text:
\section{Construction of the Flows}

In this section, we complete the proof of Theorem \ref{main} by
constructing the $\nabla f$-like vector fields appearing in part
(iii) of that theorem.  We begin with some preparations (keeping
the notation of Section 4).

Theorem \ref{main} stipulates that the vector field $V$ should be
$\cS$-preserving, i.e., weakly controlled with respect to some
control data on $(X, \cS)$.  However, by Proposition \ref{approx},
it suffices to construct, for each $\cU \supset K$, a vector field
$V$ which is $\cX$-preserving, instead.  We therefore fix some
control data on $(X, \cX)$, subject to the only condition that the
quasi-distance function $\rho_{\{ p \}}$ is the standard Euclidean
distance to $p$.  All weakly controlled vector fields in the rest
of this section will be compatible with this control data.

Let $\cX'$ be the set of all $S \in \cX$ with $S \cap B \neq
\emptyset$ and $\dim S > 0$.  For every $S \in \cX'$, with
$\dim S = i$, let $V^S$ be the unique smooth vector field on
$S \cap B$ satisfying the following properties:

(a)  the (standard, Euclidean) norm $|\!| V^S_x |\!| =
|\!| x |\!|$ for every $x \in S \cap B$;

(b)  $V^S$ preserves the projection $\pi_{i-1} :
S \cap B \to Q_{i-1}$;

(c)  $V^S$ preserves the imaginary part $\im l :
S \cap B \to \R$;

(d)  $V^S_x f > 0$ for every $x \in S \cap B$.

\noindent
For every $S \in \cX'$, we fix a continuous, controlled extension
$\tilde V^S$ of $V^S$ to some neighborhood $U_S$ of $S \cap B$.
The vector field $V$
will be constructed by ``patching together'' the $\tilde V^S$.

Without loss of generality, we may assume that $U_S \cap T =
\emptyset$, for all distinct $S, T \in \cX'$ with
$\dim T \leq \dim S$.  Let $\tilde U_S$ be the union of all
$T \in \cX'$ with $\dim T > \dim S$ or $T = S$.  We say that
$\phi : \tilde U_S \to [0,1]$ is a cut-off function for
$S \cap B$ if $\supp (\phi) \subset U_S$ and $\phi^{-1} (1)$
contains a neighborhood of $S \cap B$.  Given two cut-off
functions $\phi, \psi$ for $S \cap B$, we write $\phi \less \psi$
if $\supp(\psi) \subset \phi^{-1}(1)$.

Now, let $\Phi$ be the set of all collections $\bar \phi =
\{ \phi_S  \}_{S \in \cX'}$, where $\phi_S$ is a cut-off function
for $S \cap B$.  Given $\bar \phi, \bar \psi \in \Phi$, we write
$\bar\phi \less \bar\psi$ if $\phi_S \less \psi_S$ for all $S \in
\cX'$.  It is easy to see that $(\Phi, \less)$ is a directed set:
$$\forall \; \bar\phi_1, \bar\phi_2 \in \Phi \;\; \exists \;
\bar\phi_3 \in \Phi : \; \bar\phi_1 \less \bar\phi_3, \;
\bar\phi_2 \less \bar\phi_3.$$

Fix an ordering $\cX' = \{ S_1, \dots, S_n \}$ such that
$\dim S_i \leq \dim S_j$ for $i \leq j$.  For every $\bar \phi \in
\Phi$, we define a vector field $V = V(\bar\phi)$ on $B$ as follows.
Let $V_p = 0$, and use the formula
$$V = \phi_{S_1} \cdot \tilde V^{S_1} + (1 - \phi_{S_1}) \cdot (
\phi_{S_2} \cdot \tilde V^{S_2} + (1 - \phi_{S_2}) \cdot (
\phi_{S_3} \cdot \tilde V^{S_3} + \dots +
(1 - \phi_{S_{n-1}}) \cdot V^{S_n}) \dots )$$
on $B \setminus \{ p \}$.
This formula should be parsed left to right, and evaluation should
stop as soon as one of the expressions $(1 - \phi_{S_i})$ is found
to be zero.  In this way, we will never have to evaluate one of
the $\phi_{S_j}$ or $\tilde V^{S_k}$ at a point where it is not
defined.  Also, note the missing $\;\tilde{}\;$ over $V^{S_n}$; it
is not needed because $S_n$ is the open stratum of $\cX$.  It is
not hard to check that there is a $\bar \phi_0 \in \Phi$ such that,
for every $\bar \phi \in \Phi$ with $\bar\phi_0 \less \bar\phi$, the
vector field $V(\phi)$ is $\nabla f$-like.  Theorem \ref{main} is a
consequence of the following claim.

\vspace{.1in}

\noindent
{\bf Claim 1:}  For every open $\cU \supset K$, there exists a
$\bar \phi_0 \in \Phi$ such that, for every $\bar \phi \in \Phi$
with $\bar\phi_0 \less \bar\phi$, we have
$M^\pm_{V(\phi)}(p) \subset \cU$.

\vspace{.1in}

Claim 1, in turn, follows from the following.

\vspace{.1in}

\noindent
{\bf Claim 2:}  For every $x \in B \setminus K$, there is a
neighborhood $U_x$ of $x$ and a $\bar \phi_0 \in \Phi$ such that,
for every $\bar \phi \in \Phi$ with $\bar\phi_0 \less \bar\phi$,
we have $M^\pm_{V(\phi)}(p) \cap U_x = \emptyset$.

\vspace{.1in}

Claim 2 is readily proved by induction on the dimension of the
stratum containing $x$, using the definition of $K$ and the
following.

\vspace{.1in}

\noindent
{\bf Claim 3:}  Fix an $i \in \{ 0, \dots, d-1 \}$, and let
$A \subset B$ be a compact set such that $A \cap X_i = \emptyset$.
Then for every $\epsilon > 0$, there is a $\bar \phi_0 \in \Phi$
such that, for every $\bar \phi \in \Phi$ with $\bar\phi_0 \less
\bar\phi$ and every $a \in A$, we have the following estimates:
$$|\!| d \pi_i (V(\phi)_a) |\!| < \epsilon \cdot d f (V(\phi)_a),$$
$$| \im d l (V(\phi)_a) | < \epsilon \cdot d f (V(\phi)_a).$$

\vspace{.1in}

Lastly, Claim 3 follows from the fact that the $\tilde V^S$ in the
construction of $V(\phi)$ were chosen to be continuous extensions
of the vector fields $V^S$ satisfying conditions (b)-(d).  This
completes the proof of Theorem \ref{main}.

\vfill

config: arxiv


TITLE: Shifted sum of reciprocals of primes: $\sum \limits_{\forall p \in primes}\frac{1}{p+1}$
QUESTION [1 upvotes]: I know that the sum:
$\sum \limits_{\forall p \in primes}\frac{1}{p}$ diverges.
I was wondering if:
$\sum \limits_{\forall p \in primes}\frac{1}{p+1}$ diverges as well.
And if there is a formula for the partial sum of primes up to an upperbound $n$.
My guess is that it does but I'm not quite sure what the quick way to show it is. Any help is greatly appreciated!

REPLY [5 votes]: Let $p_n$ be the $n$-th prime. Then these two series diverge:
$$
\sum_{n=1}^{\infty} 1/p_n
$$
$$
\sum_{n=1}^{\infty} 1/p_{n+1}
$$
Since $p_{n+1} \geq p_n + 1$, we conclude that the series you described is divergent as well.


TITLE: Is there any upper limit on a particles kinetic energy?
QUESTION [5 upvotes]: While the speed of light is the "ultimate speed limit" for any particle.. Is there any upper limit to kinetic energy?

REPLY [9 votes]: In principle, no: a particle may leave a reaction with any kinetic energy you can imagine.
However in practice, there's a limit known as the GZK cutoff. A baryon (proton or neutron) with energy above about $10^{19}\,\rm eV$ will see the cosmic microwave background blue-shifted so strongly that the baryon can scatter from the background photons to produce pions. The consequence is that any particle with energy above this limit must be "local", since it has not yet lost its energy. ("Local" is relative, however; the GZK horizon contains many galaxies.)
I am not sure whether there is a GZK-like limit on the kinetic energy of photons or electrons. For macroscopic objects, Randall Munroe discusses this sort of limit for baseballs and meteorites. But in general, whether there's a limit to a particle's kinetic energy depends on how close you are to the source of your accelerated particle: close, no limit; far off, limited.


TITLE: Prove that middle cancellation implies that the group is abelian
QUESTION [7 upvotes]: Suppose that $G$ is a group with the property that for every choice of elements in $G$, $axb=cxd$ implies $ab=cd$.
Prove that $G$ is Abelian.
(Middle cancellation implies commutativity).

To be an Abelian group, it is required that $\forall a,b \in G$, $a. b=b .a$
Can I request for a hint to this question? I'm not sure how to get started on this one.

REPLY [7 votes]: Hint: Given $a$ and $b$, find an element $x$ such that $axb=bxa$. (I can think of two $x$s that would work.)


TITLE: What is the angle x?
QUESTION [1 upvotes]: We have to find the angle x

I have tried using these theorems, but failed.
1. The angle at the centre is twice the angle at the circumference.
2. The exterior angle of a triangle is equal to the sum of the two opposite interior angle.

REPLY [1 votes]: Hint:
$\triangle AOC$ is isosceles
with
\begin{align}
\angle OCA
=
\angle AOC
&=
\tfrac12\,(180^\circ-20^\circ)
=80^\circ
.
\end{align}


TITLE: Find Jordan form of matrix with parameter $m\in\mathbb R$
QUESTION [0 upvotes]: the (real) matrix is $\left[ \begin{matrix} 0& 0& 0\\ 1& 0& -m^{2}\\0& 1& m\\\end{matrix} \right]$.
Well, the characteristic polynomial is $p(t)=t(t^2-mt+m^2)$ so it doesn't have any real roots except 0, and then I got stuck.

REPLY [0 votes]: For the case in which $m \neq 0$, the matrix has three distinct eigenvalues.  So, the Jordan form is just the diagonal matrix with those eigenvalues on the diagonal.
For the case in which $m = 0$, we note that transposing the matrix produces a Jordan-form matrix.  Any matrix is similar to its transpose.


TITLE: Is it possible get an algebraic expression for $\sin 1°$ that does not contain roots of negative values?
QUESTION [7 upvotes]: Is it possible get an algebraic expression for $\sin 1°$ that does not contain roots of negative values, so that it can be evaluated entirely using only real numbers?

For example, this paper gives some algebraic expressions for it, but they all involve complex numbers.

REPLY [11 votes]: If this would be possible, you could also get an algebraic expression for $\sin(10^o)$ that doesn't contain complex numbers.  That is one of three real roots
of the cubic $8 z^3 - 6 z + 1$. Now see casus irreducibilis: given a cubic with rational coefficients that is irreducible over the rationals and has three real roots, it is impossible to express the roots using real radicals.


TITLE: Subspace topology and order topology (2)
QUESTION [7 upvotes]: Yesterday I asked a very similar question but now I am confused again...
Let $I=[0,1]$. The dictionary order on $I\times I$ is just the restriction to $I\times I$ of the dictionary order on the plane $\mathbb{R}\times\mathbb{R}$. now, how can I show that the set $\{1/2\}\times (1/2,1]$ is not open in the order topology? The maximum element of $I$ is 1. So the order topology on $I\times I$ has intervals of the form $(a,1]\times (b,1]$ as basis elements. Isn't $\{1/2\}\times (1/2,1]$ itself a basis element of the order topology?

REPLY [8 votes]: You may have somewhat misunderstood the lexicographic order, since the lexicographic order topology on $I\times I$ does not have sets of the form $(a,1]\times(b,1]$ as basis elements. To avoid confusion with the usual order $\le$ on $I$, I’ll use $\preceq$ for the lexicographic order on $I\times I$. Note also that I use angle brackets for ordered pairs, e.g., $\langle x,y\rangle$, and parentheses for open intervals.
If $p=\langle x_0,y_0\rangle\in I\times I$ and $q=\langle x_1,y_1\rangle\in I\times I$, then $p\preceq q$ iff either $x_0<x_1$, or $x_0=x_1$ and $y_0\le y_1$. Suppose that $p\prec q$. If $x_0=x_1$, so that $p$ and $q$ are both on the vertical segment $\{x_0\}\times I$, then the open interval $(p,q)$ in $\langle I\times I,\preceq\rangle$ is the vertical open interval
$$(p,q)=\{x_0\}\times(y_0,y_1)=\{\langle x,y\rangle\in I\times I:x=x_0\text{ and }y_0<y<y_1\}\;.$$
If $x_0<x_1$, it’s a little more complicated:
$$\begin{align*}
(p,q)&=\Big(\{x_0\}\times(y_0,1]\Big)\cup\Big((x_0,x_1)\times I\Big)\cup\Big(\{x_1\}\times[0,y_1)\Big)\\
&=\{\langle x,y\rangle\in I\times I:x=x_0\text{ and }y_0<y\le 1\}\\
&\qquad\cup\{\langle x,y\rangle\in I\times I:x_0<x<x_1\text{ and }y\in I\}\\
&\qquad\cup\{\langle x,y\rangle\in I\times I:x=x_1\text{ and }0\le y<y_1\}\;.
\end{align*}$$
One base for the lexicographic order topology on $I\times I$ is the collection of all of these open intervals together with all of the intervals $[\langle 0,0\rangle,p)$ and $(p,\langle 1,1\rangle]$ for $p\in(I\times I)\setminus\{\langle 0,0\rangle,\langle 1,1\rangle\}$, since these latter form local bases at the endpoints $\langle 0,0\rangle$ and $\langle 1,1\rangle$, respectively.
Let $A=\left\{\frac12\right\}\times\left(\frac12,1\right]$, the specific set about which you asked. (Note that it is not of the form $(a,1]\times(b,1]$ in the first place.) Let $p=\left\langle\frac12,\frac12\right\rangle$ and $q=\left\langle\frac12,1\right\rangle$. Then $A$ is the half-open, half-closed interval $(p,q]$, which is not open, because $q$ is not in its interior. To see this, suppose that $U$ is an open nbhd of $q$. Then there are $u,v\in I\times I$ such that $u\prec q\prec v$ and $(u,v)\subseteq U$. Say $v=\langle x_0,y_0\rangle$; then $q\prec v$ implies that $x_0>\frac12$, so there is a real number $x\in\left(\frac12,x_0\right)$, and $q\prec\langle x,0\rangle\prec v$ and hence $\langle x,0\rangle\in(u,v)\subseteq U$. That is, every open nbhd of $q$ in $I\times I$ contains points with first coordinate larger than $\frac12$, so $(p,q]$, which does not contain any such points, cannot contain any open nbhd of $q$ and therefore cannot be open.


TITLE: Ratio of force of resistance due to liquid to that due to gravity
QUESTION [0 upvotes]: What is ratio of force of resistance due to liquid to that due to gravity?
For any object for example a ball falling in a liquid with constant velocity, what will be the ratio?

