text:
\begin{document}

\subjclass[2010]{Primary 14K20 ; Secondary 53C26, 14C30, 32J27}

\begin{abstract}
As in the case of irreducible holomorphic symplectic manifolds, the period domain $Compl$ of
compact complex tori of even dimension $2n$ contains twistor lines. These are special $2$-spheres parametrizing complex tori whose complex structures arise from a given quaternionic structure. In analogy with the case of irreducible holomorphic symplectic manifolds, we show that the periods of any two complex tori can be joined by a {\em generic} chain of twistor lines. Furthermore, we show that twistor lines are holomorphic submanifolds of $Compl$, of degree $2$ in the Pl\"ucker embedding of $Compl$.
\end{abstract}
\maketitle
\tableofcontents
\section*{Introduction}

Let $M$ be a complex manifold of dimension $2n\geq 2$. Then $M$ is called {\it hyperk\"ahler} with
respect to a Riemannian metric $g$ (see \cite[p. 548]{Hitchin})
if there exist covariantly constant complex structures $I,J$ and $K$ which
are isometries of the tangent bundle $TM$ with respect to $g$, satisfying
the quaternionic relations $$I^2=J^2=K^2=-1, IJ=-JI=K.$$

We call the  ordered triple $(I,J,K)$ {\it a hyperk\"ahler
structure on $M$ compatible with $g$}.
A hyperk\"ahler structure $(I,J,K)$ gives rise to a sphere $S^2$ of complex structures on $M$:
$$S^2=\{aI+bJ+cK| a^2+b^2+c^2=1\}.$$

We call the family $\mathcal M=\{(M,\lambda)| \lambda \in S^2\}  \rightarrow S^2$ a {\it twistor family over the twistor sphere $S^2$}. The family $\mathcal M$ can be endowed with a complex structure,
so that it becomes a complex manifold and the fiber $\mathcal M_\lambda$ is biholomorphic to
the complex manifold $(M,\lambda)$, see \cite[p. 554]{Hitchin}.

The well known examples of compact hyperk\"ahler manifolds are even-dimensional complex tori and irreducible holomorphic symplectic manifolds ({\it IHS manifolds}).
We recall that an IHS manifold is a simply connected compact K\"ahler manifold $M$  with $H^0(M,{\Omega}_M^2)$ generated by
an everywhere non-degenerate holomorphic 2-form $\sigma$.

Examples of IHS manifolds include  $K3$ surfaces and, more generally, Hilbert schemes of points on $K3$
surfaces. For these examples there exist well-defined period domains, carrying the structure of a complex manifold, and every twistor family $\mathcal M$
determines an embedding of the base $S^2$ into the corresponding period domain
as a 1-dimensional complex submanifold. The image of such an embedding is called a {\it twistor line}.

It is known that in the period domain of
an IHS manifold any two periods can be connected
by a path of twistor lines, meaning an ordered sequence $S_1, \ldots , S_m$ of twistor spheres such that $S_i \cap S_{i+1}$ is non-empty if $1\leq i\leq m-1$
 (see \cite{Bea-Geometrie} or \cite{Bourbaki}). Moreover, such a path can be chosen generic, that
is, the manifolds corresponding to the periods  at intersections of successive lines in the path
have trivial N\'eron Severi groups.
Here we prove a similar result for the period domain $Compl$ of complex tori of dimension $2n$.

\begin{thmi}
\label{Theorem-tw-path-conn}
Any twistor sphere on a complex torus embeds into $Compl$ as a complex 1-dimensional submanifold.
In $Compl$ any two periods can be connected by a generic path
of twistor lines.
\end{thmi}

For the case of IHS manifolds
the proof of this fact relies on the realization of the period domain
as the grassmanian of oriented positive real 2-planes in the second cohomology, where positivity
is with respect to the Beauville-Bogomolov bilinear form, again see \cite{Bea-Geometrie} or  \cite{Bourbaki}.
This bilinear form provides a
very convenient tool for investigating the local topology of this period domain.

 For complex tori, however, we
do not have such a realization of their period domain and cannot use a similar argument. Here, instead,
we need to use the (less refined) fact that  the period domain of complex tori is a homogeneous space (which, certainly, the period
domain of an IHS manifold is, as well). Its homogeneous nature allows
us to proceed with proving the twistor path connectivity in steps that are more or less parallel to the steps
of the proof of the twistor path connectivity for the period domains of IHS manifolds.
Let $A$ be a complex torus of dimension $2n$. The period domain $Compl$ can be considered as an open subset of the
Grassmanian $G(2n,4n)$, whose points are $2n$-dimensional complex planes, realizing the real weight 1 Hodge structures on the complex
$4n$-dimensional vector space $T_{0,\RR}A \otimes \CC$.  The open subset consists of those $2n$-planes
in $T_{0,\RR}A \otimes \CC$ that do not intersect the real subspace $V_\RR:=T_{0,\RR}A \subset T_{0,\RR}A \otimes \CC$.
Explicitly, a complex structure $I \colon V_\RR \rightarrow V_\RR$ corresponds to
the point $(\mathbb{1}-iI)V_\RR \in G(2n,4n)$ where $\mathbb{1}$ denotes the identity map.

\begin{remi}
There is a relation between twistor path connectivity and rational connectedness, that is, the
connectedness of points of a complex manifold by chains of rational curves (for the latter see, for example, \cite{Kollar}).
The  grassmanian $G(2n,4n)$ being
a rational variety ($G(2n,4n) \overset{\sim}{\dashrightarrow} \PP^{4n^2}$), is certainly  rationally connected.
However, rational connectedness is  a weaker property than twistor path connectivity.
Indeed, the variety of lines in $\PP^{4n^2}$, passing through a
fixed point, has complex dimension $4n^2-1$ (and the dimension of the variety
of rational curves of degree $d>1$ in $\PP^{4n^2}$, passing through a fixed point, is even larger), thus its real dimension is $8n^2-2$.
On the other hand, our dimension count
in Corollary \ref{Corollary-M-I-dimension}
of the space of all twistor lines, passing through a fixed point in the period domain,
is $4n^2-1$. Thus, through a given point, there are half as many twistor directions than general rational curves,
and the problem of twistor path connectivity may be roughly considered as a ``sub-Riemannian''
version of rational connectedness.
\end{remi}

The plan of the paper is as follows.

In Section \ref{Space-of-twistor-lines:-dimension-count} we describe our basic set-up and show that the twistor spheres $S^2 \subset Compl$ are complex submanifolds (Corollary \ref{corSanalytic}). We define the
union $C_I$
of all twistor spheres passing through a given period $I$ and show that the stabilizer group $G_I$ of $I$ acts transitively on the set of twistor spheres containing $I$.
The sets $C_I$ will serve as the main tool in the proof of twistor path connectivity.

 In Section \ref{Twistor-path-connectivity-of-Compl} we provide an argument, illustrated by a picture, that the set of periods reachable from a given one $I$ by means of all possible triples of consecutive twistor spheres contains an open neighborhood
of the initial period. Then, the connectedness of the period domain allows us to conclude that any two periods can be connected by a path of twistor lines.
The three spheres argument is essentially due to the transversality formulated in its most general form in Proposition \ref{Proposition-general-transversality}. We also show that $C_I$ is a real analytic subset of $Compl$.

In Section \ref{Connectivity-by-generic-twistor-paths}
we prove the generic connectivity part of Theorem \ref{Theorem-tw-path-conn}.
The idea of the proof is to show that the space of triples of consecutive
twistor spheres connecting a fixed pair of periods is not the union of
of its subspaces for which the first or, respectively, the second, of the two joint points
belongs to the locus of tori with nontrivial N\'eron-Severi group in the period domain.
Again, the transversality, stated in Proposition \ref{Proposition-general-transversality},
constitutes the main tool for proving generic connectivity.

In Section \ref{Twistor-spheres-are-complex-submanifolds} we prove that the degree of twistor lines in $G(2n,4n)$
with respect  to the Pl\"ucker embedding is 2. Here we use the fact that the group $G := GL^+(V_\RR)$ acts transitively on the set of all twistor lines together with an explicit computation on an explicit example.

In Section \ref{Appendix} we gather the calculations needed in the example of Section \ref{Twistor-spheres-are-complex-submanifolds}.

The authors are indebted to Eyal Markman for suggesting the problem of twistor path connectivity
of the period domain and many useful comments that helped improve
the exposition.

\section{The space of twistor spheres}
\label{Space-of-twistor-lines:-dimension-count}
\subsection{} Let $A$ be a complex torus of complex dimension $2n$. Denote
by $V_\RR$ the real tangent space $T_{\RR,0}A$ and
by $V$ the complex tangent space $T_{\CC,0}A\subset T_{\RR,0}A \otimes \CC$,
so that $\dim_\RR V_\RR = 2\dim_\CC V=4n$.
Let $I\colon V_\RR \rightarrow V_\RR$ be the operator of the complex structure on $V_\RR$ induced
by scalar multiplication by $i$ on $V$.
Let  $G := GL^+(V_\RR)\cong GL^+_{4n}(\RR)$ be the group of
orientation-preserving automorphisms of $V_\RR$, which acts via conjugation on the set of complex structures on $V_\RR$: $g\cdot I=\, ^g\! I:=gIg^{-1}$
for $g \in G$. This action is clearly transitive.
We consider the set $Compl$ of all complex structures on $V_\RR$ as the
orbit of $I$:
$Compl=G\cdot I$, diffeomorphic to the homogeneous space
$G/GL(V) \cong GL^+_{4n}(\RR)/GL_{2n}(\CC)$.  It carries the structure of a complex manifold, see \cite[p. 31]{BirkenhakeLange99} and Proposition \ref{propCompl}.
Assume that, in addition to $I$, there is a complex structure $J$  on $V_\RR$ anticommuting with $I$.
Then $I$ and $J$ determine a twistor sphere $$S(I,J,K) :=\{aI+bJ+cK|a^2+b^2+c^2=1\},$$ where $K=IJ$.
In general, for two complex structures $I_1,I_2$, not necessarily anticommuting, such that $I_1 \neq \pm I_2$,
and such that they are contained in the same twistor sphere $S$, we will also denote
this sphere by $S(I_1,I_2)$.
Our notation is justifed by the
following lemma.

\begin{lem}\label{lempair}
Every twistor sphere $S$ is uniquely determined by any pair of non-proportional complex structures $I_1,I_2 \in S$.
\end{lem}

\begin{proof}
By definition, $S=\{aI+bJ+cK|a^2+b^2+c^2=1\}$ for some $I,J$ and $K=IJ$ satisfying the quaternionic relations.
Let the 3-dimensional real vector space $\langle I,J,K\rangle_\RR \subset End\, V_\RR$
be equipped with the inner product $(\cdot, \cdot)$ defined by requiring the basis $I,J,K$
to be orthonormal. Let $\HH := \HH (I,J) = \langle Id, I,J,K\rangle_\RR \subset End\, V_\RR$ be the subalgebra of quaternions generated by $I$ and $J$.
For arbitrary vectors $u,v \in \langle I,J,K\rangle_\RR \subset \HH$ we have
the equality in $\HH$, $u\cdot v=-(u,v)Id+u\times v$ where $\cdot$ denotes the product in $\HH$ and $u \times v$ is the ordinary cross product of vectors in $\RR^3 = \langle I,J,K\rangle_\RR$.
Note that $u$ and $v$ are orthogonal if and only if $u$
and $v$ anticommute, and the vectors of length 1 in $\langle I,J,K\rangle_\RR$ are precisely those belonging to $S$. So our inner product does not depend on the choice of the quaternionic basis $I,J,K$
spanning the sphere $S$. The complex structures $I_1,I_2$ determine a 2-plane $\langle I_1,I_2\rangle_\RR$ in $End\, V_\RR$. The intersection of this 2-plane with the set of all complex structures is the set of vectors of length $1$ for our inner product, hence equal to the great circle
$S^1=S\cap \langle I_1,I_2\rangle_\RR$ and independent of the choice of twistor sphere $S$ containing $I_1, I_2$.
Choose $u,v \in S$ to be an orthonormal basis of our plane $\langle I_1,I_2\rangle_\RR$
or, equivalently, a pair of anticommuting complex structures in $S^1$.
Then $u\cdot v=u\times v$ is again a point in $S$ whose corresponding vector is orthogonal to our 2-plane
(which is equivalent to $u\cdot v$ being a complex structure anticommuting with $I_1,I_2$).
Thus $S=\{au+bv+c \: u\cdot v| a^2+b^2+c^2=1\}$ which shows that the set $S$
is uniquely determined by $u,v$ (and $u\cdot v$) and hence it is uniquely determined by $I_1$ and $I_2$.
\end{proof}

\subsection{} By the definition of $Compl$, the group $G$ acts transitively on the space of complex structures $Compl$ on $A$:
$$g \in G \colon J \mapsto \: ^g\! J= gJg^{-1}.$$ In particular, $G$ acts on the set of all twistor spheres
$S(I,J)$:
\[
g\cdot S(I,J)=S(\,^g\! I,\,^g\! J).
\]

