# The Steenrod algebra#

AUTHORS:

• John H. Palmieri (2008-07-30): version 0.9: Initial implementation.

• John H. Palmieri (2010-06-30): version 1.0: Implemented sub-Hopf algebras and profile functions; direct multiplication of admissible sequences (rather than conversion to the Milnor basis); implemented the Steenrod algebra using CombinatorialFreeModule; improved the test suite.

This module defines the mod $$p$$ Steenrod algebra $$\mathcal{A}_p$$, some of its properties, and ways to define elements of it.

From a topological point of view, $$\mathcal{A}_p$$ is the algebra of stable cohomology operations on mod $$p$$ cohomology; thus for any topological space $$X$$, its mod $$p$$ cohomology algebra $$H^*(X,\mathbf{F}_p)$$ is a module over $$\mathcal{A}_p$$.

From an algebraic point of view, $$\mathcal{A}_p$$ is an $$\mathbf{F}_p$$-algebra; when $$p=2$$, it is generated by elements $$\text{Sq}^i$$ for $$i\geq 0$$ (the Steenrod squares), and when $$p$$ is odd, it is generated by elements $$\mathcal{P}^i$$ for $$i \geq 0$$ (the Steenrod reduced pth powers) along with an element $$\beta$$ (the mod p Bockstein). The Steenrod algebra is graded: $$\text{Sq}^i$$ is in degree $$i$$ for each $$i$$, $$\beta$$ is in degree 1, and $$\mathcal{P}^i$$ is in degree $$2(p-1)i$$.

The unit element is $$\text{Sq}^0$$ when $$p=2$$ and $$\mathcal{P}^0$$ when $$p$$ is odd. The generating elements also satisfy the Adem relations. At the prime 2, these have the form

$\text{Sq}^a \text{Sq}^b = \sum_{c=0}^{[a/2]} \binom{b-c-1}{a-2c} \text{Sq}^{a+b-c} \text{Sq}^c.$

At odd primes, they are a bit more complicated; see Steenrod and Epstein [SE1962] or sage.algebras.steenrod.steenrod_algebra_bases for full details. These relations lead to the existence of the Serre-Cartan basis for $$\mathcal{A}_p$$.

The mod $$p$$ Steenrod algebra has the structure of a Hopf algebra, and Milnor [Mil1958] has a beautiful description of the dual, leading to a construction of the Milnor basis for $$\mathcal{A}_p$$. In this module, elements in the Steenrod algebra are represented, by default, using the Milnor basis.

Bases for the Steenrod algebra

There are a handful of other bases studied in the literature; the paper by Monks [Mon1998] is a good reference. Here is a quick summary:

• The Milnor basis. When $$p=2$$, the Milnor basis consists of symbols of the form $$\text{Sq}(m_1, m_2, ..., m_t)$$, where each $$m_i$$ is a non-negative integer and if $$t>1$$, then the last entry $$m_t > 0$$. When $$p$$ is odd, the Milnor basis consists of symbols of the form $$Q_{e_1} Q_{e_2} ... \mathcal{P}(m_1, m_2, ..., m_t)$$, where $$0 \leq e_1 < e_2 < ...$$, each $$m_i$$ is a non-negative integer, and if $$t>1$$, then the last entry $$m_t > 0$$.

When $$p=2$$, it can be convenient to use the notation $$\mathcal{P}(-)$$ to mean $$\text{Sq}(-)$$, so that there is consistent notation for all primes.

• The Serre-Cartan basis. This basis consists of ‘admissible monomials’ in the Steenrod operations. Thus at the prime 2, it consists of monomials $$\text{Sq}^{m_1} \text{Sq}^{m_2} ... \text{Sq}^{m_t}$$ with $$m_i \geq 2m_{i+1}$$ for each $$i$$. At odd primes, this basis consists of monomials $$\beta^{\epsilon_0} \mathcal{P}^{s_1} \beta^{\epsilon_1} \mathcal{P}^{s_2} ... \mathcal{P}^{s_k} \beta^{\epsilon_k}$$ with each $$\epsilon_i$$ either 0 or 1, $$s_i \geq p s_{i+1} + \epsilon_i$$, and $$s_k \geq 1$$.

Most of the rest of the bases are only defined when $$p=2$$. The only exceptions are the $$P^s_t$$-bases and the commutator bases, which are defined at all primes.

• Wood’s Y basis. For pairs of non-negative integers $$(m,k)$$, let $$w(m,k) = \text{Sq}^{2^m (2^{k+1}-1)}$$. Wood’s $$Y$$ basis consists of monomials $$w(m_0,k_0) ... w(m_t, k_t)$$ with $$(m_i,k_i) > (m_{i+1},k_{i+1})$$, in left lex order.

• Wood’s Z basis. For pairs of non-negative integers $$(m,k)$$, let $$w(m,k) = \text{Sq}^{2^m (2^{k+1}-1)}$$. Wood’s $$Z$$ basis consists of monomials $$w(m_0,k_0) ... w(m_t, k_t)$$ with $$(m_i+k_i,m_i) > (m_{i+1}+k_{i+1},m_{i+1})$$, in left lex order.

• Wall’s basis. For any pair of integers $$(m,k)$$ with $$m \geq k \geq 0$$, let $$Q^m_k = \text{Sq}^{2^k} \text{Sq}^{2^{k+1}} ... \text{Sq}^{2^m}$$. The elements of Wall’s basis are monomials $$Q^{m_0}_{k_0} ... Q^{m_t}_{k_t}$$ with $$(m_i, k_i) > (m_{i+1}, k_{i+1})$$, ordered left lexicographically.

(Note that $$Q^m_k$$ is the reverse of the element $$X^m_k$$ used in defining Arnon’s A basis.)

• Arnon’s A basis. For any pair of integers $$(m,k)$$ with $$m \geq k \geq 0$$, let $$X^m_k = \text{Sq}^{2^m} \text{Sq}^{2^{m-1}} ... \text{Sq}^{2^k}$$. The elements of Arnon’s A basis are monomials $$X^{m_0}_{k_0} ... X^{m_t}_{k_t}$$ with $$(m_i, k_i) < (m_{i+1}, k_{i+1})$$, ordered left lexicographically.

(Note that $$X^m_k$$ is the reverse of the element $$Q^m_k$$ used in defining Wall’s basis.)

• Arnon’s C basis. The elements of Arnon’s C basis are monomials of the form $$\text{Sq}^{t_1} ... \text{Sq}^{t_m}$$ where for each $$i$$, we have $$t_i \leq 2t_{i+1}$$ and $$2^i | t_{m-i}$$.

• $$P^s_t$$ bases. Let $$p=2$$. For integers $$s \geq 0$$ and $$t > 0$$, the element $$P^s_t$$ is the Milnor basis element $$\mathcal{P}(0, ..., 0, p^s, 0, ...)$$, with the nonzero entry in position $$t$$. To obtain a $$P^s_t$$-basis, for each set $$\{P^{s_1}_{t_1}, ..., P^{s_k}_{t_k}\}$$ of (distinct) $$P^s_t$$’s, one chooses an ordering and forms the monomials

$(P^{s_1}_{t_1})^{i_1} ... (P^{s_k}_{t_k})^{i_k}$

for all exponents $$i_j$$ with $$0 < i_j < p$$. When $$p=2$$, the set of all such monomials then forms a basis, and when $$p$$ is odd, if one multiplies each such monomial on the left by products of the form $$Q_{e_1} Q_{e_2} ...$$ with $$0 \leq e_1 < e_2 < ...$$, one obtains a basis.

Thus one gets a basis by choosing an ordering on each set of $$P^s_t$$’s. There are infinitely many orderings possible, and we have implemented four of them:

• ‘rlex’: right lexicographic ordering

• ‘llex’: left lexicographic ordering

• ‘deg’: ordered by degree, which is the same as left lexicographic ordering on the pair $$(s+t,t)$$

• ‘revz’: left lexicographic ordering on the pair $$(s+t,s)$$, which is the reverse of the ordering used (on elements in the same degrees as the $$P^s_t$$’s) in Wood’s Z basis: ‘revz’ stands for ‘reversed Z’. This is the default: ‘pst’ is the same as ‘pst_revz’.

• Commutator bases. Let $$c_{i,1} = \mathcal{P}(p^i)$$, let $$c_{i,2} = [c_{i+1,1}, c_{i,1}]$$, and inductively define $$c_{i,k} = [c_{i+k-1,1}, c_{i,k-1}]$$. Thus $$c_{i,k}$$ is a $$k$$-fold iterated commutator of the elements $$\mathcal{P}(p^i)$$, …, $$\mathcal{P}(p^{i+k-1})$$. Note that $$\dim c_{i,k} = \dim P^i_k$$.

Commutator bases are obtained in much the same way as $$P^s_t$$-bases: for each set $$\{c_{s_1,t_1}, ..., c_{s_k,t_k}\}$$ of (distinct) $$c_{s,t}$$’s, one chooses an ordering and forms the resulting monomials

$c_{s_1, t_1}^{i_1} ... c_{s_k,t_k}^{i_k}$

for all exponents $$i_j$$ with $$0 < i_j < p$$. When $$p$$ is odd, one also needs to left-multiply by products of the $$Q_i$$’s. As for $$P^s_t$$-bases, every ordering on each set of iterated commutators determines a basis, and the same four orderings have been defined for these bases as for the $$P^s_t$$ bases: ‘rlex’, ‘llex’, ‘deg’, ‘revz’.

Sub-Hopf algebras of the Steenrod algebra

The sub-Hopf algebras of the Steenrod algebra have been classified. Milnor proved that at the prime 2, the dual of the Steenrod algebra $$A_*$$ is isomorphic to a polynomial algebra

$A_* \cong \GF{2} [\xi_1, \xi_2, \xi_3, ...].$

The Milnor basis is dual to the monomial basis. Furthermore, any sub-Hopf algebra corresponds to a quotient of this of the form

$A_* /(\xi_1^{2^{e_1}}, \xi_2^{2^{e_2}}, \xi_3^{2^{e_3}}, ...).$

The list of exponents $$(e_1, e_2, ...)$$ may be considered a function $$e$$ from the positive integers to the extended non-negative integers (the non-negative integers and $$\infty$$); this is called the profile function for the sub-Hopf algebra. The profile function must satisfy the condition

• $$e(r) \geq \min( e(r-i) - i, e(i))$$ for all $$0 < i < r$$.

At odd primes, the situation is similar: the dual is isomorphic to the tensor product of a polynomial algebra and an exterior algebra,

$A_* = \GF{p} [\xi_1, \xi_2, \xi_3, ...] \otimes \Lambda (\tau_0, \tau_1, ...),$

and any sub-Hopf algebra corresponds to a quotient of this of the form

$A_* / (\xi_1^{p^{e_1}}, \xi_2^{p^{e_2}}, ...; \tau_0^{k_0}, \tau_1^{k_1}, ...).$

Here the profile function has two pieces, $$e$$ as at the prime 2, and $$k$$, which maps the non-negative integers to the set $$\{1, 2\}$$. These must satisfy the following conditions:

• $$e(r) \geq \min( e(r-i) - i, e(i))$$ for all $$0 < i < r$$.

• if $$k(i+j) = 1$$, then either $$e(i) \leq j$$ or $$k(j) = 1$$ for all $$i \geq 1$$, $$j \geq 0$$.

(See Adams-Margolis [AM1974], for example, for these results on profile functions.)

This module allows one to construct the Steenrod algebra or any of its sub-Hopf algebras, at any prime. When defining a sub-Hopf algebra, you must work with the Milnor basis or a $$P^s_t$$-basis.

