text:
\begin{document}

\begin{abstract} We study $6$-dimensional nearly K{\"a}hler manifolds
  admitting a Killing vector field of unit length. In the compact
case it is shown that up to a finite cover there is only one geometry possible, that of the
$3$--symmetric space $S^{3} \times S^3$.

\end{abstract}

\maketitle

\section{Introduction}
Nearly K{\"a}hler geometry (shortly NK in what follows) naturally arises
as one of the sixteen classes
of almost Hermitian manifolds appearing in the celebrated
Gray--Hervella classification \cite{Hervella}. These
manifolds were studied intensively in the seventies by A. Gray
\cite{Gray}. His initial motivation was
inspired by the concept of weak holonomy \cite{Gray}, but very
recently it turned out that this concept, as defined by
Gray, does not produce any new geometric structure (see
\cite{Alexandrov}) other than those coming from a Riemannian holonomy
reduction.
One of the most important
properties of NK manifolds is that their canonical Hermitian
connection has totally skew--symmetric, parallel
torsion \cite{Kiri}. From this point of view, they naturally fit into the
setup proposed in \cite{Cleyton-Swann} towards a weakening of the
notion of Riemannian holonomy. The same property suggests that
NK manifolds might be objects of interest in string theory
\cite{friedrich2}.

The structure theory of compact NK manifolds, as developed in
\cite{Nagy} reduces their study to positive quaternionic K{\"a}hler
manifolds and nearly--K{\"a}hler manifolds of dimension $6$.
The last class of manifolds falls in the area of special metrics with
very rigid -- though not yet fully understood -- properties.

Indeed, it is known since a long time that in $6$ dimensions, a
NK metric which is not K{\"a}hler has to be Einstein of positive scalar
curvature. Moreover, such a structure is characterized by the
existence of some (at least locally defined) real Killing spinor
\cite{Grunewald}.
Combining these properties with the fact that the first Chern class
(form) vanishes \cite{Gray}, one observes that non--K{\"a}hler,
nearly K{\"a}hler
six--dimensional manifolds solve most of the type II string equations
\cite{friedrich2}. Despite of all these interesting features, very
little is known about these manifolds. In particular, apart from the
$3$--symmetric spaces
$$ S^6, S^3 \times S^3, \mathbb{C}P^3, F(1,2) $$
no compact example is available and moreover these is are the only
compact homogeneous examples \cite{Butruille}.

In a recent article \cite{Hitchin}, Hitchin shows that
nearly parallel  $G_2$--structures (\cite{fkms} for an account)
and NK manifolds of $6$ dimensions have the same variational
origins. On the other hand,
many examples of nearly parallel $G_2$--structures are available
since any $7$--dimensional, $3$--Sasakian manifold carries such a
structure \cite{fkms, GaSa} and a profusion
of compact examples of the latter were produced in \cite{Galicki}.
Since one property of $3$--Sasakian manifolds is to admit
unit Killing vector fields, one might ask whether this can happen in the
NK setting.

In the present paper we study $6$--dimensional non--K{\"a}hler,
nearly K{\"a}hler manifolds which globally admit a Killing vector field
$\xi$ of
constant length. After recalling
some elementary features of nearly K{\"a}hler geometry in
Section 2, we show in Section 3 that any Killing vector field
of unit length induces a transversal almost hyper--Hermitian structure
on the manifold.
The almost hyper--Hermitian structure is preserved by the Killing
vector field $\xi$ but not by
$J \xi$ (here $J$ denotes the almost complex structure of the
nearly K{\"a}hler structure). We measure this in the fourth section by
computing the Lie derivatives of the various geometrically significant
tensors in the direction of $J \xi$. This technical part is used
in Section 5 to perform a double reduction of the $6$--dimensional
nearly K{\"a}hler manifold. The resulting $4$--dimensional manifold
is in fact a K{\"a}hler--Einstein surface of positive scalar curvature
admitting a orthogonal almost--K{\"a}hler structure inducing the opposite
orientation.
The geometry of the situation is completely understood in terms of
this data. Moreover, if the nearly K{\"a}hler manifold is compact, a
Sekigawa--type argument from \cite{adm}
shows that the almost--K{\"a}hler structure is
actually integrable, allowing us to prove the main result of this paper.
\begin{ath}\label{main}
Let $(M^6,g,J)$ be a complete nearly K{\"a}hler manifold. If $g$ admits a unit
Killing vector field, then up to a finite cover $(M^6,g,J)$ is
isometric to $S^3 \times S^3$ endowed with its canonical NK structure.
\end{ath}

\section{Nearly K{\"a}hler Manifolds}

An almost Hermitian manifold $(M, g, J)$ is called
{\it nearly K{\"a}hler} if $(\nabla_X J) X = 0$ is satisfied
for all vector fields $X$. In other words, the covariant derivative of
$J$ (viewed as a ($3,0$)--tensor via the metric $g$)
is skew--symmetric in all three arguments, not only in the last two,
as it is the case for general almost Hermitian structures.
This is equivalent to
$d \Omega = 3 \nabla \Omega$, where $\Omega$ is the fundamental
$2$--form, {\em i.e.} $\Omega(X, Y) := g(J X, Y)$. The following
lemma summarizes some of the known identities for nearly K{\"a}hler manifolds.

\begin{elem}\label{nk}
Let $(M, g, J)$ be a nearly K{\"a}hler manifold. Then
\begin{enumerate}
\item \quad $(\nabla_X J) Y + (\nabla_Y J) X = 0 $
\item \quad $(\nabla_{JX} J) Y = (\nabla_{X} J) JY $
\item \quad
$
J((\nabla_X J) Y) = - (\nabla_X J) JY = - (\nabla_{JX} J) Y
$
\item \quad
$ g(\nabla_X Y, X) = g(\nabla_X JY, JX) $
\item \quad
$
2 g((\nabla^2_{W, X} J) Y, Z)
 = - \sigma_{X,Y,Z}   g((\nabla_W J) X,  (\nabla_Y J) JZ)
$
\end{enumerate}
where $ \sigma_{X,Y,Z}  $ denotes the cyclic sum over
the vector fields $ X, Y, Z$.
\end{elem}

A nearly K{\"a}hler manifold is called
to be of {\it constant type} $\alpha$ if
$$
\|(\nabla_X J)(Y)\|^2 = \alpha \{\|X\|^2 \|Y\|^2 - g(X, Y)^2
 - g(J X, Y)^2\}
$$
holds for any vector fields $X, Y$. A. Gray proved that a
nearly K{\"a}hler manifold of positive constant type is necessarily
$6$--dimensional (cf.~\cite{Gray}). Moreover he showed:
\begin{epr}
Let $(M, g, J)$ be a $6$--dimensional nearly K{\"a}hler, non--K{\"a}hler
manifold, then
\begin{enumerate}
\item\quad $M$ is of constant type $\alpha > 0$.
\item\quad $c_1(M) = 0$ and in particular $M$ is a spin manifold.
\item\quad $(M, g)$ is Einstein and $\Ric=5 \alpha \Id=5 \Ric^*$
\end{enumerate}
\end{epr}
Here the $\ast$--Ricci curvature $\Ric^*$ is defined as
$\Ric^*(X, Y) = \tr(Z\mapsto R(X, JZ) JY))$. From this it easily
follows that $\Ric^*(X, Y) = {\mathcal R}(\Omega)(X, JY)$, where $\mathcal R$
denotes the curvature operator on $2$--forms.

\begin{elem}
Let $(M^6, g J)$ be a nearly K{\"a}hler manifold of constant type $\alpha$, then
\bea
g((\nabla_U J) X, (\nabla_Y J) Z)
&=&
\alpha \{
g(U, Y) g(X, Z) - g(U, Z) g(X, Y)\\
&&\phantom{xxxxxxxxx} -
g(U, JY) g(X, JZ) + g(U, JZ) g(X, JY) \}
\eea
\end{elem}
\begin{ecor}\label{nabla}
Let $(M^6, g, J)$ be a nearly K{\"a}hler manifold of constant type
$\alpha=1$, then
\bea
(\nabla_X J)\circ (\nabla_X J)  Y
&=&
- |X|^2 Y\qquad \qquad\mbox{for} \quad Y \perp X, JX\\[1.5ex]
\nabla^* \nabla \Omega
&=&
\quad 4 \Omega
\eea
\end{ecor}

\begin{elem}\label{ortho}
Let $X$ and $Y$ be any vector fields on $M$, then the vector field
$(\nabla_XJ) Y$ is orthogonal to $X, JX, Y, $ and $JY$.
\end{elem}

This lemma allows us to use adapted frames $\{e_i\}$ which
are especially convenient for local calculations. Let $e_1$ and
$e_3$ be any two orthogonal vectors and define:
$$
e_2:=Je_1,\quad e_4:=Je_3,\qquad
e_5:=(\nabla_{e_1} J) e_3,\qquad
e_6:=Je_5
$$
\begin{elem}
With respect to an adapted frame $\{e_i\}$ one has
\bea
\nabla J &=& \quad e_1 \wedge e_3 \wedge e_5 -
e_1 \wedge e_4 \wedge e_6 -
e_2 \wedge e_3 \wedge e_6 -
e_2 \wedge e_4 \wedge e_5\\
\ast (\nabla J) &=& -  e_2 \wedge e_4 \wedge e_6 +
e_2 \wedge e_3 \wedge e_5 +
e_1 \wedge e_4 \wedge e_5 +
e_1 \wedge e_3 \wedge e_6\\
\Omega &=& \quad e_1 \wedge e_2 + e_3 \wedge e_4 + e_5 \wedge e_6
\eea
\end{elem}
\begin{ecor}\label{elementary}
Let $\Omega$ be the fundamental $2$--form and $X$ an arbitrary vector
field, then
\begin{enumerate}
\item\quad
$
X \i \Omega = J X^{\flat},\quad
X \i \ast \Omega = JX^{\flat} \wedge \Omega,\quad
X \i d \Omega = JX \i \ast d \Omega
$
\item\quad
$
| \Omega |^2 = 3,\quad
\ast \Omega = \frac12 \Omega \wedge \Omega,\quad
\vol = e_1 \wedge\ldots\wedge e_6 = \frac16 \Omega^3,\quad
\Omega \wedge d \Omega = 0
$
\item\quad
$
\ast X^{\flat} = \frac12 JX^{\flat} \wedge \Omega \wedge \Omega
,$

where $X^{\flat}$ denotes the $1$--form which is metric dual to $X$.
\end{enumerate}

\end{ecor}

\begin{epr}
Let $(M^6, g, J)$ be a nearly K{\"a}hler, non--K{\"a}hler manifold
with fundamental $2$--form $\Omega$ which is of constant type
$\alpha = 1$, then
$$
\Delta \Omega = 12 \Omega
$$
\end{epr}
{\sc Proof.}
To show that $\Omega$ is an eigenform of the Laplace operator
we will use the Weitzenb{\"o}ck formula on $2$--forms, {\em i.e.}
$\Delta  = \nabla^*\nabla + \frac{s}{3} \Id-2 {\mathcal R}$.

From Corollary \ref{nabla} we know that
$\nabla^*\nabla \Omega = 4 \Omega$.
Since we assume $M$ to be of constant type $1$ the scalar curvature
is $s=30$ and the $\ast$--Ricci curvature $\Ric^*$ coincides with
the Riemannian metric $g$. Hence,
$ {\mathcal R}(\Omega)(X, Y) = - Ric^*(X, J Y) = \Omega(X, Y)$.
Substituting this into the Weitzenb{\"o}ck formula yields $\Delta\Omega=12\Omega$.