Consider the subgroup $G_I$ of $G$ fixing $I$ via the above action:  $G_I=GL(V_\RR,I) \cong GL(V)$, so that $Compl \cong G/G_I$.
For $g \in G_I$ we have $g \cdot S(I,J)=S(I,\,^g\!J)$.
We have

\begin{prop}
\label{transitivity}
The group $G_I$ acts transitively on the set $N_I$ of complex structures anticommuting with $I$.
\end{prop}

\begin{proof}

Fix a hermitian form $h$ on $V$ (determining a K\"ahler class in $H^{1,1}(A)$)
and a global holomorphic form $\sigma \in H^{2,0}(A)$.
Any operator $J\colon V_\RR \rightarrow V_\RR$
defined by the equation $h(x,Jy)=Re\, \sigma(x,y)$
is non-degenerate, skew-symmetric with respect to $h$ and  anticommutes with $I$.
As $h(\cdot,\cdot)$ is positive definite,
for such $J$, its square $J^2$ is a diagonalizable
operator with negative real eigenvalues, such that the vector space $V_\RR$ is an $h$-orhogonal
sum of the eigenspaces $V_\lambda$ for $J^2$: $V_\RR=\bigoplus^\perp V_\lambda$.
Moreover, as $J^2$ commutes with $I$, we have that the $V_\lambda$ are $I$-invariant.
It follows that every $V_\lambda$ is an orthogonal sum of subspaces of the form $\langle v, Iv, Jv, Kv\rangle, K=IJ, v \neq 0$.
As $h(Jx,Jy)=Re\,\sigma(Jx,y)=-Re\,\sigma(y, Jx)=-h(y, J^2x)=h(-J^2x,y)$, we see that $J^2=-Id$ if and only if
$J$ is an isometry with respect to $h$. Let us replace $J$ on each eigenspace $V_\lambda$ by $\frac{1}{\sqrt{-\lambda}}J$.
Denoting this operator again by $J$
we obtain $J^2=-Id$, moreover the vectors $v, Iv, Jv, Kv$ for $v \in V_\lambda$
all have equal $h$-norm and form an orthogonal basis of the subspace they span.
This tells us that $V_\lambda$, and hence the whole $V_\RR$, decomposes into an orthogonal sum of subspaces
$\langle v, Iv, Jv, Kv\rangle$ such that the union of the bases of these subspaces forms an orthonormal basis of $V_\RR$.
The matrix of $h$ on this basis is the identity matrix, and the matrix of $J$ has the block-diagonal form with the following 4$\times$4 blocks on the diagonal
$$\left(\begin{array}{cc|cc} 0 & 0 &   -1 & 0\\ 0 & 0 &   0 & 1 \\
\hline
 1 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0\end{array}\right).$$

Summarizing the above observations, we see that our initial choice of a K\"{a}hler class $h \in H^{1,1}(A)$ and
nondegenerate $\sigma \in H^{2,0}(A)$ defines a complex structure $J$ which anticommutes with $I$,
and, moreover, gives, in a non-unique way, a choice of an $h-$orthonormal basis where $J$ has the canonical form above. The group $G_I \cong GL(V) < GL^+(V_\RR)=GL^+_{4n}(\RR)$ acts transitively on the set of such bases.

Conversely, given a complex structure $I$, any operator $J \colon V_\RR \rightarrow V_\RR$
 which anticommutes with $I$ and whose square is $-Id$ (but which is not a priori defined as  a skew-symmetric operator with respect to a form on $V_\RR$ induced by a hermitian form on $V$),  originates, via a procedure similar to the one described above, from
an appropriate K\"ahler class $h$ and a non-degenerate holomorphic 2-form $\sigma$.
\end{proof}

\subsection{} Therefore, given a complex structure $J\in N_I$, $N_I = G_I \cdot J \cong G_I / G_{I,J}$ is the orbit of $J$ under $G_I$, where $G_{I,J}$ is the stabilizer group of $J$ in $G_I$.
Since $G_{I,J}$ is the subgroup of elements of $G_I = GL(V)$ commuting with $J$, that is, preserving the quaternionic structure
on $V_\RR$ determined by $I$ and $J$, we have $G_{I,J} \cong GL(V, \mathbb{H})$ which we will also
denote by $G_\HH$. So $N_I \cong GL(V)/GL(V, \mathbb{H})$ and we deduce

\begin{cor}\label{corNIdim}
The set $N_I$ is a real submanifold of $Compl$ of  dimension $4n^2$.
\end{cor}
\begin{proof}
The dimension of the orbit as a complex manifold is
$\dim_\CC\, GL(V)-\dim_\CC\, GL(V, \mathbb{H})=(2n)^2-2n^2=2n^2$
(the complex dimension of the space of quaternionic $n$$\times$$n$ matrices is  $2n^2$).
The real dimension is thus
equal to $4n^2$.
\end{proof}

\subsection{}\label{subsecSMN} Let $S=S(I,J)$ for $J\in N_I$ be a twistor sphere.
Define $G_{I,S}\subset G_I$ to be the stabilizer of $S$ as a set, i.e., the set of elements $g$ of $G_I$ such that $g \cdot S \subset S$. For any $g\in G_{I,S}$, the complex
structure $^g\! J\in S$ also anticommutes with $I$, so $^g\! J$ is of the form
$aJ+bK, a^2+b^2=1$. Setting $a=\cos t, b=\sin t$ we have $aJ+bK=
e^{\frac{tI}{2}}Je^{-\frac{tI}{2}}$, where $e^{sI} = \cos s \: \mathbb{1} + \sin s \: I\in G_I$ realizes,
via the adjoint action,
the rotations of $S$ around $\{\pm I\}$. Conversely, if $g\in G_I$ and $^g\!J \in S$, then $g \in G_{I,S}$.
The set of $g \in G_{I,S}$ such that $^g\!J=J$ is the quaternionic subgroup
$G_{I,J}=G_{\mathbb H} \subset G_{I,S}$.
Explicitly, we have
$G_{I,S}=\langle e^{tI}, t\in \RR\rangle \times G_{\HH}$, where $\langle e^{tI}, t\in \RR\rangle \cong S^1$ (which is a subgroup of the center of $G_I$).
This
tells us, in particular, that $\dim_\RR\, G_{I,S}=\dim_\RR \,G_{\mathbb H}+1=4n^2+1$.

Let $M_I$ be the set of all twistor spheres in $Compl$ containing $I$. The natural map $N_I \ra M_I$ identifies two complex structures $J_1$ and $J_2$ whenever they belong to the same twistor sphere through $I$, i.e., $S(I, J_1) = S(I, J_2)$. More precisely, they belong to the great circle in $S := S(I, J_1)$ consisting of elements anticommuting with $I$.
Hence, for the $S^1$-action $J \in N_I \mapsto e^{tI}J=e^{\frac{tI}{2}}Je^{-\frac{tI}{2}}$ on $N_I$ defined above, we have $N_I / S^1 = M_I$. Therefore Corollary \ref{corNIdim} immediately implies

\begin{cor}
\label{Corollary-M-I-dimension}
The set $M_I$ is a real manifold of dimension $4n^2 -1$.
\end{cor}

\subsection{The twistor cone of $I$} Define the set $C_I := \bigcup_{S\in M_I}S \subset Compl$ as the union of all twistor spheres containing $I$.
All spheres in this union contain the complex structures $I$ and $-I$.
We will sometimes refer to the set $C_I$ as a cone. Proposition \ref{transitivity} immediately implies
\begin{cor}\label{corMItrans}
The group $G_I$ acts transitively on $M_I\cong G_I/G_{I,S}$ so that
$C_I=\bigcup_{g\in G_I}g \cdot S(I,J)$.
\end{cor}
We will
give an explicit local parametrization of $C_I$ in  the next section and prove that
the cone $C_I$ is a real-analytic subset of $Compl$ of dimension $4n^2+1$ (Proposition \ref{Prop-cone-dimension}).

\subsection{} We now describe the complex structure on the tangent bundle of the orbit
$Compl = G \cdot I$. Then we will see that the tangent bundle $TS^2$ of an arbitrary twistor sphere
$S^2 \subset Compl$
is a subbundle of the restricted tangent bundle $TCompl|_{S^2}$, invariant under the complex structure of $TCompl|_{S^2}$.
This will imply that $S^2$ is a complex submanifold in $Compl$.

\begin{prop}\label{propCompl}  The manifold $Compl$ is a complex manifold. Its complex structure $l_I$ is given by left multiplication by $I$ on $T_ICompl \subset End (V_\RR)$.
\end{prop}
\begin{proof}
In order to prove the theorem we will use the classical period matrix realizations of open charts of
the period domain. The charts carry the (compatible) complex structures, induced from
the complex affine space containing the period matrices, thus  giving a globally defined complex structure
on the period domain.
We will show that the complex structure $l_I \colon T_ICompl \rightarrow T_ICompl$ in the statement is induced by this complex structure.

The period matrix realization of the period domain of marked complex tori
uses the complex space $V$ with a fixed basis $e_1,\dots, e_{2n}$ so that an arbitrary
full rank lattice $\Gamma \subset V_\RR$, determining the complex torus $A=V_\RR/\Gamma$, can be written in terms of this basis of $V$,
$$\langle \gamma_1,\dots,\gamma_{2n}, \dots, \gamma_{4n}\rangle=(e_1,\dots , e_{2n})(Z_1,Z_2).$$
The complex affine space $Mat_{2n\times 4n}(\CC) \cong \CC^{8n^2}$ of $2n \times 4n$ complex matrices $Z=(Z_1,Z_2)$  carries
a natural complex structure, given by multiplication by $i$. The affine chart of matrices $(Z_1,Z_2)$, subject to the open condition
of the corresponding $\Gamma$ having full rank, holomorphically embeds into $G(2n,4n)$, where the embedding
sends $(Z_1,Z_2)$ to the $2n$-dimensional subspace of $V_\RR \otimes\CC$ spanned by the rows of $(Z_1,Z_2)$.
 There is a mapping $f$ from an open subspace
of $Mat_{2n\times 4n}(\CC)$ to $Compl$, namely, the matrix $Z=(Z_1,Z_2)$, whose columns generate a full rank lattice $\Gamma$ in $V_\RR$, uniquely determines
the complex structure $f(Z)=I_Z \colon V_\RR \rightarrow V_\RR$ of the torus
$A \cong V_\RR/\Gamma$. Such $I_Z$ is induced
by multiplication by $i$ on $V$, that is, the matrix of $I_Z$ in the basis
$\gamma_1,\dots,\gamma_{2n}, \dots, \gamma_{4n}$ of $\Gamma$,
denoted also by $I_Z$, satisfies
$$(e_1,\dots, e_{2n})Z I_Z = (e_1,\dots,e_{2n})iZ.$$
This can be written in short as the relation
$$Z f(Z)=i Z.$$
The mapping $f$ (certainly, not bijective) is a submersion onto $Compl$.
We want to use the differential $df|_Z$ to induce the complex structure on $T_{f(Z)}Compl$.
If we take  any small tangent vector $X \in T_Z Mat_{2n\times 4n}(\CC)$
then, writing $f(Z+X)=I_Z+ df|_{Z}(X)+\dots$ and $f(Z+iX)=I_Z+ df|_{Z}(iX)+\dots$, we have that the equalities
\[(Z+X)f(Z+X)= i(Z+X),\quad (Z+iX)f(Z+iX)= i(Z+iX)\]
imply
\[Z df(X)+XI_{Z}=iX, \quad Z df(iX)+iXI_{Z}=-X.\]
Multiplying the first of these equalities by $i$ and comparing to the second, we obtain
$i Z df(X)=Z df(iX)$. For $Z$ in the domain of $f$, this equation has
a unique solution $df(iX)=I_{Z}df(X)$, which is precisely the statement of the proposition.
\end{proof}

\begin{cor}\label{corSanalytic}
The twistor spheres $S^2 \subset Compl$ are complex submanifolds.
\end{cor}
\begin{proof}
The proof is based on the simple observation that the tangent space of $S^2=S(I,J)$
at the point $I$,
for $I,J,K=IJ$ satisfying the quaternionic identities,  is the 2-plane $\langle J,K\rangle_\RR \subset T_ICompl$ (this can be shown using the parametrization of $S^2$ by $\langle e^{sJ} e^{tK} I\mid s,t \in \RR\rangle$ as in Paragraph \ref{subsecSMN})
and this plane is obviously invariant under left multiplication by $I$. Thus, $TS^2$ is a complex subbundle of $TCompl|_{S^2}$ and
thus  $S^2 \subset Compl$ is a complex submanifold.
\end{proof}

\section{Twistor path connectivity of $Compl$}
\label{Twistor-path-connectivity-of-Compl}
The main result of this section is Theorem \ref{Theorem-connectivity}.
Before proving it we need to introduce a certain mapping and prove an important technical result
 about it (Proposition \ref {Injectivity-proposition}).

