Twistor spaces of non-flat Bochner-K\"ahler manifolds

Abstract:
  For the twistor spaces of the Bochner-K\"ahler manifold $M = H^l \times P^n$,
systems of holomorphic coordinates are constructed. As an application of them,
an explicit description of the moduli space of relative deformations of fibers
of $M$'s twistor space is given.

\section{Introduction}

It was shown by Kamishima in \cite{[Ka]} that
a Bochner-K\"ahler manifold of dimension $n>1$
 is locally isomorphic to one of
the following model spaces:
\begin{enumerate}
\item
a flat geometry $({\mathbb C}^n\times (\qopname\relax o{U}(n)\times {\mathbb R}_+), {\mathbb C}^n)$,
\item
a non-flat geometry
 $(\qopname\relax o{PU}(1,l)\times \qopname\relax o{PU}(m+1), \Hypb^l\times{\bf P}^m)$.
\end{enumerate}
This can be compared to the fact that a conformally flat
manifold of dimension $n>2$ is locally conformal to a region
of the sphere of the same dimension.
In twistor theory, it is well-known that an even dimensional
conformally flat manifold has an
integrable twistor space
(\cite{[O.R]}, \cite{[M]}, \cite{[I1]}, \cite{[I3]}).
It is interesting, as an analogy, that
a Bochner-K\"ahler manifold has integrable twistor spaces
defined by O'Brian and Rawnsley in \cite{[O.R]}.
Since it was proved by calculating
the Nijenhuis tensors, the proof does not  give holomorphic
coordinates. Hence it would be natural to ask to
``give a system of local coordinates of the
twistor spaces of the above model spaces''.
Furthermore, to study a generalization of the Penrose transform,
it is necessary to ``construct the moduli space of relative deformations
of fibers'', which we call the {\it complexification}.

In the present paper, we give answers to the above two questions.

Since the first space is flat, its twistor space as a Riemannian manifold
has already an integrable complex structure.
Hence the problems can be solved easily by restricting the Penrose
diagram as a Riemannian manifold.

In the second case,
we give local coordinates by constructing a local embedding
to a product of complex projective spaces
(Theorem \ref{hol} and Theorem \ref{embd}).
The complexification of ${\bf H}^l\times{\bf P}^m$
is $\Hypb^l\times {\bar{\bf H}}^l \times
	\left({\bf P}^m\times \bar{{\bf P}}^m \setminus F(1,m;m+1)\right)$
(Corollary \ref{cmlx}).

Let us explain briefly the contents of this paper.
In \S2, we review the definition of the twistor spaces of Hermitian
manifolds. The answers in the flat case are explained shortly.
In \S3, we construct a system of local coordinates
of the twistor spaces in the second case.
In \S4, we give an explicit description of the complexification
of the model space, by extending maps in the previous section.
In \S5, we give some examples.

\section{The twistor spaces of Hermitian manifolds}

In this section, we review the definition of the twistor spaces
of a general Hermitian manifold $M$.
Let $n$ be the complex dimension of $M$ and
 $k$  an integer between $0$ and $n$.
  The $k$-th twistor space $Z_k(M)$ of $M$
is the total space of the Grassmannian bundle $G_k(T^{(1,0)}M)$
equipped with the almost complex structure defined as follows.
By the Hermitian metric on $M$, we can define a connection on its
 frame bundle.
Since $G_k(M)$ is an associated bundle of the frame bundle,
the connection decomposes
the real tangent bundle of $Z_k(M)$ as
\begin{equation*}
	T Z_k(M) =  T_V Z_k(M) \oplus T_H Z_k(M).
\end{equation*}
Since its fibers have natural holomorphic structures,
we have a complex structure on the first component.
For $x\in M$ and $L\in G_k(T^{(1,0)}_xM)$,
decompose the second component as
$$
	T_H Z_k(M)_{(x,L)} = T^{(1,0)}_x M = L \oplus L^{\perp}.
 $$
This decomposition is compatible with the original complex structure,
since $L$ is a complex subspace of $T^{(1,0)}_xM$.
We define a new complex structure on ${T_H} Z_k(M)_{(x,L)}$
by replacing the complex structure
on $L$ with its conjugate structure.

We can write
explicitly the space of horizontal $(1,0)$-forms,
by using the Pl\"ucker coordinates
with notation of multi-indices similar to
that in \cite{[I2]} and \cite{[I3]}.
Let $(e_1,\dots,e_n)$ be a local orthonormal basis of
the space of $(1,0)$-vectors of $M$.
For $I_0=(i_1\dots i_k)$ with $i_1 < \dots < i_k$,
let $z^{I_0}$ be the coefficient of
$e_{I_0} = e_{i_1}\wedge\dots\wedge e_{i_k}$.
A multi-index $I$ is a sequence of elements of $\{1,\dots,n\}$.
The composition of the sequences $I$ and $J$ is written as $IJ$.
We put relations between $z^I$'s as:
\begin{align*}
	z^{I_1iiI_2} &= - z^{I_1I_2},\\
	z^{I_1ijI_2} &= - z^{I_1jiI_2}, \quad i\not=j.
\end{align*}
Then, for any multi-index $I$, there is a unique subsequence
$I_0$ of $(1,\dots,n)$ such that $z^I = \pm z^{I_0}$.
We denote $\abs{I}$ to be the length of $I_0$.
Then the space of horizontal $(1,0)$-forms
of the twistor space is spanned by the following forms:
\begin{equation}\label{ACS}
  \begin{split}
	\sum_{j\not\in I} z^{jI} \bar{e}^j,&\quad	\abs{I}=k-1,\\
	\sum_{i\in I}  z^{iI} e^i,&\quad \abs{I}=k+1,
  \end{split}
\end{equation}
where $(e^1,\dots,e^n)$ is the dual basis.

 The integrability condition of the almost complex structure
is given by O'Brian and Rawnsley
in \cite{[O.R]}. A Bochner-K\"ahler manifold,
which we will study in the following sections, is an example
of such manifolds.

Henceforth, we assume that the almost complex structure of $Z_k(M)$
is integrable.

We can define the complexification of
an even dimensional conformally flat manifold
by considering the relative deformations of fibers of its twistor space.
Since Grassmannian manifolds are rigid, we can
also define the {\it complexification} of $M$ in a similar way.

For $(A,B)\in \qopname\relax o{U}(k)\times\qopname\relax o{U}(n-k)$, define
\begin{align*}
	\rho(A,B): {\mathbb C}^k\oplus {\mathbb C}^{n-k} &\rightarrow {\mathbb C}^k\oplus{\mathbb C}^{n-k}\\
		    (v_1,v_2) &\mapsto (A^*v_1, Bv_2).
\end{align*}
Then, for $x\in M$, the normal bundle $N_x$
of the fiber $Z_k(M)_x$ is written as
$$
	N_{x} \simeq N = \qopname\relax o{U}(n) \times_{\rho} ({\mathbb C}^k\oplus{\mathbb C}^{n-k}).
 $$
By the theorem of Bott-Borel-Weil-Kostant, we have:
$$
	H^i(Z_k(M)_x, {\mathcal O}(N_x)) =
	  \begin{cases}
	    \Lambda^{(1,0)}_x M \oplus T^{(1,0)}_x M,&\quad \text{if $i=0$,}\\
	    0,& \quad \text{if $i>0$.}
	  \end{cases}
 $$
Hence, by the relative deformation theory of Kodaira (\cite{[Ko]}),
the moduli space
$$
	M_{\mathbb C} = \{ Z\subset Z_k(M)\mid \text{$Z$ is a deformation of
	 a fiber.} \}
 $$
is a $2n$-dimensional complex manifold, which includes $M$
as the space of {\it real} fibers.

 As an example, we describe the situation shortly when $M={\mathbb C}^n$ with
flat metric.

 Let $X= {\mathbb C}^n\times \bar{{\mathbb C}}^n	= H^0(G_{k,n}, \mathcal{O}(N))$.
Let $Y$ be the submanifold of
$N\times X$ defined by
$$
	Y = \{ (f(z),f) \mid f\in
	  X, z\in G_{k,n} \}.
 $$
We have a canonical double fibration:
\begin{equation*}
\setlength{\unitlength}{0.5mm}%
  \begin{picture}(50,30)
    \put(20,20){$Y$}
    \put(0,0){$N$}
    \put(40,0){$X.$}
    \put(18,18){\vector(-1,-1){10}}
    \put(27,18){\vector(1,-1){10}}
    \put(7,16){$\scriptstyle p_1$}
    \put(34,16){$\scriptstyle p_2$}
  \end{picture}
\end{equation*}
Then we can show easily that, for each $w\in N$, there is
a unique element $v\in {\mathbb C}^n$ such that
$w\in p_1\circ p_2^{-1}(v,\bar{v})$.
By the projection $w\mapsto v$, we can identify $N$ with
$Z_k({\mathbb C}^n)$.
The {\it complexification} of ${\mathbb C}^n$ is $X$, namely,
for $x\in X$,  the corresponding
submanifold is written as $p_1\circ p_2^{-1}(x)$.
\begin{rem}
Since the twistor space $Z({\mathbb C}^n)$ as a Riemannian manifold
satisfies the integrability condition,
this diagram can be obtained by restricting
the Penrose double fibration as a Riemannian manifold,
which is described in detail by Murray in \cite{[M]}.
\end{rem}

\section{The model spaces of non-flat Bochner-K\"ahler manifolds
and their twistor spaces}

Let $V$ be a complex vector space of dimension $l+1$  having
the indefinite metric:
$$
	\begin{pmatrix}
	  1 & 0\\
	  0 & -I_l
	\end{pmatrix}.
 $$
Then we can consider the $l$-dimensional complex hyperbolic space
$\Hypb^l = \Hypb^l(V)$ as an open submanifold of ${\bf P}^l = {\bf P}(V)$ by
$$
	\Hypb^l = \{x = (1: x_1 : \dots : x_l)\in {\bf P}^l \mid
	  \abs{x_1}^2 + \dots + \abs{x_l}^2 < 1 \}.
 $$
Let $W$ be a Hermitian vector space of dimension $m+1$
and  ${\bf P}^m={\bf P}(W)$ its projective space.
Let $y = (1:y_1:\dots:y_m)$ denote its local affine coordinates
with respect to an orthonormal basis of $W$.

We consider $\Hypb^l$ and ${\bf P}^m$	 as K\"ahler manifolds
by the well-known K\"ahler forms
\begin{align*}
	& -\sqrt{-1} \partial \bar{\partial}
	  \log \left(1- \sideset{}{_i}\sum \abs{x^i}^2\right),\\
	& \sqrt{-1} \partial \bar{\partial}
	  \log\left(1+\sideset{}{_j}\sum \abs{y^j}^2\right),
\end{align*}
respectively.
Then $M=\Hypb^l\times {\bf P}^m$ with the product metric is
an example of Bochner-K\"ahler manifolds.
Hence
the almost complex structure of $Z_k(M)$ is integrable.
By the definition of the metric,
the group $\qopname\relax o{U}(1,l) \times \qopname\relax o{U}(m+1)$ acts on $M$ isometrically.

Let $x_0=(1,0,\dots,0)$ and $y_0=(1,0,\dots,0)$.
By identifying a tangent vector with an element of
$x_0^\perp\oplus y_0^\perp$,
we can consider it as an element of $V\oplus W$.
Let $z^{IJ'}$ be the Pl\"ucker coordinate,
where $I$ (resp. $J$) is a multi-index
with respect to $x_1,\dots,x_l$ (resp. $y_1,\dots,y_m$).
Note that the local frame of the tangent bundle that
we choose is orthonormal only at $(x_0,y_0)$.
Hence \eqref{ACS} is valid only at points on the fiber over $(x_0,y_0)$.

Let $L$ be a complex vector space with metric.
For a vector $x\in L$, we regard $\bar{x}$ as
 an element of $L^*$ by the canonical
isomorphism $\bar{L} \simeq L^*$.
Then, by using the decomposition:
$$
	\Lambda^{k+1}(V\oplus W) = \bigoplus_{a+b=k+1}\Lambda^aV\otimes\Lambda^bW,
 $$
we define
\begin{align*}
  \phi_{a,b} &= \left.
  x\wedge\left(1-\frac{y\wedge i(\overline{y})}{(y,y)}\right) z +
    y\wedge\left(1-\frac{x\wedge i(\overline{x})}{(x,x)}\right) z
  \right|_{\Lambda^{a} V\otimes \Lambda^{b} W}\\
  &= \sum_{\abs{I}=a, \abs{J}=b} \phi_{I,J} e_I\wedge e_{J'},
\end{align*}
where $i$ denotes the interior multiplication.
In the following, we assume that multi-indices $I,J,\dots$
do not include the index $0$.
Then we have
\begin{equation}\label{a}
\begin{split}
	\phi_{0I,0J} &= (-1)^a
	  \left(
	    \sum_{i\not\in I}\frac{\overline{x}^i z^{iIJ'}}{(x,x)} +
	    \sum_{j\not\in J}\frac{\overline{y}^j z^{j'IJ'}}{(y,y)}
	  \right),\\
	\phi_{I,0J} &= (-1)^a z^{IJ'} - \sum_{i\in I} x^i \phi_{0iI,0J},\\
	\phi_{0I,J} &= z^{IJ'} - \sum_{j\in J} y^j \phi_{0I,0jJ},\\
	\phi_{I,J} &= - \sum_{j\in J} y^j z^{j'IJ'}
	  - \sum_{i\in I}x^i \phi_{0iI,J},
\end{split}
\end{equation}
where $\bar{x}^i$ is the complex conjugate of $x^i$.