\r

\begin{ecor}\label{djxi}
If $\xi$ is any vector field satisfying $L_\xi( \ast  d  \Omega)=0$

then
\beq\label{c29}
d (J\xi \i d\Omega)=- 12 J\xi^{\flat} \wedge \Omega
\eeq
\end{ecor}
{\sc Proof.}
We use Corollary \ref{elementary} and the relation
$L_\xi = d \circ\xi \i + \xi \i \circ d$ to obtain
$$
d(J\xi \i d\Omega)
 =
- d(\xi \i \ast d \Omega)
 =
\xi \i (d \ast d \Omega)
 =
- \xi \i (\ast d^* d \Omega)
 =
- \xi \i (\ast \Delta \Omega)
$$
Note that $d^* \Omega = 0$. Again applying Corollary \ref{elementary}
together with (\ref{c29}) we get
$$
d(J\xi \i d\Omega)
 =
- 12 \xi \i (\ast \Omega)
 =
-6 \xi \i (\Omega \wedge \Omega)
 =
- 12 (\xi \i \Omega) \wedge \Omega
 =
- 12 J\xi^{\flat}  \wedge \Omega
$$

\r

The structure group of a nearly K{\"a}hler manifold $(M^6, g, J)$
reduces to $\SU(3)$ which implies the decomposition
$\Lambda^2(TM)=\Lambda^{inv}\oplus\Lambda^{anti}$ with
\bea
\alpha\in\Lambda^{inv}
&\Leftrightarrow &
\alpha (X, Y) = \quad\alpha (JX, JY)\\
\alpha\in\Lambda^{anti}
&\Leftrightarrow &
\alpha (X, Y) = - \alpha (JX, JY)
\eea
We will denote the projection of a $2$--form $\alpha$ onto $\Lambda^{inv}$
by $\alpha^{(1,1)}$ and the projection onto $\Lambda^{anti}$ by
$\alpha^{(2,0)}$. This is motivated by the isomorphisms
$$
\Lambda^{inv}\otimes\CM \cong \Lambda^{(1,1)}(TM) \cong {\rm u}(3),
\qquad
\Lambda^{anti}\otimes\CM \cong
\Lambda^{(2,0)}(TM) \oplus \Lambda^{(0,2)}(TM)
$$
\begin{elem}
The decomposition $\Lambda^2(TM)=\Lambda^{inv}\oplus\Lambda^{anti}$
is orthogonal and the projections of a $2$--form $\alpha$ onto the
two components are given by
\bea
\alpha^{(1,1)} (X, Y)
&=&
\frac12 (\alpha (X, Y) + \alpha (JX, JY))
 =
\frac12 \Re  (\alpha([X+iJX], [Y-iJY]))\\
\alpha^{(2,0)} (X, Y)
&=&
\frac12 (\alpha (X, Y) - \alpha (JX, JY))
 =
\frac12 \Re  (\alpha([X+iJX], [Y+iJY]))
\eea
\end{elem}

Under the isomorphism $\Lambda^2(TM)\stackrel{\sim}{\rightarrow} \End_0(TM)$
every $2$--form $\alpha$ corresponds to a skew--symmetric endomorphism
$A$ which is defined by the equation $\alpha(X, Y) = g(AX, Y)$.
Note that $|A|^2 = 2 |\alpha|^2$, where $|\cdot |$ is the norm
induced from the Riemannian metric on the endomorphisms and on the
$2$--forms.

\begin{elem}\label{invanti}
Let $A^{(2,0)}$ resp. $A^{(1,1)}$ be the endomorphisms corresponding
to the components $\alpha^{(2,0)}$ resp. $\alpha^{(1,1)}$, then
\begin{enumerate}
\item \quad
$
A^{(2,0)} = \frac12 (A + J A J),\qquad
A^{(1,1)} = \frac12 (A - J A J)
$
\item \quad
$
J\circ A^{(1,1)} = A^{(1,1)}\circ J,\qquad
J\circ A^{(2,0)} = - A^{(2,0)}\circ J
$

\end{enumerate}
\end{elem}

\section{The Transversal Complex Structures}

In this section we consider  $6$--dimensional compact non--K{\"a}hler
nearly K{\"a}hler manifolds $(M, g, J)$ of constant type $\alpha
\equiv 1$, {\it i.e.} with scalar curvature normalized by
$s \equiv 30$. We will assume from now on that $ (M, g) $
is not isometric to the sphere with its standard metric.

\begin{epr}
Let $\xi$ be a Killing vector field then the Lie derivative of
the almost complex structure $J$ with respect to $\xi$ vanishes
$$
L_\xi J = 0.
$$
\end{epr}
{\sc Proof.}
Nearly K{\"a}hler structures $J$ on $(M, g)$ are in one--to--one correspondence
with Killing spinors of unit norm on $M$ (cf.~\cite{Grunewald}).
However, if  $ (M, g) $ is not isometric to the standard  sphere,
the (real) space of Killing spinors is $1$--dimensional, so there
exist exactly $2$ nearly K{\"a}hler structures compatible with $g$: $J$
and $-J$. This shows that the identity component of the isometry group
of $M$ preserves $J$, so in particular $ L_\xi J= 0 $ for every
Killing vector field $\xi$.

\r

Since a Killing vector field $ \xi $ satisfies by definition
$ L_\xi g= 0 $ we obtain that the Lie derivative $ L_\xi $
of all natural tensors constructed out of $g$ and $J$ vanishes.
In particular, we have

\begin{ecor}\label{enc}
If $\xi$ is a Killing vector field on the
  nearly K{\"a}hler manifold $(M,g,J)$ with fundamental $2$--form
  $\Omega$, then
$$
L_\xi (\Omega)  =  0,\qquad
L_\xi (d \Omega)  =  0,\qquad
L_\xi (\ast d \Omega)  =  0.\qquad
$$
\end{ecor}

In order to simplify the notations we denote by $\zeta:=\xi^{\flat}$
the dual 1--form to $\xi$.

\begin{ecor}\label{dxi}
$\qquad
d(J \zeta) = - \xi \i d\Omega
$
\end{ecor}
{\sc Proof.}
From $L_\xi \Omega = 0$ follows: $ - \xi \i d\Omega
=d (\xi \i \Omega)=d(J \zeta)$.

\r

\bigskip

\begin{elem}\label{dxi20}
Let $\xi$ be a Killing vector field with metric dual $\zeta$ and let
$d\zeta = d\zeta^{(1,1)} + d\zeta^{(2,0)}$ be the type decomposition of
$d\zeta$. Then the endomorphisms corresponding to $ d\zeta^{(2,0)} $ and
$ d\zeta^{(1,1)} $ are $ - \nabla_{J\xi}J $ and
$ 2\nabla_{\cdot}\xi +  \nabla_{J\xi}J$ respectively.
\end{elem}
{\sc Proof.}
The equation $ L_\xi J = 0$ applied to a vector field
$X$ yields $[\xi, JY]=J [\xi, Y]$, which can be written as
$$
\nabla_\xi JY - \nabla_{JY} \xi = J \nabla_\xi Y - J \nabla_{Y} \xi.
$$
From this equation we obtain
\beq\label{rt}
(\nabla_\xi J) Y =  \nabla_{JY}\xi - J \nabla_{Y} \xi.
\eeq
Let $A$ ($=2\nabla_{\cdot}\xi$) denote the skew--symmetric endomorphism
corresponding to the $2$--form $d\zeta$.
Then (\ref{rt}) can be written
$$
\nabla_\xi J = \frac12  (A\circ J -  J\circ A)
$$
From Lemma \ref{nk} we  get
$  \nabla_{J\xi} J = (\nabla_{\xi} J)\circ J$, whence
$$
\nabla_{J\xi} J
 =
\frac12  (A\circ J -  J\circ A)\circ J
 =
-  \frac12 (A +  J\circ A\circ J)
 = - A^{(2,0)}
$$
For the corresponding $2$--form we obtain:
$d \zeta^{(2,0)} = -\nabla_{J\xi} \Omega$. Finally
we compute $ A^{(1,1)} $ as
$$
A^{(1,1)}
  =
A   -  A^{(2,0)}
  =
2 \nabla_{\cdot}\xi   +  \nabla_{J\xi} J \ .
$$

\r

From now on we will mainly be interested in compact nearly K{\"a}hler
manifolds admitting a Killing vector field $\xi$ of constant length
(normalized to $1$). We start by collecting several elementary properties

\begin{elem}\label{l35}
The following relations hold:
\begin{enumerate}
\item \quad
$\nabla_{\xi} \xi =
\nabla_{J\xi} \xi =
\nabla_{\xi} J\xi =
\nabla_{J\xi} J\xi = 0,\qquad [\xi, J\xi] = 0
$
\item \quad
$\xi  \i  d\zeta = J\xi \i  d\zeta = 0 \ . $
\end{enumerate}
In particular, the distribution $ V:=\spann\{\xi, J\xi\} $ is integrable.
\end{elem}

We now define two endomorphisms which
turn out to be complex structures on the orthogonal complement
$  H:= \{\xi, J\xi\}^\perp$:
\beq\label{ik}
I  :=   \nabla_\xi J,
\qquad\mbox{and}\qquad
K  :=  \nabla_{J\xi} J \ .
\eeq

Note that $ -K $ is the endomorphism corresponding to
$ d\zeta^{(2,0)} $. Later on, we will see that the endomorphism
$ \J:=\nabla_{\cdot }\xi + \frac12 \nabla_{J\xi}J$,
corresponding to $ \frac12 d\zeta^{(1,1)}$, defines a complex
structure on $  H$, too.

\begin{elem}\label{acs}
The endomorphisms $I$ and $K$ vanish on $\spann\{\xi, J\xi\}$
and define complex structures on $ H = \{\xi, J\xi\}^\perp$
compatible with the metric $ g$.
Moreover they satisfy $ K=I \circ J, $
$ 0 = I \circ J + J \circ I $
and for any $  X, Y \in H$ the equation
$$
(\nabla_XJ)Y
  =
\la Y,   I X \ra \xi
  +
\la Y,   K X \ra   J\xi .
$$
\end{elem}

We will call endomorphisms of $TM$, which are complex
structures on $ H = \{\xi, J\xi\}^\perp$ {\it transversal
complex structures}.
The last equation shows that $ \nabla J $ vanishes in
the direction of $H$. It turns out that the  same is true
for transversal complex structures $I$ and $J$.

\begin{elem}\label{covtrans}
The transversal complex structures $  I $ and $ K $ are
parallel in the direction of the distribution $ H$, {\em i.e.}
$$
\la (\nabla_XI)Y,   Z \ra   =
\la (\nabla_XK)Y,   Z \ra   =
0
$$
holds for all vector fields $ X, Y, Z $ in $H$.
\end{elem}
{\sc Proof.}
First of all we compute for any vector fields $X, Y, Z$
the covariant derivative of $I$.
$$
\la (\nabla_X I)   Y,  Z \ra
  =
\la (\nabla_X(\nabla_\xi I)) Y,  Z \ra
  =
\la (\nabla^2_{X, \xi} I)Y,  Z \ra  +
\la (\nabla_{\nabla_X\xi}I)Y,  Z \ra .
$$
If $X$ is a vector field in $ H$, then Lemma~\ref{ortho}
implies that $ \nabla_XJ $ maps $H$ to $V$ and vice versa.

Let now $X, Y, Z$ be any vector fields in $H$. Then  both summands
in the above  formula for $\nabla_X I$ vanish, which is clear
after rewriting the first summand using formula $(5)$ of Lemma~\ref{nk}
and the second by using  formula $(1)$ of Lemma~\ref{nk}.

The proof for the transversal complex structure $K$ is similar.

\r

Our next goal is to show that $ \frac12 d\zeta^{(1,1)} $
defines an complex structure on the orthogonal
complement of $ \spann \{\xi, J\xi\}$.

\begin{elem} \label{norm}
Let $\xi$ be a Killing vector field of length $1$ then
$$
\|d \zeta^{(1,1)}\|^2 = 8 \qquad
\|d \zeta^{(2,0)}\|^2 = 2
$$
\end{elem}
{\sc Proof.}
We already know that the skew--symmetric endomorphism $K$ corresponding
to $-d \zeta^{(2,0)}$ is an complex structure on $\{\xi, J\xi\}^\perp$
with $K(\xi)=K(J\xi)=0$. Hence it has norm $4$ and $\|d \zeta^{(2,0)}\|^2
=\frac12 \|K\|^2=2$. Next, we have to compute the norm of
$d \zeta^{(1,1)}$. Since $\xi $ is a Killing vector field we have
$ \Delta \zeta = 10 \zeta$, hence
$$
\|d \zeta^{(1,1)}\|^2
 =
\|d\zeta\|^2 - \|d \zeta^{(2,0)}\|^2
 =
(\Delta \zeta, \zeta) - 2
 = 8
$$

\r

\begin{ecor}
The square norm of the endomorphism $\J$ corresponding to
$ \frac12d \zeta^{(1,1)}$ is equal to $4$.
\end{ecor}

\begin{epr}
Let $(M^6, g, J)$ be a compact nearly K{\"a}hler, non--K{\"a}hler
manifold of constant type $1$ and let $\xi$ be a Killing vector field
of constant length $1$ with dual $1$--form $\zeta$. Then
$$
d^*(J\zeta) = 0 \qquad \mbox{and} \qquad \Delta(J\zeta) = 18 J\zeta .
$$
In particular, the vector field $ J\xi $ is never a
Killing vector field.
\end{epr}
{\sc Proof.}
We start to compute the $L^2$--norm of the function $d^*(J\zeta)$:
$$
\|d^*(J\zeta)\|^2
 =
(d^* J \zeta, d^* J \zeta)
 =
(\Delta (J \zeta), J \zeta)
 - (d^* d (J \zeta), J \zeta)
$$
Since $\Delta=\nabla^*\nabla+\Ric  $  on $1$--forms and $\Ric=5 \Id$ we obtain
$$
\|d^*(J\zeta)\|^2
 =
\|\nabla (J \zeta)\|^2 + 5 \|J \zeta\|^2 - \|d J \zeta\|^2
$$
To compute the norm of $\nabla (J \zeta)$ we use the formula
$2 (\nabla X^{\flat}) = dX^{\flat} + L_Xg$ which holds for any vector
field $X$.

Note that the decomposition
$T^*M\otimes T^*M\cong \Lambda^2(TM)\oplus\Sym^2(TM)$
is orthogonal. Together with Lemma~\ref{lie} this yields
\bea
\|\nabla (J \zeta)\|^2
 =
\frac14 \left( \|d J \zeta\|^2 + \|L_{J\xi} g\|^2\right)
 =
\frac14 \left( \|d J \zeta\|^2 + 4 \|\J\|^2\right)
\eea
From Lemma \ref{norm} follows:
$\|\J\|^2 = \frac12 \|d \zeta^{(1,1)}\|^2 = 4$.
Using Corollary \ref{dxi}, the formula
$d \Omega = 3 \nabla \Omega$ and the fact that
$I=\nabla_{\xi}J$ is again  a transversal complex structure,
we find
$$
\|d J\zeta\|^2 = 9 \|\xi \i \nabla \Omega\|^2 = 36.
$$
Combining all these computations yields $\|d^*(J\zeta)\|^2 = 0$.