\subsection{} \label{subsecPhi} Let $I,J,K$ be a triple of complex structures belonging to a twistor sphere $S$.
 Consider the smooth mapping
 \[
 \begin{array}{cccc}
 \Phi \colon & G_J \times G_K & \longrightarrow & Compl,\\
& (g_1,g_2) &\longmapsto & ^{g_1g_2}\! I,
\end{array}
\]
where, as before, the action on $Compl$ is by conjugation: $g\cdot I=\,^g\!I=gIg^{-1}$. The mapping
$\Phi$ clearly sends $G_\HH \times G_\HH$ to $I$, so that
 its differential $d_{(e,e)}\Phi$ factors through
$$\widetilde{d_{(e,e)}\Phi}: T_eG_J/T_eG_{\HH} \oplus T_eG_K/T_eG_{\HH} \rightarrow T_ICompl.$$
\begin{prop}
\label{Injectivity-proposition}
Suppose $I,J,K$ is a quaternionic triple. The mapping
$$\widetilde{d_{(e,e)}\Phi}: T_eG_J/T_eG_{\HH} \oplus T_eG_K/T_eG_{\HH} \rightarrow T_ICompl$$
is injective. Hence, since the two spaces have the same dimension, it is an isomorphism.
\end{prop}
\begin{proof}
By the definition of $\widetilde{d_{(e,e)}\Phi}$, its restrictions to the above direct summands
are injective. Let us show that it is injective on the direct sum.
Consider $X\in T_eG_J, Y\in T_eG_K$ and the vector
$\widetilde{d_{(e,e)}\Phi}(X+T_eG_\HH, Y+T_eG_\HH)$, which is
 $$d_{(e,e)}\Phi(X+Y)=\left.\frac{d}{dt}\right|_{t=0}(e^{tX}e^{tY}\cdot I) = (X+Y)I-I(X+Y)\in T_ICompl.$$ Assume that this vector is zero,
that is, $X+Y$ commutes with $I$:
\begin{equation}
\label{fail-of-injectivity}
I(X+Y)=(X+Y)I.
\end{equation}
Then the conjugate $(X+Y)^J=J^{-1}(X+Y)J=X^J+Y^J=X-JYJ$ must also commute with $I$.
Using that $Y$ commutes with $K$ we obtain $$X-JYJ=X-JYKI=X-JKYI=X-IYI.$$
The commutation with $I$ is expressed now by $I(X-IYI)=(X-IYI)I$, or
$$IX+YI=XI+IY,$$
which gives $$I(X-Y)=(X-Y)I.$$
Adding the last equality to (\ref{fail-of-injectivity}) side by side gives that
$XI=IX$, hence $YI=IY$, which implies $X,Y \in T_eG_\HH$. This
proves the required injectivity of $\widetilde{d_{(e,e)}\Phi}$.
\end{proof}

\begin{cor}\label{Phisubmers}
Suppose $I,J,K$ is a quaternionic triple. The mapping $\Phi$ is a submersion at $(e,e)\in G_J \times G_K$, that is
$$d_{(e,e)}\Phi(T_eG_J\oplus T_eG_K) =T_ICompl\cong \RR^{8n^2}.$$
\end{cor}
\begin{proof}
As $\dim_\RR T_eG_J/T_eG_{\HH}=\dim_\RR T_eG_K/T_eG_{\HH}=4n^2$, the statement that
$\Phi$ is a submersion follows from the fact that $d_{(e,e)}\Phi$ factors through $\widetilde{d_{(e,e)}\Phi}$
and the fact that the mapping
$\widetilde{d_{(e,e)}\Phi}: T_eG_J/T_eG_{\HH} \oplus T_eG_K/T_eG_{\HH} \rightarrow T_ICompl$ is an isomorphism by Proposition \ref{Injectivity-proposition}.
\end{proof}

\begin{thm}
\label{Theorem-connectivity}
Given a complex structure $I\in End(V_\RR) $, there is a neighborhood of $I$
in the space of complex structures on $V_\RR$ such that, for any complex structure $I_1$
in this neighborhood, there is a twistor path consisting of three spheres joining $I$ to $I_1$.
Consequently, the space of complex structures $Compl$ is twistor path connected.
\end{thm}
\begin{proof}
Choose a complex structure $J$,  anticommuting with $I$, and consider the sphere $S=S(J,I)$
 and the cone
$C_J$.
By Lemma \ref{lempair}, the complex structures $K=IJ$ and $I$ span the sphere $S=S(K,I)=S(J,I)$. We can then form
the cone $C_{K}$ whose intersection with $C_J$ contains $S$.
See Picture 1 below where the cones $C_J$ and $C_K$ are depicted by transversal planes
and the sphere $S$ lying in their intersection is depicted by a line.

We first show that
the images of $C_{K}$ under the action of $G_J$ (``rotation of $C_K$ around $J$'') sweep out an open neighborhood of $I$ in
$Compl$.
Since
$\Phi$ is a submersion by Corollary \ref{Phisubmers}, there exist neighborhoods  $U_{e,J}\subset G_J$
and $U_{e,K} \subset G_K$ of $e$ such that the set $\Phi(U_{e,J}\times U_{e,K})$
contains an open neighborhood
of $I$.
By definition, the cone $C_K$ contains the orbit $G_K\cdot I$.
Hence the union $\bigcup_{g\in G_J}{}^gC_K$ contains the image of $\Phi$
and consequently it contains an open neighborhood of $I$.

Now the three twistor spheres connecting $I$ to an arbitrary point $I_1$ in this neighborhood are found as illustrated in the following picture.

\vspace*{0.8cm}

\input{Picture1-2.pic}

\vspace*{-12cm}
\begin{center}
Picture 1.
\end{center}
\medskip

The connectedness
of the space $Compl$ allows us to conclude, as in \cite[Prop. 3.7]{Bourbaki} that $Compl$ is twistor path connected.
\end{proof}

\subsection{} Another immediate consequence of the injectivity of $\widetilde{d_{(e,e)}\Phi}$ proved in Proposition \ref{Injectivity-proposition}
is the following

\begin{cor}\label{cortrans}
For a quaternionic triple $I,J,K$, the triple intersection of the submanifolds $G_I/G_\HH,G_J/G_\HH$ and $G_K/G_\HH$ of the homogeneous
space $G/G_\HH$ at $eG_\HH$ is transversal.
\end{cor}
The following generalization of this transversality is one of the main ingredients of the proof of connectivity by generic twistor paths in Section
\ref{Connectivity-by-generic-twistor-paths}.

\begin{prop}
\label{Proposition-general-transversality}
Let $I_1,I_2,I_3$ be complex structures belonging to the same twistor sphere $S$.
The submanifolds $G_{I_1}/G_\HH, G_{I_2}/G_\HH, G_{I_3}/G_\HH$ in $G/G_\HH$
intersect transversally (as a triple) if and only if $I_1,I_2,I_3$ are linearly independent as vectors in
$End\, V_\RR$.
\end{prop}
\begin{proof} Choose anticommuting
complex structures $I,J$ in $S$, and set $K=IJ$.
By Corollary \ref{cortrans},
\begin{equation}
\label{I-J-K-transversality}
T_eG/T_eG_\HH=V_I \oplus V_J \oplus  V_K,
\end{equation}
where we set  $V_I:=T_eG_{I}/T_eG_\HH, V_J:=T_eG_{J}/T_eG_\HH, V_K:=T_eG_{K}/T_eG_\HH$.

We shall prove that $T_eG/T_eG_\HH$ also decomposes into the direct sum of its subspaces
$V_i:=T_eG_{I_i}/T_eG_\HH, i=1,2,3.$ Put $I_i=a_iI+b_iJ+c_iK, i=1,2,3$.
Assume, on the contrary, that for certain vectors
$X \in V_1, Y \in V_2$ and $Z \in V_3$
we have $X+Y+Z=0$. Let $X:=X_I+X_J+X_K$ be the decomposition of $X$ into the sum of its components
in the respective subspaces of (\ref{I-J-K-transversality}),
and do similarly for $Y$ and $Z$.  Then for $X$ the commutation relation $[X,I_1]=0$
can be written as
$$a_1[X_J+X_K,I]+b_1[X_I+X_K,J]+c_1[X_I+X_J,K]=0.$$
Note that in the above expression, the term $[X_J,I]$, for example, anticommutes with both $I,J$, hence commutes with $K=IJ$,
and an analogous commutation relation holds for the other terms as well. Hence we can decompose the expression on the left side of the above equality
with respect to (\ref{I-J-K-transversality}):
$$(b_1[X_K,J]+c_1[X_J,K])+(a_1[X_K,I]+c_1[X_I,K])+(a_1[X_J,I]+b_1[X_I,J])=0.$$
From here we conclude that $b_1[X_K,J]+c_1[X_J,K]=0, a_1[X_K,I]+c_1[X_I,K]=0$
and $a_1[X_J,I]+b_1[X_I,J]=0$. Perturbing
the quaternionic triple $I,J,K$, we may assume that all $a_i, i=1,2,3,$ are nonzero. Then we can use
the last two equalities to express
\begin{equation}
\label{X-I-reduction}
[X_J,I]=-\frac{b_1}{a_1}[X_I,J], \quad [X_K,I]=-\frac{c_1}{a_1}[X_I,K].
\end{equation}
Note  that $F_J := [\cdot,J] \colon V_I \rightarrow V_K, F_K := [\cdot, K] \colon V_I \rightarrow V_J$
and $F_I := [\cdot, I] \colon V_J \rightarrow V_K$ are isomorphisms of the respective vector spaces. Then, using (\ref{X-I-reduction}),
we can write
$$X_J=-\frac{b_1}{a_1}F^{-1}_I \circ F_J(X_I), \quad X_K= -\frac{c_1}{a_1}F^{-1}_I \circ F_K(X_I),$$
so that
$$X=X_I +\left(-\frac{b_1}{a_1}F^{-1}_I \circ F_J(X_I)\right) +  \left(-\frac{c_1}{a_1}F^{-1}_I \circ F_K(X_I)\right).$$
Using $a_2,a_3 \neq 0$, we obtain similar expressions for $Y$ and $Z$.
Since $F_I, F_J, F_K$ are isomorphisms, the equality $X+Y+Z=0$ can now be written as
\[
\left(\begin{array}{ccc}
1 & 1 & 1\\
-\frac{b_1}{a_1} & -\frac{b_2}{a_2} & -\frac{b_3}{a_3}\\
-\frac{c_1}{a_1} & -\frac{c_2}{a_2} & -\frac{c_3}{a_3}
\end{array}\right)
\left(\begin{array}{c}
X_I \\
Y_I\\
Z_I
\end{array}\right)=
\left(\begin{array}{c}
0 \\
0\\
0
\end{array}\right).
\]
This has a nontrivial solution if and only if the columns of the matrix, i.e., $I_1,I_2,I_3$, are linearly dependent.
\end{proof}

\subsection{} We can now prove that the cone $C_I$ has a real analytic structure. Define the incidence correspondence
\[
S_I := \{ (S, J) \mid J\in S \} \subset M_I \times Compl.
\]
Then $S_I$ is an $S^2$-bundle over $M_I$ and $C_I$ is the image of $S_I$ by the projection to $Compl$:
\[
\xymatrix{
N_I \ar@{^{(}->}[r] \ar[rd] & S_I \ar[d]^{pr_1} \ar[r]^-{pr_2} & C_I \subset Compl\\
& M_I.
}
\]
The projection $S_I \ra M_I$ has two sections $\sigma_+$ and $\sigma_-$, given by $+I$ and $-I$ respectively.
\begin{prop}
\label{Prop-cone-dimension}
The map $pr_2: S_I \ra C_I$ is a diffeomorphism away from the images of $\sigma_{\pm}$ and contracts these images to points. Therefore the cone $C_I$ is a real-analytic subset of $Compl$ of dimension $4n^2+1$, smooth away from
the points $\pm I$.
\end{prop}
\begin{proof}
First note that $pr_2$ clearly contracts the images of $\sigma_{\pm}$. Also, it is injective away from $\pm I$ by Lemma \ref{lempair}. To see that it is also an immersion away from $\pm I$, let $J$ be in $ C_I\setminus\{\pm I\}$, not necessarily anticommuting with $I$.
Define the following mapping
$$\Phi\colon (T_eG_I/T_eG_\HH) \times \RR \rightarrow C_I,$$
$$(X,t) \mapsto e^Xe^{tK}Je^{-tK}e^{-X},$$
where $K \in S(I,J)\setminus S^1$ for
$S^1=\langle I,J\rangle_\RR \cap S(I,J)$.
Then the restriction of $\Phi$ on a small enough neighborhood of $(0,0) \in  (T_eG_I/T_eG_\HH) \times \RR$ defines a parametrization of  $C_I$ around $J$.

Here the subgroup $e^{tK}, t \in \RR$, rotates the sphere $S = S(I,J)$ around the axis $\{\pm K\}$
and, together with the rotation subgroup $e^{tI} \subset G_{I,S} \subset G_I$, sweeps out in $S$, via the above action,
a neighborhood of any point of $S$ other than $\pm I, \pm K$. Proposition \ref{Proposition-general-transversality} provides  that $K$ may be chosen arbitrarily in $S\setminus S^1$, which in turn gives us that $C_I$ is a manifold, smooth away
from $\pm I$,
of dimension $\dim_\RR (G_I/G_{I,S})+\dim_\RR S=(4n^2-1)+2=4n^2+1$.
The fact that the points $\pm I$ are indeed singular points of the cone $C_I$ is easy to prove.
\end{proof}

We can now use Proposition \ref{Injectivity-proposition} to also prove

\begin{cor}
For a quaternionic triple $I,J,K$, the cones $C_J$ and $C_K$ intersect transversely at $\pm I$ in $S:=S(I,J) \subset C_K \cap C_J$
and hence, by continuity, at all points in some neighborhood of $\pm I$.
\end{cor}

\begin{proof}
The transversality of intersection at $I \in S$ means
that the intersection of the quotient subspaces $W_J=T_IC_J/T_IS$ and $W_K=T_IC_K/T_IS$
in $T_ICompl/T_IS$ is zero.
The subspaces $W_J$ and $W_K$ are isomorphic images, under the quotient mapping $T_ICompl \rightarrow T_ICompl/T_IS$, of
the subspaces $T_eG_J/T_eG_{J,S}$ and $T_eG_K/T_eG_{K,S}$ respectively,
in the (non-canonical) decompositions of the transversal subspaces
$$\widetilde{d_{(e,e)}\Phi}(T_eG_J) \cong T_eG_J/T_eG_{J,S} \oplus T_eG_{J,S}/T_eG_\HH$$
and $$\widetilde{d_{(e,e)}\Phi}(T_eG_K) \cong T_eG_K/T_eG_{K,S} \oplus T_eG_{K,S}/T_eG_\HH.$$
Thus, the transversality of $W_J$ and $W_K$ follows.
\end{proof}

\section{Connectivity by generic twistor paths}
\label{Connectivity-by-generic-twistor-paths}

Recall that a period in $Compl$ is {\it generic} if the corresponding complex torus
has trivial N\'eron Severi group.
A twistor path in $Compl$
is called {\it generic}, if its successive twistor spheres intersect at generic periods.
In this section we prove the connectivity part of  Theorem \ref{Theorem-tw-path-conn}, i.e.,

\begin{prop}
\label{Proposition-generic-connectivity}
Any two periods in the  period domain $Compl$ can be connected by a generic twistor path.
\end{prop}

In this section, with the exception of Lemma \ref{Lemma-generic-period} and its proof, we do not assume that the complex structures $I,J,K$ (with or without subscripts) anticommute.