REPLY [1 votes]: If the velocity is constant then ratio is 1:1


TITLE: Generating $3\times 3$ Unitary Matrices close to the Identity
QUESTION [1 upvotes]: Is there a way to generate uniformly spaced, or better still, to parametrise Unitary Matrices close to Identity. The solution is required primarily for $SU(3)$, but also in general for $SU(n)$.

It might be relevant that I am trying to optimise a function that takes in a unitary matrix as its argument. I am aware of the fact that $SU(n)$ matrices are exponentiated hermitian traceless generators. But that does not quite help me.

The outline of the problem that I am trying to solve is as follows -

I have an equation for the evolution of some unitary matrix. The evolution is constrained by the unitarity of the matrix - and i have the fixed points of this flow by generating the entire spectrum of $SU(3)$ matrices. But now i also have to find out the stability of these fixed points. I would prefer to use a systematic root finding type algorithm to accomplish this - which is where generating unitary matrices close to the identity comes in.
After the initial brute force search in this space, i would also require increased resolution for bisection or Newton-Ralphson to work. I already have a way to generate a uniform distribution of unitary matrices RANDOMLY - refer here. But I cannot use that for the root finding bit.

Hope the explanation helps! I have heard something about using two Householder transformations to do this, but didnt find anything to back it up! Please cite references if you don't have enough time. I will complete the answers and choose the one that leads me to the correct answer.

REPLY [3 votes]: If you want to generate random elements of $SU(3)$ close to the identity, it seems to me that the most logical approach would be to generate random elements of the Lie algebra $\mathfrak{su}(3)$, and then exponentiate them.
The Lie algebra $\mathfrak{su}(3)$ consists of all $3\times 3$ matrices $A$ with trace $0$ satisfying $A^\dagger = -A$.  Such a matrix can be written as
$$
A \;=\; \begin{bmatrix}ia_1 & b_1+ic_1 & b_2+ic_2 \\ -b_1+ic_1 & ia_2 & b_3+ic_3 \\ -b_2+ic_2 & -b_3+ic_3 & ia_3\end{bmatrix}
$$
where $a_k,b_k,c_k$ are real and $a_1+a_2+a+3 = 0$.
Given a matrix $A\in\mathfrak{su}(3)$, the exponential $\exp(tA)$ is a unitary matrix for all $t\in\mathbb{R}$.
There is a standard inner product on $\mathfrak{su}(3)$, namely the negative of the Killing form, which is invariant under the adjoint action.  This is given by the formula
$$
\langle A, A'\rangle = 12\sum_{k=1}^3 \left(\frac{1}{2}a_ka_k' + b_kb_k' + c_kc_k'\right)
$$
If we drop the factor of $12$, then we can easily find an orthonormal basis for $\mathfrak{su}(3)$ with eight elements:
$$
M_1=\begin{bmatrix}1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 0\end{bmatrix}, \quad M_2=\frac{1}{\sqrt{3}}\begin{bmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 2\end{bmatrix}
$$
$$
M_3=\begin{bmatrix}0 & 1 & 0 \\ -1 & 0 & 0 \\ 0 & 0 & 0\end{bmatrix},
\quad M_4=\begin{bmatrix}0 & 0 & 1 \\ 0 & 0 & 0 \\ -1 & 0 & 0\end{bmatrix},
\quad M_5=\begin{bmatrix}0 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & -1 & 0\end{bmatrix},
$$
$$
M_6=\begin{bmatrix}0 & i & 0 \\ i & 0 & 0 \\ 0 & 0 & 0\end{bmatrix},
\quad M_7=\begin{bmatrix}0 & 0 & i \\ 0 & 0 & 0 \\ i & 0 & 0\end{bmatrix},
\quad M_8=\begin{bmatrix}0 & 0 & 0 \\ 0 & 0 & i \\ 0 & i & 0\end{bmatrix},
$$
These are known as the Gell-Mann matrices.
These eight matrices are like the standard basis of unit vectors for $\mathfrak{su}(3)$.  For example, if you want a random point on the sphere in $\mathfrak{su}(3)$ of radius $0.1$, the thing to do is to choose eight random coordinates $(x_1,\ldots,x_8)$ such that $x_1^2 + \cdots + x_8^2 = 0.1^2$, and then compute
$$
M \;=\; x_1M_1 + \cdots + x_8 M_8.
$$
This will be a random matrix at a distance of $0.1$ from the zero matrix in $\mathfrak{su}(3)$.
To get a random matrix in $SU(3)$, you simply exponentiate:
$$
\exp(M) \;=\; I + M + \frac{1}{2}M^2 + \frac{1}{3!}M^3 + \frac{1}{4!}M^4 + \cdots
$$
Then $\exp(M)$ will be a random matrix in $SU(3)$ at a distance of $0.1$ from the identity.
Note:  The only part I haven't explained yet is how to pick a random point $(x_1,\ldots,x_8)$ on the sphere of radius $0.1$ in $\mathbb{R}^8$.  Here is a standard algorithm:

Pick eight random numbers $(r_1,\ldots,r_8)$ between $-1$ and $1$.
If $r_1^2 + \cdots + r_8^2 > 1$, throw the numbers out and go back to step 1.  Keep picking until you choose one that satisfies $r_1^2 + \cdots + r_8^2 \leq 1$
Now we have a randomly chosen tuple $(r_1,\ldots,r_8)$ such that $r_1^2 + \cdots + r_8^2 \leq 1$.  Then let
$$
(x_1,\ldots,x_8) \;=\; \frac{0.1}{\sqrt{r_1^2+\cdots+r_8^2}}(r_1,\ldots,r_8)
$$


TITLE: If $a \cdot a = a$ then $a=0$ or $a=1$?
QUESTION [1 upvotes]: If $a\cdot a = a$ then $a=0$ or $a=1$?

Where $a \in \mathbb{R}.$
I try this way:
Suppose that a is different to $0$ it implies the existence of multiplicative inverse:
$$a^{-1}(a\cdot a)=a^{-1}\cdot a$$
$$(a^{-1}a)= 1~~~~~\text{(by associativity)}$$
$$1\cdot a=1~~~~~\text{(multiplicative inverse)}$$
Therefore $a=1$.
Is this even possible to negate one of my conclusion to get to the other one and then do the same to prove the other conclusion or this is terribly wrong? I appreciate every help I'm losing myself.

REPLY [2 votes]: A priori you cannot assume there exists a multiplicative inverse.
Instead, using associativity (I hope you have that, at least) rewrite $a^2=a$ as $a\cdot(a-1)=0$. Now it all depends on the fact if your ring allows zero divisors or not. If it does not, then either $a=0$ or $a=1$, otherwise not.


TITLE: Bounded self-adjoint linear operator is injective...
QUESTION [0 upvotes]: Let $T : H → H$ be a bounded self–adjoint linear operator on a Hilbert space $H$. Suppose
the range $R(T)$ is dense in $H$. Prove that $T$ is injective.

REPLY [0 votes]: We have $R(T)^{\perp}=\overline{R(H)}^{\perp}=H^{\perp}=\{0\}$, but $\ker T^{\ast}=(R(T))^{\perp}$.


text:
\begin{document}

\papernumber{003}

\twocolumn[
\begin{@twocolumnfalse}
  \maketitle

\begin{abstract}
It is well-known that the group of regular projective transformations of $\mathbb{P}^3(\mathbb{R})$ is isomorphic to the group of projective automorphisms of Klein's quadric $M_2^4\subset\mathbb{P}^5(\mathbb{R})$. We introduce the Clifford algebra $\clifford_{(3,3)}$ constructed over the quadratic space $\mathbb{R}^{(3,3)}$ and describe how points on Klein's quadric are embedded as null vectors, {\it i.e.},  grade-$1$ elements squaring to zero. Furthermore, we discuss how geometric entities from Klein's model can be transferred to this homogeneous Clifford algebra model. Automorphic collineations of Klein's quadric can be described by the action of the so called sandwich operator applied to vectors $\mathfrak{v}\in\bigwedge^1 V$. Vectors correspond to null polarities in $\mathbb{P}^3(\mathbb{R})$. We introduce a factorization algorithm. With the help of this algorithm we are able to factorize an arbitrary versor $\mathfrak{g}\in\clifford_{(3,3)}$ into a set of non-commuting vectors $\mathfrak{v}_i\in\bigwedge^1 V,\,i=1,\dots, k,\, 1\leq k\leq 6$ corresponding to null polarities with $\mathfrak{g}=\mathfrak{v}_1\dots\mathfrak{v}_k$. Thus, we present a method to factorize every collineation in $\mathbb{P}^5(\mathbb{R})$ that is induced by a projective transformation acting on $\mathbb{P}^3(\mathbb{R})$ into a set of at most six involutoric automorphic collineations of Klein's quadric corresponding to null polarities respectively skew-symmetric $4 \times 4$ matrices.
Moreover, we give an outlook for Lie's sphere geometry, {\it i.e.}, the homogeneous Clifford algebra model constructed with the quadratic form corresponding to Lie's quadric $L_1^{n+1}\subset\mathbb{P}^{n+2}(\mathbb{R})$.
\end{abstract}

  \keywords{Clifford algebra, line geometry, Klein's quadric, null polarity, factorization}
\end{@twocolumnfalse}]

\section{Introduction}
Klein's quadric denoted by $M_2^4\subset\mathbb{P}^5(\mathbb{R})$ and subspace intersections of Klein's quadric are well-known, cf \cite{potwal} or \cite{weiss:vor}. The group of regular projective transformations of $\mathbb{P}^3(\mathbb{R})$ is isomorphic to the group of automorphic collineations of Klein's quadric. Moreover, Cayley-Klein geometries can be represented by Clifford algebras, where the group of Cayley-Klein isometries is given by the Pin group of the corresponding Clifford algebra, see \cite{gunn:kinematics}. Therefore, we recall fundamental basics of Clifford algebras and recall the homogeneous Clifford algebra model corresponding to Klein's quadric. The relationsship between projective transformations and elements of the Pin group is given. The full construction can be found in \cite{klawitter_lingeo}.

\section{Geometric Algebra}
General introductions to Clifford algebras can be found for example in \cite{gallier:cliffordalgebrascliffordgroups,garling:cliffordalgebras,perwass} and \cite{porteous:cliffordalgebrasandtheclassicalgroups}.