Next we want to compute $\Delta J\zeta$. We start by using the Weitzenb{\"o}ck
formula on $1$--forms and the equation
$
\nabla^*\nabla (J\tau) = (\nabla^*\nabla J)\tau + J(\nabla^*\nabla \tau) -
2 \sum (\nabla_{e_i}J)(\nabla_{e_i}\tau)
$ for any $1$--from $\tau$.
This gives
\bea
\Delta J\zeta
&=&
\nabla^*\nabla J\zeta + \Ric(J\zeta)
 =
(\nabla^*\nabla J)\zeta + J(\nabla^*\nabla \zeta) + 5 (J\zeta)
 - 2 \sum (\nabla_{e_i}J)(\nabla_{e_i}\zeta)\\
&=&
14 J \zeta - 2 \sum (\nabla_{e_i}J)(\nabla_{e_i}\zeta)
\eea
For the last equation we used Corollary \ref{nabla} and the assumption
that $\xi$ is
a Killing vector field, hence $\nabla^*\nabla\zeta=\Delta
\zeta-\Ric(\zeta)=5 \zeta$.
Since $d^*(J\zeta) = 0$ we can compute $\Delta J\zeta$ by
\bea
\Delta J\zeta
&=&
d^*d(J\zeta)
 =
\sum e_i \i \nabla_{e_i}(\xi \i d\Omega)
 =
\sum e_i \i (\nabla_{e_i} \xi) \i d\Omega
 + e_i \i \xi \i (\nabla_{e_i} d\Omega)\qquad \\
&=&
- 3 \sum (\nabla_{e_i}\Omega)(\nabla_{e_i}\xi) + 12 J\zeta
\eea
Comparing these two equations for $\Delta J\zeta$ we get
$\sum (\nabla_{e_i}\Omega)(\nabla_{e_i}\xi) = - 2 J\zeta$,
so finally  $\Delta J\zeta = 18 J\zeta$.

Since $\Delta X=2\Ric(X)=10X$ for every Killing vector field $X$,
$J\zeta$ cannot be Killing.

\r

Notice that if the manifold $M$ is not assumed to be compact a local
calculation still shows that $d^*(J\zeta)$ is a constant.

\begin{ecor}
If $ \xi $ is a Killing vector field of unit length, then
$$
\la d\zeta,  \Omega\ra  =  \la d\zeta^{(1,1)},  \Omega\ra  =  0 .
$$
\end{ecor}
{\sc Proof.}
Using Corollary \ref{elementary} we calculate
$$
\la d\zeta, \Omega \ra   \vol
 =
d\zeta \wedge \ast \Omega
 =
\frac12 d\zeta \wedge \Omega \wedge \Omega
 =
\frac12 d(\zeta \wedge \Omega \wedge \Omega)
 =
- d(\ast J\zeta)
 =
\ast (d^*J\zeta)
 =
0
$$
This proves the corollary since the decomposition
$\Lambda^2(TM)=\Lambda^{inv}\oplus\Lambda^{anti}$
is orthogonal and $\Omega\in\Lambda^{inv}$.

\r

We are now ready to prove that the $ (1,1)$--part of $ d \zeta $ defines
a fourth complex structure on $  H =\{\xi, J\xi\}^\perp$.
Indeed we have
\begin{epr}\label{bquadrat}
Let $\xi$ be a Killing vector field of constant length $1$. Then
$$
\J
  :=
\nabla_{\cdot }\xi  +  \frac12 K
$$
defines a transversal complex structure on $ H $ which
is compatible with the metric $g$.
Moreover, $  \J $ is the skew--symmetric endomorphism
corresponding to $ \frac12d \zeta^{(1,1)} $ and it commutes with $I,\
J$ and $K$:
$$
[  \J, J  ]
  =
[  \J,   K  ]
  =
[  \J, I  ] =0.
$$
\end{epr}
{\sc Proof.}
Lemma~\ref{dxi20} shows that $ \J $ corresponds to $ \frac12
d\zeta^{(1,1)} $.
Since $\la d\zeta^{(1,1)}, \Omega\ra = 0 $ and since
$ \J $ vanishes on $\spann\{\xi, J\xi\}$,
it follows that
$ d\zeta^{(1,1)}\in\Lambda^{(1,1)}_0(H)=\Lambda^2_-(H)$.
Hence
$$
\J^2 = - \frac14 \|\J\|^2 \Id_H = - \Id_H.
$$
Finally, $ \J $, commutes with $I,\
J$ and $K$ since endomorphisms corresponding to self--dual and
anti--self--dual $2$--forms in 4 dimensions commute.

\r

\section{Projectable Tensors}

In this section we want to study which of the above defined tensors descend
to the space of leaves of the integrable distribution $V = \spann
\{\xi, J\xi\}$.
For doing this we have to compute Lie derivatives in the direction
of $\xi$ and $J\xi$. We first remark that the flow of $\xi$ preserves
both the metric $g$ and the almost complex structure $J$, thus it
also preserves $I,K,d\zeta,\J$ etc.

The situation is more complicated for $J\xi$. Since $J\xi$ is not
Killing for $g$, we have to specify, when computing the Lie
derivative of a tensor with respect to $J\xi$, whether the given
tensor is regarded as endomorphism or as bilinear form.
Here we need the following lemma, a direct consequence of the
definition
of the Lie
derivative and Proposition \ref{bquadrat}.

\begin{elem}\label{lie}
Let $\alpha\in\Gamma(T^*M\otimes T^*M)$ be a $(2,0)$--tensor and
$A$ be the corresponding endomorphism. Then the Lie derivatives of $A$
and $\alpha$
with respect to $J\xi$ are related by
\beq\label{li1}
(L_{J\xi} \alpha)(X, Y) =
g((L_{J\xi}A)X, Y) + 2 g(J  \J  A(X), Y)  .
\eeq
In particular, the Lie derivative of the Riemannian metric $g$ with
respect to $J\xi$ is:
\beq\label{li2}
L_{J\xi} g=2  g(J  \J  \cdot, \cdot).
\eeq
\end{elem}

{\sc Proof.} Taking the Lie derivative in $\a(X,Y)=g(AX,Y)$ yields
$$(L_{J\xi} \alpha)(X, Y)=(L_{J\xi} g)(AX,Y)+g((L_{J\xi}A)X, Y).$$
Thus (\ref{li2}) implies (\ref{li1}). Taking $\a=g$ in (\ref{li1})
yields (\ref{li2}), so the two assertions are equivalent.

Using Proposition \ref{bquadrat} we can write:
\bea
L_{J\xi} g(X,Y)&=&g(\n_XJ\xi,Y)+g(X,\n_YJ\xi)\\
&=&\n J(X,\xi,Y)+\n J(Y,\xi,X)+g(J\n_X\xi,Y)+g(J\n_Y\xi,X)\\
&=&g(J(\J-\frac12 K)X,Y)+g(J(\J-\frac12 K)Y,X)=2g(J\J X,Y).
\eea

\r

From Lemma \ref{l35} we see that $L_{J\xi}\xi=L_{J\xi}J\xi=0$.
Thus, if $\zeta$ and $J\zeta$ denote the metric duals of $\xi$ and $J\xi$,
(\ref{li2}) shows that $L_{J\xi}\zeta=L_{J\xi}J\zeta=0$. We thus get
\beq\label{ll1}
L_{J\xi}(d\zeta) = L_{J\xi}(d J \zeta) = 0.
\eeq

\begin{elem}\label{lieformulas}
If $\xi$ is a Killing vector field of constant length $1$, then
\beq\label{ll2}
L_{J\xi}(\Omega) = J\xi \i  d\Omega - d\zeta
 =  4\omega_K  - 2\omega_{\J}
,\qquad
L_{J\xi}(J) = 4 K,
\eeq
\beq\label{ll3}
L_{J\xi}(\omega_K) = -4\Omega  +  4 \zeta \wedge J\zeta
 = -4\omega_J
,\qquad
L_{J\xi}(K) = -4J|_H- 2 I\J
\eeq
\beq\label{ll4}
L_{J\xi}(\omega_{\J}) = -2\Omega  + 2 \zeta \wedge J\zeta,\qquad
L_{J\xi}(\J)=0
 \eeq
\beq\label{ll5}
L_{J\xi}(\omega_I) =  0,\qquad
L_{J\xi}(I) =  2  \J K
\eeq
where $ \omega_I,  \omega_K, \omega_{\J}$ are the $2$--forms
corresponding to $ I,  K, \J $ and $ \omega_J $
denotes the projection of $ \Omega $ onto
$ \Lambda^2H$, {\em i.e.} $ \omega_J = \Omega - \xi \wedge J\xi$.
\end{elem}
{\sc Proof.}
We will repeatedly use the formula
$ L_X \alpha = X \i d \alpha + d X \i \alpha$,
which holds for any vector field $X$ and any differential form
$\alpha$.

Proposition \ref{bquadrat} shows that
\beq\label{ess}
2\omega_{\J}=d\zeta+\omega_K\eeq
hence
$$
L_{J\xi} \Omega
  =
J\xi \i d\Omega  + d(J\xi \i \Omega)
  =
J\xi \i d\Omega  - d\zeta=3\omega_K-d\zeta=4\omega_K-2\omega_{\J}.
$$
The second equation in (\ref{ll2}) follows by taking $A=J$ in Lemma~\ref{lie}.

Using Corollary~\ref{djxi} we get
$$L_{J\xi}(\omega_K)=
\frac13L_{J\xi} ( J\xi \i d\Omega)
  =
\frac13J\xi \i d(J\xi \i d\Omega)
  =
-4   J\xi \i  (J\zeta \wedge \Omega)
  =
-4 \Omega  +  4 \zeta \wedge J\zeta.
$$
The second equation in (\ref{ll3})
follows directly from the first one, by taking
$A=K$ in Lemma \ref{lie}.

Using (\ref{ll1}), (\ref{ll3}) and (\ref{ess}) we obtain
$$2L_{J\xi}\omega_{\J}=L_{J\xi}(d\zeta+\omega_K)=L_{J\xi}\omega_K=
-4 \Omega  +  4 \zeta\wedge J\zeta.$$
The second part of (\ref{ll4}) follows from Lemma \ref{lie}.

Finally, in order to prove (\ref{ll5}) we use (\ref{ll1}) twice:
$$L_{J\xi}\omega_I=\frac13 L_{J\xi}(\xi\i d\Omega)=\frac13 \xi\i
L_{J\xi}(d\Omega)=-4\xi\i(J\zeta\wedge\Omega)=0,$$
and the second part follows from Lemma~\ref{lie} again.

\r

\section{The Transversal Involution}

We define a transversal orthogonal involution $  \sigma \in \End(TM)  $ by
$$
\sigma   =   K \circ \J,
$$
{\em i.e.} we have $ \sigma^2 = \Id $ on $ H $ and $ \sigma = 0 $
on $ V=\spann\{\xi, J\xi\}$. Hence, the distribution $ H $ splits into
the $(\pm 1)$--eigenspaces of $  \sigma $ and we can define a new
metric $ g_0 $ on $ M $ as
\begin{equation}\label{g0}
g_0 = g + \frac12 g(\sigma \cdot, \cdot),
\end{equation}
{\em i.e.} we have $g_0 = g$ on $V$, $g_0 = \frac12 g $ on the
$(-1)$--eigenspace of $ \sigma $ and $ g_0 = \frac32 g $ on the
$(+1)$--eigenspace of $ \sigma$. The reason for introducing $g_0$ is
the fact that, in contrast to $g$,  this new
metric is preserved by the flow of $J\xi$ (cf. Corollary \ref{cor47} below).