\subsection{Outline of the proof}\label{subsecoutline}
Define $\mathcal T$ to be the closure, in $Compl \times Compl \times Compl$, of the set of triples $(I,J,K)$ that are linearly independent and belong to the same twistor sphere. Denote by
\[
\begin{array}{cccl}
pr_1\colon & Compl\times Compl \times Compl & \longrightarrow & Compl,\\
pr_{23}\colon & Compl\times Compl \times Compl & \longrightarrow & Compl \times Compl
\end{array}
\]
the respective projections.
For $(I_1,J_1,K_1)\in \cT$, we defined, in Paragraph \ref{subsecPhi}, the mapping $\Phi_{I_1, J_1,K_1}$:
\[
\begin{array}{cccl}
\Phi_{I_1, J_1,K_1} \colon & G_{J_1} \times G_{K_1} & \longrightarrow & Compl, \\
& (g_1,g_2) & \longmapsto & g_1g_2I_1g_2^{-1}g_1^{-1}=\, ^{g_1g_2}I_1.
\end{array}
\]
Proposition \ref{Proposition-general-transversality}
tells us that, when $I_1,J_1,K_1$ are linearly independent, $\Phi_{I_1, J_1,K_1}$ is a submersion near $(e,e) \in G_{J_1} \times G_{K_1}$.
In other words, there is a
neighborhood $U_{e,G} \subset G=GL^+(V_\RR)$ of $e\in G$
such that the map $\Phi_{I_1,J_1,K_1}$ is submersive on $U_{e,J_1} \times U_{e,K_1}$, where
$U_{e,J_1}:=U_{e,G} \cap G_{J_1}$ and $U_{e,K_1}:=U_{e,G} \cap G_{K_1}$
 (and the image is, thus, a neighborhood of $I_1$ in $Compl$).

Let $I_2$ be an arbitrary point in the image of $\Phi_{I_1,J_1,K_1}$ and let $(g_1,g_2) \in U_{e,J_1} \times U_{e,K_1}$ be such that $I_2={}^{g_1g_2}I_1$.
With this notation, the three twistor spheres connecting $I_1$ to $I_2$

are: $S_1:=S(I_1,J_1,K_1)$, $S:={}^{g_1}S_1=S( {}^{g_1}I_1,{}^{g_1}\!J_1=J_1,
{}^{g_1}\!K_1)$ and $S_2:={}^{g_1g_2}S_1=S( I_2,{}^{g_1g_2}J_1,{}^{g_1g_2}K_1 = {}^{g_1}K_1)$,
with the joint points $J_1$ and ${}^{g_1}K$.

We are going to show
that,
for a fixed $I_1$, there is a neighborhood $U_{I_1}\subset Compl$ of $I_1$
such that for any $I_2\in U_{I_1}$,
we can choose a generic $J \in C_{I}$, a $K\in S(I,J)$ and find $(g_1,g_2) \in \Phi^{-1}_{I,J,K}(I_1)$
as above such that $^{g_1g_2}K$ is also generic.

We begin by proving, in Lemma \ref{Baire-lemma}, that the set of non-generic periods in $C_{I_1}$ is a countable union of proper analytic subsets, i.e., $J$ can be chosen generic.

Next, for $I_2$ close to $I_1$, and with $S_1,S,S_2$ as above, connecting $I_1$ to $I_2$, the initial sphere $S_1$ together with the choice of $J,K\in S_1$, uniquely determines the terminal sphere $S_2$ together with the pair of periods
 ${}^{g_1g_2}J,{}^{g_1g_2}K$.

To justify this uniqueness we first need to control the fibers of the maps $\Phi_{I,J,K}$ in a neighborhood of $(I_1, J_1, K_1)$, which we do in Lemma \ref{Fiber-lemma}.
This allows us to introduce, in Paragraphs \ref{subsecPsi12} and \ref{subsecPsi21}, two maps $\Psi^{I_1\ra I_2}$ and $\Psi^{I_2\ra I_1}$ which,

roughly, switch $(S_1,J,K)$ and $(S_2,{}^{g_1g_2}J,{}^{g_1g_2}K)$.

We then show in Lemma \ref{lemPsicompId}, after shrinking our various domains, that the composition of $\Psi^{I_1\ra I_2}$ and $\Psi^{I_2\ra I_1}$ is the identity. Corollary \ref{corfinal} then shows that this implies the irreducibility of the set of triples $(S_1,S,S_2)$ joining $I_1$ and $I_2$ mentioned in the introduction,
which gives that $J$ and $^{g_1g_2}K$ can both be chosen generic.

Thus the chain of three twistor spheres connecting $I_1$ to $I_2$ for every $I_2$ in some neighborhood of $I_1$
can be chosen in such a way that the periods at the intersections are generic. For arbitrary $I_1$ and $I_2$, we connect them by a path in $Compl$ consisting of generic triple subchains.

\subsection{} Let us first show that there are generic periods $J \in C_I$. Dimension-wise this is not trivial
because $\dim_\RR C_I=4n^2+1$, whereas the real dimension of the locus of, for example,
abelian varieties in $Compl$ is $4n^2+2n$. For an alternating form $\Omega$ on $V_\RR$
we denote by $Compl_\Omega$ the locus of periods in $Compl$ at which $\Omega$
represents a class of Hodge (1,1)-type, that is
$$Compl_\Omega=\{I\in Compl \,|\, \Omega(I\cdot,I\cdot ) = \Omega(\cdot,\cdot)\}.$$
If we fix a basis of $V_\RR$ and switch to matrix descriptions, then the condition
$\Omega(I\cdot,I\cdot)=\Omega(\cdot,\cdot)$ simply becomes $^tI\,\Omega I = \Omega$, where $I$ and $\Omega$ also denote the matrices of the corresponding complex structure and alternating form.
The locus of
marked complex tori with nontrivial N\'eron Severi group is
$$\mathcal L_{NS}=\bigcup_{0 \neq [\Omega] \in H^2(A,\QQ)} Compl_\Omega,$$
where $A$ is a fixed complex torus.

\begin{lem}
\label{Baire-lemma}
For every $I \in Compl$ the set of non-generic periods in $C_I$, that is $C_I \cap \mathcal L_{NS}$,
is a countable union of closed subsets of $C_I$ none of which contains an open neighborhood (in $C_I$)
of any of its points.
\end{lem}

The proof follows from the following lemmas.
\begin{lem}
\label{Lemma-open-subset-invariance}
For any alternating form $\Omega$ and any twistor sphere $S$, the intersection $S\cap Compl_\Omega$ is either finite or all of $S$.
\end{lem}

\begin{lem}
\label{Lemma-generic-period}
For any $J$ anti-commuting with $I$ and any nonzero alternating form $\Omega$ on $V_\RR$ there is a neighborhood $U_\Omega \subset C_{I}$ of
$J$ such that the locus $Compl_\Omega$ intersects $U_\Omega$
along a submanifold of positive codimension.
\end{lem}

Let us assume Lemmas \ref{Lemma-open-subset-invariance} and \ref{Lemma-generic-period} for a moment and prove Lemma \ref{Baire-lemma}.

\begin{proof}[Proof of Lemma \ref{Baire-lemma}]
Assume that some  $Compl_\Omega$
in $\mathcal L_{NS}$
contains an open neighborhood $U \subset C_I$ of a point $I_1 \in C_I$. This implies that,
for every twistor sphere $S=S(I,\cdot) \subset C_I$ intersecting $U$, the intersection $S\cap U$
contains a non-empty open open subset of $S$. Hence,
by Lemma \ref{Lemma-open-subset-invariance}, all of the periods of such $S$ are contained in
$Compl_\Omega$, including the ones anticommuting with $I$. This holds for all spheres $S$ intersecting $U$, that is, for all $g \in U_{e} \subset G_I$, where $U_e$ is a neighborhood of $e$ in $G_I$, we have
that $S(I,{}^gJ))$ is contained in $Compl_\Omega$. Thus
we have a contradiction
with the existence of $U_\Omega$ as in Lemma \ref{Lemma-generic-period}.
\end{proof}

\begin{proof}[Proof of Lemma \ref{Lemma-open-subset-invariance}]
Follows from the fact that $Compl_\Omega$ and $S$ (Corollary \ref{corSanalytic}) are complex analytic subsets of $Compl$.
\end{proof}

\begin{proof}[Proof of Lemma \ref{Lemma-generic-period}]
If $J \notin Compl_\Omega$ there is nothing to prove. Assume $J \in Compl_\Omega$.

As in the proof of Proposition \ref{transitivity}, choose a basis of $V_\RR$ in which the matrices of $J$ and $I$ are bloc diagonal with $4\times 4$ blocs
\[
\left(\begin{array}{cc|cc}
0 & -1 &   0 & 0\\
1 & 0 &   0 & 0 \\
\hline
 0 & 0 & 0 & -1 \\
0 & 0 & 1 & 0\end{array}\right),
\left(\begin{array}{cc|cc}
0 & 0 &   -1 & 0\\
0 & 0 &   0 & 1 \\
\hline
 1 & 0 & 0 & 0 \\
0 & -1 & 0 & 0\end{array}\right)
\]
respectively. Such a basis is the union of sets of vectors of the form $\{ v, Jv, Iv, JI v \}$. We denote the matrices of $I, J$ and $\Omega$ in this basis by the same letters.

Consider the orbit of $J$ under the conjugation action of  $G_I$: $G_I \cdot J \cong G_I/G_\HH$.
Let
\[
\begin{array}{cccc}
\Psi \colon & G_I & \longrightarrow & Compl,\\
& g & \longmapsto & ^g\!J=gJg^{-1},
\end{array}
\]
be the evaluation map of the action.
Put $G_\Omega :=\Psi^{-1}(Compl_\Omega)$, that is, $$G_\Omega=\{g \in G_I\,|\, ^t(^g\!J)\Omega
({}^g\!J) =\Omega\},$$
(note that $G_\Omega$ need not be a subgroup in $G_I$).
Let $g(\tau)$ be any curve in $G_\Omega$ with tangent
vector $X := g^\prime(0) \in T_eG_\Omega$ at $e = g(0) \in G_\Omega$. Then, differentiating the constant function $^t(^{g(\tau)}\!J)\Omega (^{g(\tau)}\!J)$
at $\tau=0$ we obtain
\[
-\, ^t\! X \, ^t\! J\Omega J+\, ^t\! J\, ^t\! X\Omega J +\, ^t\! J \Omega XJ -\, ^t\!J \Omega J X = 0.
\]
The left hand side may be simplified, given that $^t\! J\Omega J =\Omega$ and $^t\! J=J^{-1}=-J$, to
\[
\begin{array}{l}
-\,^t\! X \Omega +\, ^t\! J\, ^t\! X J\, ^t\! J \Omega J +\, ^t\! J \Omega J\, ^t\! J X J -\Omega X=\\
=-\, ^t\! X \Omega +\, ^t\! J\, ^t\! X J\Omega + \Omega\, ^t\! J X J -\Omega X\\
=\, ^t\! (X^J-X) \Omega + \Omega (X^J-X),
\end{array}
\]
where
\[
X^J:=J^{-1}XJ=JXJ^{-1}.
\]
So, denoting $Y:=X^J-X$, we have the equality
\begin{equation}
\label{Equation-locus}
^tY \Omega + \Omega Y=0,
\end{equation}
where $Y$ commutes with $I$ and anticommutes with $J$. Note that for any $X$, $X = \frac{1}{2} (X + X^J) + \frac{1}{2} (X - X^J)$, and $T_e G_\HH$ is the subspace of elements of $T_e G_I$ that commute with $I$. Hence, the subspace of $Y$'s
in $T_eG_I$ anticommuting with $J$ maps isomorphically onto the quotient space $V_I:= T_e{G_I}/T_e{G_\HH}$
under the quotient map $T_e{G_I} \rightarrow V_I$. So we need to check
that for a nonzero $\Omega$ the space of solutions to (\ref{Equation-locus}) has dimension
strictly less than $\dim_\RR V_I=4n^2$ (i.e., not all of the orbit $G_I\cdot J$ lies in $\mathcal L_{NS}$).

Now conjugate equation (\ref{Equation-locus}) by $I$ to obtain
$$^tY \Omega^I + \Omega^I Y=0.$$
Adding and subtracting this from (\ref{Equation-locus}) we obtain
$$^tY (\Omega +\Omega^I) + (\Omega+\Omega^I) Y=0\quad \hbox{and} \quad ^tY (\Omega -\Omega^I) + (\Omega-\Omega^I) Y=0.$$
So we may assume that $\Omega$ is either $I$-invariant
or $I$-anti-invariant in equation (\ref{Equation-locus}).