Elements of the Steenrod algebra

Basic arithmetic, $$p=2$$. To construct an element of the mod 2 Steenrod algebra, use the function Sq:

sage: a = Sq(1,2)
sage: b = Sq(4,1)
sage: z = a + b
sage: z
Sq(1,2) + Sq(4,1)
sage: Sq(4) * Sq(1,2)
Sq(1,1,1) + Sq(2,3) + Sq(5,2)
sage: z**2          # non-negative exponents work as they should
Sq(1,2,1) + Sq(4,1,1)
sage: z**0
1

>>> from sage.all import *
>>> a = Sq(Integer(1),Integer(2))
>>> b = Sq(Integer(4),Integer(1))
>>> z = a + b
>>> z
Sq(1,2) + Sq(4,1)
>>> Sq(Integer(4)) * Sq(Integer(1),Integer(2))
Sq(1,1,1) + Sq(2,3) + Sq(5,2)
>>> z**Integer(2)          # non-negative exponents work as they should
Sq(1,2,1) + Sq(4,1,1)
>>> z**Integer(0)
1


Basic arithmetic, $$p>2$$. To construct an element of the mod $$p$$ Steenrod algebra when $$p$$ is odd, you should first define a Steenrod algebra, using the SteenrodAlgebra command:

sage: A3 = SteenrodAlgebra(3)

>>> from sage.all import *
>>> A3 = SteenrodAlgebra(Integer(3))


Having done this, the newly created algebra A3 has methods Q and P which construct elements of A3:

sage: c = A3.Q(1,3,6); c
Q_1 Q_3 Q_6
sage: d = A3.P(2,0,1); d
P(2,0,1)
sage: c * d
Q_1 Q_3 Q_6 P(2,0,1)
sage: e = A3.P(3)
sage: d * e
P(5,0,1)
sage: e * d
P(1,1,1) + P(5,0,1)
sage: c * c
0
sage: e ** 3
2 P(1,2)

>>> from sage.all import *
>>> c = A3.Q(Integer(1),Integer(3),Integer(6)); c
Q_1 Q_3 Q_6
>>> d = A3.P(Integer(2),Integer(0),Integer(1)); d
P(2,0,1)
>>> c * d
Q_1 Q_3 Q_6 P(2,0,1)
>>> e = A3.P(Integer(3))
>>> d * e
P(5,0,1)
>>> e * d
P(1,1,1) + P(5,0,1)
>>> c * c
0
>>> e ** Integer(3)
2 P(1,2)


Note that one can construct an element like c above in one step, without first constructing the algebra:

sage: c = SteenrodAlgebra(3).Q(1,3,6)
sage: c
Q_1 Q_3 Q_6

>>> from sage.all import *
>>> c = SteenrodAlgebra(Integer(3)).Q(Integer(1),Integer(3),Integer(6))
>>> c
Q_1 Q_3 Q_6


And of course, you can do similar constructions with the mod 2 Steenrod algebra:

sage: A = SteenrodAlgebra(2); A
mod 2 Steenrod algebra, milnor basis
sage: A.Sq(2,3,5)
Sq(2,3,5)
sage: A.P(2,3,5)   # when p=2, P = Sq
Sq(2,3,5)
sage: A.Q(1,4)     # when p=2, this gives a product of Milnor primitives
Sq(0,1,0,0,1)

>>> from sage.all import *
>>> A = SteenrodAlgebra(Integer(2)); A
mod 2 Steenrod algebra, milnor basis
>>> A.Sq(Integer(2),Integer(3),Integer(5))
Sq(2,3,5)
>>> A.P(Integer(2),Integer(3),Integer(5))   # when p=2, P = Sq
Sq(2,3,5)
>>> A.Q(Integer(1),Integer(4))     # when p=2, this gives a product of Milnor primitives
Sq(0,1,0,0,1)


Associated to each element is its prime (the characteristic of the underlying base field) and its basis (the basis for the Steenrod algebra in which it lies):

sage: a = SteenrodAlgebra(basis='milnor').Sq(1,2,1)
sage: a.prime()
2
sage: a.basis_name()
'milnor'
sage: a.degree()
14

>>> from sage.all import *
>>> a = SteenrodAlgebra(basis='milnor').Sq(Integer(1),Integer(2),Integer(1))
>>> a.prime()
2
>>> a.basis_name()
'milnor'
>>> a.degree()
14


It can be viewed in other bases:

sage: a.milnor() # same as a
Sq(1,2,1)
Sq^9 Sq^4 Sq^1 + Sq^11 Sq^2 Sq^1 + Sq^13 Sq^1
Sq(1,2,1)

>>> from sage.all import *
>>> a.milnor() # same as a
Sq(1,2,1)
Sq^9 Sq^4 Sq^1 + Sq^11 Sq^2 Sq^1 + Sq^13 Sq^1
Sq(1,2,1)


Regardless of the prime, each element has an excess, and if the element is homogeneous, a degree. The excess of $$\text{Sq}(i_1,i_2,i_3,...)$$ is $$i_1 + i_2 + i_3 + ...$$; when $$p$$ is odd, the excess of $$Q_{0}^{e_0} Q_{1}^{e_1} ... \mathcal{P}(r_1, r_2, ...)$$ is $$\sum e_i + 2 \sum r_i$$. The excess of a linear combination of Milnor basis elements is the minimum of the excesses of those basis elements.

The degree of $$\text{Sq}(i_1,i_2,i_3,...)$$ is $$\sum (2^n-1) i_n$$, and when $$p$$ is odd, the degree of $$Q_{0}^{\epsilon_0} Q_{1}^{\epsilon_1} ... \mathcal{P}(r_1, r_2, ...)$$ is $$\sum \epsilon_i (2p^i - 1) + \sum r_j (2p^j - 2)$$. The degree of a linear combination of such terms is only defined if the terms all have the same degree.

Here are some simple examples:

sage: z = Sq(1,2) + Sq(4,1)
sage: z.degree()
7
sage: (Sq(0,0,1) + Sq(5,3)).degree()
Traceback (most recent call last):
...
ValueError: element is not homogeneous
sage: Sq(7,2,1).excess()
10
sage: z.excess()
3
sage: B = SteenrodAlgebra(3)
sage: x = B.Q(1,4)
sage: y = B.P(1,2,3)
sage: x.degree()
166
sage: x.excess()
2
sage: y.excess()
12

>>> from sage.all import *
>>> z = Sq(Integer(1),Integer(2)) + Sq(Integer(4),Integer(1))
>>> z.degree()
7
>>> (Sq(Integer(0),Integer(0),Integer(1)) + Sq(Integer(5),Integer(3))).degree()
Traceback (most recent call last):
...
ValueError: element is not homogeneous
>>> Sq(Integer(7),Integer(2),Integer(1)).excess()
10
>>> z.excess()
3
>>> B = SteenrodAlgebra(Integer(3))
>>> x = B.Q(Integer(1),Integer(4))
>>> y = B.P(Integer(1),Integer(2),Integer(3))
>>> x.degree()
166
>>> x.excess()
2
>>> y.excess()
12


Elements have a weight in the May filtration, which (when $$p=2$$) is related to the height function defined by Wall:

sage: Sq(2,1,5).may_weight()
9
sage: Sq(2,1,5).wall_height()
[2, 3, 2, 1, 1]
sage: b = Sq(4)*Sq(8) + Sq(8)*Sq(4)
sage: b.may_weight()
2
sage: b.wall_height()
[0, 0, 1, 1]

>>> from sage.all import *
>>> Sq(Integer(2),Integer(1),Integer(5)).may_weight()
9
>>> Sq(Integer(2),Integer(1),Integer(5)).wall_height()
[2, 3, 2, 1, 1]
>>> b = Sq(Integer(4))*Sq(Integer(8)) + Sq(Integer(8))*Sq(Integer(4))
>>> b.may_weight()
2
>>> b.wall_height()
[0, 0, 1, 1]


Odd primary May weights:

sage: A5 = SteenrodAlgebra(5)
sage: a = A5.Q(1,2,4)
sage: b = A5.P(1,2,1)
sage: a.may_weight()
10
sage: b.may_weight()
8
sage: (a * b).may_weight()
18
sage: A5.P(0,0,1).may_weight()
3

>>> from sage.all import *
>>> A5 = SteenrodAlgebra(Integer(5))
>>> a = A5.Q(Integer(1),Integer(2),Integer(4))
>>> b = A5.P(Integer(1),Integer(2),Integer(1))
>>> a.may_weight()
10
>>> b.may_weight()
8
>>> (a * b).may_weight()
18
>>> A5.P(Integer(0),Integer(0),Integer(1)).may_weight()
3


Since the Steenrod algebra is a Hopf algebra, every element has a coproduct and an antipode:

sage: Sq(5).coproduct()
1 # Sq(5) + Sq(1) # Sq(4) + Sq(2) # Sq(3) + Sq(3) # Sq(2) + Sq(4) # Sq(1) + Sq(5) # 1
sage: Sq(5).antipode()
Sq(2,1) + Sq(5)
sage: d = Sq(0,0,1); d
Sq(0,0,1)
sage: d.antipode()
Sq(0,0,1)
sage: Sq(4).antipode()
Sq(1,1) + Sq(4)
sage: (Sq(4) * Sq(2)).antipode()
Sq(6)
sage: SteenrodAlgebra(7).P(3,1).antipode()
P(3,1)

>>> from sage.all import *
>>> Sq(Integer(5)).coproduct()
1 # Sq(5) + Sq(1) # Sq(4) + Sq(2) # Sq(3) + Sq(3) # Sq(2) + Sq(4) # Sq(1) + Sq(5) # 1
>>> Sq(Integer(5)).antipode()
Sq(2,1) + Sq(5)
>>> d = Sq(Integer(0),Integer(0),Integer(1)); d
Sq(0,0,1)
>>> d.antipode()
Sq(0,0,1)
>>> Sq(Integer(4)).antipode()
Sq(1,1) + Sq(4)
>>> (Sq(Integer(4)) * Sq(Integer(2))).antipode()
Sq(6)
>>> SteenrodAlgebra(Integer(7)).P(Integer(3),Integer(1)).antipode()
P(3,1)


Applying the antipode twice returns the original element:

sage: y = Sq(8)*Sq(4)
sage: y == (y.antipode()).antipode()
True

>>> from sage.all import *
>>> y = Sq(Integer(8))*Sq(Integer(4))
>>> y == (y.antipode()).antipode()
True


Internal representation: you can use any element as an iterator (for x in a: ...), and the method monomial_coefficients() returns a dictionary with keys tuples representing basis elements and with corresponding value representing the coefficient of that term:

sage: c = Sq(5).antipode(); c
Sq(2,1) + Sq(5)
sage: for mono, coeff in c: print((coeff, mono))
(1, (5,))
(1, (2, 1))
sage: c.monomial_coefficients() == {(2, 1): 1, (5,): 1}
True
sage: sorted(c.monomials(), key=lambda x: tuple(x.support()))
[Sq(2,1), Sq(5)]
sage: sorted(c.support())
[(2, 1), (5,)]
sage: sorted(elt.monomials(), key=lambda x: tuple(x.support()))
[Sq^9 Sq^1, Sq^10]

sage: A7 = SteenrodAlgebra(p=7)
sage: a = A7.P(1) * A7.P(1); a
2 P(2)
2
P(2)
2 P(2)
{(0, 2, 0): 2}

>>> from sage.all import *
>>> c = Sq(Integer(5)).antipode(); c
Sq(2,1) + Sq(5)
>>> for mono, coeff in c: print((coeff, mono))
(1, (5,))
(1, (2, 1))
>>> c.monomial_coefficients() == {(Integer(2), Integer(1)): Integer(1), (Integer(5),): Integer(1)}
True
>>> sorted(c.monomials(), key=lambda x: tuple(x.support()))
[Sq(2,1), Sq(5)]
>>> sorted(c.support())
[(2, 1), (5,)]
>>> sorted(elt.monomials(), key=lambda x: tuple(x.support()))
[Sq^9 Sq^1, Sq^10]

>>> A7 = SteenrodAlgebra(p=Integer(7))
>>> a = A7.P(Integer(1)) * A7.P(Integer(1)); a
2 P(2)
2
P(2)
2 P(2)
{(0, 2, 0): 2}


The tuple in the previous output stands for the element $$\beta^0 P^2 \beta^0$$, i.e., $$P^2$$. Going in the other direction, if you want to specify a basis element by giving the corresponding tuple, you can use the monomial() method on the algebra:

sage: SteenrodAlgebra(p=7, basis='adem').monomial((0, 2, 0))
P^2
sage: 10 * SteenrodAlgebra(p=7, basis='adem').monomial((0, 2, 0))
3 P^2

>>> from sage.all import *
P^2
>>> Integer(10) * SteenrodAlgebra(p=Integer(7), basis='adem').monomial((Integer(0), Integer(2), Integer(0)))
3 P^2


In the following example, elements in Wood’s Z basis are certain products of the elements $$w(m,k) = \text{Sq}^{2^m (2^{k+1}-1)}$$. Internally, each $$w(m,k)$$ is represented by the pair $$(m,k)$$, and products of them are represented by tuples of such pairs.

sage: A = SteenrodAlgebra(basis='wood_z')
sage: t = ((2, 0), (0, 0))
sage: A.monomial(t)
Sq^4 Sq^1

>>> from sage.all import *
>>> A = SteenrodAlgebra(basis='wood_z')
>>> t = ((Integer(2), Integer(0)), (Integer(0), Integer(0)))
>>> A.monomial(t)
Sq^4 Sq^1


See the documentation for SteenrodAlgebra() for more details and examples.

sage.algebras.steenrod.steenrod_algebra.AA(n=None, p=2)[source]#

This returns the Steenrod algebra $$A$$ or its sub-Hopf algebra $$A(n)$$.