Let $p_1$ and $p_2$ be the projections:
\begin{align*}
	p_1: \Hypb^l\times {\bf P}^m &\rightarrow \Hypb^l,\\
	p_2: \Hypb^l\times {\bf P}^m &\rightarrow {\bf P}^m.
\end{align*}
Let $L_{\Hypb^l}$ and $L_{{\bf P}^m}$ be the tautological line bundles.
Let $L_{Z_k(M)}$ denote
the tautological line bundle
as a Grassmannian bundle.
Then $\phi_{a,b}$ is canonically extended to a map
$$
	\phi_{a,b}: L_{Z_k(M)}\otimes p_1^*(L_{\Hypb^l}^a)
			\otimes p_2^*(L_{{\bf P}^m}^b)
		\rightarrow
	\Lambda^a V\otimes \Lambda^b W,
 $$
which is equivariant under the
action of $\qopname\relax o{U}(1,l)\times \qopname\relax o{U}(m+1)$.
The space on the left hand side has an almost complex structure
induced by that of $Z_k(M)$ and the canonical connection
as a line bundle. Then we have the following theorem.

\begin{thm}\label{hol}
The map $\phi_{a,b}$ is holomorphic.
\end{thm}

\begin{proof}
It suffices
to calculate that $\phi_{a,b}$ is $\overline{\partial}$-closed at points on the
fiber over $(x_0,y_0)$.
Then it is immediate by the definition
of the almost complex structure.
\end{proof}

\begin{rem}
This theorem shows that the line bundle
$L_{Z_k(M)}\otimes p_1^*(L_{\Hypb^l}^a)
		\otimes p_2^*(L_{{\bf P}^m}^b)$
has a natural holomorphic structure if $a+b=k+1$.
In fact, if both $l$ and $m$ are positive,
this is necessary,
which can be shown directly by computing the integrability condition.
\end{rem}

For non-negative integers $a$ and $b$ such that $a+b=k$, put
$$
	U_{a,b} = \{ \kappa \in Z_k(M) \mid
	  \phi_{a+1,b}(\kappa) \not= 0, \phi_{a,b+1}(\kappa) \not= 0 \}.
 $$
These open sets cover $Z_k(M)$. In fact, we have the following
lemma.
\begin{lem}
Let $(x,y,z)$ be a point of $Z_k(M)$ and
 $L$ the subspace of
$x^\perp\oplus y^\perp$ corresponding to $z$. Then
$(x,y,z)\in U_{a,k-a}$ if and only if
$\dim (L\cap x^\perp) \le a \le k - \dim(L\cap y^\perp)$.
\end{lem}

\begin{proof}
By \eqref{a}, we have
\begin{align}
	\phi_{0I,0J}(x_0, y_0, z) &= 0,\label{eq1}\\
	\phi_{I,0J}(x_0, y_0, z) &= (-1)^{\abs{I}} z^{IJ'},\label{eq2}\\
	\phi_{0I,J}(x_0, y_0, z) &= z^{IJ'},\label{eq3}\\
	\phi_{I,J}(x_0, y_0, z) &= 0\label{eq4}.
\end{align}
Hence, by the Pl\"ucker relations, we have
$$
	U_{a,b} = \{ (x,y,z)\in Z_k(M) \mid
	  L |_{\Lambda^a x^{\perp}\otimes \Lambda^b y^{\perp}} \not=0 \},
 $$
from which the lemma follows immediately.
\end{proof}

Now we define a map:
\begin{align*}
	\psi_{a,b}:
	 U_{a,b} &\rightarrow {\bf P}(\Lambda^{a+1}V\otimes\Lambda^{b}W)
		\times {\bf P}(\Lambda^{a}V\otimes\Lambda^{b+1}W)\\
	 \kappa &\mapsto ([\phi_{a+1,b}(\kappa)],[\phi_{a,b+1}(\kappa)]).
\end{align*}

\begin{thm}\label{embd}
The map $\psi_{a,b}$ is an embedding.
\end{thm}

\begin{proof}
Since the map is equivariant,
to prove the injectivity,
 it suffices to prove that
if $\psi_{a,b}(x,y,w) = \psi_{a,b}(x_0,y_0,z)$ then $x=x_0$,
$y=y_0$ and $w=z$.
Suppose $(y,y_0) = 0$. Then we can assume $y = (0,1,0,\dots,0)$.
We assume, for a while, that $J$ does not include the index $1$.
We compute
$$
	\phi_{0I,0J} = 0 = w^{I0'J'}, \quad \abs{I} = a, a-1.
 $$
Hence we have
\begin{gather*}
	\phi_{I,0J} = (-1)^{\abs{I}} z^{IJ'} =
	  - \sum_{i\in I} x^i \phi_{0iI,0J} = 0,\quad \abs{I} = a,\\
	\phi_{I,01J} = (-1)^{\abs{I}}z^{I1'J'} =
	- (-1)^{\abs{I}}w^{I0'J'} = 0,\quad \abs{I} = a.
\end{gather*}
These contradict with the assumption $z\in U_{a,k-a}$.

Hence we should have $(y,y_0) \not=0$.
Then, by \eqref{eq2} and \eqref{eq3},
there exists a constant $c$ such that
$w^{IJ'} = c z^{IJ'}$ for all $(I,J)$ with $\abs{I} = a-1, a, a+1$.
Hence we have $w=z$ by the Pl\"ucker relations.
By \eqref{eq1} and  \eqref{eq4}, the equations for $x$ and $y$ become
\begin{align*}
	\sum_{i\not\in I}
	    \frac{\overline{x}^i z^{iIJ'}}{(x,x)}
	+\sum_{j\not\in J}
	    \frac{\overline{y}^j z^{j'IJ'}}{(y,y)} &= 0,\\
	 \sum_{j\in J} y^j z^{j'IJ'} + \sum_{i\in I} x^i z^{iIJ'} &= 0.
\end{align*}
Let $x'=(x^1,\dots,x^l)$ and $y'=(y^1,\dots,y^m)$. Then these equations
 mean that there is a $k$-dimensional subspace
$L$ of $x_0^\perp\oplus y_0^\perp$ such that
$$
	(x',y') \in L \subset
	  (\frac{x'}{(x,x)}, \frac{y'}{(y,y)})^\perp.
 $$
Hence we have $x=x_0$ and $y=y_0$, concluding that the map
is injective.

 Injectivity of the derivative is also computed by
simple computations at points on the fiber over $(x_0, y_0)$.
\end{proof}

\section{The complexification of the model spaces}

In this section, we give an explicit description
of the {\it complexification} $M_{\mathbb C}$
of the model space $M = \Hypb^l\times{\bf P}^m$.

Let $\xi, \zeta\in \Hypb^l$ and
$\mu, \nu\in{\bf P}^m$ such that $(\mu,\nu)\not=0$.
We construct an embedding of
$G_{k,n} = G_k(V/\vecsp{\xi} \oplus W/\vecsp{\mu})$
in $Z_k(M)$ by
generalizing $\phi_{a,b}$	in the previous section.
Let $a$ and $b$ be non-negative integers such that $a+b=k+1$.
For $z\in G_k(V/\vecsp{\xi}\oplus W/\vecsp{\mu})$, we define:
$$
	\Phi_{a,b}(z) =
	  \left.
	  \xi\wedge
	    \left(
		1 - \frac{\mu\wedge i(\bar{\nu})}{(\mu,\nu)}
	    \right)z
	  + \mu\wedge
	    \left(
		1 - \frac{\xi\wedge i(\bar{\zeta})}{(\xi,\zeta)}
	    \right)z
	 \right|_{\Lambda^aV\otimes\Lambda^bW\,.}
 $$
For non-negative integers $a$ and $b$ satisfying $a+b=k$,  put
\begin{align*}
	U'_{a,b} &= \{ z \in G_{k,n} \mid
	  \Phi_{a+1,b}(z) \not= 0, \Phi_{a,b+1}(z) \not= 0 \}\\
	&= \{ z \in G_{k,n} \mid
	  z |_{\Lambda^a V/\vecsp{\xi}\otimes \Lambda^b W/\vecsp{\mu}}
		 \not=0 \}.
\end{align*}
Then we define
\begin{align*}
	\Psi_{a,b}(\xi,\zeta,\mu,\nu)=	\Psi_{a,b}:
	 U'_{a,b} &\rightarrow {\bf P}(\Lambda^{a+1}V\otimes\Lambda^{b}W)
		\times {\bf P}(\Lambda^{a}V\otimes\Lambda^{b+1}W)\\
	 z &\mapsto ([\Phi_{a+1,b}(z)],[\Phi_{a,b+1}(z)])
\end{align*}

\begin{thm}
Let $z\in G_{k,n}$.
There exists $\kappa\in Z_k(M)$ such that
$\psi_{a,b}(\kappa) = \Psi_{a,b}(z)$
for all $(a,b)$ such that $z\in U'_{a,b}$.
Hence $\Psi_{a,b}$ induces an embedding of $G_{k,n}$ in $Z_k(M)$.
\end{thm}

\begin{proof}
Let $L$ be the subspace of $V/\vecsp{\xi}\oplus W/\vecsp{\mu}$
corresponding to $z$.
We can assume $\zeta = x_0$ and $\nu = y_0$.
By the isomorphisms
$x_0^{\perp} \simeq V/\vecsp{\xi}$ and
$y_0^{\perp} \simeq W/\vecsp{\mu}$
defined by the canonical projections,
we can consider $L$ as a subspace of $x_0^{\perp}\oplus y_0^{\perp}$.
We give a point $\kappa = (x, y, w)$ which satisfies
the property. We fix $a$ and $b$ for a while, and calculate
 the coefficients
of $\Phi_{a+1,b}(z)$ and $\Phi_{a,b+1}(z)$.
By a similar argument
in the proof of Theorem~\ref{embd}, we should have
$(y,y_0) \not= 0$. Thus, by \eqref{a},	we have
\begin{align*}
	\Phi_{0I,0J}(z) &=  0,\\
	\Phi_{I,0J}(z) &= (-1)^{\abs{I}}z^{IJ'},\\
	\Phi_{0I,J}(z) &= z^{IJ'},\\
	\Phi_{I,J}(z) &= - \sum_{j\in J} \mu^j z^{j'IJ'}
		   - \sum_{i\in I} \xi^i z^{iIJ'}.
\end{align*}
Hence, by the Pl\"ucker relations, we have $w=z$.
The coordinates of $x$ and $y$ satisfy the equation:
\begin{equation}\label{int}
	\sum_{i\not\in I} \frac{\bar{x}^i z^{iIJ'}}{(x,x)}
	  +
	\sum_{j\not\in J} \frac{\bar{y}^j z^{j'IJ'}}{(y,y)}
	  = 0.
\end{equation}
Define
\begin{align*}
	V_1 &= L \cap x_0^\perp,\\
	W_1 &= L \cap y_0^\perp.
\end{align*}
Let $V_2$ and $W_2$ be the subspaces such that
\begin{align*}
	x_0^\perp &= V_1 \oplus V_2\\
	y_0^\perp &= W_1 \oplus W_2
\end{align*}
which are orthogonal decompositions.
Then \eqref{int} means that
\begin{align*}
	x &\in \vecsp{x_0} + V_2,\\
	y &\in \vecsp{y_0} + W_2.
\end{align*}
Hence the problem is reduced to the case $V_1 = W_1 = 0$,
which implies that $\min(l,m)\ge k$.
By considering the canonical anti-holomorphic diffeomorphism:
\begin{align*}
	Z_k(M) &\rightarrow Z_{n-k}(M)\\
	L &\mapsto L^\perp,
\end{align*}
we can further assume that
\begin{gather*}
	\dim V = \dim W = k+1,\\
	z \in \bigcap_{0\le a \le k} U'_{a,k-a}.
\end{gather*}
In this case, it is easily verified that the
equations for $x$ and $y$ are independent with respect to $a$.
Hence it suffices to find $x$ and $y$ when $a=0$.

An image of the map
$$
	\Psi_{0,k}: \bigcap_{0\le a \le k} U'_{a,k-a}
	  \rightarrow {\bf P}(\qopname\relax o{Hom}(W, V))
 $$
is a projectivized invertible linear map.
Hence it suffices to find
$x\in V$ and $y\in W$ such that
\begin{align*}
	A(y) &= x,\\
	A(y^\perp) &= x^\perp,
\end{align*}
for an arbitrary invertible linear map $A: W\rightarrow V$.
Indeed, if we find them, we have
$$
	\psi_{0,k}\left (x,y, \vecsp{A(e_j) + e_j \mid j=1,\dots,k}\right)
	  = [A],
 $$
where $(e_1,\dots,e_k)$ is a basis of $y^\perp$.