\medskip
\noindent {\it Case of $I$-invariant $\Omega$.} As $\Omega$ is $J$-invariant, it determines
a skew-symmetric operator $\Omega \colon V_\RR \rightarrow V_\RR$, commuting with $J$.
So we may choose
an $\Omega$-invariant plane $P=\langle v, Jv\rangle \subset V_\RR$ corresponding to a complex eigenvector
$v-iJv$ of
$\Omega \colon V_\RR \rightarrow V_\RR$ such that the matrix of $\Omega|_P$ is
\[\left (\begin{array}{cc}
0& -\lambda \\
\lambda & 0
\end{array}\right).\]
The complex structure $I$ provides another such plane $IP=\langle Iv, JIv \rangle$, which is also $\Omega$-invariant and orthogonal to $P$, so that on $P\oplus IP=\langle v,Jv,Iv,JIv \rangle$ the matrices of
$\Omega, J$ and $I$ are 4$\times$4-block-diagonal with the following blocks on the diagonal
\[\left(\begin{array}{cc|cc}
0 & -\lambda &   0 & 0\\
\lambda & 0 &   0 & 0 \\
\hline
 0 & 0 & 0 & \lambda \\
0 & 0 & -\lambda & 0\end{array}\right),
\left(\begin{array}{cc|cc}
0 & -1 &   0 & 0\\
1 & 0 &   0 & 0 \\
\hline
 0 & 0 & 0 & -1 \\
0 & 0 & 1 & 0\end{array}\right),
\left(\begin{array}{cc|cc}
0 & 0 &   -1 & 0\\
0 & 0 &   0 & 1 \\
\hline
 1 & 0 & 0 & 0 \\
0 & -1 & 0 & 0\end{array}\right)
.\]
The condition that $Y$ commutes with $I$ and anticommutes with $J$
tells us that $Y$ has a 4$\times$4-block structure with blocks of the form
\[\left(\begin{array}{cc|cc}
a_1 & a_2 &   b_1 & b_2\\
a_2 & -a_1 &   b_2 & -b_1 \\
\hline
 -b_1 & b_2 & a_1 & -a_2 \\
b_2 & b_1 & -a_2 & -a_1\end{array}\right)
.\]

Noting that $\Omega = JD=DJ$ for a diagonal matrix $D$ commuting with $J$, we can rewrite  (\ref{Equation-locus}) as
\begin{equation}
\label{Equation-locus-short-form}
DY=\;  ^tYD, \; Y^I=Y, \; Y^J=-Y.
\end{equation}
For notational convenience we write the matrix $Y$ in terms of its 2$\times$2-blocks $Y_{k,l}$,
$Y=(Y_{k,l})$, $1\leqslant k,l \leqslant 2n$, and denote by $\mathbb{1}_{2\times 2}$ the 2$\times$2 identity
matrix.
If at least one $\lambda_i$, $1\leqslant i\leqslant n$, in $D$ is nonzero we get for all $1\leqslant j \leqslant n$ the equalities of 4$\times$4-blocks
\[
\begin{array}{l}
\left(\begin{array}{cc}
\lambda_i  {\mathbb 1_{2 \times 2}}& 0\\
0 & -\lambda_i {\mathbb 1_{2 \times 2}}    \\
\end{array}\right) \cdot
\left(\begin{array}{cc}
Y_{2i-1,2j-1} & Y_{2i-1,2j} \\
 Y_{2i,2j-1} & Y_{2i,2j}\\
\end{array}\right)=
\\
=
\left(\begin{array}{cc}
^tY_{2j-1,2i-1} & ^tY_{2j,2i-1} \\
 ^tY_{2j-1,2i} & ^tY_{2j,2i}\\
\end{array}\right) \cdot
\left(\begin{array}{cc}
\lambda_j  {\mathbb 1_{2 \times 2}}& 0\\
0 & -\lambda_j {\mathbb 1_{2 \times 2}}    \\
\end{array}\right).
\end{array}
\]
These matrix equalities
completely determine all $n-1$ off-diagonal 4$\times$4-entries of $Y$ in the $i$-th ``fat'' row of
4$\times$4-blocks
in terms of the off-diagonal 4$\times$4-entries of the $i$-th ``fat'' column,
$1\leqslant i \leqslant n$. So the codimension of the space of solutions of
(\ref{Equation-locus-short-form}) is at least
$4(n-1)$ (precise lower bound that is reached in the least restrictive case $\lambda_j=\lambda_i$ for all $j$). For the diagonal 4$\times$4-entry, $i=j$, we obtain $b_2=0$ in $Y_{2i-1,2i}$, so that the codimension
is at least $4n-3$.

\medskip
\noindent {\it Case of $I$-anti-invariant $\Omega$.}
The only difference with the previous case is that in the basis $\langle v,Jv, Iv, JIv \rangle$
as above the matrix of $\Omega$ is
\[\left(\begin{array}{cc|cc}
0 & -\lambda &   0 & 0\\
\lambda & 0 &   0 & 0 \\
\hline
 0 & 0 & 0 & -\lambda \\
0 & 0 & \lambda & 0\end{array}\right).\]
So equation (\ref{Equation-locus-short-form}), written block-wise, gives that in addition to the
$4(n-1)$ conditions for the off-diagonal entries, for the diagonal 4$\times$4-entry we have $b_1=0$, which still
results in the lower bound $4(n-1)+1=4n-3$ for the codimension of the space of solutions of (\ref{Equation-locus-short-form}).

Now, by Lemma \ref{Lemma-open-subset-invariance}, either a twistor sphere in $C_I$ entirely lies in
some $Compl_\Omega$ or
its intersection with $\mathcal L_{NS}$ contains only finitely many points of each $Compl_\Omega$.
If $I\not\in \mathcal L_{NS}$ then no twistor sphere in $C_I$ is contained in
$\mathcal L_{NS}$. The codimension estimate above then allows us to conclude that, for every nonzero $\Omega$, the subset $C_I \cap Compl_\Omega$ is of codimension
at least $(4n-3)+2=4n-1>0$ in $C_I$. If $I\in \mathcal L_{NS}$, the lower bound for the codimension is still at least
$4n-3>0$.
The proof is now complete.
\end{proof}

\subsection{} \label{subsecJ12K12}

The transversality of the triple intersection of $G_{I_1}/G_\HH, G_{J_1}/G_\HH, G_{K_1}/G_\HH$
at $eG_\HH$, which is equivalent
to the direct sum decomposition $T_eG/T_eG_\HH=T_eG_{I_1}/T_eG_\HH \oplus T_eG_{J_1}/T_eG_\HH
\oplus T_eG_{K_1}/T_eG_\HH$, is preserved if we perturb $(I_1,J_1,K_1)\in \cT$ a little. In other words, there is a compact neighborhood
$U_{I_1,J_1,K_1}\subset \mathcal T$ of $(I_1,J_1,K_1)$ and a compact neighborhood
$U_{e,G}\subset G$ such that $\Phi_{I,J,K}\colon U_{e,J} \times U_{e,K} \rightarrow Compl$
is a submersion onto its image for all $(I,J,K)\in U_{I_1,J_1,K_1}$. Moreover,
 there is a compact neighborhood $U_{I_1}\subset Compl$ of $I_1$ which is contained in the image
$\Phi_{I,J,K}(U_{e,J}\times U_{e,K})$ for all $(I,J,K)\in U_{I_1,J_1,K_1}$. We will always assume
that for each neighborhood $U_{e,G}$ we made a choice of such $U_{I_1}=U_{I_1}(U_{e,G})$.
Note that every $I_2\in U_{I_1}$ is a regular value of $\Phi_{I,J,K}$ for all
$(I,J,K)\in  U_{I_1,J_1,K_1}$.

\begin{lem}
\label{Fiber-lemma}
There exists a  neighborhood $U_{e,G}$ such that
for all $I_2 \in U_{I_1}$ and for all $(I,J,K)\in U_{I_1,J_1,K_1}$,
the full preimage $\Phi_{I,J,K}^{-1}(I_2)$
is an $8n^2$-dimensional submanifold in $U_{e,J} \times U_{e,K}$
of the form
\begin{equation}
\label{Fiber-equation}
\{(f_1h_1, h_1^{-1}f_2h_2)\mid h_1,h_2 \in G_\HH\} \cap (U_{e,J} \times U_{e,K}),
\end{equation}
where $(f_1,f_2)$ is a pair in $U_{e,J} \times U_{e,K}$ such that $\Phi_{I,J,K}(f_1,f_2)=I_2$.
\end{lem}

\begin{proof}[Proof of Lemma \ref{Fiber-lemma}] The fact that
$\Phi_{I,J,K}^{-1}(I_2) \cap(U_{e,J} \times U_{e,K})$ consists of
a finite number of $8n^2$-dimensional manifolds follows from the regularity of $I_2$.

While the part of the fiber in (\ref{Fiber-equation}) may have been easily guessed, the fact that for a small enough $U_{e,G}$ this is the whole fiber
follows from Proposition \ref{Proposition-general-transversality}.
Indeed, assuming that we have $(f_1,f_2), (g_1,g_2) \in
\Phi^{-1}_{I,J,K}(I_2)\subset G_{J}\times G_{K}$, we see that $f_2^{-1}f_1^{-1}g_1g_2 \in G_{I}$.
Setting $g_{I}=f_2^{-1}f_1^{-1}g_1g_2$ and $g_{J}=f_1^{-1}g_1 \in G_{J}$ we
have the equality
$$f_2g_{I}=g_{J}g_2.$$
The left side of the equality lies in $G_KG_{I}$ and the right side lies in $G_JG_K$.
If we restrict
ourselves to $\Phi^{-1}_{I,J,K}(I_2) \cap (U_{e,G}\times U_{e,G})$ for a small enough neighborhood $U_{e,G} \subset G$ then Proposition  \ref{Proposition-general-transversality} tells us that,
for every element  in the product $U_{e,J}U_{e,K}U_{e,I}$,
each of its three factors is uniquely determined up to a $G_\HH$-correction.

So from our equality $f_2g_{I}=g_{J}g_2$ we obtain
$g_{I}, g_{J} \in G_\HH$, which, after setting
$h_1 := g_J=f_1^{-1}g_1$ and $h_2 :=g_I$, implies
that $g_1=f_1h_1$ and $g_2=g_J^{-1}f_2g_I=h_1^{-1}f_2h_2$.

Since $U_{I_1},U_{I_1,J_1,K_1},U_{e,G}$ are compact and $U_{e,G}$ is independent of the choice of $(I,J,K)\in U_{I_1,J_1,K_1}$,
there is a universal upper bound for the number of connected components
of $\Phi_{I,J,K}^{-1}(I_2)\cap (U_{e,J} \times U_{e,K})$, for all
$I_2\in U_{I_1}$ and all $(I,J,K)\in U_{I_1,J_1,K_1}$. Therefore we can shrink the compact neighborhood
$U_{e,G}$ so that the fibers $\Phi_{I,J,K}^{-1}(I_2) \cap (U_{e,J} \times U_{e,K})$ for all
$I_2\in U_{I_1}$ and all $(I,J,K)\in U_{I_1,J_1,K_1}$ contain only the component specified in (\ref{Fiber-equation}).
\end{proof}
Regarding the proof of Lemma \ref{Fiber-lemma}, we note the following.
\begin{rem}
\label{Fiber-deformation-remark}
It is not hard to see that the  fiber $\Phi_{I,J,K}^{-1}(I_2)$ in Lemma \ref{Fiber-lemma},
as a topological subspace of $G \times G$, depends continuously on
$I_2\in U_{I_1}$ and  $(I,J,K)\in U_{I_1,J_1,K_1}$.
\end{rem}

\begin{rem}
In general,  it is possible that $g \in U_{e,G}$ is not uniquely representable as a triple product
of elements  in the larger sets $G_{J}$, $G_{K}$, $G_{I}$
and thus we cannot say if the whole fiber $\Phi^{-1}_{I,J,K}(I_2) \subset G_{J}\times G_{K}$ consists of just one
$G_\HH \times G_\HH$-orbit as in Lemma \ref{Fiber-lemma}. This is why we possibly need to shrink $U_{e,G}$.
\end{rem}

\subsection{}\label{subsecPsi12} Recall that, for any $I$, $M_I=G_{I}/G_{I,S}$ parametrizes the twistor lines through $I$ (see Paragraph \ref{subsecSMN}). For all $I$, put $U_{I_1,J_1,K_1}(I) :=pr_{23}(pr_1^{-1}(I)\cap U_{I_1,J_1,K_1})$.
Then $U_{I_1,J_1,K_1} (I_1)$ is a neighborhood of
$(J_1,K_1)$ in $C_{I_1}\times_{M_{I_1}} C_{I_1} = pr_{23} (pr_1^{-1} (I_1) \cap \cT)$.
Consider the map
\[
\begin{array}{cccl}
\Psi^{I_1 \rightarrow I_2} \colon & U_{I_1,J_1,K_1} (I_1) &
\longrightarrow & {C_{I_2}}
\times_{M_{I_2}} C_{I_2} = pr_{23} (pr_1^{-1} (I_2) \cap \cT),\\
& (S(J,K),J,K) & \longmapsto &(S({}^{f_1f_2}J, {}^{f_1f_2}K), {}^{f_1f_2}K, {}^{f_1f_2}J),
\end{array}
\]
where $ (f_1,f_2) \in \Phi^{-1}_{I_1,J,K}(I_2) \cap ({U_{e,J} \times U_{e,K}})$,
and we use, in an obvious way, the triple notation of the kind $(S(J,K),J,K)$ for the elements of the
fiber products above. Note the switched order of ${}^{f_1f_2}K, {}^{f_1f_2}J$.
The role of this change of order will be clarified later.

Lemma \ref{Fiber-lemma} guarantees that the mapping $\Psi^{I_1 \rightarrow I_2}$
is well-defined, as its value at $(S(J,K),J,K)$ is uniquely determined by the fiber
$\Phi^{-1}_{I_1,J,K}(I_2) \cap ({U_{e,J} \times U_{e,K}})$,
so it does not depend on the choice of a particular point in the fiber.