\begin{definition}
Let $V$ be a real valued vector space of dimension $n$. Furthermore, let $b:V\mapsto \mathbb{R}$ be a quadratic form on $V$. The pair $(V,b)$ is called \emph{quadratic space}.
\end{definition}
\noindent
We denote the Matrix corresponding to $b$ by $\mathrm{B}_{ij}$ with $i\leq j,j\leq n$. Therefore $b(x_i,x_j)=\mathrm{B}_{ij}$ for some basis vectors $x_i$ and $x_j$.
\begin{definition}
The Clifford algebra is defined by the relations
\begin{equation}
\label{EQ1}
x_ix_j+x_jx_i=2\mathrm{B}_{ij},\quad 1\leq i,j\leq n.
\end{equation}
\end{definition}
\noindent
Usually, the algebra is denoted by $\clifford(V,b)$.
By Silvester's law of inertia we can always find a basis $\lbrace e_1,\dots,e_n \rbrace$ of $V$ such that $e_i^2$ is either $1,-1$ or $0$.
\begin{definition}
The number of basis vectors that square to $(1,-1,0)$ is called \emph{signature} $(p,q,r)$. If $r\neq 0$ we call the geometric algebra \emph{degenerated}. We denote a Clifford algebra by $\clifford_{(p,q,r)}$.
\end{definition}
\begin{remark}
A quadratic real space with signature $(p,q,0)$ is abbreviated by $\mathbb{R}^{(p,q)}$.
\end{remark}
\noindent
With the new basis $\lbrace e_1,\dots,e_n \rbrace$ the relations \eqref{EQ1} become
\begin{equation*}
e_ie_j+e_je_i=0,\quad i\neq j.
\end{equation*}
In the remainder of this paper we shall abbreviate the product of basis elements with lists
\begin{equation*}
e_{12\dots k}:=e_1e_2\dots e_k,\, \mbox{with $0\leq k\leq n$}.
\end{equation*}
The $2^n$ monomials
\begin{equation*}
e_{i_1}e_{i_2}\dots e_{i_k},\quad 0\leq k\leq n
\end{equation*}
form the standard basis of the Clifford algebra. Furthermore, a Clifford algebra is the direct sum $\bigoplus\limits_{i=0}^n{\bigwedge}^i V$ of all exterior products ${\bigwedge}^i V$ of any grade $0\leq i\leq n$ where $e_{k_1}\dots e_{k_i},\,k_1<\dots<k_i$ form a basis of ${\bigwedge}^i V$. Therefore, a Clifford algebra is a graded algebra and its dimension is calculated by
\begin{equation*}
\dim \clifford_{(p,q,r)}=\sum\limits_{i=0}^n{\dim {\bigwedge}^i V}=\sum\limits_{i=0}^n\binom{n}{i}=2^n.
\end{equation*}
Moreover, the Clifford algebra $\clifford_{(p,q,r)}$ can be decomposed in an even and an odd part
\begin{align*}
\clifford_{(p,q,r)}&=\clifford_{(p,q,r)}^+ \oplus \clifford_{(p,q,r)}^-\\
&=\bigoplus\limits_{\substack{i=0\\i \text{ even}}}^n{{\bigwedge}^i V} \oplus \bigoplus\limits_{\substack{i=0\\i \text{ odd}}}^n{{\bigwedge}^i V}.
\end{align*}
The even part $\clifford_{(p,q,r)}^+$ is a subalgebra, because the product of two even graded monomials must be even graded since the generators cancel only in pairs. Elements contained in ${\bigwedge}^i V$ are called \emph{$i$-vectors} and the $\mathbb{R}$-linear combination of $i$-vectors $i\leq n$ is called a \emph{multi-vector}. A multi-vector $\mathfrak{A}$ is called homogeneous if $\left[ \mathfrak{A}\right]_i=\mathfrak{A},\, i\leq n$ where $\left[\cdot\right]_m\ m\in\mathbb{N}$ denotes the grade-$m$ part of the multi-vector $\mathfrak{A}$. The product of invertible vectors is called a \emph{versor} and the product of vectors where at least one vector is not invertible is called a \emph{null versor}.
\begin{definition} \label{centerclifford}
The center of a ring $\mathcal{R}$ is the set of all elements that commute with all other elements
\[\mathcal{C}(\mathcal{R}):=\left\lbrace c\in\mathcal{R}\mid cx=xc \mbox{ for all }a\in\mathcal{R}\right\rbrace.\]
\end{definition}
\noindent We are interested in the center of a Clifford algebra, see \cite[p.95]{garling:cliffordalgebras}. The center of a Clifford algebra $\clifford_{(p,q,r)}$ is
\begin{enumerate}
\item[(1)] $\mathcal{C}(\clifford_{(p,q,r)})=\lbrace\alpha e_0+\beta e_{12\ldots n}\mid\,\alpha,\beta\in\mathds{R}\rbrace$ if $n$ is odd,
\item[(2)] $\mathcal{C}(\clifford_{(p,q,r)})=\lbrace\alpha e_0\mid\,\alpha\in \mathds{R}\rbrace$ if $n$ is even.
\end{enumerate}
For the even part the center is
\begin{enumerate}
\item[(3)] $\mathcal{C}(\clifford_{(p,q,r)}^+)=\lbrace\alpha e_0\mid\,\alpha\in\mathds{R}\rbrace$ if $n$ is odd,
\item[(4)] $\mathcal{C}(\clifford_{(p,q,r)}^+)=\lbrace\alpha e_0+\beta e_{12\ldots n}\mid\,\alpha,\beta\in\mathds{R}\rbrace$ if $n$ is even.
\end{enumerate}
\subsection{Clifford Algebra Automorphisms}
\noindent For our purposes two automorphisms that exist on each Clifford algebra are interesting. The \emph{conjugation} is an \emph{anti-involution} denoted by an asterisk, see \cite{selig:geometricfundamentalsofrobotics}. Its effect on generators is given by $e_i^\ast=-e_i$. There is no effect on scalars. Extending the conjugation by using linearity yields
\begin{equation*}
(e_{i_1}e_{i_2}\dots e_{i_k})^\ast=(-1)^k e_{i_k}\dots e_{i_2}e_{i_1}
\end{equation*}
with $ 0\leq i_1<i_2<\dots <i_k\leq n$. The geometric product of a vector $\mathfrak{v}=\sum\limits_{i=1}^{n}{x_ie_i}\in{\bigwedge}^1 V$ with its conjugate results in
\begin{align*}
\mathfrak{v}\mathfrak{v}^\ast&=-x_1^2-x_2^2-\dots-x_p^2+x_{p+1}^2+\dots x_{p+q}^2\\
&=-b(v,v),
\end{align*}
where $v=(x_1,\dots x_n)^\mathrm{T}\in\mathbb{R}^n$.
\begin{definition}
The inverse element of a versor $\mathfrak{v}\in\clifford_{(p,q,r)}$ is defined by
\begin{equation*}
\mathfrak{v}^{-1}:=\frac{\mathfrak{v}^\ast}{N(\mathfrak{v})},
\end{equation*}
with $N(\mathfrak{v}):=\mathfrak{v}\mathfrak{v}^\ast$.
\end{definition}
\noindent
The map $N:\clifford_{(p,q,r)}\to\clifford_{(p,q,r)}$ is called the \emph{norm} of the Clifford algebra. For general multi-vectors $\mathfrak{M}\in\clifford_{(p,q,r)}$ inverse elements exist and are defined through the relation $\mathfrak{M}\mathfrak{M}^{-1}=\mathfrak{M}^{-1}\mathfrak{M}=1$, but the determination is more difficult and can be found in \cite{Fontijne:EfficientImplementationofGeometricAlgebra}. Note that in general not every element is invertible.
The other automorphism we are dealing with is the \emph{main involution}. It is denoted by $\alpha$ and defined by
\begin{equation*}
\alpha(e_{i_1}e_{i_2}\dots e_{i_k})=(-1)^k e_{i_1}e_{i_2}\dots e_{i_k}
\end{equation*}
for $ 0\leq i_1<i_2<\dots <i_k\leq n$. The main involution has no effect on the even subalgebra and it commutes with the conjugation, {\it i.e.}, $\alpha(\mathfrak{M}^\ast)=\alpha(\mathfrak{\mathfrak{M}})^\ast$ for arbitrary $\mathfrak{\mathfrak{M}}\in\clifford_{(p,q,r)}$.
\subsection{Clifford Algebra Products}
\noindent On vectors $\mathfrak{a},\mathfrak{b}\in{\bigwedge}^1 V$ we can write the inner product in terms of the geometric product
\begin{equation}\label{EQ10}
\mathfrak{a}\cdot \mathfrak{b}:=\frac{1}{2}(\mathfrak{a}\mathfrak{b}+\mathfrak{b}\mathfrak{a}).
\end{equation}
A generalization of the inner product to homogeneous multi-vectors can be found in \cite{hestenes:cliffordalgebra}.
For $\mathfrak{A}\in{\bigwedge}^k V,\ \mathfrak{B}\in{\bigwedge}^l V$ the generalized inner product is defined by
\begin{equation*}
\mathfrak{A}\cdot \mathfrak{B}:=\left[ \mathfrak{A}\mathfrak{B} \right]_{\vert k-l\vert}.
\end{equation*}
There is another product on vectors, {\it i.e.}, the outer (or exterior) product
\begin{equation}\label{EQ12}
\mathfrak{a}\wedge \mathfrak{b}:=\frac{1}{2}(\mathfrak{a}\mathfrak{b}-\mathfrak{b}\mathfrak{a}).
\end{equation}
This product can also be generalized to homogeneous multi-vectors, see again \cite{hestenes:cliffordalgebra}. For $\mathfrak{A}\in{\bigwedge}^k V,\ \mathfrak{B}\in{\bigwedge}^l V$ the generalized outer product is defined by
\begin{equation*}
\mathfrak{A}\wedge \mathfrak{B}:=\left[ \mathfrak{A}\mathfrak{B} \right]_{\vert k+l\vert}.
\end{equation*}
From equation \eqref{EQ10} and \eqref{EQ12} it follows that for vectors the geometric product can be written as the sum of the inner and the outer product
\begin{equation*}
\mathfrak{ab}=\mathfrak{a}\cdot \mathfrak{b}+\mathfrak{a}\wedge \mathfrak{b}.
\end{equation*}
More general this can be defined for multivectors with the commutator and the anti-commutator product, see \cite{perwass}. For treating geometric entities within this algebra context the definition of a $k$-blade, the \emph{inner product null space}, and its dual the \emph{outer product null space} is needed.
\begin{definition}
A $k$-blade is the $k$-fold exterior product of vectors $\mathfrak{v}\in \bigwedge^1 V$.
Therefore, a $k$-blade can be written as
\[\mathfrak{A}=\mathfrak{a}_1\wedge \mathfrak{a}_2 \wedge\dots \mathfrak{a}_k.\]
If the $k$-blade squares to zero, it is called a null $k$-blade.
The \emph{inner product null space} (IPNS) of a blade $\mathfrak{A}\in{\bigwedge}^k V$, cf. \cite{perwass}, is defined by
\begin{equation*}
\mathds{NI}(\mathfrak{A}):=\left\lbrace \mathfrak{v}\in{\bigwedge}^1 V\mid\mathfrak{v}\cdot \mathfrak{A}=0 \right\rbrace.
\end{equation*}
Moreover, the \emph{outer product null space} (OPNS) of a blade $\mathfrak{A}\in{\bigwedge}^k V$ is defined by
\begin{equation*}
\mathds{NO}(\mathfrak{A}):=\left\lbrace \mathfrak{v}\in{\bigwedge}^1 V\mid\mathfrak{v}\wedge \mathfrak{A}=0 \right\rbrace.
\end{equation*}
\end{definition}
\begin{remark}
The same set can be described with inner product or outer product null spaces. In the non-degenrate case the change between both representations is achieved with the \emph{pseudoscalar} $\mathfrak{J}=e_1\dots e_n$
\[\mathds{NO}(\mathfrak{A})=\mathds{NI}(\mathfrak{AJ}).\]
\end{remark}
\noindent Later on we work in a homogeneous Clifford algebra model, and therefore, it does not matter from which side we multiply with the pseudoscalar, since the inner product respectively outer product null space is not affected.
Moreover, multiplication with the pseudoscalar corresponds to the application of the polarity corresponding to the measure quadric. In the degenerate case the \emph{Poincar\'{e}-identity} has to be used, compare to \cite{gunn:kinematics}.
\subsection{Pin and Spin Groups}
\noindent With respect to the geometric product the units, {\it i.e.}, the invertible elements of a Clifford algebra denoted by $\clifford_{(p,q,r)}^\times$ form a group.
\begin{definition}
The \emph{Clifford group} is defined by
\begin{align*}
\Gamma(\clifford_{(p,q,r)})\!:=\!&\left\lbrace \mathfrak{g}\in\clifford_{(p,q,r)}^\times\mid \alpha(\mathfrak{g})\mathfrak{v} \mathfrak{g}^{-1}\in {\bigwedge}^1 V \right.  \\
 &\left. \mbox{ for all } \mathfrak{v}\in{\bigwedge}^1 V \right\rbrace.
\end{align*}
\end{definition}
\noindent
A proof that $\Gamma(\clifford_{(p,q,r)})$ is indeed a group with respect to the geometric product can be found in \cite{gallier:cliffordalgebrascliffordgroups}.
We define two important subgroups of the Clifford group.
\begin{definition}
The \emph{Pin group} is the subgroup of the Clifford group with $N(\mathfrak{g})=\pm 1$.
\begin{align*}
\mbox{Pin}_{(p,q,r)}\!:=\!&\left\lbrace \mathfrak{g}\in\clifford_{(p,q,r)}\mid \mathfrak{gg}^\ast\!=\!\pm 1\mbox{ and }\right.\\
 &\left.\alpha(\mathfrak{g})\mathfrak{v}\mathfrak{g}^\ast\in {\bigwedge}^1 V \mbox{ for all } \mathfrak{v}\in{\bigwedge}^1 V \right\rbrace.
\end{align*}
Furthermore, we define the \emph{Spin group} by $Pin_{(p,q,r)}\cap \clifford_{(p,q,r)}^+$
\begin{align*}
\mbox{Spin}_{(p,q,r)}\!:=\!&\left\lbrace \mathfrak{g}\in\clifford_{(p,q,r)}^+\mid \mathfrak{gg}^\ast\!=\!\pm 1\mbox{ and }\right.\\
&\left. \alpha(\mathfrak{g})\mathfrak{v}\mathfrak{g}^\ast\in {\bigwedge}^1 V \mbox{ for all } \mathfrak{v}\in{\bigwedge}^1 V \right\rbrace.
\end{align*}
\end{definition}
\begin{remark}
For non-degenerated Clifford algebras the Pin group is a double cover of the orthogonal group of the quadratic space $(V,b)$. Moreover, the Spin group is a double cover of the special orthogonal group of $(V,b)$, see \cite{gallier:cliffordalgebrascliffordgroups}.
\end{remark}

\section{The homogeneous Clifford Algebra model corresponding to Klein's Quadric}
To construct a homogeneous Clifford algebra model we use a vector space as model for the projective space which is the base space of the model. Thus, we use $\mathbb{R}^6$ as model space for $\mathbb{P}^5(\mathbb{R})$ together with the quadric form $b$ given by the corresponding symmetric matrix
\[\mathrm{B}=\begin{pmatrix}
\mathrm{O} & \mathrm{I}\\\mathrm{I} & \mathrm{O}
\end{pmatrix}\]
where $\mathrm{O}$ denotes the $3\times 3$ zero matrix and $\mathrm{I}$ the $3\times 3$ identity matrix. The $\bigwedge^1 V$ subspace is identified with the vector space, and therefore, with the projective space. The square of a general vector $\mathfrak{v}=\sum_ {i=1}^6{x_ie_i}$ results in
\[\mathfrak{vv}=x_1x_4+x_2x_5+x_3x_6.\]
This is the equation of Klein's quadric and a vector corresponds to a point on Klein's quadric if it is a null vector, {\it i.e.}, a vector squaring to zero. Since the Clifford algebra allows the description of subspaces with the use of blades of higher grade, we can express linear line manifolds within the homogeneous Clifford algebra model, see \cite{klawitter_lingeo, klawitter_diss}.
\subsection{Linear Line Manifolds}
Clifford algebras carry the subspace structure of Grassmann algebras together with the metric properties derived with the corresponding Cayley-Klein geometry. Therefore, we construct blades of grade $k$, $k\leq 5$ and examine the geometric inner product and geometric outer product null spaces.
\paragraph{Two-blades}
The exterior product of two vectors corresponds to a two-blade that describes a line $P_1^1$ in $\mathbb{P}^5(\mathbb{R})$. We use the outer product null space to determine the set of points on $P_1^1$ contained in $M_2^4$, and therefore, the set of lines in $\mathbb{P}^3(\mathbb{R})$ corresponding to the two-blade. If the whole line $P_1^1$ is contained in $M_2^4$ the corresponding two-blade is a null two-blade. Thus, the outer product null space corresponds to a pencil of lines in $\mathbb{P}^3(\mathbb{R})$ or to two lines in $\mathbb{P}^3(\mathbb{R})$ that may be identical. The inner product null space of a two-blade corresponds to a three-space intersection with Klein's quadric, {\it i.e.}, a linear line congruence in $\mathbb{P}^3(\mathbb{R})$.
\paragraph{Three-blades}
Three-blades can be generated as the exterior product of three vectors and describe two-spaces in $\mathbb{P}^5(\mathbb{R})$. The outer product null space of a three-blade corresponds to the two-space intersection of the two-space $P_1^2$ described by the three-blade with $M_2^4$. If the three-blade is a null three-blade $P_1^2$ is completely contained in Klein's quadric and the outer product null space corresponds to a bundle or a field of lines in $\mathbb{P}^3(\mathbb{R})$. The outer product null space of a non-null three-blade corresponds to a conic section of $M_2^4$ that defines a regulus in $\mathbb{P}^3(\mathbb{R})$. In a dual way the inner product null space of a three-blade is the intersection of the image of $P_1^2$ under the polarity corresponding to the meassure quadric with Klein's quadric.
\paragraph{Four-blades}
Outer product null spaces of four-blades correspond to three-space intersections with Klein's quadric and the corresponding set of lines in $\mathbb{P}^3(\mathbb{R})$ is a linear congruence of lines. The dual of a three-space in $\mathbb{P}^5(\mathbb{R})$ is a line. Thus, the inner product null space of a four-blade is the intersection of this line with $M_2^4$.
\paragraph{Five-blades}
Five-blades describe four-spaces in Klein's model and the intersection of a four-space with Klein's quadric results in a linear congruence of lines. The outer product null space of a five-blade corresponds to a linear congruence of lines. Moreover, the inner produt null space of a five-blade yields a vector.
\subsection{Clifford Group Action}\label{sectionCliffordgroup}
The null vectors correspond to points on Klein's quadric. Moreover, the sandwich action of an arbitrary versor on a null vector results a null vector again. Thus Klein's quadric is mapped to itself. To examine the action of the whole Clifford group, we first investigate the action of non-null vectors on null vectors.
\begin{remark}\label{remarksandwich}
Since we are working in a homogeneous Clifford algebra model multiplication with a real factor does not change the geometric meaning of an algebra element. Therefore, we use the conjugate element instead of the inverse element. Hence, the sandwich operator that we use is given by:
\[\alpha(\mathfrak{a})\mathfrak{va}^\ast=\mathfrak{aa}^\ast(\alpha(\mathfrak{a})\mathfrak{va}^{-1}).\]
This operator does not involve an inverse, and therefore, it can also be applied if the element $\mathfrak{a}$ is not invertible.
\end{remark}
\paragraph{Sandwich action of vectors}
Let $\mathfrak{a}=\sum_{i=1}^6{a_ie_i}$ and $\mathfrak{v}=\sum_{i=1}^6{x_ie_i}$

be two vectors with $\mathfrak{aa}\neq0$ and $\mathfrak{vv}=0$.

