def sum_of_partitions(self, la):
r"""
Return the sum over all sets partitions whose shape is ``la``,
scaled by `\prod_i m_i!` where `m_i` is the multiplicity
of `i` in ``la``.
INPUT:
- ``la`` -- an integer partition
OUTPUT:
- an element of ``self``
EXAMPLES::
sage: w = SymmetricFunctionsNonCommutingVariables(QQ).dual().w()
sage: w.sum_of_partitions([2,1,1])
2*w{{1}, {2}, {3, 4}} + 2*w{{1}, {2, 3}, {4}} + 2*w{{1}, {2, 4}, {3}}
+ 2*w{{1, 2}, {3}, {4}} + 2*w{{1, 3}, {2}, {4}} + 2*w{{1, 4}, {2}, {3}}
"""
la = Partition(la)
c = prod([factorial(_) for _ in la.to_exp()])
P = SetPartitions()
return self.sum_of_terms([(P(m), c) for m in SetPartitions(sum(la), la)], distinct=True)
def hall_polynomial(nu, mu, la, q=None):
r"""
Return the (classical) Hall polynomial `P^{\nu}_{\mu,\lambda}`
(where `\nu`, `\mu` and `\lambda` are the inputs ``nu``, ``mu``
and ``la``).
Let `\nu,\mu,\lambda` be partitions. The Hall polynomial
`P^{\nu}_{\mu,\lambda}(q)` (in the indeterminate `q`) is defined
as follows: Specialize `q` to a prime power, and consider the
category of `\GF{q}`-vector spaces with a distinguished
nilpotent endomorphism. The morphisms in this category shall be
the linear maps commuting with the distinguished endomorphisms.
The *type* of an object in the category will be the Jordan type
of the distinguished endomorphism; this is a partition. Now, if
`N` is any fixed object of type `\nu` in this category, then
the polynomial `P^{\nu}_{\mu,\lambda}(q)` specialized at the
prime power `q` counts the number of subobjects `L` of `N` having
type `\lambda` such that the quotient object `N / L` has type
`\mu`. This determines the values of the polynomial
`P^{\nu}_{\mu,\lambda}` at infinitely many points (namely, at all
prime powers), and hence `P^{\nu}_{\mu,\lambda}` is uniquely
determined. The degree of this polynomial is at most
`n(\nu) - n(\lambda) - n(\mu)`, where
`n(\kappa) = \sum_i (i-1) \kappa_i` for every partition `\kappa`.
(If this is negative, then the polynomial is zero.)
These are the structure coefficients of the
:class:`(classical) Hall algebra <HallAlgebra>`.
If `\lvert \nu \rvert \neq \lvert \mu \rvert + \lvert \lambda \rvert`,
then we have `P^{\nu}_{\mu,\lambda} = 0`. More generally, if the
Littlewood-Richardson coefficient `c^{\nu}_{\mu,\lambda}`
vanishes, then `P^{\nu}_{\mu,\lambda} = 0`.
The Hall polynomials satisfy the symmetry property
`P^{\nu}_{\mu,\lambda} = P^{\nu}_{\lambda,\mu}`.
ALGORITHM:
If `\lambda = (1^r)` and
`\lvert \nu \rvert = \lvert \mu \rvert + \lvert \lambda \rvert`,
then we compute `P^{\nu}_{\mu,\lambda}` as follows (cf. Example 2.4
in [Sch2006]_):
First, write `\nu = (1^{l_1}, 2^{l_2}, \ldots, n^{l_n})`, and
define a sequence `r = r_0 \geq r_1 \geq \cdots \geq r_n` such that
.. MATH::
\mu = \left( 1^{l_1 - r_0 + 2r_1 - r_2}, 2^{l_2 - r_1 + 2r_2 - r_3},
\ldots , (n-1)^{l_{n-1} - r_{n-2} + 2r_{n-1} - r_n},
n^{l_n - r_{n-1} + r_n} \right).
Thus if `\mu = (1^{m_1}, \ldots, n^{m_n})`, we have the following system
of equations:
.. MATH::
\begin{aligned}
m_1 & = l_1 - r + 2r_1 - r_2,
\\ m_2 & = l_2 - r_1 + 2r_2 - r_3,
\\ & \vdots ,
\\ m_{n-1} & = l_{n-1} - r_{n-2} + 2r_{n-1} - r_n,
\\ m_n & = l_n - r_{n-1} + r_n
\end{aligned}
and solving for `r_i` and back substituting we obtain the equations:
.. MATH::
\begin{aligned}
r_n & = r_{n-1} + m_n - l_n,
\\ r_{n-1} & = r_{n-2} + m_{n-1} - l_{n-1} + m_n - l_n,
\\ & \vdots ,
\\ r_1 & = r + \sum_{k=1}^n (m_k - l_k),
\end{aligned}
or in general we have the recursive equation:
.. MATH::
r_i = r_{i-1} + \sum_{k=i}^n (m_k - l_k).