\vspace*{1cm}
\hspace*{40mm}
\input{Picture4-2.pic}

\vspace*{-11.5cm}
\begin{center}
Picture 2: For fixed $I_1$ and $I_2$ any pair $(J,K) \in C_{I_1}\times_{M_{I_1}} C_{I_1}$ near $(J_1,K_1)$ determines a unique pair
$({}^{f_1f_2}K,\,^{f_1f_2}\!J)\in C_{I_2}\times_{M_{I_2}} C_{I_2}$.
\end{center}
\medskip
\subsection{}
\label{Inverse-preparation-subsection}
Next, for each $(I,J,K) \in U_{I_1,J_1,K_1}$  consider the mapping
$\Phi_{I,K,J}$ (note that  we switched $J$ and $K$ in the subscript).
By shrinking the original $ U_{I_1,J_1,K_1}$ and $U_{e,G}$ if needed,
we can find a compact neighborhood $V_{e,G} \subset G$ such that
 for each $(I,J,K) \in U_{I_1,J_1,K_1}$ we have

(a) $\Phi_{I,K,J}\colon V_{e,K} \times V_{e,J} \rightarrow Compl$ is a submersion
onto its image;

(b) every fiber of this mapping is of the form described in Lemma \ref{Fiber-lemma};

and

(c) the image $\Phi_{I,K,J}( V_{e,K} \times V_{e,J})$ contains
$U_{I_1} \subset  \underset{(I,J,K)\in U_{I_1,J_1,K_1}}{\bigcap}\Phi_{I,J,K}(U_{e,J}\times U_{e,K})$ (see Paragraph \ref{subsecJ12K12}).
\vskip5pt
By Lemma \ref{Fiber-lemma}, conditions (a) and (b) are satisfied. We need only to comment on (c). By Lemma \ref{Fiber-lemma}, for the original triple $(I_1,J_1,K_1) \in U_{I_1,J_1,K_1}$, we can find $V_{e,G}$ such that
$\Phi_{I_1,K_1,J_1}\colon  V_{e,K_1} \times V_{e,J_1} \rightarrow Compl$, where
$V_{e,K_1}:=V_{e,G} \cap G_{K_1},V_{e,J_1}:=V_{e,G} \cap G_{J_1}$, satisfies (a) and (b). Shrinking $U_{e,G}$
and, thus, $U_{I_1}$, if needed,
we can satisfy (c) for $\Phi_{I_1,K_1,J_1}$. Now shrinking $U_{I_1,J_1,K_1}$ and again $U_{e,G}$, if needed, we can satisfy conditions (a), (b) and (c)
for all $(I,J,K)\in U_{I_1,J_1,K_1}$.

\subsection{}
\label{Domain-preparation-subsection}
Now introduce $V_{I_1,K_1,J_1}:=\{(I,K,J)\,|\, (I,J,K)\in U_{I_1,J_1,K_1}\}$
and $V_{I_1,K_1,J_1}(I):=pr_{23}(pr_1^{-1}(I) \cap V_{I_1,K_1,J_1})$.

Then, for all $(I,K,J)$ in the interior of $V_{I_1,K_1,J_1}$, the set
$pr_1(V_{I_1,K_1,J_1})$ is a neighborhood of $I$ in $Compl$ and $V_{I_1,K_1,J_1}(I)$ is a neighborhood of $(K,J)\in C_I\times_{M_I} C_I$.
Note that due to Condition (c) in Paragraph \ref{Inverse-preparation-subsection},
for all $I \in U_{I_1}\cap pr_1(V_{I_1,K_1,J_1})$ and for all $(K,J)\in V_{I_1,K_1,J_1}(I)$,
the image $\Phi_{I,K,J}(U_{e,K}\times U_{e,J})$ contains the neighborhood $U_{I_1}$.

\subsection{}\label{subsecPsi21} Choose $I_2 \in U_{I_1}\cap pr_1(V_{I_1,K_1,J_1})$ and $K,J$ such that $(I_2,K,J) \in V_{I_1,K_1,J_1}$. Conditions (a),(b) and (c) in Paragraph \ref{Inverse-preparation-subsection} allow us to define, analogously to $\Psi^{I_1 \rightarrow I_2}$, the map
\[
\begin{array}{cccc}
\Psi^{I_2 \rightarrow I_1} \colon &
V_{I_1,K_1,J_1}(I_2) & \longrightarrow & {C_{I_1}}\times_{M_{I_1}} C_{I_1},\\
& (S(J,K),K,J) & \longmapsto & (S({}^{d_1d_2}J, {}^{d_1d_2}K),{}^{d_1d_2}J,{}^{d_1d_2}K),
\end{array}
\]
for $(d_1,d_2) \in \Phi^{-1}_{I_2,K,J}(I_1)$ (again, note the reversed order of $J$ and $K$ in the subscript).

The period $^{f_1f_2}K$ in Picture 2 above will play the role of the ``rotation center''
for  $\Phi_{I_2,\,{}^{f_1f_2}K,\,{}^{f_1f_2}J}$ (here $J,K \in C_{I_1}$),
similar to the role that $J$ plays  for $\Phi_{I_1,J,K}$.
This explains why we switched $J$ and $K$.

Below we will impose restrictions on the domain of $\Psi^{I_1 \rightarrow I_2}$
in order for the image of this map to be contained in the domain of $\Psi^{I_2 \rightarrow I_1}$,
so that we can compose them.

We begin by choosing a compact neighborhood
$U_{J_1,K_1}$ of $(J_1,K_1)$ in $U_{I_1,J_1,K_1} (I_1)$, which can at first be all of $U_{I_1,J_1,K_1} (I_1)$.
We will later modify $U_{J_1,K_1}$, without changing the original $U_{I_1,J_1,K_1}$.

\begin{lem}\label{lemPsidomains}
For fixed $V_{I_1,K_1,J_1}$, we can shrink $U_{e,G}$ and

 $U_{J_1,K_1}$ so that for arbitrary $I_2 \in U_{I_1}$,
\[
\Psi^{I_1 \rightarrow I_2}(U_{J_1,K_1}) \subset V_{I_1,K_1,J_1}(I_2).
\]
\end{lem}

\begin{proof}
As in Paragraph \ref{Inverse-preparation-subsection}, this
follows from the fact that the mapping $\Psi^{I_2\rightarrow I_1}$
depends continuously on  $I_2$ (see Remark \ref{Fiber-deformation-remark}), and that
 $$\underset{I_2\to I_1}{\lim}\Psi^{I_2\rightarrow I_1}=(12)\colon U_{J_1,K_1} \rightarrow
V_{I_1,K_1,J_1}(I_1),$$ $$(S(J,K),J,K)\mapsto (S(J,K),K,J),$$ the latter mapping is trivially defined on the whole
$U_{J_1,K_1}$,

so that the sizes of the domains $V_{I_1,K_1,J_1}(I_2)$ of $\Psi^{I_2\rightarrow I_1}$'s are bounded away from zero, when $I_2$ is close to $I_1$.

As before, we can further shrink $U_{e,G}$ (and hence $U_{I_1}$), if needed, so that
properties (a), (b), (c) in Paragraph \ref{Inverse-preparation-subsection} hold independently of the point $I_2\in U_{I_1}$.
\end{proof}

\subsection{} Possibly shrinking $U_{e,G}$, we can and will assume that it is invariant under taking inverses,
$g\mapsto g^{-1}$.

\begin{lem}\label{lemPsicompId}

Possibly further shrinking $U_{e,G}$  and $U_{J_1,K_1}$, satisfying the conclusion of Lemma \ref{lemPsidomains}, we have for all $I_2\in U_{I_1}$

\[
\Psi^{I_2\rightarrow I_1} \circ \Psi^{I_1 \rightarrow I_2} =
Id|_{U_{J_1,K_1}}.
\]
\label{lemma-composition}
\end{lem}
\begin{proof}
For all $(f_1,f_2) \in \Phi^{-1}_{I_1,J,K}(I_2)\cap (U_{e,G} \times U_{e,G})$ and all $(S(J,K),J,K) \in
U_{J_1,K_1}
$, we want the neighborhoods
$V_{e, \,{}^{f_1f_2}K}=V_{e,G} \cap G_{{}^{f_1f_2}K},
V_{e,\,{}^{f_1f_2}J}=V_{e,G} \cap G_{{}^{f_1f_2}J}$ to contain, respectively, the neighborhoods
${}^{f_1f_2}U_{e,K}=f_1f_2U_{e,K}f_2^{-1}f_1^{-1}$ and ${}^{f_1f_2}U_{e,J}=f_1f_2U_{e,J}f_2^{-1}f_1^{-1}$,
so that, in particular, $V_{e,\, {}^{f_1f_2}K} \times V_{e,\,{}^{f_1f_2}J}$
contains the pair
\[
(d_1,d_2)=(f_1f_2\cdot f_2^{-1} \cdot f_2^{-1}f_1^{-1}, f_1f_2 \cdot f_1^{-1} \cdot f_2^{-1}f_1^{-1})= (f_1f_2^{-1}f_1^{-1}, f_1f_2 \cdot f_1^{-1} \cdot f_2^{-1}f_1^{-1}).\]
Here the invariance of $U_{e,G}$ under taking inverses is used. The pair $(d_1, d_2)$
certainly belongs to the preimage $\Phi^{-1}_{I_2,\, {}^{f_1f_2}K,\,{}^{f_1f_2}J}(I_1)$ as the product
of its entries is $f_1f_2^{-1}f_1^{-1} \cdot f_1f_2 \cdot f_1^{-1} \cdot f_2^{-1}f_1^{-1}=f_2^{-1}f_1^{-1}$.

Note that,
for $U_{e,G}$ small enough, the neighborhoods ${}^{f_1f_2}U_{e,K} \times {}^{f_1f_2}U_{e,J}$
will also be uniformly small for all
$(f_1,f_2) \in \Phi_{I_1,J,K}^{-1}(I_2) \cap (U_{e,G} \times U_{e,G})$, so that
the fiber of
$\Phi_{I_2,\,{}^{f_1f_2}K,\,{}^{f_1f_2}J}^{-1}(I_1)$ in ${}^{f_1f_2}U_{e,K} \times {}^{f_1f_2}U_{e,J}$
consists of a unique connected component
of the form described in Lemma \ref{Fiber-lemma}. Then the pair $(d_1,d_2)$
is contained in this ``good'' part of the fiber $\Phi_{I_2,\,{}^{f_1f_2}K,\,{}^{f_1f_2}J}^{-1}(I_1)$ and
we can use $(d_1,d_2)$
to evaluate $\Psi^{I_2 \mapsto I_1}$ at $(S({}^{f_1f_2}J,{}^{f_1f_2}K),
{}^{f_1f_2}J, {}^{f_1f_2}K)$.
Thus $$\Phi_{I_2,\,{}^{f_1f_2}K,\,{}^{f_1f_2}J}(d_1,d_2)=I_1$$ and $${}^{d_1d_2f_1f_2}J=J,
{}^{d_1d_2f_1f_2}K=K,$$
so that $$\Psi^{I_2 \mapsto I_1}(S({}^{f_1f_2}J,{}^{f_1f_2}K),{}^{f_1f_2}K, {}^{f_1f_2}J)=(S(J,K),J,K),$$
where, certainly, $S(J,K)=S(I_1,J,K)$,
proving that the composition $\Psi^{I_2\rightarrow I_1} \circ \Psi^{I_1 \rightarrow I_2}$ is the identity
on $U_{J_1,K_1}$.

In order to ensure that $V_{e,\, {}^{f_1f_2}K} \times V_{e,\,{}^{f_1f_2}J}$ contains $(d_1,d_2)$,
we assume, shrinking $U_{e,G}$ and $U_{J_1,K_1}$
if necessary, but not changing $V_{e,G}$ and the previously fixed $V_{I_1,K_1,J_1}$, that
for all $(S(J,K),J,K) \in U_{J_1,K_1}

$ and for all points $(f_1,f_2) \in
\Phi^{-1}_{I_1,J,K}(I_2)\cap (U_{e,G} \times U_{e,G})$, the neighborhoods
$V_{e,\, {}^{f_1f_2}K}, V_{e,\,{}^{f_1f_2}J}$ contain, respectively, the neighborhoods
${}^{f_1f_2}U_{e,K}$ and ${}^{f_1f_2}U_{e,J}$.
\end{proof}

\begin{cor}\label{corfinal} Let $U_{I_1}$ be defined by $U_{e,G}$ ($U_{e,G}$ satisfying Lemma \ref{lemma-composition}).
For arbitrary $I_2 \in U_{I_1}$, both joint points $J \in C_{I_1}$ and  ${}^{f_1f_2}K\in C_{I_2}$ of a triple of twistor spheres connecting $I_1$ and $I_2$, can be chosen generic.
\end{cor}
\begin{proof}
Define
\[
\begin{array}{cccc}
pr_K \colon & V_{I_1,K_1,J_1}(I_2)
& \longrightarrow & C_{I_2} \subset Compl, \\
& (S(J,K),K,J)&\longmapsto &K.
\end{array}
\]
This projection is a submersion onto its image.
By Lemma  \ref{Baire-lemma}, the locus $\mathcal L_{NS}$ intersects $C_{I_2}$
in a countable union of closed submanifolds of positive codimension in $C_{I_2}$.
As the mapping $pr_K$
is a submersion onto its image, the preimage $pr_K^{-1}(\mathcal L_{NS}\cap C_{I_2})$ is also a countable union of
closed submanifolds of positive codimension in $V_{I_1,K_1,J_1}(I_2)$.