The action of the sandwich operator $\alpha(\mathfrak{a})\mathfrak{va}^\ast$ on the $\bigwedge^1 V$ subspace can be expressed as product of a matrix with a vector. The matrix has the form:
{\arraycolsep4pt
\begin{equation}
\mathrm{M}=\begin{pmatrix}
  k_1 & a_1a_5   & a_1a_6 & a_1a_1 & a_1a_2 & a_1a_3\\
 a_2a_4 & k_2    & a_2a_6 & a_2a_1 & a_2a_2 & a_2a_3\\
 a_3a_4 & a_3a_5 & k_3    & a_3a_1 & a_3a_2 & a_3a_3 \\
 a_4a_4 & a_4a_5 & a_4a_6 & k_4    & a_4a_2 & a_4a_3\\
 a_5a_4 & a_5a_5 & a_5a_6 & a_5a_1 & k_5    & a_5a_3\\
 a_6a_4 & a_6a_5 & a_6a_6 & a_6a_1 & a_6a_2 &  k_6
\end{pmatrix},
\end{equation}}
with
\begin{align*}
k_1&=-\!a_5a_2\!-\!a_6a_3,\quad &&k_2=-\!a_6a_3\!-\!a_4a_1,\\
k_3&=-\!a_4a_1\!-\!a_5a_2,\quad &&k_4=-\!a_5a_2\!-\!a_6a_3,\\
k_5&=-\!a_6a_3\!-\!a_4a_1,\quad &&k_6=-\!a_4a_1\!-\!a_5a_2.
\end{align*}
This involutoric automorphism of Klein's quadric corresponds to a null polarity acting on $\mathbb{P}^3(\mathbb{R})$, see \cite{klawitter_lingeo}. Furthermore, the Clifford group is generated by invertible  vectors, and therefore, all elements of the Clifford group correspond to compositions of null polarities in $\mathbb{P}^3(\mathbb{R})$. Moreover, this means that every element of $\clifford_{(3,3)}^+$ that is an even product of vectors corresponds to a collineation and every element of $\clifford_{(3,3)}^-$ that is an odd product of vectors corresponds to a correlation.

\paragraph{Collineations}
A general element $\mathfrak{g}\in\clifford_{(3,3)}^+$ corresponding to a collineation is given by
\begin{align*}
\mathfrak{g}&=g_{1}e_0\!+\!g_{2}e_{12}\!+\!g_{3}e_{13} \!+\!g_{4}e_{14}\!+\!g_{5}e_{15}\!+\!g_{6}e_{16}\\
&+\!g_{7}e_{23}\!+\!g_{8}e_{24}\!+\!g_{9}e_{25}\!+\!g_{10}e_{26}\!+g_{11}e_{34}\\
&+\!g_{12}e_{35}\!+\!g_{13}e_{36}\!+\!g_{14}e_{45}\!+\!g_{15}e_{46}\!+\!g_{16}e_{56}\\
&+\!g_{17}e_{1234}\!+\!g_{18}e_{1235}\!+\!g_{19}e_{1236}\!+\! g_{20}e_{1245}\\&
+g_{21}e_{1246}\!+\!g_{22}e_{1256} \!+\!g_{23}e_{1345}\!+\!g_{24}e_{1346}\\
&+g_{25}e_{1356}\!+\!g_{26}e_{1456}\!+\!g_{27}e_{2345}\!+\!g_{28}e_{2346}\\
&+g_{29}e_{2356}\!+\!g_{30}e_{2456}\!+\!g_{31}e_{3456}\!+\!g_{32}e_{123456}.
\end{align*}
We derive constraints to this element by $\alpha(\mathfrak{g})\mathfrak{vg}^\ast\in\bigwedge^1 V$ for all $\mathfrak{v}\in\bigwedge^1 V$. If we try to compute the corresponding collineation we have to distinguish the type of the collineation, {\it i.e.}, if it acts on points or on planes.
\paragraph{Action on Points}
First we assume $\mathfrak{g}\in\clifford_{(3,3)}^+$ corresponds to a collineation acting on points.
The automorphic collineation of Klein's quadric induced by $\mathfrak{g}$ can be transferred to a collineation that acts on $\mathbb{P}^3(\mathbb{R})$. Therefore, the element $g\in\clifford_{(3,3)}^+$ is applied to a null three-blade corresponding to a bundle of lines under the constraint that $\mathfrak{g}$ is a versor, see \cite{klawitter_lingeo}. This results in the matrix $(m_{i,j}), i,j=0,\dots,3$ for the collineation with
\begin{align*}
m_{00}&=g_{1}\!-\!g_{20}\!-\!g_{24}\!-\!g_{32}\!-\!g_{29}\!+\!g_{9}\!+\!g_{4}\!+\!g_{13},\\
m_{11}&=g_{24}\!-\!g_{9}\!+\!g_{20}\!-\!g_{13}\!-\!g_{32}\!+\!g_{1}\!+\!g_{4}\!-\!g_{29},\\
m_{22}&=g_{1}\!-\!g_{13}\!-\!g_{32}\!-\!g_{4}\!+\!g_{29}\!+\!g_{9}\!-\!g_{24}\!+\!g_{20},\\
m_{33}&=g_{24}\!+\!g_{13}\!+\!g_{29}\!+\!g_{1}\!-\!g_{4}\!-\!g_{9}\!-\!g_{20}\!-\!g_{32},
\end{align*}
and
\begin{align*}
m_{01}&=2(g_{7}+g_{17}),&&m_{02}=2(g_{18}\!-\!g_{3}),\\
m_{03}&=2(g_{19}\!+\!g_{2}),&&m_{10}=-\!2(g_{26}\!+\!g_{16}),\\
m_{12}&= 2(g_{5}\!+\!g_{25}), &&m_{13}=2(g_{6}\!-\!g_{22}),\\
m_{20}&= 2(g_{15}\!-\!g_{30}),&&m_{21}=2(g_{8}\!+\!g_{28}),\\
m_{23}&=2(g_{21}\!+\!2g_{10}),&&m_{30}=-\!2(g_{31}\!+\!g_{14}),\\
m_{31}&=2(g_{11}\!-\!g_{27}), &&m_{32}=2(g_{23}\!+\!g_{12}).
\end{align*}
The system of equations
\begin{equation}\label{systemcollineation}
(m_{i,j})=(c_{i,j}),\ i,j=0,\dots,3
\end{equation}
can be used to compute an element of $\clifford_{(3,3)}^+$ corresponding to a collineation acting on points with matrix representation $\mathrm{C}=(c_{i,j}),\ i,j=0,\dots,3$. Therefore, constraint equations derived from $\alpha(\mathfrak{g})\mathfrak{vg}^\ast\in\bigwedge^1 V$ and $\mathfrak{gg}^\ast=\pm 1$ are used.\\

\paragraph{Action on Planes} Moreover, the same automorphic collineation of Klein's quadric may correspond to a collineation in $\mathbb{P}^3(\mathbb{R})$ that maps planes to planes. This action can also be represented as matrix vector product. Therefore, the general element $\mathfrak{g}\in\clifford_{(3,3)}^+$ is applied to a null three-blade corresponding to a field of lines. The action can be expressed as collineation with matrix $(m_{i,j}),\,i,j,=0,\dots,3$
\begin{align*}
m_{00}&=g_{32}\!-\!g_{20}\!-\!g_{13}\!-\!g_{29}\!-\!g_{9}\!-\!g_{24}\!+\!g_{1}\!-\!g_{4},\\
m_{11}&=g_{1}\!+\!g_{9}\!+\!g_{20}\!+\!g_{24}\!+\!g_{13}\!-\!g_{29}\!+\!g_{32}\!-\!g_{4},\\
m_{22}&=g_{20}\!+\!g_{29}\!+\!g_{4}\!+\!g_{13}\!-\!g_{9}\!-\!g_{24}\!+\!g_{1}\!+\!g_{32},\\
m_{33}&=g_{9}\!+\!g_{24}\!+\!g_{29}\!-\!g_{13}\!+\!g_{1}\!+\!g_{4}\!-\!g_{20}\!+\!g_{32},
\end{align*}
and
\begin{align*}
m_{01}&=2(g_{16}-g_{26}),&&m_{02}=-2(g_{15}+g_{30}),\\
m_{03}&=2(g_{14}-g_{31}),&&m_{10}=2(g_{17}-g_{7}),\\
m_{12}&=2(g_{28}-g_{8}), &&m_{13}=-2(g_{27}+g_{11}),\\
m_{20}&=2(g_{3}+g_{18}) ,&&m_{21}=2(g_{25}-g_{5}),\\
m_{23}&=2(g_{23}-g_{12}),&&m_{30}=2(g_{19}-g_{2}),\\
m_{31}&=-2(g_{22}+g_{6}), &&m_{32}=2(g_{21}-g_{10}).
\end{align*}
Depending on the action of the collineation, {\it i.e.}, if it acts on points or planes the elements of the Spin group can be transferred to their matrix representations and vice versa.
\paragraph{Correlations}
General elements of the odd part $\clifford_{(3,3)}^-$ have the form
\begin{align*}
\mathfrak{h} &= h_{1}e_{1}\!+\!h_{2}e_{2}\!+\!h_{3}e_{3}\!+\!h_{4}e_{4}\!+\!h_{5}e_{5}\!+\!h_{6}e_{6}\\
&+h_{7}e_{123}\!+\!h_{8}e_{124}\!+\!h_{9}e_{125}\!+\!h_{10}e_{126}\!+\!h_{11}e_{134}\!\\
&+h_{12}e_{135}\!+\!h_{13}e_{136}\!+\!h_{14}e_{145}\!+\!h_{15}e_{146}\!+\!h_{16}e_{156}\\
&+h_{17}e_{234}\!+\!h_{18}e_{235}\!+\!h_{19}e_{236}\!+\!h_{20}e_{245}\!+\!h_{21}e_{246}\\
&+h_{22}e_{256}\!+\!h_{23}e_{345}\!+\!h_{24}e_{346}\!+\!h_{25}e_{356}\!+\!h_{26}e_{456}\\
&+h_{27}e_{12345}\!+\!h_{28}e_{12346}\!+\!h_{29}e_{12356}\!+\!h_{30}e_{12456}\\
&+h_{31}e_{13456}\!+\!h_{32}e_{23456}.
\end{align*}
Their action on points or planes of the three-dimensional projective space $\mathbb{P}^3(\mathbb{R})$ can be described by the sandwich action on null three-blades corresponding to two-spaces contained entirely in $M_2^4$ that correspond to bundles of lines or fields of lines.
\paragraph{Action on Planes}
The action of a general element $\mathfrak{h}\in\clifford_{(3,3)}^-$ applied to a null three-blade corresponding to a two-space contained in Klein's quadric that is the image of a filed of lines under the Klein map can be expressed as $4\times 4$ matrix $(m_{i,j}), i,j=0,\dots,3$ that determines a correlation with
$m_{00}\!=\!-2h_{7},\ m_{11}\!=\!-2h_{16},\ m_{22}\!=\!2h_{21},\ m_{33}\!=\!-2h_{23},$
and

\begin{align*}
m_{01}&=h_{9}+h_{13}-h_{29}+h_{1},\\
m_{02}&=h_{2}-h_{8}+h_{28}+h_{19},\\
m_{03}&=h_{3}-h_{27}-h_{18}-h_{11},\\
m_{10}&=h_{29}+h_{13}+h_{9}-h_{1},\\
m_{12}&=h_{6}-h_{22}+h_{15}+h_{30},\\
m_{13}&=h_{31}-h_{25}-h_{5}-h_{14},\\
m_{20}&=h_{19}-h_{8}-h_{2}-h_{28},\\
m_{21}&=h_{15}-h_{30}-h_{6}-h_{22},\\
m_{23}&=h_{4}-h_{20}+h_{32}+h_{24},\\
m_{30}&=h_{27}-h_{11}-h_{18}-h_{3},\\
m_{31}&=h_{5}-h_{31}-h_{25}-h_{14},\\
m_{32}&=h_{24}-h_{4}-h_{32}-h_{20}.
\end{align*}
\paragraph{Action on Points} The same element $\mathfrak{h}\in\clifford_{(3,3)}^-$ can be interpreted as correlation in $\mathbb{P}^3(\mathbb{R})$ that acts on points and maps them to planes. This action can be described by a product of a matrix $m_{(i,j)},i,j=0,\dots,3$ with a vector. This matrix can be expressed with the coefficients of $\mathfrak{h}\in\clifford_{(3,3)}^-$ and has the entries $m_{00}\!=\!2h_{26},\ m_{11}\!=\!2h_{17},\ m_{22}\!=\!-2h_{12},\ m_{33}\!=\!2h_{10},$
and

