text:
\begin{document}
\maketitle

\centerline{\scshape Susanne Pumpl\"un}
\medskip
{\footnotesize
 \centerline{School of Mathematical Sciences}
  \centerline{University of Nottingham}
   \centerline{University Park,
  Nottingham NG7 2RD, United Kingdom}
   }

\bigskip

\centerline{\scshape Thomas Unger}
\medskip
{\footnotesize
 \centerline{School of Mathematical Sciences}
  \centerline{University College Dublin}
   \centerline{Belfield, Dublin 4, Ireland}
   }

\begin{abstract}
Associative division algebras are a rich source of fully diverse space-time block codes (STBCs). In this paper the systematic construction of fully diverse  STBCs from nonassociative algebras is discussed.  As examples, families of fully diverse $2\times 2$, $2\times 4$ multiblock   and $4\x 4$ STBCs
are designed, employing nonassociative quaternion division algebras.
\end{abstract}

\section{Introduction}
Space-time coding is used for reliable high rate transmission
over wireless digital channels with multiple antennas at both the transmitter and receiver ends.
From a mathematical point of view,  a space-time block code (STBC) consists of a
family of  matrices with complex entries (the codebook)  that satisfies a number of properties which determine how well the code performs.

The first aim is to find \emph{fully diverse} codebooks, where the difference of any two
code words   has full rank. Once a fully diverse codebook is found it is then further optimized to satisfy additional design criteria (see Section~\ref{sec6}).

Using central simple associative division algebras to build
space-time block codes allows for a systematic code design
(see for instance \cite{O-R-B-V}, \cite{E-S-K}, \cite{S-R-S}, \cite{H-L-R-V}, \cite{Be-Og1}, \cite{Be-Og2}, \cite{Be-Og3} and the excellent survey \cite{Seth}).

Most of the existing  codes  are built from cyclic division algebras
over $F=\mathbb{Q}(i)$ or $F= \mathbb{Q}(\zeta_3)$
with $\zeta_3=e^{2\pi i /3}$ a third root of unity. These fields are used for
 the transmission of QAM or HEX constellations, respectively.

There are two  ways to embed an associative division algebra into a matrix algebra in order to obtain a
 codebook: the left regular
representation of the algebra and the representation over some maximal subfield. For instance, a real $4\times 4$ orthogonal design is obtained
by the left regular representation of the real quaternions $\mathbb{H}=(-1,-1)_\mathbb{R}$
(see for instance \cite[p.~1458]{T-J-C1}), whereas the  Alamouti code \cite{Al} uses the representation of $\mathbb{H}$ over its maximal
subfield $\mathbb{C}$.

In \cite[p.~2608]{S-R-S}, the authors note that ``the Alamouti code is the only rate-one STBC
which is full rank over any finite subset of $\mathbb{C}$, which is due to the fact that the set of quaternions $\mathbb{H}$
is the only division algebra which has the entire complexes as its maximal subfield.''
There are however other, \emph{nonassociative}, division algebras over $\mathbb{R}$ of dimension 4 which contain $\mathbb{C}$
as a subfield and which yield  STBCs that are full rank over any finite subset of $\mathbb{C}$:
the  \emph{nonassociative quaternion division algebras} over $\mathbb{R}$ which were classified in \cite{Al-H-K} and which we will employ here.

In this paper we show how  nonassociative division algebras can be
used  to systematically construct fully diverse linear STBCs.
If a nonassociative algebra $A$ behaves well enough,
 one can also obtain fully diverse families of matrices using subfields of $A$.
We  use 4-dimensional nonassociative quaternion division algebras
to construct new
examples of  fully diverse $2\times 2$ and $4\x 4$ space-time block codes and of $2\times 4$ multiblock space-time  codes (cf. \cite{L-M-M}). We also investigate when these codes satisfy the non-vanishing determinant property.

The paper is organized as follows: in Sections~\ref{sec2} and \ref{sec3}
we present nonassociative algebras and the Cayley-Dickson doubling process, respectively. In Section~\ref{sec4} we define nonassociative quaternion division algebras (constructed via a
generalized Cayley-Dickson doubling process). In Section~\ref{sec5} we explain the general framework for obtaining fully diverse STBCs from nonassociative division algebras.
In Section~\ref{sec6} we list the design criteria used in the construction of STBCs.
In Section~\ref{sec7} we look in more detail at the construction of fully diverse $2\x 2$ STBCs from nonassociative quaternion algebras. We also discuss the non-vanishing determinant  and
information lossless  properties. This is followed by many examples.
In Sections~\ref{sec8} and \ref{sec9} we discuss the construction of fully diverse $2\x 4$ multiblock codes and $4\x 4$ codes from nonassociative quaternion algebras, respectively. In Appendix~\ref{AppA} we collect results from algebraic number theory that are needed in this paper.

\section{Nonassociative algebras}\label{sec2}

Let $F$ be a field and let $A$ be a finite-dimensional $F$-vector space.
We call $A$ an \emph{algebra} over $F$ if there exists an
$F$-bilinear map $A\times A\to A$, $(x,y) \mapsto x \cdot y$ (also denoted simply by juxtaposition $xy$),
called  \emph{multiplication}, on $A$.  This definition does not imply that the algebra is associative; we only have
$c(xy)=(cx)y=x(cy)$ for all $c\in F$, $x,y\in A$. Hence we also call such an algebra
a \emph{nonassociative algebra}, in the sense that it is not necessarily associative.
A (nonassociative) algebra $A$ is called \emph{unital} if there is
an element in $A$ (which can be shown to be uniquely determined), denoted by 1, such that $1x=x1=x$ for all $x\in A$.
We will only consider unital nonassociative algebras.

For a nonassociative $F$-algebra $A$, associativity in $A$ is measured by the {\it
associator}
\[[x, y, z]  = (xy) z - x (yz).\]
The \emph{nucleus} of $A$ is
defined as
\[\cN(A)  = \{ x \in A \, \vert \, [x, A, A] = [A, x, A] = [A,
A, x] = 0 \}.\]
The nucleus is an associative subalgebra of $A$ (it may be
zero), and $x(yz) = (xy) z$ whenever one of the elements $x, y, z$ is in
$\cN(A)$. In other words, the nucleus of the algebra $A$ contains all the elements of $A$
which associate with every other two elements in $A$. Suppose $K\subset A$ is a subfield of $A$.
An $F$-algebra $A$ is \emph{$K$-associative} if $K$ is contained in the nucleus $\cN(A)$.
The \emph{left nucleus} of $A$ is defined as
$$\cN_\ell(A) = \{ x \in A \, \vert \, [x, A, A]  = 0 \},$$
 the \emph{middle nucleus} of $A$ is
defined as
$$\cN_m(A) = \{ x \in A \, \vert \, [A, x, A]  = 0 \}$$
 and  the \emph{right nucleus} of $A$ is
defined as
$$\cN_r(A) = \{ x \in A \, \vert \, [A,A, x]  = 0 \}.$$
Their intersection is the nucleus $\cN(A)$.

 A nonassociative algebra $A$ is called a \emph{division algebra} if for any $a\in A$, $a\not=0$,
the left multiplication  with $a$, $\lambda_a(x)=ax$,  and the right multiplication with $a$, $\rho_a(x)=xa$, are bijective. The algebra
$A$ is a division algebra if and only if $A$ has no zero divisors  \cite[pp. 15, 16]{Sch}.
Note that if the $F$-algebra $A$ is associative and finite-dimensional as an $F$-vector space, this
definition of division algebra coincides with the usual one for associative algebras.

\section{The Cayley-Dickson doubling process}\label{sec3}

The Cayley-Dickson doubling process is a well-known way to construct a new
algebra with involution from a given algebra with involution.
It can be motivated by the observation that the complex numbers can be viewed as pairs of real numbers with componentwise
addition  and a suitably defined multiplication:

\begin{example}
We define a multiplication on  $\mathbb{R}\x \mathbb{R}$ via
$$(u,v)(u',v'):=(uu' -v'v,uv'+u'v),$$
for $u,v,u',v'\in \mathbb{R}$. The unit element for this multiplication is $(1,0)$.
Let $i=(0,1)$. Then $i^2=(-1,0)$. We can now write the pair
 $(u,v)$ as $(u,v)=(u,0)+(0,1)(v,0)$ and identify it with
 the element $u+iv\in \mathbb{R}\oplus  i\mathbb{R}$.
In this way  we obtain the complex numbers
$$\mathbb{C}=\mathbb{R}\oplus  i\mathbb{R}.$$
For $x=u+iv$, $y=u'+iv'$ with $u,v,u',v'\in \mathbb{R}$, we have
$x y=(uu' -v'v)+i(v'u+vu')$.

Let $\can$ denote  complex conjugation, given by $\bar{x}=u-iv$ for $x=u+iv$. Then we can also write $\overline{(u,v)}=(u,-v)$.

The above process can be repeated with $\C$ instead of $\R$: define a multiplication on  $\mathbb{C}\x \mathbb{C}$ via
$$(u,v)(u',v'):=(uu' -v'\bar{v}, \bar{u}v'+u'v),$$
for $u,v,u',v'\in \mathbb{C}$. The unit element for this multiplication is $(1,0)$.
Let $j=(0,1)\in \C\x \C$. Then $j^2=(-1,0)$.
We identify the element $(u,v)\in \C\x\C$ with $u+jv\in \C\oplus j\C $. In this way we obtain the Hamilton quaternions
\[\mathbb{H}=\C \oplus j\C .\]
We define quaternion conjugation (again denoted $\can$) via
\[\bar{(u,v)}=(\bar{u}, -v).\]
\end{example}

Another iteration of this process, this time starting with $\mathbb{H}$, results in the (Cayley-Graves) octonion algebra $\mathbb{O}$.

\begin{remark}
In 1958 it was shown that finite-dimensional real division algebras can only
have dimension 1, 2, 4 or 8 (cf. \cite{Bo-M}).
In addition to the well-known algebras
$\mathbb{R}$, $\mathbb{C}$, $\mathbb{H}$ and $\mathbb{O}$, there exist other finite-dimensional real division algebras. The algebras
$\mathbb{R}$, $\mathbb{C}$, $\mathbb{H}$ and $\mathbb{O}$ are just  the alternative ones (see \cite[p.~48]{Sch}).
Indeed,  a complete classification
of these algebras is still far away. Only certain subclasses are understood thus far.
Moreover, the restriction on the dimension only holds for real and real closed fields (see \cite{D-D-H}).
Over number fields there exist also higher dimensional division algebras as
well as division algebras which do not appear over the
real numbers.
\end{remark}

The previous construction of the Hamilton quaternions as a double of the complex numbers serves as a motivating example for obtaining generalized
 (associative) quaternion algebras as doubles,  as we will do now.

Let $F$ be a field. Let $K$ be  a separable quadratic field extension of $F$ with non-trivial Galois automorphism $\sigma:K\to K$.
Let  $b\in F^\times:=F\setminus \{0\}$. Then the 4-dimensional $F$-vector space
 $K\x K$ can be made into a new unital associative (but not commutative) algebra over $F$ via the multiplication
$$(u,v)(u',v'):=(uu'+b v'\sigma(v),\sigma(u)v'+u'v)$$
for $u,u',v,v'\in K$. The unit element is given by $(1,0)$. The automorphism $\sigma$ induces an involution $\can$ on $K\x K$ as follows:
\[\bar{(u,v)}:= (\sigma(u), -v).\]
Let $j=(0,1)$. Then $j^2=(b,0)$. We identify $(u,v)\in K\x K$ with $u+jv$ in $K\oplus jK $.
The algebra $K\oplus jK$ is called the \emph{Cayley-Dickson double of $K$ $($with scalar $b$$)$} and denoted by $\Cay(K,b)$ (cf. \cite{A2}).

The Cayley-Dickson double of $K$ yields a quaternion algebra over $F$. If $F$ has characteristic not 2 and
$K=F(\sqrt{a})=F(i)$, $\sigma: \sqrt{a} \mapsto -\sqrt{a}$, we have
$$\Cay(K,b)\cong (a,b)_{F}.$$
The standard basis $\{1,i,j,k\}$ of the quaternion algebra $(a,b)_F$ satisfies $i^2=a$, $j^2=b$,
$k=ij$ and $ij=-ji$.

The Cayley-Dickson doubling process depends on the scalar $b$ only up to an invertible square, i.e.
$$\Cay(K,b)\cong \Cay(K,bd^2)$$
for every $d\in F^\times$. The algebra $\Cay(K,b)$ is a division algebra if and only if
$b\not \in N_{K/F}(K^\times)$, where $N_{K/F}$ is the norm of the
field extension $K/F$.

The quadratic \emph{norm} $N_A:A\to F$ of the algebra $A=\Cay(K,b)$ is given by
$$N_A(u+jv)=N_{K/F}(u)-b N_{K/F}(v)$$
for $u,v\in K$.
If $F$ has characteristic not 2, a straightforward computation shows that
$$N_A(x)=x\bar{x}=\bar{x}x=q_0^2-aq_1^2-bq_2^2+abq_3^2$$
for $x=q_0+iq_1+jq_2+jiq_3$, $q_i\in F$.

\begin{example}
Let $\mathbb{H}=(-1,-1)_\mathbb{R}$ denote Hamilton's quaternion algebra.
This is just the algebra $\Cay(\mathbb{C},-1)$, as already established above.
\end{example}

\begin{example} The algebra
$(5,i)_{\mathbb{Q}(i)}=\Cay(K,i)$ with $K=\mathbb{Q}(i,\sqrt{5})$ is the quaternion algebra used
in the construction of the
Golden code \cite{Og}. This algebra is isomorphic to the cyclic algebra $(K/F,\sigma, i)$ where $\sigma:\sqrt{5}\mapsto-\sqrt{5}$.
Since $K=\mathbb{Q}(i,\sqrt{5})\cong\mathbb{Q}(i, \theta)$, where $\theta=
\frac{1+\sqrt{5}}{2}$ is the golden number, we also have that
$\Cay(\mathbb{Q}(i,\theta),i)\cong (5,i)_{\mathbb{Q}(i)}.$
\end{example}

\begin{remark} The Cayley-Dickson doubling process can be iterated: the quaternion algebras double to \emph{octonion algebras}, which in turn double to \emph{sedenion algebras}. Continuing the doubling process results in successive \emph{generalized Cayley-Dickson algebras}.
\end{remark}

\section{Nonassociative quaternion division algebras}\label{sec4}

Let $F$ be a field of characteristic not 2.
Let $K$ be a quadratic field extension of $F$ with non-trivial Galois automorphism
$\sigma$ and let $b\in K\setminus F$. We define an algebra structure on
the $F$-vector space $K\x K$ via the multiplication
$$(u,v)(u',v'):=(uu'+b v'\sigma(v), \sigma(u)v'+u'v)$$
for $u,u',v,v'\in K$. The multiplication is thus defined just
as for  quaternion algebras with the exception that we require the scalar $b$ to lie outside of $F$. We denote the algebra again by
$\Cay(K,b)$. Its  unit element is $(1,0)$.

