (2m,\C )=\laso^* (2m)$, which is defined as follows
 \[\laso^* (2m):=\left\{\left( \begin{array}{cc} A& B\\-\ol{B}^t & \ol{A}^t\end{array}\right)\mid A\in \laso(m,\C), B=\ol{B}^t\right\}\]
(see \cite[p. 446]{helgason78}).
\eprf

\subsection{Irreducibility of  $S^1\cdot SO_0(1,n)$}
In this section we explain that $S^1\cdot SO_0(1,n)\subset U(1,n)\subset  SO(2,2n)$ acts irreducibly.

\bs
 Let $\lag\subset \laso (p,q)$ act irreducibly on $\rr^{p,q}$. This representation is of real type if and only if $\wt{\lag}:=\i\cdot \rr\+ \lag$ acts irreducibly on $\C^{p+q}$ as real representation.
 \es
 \bprf
 If the representation of $\lag$ on $\rr^{p,q}$ is of real type then its complexification is still irreducible, and so is the representation of $\wt{\lag}$ on $\C^{p+q}$. But by definition, there is no conjugation that is invariant under $\wt{\lag}$, and thus
the representation of $\wt{\lag}$ on $\C^{p+q}$ is of complex type, which means that it is still irreducible as a real representation.

On the other hand assume that   $\C^{p+q}$ is irreducible as real and therefore as a complex  representation of $\wt{\lag}$. Assume furthermore that the representation of $\lag$ on $\rr^{p,q}$ is not of real type. By the remarks in the previous section, this is equivalent to the existence of a $\lag$-invariant complex structure $J$ on $\rr^{p,q}$ and to the existence of a complex $\lag$-invariant subspace $V\subset \C^{p+q}$. Extending $J$ complex linear gives an invariant complex structure on $\C^{p+q}$. Note that $J\not= \i\cdot$Id, because otherwise $\wt{\lag}$ could no longer act irreducibly on $\C^{p,q}$. Hence, $I:= \i\cdot J $ is $\wt{\lag}$-invariant, satisfies $I^2=$ Id and is not a multiple of the identity. Hence, it has non-trivial invariant eigen spaces to the eigen values $\pm 1$. But this again contradicts to the irreducibility of $\C^{p,q}$ under $\wt{\lag}$. Therefore, $\rr^{p,q}$ must be of real type for $\lag$.
 \eprf
This gives the following conclusion in the case $p=1$.
 \bfolg
For $n>1$,  $S^1\cdot SO_0(1,n)$ is an irreducible subgroup of $U(1,n)\subset SO_0(2,n)$, is not contained in $SU(1,n)$,  and   has no further irreducible subgroups.
\efolg
\bprf From the previous section we know that irreducible representations of non-real type are unitary, but this is not possible for $\lag \subset \laso(1,n)$. In fact, there is no proper irreducibly acting subalgebra of $\laso(1,n)$, see \cite{olmos-discala01}.  But $\laso(1,n)$ is of real type, and the result follows from the proposition.

For the minimality assume that $\lag\subset\i\rr\+\laso(1,n)$ acts irreducibly. But then the projection of $\lag$ onto $\laso(1,n)$ acts irreducibly and thus has to be equal to $\laso (1,n)$. But this implies $\laso(1,n)=[\lag,\lag]\subset \lag\subset \i\rr\cdot \laso(1,n)$. Hence, $\lag=\i\rr\cdot \laso(1,n)$. That $S^1\times SO(1,n)$ is not contained in $SU(1,n)$ is obvious.
\eprf

\subsection{Reduction to simple Lie algebras}
In this section we show that an irreducible subalgebra of $\laso (2,n)$ is either contained in $\mf{u} (1,n/2)$ or simple.

\bigskip

First, we have to recall some more general facts about representations of real Lie algebras.
Let $\lag$ be a real Lie algebra and $V$ an irreducible {\em complex} representation. We have seen that $V$ is of {\em real type} if and only if there is a $\lag$-invariant existence of a  invariant conjugation squaring to one.
Furthermore, on says that a complex irreducible representation $V$ that is not self-conjugate is of  {\em of complex type}, and if $V$ is self-conjugate with a conjugation squaring to $-1$ it is called {\em of quaternionic type}.
Based on this distinction and on the description of the center in \cite{att05} we proved the following:
 \bs\label{folge3} Let $G\subset SO_0(p,q)$ a
connected Lie subgroup of $SO_0(p,q)$ which acts irreducibly. If $G$
is not semisimple, then $p$ and $q$ are even and $G$ is a subgroup
of $ U(p/2, q/2)$ with centre $U(1)$.
In particular, if $G\subset SO(2,n)$, then $G\subset U(1,n/2)$ or semi-simple.

\es

Here we will strengthen this result for the case $G\subset SO(2,n)$ by replacing ``semi-simple'' by ``simple''. This will be based on the following general fact on complex irreducible representation of semi-simple complex Lie algebras (for a reference, see for example \cite[p. 11]{onishchik04}):
\\[3mm]
{\bf Fact.}
If $\lag=\lag_1\+\lag_2$ is a semi-simple Lie algebra decomposing into non-trivial ideals $\lag_1$ and $\lag_2$.  Then $V$ is a complex irreducible representation of $\lag$ if and only if $V=V_1\otimes V_2$ where $V_i$ are irreducible representations of $\lag_i$.
\\
\blem
Let $\lag\+\lah$ be semi-simple and $W=U\otimes V$ an irreducible complex representation. Then $W$ is self-dual  if and only if both, $U$ and $V$ are  self-dual. The invariant isomorphisms are related by
 $\psi=\psi_1\otimes \psi_2$.
\elem
\bprf
The `if'-direction is obvious, $\psi=\psi_1\otimes \psi_2$ defines the required invariant isomorphism.