\begin{align*}
m_{01}&=h_{32}-h_{4}-h_{20}-h_{24},\\
m_{02}&=h_{14}-h_{31}-h_{25}-h_{5},\\
m_{03}&=h_{30}+h_{15}+h_{22}-h_{6},\\
m_{10}&=h_{4}-h_{32}-h_{24}-h_{20},\\
m_{12}&=h_{18}-h_{27}-h_{3}-h_{11},\\
m_{13}&=h_{2}+h_{8}+h_{19}-h_{28},\\
m_{20}&=h_{31}+h_{14}-h_{25}+h_{5},\\
m_{21}&=h_{3}-h_{11}+h_{27}+h_{18},\\
m_{23}&=h_{9}-h_{13}-h_{1}-h_{29},\\
m_{30}&=h_{15}-h_{30}+h_{6}+h_{22},\\
m_{31}&=h_{8}-h_{2}+h_{28}+h_{19},\\
m_{32}&=h_{1}-h_{13}+h_{29}+h_{9}.
\end{align*}
Depending on the action of the correlation, {\it i.e.}, if it acts on points or planes the elements of the Pin group can be transferred to their matrix representations and vice versa.
\section{A Factorization Algorithm and its Application}
At this point we recall a theorem from \cite{klawitter_lingeo}.
\begin{theorem}\label{theoremKlein}
Every regular collineation or correlation can be expressed as the product of six null polarities at the most.
\end{theorem}
\noindent For a proof we refer to \cite{klawitter_lingeo} or \cite{klawitter_diss}. With the help of a modified version of a factorization algorithm introduced in \cite[p. 107]{perwass} we factorize arbitrary regular $4\times 4$ matrices into the product of six skew symmetric matrices at the most. Let $\left[  \mathfrak{g}\right]_k$ denote the grade-$k$ part of the versor $\mathfrak{g}$. For example a versor $\mathfrak{g}$ with maximal grade four can be written as $\mathfrak{g}=\left[  \mathfrak{g}\right]_0+\left[  \mathfrak{g}\right]_2+\left[  \mathfrak{g}\right]_4$.The part of maximal grade $\left[  \mathfrak{g}\right]_{\mathrm{max}}$ is always a blade, see \cite{perwass}. The non-null vector $\mathfrak{v}\in\bigwedge^1 V$ that is contained in
$\mathds{NO}(\left[  \mathfrak{g}\right]_{\mathrm{max}})$ satisfies
\begin{equation}\label{gradereduction}
\mathfrak{vg}_{\mathrm{max}}=\mathfrak{v}\cdot\mathfrak{g}_{\mathrm{max}}.
\end{equation}
Hence, $\mathfrak{g}'=\mathfrak{gv}^{-1}$ reduces the grade of the maximum blade of $\mathfrak{g}$ by one. With remark \ref{remarksandwich} this can be simplified to $\mathfrak{g}'=\mathfrak{gv}^\ast$ for homogeneous Clifford algebra models. Moreover, for vectors $\mathfrak{v}\in\bigwedge^1 V$ we have $\mathfrak{v}^\ast=-\mathfrak{v}$, and therefore, we can use  $\mathfrak{g}'=\mathfrak{gv}$ to reduce the degree of the initial versor $\mathfrak{g}$. The result of this product is again a versor. Repeated application of this process results in a set of vectors $\mathfrak{v}_1,\dots,\mathfrak{v}_{\mathrm{max}}$ whose geometric product results in the versor $\mathfrak{g}$ except for a real factor. Note that this procedure may fail for null versors. We show this algorithm in an example
\begin{example}\label{ex1}
Let $\mathrm{K}\in \mathrm{PGL}(\mathbb{P}^3(\mathbb{R}))$ be a collineation of $\mathbb{P}^3(\mathbb{R})$ that acts on points
\[K=\begin{pmatrix}
1 & 0 & 3 & 0\\1 & 1 & 0 & 1\\1 & 2 & 1 & 0\\1 & 1 & 2 & 1
\end{pmatrix}.\]
To get a versor $\mathfrak{g}\in\clifford_{(3,3)}^+$ corresponding to this collineation we have to solve the system \eqref{systemcollineation}.
\begin{align*}
&2(g_{7}\!+\!g_{17})\!=\!0, & 2(g_{18}\!-\!g_{3})\!=\!3,\\
&2(g_{19}\!+\!g_{2})\!=\!0,&-2(g_{26}\!+\!g_{16})\!=\!1,\\
&2(g_{5}\!+\!g_{25})\!=\!0, & 2(g_{6}\!-\!g_{22})\!=\!1,\\
&2(g_{15}\!-\!g_{30})\!=\!1, &2(g_{8}\!+\!g_{28})\!=\!2,\\
&2(g_{21}\!+\!2g_{10})\!=\!0, &-2(g_{31}\!+\!g_{14})\!=\!1,\\
& 2(g_{11}\!-\!g_{27})\!=\!1, &2(g_{23}\!+\!g_{12})\!=\!2,
\end{align*}

\begin{align*}
&g_{1}\!-\!g_{20}\!-\!g_{24}\!-\!g_{32}\!-\!g_{29}\!+\!g_{9}\!+\!g_{4}\!+\!g_{13}\!=\!1,\\
&g_{24}\!-\!g_{9}\!+\!g_{20}\!-\!g_{13}\!-\!g_{32}\!+\!g_{1}\!+\!g_{4}\!-\!g_{29}\!=\!1,\\
&g_{1}\!-\!g_{13}\!-\!g_{32}\!-\!g_{4}\!+\!g_{29}\!+\!g_{9}\!-\!g_{24}\!+\!g_{20}\!=\!1,\\
&g_{24}\!+\!g_{13}\!+\!g_{29}\!+\!g_{1}\!-\!g_{4}\!-\!g_{9}\!-\!g_{20}\!-\!g_{32}\!=\!1.
\end{align*}
There are two possibilities to guarantee that the resulting versor is in the Spin group, {\it i.e.}, $\mathfrak{gg}^\ast=1$ or $\mathfrak{gg}^\ast=-1$. We compute both solutions and start with the constraint equations implied by eq. $\alpha(\mathfrak{g})\mathfrak{vg}^\ast\in\bigwedge^1 V$ for all $\mathfrak{v}\in\bigwedge^1 V$ and $\mathfrak{gg}^\ast=1$.
The corresponding Spin group element has the form:
\begin{align*}
\mathfrak{g_+}&={1\over 8\sqrt{2}} \big(7 e_{0}\! +\!6 e_{12}\!-\!6 e_{13}\! +\! e_{14}\!- \!2 e_{15}\!- \!6 e_{23}\\
&+6 e_{24}\! - \!e_{25}\! -\!2 e_{26}\! +\!2 e_{34}\!+\!6 e_{35}\!-\!5 e_{36}\! -\!4 e_{45}\\
&+2 e_{46}\! +\!6 e_{1234}\! -\! 4 e_{56}\! +\!6 e_{1235}\! -\!6 e_{1236}\\
&-5 e_{1245}\!+\!2 e_{1246}\!-\!4 e_{1256}\! +\!2 e_{1345}\!-\!e_{1346}\\
&+2 e_{1356}\!-\!2 e_{2345}\!+\!2 e_{2346} \!+\! e_{2356}\!-\!2 e_{2456} \\
&-e_{123456}\big).
\end{align*}
If we demand that $\mathfrak{gg}^\ast=-1$ the resulting Spin group element is computed as
\begin{align*}
\mathfrak{g}_-&={1\over 8\sqrt{2}}\big(e_{0}\!-\!6 e_{12}\!-\!6 e_{13}\!-\!e_{14}\!+\!2 e_{15}\!+\!4 e_{16}\\
&+6 e_{23}\!+\!2 e_{24}\!+\!e_{25}\!+\!2 e_{26}\!+\!2 e_{34}\!+\!2 e_{35}\!+\!5 e_{36}\\
&+2 e_{46}\!-\!6 e_{1234}\!+\!6 e_{1235}\!+\!6 e_{1236}\!+\!5 e_{1245}\\
&-2 e_{1246}\!+\!6 e_{1345}\!+\!e_{1346}\!-\!2 e_{1356}\!-\!4 e_{1456}\\
&-2 e_{2345}\!+\!6 e_{2346}\!-\!e_{2356}\!-\!2 e_{2456}\!-\!4 e_{3456}\\
&-7 e_{123456}\big).
\end{align*}
Both elements $\mathfrak{g}_+$ and $\mathfrak{g}_-$ correspond to the same collineation whose entries can be computed with the coefficients of $\mathfrak{g}_+$ and $\mathfrak{g}_-$ and $(m_{i,j}),i,j=0,\dots, 3$.
\end{example}
\noindent Depending on the matrix of the projective transformation, {\it i.e.}, if it describes a collineation or correlation that acts on points or planes the corresponding versor can be computed with the use of the corresponding systems derived in the last section.
\begin{remark}
The matrix
\[K=\begin{pmatrix}
-1 & 0 & 3 & 0\\1 & 1 & 0 & 1\\1 & 2 & 1 & 0\\1 & 1 & 2 & 1
\end{pmatrix}\]
results in a versor with complex entries. Thus, we have to work over the complex numbers.
\end{remark}
\noindent Moreover, it can be verified that $\mathfrak{g}_+\corr \mathfrak{Jg}_-$, where $\corr$ means is equal up to a scalar factor. Multiplication with the pseudoscalar does not change the action of the versor since
\begin{align*}
(\mathfrak{Jg})\mathfrak{v}(\mathfrak{Jg})^\ast&=\alpha(\mathfrak{J})\alpha(\mathfrak{g})\mathfrak{v}\mathfrak{g}^\ast\mathfrak{J}^\ast\\
&=\mathfrak{J}\mathfrak{J}^\ast\alpha(\mathfrak{g})\mathfrak{v}\mathfrak{g}^\ast\\
&\corr \alpha(\mathfrak{g})\mathfrak{v}\mathfrak{g}^\ast.
\end{align*}
\begin{remark}
Multiplication with an element of the center $\mathcal{C}(\clifford_{(3,3)})=a e_0 + b e_{123456},\ a,b\in\mathbb{C}$ has no effect on the sandwich action of a versor. Therefore, we have a group isomorphism
\begin{large}
\[\nicefrac{\mathrm{PGL(4,\mathbb{C})}}{\mathcal{C}(\mathrm{PGL(4,\mathbb{C})})}\to\nicefrac{\mathrm{Pin}(\clifford_{(3,3)})}{\mathcal{C}(\clifford_{(3,3)})}\]
\end{large}
\end{remark}
\noindent Now we factorize the element $\mathfrak{g}_+$ into vectors that correspond to null polarities. Due to the homogeneous setting we start with the numerator of $\mathfrak{g}_+$ of ex. \ref{ex1} denoted by $\mathfrak{g}\ (=8\sqrt{2}\mathfrak{g}_+)$.
\begin{example}
To factorize $\mathfrak{g}$ we first observe that $\left[ \mathfrak{g}\right]_\mathrm{max}=\left[ \mathfrak{g}\right]_6=-e_{123456}$. Multiplication with a scalar does not change the outer product null space of a $k$-blade, and therefore, it follows
\[\mathds{NO}(\left[ \mathfrak{g}\right]_\mathrm{max})=\mathds{NO}(e_{123456})=\lbrace\mathfrak{v}\in{\bigwedge}^1 V\rbrace.\]
We can now chose an arbitrary non-null vector $\mathfrak{v}_1\in\mathds{NO}(\left[ \mathfrak{g}\right]_\mathrm{max})$. Let $\mathfrak{v}_1=e_1\!+\!e_4$
be the first factor. The versor $\mathfrak{g}_1\!=\!\mathfrak{gv}_1$ has maximal grade five and the grade-$5$ part is given by
\[\left[ \mathfrak{g}_1\right]_5\! \corr \!e_{23456}\!-\!4e_{12345}\!+\!4e_{12346}\!-\!3e_{12456}\!+\!e_{13456}.\]
The outer product null space of $\left[ \mathfrak{g}_1\right]_\mathrm{5}$ is computed as
\begin{equation*}
\mathds{NO}(\left[ \mathfrak{g}_1\right]_5)=\lbrace\mathfrak{a}\in{\bigwedge}^1 V \mid \mathfrak{a \wedge \left[ \mathfrak{g}_1\right]_5}=0\rbrace.
\end{equation*}
This results in the set of all vectors $\mathfrak{a}=\sum_{i=1}^6{a_ie_i}$
satisfying $a_1\!-\!a_2\!-\!3a_3\!+\!4a_5\!+\!4a_6\!=\!0$.
We chose the non-null vector $\mathfrak{v}_2=4 e_2\!+\! e_5\in\mathds{NO}(\left[ \mathfrak{g}_1\right]_5)$
and compute the grade-four part of $\mathfrak{g}_2\!=\!\mathfrak{g}_1\mathfrak{v}_2$
\begin{align*}
\left[ \mathfrak{g}_2\right]_4&\corr e_{2356}\!-\!12e_{1234}\!+\!5e_{2345}\!-\!4e_{1236}\!+\!4e_{1235}\\
&+\!e_{1245}\!+\!e_{1345}\!-\!8e_{2346}\!+\!4e_{2456}\!-\!e_{3456}\\
&+8e_{1246}\!-\!3e_{1256}\!-\!4e_{1346}\!+\!e_{1356}\!+\!e_{1456}.
\end{align*}
Again we determine a vector contained in the outer product null space of $\left[ \mathfrak{g}_2\right]_4$. This null space is three-dimensional and spanned by the vectors
\begin{align*}
\mathfrak{b}_1&=e_1\!+\!5 e_2 \!+\! e_6, &&\mathfrak{b}_2=4e_1\!+\!8e_2\!+\!e_5\\
\mathfrak{b}_3&=e_1\!+\!e_2\!+\!e_4, &&\mathfrak{b}_4=e_1\!+\!4e_2\!-\!e_3.
\end{align*}
We chose the non-null vector $\mathfrak{b}_3$ and set $\mathfrak{v}_3=\mathfrak{b}_3$. In the next step we have to compute the outer product null space of $\left[ \mathfrak{g}_3\right]_3$ with $\mathfrak{g}_3=\mathfrak{g}_2\mathfrak{v}_3$ that is two-dimensional and spanned by:
\begin{align*}
\mathfrak{b}_1&=e_1\!+\!4e_2\!-\!3e_3,&&\mathfrak{b}_2=3e_1\!+\!7e_2\!-\!e_4\!+\!e_5,\\
\mathfrak{b}_3&=4e_2\!-\!e_4\!+\!e_6
\end{align*}
To keep the factors simple we set
\[\mathfrak{v}_4=\mathfrak{b}_1+\mathfrak{b}_3=-e_1+e_3-e_4+e_6.\]
The reduced versor $\mathfrak{g}_4=\mathfrak{g}_3\mathfrak{v}_4$ has maximal grade two and the outer product null space of the grade-two part is spanned by the two vectors:
\begin{align*}
\mathfrak{b}_1&=2e_1\!+\!3e_2\!+\!e_3\!-\!e_4\!+\!e_5,\\
\mathfrak{b}_2&=4e_2\!-\!e_4\!+\!e_6.
\end{align*}
Note that the vector $\mathfrak{b}_2$ is a null vector, and therefore, we chose $\mathfrak{v}_5=\mathfrak{b}_1$ as the next factor. The last factor $\mathfrak{v}_6$ is obtained by
\begin{align*}
\mathfrak{v}_6&=\mathfrak{g}_4\mathfrak{v}_5\\
&=8e_1\!+\!8e_2\!+\!4e_3\!-\!3e_4\!+\!4e_5\!-\!e_6.
\end{align*}
This results in the factorization
\[\mathfrak{v}_6\mathfrak{v}_5\mathfrak{v}_4\mathfrak{v}_3\mathfrak{v}_2\mathfrak{v}_1\corr\mathfrak{g}.\]
The next step is to carry over the factorization to projective transformations. Each vector corresponds to a null polarity. Therefore, we aim at a matrix product of the form
\[\mathrm{M}_6\mathrm{M}_5\mathrm{M}_4\mathrm{M}_3\mathrm{M}_2\mathrm{M}_1\corr\mathrm{K}.\]
Since we started with a collineation that maps points to points we have to ensure that the vectors are transferred to projective mappings of the right type. For example the product $\mathrm{M}_2\mathrm{M}_1$ must be the matrix corresponding to a collineation that maps points to points. Hence, $\mathrm{M}_1$ has to correspond to a null polarity acting on points and $\mathrm{M}_2$ to a null polarity acting on planes. Thus, the matrix product has to be understand as
\[\mathrm{M}_6\mathrm{M}_5^\ast\mathrm{M}_4\mathrm{M}_3^\ast\mathrm{M}_2\mathrm{M}_1^\ast\corr\mathrm{K},\]
where $\mathrm{M}_i^\ast,\ i=1,3,5$ is a null polarity acting on points and $\mathrm{M}_i,\ i=2,4,6$ is a null polarity acting on planes. We compute the matrices $\mathrm{M}_i^\ast$ and $\mathrm{M}_i$ with the matrix representations computed in section \ref{sectionCliffordgroup}. This results in the six null polarities:
{\arraycolsep3pt
\begin{align*}
\mathrm{M}_1^\ast&\!\!=\!\!\begin{small}\begin{pmatrix} 0 & \!-\!1 & 0 & 0 \\1 & 0 & 0 & 0 \\0 & 0 & 0 & \!-\!1 \\0 & 0 & 1 & 0 \end{pmatrix}\end{small}\!\!\!,
&&\mathrm{M}_2\!\!=\!\!\begin{small}\begin{pmatrix} 0 & 0 & 4 & 0 \\0 & 0 & 0 & \!-\!1 \\\!-4 & 0 & 0 & 0 \\0 & 1 & 0 & 0\end{pmatrix}\end{small}\!\!\!,\\
\mathrm{M}_3^\ast&\!\!=\!\!\begin{small}\begin{pmatrix} 0 & \!-\!1 & 0 & 0 \\1 & 0 & 0 & 1 \\0 & 0 & 0 & \!-\!1 \\0 & \!-\!1 & 1 & 0\end{pmatrix}\end{small}\!\!\!,
&&\mathrm{M}_4\!\!=\!\!\begin{small}\begin{pmatrix} 0 & \!-\!1 & 0 & 1 \\1 & 0 & 1 & 0 \\0 & \!-\!1 & 0 & \!-\!1 \\\!-\!1 & 0 & 1 & 0\end{pmatrix}\end{small}\!\!\!,\\
\mathrm{M}_5^\ast&\!\!=\!\!\begin{small}\begin{pmatrix} 0 & 1 & \!-\!1 & 0 \\\!-\!1 & 0 & \!-\!1 & 3 \\1 & 1 & 0 & \!-\!2 \\0 & \!-\!3 & 2 & 0\end{pmatrix}\end{small}\!\!\!,
&&\mathrm{M}_6\!\!=\!\!\begin{small}\begin{pmatrix} 0 & 8 & 8 & 4 \\\!-\!8 & 0 & \!-\!1 & \!-4 \\\!-\!8 & 1 & 0 & \!-\!3 \\\!-4 & 4 & 3 & 0\end{pmatrix}\end{small}\!\!\!.
\end{align*}}