Since  $b\in K\setminus F$, the multiplication of  $\Cay(K,b)$ is not associative
anymore. It is not even third power-associative, meaning that in general
$(x^2)x\not=x(x^2)$. The algebra
$\Cay(K,b)$ with $b\in K$ and not in $F$ is called a \emph{nonassociative quaternion algebra} over $F$.

\begin{remark}\label{rem4.1}
 Let $K=F(\sqrt{a})=F(i)$ be a quadratic field extension and let $b\in K\setminus F$. Let $A=\Cay(K,b)$ be a nonassociative quaternion
 division algebra. Put
 $$j=(0,1)\in \Cay(K,b).$$
Then $A$ has $F$-basis $\{1,i,j,ji\}$ such that $i^2=a$, $j^2=b$  and
 $xj=j\sigma(x)$
 for all $x\in K$ (so in particular $ij=-ji$).
\end{remark}

\begin{theorem}[{\cite{As-Pu}} or {\cite{W}}]\label{thm4}
The nonassociative quaternion algebra $\Cay (K, b) $
has nucleus $K$ and is a division algebra over $F$.
\end{theorem}

Thus products involving a factor from $K$ are still associative. Furthermore, nonassociative quaternion algebras are always division algebras, which is not the case for the usual associative quaternion algebras.

\begin{remark}\label{rem4.3}
Let $F=\mathbb{R}$ and $b,b'\in\C$. Let
$b=p+iq$. Then $\Cay (\mathbb{C}, b) \cong \Cay (\mathbb{C}, b')$
if and only if $b'=t(p\pm iq)$ for some positive $t\in \mathbb{R}$  \cite[Thm. 14]{Al-H-K}.

Over $\mathbb{Q}$, we can easily find non-isomorphic nonassociative quaternion division algebras:
it was observed in \cite{W} that two nonassociative quaternion  algebras $\Cay(K,b)$ and $\Cay(L,c)$ can only be isomorphic if $L\cong K$.
Moreover,
$$\Cay(K,b)\cong \Cay(K,c) \text{ iff } g(b)=d\sigma(d)c$$
for some automorphism $g\in \Aut(K)$ and some non-zero $d\in K$.
\end{remark}

 Nonassociative quaternion algebras provided early examples of real nonassociative division algebras which
were neither power-associative nor quadratic.
They were investigated for the first time by Dickson \cite{D} in 1935 and by Albert \cite{A1} in 1948.
In 1987, Waterhouse \cite{W} completely classified these algebras over a field of characteristic not 2.

The only division algebras which appear in the classification of 4-dimensional $K$-associative algebras, cf. \cite{Al-H-K} or \cite{W}, are the generalized quaternion division algebras and the
 nonassociative quaternion division algebras over $F$.

\section{STBCs from nonassociative division algebras: the general setup}\label{sec5}

The general setup for constructing a fully diverse  STBC from an associative division algebra $A$ is simple: associate to each nonzero element $x\in A$ a square matrix $X$ over a fixed subfield of $A$ (normally the base field, via the left regular representation, or a maximal extension of the base field). The difference of any two such matrices $X-X'$ (with $X\not=X'$) will then always be invertible.
This procedure can be adapted to work in the nonassociative case as well, as will be explained in this section.

\subsection{The left regular representation}
Let $A$ be a nonassociative division algebra over $F$ of dimension $n$ as an $F$-vector space.
Let $a$ be any element in $A$. The left multiplication $\lambda_a:A\to A$ determined by $a$ is defined
by $x\mapsto ax$ for all $x\in A$. The operator $\lambda_a$ is  linear and the set $\{\lambda_a\,|\, a\in A\}$ is a  subspace of the
associative algebra $\End_F(A)$, the algebra of $F$-linear transformations on $A$.
Consider the left regular representation
$$\lambda: A \to \End_F(A), a\mapsto \lambda_a.$$
If $\lambda_a=\lambda_b$ then $ax=bx$ for all $x\in A$, hence $(a-b)x=0$ for all $x$, which yields $a=b$
($A$ is a finite-dimensional division algebra and as such does not have zero divisors) and we have an injection.

After a choice of $F$-basis for $A$, we can embed $\End_F(A)$ into the algebra ${\rm Mat}_n(F)$ of $n\x n$-matrices with entries from $F$, where $n=\dim_F(A)$.
In this way we get an embedding $\lambda:A\hookrightarrow {\rm Mat}_n(F)$ of vector spaces.

Contrary to the situation for associative division algebras, this only
embeds the vector space $A$ into the vector space ${\rm Mat}_n(F)$; the algebra structure of $A$ is disregarded here.

Nonetheless, all non-zero elements of $A$ are invertible, hence all $\lambda_a$ with $a\not=0$ are bijective and so all
non-zero matrices in $\lambda(A)$ have non-zero determinant. Now
$X\pm Y\in \lambda(A)$ for all $X,Y\in \lambda(A)$.
Thus $\lambda(A)$ constitutes a linear codebook which in addition is fully diverse,
since the rank of the difference of two distinct codewords is maximal.

\subsection{Representation over a maximal subfield}
For coding purposes, an associative division algebra $A$ is often  considered as a vector space over some
 subfield $K$ of the algebra $A$. Usually $K$ is maximal with respect to inclusion.
Given a nonassociative $F$-algebra $A$ with a maximal subfield $K$, this is not always possible
because of the nonexistence of the associative law. So what are the minimum requirements on a nonassociative algebra in order
to have such a representation?

Let $K$ be a subfield of the $F$-algebra $A$.
We need $A$ to be a right $K$-vector space, i.e. we need
$$x(cd)=(xc)d \text{ for all }x\in A, c,d\in K.$$
This is satisfied for instance if $K \subset \cN_r(A)$ or if $ K \subset \cN_m(A) $.

We also need that left multiplication $\lambda_a$ is a linear endomorphism of the right $K$-vector space
 $A$, i.e. that $(\alpha a)x=\alpha ( ax)$ for all $\alpha\in K$, $a,x\in A$, which is equivalent to
 $K\subset \cN_\ell(A)$.
 Then
$$\lambda_{\alpha a}(x)=(\alpha a)x=\alpha ( ax)=\alpha \lambda_a(x)$$
for all $a,x\in A$, $\alpha\in K$ and $\lambda_a\in \End_K(A)$, so $\lambda: A \hookrightarrow \End_K(A), a\mapsto \lambda_a.$

Thus,  let $K$ be a subfield of $A$, maximal with respect to inclusion and assume that $K \subset \cN_r(A)\cap \cN_\ell(A)$ or $ K \subset \cN_m(A) \cap \cN_\ell(A)$.
 Consider $A$ as a right $K$-vector space.
After a choice of  $K$-basis for $A$, we can embed $\End_K(A)$ into the vector space
${\rm Mat}_r(K)$ where $r=\dim_K (A)$. In this way we get an embedding
$$\lambda:A\hookrightarrow {\rm Mat}_r(K)$$
 of vector spaces. Obviously, we have $X\pm Y\in \lambda(A)$ for all $X,Y\in \lambda(A)$.
Thus $\lambda(A)$ constitutes a linear codebook.

\begin{remark}
If we want to consider $A$
as a \emph{left} $K$-vector space, we
require $K\subset \cN_\ell(A) $ and $K\subset \cN_m(A)$ and  adjust the above construction accordingly.
\end{remark}

More generally one can do the following:
let $D$ be a subalgebra of $A$,  assume that
$D\subset \cN_r(A)\cap \cN_\ell(A)$ or that
$D \subset \cN_m(A) \cap \cN_\ell(A)$
 and suppose $A$ can be viewed as a free right $D$-module of rank $r$.
After a choice of a $D$-basis for $A$, we can embed the right $D$-module $\End_D(A)$ into the vector space
${\rm Mat}_r(D)$. In this way we get an embedding
$\lambda:A\to {\rm Mat}_r(D)$
 of $D$-modules. Obviously, we have $X\pm Y\in \lambda(A)$ for all $X,Y\in \lambda(A)$.
Thus $\lambda(A)$ is a linear codebook.

\begin{remark}
It is not known whether there exist 8-dimensional real division algebras with some (left, middle or right)
nucleus isomorphic to $\mathbb{H}$. The fact that there are no 8-dimensional real division algebras with
two associative nuclei (left, middle or right) isomorphic to $\mathbb{H}$ suggests a negative answer \cite[Proposition~3]{J-P}.
This need not be the case over other base fields, however.
\end{remark}
\medskip

\emph{In the remainder of this paper, all fields are assumed to be algebraic number fields unless stated otherwise.}

\section{STBC Design criteria}\label{sec6}

Let $\cC\subset \Mat_n(\C)$ be a space-time block code. In order for $\cC$ to perform well, it should satisfy property~(1) below (as remarked before) and as many of the other properties as possible.

 \begin{enumerate}[(1)]

\item It is \emph{fully diverse}: $\det(X-X')\not=0$ for all matrices $X\not=X',$ $X,X'\in \mathcal{C} $.
\item It has \emph{full rate}, which means that the $n^2$ degrees of freedom are used to transmit $n^2$ information symbols.
\item It has \emph{non-vanishing determinant}: the minimum determinant
of the code,
$$\delta(\mathcal{C})=\inf_{X'\not=X''\in \mathcal{C}}|\det(X'-X'')|^2,$$
 is bounded below by a constant even if the codebook $\mathcal{C}$   is infinite.
\item It has \emph{cubic shaping}: each layer of a codeword is of the form $Rv$, where $R$ is a
unitary matrix and $v$ is a vector containing the information symbols.  As a consequence it is \emph{information lossless}.

\item It induces \emph{uniform average energy per antenna}: the $i$th antenna will transmit the $i$th row of the codeword;
on average, the norms of the rows should be equal in order to have a balanced repartition of
the energy at the transmitter.
 \end{enumerate}

These properties, originally considered for codes based on associative division algebras, also make sense in the nonassociative case.
Codes that satisfy all the properties above are called \emph{perfect codes}. We refer to \cite{O-R-B-V} for more details. The Golden Code \cite{B-R-V} is the best performing $2\x 2$ perfect STBC, cf. \cite{Og}.

\section{$2\times 2$ Codebooks from nonassociative quaternion division algebras}
\label{sec7}

STBCs based on associative quaternion algebras seem to have been considered explicitly for the first time in \cite{BR}. See also \cite{U-M} for more details. In this section we  look at the construction of STBCs based on nonassociative quaternion algebras.

Roughly speaking, constructing a nonassociative quaternion division
algebra boils down to choosing the nonzero scalar $b$ in the quadratic field extension $K=F(\sqrt{a})$ of the base field
$F$, and \emph{not} in $F$ itself. In contrast, $b$ \emph{is} chosen in $F$ in
the construction of a classical generalized quaternion
 algebra $(a,b)_F$ over $F$.  This usually gives us more
freedom of choice for $b$, despite the restriction that we will still have to require $|b|^2=1$ in order to get a
balanced repartition of the energy at the transmitter.
The choice of $F=\mathbb{Q}(i)$ allows us to transmit QAM constellations.

\subsection{Fully diverse codebook construction} \label{sec6.1}
Let $K$ be a quadratic field extension of $F$ with non-trivial Galois automorphism $\sigma$.
Let $A=\Cay(K,b)$  be a nonassociative quaternion division algebra over $F$ (so $b\in K\setminus F$) with $F$-basis $\{1,i,j,ji\}$ (see Remark~\ref{rem4.1}).

The algebra  $A$ is $K$-associative by Theorem~\ref{thm4}, hence we can consider $A$ as a right vector space over the subfield $K$ of $A$. The field $K$ is maximal with respect to inclusion. For $x\in A$, the left multiplication
 $\lambda_x:A\to A$, $a\mapsto xa$, is a $K$-linear endomorphism of the right $K$-vector space $A$.
Therefore $\lambda_x\in \End_K(A)$ and we get an injective $K$-linear map
$$\lambda: A \hookrightarrow \End_K(A), x\mapsto \lambda_x.$$
Consider the $K$-basis $\{1,j\}$ of $A$. Then $\End_K(A)\cong \Mat_2(K)$ as vector spaces and
we get an embedding $\lambda:A\hookrightarrow {\rm Mat}_2(K)$ of vector spaces, which sends $x\in A$ to the matrix of $\lambda_x$ with respect to the basis
$\{1,j\}$.

\begin{lemma}\label{lem6.1}
\[\lambda(A)\cong \Biggl\{\left [\begin {array}{cc}
x_0 &  b\sigma(x_1)\\
x_1 & \sigma(x_0)
\end {array}\right ] \,\Bigg\vert\, x_0,x_1\in K \Biggr\}
 \]
\end{lemma}

\begin{proof}
Let $x=x_0+jx_1\in A$ with $x_0, x_1 \in K$. Then
\begin{align*}
\lambda_x(1)&=x_0+jx_1,\\
\lambda_x(j)&=x_0j+jx_1j=j\sigma(x_0)+j^2\sigma(x_1)=b\sigma(x_1)+j\sigma(x_0)
\end{align*}
by the rules in Remark~\ref{rem4.1}.
\end{proof}

\begin{lemma}
For any
\[0\not=X= \left[\begin {array}{cc}
x_0 &  b\sigma(x_1)\\
x_1 & \sigma(x_0)\\
\end {array}\right ]
 \]
with $x_i\in K$ for $i=1,2$,  we have
$$\det(X)=N_{K/F}(x_0)-b N_{K/F}(x_1)\not=0.$$
\end{lemma}

\begin{proof} For $b=u+\sqrt{a}v\in K$, $u,v\in F,$ $v\not=0$, we compute
$$\det(X)=x_0\sigma(x_0)- b\sigma(x_1)x_1=(N_{K/F}(x_0)-u N_{K/F}(x_1))-\sqrt{a}v  N_{K/F}(x_1).$$
Since $(x_0,x_1)\not=(0,0)$ and since $N_{K/F}(x)=0$ iff $x=0$, we get
$\det(X)\not=0$.
\end{proof}

Since $X\pm Y\in \lambda(A)$ for all $X,Y\in \lambda(A)$, the difference of any two distinct elements in $\lambda(A)$ will have non-zero determinant.
 Therefore the (infinite linear) codebook built on $A$, $\cC:=\lambda(A)$,
is fully diverse.

\subsection{Non-vanishing determinant} \label{S6.2}

We closely follow the approach in \cite[\S17]{Be-Og2}.
The minimum determinant of $\cC$ determines the coding gain and is defined as
$$\delta(\mathcal{C})=\inf_{X'\not=X''\in \mathcal{C}}|\det(X'-X'')|^2.$$
The discussion in
 \cite[p.~73]{Be-Og2} can easily be adapted to the more general set-up of  nonassociative
algebras. Since the codebook $\cC$ is linear (it is based on an algebra) we have
$$\delta(\mathcal{C})=\inf_{0\not=X\in \mathcal{C}}|\det(X)|^2.$$
Let us compute the minimum determinant  of the codebook
\[\mathcal{C}=\Biggl\{\left [\begin {array}{cc}
c+d\sqrt{a} &  b(e-f\sqrt{a})\\
e+f\sqrt{a} & c-d\sqrt{a}
\end {array}\right ]\,\Bigg|\,c,d,e,f\in F \Biggr\},
 \]
obtained from the nonassociative quaternion division algebra $A=\Cay(K,b)$ with $K=F(\sqrt{a})$ and $b\in K\setminus F$ in \S\ref{sec6.1}.
We obtain
$$\delta(\mathcal{C})=\inf_{c,d,e,f\in F}|N_{K/F}(c+d\sqrt{a})-b
N_{K/F}(e+f\sqrt{a})|^2$$
with the infimum taken over all $(c,d,e,f)\not=(0,0,0,0)$, or equivalently
$$\delta(\mathcal{C})=\inf_{c,d,e,f\in F}|c^2-ad^2-be^2+abf^2|^2.$$
Thus
 $$\delta(\mathcal{C})\in K\cap \mathbb{R}^+.$$
Since $A$ is a division algebra, $\delta(\cC)\not=0$. If the code $\cC$ is finite, i.e.
 if the information symbols $c,d,e,f$ belong to a finite constellation in $F$, then $\delta(\cC)$ is bounded below by a constant. If the constellation size increases however, $\delta(\mathcal{C})$ can get arbitrarily
close to zero (e.g. let $(c,d,e,f)=(\frac{1}{n},0,0,0)$; as $n$ increases, $\delta(\cC)$ will approach zero). This will also be the case for infinite codes.

