Conway and Ryba considered a table of bi-infinite Fibonacci sequences
and discovered new interesting patterns.
We extend their considerations to tables that are defined by the
recurrence $X_{n+1}=dX_n+X_{n-1}$ for natural numbers $d$.
In our search
for new patterns we run into a Red Wall and exotic numeration systems.

The Fibonacci sequence $(F_n)=0,1,1,2,3,5,8,13,21,34,\ldots$ and its companion,
the Lucas sequence $(L_n)=2,1,3,4,7,11,18,29,47,\ldots$
have been examined inside out.
A few years ago, John Conway and Alex Ryba managed
to find a surprising new shape in the Fibonacci recurrence $X_{n+1}=X_n+X_{n-1}$, which they called
the Empire State Building. In this paper we adopt their point of view
and look for similar shapes in the
recurrence $X_{n+1}=dX_{n}+X_{n-1}$ for general $d$.
We will see that numeration systems
with a negative base emerge naturally from our considerations.

§ THE WYTHOFF ARRAY AND THE EMPIRE STATE BUILDING

The starting point of Conway and Ryba in [5] is the following table
of Fibonacci recurrent sequences. They call it the Garden State:

\[
\begin{NiceArray}{cccccccccccc}
0&1&1&2&3&5&8&13&21&34& \cdots&  \\
1&3&4&7&11&18&29&47&76&123& \cdots  \\
2&4&6&10&16&26&42&68&110&178& \cdots \\
3&6&9&15&24&39&63&102&165&267& \cdots \\
4&8&12&20&32&52&84&136&220&356& \cdots \\
5&9&14&23&37&60&97&157&254&411& \cdots \\
6&11&17&28&45&73&118&191&309&500& \cdots \\
7&12&19&31&50&81&131&212&343&555& \cdots \\
8&	14&	22&	36&	58&	94&	152&	246&	398&	644& \cdots \\
9&	16&	25&	41&	66&	107&	173&	280&	453&	733& \cdots \\
\vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \ddots
\CodeAfter
\tikz \draw [line width=0.6 mm] (1-|1) -- (1-|3) |- (12-|3);
%  \tikz \draw [thick] (2-|1) -- (2-|2) |- (3-|3) |- (4-|6) ;
\end{NiceArray}
\]

The first ten rows of the Garden State. If we ignore the first two columns (called the
seed and the wall), then
we get the Wythoff array in which
each natural number occurs exactly once.

Table <ref> had been encountered before, without the first two columns.
Morrison  [11] found it first, modifying an earlier array of Stolarsky [15].
He proved that each Fibonacci recurrent sequence with positive terms
occurs exactly once in the array (after deleting or adding
some initial terms).
He also proved that the
rows consist of losing positions in Wythoff's game, which is why he called it
the Wythoff array.

Kimberling [10] found a neat connection between the Wythoff array and Zeckendorf numeration.
In this numeration system, each natural number $N$
is written as a sum of non-consecutive Fibonacci numbers

\[
N=\sum_{2\leq j\leq i} d_jF_j
\]

with $F_i$ the largest Fibonacci number bounded by $N$ and $d_j\in\{0,1\}$ and $d_{j-1}=0$ if $d_j=1$.
Each number can be represented by a binary word, starting with the digit of $F_2$.
We write words with least significant digit first, or lsd.
For instance, the Zeckendorf word of $15$ is $010001$.
If we arrange the words that start and end with a $1$ in increasing lexicographical order, then we get
$1, 101, 1001, 10001, 10101, \ldots$.
These are the binary words for the first column of the Wythoff array.
The second column contains words that start with $01$,
the third contains those that start with $001$, and so forth.
To locate a number in the array, simply determine its Zeckendorf expansion.
The prefix of zeros determines the column. The suffix determines the row.

Conway and Ryba extend the Wythoff array by two columns, called
the seed and the wall.
The seed $0,1,2,3,\ldots$ numbers the floors of the
Empire State Building, counting from the top.
A seed term and a wall term determine the state and the recurrent sequence
grows from there, which is why Conway and Ryba say that this table is
a garden.

If we continue the recurrence to the left we get alternating signs.
If we take absolute values $Y=|X|$,
then we retrieve the Fibonacci recurrence $Y_{-n-1}=Y_{-n}+Y_{-n+1}$,
which now proceeds to the left.
By Morrison's result, every recurrent sequence appears in the
Wythoff array and so we eventually re-encounter the wall on the left:

\[
\begin{NiceArray}{ccccccccccc}
\cdots &13&-8&5&-3&2&-1&1&0&1&\\
\cdots &18&-11&7&-4&3&-1&2&1&3\\
\cdots &10&-6&4&-2&2&0&2&2&4\\
\cdots &15&-9&6&-3&3&0&3&3&6\\
\cdots &20&-12&8&-4&4&0&4&4&8\\
\cdots &12&-7&5&-2&3&1&4&5&9\\
\cdots &17&-10&7&-3&4&1&5&6&11\\
\cdots &9&-5&4&-1&3&2&5&7&12\\
\cdots &14	&-8	&6	&-2	&4	&2	&6	&8	&14\\
\cdots &19	&-11	&8	&-3	&5	&2	&7	&9	&16\\
\iddots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots
\CodeAfter
\tikz \draw [line width=0.6 mm] (12-|11) |- (1-|11) -- (1-|8) |- (2-|6) |- (3-|4) |- (6-|2) |- (12-|2);
\end{NiceArray}
\]

Extending the recurrence to the left. The wall reappears, with positive terms, and so
does the seed in front of it, with negative terms. The order is shuffled. For instance, the seed 5 (sixth row) appears two rows down (eighth row) next to the left wall term as -5.

Conway and Ryba prove that
the number of terms between the walls is odd, which is why it is natural
to center the table around the middle term (the pillar).
The resulting figure has the outline of a skyscraper, made up of
of blocks of constant width (number of terms between the walls)
which get longer and longer as we go down the table.

If the left wall term and the right wall term are equal, then
the row is palindromic (ignoring the signs).
Such rows are evenly spaced within each block.
By underlining these rows within the walls, the structure gets even more likeliness to the Empire State Building.
Within a block, the palindromes are either
multiples of the Fibonacci numbers or of the Lucas numbers. That is why Conway and Ryba
call them Fifi blocks and Lulu blocks. The Empire State Building has lots of structure.
What types of buildings do we get for other recursions?

\[
\tiny{
\begin{NiceArray}{rrrrrrr||r||rrrrrrr}[columns-width=auto]
\cdots &-8&5&-3&2&-1&1&0&1&1&2&3&5&8&\cdots\\
\cdots &18&-11&7&-4&3&-1&2&1&3&4&7&11&18&\cdots\\
\cdots &-16&10&-6&4&-2&2&0&2&2&4&6&10&16&\cdots\\
\cdots &-24&15&-9&6&-3&3&0&3&3&6&9&15&24&\cdots\\
\cdots &-32&20&-12&8&-4&4&0&4&4&8&12&20&32&\cdots\\
\cdots &31&-19&12&-7&5&-2&3&1&4&5&9&14&23&\cdots\\
\cdots &44&-27&17&-10&7&-3&4&1&5&6&11&17&28&\cdots\\
\cdots &23&-14&9&-5&4&-1&3&2&5&7&12&19&31&\cdots\\
\cdots &36&-22&14	&-8	&6	&-2	&4	&2	&6	&8	&14&22&36&\cdots\\
\cdots &49&-30&19	&-11	&8	&-3	&5	&2	&7	&9	&16&25&41&\cdots\\
\cdots&28&	-17&	11&	-6&	5&	-1&	4&	3&	7&	10&	17&	27&	44&\cdots\\
\cdots&41&-25&16&-9&7&-2&5&3&8&11&19&30&49&\cdots\\
\cdots&54&-33&21&-12&9&-3&6&3&9&12&21&33&54&\cdots\\
\cdots&-53&33&-20&13&-7&6&-1&5&4&9&13&22&35&\cdots\\
\cdots&-74&46&-28&18&-10&8&-2&6&4&10&14&24&38&\cdots\\
\cdots&-40&25&-15&10&-5&5&0&5&5&10&15&25&40&\cdots\\
\cdots&-61&38&-23&15&-8&7&-1&6&5&11&16&27&43&\cdots\\
\cdots&-82&51&-31&20&-11&9&-2&7&5&12&17&29&46&\cdots\\
\cdots&-48&30&-18&12&-6&6&0&6&6&12&18&30&48&\cdots\\
\cdots&-69&43&-26&17&-9&8&-1&7&6&13&19&32&51&\cdots\\
\cdots&-35&22&-13&9&-4&5&1&6&7&13&20&33&53&\cdots\\
\cdots&-56&35&-21&14&-7&7&0&7&7&14&21&35&56&\cdots\\
\cdots&-77&48&-29&19&-10&9&-1&8&7&15&22&37&59&\cdots\\
\cdots&-43&27&-16&11&-5&6&1&7&8&15&23&38&61&\cdots\\
\cdots&-64&40&-24&16&-8&8&0&8&8&16&24&40&64&\cdots\\
\cdots&-85&53&-32&21&-11&10&-1&9&8&17&25&42&67&\cdots\\
\cdots&-51&32&-19&13&-6&7&1&8&9&17&26&43&69&\cdots\\
\cdots&-72&45&-27&18&-9&9&0&9&9&18&27&45&72&\cdots\\
\cdots&-38&24&-14&10&-4&6&2&8&10&18&28&46&74&\cdots\\
\cdots&-59&37&-22&15&-7&8&1&9&10&19&29&48&77&\cdots\\
\cdots&-80&50&-30&20&-10&10&0&10&10&20&30&50&80&\cdots\\
\cdots&-46&29&-17&12&-5&7&2&9&11&20&31&51&82&\cdots\\
\cdots&-67&42&-25&17&-8&9&1&10&11&21&32&53&85&\cdots\\
\cdots&-88&55&-33&22&-11&11&0&11&11&22&33&55&88&\cdots\\

\iddots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots  & \vdots & \vdots & \vdots & \vdots & \vdots &\vdots &\ddots
\CodeAfter
\tikz \draw [line width=0.6 mm](36-|15)|-(35-|14)-|(14-|14)-|(6-|13) -| (3-|12) -| (2-|11) -| (1-|10) -- (1-|7) |- (2-|6) |- (3-|5) |- (6-|4) |- (14-|3) |-(35-|3) -|(36-|2);
\tikz \draw [line width=0.1 mm] (2-|6) -- (2-|10);
\tikz \draw [line width=0.1 mm] (3-|5) -- (3-|11);
\tikz \draw [line width=0.1 mm] (4-|5) -- (4-|12);
\tikz \draw [line width=0.1 mm] (5-|5) -- (5-|12);
\tikz \draw [line width=0.1 mm] (6-|5) -- (6-|12);
\tikz \draw [line width=0.1 mm] (10-|4) -- (10-|13);
\tikz \draw [line width=0.1 mm] (14-|4) -- (14-|13);
\tikz \draw [line width=0.1 mm] (17-|3) -- (17-|14);
\tikz \draw [line width=0.1 mm] (20-|3) -- (20-|14);
\tikz \draw [line width=0.1 mm] (23-|3) -- (23-|14);
\tikz \draw [line width=0.1 mm] (26-|3) -- (26-|14);
\tikz \draw [line width=0.1 mm] (29-|3) -- (29-|14);
\tikz \draw [line width=0.1 mm] (32-|3) -- (32-|14);
\tikz \draw [line width=0.1 mm] (35-|3) -- (35-|14);
\tikz \draw [red] [line width=0.6 mm] (1-|8) |- (2-|7);
\tikz \draw [red] [line width=0.6 mm] (2-|7) |- (4-|6);
\tikz \draw [red] [line width=0.6 mm] (4-|6) |- (6-|6);
\tikz \draw [red] [line width=0.6 mm] (6-|6) |- (9-|5);
\tikz \draw [red] [line width=0.6 mm] (9-|5) |- (22-|4);
\tikz \draw [red] [line width=0.6 mm] (22-|4) |- (36-|4);
\end{NiceArray}}
\]

The top five blocks of the Empire State. A larger figure with
more details is given in [5].  The additional
red decoration is defined in the next section..

§ OSTROWSKI ARRAYS AND THE PELL TOWER

We consider what happens if Conway and Ryba's considerations are extended to
Ostrowski numeration.
In its general form, Ostrowski numeration is based on the denominators of convergents to arbitrary
irrationals. It is particularly nice for quadratic irrationals [7],
but we will limit our attention to the irrationals $\frac{d+\sqrt{d^2+4}}{2}$.
For a fixed $d$ the denominators of its convergents satisfy the recursion $X_{n+1}=dX_n+X_{n-1}$.
If $d=1$ we get the Fibonacci numbers, and Zeckendorf numeration.
If $d=2$ we
get the Pell numbers
More generally,
let $(D_n)=1,d,d^2+1,\ldots$ be the sequence of denominators for a fixed natural number $d$.
It is known [1] that
every natural number $N$ can be represented uniquely in the form

\[
N=\sum_{1\leq j\leq i} d_jD_j
\]

where $D_i$ is the largest denominator less than $N$ and the digits $d_j$ satisfy

* $0\leq d_1<d$.