For the other direction we consider the identification $\tau: U\simeq U\otimes v_0$ for a fixed $v_0\in V$. $\tau $ is not only an  isomorphism of vector spaces but also of representations of $\lag$, i.e.
\[A (\tau (u))= A (u\otimes v_0)= Au\otimes v_0= \tau (Au).\]
Let $\psi: W\simeq W^*$ be the the isomorphism yielding the self-duality of $W$.
This implies that there are $u_0,\hat{u}_0\in U$ and $v_1\in V$ such that
\[ \left[\psi ( u_0\otimes v_0)\right] (\hat{u}_0\otimes v_1)\not=0.\]
Otherwise, $u_0\otimes v_0  $ would be in the kernel of $\psi$. Hence
by defining
\[\left[ \psi_1 (u)\right](\hat{u}):= \left[\psi ( u\otimes v_0)\right] (\hat{u}\otimes v_1)\]
we obtain a $\lag$-invariant homomorphism $\psi_1:U\simeq U^*$ which is non trivial. By the Schur lemma, $\psi_1$ is an isomorphism.
Obviously, for $V$ one can proceed in the same way.
The Schur-lemma also gives the uniqueness of the invariant structures and the relation between them.
\eprf
\blem
Let $\lag\+\lah$ be semi-simple and $W=U\otimes V$ an irreducible complex self-dual representation. Then $W$ is self-conjugate if and only if both, $U$ and $V$ are  self-conjugate. The invariant isomorphisms are related by
 $\psi=\psi_1\otimes \psi_2$.
\elem
\bprf As $W$ is self-dual,  both $U$ and $V$ are self dual.
Hence, $\ol{U}\simeq \ol{U}^*$ and $\ol{V}\simeq \ol{V}^*$. If $\psi:W\simeq W^*$ and $C:W\simeq \ol{W}$, analogously as in the proof of the previous lemma, one defines  $\phi_1: U\rightarrow \ol{U}^*$ via
\[\left[ \phi_1 (u)\right](\hat{u}):= \left[\psi ( u\otimes v_0)\right] (C(\hat{u}\otimes v_1)).\]
Again, by the Schur lemma, this is an isomorphism, yielding an isomorphism $\psi_1:U\simeq \ol{U}$. All invariant structures are uniquely defined.
\eprf

\btheo
Let $\lag\subset \laso(2,n)$ be an irreducibly acting Lie algebra. Then $\lag\subset \mf{u}(1,n/2)$ or $\lag$ is simple.
\etheo
\bprf
By Proposition \ref{folge3} we can suppose that $\lag$ is semisimple and that the representation of $\lag$ on $\rr^{2,n}$ is of real type.
Assume that
$\lag=\lag_1\+\lag_2$ is not simple. Then its complexification is semisimple and not simple, and thus, the complexified representation $\C^{n+2}$ of  $\rr^{2,n}$ is a tensor product, $\C^{n+2} =V_1\otimes V_2$ of irreducible representations of $\lag_1$ and $\lag_2$. As $\C^{n+2}$ is of real type, the second lemma implies that $V_1$ and $V_2$ are either both of real type or both of quaternionic type. Since $\lag\subset \laso (n+2,\C)$, by the first lemma both are self-dual, defined by either two complex linear symmetric or symplectic forms.

Assume first that both, $V_1$ and $V_2$ are of real type, i.e. $V_i=E_i^\C$ where $E_i$ are irreducible real  representations of $\lag_i$. If $\lag_i\subset \laso (V_i)$, also both $E_i$ are orthogonal, i.e. $\lag_1\subset \laso (p,q)$ and $\lag_2\subset \laso(r,s)$ with
$2=ps+qr$. W.l.o.g. this yields two cases:
The first is $\lag_1=\laso (2)$ and $\lag_2\subset \laso (1,n/2)$ acting on $\rr^2\otimes \rr^{1,\frac{n}{2}}$, or  $\lag_1=\lag_2=\laso(1,1)$. But both cases  contradict to the assumption that $\lag$ was semisimple.

Now we consider the case where the $V_i$'s and thus both $E_i$'s are symplectic representations. In this case the defining scalar product on $\rr^{2,n}$ has neutral signature, i.e. $\lag\subset \laso (2,2)$, and $\lag_i\subset \mf{sp}(1,\rr)=\lasl (2,\rr)$ acting irreducible. Hence, $\lag_i$ either one-dimensional and therefore Abelian, two-dimensional, and thus solvable, or  equal to $\lasl(2,\rr)$. The first two possibilities are excluded by the semisimplicity assumption. We  obtaining that
 $\lag$ is equal to $\lasl(2,\rr)\+\lasl(2,\rr)=\laso (2,2)$.

Now we have to deal with the case where both representations, $V_1$ and $V_2$ are of quaternionic type. As $\lag\subset \laso (n+2,\C)$, they are  either both orthogonal or both symplectic.

Using Lemma \ref{oslemma} we can conclude the proof of the theorem:
First consider the case that $\lag=\lag_1\+\lag_2$ with $\lag_i\subset \mf{sp}(p_i,q_i)\subset \mf{u}(2p_i,2q_i)$. The tensor product of the  hermitian forms on $V_i$ defines a hermitian form of signature $(4(p_1q_2+p_2q_1), 4(p_1p_2+q_1q_2))$ on $V=E^\C$. Since $V$ is an irreducible representation of $\lag$, the space of hermitian forms on $V$ is one-dimensional. Hence, the defined hermitian form is a multiple of the hermitian form obtained by extending the signature $(2,n)$ scalar product on $E$ to $V$. But $2\not= 4(p_1q2+p_2q_1)$ which excludes this case.

For the case $\lag_i\subset \laso^*(2m_i)\subset \mf{u}(m_i,m_i)$ we obtain that $\lag\subset \mf{u}(2m_1m_2, 2m_1m_2)$, which implies $m_i=1$, $V_i=\C^2$ and $\lag_i=\laso^* (2)=\laso(2)$ and $\lag $ is no longer semisimple.
\eprf

 \subsection{Duality for symmetric spaces and consequences}
\label{duality}
For a Riemannian symmetric space $G/K$ given by the Cartan decomposition $\lag=\lak\+\lam$, i.e. by the Lie triple system $\lam$ with the Lie bracket of $\lag$ there is a corresondence between totally geodesic submanifolds and Lie subtriples,
\begin{equation}
\label{Lst}
\{\text{totally geodesic submanifolds $G'/K'$ of $G/K$}\}\simeq
\{\text{Lie subtriples $\lam'\subset \lam$}\}\end{equation}
On the other hand there is the dualtity between compact and non compact symmetric spaces. If $\lag=\lak\+\lam$ is a Cartan decomposition defining  a Riemannian symmetric space, then the one defines the Lie algebra $\lag^*:=\lak\+\lam^*$ by setting
\[\lam^*:= \lam\text{ with Lie bracket }[X,Y]^*:=-[X,Y]\]
and all the other commutators are the same as in $\lag$. In other words, $\lag^*=\lak \+ \i\cdot \lam$ with complex linearly extended commutator.  Then $\lag^*$ defines the dual symmetric space. We get the following correspondence
\be
\{\text{totally geodesic submanifolds  of $G^*/K$}\}&\simeq &\{\text{totally geodesic submanifolds  of $G/K$}\}\\[-2mm]
\turnsimeq\hspace{1cm} &&\hspace{1cm} \turnsimeq\\
\{\text{Lie subtriples in  $\lam^*$}\}&\simeq & \{\text{Lie subtriples in  $\lam$}\}.
\ee
Hence, a totally geodesic submanifold in a compact Riemannian symmetric space $G/K$ is a compact Riemannian symmetric space $H/L$, and the corresponding totally geodescic submanifold in the non-compact dual $G^*/K$ is given by $H^*/L$. For more details we refer the reader to \cite[Chapter 9]{BCO03}.

