# On 2-Near Perfect Numbers

Vedant Aryan
<EMAIL_ADDRESS>
Dev Madhavani
<EMAIL_ADDRESS>
Savan Parikh
<EMAIL_ADDRESS>
Ingrid Slattery
<EMAIL_ADDRESS>
Joshua Zelinsky
<EMAIL_ADDRESS>

###### Abstract

Let $\sigma(n)$ be the sum of the positive divisors of $n$. A number $n$ is
said to be 2-near perfect if $\sigma(n)=2n+d_{1}+d_{2}$, where $d_{1}$ and
$d_{2}$ are distinct positive divisors of $n$. We give a complete description
of those $n$ which are 2-near perfect and of the form $n=2^{k}p^{i}$ where $p$
is prime and $i\in\\{1,2\\}$. We also prove related results under the
additional restriction where $d_{1}d_{2}=n$.

##  1 Introduction

A perfect number is a positive integer that is equal to the sum of its proper
positive divisors. Equivalently, a perfect number is an $n$ such that
$\sigma(n)=2n$, where $\sigma(n)$ is the sum of all positive divisors of $n$.
Perfect numbers have been studied since antiquity. The idea of perfect numbers
has been generalized in a variety of ways. A classic generalization is the
notion of a multiply perfect number, defined as a number $n$ which satisfies
$\sigma(n)=mn$ for some other integer $m$. Sierpiñski[6] introduced the term
pseudoperfect number to mean a number $2n$ such that $n$ is the sum of some
subset of its divisors. Pollack and Shevelev [5] separated the pseudoperfect
numbers into separate types by introducing the idea of a $s$-near perfect
number. A number $n$ is $s$-near perfect if $2n$ is the sum of all its
positive divisors excepting $s$ of them. For example, while 12 is not perfect,
it is 1-near perfect, since $1+2+3+6+12=2(12)$. Here the divisor which has
been removed from the set is 4. We will refer to divisors removed from the set
as omitted divisors. Note that any pseudoperfect number is $s$-near perfect
for some $s$, and one can think of perfect numbers as $0$-near perfect
numbers. In some sense, multiply perfect numbers are a multiplicative
generalization of perfect numbers, while $s$-near perfect numbers are a more
additive generalization.

A number $n$ is said to be abundant if it satisfies $\sigma(n)>2n$. If $n$ is
$s$-near perfect for some $s>0$, then $n$ must be abundant. However, it is
possible for a number to be abundant while not being $s$-near perfect for any
$s$. An example is 70, where $\sigma(70)=144$. Numbers which are abundant but
not $s$-near perfect for any $s$ are said to be weird. A classic open problem
is whether there are any odd weird numbers.

In addition to the more general notion of $s$-near perfect numbers, Pollack
and Shevelev [5] also used the term near perfect number to mean 1-near perfect
number. Another classic open problem is whether there is any $n$ such that
$\sigma(n)=2n+1$. Such numbers are called quasiperfect numbers. Note that any
quasiperfect number is a $1$-near perfect number with omitted divisor $1$.

Pollack and Shevelev constructed the following three distinct families of
1-near perfect numbers:

  1. 1.

$2^{t-1}(2^{t}-2^{k}-1)$ where $2^{t}-2^{k}-1$ is prime. Here $2^{k}$ is the
omitted divisor.

  2. 2.

$2^{2p-1}(2^{p}-1)$ where $2^{p}-1$ is prime. Here $2^{p}(2^{p}-1)$ is the
omitted divisor.

  3. 3.

$2^{p-1}(2^{p}-1)^{2}$ where $2^{p}-1$ is prime. Here $2^{p}-1$ is the omitted
divisor.

Subsequent work by Ren and Chen [7] showed that all near perfects with two
distinct prime factors must be either 40, or one of the the three families
above.

The only known near perfect odd number is
$173369889=(3^{4})(7^{2})(11^{2})(19^{2}).$ Tang, Ma, and Feng [8] showed that
this is the only odd near perfect number with four or fewer distinct prime
divisors. Cohen, Cordwell, Epstein, Kwan, Lott, and Miller proved general
asymptotics for $s$-near perfect numbers for $s\geq 4.$ Recent work by
Hasanalizade [2] gave a partial classification of near perfect numbers which
are also Fibonacci or Lucas numbers.

Li and Liao [3] classified all even near perfects of the form
$2^{a}p_{1}p_{2}$ where $p_{1}$ and $p_{2}$ are distinct primes.

The main results of this paper are twofold. First, we give a complete
description of $2$-near perfects of the form $2^{k}p$ or $2^{k}p^{2}$ where
$p$ is prime. Second, we use these characterizations to introduce a closely
related notion of strongly 2-near perfect numbers, and give a characterization
of those of the form $2^{k}p$.

In particular, we have the following two main results.

######  Theorem 1.

Assume $n$ is a $2$-near perfect number with omitted divisors $d_{1}$ and
$d_{2}$. Assume further that $n=2^{k}p$ where $p$ is prime and $k$ is a
positive integer. Then one must have, without loss of generality, one of four
situations.

  1. 1.

$p=2^{k}-1$. Here we have $d_{1}=1$ and $d_{2}=p$.

  2. 2.

$p=2^{k+1}-2^{a}-2^{b}-1$ for some $a,b\in\mathbb{N}$. Here $d_{1}=2^{a}$ and
$d_{2}=2^{b}$.

  3. 3.

$p=\frac{2^{k+1}-2^{a}-1}{1+2^{b}}$ for some $a,b\in\mathbb{N}$. Here
$d_{1}=2^{a}$ and $d_{2}=2^{b}p$.

  4. 4.

$p=\frac{2^{k+1}-1}{1+2^{a}+2^{b}}$ for some $a,b\in\mathbb{N}$. Here
$d_{1}=2^{a}p$ and $d_{2}=2^{b}p$.

######  Theorem 2.

Assume that $n$ is a $2$-near perfect number with omitted divisors $d_{1}$ and
$d_{2}$. Assume further that $n=2^{k}p^{2}$ where $p$ is prime. Then
$n\in\\{18,36,200\\}$.

We recall the following basic facts about $\sigma(n)$ will be useful
throughout:

######  Lemma 3.

The function $\sigma(n)$ has the following properties:

  1. 1.

$\sigma(n)$ is multiplicative. That is, $\sigma(ab)=\sigma(a)\sigma(b)$
whenever $a$ and $b$ are relatively prime.

  2. 2.

For a prime $p$, $\sigma(p^{k})=p^{k}+p^{k-1}\cdots+1=\frac{p^{k+1}-1}{p-1}.$

##  2 Proof of Theorem 1.

Let us now prove Theorem 1.

###### Proof.

Assume we have a 2-near perfect number of the form $n=2^{k}p$ with two omitted
divisors $d_{1},d_{2}$, $d_{1}\neq d_{2}$ and odd prime $p$. Because $n$ is
near perfect, we have that:

$\sigma(n)=2n+d_{1}+d_{2}.$ (1)

Using Lemma 3, we then have:

$\sigma(n)=\sigma(2^{k}p)=(2^{k+1}-1)(p+1)$ (2)

So, setting equation (1) equal to equation (2), we have:

$(2^{k+1}-1)(p+1)=2n+d_{1}+d_{2}=2^{k+1}p+d_{1}+d_{2},$

and hence

$p=2^{k+1}-1-d_{1}-d_{2}.$ (3)

Because $p$ is odd, we have that $2^{k+1}-1-d_{1}-d_{2}$ is odd. Since
$2^{k+1}-1$ is always odd, we have that $-(d_{1}+d_{2})$ must be even. If
$-(d_{1}+d_{2})$ is even, $d_{1}$ and $d_{2}$ must be of the same parity. We
thus need to consider two situations: where $d_{1},d_{2}$ are both odd, and
where they are both even. We will call the first situation Case 1, and we
shall separate the second situation, Case 2, into three separate subcases
without loss of generality.

Case 1: In this case, $d_{1},d_{2}$ are both odd.

The only odd divisors of $n$ are $1$ and $p$, so we can, without loss of
generality, set $d_{1}=1$ and $d_{2}=p$ to find:

$p=2^{k+1}-1-1-p,$

and hence

$p=2^{k}-1.$

Thus, our first family of 2-near perfect numbers correspond to Mersenne primes
and have the form $2^{k}(2^{k}-1)$ (twice an even perfect number).

We now consider the situation where where $d_{1},d_{2}$ are both even. We
shall break this down into three subcases, depending on the types of values
for $d_{1}$ and $d_{2}$.

Case 2.1: In this case we have $d_{1}=2^{a},d_{2}=2^{b},$ where $0<a<b\leq k$
We can use in our definitions of $d_{1}$ and $d_{2}$ in Equation (3) to find:

$p=2^{k+1}-2^{a}-2^{b}-1$

This is our second family of 2-near perfect numbers.

Case 2.2: In this case, $d_{1},d_{2}$ are both even, and
$d_{1}=2^{a},d_{2}=2^{b}p,$ and $a,b\in(0,k]$

We use a similar strategy, and plug in our definitions of $d_{1},d_{2}$ into
Equation (3):

$p=2^{k+1}-2^{a}-2^{b}p-1,$

and so

$p(1+2^{b})=2^{k+1}-2^{a}-1,$

which becomes

$p=\frac{2^{k+1}-2^{a}-1}{1+2^{b}}.$

This is our third family of 2-near perfect numbers.

Case 2.3: In this case, $d_{1},d_{2}$ are both even and we have
$d_{1}=2^{a}p,d_{2}=2^{b}p,$ and $0<a<b\leq k$

Using the same technique, equation (3) tells us:

$p=2^{k+1}-1-2^{a}p-2^{b}p,$

and hence,

$p(1+2^{a}+2^{b})=2^{k+1}-1,$

which implies that

$p=\frac{2^{k+1}-1}{1+2^{a}+2^{b}}.$

This is the fourth and final family of 2-near perfect numbers. ∎

##  3 Proof of Theorem 2

One major technique we will use is what we call the discriminant sandwich
method: we show that a given Diophantine equation has only a restricted set of
possible solutions. We do so by showing that the equation is a quadratic
equation in one variable, and thus in order to have integer valued solutions,
the discriminant must be a perfect square. However, we will show that the
discriminant must, except in a limited set of cases, be shown to be strictly
between two consecutive perfect squares, and thus aside from those situations,
we have no solution. Discriminant sandwiching will be used extensively in what
follows.

######  Lemma 4.

Let $a$ and $k$ be positive integers such that $D=2^{2k+2}+2^{k+2}-2^{a+2}-7$.
If $0\leq a\leq k$ and $D$ is a perfect square, then $k=a=1$.

###### Proof.

Let us assume that $D$ is a perfect square. For all $a$, note that

$2^{2k+2}+2^{k+2}-2^{a+2}-7<2^{2k+2}+2^{k+2}+2=(2^{k+1}+1)^{2}$

Thus, if we have

$D=2^{2k+2}+2^{k+2}-2^{a+2}-7>2^{2k+2}=(2^{k+1})^{2}$ (4)

then the quantity in question cannot be a perfect square because it is
sandwiched between two consecutive perfect squares. So we must have that

$2^{2k+2}+2^{k+2}-2^{a+2}-7\leq 2^{2k+2}$

and therefore

$2^{k+2}-2^{a+2}\leq 7.$ (5)

If $k>a\geq 1$, then from Equation 5 we have that $2^{k+2}-2^{k+1}\leq
2^{k+2}-2^{a+2}\leq 7$. Thus,

$2^{k+2}-2^{k+1}=2^{k+1}\leq 7,$

which implies that $k\leq 1$.

However, given the conditions for this case, no solutions are possible.

Now, consider the case when $k=a\geq 1$. In this case, we have

$D=2^{2k+2}+2^{k+2}-2^{k+2}-7=2^{2k+2}-7.$

Given this, note that:

$2^{2k+2}-7<2^{2k+2}=(2^{k+1})^{2}$

Using the same logic as earlier, we see that if

$D>(2^{k+1}-1)^{2},$

then $D$ will be sandwiched between two consecutive perfect squares, and thus
will not be a square itself. Thus, we can assume:

$2^{2k+2}-7\leq(2^{k+1}-1)^{2}=2^{2k+2}-2^{k+2}+1,$

which implies that $k\leq 1$. The bounds for this case require $k\geq 1$, so
the only solution possible is $(a,k)=(1,1)$. ∎

Essentially, Lemma 4 is the sandwiching part of the discriminant sandwich we
will use in the proof of Proposition 8 below.