* $0\leq d_i\leq d $ for $i>1$.

* If $d_i=d$ then $d_{i-1}=0$.

The Ostrowski representation of a number is a word
with alphabet $\{0,1,\ldots,d\}$.
We say that $w$ is an Ostrowski word if each $d$ is preceded by $0$.
We say that it is trimmed if it cannot be written as $0v$
for an Ostrowski word $v$.
If we put the trimmed words in increasing alphabetical order, then they
represent an increasing sequence of numbers. By adding initial zeros, we get subsequent sequences,
which we can arrange in columns just like in
the Wythoff array.
We shall call this a $d$-Ostrowski array,
and shall often omit the $d$ in our notation.
We limit our attention to $d>1$, since Zeckendorf numeration
was fully covered in [5].

\[
\begin{NiceArray}{cccccccccc}
\vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \ddots
\CodeAfter
\tikz \draw [line width=0.6 mm] (1-|1) -- (1-|2) |- (11-|2);
%  \tikz \draw [thick] (2-|1) -- (2-|2) |- (3-|3) |- (4-|6) ;
\end{NiceArray}
\]

The first ten rows of the Pell array
(the $2$-Ostrowski array),
with an additional first column of wall terms. The table is seedless.

If $n$ is represented by the Ostrowski word $w$ then we define out$(n)$ to
be the number represented by $0w$.
The columns in the Ostrowski array are generated by iteration of this function.
We write $\alpha=\frac{d+\sqrt{d^2+4}}{2}$ and $\beta=\frac{d-\sqrt{d^2+4}}{2}$,
suppressing $d$ in our notation.
Note that $\alpha\beta=-1$ and $\alpha+\beta=d$.

For every natural number $n$ out$(n)=\lfloor \alpha n + \frac 1\alpha \rfloor$.

The denominators satisfy

\begin{equation}\label{eq:binet}
\end{equation}

from which $D_{n+1}-\alpha D_n=\beta^n$ follows. If $n=\sum_{1\leq j\leq i} d_jD_j$

\begin{equation}\label{eq:1}
\text{out}(n)-\alpha n=\sum_{1\leq j\leq i} d_j\beta^j.
\end{equation}

Even powers are positive, odd
powers are negative. The first digit is bounded by $d-1$ since it cannot be preceded by zero.
It follows that

\[
d \sum_{k=1}^\infty \beta^{2k-1} - \beta < \text{out}(n)-\alpha n < d \sum_{k=1}^\infty \beta^{2k},
\]

which is equal to

\[
\frac {d\beta}{1-\beta^2} - \beta < \text{out}(n)-\alpha n < \frac {d\beta^2}{1-\beta^2}.
\]

Now $\beta^2=d\beta + 1$ and $\beta = -\frac 1\alpha$. Therefore, out$(n)$ is the unique
integer in the interval $(\alpha n + \frac 1\alpha -1, \alpha n+\frac 1 \alpha)$.

We shall say that two sequences $(X_n)$ and $(Y_n)$ are tail equivalent if there exists an integer $j$ such
that $X_n=Y_{n+j}$ for sufficiently large $n$.
Following Morrison, we say that a table $(A_{m,n})$
that is made from positive recursive sequences
forms a Stolarsky array if it satisfies the following properties:

* Each row satisfies the recurrence $X_{n+1}=dX_n + X_{n-1}$.

* Each natural number occurs once in the table.

* For every positive recurrent sequence $(B_n)$ there exists an $m$
such that $(A_{m,n})$ and $(B_n)$ are tail equivalent.

Morrison's result extends to Ostrowski arrays.

The $d$-Ostrowski array is a Stolarsky array.

Properties 1 and 2 follow immediately from the definition of the $d$-Ostrowski array.
There exists $a>0$ and $b$ such that $X_n=a\alpha^n + b\beta^n$. For every $\epsilon>0$
there exists an $k$ such that $|X_{k+1}-\alpha X_k|<\epsilon$. In particular,
we may choose $\epsilon=-\beta$. By property 2 there exists an $A_{m,n}$ that is
equal to $X_k$ and by lemma <ref> $A_{m,n+1}=X_{k+1}$. Therefore, $A_{m,n}$
and $X_n$ are tail equivalent.

The first column $A_{1,n}$ of the $d$-Ostrowski array is equal to the
non-homogeneous Beatty sequence $\lfloor n\cdot\frac{\alpha}{\alpha-1}-\frac 1{\alpha(\alpha-1)}\rfloor$.

By lemma <ref> the numbers that can be written as out$(n)$
form the non-homogeneous Beatty sequence $\lfloor n\alpha+\frac 1\alpha\rfloor$.
The first column  contains the numbers that cannot be written in this form,
i.e., the complementary Beatty sequence.
The formula for the complementary sequence can be found in [8].

Expecting all first columns of Ostrowski arrays to be present in the OEIS is a bit much,
but it turns out that the number which does appear is quite limited.
For $d=1$ we have the upper Wythoff sequence A001950.
For $d=2$ the column
does not occur in the OEIS, although it is very close to A081031, the positions of the white keys on a
piano keyboard, given by $\lfloor \frac {12 n - 3}7\rfloor$. The reason is that the fraction
$\frac {12}7$ is a convergent of $\frac{\alpha}{\alpha - 1}$.
The complementary Beatty sequence of the first column appears in the OEIS for
$d=1$ (lower Wythoff) and for $d=2$, sequence A082845.

We added a column $A_{m,0}$ of wall terms to the Ostrowski array.
If $w=jv$ represents $A_{m,1}$, then $v$ represents $A_{m,0}$.
It may not be an Ostrowski word.

For $d>1$ the sequence $A_{m,0}$ of wall terms is equal to
$\lfloor \frac{m\alpha}{\alpha+1}\rfloor$.

A wall term is represented by the word that labels its row, with the initial digit deleted.
This is either
a legal Ostrowski word (indeed, all of them occur) or an illegal word with prefix $d$.
Therefore, the wall terms contain all non-negative integers with some repetitions
because of the illegal words.
These repetitions can be obtained from a legal $0w$ by replacing the initial digit zero
by $d$, if we allow the empty word for the first row.
The repetitions occur at out$(w)+d$, which by lemma <ref> is equal to

\[
\left\lfloor (n-1)\alpha+\frac 1\alpha\right\rfloor+d=\lfloor n\alpha\rfloor.
\]

Here we put $n-1$ to start the count at zero, to include the empty word $w$.
The number of repetitions up to but not including
$k$ is equal to $\lfloor \frac{k}{\alpha}\rfloor$.
Therefore, at index $m=k+\lfloor \frac{k}{\alpha}\rfloor$ we have $A_{m,0}=k$.
Expressing $k$ in terms of $m$ we find that
k=\frac {(m+\epsilon)\alpha}{1+\alpha}
for some $0<\epsilon<1$. Thus, if $k$ occurs first at index $m$, then

\[
A_{m,0}=\left\lfloor \frac {m\alpha}{\alpha+1}\right\rfloor.
\]

If $k$ repeats at the next index $m+1$, then $k=n\alpha-\epsilon$ for some $n$ and $0<\epsilon<1$.
Therefore $\lfloor \frac k\alpha\rfloor=n-1=\frac k\alpha+\frac \epsilon\alpha -1$.
Now by the same argument as above, $m+1=k+\lfloor \frac{k}{\alpha}\rfloor+1=k+\frac k\alpha+\frac\epsilon\alpha$.
We get $A_{m+1,0}=\left\lfloor \frac {(m+1)\alpha}{\alpha+1}\right\rfloor.$

For $d=2$ we have sequence A049472. For $d=3$ the sequence is very similar to
A093700, which is $\lfloor n\gamma\rfloor$ for $\gamma=-\log_{10}{(3-\sqrt 8)}$.
This is because $\alpha/(\alpha+1)$ is very close to $\gamma$.

Let $A_{m,n}$ be an Ostrowski array. For a fixed $k$ and running index $n$, the differences $A_{k,n+1}-\alpha A_{k,n}$ have alternating signs and diminish by a factor $\frac 1\alpha$.

Let $n$ be a number in that row and write $r=\text{out}(n)-\alpha n$. By equation <ref>, the consecutive difference
in the row satisfies $\text{out}^2(n)-\alpha\cdot\text{out}(n)=\beta r$. The sign of $\beta$ is negative
and it is equal to $\frac 1\alpha$ in absolute value.

With the formulas for the wall and the first column in hand, it
is easy to extend the Ostrowski array to the left.
By theorem <ref> the wall reappears (only once)
but it does not have the regular structure of the Empire State.
The extension of the Pell array produces a terrace building which we call the Pell Tower.

\[
\begin{NiceArray}{rrrrrrrrrrl}[columns-width=auto]

\cdots&-70&29 &-12 &5 &-2 &1 &0 &1 &\cdots&1\\
\cdots&-41&17 &-7 &3 &-1 &1 &1 &3 &\cdots&11\\
\cdots&58&-24 &10 &-4 &2 &0 &2 &4 &\cdots&02\\
\cdots&-82&34 &-14 &6 &-2 &2 &2 &6 &\cdots&101\\
\cdots&-53&22 &-9 &4 &-1 &2 &3 &8 &\cdots&111\\
\cdots&46&-19 &8 &-3 &2 &1 &4 &9 &\cdots&021\\
\cdots&-94&39 &-16 &7 &-2 &3 &4 &11 &\cdots&102\\
\cdots&-65&27 &-11 &5 &-1 &3 &5 &13 &\cdots&1001\\
\cdots&-36&15 &-6 &3 &0 &3 &6 &15 &\cdots&1101\\
\cdots&-63&-26 &11 &-4 &3 &2 &7 &16 &\cdots&0201\\
\cdots&-77&32 &-13 &6 &-1 &4 &7 &18 &\cdots&1011\\
\cdots&-48&20 &-8 &4 &0 &4 &8 &20 &\cdots&1111\\
\cdots&51&-21 &9 &-3 &3 &3 &9 &21 &\cdots&0211\\
\cdots&-89&37 &-15 &7 &-1 &5 &9 &23 &\cdots&1021\\
\cdots&-60&25 &-10 &5 &0 &5 &10 &25 &\cdots&1002\\
\cdots&-31&13 &-5 &3 &1 &5 &11 &27 &\cdots&1102\\
\cdots&68&-28 &12 &-4 &4 &4 &12 &28 &\cdots&0202\\
\cdots&-72&30 &-12 &6 &0 &6 &12 &30 &\cdots&10001\\
\cdots&-43&18&-7&4&1&6&13&32&\cdots&11001\\
\cdots&56&-23&10&-3&4&5&14&33&\cdots&02001\\
\cdots&-84&35&-14&7&0&7&14&35&\cdots&10101\\
\cdots&-55&23&-9&5&1&7&15&37&\cdots&11101\\
\cdots&44&-18&8&-2&4&6&16&38&\cdots&02101\\
\cdots&-96&40&-16&8&0&8&16&40&\cdots&10201\\
\cdots&-67&28&-11&6&1&8&17&42&\cdots&10011\\
\cdots&-38&16&-6&4&2&8&18&44&\cdots&11011\\
\cdots&61&-25&11&-3&5&7&19&45&\cdots&02011\\
\cdots&79&33&-13&7&1&9&19&47&\cdots&10111\\
\cdots&-50&21&-8&5&2&9&20&49&\cdots&11111 \\
\cdots&49& -20&9&-2&5&8&21&50&\cdots&02111 \\
\cdots&-83& 38&-15&8&1&10&21&52&\cdots&10211 \\
\cdots&-62& 26&-10&6&2&10&22&54&\cdots&10021 \\
\cdots&-33& 14&-5&4&3&10&23&56&\cdots&11021 \\
\cdots&66& -27&12&-3&6&9&24&57&\cdots&02021 \\
\cdots&-74& 31&-12&7&2&11&24&59&\cdots&10002 \\
\cdots&-45& 19&-7&5&3&11&25&61&\cdots&11002\\
\cdots&54& -22&10&-2&6&10&26&62&\cdots&02002 \\
\cdots&-86& 36&-14&8&2&12&26&64&\cdots&10102\\
\cdots&-57& 24&-9&6&3&12&27&66&\cdots&11102 \\
\cdots&42& -17&8&-1&6&11&28&67&\cdots&02102 \\
\cdots&-98& 41&-16&9&2&13&28&69&\cdots&10202\\
\cdots&-69&29&	-11&	7&	3&	13&	29&	71&\cdots&100001\\

\iddots&\vdots & \vdots & \vdots  & \vdots & \vdots & \vdots & \vdots & \vdots &\ddots
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The Pell Tower and its irregular left wall. On the right
we added a final column that
contains the word $w$ that generates the row. Palindromic rows are underlined and they clearly
do not occur at regular distances as they do in the Empire State Building, although their
number of occurrences in blocks can be specified.
The distance between the right wall and the red wall is equal to $|w|$.