INPUT:

• $$n$$ – non-negative integer, optional (default None)

• $$p$$ – prime number, optional (default 2)

OUTPUT:

If $$n$$ is None, then return the full Steenrod algebra. Otherwise, return $$A(n)$$.

When $$p=2$$, $$A(n)$$ is the sub-Hopf algebra generated by the elements $$\text{Sq}^i$$ for $$i \leq 2^n$$. Its profile function is $$(n+1, n, n-1, ...)$$. When $$p$$ is odd, $$A(n)$$ is the sub-Hopf algebra generated by the elements $$Q_0$$ and $$\mathcal{P}^i$$ for $$i \leq p^{n-1}$$. Its profile function is $$e=(n, n-1, n-2, ...)$$ and $$k=(2, 2, ..., 2)$$ (length $$n+1$$).

EXAMPLES:

sage: from sage.algebras.steenrod.steenrod_algebra import AA as A
sage: A()
mod 2 Steenrod algebra, milnor basis
sage: A(2)
sub-Hopf algebra of mod 2 Steenrod algebra, milnor basis,
profile function [3, 2, 1]
sage: A(2, p=5)
sub-Hopf algebra of mod 5 Steenrod algebra, milnor basis,
profile function ([2, 1], [2, 2, 2])

>>> from sage.all import *
>>> from sage.algebras.steenrod.steenrod_algebra import AA as A
>>> A()
mod 2 Steenrod algebra, milnor basis
>>> A(Integer(2))
sub-Hopf algebra of mod 2 Steenrod algebra, milnor basis,
profile function [3, 2, 1]
>>> A(Integer(2), p=Integer(5))
sub-Hopf algebra of mod 5 Steenrod algebra, milnor basis,
profile function ([2, 1], [2, 2, 2])

sage.algebras.steenrod.steenrod_algebra.Sq(*nums)[source]#

Milnor element Sq(a,b,c,…).

INPUT:

• a, b, c, … – non-negative integers

OUTPUT: element of the Steenrod algebra

This returns the Milnor basis element $$\text{Sq}(a, b, c, ...)$$.

EXAMPLES:

sage: Sq(5)
Sq(5)
sage: Sq(5) + Sq(2,1) + Sq(5)  # addition is mod 2:
Sq(2,1)
sage: (Sq(4,3) + Sq(7,2)).degree()
13

>>> from sage.all import *
>>> Sq(Integer(5))
Sq(5)
>>> Sq(Integer(5)) + Sq(Integer(2),Integer(1)) + Sq(Integer(5))  # addition is mod 2:
Sq(2,1)
>>> (Sq(Integer(4),Integer(3)) + Sq(Integer(7),Integer(2))).degree()
13


Entries must be non-negative integers; otherwise, an error results.

This function is a good way to define elements of the Steenrod algebra.

sage.algebras.steenrod.steenrod_algebra.SteenrodAlgebra(p=2, basis='milnor', generic='auto', **kwds)[source]#

The mod $$p$$ Steenrod algebra

INPUT:

• p – positive prime integer (default: 2)

• basis – string (default: 'milnor')

• profile – a profile function in form specified below (default: None)

• truncation_type – 0 or $$\infty$$ or ‘auto’ (default: 'auto')

• precision – integer or None (default: None)

• generic – (default: 'auto')

OUTPUT:

mod $$p$$ Steenrod algebra or one of its sub-Hopf algebras, elements of which are printed using basis

See below for information about basis, profile, etc.

EXAMPLES:

Some properties of the Steenrod algebra are available:

sage: A = SteenrodAlgebra(2)
sage: A.order()
+Infinity
sage: A.is_finite()
False
sage: A.is_commutative()
False
sage: A.is_noetherian()
False
sage: A.is_integral_domain()
False
sage: A.is_field()
False
sage: A.is_division_algebra()
False
sage: A.category()
Category of supercocommutative super Hopf algebras
with basis over Finite Field of size 2

>>> from sage.all import *
>>> A = SteenrodAlgebra(Integer(2))
>>> A.order()
+Infinity
>>> A.is_finite()
False
>>> A.is_commutative()
False
>>> A.is_noetherian()
False
>>> A.is_integral_domain()
False
>>> A.is_field()
False
>>> A.is_division_algebra()
False
>>> A.category()
Category of supercocommutative super Hopf algebras
with basis over Finite Field of size 2


There are methods for constructing elements of the Steenrod algebra:

sage: A2 = SteenrodAlgebra(2); A2
mod 2 Steenrod algebra, milnor basis
sage: A2.Sq(1,2,6)
Sq(1,2,6)
sage: A2.Q(3,4)  # product of Milnor primitives Q_3 and Q_4
Sq(0,0,0,1,1)
sage: A2.pst(2,3)  # Margolis pst element
Sq(0,0,4)
sage: A5 = SteenrodAlgebra(5); A5
mod 5 Steenrod algebra, milnor basis
sage: A5.P(1,2,6)
P(1,2,6)
sage: A5.Q(3,4)
Q_3 Q_4
sage: A5.Q(3,4) * A5.P(1,2,6)
Q_3 Q_4 P(1,2,6)
sage: A5.pst(2,3)
P(0,0,25)

>>> from sage.all import *
>>> A2 = SteenrodAlgebra(Integer(2)); A2
mod 2 Steenrod algebra, milnor basis
>>> A2.Sq(Integer(1),Integer(2),Integer(6))
Sq(1,2,6)
>>> A2.Q(Integer(3),Integer(4))  # product of Milnor primitives Q_3 and Q_4
Sq(0,0,0,1,1)
>>> A2.pst(Integer(2),Integer(3))  # Margolis pst element
Sq(0,0,4)
>>> A5 = SteenrodAlgebra(Integer(5)); A5
mod 5 Steenrod algebra, milnor basis
>>> A5.P(Integer(1),Integer(2),Integer(6))
P(1,2,6)
>>> A5.Q(Integer(3),Integer(4))
Q_3 Q_4
>>> A5.Q(Integer(3),Integer(4)) * A5.P(Integer(1),Integer(2),Integer(6))
Q_3 Q_4 P(1,2,6)
>>> A5.pst(Integer(2),Integer(3))
P(0,0,25)


You can test whether elements are contained in the Steenrod algebra:

sage: w = Sq(2) * Sq(4)
sage: w in SteenrodAlgebra(2)
True
sage: w in SteenrodAlgebra(17)
False

>>> from sage.all import *
>>> w = Sq(Integer(2)) * Sq(Integer(4))
>>> w in SteenrodAlgebra(Integer(2))
True
>>> w in SteenrodAlgebra(Integer(17))
False


Different bases for the Steenrod algebra:

There are two standard vector space bases for the mod $$p$$ Steenrod algebra: the Milnor basis and the Serre-Cartan basis. When $$p=2$$, there are also several other, less well-known, bases. See the documentation for this module (type sage.algebras.steenrod.steenrod_algebra?) and the function steenrod_algebra_basis for full descriptions of each of the implemented bases.

This module implements the following bases at all primes:

• ‘milnor’: Milnor basis.

• ‘pst’, ‘pst_rlex’, ‘pst_llex’, ‘pst_deg’, ‘pst_revz’: various $$P^s_t$$-bases.

• ‘comm’, ‘comm_rlex’, ‘comm_llex’, ‘comm_deg’, ‘comm_revz’, or these with ‘_long’ appended: various commutator bases.

It implements the following bases when $$p=2$$:

• ‘wood_y’: Wood’s Y basis.

• ‘wood_z’: Wood’s Z basis.

• ‘wall’, ‘wall_long’: Wall’s basis.

• ‘arnon_a’, ‘arnon_a_long’: Arnon’s A basis.

• ‘arnon_c’: Arnon’s C basis.

When defining a Steenrod algebra, you can specify a basis. Then elements of that Steenrod algebra are printed in that basis:

sage: adem = SteenrodAlgebra(2, 'adem')
sage: x = adem.Sq(2,1)  # Sq(-) always means a Milnor basis element
sage: x
Sq^4 Sq^1 + Sq^5
sage: y = Sq(0,1)    # unadorned Sq defines elements w.r.t. Milnor basis
sage: y
Sq(0,1)
Sq^2 Sq^1 + Sq^3
P^10 P^2 + 4 P^11 P^1 + P^12

>>> from sage.all import *
>>> x = adem.Sq(Integer(2),Integer(1))  # Sq(-) always means a Milnor basis element
>>> x
Sq^4 Sq^1 + Sq^5
>>> y = Sq(Integer(0),Integer(1))    # unadorned Sq defines elements w.r.t. Milnor basis
>>> y
Sq(0,1)
Sq^2 Sq^1 + Sq^3
P^10 P^2 + 4 P^11 P^1 + P^12


If you add or multiply elements defined using different bases, the left-hand factor determines the form of the output:

sage: SteenrodAlgebra(basis='adem').Sq(3) + SteenrodAlgebra(basis='pst').Sq(0,1)
Sq^2 Sq^1
sage: SteenrodAlgebra(basis='pst').Sq(3) + SteenrodAlgebra(basis='milnor').Sq(0,1)
P^0_1 P^1_1 + P^0_2
sage: SteenrodAlgebra(basis='milnor').Sq(2) * SteenrodAlgebra(basis='arnonc').Sq(2)
Sq(1,1)

>>> from sage.all import *
Sq^2 Sq^1
>>> SteenrodAlgebra(basis='pst').Sq(Integer(3)) + SteenrodAlgebra(basis='milnor').Sq(Integer(0),Integer(1))
P^0_1 P^1_1 + P^0_2
>>> SteenrodAlgebra(basis='milnor').Sq(Integer(2)) * SteenrodAlgebra(basis='arnonc').Sq(Integer(2))
Sq(1,1)


You can get a list of basis elements in a given dimension:

sage: A3 = SteenrodAlgebra(3, 'milnor')
sage: A3.basis(13)
Family (Q_1 P(2), Q_0 P(3))

>>> from sage.all import *
>>> A3 = SteenrodAlgebra(Integer(3), 'milnor')
>>> A3.basis(Integer(13))
Family (Q_1 P(2), Q_0 P(3))


Algebras defined over different bases are not equal:

sage: SteenrodAlgebra(basis='milnor') == SteenrodAlgebra(basis='pst')
False

>>> from sage.all import *
>>> SteenrodAlgebra(basis='milnor') == SteenrodAlgebra(basis='pst')
False


Bases have various synonyms, and in general Sage tries to figure out what basis you meant:

sage: SteenrodAlgebra(basis='MiLNOr')
mod 2 Steenrod algebra, milnor basis
sage: SteenrodAlgebra(basis='MiLNOr') == SteenrodAlgebra(basis='milnor')
True
mod 2 Steenrod algebra, serre-cartan basis
'serre-cartan'
sage: SteenrodAlgebra(basis='wood---z---').basis_name()
'woodz'

>>> from sage.all import *
>>> SteenrodAlgebra(basis='MiLNOr')
mod 2 Steenrod algebra, milnor basis
>>> SteenrodAlgebra(basis='MiLNOr') == SteenrodAlgebra(basis='milnor')
True
mod 2 Steenrod algebra, serre-cartan basis
'serre-cartan'
>>> SteenrodAlgebra(basis='wood---z---').basis_name()
'woodz'


As noted above, several of the bases (‘arnon_a’, ‘wall’, ‘comm’) have alternate, sometimes longer, representations. These provide ways of expressing elements of the Steenrod algebra in terms of the $$\text{Sq}^{2^n}$$.

sage: A_long = SteenrodAlgebra(2, 'arnon_a_long')
sage: A_long(Sq(6))
Sq^1 Sq^2 Sq^1 Sq^2 + Sq^2 Sq^4
sage: SteenrodAlgebra(2, 'wall_long')(Sq(6))
Sq^2 Sq^1 Sq^2 Sq^1 + Sq^2 Sq^4
sage: SteenrodAlgebra(2, 'comm_deg_long')(Sq(6))
s_1 s_2 s_12 + s_2 s_4

>>> from sage.all import *
>>> A_long = SteenrodAlgebra(Integer(2), 'arnon_a_long')
>>> A_long(Sq(Integer(6)))
Sq^1 Sq^2 Sq^1 Sq^2 + Sq^2 Sq^4
>>> SteenrodAlgebra(Integer(2), 'wall_long')(Sq(Integer(6)))
Sq^2 Sq^1 Sq^2 Sq^1 + Sq^2 Sq^4
>>> SteenrodAlgebra(Integer(2), 'comm_deg_long')(Sq(Integer(6)))
s_1 s_2 s_12 + s_2 s_4


Sub-Hopf algebras of the Steenrod algebra:

These are specified using the argument profile, along with, optionally, truncation_type and precision. The profile argument specifies the profile function for this algebra. Any sub-Hopf algebra of the Steenrod algebra is determined by its profile function. When $$p=2$$, this is a map $$e$$ from the positive integers to the set of non-negative integers, plus $$\infty$$, corresponding to the sub-Hopf algebra dual to this quotient of the dual Steenrod algebra:

$\GF{2} [\xi_1, \xi_2, \xi_3, ...] / (\xi_1^{2^{e(1)}}, \xi_2^{2^{e(2)}}, \xi_3^{2^{e(3)}}, ...).$

The profile function $$e$$ must satisfy the condition

• $$e(r) \geq \min( e(r-i) - i, e(i))$$ for all $$0 < i < r$$.