######  Lemma 5.

Let $b$ and $k$ be non-negative integers and $p$ be an odd number such that

$(2^{k+1}-1)(p^{2}+p+1)=2^{k+1}p^{2}+2^{b}p+1.$ (6)

then $p|2^{k}-1$, and $p+1|2^{b}-2$.

###### Proof.

Assume one has a solution to Equation 6. Then, if we take the equation modulo
$p$, we get that

$2^{k+1}-1\equiv 1\mod p,$

and hence $p|2^{k+1}-2=2(2^{k}-1)$. Since $p$ is odd, we have $p|2^{k}-1$. To
prove the second half, observe that we can rewrite our initial equation as

$2^{k+1}(p+1)=2^{b}p+p^{2}+p+2,$

which implies that $p+1|2^{b}p+p^{2}+p+2$. Hence $p+1|2^{b}p+p^{2}-p$, and so
$p+1|p(2^{b}+p-1)$. Since $p$ and $p+1$ are relatively prime, we have then
that $p+1|2^{b}+p-1$ Finally, we take away another multiple of $p+1$ to get
$p+1|2^{b}-2$. ∎

######  Lemma 6.

The equation

$(2^{k+1}-1)(p^{2}+p+1)=2^{k+1}p^{2}+2^{b}p+1$ (7)

has no solutions where $p$ is odd, and $2\leq b\leq k-1$.

###### Proof.

Assume we have a solution to the equation. From Lemma 5, we choose an integer
$x$ such that $x(p+1)=2^{b}-2$. Note that $x$ must be odd since $p+1$ is even,
and $2^{b}-2$ is not divisible by 4. We then have $2^{b}=xp+x+2$. When we
substitute this back into Equation 7, and solve for $2^{k+1}$, we get that
$2^{k+1}=xp+p+2$. By taking the difference of these two expressions, we
obtain:

$2^{k+1}-2^{b}=(xp+p+2)-(xp+x+2)=p-x.$

Because $b\leq k-1$, $2^{k+1}-2^{b}>2^{k}$, we have $p-x>2^{k}$, and hence
$p>2^{k}+1$. But this contradicts Lemma 5, since we must have $p|2^{k}-1$. ∎

######  Lemma 7.

Let $n$ be a 2-near perfect number of the form $n=2^{k}p^{2}$ where $p$ is an
odd prime. Assume further that the omitted divisors of $n$ are $d_{1}$ and
$d_{2}$. Then, we have

$d_{1}+d_{2}=-p^{2}+(2^{k+1}-1)p+(2^{k+1}-1),$ (8)

and $d_{1}$ and $d_{2}$ are of opposite parity.

###### Proof.

Assume as given. Then we have

$\sigma(n)-d_{1}-d_{2}=2n$
$\sigma(2^{k}p^{2})-d_{1}-d_{2}=2(2^{k}p^{2})=2^{k+1}p^{2}$
$(2^{k+1}-1)(p^{2}+p+1)-d_{1}-d_{2}=2^{k+1}p^{2}$
$d_{1}+d_{2}=-2^{k+1}p^{2}+(2^{k+1}-1)(p^{2}+p+1).$

which is equivalent to Equation 8. Since the right-hand side of Equation 8 is
odd, $d_{1}$ and $d_{2}$ must be of opposite parity. ∎

Let’s look at the possible divisors of $n$, assuming $n=2^{k}p^{2}$. Every
possible divisor can be of one of three types. Type I divisors are powers of
2, that is $d=2^{a}$ for some $0\leq a\leq k$. Type II divisors are $p$ or
$p^{2}$. That is $d=p^{m}$ where $m\in\\{1,2\\}$. Type III divisors are of the
form $d=2^{b}p^{j}$ where $0<b\leq k$ and $j\in\\{1,2\\}$.

We may then, without loss of generality, break down our situation into the
following six cases depending on all the possible combinations of omitted
divisor types, $d_{1}$ and $d_{2}$.

Case | $d_{1}$ | $d_{2}$
---|---|---
1 | I | I
2 | I | II
3 | I | III
4 | II | II
5 | II | III
6 | III | III

We will handle each of these six cases separately. But before we do, we will
observe that Cases 4 and 6 are both trivial since they require that both
$d_{1}$ and $d_{2}$ of the same parity, which contradicts Lemma 7. We thus
only consider Cases 1, 2, 3, and 5.

######  Proposition 8.

If $n$ is a 2-near perfect number of the form $n=2^{k}p^{2}$, where $p$ is an
odd prime, with omitted divisors of Case 1 form, then $n=18$, and the omitted
divisors are $1$ and $2$.

###### Proof.

Assume we are in Case 1. In this case, both $d_{1}$ and $d_{2}$ must be
distinct powers of 2. Since $d_{1}+d_{2}$ is odd, one of the omitted divisors
must be odd (and hence equal to 1). Without loss of generality, we will set
$d_{1}=1$, and $d_{2}=2^{a}$ where $1\leq a\leq k$. Putting this into Equation
8, we get that

$p^{2}-(2^{k+1}-1)p-(2^{k+1}-2^{a}-2)=0.$ (9)

Equation 9 is a quadratic equation in $p$. Thus, in order to have a solution,
its discriminant, defined as

$D=2^{2k+2}+2^{k+2}-2^{a+2}-7$

must be a perfect square. From Lemma 4, $D$ is only a perfect square if
$k=a=1$. In this case, Equation 9 becomes just $p^{2}-3p=0$. Thus, one must
have $p=3$, and so $n=18$ with $d_{1}=1,d_{2}=2$. One can verify this result:
$2(18)=\sigma(18)-(1+2)$.

∎

######  Proposition 9.

There are no 2-near perfect numbers of the form $2^{k}p^{2}$ with omitted
divisors of the Case 2.

###### Proof.

We will apply the discriminant sandwich method to this situation. Again we are
working with $n=2^{k}p^{2}$, but we now have omitted divisors of the following
form $d_{1}=2^{a}$, $d_{2}=p^{m}$ where $a\in(0,k]$ and $m\in[1,2].$ We will
break Case 2 down into two subcases, depending on whether $m=1$ or $m=2$.

We first consider the situation where $m=1$. Applying this to Equation 8, we
obtain

$0=p^{2}-(2^{k+1}-2)p-(2^{k+1}-2^{a}-1).$ (10)

Equation 10 has an even discriminant $D$, which means that if $D$ is a perfect
square, it must be divisible by 4. Thus, we can define

$D^{\prime}=\frac{D}{4}=2^{2k}-2^{a},$

and just as well assume that $D^{\prime}$ is a perfect square. Note that
$D^{\prime}$ is still even, so we can skip over checks against odd squares. We
have that $2^{2k}-2^{a}<(2^{k})^{2}$, and so

$2^{2k}-2^{a}\leq(2^{k}-2)^{2}.$

With a little algebra we then obtain that

$2^{k+2}-2^{a}\leq 4$

Without loss of generalization, let’s say $k=a+m,m\in[0,k),a>0$. We then have
that

$2^{a}(2^{m+2}-1)=2^{a+m+2}-2^{a}\leq 4.$

It is evident after plugging in the minimum values for $a$ and $m$ that no
solution exists. We thus have shown that when $m=1$, no solution exists.

We now consider the case when $m=2.$ We then obtain from Equation 8,

$0=2p^{2}-(2^{k+1}-1)p-2^{k+1}+2^{a}+1.$ (11)

We then need that the discriminant D, defined as

$D=2^{2k+2}-2^{k+2}+2^{k+4}-2^{a+3}-7.$

is a perfect square. We thus must have

$D<(2^{k+1}+3)^{2}.$

Since $D$ is odd, it cannot be equal to the next smallest square, which is
even. So,

$D\leq(2^{k+1}+1)^{2}.$

We thus have

$2^{2k+2}-2^{k+2}+2^{k+4}-2^{a+3}-7\leq 2^{2k+2}+2^{k+2}+1,$

which implies that

$2^{k}-2^{a}\leq 1.$

Thus, we can only have a solution when $k=a$ or we have $k=1$ and $a=0$.
However, since $d_{1}$ and $d_{2}$ must be of opposite parity, we cannot have
$a=0$. Thus, we need consider only the case when $k=a$. Our expression for $D$
simplifies so that we have $D=2^{2k+2}+2^{k+2}-7$. But this quantity cannot be
a perfect square since

$(2^{k+1})^{2}<2^{2k+2}+2^{k+2}-7<(2^{k+1}+1)^{2},$

and so $D$ is again sandwiched between two consecutive perfect squares. Thus,
there are no solutions for Case 2 when $m=2$. Since no solutions exist for all
possible cases for $m$, Proposition 9 is proven. ∎

######  Lemma 10.

If $p$ is an odd number such that

$(2^{k+1}-1)(p^{2}+p+1)=2^{k+1}p^{2}+2^{b}p^{2}+1$ (12)

where $k$ and $b$ are positive integers, then $p|2^{k}-1$.

###### Proof.

Assume one has a solution to Equation 12. Then, if we take the equation modulo
$p$, we get that

$2^{k+1}-1\equiv 1\mod p,$

and hence $p|2^{k+1}-2=2(2^{k}-1)$. Since $p$ is odd, we have $p|2^{k}-1$. ∎

######  Lemma 11.

If $p$ is an odd number which is a solution to Equation 12, then
$p+1|2^{b}+2$.

###### Proof.

Assume one has a solution to Equation 12. We can rewrite this as

$2^{k+1}(p+1)=(2^{b}+1)p^{2}+p+2$

which implies that

$p+1|(2^{b}+1)p^{2}+p+2,$

and hence

$p+1|(2^{b}+1)p^{2}-p=p((2^{b}+1)p-1).$

Since $p$ and $p+1$ are relatively prime,

$p+1|(2^{b}+1)p-1,$

and so by similar logic,

$p+1|2^{b}+2.$

∎

Note that Lemma 11 is distinct from Lemma 5, since the equations needed are
different, and one has a $+2$, and the other a $-2$ on the right-hand side.

######  Proposition 12.

Let $n=2^{k}p^{2}$ be a 2-near perfect number with omitted divisors of the
Case 3 form with omitted divisors $1$ and $2^{b}p^{2}$. Then $n=36$, and our
omitted divisors are $1$ and $18.$

###### Proof.

Assume as given. So from Equation 8, we have some $b$ such $b\leq k$, and $p$
prime such that

$(2^{k+1}-1)(p^{2}+p+1)=2^{k+1}p^{2}+2^{b}p^{2}+1.$ (13)

By Lemma 10 and 11, we have

$p|2^{k}-1$

and

$p+1|2^{b}+2.$

Thus, there is a positive integer $z$ such that

$z(p+1)=2^{b}+2,$

and

$z(p+1)-2=2^{b}.$

If we take Equation 13 modulo $2^{b}$ we also get that $2^{b}|p^{2}+p+2$.

We also have

$z(p+1)-2|p^{2}+p+z(p+1)$ $z(p+1)-2|p(p+1)+z(p+1)$ $z(p+1)-2|(p+1)(p+z).$ (14)

Let $Q$ be some integer such that $Q|z(p+1)-2$ and $Q|(p+1)$. Then Q will
divide any linear combination of those terms. Thus we have

$Q|z(p+1)-2-z(p+1)=-2.$

Thus, the only possible common factors of $z(p+1)-2$ and $(p+1)$ are 1 and 2.
Hence Equation 14 may be strengthened to:

$z(p+1)-2|2(p+z)$ (15)

Therefore, we know that

$z(p+1)-2\leq 2(p+z),$

which implies that

$z\leq\frac{2p+2}{p-1}=2+\frac{4}{p-1}.$

Since $p\geq 3$, we have $z\leq 3$, and hence have only three cases, $z=1$,
$z=2$ or $z=3$. One can easily check that if $p=3$ then the only case which
leads to integer values is when $z=1$ and $b=1$. Here $n=36$, and our omitted
divisors are $d_{1}=1,d_{2}=18$. Thus, we may assume that $p>3$, which implies
$z=1$ or $z=2$. However, if $b=1$, we get a contradiction if $p>3$. Thus, may
assume that $b>1$ which forces $z$ to be odd, and so $z=1$.