An inspection of this building suggests various

* The distance between the walls is either $|w|$ or $|w|+1$, where $|w|$ denotes the length
of the word that generates the row. Distance $|w|+1$ appears to be prevalent. This is illustrated by the
red wall at distance $|w|$ from the right wall, where we chose red since this is the color for
negative numbers.

* Columns on the left contain positive and negative numbers. The sign depends on whether $w$ starts with $02$ or not.

* All integers (positive and negative, but not zero) appear to the left of the red wall.

We shall see that these observations can be substantiated by properties of the negative Pell numbers $P_{-n}$ for $n\geq 1$.
It turns out that the $P_{-n}$, or more generally the denominators $D_{-n}$,
constitute a basis of a numeration system.
We refer to these numbers as negative denominators, meaning that their index is negative.
Negative denominators have alternating signs, just like $(-b)^n$ for a negative
base $-b$.
Systems with negative bases have been extensively studied, see [9] and references therein.
The following result appears to be new, but we will see further
down that it can be connected to the numeration system with base $-\alpha$.
It complements a result of Bunder who proved that the negative Fibonacci numbers
$F_{-1}, F_{-2},\ldots$ form a numeration system [3].

Let $d>1$ be fixed.
Every integer $N$ (positive or negative) can be represented uniquely in the form

\begin{equation}\label{eq:2}
N=\sum_{1\leq j\leq i} d_jD_{-j}
\end{equation}

with digits $d_j\in \{0,1,\ldots, d\}$ such that $d_{i+1}=0$ if $d_{i}=d$.
If $N$ is represented by $w$ then the sign of $N$ is equal to $(-1)^{|w|+1}$.

Suppose that $|w|=2j+1$ is odd. We will prove that $D_{2j}+1\leq N\leq D_{2j+2}$, that
there are $D_{2j+2}-D_{2j}$ words of this length, and that none of them represents
the same number. This will settle the case of words of odd length. The words of even
length are left to the reader.

The even indices correspond to the negative denominators.
Under the fixed word length $2j+1$, the largest number is represented by $d0d0\cdots 0d$
and the smallest number is represented by $0d0d0\cdots (d-1)1$.
The largest number is

\[dD_1+dD_3+\cdots + dD_{2j+1}=D_2+dD_3+\cdots+dD_{2j+1}=D_4+dD_5+\cdots dD_{2j+1}.\]

By a cascade of recursions this works out to $D_{2j+2}$. Similarly, the smallest number is

\[-dD_2-dD_4-\cdots - dD_{2j}+D_{2j+1}+D_{2j},\]

which is equal to

\[D_1-D_3-dD_4-\cdots +D_{2j+1}-D_{2j}=D_1-D_5-dD_6-\cdots +D_{2j+1}+D_{2j}.\]

By a cascade of recursions this works out to $D_1+D_{2j}$. We conclude that $D_{2j}+1\leq N\leq D_{2j+2}$.
We leave it to the reader to check that for even word length $2j$ we find $-D_{2j+1}+1\leq N\leq -D_{2j-1}$.
In particular, words of different lengths represent different numbers.

Let $v$ be any admissible word of length $n$ and let $x$ be any non-zero digit. Then $vx$ is an admissible
word of length $n+1$ unless the final digit of $v$ is equal to $d$. In this case, we replace $v=ud$ by $\bar v=u0$,
which has suffix $00$. The words $\bar v x$ cover all words of length $n+1$ except those of the form $t0v$ for
words $t$ of length $n-1$. Therefore $W_{n+1}=dW_n+W_{n-1}$ where $W_n$ is the number of words of length $n$.
We have $W_0=0=D_1-D_{-1}$ and $W_1=d=D_2-D_0$. Therefore $W_n=D_{n+1}-D_{n-1}$.

Finally, suppose that different admissible words $v$ and $w$ represent the same number.
By subtracting an equal amount from their digits $v_i$ and $w_i$, we may suppose that at least
one of these digits is equal to zero. This implies that $v$ and $w$ have different lengths,
but we already found that this implies that they represent different numbers.

Note that in these words each $d$ is followed by a zero.
In an Ostrowski word, each $d$ is
preceded by a zero. If we take the usd instead of
the lsd representation, then we get
an Ostrowski word $w$, unless it starts with a $d$. However,
we can replace $w$ by $0w$, because in usd it represents the same number
anyway. From now on, if we speak about the usd representation, we assume
that its initial digit is either equal to $0<j<d$, or its initial two
digits are $0d$. In this way, the words that label the rows in table <ref>
can also be read as usd representations of terms starting from the red wall.

Each integer occurs exactly once to the left of the red wall.

The rows are labelled by Ostrowski words $w$. Each integer has a unique usd representation
$w0^k$. To get to the red wall, we need to take $|w|$ steps to the left of the right wall.
The number immediately to the left of the red wall has usd representation $w$. If we take
$k$ further steps, we get to $w0^k$. The length of the word determines the sign.
This substantiates observations 2 and 3.

The wall terms of the Ostrowski array are the numbers $A_{m,0}$.
The red wall terms are the numbers $A_{m,n}$ that occur right outside the red wall (as viewed from within the building,
taking the terrace to be outside).
We represent them as $A_{m,r}$, suppressing $m$ in our notation. The number of terms within the building
is $1-r$.
We call the array to the left of the red wall the negative Ostrowski array
\[\bar A_{m,n}=A_{m,r-n}.\]

It is the natural counterpart of
the Ostrowski array. Inhabitants of the Pell Tower enjoy the view of these two arrays.
The operation $n\to\text{out}(n)$ moves one step to the right in the Ostrowski array.
Its counterpart $n\to\text{nut}(n)$ takes one step to the left in
the negative Ostrowski array.

For any integer $n$ we have nut$(n)=\lceil -n \alpha \rceil$.

As in the proof of lemma <ref> we compute $D_{-n-1}+\alpha D_{-n}=\alpha^{-n}$.
If $n=\sum_{1\leq j\leq i} d_jD_{-j}$
then nut$(n)+\alpha n=\sum_{1\leq j\leq i} d_j\alpha^{-j}>0$.
The sum $\sum_{1\leq j\leq i} d_j\alpha^{-j}$ is maximized by
taking all digits equal to $d$ at odd indices and zero at even indices.
Since $d(\alpha^{-1}+\alpha^{-3}+\cdots)=1$ we get
that nut$(n)+\alpha n<1$.

Surprisingly, the negative base representation is even easier to compute than the
ordinary Ostrowski representation, thanks to its connection with base $-\alpha$
through lemma <ref>.
To find the lsd representation of $N_1$, compute $N_2=\lceil -N_1/\alpha\rceil$ and put
$x_1=N_1-\lceil -N_2\alpha\rceil$.
It follows from the lemma that $N_1-x_1$ is represented by $\text{nut}(N_2)$,
hence the representation of $N_1$ has first digit $x_1$. Continue with $N_2$ to find digit $x_2$, etc.
Terminate as soon as $0< N_k \leq d$ and put $x_k=N_k$. This is a standard digit generating
known as the greedy expansion [6]. To see that it produces Ostrowski words,
observe that digit $d$ occurs only if $N_1$ is at the end of $(-N_2\alpha, -N_2\alpha+\alpha)$
which has length $\alpha-d=\frac 1\alpha$. In particular $-N_2\alpha=N_1-d-\epsilon$
for some $\epsilon<\frac 1\alpha$. Therefore

\[
\frac {-N_2}{\alpha}=-N_2\alpha+dN_2=N_1-d-\epsilon + dN_2,
\]

which rounds up to $N_3=N_1-d+dN_2$ with digit

\[
x_2=N_2-\lceil -N_3\alpha\rceil<\epsilon \alpha
\]

If $x_1=d$ then $x_2=0$. The greedy expansion produces Ostrowski words.

Fix $m$ and consider the row $A_{m,n}$ in the Ostrowski array. There exists a
row $\bar A_{k,n}$ in the negative Ostrowski array and a number $i\in\{0,1\}$
such that $A_{m,n}=|\bar A_{k,i+n}|$.

Let $N=A_{m,1}$ be the first term of the sequence. Then both $-N$ and $N$ occur in the negative Ostrowski array.
Their consecutive terms are, respectively, $\lfloor N\alpha\rfloor$ and $\lfloor -N\alpha \rfloor$.
In absolute value these terms are $\lfloor N\alpha\rfloor$ and
$\lfloor N\alpha \rfloor$. One of these two absolute values is equal to $A_{m,2}$.
Since two recursive sequences are equal if they have two identical terms, we conclude that $A_{m,n}$
occurs (possibly as a tail) in the negative Ostrowski array.

By lemma <ref> we have $||\bar A_{k,2}|-|\alpha \bar A_{k,1}||<1$.
In analogy of equation <ref> we get that consecutive differences $\text{nut}(n)+\alpha n$
diminish by $\frac 1\alpha$.
The next difference will be $<\frac 1 \alpha$. Hence, $|\bar A_{k,2}|$ and $|\bar A_{k,3}|$
are consecutive terms in the Ostrowski array.
Therefore, the row $A_{m,n}$ starts either at the first or second index of $\bar A_{k,n}$.
This substantiates a part of the first observation.

Let $(C_n)$ be.the absolute values of the terms $A_{m,b}$ where the red wall and the left wall coincide.
The complementary series of $(C_n)$ is given by $\lfloor n\alpha \rfloor \cup \lceil n\alpha\rceil$.
These are the absolute values of the numbers on the terraces.

The terms in the Ostrowski array make up $\mathbb N$. The terms in the negative array
make up $-\mathbb N\cup\mathbb N$.
Every row in the Ostrowski array occurs once in the negative array, possibly skipping
the first term.
The absolute values of the terms that are skipped form the natural numbers,
and they are all in the first column of the negative array.
Therefore, $|\bar A(m,1)|$ contains all natural numbers, with some doubles.

A row in the Ostrowski array skips the first term if, and only if,
$A(m,1)=\lceil -n \alpha \rceil$ for some $n$ by lemma <ref>.
The doubles in $|\bar A(m,1)|$ are all natural numbers that are
not of this form. In other words, the doubles are unequal to $n\alpha$
rounded up or down.
The density of the sequence $(C_n)$ for $d>1$ is equal to $1-\frac 2\alpha$ which is approximately
$0.172$ if $d=2$. This substantiates the remaining part of the first observation.

For the Pell array $(C_n)$ is given by A276879 in the OEIS. For larger
values of $d$, the density of $(C_n)$ increases to one and almost half of the red wall
coincides with the left wall.

We extend the notion of a Stolarsky array to include recursive sequences that contain negative numbers:

* Each row satisfies the recurrence $X_{n+1}=-dX_n + X_{n-1}$.

* Each non-zero integer occurs once in the table.

* For every recurrent sequence $(B_n)$ there exists an $m$
such that $(A_{m,n})$ is tail equivalent to $(B_n)$ or to $(-B_n)$.

Our previous results imply:

The negative Ostrowski array is a Stolarsky array.

The sequence of denominators $(D_n)$ is palindromic and so is the sequence
$(E_n)$ given by $\cdots,d^2+2,-d,2,d,d^2+2,\cdots$.
These are the so-called companion numbers [2].
It is not hard to prove that
all palindromic sequences are multiples of $(D_n)$ or $(E_n)$, if
we allow multiples to be halves if $d$ is even.
Following the Fifis and Lulus from [5],
let's call the multiples of $(D_n)$ Deedees and call the multiples of $(E_n)$

The Empire State Building is divided in blocks, counting from zero,
where block $k$ consists
of all rows that are labelled by words of length
$|w|=2k$ or $|w|=2k+1$.
Fifis occur in the even blocks and Lulus occur in the odd blocks.
We modify this definition for
the Pell Tower and define block $k$ to contain the
rows labelled by
words of length $|w|=2k-1$ or $|w|=2k$.
The distribution of Deedees and Edees over these blocks is not
as nice as for the Empire State Building, but we can still locate
how many there are per block.