This, combined with the condition that `r_0 = r`, determines the
`r_i` uniquely (recursively). Next we define
.. MATH::
t = (r_{n-2} - r_{n-1})(l_n - r_{n-1})
+ (r_{n-3} - r_{n-2})(l_{n-1} + l_n - r_{n-2}) + \cdots
+ (r_0 - r_1)(l_2 + \cdots + l_n - r_1),
and with these notations we have
.. MATH::
P^{\nu}_{\mu,(1^r)} = q^t \binom{l_n}{r_{n-1}}_q
\binom{l_{n-1}}{r_{n-2} - r_{n-1}}_q \cdots \binom{l_1}{r_0 - r_1}_q.
To compute `P^{\nu}_{\mu,\lambda}` in general, we compute the product
`I_{\mu} I_{\lambda}` in the Hall algebra and return the coefficient of
`I_{\nu}`.
EXAMPLES::
sage: from sage.combinat.hall_polynomial import hall_polynomial
sage: hall_polynomial([1,1],[1],[1])
q + 1
sage: hall_polynomial([2],[1],[1])
1
sage: hall_polynomial([2,1],[2],[1])
q
sage: hall_polynomial([2,2,1],[2,1],[1,1])
q^2 + q
sage: hall_polynomial([2,2,2,1],[2,2,1],[1,1])
q^4 + q^3 + q^2
sage: hall_polynomial([3,2,2,1], [3,2], [2,1])
q^6 + q^5
sage: hall_polynomial([4,2,1,1], [3,1,1], [2,1])
2*q^3 + q^2 - q - 1
sage: hall_polynomial([4,2], [2,1], [2,1], 0)
1
TESTS::
sage: hall_polynomial([3], [1], [1], 0)
0
"""
if q is None:
q = ZZ['q'].gen()
R = q.parent()
# Make sure they are partitions
nu = Partition(nu)
mu = Partition(mu)
la = Partition(la)
if sum(nu) != sum(mu) + sum(la):
return R.zero()
if all(x == 1 for x in la):
r = [len(la)] # r will be [r_0, r_1, ..., r_n].
exp_nu = nu.to_exp() # exp_nu == [l_1, l_2, ..., l_n].
exp_mu = mu.to_exp() # exp_mu == [m_1, m_2, ..., m_n].
n = max(len(exp_nu), len(exp_mu))
for k in range(n):
r.append(r[-1] + sum(exp_mu[k:]) - sum(exp_nu[k:]))
# Now, r is [r_0, r_1, ..., r_n].
exp_nu += [0] * (n - len(exp_nu)) # Pad with 0's until it has length n
# Note that all -1 for exp_nu is due to indexing
t = sum((r[k - 2] - r[k - 1]) * (sum(exp_nu[k - 1:]) - r[k - 1])
for k in range(2, n + 1))
if t < 0:
# This case needs short-circuiting, since otherwise q**-t
# might throw an exception if q is non-invertible.
return R.zero()
return q**t * q_binomial(exp_nu[n-1], r[n-1], q) \
* prod([q_binomial(exp_nu[k-1], r[k-1] - r[k], q)
for k in range(1, n)], R.one())
from sage.algebras.hall_algebra import HallAlgebra
H = HallAlgebra(R, q)
return (H[mu] * H[la]).coefficient(nu)
def hall_polynomial(nu, mu, la, q=None):
r"""
Return the (classical) Hall polynomial `P^{\nu}_{\mu,\lambda}`
(where `\nu`, `\mu` and `\lambda` are the inputs ``nu``, ``mu``
and ``la``).
Let `\nu,\mu,\lambda` be partitions. The Hall polynomial
`P^{\nu}_{\mu,\lambda}(q)` (in the indeterminate `q`) is defined
as follows: Specialize `q` to a prime power, and consider the
category of `\GF{q}`-vector spaces with a distinguished
nilpotent endomorphism. The morphisms in this category shall be
the linear maps commuting with the distinguished endomorphisms.
The *type* of an object in the category will be the Jordan type
of the distinguished endomorphism; this is a partition. Now, if
`N` is any fixed object of type `\nu` in this category, then
the polynomial `P^{\nu}_{\mu,\lambda}(q)` specialized at the
prime power `q` counts the number of subobjects `L` of `N` having
type `\lambda` such that the quotient object `N / L` has type
`\mu`. This determines the values of the polynomial
`P^{\nu}_{\mu,\lambda}` at infinitely many points (namely, at all
prime powers), and hence `P^{\nu}_{\mu,\lambda}` is uniquely
determined. The degree of this polynomial is at most
`n(\nu) - n(\lambda) - n(\mu)`, where
`n(\kappa) = \sum_i (i-1) \kappa_i` for every partition `\kappa`.
(If this is negative, then the polynomial is zero.)
These are the structure coefficients of the
:class:`(classical) Hall algebra <HallAlgebra>`.
If `\lvert \nu \rvert \neq \lvert \mu \rvert + \lvert \lambda \rvert`,
then we have `P^{\nu}_{\mu,\lambda} = 0`. More generally, if the
Littlewood-Richardson coefficient `c^{\nu}_{\mu,\lambda}`
vanishes, then `P^{\nu}_{\mu,\lambda} = 0`.