We have from Equation 15, $p-1|2(p+1)$. So there is some $m$ such that
$m(p-1)=2(p+1)$. If $m=1$ then we get a negative value for $p$ and if $m=2$,
we immediately get a contradiction. So we may assume that $m\geq 3$. If $m=3$,
then we have $3(p-1)=2(p+1)$, which yields $p=5$ which quickly leads to a
contradiction. We thus must have $m\geq 4$. However, if $4(p-1)\leq 2(p+1)$,
then one must have $p=3$, but we are in the situation where $p>3$.

Thus, our only possibility is when $n=36$. ∎

######  Proposition 13.

If $n$ is a 2-near perfect number of the form $n=2^{k}p^{2}$, where $p$ is an
odd prime, with omitted divisors of Case V form, then $n=200$.

###### Proof.

Assume we have such an $n$, with omitted divisors $d_{1}$ and $d_{2}$. Then,
without loss of generality, we may assume that $d_{1}=p^{j}$ for $j\in[1,2]$
and $d_{2}=2^{b}p^{g}$ for some $g\in[1,2]$, and $1\leq b\leq k$. We may break
this down into four cases:

Case | $d_{1}$ | $d_{2}$
---|---|---
1 | $p$ | $2^{b}p$
2 | $p$ | $2^{b}p^{2}$
3 | $p^{2}$ | $2^{b}p$
4 | $p^{2}$ | $2^{b}p^{2}$

Case 1 can be handled by the discriminant sandwich technique.

For Case 1, Equation 8 becomes

$p+2^{b}p=-p^{2}+(2^{k+1}-1)p+(2^{k+1}-1).$ (16)

The relevant discriminant from Equation 16 is

$D=2^{2k+2}-2^{k+b+2}+2^{2b}+2^{b+2}.$

We have then

$(2^{k+1}-2^{b})^{2}<D<(2^{k+1}-2^{b}+1)^{2},$

so $D$ cannot be a perfect square. Thus the equation has no solutions.
Now, consider Case 2. In this case Equation 8 becomes

$p+2^{b}p^{2}=-p^{2}+(2^{k+1}-1)p+(2^{k+1}-1).$ (17)

The relevant discriminant from Equation 17 is

$D=2^{2k+2}+2^{k+b+3}-2^{b+2}.$ (18)

We wish to show that such a $D$ is not a perfect square. We do so by splitting
into two subcases, when $b$ is even and when $b$ is odd. In both cases we will
get a contradiction from assuming that $D$ is a perfect square.

First, assume that $b$ is even. Then $2^{b+2}$ is a perfect square, and so if
$D$ is a perfect square, we may factor out $2^{b+2}$ from Equation 18. In that
case, we have then that

$D_{0}=2^{2k-b}+2^{k+1}-1$

must be a perfect square. However, if $b<k$, then $D_{0}\equiv 3$ (mod 4), and
thus $D_{0}$ cannot be a pefect square. Thus, we must have $b=k$. In that
situation we have $D_{0}=2^{k}+2^{k+1}-1$. However, since $b$ is even and
greater than $1$, we must then have $k\geq 2$. Thus, we still have
$D_{0}\equiv 3$ (mod 4).

Now, for our second subcase, assume that $b$ is odd. Then $2^{b+1}$ is a
perfect square, and so we may factor that quantity out of $D$ and still have a
perfect square. We thus have that $D_{1}=2^{2k-b+1}+2^{k+2}-2$ must be a
perfect square. However, we have then that $D_{1}\equiv 2$ (mod 4), and so we
have again reached a contradiction.

Since both cases lead to a contradiction, we conclude that the relevant
discriminant is never a perfect square, and thus the equation has no
solutions.

We now consider Case 3, where $d_{1}=p^{2}$ and $d_{2}=2^{b}p^{2}$. Equation 8
then becomes:

$p^{2}+2^{b}p=-p^{2}+(2^{k+1}-1)p+2^{k+1}-1,$

and hence

$-2p^{2}+(2^{k+1}-2^{b}-1)p+2^{k+1}-1=0.$ (19)

This has a corresponding discriminant value given by

$D=x^{2}-2x(2^{b}-3)+2^{2b}+2^{b+1}-7,$ (20)

where $x=2^{k+1}$. We note that if $b=1$, then we get that either $p=-1$ or
$p=\frac{2^{k+1}-1}{2}$, neither of which is a prime. Thus, we may assume that
$b>1$. Since $b>1,$ we have

$2^{b+3}>16,$

which implies that

$2^{b+1}-7>-6\cdot 2^{b}+9.$

We then obtain that

$x^{2}-2x(2^{b}-3)+2^{2b}+2^{b+1}-7>x^{2}-2x(2^{b}-3)+2^{2b}-6\cdot 2^{b}+9,$

which implies that

$D>x^{2}-2x(2^{b}-3)+(2^{b}-3)^{2}=(x-(2^{b}-3))^{2}.$

If we have a solution to our original equation, $D$ must be a perfect square.
Equation 20 also shows that $D$ must be odd. Thus, we cannot have
$D=(x-(2^{b}-2))^{2},$ and thus we have

$D\geq(x-(2^{b}-1))^{2}$ (21)

$3\cdot 2^{b}\geq 2^{k+1}+8,$ (22) $4\cdot 2^{b}>2^{k+1}.$ (23)

Inequality 23 implies that $b>k-1$. Since we have that $b\leq k$, and $b$ is a
natural number, we conclude that $b=k$. We thus may replace $k$ with $b$ in
Equation 19 to obtain

$p^{2}+2^{b}p=-p^{2}+(2^{b+1}-1)p+2^{b+1}-1,$

which is equivalent to

$2^{b}(p+2)=2p^{2}+p+1.$ (24)

Thus, we have $p+2|2p^{2}+p+1$. We then have:

$p+2|(2p^{2}+p+1)+(3-2p)(p+2)=7$

Since $p+2|7$, we must have $p=5$. We then can solve to get that $b=k=3$. This
yields $n=(2^{3})(5^{2})=200$, which is in fact a 2-near perfect number of the
desired form. Here our omitted divisors are $25$ and $40$.

We now turn our attention to Case 4. That is, we have $d_{1}=p^{2}$ and
$d_{2}=2^{b}p^{2}$. In this situation Equation 8 becomes

$(2^{k+1}-1)(p^{2}+p+1)-2^{k+1}p^{2}=p^{2}+2^{b}p^{2},$

which can be rewritten as

$2^{k+1}(p+1)=2p^{2}+2^{b}p^{2}+p+1.$ (25)

Therefore, we have:

$(p+1)|(2p^{2}+2^{b}p^{2}+p+1),$
$(p+1)|(2p^{2}+2^{b}p^{2}+p+1)-(p+1)=2p^{2}+2^{b}p^{2},$

and thus

$(p+1)|p^{2}(2+2^{b}).$

Since $p+1$ and $p^{2}$ are relatively prime this implies

$(p+1)|(2+2^{b}).$

Thus, there exists a positive integer $z$ such that $z(p+1)=2+2^{b},$ and
hence

$2^{b}=z(p+1)-2.$ (26)

We note that $p+1$ is even and the only way which $2+2^{b}$ can be divisible
by 4 is if $b=1$ (which does not lead to a solution). Thus, $z$ is odd.

If we take Equation 25 modulo $2^{b}$, we obtain that

$2^{b}|2p^{2}+p+1.$ (27)

We note that Equation 27 allows one to obtain a finite set of possible values
$b$ for any given fixed choice of $p$, and then use each to solve for $k$. We
may thus with only a small amount of effort verify that we must have $p>23$.

We may combine Equation 27 with Equation 26 to obtain:

$z(p+1)-2|2p^{2}+p+1.$

We then have:

$zp+z-2|2p^{2}+p+1$ $zp+z-2|2z(2p^{2}+p+1)-(4p-2)(zp+z-2)=4z+8p-4$

and so

$zp+z-2|4(z+2p-1).$ (28)

Consider now the situation where we have equality in the relationship in
Equation 28. Then we have

$zp+z-2=4(z+2p-1),$

which is equivalent to $z(p+1)=8p+2$. Thus,

$p+1|8p+2,$

and hence

$p+1|6p.$

Since $p+1$ and $p$ are relatively prime, this forces us to have $p+1|6$, and
hence we must have $p=5$, which we can verify does not work. Thus, we must
have some integer $m\geq 2$ such that

$m(zp+z-2)=4(z+2p-1),$

and thus we have

$zp+z-2\leq 2(z+2p-1),$

which is equivalent to

$z\leq\frac{4p}{p-1}.$ (29)

We have that $p\geq 7$, and thus, Inequality 29 implies that $z<6$. Since $z$
is odd, we must then have $z=1$, $z=3$ or $z=5$. If we have $z=1$, then
Equation 28 implies that $p-1|8p.$ But since $p-1$ is relatively prime to $p$,
we must have $p-1|8,$ which is impossible since we know that $p>23$.

Using similar logic, for $z=3$, we obtain from Equation 28 that

$3p+1|4(2p+2)=8(p+1)$ $3p+1|8p+8-8(3p+1)=-16p.$

Thus, $3p+1$ is relatively prime to $p$, so we must have that

$3p+1|16.$

But, once again, we have $p>23$, so there cannot be any solution.

Finally, for $z=5$, we obtain, from Equation 28, that

$5p+3|4(2p+4)=8p+16,$

and thus

$5p+3|3(8p+16)=24p+48$ $5p+3|24p+48-16(5p+3)=-56p.$

Since $p\neq 3$, $5p+3$ and $p$ are relatively prime, so we must have

$5p+3|56.$

But, once again this contradicts that $p>23$.

∎

Thus, we have completed the proof of Theorem 2.

##  4 Strongly 2 near perfect numbers

A slightly different way of defining a number to be pseudoperfect is to say
that a number $n$ is pseudoperfect if there is a set $S$ which is a subset of
the positive divisors of $n$ such that the sum of the elements in $S$ sums to
$2n$. The last author and Tim McCormack [4] studied what they called strongly
pseudoperfect numbers. A number $n$ is said to be strongly pseudoperfect if
there is a subset $S$ of divisors of $n$ where the sum of the elements sums to
$2n$, and where we also have the property that $d\in S$ if and only if
$\frac{n}{d}\in S$. It is natural to combine the notion of 2-near perfect and
strongly pseudoperfect as follows: We say that a number $n$ is strongly 2-near
perfect if $n$ is strongly pseudoperfect and also 2-near perfect. Note that
this is equivalent to $n$ having a divisor $d$ such that

$\sigma(n)-d-\frac{n}{d}=2n.$

The following table gives all seven strongly 2-near perfect numbers less than
one million:

$n$ | $\sigma(n)$ | $d_{1}$ | $d_{2}$
---|---|---|---
156 | 392 | 2 | 78
352 | 756 | 8 | 44
6832 | 15376 | 4 | 1708
60976 | 122512 | 148 | 412
91648 | 184140 | 128 | 716
152812 | 306432 | 302 | 506
260865 | 539136 | 15 | 17391

In this section, we will give a description of all strongly 2-near perfect
numbers $n$ of the form $n=2^{k}p$ for a prime $p$.

######  Lemma 14.

If $n$ is a strong 2-near perfect number of the form $2^{k}p$ for some odd
prime $p$ and natural number $k$, then $p=\frac{2^{k+1}-2^{a}-1}{1+2^{k-a}}$.

###### Proof.

Assume we have a strong 2-near perfect number. By looking at our four families
of numbers which arise from Theorem 1, we can see that only numbers in the
third family might possibly be strongly 2-near perfect. In the first family,
the product of omitted divisors $d_{1}d_{2}$ is odd, so one cannot have
$d_{1}d_{2}=n$ since $n$ is even. In the second family, we have $d_{1}d_{2}$
is a power of 2, and thus is not $n$. In our fourth family, we have
$p^{2}|d_{1}d_{2}$ so $d_{1}d_{2}\neq n$.