Codes whose  minimum determinant is bounded below by a constant which is independent of the size of the constellation from which the information symbols are chosen are said to satisfy  the \emph{non-vanishing determinant} (NVD) property, cf. Section~\ref{sec6}.

For associative division algebras over a number field $F$ and with maximal subfield $K$ infinite codes that satisfy
 the NVD property can often be obtained by restricting the entries in the codebook to the ring of integers $\mathcal{O}_K$. If $F=\Q$ or $F$ is  quadratic imaginary, then the resulting code will still be infinite, and its minimum determinant is guaranteed to be bounded
 away from zero, cf. \cite[Cor.~17.8]{Be-Og2}.

 Let us look at what happens for a code $\cC$, based on a nonassociative quaternion division algebra.

\begin{proposition}\label{prop6.2}
 Let $F$ be a number field and let $K=F(\sqrt{a})$ for some nonzero square-free
$a\in \mathcal{O}_F$. Let $b\in K\setminus F$.
Let $\cC=\lambda(\Cay(K,b))$ and let $\mathcal{C}_{\mathcal{O}_K}$ denote the code  obtained from $\cC$
by restricting the elements of $K$ to elements of $\mathcal{O}_K$.
Then there exists a constant $c>0$ such that
\[\delta(\mathcal{C}_{\mathcal{O}_K})\in \frac{1}{c} \cO_K\cap \mathbb{R}^+.\]
If $K$ is quadratic imaginary, then there exists an integer $d>0$ such that
\[\delta(\mathcal{C}_{\mathcal{O}_K})\geq \frac{1}{d}\]
$($and so $\mathcal{C}_{\mathcal{O}_K}$ satisfies the NVD property$)$, otherwise $\delta(\mathcal{C}_{\mathcal{O}_K})$ can become arbitrarily small.
\end{proposition}

 \begin{proof} Write $b$ as a fraction $b=\frac{b_n}{b_d}$ with $b_n,b_d\in \mathcal{O}_K$
(not necessarily unique)
 and $b_d\not=0$. A codeword of $\mathcal{C}_{\mathcal{O}_K}$ is of the form
\[ \left[\begin {array}{cc}
u &  b\sigma(v)\\
v & \sigma(u)\\
\end {array}\right ]
 \]
with $u,v\in \cO_K$. Thus we have
\begin{align*}
\delta(\mathcal{C}_{\mathcal{O}_K})&=\inf_{u,v\in \mathcal{O}_K}|N_{K/F}(u)-b N_{K/F}(v)|^2\\
&=\inf_{u,v\in \mathcal{O}_K} \frac{1}{|b_d|^2}
|b_d N_{K/F}(u)-b_n N_{K/F}(v)|^2\in \frac{1}{|b_d|^2}  \mathcal{O}_K\cap \mathbb{R}^+
\end{align*}
with the infimum taken over all $(u,v)\not=(0,0)$.  Taking $c=|b_d|^2$ establishes the first part of the proposition.

Assume that $K$ is  quadratic imaginary (i.e. $F=\Q$ and  $a<0$). Then  it follows from Proposition~\ref{propA1} that   $\cO_K \cap \mathbb{R}^+=\mathbb{N}$ and that
$|b_d|^2=N_{K/\Q}(b_d)$ is a positive integer. Thus,   among all possible pairs
 $(b_n, b_d) \in \cO_K^2$ (with $b_d\not=0$) that satisfy $b=\frac{b_n}{b_d}$,
we can choose a pair $(b_n,b_d)$ in such a way that $|b_d|^2$ is minimal. Let $d=|b_d|^2$.

If $K$ is  not quadratic imaginary it follows from the Dirichlet Unit Theorem (Proposition~\ref{propA8}) that $\cO_K$ contains units $u$ such that $|u|^2$ is arbitrarily large or small,
so that
$\delta(\mathcal{C}_{\mathcal{O}_K})$ can become arbitrarily small.
\end{proof}

\begin{remark}
 It follows from the proposition that codes based on nonassociative quaternion algebras only satisfy the NVD property  if we assume that
$K$ is a quadratic imaginary number field.

Assume that $K$ is  quadratic imaginary.   If we assume in addition that $\cO_K$ is a unique factorization domain (or, equivalently, a principal ideal domain; cf.  Appendix~\ref{AppA}), we can write $b$ as an irreducible fraction $b=b_n/b_d$ and $b_d$ will be unique up to multiplication by a unit. By the Dirichlet Unit Theorem the only units in $\cO_K$ are roots of unity, so that $|b_d|^2$ is unique. For information symbols $u$, $v$ taken from a QAM constellation we use the field $K=\Q(i)$ which is quadratic imaginary and whose ring of integers $\cO_K=\Z[i]$ is a unique factorization domain.
\end{remark}

\subsection{Information lossless encoding} A code $\cC$ is \emph{information lossless} if
it is obtained from information symbols in such a way that the energy needed to transmit them is the same as the energy needed to transmit the information symbols without encoding. By \cite[Prop. 3.5]{O-B-V} it suffices to construct the layers  of each codeword from the information symbols vector by applying a unitary matrix. This procedure is called \emph{cubic shaping} (cf. \cite[p.~3886]{O-R-B-V}), as it corresponds to an isometry transformation of the cubic lattice $\mathbb{Z}[i]^n$ in the case of information symbols taken from a QAM constellation. The term \emph{good shaping} is also used.

The energy needed to transmit  a complex number $z$ is determined by $|z|^2$. The energy needed to transmit a codeword $X=[x_{i,j}]\in \cC$ is determined by  its squared Frobenius norm $\Vert X\Vert^2 = \sum_{i,j} |x_{i,j}|^2$.

Information lossless encoding of information symbols $c,d,e,f \in F$ into a codeword
\[ X=
\left [\begin {array}{cc}
x_0 &  b\sigma(x_1)\\
x_1 & \sigma(x_0)
\end {array}\right ]
\]
of $\cC$ can be done as follows. Let $\{u_0,u_1\}$ be an $F$-basis of $K$. Let
\[ G=
\left [\begin {array}{cc}
u_0 &  u_1\\
\sigma(u_0) & \sigma(u_1)
\end {array}\right ]
\]
be the matrix of the embeddings of  the basis. Let $x_0=cu_0+du_1$ and $x_1=eu_0+fu_1$ be elements of $K$ and consider the vectors $\mathbf{x}_0=(c,d)^T$, $\mathbf{x}_1=(e,f)^T$, each containing two information symbols.
Then
$$G \mathbf{x}_0=(x_0,\sigma(x_0))^T, \quad G \mathbf{x}_1=(x_1,\sigma(x_1))^T.$$
Let $\Gamma_1=I_2$ be the identity matrix and
\[ \Gamma_2=
\left [\begin {array}{cc}
0 &  b\\
1 & 0
\end {array}\right ].
\]
Then we can write $X$ as the sum of its layers,
$$X=\Gamma_1{\rm diag}(G\mathbf{x}_0)+\Gamma_2{\rm diag}(G\mathbf{x}_1)=\left [\begin {array}{cc}
x_0 &  0\\
0 & \sigma(x_0)
\end {array}\right ] +
\left [\begin {array}{cc}
0 &  b\sigma(x_1)\\
x_1 & 0
\end {array}\right ].$$
In order for the encoding to be information lossless, we want $\Gamma_2$ and $G$ to be unitary. Note that the matrix $\Gamma_2$ is unitary if and only if $|b|^2=1$.

In order to find a good unitary matrix $G$ (if one exists) and also to satisfy the NVD property, one usually restricts $x_0$ and $x_1$ to the ring of integers $\cO_K$ or an ideal $I$ of $\cO_K$ with ``good'' properties. See \cite[\S4.3-4.4]{O-B-V} for more details.
\bigskip

In the rest of this section we construct fully diverse $2\x 2$ codebooks $\lambda(A)$, based on nonassociative quaternion algebras $A$. From these we construct codebooks $\cC$ that satisfy the NVD and/or cubic shaping properties in certain cases. The codebooks are all infinite. When restricting entries of the codewords to the ring of integers $\cO_K$, we indicate this by writing $\cC_{\cO_K}$.

\subsection{Nonassociative Alamouti codes}

Let $F=\mathbb{R}$, $K=\mathbb{C}$ and let $\sigma=\can$ be  complex
conjugation. Recall \cite{Al} that the Alamouti Code is obtained from the quaternion division algebra $\mathbb{H}=(-1,-1)_\mathbb{R}$ over
$\mathbb{R}$ and yields codewords of the form
\[\left [\begin {array}{cc}
c+id &  -e+if\\
e+if & c-id
\end {array}\right ],
\]
with  $c+id,e+if$ the information symbols  ($c,d,e,f\in \mathbb{R}$). Used with QAM symbols it achieves the
diversity-multiplexing gain trade-off (DMT) of a MISO channel with 2 transmit antennas and 1 receive antenna.

\begin{example} \label{ex7.1} (i)
 Let $A=\Cay(\mathbb{C},i)$.  Then
\[\lambda(A)= \Biggl\{\left [\begin {array}{cc}
c+id &  f+ie\\
e+if & c-id
\end {array}\right ]\,\Bigg|\, c+id,e+if\in \mathbb{C} \Biggr\}.
\]
The codebook obtained from $\Cay(\mathbb{C},i)$ closely resembles the Alamouti Code.

Consider also  $B=\Cay(\mathbb{C},-i)$.  Then
\[\lambda(B)= \Biggl\{\left [\begin {array}{cc}
c+id &  -(f+ie)\\
e+if & c-id
\end {array}\right ]\,\Bigg|\, c+id,e+if\in \mathbb{C} \Biggr\}
\]
is another Alamouti-like code. Note that the algebras $A$ and $B$ are isomorphic by Remark~\ref{rem4.3}.

Let us now consider QAM symbols only.

(ii) Let  $A=\Cay(\mathbb{Q}(i),i)$ over $\mathbb{Q}$ and restrict the entries in $\lambda(A)$ to $\Z[i]$.
Then
\[\lambda(A)_{\Z[i]}= \Biggl\{\left [\begin {array}{cc}
c+id &  f+ie\\
e+if & c-id
\end {array}\right ]\,\Bigg|\, c+id,e+if\in \mathbb{Z}[i] \Biggr\}.
\]
From $\lambda(A)_{\Z[i]}$ we obtain a code $\cC_A$ with good shaping as follows: $\{1,i\}$ is a $\mathbb{Z}$-basis of $\mathbb{Z}[i]$ and
\[ \frac{1}{\sqrt{2}}G=\frac{1}{\sqrt{2}}
\left [\begin {array}{cc}
1 & i\\
1 & -i
\end {array}\right ]
\]
is a unitary matrix. So a codeword of $\cC_A$ is given by
\[\frac{1}{\sqrt{2}} \left [\begin {array}{cc}
c+id &  f+ie\\
e+if & c-id
\end {array}\right ],
\]
$c,d,e,f\in\mathbb{Z}$.

(iii) Similarly, $B=\Cay(\mathbb{Q}(i),-i)$ yields
\[\lambda(B)_{\Z[i]}= \Biggl\{\left [\begin {array}{cc}
c+id &  -(f+ie)\\
e+if & c-id
\end {array}\right ]\,\Bigg|\, c+id,e+if\in \mathbb{Z}[i] \Biggr\},
\]
resulting in a shaped code  $\cC_B$ with codewords
\[\frac{1}{\sqrt{2}} \left [\begin {array}{cc}
c+id &  -(f+ie)\\
e+if & c-id
\end {array}\right ],
\]
$c,d,e,f\in\mathbb{Z}$.

In examples  (ii) and (iii), $F=\mathbb{Q}$, $K=\mathbb{Q}(i)$ is a quadratic imaginary number field, $b=\pm i$ and $\cO_K=\mathbb{Z}[i]$ is a principal ideal domain.
Hence, before shaping, the minimum determinant of each code is bounded below by $1$
by Proposition~\ref{prop6.2}. Thus the minimum determinant of both shaped codes is lower bounded by the constant~${1}/{2}$.

To summarize: the codes in (ii) and (iii) are fully diverse, satisfy the NVD property, have good shaping and clearly also satisfy the  uniform average transmitted energy per antenna property. Used for a $2\times 2$ MIMO channel
they are only half-rate though, since 4 transmitted signals are used to transmit 2 QAM  information symbols.
\end{example}

\subsection{Nonassociative Golden Codes}\label{S7.2}

  Let $F=\mathbb{Q}(i)$ and let $K= \mathbb{Q}(i)(\sqrt{5})$.  Then $\mathcal{O}_K=\mathbb{Z}[i][\frac{1+\sqrt{5}}{2}]$.
The Golden Code \cite{B-R-V} uses the maximal $\mathbb{Z}[i]$-order  in the quaternion division algebra
$(5,i)_{\mathbb{Q}(i)}=\Cay\bigl( \mathbb{Q}(i)(\sqrt{5}),i\bigr)$ which can be described
by the Cayley-Dickson doubling $\Cay\Bigl(\mathbb{Z}[i][\frac{1+\sqrt{5}}{2}],i\Bigr)$, defined in the obvious way. (Note that associative quaternion algebras are precisely the cyclic algebras of dimension $4$.)
After
 shaping by
$\frac{1}{\sqrt{5}}$, a codeword of $\cC$ is thus of the form
\[ X=\frac{1}{\sqrt{5}}
\left [\begin {array}{cc}
c+d\theta &  e+f\theta\\
i(e+f\sigma(\theta) ) & c+d\sigma({\theta})
\end {array}\right ]
\]
with $\theta=\frac{1+\sqrt{5}}{2}$ the golden number, $\sigma:\mathbb{Q}(i)(\sqrt{5})\to \mathbb{Q}(i)(\sqrt{5})$,
$\sigma(i)= i$, $\sigma(\sqrt{5})= -\sqrt{5}$ and $c,d,e,f\in \mathbb{Z}[i]$.
To obtain an energy-efficient code, the entries in the codewords are then restricted to elements in the principal
ideal $I$ in $\mathcal{O}_K$ of norm 5 in $\mathbb{Q}$ which is generated by $\alpha=1+i-i\theta$. So, finally the Golden Code is
given by the codewords
\[ X=\frac{1}{\sqrt{5}}
\left [\begin {array}{cc}
\alpha(c+d\theta) &  \alpha(e+f\theta)\\
i\sigma(\alpha)(e+f\sigma(\theta)) & \sigma(\alpha) \sigma(c+d\sigma(\theta))
\end {array}\right ]
\]
with $c,d,e,f\in \mathbb{Z}[i]$ and has minimum determinant $1/5$. We refer to \cite{B-R-V} for the details.

\begin{example} \label{ex7.3}
Consider the nonassociative quaternion division algebra
$$A=\Cay\Bigl( \mathbb{Q}(i)(\sqrt{5}),\frac{i+\sqrt{5}}{i-\sqrt{5}}\Bigr)$$
 over
$\mathbb{Q}(i)$, where $|\frac{i+\sqrt{5}}{i-\sqrt{5}}|^2=1$ guarantees that $\Gamma_2$ is unitary.