The Hall polynomials satisfy the symmetry property
`P^{\nu}_{\mu,\lambda} = P^{\nu}_{\lambda,\mu}`.
ALGORITHM:
If `\lambda = (1^r)` and
`\lvert \nu \rvert = \lvert \mu \rvert + \lvert \lambda \rvert`,
then we compute `P^{\nu}_{\mu,\lambda}` as follows (cf. Example 2.4
in [Schiffmann]_):
First, write `\nu = (1^{l_1}, 2^{l_2}, \ldots, n^{l_n})`, and
define a sequence `r = r_0 \geq r_1 \geq \cdots \geq r_n` such that
.. MATH::
\mu = \left( 1^{l_1 - r_0 + 2r_1 - r_2}, 2^{l_2 - r_1 + 2r_2 - r_3},
\ldots , (n-1)^{l_{n-1} - r_{n-2} + 2r_{n-1} - r_n},
n^{l_n - r_{n-1} + r_n} \right).
Thus if `\mu = (1^{m_1}, \ldots, n^{m_n})`, we have the following system
of equations:
.. MATH::
\begin{aligned}
m_1 & = l_1 - r + 2r_1 - r_2,
\\ m_2 & = l_2 - r_1 + 2r_2 - r_3,
\\ & \vdots ,
\\ m_{n-1} & = l_{n-1} - r_{n-2} + 2r_{n-1} - r_n,
\\ m_n & = l_n - r_{n-1} + r_n
\end{aligned}
and solving for `r_i` and back substituting we obtain the equations:
.. MATH::
\begin{aligned}
r_n & = r_{n-1} + m_n - l_n,
\\ r_{n-1} & = r_{n-2} + m_{n-1} - l_{n-1} + m_n - l_n,
\\ & \vdots ,
\\ r_1 & = r + \sum_{k=1}^n (m_k - l_k),
\end{aligned}
or in general we have the recursive equation:
.. MATH::
r_i = r_{i-1} + \sum_{k=i}^n (m_k - l_k).
This, combined with the condition that `r_0 = r`, determines the
`r_i` uniquely (recursively). Next we define
.. MATH::
t = (r_{n-2} - r_{n-1})(l_n - r_{n-1})
+ (r_{n-3} - r_{n-2})(l_{n-1} + l_n - r_{n-2}) + \cdots
+ (r_0 - r_1)(l_2 + \cdots + l_n - r_1),
and with these notations we have
.. MATH::
P^{\nu}_{\mu,(1^r)} = q^t \binom{l_n}{r_{n-1}}_q
\binom{l_{n-1}}{r_{n-2} - r_{n-1}}_q \cdots \binom{l_1}{r_0 - r_1}_q.
To compute `P^{\nu}_{\mu,\lambda}` in general, we compute the product
`I_{\mu} I_{\lambda}` in the Hall algebra and return the coefficient of
`I_{\nu}`.
EXAMPLES::
sage: from sage.combinat.hall_polynomial import hall_polynomial
sage: hall_polynomial([1,1],[1],[1])
q + 1
sage: hall_polynomial([2],[1],[1])
1
sage: hall_polynomial([2,1],[2],[1])
q
sage: hall_polynomial([2,2,1],[2,1],[1,1])
q^2 + q
sage: hall_polynomial([2,2,2,1],[2,2,1],[1,1])
q^4 + q^3 + q^2
sage: hall_polynomial([3,2,2,1], [3,2], [2,1])
q^6 + q^5
sage: hall_polynomial([4,2,1,1], [3,1,1], [2,1])
2*q^3 + q^2 - q - 1
sage: hall_polynomial([4,2], [2,1], [2,1], 0)
1
"""
if q is None:
q = ZZ['q'].gen()
R = q.parent()
# Make sure they are partitions
nu = Partition(nu)
mu = Partition(mu)
la = Partition(la)
if sum(nu) != sum(mu) + sum(la):
return R.zero()
if all(x == 1 for x in la):
r = [len(la)] # r will be [r_0, r_1, ..., r_n].
exp_nu = nu.to_exp() # exp_nu == [l_1, l_2, ..., l_n].
exp_mu = mu.to_exp() # exp_mu == [m_1, m_2, ..., m_n].
n = max(len(exp_nu), len(exp_mu))
for k in range(n):
r.append(r[-1] + sum(exp_mu[k:]) - sum(exp_nu[k:]))
# Now, r is [r_0, r_1, ..., r_n].
exp_nu += [0]*(n - len(exp_nu)) # Pad with 0's until it has length n
# Note that all -1 for exp_nu is due to indexing
t = sum((r[k-2] - r[k-1])*(sum(exp_nu[k-1:]) - r[k-1]) for k in range(2,n+1))
if t < 0:
# This case needs short-circuiting, since otherwise q**-t
# might throw an exception if q is non-invertible.
return R.zero()
return q**t * q_binomial(exp_nu[n-1], r[n-1], q) \
* prod([q_binomial(exp_nu[k-1], r[k-1] - r[k], q)
for k in range(1, n)], R.one())
from sage.algebras.hall_algebra import HallAlgebra
H = HallAlgebra(R, q)
return (H[mu]*H[la]).coefficient(nu)