This correspondence will enable us to describe totally geodesic submanifolds in the Lie Ball $SO_0(2,n)/ SO(2)\times SO(n)$ with the help of totally geodesic submanifolds in the complex quadric $Q^n= SO(n+2)/SO(2)\times SO(n)$.

We can now apply Karpelevich's Theorem \ref{Karp} to what we have obtained so far.
\btheo \label{conseq}
Let $G\subset SO_0(2,n)$ be a connected irreducibly acting subgroup. Then $G\subset U(1,n)$ or $G$ is simple and equal to the effectively acting isometry group of a totally geodesic submanifold in the non-compact symmetric space $SO_0(2,n)/SO(2)\times SO(n)$.

\etheo
\bprf
Let $G\subset SO_0(2,n)$ but $G\not\subset U(1,n)$. From the previous section we know that $G$ is simple. By Karpelvich's Theorem \ref{Karp} it follows that $G$ has a totally geodesic orbit $\cal T$  in the non-compact symmetric space $\cal L^n:= SO_0(2,n)/SO(2)\times SO(n)$. The subgroup
\[ I(\cal T):=\{ A\in G\mid Ap=p\text{ for all } p\in \cal T\}\]
is a normal subgroup in $G$. As $G$ is simple, $I(T)$ is trivial and $G$ acts effectively on $\cal T$. Hence, $\cal T= G/K\subset \cal L^n$ is a non-compact symmetric space with $K\subset G$ maximally compact.
\eprf
In the next section we will determine all irreducibly acting groups $G\subset U(1,n)$ by applying Karpelevich's theorem to the complex projective space. In the last section we will then use a classification of totally geodesic submanifolds in $Q^n= SO(n+2)/SO(2)\times SO(n)$ by \cite{klein08} and the just explained duality to determine the remaining $G$'s.

\section{Irreducible subgroups of $U(1,n)$ and complex hyperbolic space}

Using Karpelevich's Theorem in this section we will proof the following statement.

\btheo \label{u1n} Let $G\subset U(1,n)\subset SO(2,2n)$ be a connected subgroup that acts irreducibly on $\rr^{2,2n}$. Then $SU(1,n)\subset G$ or $G=S^1\cdot SO_0(1,n)$.
\etheo
To this end we consider the complex vector space
$\C^{n+1} =: \C^{n,1}$  endowed with the Hermitian form $Q$:

\[ Q =  -|z_0|^2 + |z_1|^2 + |z_2|^2 + \cdots + |z_n|^2 \, . \]

Let us denote by $U(1,n) \subset GL(n+1, \C)$ the subgroup that preserves $Q$.\\

Let $\mathcal{N} := \{ p \in \mathbb{C}^{n,1} : Q(p) < 0 \}$ be the set of negative points. Notice
that $\mathcal{N}$ is a cone preserved by the $U(1,n)$-action. Let us call $\mathbb{C}H^n$ the projectivization of $\mathcal{N}$. Thus, by taking $z_{0} = 1$ we can see that $\mathbb{C}H^n$ is identified with the unit ball of $\mathbb{C}^n$. Namely, \[ \mathbb{C}H^n \cong \{ Z \in \C^n \, : \, |Z|^2 < 1 \} \, \, . \]

It is standard to see that the Hermitian form $Q$ induces on $\ccc H^n$ a $U(1,n)$-invariant Riemannian metric of constant holomorphic curvature. Indeed, we get $\ccc H^n \cong SU(1,n)/U(n)$ as symmetric space of rank one. Notice that the $U(1,n)$-action on $\C H^n$ is not effective since the matrices $e^{i \theta} Id \in U(1,n)$ leaves invariant any complex line. Recall also that the presentation $\ccc H^n \cong SU(1,n)/U(n)$ as symmetric quotient is unique. Namely, if  $\ccc H^n \cong G/K$ where $G$ is semisimple (connected and simply connected) and $K \subset G$ maximal compact then $G = SU(1,n)$ and $K = U(n)$.

The following fact about totally geodesic submanifolds of $\ccc H^n$ can be found in  \cite[pp. 74]{goldman99}, for example.

\bs \label{totalgeo} Let $\mathcal{T} \subset \ccc H^n$ be a complete totally geodesic submanifold. Then $\cal T$ is either a totally real submanifold or a complex submanifold. In the totally real case $\cal T$ is isometric to real hyperbolic space, otherwise $\cal T$ is biholomorphic and isometric to a lower dimensional complex hyperbolic space. In particular,
there exists a real vector subspace $V \subset \ccc^{n,1}$ such that $ \mathcal{T} = V \bigcap \ccc H^n $.
\es

Now we are ready to deduce Theorem \ref{u1n} from Karpelevich's Theorem.

\bprf[Proof of Theorem \ref{u1n}]
Let $H\subset U(1,n)$ be connected and acting irreducibly on $\rr^{2,2n}$ then
 $H$ is reductive, i.e.
 $H=Z\cdot S$ where $Z$ is the centre and $S$ semisimple.

 According with Proposition \ref{folge3} we know that the centre
 $Z$ is trivial or equal to $S^1$.

Hence, the semisimple part $S$ cannot be trivial.

Now,  according to Karpelevich's Theorem $S$ has a totally geodesic orbit $T$ of $\C H^n$. If $T$ is a complex submanifold then Proposition \ref{totalgeo} implies that $S$ must be transitive on $\C H^n$ since otherwise the complex subspace $V$ associated to $T$ is invariant by $S$ and  $Z=S^1$. Thus $H$ can not be irreducible. So $S$ is transitive and we get by the uniqueness of the representation of the symmetric quotient that $SU(1,n) = S$.