The codebook based on $A$ is
\[\lambda(A)=\Biggl\{\left [\begin {array}{cc}
c+d\theta &  e+f\theta\\
\frac{i+\sqrt{5}}{i-\sqrt{5}}(e+f\sigma(\theta) ) & c+d\sigma(\theta)
\end {array}\right ]\,\Bigg|\,c,d,e,f\in \mathbb{Q}(i) \Biggr\}.
 \]
(Compared to the general code construction in Lemma~\ref{lem6.1}, we are transposing the matrices here in order  to better compare them  with the Golden Code matrices above. This does not influence the  behaviour of the code.) The code has full diversity and uniform average transmitted energy  per antenna. Now
$\{1,\theta\}$ is a $\mathbb{Q}(i)$-basis of $\mathbb{Q}(i)(\sqrt{5})$, but
\[ G=
\left [\begin {array}{cc}
1 &  \theta\\
1 & \sigma(\theta)
\end {array}\right ]
\]
is not a unitary matrix, so we have an energy shaping loss.  To obtain an energy-efficient code,  restrict
the entries in the codeword again to elements in the principal
ideal $I$ in $\mathcal{O}_K$ generated by $\alpha=1+i-i\theta$. Then a nonassociative Golden Code is
given by the codewords
\[ X=\frac{1}{\sqrt{5}}
\left [\begin {array}{cc}
\alpha(c+d\theta) &  \alpha(e+f\theta)\\
\frac{i+\sqrt{5}}{i-\sqrt{5}}\sigma(\alpha)(e+f\sigma(\theta)) & \sigma(\alpha) \sigma(c+d\sigma(\theta))
\end {array}\right ]
\]
with $c,d,e,f\in \mathbb{Z}[i]$. The choice of the ideal $I$ is optimal here for the exact same reasons as the ones
given in \cite[p.~1433]{B-R-V} and yields good shaping. The code is also
 full rate, fully diverse and has uniform average transmitted energy per antenna.
\end{example}

With the same arguments codes can be constructed using  any of the infinitely many scalars $b\in  \mathbb{Q}(i)(\sqrt{5})\setminus \Q(i)$ with $|b|^2=1$. All these codes, however, have vanishing determinant by Proposition~\ref{prop6.2}. We give another example:

\begin{example} \label{ex7.4}
Let $b=\frac{2i+\sqrt{5}}{3}$ and consider
the nonassociative quaternion algebra
$$\Cay\Bigl( \mathbb{Q}(i)(\sqrt{5}),\frac{2i+\sqrt{5}}{3}\Bigr)$$
 over $\mathbb{Q}(i)$. Again $|\frac{2i+\sqrt{5}}{3}|^2=1$ and we obtain another
  code which has full diversity
and uniform average transmitted energy per antenna. To obtain an energy-efficient code, we restrict
the entries in the codeword again to elements in the principal
ideal $I$ in $\mathcal{O}_K$ generated by $\alpha=1+i-i\theta$. Then another nonassociative Golden Code with good shaping is
given by the codewords
\[ X=\frac{1}{\sqrt{5}}
\left [\begin {array}{cc}
\alpha(c+d\theta) &  \alpha(e+f\theta)\\
\frac{2i+\sqrt{5}}{3}\sigma(\alpha)(e+f\sigma(\theta)) & \sigma(\alpha) \sigma(c+d\sigma(\theta))
\end {array}\right ]
\]
with $c,d,e,f\in \mathbb{Z}[i]$.
\end{example}

\subsection{Optimality of the Golden Code}
Oggier \cite{Og} shows that the Golden Code is optimal inside the class of cyclic algebra based $2\times 2$ codes built over fields $K=\Q(i)(\sqrt{d})$    in the following sense: the minimum determinant of such codes is inversely proportional to $|d_{K/\Q(i)}|$, where $d_{K/\Q(i)}$ denotes the relative discriminant of $K/\Q(i)$.
For the Golden Code $|d_{K/\Q(i)}|=5$.
While it is possible to consider fields $K=\Q(i)(\sqrt{d})$ with $|d_{K/\Q(i)}|<5$, Oggier shows that  the resulting codes are no longer fully diverse \cite[III]{Og}.

This problem does not occur in the nonassociative case by Theorem~\ref{thm4}. It is possible to construct fully diverse nonassociative codes over fields $K=\Q(i)(\sqrt{d})$ with $|d_{K/\Q(i)}|<5$,  but by Proposition~\ref{prop6.2} these codes do not satisfy the NVD property.  In the examples below we will consider the cases $|d_{K/\mathbb{Q}(i)}|=4$ and $|d_{K/\mathbb{Q}(i)}|=3$. The case  $|d_{K/\mathbb{Q}(i)}|=2$ does not exist, cf. Proposition~\ref{propA5}.

\begin{example}\label{ex7.5}
Let  $K=\mathbb{Q}(i)(\sqrt{2})$.
Then $|d_{K/\mathbb{Q}(i)}|=4$ and $\sigma(\sqrt{2})=-\sqrt{2}$. Moreover, $K= \mathbb{Q}(i)(\zeta_8)$ where $\zeta_8=\frac{1+i}{\sqrt{2}}$
is an 8th root of unity and $\sigma(\zeta_8)=-\zeta_8$.  We have that $\{1,\zeta_8\}$ is a $\mathbb{Z}[i]$-basis for the ring of integers
$\mathcal{O}_K=\mathbb{Z}[i][\zeta_8]$.

 Consider the nonassociative quaternion division algebra
$A=\Cay( \mathbb{Q}(\zeta_8),\zeta_8)$
over $ \mathbb{Q}(i)$.
The choice of $b=\zeta_8$  guarantees that $\Gamma_2$ is unitary, since $|\zeta_8|^2=1$.
We obtain the codebook
\begin{align*}
\lambda(A)&=\Biggl\{
\left [\begin {array}{cc}
x_0 &  \zeta_8\sigma(x_1)\\
x_1 & \sigma(x_0)
\end {array}\right ]\, \Bigg|\, x_0, x_1\in \mathbb{Q}(i)(\zeta_8) \Biggr\} \\
&=\Biggl\{
\left [\begin {array}{cc}
u_0+\zeta_8 w_0 &  \zeta_8(u_1-\zeta_8 w_1)\\
u_1+\zeta_8 w_1 & u_0-\zeta_8 w_0
\end {array}\right ] \,\Bigg|\, u_0, u_1,w_0, w_1\in \mathbb{Q}(i) \Biggr\}.
\end{align*}
Now $\{1,\zeta_8\}$ is a $\mathbb{Q}(i)$-basis of $\mathbb{Q}(i)(\zeta_8)$ and
\[ \frac{1}{\sqrt{2}}G=  \frac{1}{\sqrt{2}}
\left [\begin {array}{cc}
1 &  \zeta_8\\
1 & \sigma(\zeta_8)
\end {array}\right ]
=  \frac{1}{\sqrt{2}}
\left [\begin {array}{cc}
1 &  \zeta_8\\
1 & -\zeta_8
\end {array}\right ]
\]
is a unitary matrix. So after multiplying the matrices in the codebook by $ \frac{1}{\sqrt{2}}$ and restricting the information symbols to $\Z[i]$, we obtain a code that has good
shaping:
\[ \mathcal{C}=
\Biggl\{ \frac{1}{\sqrt{2}}
\left [\begin {array}{cc}
u_0+\zeta_8 w_0 &  \zeta_8(u_1-\zeta_8 w_1)\\
u_1+\zeta_8 w_1 & u_0-\zeta_8 w_0
\end {array}\right ] \,\Bigg|\, u_0, u_1,w_0, w_1\in \mathbb{Z}[i] \Biggr\}.
\]
The code $\cC$ is full rate, has full diversity and good shaping.
The factor $\zeta_8$ in the
first row of the codeword guarantees uniform average transmitted energy per antenna since $|\zeta_8|^2=1$.
This code does not satisfy the NVD property however by Proposition~\ref{prop6.2}.
\end{example}

\begin{example}
Let $K=\mathbb{Q}(i)(\sqrt{3})$. Then $|d_{K/\mathbb{Q}(i)}|=3$
 and $\sigma(\sqrt{3})=-\sqrt{3}$.
Moreover, $K=\mathbb{Q}(i)(\zeta_3)$ where $\zeta_3=e^{2\pi i /3}=\frac{-1+i\sqrt{3}}{2}$ is a third root of unity. We have $\sigma(\zeta_3)=\frac{-1-i\sqrt{3}}{2}=\overline{\zeta_3}$.
We know that $\{1,\zeta_3\}$ is a $\mathbb{Z}[i]$-basis for the ring of integers
$\mathcal{O}_K=\mathbb{Z}[i][\zeta_3]$.

Consider the nonassociative quaternion division algebra
$A=\Cay(\mathbb{Q}(i)(\zeta_3),\zeta_3)$ over $\mathbb{Q}(i)$. We obtain the codebook
\[\lambda(A)=\Biggl\{\left [\begin {array}{cc}
u_0+\zeta_3 w_0 &  \zeta_3(u_1+\overline{\zeta_3} w_1)\\
u_1+\zeta_3 w_1& u_0+\overline{\zeta_3}w_0
\end {array}\right ]\,\Bigg|\,u_0, u_1,w_0, w_1\in \mathbb{Q}(i) \Biggr\}.
 \]
 This  time the matrix $G$ (up to scaling) is not unitary. Thus the energy required to send the linear combination of the information symbols
 on each layer is higher than the energy needed
to send the information symbols themselves and we would
 still have to optimize for energy efficiency. In addition the discriminant of this code is not bounded away from  zero by Proposition~\ref{prop6.2}.
\end{example}

\section{$2\times 4$ Multiblock space-time codes from nonassociative quaternion algebras}\label{sec8}

Let $F$  be a number field, let $a\in F^\x$ and
let $K=F(\sqrt{a})$ be a quadratic field extension of $F$ with non-trivial Galois automorphism $\sigma$
and norm $N_{K/F}(x)=x\sigma(x)$.  Let $b\in K\setminus F$, so that
 $A=\Cay(K,b)$ is a nonassociative quaternion division algebra.
 A $2\x 4$  \emph{multiblock space-time code} based on $A$ is a set of matrices of the form $Y=[X |\sigma(X)]$ where $X\in \lambda(A)$
 (see \cite{L-M-M} for a more general construction in the associative case).

This yields a codebook $\cC$ consisting of  matrices of the form
\[Y= [X |\sigma(X)]= \left [\begin {array}{cc|cc}
x_0 &  b\sigma(x_1) &    \sigma(x_0) &  \sigma(b) x_1\\
x_1 & \sigma(x_0) &  \sigma(x_1) &  x_0   \\
\end {array}\right ]
 \]
with $x_0, x_1\in K$. They have full rank since $X$ comes from the division algebra $A$. In this set-up, we want the code to satisfy a \emph{generalized} NVD property (see \cite{Lu}) which can be achieved by bounding
the \emph{generalized minimum determinant}
$$\delta_g(\mathcal{C})=\inf_{0\not=X \in \lambda(A)}|\det(X) \det(\sigma(X))|$$
away from zero.

\begin{proposition}
Let $F$ be a number field and let $K=F(\sqrt{a})$ for some nonzero square-free
$a\in \mathcal{O}_F$. Let $b\in K\setminus F$.
Let $A=\Cay(K,b)$ and let
\[\mathcal{C}_{\mathcal{O}_K}=\{[X|\sigma(X)] \mid X\in \lambda(A)_{\cO_K}\}\]
denote the $2\x 4$ multiblock code  obtained from $\cC$
by restricting the elements of $K$ to elements of $\mathcal{O}_K$.
Then there exists a constant $c>0$ such that
\[\delta_g(\mathcal{C}_{\mathcal{O}_K})\in \frac{1}{c} \cO_F\cap \mathbb{R}^+.\]
If $F=\mathbb{Q}$ or $F$ is  quadratic imaginary, then there exists an integer $d>0$ such that
\[\delta_g(\mathcal{C}_{\mathcal{O}_K})\geq \frac{1}{\sqrt{d}}\]
$($and so $\mathcal{C}_{\mathcal{O}_K}$ satisfies the generalized NVD property$)$, otherwise $\delta_g(\mathcal{C}_{\mathcal{O}_K})$ can become arbitrarily small.
\end{proposition}

\begin{proof} Write $b$ as a fraction $b=\frac{b_n}{b_d}$ with $b_n,b_d\in \cO_K$ (not necessarily unique) and $b_d\not=0$.
We have
\begin{align*}
\delta_g(\mathcal{C}_{\mathcal{O}_K})&=\inf_{0\not=X \in \lambda(A)_{\mathcal{O}_K}}|\det(X) \det(\sigma(X))|\\
&=\inf_{0\not=X \in \lambda(A)_{\mathcal{O}_K}}|\det(X) \sigma(\det(X))|\\
&=\inf_{0\not=X \in\lambda(A)_{\mathcal{O}_K}}|N_{K/F}(\det(X))|\\
&=\mathop{\inf_{x_0,x_1\in \mathcal{O}_K}}_{(x_0,x_1)\not=(0,0)  }|N_{K/F}(N_{K/F}(x_0)-bN_{K/F}(x_1))|\\
&=\mathop{\inf_{x_0,x_1\in \mathcal{O}_K}}_{(x_0,x_1)\not=(0,0)  }  \frac{1} {|N_{K/F}(b_d)|}
|N_{K/F}(b_dN_{K/F}(x_0)-b_nN_{K/F}(x_1))|\\
& \in \frac{1}{|N_{K/F}(b_d)|} \cO_F \cap \R^+
\end{align*}
since $N_{K/F}(\mathcal{O}_K)\subset \mathcal{O}_F$. Taking $c= |N_{K/F}(b_d)|$ establishes the first part of the proposition.

Assume that $F=\Q$. Then $|N_{K/F}(b_d)|$ is a positive integer. Thus,   among all possible pairs
 $(b_n, b_d) \in \cO_K^2$ (with $b_d\not=0$) that satisfy $b=\frac{b_n}{b_d}$,
we can choose a pair $(b_n,b_d)$ in such a way  that $|N_{K/F}(b_d)|$ is minimal . Furthermore, $\cO_F\cap \R^+ =\N$. We let $d= |N_{K/F}(b_d)|^2$ in this case.

Next assume that $F$ is  quadratic imaginary (i.e. $F=\Q(\sqrt{m})$ and  $m<0$). Then it follows
from Proposition~\ref{propA1} that $\cO_F \cap \mathbb{R}^+=\mathbb{N}$ and that $|N_{K/F}(b_d)|^2=N_{F/\Q}(N_{K/F}(b_d))$ is a positive integer.
Thus,   among all possible pairs
 $(b_n, b_d) \in \cO_K^2$ (with $b_d\not=0$) that satisfy $b=\frac{b_n}{b_d}$,
we can choose a pair $(b_n,b_d)$ in such a way that
 $|N_{K/F}(b_d)|^2$ is minimal. Let $d=|N_{K/F}(b_d)|^2$.

If $F$ is  not $\Q$ or not quadratic imaginary it follows from the Dirichlet Unit Theorem (Proposition~\ref{propA8}) that $\cO_F$ contains units $u$ such that $|u|^2$ is arbitrarily large or small,  so that
$\delta_g(\mathcal{C}_{\mathcal{O}_K})$ can become arbitrarily small.
\end{proof}

\begin{remark}
If $F=\mathbb{Q}$ or $F$ is  quadratic imaginary, the generalized minimum determinant  of $\cC_{\cO_K}$ is lower bounded by a positive constant and the generalized NVD property is satisfied. As a consequence the code will achieve the diversity-multiplexing gain trade-off, as explained in \cite[p.~5232]{L-M-M}.