We denote the indefinite metric on $V$ as $(,)_G$.
By the automorphism $A$, we define a Hermitian metric $(,)_H$ on $V$.
For a positive number $r$, let $S_r$ be the real hypersurface of $V$
defined as
$$
	S_r = \{ v\in V \mid (v,v)_H = r \}.
 $$
Let $H$ be the real hypersurface of $V$ defined as
$$
	H = \{ v\in V \mid (v,v)_G = 1 \}.
 $$
Let $r_0$ be the minimum number such that $S_r\cap H \not = \emptyset$.
Then we can show easily that
$S_{r_0}$ and $H$ intersect at one point $x$.
Then $x$ and $y = A^{-1}(x)$ satisfy
the desired property.
\end{proof}

Let $(\xi,\zeta,\mu,\nu)$ be as above.
We denote the corresponding submanifold of $Z_k(M)$ as
$Z_k(\xi,\zeta,\mu,\nu)$.
The intersection of two such submanifolds
can be completely determined as follows.

\begin{thm}
Two submanifolds $Z_k(\xi_i,\zeta_i,\mu_i,\nu_i)$, $i=1,2$
intersect if and only if
$$
      \frac{(\xi_1,\zeta_2)(\xi_2,\zeta_1)}{(\xi_1,\zeta_1)(\xi_2,\zeta_2)}
      = \frac{(\mu_1,\nu_2)(\mu_2,\nu_1)}{(\mu_1,\nu_1)(\mu_2,\nu_2)}.
 $$
When this condition is satisfied, the intersection is
one of the following:
\begin{align*}
	G_{k,n-1},& \quad \text{if $\xi_1=\xi_2$, $\mu_1=\mu_2$,}\\
	G_{k-1,n-1},& \quad \text{if $\zeta_1=\zeta_2$, $\nu_1=\nu_2$,}\\
	G_{k-1,n-2},& \quad \text{otherwise.}
\end{align*}
\end{thm}

\begin{proof}
We can put $\zeta_2 = x_0$ and $\nu_2 = y_0$.
By a similar argument in the proof of Theorem \ref{embd},
we can show that two submanifolds are disjoint if
$(\mu_1,y_0) = 0$.
Hence we can assume that $(\mu_1,y_0) \not= 0$.
Let $z$ and $w$ be elements of $G_k(x_0^\perp\oplus y_0^\perp)$
such that
$\Psi_{a,b}(\xi_1,\zeta_1,\mu_1,\nu_1)(z) =
\Psi_{a,b}(\xi_2, x_0, \mu_2, y_0)(w)$.
Then, since $\Phi_{0I,0J}(z) = 0$ for $\abs{I} = a, a-1$, we have $z = w$.
Let $\xi_1 = (1,\xi_1^1,\dots,\xi_1^l)$ and
$\mu_1 = (1,\mu_1^1,\dots,\mu_1^m)$. Then
the equations are written as
\begin{equation}\label{intcon}
\begin{split}
	\frac{\sum_{i\not\in I} {\bar\zeta}_1^i z^{iIJ'}}
	    { {\bar\zeta}_1^0 - \sum_i \xi_1^i {\bar\zeta}_1^i } +
	  \frac{\sum_{j\not\in J} {\bar\nu}_1^j z^{j'IJ'}}
	    { {\bar\nu}_1^0 + \sum_j \mu_1^j {\bar\nu}_1^j} &= 0,\\
	\sum_{i\in I} (\xi_1^i - \xi_2^i)z^{iIJ'} +
	  \sum_{j\in J} (\mu_1^j - \mu_2^j)z^{j'IJ'} &= 0.
\end{split}
\end{equation}
Let $L$ be the subspace of $x_0^\perp\oplus y_0^\perp$ corresponding
to $z$ and $\xi_i' = (0,\xi_i^1,\dots,\xi_i^l)$,
$\mu_i' = (0,\mu_i^1,\dots,\mu_i^m)$, $i=1,2$.
Then \eqref{intcon} means that
$$
	(\xi_1'-\xi_2', \mu_1'-\mu_2') \in L \subset
	(\frac{\zeta_1}{(\zeta_1, \xi_1)},
	   \frac{\nu_1}{(\nu_1, \mu_1)})^\perp.
 $$
Then the theorem follows immediately.
\end{proof}

\begin{cor}\label{cmlx}
 Let $M=\Hypb^l\times {\bf P}^m$. Then we have
$$
	M_{\mathbb C} = \Hypb^l\times {\bar{\bf H}}^l \times
	  \left({\bf P}^m\times \bar{{\bf P}}^m \setminus F(1,m;m+1)\right).
 $$
\end{cor}

\section{Some examples}
\begin{enumerate}
\item $M={\bf P}(W)={\bf P}^n$:
We can write $Z_k({\bf P}^n) = F(k,k+1;n+1)$.
The projection map is written as
\begin{equation*
\begin{split}
	F(k,k+1;n+1) &\rightarrow {\bf P}^n\\
	(U,U') &\mapsto U^\perp \cap U'
\end{split}
\end{equation*}
The complexification $M_{\mathbb C}$ is ${\bf P}^n \times \bar{{\bf P}}^n
\setminus F(1,n;n+1)$.
For $(\mu,\nu)\in M_{\mathbb C}$, the corresponding submanifold is written as
\begin{align*}\label{cmplx}
	\{(U, U') \mid U \subset \nu, \mu \subset U'\}.
\end{align*}
This is a straightforward generalization of the
well-known description of the twistor space of ${\bf P}^2$.

\item $M={\bf H}(V) = {\bf H}^n$:
For a subspace $U$ of $V$, we write
$$
	U = \begin{cases}
	  U_{\pm},&\quad \text{if $U$ has a vector with positive
		norm},\\
	  U_{-},&\quad \text{if $G|_U$ is negative definite}.
	\end{cases}
 $$
Then we have
$$
	Z_k({\bf H}^n) =
	  \{ (U_-, U'_{\pm}) \in F(k,k+1;V) \}
 $$
The projection map is written as:
\begin{align*}
	Z_k({\bf H}^n) &\rightarrow {\bf H}^n\\
	(U, U') &\mapsto U^\perp \cap U'
\end{align*}
$M_{\mathbb C}$ is ${\bf H}^n \times \bar{{\bf H}}^n$.
For $(\xi,\zeta)\in M_{\mathbb C}$, the corresponding submanifold
is written as:
$$
	\{(U,U')\mid U\subset \zeta, \xi \subset U' \}
 $$
\item $Z_1(\Hypb^l \times {\bf P}^m)$ with $l,m\ge 1$:
The restriction of $\phi_{1,1}$ to $U_{1,0}\cap U_{0,1}$
is an isomorphism to
the projectivization of
the space of rank 2 elements of $V\otimes W$.
The boundary of the image is identified with
${\bf P}(V)\times{\bf P}(W)$ by the Segre embedding.
The space ${\bf P}(V)$ has a decomposition:
$$
	{\bf P}(V) = {\bf H}_+ \cup S^{2l-1} \cup {\bf H}_-,
 $$
by the sign of the norm.
Then we obtain $Z_1(\Hypb^l \times {\bf P}^m)$ by
changing the boundary as follows:
\begin{enumerate}
\item remove $S^{2l-1}\times {\bf P}^m$,
\item replace
  ${\bf H}_+  \times {\bf P}^m$ with
    ${\bf P}(T^{(1,0)}{\bf H}_+)\times {\bf P}^m$,
\item replace
${\bf H}_- \times {\bf P}^m$ with
${\bf H}_-\times {\bf P}(T^{(1,0)}{\bf P}^m)$.
\end{enumerate}
\end{enumerate}


L.A. Ferreira, D.I. Olive and M.V. Saveliev

Orthogonal Decomposition of Some Affine Lie Algebras in Terms of their
  Heisenberg Subalgebras

Abstract:
  In the present note we suggest an affinization of a theorem by Kostrikin
et.al. about the decomposition of some complex simple Lie algebras ${\cal G}$
into the algebraic sum of pairwise orthogonal Cartan subalgebras. We point out
that the untwisted affine Kac-Moody algebras of types $A_{p^m-1}$ ($p$ prime,
$m\geq 1$), $B_r, \, C_{2^m}, D_r,\, G_2,\, E_7,\, E_8$ can be decomposed into
the algebraic sum of pairwise or\-tho\-go\-nal Heisenberg subalgebras. The
$A_{p^m-1}$ and $G_2$ cases are discussed in great detail. Some possible
applications of such decompositions are also discussed.

\section{Introduction}

In the early eighties A. I. Kostrikin, I. A. Kostrikin and V. A. Ufnarovskii
 proved a theorem \cite{KKU81}, see
also \cite{KKU811} and \cite{KKU82}, that the
complex Lie algebras ${\cal G}$ of types
$A_{p^m-1}$ ($p$ prime,
$m\geq 1$), $B_r, \, C_{2^m}, D_r,\, G_2,\, E_7,\, E_8$ can be decomposed into
the algebraic sum of Cartan subalgebras, pairwise orthogonal with respect
to the Killing form; i.e.
\begin{equation}
{\cal G} = {\cal H}_0 \, \oplus \, {\cal H}_1 \, \oplus \, \cdots \, \oplus \, {\cal H}_h
\lab{decom}
\end{equation}
with
\begin{equation}
\mathop{\rm Tr} \( {\cal H}_i \, {\cal H}_j \) = 0 \, \, \, \, \mbox{\rm for $i \neq j$},
\lab{otho}
\end{equation}
where $h$ is the Coxeter number of ${\cal G}$. The fact that the
number of Cartan subalgebras in the above decomposition \rf{decom} is $h+1$
accords with the well-known observation that for all semisimple algebras
$h+1 = \mbox{\rm dim ${\cal G}$}/r,\; r=\mbox{\rm rank ${\cal G}$}$.

Here we show that this theorem can be extended to many of the
corresponding affine Lie algebras ${\hat {\cal G}}$, in the sense that the latter may
be decomposed  into the algebraic  sum of pairwise orthogonal Heisenberg
subalgebras. Those  subalgebras are in fact the affinization of the Cartan
subalgebras in the  corresponding decomposition of the finite--dimensional
algebras ${\cal G}$. The problem of extending the results to the affine case
consists in identifying the gradations of the Kac-Moody algebra ${\hat {\cal G}}$
which respect the decomposition \rf{decom}.

The integral gradations of an affine Kac-Moody algebra ${\hat {\cal G}}$ are
provided by the grading
operators \cite{KP85}
\begin{equation}
Q = \sum_{a=1}^{r} s_a \lambda^v_a \cdot H + N d,
\lab{gradop}
\end{equation}
where $(s_0, s_1, \cdots ,s_r)$ is a vector of
non negative, relatively prime  integers, $\lambda^v_a \equiv 2
\lambda_a/\alpha_a^2$
with $\lambda_a$ and $\alpha_a$ being  the fundamental weights and simple roots
of ${\cal G}$ respectively. Furthermore,
\begin{equation}
N = \sum_{i=0}^{r} s_i \, m_i^{\psi}, \qquad
\psi = \sum_{a=1}^{r}m_a^{\psi}\,  \alpha_a , \qquad
m_0^{\psi} \equiv 1
\lab{grad}
\end{equation}
and $\psi$ is the maximal root of ${\cal G}$. Two gradations are counted as
equivalent if the corresponding vectors $(s_0,s_1,\cdots ,s_r)$ and $(s'_0,
s'_1, \cdots s'_r)$ are related by a symmetry of the extended Dynkin diagram
 of ${\cal G}$.

Given a finite--dimensional complex simple Lie algebra ${\cal G}$ with
decomposition \rf{decom}, we look
for a gradation such that the corresponding affine Kac--Moody algebra ${\hat {\cal G}}$
can be written as
\begin{equation}
{\hat {\cal G}} = {\hat {\cal H}}_0 \, \oplus \, {\hat {\cal H}}_1 \, \oplus \, \cdots \, \oplus \, {\hat {\cal H}}_h,
\lab{decomkm}
\end{equation}
where each subspace ${\hat {\cal H}}_i$ is a Heisenberg subalgebra of ${\hat {\cal G}}$  with
generators being eigenvectors of the corresponding grading operator $Q$. In
addition, under the restriction of ${\hat {\cal G}}$ to ${\cal G}$, the elements of ${\hat {\cal H}}_i$
become those of ${\cal H}_i$.

For the case of the homogeneous gradation it is quite trivial to make
the extension of the decomposition \rf{decom} with the orthogonality
property \rf{otho} to the affine case. In this case
$s_0=1$ and $s_a =0$ for $a=1,2\cdots ,r$, and so $Q=d$. This gradation is
insensitive to the structure of the root system of ${\cal G}$  and therefore it does
not matter what linear combinations of the Weyl--Cartan basis elements the
generators of ${\cal H}_i$ are. We construct the generators of ${\hat {\cal H}}_i$ by
replacing in ${\cal H}_i$ the elements $H_a$ by $H_a^n$ and $E_{\alpha}$ by
$E_{\alpha}^n$, with $n$
being the integral eigenvalues of $d$; in other words $\lbrack d , H_a^n \rbrack =
n H_a^n$, and
$\lbrack d , E_{\alpha}^n \rbrack = n E_{\alpha}^n$. The orthogonality condition is preserved
since if, say $S$ and $T$ are two orthogonal elements of ${\cal G}$, so are the
elements  $S^m$ and $T^n$ of ${\hat {\cal G}}$; i.e., $\mathop{\rm Tr} ( S^m T^n ) = \mathop{\rm Tr} ( S T )
\, \delta_{m+n,0}$.