The number of Deedees in block $k$ is equal to the number of times $k$
occurs in the sequence $\lceil\log_{\alpha}(n)\rceil$.
Let $D=d^2+4$ for
odd $d$ and $D=(d/2)^2+1$ for even $d$.
The number of
Edees in block $k$ is equal to the number of times $k$ occurs in

Consider $(jD_n)$ for some fixed $j$.
The first term of this sequence in the negative Ostrowski array
occurs at index $-i$ such that
$jD_{-i-1}+\alpha jD_{-i}<1< jD_{-i}+\alpha jD_{-i+1}$.
A straightforward computation using equation <ref> gives
\[\alpha^{i-1}<j<\alpha^i,\]

and so $i=\lceil\log_{\alpha}(j)\rceil$.
The first term of $(jD_n)$ that is in the (positive) Ostrowski array is either
$jD_i$ or $jD_{i+1}$. We conclude that the number of terms in between the blue wall and the
right wall is either $2i-1$ or $2i$. Hence $(jD_n)$ is in block $\lceil \log_{\alpha}(j)\rceil$.

The companion numbers are given by

\begin{equation}\label{eq:comp}
\end{equation}

The first term of the sequence $(jE_n)$ that occurs in the negative array is at $-i$ such that
$jE_{-i-1}+\alpha jE_{-i}<1< jE_{-i}+\alpha jE_{-i+1}$. A straightforward computation using
equation <ref> gives

\[\alpha^{i-1}<j\sqrt{\Delta}<\alpha^i,\]

where $\Delta=d^2+4$,
and so $i=\lceil\log_{\alpha}(j\sqrt{\Delta})\rceil$.
If $d$ is even, then the multiples are halves and we
need to adjust by dividing $\Delta$ by four.

We conclude this section with some remarks on the case
of Fibonacci numbers.
We do not supply proofs, as they are either very similar to
the proofs above or they are consequences of the results
of Conway and Ryba.
Bunder proved that each integer can be written as $N=\sum_{1\leq j\leq i} d_jF_{-j}$
for digits $d_j\in\{0,1\}$ such that $d_{j+1}=0$ if $d_j=1$.
This is the analogue of our theorem <ref>.
Bunder's algorithm Z to determine the expansion is not very complicated, but it
involves a few different operations and the proof of its correctness requires a bit of work.
There is a simpler algorithm!
The analogue of our lemma <ref> holds for the negative base $-\gamma$,
where $\gamma=\frac{1+\sqrt 5}2$ is the golden ratio.
To determine the negative Zeckendorf representation, divide and round by $-\gamma$ and terminate at one.
We can also put a red wall
within the Empire State Building. Since the array involves a seed, which is absent
for Ostrowski arrays for $d>1$, the distance between the red wall and the left wall
is either 1 or 2. The $n$-th block (counting blocks from zero)
is divided into two parts, starting with $F_{2n}$ rows
of distance 2 and ending with $F_{2n+2}$ rows of distance 1.
The negative Wythoff array is a Stolarsky array.

§ OSTRONOMETRY

The Fibonacci numbers and the Lucas numbers satisfy many interesting
identities. The  oldest seems to
be Cassini's identity from 1680, if not earlier:

\[
\]

This was only the start of an ever growing list.
Bicknell [2] observed that these identities all
extend to our denominators $D_n$, and we retrace her footsteps. For instance, Cassini's identity extends to

\[
\]

In these generalized identities the companion numbers $E_n$
take the role of the Lucas numbers.

This conversion depends on a trick involving trigonometry,
which is why Conway and Ryba call it Fibonometry [4].
The trick is apparently due to Vajda [16] and extends to $D_n$ and $E_n$
and it seems natural to call this Ostronometry.
By definition

\[
\sin(nz)=\frac{e^{inz}-e^{-inz}}{2i}\ ,\ \ \ \cos(nz)=\frac{e^{inz}+e^{-inz}}2
\]

The recursive sequence
that starts with $2$ and $d$ produces the companion
numbers $(E_n)$. We write $\Delta=d^2+4$. Vajda's trick
puts $z=\frac {\pi} 2 - i\log(\alpha)$.
By equations <ref> and <ref>

\begin{equation}\label{eq:trick}
\sin(nz) = \frac {D_n \sqrt{\Delta}}{2}\cdot i^{n-1} \ ,\ \ \ \cos(nz)=\frac{E_n}{2}\cdot i^n.
\end{equation}

If $d=1$ then $D_n$ and $E_n$ are the Fibonacci and Lucas numbers.
Now trigonometric equations transform into Cassini-like identities.
Fibonacci identities transform into Ostrowski identities under
$F_n\to D_n\cdot\sqrt{d^2+4}/\sqrt{5}$ and $L_n\to E_n$.
For instance, the fundamental equation $\cos^2(nz)+\sin^2(nz)=1$ becomes

\[
E_n^2-\Delta D_n^2=(-1)^n 4,
\]

and we see that $(D_n,E_n)$ solves a Pell equation. Cassini's identity
follows from

\begin{equation*}
\sin^2(x) -  \sin^2(y) = \sin(x + y) \sin (x - y),
\end{equation*}

if $x=nz$ and $y=z$.
The trigonometric Jacobi identity from [5]
is equal to

\[
\sin(a)\sin(b-c)+\sin(b)\sin(c-a)+\sin(c)\sin(a-b)=0.
\]

It should be read in terms of $az, bz, cz$ and transforms to

\[
\]

Cassini's identity is the special case in which $a,b,c$ is equal to $n+1,n,1$.
A generalization of Cassini's identity, which is sometimes named after d'Octagne, is

\[
\]

It is another consequence of the trigonometric Jacobi equation. By Ostronometry we get

\begin{equation}\label{eq:3}
\end{equation}

Ostronometry can also be used to demonstrate divisibility properties of
the denominators. It is well known that $F_d$ divides $F_n$ if
$d$ divides $n$. By Fibonometry, this follows from the
fact that by De Moivre $\sin(nz)$ is a sum of $\sin^j(z)\cos^{n-j}(z)$ for odd $j$.
More can be said. If we take $m=n+1$ in equation <ref> then we find that
two consecutive denominators
are relatively prime. Therefore $\gcd(D_m,D_n)=\gcd(D_{m-n},D_n)$ and by a run
of the lazy Euclidean algorithm, which subtracts one at a time, we conclude that
$\gcd(D_m,D_n)=D_{\gcd(m,n)}$. In particular, $D_n$ divides $D_m$ if and only
if $n$ divides $m$, which is a well known fact for Fibonacci numbers.
Another fun fact is Carmichael's theorem, which says that the product of any $n$ consecutive Fibonacci
numbers is divisible by $F_1\cdots F_n$. It is a consequence of d'Octagne's
identity <cit.> and hence extends to $D_n$.

Vajda's trick replaces $e^{iz}$ by a fundamental
solution of the Pell equation $|X^2-\Delta Y^2|=1$.

\[
\cos(z)+i \sin(z)\longleftrightarrow \frac {E_1 + D_1\sqrt {\Delta}}{2}\cdot i.
\]

The reason why this works is its stability under $n$-th powers

\[
\cos(nz)+i \sin(nz)\longleftrightarrow \frac {E_n + D_n\sqrt {\Delta}}{2}\cdot i^n.
\]

The left-hand side follows from De Moivre's identity. The right-hand side
follows from the fact that solutions of
the Pell equation form a cyclic group.
Vajda uses the hyperbolic sine and cosine, which give cleaner formulas,
but their identities are less familiar.

It is possible to adapt Vajda's trick to other rows in the Ostrowski table, although
it gets a little cumbersome. Pick a row $m$ in the Ostrowski table
and write $Y_n=A_{m,n}$. The generalized Binet formula gives

\[
Y_n = \left(a+\frac{b}{\sqrt{\Delta}}\right)\alpha^n\ +\ \left(a-\frac{b}{\sqrt{\Delta}}\right)\beta^n
\]

for $2a=Y_0$ and $ad+b=Y_1$. The numbers $Y_n$ need companions $X_n$ to solve a Pell
equation $X_n^2 - \Delta Y_n^2 = (-1)^{n}C$ for some constant $C$. Asymptotically, $X_n$ needs to be
equal to $\sqrt{\Delta}Y_n$ and so

\[
X_n = \left(a\sqrt{\Delta}+b\right)\alpha^n\ +\ \left(-a\sqrt{\Delta}+b\right)\beta^n.
\]

We have that

\begin{equation}\label{eq:norm}
\end{equation}

By taking the norm it follows that the pairs $(X_n,Y_n)$ solve the Pell equation above with $C=X_0^2-Y_0^2\Delta$.
Choose $\phi$ such that

\[
\sin(\phi) = \frac { Y_0 \sqrt{\Delta}}{i\sqrt{C}} \ ,\ \ \ \cos(\phi)=\frac{X_0}{\sqrt{C}}.
\]

Vajda's trick in equation <ref> combined with equation <ref> gives

\[
\frac{X_n+Y_n\sqrt{\Delta}}{\sqrt C}=(\cos(\phi)+i\sin(\phi))(\cos(nz)+i\sin(nz))i^{-n}.
\]

According to equation <ref> we have $Y_n=X_0D_n+Y_0E_n$, which corresponds to
the computation of the imaginary part of $(\cos(\phi)+i\sin(\phi))(\cos(nz)+i\sin(nz))$.
From this we find $Y_n$ and hence $X_n$.
Vajda's trick for Ostrowski rows is

\begin{equation}\label{eq:modi}
\sin(nz+\phi) = \frac {Y_n \sqrt{\Delta}}{\sqrt{C}}\cdot i^{n-1} \ ,\ \ \ \cos(nz+\phi)=\frac{X_n}{\sqrt{C}}\cdot i^n.
\end{equation}

It is a little more cumbersome since it has an extra angle $\phi$ and a constant $C$. For instance,
if we read $a$ as $az+\phi$, $b$ as $bz+\phi$ and $c$ as $z$, then Jacobi's trigonometric identity transforms to

\[
Y_bY_{a-1}-Y_aY_{b-1}=(-1)^bD_{a-b}\cdot \frac C4.
\]

§ CONCLUDING REMARKS

The notion that negative numbers were once considered satanic is a modern myth [12],
but it is fair to say that they have not received as much attention as positive numbers.
We should treat all numbers, negative and positive, odd and even, with equal respect, regardless
of orientation or parity.
The results in this paper underline this.

It would be nice if the results in this paper can be extended, in one form or another,
to Ostrowski numeration systems for convergents of quadratic irrationals.
In particular, the irrationals with a purely periodic continued fraction
deserve attention.

Our understanding of the combinatorial and dynamical properties of
negative beta-expansions has progressed tremendously
thanks to works of Charlier, Frougny, Ito, Pelantová, Steiner, and many others.
An overview of the literature with open problems is given in <cit.>.
Negative bases have recently been implemented in
the automatic theorem prover Walnut, which already devoured its first
conjectures [13].

Overt nationalism has regained respectability once again, so
let me highlight the abundance of Dutch mathematicians  in this
paper. Maarten Bunder, Gerrit Lekkerkerker
(who preceded Edouard Zeckendorf), John Pell, and Willem Wythoff all studied or
worked at the
University of Amsterdam, just like me.
Edouard Zeckendorf grew up near Liège, where he studied medicin, but his parents were
from Amsterdam.
This city is infamous for various reasons, yet its uncanny
connection with recursion has so far gone unnoticed.

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J.-P. Allouche and J. Shallit, Automatic Sequences, Cambridge
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[2]
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[3]
M. Bunder, Zeckendorf representations using negative fibonacci
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[4]
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[5]
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[6]
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[7]
C. Epifanioa, C. Frougny, A. Gabriele, F. Mignosi, and J. Shallit, Sturmian graphs and integer representations over numeration systems,
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[8]
A. S. Fraenkel, The bracket function and complementary sets of
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[9]
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[10]
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Contravariant Form on Tensor Product of Highest Weight Modules

Contravariant Form on Tensor Product
of Highest Weight Modules

Andrey I. MUDROV

A.I. Mudrov

Department of Mathematics, University of Leicester, University Road, LE1 7RH
Leicester, UK<EMAIL_ADDRESS>
Received August 23, 2018, in final form March 25, 2019; Published online April
07, 2019

We give a criterion for complete reducibility of tensor product $V\otimes Z$
of two irreducible highest weight modules $V$ and $Z$ over a classical or
quantum semi-simple group in terms of a contravariant symmetric bilinear form
on $V\otimes Z$. This form is the product of the canonical contravariant forms
on $V$ and $Z$. Then $V\otimes Z$ is completely reducible if and only if the
form is non-degenerate when restricted to the sum of all highest weight
submodules in $V\otimes Z$ or equivalently to the span of singular vectors.

highest weight modules; contravariant form; tensor product; complete
reducibility

17B10; 17B37

In memoriam Ludwig Faddeev

##  1 Introduction

Consider a semi-simple Lie algebra over the complex field and its classical or
quantum universal enveloping algebra (assuming $q$ not a root of unity in the
latter case). Fix a pair of irreducible highest weight modules. The question
when their tensor product is completely reducible is a fundamental problem of
representation theory. That is always the case when the modules are finite-
dimensional, due to the Weyl theorem, cf. [6] for classical and [8] for
quantum groups. However the problem is a challenge if one of them is infinite-
dimensional. We give a criterion of complete reducibility in terms of a
canonical contravariant form that is the product of the Shapovalov forms on
the factors [14]. Quite amazingly, this result appears to be new.