\begin{elem} If $A^{\flat}$ denotes the $(2,0)$--tensor corresponding
to an endomorphism $A$, and $\alpha^{\sharp}$ denotes the endomorphism
corresponding to a $(2,0)$--tensor $\alpha$ with respect to the metric
$g$, then
$$
L_{J\xi}   \sigma^{\flat}=  - 4 (J   \J)^{\flat}
$$
\end{elem}
{\sc Proof.}
The right parts of (\ref{ll3}) and (\ref{ll4}) read
$$L_{J\xi}(K) = -4J  +  4 (\zeta\wedge\J\zeta)^{\sharp}- 2 I\J ,
\qquad
L_{J\xi}(\J)=2 (\zeta\wedge\J\zeta)^{\sharp}.$$
We clearly have $(\zeta\wedge\J\zeta)^{\sharp} \circ K=K \circ
(\zeta\wedge\J\zeta)^{\sharp}=(\zeta\wedge\J\zeta)^{\sharp} \circ
\J=\J \circ(\zeta\wedge\J\zeta)^{\sharp}=0$,
therefore
$$
L_{J\xi}   \sigma
  =
(L_{J\xi}  K)  \J   +   K  (L_{J\xi}  \J)
  =
(- 4 J +4(\zeta\wedge\J\zeta)^{\sharp} - 2 J \J K)  \J
  =
-4 J \J - 2 I.
$$
Thus Lemma~\ref{lie} gives
$$
L_{J\xi}   \sigma^{\flat}
  =
(-4 J \J - 2 I)^{\flat}   +   2 (J \J ( K \J))^{\flat}
  =
- 4 (J   \J)^ {\flat}.
$$
\r

\begin{ecor}\label{cor47} The metric $g_0$ is preserved by the flow of
  $J\xi$:
$$
L_{J\xi}   g_0=0.
$$
\end{ecor}
{\sc Proof.}
Direct consequence of (\ref{li2}):
$$L_{J\xi}   g_0
  =   L_{J\xi} (g + \frac12  \sigma^{\flat})=2(J\J)^{\flat}+\frac12(- 4 (J
  \J)^{\flat})=0.
$$
\r

\begin{epr}\label{lev}
For every horizontal vector fields $X,\ Y,\ Z\in H$, the Levi--Civita
connection $ \nabla^{g_0} $ of $ g_0 $
is related to the Levi--Civita
connection $ \nabla$ of $g$ by the formula
\beq\label{lv}
g_0(\nabla^{g_0}_X Y,   Z)
  =
g_0(\nabla_X Y,  Z)
  +
\frac13
g_0((1 - \tfrac{1}{2} \sigma)
[(\nabla_X \sigma)Y  +  (\nabla_{KX}\J)Y], Z) .
\eeq
\end{epr}
{\sc Proof.}
Since the expression $g_0(\nabla^{g_0}_X Y,   Z)-g_0(\nabla_X Y,  Z)$
is tensorial, we may suppose that $X,Y,Z$ are $\n$--parallel at some
point where the computation is performed.
The Koszul formula for $ g_0(\nabla^{g_0}_X Y,   Z) $ yields directly
\begin{equation}\label{g5}
2 g_0(\nabla^{g_0}_X Y,   Z)
  =
2 g_0(\nabla_X Y,   Z)
  +  \frac{1}{2}
\left[
\la (\nabla_X \sigma ) Y,  Z\ra
  +
\la (\nabla_Y \sigma ) X,  Z \ra
  -
\la (\nabla_Z \sigma ) X,   Y\ra
\right] \ ,
\end{equation}
where $ \la \cdot, \cdot \ra $ denotes the metric $ g$.
Since $ \sigma = K \circ \J = \J \circ K $ and since -- according to
Lemma~\ref{covtrans} -- $ K $ is $\n$--parallel
in direction of $ H $,  we obtain
$
\la (\nabla_X \sigma ) Y,  Z\ra
  =
\la (\nabla_X \J)   K   Y,  Z\ra .
$
Now, (\ref{ess}) shows that
$ 2\omega_{\J} - \omega_K = d\xi $ is a closed $2$--form. Hence,
$$
\la (\nabla_{X_1} \J) X_2,  X_3\ra
  +
\la (\nabla_{X_2} \J) X_3,  X_1\ra
  +
\la (\nabla_{X_3} \J) X_1,  X_2\ra
  =   0
$$
for all vectors $X_i$.
Using this equation for $ X_1 = Y, X_2 = KX $ and
$ X_3 = Z $ we obtain
\bea
\firstline{1cm}{
\la (\nabla_X \sigma ) Y,  Z\ra
  +
\la (\nabla_Y \sigma ) X,  Z \ra
  -
\la (\nabla_Z \sigma ) X,   Y\ra
  =  }
\\[1.5ex]
&=&
\la (\nabla_X \J) K Y,  Z\ra
  +
\la (\nabla_Y \J) K X,  Z \ra
  -
\la (\nabla_Z \J) K X,   Y\ra
\\[1.5ex]
&=&
\la (\nabla_X \J) K Y    +    (\nabla_{KX} \J) Y,   Z \ra
  =
\la (\nabla_X \sigma)  Y    +    (\nabla_{KX} \J) Y,   Z \ra
\eea
The desired formula then follows from (\ref{g0}) and (\ref{g5})
using
\begin{equation}\label{idh}
(\id_H + \frac12 \sigma)(\id_H - \frac12 \sigma ) = \frac34 \id_H.
\end{equation}

\r

We consider the space of leaves, denoted by $N$, of the integrable
distribution $V = \spann \{\xi, J\xi\}$. The $4$--dimensional
manifold $N$ is {\em a priori} only locally defined. It can be thought
of as the base space of
a locally defined principal torus bundle $\mathbb{T}^2 \hookrightarrow
M \to N$. The local
action of the torus is obtained by integrating the vector fields $\xi$ and
$\xi^{\prime}=\frac{1}{2\sqrt{3}}J\xi$. Moreover, if one considers the
$1$--forms $\zeta$ and $\zeta'$ on $M$
associated via the metric $g$ to the vector fields $\xi$ and
$2\sqrt{3} J\xi$ it follows that
$\zeta(\xi)=\zeta'(\xi^{\prime})=1$ and the Lie derivatives of
$\zeta$ and $\zeta'$ in the directions
of $\xi$ and $\xi^{\prime}$ vanish by (\ref{ll1}).
Therefore $\zeta$ and $\zeta'$ are principal connection
$1$--forms in the torus bundle $\mathbb{T}^2 \hookrightarrow M \to N$.

A tensor field on
$M$ projects to $N$ if and only if  it is horizontal and its Lie
derivatives with respect to $\xi$ and
$J\xi$ both vanish. All horizontal tensors defined above have vanishing Lie
derivative with respect to $\xi$. Using (\ref{ll5}) together with
Corollary \ref{cor47} we see that $\o_I$ and $g_0$ project down to
$N$. Moreover, $\o_I$ is compatible with $g_0$ in the sense that
$$ \omega_I(X,Y)=\frac{2}{\sqrt{3}}g_0(I_0 X, Y),\qquad\forall\
X,Y\in H,$$
where $I_0$ is the $g_0$--compatible complex structure on $H$ given by
$$
I_0=\frac{2}{\sqrt{3}}(I-\frac{1}{2}\sigma I).
$$
This follows directly from (\ref{g0}) and (\ref{idh}). Keeping the
same notations for the projections on $N$ of projectable tensors (like
$g_0$ or $I_0$) we now prove

\begin{ath} \label{mt}
$(N^4,g_0,I_0)$ is a K{\"a}hler manifold.
\end{ath}
{\sc Proof.}
In order to simplify notations we will denote by $\t\nabla$ and
$\t\nabla^{g_0}$ the
partial connections on the distribution $H$ given by the
$H$--projections of the Levi--Civita
connections $\nabla$ and $\nabla^{g_0}$.

Then Proposition \ref{lev} reads
\beq\label{lle1}
\t\nabla^{g_0}_X = \t\nabla_X +
\frac13 (\id_H-\frac12\sigma)(\t\n_X\sigma+\t\n_{KX}\J).
\eeq

We have to check that $\t\nabla^{g_0}_XI_0=0$
for all $X$ in $H$. We first notice the tautological relation
$\t\nabla_X\id_H =0$.
From Lemmas \ref{acs} and \ref{covtrans} we have
$\t\nabla_XI=\t\nabla_XJ=\t\nabla_XK=0$
for all $X$ in $H$. Moreover, the fact that
$I,\ J$ and $K$ commute with $\J$ and the relation $\J^2=\id_H$ easily
show that $\t\n_X\J$ commutes with $I,\ J,\ K$ and anti--commutes with $\J$
and $\s$. Consequently, $\t\n_X\s\ (=K\t\n_X\J)$ commutes with $K$ and
anti--commutes with $I,\ J,\ \J$ and $\s$ for all $X\in H$.

We thus get
\beq\label{ji1}
 \t\nabla_X I_0=\t\nabla_X\frac{2}{\sqrt{3}}(I-\frac{1}{2}\sigma I)
=-\frac{1}{\sqrt{3}}(\t\nabla_X\s)I.
\eeq
On the other hand, the commutation relations above show immediately
that the endomorphism $I_0$ commutes with
$(\id_H-\frac12\sigma)\t\n_{KX}\J$ and
anti--commutes with $(\id_H-\frac12\sigma)\t\n_X\sigma$. Thus
the endomorphism $\frac13
(\id_H-\frac12\sigma)(\t\n_X\sigma+\t\n_{KX}\J)$
 acts on $I_0$ by
\bea\frac13
(\id_H-\frac12\sigma)(\t\n_X\sigma+\t\n_{KX}\J)(I_0)&=&
2\frac13
(\id_H-\frac12\sigma)(\t\n_X\sigma)I_0\\
&=&\frac{4}{3\sqrt3}(\id_H-\frac12\sigma)
\t\n_X\sigma(\id_H-\frac12\sigma)I\\
&=&\frac{4}{3\sqrt3}(\id_H-\frac12\sigma)(\id_H+\frac12\sigma)
(\t\n_X\sigma)I\\
&=&\frac{1}{\sqrt{3}}(\t\nabla_X\s)I.
\eea

This, together with (\ref{lle1}) and (\ref{ji1}), shows that
$\t\nabla^{g_0}_XI_0=0$.

\r

We will now look closer at the structure of the metric $g$. Since
$$g=g_0+\frac12 \omega_K(\J\cdot, \cdot),$$
the geometry of $N$, together with the form $\omega_K$ and the almost
complex structure $\J$ determine
completely the nearly K{\"a}hler metric $g$. But the discussion below will
show that $\omega_K$ depends also in an explicit way on the geometry of
the K{\"a}hler surface $(N^4,g_0, I_0)$.

If $\alpha$ is a $2$--form on $H$ we shall denote by $\alpha^{\prime}$,
resp. $\alpha^{\prime \prime}$ the invariant, resp. anti--invariant parts of
$\alpha$ with respect to the almost complex structure $I_0$. An easy algebraic
computation shows that $\omega_J$ is $I_0$--anti--invariant whilst
\begin{equation}\label{52}
 \omega_K^{\prime}=-\frac{1}{3}(\omega_K-2\omega_{\J}), \
\mbox{and} \ \omega_K^{\prime \prime}=\frac{2}{3}(2
\omega_K-\omega_{\J}).
\end{equation}
Consider now the complex
valued $2$-form of $H$ given by
\begin{equation}
 \Psi=\sqrt{3}\omega_K^{\prime \prime}+2i\omega_J.
\end{equation}
It appears then from Lemma \ref{lieformulas} and (\ref{52}) that
\begin{equation}\label{60}
 L_{\xi^{\prime}}\Psi=i\Psi.
\end{equation}

Thus $\Psi$ is not projectable on $N$, but it can be interpreted as a
$\L$--valued $2$--form on $N$, where $\L$ is the complex line bundle
over $N$ associated
to the (locally defined) principal $S^1$--bundle
$$M/\{\xi\}\to N:=M/\{\xi,\xi'\}$$
with connection form $\zeta'$.

Corollary \ref{dxi}, together with
(\ref{ik}) implies that the curvature form of $\L$ equals
\beq\label{dz}
d \zeta'=-6\sqrt{3} \omega_I=-12g_0(I_0
\cdot, \cdot).
\eeq
Notice that, since the curvature form of $\L$ is of type $(1,1)$, the
Koszul--Malgrange theorem implies that $\L$ is holomorphic.

The following proposition computes the Ricci curvature of the K{\"a}hler
surface $(N^4,g,I_0)$ by identifying the line bundle $\L$ with the
anti--canonical bundle of $(N,I_0)$.

\begin{epr}\label{kahl}
$(N^4,g,I_0)$ is a K{\"a}hler-Einstein surface with Einstein constant equal to
$12$. Moreover, $\L$ is isomorphic to the anti--canonical line bundle
$\K$ of $(N^4,g,I_0)$.
 \end{epr}
{\sc Proof.}
We first compute $\omega_J(I_0 \cdot, \cdot)=-\frac{\sqrt{3}}{2}
\omega_K^{\prime \prime}$ and $(\omega_K^{\prime \prime})(I_0 \cdot, \cdot)=
\frac{2}{\sqrt{3}} \omega_J$. These lead to
$$ \Psi(I_0 \cdot, \cdot)=-i \Psi $$
in other words $\Psi$ belongs to $\Lambda^{0,2}_{I_0}(H, \mathbb{C})$.
We already noticed that by (\ref{60}), $\Psi$ defines a
section of the holomorphic line bundle
\begin{equation}\label{59}
 \Lambda^{0,2}_{I_0}(N)\otimes\L=\K^{-1} \otimes\L.
\end{equation}
Since $\Psi$ is non--vanishing, this section induces an isomorphism
$\Psi: \K\to\L$. We
now show that $\Psi$ is in fact $\t\n^{g_0}$--{\it{parallel}}.

Notice first that $K$ commutes with
$(\id_H-\frac12\sigma)(\t\n_X\sigma+\t\n_{KX}\J)$ (it actually
commutes with each term of this endomorphism), and $\t\n K=0$ by Lemma
\ref{covtrans}. Thus (\ref{lle1}) shows that
\beq\label{t1}\t\n^{g_0}K=0.
\eeq
Furthermore, using the relation
$$g(\cdot,\cdot)=\frac43
g_0((1-\frac{\sigma}2)\cdot,\cdot),$$
 $\Psi$ can be expressed as
\beq\label{t2}\Psi=\frac{4}{\sqrt 3}g_0((K-iI_0 K)\cdot,\cdot).\eeq
Since $\t\n^{g_0} g_0=0$ and $\t\n^{g_0}I_0=0$ (by Theorem \ref{mt}),
(\ref{t1}) and (\ref{t2}) show
that $\Psi$ is $\t\n^{g_0}$--parallel.