For other gradations the extension is more delicate, and we discuss here
the case of the Lie algebras $A_{p^m-1}$ and $G_2$ only. The cases of
gradations which are not of type \rf{gradop} deserve further study. At the
end of section \ref{sec2} we comment on a special gradation of $A_2$.

\section{The case $A_{p-1}$ with $p$ prime}
\label{sec2}

In refs. \cite{KKU81}, \cite{KKU811} (see also \cite{PZ81}) the algebra
$A_{p-1}$ was decomposed as in
\rf{decom} using some special properties of the defining  representation.
Consider the $p\times p$ matrices
\begin{equation}
g=\(
\begin{array}{cccccc}
0&x_1&0&0&\cdots &0\\
0&0&x_2&0&\cdots &0\\
0&0&0&x_3&\cdots &0\\
\vdots&\vdots&\vdots&\vdots&\vdots&\vdots\\
0&0&0&0&\cdots &x_{p-1}\\
x_0&0&0&0&\cdots &0
\end{array}\),
\qquad
D=\(
\begin{array}{cccccc}
1&0&0&0&\cdots &0\\
0&\omega&0&0&\cdots &0\\
0&0&\omega^2&0&\cdots &0\\
\vdots&\vdots&\vdots&\vdots&\vdots&\vdots\\
0&0&0&\cdots &\omega^{p-2}&0\\
0&0&0&0&\cdots &\omega^{p-1}
\end{array}\),
\lab{gd}
\end{equation}
where $x_i$, $i=0,1,2\ldots p-1$ are arbitrary non vanishing constants,
$\omega$ is the $p$-th root of unit ($\omega =\exp (2\pi
i/p), \, \omega^p=1$), and so
\begin{equation}
g \, D = \omega \, D \, g,  \qquad  g^p =x_0 x_1 \ldots x_{p-1}\hbox{{1}\kern-.25em\hbox{l}} , \qquad
D^p=\hbox{{1}\kern-.25em\hbox{l}} .
\lab{prop}
\end{equation}
The subspaces in \rf{decom} are defined as
\begin{eqnarray}
{\cal H}_0 &=& \{ D^k \, , \, k=1,2,3, \ldots ,p-1\};\nonumber\\
{\cal H}_{l+1} &=& \{ \( D^{l} g\)^k \, , \, k=1,2,3, \ldots ,p-1\},
\qquad l=0,1,2 ,\ldots ,p-1;
\lab{subsp}
\end{eqnarray}
and the Coxeter number $h$ of $A_{p-1}$ equals $p$.

The  subspaces \rf{subsp} are obviously abelian. However, the linear
independency and  orthogonality of them requires $p$ to be a prime number.
Using \rf{prop} one observes that
\begin{equation}
\( D^{l} g\)^k \sim  \, D^{lk} g^k,
\end{equation}
where here, and below, the symbol $\sim $ means the two quantities are
proportional (the proportionality constant being $\omega$ dependent).
Therefore, if for given $l$ and $l^{\prime}$ one has $lk = l^{\prime} k + mp$,
with $m$
integer, then $\( D^{l} g\)^k \sim \( D^{l^{\prime}} g\)^k$. Now, if $p$ is prime,
either $k$ or $(l- l^{\prime})$ will have to have $p$ in its expansion as a
product of primes. But this can not happen since they can be at most $(p-1)$.
Therefore  the subspaces \rf{subsp} are linearly independent for $p$ prime. As
for the orthogonality, consider $\mathop{\rm Tr} \( \( D^{l} g\)^k \( D^{l^{\prime}}
g\)^{k^{\prime}} \) \sim \mathop{\rm Tr} \( D^{lk+l^{\prime}k^{\prime}} g^{k+k^{\prime}}\)$. That does
not vanish for $l\neq  l^{\prime}$ only if $k+k^{\prime}= mp$ and
$lk+l^{\prime}k^{\prime}=np$, with $m$ and $n$ integers, and so for
$(l-l^{\prime})k=(n-ml^{\prime})p$, which is impossible for $p$ prime.

Now to extend the consideration to the affine case, we have to select a
gradation which respects
the decomposition written above. As we have already said, the homogenous
gradation fits our
needs. However, in this particular case the principal gradation also does, and
we find it more interesting and important for applications (see comments at
the end of this section on an additional gradation for $A_2$).

In fact the homogeneous and principal gradations are the only integer
gradations which respect the decomposition above. In order to see that, one
observes that the elements $D^lg$ are linear combinations of the step operators
for simple roots $E_{\alpha_a}$, $a=1,2,\ldots ,p-1$, and $E_{-\psi}$ where $\psi$
is the highest root. Now when we affinize the algebra we will need to find
integers $n_a$ and $n_b$ such that $E_{\alpha_a}^{n_a}$ and $E_{\alpha_b}^{n_b}$
have  the same grade under the grading operator
\rf{gradop}. Therefore we need  $n_b  = n_a +{s_a - s_b\over N}$. But since $s_a$
and $s_b$ are non negative it follows that $\mid s_a - s_b\mid$ is smaller than
$s_a$ and/or $s_b$. On the other hand, from \rf{grad} we see that $N$ is
greater
than  $s_a$ and $s_b$. So it is impossible to satisfy the above equality with
$n_a$ and $n_b$ integers unless
$s_a = s_b$ for all $a$ and $b$. Now we also need $E_{\alpha_a}^{n_a}$ and
$E_{-\psi}^{n_0}$ to have the same grade, and so $s_a + N n_a =
-\sum_{b=1}^{p-1} s_b + N n_0$ (since $\psi = \sum_{b=1}^{p-1} \alpha_b$).
Therefore we need $n_a = n_0 - {p s \over {(p-1)s + s_0}}$, where we used the fact
that all $s_a$'s are equal and have set $s_a \equiv s$ for any $a$. So we have
to have $s=0$ (homogeneous) or $s=s_0$ (principal).

We then consider the principal gradation and  define the following generators
of
the affine Kac--Moody algebra
$\hat{sl}(p, \rlx\hbox{\,$\inbar\kern-.3em{\rm C}$} )$:
\begin{eqnarray}
{\hat {\cal H}}^{0}_{k,pn} \equiv \omega_k H_1^n + \omega_k^2 H_2^n + \ldots
+ \omega_k^{p-1} H_{p-1}^n +  H_0^n,
\lab{h0}
\end{eqnarray}
with $k=1,2, \ldots ,p-1$; and also
\begin{eqnarray}
{\hat {\cal H}}^{l+1}_{1,pn+1} &\equiv& \omega_l E_{\alpha_1}^n + \omega_l^2 E_{\alpha_2}^n +
\omega_l^3  E_{\alpha_3}^n + \ldots + \omega_l^{p-1} E_{\alpha_{p-1}}^n +
 E_{\alpha_{0}}^{n+1};\\
{\hat {\cal H}}^{l+1}_{2,pn+2} &\equiv& \omega_{2l} E_{\alpha_1+\alpha_2}^n + \omega_{2l}^2
E_{\alpha_2+\alpha_3}^n +
\omega_{2l}^3  E_{\alpha_3+\alpha_4}^n + \ldots \nonumber\\
&+& \omega_{2l}^{p-2} E_{\alpha_{p-2}+\alpha_{p-1}}^{n}
+ \omega_{2l}^{p-1} E_{\alpha_{p-1}+\alpha_0}^{n+1} + E_{\alpha_{0}+\alpha_1}^{n+1};\\
{\hat {\cal H}}^{l+1}_{3,pn+3} &\equiv& \omega_{3l} E_{\alpha_1+\alpha_2+\alpha_3}^n + \omega_{3l}^2
E_{\alpha_2+\alpha_3+\alpha_4}^n + \omega_{3l}^3  E_{\alpha_3+\alpha_4+\alpha_5}^n + \ldots \nonumber\\
&+& \omega_{3l}^{p-3} E_{\alpha_{p-3}+\alpha_{p-2}+\alpha_{p-1}}^{n}
+ \omega_{3l}^{p-2} E_{\alpha_{p-2}+\alpha_{p-1}+\alpha_0}^{n+1} \nonumber\\
&+& \omega_{3l}^{p-1} E_{\alpha_{p-1}+\alpha_0+\alpha_1}^{n+1}
+ E_{\alpha_{0}+\alpha_1+\alpha_2}^{n+1};\\
\vdots &\vdots& \vdots \nonumber\\
{\hat {\cal H}}^{l+1}_{k,pn+k} &\equiv& \omega_{kl} E_{\alpha_1+\ldots +\alpha_k}^n + \omega_{kl}^2
E_{\alpha_2+\ldots +\alpha_{k+1}}^n + \omega_{kl}^3  E_{\alpha_3+\ldots +\alpha_{k+2}}^n + \ldots
\nonumber\\
&+& \omega_{kl}^{p-k} E_{\alpha_{p-k}+\ldots +\alpha_{p-1}}^{n}
+ \omega_{kl}^{p-k+1} E_{\alpha_{p-k+1}+\ldots +\alpha_{p-1}+\alpha_0}^{n+1} + \ldots \nonumber\\
&+& \omega_{kl}^{p-1} E_{\alpha_{p-1}+\alpha_0+\alpha_1+\ldots + \alpha_{k-2}}^{n+1}
+ E_{\alpha_{0}+\alpha_1+\ldots +\alpha_{k-1}}^{n+1};\\
\vdots &\vdots& \vdots \nonumber\\
{\hat {\cal H}}^{l+1}_{p-1,pn+p-1} &\equiv& \omega_{(p-1)l} E_{\alpha_1+\alpha_2+\ldots +\alpha_{p-1}}^n
+
\omega_{(p-1)l}^2 E_{\alpha_2+\alpha_3+\ldots +\alpha_{p-1} + \alpha_0}^{n+1}\nonumber\\
& +&
\omega_{(p-1)l}^3  E_{\alpha_3+\ldots +\alpha_{p-1}+\alpha_0 + \alpha_1}^{n+1} +
\ldots
\nonumber\\
&+& \omega_{(p-1)l}^{p-1} E_{\alpha_{p-1}+\alpha_{0}+\alpha_{1}+\ldots +\alpha_{p-3}}^{n+1}
+ E_{\alpha_{0}+\alpha_1+\ldots +\alpha_{p-2}}^{n+1},
\lab{step}
\end{eqnarray}
with $l=0,1,2, \ldots ,p-1$. Here we have denoted $\omega_k \equiv \exp (2 \pi
i k/p)$, and $\alpha_i$, $i=0,1,2, \ldots ,p-1$ are the simple roots of
$\hat{sl}(p, \rlx\hbox{\,$\inbar\kern-.3em{\rm C}$} )$ with $\alpha_0 = - \psi$; $\ \psi \equiv$ is the maximal
root of
$\hat{sl}(p, \rlx\hbox{\,$\inbar\kern-.3em{\rm C}$} )$.  The elements $H_i^n \equiv 2 \alpha_i \cdot H^n/\alpha_i^2 +
\delta_{i,0} C$ with
$C$ being the centre of $\hat{sl}(p, \rlx\hbox{\,$\inbar\kern-.3em{\rm C}$} )$,
$i=0,1,2, \ldots ,p-1$ and $E_{\alpha}^n$ are the generators of  $\hat{sl}(p,
\rlx\hbox{\,$\inbar\kern-.3em{\rm C}$} )$ in
the Chevalley basis. They satisfy the commutation relations
\begin{eqnarray}
\lbrack H_a^m \, , \, H_b^n \rbrack &=& C {2\over \alpha_a^2} K_{ab} m \delta_{m+n,0} \nonumber\\
\lbrack H_a^m \, , \, E_{\pm \alpha}^n \rbrack &=& \pm K_{\alpha a} E_{\pm \alpha}^{m+n} \nonumber \\
\lbrack E_{\alpha}^m \, , \, E_{-\alpha}^n \rbrack &=&  l_a^{\alpha} H_a^{m+n} +
C {2\over \alpha^2} m \delta_{m+n,0} \nonumber\\
\lbrack E_{\alpha}^m \, , \, E_{\beta}^n \rbrack &=& \epsilon (\alpha , \beta )
E_{\alpha + \beta}^{m+n}; \, \, \, \qquad \mbox{\rm if $\alpha + \beta$ is a root of ${\cal G}$}
\nonumber\\
\lbrack d \, , \, H_b^m \rbrack &=& m H_b^m \nonumber\\
\lbrack d \, , \, E_{\alpha}^m \rbrack &=& m  E_{\alpha}^m
\lab{km}
\end{eqnarray}
with $a,b=1,2, \ldots ,p-1$;  $K_{\alpha a}= 2 \alpha \cdot \alpha_a / \alpha_a^2 = n_b^{\alpha}
K_{ba}$ with $K_{ab}$ being the Cartan matrix of ${\cal G}$;
$n_a^{\alpha}$ and $l^{\alpha}_a$ are the integers in the expansions
$\alpha = \sum_{a=1}^r n^{\alpha}_a \alpha_a$ and $\alpha /\alpha^2 = \sum_{a=1}^r l^{\alpha}_a \alpha_a /
\alpha_a^2$, and
$\epsilon (\alpha , \beta )$ are structure constants.