This is specified via profile, and optionally precision and truncation_type. First, profile must have one of the following forms:

• a list or tuple, e.g., [3,2,1], corresponding to the function sending 1 to 3, 2 to 2, 3 to 1, and all other integers to the value of truncation_type.

• a function from positive integers to non-negative integers (and $$\infty$$), e.g., lambda n: n+2.

• None or Infinity – use this for the profile function for the whole Steenrod algebra.

In the first and third cases, precision is ignored. In the second case, this function is converted to a tuple of length one less than precision, which has default value 100. The function is truncated at this point, and all remaining values are set to the value of truncation_type.

truncation_type may be 0, $$\infty$$, or ‘auto’. If it’s ‘auto’, then it gets converted to 0 in the first case above (when profile is a list), and otherwise (when profile is a function, None, or Infinity) it gets converted to $$\infty$$.

For example, the sub-Hopf algebra $$A(2)$$ has profile function [3,2,1,0,0,0,...], so it can be defined by any of the following:

sage: A2 = SteenrodAlgebra(profile=[3,2,1])
sage: B2 = SteenrodAlgebra(profile=[3,2,1,0,0]) # trailing 0's ignored
sage: A2 == B2
True
sage: C2 = SteenrodAlgebra(profile=lambda n: max(4-n, 0), truncation_type=0)
sage: A2 == C2
True

>>> from sage.all import *
>>> A2 = SteenrodAlgebra(profile=[Integer(3),Integer(2),Integer(1)])
>>> B2 = SteenrodAlgebra(profile=[Integer(3),Integer(2),Integer(1),Integer(0),Integer(0)]) # trailing 0's ignored
>>> A2 == B2
True
>>> C2 = SteenrodAlgebra(profile=lambda n: max(Integer(4)-n, Integer(0)), truncation_type=Integer(0))
>>> A2 == C2
True


In the following case, the profile function is specified by a function and truncation_type isn’t specified, so it defaults to $$\infty$$; therefore this gives a different sub-Hopf algebra:

sage: D2 = SteenrodAlgebra(profile=lambda n: max(4-n, 0))
sage: A2 == D2
False
sage: D2.is_finite()
False
sage: E2 = SteenrodAlgebra(profile=lambda n: max(4-n, 0), truncation_type=Infinity)
sage: D2 == E2
True

>>> from sage.all import *
>>> D2 = SteenrodAlgebra(profile=lambda n: max(Integer(4)-n, Integer(0)))
>>> A2 == D2
False
>>> D2.is_finite()
False
>>> E2 = SteenrodAlgebra(profile=lambda n: max(Integer(4)-n, Integer(0)), truncation_type=Infinity)
>>> D2 == E2
True


The argument precision only needs to be specified if the profile function is defined by a function and you want to control when the profile switches from the given function to the truncation type. For example:

sage: D3 = SteenrodAlgebra(profile=lambda n: n, precision=3)
sage: D3
sub-Hopf algebra of mod 2 Steenrod algebra, milnor basis, profile function [1, 2, +Infinity, +Infinity, +Infinity, ...]
sage: D4 = SteenrodAlgebra(profile=lambda n: n, precision=4); D4
sub-Hopf algebra of mod 2 Steenrod algebra, milnor basis, profile function [1, 2, 3, +Infinity, +Infinity, +Infinity, ...]
sage: D3 == D4
False

>>> from sage.all import *
>>> D3 = SteenrodAlgebra(profile=lambda n: n, precision=Integer(3))
>>> D3
sub-Hopf algebra of mod 2 Steenrod algebra, milnor basis, profile function [1, 2, +Infinity, +Infinity, +Infinity, ...]
>>> D4 = SteenrodAlgebra(profile=lambda n: n, precision=Integer(4)); D4
sub-Hopf algebra of mod 2 Steenrod algebra, milnor basis, profile function [1, 2, 3, +Infinity, +Infinity, +Infinity, ...]
>>> D3 == D4
False


When $$p$$ is odd, profile is a pair of functions $$e$$ and $$k$$, corresponding to the quotient

$\GF{p} [\xi_1, \xi_2, \xi_3, ...] \otimes \Lambda (\tau_0, \tau_1, ...) / (\xi_1^{p^{e_1}}, \xi_2^{p^{e_2}}, ...; \tau_0^{k_0}, \tau_1^{k_1}, ...).$

Together, the functions $$e$$ and $$k$$ must satisfy the conditions

• $$e(r) \geq \min( e(r-i) - i, e(i))$$ for all $$0 < i < r$$,

• if $$k(i+j) = 1$$, then either $$e(i) \leq j$$ or $$k(j) = 1$$ for all $$i \geq 1$$, $$j \geq 0$$.

Therefore profile must have one of the following forms:

• a pair of lists or tuples, the second of which takes values in the set $$\{1,2\}$$, e.g., ([3,2,1,1], [1,1,2,2,1]).

• a pair of functions, one from the positive integers to non-negative integers (and $$\infty$$), one from the non-negative integers to the set $$\{1,2\}$$, e.g., (lambda n: n+2, lambda n: 1 if n<3 else 2).

• None or Infinity – use this for the profile function for the whole Steenrod algebra.

You can also mix and match the first two, passing a pair with first entry a list and second entry a function, for instance. The values of precision and truncation_type are determined by the first entry.

More examples:

sage: E = SteenrodAlgebra(profile=lambda n: 0 if n<3 else 3, truncation_type=0)
sage: E.is_commutative()
True

sage: A2 = SteenrodAlgebra(profile=[3,2,1]) # the algebra A(2)
sage: Sq(7,3,1) in A2
True
sage: Sq(8) in A2
False
sage: Sq(8) in SteenrodAlgebra().basis(8)
True
sage: Sq(8) in A2.basis(8)
False
sage: A2.basis(8)
Family (Sq(1,0,1), Sq(2,2), Sq(5,1))

sage: A5 = SteenrodAlgebra(p=5)
sage: A51 = SteenrodAlgebra(p=5, profile=([1], [2,2]))
sage: A5.Q(0,1) * A5.P(4) in A51
True
sage: A5.Q(2) in A51
False
sage: A5.P(5) in A51
False

>>> from sage.all import *
>>> E = SteenrodAlgebra(profile=lambda n: Integer(0) if n<Integer(3) else Integer(3), truncation_type=Integer(0))
>>> E.is_commutative()
True

>>> A2 = SteenrodAlgebra(profile=[Integer(3),Integer(2),Integer(1)]) # the algebra A(2)
>>> Sq(Integer(7),Integer(3),Integer(1)) in A2
True
>>> Sq(Integer(8)) in A2
False
>>> Sq(Integer(8)) in SteenrodAlgebra().basis(Integer(8))
True
>>> Sq(Integer(8)) in A2.basis(Integer(8))
False
>>> A2.basis(Integer(8))
Family (Sq(1,0,1), Sq(2,2), Sq(5,1))

>>> A5 = SteenrodAlgebra(p=Integer(5))
>>> A51 = SteenrodAlgebra(p=Integer(5), profile=([Integer(1)], [Integer(2),Integer(2)]))
>>> A5.Q(Integer(0),Integer(1)) * A5.P(Integer(4)) in A51
True
>>> A5.Q(Integer(2)) in A51
False
>>> A5.P(Integer(5)) in A51
False


For sub-Hopf algebras of the Steenrod algebra, only the Milnor basis or the various $$P^s_t$$-bases may be used.

sage: SteenrodAlgebra(profile=[1,2,1,1], basis='adem')
Traceback (most recent call last):
...
NotImplementedError: for sub-Hopf algebras of the Steenrod algebra, only the Milnor basis and the pst bases are implemented

>>> from sage.all import *
Traceback (most recent call last):
...
NotImplementedError: for sub-Hopf algebras of the Steenrod algebra, only the Milnor basis and the pst bases are implemented


The generic Steenrod algebra at the prime $$2$$:

The structure formulas for the Steenrod algebra at odd primes $$p$$ also make sense when $$p$$ is set to $$2$$. We refer to the resulting algebra as the “generic Steenrod algebra” for the prime $$2$$. The dual Hopf algebra is given by

$A_* = \GF{2} [\xi_1, \xi_2, \xi_3, ...] \otimes \Lambda (\tau_0, \tau_1, ...)$

The degree of $$\xi_k$$ is $$2^{k+1}-2$$ and the degree of $$\tau_k$$ is $$2^{k+1}-1$$.

The generic Steenrod algebra is an associated graded algebra of the usual Steenrod algebra that is occasionally useful. Its cohomology, for example, is the $$E_2$$-term of a spectral sequence that computes the $$E_2$$-term of the Novikov spectral sequence. It can also be obtained as a specialisation of Voevodsky’s “motivic Steenrod algebra”: in the notation of [Voe2003], Remark 12.12, it corresponds to setting $$\rho = \tau = 0$$. The usual Steenrod algebra is given by $$\rho = 0$$ and $$\tau = 1$$.

In Sage this algebra is constructed using the ‘generic’ keyword.

Example:

sage: EA = SteenrodAlgebra(p=2,generic=True) ; EA
generic mod 2 Steenrod algebra, milnor basis
sage: EA[8]
Vector space spanned by (Q_0 Q_2, Q_0 Q_1 P(2), P(1,1), P(4))
over Finite Field of size 2

>>> from sage.all import *
>>> EA = SteenrodAlgebra(p=Integer(2),generic=True) ; EA
generic mod 2 Steenrod algebra, milnor basis
>>> EA[Integer(8)]
Vector space spanned by (Q_0 Q_2, Q_0 Q_1 P(2), P(1,1), P(4))
over Finite Field of size 2

class sage.algebras.steenrod.steenrod_algebra.SteenrodAlgebra_generic(p=2, basis='milnor', **kwds)[source]#

The mod $$p$$ Steenrod algebra.

Users should not call this, but use the function SteenrodAlgebra() instead. See that function for extensive documentation.

EXAMPLES:

sage: sage.algebras.steenrod.steenrod_algebra.SteenrodAlgebra_generic()
mod 2 Steenrod algebra, milnor basis
sage: sage.algebras.steenrod.steenrod_algebra.SteenrodAlgebra_generic(5)
mod 5 Steenrod algebra, milnor basis
mod 5 Steenrod algebra, serre-cartan basis

>>> from sage.all import *
>>> sage.algebras.steenrod.steenrod_algebra.SteenrodAlgebra_generic()
mod 2 Steenrod algebra, milnor basis
>>> sage.algebras.steenrod.steenrod_algebra.SteenrodAlgebra_generic(Integer(5))
mod 5 Steenrod algebra, milnor basis
mod 5 Steenrod algebra, serre-cartan basis

class Element[source]#

Class for elements of the Steenrod algebra. Since the Steenrod algebra class is based on CombinatorialFreeModule, this is based on IndexedFreeModuleElement. It has new methods reflecting its role, like degree() for computing the degree of an element.