Hence the Ricci form of $(N^{4},g_0,I_0)$ is opposite to the
curvature form of $\L$. From (\ref{dz}) we obtain
$Ric_{g_0}=12g_0$, thus finishing the proof.

\r

\begin{epr} The almost complex structure $\J$ on $H$ is projectable and
  defines an almost K{\"a}hler structure on $(N,g_0)$ commuting with $I_0$.
\end{epr}

{\sc Proof.}
Lemma \ref{lieformulas} shows that $\J$ is projectable onto $N$. Let
us denote the
associated $2$--form with respect to $g_0$ by $\o^0_{\J}$.
Identifying forms and endomorphisms via the metric $g$ we can write
$$\o^0_{\J}:=g_0(\J\cdot,\cdot)
=(1+\frac12\sigma)\J=(1+\frac12 K\J)\J=\J-\frac12 K=\frac12d\zeta.$$
This shows that $\o^0_{\J}$ is closed, so the projection of
$(g_0,\J)$ onto $N$ defines an almost
K{\"a}hler structure.

\r

Together with Proposition \ref{kahl}, we see
that the locally defined manifold $N$ carries
a K{\"a}hler structure $(g_0,I_0)$ and an almost K{\"a}hler
structure $(g_0,\J)$, both obtained by projection from $M$.
Moreover $g_0$ is Einstein with positive scalar curvature. If $N$ were
compact, we could directly apply Sekigawa's proof of the Goldberg
conjecture in the positive curvature case in order to conclude that
$(g_0,\J)$ is K{\"a}hler. As we have no information on
the global geometry of $N$, we use the following idea.
On any almost K{\"a}hler Einstein manifold, a Weitzenb{\"o}ck--type formula was
obtained in \cite{adm}, which in the compact case shows by integration
that the manifold is actually K{\"a}hler provided the Einstein constant
is non--negative. In the present situation, we simply interpret on $M$
the corresponding formula on $N$, and after integration over $M$ we prove a
pointwise statement which down back on $N$ just gives the
integrability of the almost K{\"a}hler structure.

The following result is a particular case (for Einstein metrics) of
Proposition 2.1 of \cite{adm}:

\begin{epr} \label{prop2} For any almost K{\"a}hler Einstein manifold
$(N^{2n},g_0,J,\O)$ with covariant derivative denoted by $\n$ and
curvature tensor $R$, the following pointwise relation holds:

\begin{equation}\label{la1}
\Delta^{N} s^*-
8 \delta^{N} (\langle \rho^* , \n_{\cdot}  \O\rangle) = -
8|R''|^2  - |\n^*\n \O |^2 - |\phi|^2
-\frac{s}{2n}|\n\O|^2,
\end{equation}
where $s$ and $s^*$ are respectively the scalar and $*$--scalar
curvature, $\rho ^*:=R(\O)$ is the $*$--Ricci form,
$\phi(X,Y) = \langle \n_{JX} \O,\n_Y \O \rangle $,
and $R''$ denotes the projection of the curvature tensor on the
space of endomorphisms of $[\Lambda ^{2,0}N]$ anti--commuting with $J$.
\end{epr}

We apply this formula to the (locally defined)
almost K{\"a}hler Einstein manifold
$(N,g_0,\J)$ with Levi--Civita covariant derivative denoted $\n^0$ and
almost K{\"a}hler form $\hat\O$ and obtain
\begin{equation}\label{fdn}F+\delta ^N \alpha =0,
\end{equation}
where
$$F:=8|R''|^2+|(\n^0)^*\n^0 \hat\O
|^2+|\phi|^2+\frac{s}{4}|\n^0\hat \O|^2$$
is a non--negative function on $N$ and
$$\alpha:=ds^*-8g_0(\rho^*,
\n^0_{\cdot}\hat \O)$$
is a 1--form, both $\alpha$ and $F$ depending in an explicit way on
the geometric data $(g_0,\J)$. Since the Riemannian submersion
$\pi:(M,g_0)\to N$ has minimal (actually totally geodesic) fibers, the
codifferentials on $M$ and $N$ are related by $\delta^M(\pi ^*\a)=\pi ^*
\delta^N\a$ for every 1--form $\a$ on $N$. Thus (\ref{fdn}) becomes
\begin{equation}\label{fdn1}\pi ^*F+\delta ^M (\pi ^*\alpha) =0.
\end{equation}
Notice that the function $\pi ^*F$ and the $1$--form $\pi ^*\alpha$ are
well--defined {\em global} objects on $M$, even though $F$, $\a$ and
the manifold $N$ itself
are just local. This follows from the fact that $F$ and
$\a$ only depend on the geometry of $N$, so $\pi ^*F$ and $\pi
^*\alpha$ can be explicitly defined in terms of $g_0$ and $\J$ on $M$.

When $M$ is {\it{compact}}, since $\pi ^*F $ is non--negative,
(\ref{fdn1}) yields, after
integration over $M$, that $\pi ^*F=0$. Thus $F=0$ on $N$ and this
shows, in particular, that $\phi=0$, so $\J$ is parallel on $N$.

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

By the discussion above, when $M$ is compact, $\J$ is parallel on $N$
with respect to the Levi--Civita
connection of the metric $g_0$, so $\J$ is
$\t\n^{g_0}$--parallel on $H$.

\begin{elem} \label{l61} The involution $\s$ is $\t\n$--parallel.
\end{elem}
{\sc Proof.} Since $\s=\J K$, (\ref{t1}) shows that
$\tilde{\nabla}^{g_0}\s=0$. Using (\ref{lle1}) and the fact that $\s$
anti--commutes with  $\tilde{\nabla}_X\sigma$ and $\tilde{\nabla}_X\J$
for every $X\in H$, we obtain
$$ \tilde{\nabla}_X\s+\frac{2}{3}
(id_H-\frac{1}{2}\sigma)(\tilde{\nabla}_X\sigma+\tilde{\nabla}_{KX}\J)
\s=0$$
for all $X$ in $H$. Since $I$ commutes with $\tilde{\nabla}_X\J$ and
anti--commutes with $\s$ and $\tilde{\nabla}_X\sigma$, the
$I$--invariant part of the above equation reads
\beq\label{pp}\frac23
(\tilde{\nabla}_X\sigma)\s+\frac13\tilde{\nabla}_{KX}\J=0.\eeq
But $\s=\J K$ and $\t\n K=0$, so from (\ref{pp}) we get
$$2(\t\n_X\J)\J=\tilde{\nabla}_{KX}\J.$$
Replacing $X$ by $KX$ and applying this formula twice yields
$$\tilde{\nabla}_{X}\J=-2(\t\n_{KX}\J)\J=4\t\n_X\J,$$
thus proving the lemma.

\r

We now recall that the first canonical Hermitian connection of the
NK structure $(g,J)$ is given by
$$ \overline{\nabla}_U=\nabla_U+\frac{1}{2}(\nabla_U J)J $$
whenever $U$ is a vector field on $M$. We will show that $(M^6,g)$ is
a homogeneous space actually by showing that
$\overline{\nabla}$ is a Ambrose--Singer connection, that is
$\overline{\nabla} \bar T=0$ and $\overline{\nabla} \bar{R}=0$, where
$\bar{T}$  and $\bar{R}$ denote the torsion and
curvature tensor of the canonical connection $\overline{\nabla}$.

Let $H_{\pm}$ be the eigen--distributions of the involution
$\sigma$ on $H$, corresponding to the eigenvalues $\pm 1$. We define the new
distributions
$$ E=<\xi> \oplus H_{+} \ \mbox{and} \ F=<J\xi> \oplus H_{-}.$$
Obviously, we have a $g$--orthogonal splitting $TM=E \oplus F$, with
$F=JE$.

\begin{elem} The splitting $TM=E \oplus F$ is parallel with respect to
  the first canonical connection.
\end{elem}
{\sc Proof. }
For every tangent vector $U$ on $M$ we can write
$$ \overline{\nabla}_U\xi=\nabla_U \xi+\frac{1}{2}(\nabla_UJ)J\xi
=\J U-\frac{1}{2}KU+\frac{1}{2}JIU= (\sigma+1)\J U, $$
showing that $\overline{\nabla}_U\xi$ belongs to $E$ (actually to
$H_+$) for all $U$ in $TM$.

Let now $Y_{+}$ be a local section of $H_{+}$. We have to consider three
cases. First,
$$\overline{\nabla}_{\xi} Y_{+}=\n_{\xi}Y_++\frac12 (\n_{\xi}J)JY_+
=L_{\xi}Y_++\n_{Y_+}\xi+\frac12 IJY_+=L_{\xi}Y_++\J Y_+$$
belongs to $H_+$ since $L_\xi$ and $\J$ both preserve $H_+$. Next,
if $X$ belongs to $H $ then:
$$
\overline{\nabla}_{X} Y_{+}=\tilde{\nabla}_XY_{+}+\la
\overline{\nabla}_XY_{+},\xi\ra \xi+\la
\bar{\nabla}_XY_{+},J\xi\ra J\xi $$
But $< \overline{\nabla}_XY_{+},J\xi> =<JY_{+},
\bar{\nabla}_X\xi>=0$ by the above discussion
and the fact that $JY_{+}$ is in $H_{-}$, and
$\t{\nabla}_XY_{+}$ is an element of $H_+$ by Lemma \ref{l61}. Thus
$\overline{\nabla}_{X} Y_{+}$ belongs to $E$.

The third case to consider is
\bea\overline{\nabla}_{J\xi} Y_{+}&=&{\nabla}_{J\xi} Y_{+}+\frac12
(\n_{J\xi}J)JY_+ =L_{J\xi} Y_{+}+\n_{Y_+}J\xi-\frac12\n_{\xi}Y_+\\
&=&L_{J\xi} Y_{+}+(\n_{Y_+}J)\xi+J\n_{Y_+}\xi-\frac12 IY_+\\
&=&L_{J\xi} Y_{+}-IY_++J(\J Y_+-\frac12 KY_+)-\frac12 IY_+\\
&=&L_{J\xi} Y_{+}-2IY_++J\J Y_+.
\eea
On the other hand
$$L_{J\xi}Y_+=L_{J\xi}\s Y_+=\s L_{J\xi}Y_++(L_{J\xi}\s) Y_+,$$
so the $H_-$--projection of $L_{J\xi}Y_+$ is
$$\pi_{H_-}L_{J\xi}Y_+=\frac{1-\s}{2}L_{J\xi}Y_+=\frac12 (L_{J\xi}\s)
Y_+.$$
Using (\ref{ll3}) and (\ref{ll4}) and the previous calculation we get
\bea\pi_{H_-}(\overline{\nabla}_{J\xi} Y_{+})
&=&\pi_{H_-}(L_{J\xi} Y_{+}-2IY_++J\J Y_+)=\pi_{H_-}(\frac12 (L_{J\xi}\s)
Y_+-2IY_++J\J Y_+)\\
&=&\pi_{H_-}((I-2J\J-2I+J\J)Y_+)=-\pi_{H_-}((1+\s)(Y_+))=0.
\eea
Thus $E$ is $\overline{\nabla}$--parallel, and since $F=JE$ and
$\overline{\nabla}J=0$ by definition, we see that
$F$ is $\overline{\nabla}$--parallel, too.

\r

Therefore the canonical Hermitian connection of $(M^6,g,J)$ has
reduced holonomy, more precisely complex
irreducible but real reducible. Using Corollary 3.1, page 487 of
\cite{Nagy} we obtain that $\overline{\nabla} \bar{R}=0$.
Moreover, the condition $\overline{\nabla} \bar{T}=0$ is always
satisfied on a NK manifold (see \cite{bm}, lemma 2.4 for instance). The
Ambrose--Singer theorem shows that if $M$ is simply
connected, then it is a homogeneous space. To conclude
that $(M,g,J)$ is actually $S^3 \times S^3$ we use the fact that the
only homogeneous NK manifolds
are $S^6, S^3 \times S^3, \mathbb{C}P^3, F(1,2)$ (see
\cite{Butruille}) and
among these spaces only $S^3 \times S^3$ has vanishing Euler
characteristic. If $M$ is not simply connected, one applies the
argument above to the universal cover of $M$ which is compact and finite by
Myers' theorem. The proof of Theorem \ref{main} is now complete.

\r

\section{The Inverse Construction}

The construction of the (local) torus bundle $M^6\to N^4$ described in
the previous sections gives rise to the following Ansatz for constructing local
NK metrics.

Let $(N^4,g_0,I_0)$ be a (not necessarily complete)
K{\"a}hler surface with $Ric=12g_0$ and assume
that $g_0$ carries a compatible almost--K{\"a}hler
structure $\J$ which commutes with $I_0$. Let $\L \to N$ be
the anti--canonical line bundle of $(N^4,g_0,I_0)$
and let $ \pi_1 : M_1 \to N$ be the associated principal circle
bundle. Fix
a principal connection form $\theta$ in $M_1$ with curvature
$-12\omega_{(g_0,I_0)}$.
Let  $H$ be the horizontal distribution of this connection and let  $\Phi$ in
$\Lambda^{0,2}_{I_0}(H, \mathbb{C})$ be the "tautological"
$2$--form obtained by the lift of the identity
map $1_{\L^{-1}} : \L^{-1} \to \L^{-1}$.