Notice all the generators defined above are eigenvectors of the principal
gradation operator, see \rf{gradop},
\begin{equation}
 Q_{\mbox{\rm pr.}} = \sum_{a=1}^{p-1} \lambda^v_a \cdot H + p d,
\end{equation}
i.e.
\begin{eqnarray}
\lbrack Q_{\mbox{\rm pr.}} \, , \, {\hat {\cal H}}_{k,pn}^0 \rbrack &=& pn \, {\hat {\cal H}}_{k,pn}^0, \\
\lbrack Q_{\mbox{\rm pr.}} \, , \, {\hat {\cal H}}_{k,pn+k}^{l+1} \rbrack &=&
(pn+k) \, {\hat {\cal H}}_{k,pn+k}^{l+1}.
\end{eqnarray}

We then define the subspaces in \rf{decomkm} as
\begin{eqnarray}
{\hat {\cal H}}_0 &\equiv & \{ {\hat {\cal H}}_{k,pn}^0 \, , \, k=1,2,\ldots ,p-1 \, , \,
 n \in \rlx\hbox{\sf Z\kern-.4em Z} \};\nonumber\\
{\hat {\cal H}}_{l+1} &\equiv & \{ {\hat {\cal H}}_{k,pn+k}^{l+1} \, , \, k=1,2,\ldots ,p-1 \, , \,
 n \in \rlx\hbox{\sf Z\kern-.4em Z} \} \qquad l=0,1, \ldots ,p-1.
\lab{kmsubspaces}
\end{eqnarray}

Each one of these subspaces constitutes a Heisenberg
subalgebra of $\hat{sl}(p, \rlx\hbox{\,$\inbar\kern-.3em{\rm C}$} )$. The first one, ${\hat {\cal H}}_0$, is a homogeneous
Heisenberg subalgebra and the others,  ${\hat {\cal H}}_{l+1}$, are principal ones.
The proof that this is indeed so, can be obtained by realizing the
Kac-Moody algebra \rf{km} in the following way.

Introduce the bracket
\begin{equation}
\lbrack A(z)\, , \, B(z) \rbrack \equiv \lbrack A(z)\, , \, B(z) \rbrack_{\mbox{\rm o.c.}} +
C \hbox{{1}\kern-.25em\hbox{l}} \oint {dz \over 2 \pi i } \mathop{\rm Tr} ( B(z) {d\, \over dz}A(z))
\lab{bracket}
\end{equation}
where $\lbrack \, , \,  \rbrack_{\mbox{\rm o.c.}}$ is the ordinary commutator between
matrices, i.e. $\lbrack X\, , \, Y \rbrack_{\mbox{\rm o.c.}} \equiv  X Y - Y X$. The
commutation relations \rf{km} can be obtained from this bracket by writing the
Kac-Moody generators as the matrices:
\begin{eqnarray}
\( H_a^{n}\)_{ij}&\equiv& z^n (\delta_{i,a}\delta_{j,a} - \delta_{i,a+1}\delta_{j,a+1}),\nonumber\\
\( H_0^{n}\)_{ij}&\equiv& z^n (-\delta_{i,1}\delta_{j,1} + \delta_{i,p}\delta_{j,p}) + C
\delta_{n,0}\,\delta_{ij},
\nonumber\\
\( E_{\alpha_a}^n\)_{ij}&\equiv& z^n \delta_{i,a} \delta_{j,a+1}, \nonumber\\
\( E_{\alpha_0}^n\)_{ij}&\equiv& z^n \delta_{i,p} \delta_{j,1}, \nonumber\\
\( E_{\alpha_a+\alpha_{a+1}+\ldots +\alpha_{a+k}}^n\)_{ij} &\equiv& z^n \delta_{i,a}
\delta_{j,a+k+1}
\lab{defrep}
\end{eqnarray}
with $i,j=1,2,\ldots p$; $a=1,2,\ldots p-1$ and $k=1,2,\ldots p-a-1$. The
negative root step operators are obtained from the positives ones by
transposing the matrices and inverting the power of $z$, i.e.
$E_{-\alpha}^n =\( E_{\alpha}^{-n}\)^{\dagger}$
(notice that for $k>l$, $\alpha_k + \alpha_{k+1}+\ldots +\alpha_{p-1}+\alpha_0+\alpha_1+\alpha_2+\ldots
+\alpha_l=-\alpha_{l+1}-\alpha_{l+2}-\ldots -\alpha_{k-1}$). In addition we have $d\equiv \hbox{{1}\kern-.25em\hbox{l}}
\,  z {d\, \over dz}$.

By introducing the matrix
\begin{equation}
g(z) \equiv  g\mid_{x_0=z,x_a=1} \equiv E_{\alpha_1}^0  + E_{\alpha_2}^0 + \ldots
+ E_{\alpha_{p-1}}^0 + E_{\alpha_0}^1
\lab{gz}
\end{equation}
with $a=1,2,\ldots p-1$ and $g$ given in \rf{gd} ($g(z)^p=z \hbox{{1}\kern-.25em\hbox{l}}$), we can
then write the generators \rf{h0}-\rf{step} as
\begin{eqnarray}
{\hat {\cal H}}^{0}_{k,pn} &\equiv& z^n \( \omega_k-1\) D^k + \delta_{n,0} C \hbox{{1}\kern-.25em\hbox{l}}
\nonumber\\
{\hat {\cal H}}^{l+1}_{k,pn+k} &\equiv& z^n \omega_{kl} D^{kl} (g(z))^k \sim z^n \( D^{l}
g(z)\)^k
\lab{basis}
\end{eqnarray}
with $l=0,1,\ldots p-1$ and $k=1,2, \ldots p-1$.

The orthogonality of the subspaces \rf{kmsubspaces} is shown by using the same
arguments as the ones given above in the finite dimensional case, but now
using the  bilinear form
\begin{equation}
\langle A(z) \mid B(z) \rangle \equiv
\oint {dz \over 2 \pi i z} \mathop{\rm Tr} (A(z)  B(z))
\end{equation}

Using \rf{bracket} one obtains
\begin{eqnarray}
\lbrack {\hat {\cal H}}^{0}_{j,pm} \, ,\, {\hat {\cal H}}^{0}_{k,pn}\rbrack &=& C\,
\,(\omega_{j}-1)(\omega_{p-j}-1)\,  p m\, \delta_{n+m,0} \delta_{k,p-j}
\lab{heis1}\\
\lbrack {\hat {\cal H}}^{l+1}_{j,pm+j} \, ,\, {\hat {\cal H}}^{l+1}_{k,pn+k}\rbrack &=& C\, \omega_l^{k+j+kj} \,
(pm+j)\, \delta_{m+n+1,0} \delta_{k,p-j}
\lab{heis2}
\end{eqnarray}
which constitute a set of $p+1$ orthogonal Heisenberg subalgebras.

We now comment on a connection between the decomposition above and a special
basis for the affine Kac-Moody algebras. Such a basis \cite{kklw,kac} is a
generalization of the basis constructed by Kostant \cite{kostant} for finite
dimesional simple Lie algebras. Consider the principal gradation of an affine
Kac-Moody algebra ${\hat {\cal G}}$
\begin{equation}
{\hat {\cal G}} = \oplus \, {\hat {\cal G}}_n
\end{equation}
with $n\in \rlx\hbox{\sf Z\kern-.4em Z}$ and
\begin{equation}
\lbrack Q_{\mbox{\rm pr.}} \, , \, {\hat {\cal G}}_n \rbrack = n \,{\hat {\cal G}}_n
\end{equation}
and $Q_{\mbox{\rm pr.}}$ being the grading operator \rf{gradop} corresponding
to  all $s_i$'s equal to $1$. Let ${\hat {\cal H}} \subset {\hat {\cal G}}$ be a principal Heisenberg
subalgebra with generators $E_M \in {\hat {\cal G}}_M$, where $M$ are the exponents of
${\hat {\cal G}}$, and satisfying
\begin{equation}
\lbrack E_M \, , \, E_N \rbrack = {C \over h} \mathop{\rm Tr} (E_M E_N ) M \delta_{M+N,0}
\end{equation}
These exponents have a period equal to the Coxeter number $h$ of the finite
simple Lie algebra ${\cal G}$ (of which ${\hat {\cal G}}$ is the affinization of), i.e. they
have the form $M=m_a + nh$ where $n$ is an integer and $m_a$, $a=1,2,\ldots r$
are the exponents of ${\cal G}$ \cite{kac}. In particular, unity is always an
exponent and it follows that $E_1$ is a linear combination of the simple root
step operators
$E_{\alpha_a}^0$ and $E_{\alpha_0}^1$, $a=1,2, \ldots r$. The complement ${\cal F}$ of
${\hat {\cal H}}$ in ${\hat {\cal G}}$ is such that the dimension of ${\cal F}_m \subset {\hat {\cal G}}_m$ for $m
\ne 0$ is equal to the rank of ${\cal G}$ ($\equiv r$). The subspace ${\cal F}_0$ has
dimension $r+2$ and is generated by $C$, $D$ and $H_a^0$, $a=1,2, \ldots, r$.
The complement ${\cal F}$ ad-diagonalize ${\hat {\cal H}}$ and except for the extra
two generators of ${\cal F}_0$, the basis, $F^a_m$, of
${\cal F}_m$  can be chosen such that
\begin{equation}
\lbrack E_M \, , \, F^a_n \rbrack \sim F^a_{M+n} \lab{ef}
\end{equation}

We point out that for the affine algebras $\hat{sl}(p, \rlx\hbox{\,$\inbar\kern-.3em{\rm C}$} )$ with $p$ prime,
the complement ${\cal F}$ is in fact an algebraic sum of $p$ Heisenberg
subalgebras, with $p-1$ of them being principal and $1$ being homogeneous. In
addition, the role played by ${\hat {\cal H}}$ can be replaced by any one of these $p-1$
principal Heisenberg subalgebras, with ${\hat {\cal H}}$ being now part of the new
complement ${\cal F}$.

To establish this fact we show that by choosing one the principal Heisenberg
subalgebras ${\hat {\cal H}}_{l+1}$ in \rf{kmsubspaces} to play the role of ${\hat {\cal H}}$ as
above, then the basis elements of the remaining Heisenberg subalgebras
ad-diagonalize ${\hat {\cal H}}$.

Using \rf{bracket} and \rf{basis} one gets
\begin{equation}
\lbrack {\hat {\cal H}}_{j,pm+j}^{l+1} \, , \, {\hat {\cal H}}_{k,pn}^0 \rbrack = (\omega_{kj}-1)(1-\omega_{p-k})
{\hat {\cal H}}_{j,p(m+n)+j}^{l^{\prime}+1}
\lab{comp1}
\end{equation}
where $l^{\prime}$ is the integer such that $(j(l^{\prime}-l)-k)$ is a multiple of
$p$. Notice that $l^{\prime}$ and $l$ are never equal since  $k$ is not allowed to
be a multiple of $p$ (see \rf{basis}). In addition one gets ($l\neq l^{\prime}$)
\begin{eqnarray}
\lbrack {\hat {\cal H}}_{j,pm+j}^{l+1} \, , \, {\hat {\cal H}}_{k,pn+k}^{l^{\prime}+1} \rbrack =
(\omega_{l^{\prime}}^{jk}-\omega_l^{jk}){\hat {\cal H}}_{k+j,p(m+n)+k+j}^{l^{\prime\pr}+1}
\qquad \mbox{\rm if $k+j\neq p$}
\lab{comp2}
\end{eqnarray}
where $l^{\prime\pr}$ is the integer such that $k(l^{\prime\pr}-l^{\prime}) +
j(l^{\prime\pr}-l)$ is a multiple of $p$, and (again $l\neq l^{\prime}$)
\begin{eqnarray}
\lbrack {\hat {\cal H}}_{j,pm+j}^{l+1} \, , \, {\hat {\cal H}}_{k,pn+k}^{l^{\prime}+1} \rbrack =
{\omega_{j(l-l^{\prime})}(\omega_{l^{\prime}}^{-j^2}-\omega_l^{-j^2})\over
(\omega_{j(l-l^{\prime})}-1)} {\hat {\cal H}}_{j(l-l^{\prime}),p(m+n+1)}^0
\qquad \mbox{\rm if $k+j= p$}
\lab{comp3}
\end{eqnarray}
Notice that in \rf{comp2} $l^{\prime\pr}$ can not be equal to $l$ or $l^{\prime}$,
since if it is equal, let us say, to $l$ then either $k$ or
$(l^{\prime\pr}-l^{\prime})$ would have to have $p$ in its expansion as a product of
primes, and that is not allowed. We point out that the term $\delta_{n,0}\, C\,
\hbox{{1}\kern-.25em\hbox{l}}$ in the expression of ${\hat {\cal H}}_{k,pn}^0$ given by \rf{basis}, was not
important in obtaining
\rf{heis1} or the orthogonality property. However it was crucial in
establishing \rf{comp3}.

Therefore the elements of the $p+1$ Heisenberg subalgebras \rf{kmsubspaces}
ad-diagonalize any one of the $p$ principal Heisenberg subalgebras
${\hat {\cal H}}_{l+1}$, $l=0,1, \ldots p-1$. They therefore constitute a basis for the
complement ${\cal F}$ discussed above. Notice however that the basis elements
$F^a_n$ for a fixed $a$ are not all in the same Heisenberg subalgebra.
In addition, the results of brackets between elements of two given Heisenberg
subalgebras  do not belong necessarily to a fixed third Heisenberg subalgebra,
in other words the decomposition is not multiplicative orthogonal in
the sense of ref. \cite{KKU82}.