Tensor products of representations of simple Lie algebras have been studied in
detail in [2, 10], where the main tool was the action of the center of their
universal enveloping algebras. Our approach based on contravariant forms
displays a relation with extremal projectors [1, 9] and dynamical Weyl group
[7]. Unlike [2], we allow for arbitrary dimensions of the tensor factors. We
simultaneously address the classical and quantum cases, although the
exposition is provided for quantum groups. The classical case is a
simplification that can be obtained by setting the deformation parameter $q\to
1$.

We prove that the tensor product of two irreducible modules $V\otimes Z$ of
highest weight is completely reducible if and only if the canonical
contravariant form is non-degenerate when restricted to the sum of all highest
weight submodules in $V\otimes Z$. It is easy to see that the necessary
condition here comes for free. Conversely, if the form is non-degenerate on
the sum of highest weight submodules, they are all irreducible. It is then
sufficient to prove that the sum exhausts all of $V\otimes Z$, which takes a
certain effort.

Equivalently, $V\otimes Z$ is completely reducible if and only if the form is
non-degenerate on the subspace of singular vectors in $V\otimes Z$, i.e.,
annihilated by positive Chevalley generators. Having parameterized them with
elements from either of reciprocal “extremal subspaces” $V^{+}_{Z}\subset V$
and $Z^{+}_{V}\subset Z$, we construct operators $\theta_{V,Z}$,
$\theta_{Z,V}$ sending $V^{+}_{Z}$ and $Z^{+}_{V}$ to their duals under the
Shapovalov forms on $V$ and $Z$, respectively. These parametrisations link the
canonical form with Zhelobenko cocycles [15] and lay a ground for developing
computational methods.

One application of the presented result is quantization of equivariant vector
bundles over semi-simple conjugacy classes and coadjoint orbits regarded as
projective modules over quantized function algebras. The infinite-dimensional
modules of interest are parabolic or generalized parabolic modules generating
module categories over the category of finite-dimensional representations. A
quantim conjugacy class or a coadjoint orbit is realized by linear
endomorphisms on such a module whose tensor product with finite-dimensional
modules supports “representations” of a trivial bundle. Its non-trivial sub-
bundles correspond to invariant idempotents that project the tensor product
onto submodules [12, 13].

##  2 Preliminaries

Suppose that $\mathfrak{g}$ is a complex semi-simple Lie algebra and
$\mathfrak{h}\subset\mathfrak{g}$ its Cartan subalgebra. Fix a triangular
decomposition
$\mathfrak{g}=\mathfrak{g}_{-}\oplus\mathfrak{h}\oplus\mathfrak{g}_{+}$ with
nilpotent Lie subalgebras $\mathfrak{g}_{\pm}$. Denote by $\mathrm{R}$ the
root system of $\mathfrak{g}$, and by $\mathrm{R}^{+}$ the subset of positive
roots with basis $\Pi^{+}$ of simple roots. Choose an inner product
$(\cdot,\cdot)$ on $\mathfrak{h}$ as a multiple of the restricted Killing form
and transfer it to the dual space $\mathfrak{h}^{*}$. For all
$\lambda\in\mathfrak{h}^{*}$ denote by $h_{\lambda}$ the element of
$\mathfrak{h}$ such that $\mu(h_{\lambda})=(\mu,\lambda)$ for all
$\mu\in\mathfrak{h}^{*}$.

By $U_{q}(\mathfrak{g})$ we understand the standard quantum group [3, 4], that
is a complex unital algebra with the set of generators $e_{\alpha}$,
$f_{\alpha}$, and $q^{\pm h_{\alpha}}$ satisfying

$\displaystyle q^{\pm
h_{\alpha}}e_{\beta}=q^{\pm(\alpha,\beta)}e_{\beta}q^{\pm
h_{\alpha}},\qquad[e_{\alpha},f_{\beta}]=\delta_{\alpha,\beta}[h_{\alpha}]_{q},$
$\displaystyle q^{\pm
h_{\alpha}}f_{\beta}=q^{\mp(\alpha,\beta)}f_{\beta}q^{\pm
h_{\alpha}},\qquad\alpha\in\Pi^{+},$

plus the Serre relations among $\\{e_{\alpha}\\}_{\alpha\in\Pi^{+}}$ and
$\\{f_{\alpha}\\}_{\alpha\in\Pi^{+}}$, see [3] for details. Here
$[h_{\alpha}]_{q}=\frac{q^{h_{\alpha}}-q^{-h_{\alpha}}}{q-q^{-1}}$, and
$q^{h_{\alpha}}q^{-h_{\alpha}}=1=q^{-h_{\alpha}}q^{h_{\alpha}}$. The complex
parameter $q\not=0$ is assumed not a root of unity.

Fix the comultiplication in $U_{q}(\mathfrak{g})$ as

$\displaystyle\Delta(f_{\alpha})=f_{\alpha}\otimes 1+q^{-h_{\alpha}}\otimes
f_{\alpha},\qquad\\!\\!\Delta\big{(}q^{\pm h_{\alpha}}\big{)}=q^{\pm
h_{\alpha}}\otimes q^{\pm
h_{\alpha}},\qquad\\!\\!\Delta(e_{\alpha})=e_{\alpha}\otimes
q^{h_{\alpha}}+1\otimes e_{\alpha}.$

Then the antipode acts on the generators by the assignment
$\gamma(f_{\alpha})=-q^{h_{\alpha}}f_{\alpha}$, $\gamma\big{(}q^{\pm
h_{\alpha}}\big{)}=q^{\mp h_{\alpha}}$,
$\gamma(e_{\alpha})=-e_{\alpha}q^{-h_{\alpha}}$. The counit homomorphism
returns $\epsilon(e_{\alpha})=\epsilon(f_{\alpha})=0$, and
$\epsilon\big{(}q^{\pm h_{\alpha}}\big{)}=1$.

Denote by $U_{q}(\mathfrak{h})$, $U_{q}(\mathfrak{g}_{+})$,
$U_{q}(\mathfrak{g}_{-})$ the subalgebras in $U_{q}(\mathfrak{g})$ generated
by $\big{\\{}q^{\pm
h_{\alpha}}\big{\\}}_{\alpha\in\Pi^{+}}$,$\\{e_{\alpha}\\}_{\alpha\in\Pi^{+}}$,
and $\\{f_{\alpha}\\}_{\alpha\in\Pi^{+}}$, respectively. The quantum Borel
subgroups
$U_{q}(\mathfrak{b}_{\pm})=U_{q}(\mathfrak{g}_{\pm})U_{q}(\mathfrak{h})$ are
Hopf subalgebras in $U_{q}\\!(\mathfrak{g})$. We introduce a grading on
$U_{q}\\!(\mathfrak{g}_{\pm}\\!)$ by setting
$\deg(e_{\alpha})\\!=\\!1\\!=\\!\deg(f_{\alpha})$.

We will need the following involutive maps on $U_{q}(\mathfrak{g})$. The
assignment

$\displaystyle\sigma\colon\ e_{\alpha}\mapsto f_{\alpha},\qquad\sigma\colon\
f_{\alpha}\mapsto e_{\alpha},\qquad\sigma\colon\ q^{h_{\alpha}}\mapsto
q^{-h_{\alpha}}$

gives rise to an algebra automorphism of $U_{q}(\mathfrak{g})$. It is also a
coalgebra anti-automorphism. The assignment

$\displaystyle\tilde{\omega}\colon\ e_{\alpha}\mapsto
f_{\alpha},\qquad\tilde{\omega}\colon\ f_{\alpha}\mapsto
e_{\alpha},\qquad\tilde{\omega}\colon\ q^{h_{\alpha}}\mapsto q^{h_{\alpha}}$

for all $\alpha\in\Pi^{+}$ extends to an algebra anti-automorphism of
$U_{q}(\mathfrak{g})$. This map flips the subalgebras
$U_{q}(\mathfrak{g}_{+})$ and $U_{q}(\mathfrak{g}_{-})$ but it is not
compatible with the comultiplication.

We will also work with the involutive map $\omega=\gamma^{-1}\circ\sigma$,
which is an algebra anti-automorphism and preserves comultiplication. However,
it does not exchange $U_{q}(\mathfrak{g}_{+})$ and $U_{q}(\mathfrak{g}_{-})$
as $\tilde{\omega}$ does. To fix this disadvantage, we have to pass to the
Borel subalgebras. Let us show that replacement of $\tilde{\omega}$ with
$\omega$ does not affects certain left/right ideals in
$U_{q}(\mathfrak{b}_{\pm})$ which are of interest for the present exposition.

An element $\psi\in U_{q}(\mathfrak{g})$ is said to be of weight
$\lambda\in\mathfrak{h}^{*}$ if $q^{h_{\alpha}}\psi
q^{-h_{\alpha}}=q^{(\lambda,\alpha)}\psi$. Such weights are linear
combinations of simple roots with integer coefficients.

######  Lemma 2.1.

The map $f_{\alpha}\mapsto cf_{\alpha}q^{\pm h_{\alpha}}$, with non-zero
$c\in\mathbb{C}$, extends to an automorphism of the Borel subalgebra
$U_{q}(\mathfrak{b}_{-})$. Under this map, any element $\psi\in
U_{q}(\mathfrak{b}_{-})$ of a given weight is sent to $\psi\phi$ for some
$\phi\in U_{q}(\mathfrak{h})$.

###### Proof.

Since the Serre relations are homogeneous, cf. [3], it is sufficient to assume
that $\psi$ is a monomial in the Chevalley generators and check that $\phi$
depends only on the weight of $\psi$. Let $\beta_{1},\ldots,\beta_{m}$ be a
sequence of simple positive roots and take $\psi=f_{\beta_{1}}\cdots
f_{\beta_{m}}$. Then
$\phi=c^{m}q^{\mp\sum_{i<j}(\beta_{i},\beta_{j})}q^{\sum_{i=1}^{m}h_{\beta_{i}}}$.
∎

######  Corollary 2.2.

Let $I\in U_{q}(\mathfrak{g}_{+})$ be a $U_{q}(\mathfrak{h})$-graded left
ideal. Then the right ideals in $U_{q}(\mathfrak{b}_{-})$ generated by
$\tilde{\omega}(I)$ and by $\omega(I)$ coincide.

###### Proof.

A $U_{q}(\mathfrak{h})$-graded left ideal is generated by elements of certain
weights. Now the proof follows from Lemma 2.1. ∎

##  3 Extremal vectors in tensor product

In this preparatory section we fix some notation and describe two
parametrizations of singular vectors in $V\otimes Z$ by hom-sets relative to
the subalgebra $U_{q}(\mathfrak{g}_{+})$. We will use those parameterizations
for constructing pullbacks of a contravariant form on $V\otimes Z$.

For every $U_{q}(\mathfrak{h})$-module $V$ we denote by $\Lambda(V)$ the set
of its weights, i.e., $\mu\in\Lambda(V)$ if there is a non-zero $v\in V$ such
that $q^{h_{\alpha}}v=q^{(\alpha,\mu)}v$, $\forall\,\alpha$. The set of $v\in
V$ that satisfy this condition (including the zero vector) is denoted by
$V[\mu]$.

Let $\Gamma=\mathbb{Z}\Pi^{+}$ denote the root lattice and set
$\Gamma_{+}=\mathbb{Z}_{+}\Pi^{+}$ ($\mathbb{Z}_{+}$ stands for the monoid of
non-negative integers). For $\alpha,\beta\in\Lambda(V)$ we write
$\alpha\prec\beta$ or $\beta\succ\alpha$ if
$\beta-\alpha\in\Gamma_{+}\backslash\\{0\\}$.

$V$ is called a highest weight module if there is a vector $1_{\nu}$ of weight
$\nu\in\Lambda(V)$ such that $V$ is cyclically generated by $v$ and
$e_{\alpha}1_{\nu}=0$ for all $\alpha\in\Pi$. Then $1_{\nu}$ is called highest
vector and $\nu$ the highest weight. $V$ is also cyclicly generated by
$1_{\nu}$ as a module over $U_{q}(\mathfrak{g}_{-})$. It is a quotient of the
Verma module of the same highest weight $\nu$, which is freely generated over
$U_{q}(\mathfrak{g}_{-})$ by its highest vector. Clearly $\mu\prec\nu$ for all
$\mu\in\Lambda(V)$ distinct from $\nu$.

Recall that a non-zero weight vector of a $U_{q}(\mathfrak{g})$-module $V$ is
called singular (extremal) if it is killed by all positive generators of
$U_{q}(\mathfrak{g})$. A singular vector $u$ of weight $\lambda$ defines a
homomorphism of the corresponding Verma module extending the assignment
$1_{\lambda}\mapsto u$ on its highest vector. The submodule in $V$ generated
by a singular vector $u$ is a highest weight module with the highest vector
$u$. We call it a highest weight submodule.