Give $M_1$ the Riemannian metric
\begin{equation}
g_1=\theta\otimes\theta+\frac{2}{3}
\pi_1^{\star}g_0-\frac{1}{2\sqrt{3}} (Re \Phi)(\J \cdot,
\cdot).
\end{equation}
Let now $M$ denote the principal $S^1$--bundle $ \pi : M \to M_1 $
with first Chern class represented by the closed $2$--form
$\Omega=2\pi_1^{\star}g_0(\J \cdot, \cdot)$. Since we work locally
we do not have to worry about integrability matters. Let $\mu$ be a connection
$1$--form in $M$ and give $M$ the Riemannian metric
$$g=\mu^2+\pi^{\star}g_1.$$
We consider on $M$ the $2$--form
\begin{equation}
\omega=\frac{1}{2\sqrt{3}} \mu \wedge
\pi^{\star}\theta+\frac{1}{2}\pi^{\star}(Im\Phi).
\end{equation}

By a careful inspection of the discussion in the previous sections we
obtain:
\begin{epr} $(M^6,g, \omega)$ is a nearly K{\"a}hler manifold of constant
type equal to $1$.
Moreover, the vector field dual to $\mu$ is a unit Killing
vector field.
\end{epr}

Notice that the only {\em compact} K{\"a}hler--Einstein surface
$(N^4,g_0,I_0)$ with $Ric=12g_0$ possessing an almost K{\"a}hler
structure commuting with $I_0$ is the product of two spheres of radius
$\frac1{2\sqrt3}$ (see \cite{adm}), which corresponds, by the above
procedure, to the nearly K{\"a}hler structure on $S^3\times S^3$. Thus
the new NK metrics provided by our Ansatz cannot be compact, which is
concordant with Theorem \ref{main}.

\labelsep .5cm


TITLE: (EDITED) Show that $\displaystyle\sum_{n=0}^{\infty} {2n \choose n } \frac{2^n}{(n+1)\cdot 3^{2n+1}} =\frac{1}{2}$
QUESTION [0 upvotes]: I have a series which I have shown converges with the ratio test. I've observed that the partial sums appear to be approaching $\frac{1}{2}$.
How would I go about proving this?  I have considered using the Squeeze theorem, but I can't find any information on using that theorem to show convergence to a specific limit, and I currently have some misgivings that it is rigorous.
Thank you in advance for your help!
EDIT: The series of interest is
$$\sum_{n=0}^{\infty} {2n \choose n } \frac{2^n}{(n+1)\cdot 3^{2n+1}}$$

REPLY [3 votes]: Rewrite $$S=\sum_{n=0}^{\infty} {2n \choose n } \frac{2^n}{(n+1)\cdot 3^{2n+1}}=\frac 13\sum_{n=0}^{\infty} {2n \choose n } \frac{x^n}{(n+1)}$$ where $x=\frac 29$ that is to say
$$S=\frac 1{3x}\sum_{n=0}^{\infty} {2n \choose n } \frac{x^{n+1}}{(n+1)}$$
$$T=\sum_{n=0}^{\infty} {2n \choose n } \frac{x^{n+1}}{(n+1)}\implies T'=\sum_{n=0}^{\infty} {2n \choose n }x^n =\frac{1}{\sqrt{1-4 x}}$$
$$T=\int \frac{dx}{\sqrt{1-4 x}}=-\frac{1}{2} \sqrt{1-4 x}+C$$ and $$C=\lim_{x\to 0} \, \frac{1}{2}\frac{1}{\sqrt{1-4 x}}=\frac 12$$ All of the above make
$$S=\frac 1{3x}\frac{1-\sqrt{1-4 x} }2=\frac{1-\sqrt{1-4 x} }{6x}$$ Now, make $x=\frac 29$.


TITLE: using half identities to find exact value of each trigonometric expression
QUESTION [2 upvotes]: a) $\cos{({105}^{°})}$
b) $\sin{(\frac{3\pi}{8})}$
c) $\cot{({67.5}^{°})}$
please do explain how you are able to get the answer as I'm still confused about this topic... Thank you

REPLY [0 votes]: $(1)$
$$\cot A=\dfrac{\cos A}{\sin A}=\dfrac{2\cos^2A}{2\cos A\sin A}=\dfrac{1+\cos2A}{\sin2A}$$
or $$\cot A=\cdots=\dfrac{\sin2A}{1-\cos2A}$$

$(2)$
$$\cos2B=2\cos^2B-1=1-2\sin^2B$$
$$\cos^2B=\dfrac{1+\cos2B}2,\cos B=\pm\sqrt{\dfrac{1+\cos2B}2}$$
Now use All Sin Tan Cos Rule


TITLE: Can a wormhole be created if it has not always existed?
QUESTION [36 upvotes]: I know there are solutions to Einstein's field equations that give a wormhole geometry. But they are time independent. They are static. Is there a process where empty flat spacetime can evolve into a wormhole by an appropriate flow of matter and energy and negative energy?
If so, it would change the topology of spacetime. Does General Relativity permit this? How would a hole in spacetime form? What determines where the other mouth of the wormhole would be located?

REPLY [44 votes]: The current state of knowledge is basically that we don't know.
The topology change is contentious - it is unclear if it is permitted, but there are also disagreements about why and how. The standard approach is to shout "quantum gravity!" and escape in the confusion.
One classical argument for why making a wormhole would be problematic is the topology censorship theorem. It states:

Every causal curve extending from past null infinity to future null infinity
  can be continuously deformed to a  curve near
  infinity.
Roughly speaking, this says that an observer, whose trip begins and ends near
  infinity,  and who thus remains outside all black holes, is unable to probe
  any nontrivial topological structures.

Now, there are issues with the theorem (since it assumes the null energy condition that wormholes and quantum fields often break, and some topological assumptions). But it seems to be a good reason to suspect wormhole formation or existence is not allowed unless it gets hidden by a topologically spherical event horizon. Except that general relativity on its own seems to be too much of a local theory to be really good defence against non-trivial topology.
There are quantum gravity papers arguing that a cosmic string breaking by tunnelling can produce traversable wormholes and gleefully break the (classical) topological censorship. Here the wormholes show up at the ends of the string, initially next to each other.

REPLY [7 votes]: Making a solution of the formation of a wormhole isn't too hard (I'm lying, it is hard), but whether or not that's a reasonable solution is a different matter.
The simplest case is to just consider the case of the collapse of a wormhole in reverse. Take a spacetime that is originally just two copies of $\mathbb{R}^3$, and at some point, remove a point from both copies. That point will then grow in size, to an open ball, on which you can identify the boundaries (this will only work for a ball of radius $> 0$, so the initial singularity is indeed a singularity).
This is entirely artificial, but it does illustrate some of the problems. While there is nothing fundamentally wrong with this solution, you can just ask why the singularity would develop in such a way that the edges would identify, instead of simply remaining a singularity, or even appearing at all.
By the various theorems relating to spacetime topology, any such change in topology will be in some way unpalatable. It will either involve closed timelike curves or singularities in some way. In particular, by Geroch's theorem, it cannot be globally hyperbolic. If your spacetime isn't globally hyperbolic, there is also no guarantee of the uniqueness of development. There exists a theorem saying that any spacetime developping closed timelike curve also has a possible development without them, and I suspect the same may be true for wormholes.
A possibility that goes back to the very origin of wormholes in the 1950's is the microscopic structure of spacetime. Some theories have it that quantum gravity, along with having a sum over every possible metric (in the path integral formalism), there may also be a sum over topologies, and spacetime may simply evolve wormholes at that scale naturally. In other words, we may get something like
\begin{equation}
Z = \sum_{M \in \mathrm{Top}} \int \mathcal{D}g \exp[i \left(\int_M (R_g + L_M)  d\mu[g]\right)]
\end{equation}
Whether that is accurate, and whether this allows for the production of traversable macroscopic wormholes is another matter.

REPLY [5 votes]: Unfortunately the answer is that we don't know, and most likely a classical theory like General Relativity that concerns itself mostly with local properties of geometry is inherently unfit to aboard the problem of global topology
The clues we get from General Relativity is that some static adiabatic (i.e. reversible) solutions with spacetimes topologically inequivalent to Minkowski do exist, but they require unphysical negative energy.
If we relax the reversibility requirement on trajectories, we might consider static Kerr geometries as unidirectional wormholes, which can exist without any negative energy required
Intuitively, a topology change of spacetime cannot (shouldn't?) occur without at least a transient singularity occurring somewhere in the spacetime, but no one has ever obtained the equivalent of a maximally extended spacetime from a finite-time collapse (see this answer)


TITLE: What's the point of the subscripts in the triple product rule?
QUESTION [1 upvotes]: $$\left ( \frac{\partial x}{\partial  y}\right )_z \left ( \frac{\partial y}{\partial  z}\right )_x\left ( \frac{\partial z}{\partial  x}\right )_y = -1$$

I don't understand what the subscripts are there for. I was thinking it meant that the partial derivative is specified on a particular contour of, say, $z$. This seems to turn the partial derivative from a function of two variables to a function of one variable? But why is this necessary? Why not just write

$$\left ( \frac{\partial x}{\partial  y}\right )\left ( \frac{\partial y}{\partial  z}\right )\left ( \frac{\partial z}{\partial  x}\right )= -1$$

REPLY [1 votes]: The Wikipedia article Triple product rule explicitly explains the notation:

Note: In each factor the variable in the numerator is considered to be an implicit function of the other two. In each factor the subscripted variable is being held constant.


TITLE: Diagonalization:Algebraic and Geometric multiplicity
QUESTION [4 upvotes]: How could I prove that if the geometric multiplicity of all eigen values of a matrix equals its algebraic multiplicity, then the matrix is diagonalizable?

REPLY [5 votes]: Well, that's obvious: if the geometric multiplicities of the eigenvalues are equal to their algebraic multiplicities, as the sum of the algebraic multiplicities is the dimension $n$ of the vector space, it means we have a set of $n$ linearly independent eigenvectors, which are therefore a basis of eigenvectors.


TITLE: Calculating $\lim_{n\to\infty}\left( \frac{4n^2+5n-6}{4n^2+3n-10}\right)^{3-4n}$
QUESTION [1 upvotes]: Calculate $$\lim_{n\to\infty}\left( \frac{4n^2+5n-6}{4n^2+3n-10}\right)^{3-4n}$$
Here is my attempt:
$$\lim_{n\to\infty}\left( \frac{4n^2+5n-6}{4n^2+3n-10}\right)^{3-4n}= \left(\frac{4\infty^2+5\infty-6}{4\infty^2+3\infty-10}\right)^{3-4\infty}$$
$$=\left(\frac{\infty(4\infty+5)}{\infty(4\infty+3)}\right)^{-4\infty}=\left(\frac{4\infty}{4\infty}\right)^{-4\infty} = 1^{-4\infty} = \boxed{1}$$
However, when I try to graph the function, I can't reliably get my answer due to precision limitations, and I feel that this method of calculating limits is less than ideal. How can I confirm that this is indeed the limit?

REPLY [1 votes]: In addition to all the other answers, that give you the correct solution,
let me phrase in no uncertain terms that your solution attempt is very misguided and will lead you to wrong results in many cases.
There are reasons why you almost never see a matematician use thy $\infty$-symbol in a formula like $4\infty+5$. The main reason is that many properties about numbers that we are tought and use intuitively are not true if you think about limits reaching an actual 'number' $\infty$.
If you have two sequences $a_n$ and $b_n$ with the limits $$\lim_{n\to \infty}a_n=\infty,\quad \lim_{n\to \infty}b_n=\infty,$$
then
$$\lim_{n\to \infty}(a_n-b_n) \text{ could be anything from }+\infty, 2019, 0, -\infty, \text {any other real number or not existing at all.}$$
 You cannot model that with the formula $\lim_{n\to \infty}(a_n-b_n)=\infty - \infty$. Once you use $\infty$ as an actual number, you have lost all the information how the series converges to $\infty$, which means you cannot tell which of the cases I described is actually correct for two given series $a_n$ and $b_n$.
The same is true for $\lim_{n\to \infty}\frac{a_n}{b_n}$, which you used in your formulas.
The way you handled this shows that you have some correct intuition about limits, but unfortunately at the crucial moment, when you are dealing with
$$\left(\frac{4\infty}{4\infty}\right)^{-4\infty}$$
you are now a victim of what you did before, because you have now no idea how either of the terms
$$ \frac{4n^2+5n-6}{4n^2+3n-10} \text{ and } 3-4n$$
converges to $1$ or $-\infty$, resp., which is important to know (if you look at the real solutions).


text:
\section{Case Study: Twitter Conversations}
\label{sec:results}

This section presents the results of applying the proposed growth
methods to predict evolution of the mention networks between Twitter
users. First, the networks are described and the assumptions of the
spectral evolution model are verified. Second, the performance of the
extended model is analyzed and compared to the performance of the
using common methods from the original spectral evolution model.