\noindent With these six skew symmetric matrices it can be verified that
\[\mathrm{M}_6\mathrm{M}_5^\ast\mathrm{M}_4\mathrm{M}_3^\ast\mathrm{M}_2\mathrm{M}_1^\ast=-{1\over 4}\mathrm{K}.\]
Thus, we have found one possible factorization of the initial projective mapping into six null polarities.
\end{example}
\noindent It is worth to check the maximal grade of a versor that is solution of the other systems that were derived in section \ref{sectionCliffordgroup} if a factorization of a $4\times 4$ matrix is searched. The maximal grade of the versor determines the number of skew-symmetric matrices that are necessary to factorize the given matrix and for correlations the maximal possible grade is five.
\section{The homogeneous Clifford Algebra Model of Lie's Sphere Geometry}

The same construction that we described for Klein's quadric can be applied to any quadric. As a second example that shall demonstrate the power of this calculus we examine Lie sphere geometry in a Clifford algebra context.
Lie sphere geometry is the geometry of oriented spheres. Especially, for the three-dimensional case the set of oriented spheres can be mapped  to a hyperquadric $L^4_1$ in five-dimensional projective space $\mathbb{P}^5(\mathbb{R})$. The construction goes back to S. \textsc{Lie} and was treated again by W. \textsc{Blaschke}, cf. \cite{Blaschke:VorlesungenuberDifferentialgeometrieundGeometrischeGrundlagenvonEinsteinsRelativitatstheorieBd.3}. A modern treatment of this topic can be found in \cite{Cecil:Liespheregeometry}. Moreover, the Lie construction can be achieved for arbitrary dimension.

\subsection{Lie's Quadric}\index{Lie's quadric}\label{subsectionLiequadrik}
A point model for the set of oriented hyperspheres, hyperplanes, and points (considered as spheres of radius $0$) of $\mathbb{R}^n$ is given by the projective hyperquadric
\[L_1^{n+1}:\,-x_0^2+x_1^2+\ldots+x_{n+1}^2-x_{n+2}^2=0.\]

For our purposes it is convenient that we restrict ourselves to the case of oriented spheres in three-dimensional Euclidean space. Nevertheless, we formulate the calculus for arbitrary dimensions. The quadric $L^{n+1}_1\subset\mathbb{P}^{n+2}(\mathbb{R})$ is of dimension $n\!+\!1$, degree $2$, and is called Lie's quadric. The maximal dimension of subspaces contained by $L_1^{n+1}$ is $1$, and therefore, there are no two-spaces contained entirely in $L_1^{n+1}$. Oriented hyperspheres, hyperplanes, and points are represented in Lie coordinates as shown in Table \ref{talbeliecoords}.

It is not difficult to recover the Euclidean representation from Lie coordinates. If $x_0+x_1=0$ and if $x_{n+2}=0$ we have the point at infinity. If $x_{n+2}\neq 0$ we bring the point to the form $ \left( h,-h,N,1\right)^\mathrm{T}\mathbb{R} $ by dividing by $x_{n+2}$. If $x_0+x_1\neq 0$ and if $x_{n+2}=0$, we have a proper point. We obtain its normal form by dividing by $x_0+x_1$. The last case is if $x_{n+2}\neq 0$. In this case we have an oriented sphere. Again we get its normal form through division by $x_0+x_1$.\\

{\linespread{1.4}
\begin{table*}[hbt]
\begin{center}
\begin{tabular}{|c|c|}
\hline
\textbf{Euclidean}&\textbf{Lie}\\
\hline
points: $u\in\mathbb{R}^n$ & $\left( \frac{1+u\cdot u}{2},\frac{1-u\cdot u}{2},u_1,\dots,u_n,0  \right)^\mathrm{T} \mathbb{R}$\\
$\infty$ &  $\left( 1,-1,0,\dots,0,0\right)^\mathrm{T} \mathbb{R}$ \\
sphere: center $p\in\mathbb{R}^n$, signed radius $r$ &  $\left( \frac{1+p\cdot p-r^2}{2},\frac{1-p\cdot p+r^2}{2},p_1,\dots,p_n,r \right)^\mathrm{T} \mathbb{R}$ \\
planes: $u\cdot N=h$, unit normal $N\in\mathbb{R}^n$ &  $\left( h,-h,N_1,\dots,N_n,1\right)^\mathrm{T} \mathbb{R} $\\
\hline
\end{tabular}
\caption{Correspondence between Euclidean entities and Lie-coordinates.}
\label{talbeliecoords}
\end{center}
\end{table*}}
\noindent The fundamental invariant of Lie sphere geometry is the oriented contact of spheres. It is not difficult to show that two spheres are in oriented contact if, and only if, their Lie coordinates $s_1,s_2\in L_1^{n+1}$ satisfy $\ell(s_1,s_2) =0$, where $\ell(\cdot,\cdot)$ denotes the bilinear form corresponding to $L_1^{n+1}$.\\

\noindent Especially for $n=3$ the lines on $L^4_1$ correspond to so called parabolic pencils of spheres. These pencils consist of all oriented spheres with one common point of contact. Furthermore, each parabolic pencil contains exactly one point, {\it i.e.}, sphere of radius $0$. If this point sphere is not $\infty$ the pencil contains exactly one oriented hyperplane $\Sigma$.
\begin{remark}
Conics on Lie's quadric correspond to Dupin cyclides, that are the envelopes of two one-parameter families of spheres.
\end{remark}
\noindent The group of Lie transformations shows up as the group of projective automorphisms of $L^{n+1}_1$. This group is isomorphic to $\mathrm{O}(n+1,2)/\pm 1$, see \textsc{Cecil} \cite{Cecil:Liespheregeometry}. Since the Pin group of the Clifford algebra $\clifford_{(n+1,2,0)}$ is a double cover of $\mathrm{O}(n+1,2)$ we can use this group to describe Lie transformations.

\subsection{The homogeneous Clifford Algebra Model corresponding to Lie Sphere Geometry}\label{subsectionLieClifford}
In this section we discuss the Clifford algebra model for Lie Sphere Geometry in the three-dimensional case. Therefore, the projective space we are dealing with is a five-dimensional space $\mathbb{P}^5(\mathbb{R})$. The homogeneous Clifford algebra model is obtained with the six-dimensional real vector space $\mathbb{R}^6$ as a model for the projective image space together with the quadratic form of Lie's quadric
\[\mathrm{Q}=\begin{pmatrix}
-1 & 0 &0 & 0 &0 &0\\
 0 & 1 &0 & 0 &0 &0\\
 0 & 0 &1 & 0 &0 &0\\
 0 & 0 &0 & 1 &0 &0\\
 0 & 0 &0 & 0 &1 &0\\
 0 & 0 &0 & 0 &0 &-1\\
\end{pmatrix}.\]
This algebra has signature $(p,q,r)=(4,2,0)$ and is of dimension $2^6=64$. Again the advantage of the Clifford algebra lies in the common description of the application of Lie transformations. Arbitrary projective subspaces of $\mathbb{P}^5(\mathbb{R})$ are transformed by the sandwich operator. As an example we determine all Lie inversions that leave the point at infinity fixed, {\it i.e.}, the subgroup of \emph{Laguerre} transformations. The point at infinity has the form $\mathfrak{p}=e_1\!-\!e_2$, compare to Table \ref{talbeliecoords}. A general invertible vector is given by $\mathfrak{a}=\sum_{i=1}^6{a_ie_i}$

with $\mathfrak{aa}\neq 0$. The application of the sandwich operator to $\mathfrak{p}$ results in
\begin{align*}
\alpha(\mathfrak{a})\mathfrak{pa}^\ast&=-2(a_1\!+\!a_2)a_3e_3\!-\!2(a_1\!+\!a_2)a_4e_4\\
-&(a_1^2\!+\!a_2^2\!+\!a_3^2\!+\!a_4^2\!+\!a_5^2\!-\!a_6^2\!+\!2a_2a_1)e_1\\
-&(a_1^2\!+\!a_2^2\!-\!a_3^2\!-\!a_4^2\!-\!a_5^2\!+\!a_6^2\!+\!2a_2a_1)e_2 \\
-&2(a_1\!+\!a_2)a_5e_5\!-\!2(a_1\!+\!a_2)a_6e_6.
\end{align*}
To guarantee that this entity represents the point at infinity, we first see that $a_1\!+\!a_2\!=\!0$. With this condition the coefficients of $e_3,e_4,e_5$, and $e_6$ vanish. Moreover, the sum of the coefficients $e_1$ and $e_2$ has to vanish. This results in
\[-\!2a_1^2\!-\!2a_2^2\!-\!4a_2a_1\!=\!-\!2(a_1\!+\!a_2)^2\!=\!0.\]
Therefore, the only condition to a Lie inversion that it represents a Laguerre transformation is given by $a_1\!+\!a_2\!=\!0$ and the subgroup of Laguerre transformations is generated by all vectors with $a_1\!+\!a_2\!=\!0$.
\begin{remark}
Analogue to Th. \ref{theoremKlein} we can formulate a similar theorem for Lie sphere geometry in arbitrary dimensions. Since the projective model space for $n$-dimensional Lie sphere geometry has dimension $n+2$, the vector space for the homogeneous Clifford algebra model has dimension $n+3$. That means the highest grade is equal to $n+3$, and therefore, every group element can be written as the composition of $n+3$ vectors at the most. Especially for the case $n=3$, we have similar results as for Klein's quadric. In this case six involutions are necessary to generate the whole group.
\end{remark}
\noindent Let us reformulate this remark as theorem.
\begin{theorem}
Every Lie transformation in $n$-dimensional space is the composition of $n\!+\!3$ involutions, that correspond to the sandwich action of vectors.
\end{theorem}

\section{Conclusion}
With the help of the homogeneous Clifford algebra model corresponding to line geometry we introduced null polarities as fundamental involutions. After we transferred linear line manifolds to this model, we derived the correspondence between regular projective transformations and versors. Every versor can be factorized into vectors that correspond to null polarities. Therefore, every $4\times 4$ matrix corresponding to a projective transformation can be expressed as the product of six skew symmetric matrices at the most. Moreover, we presented a homogeneous Clifford algebra model corresponding to Lie sphere geometry.
\section*{Acknowledgments}

This work was supported by the research project "Line Geometry for Lightweight Structures", funded by the DFG
(German Research Foundation) as part of the SPP 1542.

\setlength{\bibsep}{2\parskip}


TITLE: $u_t+uu_x+cu=0$ with initial condition
QUESTION [2 upvotes]: I'm trying to solve $u_t+uu_x=-cu$ with $u(x,0)=u_0(x)$
Using the method of characteristics we get $\frac{dt}{ds}=1$ or $t=s$ likewise, $\frac{dx}{ds}=u$ and $\frac{du}{ds}=-cu$. The last equations implies that $u=u_0(x)e^{ct}$ so the second implies
$\frac{dx}{ds}=u=u_0(x)e^{ct}$ How do I finish this off?
I stumbled across this link but I don't understand the accepted solution after the part where the person says let $x(0)=f(u_0)$ and everything after that.
From $u(x,0)=u_0(x)$ isn't $x(0)$ just $x$ or $s$ or am I doing something wrong?
Anyways how would I finish off the solution?

REPLY [1 votes]: $$u_t+u\:u_x=-cu\tag 1$$
$$ds=\frac{dt}{1}=\frac{dx}{u}=\frac{du}{-c}$$
A first characteristic equation comes from solving $\frac{dt}{1}=\frac{du}{-c}$ :
$$u\,e^{ct}=c_1$$
A second characteristic equation comes from $\frac{dx}{u}=\frac{du}{-c}$:
$$u+c\,x=c_2$$
The general solution of the PDE expressed on the form of implicit equation $c_2=F(c_1)$ with arbitrary function $F$ is :
$$\boxed{u+c\,x=F\left(u\,e^{ct}\right)}\tag 2$$
CONDITION :
$u(x,0)=u_0(x)$ with known (given) function $u_0(x)$ .
$$u_0(x)+c\,x=F\left(u_0(x)\right)$$
Consider the inverse function of $X=u_0(x)$ say $x(X)=u_0^{(-1)}(X)$
The superscript $(-1)$ denotes the inverse function, not a power $-1$. Take care to not confuse.
$$X+c\,u_0^{(-1)}(X)=F\left(X\right)$$
Since $u_0(x)$ is a known function we can consider that the inverse function $u_0^{(-1)}(X)$ is known in general. Thus the function $F(X)$ is known now :
$$F\left(X\right)=X+c\,u_0^{(-1)}(X)$$
We put it into the above general solution $(2)$ where $X=u\,e^{ct}$ :
$$\boxed{u+cx=u\,e^{ct}+c\,u_0^{(-1)}\left(u\,e^{ct}\right)}\tag 3$$
In this implicit equation, $u(x,t)$ is the particular solution of the PDE which agrees with the specified condition.
When it is possible, solving the implicit equation for $u$ would give the solution on explicit form. Of course this is not possible if $u_0(x)$ is not given explicitly. But even if $u_0(x)$ is explicitly given  it is not always possible with standard functions. In those cases one consider the implicit solution $(3)$ as the final solution of the problem.