By $V^{*}$ we mean the $U_{q}(\mathfrak{h})$-graded restricted (left) dual to
a highest weight $U_{q}(\mathfrak{g})$-module $V$. It is isomorphic to the
right dual ${}^{*}\\!V$ since $U_{q}(\mathfrak{g})$ is quasi-triangular. If
$V$ is irreducible, then $V^{*}$ is cyclically generated by a vector of lowest
weight, i.e., the one annihilated by all $f_{\alpha}$, $\alpha\in\Pi$.

In what follows, $V$ and $Z$ are highest weight modules. They are assumed
irreducible unless otherwise is explicitly stated. We reserve notation $\nu$
for the highest weight of $V$ and $\zeta$ for the highest weight of $Z$. By
$1_{\nu}$ and $1_{\zeta}$ we denote their highest weight generators. The
vector subspace in $V\otimes Z$ spanned by singular vectors is denoted by
$(V\otimes Z)^{+}$.

Denote by $V\circledast Z$ the sum of all highest weight submodules in
$V\otimes Z$; it is generated by $(V\otimes Z)^{+}$. Let
$u=\sum_{i}v_{i}\otimes z_{i}\in(V\otimes Z)^{+}$ be a singular vector, where
$v_{i}$ and $z_{i}$ are weight vectors. Assuming the set of $v_{i}$
independent, we call $\\{z_{i}\\}$ $Z$-coefficients of $u$, with the lowest
weight coefficient corresponding to $1_{\nu}$. We define the $V$-coefficients
of $u$ similarly.

######  Lemma 3.1.

Singular vectors in $V\otimes Z$ are in bijection with
$\mathrm{Hom}_{U_{q}(\mathfrak{g}_{+})}(V^{*},Z)$ and, alternatively, with
$\mathrm{Hom}_{U_{q}(\mathfrak{g}_{+})}({}^{*}\\!Z,V)$.

###### Proof.

Suppose that $\\{v_{i}\\}$ form a basis in $V$. Let $\\{v^{i}\\}\subset V^{*}$
be the dual basis and $\\{z_{i}\\}\subset Z$ the set of coefficients for a
given singular vector $u$. Then the map $V^{*}\to Z$, $v^{i}\mapsto z_{i}$, is
a $U_{q}(\mathfrak{g}_{+})$-module homomorphisms. It is clear that every
singular vector is obtained this way. The case of
$\mathrm{Hom}_{U_{q}(\mathfrak{g}_{+})}({}^{*}\\!Z,V)$ is similar. ∎

Obviously a $Z$-coefficient with a maximal weight in the expansion of $u$ is
killed by $\\{e_{\alpha}\\}_{\alpha\in\Pi^{+}}$ and is singular in $Z$.
Therefore, for irreducible $Z$, such a coefficient is equal to $1_{\zeta}$, up
to a non-zero scalar factor. Reciprocally, the $V$-coefficient of a maximal
weight is singular in $V$ and therefore is a non-zero scalar multiple of
$1_{\nu}$.

Let $I^{-}_{Z}\subset U_{q}(\mathfrak{g}_{-})$ be the annihilator of
$1_{\zeta}\in Z$ and put $I^{+}_{Z}=\sigma(I^{-}_{Z})\subset
U_{q}(\mathfrak{g}^{+})$. We define left ideals $I^{\pm}_{V}\subset
U_{q}(\mathfrak{g}^{\pm})$ similarly. Denote by $V^{+}_{Z}\subset V$ and
$Z^{+}_{V}\subset Z$ the kernels of $I^{+}_{Z}$ and $I^{+}_{V}$, respectively.

######  Proposition 3.2.

The vector subspace $V^{+}_{Z}$ is isomorphic to $Z^{+}_{V}$ via
$V^{+}_{Z}\simeq(V\otimes Z)^{+}\simeq Z^{+}_{V}$.

###### Proof.

Since $V$ and $Z$ are irreducible, their dual modules are cyclicly generated
by the lowest vectors. Therefore
$\mathrm{Hom}_{U_{q}(\mathfrak{g}_{+})}({}^{*}\\!Z,V)=V^{+}_{Z}$ and
$\mathrm{Hom}_{U_{q}(\mathfrak{g}_{+})}({}^{*}\\!V,Z)=Z^{+}_{V}$, so one is
left to apply Lemma 3.1. ∎

Observe that the isomorphisms are implemented by the assignments $u\mapsto
u_{1}(u_{2},{}^{*}\\!1_{\zeta})\in V^{+}_{Z}$, and
$u\mapsto(u_{1},{}^{*}\\!1_{\nu})u_{2}\in Z^{+}_{V}$, where $u=u^{1}\otimes
u^{2}\in(V\otimes Z)^{+}$ in the Sweedler notation. These assignments involve
the invariant pairing with the lowest vectors ${}^{*}\\!1_{\nu}$,
${}^{*}\\!1_{\zeta}$ (of weights $-\nu$ and $-\zeta$) of the dual modules and
return the coefficients of lowest weights. We call them leading coefficients.
Other coefficients in the expansion of a given singular vectors are obtained
from them by the action of $U_{q}(\mathfrak{g}_{+})$.

We denote the inverse isomorphisms by $\delta_{l}\colon V^{+}_{Z}\to(V\otimes
Z)^{+}$ and $\delta_{r}\colon Z^{+}_{V}\to(V\otimes Z)^{+}$.

######  Example 3.3.

Suppose $\mathfrak{k}$ is a Levi subalgebra in $\mathfrak{g}$ and let
$\Pi^{+}_{\mathfrak{k}}\subset\Pi^{+}_{\mathfrak{g}}$ be its basis of simple
roots. Consider $Z$ to be a scalar parabolic Verma module corresponding to a
one-dimensional representation of $U_{q}(\mathfrak{k})$. The left ideal
$I^{+}_{Z}$ is then generated by
$\\{e_{\alpha}\\}_{\alpha\in\Pi^{+}_{\mathfrak{k}}}$, and $V^{+}_{Z}$ becomes
the subspace of $U_{q}(\mathfrak{k})$-singular vectors in $V$.

##  4 Contravariant forms

Let $V$ be a $U_{q}(\mathfrak{g})$-module. A symmetric bilinear form
$\langle\cdot,\cdot\rangle$ on $V$ is called contravariant if for all $v,w\in
V$ and all $x\in U_{q}(\mathfrak{g})$ it satisfies $\langle
xv,w\rangle=\langle v,\omega(x)w\rangle$. The following properties of
contravariant forms readily follow.

######  Proposition 4.1.

  1. $i)$

The kernel of the form is a submodule in $V$.

  2. $ii)$

Weight subspaces $V[\mu]$ and $V[\nu]$ for different $\mu,\nu\in\Lambda(V)$
the are orthogonal.

  3. $iii)$

Two highest weight submodules are orthogonal if and only if their canonical
generators are orthogonal.

It is known that every module of highest weight is equipped with a unique, up
to a scalar multiplier, contravariant form, which is non-degenerate if and
only if the module is irreducible. When $V$ is a Verma module, it is called
Shapovalov form [14], which name we apply to all highest weight modules. The
Shapovalov form is equivalent to an invariant pairing
$V\otimes{}^{*}\\!V\to\mathbb{C}$ via the linear isomorphism $\sigma\colon
V\to{}^{*}\\!V$. We will work with both contravariant forms and invariant
pairings.

Now we return to the settings of the previous section.

######  Definition 4.2.

Let $V$ and $Z$ be two highest weight modules. By canonical contravariant form
on $V\otimes Z$ we mean the product of the Shapovalov forms on $V$ and $Z$.

Denote by $V^{\perp}_{Z}\subset V$ the vector subspace $\omega(I^{+})V$, i.e.,
the image of the right ideal $\omega(I^{+}_{Z})\subset
U_{q}(\mathfrak{b}_{-})$. By Lemma 2.2, it is equal to image of the right
ideal $\tilde{\omega}(I^{+}_{Z})\subset U_{q}(\mathfrak{g}_{-})$. It is
exactly the annihilator of $V^{+}_{Z}$, so the quotient
${}^{+}\\!V_{Z}=V/V^{\perp}_{Z}$ is isomorphic to the dual vector space to
$V^{+}_{Z}$ with respect to the form. We identify ${}^{+}\\!V_{Z}$ with a
fixed vector subspace in $V$ that is transversal to $V^{\perp}_{Z}$.

Define extremal twist operator $\theta_{V,Z}\colon V^{+}_{Z}\to{}^{+}\\!V_{Z}$
as follows. For $v\in V^{+}_{Z}[\xi]$ write $u=\delta_{l}(v)$ as
$u=\sum_{\alpha\in\Gamma^{+}}\sum_{i}v_{\alpha,i}\otimes
f_{\alpha,i}1_{\zeta}$, where $v_{\alpha,i}\in V$ are vectors of weight
$\xi+\alpha$ and $f_{\alpha,i}\in U_{q}(\mathfrak{g}_{-})$ are some elements
of weight $-\alpha$. Then put $\theta_{V,Z}(v)$ equal to the image of
$\sum_{\alpha\in\Gamma^{+}}\sum_{i}\gamma^{-1}(f_{\alpha,i})v_{\alpha,i}$ in
${}^{+}\\!V_{Z}$ projected along $V^{\perp}_{Z}$. This image is independent of
the choice of $f_{\alpha,i}$. Indeed, the if $f^{\prime}_{\alpha,i}\in
U_{q}(\mathfrak{g}_{-})$ is another element such that
$f_{\alpha,i}^{\prime}1_{\zeta}=f_{\alpha,i}1_{\zeta}$, then
$\psi_{\alpha,i}=f_{\alpha,i}^{\prime}-f_{\alpha,i}$ belongs to $I^{-}_{Z}$,
and
$\gamma^{-1}(\psi_{\alpha,i})=(\omega\circ\sigma)(\psi_{\alpha,i})\in\omega(I^{+}_{Z})$.
Therefore $f_{\alpha,i}^{\prime}v_{i}=f_{\alpha,i}v_{i}\mod V^{\perp}_{Z}$ as
required. Further on we will suppress the index $i$ replacing
$\sum_{i}v_{\alpha,i}\otimes f_{\alpha,i}$ with $v_{\alpha}\otimes f_{\alpha}$
in the Sweedler-like fashion.

Recall that there are natural $U_{q}(\mathfrak{g}_{\pm})$-module epimorphisms
$U_{q}(\mathfrak{g}_{+})\to{}^{*}\\!Z$ and $U_{q}(\mathfrak{g}_{-})\to Z$,
where the algebras act on themselves by left multiplication. Extend the tensor
product ${}^{*}\\!Z\otimes Z$ by infinite formal sums
$\sum_{\xi\in\Lambda({}^{*}\\!Z),\zeta\in\Lambda(Z)}{}^{*}\\!Z[\xi]\hat{\otimes}Z[\zeta]$.
Let $\mathcal{S}$ denote the $U_{q}(\mathfrak{g})$-invariant image of $1$
under the coevaluation morphism $\mathbb{C}\to{}^{*}\\!Z\hat{\otimes}Z$. Using
a linear section of the module map
$U_{q}(\mathfrak{g}_{+})\hat{\otimes}U_{q}(\mathfrak{g}_{-})\to{}^{*}\\!Z\hat{\otimes}Z$,
we pick up a lift $\mathcal{F}\in
U_{q}(\mathfrak{g}_{+})\hat{\otimes}U_{q}(\mathfrak{g}_{-})$ of $\mathcal{S}$.

######  Lemma 4.3.

The map $\delta_{l}\colon V^{+}_{Z}\to(V\otimes Z)^{+}$ acts by the assignment
$v\mapsto\mathcal{F}_{1}v\otimes\mathcal{F}_{2}1_{\zeta}$, in the Sweedler
symbolic notation for tensor factors.

###### Proof.

Indeed, ${}^{*}\\!Z\hat{\otimes}Z$ is a $U_{q}(\mathfrak{g})$-module, and the
inverse form $\mathcal{S}\in{}^{*}\\!Z\hat{\otimes}Z$ is
$U_{q}(\mathfrak{g})$-invariant. The homomorphism ${}^{*}\\!Z\otimes Z\to
V\otimes Z$ of $U_{q}(\mathfrak{g}_{+})$-modules determined by $v\in
V^{+}_{Z}$ extends to a homomorphism ${}^{*}\\!Z\hat{\otimes}Z\to V\otimes Z$
that sends $\mathcal{S}$ to
$u=\mathcal{F}_{1}v\otimes\mathcal{F}_{2}1_{\zeta}$. Therefore $u$ is
$U_{q}(\mathfrak{g}_{+})$-invariant and hence singular. Evaluating the
invariant inner product of the $Z$-component of
$\mathcal{F}_{1}v\otimes\mathcal{F}_{2}1_{\zeta}$ with ${}^{*}\\!1_{\zeta}$ we
get $v$, which proves the statement. ∎

The element $\Upsilon_{Z}=\gamma^{-1}(\mathcal{F}_{2})\mathcal{F}_{1}$ belongs
to a certain extension of $U_{q}(\mathfrak{g})$ and defines a
$U_{q}(\mathfrak{h})$-linear map $V^{+}_{Z}\to V$. Remark that $\Upsilon_{Z}$
controls the transformation of the antipode under the quasi-Hopf twist of
$U_{q}(\mathfrak{g})$ by $\mathcal{F}$ [5]. These elements for $Z$ a Verma
module were introduced in [7] where they participated in construction of
dynamical Weyl group. For the case of parabolic modules they appeared in [11]
in connection with dynamical twist over a non-abelian base.