Thus, we may assume that we have a number arising from the third family. In
that situation, from $d_{1}d_{2}=n$ we get that $a+b=k$, from which the result
follows. ∎

######  Lemma 15.

Assume that $n$ is a strong $2$-near perfect number of the form $n=2^{k}p$
with $p=\frac{2^{k+1}-2^{a}-1}{1+2^{k-a}}$. Then $k<2a$.

###### Proof.

Assume that $n$ is a strong $2$-near perfect number of the form $n=2^{k}p$
with $p=\frac{2^{k+1}-2^{a}-1}{1+2^{k-a}}$, and that $k\geq 2a$. Thus, we have

$1+2^{k-a}|2^{k+1}-2^{a}-1,$

which implies that

$1+2^{k-a}|2^{k+1}-2^{a}-1+(1+2^{k-a})=2^{k+1}-2^{a}+2^{k-a}=2^{a}(2^{k+1-a}+2^{k-2a}-1).$

Since $k\geq 2a$, $2^{k+1-a}+2^{k-2a}-1$ is a positive integer. We also have
that $(1+2^{k-a},2^{a})=1$, so we have then that

$1+2^{k-a}|2^{k+1-a}+2^{k-2a}-1.$ (30)

Consider the situation where $k=2a$. Then Equation 30, becomes

$1+2^{a}|2^{a+1},$

which has no solutions. So, we may assume that $k>2a$.

We have from Equation 30 that there is some $m$ such that

$m(1+2^{k-a})=2^{k+1-a}+2^{k-2a}-1.$ (31)

Note that the right-hand side of the equation is odd, so $m$ must be odd. If
$m=1$ then we have

$(1+2^{k-a})=2^{k+1-a}+2^{k-2a}-1,$

which implies that

$1+2^{k-a-1}=2^{k-a}-2^{k-2a-1}.$ (32)

The left-hand side of Equation 32 is odd, and the only way for the right-hand
side to be odd is if $k-2a-1=0$. The only solution of this system of equations
is when $k=3$ and $a=1$, which forces $p=\frac{13}{5}$ which is not an
integer. Thus, we have $m\neq 1$, and so $m\geq 3$.

We thus have $3(1+2^{k-a-1})\leq 2^{k-a}-2^{k-2a-1}$, which is impossible.

∎

######  Proposition 16.

Assume that $n$ is a strong $2$-near perfect number of the form $n=2^{k}p$
with $p=\frac{2^{k+1}-2^{a}-1}{1+2^{k-a}}$. Then $k=a+2$, and the omitted
divisors are $d_{1}=2^{a}$ and $d_{2}=4p$, with $p=\frac{2^{a+3}-2^{a}-1}{5}$,
and $a\equiv 3$ (mod 4).

First, we need to prove the following lemma.

######  Lemma 17.

If $2^{b}+1|3(2^{a})+1$, then $b=2$.

###### Proof.

Assume that $2^{b}+1|3(2^{a})+1$. We thus have for some positive integer $m$,

$m(2^{b}+1)=3(2^{a})+1.$ (33)

Notice that $3(2^{a})+1$ is never divisible by 3, and thus $b$ must be even,
and $m$ cannot be divisible by 3. If $m=1$ then we have $2^{b}+1=3(2^{a})+1$
which would imply we would have $3|2^{b}$, which cannot happen. Thus, we may
assume that $m\geq 5$. This implies that $a>b$.

Set $a=bq+r$ where $0\leq r<b$.

We have

$(3)(2^{qb+r}+1)=(3)((2^{b})^{q}2^{r}+1)\equiv
3((-1)^{q}2^{r}+1)\pmod{2^{b}+1}\equiv 0.$ (34)

Now, we separate into two cases, depending on whether $q$ is even or odd. If
$q$ is even, then Equation 34 yields that

$3(2^{r})+1\equiv 0\pmod{2^{b}+1},$ (35)

and so

$2^{b}+1|3(2^{r})+1$ (36)

Equation 36 implies that $r\geq b-1$. We thus have $r=b-1$. Thus,

$2^{b}+1|3(2^{b-1})+1=2^{b}+1+2^{b-1}$

which is impossible.

We now consider the case where $q$ is odd. Then Equation 34 implies that

$-3(2^{r})+1\equiv 0\pmod{2^{b}+1}.$ (37)

and thus,

$2^{b}+1|3(2^{r})-1,$

which similarly leads to a contradiction, unless $b=2$ and $r=1$.

∎

One might wonder if Lemma 17 can be strengthened to conclude that if $p$ is a
prime where $p|2^{b}+1$ for some even $b$ and $p|3(2^{a})+1$ for some $a$,
then one must have $p=5$. However, this is in fact not true. In particular,
note that $29|2^{14}+1$, but it is also true that $29|3(2^{9})+1.$

We now prove Proposition 16.

###### Proof.

Assume that $n$ is a strongly $2$-near perfect number of the form $n=2^{k}p$
with $p=\frac{2^{k+1}-2^{a}-1}{1+2^{k-a}}$. Our proof is complete if we can
show that we must have $k=a+2$. If we have $k=a+b$, then this is the same as
$2^{b}+1|2^{a+b+1}-2^{a}-1$, which implies that
$2^{b}+1|2^{a+b+1}-2^{a}+2^{b}$. We have

$2^{b}+1|2^{a+b+1}-2^{a}-1-2^{a+1}(1+2^{b})=-3(2^{a})-1$

and so $2^{b}+1|3(2^{a})+1$, which allows us to apply Lemma 17, to conclude
that $b=2$, and the rest follows simply from noting that all 2-near perfect of
this form are in Case 3. ∎

We list below the first few values of $a$ where $\frac{2^{a+3}-2^{a}-1}{5}$ is
prime, and its corresponding prime $p$, each of which corresponds to a strong
2-near perfect number of the form $2^{a+2}p$. We do not include the last two
primes as they are too big to fit on one line.

$a$ | $p$
---|---
3 | 11
7 | 179
19 | 734003
27 | 187904819
31 | 3006477107
39 | 769658139443
151 | 3996293539576687666963200714458586381871690547
1 99 | –
451 | –

Standard heuristic arguments suggest that there should be infinitely many
primes of the form $\frac{2^{a+3}-2^{a}-1}{5}$.

##  5 Open problems

One obvious direction is try to extend the classification of 2-near perfect
numbers to classify all of the form $2^{k}p^{m}$ where $m\geq 3$.

######  Conjecture 18.

There are only finitely many 2-near perfect numbers of the form $2^{k}p^{m}$
where $m\geq 2$.

A slightly weaker conjecture is the following.

######  Conjecture 19.

For any fixed $m\geq 2$, there are only finitely many 2-near perfect numbers
of the form $2^{k}p^{m}$.

Another direction to go in is to change the signs in the relationship
$\sigma(n)=2n+d_{1}+d_{2}$. The two other options are
$\sigma(n)=2n-d_{1}-d_{2}$ and $\sigma(n)=2n+d_{1}-d_{2}$. It seems likely
that the main method used in this paper, including the discriminant sandwich
would be successful for the first of these two situations, but the situation
with mixed signs on the divisors may be more difficult.

##  6 Acknowledgements

This paper was written as part of the Hopkins School Mathematics Seminar
2022-2023. Steven J. Miller suggested the problem of changing signs as
discussed in the final section.

## References

  * [1] P. Cohen, K. Cordwell, A. Epstein, C. Kwan, A. Lott, S. J. Miller On near perfect numbers, Acta Arith. 194 (2020), 341-366
  * [2] Hasanalizade, E. On near-perfect numbers of special forms. Bull. Aust. Math. Soc. 108, No. 3, 366-372 (2023).
  * [3] Y. Li and Q. Liao, A class of new near perfect numbers, J. Korean Math. Soc., Volume, 52 Issue 4, pg. 751-763 (2015).
  * [4] T. McCormack, J. Zelinsky, Weighted Versions of the Arithmetic-Geometric Mean and Zaremba’s Function, to appear.
  * [5] P. Pollack, V. Shevelev, On perfect and near perfect numbers, J. Number Theory 132, No. 12, 3037-3046 (2012).
  * [6] W. Sierpiñski, Sur les nombres pseudoparfaits, Matematiĉki Vesnik, Vol. 2 (17), No. 33, pg. 212-213 (1965).
  * [7] X. Ren, Y. Chen, On near perfect numbers with two distinct prime factors, Bull. Aust. Math. Soc. 88, No. 3, 520-524 (2013).
  * [8] M. Tang, X. Ma, M. Feng, On near perfect numbers, Colloq. Math. 144, No. 2, 157-188 (2016)


#  To the memory of A.M. Vinogradov

Quantum vector fields via quantum doubles and their applications

Dimitry Gurevich
Institute for Information Transmission Problems
Bolshoy Karetny per. 19, Moscow 127051, Russian Federation
Pavel Saponov
National Research University Higher School of Economics,
20 Myasnitskaya Ulitsa, Moscow 101000, Russian Federation
and
Institute for High Energy Physics, NRC ”Kurchatov Institute”
Protvino 142281, Russian Federation  gurevich<EMAIL_ADDRESS>
###### Abstract

By treating generators of the reflection equation algebra corresponding to a
Hecke symmetry as quantum analogs of vector fields, we exhibit the
corresponding Leibniz rule via the so-called quantum doubles. The role of the
function algebra in such a double is attributed to another copy of the
reflection equation algebra. We consider two types of quantum doubles: these
giving rise to the quantum analogs of left vector fields acting on the
function algebra and those giving rise to quantum analogs of the adjoint
vector fields acting on the same algebra. Also, we introduce quantum partial
derivatives in the generators of the reflection equation algebra and then at
the limit $q\rightarrow 1$ we get quantum partial derivatives on the
enveloping algebra $U(gl_{N})$ as well as on a certain its extension.

##  1 Introduction

The central problem of any attempt to generalize the notion of a vector field
on a manifold is the definition of the corresponding generalized Leibniz rule.
Thus, the Leibniz rule for super-fields acting on a super-variety should take
into account the parity of all elements involved. Such super-fields generate
super-Lie algebras. In the late 80’s one of the authors introduced analogs of
Lie algebras, associated with involutive symmetries, that is
braidings111Recall that by a braiding we mean a linear operator $R:V^{\otimes
2}\to V^{\otimes 2}$ subject to the so-called braid relation  $(R\otimes
I)(I\otimes R)(R\otimes I)=(I\otimes R)(R\otimes I)(I\otimes R).$
$R:V^{\otimes 2}\to V^{\otimes 2}$, such that $R^{2}=I\otimes I$ (see [G] and
references therein). Hereafter, $V$ is a vector space of dimension $N$ over
the field ${{C}}$ and $I$ is the identity operator in $V$ or the $N\times N$
unit matrix. The Leibniz rule for the corresponding vector fields was also
introduced via such symmetries.

Latter it became clear that many aspects of this approach can be formulated in
terms of the so-called reflection equation (RE) algebras, associated with
Hecke symmetries, that is braidings $R$ such that

$R^{2}=I\otimes I+(q-q^{-1})R,\qquad q\in{{C}}\setminus\\{\pm 1,0\\}.$ (1.1)

With a given Hecke symmetry $R$ we associate two forms of the RE algebra: the
non-modified ${\cal{L}}(R)$ and modified $\hat{{\cal L}}(R)$ ones. They are
respectively generated by entries of the matrices $L=(l_{i}^{j})_{1\leq
i,j\leq N}$ and $\hat{L}=(\hat{l}_{i}^{j})_{1\leq i,j\leq N}$, which are
subject to the systems of relations

$R_{12}L_{1}R_{12}L_{1}-L_{1}R_{12}L_{1}R_{12}=0,$ (1.2)
$R_{12}\hat{L}_{1}R_{12}\hat{L}_{1}-\hat{L}_{1}R_{12}\hat{L}_{1}R_{12}=R_{12}\hat{L}_{1}-\hat{L}_{1}R_{12},$
(1.3)

where $L_{1}=L\otimes I$ and similarly for all other matrices.