EXAMPLES:

Since this class inherits from IndexedFreeModuleElement, elements can be used as iterators, and there are other useful methods:

sage: c = Sq(5).antipode(); c
Sq(2,1) + Sq(5)
sage: for mono, coeff in c: print((coeff, mono))
(1, (5,))
(1, (2, 1))
sage: c.monomial_coefficients() == {(2, 1): 1, (5,): 1}
True
sage: sorted(c.monomials(), key=lambda x: tuple(x.support()))
[Sq(2,1), Sq(5)]
sage: sorted(c.support())
[(2, 1), (5,)]

>>> from sage.all import *
>>> c = Sq(Integer(5)).antipode(); c
Sq(2,1) + Sq(5)
>>> for mono, coeff in c: print((coeff, mono))
(1, (5,))
(1, (2, 1))
>>> c.monomial_coefficients() == {(Integer(2), Integer(1)): Integer(1), (Integer(5),): Integer(1)}
True
>>> sorted(c.monomials(), key=lambda x: tuple(x.support()))
[Sq(2,1), Sq(5)]
>>> sorted(c.support())
[(2, 1), (5,)]


See the documentation for this module (type sage.algebras.steenrod.steenrod_algebra?) for more information about elements of the Steenrod algebra.

The additive order of any nonzero element of the mod p Steenrod algebra is p.

OUTPUT: 1 (for the zero element) or p (for anything else)

EXAMPLES:

sage: z = Sq(4) + Sq(6) + 1
2
1

>>> from sage.all import *
>>> z = Sq(Integer(4)) + Sq(Integer(6)) + Integer(1)
2
1

basis_name()[source]#

The basis name associated to self.

EXAMPLES:

sage: a = SteenrodAlgebra().Sq(3,2,1)
sage: a.basis_name()
'milnor'
'serre-cartan'
sage: a.change_basis('wood____y').basis_name()
'woody'
sage: b = SteenrodAlgebra(p=7).basis(36)[0]
sage: b.basis_name()
'milnor'
'serre-cartan'

>>> from sage.all import *
>>> a = SteenrodAlgebra().Sq(Integer(3),Integer(2),Integer(1))
>>> a.basis_name()
'milnor'
'serre-cartan'
>>> a.change_basis('wood____y').basis_name()
'woody'
>>> b = SteenrodAlgebra(p=Integer(7)).basis(Integer(36))[Integer(0)]
>>> b.basis_name()
'milnor'
'serre-cartan'

change_basis(basis='milnor')[source]#

Representation of element with respect to basis.

INPUT:

• basis – string, basis in which to work.

OUTPUT: representation of self in given basis

The choices for basis are:

• ‘milnor’ for the Milnor basis.

• ‘wood_y’ for Wood’s Y basis.

• ‘wood_z’ for Wood’s Z basis.

• ‘wall’ for Wall’s basis.

• ‘wall_long’ for Wall’s basis, alternate representation

• ‘arnon_a’ for Arnon’s A basis.

• ‘arnon_a_long’ for Arnon’s A basis, alternate representation.

• ‘arnon_c’ for Arnon’s C basis.

• ‘pst’, ‘pst_rlex’, ‘pst_llex’, ‘pst_deg’, ‘pst_revz’ for various $$P^s_t$$-bases.

• ‘comm’, ‘comm_rlex’, ‘comm_llex’, ‘comm_deg’, ‘comm_revz’ for various commutator bases.

• ‘comm_long’, ‘comm_rlex_long’, etc., for commutator bases, alternate representations.

See documentation for this module (by browsing the reference manual or by typing sage.algebras.steenrod.steenrod_algebra?) for descriptions of the different bases.

EXAMPLES:

sage: c = Sq(2) * Sq(1)
sage: c.change_basis('milnor')
Sq(0,1) + Sq(3)
sage: c.change_basis('serre-cartan')
Sq^2 Sq^1
sage: d = Sq(0,0,1)
sage: d.change_basis('arnonc')
Sq^2 Sq^5 + Sq^4 Sq^2 Sq^1 + Sq^4 Sq^3 + Sq^7

>>> from sage.all import *
>>> c = Sq(Integer(2)) * Sq(Integer(1))
>>> c.change_basis('milnor')
Sq(0,1) + Sq(3)
>>> c.change_basis('serre-cartan')
Sq^2 Sq^1
>>> d = Sq(Integer(0),Integer(0),Integer(1))
>>> d.change_basis('arnonc')
Sq^2 Sq^5 + Sq^4 Sq^2 Sq^1 + Sq^4 Sq^3 + Sq^7

coproduct(algorithm='milnor')[source]#

The coproduct of this element.

INPUT:

• algorithmNone or a string, either ‘milnor’ or ‘serre-cartan’ (or anything which will be converted to one of these by the function get_basis_name). If None, default to ‘serre-cartan’ if current basis is ‘serre-cartan’; otherwise use ‘milnor’.

See SteenrodAlgebra_generic.coproduct_on_basis() for more information on computing the coproduct.

EXAMPLES:

sage: a = Sq(2)
sage: a.coproduct()
1 # Sq(2) + Sq(1) # Sq(1) + Sq(2) # 1
sage: b = Sq(4)
sage: (a*b).coproduct() == (a.coproduct()) * (b.coproduct())
True

1 # Sq^2 + Sq^1 # Sq^1 + Sq^2 # 1
1 # Sq^2 + Sq^1 # Sq^1 + Sq^2 # 1

sage: d = a.change_basis('comm_long'); d.coproduct()
1 # s_2 + s_1 # s_1 + s_2 # 1

sage: A7 = SteenrodAlgebra(p=7)
sage: a = A7.Q(1) * A7.P(1); a
Q_1 P(1)
sage: a.coproduct()
1 # Q_1 P(1) + P(1) # Q_1 + Q_1 # P(1) + Q_1 P(1) # 1
1 # Q_1 P(1) + P(1) # Q_1 + Q_1 # P(1) + Q_1 P(1) # 1

>>> from sage.all import *
>>> a = Sq(Integer(2))
>>> a.coproduct()
1 # Sq(2) + Sq(1) # Sq(1) + Sq(2) # 1
>>> b = Sq(Integer(4))
>>> (a*b).coproduct() == (a.coproduct()) * (b.coproduct())
True

1 # Sq^2 + Sq^1 # Sq^1 + Sq^2 # 1
1 # Sq^2 + Sq^1 # Sq^1 + Sq^2 # 1

>>> d = a.change_basis('comm_long'); d.coproduct()
1 # s_2 + s_1 # s_1 + s_2 # 1

>>> A7 = SteenrodAlgebra(p=Integer(7))
>>> a = A7.Q(Integer(1)) * A7.P(Integer(1)); a
Q_1 P(1)
>>> a.coproduct()
1 # Q_1 P(1) + P(1) # Q_1 + Q_1 # P(1) + Q_1 P(1) # 1
1 # Q_1 P(1) + P(1) # Q_1 + Q_1 # P(1) + Q_1 P(1) # 1


Once you have an element of the tensor product, you may want to extract the tensor factors of its summands.

sage: b = Sq(2).coproduct()
sage: b
1 # Sq(2) + Sq(1) # Sq(1) + Sq(2) # 1
sage: supp = sorted(b.support()); supp
[((), (2,)), ((1,), (1,)), ((2,), ())]
sage: Sq(*supp[0][0])
1
sage: Sq(*supp[0][1])
Sq(2)
sage: [(Sq(*x), Sq(*y)) for (x,y) in supp]
[(1, Sq(2)), (Sq(1), Sq(1)), (Sq(2), 1)]

>>> from sage.all import *
>>> b = Sq(Integer(2)).coproduct()
>>> b
1 # Sq(2) + Sq(1) # Sq(1) + Sq(2) # 1
>>> supp = sorted(b.support()); supp
[((), (2,)), ((1,), (1,)), ((2,), ())]
>>> Sq(*supp[Integer(0)][Integer(0)])
1
>>> Sq(*supp[Integer(0)][Integer(1)])
Sq(2)
>>> [(Sq(*x), Sq(*y)) for (x,y) in supp]
[(1, Sq(2)), (Sq(1), Sq(1)), (Sq(2), 1)]


The support of an element does not include the coefficients, so at odd primes it may be better to use monomial_coefficients:

sage: A3 = SteenrodAlgebra(p=3)
sage: b = (A3.P(1)**2).coproduct()
sage: b
2*1 # P(2) + 2*P(1) # P(1) + 2*P(2) # 1
sage: sorted(b.support())
[(((), ()), ((), (2,))), (((), (1,)), ((), (1,))), (((), (2,)), ((), ()))]
sage: b.monomial_coefficients()
{(((), ()), ((), (2,))): 2,
(((), (1,)), ((), (1,))): 2,
(((), (2,)), ((), ())): 2}
sage: mc = b.monomial_coefficients()
sage: sorted([(A3.monomial(x), A3.monomial(y), mc[x,y]) for (x,y) in mc])
[(1, P(2), 2), (P(1), P(1), 2), (P(2), 1, 2)]

>>> from sage.all import *
>>> A3 = SteenrodAlgebra(p=Integer(3))
>>> b = (A3.P(Integer(1))**Integer(2)).coproduct()
>>> b
2*1 # P(2) + 2*P(1) # P(1) + 2*P(2) # 1
>>> sorted(b.support())
[(((), ()), ((), (2,))), (((), (1,)), ((), (1,))), (((), (2,)), ((), ()))]
>>> b.monomial_coefficients()
{(((), ()), ((), (2,))): 2,
(((), (1,)), ((), (1,))): 2,
(((), (2,)), ((), ())): 2}
>>> mc = b.monomial_coefficients()
>>> sorted([(A3.monomial(x), A3.monomial(y), mc[x,y]) for (x,y) in mc])
[(1, P(2), 2), (P(1), P(1), 2), (P(2), 1, 2)]

degree()[source]#

The degree of self.

The degree of $$\text{Sq}(i_1,i_2,i_3,...)$$ is

$i_1 + 3i_2 + 7i_3 + ... + (2^k - 1) i_k + ....$

At an odd prime $$p$$, the degree of $$Q_k$$ is $$2p^k - 1$$ and the degree of $$\mathcal{P}(i_1, i_2, ...)$$ is

$\sum_{k \geq 0} 2(p^k - 1) i_k.$

ALGORITHM: If is_homogeneous() returns True, call SteenrodAlgebra_generic.degree_on_basis() on the leading summand.

EXAMPLES:

sage: Sq(0,0,1).degree()
7
sage: (Sq(0,0,1) + Sq(7)).degree()
7
sage: (Sq(0,0,1) + Sq(2)).degree()
Traceback (most recent call last):
...
ValueError: element is not homogeneous

sage: A11 = SteenrodAlgebra(p=11)
sage: A11.P(1).degree()
20
sage: A11.P(1,1).degree()
260
sage: A11.Q(2).degree()
241

>>> from sage.all import *
>>> Sq(Integer(0),Integer(0),Integer(1)).degree()
7
>>> (Sq(Integer(0),Integer(0),Integer(1)) + Sq(Integer(7))).degree()
7
>>> (Sq(Integer(0),Integer(0),Integer(1)) + Sq(Integer(2))).degree()
Traceback (most recent call last):
...
ValueError: element is not homogeneous

>>> A11 = SteenrodAlgebra(p=Integer(11))
>>> A11.P(Integer(1)).degree()
20
>>> A11.P(Integer(1),Integer(1)).degree()
260
>>> A11.Q(Integer(2)).degree()
241

excess()[source]#

Excess of element.

OUTPUT: excess – non-negative integer

The excess of a Milnor basis element $$\text{Sq}(a,b,c,...)$$ is $$a + b + c + \cdots$$. When $$p$$ is odd, the excess of $$Q_{0}^{e_0} Q_{1}^{e_1} \cdots P(r_1, r_2, ...)$$ is $$\sum e_i + 2 \sum r_i$$. The excess of a linear combination of Milnor basis elements is the minimum of the excesses of those basis elements.

See [Kr1971] for the proofs of these assertions.