TITLE: Map between Stiefel manifold and the Grassmannian
QUESTION [3 upvotes]: I'm working on problem 2-7 in Lee's introduction to Riemannian Manifolds and am having trouble on part (b):
Let $V_k(\mathbb{R}^n)$ be the Stiefel manifold and $G_k(\mathbb{R}^n)$ denote the Grassmannian. I want to show that the map
$$ \pi : V_k(\mathbb{R}^n) \rightarrow G_k(\mathbb{R}^n)$$
that sends a k-tuple to its span is a surjective smooth submersion.
I have already shown that $\pi$ is surjective and that both $V_k(\mathbb{R}^n)$ and $G_k(\mathbb{R}^n)$ are smooth manifolds. Hence if I can show that $\pi$ is a smooth map of constant rank, the result should follow. I appreciate any help.

REPLY [3 votes]: Here's a proof that $\pi$ has constant rank.
Given a point $p \in V_k(\mathbb{R}^n)$, I'll write it as $p = \{v_1,...,v_k\}$, with the $v_i \in \mathbb{R}^n$ orthornomal.
Set $G = O(n) = \{A\in M_n(\mathbb{R}): AA^t = I\}$ and note that there is a natural smooth action of $O(n)$ on $V_k(\mathbb{R}^n)$, defined by $A\ast p = \{Av_1, ..., Av_k\}$.  There is likewise an action of $G$ on $G_k(\mathbb{R}^n)$ defined by as follows.  Given $P\in G_k(\mathbb{R}^n)$, choose a basis for $P$, apply $A$ to each basis element, and then take the span.
Proposition:  The $G$-action on both $V_k(\mathbb{R}^n)$ and $G_k(\mathbb{R}^n)$ is transitive.  Morever, $\pi$ is $G$-equivariant: for $p\in V_k(\mathbb{R}^n)$, $\pi(A\ast p) = A\ast \pi(p)$.
Proof:  We first show that $G$-action on $V_k$ is transitive.  It is enough to show that this is true in the special case that $p = \{e_1,..., e_k\}$ with $e_i\in \mathbb{R}^n$ the standard orthonormal basis.
To that end, given $p = \{v_1,...,v_k\}$, extend this to an orthonormal basis $\{v_1,...,v_n\}$.  Define $A$ to the matrix in $O(n)$ whose $i$-th column if $v_i$.  The fact that $\{v_1,...,v_n\}$ is an orthonormal basis of $\mathbb{R}^n$ implies that $A$ really is an element of $O(n)$.  Then simply note that $Ae_i = v_i$, so $A\{e_1,...,e_k\} = p$.
The fact that $\pi$ is $G$-equivariant is easy:  $$\pi(A\ast \{v_1,...,v_k\}) = \operatorname{span}\{ Av_1,...,A v_k\} = A\ast \operatorname{span}\{v_1,...,v_k\} = A\ast \pi(\{v_1,...,v_k\}).$$
From these two facts, and surjectivity of $\pi$, it follows that the $G$-action on $G_k(\mathbb{R}^n)$ is transitive.  Indeed, given planes $P,Q\in G_k(\mathbb{R}^n)$, pick preimages $p,q\in V_k(\mathbb{R}^n)$.  Picking $A\in G$ with $Ap= q$, it follow that $$A\ast P = A\ast \pi(p) = \pi(A\ast p) = \pi(q) = Q.$$  This completes the proof.  $\square$
Proposition:  The map $\pi$ has constant rank.
Proof:  Given $p,q\in V_k(\mathbb{R}^n)$, we need to show that the rank of $d_p \pi$ is equal to the rank of $d_q \pi$.
To that end, let $A\in O(n)$ with $A\ast p = q$.  Notice that the map $f_A:V_k(\mathbb{R}^n)\rightarrow V_k(\mathbb{R}^n)$ defined by $f_A(x) = Ax$ is a diffeomorphism with inverse $(f_A)^{-1} = f_{A^{-1}} = f_{A^t}$.  Likewise the induced map $g_A:G_k(\mathbb{R}^n)\rightarrow G_k(\mathbb{R}^n)$ is a diffeomorphism.
The previous proposition establishes that $\pi \circ f_A = g_A \circ \pi$.  Taking the differential of this equation at the point $p\in V_k(\mathbb{R}^n)$ via the chain rule, we find that $$d_{f_A(p)} \pi \circ d_p f_A = d_{\pi(p)} g_A \circ d_p \pi.$$  Since both $d_{f_A(p)}$ and $d_{\pi(p)} g_A$ are isomorphisms, this equation tells us that the rank of $d_{f_A(p)} \pi$ is equal to the rank of $d_p \pi$.  Since $f_A(p) = q$, this is exactly what we wanted to show.  $\square$.


TITLE: Expectation of minimum of normally distributed random variables
QUESTION [2 upvotes]: Let $(X,Y)$ be normally distributed and such that
$\;\;\;\;\mathrm{Cov}(X,Y)=\varrho$, and $\mathrm{Var}(X)=\mathrm{Var}(Y)=1$.
For which $\varrho$ does the following equality hold?
$\;\;\;\;\min (E(X),E(Y)) = E(\min(X,Y))$
For $\varrho=1$? For $\varrho=0$? For $\varrho=-1$? For any other $\varrho$?

REPLY [0 votes]: Let's enter random variable $T=X-Y$,for any 2-D distribution occurs: $$V(X-Y)=V(X)+V(Y)-2\cdot COV(X,Y)$$
Remembering that $V(X)=V(Y)=1$ we obtain: $$V(T)=V(X-Y)=2\cdot(1-\rho)$$
It's obvious that if $\rho=1$ then $V(T)=0$ , which means that $T$ is no more random variable, but deterministic value equal to $E(T)$. It results even if Bienaymé-Chebyshev inequality ,described here and here and is truth for any distribution with real valued random variable (there is no sense defining function $\min$ or $\max$ for numbers which are not real). Let $E(T)=C$, in such a case, we have to find such a $C$ that:$$\min(E(X),E(X-C))=E(\min(X,X-C))$$
$E(X-C)=E(X)-C$ for any distribution. If we want to find what is greater $X$ , or $X-C$ (we know the value of $C$ , it's our parameter),then there is no need to know value of $X$ . So, after short consideration there is easy to check that: $\min(E(X),E(X-C))$ is equal to $E(\min(X,X-C))$ for any distribution with real valued random variable $X$ and any real parameter $C$. Accepting $\rho=1$, we find solution.
Let's consider other values of $\rho$, $-1\leq\rho<1$ (*). Let $f(x,y)$ be probability density function of random variables $X$,$Y$ and $g(t,x)$ probability density function for variables $T$,$X$.
$$E(\min(X,Y))=\int _{-\infty }^{+\infty }\int _{y\leq x}y f(x,y)dydx+\int _{-\infty }^{+\infty }\int _{y>x}x f(x,y)dydx$$
$$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\int _0^{+\infty }\int _{-\infty }^{+\infty }(x-t) g(t,x)dxdt+\int _{-\infty }^0\int _{-\infty }^{+\infty }x g(t,x)dxdt$$
$$=E(X)-\int_0^{+\infty } t f_T(t) \, dt\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ $$
According to definition $f_T(t)=\int_{-\infty}^{+\infty}g(t,x)dx$ is probability density function of normal distribution ($T=X+(-1)\cdot Y$) . Notice that $\int_0^{+\infty } t f_T(t) \, dt\neq 0$, ($\rho\neq 1\ \rightarrow V(T)\neq 0$), so it must be: $\min(E(X),E(Y))=E(Y)$,which implies
$$E(T)=\int_0^{+\infty } t f_T(t) \, dt$$
and is impossible when $V(T)\neq 0$.


text:
\subsection{Discussion for the Best Arm Identification Problem}\label{appsec:best_arm}
In Section~\ref{sec:bandit}, we mentioned that a confidence sequence for which the radius $R_{\alpha}$ stays constant for a stretch of samples yields a larger sample complexity. We present here more experimental details regarding this behavior.

In the following, we experiment with a single instance of best arm identification problem
where the number of arms is 2 (i.e., $K=2$). The expected rewards are generated as the same as in Section~\ref{sec:bandit}, so that Arm 0 has mean $\mu_0 = 1$ is the best arm, and Arm 1 has mean $\mu_1 \approx 0.34$. For all the methods, we use the same data.

\begin{figure}[H]
\centering
\begin{subfigure}{.49\textwidth}
\centering
    \includegraphics[width=0.95\textwidth]{figures/best_arm_confseq_ah.pdf}
\caption{Confidence intervals of \ah for two arms.}
\label{fig:best_arm_confseq_ah}
\end{subfigure}
\begin{subfigure}{.49\textwidth}
    \centering
    \includegraphics[width=0.95\textwidth]{figures/best_arm_confseq_ah_truncated.pdf}
    \caption{Confidence intervals of truncated \ah.}
\label{fig:best_arm_confseq_ah_truncated}
\end{subfigure}
\begin{subfigure}{.49\textwidth}
    \centering
    \includegraphics[width=0.95\textwidth]{figures/best_arm_radius_ah.pdf}
    \caption{Radius of \ah (original and truncated) for two arms.}
\label{fig:best_arm_radius_ah.pdf}
\end{subfigure}
\begin{subfigure}{.49\textwidth}
    \centering
    \includegraphics[width=0.95\textwidth]{figures/best_arm_radius_diff_ah.pdf}
    \caption{The difference of the radius for two arms: $R_0 - R_1$. For positive difference value, Arm 0 will be pulled. For negative difference
    value, Arm 1 will be pulled.}
\label{fig:best_arm_radius_diff_ah}
\end{subfigure}
\begin{subfigure}{.49\textwidth}
    \centering
    \includegraphics[width=0.95\textwidth]{figures/best_arm_pulled_arm_ah.pdf}
    \caption{The arms pulled at each iteration. `$\cdot$': Arm 0 is pulled. `+': Arm 1 is pulled. \ah and truncated \ah are marked in
    red and blue, respectively. }
\label{fig:best_arm_pulled_arm_ah}
\end{subfigure}
\caption{Identify the best arm out of two using \ah and its truncated variant.}
\label{fig:best_arm_ah}

\end{figure}

We first explain this phenomenon using \ah and its truncated variant.

\ah can result in confidence intervals that are larger than $[0, 1]$. In the truncated version of \ah, the upper confidence term of a confidence interval
will be capped at $1$, and the lower confidence term will be cut at $0$, so that all the confidence intervals stay in $[0, 1]$ throughout the experiment.  We shall see that the truncated variant
would result in stationary radius and yield larger sample complexity compared with \ah.

Figures~\ref{fig:best_arm_confseq_ah} and~\ref{fig:best_arm_confseq_ah_truncated} show the confidence intervals of each arm at each iteration, when \ah and truncated \ah are plugged into Algorithm~\ref{alg:bestarm}.
The algorithm will stop when the confidence intervals of the two arms completely separate (i.e., the lower bound of Arm 0 goes above the upper bound of Arm 1). Figure~\ref{fig:best_arm_confseq_ah} and~\ref{fig:best_arm_confseq_ah_truncated} show that
\ah used 107 iterations, while the truncated \ah used 132 iterations.
One can observe that in the initial stage of the algorithm, the confidence interval, without truncation, will likely get updated once a sample adds in, which does not hold for the truncated version; compare the first 15 iterations in Figures~\ref{fig:best_arm_confseq_ah} and~\ref{fig:best_arm_confseq_ah_truncated}.
Therefore, the radius will not get updated for truncated \ah, as shown in Figure~\ref{fig:best_arm_radius_ah.pdf}.
Recall that  Algorithm~\ref{alg:bestarm} samples an arm with largest radius; when both radii are same, we sample the arm with smaller empirical mean.
Due to the stationary radius,  in those iterations, truncated \ah keeps sampling the same arm
till an update happens.

In Figure~\ref{fig:best_arm_radius_diff_ah}, we plot the difference between the radius for Arm 0 and Arm 1: $R_0 - R_1$.  Arm 0 will be sampled if this value is positive and vice versa. Again, if $R_0$ is equal to $R_1$, we shall sample the arm with lower empirical mean.
We can see the difference fluctuates evenly for \ah, so that \ah almost alternatively samples each arm, and the confidence intervals of both arms gets updated alternatively as shown in Figure~\ref{fig:best_arm_confseq_ah}. In contrast,
for truncated \ah, the difference consistently stays above or below zero for some time, which means the same arm gets sampled. See Figure~\ref{fig:best_arm_pulled_arm_ah} for the arms pulled at each iteration; the `+' and `$\cdot$' appear almost side-by-side with \ah and they appear disproportionately with truncated \ah.

As mentioned, Algorithm~\ref{alg:bestarm} stops when the two confidence intervals separate, and it is not crucial for those intervals to be shorter.
Hence, it will stop fast if (i) the confidence interval gets updated by every sample and (ii) the updates are significant for
small number of samples (the early stage). Truncated \ah underperforms in both aspects. This is also the reason why the Berstein type of confidence sequences underperforms \ah in this problem (c.f. Section~\ref{sec:bandit}). Even though they are shorter for larger samples; \ah is better with smaller samples.

Next, we investigate the performance for Bentkus type of methods.

We write \bk to be the variant from Section~\ref{sec:bandit}, that is,
we output confidence interval $\{[\mu_n^{\low*}, \mu_n^{\up*}], n\ge 1\}$ as in Theorem~\ref{thm:empirical-Bentkus-arbitrary-mean},
but output radius $R_n = \mu_n^{\up} - \mu_n^{\low}$.