Note that these algebras are isomorphic to each other. Their isomorphism can
be established by the following relation on the generating matrices $L$ and
$\hat{L}$:

$L=I-(q-q^{-1})\,\hat{L}.$ (1.4)

Thus, in fact, we are dealing with one algebra realized in two different sets
of generators: $\\{l_{i}^{j}\\}$ and $\\{\hat{l}_{i}^{j}\\}$. Observe that the
isomorphism (1.4) fails at $q=\pm 1$. If a Hecke symmetry $R$ is a deformation
of the usual flip $P$, then at the limit $q\to 1$ the algebra ${\cal{L}}(R)$
tends to $\mathrm{Sym}(gl_{N})$, whereas $\hat{{\cal L}}(R)$ tends to the
algebra $U(gl_{N})$.

The generators of the RE algebra (modified or not) are usually considered as
quantum (or braided) analogs of the classical vector fields. Habitually, the
role of the function algebra is played by the corresponding RTT algebra (see
[IP] and references therein). In [GPS2, GPS3] we suggested a different version
of the quantum calculus, in which the role of the function algebra was played
by another copy of the RE algebra or by an algebra, generated by the space $V$
or its dual $V^{*}$.

All these examples can be included into the family of the so-called quantum
doubles. By a quantum double we mean a couple of associative algebras $(A,B)$
endowed with a permutation map ${\sigma}:A\otimes B\to B\otimes A$, satisfying
a set of certain axioms (see the next section). On the base of this map one
can introduce the structure of a unital associative algebra on the linear
space $B\otimes A$. If the algebra $A$ is endowed with an algebra homomorphism
$\varepsilon_{A}:A\to{{C}}$, then it is possible to define an action of the
algebra $A$ onto $B$ (see formula (2.2) below) and this action will be
compatible with the algebraic structures of both components.

The objective of this article is two-folded. First, we compare two quantum
doubles, giving rise to quantum analogs of the left vector fields and to
quantum analogs of the adjoint vector fields. These quantum vector fields act
on the RE algebra, which plays the role of the function algebra
$\mathrm{Sym}(gl_{N})$. This construction is exhibited in Section 2 in terms
of quantum doubles.

Second, we discuss some applications of these two classes of quantum vector
fields. Namely, in Section 3 we consider invariant differential operators,
corresponding to the central elements of the RE algebra and represented by the
quantum left vector fields, and formulate a conjecture on their spectrum.

In Section 4 we establish a series of matrix Capelli identities which are
quantum generalizations of results by A.Okounkov (see [Ok1, Ok2]). These
generalizations are based on the quantum analogs of the partial derivatives in
the generators of an RE algebra, defined in Section 2. Moreover, in Section 6
we get analogs of the partial derivatives on the enveloping algebra
$U(gl_{N})$ by passing to the limit $q\rightarrow 1$ in the corresponding
structures of the RE algebra. As for the quantum adjoint vector fields, we
show that they can be reduced to the so-called “quantum orbits”, which are
some quotients of the RE algebras. This is done in Section 5.

##  2 Quantum doubles and partial derivatives

###### Definition 1

A couple $(A,B)$ of unital associative algebras $A$ and $B$ is called an
associative double if it is equipped with a linear invertible permutation map

$\sigma:A\otimes B\to B\otimes A,$

which meets the following requirements:

$\sigma\circ(\mu_{A}\otimes\mathrm{id}_{B})=(\mathrm{id}_{B}\otimes\mu_{A})\circ\sigma_{12}\circ\sigma_{23}\quad\mathrm{on}\quad
A\otimes A\otimes B,$
$\sigma\circ(\mathrm{id}_{A}\otimes\mu_{B})=(\mu_{B}\otimes\mathrm{id}_{A})\circ\sigma_{23}\circ\sigma_{12}\quad\mathrm{on}\quad
A\otimes B\otimes B,$ $\sigma(1_{A}\otimes b)=b\otimes
1_{A},\quad\sigma(a\otimes 1_{B})=1_{B}\otimes a\qquad\forall\,a\in
A,\,\forall\,b\in B.$

Here $\mu_{A}:A\otimes A\to A$ is the multiplication in the algebra $A$,
$1_{A}$ is its unit element, and similarly for $B$.

In general, if the algebras $A$ and $B$ are introduced via relations on their
generators, the verification of the above properties reduces to checking that
the map ${\sigma}$ preserves the ideals defining these algebras. In this sense
we say that the defining relations of both algebras are compatible with the
permutation map ${\sigma}$.

Note, that an obvious and well-known example of such a permutation map is the
flip $\sigma=P$:

$P(a\otimes b)=b\otimes a\quad\forall\,a\in A,\,b\in B.$

If the permutation map $\sigma$ is not a flip (or a super flip), we call the
corresponding double a quantum double (QD).

By means of the permutation map ${\sigma}$ we can introduce a unital
associative algebra which is isomorphic as a vector space to the tensor
product $B\otimes A$. For this we consider a free tensor algebra
$\mathrm{T}(A\oplus B)$ and take its quotient over an ideal generated by the
multiplication rules in $A$, $B$ and by elements $a\otimes b-\sigma(a\otimes
b)$, $\forall\,a\in A$, $\forall\,b\in B$. So, in the quotient algebra the
following relations take place:

$a\otimes b=\sigma(a\otimes b).$ (2.1)

Below the equalities (2.1) will be referred to as the permutation relations.

Moreover, if the algebra $A$ is equipped with an algebra homomorphism
$\varepsilon_{A}:A\to{{C}}$, then it becomes possible to define a linear
action $\triangleright:A\otimes B\rightarrow B$ of the algebra $A$ onto $B$:

$a\triangleright b:=(\mathrm{id}_{B}\otimes\varepsilon_{A}){\sigma}(a\otimes
b),\quad\forall\,\,a\in A,\,b\in B.$ (2.2)

Note, that due to requirements imposed on the map $\sigma$ this action defines
a representation of the algebra $A$ in the algebra $B$.

Consider now two examples of the QD which are of the main interest for us. As
the first example we consider the QD $(A,B)$ of two RE algebras $A=\hat{{\cal
L}}(R)$ and $B={\cal{M}}(R)$ equipped with the following permutation map

$\sigma:\,R_{12}\hat{L}_{1}R_{12}\otimes M_{1}\to M_{1}\otimes
R_{12}\hat{L}_{1}R^{-1}_{12}+R_{12}M_{1}\otimes 1_{A}.$ (2.3)

Consequently, the corresponding permutation relations are (from now on we omit
the tensor product signs and the unit elements):

$R_{12}\hat{L}_{1}R_{12}M_{1}=M_{1}R_{12}\hat{L}_{1}R^{-1}_{12}+R_{12}M_{1}.$
(2.4)

This QD was introduced in [GPS2, GPS3]. The modified RE algebra $A=\hat{{\cal
L}}(R)$ admits a homomorphism $\varepsilon:\hat{{\cal L}}(R)\rightarrow{{C}}$
which on generators of $\hat{{\cal L}}(R)$ is defined as
$\varepsilon(\hat{L})=0$, $\varepsilon(1_{A})=1$. Then, in accordance with
(2.2) we get the action of the algebra $\hat{{\cal L}}(R)$ onto
${\cal{M}}(R)$:

$\hat{L}_{1}R_{12}\triangleright M_{1}=M_{1}.$

Note, that if we replace the Hecke symmetry $R$ to the flip $P$ in the
permutation map (2.3), then the above action acquires the meaning of the left
action of the enveloping algebra222We assume the realization of $U(gl_{N})$ as
an algebra of right invariant differential opereators on
$\mathrm{Sym}(gl_{N})$. $U(gl_{N})$ onto the commutative algebra
$\mathrm{Sym}(gl_{N})$.

Using the change (1.4) of the generating matrices, we can pass from the QD
$(\hat{{\cal L}}(R),{\cal{M}}(R))$ to that $({\cal{L}}(R),{\cal{M}}(R))$. The
permutation relations in the latter QD take the form

$R_{12}L_{1}R_{12}M_{1}=M_{1}R_{12}L_{1}R^{-1}_{12}.$ (2.5)

Since $\varepsilon(L)=I$ we get the action

$L_{1}R_{12}\triangleright M_{1}=R^{-1}_{12}M_{1}.$ (2.6)

From the technical point of view this form of the QD is easier to deal with,
so we use it below for a study of invariant operators.

As the second example consider the QD $(\hat{{\cal L}}(R),{\cal{M}}(R))$
composed of the same algebras but equipped with another permutation map
$\tilde{\sigma}$ leading to the following permutation relations:

$R_{12}\hat{L}_{1}R_{12}M_{1}-M_{1}R_{12}\hat{L}_{1}R_{12}=R_{12}M_{1}-M_{1}R_{12}.$
(2.7)

With the homomorhism $\varepsilon(\hat{L})=0$ we get the action of the form:

$\hat{L}_{1}R_{12}\triangleright M_{1}=M_{1}-R^{-1}_{12}M_{1}R_{12}.$

Now, if we pass to the classical limit, assuming that $R$ tends to $P$, the
above action again turns into the action of the enveloping algebra $U(gl_{N})$
onto commutative algebra $\mathrm{Sym}(gl_{N})$ but the generators of
enveloping algebra should be realized as the adjoint vector fields.

Here, we can also express $\hat{L}$ via $L$ by means of the relation (1.4).
Then we get a QD $({\cal{L}}(R),{\cal{M}}(R))$, equipped with the permutation
relations of the form:

$R_{12}L_{1}R_{12}M_{1}=M_{1}R_{12}L_{1}R_{12}.$ (2.8)

As for the homomorphism $\varepsilon$, we always put $\varepsilon(L)=I$ for
the generating matrix of the RE algebra ${\cal{L}}(R)$ and
$\varepsilon(\hat{L})=0$ for that of the modified RE algebra $\hat{{\cal
L}}(R)$.

Now, we go back to the QD $(\hat{{\cal L}}(R),{\cal{M}}(R))$ equipped with the
permutation relations (2.4). As was shown in [GPS3], the matrix
$D=\|{\partial}_{i}^{j}\|$, defined by $D=M^{-1}\hat{L}$ satisfies the matrix
relations:

$R^{-1}_{12}D_{1}R^{-1}_{12}D_{1}=D_{1}R^{-1}_{12}D_{1}R^{-1}_{12}$

and

$D_{1}R_{12}M_{1}R_{12}=R_{12}M_{1}R^{-1}_{12}D_{1}+R_{12}.$ (2.9)

The entries of the matrix $D$ generate an RE algebra333Observe that $R^{-1}$
is a braiding, subject to the Hecke condition with $q$ replaced by $q^{-1}$.
${\cal D}(R^{-1})$ which forms a QD with the algebra ${\cal{M}}(R)$, equipped
with the permutation relations (2.9). To define the action of ${\cal
D}(R^{-1})$ onto ${\cal{M}}(R)$ we put $\varepsilon(D)=0$.

We treat the entries of the matrix $D$ as quantum analogs of the usual partial
derivatives and call them quantum partial derivatives (QPD). This treatment is
motivated by the fact that if $R=P$ the generators $m_{i}^{j}$ become
commutative and the entries of the matrix $D$ become the partial derivatives
${\partial}_{i}^{j}=\partial/\partial m_{j}^{i}$ while the permutation
relations (2.9) turn into the classical Leibniz rules. The QD $({\cal
D}(R^{-1}),{\cal{M}}(R))$ becomes the Heisenberg-Weyl algebra with the
coordinates $(m_{i}^{j})_{1\leq i,j\leq N}$ and the corresponding partial
derivatives $(\partial_{i}^{j})_{1\leq i,j\leq N}$.

In the last section the QD $({\cal D}(R^{-1}),{\cal{M}}(R))$ will be used for
a definition of analogs of the partial derivatives on the enveloping algebra
$U(gl_{N})$.