We end this section by making a comment on a fine grading of $A_2$ \tilde c{PZ81}.
It is a $\rlx\hbox{\sf Z\kern-.4em Z}_3\times \rlx\hbox{\sf Z\kern-.4em Z}_3$ gradation where the generators have the following
gradings:
$D=(1,0)$, $D^2=(2,0)$, $g=(0,2)$, $g^2=(0,1)$, $Dg=(1,2)$, $(Dg)^2=(2,1)$,
$D^2g=(2,2)$,and $(D^2g)^2=(1,1)$, with $D$ and $g$ being the $3\times 3$
matrices defined in \rf{gd}. The addition of the gradings is given by the sum
mod $3$. The operator with grade $(0,0)$ is the identity matrix, which is not
part of the algebra $A_2$ (one has to extend to $gl(3)$ to incorporate it).
Since this gradation respects the  decomposotion \rf{decom} for $A_2$ one can
elevate it to the corresponding affine algebra. One can just attach an integer
index $n$ to the generators (eigenvalue of the derivation $d$, see \rf{km}) and
get a gradation $\rlx\hbox{\sf Z\kern-.4em Z}_3\times \rlx\hbox{\sf Z\kern-.4em Z}_3 \times \rlx\hbox{\sf Z\kern-.4em Z}$ of the Kac-Moody algebra
$A_2^{(1)}$. Such a gradation is not of the type \rf{gradop}.  However, the
decomposition in terms of Heisenberg subalgebras
\rf{decomkm} we get in this case is the same as the one we would get with the
homogeneous gradation.

\section{The case $A_{p^m-1}$ with $p$ prime and $m>1$}

In refs. \cite{KKU81}, \cite{KKU811} a construction of the orthogonal
decompositions for the algebras $A_{p^m-1}$ with $p$ prime and $m>1$ was also
performed.  In the case $m=1$ (see previous section) the basic idea was that
matrices made of powers of a given matrix $M$ commute among themselves. For the
case $m>1$ one uses the fact that, given the matrices $M_i$, $i=1,2,\ldots ,m$,
then the quantities $M_1^{n_1}\otimes  M_2^{n_2}\otimes \ldots \otimes
M_m^{n_m}$, commute among themselves under the usual commutator in the
tensor product
\begin{eqnarray}
 \lbrack A_1 \otimes  \ldots \otimes A_m \, ,\, B_1 \otimes  \ldots
\otimes B_m \rbrack \equiv
 A_1 B_1 \otimes \ldots \otimes A_m B_m -
B_1 A_1 \otimes \ldots \otimes B_m A_m
\lab{tensorcom}
\end{eqnarray}
Therefore for the complex Lie algebra $A_{p^m-1}$ with $p$ prime and $m>1$ we
introduce the following subspaces
\begin{eqnarray}
{\cal H}_0 &\equiv& \{ D^{k_1}\otimes D^{k_2}\otimes \ldots \otimes D^{k_m}\, \mid
\,  k_i=0,1,2,\ldots ,p-1  \} \nonumber\\
{\cal H}_{\bf L}&\equiv& \{ (D^{l_1}g)^{k_1}\otimes (D^{l_2}g)^{k_2}\otimes \ldots
\otimes (D^{l_m}g)^{k_m}\, \mid \, k_i=0,1,2,\ldots ,p-1 \}
\lab{subpm}
\end{eqnarray}
where $D$ and $g$ are the $p\times p$ matrices defined in \rf{gd}, and ${\bf L}
\equiv (l_1,l_2,\ldots l_{m})$, with $l_i=0,1,2,\ldots ,p-1$ . In addition the
$k_i$'s are not all allowed to have the value zero.

Notice that there are $p^m$ possible vectors ${\bf L}$ and that the dimension
of ${\cal H}_0$ and ${\cal H}_{\bf a}$ is $p^m-1$. Therefore we have $(p^m-1)(p^m+1)$
generators in those subspaces, which is the  dimension of $A_{p^m-1}$. Each
one the above subspaces is obviously abelian. For $p$ being a prime number
those generators are linearly independent and constitute a basis for
$A_{p^m-1}$. The proof for that and the orthogonality property is very similar
for the case $m=1$, given in the previous section, and for more details we
refer
to \cite{KKU811}.

In order to extend the above decomposition to the corresponding affine
Kac-Moody algebra we have to find a gradation which respects the decomposition
in terms of \rf{subpm}. We now show that the only two allowed gradations are
given by
\begin{eqnarray}
{\bf s}_{\rm hom} &=& \( 1,0,0, \ldots ,0\) , \qquad \mbox{\rm homogeneous
gradation};
\nonumber\\
{\bf s}_{\rm p} &=& \( 1,\underbrace{0, \ldots ,0,1}_{p{\rm\; entries}},
\underbrace{0, \ldots ,0,1}_{p{\rm\; entries}},\ldots\ldots
\underbrace{0, \ldots ,0,1}_{p{\rm\; entries}},
\underbrace{0, \ldots ,0}_{p-1{\rm\; entries}} \),
\lab{allgrad}
\end{eqnarray}
where we have denoted the integers $s_a$ in \rf{gradop} as the components of
the vector ${\bf s} \equiv \( s_0, s_1, \ldots ,s_{p^m-1}\)$.

To show that, it is better to express the elements of the subspaces
\rf{subpm} in terms of step operators. The elements of ${\cal H}_0$ are diagonal and
therefore are linear combinations of the Cartan subalgebra generators only.
Therefore, under \rf{gradop}, they will be sensitive to the $d$ term only.
Consequently, when affinizing the algebra what one has to do is to make sure
all Cartan subalgebras generators appearing in the expansion of the
affinized version of a given element of ${\cal H}_0$ have the same $d$-eigenvalue.
Therefore ${\cal H}_0$ does not impose any restrictions on the possible gradations.

As for the elements of ${\cal H}_{\bf L}$ we can write, using \rf{prop},
$(D^{l_1}g)^{k_1}\otimes (D^{l_2}g)^{k_2}\otimes \ldots \otimes
(D^{l_m}g)^{k_m}
\sim (D^{l_1 k_1}\otimes D^{l_2 k_2}\otimes \ldots \otimes D^{l_m k_m})
(g^{k_1}\otimes  g^{k_2}\otimes \ldots \otimes g^{k_m})$. When expanding such
an element in the Chevalley basis, the step operator content is determined by
the
term $(g^{k_1}\otimes  g^{k_2}\otimes \ldots \otimes g^{k_m})$, since the other
one just dictates which linear combination of these step operators it
corresponds to. Therefore in order to determine the gradings respecting the
decomposition \rf{subpm}, one has to consider only the $g$ part of those
elements only.

Consider the quantity
$\hbox{{1}\kern-.25em\hbox{l}} \otimes \hbox{{1}\kern-.25em\hbox{l}} \otimes \ldots \otimes\hbox{{1}\kern-.25em\hbox{l}} \otimes g$, which, written in
terms of $p^m\times p^m$ matrices,  has the form
\begin{eqnarray}
\hbox{{1}\kern-.25em\hbox{l}} \otimes \hbox{{1}\kern-.25em\hbox{l}} \otimes\ldots \otimes\hbox{{1}\kern-.25em\hbox{l}} \otimes g = \(
\begin{array}{ccccc}
g&0&0&\cdots&0\\
0&g&0&\cdots&0\\
\vdots&\vdots&\vdots&\vdots&\vdots\\
0&0&0&\cdots&g
\end{array}\)
\lab{11g}
\end{eqnarray}
where $g$ is the matrix given in \rf{gd}. Using the matrices for step
operators in the defining representation of $A_{p^m-1}$, given in \rf{defrep}
(excluding the $z$ terms), one concludes that the above element contains in
its expansion all the step operators for simple roots $E_{\alpha_a}$, except
those corresponding to $a=k p$, with $k=1,2,\ldots p^{m-1}-1$. When affinizing
such an element we will need to find integers $n_a$ and $n_b$ such that
$E_{\alpha_a}^{n_a}$ and $E_{\alpha_b}^{n_b}$ have the same grade under the
grading operator \rf{gradop}. Using arguments similar to those of the previous
section (see paragraph before eq. \rf{h0}) one then concludes that the gradings
respecting the decomposition \rf{subpm} are such that $s_a = s_b$ for all $a$
and $b$ which are not multiples of $p$.

{}From \rf{11g}, \rf{gd} and \rf{defrep} one observes that $\hbox{{1}\kern-.25em\hbox{l}} \otimes \hbox{{1}\kern-.25em\hbox{l}}
\otimes \ldots \otimes\hbox{{1}\kern-.25em\hbox{l}} \otimes g$ also contains in its expansion step
operators corresponding to the lowest root of each block, namely
$E_{-\alpha_{jp+1}-\alpha_{jp+2}-\ldots -\alpha_{jp+p-1}}$, $j=0,1,\ldots p^{m-1}-1$.
Therefore we have to find integers $n$ and ${\bar n}$ such that
$E_{\alpha_{jp+s}}^n$ ($s=1,2,\ldots p-1$) and $E_{-\alpha_{jp+1}-\alpha_{jp+2}-\ldots
-\alpha_{jp+p-1}}^{{\bar n}}$  have the same grade under the grading operator
\rf{gradop}. Therefore we need $s_{jp+s} + N n = - \sum_{q=1}^{p-1} s_{jp+q} +
N {\bar n}$, or ${\bar n} = n + {p s_{jp+s}\over N}$, where we have used the fact
that, according to the reasoning above, all $s_{jp+q}$'s ($q=1,2,\ldots p-1$)
are equal. But, by the same reasons, we have $N= s_0 + p^{m-1}(p-1) s_{jp+s} +
\sum_{k=1}^{p^{m-1}-1} s_{kp}$. Therefore, since all $s_i$'s are non negative,
it follows that ${p s_{jp+s}\over N}$ can not be an integer unless
\begin{equation}
s_{jp+s}=0, \qquad \mbox{\rm for $j=0,1,\ldots p^{m-1}-1$ and $s=1,2,\ldots
p-1$}
\lab{first}
\end{equation}
The case $p=2$ with $m=2$ may lead to other possibilities, but one can check
that due to the symmetries of the $A_{3}$ Dynkin diagram they do not lead to
gradings inequivalent to those given in \rf{allgrad}.

Consider now the quantity
\begin{eqnarray}
\hbox{{1}\kern-.25em\hbox{l}} \otimes \hbox{{1}\kern-.25em\hbox{l}} \otimes\ldots \otimes\hbox{{1}\kern-.25em\hbox{l}} \otimes g \otimes g = \(
\begin{array}{ccccc}
0&g&0&\cdots&0\\
0&0&g&\cdots&0\\
\vdots&\vdots&\vdots&\vdots&\vdots\\
0&0&0&\cdots&g\\
g&0&0&\cdots&0
\end{array}\)
\lab{gg}
\end{eqnarray}
where again $g$ is the matrix given in \rf{gd}, but now to simplify we take
all $x_i$'s  equal to unity. Using \rf{defrep} one observes that such an
element
contains in its expansion all step operators for simple roots not appering in
\rf{11g}, namely $E_{\alpha_{kp}}$ with $k=1,2,\ldots p^{m-1}-1$. Therefore we need
to find integers $n_k$ and $n_{\bar k}$ such that $E_{\alpha_{kp}}^{n^k}$ and
$E_{\alpha_{{\bar k}p}}^{n_{\bar k}}$ have the same grade under the grading
operator \rf{gradop}. Using the same arguments as above one concludes that all
$s_{kp}$'s ($k=1,2,\ldots p^{m-1}-1$) must be equal.

One observes that the step operator for the  lowest root of $A_{p^m-1}$ also
appears in \rf{gg}. Therefore we need integers $n_k$ and $n_{\psi}$ such that
$E_{\alpha_{kp}}^{n_k}$ and $E_{-\psi}^{n_{\psi}}$ ($\psi = \sum_{a=1}^{p^m-1}
\alpha_{a}$) have the same grade under the grading operator \rf{gradop}.  So, we
need $n_{\psi} = n_k + {p^{m-1} s_{kp}\over N}$, where $N=s_0 +
(p^{m-1}-1)s_{kp}$, and where we have used \rf{first} and the fact that all
$s_{kp}$'s are equal.One then concludes there are only two ways for the
ratio ${p^{m-1} s_{kp}\over N}$ to be an integer, namely
\begin{equation}
s_{kp}=0 \qquad \mbox{\rm for $k=1,2,\ldots p^{m-1}-1$}
\end{equation}
or
\begin{equation}
s_{kp}=s_0 \qquad \mbox{\rm for $k=1,2,\ldots p^{m-1}-1$}
\end{equation}
Again the case $p=2$ with $m=2$ leads to other possibilities but
they do not give inequivalent gradations. But those possibilities correspond
exactly to the two  gradations given in \rf{allgrad}. One can check that by
considering the other elements of ${\cal H}_{\bf L}$ one does not get any new
restrictions on the allowed gradations, showing that \rf{allgrad} are the only
two possibilities.