EXAMPLES:

sage: a = Sq(1,2,3)
sage: a.excess()
6
sage: (Sq(0,0,1) + Sq(4,1) + Sq(7)).excess()
1
sage: elt = Sq(0,0,1) + Sq(4,1) + Sq(7)
sage: M = sorted(elt.monomials(), key=lambda x: tuple(x.support()))
sage: [m.excess() for m in M]
[1, 5, 7]
sage: [m for m in M]
[Sq(0,0,1), Sq(4,1), Sq(7)]
sage: B = SteenrodAlgebra(7)
sage: a = B.Q(1,2,5)
sage: b = B.P(2,2,3)
sage: a.excess()
3
sage: b.excess()
14
sage: (a + b).excess()
3
sage: (a * b).excess()
17

>>> from sage.all import *
>>> a = Sq(Integer(1),Integer(2),Integer(3))
>>> a.excess()
6
>>> (Sq(Integer(0),Integer(0),Integer(1)) + Sq(Integer(4),Integer(1)) + Sq(Integer(7))).excess()
1
>>> elt = Sq(Integer(0),Integer(0),Integer(1)) + Sq(Integer(4),Integer(1)) + Sq(Integer(7))
>>> M = sorted(elt.monomials(), key=lambda x: tuple(x.support()))
>>> [m.excess() for m in M]
[1, 5, 7]
>>> [m for m in M]
[Sq(0,0,1), Sq(4,1), Sq(7)]
>>> B = SteenrodAlgebra(Integer(7))
>>> a = B.Q(Integer(1),Integer(2),Integer(5))
>>> b = B.P(Integer(2),Integer(2),Integer(3))
>>> a.excess()
3
>>> b.excess()
14
>>> (a + b).excess()
3
>>> (a * b).excess()
17

is_decomposable()[source]#

Return True if element is decomposable, False otherwise.

That is, if element is in the square of the augmentation ideal, return True; otherwise, return False.

OUTPUT: boolean

EXAMPLES:

sage: a = Sq(6)
sage: a.is_decomposable()
True
sage: for i in range(9):
....:     if not Sq(i).is_decomposable():
....:         print(Sq(i))
1
Sq(1)
Sq(2)
Sq(4)
Sq(8)
sage: [A3.P(n) for n in range(30) if not A3.P(n).is_decomposable()]
[1, P^1, P^3, P^9, P^27]

>>> from sage.all import *
>>> a = Sq(Integer(6))
>>> a.is_decomposable()
True
>>> for i in range(Integer(9)):
...     if not Sq(i).is_decomposable():
...         print(Sq(i))
1
Sq(1)
Sq(2)
Sq(4)
Sq(8)
>>> [A3.P(n) for n in range(Integer(30)) if not A3.P(n).is_decomposable()]
[1, P^1, P^3, P^9, P^27]

is_homogeneous()[source]#

Return True iff this element is homogeneous.

EXAMPLES:

sage: (Sq(0,0,1) + Sq(7)).is_homogeneous()
True
sage: (Sq(0,0,1) + Sq(2)).is_homogeneous()
False

>>> from sage.all import *
>>> (Sq(Integer(0),Integer(0),Integer(1)) + Sq(Integer(7))).is_homogeneous()
True
>>> (Sq(Integer(0),Integer(0),Integer(1)) + Sq(Integer(2))).is_homogeneous()
False

is_nilpotent()[source]#

True if element is not a unit, False otherwise.

EXAMPLES:

sage: z = Sq(4,2) + Sq(7,1) + Sq(3,0,1)
sage: z.is_nilpotent()
True
sage: u = 1 + Sq(3,1)
sage: u == 1 + Sq(3,1)
True
sage: u.is_nilpotent()
False

>>> from sage.all import *
>>> z = Sq(Integer(4),Integer(2)) + Sq(Integer(7),Integer(1)) + Sq(Integer(3),Integer(0),Integer(1))
>>> z.is_nilpotent()
True
>>> u = Integer(1) + Sq(Integer(3),Integer(1))
>>> u == Integer(1) + Sq(Integer(3),Integer(1))
True
>>> u.is_nilpotent()
False

is_unit()[source]#

True if element has a nonzero scalar multiple of P(0) as a summand, False otherwise.

EXAMPLES:

sage: z = Sq(4,2) + Sq(7,1) + Sq(3,0,1)
sage: z.is_unit()
False
sage: u = Sq(0) + Sq(3,1)
sage: u == 1 + Sq(3,1)
True
sage: u.is_unit()
True
sage: A5 = SteenrodAlgebra(5)
sage: v = A5.P(0)
sage: (v + v + v).is_unit()
True

>>> from sage.all import *
>>> z = Sq(Integer(4),Integer(2)) + Sq(Integer(7),Integer(1)) + Sq(Integer(3),Integer(0),Integer(1))
>>> z.is_unit()
False
>>> u = Sq(Integer(0)) + Sq(Integer(3),Integer(1))
>>> u == Integer(1) + Sq(Integer(3),Integer(1))
True
>>> u.is_unit()
True
>>> A5 = SteenrodAlgebra(Integer(5))
>>> v = A5.P(Integer(0))
>>> (v + v + v).is_unit()
True

may_weight()[source]#

May’s ‘weight’ of element.

OUTPUT: weight – non-negative integer

If we let $$F_* (A)$$ be the May filtration of the Steenrod algebra, the weight of an element $$x$$ is the integer $$k$$ so that $$x$$ is in $$F_k(A)$$ and not in $$F_{k+1}(A)$$. According to Theorem 2.6 in May’s thesis [May1964], the weight of a Milnor basis element is computed as follows: first, to compute the weight of $$P(r_1,r_2, ...)$$, write each $$r_i$$ in base $$p$$ as $$r_i = \sum_j p^j r_{ij}$$. Then each nonzero binary digit $$r_{ij}$$ contributes $$i$$ to the weight: the weight is $$\sum_{i,j} i r_{ij}$$. When $$p$$ is odd, the weight of $$Q_i$$ is $$i+1$$, so the weight of a product $$Q_{i_1} Q_{i_2} ...$$ equals $$(i_1+1) + (i_2+1) + ...$$. Then the weight of $$Q_{i_1} Q_{i_2} ...P(r_1,r_2, ...)$$ is the sum of $$(i_1+1) + (i_2+1) + ...$$ and $$\sum_{i,j} i r_{ij}$$.

The weight of a sum of Milnor basis elements is the minimum of the weights of the summands.

When $$p=2$$, we compute the weight on Milnor basis elements by adding up the terms in their ‘height’ - see wall_height() for documentation. (When $$p$$ is odd, the height of an element is not defined.)

EXAMPLES:

sage: Sq(0).may_weight()
0
sage: a = Sq(4)
sage: a.may_weight()
1
sage: b = Sq(4)*Sq(8) + Sq(8)*Sq(4)
sage: b.may_weight()
2
sage: Sq(2,1,5).wall_height()
[2, 3, 2, 1, 1]
sage: Sq(2,1,5).may_weight()
9
sage: A5 = SteenrodAlgebra(5)
sage: a = A5.Q(1,2,4)
sage: b = A5.P(1,2,1)
sage: a.may_weight()
10
sage: b.may_weight()
8
sage: (a * b).may_weight()
18
sage: A5.P(0,0,1).may_weight()
3

>>> from sage.all import *
>>> Sq(Integer(0)).may_weight()
0
>>> a = Sq(Integer(4))
>>> a.may_weight()
1
>>> b = Sq(Integer(4))*Sq(Integer(8)) + Sq(Integer(8))*Sq(Integer(4))
>>> b.may_weight()
2
>>> Sq(Integer(2),Integer(1),Integer(5)).wall_height()
[2, 3, 2, 1, 1]
>>> Sq(Integer(2),Integer(1),Integer(5)).may_weight()
9
>>> A5 = SteenrodAlgebra(Integer(5))
>>> a = A5.Q(Integer(1),Integer(2),Integer(4))
>>> b = A5.P(Integer(1),Integer(2),Integer(1))
>>> a.may_weight()
10
>>> b.may_weight()
8
>>> (a * b).may_weight()
18
>>> A5.P(Integer(0),Integer(0),Integer(1)).may_weight()
3

milnor()[source]#

Return this element in the Milnor basis; that is, as an element of the appropriate Steenrod algebra.

This just calls the method SteenrodAlgebra_generic.milnor().

EXAMPLES:

sage: Adem = SteenrodAlgebra(basis='adem')
Sq^3 Sq^1
sage: a.milnor()
Sq(1,1)

>>> from sage.all import *
Sq^3 Sq^1
>>> a.milnor()
Sq(1,1)

prime()[source]#

The prime associated to self.

EXAMPLES:

sage: a = SteenrodAlgebra().Sq(3,2,1)
sage: a.prime()
2
2
sage: b = SteenrodAlgebra(p=7).basis(36)[0]
sage: b.prime()
7
3

>>> from sage.all import *
>>> a = SteenrodAlgebra().Sq(Integer(3),Integer(2),Integer(1))
>>> a.prime()
2
2
>>> b = SteenrodAlgebra(p=Integer(7)).basis(Integer(36))[Integer(0)]
>>> b.prime()
7
3

wall_height()[source]#

Wall’s ‘height’ of element.

OUTPUT: list of non-negative integers

The height of an element of the mod 2 Steenrod algebra is a list of non-negative integers, defined as follows: if the element is a monomial in the generators $$\text{Sq}(2^i)$$, then the $$i^{th}$$ entry in the list is the number of times $$\text{Sq}(2^i)$$ appears. For an arbitrary element, write it as a sum of such monomials; then its height is the maximum, ordered right-lexicographically, of the heights of those monomials.

When $$p$$ is odd, the height of an element is not defined.

According to Theorem 3 in [Wal1960], the height of the Milnor basis element $$\text{Sq}(r_1, r_2, ...)$$ is obtained as follows: write each $$r_i$$ in binary as $$r_i = \sum_j 2^j r_{ij}$$. Then each nonzero binary digit $$r_{ij}$$ contributes 1 to the $$k^{th}$$ entry in the height, for $$j \leq k \leq i+j-1$$.

EXAMPLES:

sage: Sq(0).wall_height()
[]
sage: a = Sq(4)
sage: a.wall_height()
[0, 0, 1]
sage: b = Sq(4)*Sq(8) + Sq(8)*Sq(4)
sage: b.wall_height()
[0, 0, 1, 1]
sage: Sq(0,0,3).wall_height()
[1, 2, 2, 1]

>>> from sage.all import *
>>> Sq(Integer(0)).wall_height()
[]
>>> a = Sq(Integer(4))
>>> a.wall_height()
[0, 0, 1]
>>> b = Sq(Integer(4))*Sq(Integer(8)) + Sq(Integer(8))*Sq(Integer(4))
>>> b.wall_height()
[0, 0, 1, 1]
>>> Sq(Integer(0),Integer(0),Integer(3)).wall_height()
[1, 2, 2, 1]

P(*nums)[source]#

The element $$P(a, b, c, ...)$$

INPUT:

• a, b, c, … – non-negative integers

OUTPUT:

element of the Steenrod algebra given by the Milnor single basis element $$P(a, b, c, ...)$$

Note that at the prime 2, this is the same element as $$\text{Sq}(a, b, c, ...)$$.

EXAMPLES:

sage: A = SteenrodAlgebra(2)
sage: A.P(5)
Sq(5)
sage: B = SteenrodAlgebra(3)
sage: B.P(5,1,1)
P(5,1,1)
sage: B.P(1,1,-12,1)
Traceback (most recent call last):
...
TypeError: entries must be non-negative integers

sage: SteenrodAlgebra(basis='serre-cartan').P(0,1)
Sq^2 Sq^1 + Sq^3
sage: SteenrodAlgebra(generic=True).P(2,0,1)
P(2,0,1)

>>> from sage.all import *
>>> A = SteenrodAlgebra(Integer(2))
>>> A.P(Integer(5))
Sq(5)
>>> B = SteenrodAlgebra(Integer(3))
>>> B.P(Integer(5),Integer(1),Integer(1))
P(5,1,1)
>>> B.P(Integer(1),Integer(1),-Integer(12),Integer(1))
Traceback (most recent call last):
...
TypeError: entries must be non-negative integers

>>> SteenrodAlgebra(basis='serre-cartan').P(Integer(0),Integer(1))
Sq^2 Sq^1 + Sq^3
>>> SteenrodAlgebra(generic=True).P(Integer(2),Integer(0),Integer(1))
P(2,0,1)

Q(*nums)[source]#

The element $$Q_{n0} Q_{n1} ...$$ , given by specifying the subscripts.

INPUT:

• n0, n1, … – non-negative integers

OUTPUT: The element $$Q_{n0} Q_{n1} ...$$

Note that at the prime 2, $$Q_n$$ is the element $$\text{Sq}(0,0,...,1)$$ , where the 1 is in the $$(n+1)^{st}$$ position.

Compare this to the method Q_exp(), which defines a similar element, but by specifying the tuple of exponents.