Similarly, for
\[
\begin{array}{cccc}
pr_J \colon &

U_{J_1,K_1} & \longrightarrow & C_{I_1}\subset Compl, \\
& (S(J,K),J,K) & \longmapsto & J,
\end{array}
\]
$pr_J^{-1}(\mathcal L_{NS}\cap C_{I_1}) \subset

U_{J_1,K_1}$
is a countable union of closed submanifolds of positive codimension.
The mapping $\Psi^{I_2\to I_1}$ is real-analytic, so the closure of
$\Psi^{I_2\to I_1} (pr_J^{-1}(\mathcal L_{NS}\cap C_{I_2}))$ in $U_{J_1,K_1}$
does not contain interior points.
Therefore
\begin{equation}
\label{inequality-irreducibility}

U_{J_1,K_1}
\neq pr_J^{-1}(\mathcal L_{NS}\cap C_{I_1}) \cup \Psi^{I_2 \rightarrow I_1}(pr_K^{-1}
(\mathcal L_{NS}\cap C_{I_2})).
\end{equation}
Since, by Lemma \ref{lemma-composition}, $\Psi^{I_2\to I_1}\circ \Psi^{I_1\to I_2}=Id|_{U_{J_1,K_1}}$,
the inequality (\ref{inequality-irreducibility}) tells us that
the image of the mapping $\Psi^{I_1\to I_2}$ is not contained in $pr_K^{-1}(\mathcal L_{NS}\cap C_{I_2})$.
Thus we may find
a pair $(J,K)\in

U_{J_1,K_1}$ such that $J=pr_J(S(J,K),J,K)\notin \mathcal L_{NS}\cap C_{I_1}$ and
${}^{f_1f_2}K=pr_K(\Psi^{I_1 \rightarrow I_2}(S(J,K),J,K)) \notin \mathcal L_{NS}\cap C_{I_2}$,
that is, both periods are generic.
\end{proof}

\section{The degree of twistor lines}
\label{Twistor-spheres-are-complex-submanifolds}

In this section we show that, as in the case of K3 surfaces, twistor lines in $Compl$ have degree $2$ in the Pl\"ucker embedding. We first show that the group $G=GL^+(V_\RR)$ acts transitively on the set of twistor lines in $Compl$ and then compute the degree of an explicit twistor line.

\begin{lem}\label{lemtrans}
The group $G=GL^+(V_\RR)$ acts transitively on the set of twistor lines in $Compl$.
\end{lem}

\begin{proof}
Given two twistor spheres $S_1 = S(I_1, J_1)$ and $S_2= S(I_2, J_2)$, there is an element $g\in G$ sending $I_1$ to $I_2$, hence sending $S_1$ to a twistor sphere through $I_2$. The lemma now follows from Corollary \ref{corMItrans}.
\end{proof}

\subsection{} To construct our example, consider the affine chart in the Grassmannian $G(2n,4n)$
of normalized period matrices $(\mathbb{1}| Z)$, where $\mathbb{1}$ is, in general, the $2n\times 2n$
identity matrix and $Z$ now denotes a non-degenerate $2n\times 2n$ complex matrix.
Let us fix a basis of $V_\RR$ and write the matrix of an arbitrary complex structure
$I\colon V_\RR \rightarrow V_\RR$ in the following block form
\[I=\left(\begin{array}{cc}
A & B \\
C & D
\end{array}\right),\]
for $2n\times2n$ real matrices $A,B,C,D$.
Then the relation
$$(\mathbb{1}\,|\,Z)I=(i\mathbb{1}\,|\,i Z),$$
gives the matrix equations
\[ A+Z C=i\mathbb{1}, \quad B+Z D=i Z.
\]
Assume that $C$ is invertible so that the first equation
allows us to write $Z =(i\mathbb{1}-A)C^{-1}$.
The condition that $I$ is a complex structure will then guarantee that the second
equation is automatically satisfied.
\medskip

\subsection{The case $n=1$}\label{subsecn=1} Momentarily assume $n=1$ and consider the twistor sphere $S = S(I,J)$ where $I$ and $J$ have the respective matrices
\[
\left(\begin{array}{cc|cc}
0 & -1 &   0 & 0\\
1 & 0 &   0 & 0 \\
\hline
 0 & 0 & 0 & -1 \\
0 & 0 & 1 & 0\end{array}\right),
\left(\begin{array}{cc|cc}
0 & 0 &   -1 & 0\\
0 & 0 &   0 & 1 \\
\hline
 1 & 0 & 0 & 0 \\
0 & -1 & 0 & 0\end{array}\right)
\]
and put $K = IJ$.
So for $\lambda \in S$,
\[
\lambda =aI+bJ+cK=
\left(\begin{array}{cc|cc}
0 & -a &   -b & -c\\
a & 0 &   -c & b \\
\hline
 b & c & 0 & -a \\
c & -b & a & 0
\end{array}\right).
\]
Assume additionally that $b^2+c^2 \neq 0$, that is, $\lambda \in S\setminus\{\pm I\}$.
Here
\[
A=
\left(\begin{array}{cc}
 0 & -a\\
 a & 0 \\
\end{array}\right),
C=
\left(\begin{array}{cc}
 b & c\\
 c & -b \\
\end{array}\right),
C^{-1}=
\frac{1}{b^2+c^2}
\left(\begin{array}{cc}
 b & c\\
 c & -b \\
\end{array}\right).
\]
Then
\[ Z = (i\mathbb{1}-A)C^{-1}=
\frac{1}{b^2+c^2}
\left(\begin{array}{cc}
ac+ib & -ab+ic\\
 -ab+ic & -ac-ib \\
\end{array}\right)=
\left(\begin{array}{cc}
 z_1 & z_2\\
 z_3 & z_4 \\
\end{array}\right),
\]
where $Z$ clearly satisfies the equations $$z_1+z_4=0,\,\,z_2-z_3=0,\,\,\det \,Z=z_1z_4-z_2z_3=1.$$

\subsection{} Now, for a general $n$, we can
construct a twistor line in the period domain of complex $2n$-dimensional tori,
which, in the affine chart of $G(2n,4n)$ above, corresponds to the
locus of matrices $(\mathbb{1}| Z)$ where $Z$ is the bloc diagonal matrix with the same $2\times 4$-block
\[\left(\begin{array}{cc|cc}
 1 & 0 &  u & v\\
 0 & 1 & v & -u \\
\end{array}\right), u^2+v^2=-1,
\]
on the diagonal.

\subsection{}\label{pardegree}
The degree of the curve in the example is $2$ in the Pl\"ucker embedding
$G(2n,4n) \hookrightarrow \mathbb{P}^{{4n \choose{2n}}-1}$.
Indeed, the Pl\"ucker coordinates in the above affine chart are given by the maximal minors of the matrix $(\mathbb{1}| Z)$. The twistor line $S$ in our example is contained in the plane $P(S)$ with parameters $u,v$. The coordinates of the restriction of the Pl\"ucker embedding are $0,1,\pm u, \pm v, u^2 + v^2$, or products of such, with the highest degree coordinate equal to $(u^2+v^2)^n$. The image of $P(S)$ by the Pl\"ucker embedding is therefore the intersection of the image of the Grassmannian with a linear space and has degree $2n$ in that linear space since its highest degree coordinate is $(u^2+v^2)^n$. It is now immediate that the image of $S$ is the intersection of the image of $P(S)$ with a linear space given by setting all the coordinates that are powers of $u^2+v^2$ equal to $1$ (and some of the other coordinates equal to each other). This shows that the image of $S$ by the Pl\"ucker embedding is contained in a plane and is a conic is that plane.

\begin{cor}
Twistor lines have degree $2$ in the Pl\"ucker embedding of $Compl$.
\end{cor}
\begin{proof}
Follows from Lemma \ref{lemtrans} and Paragraph \ref{pardegree}.
\end{proof}

\section{Appendix}
\label{Appendix}

Let $A$ be a 2-dimensional complex torus with period belonging to the
twistor line $S$ constructed in Paragraph \ref{subsecn=1}.

For any $\Omega \in Hom(\wedge^2\Gamma,\QQ)$
such that $S\cap Compl_\Omega$ is infinite, by Lemma
\ref{Lemma-open-subset-invariance}, the whole twistor line $S$ is contained in $Compl_\Omega$. Below we determine all $\Omega$ such that $S\subset Compl_\Omega$:
these are specified by the invariance conditions
$\Omega(I\cdot,I\cdot)=\Omega(J\cdot,J\cdot)=\Omega(\cdot,\cdot)$ and form a 3-dimensional subspace in $Hom(\wedge^2\Gamma,\QQ)$. The invariance conditions mean that the first Riemann bilinear relation is satisfied.

On the other hand, the second Riemann bilinear relation does not hold: these $\Omega$ determine $(1,1)$-classes
in the cohomology of tori in this twistor line whose hermitian forms are always indefinite.
Thus, none of the classes
determined by these $\Omega$ is K\"ahler.
For the formulation of the Riemann bilinear relations see, for example,
\cite[Ch. 2]{GrHarr}.

\subsection{The first bilinear relation} Let $Q$ be the matrix of the alternating form corresponding to an $I,J$-invariant cohomology class in $H^2(A,\QQ)$, written in the basis in which the matrices of $I, J$ are as in the previous section.
The $I,J$-invariance then translates into the commutation relations $QI=IQ$ and $QJ=JQ$. A general skew-symmetric such $Q$ has the form
\[Q=\left(
\begin{array}{cccc}
0 & -b & c & -d \\
b & 0 & d & c \\
-c & -d & 0 & b \\
d & -c & -b & 0\\
\end{array}\right), \quad b,c,d \in \QQ.
\]

Such $Q$, by definition, determines a rational class of Hodge type $(1,1)$ for all tori with periods in $S(I,J)$,
so that the first bilinear relation $\Omega {Q^{-1}} \, ^t\Omega=0$ is automatically guaranteed
by the choice of $Q$.

\bigskip

\subsection{The second bilinear relation $-i\Omega {Q^{-1}} \, ^t\overline{\Omega}>0$}
For $Q$ as above it is easy to find $Q^{-1}$. Indeed, note that $$Q^2=-(b^2+c^2+d^2)\mathbb{1}_{4\times 4},$$
where $\mathbb{1}_{4\times 4}$ is the $4\times4$ identity matrix, so that
$Q^{-1}=-\frac{1}{(b^2+c^2+d^2)}Q$ and  $-i\Omega {Q^{-1}} \, ^t\overline{\Omega}>0$ is equivalent to
$i\Omega Q \, ^t\overline{\Omega}>0$. We have

\[\Omega Q=
\left(\begin{array}{cccc}
-uc+vd & -b-ud-vc & c-vb & -d+ub\\
b-vc-ud & -vd+uc & d+ub & c+vb \\
\end{array}\right),
\]
so that $\Omega Q \, ^t\overline{\Omega}$ is equal to

\[
\left(\begin{array}{cc}
(\overline{u}-u)c+(v-\overline{v})d+(u\overline{v}-\overline{u}v)b &
(\overline{u}-u)d+(\overline{v}-v)c-(1+|u|^2+|v|^2)b \\
(\overline{u}-u)d +(\overline{v}-v)c+(1+|u|^2+|v|^2)b & (u-\overline{u})c+(\overline{v}-v)d+(u\overline{v}-\overline{u}v)b\\
\end{array}\right).
\]
Setting $u=u_1+iu_2$ and $v=v_1+iv_2$, we compute
\[
i\Omega Q \, ^t\overline{\Omega} = \left(\begin{array}{cc}
2(u_2c-v_2d)+2(u_1v_2-u_2v_1)b &
2(u_2d+v_2c)-i(1+|u|^2+|v|^2)b \\
2(u_2d +v_2c)+i(1+|u|^2+|v|^2)b & -2(u_2c-v_2d)+2(u_1v_2-u_2v_1)b\\
\end{array}\right).
\]
Now the determinant $\det\, i\Omega Q \, ^t\overline{\Omega}$ is
$$\det\, i\Omega Q \, ^t\overline{\Omega}=(4b^2(u_1v_2-u_2v_1)^2-4(u_2c-v_2d)^2)-
(b^2(1+|u|^2+|v|^2)^2+4(v_2c+u_2d)^2)=$$
$$=b^2(4(u_1v_2-u_2v_1)^2-(1+|u|^2+|v|^2)^2)-4(u_2c-v_2d)^2-4(v_2c+u_2d)^2.$$
Let us show that indeed
$4(u_1v_2-u_2v_1)^2-(1+|u|^2+|v|^2)^2 \leqslant 0$ for all $b,c,d \in \RR$
and $u,v \in \CC$ such that $u^2+v^2=-1$. This would prove that
$ i\Omega Q \, ^t\overline{\Omega}>0$ never holds for the periods in our twistor line.

The complex equation $u^2+v^2=-1$ is equivalent to the two real equations
$u^2_2+v^2_2=u^2_1+v^2_1+1$ and $u_1u_2+v_1v_2=0$.
Introducing the vectors

\[X=\left(\begin{array}{c}
u_1 \\
v_1 \\
\end{array}\right),
Y=\left(\begin{array}{c}
u_2 \\
v_2 \\
\end{array}\right).
\]
these equalities can be written as $|Y|^2=|X|^2+1$ and $X \perp Y$.
The term $(u_1v_2-u_2v_1)^2$ above is the square of the dot product of $X$ and
the result of rotation of $Y$
by $\frac{\pi}{2}$, so that  $X \perp Y$ implies that $(u_1v_2-u_2v_1)^2=|X|^2|Y|^2$.
Now the equality $|Y|^2=|X|^2+1$ allows us to write $1+|u|^2+|v|^2=|X|^2+|Y|^2$
and we have
$$4|X|^2|Y|^2-(|X|^2+|Y|^2)^2=-(|X|^2-|Y|^2)^2 =-1<0.$$
So, finally we obtain $\det\, i\Omega Q \, ^t\overline{\Omega}<0$ and none of the $Q$
above determines a K\"ahler class.