Assume now that $T$ is not a complex submanifold. Then, the classification of totally geodesic submanifolds of $\ccc H^n$ imply $T \cong  \R H^n $. Otherwise $T$ is contained in a proper complex totally geodesic submanifold of $\C H^n$ and this imply that $H$ is not irreducible as above. Thus, $T$ is a totally real totally geodesic submanifold.
Without lost of generality we can assume that $T = \R H^n $ where $\R H^n = \{ Z \in \R^n \subset \C^n \, : \, |Z|^2 < 1 \} $. Notice that the Lie algebra of the group $I(T)$ is trivial.
Indeed, if $u \in Lie(I(T))$ then the tangent space to $T$ at $0 \in \R^n \subset \C^n$ is contained in the kernel of $u$. Since $T$ is totally real and $u \in \mf{u}(1,n)$ we get that also the normal space of $T$ at $0$ is contained in the kernel of $u$. Thus $u$ vanish. Since $I(T)$ is trivial we get that $S= SO_0(1,n) \subset SU(1,n)$. Now the center must be $S^1$ and so $H = S^1 \cdot SO_0(1,n) $.
\eprf

\section{The Lie ball and its totally geodesic submanifolds}

 \subsection{The projective model of the Lie ball.}

Let $\R^{2,n}$ the real vector space $\R^{n+2}$ endowed with the
quadratic form
\[q(X,Y) := \langle X, Y \rangle := -x_{0}y_{0} - x_{1}y_{1}+ \sum_{j=2}^{n+1}
x_jy_j, \] where $X = (x_0, \cdots,
x_{n+1})$ and $Y = (y_0, \cdots, y_{n+1})$.
Let $\Pi \subset \R^{2,n}$ be a $2$-dimensional subspace. The
$2$-plane $\Pi$ is called {\it negative} \rm if $q|_{\pi}$ is negative definite.
Let us define the Lie ball ${\cal L}^n$ as one connected component of the set of oriented negative
definite $2$-planes of $\R^{2,n}$. For more details about this model see \cite[p. 285, \S 6]{satake80} or \cite[p. 347]{wolf72}.
Note that $SO(2,n)$ acts transitively on the
oriented negative
definite $2$-planes, and that $SO_0(2,n)$ acts transitively on
${\cal L}^n$.

Let $\C^{2,n}$ be the complexification of the $\R^{2,n}$, i.e. $q$
becomes \[ q(Z,W) = -z_{0}\overline {w_{0}} - z_{1}\overline{w_{1}} + \sum_{j=2}^{n+1} z_j\overline{w_j}
 ,\] where $Z = (z_0, \cdots, z_{n+1})$ and $W = (w_0, \cdots,w_{n+1})$.
Let $\Pi = span_{\R}\{ A,B\} \subset \R^{2,n}$, $A,B \in
\R^{2,n}$, be an oriented negative definite $2$-plane. We can assume that
$\langle A, B \rangle = 0$ and $q(A,A) = q(B,B) < 0$.
Put $Z = A + iB \, \in \C^{2,n}$ . Then it is not difficult to see
that
\[Z \in Q^{2,n} := \{ Z=(z_0, \cdots , z_{n+1}) \in \C^{2,n} \mid
- z_{0}^2 - z_{1}^2 + \sum_{j=2}^{n+1}z_j^2  = 0 \}\] and that $q(Z,Z) < 0$.
Call $Q^{2,n}_+$ the subset of $Q^{2,n}$ of negative points, i.e.
\[Q_+^{2,n}=\{ Z=(z_0, \cdots , z_{n+1}) \in \C^{2,n} \mid
 - z_{0}^2 - z_{1}^2+\sum_{j=2}^{n+1}z_j^2 = 0\text{ and }q(Z,Z) < 0\}.\]
It follows that we can identify the Lie ball ${\cal
 L}^n$ with a subset of the projective space $\C P^{n,1}$, namely, with a connected component of the image

of the canonical projection
 $\pi : \C^{n+2} \setminus 0 \rightarrow \C P^{n,1}$.
Thus, we have homogeneous coordinates\footnote{In the appendix an explicit bi-holomorphism with the classical Cartan's domain of type $IV$ is given.} $[z_0 : z_1 : \cdots : z_{n+1} ]$ to work with the Lie ball ${\cal L}^n$.

Let $\Pi_0 = span_{\R} \{ e_{0}, e_{1} \}$ be the ``canonical'' negative definite $2$-plane.
 From now on we will assume that the Lie ball $\cal L^n$
is the connected component of $\Pi_0$.
Then $\Pi_0$ corresponds to the point $Z_0 = e_0 + \i e_{1} = (1,\i, 0,\ldots,0)$.
Thus $\Pi_0 \cong [1: \i: 0:\ldots:0] \in \pi (Q_+^{2,n}) \cong {\cal L}^n $. The isotropy group at $\Pi_0$ is $  SO(2) \times SO(n) $.\\

\subsection{Lifting submanifolds, full submanifolds, and irreducible actions}

Let $M \subset {\cal L}^n$ be a subset. We will denote $L(M) \subset \R^{2,n}$ the subset defined as follows:
\[ L(M) := \bigcup_{ \Pi \in M} \{ p \in \Pi \} \, \, \]
We call $L(M)$ the {\it lift} \rm of $M$.
Let $M \subset \R^n$ be a submanifold. The $M$ is said to be {\it full} \rm if $M$ is not contained in a proper affine subspace of $\R^n$. A submanifold $M \subset {\cal L}^n$ of the Lie ball is called {\it full} \rm if its lift $L(M)$ is full in $\R^{2,n}$.
The following is a well-known property.
\bs Let $G \subset GL(\R,n)$ be a connected Lie subgroup. Assume that $G$ acts irreducibly on $\R^n$.
Then any $G$-orbit $G.p$, $p \neq 0$, is a full submanifold of $\R^n$.
\es

\bprf Let $S_p = span_{\R} \{ G.p \} \subset \R^{n}$ be the linear span of a $G$-orbit $G.p$, $p \neq 0$.

If $S_p$ is not full from some $p$ then $S_p$ lies inside of hyperplane $\langle \cdot, v \rangle = const.$, for some $v \neq 0$. By taking derivatives we get that the Lie algebra $\frak{g}:=Lie(G)$ leaves the subspace $\W = \{ x : \langle x , v \rangle = 0 \}$ invariant. Since $G$ is connected we get that $\W$ is $G$-invariant. \eprf

The following application is also interesting.

\bs \label{full} Let $ G \subset SO(2,n)$ be a connected Lie subgroup and let $\Pi \in {\cal L}^n$ be a point.
Let $G. \Pi$ be the orbit of $G$ through $\Pi$ in the Lie ball.
If the lift $L(G.\Pi)$ is not full then $G$ do not acts irreducibly on $\R^{2,n}$.
\es

\bprf Just notice that the $G$-orbit of any point $p \in \Pi$ is contained in the same proper affine subspace that the lift $L(G.\Pi)$. Then apply the above proposition. \eprf

In the following we will classify full totally geodesic submanifolds in the Lie ball.
This is (almost) equivalent to the classification of {\it maximal} \rm totally geodesic submanifolds in the classical sense (see Sebastian Klein's table at page 11 of \cite{klein08}). We will then check whether the corresponding isometry groups are in our list of irreducible subgroups of $SO_0(2,n)$, respectively, weather or not they are simple (which is the remaining possibility after the previous sections. But first we have to recall the classification of totally geodesic submanifolds in the compact situation, i.e. for the complex quadric $Q^n$.