EXAMPLES:

sage: A2 = SteenrodAlgebra(2)
sage: A2.Q(2,3)
Sq(0,0,1,1)
sage: A5 = SteenrodAlgebra(5)
sage: A5.Q(1,4)
Q_1 Q_4
sage: A5.Q(1,4) == A5.Q_exp(0,1,0,0,1)
True
sage: H = SteenrodAlgebra(p=5, profile=[[2,1], [2,2,2]])
sage: H.Q(2)
Q_2
sage: H.Q(4)
Traceback (most recent call last):
...
ValueError: Element not in this algebra

>>> from sage.all import *
>>> A2 = SteenrodAlgebra(Integer(2))
>>> A2.Q(Integer(2),Integer(3))
Sq(0,0,1,1)
>>> A5 = SteenrodAlgebra(Integer(5))
>>> A5.Q(Integer(1),Integer(4))
Q_1 Q_4
>>> A5.Q(Integer(1),Integer(4)) == A5.Q_exp(Integer(0),Integer(1),Integer(0),Integer(0),Integer(1))
True
>>> H = SteenrodAlgebra(p=Integer(5), profile=[[Integer(2),Integer(1)], [Integer(2),Integer(2),Integer(2)]])
>>> H.Q(Integer(2))
Q_2
>>> H.Q(Integer(4))
Traceback (most recent call last):
...
ValueError: Element not in this algebra

Q_exp(*nums)[source]#

The element $$Q_0^{e_0} Q_1^{e_1} ...$$ , given by specifying the exponents.

INPUT:

• e0, e1, … – sequence of 0s and 1s

OUTPUT: The element $$Q_0^{e_0} Q_1^{e_1} ...$$

Note that at the prime 2, $$Q_n$$ is the element $$\text{Sq}(0,0,...,1)$$ , where the 1 is in the $$(n+1)^{st}$$ position.

Compare this to the method Q(), which defines a similar element, but by specifying the tuple of subscripts of terms with exponent 1.

EXAMPLES:

sage: A2 = SteenrodAlgebra(2)
sage: A5 = SteenrodAlgebra(5)
sage: A2.Q_exp(0,0,1,1,0)
Sq(0,0,1,1)
sage: A5.Q_exp(0,0,1,1,0)
Q_2 Q_3
sage: A5.Q(2,3)
Q_2 Q_3
sage: A5.Q_exp(0,0,1,1,0) == A5.Q(2,3)
True
sage: SteenrodAlgebra(2,generic=True).Q_exp(1,0,1)
Q_0 Q_2

>>> from sage.all import *
>>> A2 = SteenrodAlgebra(Integer(2))
>>> A5 = SteenrodAlgebra(Integer(5))
>>> A2.Q_exp(Integer(0),Integer(0),Integer(1),Integer(1),Integer(0))
Sq(0,0,1,1)
>>> A5.Q_exp(Integer(0),Integer(0),Integer(1),Integer(1),Integer(0))
Q_2 Q_3
>>> A5.Q(Integer(2),Integer(3))
Q_2 Q_3
>>> A5.Q_exp(Integer(0),Integer(0),Integer(1),Integer(1),Integer(0)) == A5.Q(Integer(2),Integer(3))
True
>>> SteenrodAlgebra(Integer(2),generic=True).Q_exp(Integer(1),Integer(0),Integer(1))
Q_0 Q_2

algebra_generators()[source]#

Family of generators for this algebra.

OUTPUT: family of elements of this algebra

At the prime 2, the Steenrod algebra is generated by the elements $$\text{Sq}^{2^i}$$ for $$i \geq 0$$. At odd primes, it is generated by the elements $$Q_0$$ and $$\mathcal{P}^{p^i}$$ for $$i \geq 0$$. So if this algebra is the entire Steenrod algebra, return an infinite family made up of these elements.

For sub-Hopf algebras of the Steenrod algebra, it is not always clear what a minimal generating set is. The sub-Hopf algebra $$A(n)$$ is minimally generated by the elements $$\text{Sq}^{2^i}$$ for $$0 \leq i \leq n$$ at the prime 2. At odd primes, $$A(n)$$ is minimally generated by $$Q_0$$ along with $$\mathcal{P}^{p^i}$$ for $$0 \leq i \leq n-1$$. So if this algebra is $$A(n)$$, return the appropriate list of generators.

For other sub-Hopf algebras: return a non-minimal generating set: the family of $$P^s_t$$’s and $$Q_n$$’s contained in the algebra.

EXAMPLES:

sage: A3 = SteenrodAlgebra(3, 'adem')
sage: A3.gens()
Lazy family (<bound method SteenrodAlgebra_generic.gen of mod 3 Steenrod algebra,
serre-cartan basis>(i))_{i in Non negative integers}
sage: A3.gens()[0]
beta
sage: A3.gens()[1]
P^1
sage: A3.gens()[2]
P^3
sage: SteenrodAlgebra(profile=[3,2,1]).gens()
Family (Sq(1), Sq(2), Sq(4))

>>> from sage.all import *
>>> A3.gens()
Lazy family (<bound method SteenrodAlgebra_generic.gen of mod 3 Steenrod algebra,
serre-cartan basis>(i))_{i in Non negative integers}
>>> A3.gens()[Integer(0)]
beta
>>> A3.gens()[Integer(1)]
P^1
>>> A3.gens()[Integer(2)]
P^3
>>> SteenrodAlgebra(profile=[Integer(3),Integer(2),Integer(1)]).gens()
Family (Sq(1), Sq(2), Sq(4))


In the following case, return a non-minimal generating set. (It is not minimal because $$\text{Sq}(0,0,1)$$ is the commutator of $$\text{Sq}(1)$$ and $$\text{Sq}(0,2)$$.)

sage: SteenrodAlgebra(profile=[1,2,1]).gens()
Family (Sq(1), Sq(0,1), Sq(0,2), Sq(0,0,1))
sage: SteenrodAlgebra(p=5, profile=[[2,1], [2,2,2]]).gens()
Family (Q_0, P(1), P(5))
sage: SteenrodAlgebra(profile=lambda n: n).gens()
Lazy family (<bound method SteenrodAlgebra_generic.gen of sub-Hopf algebra
of mod 2 Steenrod algebra, milnor basis, profile function [1, 2, 3, ...,
98, 99, +Infinity, +Infinity, +Infinity, ...]>(i))_{i in Non negative integers}

>>> from sage.all import *
>>> SteenrodAlgebra(profile=[Integer(1),Integer(2),Integer(1)]).gens()
Family (Sq(1), Sq(0,1), Sq(0,2), Sq(0,0,1))
>>> SteenrodAlgebra(p=Integer(5), profile=[[Integer(2),Integer(1)], [Integer(2),Integer(2),Integer(2)]]).gens()
Family (Q_0, P(1), P(5))
>>> SteenrodAlgebra(profile=lambda n: n).gens()
Lazy family (<bound method SteenrodAlgebra_generic.gen of sub-Hopf algebra
of mod 2 Steenrod algebra, milnor basis, profile function [1, 2, 3, ...,
98, 99, +Infinity, +Infinity, +Infinity, ...]>(i))_{i in Non negative integers}


You may also use algebra_generators instead of gens:

sage: SteenrodAlgebra(p=5, profile=[[2,1], [2,2,2]]).algebra_generators()
Family (Q_0, P(1), P(5))

>>> from sage.all import *
>>> SteenrodAlgebra(p=Integer(5), profile=[[Integer(2),Integer(1)], [Integer(2),Integer(2),Integer(2)]]).algebra_generators()
Family (Q_0, P(1), P(5))

an_element()[source]#

An element of this Steenrod algebra.

The element depends on the basis and whether there is a nontrivial profile function. (This is used by the automatic test suite, so having different elements in different bases may help in discovering bugs.)

EXAMPLES:

sage: SteenrodAlgebra().an_element()
Sq(2,1)
Sq^4 Sq^2 Sq^1
sage: SteenrodAlgebra(p=5).an_element()
4 Q_1 Q_3 P(2,1)
sage: SteenrodAlgebra(basis='pst').an_element()
P^3_1
sage: SteenrodAlgebra(basis='pst', profile=[3,2,1]).an_element()
P^0_1

>>> from sage.all import *
>>> SteenrodAlgebra().an_element()
Sq(2,1)
Sq^4 Sq^2 Sq^1
>>> SteenrodAlgebra(p=Integer(5)).an_element()
4 Q_1 Q_3 P(2,1)
>>> SteenrodAlgebra(basis='pst').an_element()
P^3_1
>>> SteenrodAlgebra(basis='pst', profile=[Integer(3),Integer(2),Integer(1)]).an_element()
P^0_1

antipode_on_basis(t)[source]#

The antipode of a basis element of this algebra

INPUT:

• t – tuple, the index of a basis element of self

OUTPUT:

the antipode of the corresponding basis element, as an element of self.

ALGORITHM: according to a result of Milnor’s, the antipode of $$\text{Sq}(n)$$ is the sum of all of the Milnor basis elements in dimension $$n$$. So: convert the element to the Serre-Cartan basis, thus writing it as a sum of products of elements $$\text{Sq}(n)$$, and use Milnor’s formula for the antipode of $$\text{Sq}(n)$$, together with the fact that the antipode is an antihomomorphism: if we call the antipode $$c$$, then $$c(ab) = c(b) c(a)$$.

At odd primes, a similar method is used: the antipode of $$P(n)$$ is the sum of the Milnor P basis elements in dimension $$n*2(p-1)$$, multiplied by $$(-1)^n$$, and the antipode of $$\beta = Q_0$$ is $$-Q_0$$. So convert to the Serre-Cartan basis, as in the $$p = 2$$ case. Note that in the odd prime case, there is a sign in the antihomomorphism formula: $$c(ab) = (-1)^{\deg a \deg b} c(b) c(a)$$.

EXAMPLES:

sage: A = SteenrodAlgebra()
sage: A.antipode_on_basis((4,))
Sq(1,1) + Sq(4)
sage: A.Sq(4).antipode()
Sq(1,1) + Sq(4)
Sq^3 Sq^1 + Sq^4
sage: SteenrodAlgebra(basis='pst').Sq(3).antipode()
P^0_1 P^1_1 + P^0_2
sage: a = SteenrodAlgebra(basis='wall_long').Sq(10)
sage: a.antipode()
Sq^1 Sq^2 Sq^4 Sq^1 Sq^2 + Sq^2 Sq^4 Sq^1 Sq^2 Sq^1 + Sq^8 Sq^2
sage: a.antipode().antipode() == a
True

sage: SteenrodAlgebra(p=3).P(6).antipode()
P(2,1) + P(6)
sage: SteenrodAlgebra(p=3).P(6).antipode().antipode()
P(6)

>>> from sage.all import *
>>> A = SteenrodAlgebra()
>>> A.antipode_on_basis((Integer(4),))
Sq(1,1) + Sq(4)
>>> A.Sq(Integer(4)).antipode()
Sq(1,1) + Sq(4)
Sq^3 Sq^1 + Sq^4
>>> SteenrodAlgebra(basis='pst').Sq(Integer(3)).antipode()
P^0_1 P^1_1 + P^0_2
>>> a = SteenrodAlgebra(basis='wall_long').Sq(Integer(10))
>>> a.antipode()
Sq^1 Sq^2 Sq^4 Sq^1 Sq^2 + Sq^2 Sq^4 Sq^1 Sq^2 Sq^1 + Sq^8 Sq^2
>>> a.antipode().antipode() == a
True

>>> SteenrodAlgebra(p=Integer(3)).P(Integer(6)).antipode()
P(2,1) + P(6)
>>> SteenrodAlgebra(p=Integer(3)).P(Integer(6)).antipode().antipode()
P(6)

basis(d=None)[source]#

Return basis for self, either the whole basis or the basis in degree $$d$$.

INPUT:

• $$d$$ – integer or None, optional (default None)

OUTPUT:

If $$d$$ is None, then return a basis of the algebra. Otherwise, return the basis in degree $$d$$.