TITLE: Understanding inequalities and how not to blindly apply famous inequalities
QUESTION [5 upvotes]: If $x \in (0, \pi/2)$, prove that
$$ 2 (\sin x )^{(1-\sin x)} \cdot (1-\sin x)^{\sin x} \leq 1 $$

Attempt:
We use Youngs inequality: if $\alpha+\beta = 1$ and $x,y $ are positive, then $x^{\alpha} + y^{\beta} \leq \alpha x + \beta y $.
In our problem, both $\sin x$ and $1 - \sin x$ are positive thus,
$$ 2 (\sin x )^{(1-\sin x)} \cdot (1-\sin x)^{\sin x} \leq 4(1 - \sin x) \sin x   \;\;\;\;\;\;\; (*)$$
If I let $f(x) = 4 (1-\sin x) \sin x$ then $f'(x)=0$ implies that $x= \pi/6$ is a max and since $f(\pi/6) = 1/4$, we are done.
Is this a correct argument?
MY question is: We could have also use the AM-GM inequality in (*) to obtain that
$$ f(x) \leq 8 ( (1-\sin x)^2 + \sin^2 x ) $$
But this would us lead us nowhere since the above is bounded above by $40$. Why does the AM-GM inequality gives such large bound for $f(x)$? How do we know when to apply AM-GM inequality properly?

REPLY [2 votes]: Your first solution is correct. It becomes a bit simpler to read with the substitution $u = \sin(x) \in (0, 1)$:
$$
 2u^{1-u} (1-u)^u \le 4 u(1-u) \le 1 \, .
$$
The first estimate is Young's inequality (as you noticed), this is also a special case of the Weighted AM–GM inequality.
For the second estimate you determined the maximum by computing the derivative, but actually it is just the AM-GM inequality: $\sqrt{u(1-u)} \le \frac{u + (1-u)}{2}$.
Since equality holds for $u=1/2$ (corresponding to $x = \pi/6$) the inequality is sharp.

Your second approach has two errors: First, AM-GM in the form $2ab \le a^2 + b^2$ gives
$$
 4 u (1-u) \le 2 \left( (1-u)^2 + u^2 \right)
$$
with the factor $2$ on the right-hand side, not $8$. Second, you did not consider that $0 \le u = \sin(x) \le 1$. In this range we have
$$
 2 \left( (1-u)^2 + u^2 \right) = 4 \left( u - \frac 12 \right)^2 + 1 \le 2
$$
with equality for $u=0$ and $u=1$. This gives the  estimate
$$
 4 u (1-u) \le 2 \left( (1-u)^2 + u^2 \right) \le 2
$$
which is better than your result but not sharp, because equality cannot hold simultaneously in both estimates.
Generally, when applying multiple inequalities, it is good to keep track of when equality holds. The end result is only sharp when equality can hold in all estimates simultaneously.


TITLE: Height unmixed ideal
QUESTION [0 upvotes]: Suppose $R$ is a regular local ring and $I$ is a non-zero ideal such that $I$ is a radical ideal and $I$ is height unmixed. Suppose $J$ is any radical ideal contained in $I$ and with the same height as $I$. Can we always find a prime $P\in \operatorname{Ass}(R/J)$ such that $P$ is not minimal.

REPLY [0 votes]: $I=(x) \cap (y)$, $J = (x) \cap (y) \cap (z)$.


TITLE: For $x \in \mathbb{N}$ and $1\le y \le x$, $g(x,y)=\frac{1}{2}x(x-1) + y$ is injective.
QUESTION [2 upvotes]: For $x \in \mathbb{N}$ and $1\le y \le x$, define $g(x,y)=\frac{1}{2}x(x-1) + y$.
Then $g$ is injective.
I'm trying to show this. So let $1 \le y' \le x'$ and $g(x,y)=g(x',y')$. I need to show that $x=x'$ in this case, then automatically $y=y'$.
What I have so far is
$g(x,1) \le g(x,y) < g(x+1,1)\le g(x+1,y)$.
But I can't quite figure out how to show that $x=x'$. How can we show this?

REPLY [1 votes]: What I have so far is
  $g(x,1) \le g(x,y) < g(x+1,1)\le g(x+1,y)$.

You are basically done and haven't noticed, because your argument can be extended by induction to

$g(x,1) \le g(x,y) < g(x+n,1)\le g(x+n,k)$ for $n,k\in\mathbb{N}$ and $k\leq x+n$

Now let $x,x',y,y'\in\mathbb{N}$ with $y\leq x$ and $y'\leq x'$. Assume $g(x,y)=g(x',y')$. We want to show $x=x'$ and $y=y'$.
Assume that this wouldn't be the case. Consider $x=x'$ and $y\neq y'$, that immediately leads to contradiction (because $g(x,y)\neq g(x',y')$ contradicting our initial assumption). Then because of symmetry of the following argument we only need to consider $x'<x$ and $y'\leq y$. But then we get
$$g(x',y')<g(x,1)\leq g(x,y)$$ with the inductive version of your argument, leading to the contradiction $g(x',y')< g(x,y)$.


TITLE: Let $\displaystyle f$ be a continuous real valued function on the metric space $\displaystyle (X,d)$.
QUESTION [1 upvotes]: Let $f$ be a continuous real valued function on the metric space $(X,d)$.  Let $A$ be the set of all $x\in M$ such that $f(x)\geq 0$.  Show that $A$ is closed.
$(1) Proof:$  To show that $A$ is closed, it suffices to show that $A^c$ is open.  That is, we must show $A^c=\{x\in M : f(x)<0 \}$ is open.  Let $y\in A^c$.  Then $f(y)<0$.  Because $f$ is continuous, for any given $\epsilon>0$ there exists a $\delta$ (dependent on $\epsilon$) such that $|f(x)-f(y)|<\epsilon$ whenever $d(x,y)<\delta$ and $x\in M$.  Let $\epsilon=\frac{f(y)}{2}$.  Then there exists a $\delta^*$ such that $|f(x)-f(y)|<\frac{f(y)}{2}$ whenever $d(x,y)<\delta^*$ and $x\in M$.  I claim that the $\delta^*$-neighborhood about $y$ is completely contained within $A^c$.  Let $z\in N_{\delta^*}(y)$.  Then $|f(z)-f(y)|<\frac{f(y)}{2}$, and thus $f(z)\in(f(y)-\frac{f(y)}{2},f(y)+\frac{f(y)}{2})\subset(\frac{3f(y)}{2},0)$, and thus $z\in A^c$.  Thus, $A^c$ is open, and therefore $A$ is closed.
$(2) Proof:$  To show that $A$ is closed, it suffices to show that $A$ contains all of its limits points.  Let $x$ be a limit point of $A$.  Then for any $\delta>0$, $N_{\delta}^*(x) \cap A \neq \emptyset$.  That is, there exists an element $r\in M$ such that $r\in N_{\delta}^*(x)$ and $r\in A$.  Because $r\in N_{\delta}^*(x)$, we know $d(x,r)<\delta$.  Because $r\in A$, we know $f(r)\geq 0$.  Because $f$ is continuous, we can choose an appropriate $\delta$ such that $d(r,x)<\delta$ and $d(f(r),f(x))<\epsilon$ for any given $\epsilon>0$.  If $f(r)>0$, then because the continuity of $f$ and the fact that $x$ is a limit point implies that $f(x)\geq 0$, and thus $x\in A$.  If $f(r)=0$...?
I'm comfortable with proof 1 (but if you have any comments, see anything wrong with it, or think it can be made more concise, please speak up).  However, as an exercise, I thought I would try and construct another proof.  Proof 2 is that other proof.  I've broken it down into two cases: $f(r)>0$ and $f(r)=0$.  I'm stuck with the second case.  Any suggestions?

REPLY [1 votes]: Proof 2 is good up almost until the end: "...for any given $\varepsilon > 0$." From there, you should finish the proof as follows:
Since $r \in A$, $f(r) \geq 0$, and it follows that $f(x) > f(r) - \varepsilon \geq -\varepsilon$. But $\varepsilon$ was arbitrary, and so this implies that $f(x) \geq 0$.


TITLE: Show: $f_n\geqq 0$ with $f_n\nearrow f\text{ a.s.}\implies\int f_n\, d\mu\nearrow\int f\, d\mu$
QUESTION [2 upvotes]: Let $(\Omega,\mathfrak{A},\mu)$ be a measure space and $f,f_n\colon (\Omega,\mathfrak{A})\to (\overline{\mathbb{R}},\overline{\mathcal{B}})$ be measurable functions. Show: If $f_n\geq 0$ and $f_n\nearrow f$ a.s. then
    $$
\int f_n\, d\mu\nearrow\int f\, d\mu.
$$

My idea to prove that is to use two statements we had in recent lectures:

Statement 1 Let $f,f_n$ be non-negative, measurable functions. Then from $0\leq f_n\nearrow f$ it follows $0\leq\int f_n\, d\mu\nearrow\int f\, d\mu$.

and additionally

Statement 2 Let $f,g$ be two measurable functions. If $f=g$ a.s. and $f\in\mathcal{L}_{\mu}^1$, then $g\in\mathcal{L}_{\mu}^1$ and $\int f\, d\mu=\int g\, d\mu$.

Proof:
Consider $N:=\left\{\omega\in\Omega: f_n(\omega)\not\to f(\omega)\right\}$. Then by statement 1, because of
$$
0\leq 1_{N^C}f_n\nearrow 1_{N^C}f
$$
it is
$$
\int 1_{N^C}f_n\, d\mu\nearrow\int 1_{N^C}f\, d\mu.~~~~~(*)
$$
Moreover it is
$$
f_n=1_{N^C}f_n\text{ a.s.},~~~~~~~~~f1_{N^C}=f\text{ a.s.}.
$$
Here is one point I am not totally sure about: Can $f_n1_{N^C}$ and $f1_{N^C}$ now be considered as integrable functions? I mean: The statement 1 says (*) and writes down the integrals... but does this automatically mean that the functions are integrable? I think it only means that they are quasi-integrable! Does statement 2 hold for quasi-integrable functions, too?
If yes, I can use statement 2, getting
$$
\int f_n\, d\mu=\int 1_{N^C}f_n\, d\mu\nearrow\int 1_{N^C}f_n\, d\mu=\int f\, d\mu.
$$
Greetings
Miro

REPLY [0 votes]: I think you don't really need statement 2. You already proved that $$\int 1_{N^C}f_n\, d\mu\nearrow\int 1_{N^C}f\, d\mu,$$ but since $\int 1_Nf_n\, d\mu$ and $\int 1_Nf\, d\mu$ equal $0$, you can safely say that
$$\int 1_{N^C}f_n\, d\mu  +\int 1_Nf_n\, d\mu \nearrow \int 1_{N^C}f\, d\mu+\int 1_Nf\, d\mu.$$
The left hand side equals $\int f_n\, d\mu$ (by linearity and the fact that $1_N+1_{N^C}=1$), the right hand side equals $\int f\, d\mu$, and the result follows.


TITLE: Compactness for initial-to-final map for heat equation
QUESTION [2 upvotes]: Let $M$ be a compact smooth manifold without boundary. Let $T>0$ and let $g$ be a smooth Riemannian metric on $M$. Given any $f \in L^2(M)$ let $u$ be the unique solution to the equation
$$\partial_t u -\Delta_gu=0 \quad \text{on $(0,T)\times M$},$$
subject to initial data $f$ at times $t=0$. Let the map
$$ G: L^2(M) \to L^2(M)$$
be defined via $Gf=u(T,\cdot)$ on $M$.
Is it true that $G$ is compact?

REPLY [6 votes]: Parabolic regularity show that $u$ is regular for all positive times; in particular $u(t,\cdot) \in W^{1,2}(M)$ for all $t > 0$. Interior parabolic estimates additionally show that there is a constant $C = C(g,T) > 0$ so that
\begin{equation}
\lvert u(t,\cdot) \rvert_{W^{1,2}} \leq C \lvert f \rvert_{L^2}
\quad \text{ for all $t \in [T/2,2T]$.}
\end{equation}
Rellich–Kondrachov implies that the map $G: f \in L^2(M) \mapsto u(T,\cdot) \in L^2(M)$ is compact.


TITLE: Irreducible representation decomposition
QUESTION [3 upvotes]: Any faithful finite group representation can be written as a sum of irreducible representations $\rho = \oplus_{i} a_i \rho_i$ such that $Ker(\rho)=0=\bigcap_i Ker(\rho_i)$ - is this sufficient to give us the smallest (degree) faithful representation? If $\bigcap_i Ker(\rho_i)=0$, I think we can not deduce $\forall i, Ker(\rho_i)=0$: my friend said to look at the kernels of the irreducible characters to find the smallest such $\rho$, but how to we achieve this? I can't see how knowing the kernels of the individual irreducible characters would be sufficient to find the smallest faithful representation, but maybe I'm missing some condition which follows from orthogonality or something similar. Thanks for the help!

REPLY [3 votes]: I think that the point of your friends comment is the following: as you wrote,
if $\rho = \oplus_i \rho_i,$ then $ker(\rho) = \cap_i ker(\rho_i)$.  So you need
to choose $\rho_i$ so that $\cap_i \ker(\rho_i)$ is trivial.  If you know the
kernels of the various irreps. $\rho_i$ in some explicit form, presumably you can
choose a minimal collection of them so that $\cap_i \ker(\rho_i)$ is trivial.
(At least if we are talking about representations of a finite group over a field
of characteristic zero, this is possible, since general theory assures us
that a group has a faithful semi-simple representation, e.g. its regular representation.)