 \subsection{The complex quadric and its totally geodesic submanifolds}

 The complex quadric $Q^n=SO(n+2)/SO(2)\times SO(n)$ can be viewed in to ways. First, as the Grassmannian of Z(oriented $2$-planes in $\rr^{n+2}$. Secondly, taking into account its complex nature, one can view it as a complex hypersurface in complex projective space, namely as
\[Q^n:=\left\{ [z_0:\ldots : z_{n+1}]\in \C P^{n+1} \mid \sum_{k=0}^{n+1}(z_k)^2=0\right\}.\]
 The subgroup of $SU(n+2)$ acting on $\C^{n+2}$ and thus on $\C P^{n+1}$ that leaves invariant $Q^n$ is $SO(n+2)$ with isotropy group $SO(2)\times SO(n)$. The correspondence to the Grassmannian is given by
\[ P=\span (x,y)\ \mapsto\  \pi (x+iy) \in \C P^{n+1}\]
where $\pi: \C^{n+2}\rightarrow \C P^{n+1}$ is the canonical projection.

Now we will list the totally geodesic submanifolds  in $Q^n$ and their isometry groups as classified in \cite{chen-nagano77} and \cite[Theorem 4.1 and Section 5]{klein08}. Apart from geodesics, there are the following types:

\begin{description}
\item[(I1,\boldmath $k$)] for $1\le k\le n/2 $: This orbit is defined by the following totally geodesic isometric embedding
\[ \C P^k \ni [z_0:\ldots :z_{k}]\ \mapsto\   [z_0:\ldots :z_{k}: \i  z_0:\ldots : \i z_{k}:0:\ldots :0]\in Q^n.\]
Its image is a {\em maximal} totally geodesic submanifolds if $2k=n$ and $n\ge 4$. Its isometry group is $SU(k+1)$ and the totally geodesic submanifold is isometric to $SU(k+1)/U(k)$.
\item[(I2,\boldmath $k$)] for $1\le k\le n/2 $: Here the embedding is give by the restriction of the map for type (I1,k) to real projective space $\rr P^k$ in $\C P^k$. Hence, it is never maximal. Nevertheless, it will be interesting  for our purposes. It is isometric to $O(k+1)/O(k)$.

\item[(G1,\boldmath $k$)] for $1\le k\le n-1$:
This is the embedding of a lower dimensional quadric $Q^k$ into $Q^n$,
\[ Q^k \ni [z_0:\ldots :z_{k+1}]\ \mapsto\   [z_0:\ldots :z_{k+1}:0:\ldots :0]\in Q^n.\]
It is maximal for $k=n-1\ge 2$. Its isometry group is $SO(k+2)$ and it is isometric to $SO(k+2)/SO(2)\times SO(k)$.
\item[(G2,\boldmath $k_1$,$k_2$)] for $ 1\le k_1+k_2 \le n$: This is a totally geodesic isometric embedding of a product of two spheres with radius $1/\sqrt{2}$ and of dimension $k_1$ and $k_2$ given by
\[
\left((x_0,\ldots , x_{k_1}) ,  (y_0,\ldots , y_{k_2})\right) \mapsto
 \left[ x_0:\ldots :x_{k_1}: \i  y_0:\ldots : \i y_{k_2}:0:\ldots :0\right] \in Q^n
 \]
 This orbit is maximal for $k_1+k_2=n\ge 3$. Its isometry group is given by
 $SO(k_1+1)\times SO(k_2+1)$.
 \item[(G3)] The quadric $Q^2$ is isometric to $\C P^1\times \C P^1$ i.e., $\C P^1\times \C P^1 \equiv Q^2  $ .
 Let $C=\mathbb{R}P^1 \subset \C P^1$ be the trace of a closed geodesic in $\C P^1$.
 Then the map

 \[ \C P^1\times C \rightarrow  \C P^1\times \C P^1 \equiv Q^2 \rightarrow Q^n \, \, \]

 where the last embedding represents the embedding of type (G1,2) described above. So the embedding $\C P^1\times C \hookrightarrow Q^m $ is maximal only for $n=2$.

  \item[(P1,\boldmath $k$)] for $1\le k\le n$. This is given as the embedding of type (G2, $k_1$, $k_2$) for $k_1$ or $k_2$ equal to zero. Its image is maximal for $k=n$. The isometry group is given as $SO(k+1)$.
  \item[(P2)] This is the embedding of type (G1, $k$) for $k=1$. It is maximal only for $n=2$ and its isometry group is $SO(3)$.
 \item[(A)] The totally geodesic submanifold is isometric to the $2$-sphere of radius $\sqrt{10}/2 $. It is maximal only for $n=3$ and its isometry group is given by $SO(3)$.
\end{description}

\subsection{Totally geodesic submanifolds of the Lie ball} \label{type}

We will now use Cartan's duality (as explained in Section \ref{duality}) and Klein's classification as listed in the previous section.
In the following, the immersions $u$ will be equivariant.
So they are useful to compute the corresponding immersion of the group into $SO(2,n)$.

\paragraph{Type (I1,k)} Here we have $1\le k\le n/2$.
Let us consider the following map,
 \[ u\ :\ [z_0 : \ldots : z_k] \longrightarrow [z_0 : \i z_0 : \ldots  : z_k : \i z_k:0:\ldots :0]. \]
The image of $u$ is contained in $\pi(Q^{2,n})$. In order to see which point is taken by $u$  to ${\cal L}^n $ it is enough to see that
\[ -|z_0|^2 - |\i z_0|^2 + \sum_{i=1}^{n+1}\left( |z_i|^2 + |\i z_i|^2\right)  = 2  \left(- |z_0|^2 + \sum_{i=1}^{n+1} |z_i|^2 \right) \]
Thus, $u[z_0 :\ldots :z_k:0:\ldots :0] \in {\cal L}^n $ if and only if $-|z_0|^2 + \sum_{i=1}^{n+1} |z_i|^2 < 0$.
Hence,  $u$ gives an holomorphic immersion from the complex hyperbolic space $\C H^k$ into our Lie ball ${\cal L}^n$. Namely, $\C H^k$ is regarded as the projective submanifold of $\C P^{k,1}$ with $-|z_0|^2 + \sum_{i=1}^{n+1} |z_i|^2 < 0$\footnote{To see $\C H^k$ as a ball in $\C^k$ just take $z_0 = 1$, i.e. the affine chart.}.