Therefore, if we use the homogeneous gradation ${\bf s}_{\rm hom}$ of
\rf{allgrad}, the subspaces \rf{subpm} will give rise to $p^m+1$ homogeneous
Heisenberg subalgebras. On the other hand if we use the gradation ${\bf s}_{\rm
p}$, the subspace ${\cal H}_0$ will lead to a homogeneous Heisenberg subalgebra,
and the subspaces ${\cal H}_{{\bf L}}$ to $p^m$ Heisenberg subalgebras associated
to the grading ${\bf s}_{\rm p}$.

The process of writing the elements of the subspaces \rf{subpm} in the
Chevalley basis is quite involved. However, there are some general patterns
which are useful. As we commented in the paragraph before \rf{11g}, the step
operator content of those elements is determined by the terms $g^{k_1}\otimes
g^{k_2}\otimes \ldots \otimes g^{k_m}$. These, when written in terms of
$p^m\times p^m$ matrices, will be made of blocks which are powers of  the
$p\times p$ matrix $g$ introduced in \rf{gd}. Using \rf{defrep} one can
check that such a structure implies the step operators appearing in the
expasion of a given element of the subspaces ${\cal H}_{{\bf L}}$ in
\rf{subpm} can be arranged in sets of $p$ step operators such that the
corresponding roots add up  to zero. Below we discuss the cases
$A_{p^m-1}$ with $m=2$ and $p=2,3$.

\subsection{The case $A_{3}$}

We denote the three simple roots of $A_{3}$ as $\alpha_a$, $a=1,2,3$, such that
$2\alpha_1 \cdot \alpha_2/ \alpha_2^2 = 2\alpha_3 \cdot \alpha_2/ \alpha_2^2=-1$ and $\alpha_1 \cdot
\alpha_3 =0$. We consider here the gradation ${\bf s}_{\rm p}$ in \rf{allgrad},
i.e.
\begin{equation}
{\bf s}_{\rm p}=(1,0,1,0)
\end{equation}
The corresponding results for the homogeneous gradation can be easily obtained
from those for  ${\bf s}_{\rm p}$. The generators for the affine subspace
${\cal H}_0$ in \rf{subpm} are obviously $H_a^n$, $a=1,2,3$.
The commutativity of the subspaces ${\cal H}_{{\bf L}}$ and their pairwise
orthogonality condition impose very rigid constraints on the
allowed combinations of step operators. Now it is a quite simple algebraic task
to convince oneself that the remaining $4$ affine subspaces ${\cal H}_{{\bf L}}$ are
\begin{eqnarray}
{\hat {\cal H}}_1 &=&
\{ (E_{\alpha_1}^n+E_{-\alpha_1}^n)+
(E_{\alpha_3}^n+E_{-\alpha_3}^n),\;
(E_{\alpha_1+\alpha_2}^n+E_{-\alpha_1-\alpha_2}^{n+1})+
(E_{\alpha_2+\alpha_3}^n+E_{-\alpha_2-\alpha_3}^{n+1}),\;\nonumber\\
& &(E_{\alpha_1+\alpha_2+\alpha_3}^n+E_{-\alpha_1-\alpha_2-\alpha_3}^{n+1})+
(E_{\alpha_2}^n+E_{-\alpha_2}^{n+1})\} ;\nonumber \\
{\hat {\cal H}}_2 &=&
\{ (E_{\alpha_1}^n+E_{-\alpha_1}^n)-
(E_{\alpha_3}^n+E_{-\alpha_3}^n),\;
(E_{\alpha_1+\alpha_2}^n-E_{-\alpha_1-\alpha_2}^{n+1})+
(E_{\alpha_2+\alpha_3}^n-E_{-\alpha_2-\alpha_3}^{n+1}),\;\nonumber\\
& &(E_{\alpha_1+\alpha_2+\alpha_3}^n-E_{-\alpha_1-\alpha_2-\alpha_3}^{n+1})+
(E_{\alpha_2}^n-E_{-\alpha_2}^{n+1})\} ;\nonumber \\
{\hat {\cal H}}_3 &=&
\{ (E_{\alpha_1}^n-E_{-\alpha_1}^n)+
(E_{\alpha_3}^n-E_{-\alpha_3}^n),\;
(E_{\alpha_1+\alpha_2}^n+E_{-\alpha_1-\alpha_2}^{n+1})-
(E_{\alpha_2+\alpha_3}^n+E_{-\alpha_2-\alpha_3}^{n+1}),\;\nonumber\\
& &(E_{\alpha_1+\alpha_2+\alpha_3}^n-E_{-\alpha_1-\alpha_2-\alpha_3}^{n+1})-
(E_{\alpha_2}^n-E_{-\alpha_2}^{n+1})\} ;\nonumber \\
{\hat {\cal H}}_4 &=&
\{ (E_{\alpha_1}^n-E_{-\alpha_1}^n)-
(E_{\alpha_3}^n-E_{-\alpha_3}^n),\;
(E_{\alpha_1+\alpha_2}^n-E_{-\alpha_1-\alpha_2}^{n+1})-
(E_{\alpha_2+\alpha_3}^n-E_{-\alpha_2-\alpha_3}^{n+1}),\;\nonumber\\
& &(E_{\alpha_1+\alpha_2+\alpha_3}^n+E_{-\alpha_1-\alpha_2-\alpha_3}^{n+1})-
(E_{\alpha_2}^n+E_{-\alpha_2}^{n+1})\} .
\label{A3}
\end{eqnarray}

The corresponding grading operator is (see \rf{gradop})
\begin{equation}
Q_{{\bf s}_{\rm p}}= {2\lambda_2\over \alpha_2^2}\cdot H + 2 d
\end{equation}
and the generators are split into odd and even eigenvalues, i.e. introducing
the subspaces
\begin{eqnarray}
{\hat {\cal G}}_{2n} &\equiv &
\{ E_{\alpha_1}^n, E_{-\alpha_1}^n; E_{\alpha_3}^n, E_{-\alpha_3}^n;
H_a^n \, \, (a=1,2,3)\} ,
\label{A3'}\\
{\hat {\cal G}}_{2n+1} &\equiv &
\{ E_{\alpha_1+\alpha_2}^n, E_{-\alpha_1-\alpha_2}^{n+1};
E_{\alpha_2+\alpha_3}^n, E_{-\alpha_2-\alpha_3}^{n+1} ,
E_{\alpha_1+\alpha_2+\alpha_3}^n, E_{-\alpha_1-\alpha_2-\alpha_3}^{n+1};
E_{\alpha_2}^n, E_{-\alpha_2}^{n+1}\} ,
\nonumber
\end{eqnarray}
one has
\begin{eqnarray}
\lbrack Q_{{\bf s}_{\rm p}} \, , \,  {\hat {\cal G}}_{2n} \rbrack
&=& 2n\, {\hat {\cal G}}_{2n} \nonumber\\
\lbrack Q_{{\bf s}_{\rm p}} \, , \, {\hat {\cal G}}_{2n+1} \rbrack
&=& (2n+1)\, {\hat {\cal G}}_{2n+1}
\end{eqnarray}

\subsection{The case $A_{8}$}

As usual we denote the simple roots of $A_{8}$ as $\alpha_a$, $a=1,2,\ldots,8$,
such that $\alpha_1$ and $\alpha_8$ correspond to the end points of the Dynkin diagram
of $A_{8}$. Again we consider the gradation ${\bf s}_{\rm p}$ in
\rf{allgrad}, i.e.
\begin{equation}
{\bf s}_{\rm p}=(1,0,0,1,0,0,1,0,0)
\end{equation}
The corresponding grading operator \rf{gradop} is then
\begin{equation}
Q_{{\bf s}_{\rm p}}= {2\lambda_3\over \alpha_3^2}\cdot H +
{2\lambda_6\over \alpha_6^2}\cdot H + 3 d
\lab{grada8}
\end{equation}

The affine subspace ${\cal H}_0$ in \rf{subpm} is again generated by
$H_a^n$, $a=1,2,\ldots ,8$, and the generators  of the $9$ affine subspaces
${\cal H}_{{\bf L}}$ are linear combinations of the following sets of step operators
\begin{eqnarray}
{\hat {\cal G}}^{(1)}_{3n}&=&
\{ E_{\alpha_1}^n, E_{\alpha_2}^n, E_{-\alpha_1-\alpha_2}^n; E_{\alpha_4}^n, E_{\alpha_5}^n,
E_{-\alpha_4-\alpha_5}^n;  E_{\alpha_7}^n, E_{\alpha_8}^n, E_{-\alpha_7-\alpha_8}^n\} ,
\nonumber\\
{\hat {\cal G}}^{(2)}_{3n+1}&=&
\{ E_{\alpha_2+\alpha_3}^n, E_{\alpha_4+\alpha_5+\alpha_6+\alpha_7+\alpha_8}^n,
E_{-\alpha_2-\alpha_3-\alpha_4-\alpha_5-\alpha_6-\alpha_7-\alpha_8}^{n+1};
\nonumber\\
& &
E_{\alpha_3+\alpha_4}^n, E_{\alpha_5+\alpha_6}^n, E_{-\alpha_3-\alpha_4-\alpha_5-\alpha_6}^{n+1};
\nonumber\\
& &
E_{\alpha_6+\alpha_7}^n,
E_{\alpha_1+\alpha_2+\alpha_3+\alpha_4+\alpha_5}^n, E_{-\alpha_1-\alpha_2-\alpha_3-\alpha_4-\alpha_5-\alpha_6-
\alpha_7}^{n+1}\} ,
\nonumber\\
{\hat {\cal G}}^{(3)}_{3n+1}&=&
\{ E_{\alpha_1+\alpha_2+\alpha_3}^{n}, E_{\alpha_4+\alpha_5+\alpha_6}^{n},
E_{-\alpha_1-\alpha_2-\alpha_3-\alpha_4-\alpha_5-\alpha_6}^{n+1};
\nonumber\\
& &
E_{\alpha_2+\alpha_3+\alpha_4}^{n},
E_{\alpha_5+\alpha_6+\alpha_7}^{n}, E_{-\alpha_2-\alpha_3-\alpha_4-\alpha_5-\alpha_6-\alpha_7}^{n+1};
\nonumber\\
& &
E_{\alpha_3+\alpha_4+\alpha_5}^{n}, E_{\alpha_6+\alpha_7+\alpha_8}^{n},
E_{-\alpha_3-\alpha_4-\alpha_5-\alpha_6-\alpha_7-\alpha_8}^{n+1} \} ,
\nonumber\\
{\hat {\cal G}}^{(4)}_{3n+1}&=&
\{ E_{\alpha_1+\alpha_2+\alpha_3+\alpha_4}^{n}, E_{\alpha_5+\alpha_6+\alpha_7+\alpha_8}^{n},
E_{-\alpha_1-\alpha_2-\alpha_3-\alpha_4-\alpha_5-\alpha_6-\alpha_7-\alpha_8}^{n+1};
\nonumber\\
& &
E_{\alpha_2+\alpha_3+\alpha_4+\alpha_5}^{n}, E_{\alpha_6}^{n}, E_{-\alpha_2-\alpha_3-\alpha_4-\alpha_5-\alpha_6}^{n+1};
\nonumber\\
& &
E_{\alpha_3}^{n}+E_{\alpha_4+\alpha_5+\alpha_6+\alpha_7}^{n}, E_{-\alpha_3-\alpha_4-\alpha_5-\alpha_6-
\alpha_7}^{n+1}\} ,
\end{eqnarray}
and $4$ other sets of elements which are conjugated to these by
$E_{\pm\alpha}^n \rightarrow E_{\mp \alpha}^{-n}$. The lower indices on ${\hat {\cal G}}$'s are
the eigenvalues w.r.t. the grading operator \rf{grada8}.
Notice that each of the $8$ subspaces ${\hat {\cal G}}$'s contains $9$ elements. The
generators of the $9$ Heisenberg subalgebras ${\cal H}_{{\bf L}}$ will be linear
combinations of the $9$ elements of each one of those subspaces. And  each one
of the $8$ elements of a given Heisenberg subalgebra comes from a
different subspace ${\hat {\cal G}}^{(i)}$.

\section{The case of $G_2$}

The authors of ref. \cite{KKU82} considered also what they called {\em a root
orthogonal decomposition} ${\cal R}{\cal O}{\cal D}$. In this case the algebra
is decomposed as in \rf{decom} with the subspace ${\cal H}_0$ being the usual Cartan
subalgebra and the remaining ones being generated by operators of the form
$E_{\alpha}\pm E_{-\alpha}$, with $\alpha$ being a positive root of ${\cal G}$. Clearly any two
roots, $\alpha$ and $\beta$ appearing in a given subspace ${\cal H}_i$, must be such that
$\pm\alpha\pm \beta$ are not  roots.