 If we assume in addition that $\cO_K$ is a unique factorization domain (or, equivalently, a principal ideal domain; cf.  Appendix~\ref{AppA}), we can write $b$ as an irreducible fraction $b=b_n/b_d$ and $b_d$ will be unique up to multiplication by a unit. By the Dirichlet Unit Theorem the only units in $\cO_K$ are roots of unity, so that $ |N_{K/F}(b_d)|^2$ is unique.

For QAM constellations we take $F=\Q(i)$, so that $K=\Q(i)(\sqrt{m})$ for some square-free non-zero integer $m$. Proposition~\ref{propA6} lists the fields $K$ whose ring of integers $\cO_K$ is a unique factorization domain.
\end{remark}

In the setting of multiblock space-time codes it is again  natural to ask that
$|b|^2=1$, cf. \cite[p.~5232]{L-M-M}.

\begin{example} Let $F=\Q(i)$ and
$K=\mathbb{Q}(i)(\sqrt{5})$. Let $\theta=\frac{1+\sqrt{5}}{2}$ be the golden number, $\sigma:\mathbb{Q}(i)(\sqrt{5})\to \mathbb{Q}(i)(\sqrt{5})$,
$\sigma(i)= i$, $\sigma(\sqrt{5})= -\sqrt{5}$, $b=\frac{i+\sqrt{5}}{i-\sqrt{5}}$ and
$A=\Cay( \mathbb{Q}(i)(\sqrt{5}),b)$
over $\mathbb{Q}(i)$ as in Example~\ref{ex7.3}. In order to obtain an energy-efficient code, we restrict
the entries in the codewords  to elements in the principal
ideal $I$ in $\mathcal{O}_K$,  generated by $\alpha=1+i-i\theta$. Then
\[ X=\frac{1}{\sqrt{5}}
\left [\begin {array}{cc}
\alpha(c+d\theta) & b\sigma(\alpha)(e+f\sigma(\theta)) \\
\alpha(e+f\theta) & \sigma(\alpha) (c+d\sigma(\theta))
\end {array}\right ]
\]
with $c,d,e,f\in \mathbb{Z}[i]$ and the code $\cC_{\cO_K}$ consists of  block matrices of the form
\[Y=\frac{1}{\sqrt{5}} \left [\begin {array}{cc|cc}
\alpha(c+d\theta) &  b\sigma(\alpha)(e+f\sigma(\theta)) &  \sigma(\alpha)(c+d\theta) &   \sigma(b)\alpha(e+f \theta)    \\
\alpha(e+f\theta) &  \sigma(\alpha) (c+d\sigma(\theta))   &   \sigma(\alpha)  (e+f\sigma(\theta)) &   \alpha (c+d \theta)\\
\end {array}\right ].
 \]
Note that
$$\delta_g(\mathcal{C}_{\mathcal{O}_K})\geq \frac{1}{|N_{K/F}(\sqrt{5}(i-\sqrt{5}   )  )|}=\frac{1}{30},$$
 guaranteeing that the code satisfies the generalized NVD property.
\end{example}

\begin{example} Replacing $b$ by $\frac{2i+\sqrt{5}}{3}$ (cf. Example~\ref{ex7.4})
in the previous example results in a code $\cC_{\cO_K}$ such that
$$\delta_g(\mathcal{C}_{\mathcal{O}_K})\geq \frac{1}{|N_{K/F}(3\sqrt{5}  )|}=\frac{1}{45},$$
 guaranteeing that the code satisfies the generalized NVD property.
\end{example}

\begin{example} Let  $F=\Q(i)$ and
$K=\mathbb{Q}(i)(\sqrt{2})= \mathbb{Q}(i)(\zeta_8)$ where $\zeta_8=\frac{1+i}{\sqrt{2}}$
is an 8th root of unity as in Example~\ref{ex7.5}. Let $A=\Cay( \mathbb{Q}(\zeta_8),\zeta_8)$. Then
\[X=
\frac{1}{\sqrt{2}}
\left [\begin {array}{cc}
u_0+\zeta_8 w_0 &  \zeta_8(u_1-\zeta_8 w_1)\\
u_1+\zeta_8 w_1 & u_0-\zeta_8 w_0
\end {array}\right ]\]
with $ u_0, u_1,w_0,w_1\in \mathbb{Z}[i] $ and the code $\cC_{\cO_K}$ consists of block matrices of the form
\[Y=\frac{1}{\sqrt{2}} \left [\begin {array}{cc|cc}
u_0+\zeta_8 w_0 &  \zeta_8(u_1-\zeta_8 w_1) &  u_0-\zeta_8 w_0 &  -\zeta_8
(u_1+\zeta_8 w_1)\\
u_1+\zeta_8 w_1 & u_0-\zeta_8 w_0   &    u_1-\zeta_8w_1 &  u_0+\zeta_8 w_0\\
\end {array}\right ].
 \]
We have
$$\delta_g(\mathcal{C}_{\mathcal{O}_K})\geq \frac{1}{|N_{K/F}((\sqrt{2})^2  )|}=\frac{1}{4},$$
 guaranteeing that the code satisfies the generalized NVD property.
\end{example}

\section{$4\times 4$ Codebooks from nonassociative quaternion division algebras}
\label{sec9}

Let $\mathbb{H}=(-1,-1)_\mathbb{R}$ be Hamilton's quaternion division algebra.
Its left regular representation with respect to the  basis $\{1,i,j,-ij\}$  consists of matrices of the form
\[ \left [\begin {array}{crrr}
x_0 &  -x_1 & -x_2 & -x_3 \\
x_1 & x_0  &  x_3 & - x_2 \\
x_2 & - x_3 & x_0 & x_1 \\
x_3 & x_2  & -x_1 & x_0 \\
\end {array}\right ]
 \]
with $x_\ell\in\mathbb{R}$, $\ell=0,\ldots, 3$ (cf. \cite[Example~8]{S-R-S}). This is exactly the
four-dimensional real
orthogonal design from \cite[Section III-A]{T-J-C1}.

Let us look at the left regular representation of a nonassociative
quaternion algebra $A$, this time over its base field rather than over a maximal subfield.

\subsection{Fully diverse codebook construction}

Let $F$ be a number field and let
$K=F(\sqrt{a})=F(i)$ with $i^2=a\in F^\x$ be a quadratic field extension with non-trivial Galois automorphism $\sigma:\sqrt{a}\mapsto -\sqrt{a}$.
Let $A=\Cay(K,b)$ be a nonassociative quaternion division algebra over $F$ with
$b=p+qi\in K\setminus F$, so $p,q\in F$ with $q\not=0$.
For the basis  $\{1,i,j,-ij\}$ of $A$ over $F$ the matrix representation of left multiplication with $x=x_0+x_1i+x_2j-x_3ij$
 yields the fully diverse  $4\times 4$ space-time
block code
\[ \mathcal{C}=\lambda(A)=\left.\left\{\left [\begin {array}{cccc}
x_0 &  ax_1 & px_2 -aqx_3 & aqx_2-apx_3 \\
x_1 & x_0  & qx_2-p x_3 & p x_2-aqx_3 \\
x_2 & a x_3 & x_0 & -ax_1 \\
x_3 & x_2  & -x_1 & x_0 \\
\end {array}\right ]\,\right|\, x_0,x_1,x_2,x_3\in F \right\}.
\]

 \begin{example}\label{ex9.1} Let $i^2=-1$.
 The $\mathbb{R}$-algebra $A=\Cay(\mathbb{C},i)$  yields the fully diverse $4\times 4$ space-time block code
\[\left.\left\{\left [\begin {array}{crrr}
x_0 &  -x_1 & x_3 & -x_2 \\
x_1 & x_0  & x_2 & x_3 \\
x_2 & - x_3 & x_0 & x_1 \\
x_3 & x_2  & -x_1 & x_0 \\
\end {array}\right ]\,\right|\, x_0,x_1,x_2,x_3\in \mathbb{R}\right\}.
 \]
Its matrices are not orthogonal, but their first two column vectors and, respectively,
their last two, are orthogonal to each other.
 \end{example}

\subsection{Non-vanishing determinant} Let
 $X\in \mathcal{C}$. Then
\[\det(X)=
[(x_0^2-ax_1^2)-p(x_2^2-ax_3^2)]^2-aq^2(x_2^2-ax_3^2)^2 \in F.
\]
Since the codebook is based on a division algebra, its minimum determinant equals
$$\delta(\mathcal{C})=\inf_{0\not=X\in \mathcal{C}}|\det(X)|^2$$
and is non-zero.
If the information symbols $x_0,x_1,x_2,x_3$ belong to a finite constellation in $F$,
 then $\delta(\mathcal{C})$ is bounded below by a constant which depends on the constellation size. If the constellation size increases, $\delta(\mathcal{C})$ can get arbitrarily close to zero. By restricting the entries in $\cC$ to the ring of integers $\cO_F$ we obtain for certain number fields $F$ infinite codes that satisfy the NVD property:

\begin{proposition}
 Let $F$ be a number field and let $K=F(\sqrt{a})$ for some nonzero square-free
$a\in \mathcal{O}_F$. Let $b=p+q\sqrt{a}\in K\setminus F$ with $p,q\in F$
$($so that $q\not=0$$)$.
Let $\cC=\lambda(\Cay(K,b))$ and let $\mathcal{C}_{\mathcal{O}_F}$ denote the code  obtained from $\cC$
by restricting the elements of $F$ to elements of $\mathcal{O}_F$.
Then there exists a constant $c>0$ such that
\[\delta(\mathcal{C}_{\mathcal{O}_F})\in \frac{1}{c} \cO_F\cap \mathbb{R}^+.\]
If $F=\Q$ or $F$ is quadratic imaginary, then there exists an integer $d>0$ such that
\[\delta(\mathcal{C}_{\mathcal{O}_F})\geq \frac{1}{d}\]
$($and so $\mathcal{C}_{\mathcal{O}_F}$ satisfies the NVD property$)$, otherwise $\delta(\mathcal{C}_{\mathcal{O}_F})$ can become arbitrarily small.
\end{proposition}

\begin{proof} Write $p=\frac{p_n}{p_d}$, $q=\frac{q_n}{q_d}$
with $p_n,p_d, q_n,q_d\in \mathcal{O}_F$ (not necessarily unique) and $p_d\not=0$, $q_d\not=0$.
 An easy calculation confirms that
$$
\delta(\mathcal{C}_{\mathcal{O}_F})\in \frac{1}{|q_d|^4} \mathcal{O}_F\cap \mathbb{R}^+ \text{ if } p=0,\quad
\delta(\mathcal{C}_{\mathcal{O}_F})\in \frac{1}{|p_dq_d|^4} \mathcal{O}_F\cap \mathbb{R}^+ \text{ if } p\not=0.$$
Letting $c=|q_d|^4$ if $p=0$ and  $c=|p_dq_d|^4$ if $p\not=0$ establishes the first part of the proposition.

Assume for the sake of argument that $p=0$. The case $p\not=0$ can be settled in a similar manner.

Assume that $F=\Q$.  Then $|q_d|^4$  is a positive integer.
Thus,   among all possible pairs
 $(q_n, q_d) \in \cO_F^2=\Z^2$ (with $q_d\not=0$) that satisfy $q=\frac{q_n}{q_d}$,
we can choose a pair $(q_n,q_d)$ in such a way that
 $|q_d|^4$ is minimal. Furthermore, $\cO_F\cap \R^+ =\N$.
We let $d=|q_d|^4$ in this case.

Next assume that  $F$ is  quadratic imaginary (i.e. $F=\Q(\sqrt{m})$ and  $m<0$). Then it follows from Proposition~\ref{propA1} that
$\cO_F \cap \mathbb{R}^+=\mathbb{N}$ and that
$|q_d|^4=N_{F/\Q}(q_d)^2$ is a positive integer.
Thus,   among all possible pairs
 $(q_n, q_d) \in \cO_F^2$ (with $q_d\not=0$) that satisfy $q=\frac{q_n}{q_d}$,
we can choose a pair $(q_n,q_d)$ in such a way that
$|q_d|^4$ is minimal. Let $d=|q_d|^4$.

If $F$ is  not $\Q$ or not quadratic imaginary it follows from the Dirichlet Unit Theorem (Proposition~\ref{propA8}) that $\cO_F$ contains units $u$ such that $|u|^2$ is arbitrarily large or small,  so that
$\delta(\mathcal{C}_{\mathcal{O}_F})$ can become arbitrarily small.
\end{proof}

\begin{remark}
If $F=\mathbb{Q}$ or $F$ is  quadratic imaginary, the minimum determinant  of $\cC_{\cO_F}$ is lower bounded by a positive constant
and the NVD property is satisfied.

If we assume in addition that $\cO_F$ is a unique factorization domain (or, equivalently, a principal ideal domain; cf.  Appendix~\ref{AppA}), we can write
$p$ and $q$ as  irreducible fractions $p=\frac{p_n}{p_d}$, $q=\frac{q_n}{q_d}$
and $p_d$, $q_d$ will be unique up to multiplication by a unit. By the Dirichlet Unit Theorem the only units in $\cO_F$ are roots of unity, so that $|q_d|^4$, resp.
$|p_dq_d|^4$, is unique.
\end{remark}

 \begin{example} \label{ex9.2}
The $\mathbb{Q}$-algebra $A=\Cay(\mathbb{Q}(i),i)$ yields the fully diverse $4\times 4$
space-time block code
\[ \mathcal{C}_\mathbb{Z}=\left.\left\{\left [\begin {array}{cccc}
x_0 &  -x_1 & x_3 & -x_2 \\
x_1 & x_0  & x_2 & x_3 \\
x_2 & - x_3 & x_0 & x_1 \\
x_3 & x_2  & -x_1 & x_0 \\
\end {array}\right ]\,\right|\, x_0, x_1, x_2, x_3\in \mathbb{Z}\right\}.
 \]
We obtain
$$\delta(\mathcal{C}_{\mathbb{Z}})=
\mathop{\inf_{x_0, x_1, x_2, x_3\in \mathbb{Z}}}_{
(x_0, x_1, x_2, x_3)\not=(0,0,0,0)}
|(x_0^2+x_1^2)^2+(x_2^2+x_3^2)^2 |^2=1.$$
\end{example}

\begin{example}\label{ex9.2bis}
The $\mathbb{Q}$-algebra $A=\Cay(\mathbb{Q}(i),-i)$ yields the fully diverse $4\times 4$ space-time block code
\[ \mathcal{C}_\mathbb{Z}=\left.\left\{\left [\begin {array}{cccc}
x_0 &  -x_1 & -x_3 & x_2 \\
x_1 & x_0  & -x_2 & -x_3 \\
x_2 & - x_3 & x_0 & x_1 \\
x_3 & x_2  & -x_1 & x_0 \\
\end {array}\right ]\,\right|\, x_0, x_1, x_2, x_3\in \mathbb{Z}\right\},
 \]
 again with minimum determinant $1$.
 \end{example}

In the previous two examples the first two column vectors of any codeword  and, respectively, the last two, are orthogonal to each other.