The group of isometries of $\C H^k$ is $SU(1,k) \subset SO(2,n)$ which acts irreducibly on $\R^{2,n}$ only for $k=n/2$.
To see this it is enough to identify $\R^{2,n}$ with $\C^{1,n/2}$ endowed with the quadratic form $-|w_0|^2 + \sum_{i=1}^{n/2}|w_i|^2$. The action of $SU(1,n/2)$ is transitive on the set of negative $2$-planes of $\C^{1,n/2}$ given by complex lines. For example, the complex line generated by the vector $(1,0,\ldots , 0) \in \C^{1,n/2}$ is a negative definite $2$-plane\footnote{Actually, such a complex line is the $2$-plane which we called $\Pi_0$ in the first section.} of $\R^{2,n}$.
Let $w = (w_0, \ldots ,w_{n/2}) \in \C^{1,n/2}$ be a vector. Then the $2$-plane generated by $w$, i.e. the complex line, is given by the homogeneous coordinates $[ \overline{w_0} : \i \overline{w_0} : \ldots  : \overline{w}_{n/2} : \i \overline{w}_{n/2} ]$. This show that the image of our map $u$ is the set of $2$-planes coming from complex lines of $\C^{1,n/2}$. Thus, the image $u(\C H^{n/2})$ is the orbit of $SU(1,n/2)$ through $\Pi_0$.

Notice that the lift $L(u(\C H^{n/2}))$ of the totally geodesic submanifold $u(\C H^{n/2})$ is
the union of the points in all negative complex lines. Thus, such a subset is full in $\R^{2,n}$ and
this is consistent (indeed equivalent) to the fact that $SU(1,n/2)$ acts irreducibly on $\R^{2,n}$.

\paragraph{Type  (I2,\boldmath $k$)} Here it is  $1\le k\le n/2 $.
The map $u$ is the ``real form'' of the above map:

 \[ [x_0 : \ldots  : x_k] \stackrel{u}{\rightarrow} [x_0 : \i x_0 :\ldots  : x_k : \i x_k:0:\ldots :0] \]

Thus we get an embedding  of $\R H^k$ (in the projective Klein model\footnote{Here we refer to Felix Klein.}) into the Lie ball.
Notice that the subgroup $SO(1,k) \subset SU(1,n/2)$ acts reducibly on $\R^{2,n}$, even for $k=n/2$. In the light of Theorem \ref{conseq} we do not get another irreducible subgroup of $SO(2,n)$.
But we should point out that
the group $I(u(\R H^k))$ i.e. the isometries that fix all points of the image of $u$
is given as  $I(u(\R H^k)) = SO(2)$ acting diagonally, i.e. $SO(2) \cong e^{\operatorname{i} \theta} Id$. For $k=n/2$  this group makes $G=I(u(\R H^k))\cdot SO(1,n/2)$ act irreducibly on $\rr^{2,n}$. $G$ was already on our list.

\paragraph{Type (G1,\boldmath $k$)} This is the embedding of a lower dimensional Lie ball. Its isometry group is given by $SO(2,k)$, which does not act irreducibly on $\rr^{2,n}$.

\paragraph{Type (G2,\boldmath $k_1,k_2$)} In this case  $1\le k_1+k_2 \le n$ and the map $u$ is given by:
\[ \left( [x_0:\ldots : x_{k_1}],[y_0:\ldots :y_{k_2}]\right) \stackrel{u}{\mapsto} [x_0 : \i y_0 : x_1:\ldots : x_{k_1} :\i y_1 :\ldots : \i y_{k_2} :0 : \ldots :0] \]
The image lies in ${\cal L}^n$ if and only if:
\[ -x_0^2 - y_0^2 +\sum_{i=1}^{k_1}x_i^2 + \sum_{j=1}^{k_2}y_j^2 < 0 \]
and
\[- x_0^2 + y_0^2 +\sum_{i=1}^{k_1}x_i^2 - \sum_{j=1}^{k_2}y_j^2=0. \]
Since the map is given in homogeneous coordinates we can assume that
$-x_0^2 +\sum_{i=1}^{k_1}x_i^2 = - y_0^2 + \sum_{j=1}^{k_2}y_j^2$ which shows that
the image of $u$ is in the Lie ball if and only if
$[x_0:\ldots : x_{k_1}]$ and $[y_0:\ldots :y_{k_2}]$ lie in the real hyperbolic spaces of dimensions $k_1$ and $k_2$. Hence, $u$ is an embedding of $\R H^{k_1}\times \R H^{k_2}$ into the Lie ball. The isometry groups is given by $SO(1,k_1)\times SO(1,k_2) \subset SO(2,k_1+k_2)\subset SO(2,n)$.  Thus, the isometry group of this totally geodesic submanifold does not act irreducibly, since it fixes $\R^{1,k_1}$ and $\R^{1,k_2}$.

\paragraph{Type (P1,\boldmath $k$)} Here it it $1\le k\le n$ and the embedding is given by the one of type (G2, $k_1$, $k_2$) for $k_1$ or $k_2$ equal to zero. Hence, we can write it as
\[[ x_0:\ldots : x_k ] \rightarrow [\i: x_0:\ldots : x_k:0\ldots :0] \]
Thus we get an immersion from $\R H^k$ (as in the usual Lorentzian model) into the Lie ball ${\cal L}^n$. The isometry group of the totally geodesic submanifold is $SO(1,k)$ acting reducibly even for $k=n$ by fixing the first basis vector $e_0$.

\paragraph{Type (P2)} This is the embedding of type (G1, $k$) for $k=1$ and thus  the isometry group of the totally geodesic submanifolds is given as $SO(2,1)\subset SO(2,n)$ acting reducibly by fixing $e_3, \ldots, e_{n+1}$.

\paragraph{Type (G3)} This totally geodesic submanifold is a Riemannian product. Then its isometry group $G$ is not simple. Thus this case reduce to the case of  $G \subset U(1,n)$.

\paragraph{Type (A)} Here it is $n\ge 3$. This is an embedding of $3$-dimensional real hyperbolic space into the Lie ball. The only irreducible acting (simple) subgroup of $SO(2,n)$ which did not  appear as isometry group of a totally geodesic orbit in the Lie ball is $SO(1,2)\subset SO(2,3)$. Thus we conclude that this embedding of $SO(1,2) $ gives the isometry group of a totally geodesic orbit of type $(A)$ for $n=3$. For $n>3$ it is reducible, of course. For details on this case please refer to Appendix \ref{so12}.