Observe that right hand side of the equality
$\delta_{l}(v)=\mathcal{F}(v\otimes 1_{\zeta})$, $v\in V^{+}_{Z}$, is
independent of the choice of $\mathcal{F}$. Any other lift differs from
$\mathcal{F}$ by an element from $U_{q}(\mathfrak{g})\otimes
I_{Z}^{-}+I^{+}_{Z}\otimes U_{q}(\mathfrak{g})$, which kills $V^{+}_{Z}\otimes
1_{\zeta}$. For all $v,w\in V^{+}_{Z}$, one has
$\langle\theta_{V,Z}(v),w\rangle=\langle\Upsilon_{Z}(v),w\rangle$. The right-
hand side here is independent of a particular lift, because
$U_{q}(\mathfrak{g})I^{+}_{Z}$ kills $V^{+}_{Z}$ and
$\gamma^{-1}(I^{-}_{Z})U_{q}(\mathfrak{g})V=\omega(I^{+}_{Z})V$ is orthogonal
to $V^{+}_{Z}$.

######  Proposition 4.4.

The form $\langle\theta_{V,Z}(\cdot),\cdot\rangle$ is the pullback of the
canonical form restricted to $(V\otimes Z)^{+}$ under the isomorphism
$V^{+}_{Z}\to(V\otimes Z)^{+}$.

###### Proof.

Choose extremal vectors $u,\tilde{u}\in(V\otimes Z)^{+}$ of weight
$\eta+\zeta$ and $\xi+\zeta$, respectively. Expanding them over the positive
hull of the root lattice we write $u=\sum_{\alpha\geqslant
0}v_{\eta+\alpha}\otimes f_{\alpha}1_{\zeta}=\delta_{l}(v_{\eta})$ and
$\tilde{u}=\sum_{\beta\geqslant
0}\tilde{v}_{\xi+\beta}\otimes\tilde{z}_{\beta-\zeta}=\delta_{l}(\tilde{v}_{\xi})$
using the Sweedler-like convention for suppressing summation. Then

$\displaystyle\langle u,\tilde{u}\rangle=\sum_{\alpha,\beta\geqslant 0}\langle
v_{\eta+\alpha},\tilde{v}_{\xi+\beta}\rangle\langle
f_{\alpha}1_{\zeta},\tilde{z}_{\beta-\zeta}\rangle=\sum_{\alpha,\beta\geqslant
0}\langle
v_{\eta+\alpha},\tilde{v}_{\xi+\beta}\rangle\big{\langle}1_{\zeta},\gamma^{-1}\circ\sigma(f_{\alpha})\tilde{z}_{\beta-\zeta}\big{\rangle}$
$\displaystyle\hphantom{\langle
u,\tilde{u}\rangle}{}=\sum_{\alpha,\beta\geqslant 0}\langle
v_{\eta+\alpha},\sigma(f_{\alpha})\tilde{v}_{\xi+\beta}\rangle\langle
1_{\zeta},\tilde{z}_{\beta-\zeta}\rangle=\left\langle\sum_{\alpha\geqslant
0}\gamma^{-1}(f_{\alpha})v_{\eta+\alpha},\tilde{v}_{\xi}\right\rangle=\langle\theta_{V,Z}(v_{\eta}),\tilde{v}_{\xi}\rangle$

as required. ∎

##  5 Height filtration on $\boldsymbol{V\otimes Z}$

Since all weights in a module $V$ of highest weight $\nu$ belong to
$\nu-\Gamma^{+}$ we can speak of height $\mathrm{ht}(\xi)\in\mathbb{Z}_{+}$ of
$\xi\in\Lambda(V)$ defining it as the sum of coordinates of $\nu-\xi$ with
respect to the basis $\Pi^{+}$. We define $\mathrm{ht}(\mu)$ for a weight
$\mu\in\Lambda(V\otimes Z)$ similarly by setting it equal to the sum of
coordinates of $\nu+\zeta-\mu\in\Gamma_{+}$. A vector is said to be of height
$k$ if it carries a weight of the corresponding height. The assignment $V\ni
v\mapsto v\otimes 1_{\zeta}\in V\otimes Z$ is height-preserving.

For $k\in\mathbb{Z}_{+}$ denote by $(V\otimes Z)_{k}$ the submodule in
$V\otimes Z$ generated by vectors of height $\leqslant k$ from $V\otimes
1_{\zeta}$. The submodules $(V\otimes Z)_{k}$ form an increasing filtration of
$V\otimes Z$.

######  Lemma 5.1.

Fix $\xi\in\Lambda(V)$, $\eta\in\Lambda(Z)$, $\beta\in-\Gamma_{+}$ and put
$\mu=\xi+\beta+\eta\in\Lambda(V\otimes Z)$. Then for every $\psi\in
U_{q}(\mathfrak{g}_{-})[\beta]$, $v_{\xi}\in V[\xi]$, and $z_{\eta}\in
Z[\eta]$, one has $\gamma^{-1}(\psi)v_{\xi}\otimes z_{\eta}=v_{\xi}\otimes\psi
z_{\eta}\mod(V\otimes Z)_{k-1}$ with $k=\mathrm{ht}(\mu)$.

###### Proof.

Remark that, in particular, this implies $v_{\xi}\otimes\psi
1_{\zeta}\in(V\otimes Z)_{k}$. So we can consider a monomial $\psi$ in
negative Chevalley generators and do induction on the degree of $\psi$.

For $\psi=1$ the statement is trivial. Suppose it is true for all monomials of
degree $m-1>0$ and present $\psi$ of degree $m$ as
$\psi=f_{\alpha}\psi^{\prime}$ for some $\alpha\in\Pi^{+}$. Here
$\psi^{\prime}\in U_{q}(\mathfrak{g}_{-})$ has degree $m-1$. Then
$v_{\xi}\otimes\psi^{\prime}1_{\zeta}\in(V\otimes Z)_{k-1}$ by the induction
assumption and the above remark. The action of
$\Delta(f_{\alpha})=-\gamma^{-1}(f_{\alpha})q^{-h_{\alpha}}\otimes
1+q^{-h_{\alpha}}\otimes f_{\alpha}$ on the tensor
$v_{\xi}\otimes\psi^{\prime}1_{\zeta}\in(V\otimes Z)_{k-1}$ yields

$\displaystyle v_{\xi}\otimes\psi 1_{\zeta}=v_{\xi}\otimes
f_{\alpha}\psi^{\prime}1_{\zeta}=\gamma^{-1}(f_{\alpha})v_{\xi}\otimes\psi^{\prime}1_{\zeta}\qquad\mod(V\otimes
Z)_{k-1}.$

The rightmost term transforms to
$\gamma^{-1}(\psi^{\prime})\gamma^{-1}(f_{\alpha})v_{\xi}\otimes
1_{\zeta}=\gamma^{-1}(\psi)v_{\xi}\otimes 1_{\zeta}\mod(V\otimes Z)_{k-1}$, by
the induction assumption. This completes the proof. ∎

Note that this lemma is true for arbitrary highest weight modules $V$ and $Z$,
not necessarily irreducible.

######  Corollary 5.2.

  1. $i)$

All vectors in $V\otimes Z$ of height $\leqslant k$ belong to $(V\otimes
Z)_{k}$.

  2. $ii)$

Vectors from $V^{\perp}_{Z}\otimes 1_{\zeta}$ of height $k$ belong to
$(V\otimes Z)_{k-1}$.

  3. $iii)$

Vectors from ${}^{+}\\!V_{Z}\otimes 1_{\zeta}$ of height $k$ generate
$(V\otimes Z)_{k}$ modulo $(V\otimes Z)_{k-1}$.

  4. $iv)$

For all $v\in V^{+}_{Z}$ of height $k$, $\theta_{V,Z}(v)\otimes
1_{\zeta}=\delta_{l}(v)$ modulo $(V\otimes Z)_{k-1}$.

###### Proof.

(i) readily follows from Lemma 5.1. To prove (ii), observe that, for a vector
$v\in V$ of height $k$ one has $\omega(I^{+}_{Z})v\otimes
1_{\zeta}=v\otimes\sigma(I^{+}_{Z})1_{\zeta}=v\otimes I^{-}_{Z}1_{\zeta}=0$
modulo $(V\otimes Z)_{k-1}$. Clearly (iii) follows from (ii). To prove (iv),
pick up $v\in V^{+}_{Z}$ of height $k$ and present $\theta_{V,Z}(v)$ as
$\Upsilon_{Z}(v)\mod V^{\perp}_{Z}$. Then modulo $(V\otimes Z)_{k-1}$ we get
the required:

$\displaystyle\theta_{V,Z}(v)\otimes 1_{\zeta}=\Upsilon_{Z}(v)\otimes
1_{\zeta}=\mathcal{F}(v\otimes 1_{\zeta})=\delta_{l}(v).$

Here the left equality follows from (ii), while the middle one is a
consequence of Lemma 5.1. ∎

##  6 Complete reducibility of $\boldsymbol{V\otimes Z}$

We derive a complete reducibility criterion from the following two lemmas
capturing properties of $\theta_{V,Z}$ and therefore of the canonical form on
$V\otimes Z$.

######  Lemma 6.1.

Suppose that $\theta_{V,Z}$ is surjective. Then $V\circledast Z=V\otimes Z$.

###### Proof.

It is sufficient to prove the equality on weight subspaces. The statement is
obvious for subspace of maximal weight $\nu+\zeta\in\Lambda(V\otimes Z)$. Pick
up an arbitrary $\mu$ and suppose the statement is proved for all weights of
height less than $k=\mathrm{ht}(\mu)$. Corollary 5.2 implies modulo $(V\otimes
Z)_{k-1}[\mu]$:

$\displaystyle(V\otimes Z)[\mu]=(V\otimes
Z)_{k}[\mu]={}^{+}\\!V_{Z}[\mu-\zeta]\otimes
1_{\zeta}=\theta_{V,Z}V^{+}_{Z}[\mu-\zeta]\otimes 1_{\zeta}$
$\displaystyle\hphantom{(V\otimes Z)[\mu]}{}=(V\otimes
Z)^{+}[\mu]=(V\circledast Z)[\mu].$

Here we consecutively used Lemma 5.2(i), (iii), the surjectivity assumption,
Lemma 5.2(iv), and, finally, (i) along with the induction assumption. This
proves the statement for all weights of height $k$ and hence for all weights,
by induction. ∎

######  Lemma 6.2.

Suppose that $\theta_{V,Z}$ is injective. Then all submodules of highest
weight in $V\otimes Z$ are irreducible.

###### Proof.

Since $\theta_{V,Z}$ is injective, it is an isomorphism between $V^{+}_{Z}$
and ${}^{+}\\!V_{Z}$. Then the canonical form is non-degenerate on
$V\circledast Z$. If $X\subset Y$ are submodules of highest weight, then $X$
is orthogonal to all submodules in $V\circledast Z$ of the same weight, by
Proposition 4.1(iii), (ii). Therefore $X$ is orthogonal to all $V\circledast
Z$, which is a contradiction. ∎

Now we can prove the main result of the paper.

######  Theorem 6.3.

The module $V\otimes Z$ is completely reducible if and only if the canonical
form is non-degenerate on $(V\otimes Z)^{+}$.

###### Proof.

We use the fact that the form is non-degenerate if and only if $\theta_{V,Z}$
is an isomorphism. If $V\otimes Z$ is completely reducible, then it is a sum
of highest weight modules. The canonical form is non-degenerate on
$V\circledast Z$ and therefore on $(V\otimes Z)^{+}$, by Proposition 4.1(iii).

Conversely, if the form is non-degenerate, then all highest weight submodules
are irreducible by Lemma 6.2. Since $\theta_{V,Z}$ is surjective, their sum
$V\circledast Z$ exhausts all of $V\otimes Z$ by Lemma 6.1. ∎

As an application of this criterion, we derive the following sufficient (and
necessary) condition for complete reducibility.

######  Corollary 6.4.

Suppose that all submodules in $V\otimes Z$ of highest weight are irreducible.
Then $V\otimes Z$ is completely reducible.