We write original \bk to be the one directly from Theorem~\ref{thm:empirical-Bentkus-arbitrary-mean}, i.e.,
we output confidence interval $\{[\mu_n^{\low*}, \mu_n^{\up*}], n\ge 1\}$ and radius $R_n = \mu_n^{\up*} - \mu_n^{\low*}$.
Note that $\mu_n^{\up*} = \min_{1 \leq i \leq n} \mu_i^{\up}$ is the cumulative minimum, which essentially serves as the truncation
of the upper confidence term, and similarly does the $\mu_n^{\low*}$.
We refer the readers to Theorem~\ref{thm:empirical-Bentkus-arbitrary-mean} for the details.
Similar to the previous experiment, we shall see
that the original \bk results in a larger sample complexity than \bk. Figure~\ref{fig:best_arm_confseq_bk} presents the results.

 \begin{figure}[H]
 \centering
 \begin{subfigure}{.49\textwidth}
     \centering
     \includegraphics[width=0.95\textwidth]{figures/best_arm_confseq_bk.pdf}
     \caption{Confidence intervals of \bk (variant) for two arms.}
     \label{fig:best_arm_confseq_bk}
 \end{subfigure}
 \begin{subfigure}{.49\textwidth}
     \centering
     \includegraphics[width=0.95\textwidth]{figures/best_arm_confseq_bk_truncated.pdf}
     \caption{Confidence intervals of \bk (original) for two arms.}
     \label{fig:best_arm_best_arm_confseq_bk_truncated}
 \end{subfigure}

 \begin{subfigure}{.49\textwidth}
 \centering
     \includegraphics[width=0.95\textwidth]{figures/best_arm_radius_bk.pdf}
     \caption{Radius of \bk (original and variant) for two arms.}
     \label{fig:best_arm_raidus_bk}
 \end{subfigure}
 \begin{subfigure}{.49\textwidth}
     \centering
     \includegraphics[width=0.95\textwidth]{figures/best_arm_radius_diff_bk.pdf}
     \caption{The difference of the radius for two arms: $R_0 - R_1$. For positive difference value, Arm 0 will be pulled. For negative difference
    value, Arm 1 will be pulled.}
     \label{fig:best_arm_raidus_diff_bk}
 \end{subfigure}

 \begin{subfigure}{.49\textwidth}
     \centering
     \includegraphics[width=0.95\textwidth]{figures/best_arm_pulled_arm_bk.pdf}
     \caption{The arms pulled at each iteration. `$\cdot$': Arm 0 is pulled. `+': Arm 1 is pulled. \bk and original \bk are marked in
   red and blue, respectively. }
     \label{fig:best_arm_pulled_arm_bk}
 \end{subfigure}
 \caption{Identify the best arm out of two using original \bk and the variant introduced in Section~\ref{sec:bandit}.}
 \label{fig:best_arm_bk}
\end{figure}

Patterns similar to the \ah and its truncated version happen here too. Although \bk keeps sampling the same arm in the beginning phase, it alternates the samples in the later stage. Comparing Figures~\ref{fig:best_arm_pulled_arm_ah} (\ah) and~\ref{fig:best_arm_pulled_arm_bk} (\bk), the sampling pattern of \ah is more uniform, however, \bk still outperforms \ah due to its fast convergence.


TITLE: Are there such things as non-trivial entire semigroups?
QUESTION [7 upvotes]: I'll state the theorem I am posing up front, and then explain why I think this theorem appears to be true. I am asking if anyone can prove it, or knows references to where it is proved. Please, forgive me if it is trivial, or if a counter example is trivial too.

If $\phi(s,z) : \mathbb{C}_{\Re(s)>0} \times \mathbb{C} \to
\mathbb{C}$ is holomorphic in both variables, and
  $$\phi(s_0,\phi(s_1,z)) = \phi(s_0 + s_1,z)$$ then necessarily $\frac{\partial^2}{\partial z^2} \phi(s,z) = 0$, so that $$\phi(s,z) =
e^{qs}(z-z_0) + z_0$$ for $q,z_0 \in \mathbb{C}$
  or $$\phi(s,z) = z+cs$$ for $c \in \mathbb{C}$

I ask this question because
A) I've never encountered a counter example in an extensive study of semigroups.
B) I can prove on a whole bunch of occasions if $\phi(a,z) = f(z)$; for $\Re(a) > 0$, for specific $f$; implies $\phi(s,z)$ cannot exist.
Examples include $f = \sin,\cos,\exp,\exp(p(z)), p(\exp(z))$ where $p$ is an arbitrary polynomial. If $f$ has a super attracting fixed point, no such $\phi$ exists. If $f^{\circ n}$ has fixed points $f$ doesn't have, then no such $\phi$ exists. In all these cases, no such solution exists for the orbits of $f$ either. I'm wondering if this is a universal trait. That, in some sense, the theorem above can be a kind of Liouville theorem.
Let's say that $\phi$ satisfies $f$, if $\phi(a,z) = f(z)$ for some $\Re(a) > 0$. To highlight the similarity to Liouville's theorem, the original theorem can be restated as

If $f:\mathbb{C} \to \mathbb{C}$ and some semigroup $\phi$ satisfies
  $f$, then $f= mz+b$ for some $m,b \in \mathbb{C}$.

Supposing the pair $f(\cdot)$ and $\phi(s,\cdot)$ don't have to map $\mathbb{C}$ to itself, that $\mathbb{C}$ is weakened to the unit disk $\mathbb{D}$, and $f(0) = 0$ with $|f'(0)| \neq 0,1$, then there always exists a $\phi$ satisfying $f$. It seems though, as soon as we lift from $\mathbb{D}$ (or any simply connected domain biholomorphic to $\mathbb{D}$) to $\mathbb{C}$ it fails.
I can show the result in a restricted form, which I also think is interesting

If $p:\mathbb{C} \to \mathbb{C}$ is a polynomial, and there exists
  a semigroup $\phi$ satisfying $p$ then $p(z) = mz+b$ for some $m,b \in \mathbb{C}$

My main avenue of approach for deriving a contradiction has been to consider the Weierstrass product. It follows that if $\phi(s_0,z_n) = z_n$ then $\phi(s,z_n) = z_n$, so $\phi$ has a countable list of fixed points invariant on our choice of $s$, allowing us to say
$$\phi(s,z) = z + e^{g(s,z)}\prod_{n=0}^{\infty}(1-\frac{z}{z_n})e^{-p_n(z/z_n)}$$
where $p_n(z) = \sum_{j=1}^n \frac{z^j}{j}$. I've been fiddling around with this as it seems like a smart idea. Since $\lim_{s\to 0}\phi(s,z) = z$, and we have a semigroup property, it seems reasonable to think we can show that for all $\epsilon >0$ there exists a $\delta > 0$ such that $|\phi(\delta,z)-z| < C_\delta e^{|z|^{\epsilon}}$ then by Hadamard
$$\sum_{n=0}^\infty \frac{1}{|z_n|} < \infty$$
which greatly reduces the candidates for $f$ that can be satisfied by some $\phi$—$f$'s fixed points have to be sufficiently spaced out, so to speak.
Overall, I'm lost on the general case, and seem to be only able to prove it in specific circumstances. I think it says something rather profound about multiplication and addition, that, for another reason, they are incredibly special. Help, comments, suggestions, edits, anything is welcome, thanks.

REPLY [6 votes]: Set $f_t(z)=\phi(t,z)$.
Notice that entire functions $z\mapsto f_t(z)$ all commute with each other.
I. N. Baker proved that if $f$ is a non-affine entire function with a repelling fixed point
then the set of entire functions with commute with it is countable.
MR0147650
Baker, Irvine Noel
Permutable entire functions.
Math. Z. 79 1962 243–249.
On the other hand, in a later paper he proved that for every non-affine entire function some iterate has a repelling fixed point.
MR0226009
Baker, I. N.
Repulsive fixpoints of entire functions.
Math. Z. 104 1968 252–256.
Combination of these two theorems settles your question. Indeed, start
with $f_1$. Then $f_n$ with some positive integer $n$ has a repelling fixed point. Then the set of entire functions which commute with $f_n$ is at most
countable, contradiction.
Once we know that all $f_t$  are affine with respect to $z$, the rest is easy:
commutative sub-semigroups of the affine group are just what you listed.
EDIT. Second theorem of Baker was based on the Ahlfors Islands Theorem, which was considered "heavy machinery" at that time. Since then both Baker's proof and the Ahlfors Islands theorem were very much simplified. For an elementary proof of the Baker theorem see, for example,
MR1782673 Berteloot, François; Duval, Julien Une démonstration directe de la densité des cycles répulsifs dans l'ensemble de Julia. Complex analysis and geometry (Paris, 1997), 221–222, Progr. Math., 188, Birkhäuser, Basel, 2000.


TITLE: Proof in vector spacee
QUESTION [1 upvotes]: I have problem with proofs in vector space. First is
$\vec x+(-(\vec y+\vec z))=(\vec x+(-\vec y))+(-\vec z)$
and the second
$a\cdot \vec x+b \cdot \vec y=b \cdot \vec x+a\cdot \vec y \Leftrightarrow a=b \vee \vec x=\vec y $
Could anyone help me with this? I'm sorry for my bad english.
In the second task I have:
$a\cdot \vec x+b \cdot \vec y  $
I have sentece that $ \vec x = \vec y$ or $a=b$
So i'm changing $ \vec y $ on $ \vec x$
$a\cdot \vec x+b \cdot \vec x = (a+b)\vec x = (a+b) \vec y =...$
And i don't know what can i do next.

REPLY [1 votes]: Hints:
For the first one show that, by definition, $$-(v_{1}+v_{2})=-v_{1}-v_{2}$$
where $v_{i}$ are vectors.
For the second use $$\alpha v-\beta v=(\alpha-\beta)v$$
where $\alpha,\beta$ are scalars, $v$ is a vector.
Also use $a-b=-(b-a)$


TITLE: Finding equivalence class with a binary set
QUESTION [0 upvotes]: I'm new to discrete math so there might be problems with this solution.
Prompt is to find at least one equivalence class if it is an equivalence relation.
$$X = R^2, (x_1, y_1) \sim (x_2, y_2) \iff y_1 = y_2$$
1. Reflexivity
$$(x_1, y_1) \sim (x_1, y_1) \iff y_1 = y_1$$
Since $y_1 = y_1$, it is reflexive.
2. Symmetry
$$(x_1, y_1) \sim (x_2, y_2) \iff y_1 = y_2$$
Assuming $y_1 = y_2$ (eqn1),
$$(x_2, y_2) \sim (x_1, y_1) \iff y_2 = y_1$$
from eqn 1, $y_2 = y_1$, so it is a symmetry.
3. Transitivity
$$(x_1, y_1) \sim (x_2, y_2) \iff y_1 = y_2$$
$$(x_2, y_2) \sim (x_3, y_3) \iff y_2 = y_2$$
Assuming $y_1 = y_2$ and $y_2 = y_3$, we get $y_1 = y_3$.
$$(x_1, y_1) \sim (x_3, y_3) \iff y_1 = y_3,$$
since $y_1 = y_3$, so it is transitive.
Is this the correct way to solve the problem? Also how to find the equivalence classes for the same?

REPLY [0 votes]: Yes, your proof is correct. To construct an equivalence class, pick an element of the set and find (describe) all elements that are equivalent to it. More formally: if $R$ is an equivalence relation on a set $X$, then for an $x\in X$, its equivalence class is $[x]=\{y\in X\,\colon\,yRx\}$. In this example: pick an arbitrary $(x,y)\in\mathbb{R}^2$. By the given definition, any other $(x_1,y_1)\in\mathbb{R}^2$ is equivalent to it, $(x_1,y_1)\sim(x,y)$, iff $y_1=y$, so $(x_1,y_1)=(x_1,y)$. Note that there are no constraints on $x_1$, so it can be any real. Thus the equivalence class of $(x,y)$ is $\{(x_1,y)\in\mathbb{R}^2\,\colon\,x_1\in\mathbb{R}\}$ (where $y$ is fixed). Geometrically, they are horizontal lines $y=\text{const}$.


TITLE: Does $f$ continuous in the Zariski topology imply that $\underset{x \to x_0}{\lim}f(x)=f(x_0)$?
QUESTION [0 upvotes]: Does $f$ continuous in the Zariski topology imply that $\underset{x \to x_0}{\lim}f(x)=f(x_0)$?

I used the above argument in a proof, but I found it really suspect and questionable and would like to know whether it is correct or if (and how) to replace it with a correct argument.
If it is correct, would it be possible to sketch a proof, or to point to a reference giving the proof?
Context: Given a projective variety $V$ in $\mathbb{CP}^n$, and two rational functions $\frac{f_1}{g_1}$ and $\frac{f_2}{g_2}$ which are equal everywhere in their common domain of definition, i.e. $\operatorname{Dom}\left(\frac{f_1}{g_1}\right) \cap \operatorname{Dom}\left(\frac{f_2}{g_2}\right)$.
I want to show that this implies that $f_1g_2 - f_2g_1$ equals $0$ on every point of $V$.
Obviously this holds everywhere on $\operatorname{Dom}\left(\frac{f_1}{g_1}\right) \cap \operatorname{Dom}\left(\frac{f_2}{g_2}\right)$. Then my suspicious argument goes that: this set is Zariski open, hence Zariski dense, in $V$, and $f_1g_2 - f_2g_1$ is a Zariski-continuous function which equals $0$ everywhere in this Zariski dense subset. Thus, because of continuity (the claim in yellow above), it also equals zero on $V \setminus \left[\operatorname{Dom}\left(\frac{f_1}{g_1}\right) \cap \operatorname{Dom}\left(\frac{f_2}{g_2}\right)\right]$, thus on all of $V$.
I wanted to argue this more directly, but this was the only idea I could think of at the time.
Note: this corresponds to one direction of exercise 5.4.4. in this book. Since this problem is so basic, I am quite concerned about my current lack of understanding.

REPLY [2 votes]: As per the remarks, it is not always true that one may find unique limits, depending on the topology one chooses. I'd like to present a reworking of your argument:
The locus where $f_1g_2-f_2g_1$ vanishes determines a closed subset of $\Bbb P^n$. Since $f_1g_2-f_2g_1$ is zero on $U:=\operatorname{Dom}\left(\frac{f_1}{g_1}\right) \cap \operatorname{Dom}\left(\frac{f_2}{g_2}\right)$, the function $f_1g_2-f_2g_1$ must be zero on the closure of $U$, since the closure of $U$ is the smallest closed set containing $U$. But $\overline{U}=V$, and so $f_1g_2-f_2g_1$ is zero on the whole of $V$.
This belies a fun exercise you will probably meet later in life once you know of the word "scheme" and a few adjectives about schemes: any two maps from a reduced scheme to a separated scheme which agree on a dense open are actually equal. The technique of proof is similar: show some locus is closed and contains a dense open.


TITLE: Prove or disprove this statement: For any $y \in \mathbb{Z}$ and $y \neq \pm 1$, $(y+1)^{2}$ is not divisible by $y$.
QUESTION [0 upvotes]: I want to prove or disprove this statement:

For any $y \in \mathbb{Z}$ and $y \neq \pm 1$, $(y+1)^{2}$ is not divisible by $y$.

The case where $y$ is even can be easily proved. However, I am stuck at the case where $y$ is odd. I try to start with the fact that "$y+1$ is not divisible by $y$" and expect to use something like the Euclid's Lemma. Any idea on how to proceed?

REPLY [0 votes]: Since you tagged it "polynomials" we highlight their key role here.
Note that $\ y\mid y^2\!+\!1\iff y\mid (\overbrace{y^2\!+\!1\bmod y}^{\large \color{#c00}1})\!\iff y\mid \color{#c00}1$
Generally $\ y\mid f(y)\iff y\mid \overbrace{f(y)\bmod y}^{\large \color{#c00}{f(0)}}\iff y\mid \color{#c00}{f(0)}\,$ via Polynomial $\rm\color{#c00}{Remainder}$ Theorem, where $\,f(x)\,$ is any polynomial with integer coefficvients.