\subsection{Data Description}
\label{sec:data}

The dataset consists of $31$ mention networks on Twitter users with a
profile location in Colombia. These networks capture conversations
around a set of hashtags $H$ related to popular political topics
between August $2017$ and August $2018$. Users are represented by the
vertex set $V$. The set of edges is denoted by $E$. There exists an
edge $(u,v)$ between users $u$ and $v$, if $u$ identifies a message
with a political hashtag in $H$ (e.g., \#eleccionesseguras) and
mentions~$v$ (via @username). The mention network $G=(V,E)$ is a
simple graph (without self-loops), which means that there is at most
one edge between any pair of users. Moreover, every edge $(u,v)\in E$
is associated to a timestamp representing the time at which the edge
was created.

The analysis presented in this section is based on the largest
connected component of $G$, denoted by $G_c=(V_c,E_c)$. Networks $G$
and $G_c$ are built for each hashtag $h\in
H$. Table~\ref{tab:networks} presents a brief description of each
network considered in this study, including its corresponding hashtag,
and number of vertices and edges ($|V|$ and $|E|$).

\begin{table}[htbp]
	\centering
	\begin{tabularx}{\textwidth}{@{}Ll RRRR @{}}

		& \multirow{2}*{Set of hashtags $H$} & \multicolumn{2}{c}{\;\;\;\;\;\;\;$G$} &  \multicolumn{2}{c}{\;\;\;\;\;\;\;$G_c$}  \\
		& & $|V|$ & $|E|$ & $|V_c|$ &  $|E_c|$\\[1ex]
		\hline\hline
		1  & abortolegalya & 2235 & 2202 & 1282 & 1538\\
		2  & alianzasporlaseguridad & 176 & 1074 & 150 & 351\\
		3  & asiconstruimospaz & 2514 & 14055 & 2405 & 6950\\
		4  & colombialibredefracking & 1606 & 3483 & 1476 & 3127\\
		5  & colombialibredeminas & 707 & 2685 & 655 & 1421\\
		6  & dialogosmetropolitanos & 959 & 18340 & 932 & 4134\\
		7  & edutransforma & 166 & 1296 & 161 & 404\\
		8  & eleccionesseguras & 3035 & 17922 & 2634 & 7969\\
		9  & elquedigauribe & 2375 & 6933 & 2052 & 5272\\
		10  & frutosdelapaz & 1671 & 6960 & 1479 & 3468\\
		11 & garantiasparatodos & 388 & 814 & 340 & 563\\
		12 & generosinideologia & 639 & 914 & 615 & 805\\
		13 & hidroituangoescololombia & 1028 & 3362 & 883 & 2252\\
		14 & horajudicialur & 2250 & 23647 & 2187 & 6756\\
		15 & lafauriecontralor & 2154 & 7082 & 1999 & 5309\\
		16 & lanochesantrich & 1518 & 6946 & 1444 & 3567\\
		17 & lapazavanza & 2949 & 8288 & 2775 & 6569\\
		18 & libertadreligiosa & 1584 & 13443 & 1395 & 6856\\
		19 & manifestacionpacifica & 211 & 274 & 112 & 151\\
		10 & plandemocracia2018 & 3090 & 20955 & 2962 & 7996\\
		21 & plenariacm & 1504 & 19866 & 1460 & 4782\\
		22 & proyectoituango & 1214 & 3086 & 1186 & 1891\\
		23 & reformapolitica & 2714 & 8385 & 2608 & 5928\\
		24 & rendiciondecuentas & 5103 & 25479 & 4401 & 10308\\
		25 & rendiciondecuentas2017 & 1711 & 12441 & 998 & 2933\\
		26 & resocializaciondigna & 503 & 4054 & 496 & 1171\\
		27 & salariominimo & 2494 & 7041 & 2079 & 5016\\
		28 & semanaporlapaz & 1988 & 8103 & 1732 & 4860\\
		29 & serlidersocialnoesdelito & 530 & 861 & 439 & 697\\
		30 & vocesdelareconciliacion & 161 & 1500 & 158 & 405\\
		31 & votacionesseguras & 2748 & 13307 & 2439 & 5338\\
		\hline
	\end{tabularx}
	\caption{Description of the mention networks with political hashtags. The analysis is based on the largest connected component of each network, $G_c$.}
	\label{tab:networks}
\end{table}

\subsection{Verifying the Spectral Evolution Model Assumptions}
\label{sec:spmverif}

To apply the original and extended versions of the spectral evolution
model, the assumptions on the spectrum and eigenvectors need to be
verified.  Let $t \in \nats$ be a positive number. The set of edges
$E$ is sorted by time stamps and then split into $t$ disjoint subsets
of equal size ($|E|/t$). Next, consider a total of $t$ time steps,
created in a way that each time $i$, $1\leq i\leq t$, contains the
edges of the first $i$ subsets of edges. Note that each time step $i$
represents a cumulative graph that is associated to an adjacency
matrix $\mathbf{A}_i$ and its spectral decomposition
$\mathbf{X}_i\mathbf{\Lambda}_i\mathbf{X}^T_i$.  Note also that
$\mathbf{A}_1,\mathbf{A}_2,\dots,\mathbf{A}_t$ represents the
evolution of the network over time and $\mathbf{A}_t=\mathbf{A}$
represents the adjacency matrix of the complete network.  Overall,
this process takes $O(tn^3)$ time with $n = |V|$:

each spectral decomposition takes $O(n^3)$

and there are $t$ of such decompositions to compute.
Figure~\ref{fig:steps} summarizes the process of defining a sequence
of networks for a particular mention network.

\begin{figure}[htbp]
  \centering
  \includegraphics[width=0.7\textwidth]{fig/VocesDeLaReconciliacion_bins-eps-converted-to.pdf}
  \caption{Process of defining a sequence of networks for the mention
    network \#vocesdelareconciliacion based on $t$ time steps. In
    general, for a network $G=(V,E)$ with adjacency matrix
    $\mathbf{A}$, the edge set $E$ is split into $t$ disjoint subsets
    according to their associated time stamps. Time step $i$ contains
    all the edges of the first $i$ subsets of edges. Note that each
    time $i$ is associated to an adjacency matrix $\mathbf{A}_i$ and
    $\mathbf{A}_t=\mathbf{A}$.}
  \label{fig:steps}
\end{figure}

\subsubsection{Spectral and Eigenvector Evolution}

Figure~\ref{fig:specevol} identifies the largest eigenvalues (by
absolute value) for two mention networks: namely,
\#educationtransforms and \#howwebuildpeace.  Note that for each of
these two networks, the spectra grows irregularly.  Consider the
adjacency matrix $\mathbf{A}_1$ for time $t_1$ satisfying $1\leq
t_1<t$. The eigenvectors ${(\mathbf{X}_t)}_j$ corresponding to the
largest eigenvalues are compared to the eigenvectors
${(\mathbf{X}_1)}_j$at time $t$ using the cosine distance as a
similarity measure, for each latent dimension $j$.

\begin{figure}[htbp]
  \centering
  \begin{subfigure}[b]{0.5\textwidth}
    \centering
    \includegraphics[width=\textwidth]{fig/EduTransforma_specevol.pdf}

    \label{fig:specevol1}
  \end{subfigure}
  \hspace{-0.4cm}
  \begin{subfigure}[b]{0.5\textwidth}
    \centering
    \includegraphics[width=\textwidth]{fig/AsiConstruimosPaz_specevol.pdf}

    \label{fig:specevol2}
  \end{subfigure}
  \vspace{-0.6cm}
  \caption{Spectral evolution for two mention networks,
    \#edutransforma (left) and \#asiconstruimospaz (right). Plots
    illustrate the top $8\%$ of the largest eigenvalues by absolute
    value at each time step.}
  \label{fig:specevol}
\end{figure}

Figure~\ref{fig:vecevol} shows that, during the evolution of the
network, certain eigenvectors have a similarity value close to
one. They correspond to the eigenvectors associated to the largest
eigenvalues. Note also that at points the similarity between some
eigenvectors drops to zero, which can be explained by eigenvectors
changing their order during the spectral decomposition.

\begin{figure}[htbp]
  \centering
  \begin{subfigure}[b]{0.5\textwidth}
    \centering
    \includegraphics[width=\textwidth]{fig/EduTransforma_eigvevol.pdf}

    \label{fig:vecevol1}
  \end{subfigure}
  \hspace{-0.45cm}

  \begin{subfigure}[b]{0.5\textwidth}
    \centering
    \includegraphics[width=\textwidth]{fig/VocesDeLaReconciliacion_eigvevol.pdf}

    \label{fig:vecevol2}
  \end{subfigure}
  \vspace{-0.6cm}
  \caption{Eigenvector evolution shows a one by one eigenvector
    comparison between $t_1$ and $t$ using the cosine similarity. Each
    line represents a latent dimension, i.e., a pair of eigenvectors
    being compared, with brighter colors for eigenvectors
    corresponding to the largest eigenvalues. Eigenvector evolution for
    two mention networks, \#edutransforma (left) and
    \#vocesdelareconciliacion (right).}
  \label{fig:vecevol}
\end{figure}

\subsubsection{Eigenvector Stability}
\label{sec:eigenstab}

Let $t_1$ and $t$ denote the times at which $75\%$ and $100\%$ of all
edges in a given network $G$ are present in its adjacency matrices
$\MAT{A}_1$ and $\MAT{A}_t$ at times $t_1$ and $t$,
respectively. Their spectral decomposition is
$\mathbf{A}_1=\mathbf{X}_1\mathbf{\Lambda}_1\mathbf{X}^T_1$ and
$\mathbf{A}_t=\mathbf{X}_t\mathbf{\Lambda}_t\mathbf{X}^T_t$. Similarity
values are computed for every pair of eigenvectors $\MAT{x}_i$ and
$\MAT{x}_j$ by:

\[\operatorname{sim}_{ij}(t_1,t)=\left|{(\mathbf{X}_1)}^T_i\cdot{(\mathbf{X}_t)}_j\right|.\]

Figure~\ref{fig:stab} summarizes the resulting values, plotted as a
heatmap. White cells represent a value of zero and black cells a value
of one. The more the heatmap approximates a diagonal matrix, the fewer
eigenvector permutations there are (i.e., the eigenvectors are
preserved over time). Note that the value of certain cells in
Figure~\ref{fig:stab} is between zero and one. These cells result
either from an exchange in the location of eigenvectors that have
eigenvalues very close in magnitude or from comparing eigenvalues
${({\lambda}_1)}_i$ and ${({\lambda}_t)}_i$ that do not correspond to
each other.

\begin{figure}[htbp!]
  \centering
  \includegraphics[width=0.6\textwidth]{fig/plandemocracia2018_eigvstab.pdf}
  \vspace{-0.5cm}
  \caption{Eigenvector stability for the mention network
    \#plandemocracia2018. Similarity measure for every pair of
    eigenvectors between $t_1$ and $t$. A similarity value of zero is
    represented by a white square and a similarity of one by a black
    square. The matrix illustrates the permutations of the
    eigenvectors. A diagonal matrix represents the case where
    eigenvectors are preserved over time.}
  \label{fig:stab}
\end{figure}

\subsubsection{Spectral Diagonality Test}

Recall the spectral decomposition $(\MAT{X}_1, \MAT{\Lambda}_1)$ of
$\MAT{A}_1$ from Section~\ref{sec:eigenstab}, representing the network
$G$ at time $t_1$. At any time $t_2>t_1$, the adjacency matrix
$\MAT{A}_2$ is expected to have the form
$\mathbf{A}_2=\mathbf{X}_1(\mathbf{\Lambda}_1 +
\mathbf{\Delta})\mathbf{X}^T_1$, where $\mathbf{\Delta}$ is a diagonal
matrix. The diagonal matrix $\mathbf{\Delta}$ can be used to infer if
the network's growth is related to the eigenvalues. Technically, by
using least-squares, the matrix $\mathbf{\Delta}$ can be derived as
$\mathbf{\Delta}=\mathbf{X}_1(\mathbf{A}_2-\mathbf{A}_1)\mathbf{X}^T_1$. If
$\mathbf{\Delta}$ is diagonal, then the growth between $t_1$ and $t_2$
is spectral.  Figure~\ref{fig:diag} shows the diagonality test for the
mention network \#plandemocracia2018. It has been found that most of
the networks in the dataset show irregular behavior in the spectrum
evolution. Note that the matrix $\mathbf{\Delta}$ (spectral
diagonality) is almost diagonal for all mention networks.

\begin{figure}[htbp]
  \centering
  \includegraphics[width=0.55\textwidth,center]{fig/plandemocracia2018_diag-eps-converted-to.pdf}
  \vspace{-0.5cm}
  \caption{The spectral diagonality test shows to what extent the
    network evolution is explained by the eigenvalues. If network
    growth is related to spectrum evolution, then the matrix should be
    approximately diagonal.  Spectral diagonality test for the mention
    network \#plandemocracia2018.}
  \label{fig:diag}
\end{figure}