We complete this section with one more example of a QD. Let us take
$A={\cal{L}}(R)$ again. As the algebra $B$ we choose the free tensor algebra
$T(V)$ of the space $V$. Define the permutation relations as follows:

$R_{12}L_{1}R_{12}\,x_{1}=x_{1}L_{2}.$ (2.10)

On taking $\varepsilon(L)=I$, we get the action $L_{1}R_{12}\triangleright
x_{1}=R_{12}^{-1}\,x_{1}$. Note that the permutation relations (2.10) can be
reduced to the $R$-symmetric and $R$-skew-symmetric algebras of $V$ defined
respectively by

$\mathrm{Sym}_{R}(V)=T(V)/\langle\mathrm{Im}(qI-R)\rangle,\qquad\Lambda_{R}(V)=T(V)/\langle\mathrm{Im}(q^{-1}I+R)\rangle.$

##  3 Quantum left vector fields and the invariant operators

Let $\hat{L}$ be the generating matrix of the enveloping algebra $U(gl_{N})$,
that is the matrix $\hat{L}$ meets the relation (1.3) with $R=P$. It is well
known that for all integers $k\geq 1$ the power sums $\mathrm{Tr}\hat{L}^{k}$
belong to the center of $U(gl_{N})$. Therefore, by the Schur lemma in any
irreducible finite dimensional $U(gl_{N})$-module $V_{\lambda}$ the images of
$\mathrm{Tr}\hat{L}^{k}$ are scalar operators. Note that any such a module
$V_{\lambda}$ is labelled by a partition
${\lambda}=({\lambda}_{1}\geq{\lambda}_{2}\geq\dots\geq{\lambda}_{N})$. The
eigenvalues of the mentioned scalar operators were computed in [PP].

Let us recall a way of constructing the $U(gl_{N})$-modules $V_{\lambda}$. We
start from the finite dimensional complex vector space $V$, $\dim_{\,{C}}V=N$.
In this space the standard irreducible fundamental vector representation is
realized.

Then for any integer $k\geq 1$ we consider the tensor power $V^{\otimes k}$.
This space is decomposed into the direct sum of irreducible
$U(gl_{N})$-modules:

$V^{\otimes k}=\bigoplus_{\lambda\vdash
k}\bigoplus_{T_{\lambda}}P_{T_{\lambda}}V^{\otimes k}.$

Here $T_{\lambda}$ denotes a standard Young table of the Young diagram,
corresponding to a partition $\lambda\vdash k$. The orthonormal projection
operators $P_{T_{\lambda}}$ are images of primitive idempotents
$E_{T_{\lambda}}$ of the group algebra ${{C}}[S_{k}]$ of the symmetric group
$S_{k}$ under the representation of $S_{k}$ in $V^{\otimes k}$ where elements
of the group $S_{k}$ act by permutations of factors of the tensor product
$V^{\otimes k}$.

The subspaces $V_{T_{\lambda}}=P_{T_{\lambda}}V^{\otimes k}$ are irreducible
$U(gl_{N})$-modules. For different tables $T_{\lambda}$ and
$T^{\prime}_{\lambda}$ of the same diagram $\lambda$ these modules are
isomorphic to each other and can be treated as different embeddings of the
module $V_{\lambda}$ into the space $V^{\otimes k}$:

$V_{T_{\lambda}}\simeq V_{T^{\prime}_{\lambda}}\simeq V_{\lambda}.$

In what follows we assume the Hecke symmetry $R$ to be a deformation of the
usual flip. In this case the corresponding modified RE algebra $\hat{{\cal
L}}(R)$ is a deformation of the enveloping algebra $U(gl_{N})$. Moreover, the
category of finite dimensional $U(gl_{N})$-modules can be also deformed into
that of $\hat{{\cal L}}(R)$-modules. Irreducible objects of this category can
be constructed in the same way with the use of the Hecke algebras $H_{k}(q)$
instead of ${{C}}[S_{k}]$. Namely, for a generic $q$ in the Hecke algebra
$H_{k}(q)$ there are analogous idempotents $E_{T_{\lambda}}$ (we keep the same
notation for them) such that $P_{T_{\lambda}}(R)V^{\otimes k}$ are irreducible
$\hat{{\cal L}}(R)$-modules. Here
$P_{T_{\lambda}}(R)=\rho_{R}(E_{T_{\lambda}})$ and
$\rho_{R}:H_{k}(q)\rightarrow\mathrm{End}(V^{\otimes k})$ is an $R$-matrix
representation of the Hecke algebra in the space $V^{\otimes k}$. This
representation sends an Artin generator $\tau_{i}$ of the Hecke algebra to
$R_{i}=I^{\otimes(i-1)}\otimes R\otimes I^{\otimes(k-i+1)}$. We refer the
reader to ([GPS1]) for detail and to [OP] for an explicit construction of the
primitive idempotents $E_{T_{\lambda}}$ in the Hecke algebra.

An important role in the theory of the Hecke algebras belongs to the so-called
Jusys-Murphy elements which generates the maximal commutative subagebra in the
Hecke algebra $H_{k}(q)$. In the $R$-matrix representation they form a set of
mutually commutative linear operators $J_{i}$, $1\leq i\leq k$, in the space
$V^{\otimes k}$:

$J_{1}=\mathrm{Id}_{V^{\otimes k}},\quad J_{i}=R_{i-1}J_{i-1}R_{i-1},\quad
2\leq i\leq k.$

Introduce also the notation for some “matrix copies” of the generating matrix
$L$:

$L_{\overline{1}}=L_{\underline{1}}=L_{1},\quad
L_{\overline{k}}=R_{k-1}L_{\overline{k-1}}\,R^{-1}_{k-1},\quad
L_{\underline{k}}=R^{-1}_{k-1}L_{\underline{k-1}}R_{k-1}.$

With the above notation we can write the action (2.6) in the form:

$L_{\underline{2}}\triangleright M_{1}=J_{2}^{-1}M_{1}.$ (3.1)

On applying the permutation relations (2.5) one can generalize this formula on
arbitraty “monomials” in generators of ${\cal{M}}(R)$:

$L_{\underline{k+1}}\triangleright M_{1}\dots
M_{\overline{k}}=J_{k+1}^{-1}M_{1}\dots M_{\overline{k}}.$ (3.2)

Calculating the $R$-trace at the $(k+1)$-th space and taking into account the
formula

$\mathrm{Tr}_{R(k+1)}X_{\overline{k+1}}=\mathrm{Tr}_{R(k+1)}X_{\underline{k+1}}=I_{1\dots
k}\mathrm{Tr}_{R}\,X,$ (3.3)

where $X$ is an arbitrary $N\times N$ matrix, we get the following claim (see
[S]).

###### Proposition 2

Let $\lambda\vdash k$ be a partition and $T_{\lambda}$ be a standard Young
table corresponding to the Young diagram of the partition $\lambda$. Then the
following relation holds

$\mathrm{Tr}_{R}L\triangleright P_{T_{\lambda}}(R)M_{1}M_{\overline{2}}\dots
M_{\overline{k}}=\chi_{{\lambda}}(\mathrm{Tr}_{R}L)P_{T_{\lambda}}(R)M_{1}M_{\overline{2}}\dots
M_{\overline{k}},$ (3.4)

where

$\chi_{{\lambda}}(\mathrm{Tr}_{R}L)=\frac{N_{q}}{q^{N}}-\frac{\nu}{q^{2N}}\sum_{i=1}^{k}\,q^{-2c(i)},\quad\nu=q-q^{-1},$
(3.5)

and the sum is taken over all boxes of the table $T_{\lambda}$. Here
$c(i)=n-m$ is the content of the box in which the integer $i$ is located, that
is $n$ (respectively $m$) is the number of the corresponding column (row).

Note that the symbol of the table $T$ in the notation $\chi_{\lambda}$ would
be irrelevant since the right hand side of (3.5) actually depens only on the
diagram $\lambda$ but not on the table.

###### Remark 3

Let us point out that formula (3.2) is the matrix form of the action of the
left quantum vector fields on the algebra ${\cal{M}}(R)$. In fact, it is
similar to the action of the algebra ${\cal{L}}(R)$ onto the free tensor
algebra $T(V)$ considered at the end of the previous section. Namely, using
the permutation relations (2.10) and the homomorphism $\varepsilon(L)=I$ we
can get the action

$L_{\underline{k+1}}\triangleright x_{1}\otimes\dots\otimes
x_{k}=J_{k+1}^{-1}x_{1}\otimes\dots\otimes x_{k}.$ (3.6)

Evidently, a formula similar to (3.4) is also valid in this case.

We are interested in the spectral analysis of the invariant operators, that is
these corresponding to the central elements of the algebra ${\cal{L}}(R)$
acting onto ${\cal{M}}(R)$. The proposition 2 describes the spectrum of the
operator $\mathrm{Tr}_{R}L$, corresponding to the lowest power sum.

Below, we consider some other examles of central elements and describe the
spectrum of the corresponding operators in invariant subspaces
$\mathrm{Im}(P_{T_{\lambda}})$ extracted by the projection operators
$P_{T_{\lambda}}(R)$. Besides the higher power sums $\mathrm{Tr}_{R}L^{k}$,
$k\geq 2$, we are interested in the so-called quantum elementary symmetric
polynomials $e_{k}(L)$, which are defined by the formula:

$e_{k}(L)=\mathrm{Tr}_{R(1\dots k)}(A^{(k)}L_{1}L_{\overline{2}}\dots
L_{\overline{k}}),\quad k\geq 1.$

Here $A^{(k)}$ are the $R$-skew-symmetrizers introduced by the following
recursion formula:

$A^{(1)}=I,\quad
A^{(k)}=\frac{1}{k_{q}}A^{(k-1)}\left(q^{k-1}I-(k-1)_{q}\,R_{k-1}\right)A^{(k-1)},\quad
k\geq 2,$

where $k_{q}=(q^{k}-q^{-k})/(q-q^{-1})$ is a $q$-integer. Below we assume
$q^{2k}\not=1$ for all integer $2\leq k\leq N$, and, therefore, $k_{q}\not=0$.
Any $R$-skew-symmetrizer $A^{(k)}(R)$ is a particular case of the above
idempotents $P_{T_{\lambda}}(R)$, corresponding to one-column diagram
${\lambda}=(1^{k})$.

The elementary symmetric polynomials possess a natural and convenient
parameterization in terms of the quantum eigenvalues $\\{\mu_{i}\\}_{1\leq
i\leq N}$ of the generating matrix $L$. These eigenvalues belong to an
algebraic extension of the center of the RE algebra ${\cal{L}}(R)$. They are
defined via the quantum Cayley-Hamilton identity on the matrix $L$:

$L^{N}-qe_{1}(L)L^{N-1}+q^{2}e_{2}(L)L^{N-2}+\dots+(-q)^{N}e_{N}(L)I=0.$ (3.7)

Let us introduce $N$ elements $\mu_{i}$, $1\leq i\leq N$, which satisfy a
system of $N$ polynomial equations with central elements on the left hand
side:

$q^{k}e_{k}(L)=\sum_{1\leq i_{1}<\dots<i_{k}\leq
N}\mu_{i_{1}}\dots\mu_{i_{k}},\quad 1\leq k\leq N.$ (3.8)

The elements $\mu_{i}$ are also assumed to be central. They are interpreted as
quantum analogs of eigenvalues of the matrix $L$ since in virtue of (3.8) the
Cayley-Hamilton identity (3.7) can be rewritten in the following factorized
form:

$\prod_{i=1}^{N}(L-\mu_{i}I)=0.$ (3.9)

In terms of the quantum eigenvalues $\mu_{i}$ one can express all central
symmetric polynomials of the RE algebra — the power sums, the quantum Schur
functions and so on (see [GPS1, GPS4] for more detail). For example, the power
sums in terms of quantum spectrum read:

$\mathrm{Tr}_{R}L^{k}=\sum\mu_{i}^{k}\,d_{i},\qquad
d_{i}=q^{-1}\prod_{j\not=i}^{N}\frac{\mu_{i}-q^{-2}\mu_{j}}{\mu_{i}-\mu_{j}}.$
(3.10)

Since the elements $\mu_{i}$ are central, then in any irreducible
${\cal{L}}(R)$-module they are represented by scalar operators. These
operators must be compatible with relations (3.8). In the paper [GPeS1] we
suggested the following conjecture444In the cited paper the conjecture was
formulated for a wider class of Hecke symmetries, namely those of the bi-rank
$(m|0)$. (For the notion of the bi-rank the reader is referred to [GPS1].) The
Hecke symmetries we are dealing with in the present paper is of the bi-rank
$(N|0)$..