\bigskip

We conclude that the only irreducibly acting simple proper subgroups of $SO(2,n)$ that appear as isometry group of a totally geodesic submanifold in the Lie ball are the following
\[  SU(1,n/2) \text{ and } SO(1,2)\subset SO(2,3).\]

\section{Proof of  Theorem \ref{list}}

Let $G \subset SO(2,n)$ be a connected subgroup whose action on $\R^{2,n}$ is irreducible. Assume that $G \neq SO_0(2,n)$.
If $G$ is not simple then Theorems \ref{conseq} and \ref{u1n} imply that $G$ is one of the groups in our list. Namely, $n=2p$ and either $G = U(1,p)$ or $G=S^1\cdot SO_0(1,p)$ if $p>1$. If $G$ is simple then (up to conjugation) $G$ is the group of isometries of one of the totally geodesic submanifolds of the Lie ball ${\cal L}^n$ listed in the previous section. Thus, either $G = SU(1,n/2)$ or $ G = SO_0(1,2)\subset SO(2,3)$. This completes the proof.\hfill $\Box$

\begin{appendix}

\section{Appendix}

\subsection{\boldmath $\laso(1,2)$ acting irreducibly on \boldmath $\rr^{2,3}$ and orbits of type (A)}
\label{so12}
In this section we want to describe the irreducible injections of $\laso (3)$ and $\laso (1,2)$ into $\laso (5)$ and $\laso (2,3)$, respectively, and to describe the Lie subtriples in the complex quadric and the Lie ball corresponding to them.

The irreducible injection of $\laso (3)\subset \laso (5)$ corresponds to an irreducible symmetric space  of type $AI$ for $n=3$, which we will describe in the non-compact setting. To this end split $\lasl (3,\rr)$ as
\[\lasl (3,\rr)=\laso (3)\+ sym_0(3)\]
where $sym_0(3)$ denotes the trace free, symmetric $3\times 3$ matrices. This splitting provides symmetric data, since
\[ [\laso(3),\laso(3)]\subset \laso(3),\  [\laso(3),sym_0(3)]\subset sym_0(3),\ [sym_0(3),sym_0(3)]\subset \laso(3).\]
The adjoint representation of $\laso (3) $ on the five-dimensional space $sym_0(3)$ is orthogonal with respect to the Killing form of $\lasl(3,\rr)$, which is positive definite on $sym_0(3)$. Hence, the above splitting defines an irreducible $5$-dimensional Riemannian symmetric space with isotropy group $\laso (3)\subset \laso(5)$, irreducibly.

In order to write down $\laso (3)$ in $5\times 5$ matrices,
we identify the standard basis in $\rr^5$ with the following matrices which are orthogonal under the Killing form of $\lasl (3,\rr)$:
\[e_1= S_{12},\ e_2= S_{13},\ e_3=\mathrm{diag}(-1,1,0),\  e_4=S_{23},\ e_5=\frac{1}{\sqrt{3}}\ \mathrm{ diag}(-1,-1,2),\]
where $S_{ij}$ denotes the symmetric matrix with $1$ at the $(i,j)$-th spot.
Acting via the adjoint representation, the standard basis of $\laso (3)$
\[ U=D_{12}\ ,\ \ V=D_{13}\ ,\text{ and } W=D_{23}\]
is given as follows
\[
U=
\left(\begin{array}{c|c} 0&\begin{array}{ccc} -2&0&0\\0&-1&0\end{array}\\ \hline
 \begin{array}{cc}2&0\\
 0&1\\
 0&0\end{array} &0
 \end{array}\right) \text{, }
V=
\left(\begin{array}{c|c} 0&\begin{array}{ccc} 0&-1&0\\-1&0&-\sqrt{3}\end{array}\\ \hline
 \begin{array}{cc}0&1\\
 1&0\\
 0&\sqrt{3}\end{array} &0
 \end{array}\right)
\]
and
\[
W\ =\  [U,V]\ =\
\left(\begin{array}{c|c}
\begin{array}{cc}
0&-1\\
1&0
\end{array}
&0 \\ \hline
0&
 \begin{array}{ccc}0&-1&0\\
 1&0&-\sqrt{3}\\
 0&\sqrt{3}&0\end{array}
 \end{array}\right)
\]
One verifies that
\be
[V,W]= U&\text{ and } &[W,U]=V,\ee
which are the commutator relations of $\laso (3)$.

Regarding the complex quadric $Q^n=SO(n+2)/SO(n)$,
the Cartan decomposition of the Lie algebra $\laso (n+2)$ is given by
 \be
\lak\ :=\  \laso(2)\+\laso(n)&=&\left\{\left(\begin{array}{c|c} A&0\\ \hline 0&B\end{array}\right)\mid A\in \laso(2), B\in \laso(n)\right\}\\
 \lam&:=&\left\{\left(\begin{array}{c|c} 0&\begin{array}{c} -u^t\\-v^t\end{array}\\ \hline
 \begin{array}{cc}u&v\end{array} &0
 \end{array}\right)\mid u,v\in \rrn\right\}
 \ee
Now, for $n\ge 3$, $\lan =\span (U,V)$ defines a Lie sub-triple of the Lie triple corresponding to the complex quadric, with isometry algebra $\laso (3)= [\lan,\lan]\+\lan\subset \laso (5)\subset \laso (n+2)$. The totally geodesic orbit corresponding to this Lie subtriple is of type (A), see \cite[page 85]{klein08}.

\bigskip

Now we transfer this situation to the Lie ball with Cartan decomposition
\[\laso(2,n)=\lak \+\lam^*\ \text{ with }\
 \lam^*:=\left\{\left(\begin{array}{c|c} 0&\begin{array}{c} u^t\\v^t\end{array}\\ \hline
 \begin{array}{cc}u&v\end{array} &0
 \end{array}\right)\mid u,v\in \rrn\right\}
\]
Here the totally geodesic orbits of type (A) are given by the Lie subtriple
\[\lan^*= \span (U^*,V^*)\]
where $U^*$ and $V^*$ are as above, but symmetric instead of skew symmetric. Here we have that
\[ -W=[U^*,V^*]\]
 and again
\[ [V^*,W]= U^*\ \text{ and }\  [W,U^*]=V^*.\]
These are the commutator relations of $\laso(1,2)\simeq \lasl(2,\rr)$. Again, the irreducible representation $\laso(1,2)\subset \laso (2,3)$ comes from the irreducible symmetric space of signature $(2,3)$ given by
\[\lasl (3,\rr)=\laso(1,2)\+ \mathfrak{t} \]
where $\mathfrak{t} $ is a five-dimensional complement of $\laso(1,2) $ in $\lasl(3,\rr)$, consisting of  trace-free matrices with the right symmetries.