###### Proof.

It is sufficient to prove that $\theta_{V,Z}$ is injective and therefore
surjective. That is obvious for its restriction to $\mathbb{C}1_{\nu}\subset
V^{+}_{Z}$, and further we proceed by induction on height. Suppose that $v\in
V^{+}_{Z}$ is a non-zero vector of height $k$ killed by $\theta_{V,Z}$. Then
by Corollary 5.2(iv), $\delta_{l}(v)\in(V\otimes Z)_{k-1}$ along with the
submodule it generates. But that is impossible since, by induction assumption,
$(V\otimes Z)_{k-1}$ is a direct sum of irreducible modules whose canonical
generators have heights $<k$. Therefore $\theta_{V,Z}$ is an isomorphism on
$V^{+}_{Z}$ of height $k$, and $(V\otimes Z)_{k}$ is a sum of highest weight
submodules, by Corollary 5.2(iii) and (iv). This facilitates the induction
transition and proves the assertion. ∎

##  7 Special case of finite-dimensional modules

Since finite-dimensional modules are completely reducible, so is $V\otimes Z$
once $\dim V<\infty$ and $\dim Z<\infty$. Although this case is not
particularly interesting for this exposition, we include it for completeness
and give an independent proof of complete reducibility via a sort of
“unitarian trick”.

Recall that a complex conjugation on a complex vector space $V$ is an
$\mathbb{R}$-linear involution $V\to V$, $v\mapsto\bar{v}$ satisfying
$\overline{cv}=\bar{c}\bar{v}$ for all $c\in\mathbb{C}$ and $v\in V$. It
defines a real structure on $V$, i.e., an $\mathbb{R}$-subspace
$V_{\mathbb{R}}$ of stable vectors, such that
$V=V_{\mathbb{R}}\otimes_{\mathbb{R}}\mathbb{C}$.

A bilinear form is called real if $\overline{\langle
v,w\rangle}=\langle\bar{v},\bar{w}\rangle$ for all $v,w\in V$. A real
structure on $V$ establishes a bijection between bilinear and sesquilinear
forms. For any bilinear form $\langle\cdot,\cdot\rangle$ the form $v\otimes
w\mapsto\langle\bar{v},w\rangle=(v,w)$ is anti-linear in the first argument
and linear in the second. For a real symmetric form
$\langle\cdot,\cdot\rangle$ the form $(\cdot,\cdot)$ is Hermitian. Clearly
they both are degenerate and non-degenerate simultaneously.

Suppose that $q\in\mathbb{R}$ and define a complex conjugation on
$U_{q}(\mathfrak{g})$ as an involutive antilinear Hopf algebra automorphism
acting on the generators by

$\displaystyle\overline{q^{\pm h_{\alpha}}}=q^{\pm
h_{\alpha}},\qquad\bar{f}_{\alpha}=-f_{\alpha},\qquad\bar{e}_{\alpha}=-e_{\alpha}.$

We call a representation on a vector space $V$ with a real structure real if
$\overline{xv}=\bar{x}\bar{v}$, for all $v\in V$ and all $x\in
U_{q}(\mathfrak{g})$. Since $U_{q}(\mathfrak{b}_{+})$ is invariant under the
complex conjugation, it readily follows that a Verma module $M_{\lambda}$ is
real if and only if the highest weight $\lambda$ is in the real span of
$\Pi^{+}$. Then for integral dominant $\lambda$ all the submodules in
$M_{\lambda}$ are real, as well as the irreducible finite-dimensional quotient
of $M_{\lambda}$.

The compact real form on $U_{q}(\mathfrak{g})$ is an antilinear involutive
algebra anti-automorphism and coalgebra automorphism $*$. It is defined on the
generators by

$\displaystyle(q^{\pm h_{\alpha}})^{*}=q^{\pm h_{\alpha}},\qquad
f_{\alpha}^{*}=q^{-h_{\alpha}}e_{\alpha},\qquad
e_{\alpha}^{*}=f_{\alpha}q^{h_{\alpha}}.$

One can check that $\omega\circ*$ is the complex conjugation introduced above.

A representation on a module $V$ with a Hermitian form $(\cdot,\cdot)$ is
called unitary if $(xv,w)=(v,x^{*}w)$ for all $v,w\in V$ and $x\in
U_{q}(\mathfrak{g})$. In what follows, the Hermitian form $(\cdot,\cdot)$ on a
module with real structure is obtained from a $\omega$-contravariant form by
the complex conjugation of the first argument. Then the representation
automatically becomes unitary.

Consider an example of $(n+1)$-dimensional irreducible representation of
$U_{q}(\mathfrak{sl}(2))$ with the highest vector $1_{n}$. Then the Hermitian
form is positive definite. Indeed,

$\displaystyle\big{(}f^{m}1_{n},f^{m}1_{n}\big{)}=\bigl{(}1_{n},\big{(}q^{-h}e\big{)}^{m}f^{m}1_{n}\bigr{)}$
$\displaystyle\hphantom{\big{(}f^{m}1_{n},f^{m}1_{n}\big{)}}{}=\bigl{(}1_{n},[m]_{q}![h]_{q}[h-1]_{q}\cdots[h-m+1]_{q}q^{-mh+m(m+1)}1_{n}\bigr{)}.$

It is equal to
$[n]_{q}[m-1]_{q}\cdots[n-m+1]_{q}q^{-mn+m(m+1)}[m]_{q}!(1_{n},1_{n})>0$ since
$m\leqslant n$. In particular, the Hermitian operator $q^{-h}ef$ is non-
negative and turns zero only on the lowest vector, that is, on $\ker f$.

######  Lemma 7.1.

For any finite-dimensional $U_{q}(\mathfrak{g})$-module $V$, the Hermitian
contravariant form on $V$ relative to the compact real form on
$U_{q}(\mathfrak{g})$ is positive definite.

###### Proof.

Assume for simplicity that $V$ is irreducible. Let $V_{k}\subset V$ denote the
vector subspace spanned by vectors of height $\leqslant k$. We need to prove
that the form is positive definite on all $V_{k}$. It is so on $V_{0}$ by
construction as it is normalized to $(1_{\nu},1_{\nu})=1$ on the highest
vector. Assume the lemma proved for $V_{k}$ with $k\geqslant 0$ and pick up
some non-zero $v$ of height $k+1$. Then $v=f_{\alpha}w\not=0$ for some
$\alpha\in\Pi^{+}$ and $w\in V_{k}$. By induction assumption, the form is
positive definite on $V_{k}$, so we can assume that $w$ is orthogonal to $\ker
f_{\alpha}\cap V_{k}$. As $q^{-h_{\alpha}}e_{\alpha}f_{\alpha}$ is positive on
the span of such vectors,
$(f_{\alpha}w,f_{\alpha}w)=(w,q^{-h_{\alpha}}e_{\alpha}f_{\alpha}w)>0$.
Therefore the form is positive definite on $V_{k+1}$ and on all $V$, by the
induction argument. ∎

If the modules $V$ and $Z$ are equipped with a real structure, then so is
$V\otimes Z$, and $\overline{(V\otimes Z)^{+}}=(V\otimes Z)^{+}$. This implies
that the restriction of the Hermitian form to $(V\otimes Z)^{+}$ is obtained
from the restriction of the canonical form by the conjugation of the first
argument.

######  Proposition 7.2.

Suppose that highest weight modules $V$ and $Z$ are finite-dimensional. Then
the canonical form on $V\otimes Z$ is non-degenerate on $(V\otimes Z)^{+}$.

###### Proof.

The corresponding Hermitian form is positive definite on $V\otimes Z$ and
therefore non-degenerate on $(V\otimes Z)^{+}$. Therefore the canonical
bilinear form is non-degenerate on $(V\otimes Z)^{+}$ as well. ∎

##  8 On computing extremal twist

Observe that the modules $V$ and $Z$ enter the tensor product in a symmetric
way. Therefore one can consider the operator
$\theta_{Z,V}\colon\to{}^{+}\\!Z_{V}$, which comes from the alternative
parametrization of singular vectors with elements of $Z^{+}_{V}$. The
operators $\theta_{V,Z}$ and $\theta_{Z,V}$ are equivalent in the sense that
they are simultaneously invertible. However, from the technical point of view,
one of them may be easier to work with than the other. Such an example can be
found in [12], where one of the spaces $V^{+}_{Z}$, $Z^{+}_{V}$ is easy to
identify while the other is unknown.

The canonical form on $V\otimes Z$ translates the question of complete
reducibility to non-degeneracy of a matrix, which is finite-dimensional when
restricted to each weight subspace. For computational purposes, one can work
either with $\theta_{V,Z}$ or with $\theta_{Z,V}$, depending on a particular
situation. For a special case when $Z$ is a Verma module, $\theta^{-1}_{V,Z}$
is expressed in [7] through the Zhelobenko cocycle [15] via dynamical twist,
which is closely related to $\mathcal{F}$. In this section we will show how
the result of [7] can be used in a more general situation.

Observe that calculation of $\theta_{V,Z}$ through $\Upsilon_{Z}$ involves
more information than required for a particular $V$. In fact, given a singular
vector $u=\sum_{i}v_{i}\otimes z_{i}$, we need to lift only its
$Z$-coefficients: $z_{i}\mapsto f_{i}\in U_{q}(\mathfrak{g}_{-})$. Then
$\theta_{V,Z}(v)=\sum_{i}\gamma^{-1}(f_{i})v_{i}$, where $v\in V$ is the
leading coefficient. This observation leads to the following method of finding
$\theta_{V,Z}$.

For a Verma module $\hat{M}_{\zeta}$, the operator
$\theta_{V,\hat{M}_{\zeta}}$ is a rational trigonometric function of $\zeta$
and may have poles when $\hat{M}_{\zeta}$ becomes reducible. Still we can
consider it on a subspace in $V$, where it is regular. Then we can use
$\theta_{V,\hat{M}_{\zeta}}$ to calculate $\theta_{V,Z}$ through a limit
procedure, as explained below.

Let $V^{+}_{\zeta}$ denote the vector space spanned by leading
$V$-coefficients of singular vectors in
$\big{(}V\otimes\hat{M}_{\zeta}\big{)}^{+}$. Chose a section
$\delta^{\prime}_{l}$ of the map $\big{(}V\otimes\hat{M}_{\zeta}\big{)}^{+}\to
V^{+}_{\zeta}$ assigning the leading coefficient to a singular vector. For
$v\in V^{+}_{\zeta}$ pick up a presentation $u=\sum_{i}v_{i}\otimes
f_{i}1_{\zeta}=\delta^{\prime}_{l}(v)\in V\otimes\hat{M}_{\zeta}$. We define
$\theta_{V,\hat{M}_{\zeta}}\colon V^{+}_{\zeta}\to{}^{+}\\!V_{Z}$ by
projecting $\sum_{i}\gamma^{-1}(f_{i})v_{i}$ to ${}^{+}\\!V_{Z}$ along
$V^{\perp}_{Z}$ and assigning the result to $v$. Note that it is independent
of the choice of $\delta^{\prime}_{l}$. Indeed, $\delta^{\prime}_{l}$ delivers
singular vectors in $V\otimes\hat{M}_{\zeta}$ that survive in $V\otimes Z$. If
a singular vector $u=\sum_{i}v_{i}\otimes f_{i}1_{\zeta}$ vanishes in
$V\otimes Z$, then all $f_{i}1_{\zeta}$ belong to $I^{-}_{Z}1_{\zeta}$. But
then $\sum_{i}\gamma^{-1}(f_{i})v$ is killed in projection to
${}^{+}\\!V_{Z}$.

######  Proposition 8.1.

Suppose that all singular vectors in $V\otimes Z$ are images of singular
vectors in $V\otimes\hat{M}_{\zeta}$. Then $V^{+}_{Z}=V^{+}_{\zeta}$ and
$\theta_{V,Z}=\theta_{V,\hat{M}_{\zeta}}$.

###### Proof.

The image of $\delta_{l}^{\prime}$ consists of singular vectors that survive
in projection $V\otimes\hat{M}_{\zeta}\to V\otimes Z$, hence the equality
$V^{+}_{Z}=V^{+}_{\zeta}$.

Let $u\in V\otimes Z$ be a singular vector with leading coefficient $v\in
V^{+}_{Z}$. By the hypothesis, it is the image of a singular vector
$\hat{u}\in V\otimes\hat{M}_{\zeta}$ with the same leading coefficient. But
then $\theta_{V,\hat{M}_{\zeta}}(v)=\theta_{V,Z}(v)$, because a lift of
$\hat{u}$ is a lift of $u$. ∎

### Acknowledgements

We are grateful to Joseph Bernstein for useful discussions. We are also
indebted to the anonymous referees for careful reading of the text and
valuable remarks. This study was supported by the RFBR grant 15-01-03148.

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