###### Conjecture 4

In ${\cal{L}}(R)$-invariant subspaces
$V_{T_{\lambda}}=\mathrm{Im}P_{T_{\lambda}}(R)$ the central elements $\mu_{i}$
$1\leq i\leq N$ are represented (after a proper ordering) by the following
scalar operators:

$\mu_{i}\mapsto\chi_{\lambda}(\mu_{i})\,\mathrm{Id}_{V_{T_{\lambda}}},\qquad\chi_{\lambda}(\mu_{i})=q^{-2({\lambda}_{i}+N-i)},$
(3.11)

where $\lambda_{i}$ are elements of a partition
$\lambda=(\lambda_{1},\lambda_{2},\dots,\lambda_{N})$.

The representation (3.11) is compatible with the scalar operators
corresponding to the quantum elementary symmetric polynomials
$e_{k}(L)\mapsto\chi_{\lambda}(e_{k})\,\mathrm{Id}_{V_{T_{\lambda}}}$, that
is:

$q^{k}\,\chi_{{\lambda}}(e_{k})=\sum_{1\leq i_{1}<\dots<i_{k}\leq
N}\chi_{{\lambda}}(\mu_{i_{1}})\dots\chi_{{\lambda}}(\mu_{i_{k}}).$

For the lowest elementary symmetric polynomial
$e_{1}(L)=\mathrm{Tr}_{R}(L)=q^{-1}\sum_{i}\mu_{i}$ this conjecture is easily
verified with the use of (3.5). We have also checked it for $e_{2}(L)$ by a
direct computation.

Turn now to the modified RE algebra $\hat{{\cal L}}(R)$ whose generators
$\hat{l}_{i}^{j}$ are treated as quantum analogs of the right-invariant vector
fields. The corresponding Casimir operators are $\mathrm{Tr}_{R}\hat{L}^{k}$.
The most convenient way to perform their spectral analysis is to introduce the
quantum eigenvalues $\hat{\mu}_{i}$ of the generating matrix $\hat{L}$. Taking
into account relation (1.4), we can rewrite the factorized Cayley-Hamilton
identity (3.9) in the form of the matrix identity for $\hat{L}$:

$\prod_{i=1}^{N}(L-\mu_{i}I)=0\qquad\Rightarrow\qquad\prod_{i=1}^{N}(\hat{L}-\hat{\mu}_{i}I)=0,$

where the quantum eigenvalues $\hat{\mu}_{i}$ of the generating matrix
$\hat{L}$ are connected with those of $L$ by a linear shift:

$\mu_{i}=1-\nu\hat{\mu}_{i},\qquad\nu=q-q^{-1}.$ (3.12)

Here the symbol $1$ stands for the unit element of the RE algebra.

As a direct consequence of (3.12) we obtain:

$\chi_{\lambda}({\hat{\mu}_{k}})=\frac{1-q^{-2\,({\lambda}_{k}+N-k)}}{q-q^{-1}}=q^{-(\lambda_{k}+N-k)}(\lambda_{k}+N-k)_{q},\quad
1\leq k\leq N.$

By taking the limit $q\rightarrow 1$, we get the spectral values of
$U(gl_{N})$ generating matrix $\hat{L}$ in the irreducible module
$V_{\lambda}$:

$\chi_{\lambda}({\hat{\mu}_{k}})={\lambda}_{k}+N-k,\quad 1\leq k\leq N.$

Note that this is in the full agreement with the results of [PP].

##  4 Different forms of the Capelli identity

Consider the QD $({\cal D}(R^{-1}),{\cal{M}}(R))$ of two RE algebras generated
by the enties of matrices $D$ and $M$ and equipped with the permutation
relations (2.9). Introduce the matrix $\hat{L}=MD$. As was argued in Section
2, its entries generate the modified RE algebra $\hat{{\cal L}}(R)$ and the
permutation relations with entries of the matrix $M$ are given by (2.4).

One of the remarkable properties of the QD under consideration are the
following quantum matrix Capelli identities555If $R$ is a deformation of the
flip $P$ then at the classical limit $q\rightarrow 1$ these identities
transform into those presented in [Ok2]..

###### Theorem 5

([GPeS2])  For $\forall\,k\geq 1$ the following matrix identity takes place:

$A^{(k)}\hat{L}_{\overline{1}}\,(\hat{L}_{\overline{2}}\,+qI)\dots(\hat{L}_{\overline{k}}\,+q^{k-1}(k-1)_{q}I\,)\,A^{(k)}=q^{k(k-1)}A^{(k)}M_{\overline{1}}\dots
M_{\overline{k}}\,D_{\overline{k}}\dots D_{\overline{1}}.$ (4.1)

Let us point out that identities (4.1) are valid for a wide class of the Hecke
symmetries: to prove (4.1) one needs only the braid relation and the Hecke
condition (1.1) on $R$. If we impose some additional restrictions on $R$ we
can obtain further consequences of (4.1), namely, the quantum version of the
determinant Capelli identity.

Thus, assume $R$ to be a deformation of the usual flip $P$, then the $R$-skew-
symmetrizer $A^{(N)}$ is a unit rank projector

$\mathrm{dim}\,\mathrm{Im}\,A^{(N)}=1$ (4.2)

and $A^{(N+1)}\equiv 0$ (recall that $N=\dim\,V$). A direct consequence of
(4.2) is the existence of two tensors
$|u\rangle=\|u_{i_{1}i_{2}\dots\,i_{N}}\|$ and $\langle
v|=\|v^{i_{1}i_{2}\dots\,i_{N}}\|$ such that

$A^{(N)}=|u\rangle\otimes\langle v|,\qquad\langle
v|u\rangle=\sum_{\\{i\\}}v^{i_{1}i_{2}\dots\,i_{N}}u_{i_{1}i_{2}\dots\,i_{N}}=1.$

The structure tensors $|u\rangle$ and $\langle v|$ allow one to define a
quntum analog of the determinant for the generatig matrix of the RE algebra.
Namely, we set

${\det}_{R}\,M=\langle v|M_{1}M_{\overline{2}}\dots
M_{\overline{N}}|u\rangle,\qquad{\det}_{R^{-1}}\,D=\langle
v|D_{\overline{N}}\dots D_{\overline{2}}D_{1})|u\rangle.$

With these definitions we can formulate the following corollary of Theorem 5.

###### Corollary 6

If $R$ is a deformation of the flip $P$, then the following quantum Capelli
identity takes place:

$\mathrm{Tr}_{R(1\dots
N)}A^{(N)}\hat{L}_{\overline{1}}\,(\hat{L}_{\overline{2}}\,+qI)\dots(\hat{L}_{\overline{N}}\,+q^{N-1}(N-1)_{q}I\,)=q^{-N}{\det}_{R}\,M\,{\det}_{R^{-1}}\,D.$
(4.3)

###### Remark 7

Note that this corollary remains valid mutatis mutandis if the initial Hecke
symmetry $R$ is of bi-rank $(m|0)$, $m\leq N$, the corresponding examples were
constructed in [G]. This condition means that $\mathrm{rank}\,A^{(m)}=1$ and
$A^{(m+1)}\equiv 0$. In this case we only have to replace $N$ by $m$ in
formula (4.3), while the structure tensors $|u\rangle$ and $\langle v|$ should
be extracted from the projector $A^{(m)}$.

In conclusion of the section we note that another version of the quantum
Capelli identity was suggested in [NUW]. This version is related to the RTT
algebra and is valid for $R$ coming from the QG $U_{q}(sl_{N})$.

##  5 Quantum adjoint vector fields and quantum orbits

In this section we deal with the QD $(\hat{{\cal L}}(R),{\cal{M}}(R))$,
endowed with the permutation relations (2.7). Our main objective here is to
show that the action of the modified RE algebra $\hat{{\cal L}}(R)$ on
${\cal{M}}(R)$ can be reduced onto some quotients of the algebra
${\cal{M}}(R)$. These quotient algebras are treated as quantum analogs of the
coordinate rings of the $GL(N)$-orbits in $gl_{N}^{*}$. Observe that we employ
the term “orbit” in a loose sense since even in the classical case these
quotients are “orbits” for generic matrices666This means that the eigenvalues
of the matrix are pairwise distinct. generating the orbit. The question which
quantum orbits can be considered as “generic” is discussed below. The
generators $\hat{l}_{i}^{j}$ are treated as quantum adjoint vector fields.

###### Proposition 8

In the QD $(\hat{{\cal L}}(R),{\cal{M}}(R))$ the following relations hold for
any integer $k\geq 1$

$\hat{L}\,\mathrm{Tr}_{R}(M^{k})=\mathrm{Tr}_{R}(M^{k})\hat{L}$

and consequently

$\hat{L}\triangleright\mathrm{Tr}_{R}M^{k}=0.$ (5.1)

Proof. Successively multiplying the both sides of matrix equality (2.7) by the
matrix $M_{1}$ from the right and transforming the term
$M_{1}R_{12}\hat{L}_{1}R_{12}M_{1}$ with the use of (2.7) we come to the
result:

$R_{12}\hat{L}_{1}R_{12}M^{k}_{1}-M_{1}^{k}R_{12}\hat{L}_{1}R_{12}=R_{12}M_{1}^{k}-M_{1}^{k}R_{12}.$

Then we multiply this equality by $R^{-1}_{12}$ from the left and right and
apply the $R$-trace in the second space. These operations lead to the desired
result in virtue of (3.3):

$\hat{L}_{1}\mathrm{Tr}_{R}M^{k}-\mathrm{Tr}_{R}M^{k}\hat{L}_{1}=0.$

To get the action (5.1) we apply the homomorphism $\varepsilon(\hat{L})=0$ to
the second term.

So, the quantum adjoint vector fields kill all elements
$\mathrm{Tr}_{R}M^{k}$. Note that these elements are central in the RE algebra
${\cal{M}}(R)$.

Now, consider the following quotient algebra:

$\mathcal{O}({\alpha}_{1},\dots,{\alpha}_{N})={\cal{M}}(R)/\langle\mathrm{Tr}_{R}M-{\alpha}_{1},\,\mathrm{Tr}_{R}M^{2}-{\alpha}_{2},\dots,\mathrm{Tr}_{R}M^{N}-{\alpha}_{N}\rangle,$

where ${\alpha}_{1},\dots,{\alpha}_{N}$ are some complex constants and the
notation $\langle J\rangle$ stands for the ideal, generated by a subset $J$ of
a given algebra. We call these quotients the “quantum orbits”.

The quantities ${\alpha}_{i}$ can be expressed in terms of the eigenvalues
$\mu_{i}$ of the matrix $M$ since for this matrix formula (3.10) is also
valid. In the paper [GS1] we showed (up to a conjecture on the quantum de Rham
complex) that the quantum orbits are generic if

$\mu_{i}\not=q^{2}\mu_{j},\quad\forall\,i,j.$ (5.2)

Observe that by definition a quantum orbit is generic if its cotangent module
is projective. Thus, as was shown in [GS1], under the condition (5.2) and
under the assumption that the mentioned conjecture is true the cotangent
module on the quantum orbit $\mathcal{O}({\alpha}_{1},\dots,{\alpha}_{N})$ is
projective.

##  6 Quantum partial derivatives on $U(gl_{N})$ background

In this section we consider one more application of the quantum left vector
fields. Namely, we show how to get analogs of the partial derivatives on the
enveloping algebra $U(gl_{N})$ via the QD introduced at the end of Section 2.