Finally, in order to verify a remark in the introduction, we want to show that $SO_0(1,2)$ does not act locally transitively on $S^{1,2}=SO_0(2,3)/P$  where $P$ is the parabolic subgroup given as the stabiliser of a light-like line. For general $(p,q)$, the Lie algebra of $P$ is given as $\lap=(\rr\+\laso(p,q))\ltimes \rr^{p,q}\subset \laso(p+1,q+1)$. A group $G\subset SO(p+1,q+1)$ acts locally transitively on $S^{p,q}=SO(p+1,q+1)/P$ if
\[\laso (p+1,q+1)=\lag + \lap\]
where $\lag\subset \laso(p+1,q+1)$ is the Lie algebra of $G$ (for details see \cite{alt08}). For $\lag=\laso (1,2)\subset \laso(2,3)$, irreducibly, this sum has to be direct since the parabolic $\lap$ is $7$-dimensional in this case. But $U^*\in \laso(1,2) $ fixes the line spanned by the light-like vector vector $e_1+e_3$. Hence, $\laso(1,2)\cap\lap \not=\{0\}$ and thus, this action of $SO(1,2)$  on $S^{1,2}$ is not  locally transitively.

\subsection{A biholomorphism between the Lie ball and  Cartan's bounded domain of type $IV$}
In this appendix we give an explicit bi-holomorphism between the Lie ball ${\cal L}^n$ and (the classical) Cartan's bounded domain of type $IV$ in $\C^n$.
Let $f$ be the map given as follows:

\[ f(z_1, \cdots, z_n) = [i(\Lambda - 1) : \Lambda + 1 : 2z_1 : \cdots, 2z_n  ] \, \, ,\]

where $\Lambda := z_1^2 + \cdots + z_n^2$. Thus, $f: \C^n \rightarrow \C P^{1,n}$. Let us show that $f(\C^n) \subset Q^{2,n}$. Indeed,  \[ (2z_1)^2 + \cdots + (2z_n)^2 - (i(\Lambda - 1))^2 - (\Lambda + 1)^2 = \]
\[ 4 \Lambda + (\Lambda -1)^2 - (\Lambda + 1)^2 = 4 \Lambda + (\Lambda - 1 - \Lambda - 1)(\Lambda - 1 + \Lambda + 1) = \] \[ = 4 \Lambda - 4 \Lambda = 0 \]

Let us characterize when a point $z = (z_1, \cdots, z_n)$ is take by $f$ to a point in the Lie ball ${\cal L}$.
By definition $f(z) \in {\cal L}^n$ if and only if the following condition holds:

\[ |2z_1|^2 + \cdots + |2z_n|^2 - |i(\Lambda - 1)|^2 - |\Lambda + 1|^2 < 0 \]

Notice that $f(0) = [ -i : 1: 0: \cdots : 0 ] = [1 : i: 0: \cdots : 0 ] = \Pi_0$ satisfy such a condition. Thus, we are interested in the connected component of $0 \in \C^n$ satisfying such a condition. A simple calculation shows that the above condition is equivalent with:

\[ 2(|z_1|^2 + \cdots + |z_n|^2) - |z_1^2 + \cdots + z_n^2|^2 - 1 < 0 \]

Notice that $|z_1^2 + \cdots + z_n^2|^2 \leq (|z_1|^2 + \cdots + |z_n|^2)^2$. So let $z(t)$ be a continuous curve starting at $0$ (i.e., $z(0)=0$) and satisfying the above condition. Then at each $t$ we get
\[ 2(|z_1|^2 + \cdots + |z_n|^2)  - 1 <  |z_1^2 + \cdots + z_n^2|^2 \leq (|z_1|^2 + \cdots + |z_n|^2)^2 \]
So
\[ 2(|z_1|^2 + \cdots + |z_n|^2)  - 1 <  (|z_1|^2 + \cdots + |z_n|^2)^2 \]
\[ - (|z_1|^2 + \cdots + |z_n|^2  - 1)^2 < 0 \]

Thus the curve $z(t)$ remains inside the unit ball $|z| < 1$.\\

This shows that $f$ takes the bounded domain $\Omega \subset \C^n$ given by the inequalities:

\[ |z| < 1 \]

and

\[ 2(|z_1|^2 + \cdots + |z_n|^2) - |z_1^2 + \cdots + z_n^2|^2 - 1 < 0 \]

into the Lie ball ${\cal L}^n$. Notice that $\Omega$ is indeed the classical Cartan's bounded domain of type $IV$.
Actually, $f$ is just the stereographic projection of the quadric $- w_{1}^2 - w_{0}^2 + w_1^2 + \cdots + w_n^2  = 0$ regarded as an affine sphere in the chart $w_{0} = 1$ of $\C P^{1,n}$.
\end{appendix}

\def\cprime{$'$}


TITLE: Exponential distribution and poisson process
QUESTION [1 upvotes]: Consider a post office with two clerks. Three people, A, B, and C,
  enter simultaneously. A and B go directly to the clerks, and C waits
  until either A or B leaves before he begins service. What is the
  probability that A is still in the post office after the other two
  have left when
$i)$the service time for each clerk is exactly(nonrandom) ten minutes?
$ii)$the service times are $i$ with probability $\frac{1}{3}$,
  $i=1,2,3$?
$iii)$the service times are exponential with mean $\frac{1}{\mu}$?

Let $T_1$,$T_2$ and $T_3$ the service times for $A,B,C$
$i)$ $P(T_1>t_2+t_3|t_1=t_2=t_3=10)=0$
I did not have to do this but that's okay, since this is quite intuitive
$ii)$Let $\Omega$ the space of possible values of $t_1,t_2,t_3$, then
$$t_1>t_2+t_3\Leftrightarrow t_1=3,t_2=1,t_3=1$$
$$P(T_1>t_2+t_3)=P(T_1=3)P(T_2=1)P(T_3=1)=\frac{1}{3}\frac{1}{3}\frac{1}{3}=\frac{1}{27}$$
$iii)$ Here I'm confuse anyone can help?
EDIT: The answer of $iii)$ is $\frac{1}{4}$ but I don't know how to get it

REPLY [0 votes]: The problem makes no mention of independence, so I'd be careful making that assumption. Let $S_A, S_B, S_C$ be the service times of each of the customers. Since they follow an exponential distribution with mean $\frac{1}{\mu}$, then the rate is simply $\mu$. You want to know:
$$
P(S_A > S_B + S_C)
$$
Condition on one of those, say like this:
$$
P(S_A > S_B + S_C) = P(S_A > S_B + S_C | S_A > S_B) \cdot P(S_A>S_B) + P(S_A > S_B + S_C | S_A < S_B) \cdot P(S_A<S_B)
$$
Then some simplification:
\begin{align*}
P(S_A > S_B + S_C) &= P(S_A > S_C) \cdot P(S_A > S_B) + 0 \\
&= P(S_C < S_A) \cdot P(S_B < S_A) \\
&= \frac{\mu}{\mu + \mu} \cdot \frac{\mu}{\mu + \mu} \\
&= \frac{\mu}{2\mu} \cdot \frac{\mu}{2\mu} \\
&= \frac{1}{4}
\end{align*}