 \begin{example} \label{ex9.3}
Consider the  $ \mathbb{Q}(i)$-algebra
$A=\Cay( \mathbb{Q}(i)(\zeta_8),\zeta_8)$  where
 $\zeta_8=\frac{1+i}{\sqrt{2}}$ is an 8th root of unity. Note that $a=\zeta_8^2=i$ and $b=\zeta_8$ (so $p=0$ and $q=1$).
 We obtain the fully diverse codebook
\[ \mathcal{C}_{\mathbb{Z}[i]}=\left.\left\{\left [\begin {array}{cccc}
x_0 &  ix_1 & -ix_3 & ix_2 \\
x_1 & x_0  &  x_2 &  -ix_3 \\
x_2 & i x_3 & x_0 & -ix_1 \\
x_3 & x_2  & -x_1 & x_0 \\
\end {array}\right ]\,\right|\, x_0, x_1, x_2, x_3\in \mathbb{Z}[i]\right\}
 \]
 whose minimum determinant is
$$\delta(\mathcal{C}_{\mathbb{Z}[i]})=
\mathop{\inf_{x_0,x_1,x_2,x_3\in \mathbb{Z}[i]}}_{(x_0, x_1, x_2, x_3)\not=(0,0,0,0)}|(x_0^2-ix_1^2)^2-i(x_2^2-ix_3^2)^2|^2=1.$$

\end{example}

\subsection{Information lossless encoding}

For a (transposed) matrix
\[ X=
\left [\begin {array}{cccc}
x_0 &  x_1 & x_2 & x_3 \\
ax_1 & x_0  & a x_3 &  x_2 \\
 px_2 -aqx_3 & qx_2-p x_3  & x_0 & -x_1 \\
aqx_2-apx_3  &  p x_2-aqx_3  & -ax_1 & x_0 \\
\end {array}\right ]
\]
in $\cC$
we use the following encoding: let $I_4$ be the identity matrix and let
\[ \Gamma_1=
\left [\begin {array}{cccc}
0 & 1 & 0 & 0\\
a & 0 & 0 & 0 \\
0 & 0 & 0 & -1 \\
0 & 0 & -a & 0 \\
\end {array}\right],\
\Gamma_2=
\left [\begin {array}{cccc}
0 & 0 & 1 & 0\\
0 & 0 & 0 & 1 \\
p & q & 0 & 0 \\
aq & p & 0 & 0 \\
\end {array}\right ],\
\Gamma_3=
\left [\begin {array}{cccc}
0 & 0 & 0 & 1\\
0 & 0 & a & 0 \\
-aq & -p & 0 & 0 \\
-ap & -aq & 0 & 0 \\
\end {array}\right ].
\]
The codeword $X$ is encoded as
$$X=I_4 {\rm diag}(x_0)+\Gamma_1{\rm diag}(x_1)+\Gamma_2{\rm diag}(x_2) +
\Gamma_3{\rm diag}(x_3),$$
where, for $\ell=0,\ldots,3$,
\[ {\rm diag}(x_\ell)=
\left [\begin {array}{cccc}
x_\ell & 0 & 0 & 0\\
0 & x_\ell & 0 & 0 \\
0 & 0 & x_\ell & 0 \\
0 & 0 & 0 & x_\ell \\
\end {array}\right ].
\]
The matrix $\Gamma_3$ is unitary if and only if $|a|^2=1$, $ aq\bar{p}+p\bar{q}=0$ and $|p|^2+|q|^2=1$. The matrix
$\Gamma_2$ is unitary if and only if $|p|^2+|q|^2=1$, $|a|^2|q|^2+|p|^2=1$ and $ aq\bar{p}+p\bar{q}=0$.
The matrix $\Gamma_1$ is unitary if and only if $|a|^2=1$.

Thus,  $\cC$ is  information lossless
if $|a|^2=|p|^2+|q|^2=1$ and $p\bar q+aq\bar p=0$.

It is not difficult to verify that the codes in Examples~\ref{ex9.2}, \ref{ex9.2bis} and \ref{ex9.3}
are all information lossless.

\appendix

\section{Facts from Number Theory}\label{AppA}

In this appendix we collect some results from algebraic number theory for the convenience of the reader.

Let $K$ be a number field.  The ring of integers $\cO_K$ of $K$ is a Dedekind domain \cite[I(3.1)]{Neu}.

Let $d_K$ denote the discriminant of $K$.

\begin{proposition}[{\cite[p.15]{Neu}}] \label{propA1}
Let $m\not=0$  be a square-free integer  and let $K=\Q(\sqrt{m})$. Then
\[d_K=\begin{cases}
4m & \text{if } m\equiv 2,3 \bmod 4\\
m &\text{if } m\equiv 1 \bmod 4
\end{cases}.\]
An integral basis of $K$ is given by $\{1,\sqrt{m}\}$ in the first case, by $\{1, \frac{1}{2} (1+\sqrt{m})\}$ in the second case and by  $\{1, \frac{1}{2} (m+\sqrt{m})\}$ in both cases. Thus
\[\cO_K=\begin{cases}
\Z[\sqrt{m}] & \text{if } m\equiv 2,3 \bmod 4\\
\Z\bigl[\frac{1+\sqrt{m}}{2}\bigr] &\text{if } m\equiv 1 \bmod 4
\end{cases}.\]
\end{proposition}

Let $h_K$ denote the class number of $K$, then $h_K=1$ if and only if $\cO_K$ is a principal ideal domain \cite[I,\S6]{Neu} if and only if $\cO_K$ is a unique factorization domain (since $\cO_K$ is a Dedekind domain \cite[Prop. 3.18]{Mil}).

\begin{proposition}[{\cite[p. 48]{Mil}}] Let $m$ be a positive square-free integer and let $K=\Q(\sqrt{-m})$. Then $h_K=1$ if and only if $m\in\{1,2,3,7,11,19,43,67,163\}$.
\end{proposition}

For an extension of number fields $K/F$, let $d_{K/F}$ denote the relative discriminant of $K$ over $F$.

\begin{proposition}[{\cite[p. 443]{Hasse}}] \label{propA3}
Let $L\supset K \supset F$ be a chain of number fields, then
\[d_{L/F}=N_{K/F}(d_{L/K})d^n_{K/F},\]
where $n=[K:F]$.
\end{proposition}

Let $i=\sqrt{-1}$. We collect some useful facts about quadratic extensions of $\Q(i)$.

\begin{proposition}[{\cite[Satz 2.1]{Schmal}}]  \label{propA4}
Let $m\not=0$  be a square-free integer  and let $K=\Q(i)(\sqrt{m})$. Then
\[d_K=\begin{cases}
16m^2 & \text{if } m\equiv 1,3 \bmod 4\\
64m^2 &\text{if } m\equiv 2 \bmod 4
\end{cases}.\]
\end{proposition}

\begin{proposition}\label{propA5}
Let $m\not=0$  be a square-free integer  and let $K=\Q(i)(\sqrt{m})$. Then there exists a relative integral basis of $K$ over $\Q(i)$. Furthermore,
\[|d_{K/\Q(i)}|=\frac{1}{4} \sqrt{d_K}.\]
In particular,
\[|d_{K/\Q(i)}|=\begin{cases}
|m| & \text{if } m\equiv 1,3 \bmod 4\\
2|m| &\text{if } m\equiv 2 \bmod 4
\end{cases}.\]
\end{proposition}

\begin{proof} The existence of a relative integral basis follows from \cite[Satz 4.2a]{Schmal}.  Consider the chain $K\supset \Q(i)\supset \Q$.
From Proposition~\ref{propA1} it follows that $d_{\Q(i)}=-4$. Thus,
$d_K=|d_{K/\Q(i)}|^2(-4)^2$ by Proposition~\ref{propA3}, which shows that $|d_{K/\Q(i)}|=\frac{1}{4} \sqrt{d_K}$. We conclude with Proposition~\ref{propA4}.
\end{proof}

\begin{proposition}[{\cite[pp. 915--916]{Yam}}] \label{propA6}
Let $m$ be a positive square-free integer and let
$K=\Q(i)(\sqrt{m})$. Then $h_K=1$ if and only if $m\in\{2,3,5,7,11,13,19,37,43,67,163\}$.
\end{proposition}

\begin{remark}  Note that $\Q(i)(\sqrt{m})=\Q(i)(\sqrt{-m})$. Also note that $\Q(i)(\sqrt{2})=\Q(\zeta_8)$ and $\Q(i)(\sqrt{3})=\Q(\zeta_{12})$, where $\zeta_n$ denotes a primitive $n$-th root of unity.
\end{remark}

\begin{proposition}[Dirichlet's Unit Theorem {\cite[I(7.4)]{Neu}}]\label{propA8}
Let $K$ be a number field with ring of integers $\cO_K$. Let $r$ be the number of real embeddings of $K$ and $s$ the number of pairs of complex conjugate embeddings of $K$. Let $\mu(K)$ denote the finite cyclic group of roots of unity that lie in $K$. The group of units of $\cO_K$  is the direct product of $\mu(K)$ and a free abelian group of rank $r+s-1$.
\end{proposition}

\section*{Acknowledgements} The authors wish to thank CIRM (Centro Internazionale per la Ricerca Matematica) and the Fondazione Bruno Kessler for financial support via the Research in Pairs programme and hospitality  during the period December 13--21, 2008. They warmly thank Augusto Micheletti for facilitating their visit.  The second author also wishes to thank  Sandro Mattarei and the University of Trento for their hospitality and support during the same period. The authors wish to thank
Fr\'ed\'erique Oggier and
the referees for their constructive criticism of an earlier version of this paper.


TITLE: Submanifolds of oriented manifolds
QUESTION [7 upvotes]: Let $M$ be an $n$-dimensional oriented manifold. Let $f:M\to\mathbb{R}$ be a smooth function. Suppose $c$ is a regular value of $f$ with $f^{-1}(c)$ nonempty. Show that $f^{-1}(c)$ is an oriented regular submanifold of $M$.
By constant rank thoerem, I know that $f^{-1}(c)$ is an $(n-1)$-dimensional regular submafold of $M$, but how to show that it is orientable?

REPLY [3 votes]: Suppose $S$ is the manifold $f^{-1}(c)$. Prove that $\triangledown f$ is normal to $S$ in $M$. Now if $S$ is non-orientable then there exists a continuous  choice of collection of ordered basis at each point of $S$. Attach $\triangledown f$ with it. This will show that $M$ is non-orientable, a contradiction.


TITLE: It is possible to "compact" this notation?
QUESTION [1 upvotes]: I need some help with notations. Suppose that the inequality
$$ 0\leq s_i\leq p_i p_j^{\ast} \quad\mbox{ for } i, j =1, 2 \mbox{ and } i\neq j$$
holds, where $p^{\ast}$ denote the Sobolev critical exponent. I need to write in "compact form" (I mean as in the inequality above) the following inequalities
$$ 1<s_3<p_1^{\ast}, \quad 1<\frac{s_1 s_3}{s_3-1}<p_2^{\ast}, \quad 1< s_4 <p_2^{\ast},\quad 1<\frac{s_2 s_4}{s_4-1}<p_1^{\ast}.$$
Could anyone please help me?
Thank you in advance!
EDIT: for $i, j = 1, 2, 3$ with $i\neq j$ it is $0\leq s_i\leq p_i\min(p_j^{\ast})$, I need to compact the expression:
$$1 <s_4 <p_1^{\ast}, \quad 1< s_5<p_2^{\ast}, \quad 1< s_6 <p_3^{\ast}$$
and
$$1<\frac{s_1 s_4}{s_4 -1}<\min(p_2^{\ast}, p_3^{\ast},\quad 1<\frac{s_2 s_5}{s_5-1}<\min(p_1^{\ast}, p_3^{\ast}), \quad 1<\frac{s_3 s_6}{s_6 -1}<\min(p_1^{\ast}, p_2^{\ast}).$$

REPLY [1 votes]: Assuming the following relationship is valid for $j=4$:
\begin{align*}
&1 <s_5<p_1^{\ast}, \quad 1< s_6<p_2^{\ast}, \quad 1< s_7 <p_3^{\ast}, \quad 1< s_8 <p_4^{\ast}\\
\\
&1<\frac{s_1 s_5}{s_5 -1}<\min\{p_2^{\ast}, p_3^{\ast}, p_4^{\ast}\},\quad
1<\frac{s_2 s_6}{s_6-1}<\min\{p_1^{\ast}, p_3^{\ast}, p_4^{\ast}\}\\
\\
&1<\frac{s_3 s_7}{s_7-1}<\min\{p_1^{\ast}, p_2^{\ast}, p_4^{\ast}\},\quad
1<\frac{s_4 s_8}{s_8-1}<\min\{p_1^{\ast}, p_2^{\ast}, p_3^{\ast}\}\\
\end{align*}

We might conclude for general $j\geq1$:
\begin{align*}
&1 <s_{j+1}<p_1^{\ast}, \quad 1< s_{j+2}<p_2^{\ast}, \quad \cdots, \quad 1< s_{2j} <p_j^{\ast}\\
\\
&1<\frac{s_1 s_{j+1}}{s_{j+1} -1}<\min\{p_2^{\ast},p_3^{\ast},\ldots,  p_j^{\ast}\},\quad
1<\frac{s_2 s_{j+2}}{s_{j+2}-1}<\min\{p_1^{\ast}, p_3^{\ast},\ldots, p_j^{\ast}\}\\
&\qquad\qquad\cdots\\
&1<\frac{s_j s_{2j}}{s_{2j}-1}<\min\{p_1^{\ast}, p_2^{\ast}, \ldots,p_{j-1}^{\ast}\}\\
\end{align*}

or more compactly for general $j\geq 1$ and $1\leq k\leq j$:

\begin{align*}
&\color{blue}{1 <s_{j+k}<p_k^{\ast},} \\
\\&\color{blue}{1<\frac{s_k s_{j+k}}{s_{j+k} -1}<\min\{p_1^{\ast},\ldots,  p_j^{\ast}\}\setminus\{p_k^{\ast}\}}
\end{align*}


TITLE: Determinant of Circulation Matrix or Vandermonde Alteration?
QUESTION [2 upvotes]: I'm stuck on this problem:
Given $n$ complex numbers $\alpha_\nu$, show that the determinant of
$$ \begin{vmatrix}
\alpha_1 & \alpha_2 & \cdots & \alpha_{n-1} & \alpha_n \\
\alpha_2 & \alpha_3 & \cdots & \alpha_n & \alpha_1 \\
\vdots \\
\alpha_n & \alpha_1 & \cdots & \alpha_{n-2} & \alpha_{n-1}
\end{vmatrix}
$$
is equivalent to $(-1)^{\frac{n(n-1)}{2}}\beta\cdots\beta_n$ where $$\beta_k=\sum_\nu \epsilon^\nu_k \alpha_\nu$$ and $$\epsilon_k=cos(\frac{2\pi k}{n})+isin(\frac{2\pi k}{n})$$.
Hint: Multiply the above matrix by
$$ \begin{pmatrix}
\epsilon_1 & \cdots & \epsilon_n\\
\vdots\\
\epsilon_1^n & \cdots & \epsilon_n^n
\end{pmatrix}
$$
I didn't know exactly how to multiply out that matrix, but I think I may have found another way to solve this. So, what I've done is first notice that this matrix permutes backwards every row down. I wanted to proceed by induction. I let my base case be $n=2$ and I came up with
$$ \begin{vmatrix}
\alpha_1 & \alpha_2\\
\alpha_2 & \alpha_1
\end{vmatrix}
=\alpha_1^2-\alpha_2^2
$$
Then, for $n=2$, I computed $\beta_1$ and $\beta_2$ like so:
$$\beta_1=\epsilon_1^1 \alpha_1+\epsilon_1^2 \alpha_2=-\alpha_1+\alpha_2$$
and
$$\beta_2=\epsilon_2^1 \alpha_1+\epsilon_2^2 \alpha_2=\alpha_1+\alpha_2$$.
When I plug in $n=2$ into our equivalence term, I see that this is just $-1 \beta_1 \beta_2$. So this works.
I have no clue how to even start the induction step. So here are my questions:
1. Could someone possibly give me a hint on how to start the induction step on this problem? We are using leibniz notation for determinants.
2. How do I multiply this matrix out like the hint says?
3. Would multiplying this matrix out make a big difference? Any help is greatly appreciated! Thanks!