To this purpose we return to the QD $({\cal D}(R^{-1}),{\cal{M}}(R))$, endowed
with the permutation relations (2.9). Let us introduce a new generating matrix
$N$ of the RE algebra ${\cal{M}}(R)$ by the shift analogous to (1.4):

$M=h\,I-(q-q^{-1})\,N,\qquad h\in{{C}}.$

The full list of the relations in the QD $({\cal D}(R^{-1}),\hat{\cal
N}_{h}(R))$, generated by the entries of matrices $D$ and $N$, is as follows

$R_{12}N_{1}R_{12}N_{1}-N_{1}R_{12}N_{1}R_{12}=h(R_{12}N_{1}-N_{1}R_{12}),$
$R^{-1}_{12}D_{1}R^{-1}_{12}D_{1}=D_{1}R^{-1}_{12}D_{1}R^{-1}_{12},$
$D_{1}R_{12}N_{1}R_{12}-R_{12}N_{1}R^{-1}_{12}D_{1}=R_{12}+hD_{1}R_{12}.$
(6.1)

We introduce the deformation parameter $h$ in order to present the modified RE
algebra $\hat{\cal N}_{h}(R)$ as a deformation of the RE algebra
${\cal{M}}(R)$, which corresponds to the value $h=0$.

The entries of the matrix $D=\|{\partial}_{i}^{j}\|_{1\leq i,j\leq N}$ can be
treated as analogs of partial derivatives in $n_{i}^{j}$ and we call them the
quantum partial derivatives (QPD). Using the homomorphism $\varepsilon(D)=0$,
we get the action of the QPD on the generators of $\hat{\cal N}_{h}(R)$:

$D_{1}\triangleright N_{\overline{2}}=R_{12}^{-1}.$ (6.2)

The relation (6.1) is a substitution of the Leibniz rule allowing one to
extend the action of QPD (6.2) from generators onto the whole algebra
$\hat{\cal N}_{h}(R)$.

If $R$ is a deformation of the flip $P$ then at the limit $q\rightarrow 1$
relations (6.2) turn into the classical action of the partial derivatives on
the generators of the commutative algebra777However, the action of the QPD on
higher polynomials in $n_{k}^{l}$ differs from the classical formulae and
contains corrections depending on $h$. The full classical picture restores
after specializing $h=0$. $\mathrm{Sym}(gl_{N})$:

$\partial_{i}^{j}\triangleright n_{k}^{s}=\delta_{i}^{s}\,\delta_{k}^{j}.$

This is the reason for treating the QPD $\partial_{i}^{j}$ as “derivatives” in
generators $n_{j}^{i}$.

In general, if the symmetry $R$ is involutive, it is useful to introduce the
matrix $\hat{D}=D+h^{-1}\,I$. In terms of this matrix the permutation
relations (6.1) take the form:

$\hat{D}_{1}R_{12}N_{1}R_{12}-R_{12}N_{1}R_{12}\hat{D}_{1}=h\hat{D}_{1}R_{12}$

We call the entries of the matrix $\hat{D}=\|\hat{{\partial}}_{i}^{j}\|$ the
shifted QPD888 In fact, only the diagonal elements of the matrix $D$ are
shifted..

From now on we assume that $R=P$. In this case the permutation relations for
the generators of ${\cal N}_{h}(R)$ turn into those of the universal
enveloping algebra $U(gl(N)_{h})$:

$N_{2}N_{1}-N_{1}N_{2}=h(P_{12}N_{1}-N_{1}P_{12}).$

The permutation relations (6.1) give rise to the Leibnitz rule for the QPD,
expressed via the coproduct defined on the shifted QPD

$\Delta(\hat{{\partial}}_{i}^{j})=\hat{{\partial}}_{k}^{j}\otimes\hat{{\partial}}_{i}^{k},$

and completed with
$\varepsilon(\hat{{\partial}}_{i}^{j})=h^{-1}\,\delta_{i}^{j}$.

Let us consider the example $N=2$ in more detail. It is convenient to pass to
the compact form $u(2)_{h}$ of the algebra $gl(2)_{h}$. The Lie algebra
$u(2)_{h}$ is generated by 4 elements $x$, $y$, $z$ and $t$ with the following
Lie brackets:

$[x,\,y]=hz,\qquad[y,\,z]=hx,\qquad[z,\,x]=hy,\qquad[t,\,x]=[t,\,y]=[t,\,z]=0.$

The corresponding QPD ${\partial}_{x}$, ${\partial}_{y}$, ${\partial}_{z}$ and
$\hat{\partial}_{t}$ (we use the “shifted” derivative in $t$:
$\hat{{\partial}}_{t}={\partial}_{t}+2/h$) commute with each other, while the
permutation relations with the $u(2)_{h}$ generators read:

$\begin{array}[]{l@{\quad}l@{\quad}l@{\quad}l}\hat{{\partial}}_{t}\,t-t\,\hat{{\partial}}_{t}=\frac{h}{2}\,\hat{{\partial}}_{t}&\hat{{\partial}}_{t}\,x-x\,\hat{{\partial}}_{t}=-\frac{h}{2}\,{\partial}_{x}&\hat{{\partial}}_{t}\,y-y\,\hat{{\partial}}_{t}=-\frac{h}{2}\,{\partial}_{y}&\hat{{\partial}}_{t}\,z-z\,\hat{{\partial}}_{t}=-\frac{h}{2}\,{\partial}_{z}\\\
\rule{0.0pt}{19.91692pt}{\partial}_{x}\,t-t\,{\partial}_{x}=\frac{h}{2}\,{\partial}_{x}&{\partial}_{x}\,x-x\,{\partial}_{x}=\frac{h}{2}\,\hat{{\partial}}_{t}&{\partial}_{x}\,y-y\,{\partial}_{x}=\frac{h}{2}\,{\partial}_{z}&{\partial}_{x}\,z-z\,{\partial}_{x}=-\frac{h}{2}\,{\partial}_{y}\\\
\rule{0.0pt}{19.91692pt}{\partial}_{y}\,t-t\,{\partial}_{y}=\frac{h}{2}\,{\partial}_{y}&{\partial}_{y}\,x-x\,{\partial}_{y}=-\frac{h}{2}\,{\partial}_{z}&{\partial}_{y}\,y-y\,{\partial}_{y}=\frac{h}{2}\,\hat{{\partial}}_{t}&{\partial}_{y}\,z-z\,{\partial}_{y}=\frac{h}{2}\,{\partial}_{x}\\\
\rule{0.0pt}{19.91692pt}{\partial}_{z}\,t-t\,{\partial}_{z}=\frac{h}{2}\,{\partial}_{z}&{\partial}_{z}\,x-x\,{\partial}_{z}=\frac{h}{2}\,{\partial}_{y}&{\partial}_{z}\,y-y\,{\partial}_{z}=-\frac{h}{2}\,{\partial}_{x}&{\partial}_{z}\,z-z\,{\partial}_{z}=\frac{h}{2}\,\hat{{\partial}}_{t}.\end{array}$
(6.3)

Introduce the matrix $\hat{\Theta}$, composed of the partial derivatives:

${\hat{\Theta}}=i\mathchar
1406\relax\left(\begin{array}[]{rrrr}\hat{{\partial}}_{t}&{\partial}_{x}&{\partial}_{y}&{\partial}_{z}\\\
-{\partial}_{x}&\hat{{\partial}}_{t}&-{\partial}_{z}&{\partial}_{y}\\\
-{\partial}_{y}&{\partial}_{z}&\hat{{\partial}}_{t}&-{\partial}_{x}\\\
-{\partial}_{z}&-{\partial}_{y}&{\partial}_{x}&\hat{{\partial}}_{t}\end{array}\right),$
(6.4)

where $\mathchar 1406\relax=h/2i$.

By $\hat{\Theta}(a)$ we denote the matrix whose entries result from applying
the corresponding partial derivatives to an element $a\in U(u(2)_{h})$. Thus,
we get a linear map of associative algebras:

$U(u(2)_{h})\to\mathrm{Mat}_{4}(U(u(2)_{h}))=\mathrm{End}({{C}}^{\,4})\otimes
U(u(2)_{h})$

which will be also denoted by the symbol $\hat{\Theta}$:

$\hat{\Theta}:\,a\mapsto\hat{\Theta}(a)\quad\,\forall\,a\in U(u(2)_{h}).$
(6.5)

We state (see [GS3]) that the map $\hat{\Theta}$ is actually a homomorphism of
associative algebras, that is

$\hat{\Theta}(ab)=\hat{\Theta}(a)\cdot\hat{\Theta}(b),\quad\forall\,a,b\in
U(u(2)_{h}).$ (6.6)

Consequently, the map $\hat{\Theta}$ defines a representation of the algebra
$U(u(2)_{h})$:

$\hat{\Theta}(x)\cdot\hat{\Theta}(y)-\hat{\Theta}(y)\cdot\hat{\Theta}(x)=h\hat{\Theta}(z)$

and so on. In virtue of the above properties the map $\hat{\Theta}$ is treated
as the Leibniz rule for the QPD on the algebra $U(u(2)_{h})$.

We want to stress the usefulness of this form of the Leibniz rule. So far, the
QPD were defined on the algebra of polynomials in noncommutative generators.
Our next aim is to extend their action onto some elements of a central
extension of this algebra. Namely, consider the so-called quantum radius
$r_{\mathchar 1406\relax}$:

$r_{\mathchar 1406\relax}^{2}=x^{2}+y^{2}+z^{2}+\mathchar 1406\relax^{2},$

which is an element of a central extension of the algebra $U(u(2)_{h})$. We
want to extend the action of the QPD onto $r_{\mathchar 1406\relax}$. As a
reasonable criterium of such an extension we require that the property (6.6)
would be preserved on the extended algebra. Namely, we require the relation

$\hat{\Theta}(r_{\mathchar 1406\relax})^{2}=\hat{\Theta}(r_{\mathchar
1406\relax}^{2})=\hat{\Theta}(x^{2}+y^{2}+z^{2}+\mathchar 1406\relax^{2})$

to be fulfilled.

As follows from computation of [GS3], the result of applying the map
$\hat{\Theta}$ to the quantum radius $r_{\mathchar 1406\relax}$ is

${\hat{\Theta}}(r_{\mathchar 1406\relax})=\frac{r_{\mathchar
1406\relax}^{2}+\mathchar 1406\relax^{2}}{r_{\mathchar
1406\relax}}\,I+\frac{i\mathchar 1406\relax}{r_{\mathchar 1406\relax}}\,M,$

where the matrix $M$ is of the form:

$M=\left(\\!\\!\begin{array}[]{cccc}0&x&y&z\\\ -x&0&-z&y\\\ -y&z&0&-x\\\
-z&-y&x&0\end{array}\right).$

Now we are able to compute the action of the QPD on the quantum radius:

${\partial}_{t}r_{\mathchar 1406\relax}=-\frac{i\mathchar
1406\relax}{r_{\mathchar 1406\relax}},\qquad{\partial}_{x}r_{\mathchar
1406\relax}=\frac{x}{r_{\mathchar
1406\relax}},\qquad{\partial}_{y}r_{\mathchar
1406\relax}=\frac{y}{r_{\mathchar
1406\relax}},\qquad{\partial}_{z}r_{\mathchar
1406\relax}=\frac{z}{r_{\mathchar 1406\relax}}.$

Observe that these formulae possess the usual classical limit as $\mathchar
1406\relax\to 0$ and, consequently, $r_{\mathchar 1406\relax}\to r$.

Also, in [GS4] the following problem was discussed: how to extend the action
of the QPD on the skew-field $B[B^{-1}]$, where $B$ is a central extension of
the algebra $U(u(2)_{h})$. Assuming that the above Leibniz rule is still valid
on this skew-field, we have:

$\hat{\Theta}(a)\hat{\Theta}(a^{-1})=\hat{\Theta}(e)=I.$

Thus, in order to extend the action of the partial derivatives on the skew-
field $B[B^{-1}]$ it suffices to invert the matrix $\hat{\Theta}(a)$. In [GS4]
we exhibited an example, in which such inverting was performed with the use of
the Cayley-Hamilton identity for the matrix $\hat{\Theta}(a)$.

In conclusion, we note that defining QPD on the extended algebra $U(u(2)_{h})$
enables us to introduce some dynamical models on this algebra. Certain of them
were considered in [GS2, GS3].

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