The complex Lie algebra $G_2$ was shown, in ref. \cite{KKU82}, to
possess ${\cal R}{\cal O}{\cal D}$. Denote the roots of $G_2$ as $\alpha_1$,
$\alpha_2$, $\alpha_1+\alpha_2$, $2\alpha_1+\alpha_2$, $3\alpha_1+\alpha_2$ and $3\alpha_1+2\alpha_2$, with $\alpha_1$
and
$\alpha_2$ being the simple roots. The seven subspaces in the ${\cal R}{\cal
O}{\cal
D}$ are
\begin{eqnarray}
{\cal H}_0 &=& \{ H_1 \, , \, H_2 \}\nonumber\\
{\cal H}_1^{\pm} &=& \{ E_{\alpha_1}\pm E_{-\alpha_1} \, , \, E_{3\alpha_1+2\alpha_2}\pm
E_{-3\alpha_1-2\alpha_2} \}\nonumber\\
{\cal H}_2^{\pm} &=& \{ E_{\alpha_1+\alpha_2}\pm E_{-\alpha_1-\alpha_2} \, , \, E_{3\alpha_1+\alpha_2}\pm
E_{-3\alpha_1-\alpha_2} \}\nonumber\\
{\cal H}_3^{\pm} &=& \{ E_{\alpha_2}\pm E_{-\alpha_2} \, , \, E_{2\alpha_1+\alpha_2}\pm
E_{-2\alpha_1-\alpha_2} \}
\end{eqnarray}
where $H_a$, $a=1,2$, and $E_{\pm \alpha}$ are the Cartan subalgebra generators and
step operators respectivally in the Chevalley basis. Those subspaces are
clearly abelian and orthogonal. In addition, that decomposition is also {\em
multiplicative orthogonal} in the sense that for any $i$ and $j$ there exists a
$k$ such that $\lbrack {\cal H}_i \, , \, {\cal H}_j \rbrack \subset {\cal H}_k$.

In order to extend this decomposition to the affine Kac-Moody algebra ${\hat
G_2}$ we have to find the gradings that respect the decompostion. Since the
subspaces are made of linear combinations of positive and negative step
operators associated to each root, the grading has to be such that for every
positive root $\alpha$ there are integers $n_{\pm}$ such that $E_{\alpha}^{n_+}$ and
$E_{-\alpha}^{n_-}$ have the same grades. From \rf{gradop} and \rf{grad} we see
that for a root $\alpha = m_1\alpha_1 + m_2 \alpha_2$ we need $(n_- - n_+) = 2(m_1 s_1   +
m_2 s_2)/N \equiv \mbox{\rm integer}$, where $N=s_0+3s_1+2s_2$ (since $\psi =
3\alpha_1 + 2\alpha_2$). For $\alpha \equiv \alpha_1$ we observe that this is impossible if
$s_1\neq 0$. Now taking $s_1=0$, we get for $\alpha \equiv m_1 \alpha_1 + \alpha_2$ that
$(n_- - n_+)$ is  integer only if $s_0=0$ or then if $s_2=0$. For $\alpha \equiv 3
\alpha_1 + 2 \alpha_2$ these two possibilities are also allowed. So there are two
gradations which respect the decomposition above, namely,
\begin{eqnarray}
(s_0,s_1,s_2)&=&(1,0,0) \qquad\mbox{\rm  homogeneous gradation} \nonumber\\
(s_0,s_1,s_2)&=&(0,0,1)
\end{eqnarray}
which are not equivalent, since there are no symmetries of the Dynkin diagram
relating them. We shall refer to the second gradation above as $Q_2$
gradation.

We will consider here the gradation $Q_2$, since the same results can be easily
translated to the homogeneous gradation. We define the generators
\begin{eqnarray}
{\hat {\cal H}}_{0,1}^{(2n)} &\equiv& (2 H_1^n+3 H_2^n)/\sqrt{3}  \nonumber\\
{\hat {\cal H}}_{0,2}^{(2n)} &\equiv& H_2^n  \nonumber\\
{\hat {\cal H}}_{1,1}^{\pm,(2n)} &\equiv& (E_{\alpha_1}^n \pm E_{-\alpha_1}^n)/i^{\sigma}\sqrt{3}
\nonumber\\
{\hat {\cal H}}_{1,2}^{\pm,(2n)} &\equiv& (E_{3\alpha_1+2\alpha_2}^{n-1} \pm
E_{-3\alpha_1-2\alpha_2}^{n+1})/i^{\sigma}
\nonumber\\
{\hat {\cal H}}_{2,1}^{\pm,(2n+1)} &\equiv& (E_{\alpha_1+\alpha_2}^{n} \pm
E_{-\alpha_1-\alpha_2}^{n+1})/i^{\sigma}\sqrt{3}
\nonumber\\
{\hat {\cal H}}_{2,2}^{\pm,(2n+1)} &\equiv& (E_{3\alpha_1+\alpha_2}^{n} \pm
E_{-3\alpha_1-\alpha_2}^{n+1})/i^{\sigma}
\nonumber\\
{\hat {\cal H}}_{3,1}^{\pm,(2n+1)} &\equiv& (E_{\alpha_2}^{n} \pm E_{-\alpha_2}^{n+1})/i^{\sigma}
\nonumber\\
{\hat {\cal H}}_{3,2}^{\pm,(2n+1)} &\equiv& (E_{2\alpha_1+\alpha_2}^{n} \pm
E_{-2\alpha_1-\alpha_2}^{n+1})/i^{\sigma}\sqrt{3}
\lab{basisg2}
\end{eqnarray}
where $\sigma = 0$ for the $+$ sign and $\sigma = 1$ for the $-$ sign. These
generators  are all eigenvectors of the grading operator (see \rf{gradop})
\begin{equation}
Q_2 \equiv \lambda^v_2 \cdot H + 2 d
\end{equation}
and the eigenvalues are the numbers between parentheses in the upper indices in
\rf{basisg2}.

The affine Kac-Moody algebra ${\hat G_2}$ can then be decomposed
into
\begin{equation}
{\hat G_2}={\hat {\cal H}}_0\oplus {\hat {\cal H}}_1^+\oplus {\hat {\cal H}}_1^-\oplus
{\hat {\cal H}}_2^+\oplus{\hat {\cal H}}_2^-\oplus{\hat {\cal H}}_3^+\oplus{\hat {\cal H}}_3^-
\end{equation}
with
\begin{eqnarray}
{\hat {\cal H}}_0 &\equiv& \{ {\hat {\cal H}}_{0,a}^{(2n)} \, , \, a=1,2 \, , \, n \in \rlx\hbox{\sf Z\kern-.4em Z}\} \nonumber\\
{\hat {\cal H}}_1^{\pm}&\equiv& \{ {\hat {\cal H}}_{1,a}^{\pm,(2n)} \, , \, a=1,2\, , \, n \in \rlx\hbox{\sf Z\kern-.4em Z}\}
\nonumber\\
{\hat {\cal H}}_2^{\pm}&\equiv& \{ {\hat {\cal H}}_{2,a}^{\pm,(2n+1)} \, , \, a=1,2\, , \, n \in
\rlx\hbox{\sf Z\kern-.4em Z}\}
\nonumber\\
{\hat {\cal H}}_3^{\pm}&\equiv& \{ {\hat {\cal H}}_{3,a}^{\pm,(2n+1)} \, , \, a=1,2\, , \, n \in
\rlx\hbox{\sf Z\kern-.4em Z}\}
\end{eqnarray}

Using the bilinear form
\begin{eqnarray}
\mathop{\rm Tr} ( H_a^m \, H_b^n) = \eta_{ab} \delta_{m+n,0} \qquad
\mathop{\rm Tr} ( E_{\alpha}^m \, E_{\beta}^n) = {2\over \alpha^2} \delta_{\alpha +\beta ,0}\delta_{m+n,0}
\end{eqnarray}
where
\begin{eqnarray}
\eta_{ab} \equiv {2\over \alpha_a^2} K_{ab} =
\(
\begin{array}{cc}
6&-3\\
-3&2
\end{array}\)
\end{eqnarray}
and where we have normalized the roots as $\alpha_1^2 = 2/3$ and $\alpha_2^2=2$, one
can check that the above subspaces are orthogonal.

Using \rf{km} one can check that
\begin{eqnarray}
\lbrack {\hat {\cal H}}_{0,a}^{(2m)} \, , \, {\hat {\cal H}}_{0,b}^{(2n)} \rbrack &=& C\, 2m\,
\delta_{a,b}\,\delta_{m+n,0}\nonumber\\
\lbrack {\hat {\cal H}}_{1,a}^{\epsilon ,(2m)} \, , \, {\hat {\cal H}}_{1,b}^{\epsilon ,(2n)} \rbrack &=& C\,
2m\, \delta_{a,b}\,\delta_{m+n,0}\nonumber\\
\lbrack {\hat {\cal H}}_{2,a}^{\epsilon ,(2m+1)} \, , \, {\hat {\cal H}}_{2,b}^{\epsilon ,(2n+1)} \rbrack &=&
C\, (2m+1)\, \delta_{a,b}\,\delta_{m+n+1,0}\nonumber\\
\lbrack {\hat {\cal H}}_{3,a}^{\epsilon ,(2m+1)} \, , \, {\hat {\cal H}}_{3,b}^{\epsilon ,(2n+1)} \rbrack &=&
C\, (2m+1)\, \delta_{a,b}\,\delta_{m+n+1,0}
\end{eqnarray}
where $\epsilon \equiv \pm$ and $a,b=1,2$. Therefore we have seven orthogonal
Heisenberg  subalgebras.

In addition, using the relations \rf{km} with the fact that $\epsilon (\alpha \, ,
\, \beta ) = - \epsilon (-\alpha \, ,\, -\beta )$, one can check that
\begin{eqnarray}
\lbrack {\hat {\cal H}}_i^{+} \, , \, {\hat {\cal H}}_i^{-} \rbrack &\subset& {\hat {\cal H}}_0  \nonumber\\
\lbrack {\hat {\cal H}}_0 \, , \, {\hat {\cal H}}_i^{\pm} \rbrack &\subset& {\hat {\cal H}}_i^{\mp}  \nonumber\\
\lbrack {\hat {\cal H}}_i^{\epsilon_1} \, , \, {\hat {\cal H}}_j^{\epsilon_2} \rbrack &\subset&
{\hat {\cal H}}_k^{-\epsilon_1\epsilon_2}
\end{eqnarray}
where $i,j,k=1,2,3$ and in the last relation $i\neq j\neq k$; $\epsilon_1 ,
\epsilon_2 = \pm$.

\section{Conclusion}

In the present paper we studied the decomposability problem
of the affine Lie algebras into the algebraic sum of pairwise
orthogonal Heisenberg subalgebras. In particular, the
Kostrikin--Kostrikin--Ufnarovskii theorem about the orthogonal
decomposition of some complex simple Lie algebras can be extended
to the case of the corresponding affine Lie algebras endowed with
the homogeneous gradation. However, the most interesting situation
arises when other gradations also respect such a rigid decomposition,
and we investigated this possibility in detail by the examples of the Lie
algebras of type $A_r^{(1)}$ and $G_2^{(1)}$. Unfortunately, as in the
finite--dimensional case, the consideration itself crucially depends
on an explicit realization of the algebras in question, and is performed
separately for various types of them.
So, it is very interesting to find an uniform formulation for all
algebras which admit the orthogonal (and multiplicative) decomposition.

An interesting issue stimulated by the existence of such decompositions
 concerns the representation
theory of these special affine Kac-Moody algebras. As  is well known, the
Heisenberg subalgebra is the algebra of oscillators in quantum field theory.
So, one could conjecture that representations for these algebras could be
constructed introducing a set of oscillators for each Heisenberg subalgebra in
the decomposition. These oscillators however could not be independent since the
Heisenberg subalgebras do not commute among themselves. Their commutation
relations, like the ones we calculated in the case of $A_{p-1}$ and $G_2$,
could give hints on how to relate those oscillators. These considerations could
be related to previous attempts to use division algebras, Jordan algebras,
dependent fermions etc, in the construction of representations of Kac-Moody
algebras, see e.g. \cite{jordan}.

We suspect that the orthogonal decomposition presented in section
\ref{sec2} is relevant for studying the completeness problem of the set
of the soliton  solutions to the affine Toda theories obtained in the framework
of the  technique discovered in \cite{OTU93}, especially for their nonabelian
versions discussed in \cite{OSU93}.
Moreover, the orthogonal decomposition being closely related to the
finest gradation of the algebras, could be relevant for a
construction of some multidimensional integrable nonlinear
systems in the framework of the Lie group algebraic approach \cite{ls}.
It would be also interesting to study the analogue of this
decomposition for the algebras of diffeomorphisms groups, as e.g. the
continuous limit of the $A_r$ algebra in the cyclotomic basis to that of the
area preserving diffeomorphisms on a two dimensional manifold. That could be
relevant for some problems related to an affinization of the membrane manifolds
in the context of continuous limits of the gauge theories, 3-d gravity, etc.

\vspace{.5 cm}

{\bf Acknowledgements}

\noindent The authors are grateful to  J.-L. Gervais,
 J. L. Miramontes and J.S\'anchez Guill\'en  for useful discussions;  and to A.
A. Kirillov and A. I. Kostrikin for some important comments.
We also would like to thank the Department of Physics of the
University of Wales, Swansea, where our studies of this
problem  started,  for  kind hospitality, and HEFCW for financial assistance.
M.V.S. was partially supported by the Russian
Fund for Fundamental Research, Grant \# 94-01-01585-A; and ISF, Grant  \# RMO
000; and L.A.F. was partially supported by Ministerio de Educaci\'on y Ciencia
(Spain) and FAPESP (Brazil).