REPLY [0 votes]: Doing row/column expansion and arguing by induction seems to be quite complicated so let me follow the hint given to you. First, note that the expression you are given for the determinant has $(-1)^{\frac{n(n-1)}{2}}$ inside which suggests that we should perform $\frac{n(n-1)}{2}$ row or column exchanges in order to get rid of the sign. Moving the $n$-th row to the first row, the $n-1$-th row to the second row and so on we perform $(-1)^{\frac{n(n-1)}{2}}$ row exchanges resulting in the following matrix:
$$ A = \begin{pmatrix}
\alpha_n & \alpha_1 & \alpha_2 & \cdots & \alpha_{n-2} & \alpha_{n-1} \\
\alpha_{n-1} & \alpha_n & \alpha_1 & \cdots & \alpha_{n-3} & \alpha_{n-2} \\
\vdots \\
\alpha_{2} & \alpha_3 & \alpha_4 & \cdots & \alpha_{n} & \alpha_1 \\
\alpha_1 & \alpha_2 &  \alpha_3 & \cdots & \alpha_{n-1} & \alpha_{n}
\end{pmatrix}. $$
The matrix $A$ is called a Circulant matrix and has the nice property that each diagonal is constant and each row is a right shift of the previous one. Following the hint, let us denote by $v_k = (\epsilon_k, \epsilon_k^2, \ldots, \epsilon_k^n)^T$ and $B = (v_1 | \ldots | v_n)$ the matrix you were suggested to multiply $A$ by. Calculating $Av_k$, we have:
$$ Av_k = \begin{pmatrix}
\alpha_n \epsilon_k + \alpha_1 \epsilon_k^2 + \alpha_2 \epsilon_k^3 + \cdots + \alpha_{n-2} \epsilon_k^{n-1} + \alpha_{n-1} \epsilon_k^n \\
\alpha_{n-1} \epsilon_k + \alpha_n \epsilon_k^2 + \alpha_1 \epsilon_k^3 + \cdots + \alpha_{n-3} \epsilon_k^{n-1} + \alpha_{n-2} \epsilon_k^n \\
\vdots \\
\alpha_{2} \epsilon_k + \alpha_3 \epsilon_k^2 + \alpha_4 \epsilon_k^3 + \cdots + \alpha_{n} \epsilon_k^{n-1} + \alpha_1 \epsilon_k^n \\
\alpha_1 \epsilon_k + \alpha_2 \epsilon_k^2 +  \alpha_3 \epsilon_k^3 + \cdots + \alpha_{n-1} \epsilon_k^{n-1} + \alpha_{n} \epsilon_k^n
\end{pmatrix} = (\alpha_n + \alpha_1 \epsilon_k + \ldots + \alpha_{n-2} \epsilon_k^{n-2} + \alpha_{n-1} \epsilon_k^{n-1}) \begin{pmatrix} \epsilon_k \\ \epsilon_k^2 \\ \epsilon_k^3 \\ \vdots \\ \epsilon_k^{n-1} \\ \epsilon_k^n \end{pmatrix} = (\alpha_1 \epsilon_k + \ldots + \alpha_{n-2} \epsilon_k^{n-2} + \alpha_{n-1} \epsilon_k^{n-1} + \alpha_n \epsilon_k^n) v_k = \beta_k v_k.  $$
Here, we used the fact that $\epsilon_k$ is a $n$-th root of unity and so $\epsilon_k^n = 1$ and $\epsilon_k^{n+j} = \epsilon_k^j$.
If you are familiar with the notion of eigenvalues, then this shows that $v_k$ is an eigenvector of $A$ corresponding to the eigenvalue $\beta_k$. The vectors $v_k$ are linearly independent as can be seen from the fact that $\epsilon_1, \ldots, \epsilon_n$ are all distinct and thus $\det B \neq 0$ (being a Vandermonde determinant). Thus, we have found all the eigenvalues of $A$ and so we must have $\det(A) = \prod_{j=1}^n \beta_j$ as required.
If you are not familiar with the notion of eigenvalues, we can finish the argument directly. We have
$$ A \cdot B = (Av_1 | \ldots | Av_n) = (\beta_1 v_1 | \ldots | \beta_n v_n) = B  \cdot \mathrm{diag}(\beta_1, \ldots, \beta_n). $$
Taking the determinant of both sides and using the multiplicativity property, we have $\det(A) \det(B) = \det(B) \prod_{j=1}^n \beta_j$. Since $\det(B) \neq 0$, we have $\det(A) = \prod_{j=1}^n \beta_j$.


TITLE: Show that $ \sum_{n\in \mathbb {S}} \frac{1}{n} $ is convergent
QUESTION [8 upvotes]: Let $\mathbb {S} =\left \{ 1,2,3,...,9,11,12,...,19,21,...99,111,112,113... \right \} $
i.e, the positive integers set which contain zero digit is omitted.
Now show that $ \sum_{n\in \mathbb {S}}  \frac{1}{n} $ is convergent .
I really don't have no idea about how to prove this

REPLY [2 votes]: Note that \begin{eqnarray} \sum_{n\in \mathbb{S}\cap\{10^n, \cdots, 10^{n+1}-1\}}\frac{1}{n}<\frac{9^{n+1}}{10^n}\end{eqnarray} conclude that any partial sum is bounded and thus the series converges.


TITLE: Projective Curves which are Principal Bundles
QUESTION [2 upvotes]: I have a very specific question: does anyone know of a (non-trivial) example of a projective curve which is also a homogenous space (or just a principal bundle)? The trivial example being CP^1 = SU(2)/U(1).

REPLY [4 votes]: CP^2 is not a curve. So you may have misstated your question. Nonetheless, here is my answer:
Every curve of genus 1 is a principal homogenous space for its Jacobian. Over an algebraically closed field, a principal homogenous space is just the group itself, and that is what happens in this case.
For genus g >= 2, no algebraic curve has more than 84(g-1) algebraic automorphisms. In particular, no curve can be a homogenous space.
EDIT The comment about CP^2 refers to an earlier version of the question.


TITLE: Calculating the accurate emf of an electric generator
QUESTION [1 upvotes]: Let's consider the following simple model of an electric generator:

Many textbook say the following:

The magnetic flux through the coil is $\Phi=BA\sin \theta$, where $\theta=\omega t+\phi$ is the angle the plane on which the coil lies makes with the upward vertical. $A$ denotes the area bounded by the coil, and $B$ the magnetic field strength. The emf induced, according to Farady's Law, is therefore $\mathcal E=\frac{d\Phi}{dt}=\omega BA \cos (\omega t+\phi)$.

This is, or course, under the assumption that the magnetic field is uniform and constant. However, the current in the coil produces some magnetic field, according to Ampere's Law, so the magnetic field should not be constant in general.
Question: Is the emf $\mathcal E$ still a sine wave if the change in magnetic field caused by current in the coil is taken into account? Of course, the answer depends on the type of load the generator is connected to, but now let's assume the load is a resistor. How will the waveform be different form $\mathcal E=\omega BA \cos (\omega t+\phi)$? Can I find an analytical expression for $\mathcal E$ in this case?

REPLY [0 votes]: The effect that you are worried about is self-induction. Given circuit (including the coil and the rest of wires) can be assigned quantity called self-inductance, denoted usually as $L$. The meaning of self-inductance is that magnetic flux through the circuit due to its own current is
$$
\Phi_{self}  = LI.
$$
To get total induced emf, one has to calculate total flux and negative of rate of change of this total flux:
$$
emf = -\frac{\Phi_{external}+\Phi_{self}}{dt},
$$
$$
emf = - \frac{d}{dt}\left(BA\sin (\omega t +\phi)  + LI\right),
$$
$$
emf = -\omega BA\cos (\omega t +\phi)  - L\frac{dI}{dt}.
$$


TITLE: Volume of a solid of revolution: $y = x^3$, $y = x^{1/3}$, $x \geq 0$ rotated about $y$-axis
QUESTION [2 upvotes]: I am trying to find the volume:

Rotate about $y$.
  $$y = x^3,\quad y = x^{1/3},\quad x \geq 0$$

Simple enough.
$x = y^3 \implies x = y^{\frac{1}{3}}$
$$\pi \cdot \int_0^1 y^{(1/3)^2} - y^{3^2}dy$$
$$\pi\cdot \left(\frac{3}{5} - \frac{1}{7}\right)$$
Of course that is wrong, why?

REPLY [2 votes]: Edit: Since you're using the disk method, your solution is correct, given the problem as stated. But be careful with notation: note the difference between the squaring of the functions as shown below. You evaluated as it should be evaluated, so the result is correct.
$$\pi \cdot \int_0^1 \left(y^{1/3}\right)^2 - \left(y^{3}\right)^2\,dy = \pi \int_0^1 y^{2/3} - y^6\,dy$$
Which, integrating and evaluating, gives you
$$\pi\cdot \left(\frac{3}{5} - \frac{1}{7}\right) = \frac{16\pi}{35}$$


TITLE: What is the precise relationship between pyknoticity and cohesiveness?
QUESTION [26 upvotes]: Pyknotic and condensed sets have been introduced recently as a convenient framework for working with topological rings/algebras/groups/modules/etc. Recently there has been much (justified) excitement about these ideas and the theories coming from them, such as Scholze's analytic geometry. (Small note: the difference between pyknotic and condensed is essentially set-theoretic, as explained by Peter Scholze here.)
On the other side, cohesion is a notion first introduced by Lawvere many years ago that aims to axiomatise what it means to be a category of "spaces". It has been developed further by Schreiber in the context of synthetic higher differential geometry (and also by Shulman in cohesive HoTT and by Rezk in global homotopy theory, to give a few other names in this direction).
Recently, David Corfield started a very interesting discussion on the relation between these two notions at the $n$-Category Café. The aim of this question is basically to ask what's in the title:

What is the precise relation between pyknoticity and cohesiveness?

Along with a few subquestions:

(On algebraic cohesion) It seems to me that the current notion of cohesion only  works for smooth, differential-geometric spaces: we don't really have a good notion of algebraic cohesion (i.e. cohesion for schemes/stacks/etc.) or $p$-adic variants (rigid/Berkovich/adic/etc. spaces). Is this indeed the case?
(On the relevance of cohesion to AG and homotopy theory) Despite its very young age, it's already clear that condensed/pyknotic technology is very useful and is probably going to be fruitfully applied to problems in homotopy theory and algebraic geometry. Can the same be said of cohesion?
(On "condensed cohesion") Cohesion is a relative notion: not only do we have cohesive topoi but also cohesive morphisms of topoi, which recover the former in the special case of cohesive morphisms to the punctual topos. Scholze has suggested in the comments of the linked $n$-CatCafé discussion that we should not only consider cohesion with respect to $\mathrm{Sets}$, but also to condensed sets. What benefits does this approach presents? Is this (or some variant of this idea) a convenient notion of "cohesion" for algebraic geometry?

REPLY [6 votes]: We have a case of relative cohesion used in an algebraic geometric setting discussed at the nLab. The entry for differential algebraic K-theory interprets

Ulrich Bunke, Georg Tamme, Regulators and cycle maps in higher-dimensional differential algebraic K-theory (arXiv:1209.6451)

via cohesion over the base   $Sh_\infty\left(Sch_{\mathbb{Z}}\right)$, ∞-stacks over a site of arithmetic schemes.
See also Urs Schreiber's entry, differential cohesion and idelic structure, and arithmetic elements of his research proposal, Higher theta functions and higher CS-WZW holography.


TITLE: Argument principle and the residue
QUESTION [0 upvotes]: How to prove $\int_{\gamma}f'(z)g(z)dz/f(z)=2\pi i \sum^N_{j=1}g(\alpha_j)$, where f and g are holomorphic functions and $\gamma$ is a simple closed curve such that no zero of f lies on $\gamma$. And suppose $\alpha_1, \alpha_2,... \alpha_N$ are the zeroes of f in $Int(\gamma)$.
I think this can be considered as the generalization of the argument principle, but don't know how to prove this problem. Thanks.

REPLY [0 votes]: Hint: Around a zero of order $k$, $f$ can be represented as $$(z-z_0)^kh(z)$$ with $h$ holomorphic and $h(z_0)\neq 0$. Noe apply the product rule and Cauchy's theorem.
EDIT: let $f(z)=(z-\alpha_i)^kh(z)$ as in the hint. Then $f'(z)=k(z-\alpha_i)^{k-1}h(z)+z-\alpha_i)^kh'(z)$, and hence
$$\frac{f'}{f}=\frac{k(z-\alpha_i)^{k-1}h(z)+z-\alpha_i)^kh'(z)}{(z-\alpha_i)^kh(z)}=\frac{k}{z-\alpha_i}+\frac{h'}{h}.$$
The first summand has a simple pole, the second is holomorphic.
By Cauchy's theorem one has, for any holomorphic function $g$ and any simple closed path around $z_0$
$$\frac{1}{2\pi i}\int_{\gamma}\frac{g}{z-z_0}dz=g(z_0,)$$
And for any closed path
$$\int_\gamma gdz=0.$$
Conclude


TITLE: Bounded Sequences
QUESTION [4 upvotes]: I came across the following problems during the course of my self-study of real analysis:

Show that the sequence $(x_n)$ defined by $x_n = 1+ \frac{1}{2} + \frac{1}{3} + \dots + \frac{1}{n}$ is unbounded.

I know a sequence $(x_n)$ is bounded if there exists a positive number $K$ such that $|x_n| \leq K$ for all $n$. So suppose for contradiction that it is bounded. Maybe we can define sequences $a_n = x_n-1$, $b_n = a_n-\frac{1}{2}$, $c_n = b_n- \frac{1}{3} \dots$ and try to come up with a contradiction?

Show that the sequence $(x_n)$ defined by $x_1 = x$,  $x_{n+1} = x_{n}+ 1/x_n$ is unbounded.

Suppose for contradiction that $(x_n)$ is bounded by $K$ for all $n$. Then show that there is some $K' < K$ which is also an upper bound?

Show that the sequence $(x_n)$ defined by $x_n = 1+ \frac{1}{2!}+ \frac{1}{3!} + \dots + \frac{1}{n!}$ is bounded above by $2$.

So there is some relationship between $n!$ and $2^{n-1}$. I think $n! \geq 2^{n-1}$ and we can prove this by induction on $n$? So $x_n \leq 1+1+ \frac{1}{8} + \dots + \frac{1}{2^{n-1}}$?