\subsection{Prediction Performance}
\label{sec:perform}

First, the prediction performance of the proposed approach
(Section~\ref{sec:extmodel}) is compared to that of the spectral
evolution model (Section~\ref{sec:model}). In each case, the
performance of the prediction is measured by the AUC ROC score.
Figure~\ref{fig:aucroc75} summarizes the performance comparison of the
growth methods when exact spectral decomposition and trajectories
approximations with the Rayleigh quotient. Note that for most of the
networks and growth methods, the performance of the two approaches is
highly similar. One conclusion to draw is that it is possible to
replicate the prediction of the exact method with the more efficient
method based on approximations. A $75\%$-$25\%$ train-test split was
used for the analysis depicted in Figure~\ref{fig:aucroc75}. This
ratio can be interpreted as $75\%$ of edges of the network (by
creation time) are used to predict the remaining $25\%$. The proposed
approach is used for the remainder of the analysis in this section.

\begin{figure}[htbp]
  \centering
  \includegraphics[width=.75\textwidth]{fig/auc_comp75-eps-converted-to.pdf}
  \caption[Performance evaluation]{Performance comparison of using
    exact spectral decomposition and the proposed approach where
    spectrum trajectories are approximated using the Rayleigh
    quotient. The comparison is measured by the AUC ROC score. Both
    methods seem to be consistent, i.e., the performance of the
    prediction is replicated using the Rayleigh quotient method.}
  \label{fig:aucroc75}
\end{figure}

Reducing the number of steps to be predicted while avoiding
overfitting, can bring performance improvement of the
prediction. Figure~\ref{fig:aucall} shows a comparison of the
performance for the $75\%$ and $80\%$ split ratios.  Note that
reducing the size of test set (and increasing train set) comes with
noticeable improvement in the performance of the methods, in
general. It has been verified that the growth of the spectrum is
irregular for most of the networks. Therefore, it is expected that
methods based on extrapolation, and linear and quadratic regression,
outperform the ones based on graph kernels.

\begin{figure}[htbp]
  \centering
  \includegraphics[width=.75\textwidth]{fig/auc_comp80-eps-converted-to.pdf}
  \caption[Performance evaluation]{Comparison of the prediction
    performance based on AUC ROC score for $75\%$ and $80\%$
    train-test ratios. Increasing the train set size improves the
    overall performance of the growth methods.}
  \label{fig:aucall}
\end{figure}

\begin{figure}[htbp]
  \centering
  \includegraphics[width=.6\textwidth]{fig/bestfit_auc-eps-converted-to.pdf}
  \caption[Performance evaluation]{With a $75\%$-$25\%$ train-test
    ratio for the prediction, the extrapolation method outperforms the
    regression methods for 24 out of the 31 analyzed networks. When
    the train set size increases to $80\%$, the regression methods
    outperform the extrapolation methods for 16 networks.}
  \label{fig:aucbest}
\end{figure}

It is important to include the AUC ROC score to evaluate the
methods. For instance, if the initial train-test ratio for both
methods is considered, the extrapolation method seems to outperform
the other methods in many cases. Its score is the best 24 times in
total. But the situation changes when the size of the training set
size increases: if the ratio is $80\%$, the regression methods get
better score for 16 out of 31 networks (9 and 7 for the linear and
quadratic regressions, respectively). As depicted in
Figure~\ref{fig:aucbest}, the distribution of the best method changes
alongside the modification of the train-test ratio. Improvement in the
regression methods is explained by the addition of new data available
for the prediction.  Figure~\ref{fig:aucper} shows the improvement of
the AUC ROC score in percentage with respect to the $75\%$-$25\%$
ratio.

\begin{figure}[htb!]
  \centering
  \includegraphics[width=.85\textwidth]{fig/auc_percomp-eps-converted-to.pdf}
  \caption[Performance evaluation]{Improvement of the AUC ROC score in
    percentage computed with the Rayleigh quotient considering
    different train-test ratios for extrapolation (top), linear
    (middle) and quadratic (bottom) growth methods w.r.t. the spectral
    decomposition method. Performance of the regression methods
    improve more than $10\%$ on average.}
  \label{fig:aucper}
\end{figure}

Finally, execution times of both approaches are analyzed.
Figure~\ref{fig:auctime} (top) shows the cumulative execution time of
using the spectral evolution model with exact spectral decomposition
($75\%$) and the extended spectral evolution model with approximated
spectra ($75\%$, $80\%$) when used on all 31 mention networks. Note
that there is noticeable difference (more than $\numprint{10000}$
seconds) between the execution time of the two approaches. Considering
the difference between the methods for each network, the average
improvement in favor of the extended evolution model is $71.5\%$.  The
bottom plot in Figure~\ref{fig:auctime} summarizes the execution time
improvement in percentage for each network, ranging from $45\%$ to
$80\%$. Note that the spectral extrapolation, linear and quadratic
regression outperform the graph kernels. Better prediction in the
extrapolation-based approaches seems to be consequence of their
ability to capture the irregular evolution of the eigenvalues. Also,
linear and quadratic regressions improve their performance when the
size of the train test increases.

\begin{figure}[htbp]
  \centering
  \includegraphics[width=.85\textwidth]{fig/auc_time-eps-converted-to.pdf}
  \caption[Performance evaluation]{Cumulative execution time of using
    the spectral evolution model with exact spectral decomposition
    ($75\%$) and the extended spectral evolution model with
    approximated spectra ($75\%$, $80\%$) when used on all 31 mention
    networks (top) and average execution time improvement (bottom).}
  \label{fig:auctime}
\end{figure}

config: arxiv


TITLE: Alternative definition of Euclidean operator norm
QUESTION [0 upvotes]: Given $A\in \mathbb{C}^{n\times n}$, $\|.\|$ the Euclidean operator norm, and $\rho(A)$ the spectral radius of A, how to show that
$$ \|A\| = \sup\{\rho(AB):B\in \mathbb{C}^{n\times n}, \|B\|=1\} $$

REPLY [1 votes]: Hints: For "$\ge$", use that $\rho(A) \le \|A\|$.
For "$\le$", use that $\rho(XY) = \rho(YX)$, and then use the SVD to find a unitary $B$ such that $\rho(AB) = \|A\|$.


TITLE: Favorite problems that lead to interesting diophantine equations?
QUESTION [6 upvotes]: I am looking for interesting problems (in number theory, or otherwise) that lead to interesting diophantine equations. The solution to the problem may be known, or it may be open... I just care for connections between problems and equations that one can use to motivate the study of diophantine equations, arithmetic geometry, and so on. I am more interested in problems that can be stated in elementary terms (that an undergraduate can understand), but I'll be happy to see any problems that you think fit the bill.
I'll start with one of my favorites: the congruent number problem leads to the study of elliptic curves of the form $y^2=x^3-n^2x$.
PS: the problem does not need to be a famous problem, any problem that is interesting, cute, entertaining, and leads to a diophantine equation also works!

REPLY [7 votes]: The triangular-square problem, which marlu posted, leads to a concrete instance of Pell's equation. There are quite a few elementary problems leading to Pell's equation for small values of $d$: see Barbeau's book "Pell's Equation". These are nice problems, but because $d$ turns out to be a small explicit number the fact that a general $x^2 - dy^2 = 1$ with nonsquare $d > 1$ is nontrivially solvable is not directly relevant. Here is an example of a problem that involves infinitely many instances of Pell's equation, due to H. Alder and W. Simons ($n$ and $n+1$ Consecutive Integers with Equal Sums of Squares, Amer. Math. Monthly 74 (1967), 28--30.)
For a positive integer $n$ we would like to find a set of $n$ consecutive squares and $n+1$ consecutive squares whose sums are equal:
$$
x^2 + (x+1)^2 + \cdots + (x+n-1)^2 = y^2 + (y+1)^2 + \cdots + (y+n)^2
$$
for some integers $x$ and $y$. (When $n = 1$ this is the problem of finding a Pythagorean triple with consecutive legs: 3,4,5; 20,21,29; 119,120,169;....)
Writing $z=x-y$, after some tedious algebra the above equation is the same as
$$
(y+n(1-z))^2 = n(n+1)z(z-1).
$$
Let $a^2$ be the largest square factor of $n(n+1)$: $n(n+1) = a^2b$, where $b$ is squarefree.  Then $a^2b$ is a factor of $(y+n(1-z))^2$, so (!) $ab$ is a factor of $y+n(1-z)$. Set $y+n(1-z) = abv$, with $v$ an integer.  Then the above equation is the same as
$$
a^2b^2v^2 = a^2bz(z-1) \Longleftrightarrow (2z-1)^2 - 4bv^2 = 1.
$$
We are thus reduced to solving $u^2 - 4bv^2 = 1$ for integers $u$ and $v$ (necessarily $u$ is odd), and $4b$ is not a square. Aside from the factor of 4, which could be absorbed into $v$ by insisting $v$ be even, this is a fairly general Pell equation with squarefree $b$. (One solution is $u=2n+1$ and $v = a$, leading to $x = 2n^2 + 2n+1$ and $y = 2n^2 + n$, but that is not always the first solution: try $n=8$.)
Moreover, every squarefree $b > 1$ does arise in such a problem, because there is always an integer $n$ such that $n(n+1) = a^2b$ for some integer $a$: finding $n$ and $a$ such that $n(n+1) = a^2b$ is equivalent to $(2n+1)^2 - 4ba^2 = 1$, and the equation $X^2 - 4bY^2 = 1$ does have some integral solution $(X,Y)$ by the general theory of Pell's equation, with $X$ necessarily odd.


TITLE: Rank of fundamental groups of open subsets.
QUESTION [0 upvotes]: Let $X$ be an open connected subset of the real plane. Then it is known that $\pi_1(X)$ is a free group. Is there a useful formula for the rank of $\pi_1(X)$? I suspect that the rank should be $b-1$, where $b$ is the number of boundary components. is this true? (i suspect that $X$ is homootopy equivalent to a bouqet of cicles.

REPLY [0 votes]: Studiosus's answer explains how to get the result you're looking for. This post is orthogonal: it's about how to understand the homeomorphism type of connected open subsets of the plane.
Every noncompact manifold without boundary has an exhaustion by compact submanifolds: $M = \bigcup M_i$, where $M_1 \subset M_{2} \subset \dots$ are compact manifolds contained in the interior of the next. (You can prove this by first demonstrating the existence of a proper Morse function on $M$, but it might be best to just take it for granted.)
So let $M$ be your open subset of $\Bbb R^2$. What can $M_0$ be? By the classification of compact surfaces, and the fact that no surface of positive genus (with boundary or not) can embed in $\Bbb R^2$, it has to be $S^2$ with a few open discs deleted. The same is true for all of the $M_{i+1} \setminus \text{int } M_i$.
First, a quick calculation: by the Mayer-Vietoris sequence you can see that $\chi(M_{i+1}) = \chi(M_i) + \chi(M_{i+1} \setminus \text{int }M_i).$ (This is essentially because $\chi(S^1) = 0$.)
Let $\Sigma_{0,n}$ be $S^2$ minus $n$ open discs. Then $\chi(\Sigma_{0,n}) = 2-n$. The only $\Sigma_{0,n}$ with positive Euler characteristic and with boundary are the disc $\Sigma_{0,1}$ and the annulus $\Sigma_{0,2}$. Gluing on a disc reduces the number of boundary components by one, and gluing on an annulus doesn't change the homeomorphism type of your manifold. Thus, putting together everything we've said above, because $\chi(M) = 1 - \text{rank}(\pi_1)$ if $\text{rank}(\pi_1)$ is finite, $H_k(M) = \lim_{i \to \infty} H_k(M_i)$, and our assumption that $\text{rank}(\pi_1)$ was finite, we see that the homeomorphism type of an open connected subset of the plane with $\pi_1(M)$ of finite rank is a sphere punctured $\text{rank } \pi_1(M)+1$ times. That is, it's homeomorphic to $\Bbb R^2$ with $\text{rank } \pi_1(M)$ points deleted.


TITLE: If an object moves at constant speed, does it necessarily have constant velocity?
QUESTION [1 upvotes]: If an object moves at constant speed, does it necessarily have constant velocity?

REPLY [0 votes]: If an object moving at a constant speed, It is not necessarily that it should move with constant linear velocity because Linear velocity is speed along with direction. so if the direction is not constant then Linear velocity will vary.
e.g - An object moving with constant speed in a circular path certainly doesn't have constant linear velocity because its direction keeps on changing at every instant tangentially. However, Its angular velocity will be constant.
Please note that the only condition the speed and the linear velocity be constant if the object travels in a straight line path without u-turn


TITLE: De Moivre's Theorem to calculate the fourth roots of 8
QUESTION [2 upvotes]: I was given this question:
Use de Moivre's theorem to derive a formula for the $4^{th}$ roots of 8.
As far as my understanding of this theorem goes, it is only applicaple to complex numbers. How am I supposed to use it for 8?
My initial thought was use 8 to make $z = 8 + i0$. Determining the polar form:
$$z = 8(\cos \pi + i \sin \pi )$$
from the polar form a I then get that
$$z^4 = 8^4 (\cos (4\pi) + i \sin(4\pi))$$
This is as far as I can go. I only recently started working with de Moivre's theorem so I'm not sure about my calculations and would appreciate any clarification on how to go about answering this question.

REPLY [2 votes]: Even though you were told to use the estimable Mr. de Moivre, it’s easiest without:
You know that if $\rho$ is one fourth root of a number, the others are $-\rho$ and $\pm i\rho$. Since one fourth root of $8$ is the real number $2^{3/4}$, you have them all.