REPLY [5 votes]: The simplest way to show that a sequence is unbounded is to show that for any $K\gt 0$ you can find $n$ (which may depend on $K$) such that $x_n\geq K$.
The simplest proof I know for this particular sequence is due to one of the Bernoulli brothers Oresme. I'll get you started with the relevant observations and you can try to take it from there:
Notice that $\frac{1}{3}$ and $\frac{1}{4}$ are both greater than or equal to $\frac{1}{4}$, so
$$\frac{1}{3}+\frac{1}{4}\geq \frac{1}{4}+\frac{1}{4} = \frac{1}{2}.$$
Likewise, each of $\frac{1}{5}$, $\frac{1}{6}$, $\frac{1}{7}$, and $\frac{1}{8}$ is greater than or equal to $\frac{1}{8}$, so
$$\frac{1}{5}+\frac{1}{6}+\frac{1}{7} + \frac{1}{8} \geq \frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8} = \frac{1}{2}.$$
Now look at the fractions $\frac{1}{n}$ with $n=9,\ldots,16$; compare them to $\frac{1}{16}$; then compare the fractions $\frac{1}{n}$ with $n=17,\ldots,32$ to $\frac{1}{32}$. And so on.
See what this tells you about $x_1$, $x_2$, $x_4$, $x_8$, $x_{16}$, $x_{32}$, etc.
Your proposal does not work as stated. For example, the sequence $x_n = 1+\frac{1}{2}+\frac{1}{4}+\cdots+\frac{1}{2^{n-1}}$ is bounded by $K=10$; but it's also bounded by $K=5$. Just because you can find a better bound to some proposed upper bound doesn't tell you the proposal is contradictory. It might, if you specify that you want to take $K$ to be the least upper bound of the sequence. Even so, it's hard to establish that a sequence is unbounded that way. (Note also that you haven't really defined the sequence very well: it is undefined for $x=0$, though that is the only problem.)
To get you started: Show that if you start the sequence with $-x$ instead of $x$, then you just get the same sequence multiplied by $-1$. That is, if you fix $x\neq 0$, and you let $y_1=-x$, $y_{n+1}= y_n + (1/y_n)$, then $y_k = -x_k$; so the sequence $(x_n)$ is bounded if and only if the sequence $y_k$ is bounded, and so you may assume $x\gt 0$.
Then show that if $0\lt x\lt 1$, and you let $y_1 = \frac{1}{x}$, $y_{n+1} = y_n+(1/y_n)$, then $y_k=x_k$ for $k\geq 2$; so you may assume that $x\geq 1$.
Now you have that the sequence is increasing. If it were bounded, it would converge, say to $L\gt 0$. Then
$$L = \lim_{n\to\infty}x_n = \lim_{n\to\infty}x_{n+1} = \lim_{n\to\infty}\left(x_n + \frac{1}{x_n}\right) = \lim_{n\to\infty}x_n + \frac{1}{\lim_{n\to\infty}x_n} = L+\frac{1}{L}.$$
I think that's a very big problem for $L$...
Yes, if you can prove that $n!\gt 2^{n-1}$ for all $n\geq 1$, then you can bound your sequence by a sequence of powers of $\frac{1}{2}$; if you can show that sequence is bounded, you'll be done. And, yes, you can prove the inequality in question by induction on $n$. It's very simple to do. But you messed up your computations later (that second $1$ should be a $\frac{1}{2}$). If $n!\geq 2^{n-1}$, then
$$\begin{align*}
x_n &= 1 + \frac{1}{2!} + \frac{1}{3!} + \frac{1}{4!} + \cdots + \frac{1}{n!} \\
&\leq 1 + \frac{1}{2^1} + \frac{1}{2^2} + \frac{1}{2^3} + \cdots + \frac{1}{2^{n-1}}\\
 &= \frac{1 - \frac{1}{2^n}}{1 - \frac{1}{2}}\\
 &= \frac{2^n-1}{2^{n-1}}\\
& = 2 - \frac{1}{2^{n-1}}.\end{align*}$$
Almost done!


TITLE: A sequence, whose all possible finite sub sums are bounded, converges
QUESTION [0 upvotes]: I have the following problem I can't solve.
Your help would be most appreciated!

A sequence $(a_n)$ fulfills the following condition:
the set of all possible finite sub sums of $(a_n)$ is bounded. that is there exists an $M$ such that $|\sum_k a_{n_k}| < M$ ($k$ is taken from any sub sequence of the natural numbers).
  Prove that the series $\sum_n a_n$ converges absolutely.

REPLY [1 votes]: Let us define two sequences :

$a^{+}_{n}=a_n$ if $a_n\geq 0$ and $0$ otherwise.
$a^{-}_{n}=a_n$ if $a_n\leq 0$ and $0$ otherwise.

Then we have $|a_n|=a_n^{+}-a_n^{-}$. Moreover, $a_n^{+ }\geq 0$ for all $n$, and $a_n^{- }\leq 0$ for all $n$ ; and since any subsequence of $a_n^{\pm}$ is also a subsequence of $a_n$, the partial sums of $a_n^{\pm}$ are bounded, therefore $\sum_n a_n^{\pm}$ converges. Now $\sum_n |a_n|$ is the sum of two convergent series, and is therefore convergent.


TITLE: Probability of getting the third ace in the j'th draw
QUESTION [0 upvotes]: So the problem is the following:
Given a complete deck of 52 cards, 9 cards are drawn one by one with reposition.
What is the probability that in a hand (of 9 cards) in which 3 aces come out, the 3rd will come out at the j'th draw?
My answer:
I was able to reach the correct answer (according to the solutions to the exercise) for the 9'th draw, which is
$$\binom{8}{2}\left(\frac{4}{52}\right)^3\left(\frac{48}{52}\right)^6$$
My thinking being: Probability of getting 2 aces in the first 8 cards times the probability of getting an ace in the 9'th (and last) card.
For the j'th draw I ended up with the following answer:
$$\binom{j-1}{2}\left(\frac{4}{52}\right)^3\left(\frac{48}{52}\right)^6$$
The thinking being very similar to the one I used before. According to the solution this is wrong. The correct answer should be:
$$\binom{j-1}{2}\left(\frac{4}{52}\right)^3\left(\frac{48}{52}\right)^{j-3}$$
Where did I go wrong? Did they actually mean "at least 3 aces"?

REPLY [3 votes]: Since there is exactly 3 aces in the 9 cards you can calculate the number of different positions for the aces, i.e.
AAA------  // ace in position 1, 2, 3
AA-A-----  // ace in position 1, 2, 4
A-AA-----  // ace in position 1, 3, 4
-AAA-----  // ace in position 2, 3, 4
AA--A----  // ace in position 1, 2, 5
... etc ...

That is $$\binom{9}{3} = 84$$
combinations in total.
The last ace can be in position 3, 4, ... 9
If the last ace is in position 3 there need to be exactly 2 aces in the 2 positions before position 3. Now calculate in how many ways that can be done:
$$\binom{2}{2} = 1$$
If the last ace is in position 4 there need to be exactly 2 aces in the 3 positions before position 4. Now calculate in how many ways that can be done:
$$\binom{3}{2} = 3$$
If the last ace is in position 5 there need to be exactly 2 aces in the 4 positions before position 5. Now calculate in how many ways that can be done:
$$\binom{4}{2} = 6$$
And so on...
This gives
Last ace in position 3 :  1 combination  --> probabilty  1/84
Last ace in position 4 :  3 combinations --> probabilty  3/84
Last ace in position 5 :  6 combinations --> probabilty  6/84
Last ace in position 6 : 10 combinations --> probabilty 10/84
Last ace in position 7 : 15 combinations --> probabilty 15/84
Last ace in position 8 : 21 combinations --> probabilty 21/84
Last ace in position 9 : 28 combinations --> probabilty 28/84

So in general
$$\binom{j - 1}{2} / 84$$


TITLE: Extreme lat and lon for ellipse on sphere
QUESTION [3 upvotes]: I have 2 focal points of ellipse defined on a sphere: $F_1 = (q_1, p_1)$ and $F_2 = (q_2, p_2)$ and length of major axis $2a$. $R$ of a sphere is 1.
$q_1, q_2$ are latitudes
$p_1, p_2$ are longitudes
I need to find 4 points:

point on ellipse with max latitude
point on ellipse with min latitude
point on ellipse with max longitude
point on ellipse with min longitude

I tried to convert points to vectors and use a property that sum of angles
$$\sphericalangle F_1OP + \sphericalangle F_2OP$$ is constant for each $P$ on ellipse, but it lead me nowhere. Any idea how it can be solved?

REPLY [1 votes]: A partial answer based on a comment:
Let $A, B$ be two points on the sphere (Earth) with radius taken as unity.  Define $p(A), q(A)$ as the latitude and longitude of $A$ on the Earth, analogously for other points.  Define $N$ as the North Pole with $p(N)=+\pi/2, q(N)$ arbitrary.
Draw $\triangle ABN$.  Then $|AN|=\pi/2-p(A), |BN|=\pi/2-p(B), |\angle N|=\pm (q(A)-q(B))$.  Now from the Law of Cosines
$\cos(|AB|)=\sin(p(A))\sin(p(B))+\cos(p(A))\cos(p(B))\cos(q(A)-q(B))$.
(As an exercise you may want to derive the same formula replacing $N$ with $S$ where$p(S)=-\pi/2$. Use the identity $\sin(\pi/2+\theta)=\sin(\pi/2-\theta)$.) Use this formula for the distance from any point on the ellipse to the foci, adding up the inverse cosines to equal the major axis.
Good luck!


TITLE: Diffeomorphism between an ellipse and unit circle?
QUESTION [2 upvotes]: I'm trying to learn about diffeomorphism and an example asks for a diffeomorphism between an ellipse and an unit circle.
How does one construct such?

REPLY [0 votes]: Hint: Define $T(x,y) = (x,2y).$ Verify that $T$ maps the unit circle bijectively onto the ellipse $u^2 + v^2/4=1.$


TITLE: Sections of pullback line bundle via cyclic branched cover
QUESTION [2 upvotes]: It is a basic question and I would be happy to be directed to some reference for it.
Let $f\colon X\to Y$ be a finite branched cover of smooth projective varieties, $M$ a line bundle on $Y$ and $L=f^\ast M$ its pullback to $X$. For simplicity, let's assume $\deg f=2$. Then $L$ is invariant by the involution defined by the double cover and one has a decomposition
$H^0(L) = f^\ast H^0(M) \oplus V$
where $V$ is the eigenspace corresponding to $-1$ for the action of the involution on $H^0(L)$.
Questions: what are the sections in $V$? I guess it should be related to the vanishings along the branch locus, how? What happens for higher degrees (I guess the existence of the automorphism is then not automatic, but if we have it then how do we interpret the eigenspaces)?

REPLY [6 votes]: A good reference for cyclic covers, which includes your degree 2 case, is "Lectures on Vanishing theorems" by Esnault and Viehweg. Briefly, assuming characteristic different from 2, $f_*O_X$ decomposes as $O_Y\oplus R^{-1}$, where $R$ is line bundle such that $R^{\otimes 2}=O_Y(B)$, where $B$ is the branch locus. By the projection formula $V= H^0(R^{-1}\otimes M)$.
In general, for a Galois cover, you could decompose spaces/sheaves using characters of the group. But it wouldn't be as explicit.


TITLE: Show $\lim_{x \rightarrow \infty}x^{\frac{2-n}{2}}(\ln x)^{-\frac{2+n}{2}}=0$ for $n\geq 3$
QUESTION [1 upvotes]: Show $\lim_{x \rightarrow \infty} (\frac{1}{1+x})^{\frac{n+2}{2}}x^2=\lim_{x \rightarrow \infty}x^{\frac{2-n}{2}}(\ln x)^{-\frac{2+n}{2}}=0$ for $n\geq 3$.
The first limit is zero for sure since the highest power of the denominator is higher than numerator. But why are the two limits equal and why is the second limit zero?

REPLY [0 votes]: To show that the second limit is $0$, it is enough to show that,
$$\lim_{x\rightarrow\infty}x^{\frac{n}{2}-1}(\ln x)^{\frac{n}{2}+1}= \infty \ \ \ \forall \ n\geq 3 $$
Now note that, $(\ln x)^{\frac{n}{2}+1}\geq (\ln x)^2 $ for $n\geq 3$ and $x\geq e$
Also, $x^{\frac{n}{2}-1}\geq \sqrt{x} $ for $x\geq 1$ so each of this individual terms converge to infinity separetly. Hence so does their product.


TITLE: Existence of non-trivial linear functional on any vector space
QUESTION [6 upvotes]: For every vector space $V$ does there exist a linear functional $f$ ( a linear map from $V$ to $F$ the underlying field ) such that for some $ \vec v \in V$ , $f(\vec v) \ne 0$ ? If it does exist , can we prove the existence without the "axiom of choice " ? Is the existence equivalent to axiom of choice ?

REPLY [4 votes]: No, you can't prove this without the axiom of choice.
Given any field, $F$, we can extend the universe of set theory, such that over $F$ there is a vector space which is not finitely generated, but every subspace is finitely generated. In particular, this means that there are no linear functionals from this vector space to the field (because a linear functional would have a kernel whose codimension is $1$).
But we can do even more. We can even show that there might not be a linear functional from $\Bbb R$ to $\Bbb Q$ in some models where the axiom of choice fails. In particular in models where we have automatic continuity (e.g. models where all sets of reals are Lebesgue measurable; or all sets of reals have the Baire property). In such models if $\varphi\colon\Bbb R\to\Bbb Q$ is a group homomorphism then it has to be continuous, and therefore $0$.
Finally, whether or not the existence of nontrivial functionals may or may not imply the axiom of choice. To my knowledge this is still open. I'd imagine that the answer would be "almost positive", in the sense that it's negative but some augmentation of this (e.g. every non-zero vector has a linear functional mapping it to $1$) might turn out equivalent to the axiom of choice.


TITLE: Convergence in the Caratheodory sense and Hausdorff sense
QUESTION [6 upvotes]: Among Jordan domains, I understand that Caratheodory convergence is weaker than Hausdorff convergence.
But if a sequence of Jordan domains all have rectifiable boundary whose arc length are all $L$, and their Caratheodory limit is also a Jordan domain with the same boundary arc length $L$, does this necessarily imply this sequence converge in the Hausdorff metric?
To my knowledge, examples of sequences converging in the Caratheodory sense but not in the Hausdorff sense do not preserve arc length.
-----Edit-----
I realize that Caratheodory convergence is a less commonly known concept, so here I am updating with two equivalent definitions.
A sequence of pointed domains $(\Omega_n,x_n)$ is said to converge to $(\Omega,x)$ in the Caratheodory sense if

$x_n \to x$,
for all compact $K \subseteq \Omega$, we have $K \subseteq \Omega_n$ for every $n$ sufficiently large, and
for all open connected $U$ containing $x$, if $U \subseteq \Omega_n$ for infinitely many $n$, then $U \subseteq \Omega$.

The pointed domains $(\Omega_n,x_n)$ converge to $(\Omega,x)$ in the Caratheodory sense if and only if the harmonic measures $\omega(\Omega_n,x_n)$ converge weakly to the harmonic measure $\omega(\Omega,x)$.
The definition of Hausdorff metric can be found at https://en.m.wikipedia.org/wiki/Hausdorff_distance?wprov=sfla1

REPLY [2 votes]: The answer is yes.
The boundaries $\partial\Omega_n$ are uniformly 1-Lipschitz on $[0;L]$ under the arc-length parametrization, hence, by Arzela—Ascoli,  you can extract a uniformly convergent sub-sequence, say $\partial\Omega_{n_k}\to\gamma$, where $\gamma$ also has a 1-Lipschitz parametrisation on $[0;L]$. Also, since $\Omega_{n_k}\to\Omega$ in Carathéodory sense, we must have $\partial\Omega\subset \gamma$. So, you have a rectifiable Jordan curve $\partial \Omega$ of length $L$ and also another curve $\gamma$ of length $\leq L$ that contains $\partial \Omega$. This certainly implies $\gamma=\partial\Omega$, say, by interpreting the length as the Hausdorff measure of dimension 1. So, any sub-sequence of your domains has a further sub-sequence whose boundaries uniformly converge to $\partial\Omega$, which implies the claim.