EXAMPLES:

sage: A3 = SteenrodAlgebra(3)
sage: A3.basis(13)
Family (Q_1 P(2), Q_0 P(3))
Family (Sq^12, Sq^11 Sq^1, Sq^9 Sq^2 Sq^1, Sq^8 Sq^3 Sq^1, Sq^10 Sq^2, Sq^9 Sq^3, Sq^8 Sq^4)

sage: A = SteenrodAlgebra(profile=[1,2,1])
sage: A.basis(2)
Family ()
sage: A.basis(3)
Family (Sq(0,1),)
sage: SteenrodAlgebra().basis(3)
Family (Sq(0,1), Sq(3))
sage: A_pst = SteenrodAlgebra(profile=[1,2,1], basis='pst')
sage: A_pst.basis(3)
Family (P^0_2,)

sage: A7 = SteenrodAlgebra(p=7)
sage: B = SteenrodAlgebra(p=7, profile=([1,2,1], [1]))
sage: A7.basis(84)
Family (P(7),)
sage: B.basis(84)
Family ()
sage: C = SteenrodAlgebra(p=7, profile=([1], [2,2]))
sage: A7.Q(0,1) in C.basis(14)
True
sage: A7.Q(2) in A7.basis(97)
True
sage: A7.Q(2) in C.basis(97)
False

>>> from sage.all import *
>>> A3 = SteenrodAlgebra(Integer(3))
>>> A3.basis(Integer(13))
Family (Q_1 P(2), Q_0 P(3))
Family (Sq^12, Sq^11 Sq^1, Sq^9 Sq^2 Sq^1, Sq^8 Sq^3 Sq^1, Sq^10 Sq^2, Sq^9 Sq^3, Sq^8 Sq^4)

>>> A = SteenrodAlgebra(profile=[Integer(1),Integer(2),Integer(1)])
>>> A.basis(Integer(2))
Family ()
>>> A.basis(Integer(3))
Family (Sq(0,1),)
>>> SteenrodAlgebra().basis(Integer(3))
Family (Sq(0,1), Sq(3))
>>> A_pst = SteenrodAlgebra(profile=[Integer(1),Integer(2),Integer(1)], basis='pst')
>>> A_pst.basis(Integer(3))
Family (P^0_2,)

>>> A7 = SteenrodAlgebra(p=Integer(7))
>>> B = SteenrodAlgebra(p=Integer(7), profile=([Integer(1),Integer(2),Integer(1)], [Integer(1)]))
>>> A7.basis(Integer(84))
Family (P(7),)
>>> B.basis(Integer(84))
Family ()
>>> C = SteenrodAlgebra(p=Integer(7), profile=([Integer(1)], [Integer(2),Integer(2)]))
>>> A7.Q(Integer(0),Integer(1)) in C.basis(Integer(14))
True
>>> A7.Q(Integer(2)) in A7.basis(Integer(97))
True
>>> A7.Q(Integer(2)) in C.basis(Integer(97))
False


With no arguments, return the basis of the whole algebra. This does not print in a very helpful way, unfortunately:

sage: A7.basis()
Lazy family (Term map from basis key family of mod 7 Steenrod algebra, milnor basis
to mod 7 Steenrod algebra, milnor basis(i))_{i in basis key family
of mod 7 Steenrod algebra, milnor basis}
sage: for (idx,a) in zip((1,..,9),A7.basis()):
....:      print("{} {}".format(idx, a))
1 1
2 Q_0
3 P(1)
4 Q_1
5 Q_0 P(1)
6 Q_0 Q_1
7 P(2)
8 Q_1 P(1)
9 Q_0 P(2)
sage: D = SteenrodAlgebra(p=3, profile=([1], [2,2]))
sage: sorted(D.basis())
[1, P(1), P(2), Q_0, Q_0 P(1), Q_0 P(2), Q_0 Q_1,
Q_0 Q_1 P(1), Q_0 Q_1 P(2), Q_1, Q_1 P(1), Q_1 P(2)]

>>> from sage.all import *
>>> A7.basis()
Lazy family (Term map from basis key family of mod 7 Steenrod algebra, milnor basis
to mod 7 Steenrod algebra, milnor basis(i))_{i in basis key family
of mod 7 Steenrod algebra, milnor basis}
>>> for (idx,a) in zip((ellipsis_iter(Integer(1),Ellipsis,Integer(9))),A7.basis()):
...      print("{} {}".format(idx, a))
1 1
2 Q_0
3 P(1)
4 Q_1
5 Q_0 P(1)
6 Q_0 Q_1
7 P(2)
8 Q_1 P(1)
9 Q_0 P(2)
>>> D = SteenrodAlgebra(p=Integer(3), profile=([Integer(1)], [Integer(2),Integer(2)]))
>>> sorted(D.basis())
[1, P(1), P(2), Q_0, Q_0 P(1), Q_0 P(2), Q_0 Q_1,
Q_0 Q_1 P(1), Q_0 Q_1 P(2), Q_1, Q_1 P(1), Q_1 P(2)]

basis_name()[source]#

The basis name associated to self.

EXAMPLES:

sage: SteenrodAlgebra(p=2, profile=[1,1]).basis_name()
'milnor'
sage: SteenrodAlgebra(basis='serre-cartan').basis_name()
'serre-cartan'
'serre-cartan'

>>> from sage.all import *
>>> SteenrodAlgebra(p=Integer(2), profile=[Integer(1),Integer(1)]).basis_name()
'milnor'
>>> SteenrodAlgebra(basis='serre-cartan').basis_name()
'serre-cartan'
'serre-cartan'

coproduct(x, algorithm='milnor')[source]#

Return the coproduct of an element x of this algebra.

INPUT:

• x – element of self

• algorithmNone or a string, either 'milnor' or 'serre-cartan' (or anything which will be converted to one of these by the function get_basis_name. If None, default to 'serre-cartan' if current basis is 'serre-cartan'; otherwise use 'milnor'.

This calls coproduct_on_basis() on the summands of x and extends linearly.

EXAMPLES:

sage: SteenrodAlgebra().Sq(3).coproduct()
1 # Sq(3) + Sq(1) # Sq(2) + Sq(2) # Sq(1) + Sq(3) # 1

>>> from sage.all import *
>>> SteenrodAlgebra().Sq(Integer(3)).coproduct()
1 # Sq(3) + Sq(1) # Sq(2) + Sq(2) # Sq(1) + Sq(3) # 1


The element $$\text{Sq}(0,1)$$ is primitive:

sage: SteenrodAlgebra(basis='adem').Sq(0,1).coproduct()
1 # Sq^2 Sq^1 + 1 # Sq^3 + Sq^2 Sq^1 # 1 + Sq^3 # 1
sage: SteenrodAlgebra(basis='pst').Sq(0,1).coproduct()
1 # P^0_2 + P^0_2 # 1

sage: SteenrodAlgebra(p=3).P(4).coproduct()
1 # P(4) + P(1) # P(3) + P(2) # P(2) + P(3) # P(1) + P(4) # 1
sage: SteenrodAlgebra(p=3).P(4).coproduct(algorithm='serre-cartan')
1 # P(4) + P(1) # P(3) + P(2) # P(2) + P(3) # P(1) + P(4) # 1
sage: SteenrodAlgebra(p=3, basis='serre-cartan').P(4).coproduct()
1 # P^4 + P^1 # P^3 + P^2 # P^2 + P^3 # P^1 + P^4 # 1
sage: SteenrodAlgebra(p=11, profile=((), (2,1,2))).Q(0,2).coproduct()
1 # Q_0 Q_2 + Q_0 # Q_2 + Q_0 Q_2 # 1 + 10*Q_2 # Q_0

>>> from sage.all import *
1 # Sq^2 Sq^1 + 1 # Sq^3 + Sq^2 Sq^1 # 1 + Sq^3 # 1
>>> SteenrodAlgebra(basis='pst').Sq(Integer(0),Integer(1)).coproduct()
1 # P^0_2 + P^0_2 # 1

>>> SteenrodAlgebra(p=Integer(3)).P(Integer(4)).coproduct()
1 # P(4) + P(1) # P(3) + P(2) # P(2) + P(3) # P(1) + P(4) # 1
>>> SteenrodAlgebra(p=Integer(3)).P(Integer(4)).coproduct(algorithm='serre-cartan')
1 # P(4) + P(1) # P(3) + P(2) # P(2) + P(3) # P(1) + P(4) # 1
>>> SteenrodAlgebra(p=Integer(3), basis='serre-cartan').P(Integer(4)).coproduct()
1 # P^4 + P^1 # P^3 + P^2 # P^2 + P^3 # P^1 + P^4 # 1
>>> SteenrodAlgebra(p=Integer(11), profile=((), (Integer(2),Integer(1),Integer(2)))).Q(Integer(0),Integer(2)).coproduct()
1 # Q_0 Q_2 + Q_0 # Q_2 + Q_0 Q_2 # 1 + 10*Q_2 # Q_0

coproduct_on_basis(t, algorithm=None)[source]#

The coproduct of a basis element of this algebra

INPUT:

• t – tuple, the index of a basis element of self

• algorithmNone or a string, either ‘milnor’ or ‘serre-cartan’ (or anything which will be converted to one of these by the function get_basis_name. If None, default to ‘milnor’ unless current basis is ‘serre-cartan’, in which case use ‘serre-cartan’.

ALGORITHM: The coproduct on a Milnor basis element $$P(n_1, n_2, ...)$$ is $$\sum P(i_1, i_2, ...) \otimes P(j_1, j_2, ...)$$, summed over all $$i_k + j_k = n_k$$ for each $$k$$. At odd primes, each element $$Q_n$$ is primitive: its coproduct is $$Q_n \otimes 1 + 1 \otimes Q_n$$.

One can deduce a coproduct formula for the Serre-Cartan basis from this: the coproduct on each $$P^n$$ is $$\sum P^i \otimes P^{n-i}$$ and at odd primes $$\beta$$ is primitive. Since the coproduct is an algebra map, one can then compute the coproduct on any Serre-Cartan basis element.

Which of these methods is used is controlled by whether algorithm is ‘milnor’ or ‘serre-cartan’.

OUTPUT:

the coproduct of the corresponding basis element, as an element of self tensor self.

EXAMPLES:

sage: A = SteenrodAlgebra()
sage: A.coproduct_on_basis((3,))
1 # Sq(3) + Sq(1) # Sq(2) + Sq(2) # Sq(1) + Sq(3) # 1

>>> from sage.all import *
>>> A = SteenrodAlgebra()
>>> A.coproduct_on_basis((Integer(3),))
1 # Sq(3) + Sq(1) # Sq(2) + Sq(2) # Sq(1) + Sq(3) # 1

counit_on_basis(t)[source]#

The counit sends all elements of positive degree to zero.

INPUT:

• t – tuple, the index of a basis element of self

EXAMPLES:

sage: A2 = SteenrodAlgebra(p=2)
sage: A2.counit_on_basis(())
1
sage: A2.counit_on_basis((0,0,1))
0
sage: parent(A2.counit_on_basis((0,0,1)))
Finite Field of size 2
sage: A3 = SteenrodAlgebra(p=3)
sage: A3.counit_on_basis(((1,2,3), (1,1,1)))
0
sage: A3.counit_on_basis(((), ()))
1
sage: A3.counit(A3.P(10,5))
0
sage: A3.counit(A3.P(0))
1

>>> from sage.all import *
>>> A2 = SteenrodAlgebra(p=Integer(2))
>>> A2.counit_on_basis(())
1
>>> A2.counit_on_basis((Integer(0),Integer(0),Integer(1)))
0
>>> parent(A2.counit_on_basis((Integer(0),Integer(0),Integer(1))))
Finite Field of size 2
>>> A3 = SteenrodAlgebra(p=Integer(3))
>>> A3.counit_on_basis(((Integer(1),Integer(2),Integer(3)), (Integer(1),Integer(1),Integer(1))))
0
>>> A3.counit_on_basis(((), ()))
1
>>> A3.counit(A3.P(Integer(10),Integer(5)))
0
>>> A3.counit(A3.P(Integer(0)))
1

degree_on_basis(t)[source]#

The degree of the monomial specified by the tuple t.

INPUT:

• t – tuple, representing basis element in the current basis.

OUTPUT: integer, the degree of the corresponding element.

The degree of $$\text{Sq}(i_1,i_2,i_3,...)$$ is

$i_1 + 3i_2 + 7i_3 + ... + (2^k - 1) i_k + ....$

At an odd prime $$p$$, the degree of $$Q_k$$ is $$2p^k - 1$$ and the degree of $$\mathcal{P}(i_1, i_2, ...)$$ is

$\sum_{k \geq 0} 2(p^k - 1) i_k.$

ALGORITHM: Each basis element is represented in terms relevant to the particular basis: ‘milnor’ basis elements (at the prime 2) are given by tuples (a,b,c,...) corresponding to the element $$\text{Sq}(a,b,c,...)$$, while ‘pst’ basis elements are given by tuples of pairs ((a, b), (c, d), ...), corresponding to the product $$P^a_b P^c_d ...$$. The other bases have similar descriptions. The degree of each basis element is computed from this data, rather than converting the element to the Milnor basis, for example, and then computing the degree.

EXAMPLES:

sage: SteenrodAlgebra().degree_on_basis((0,0,1))
7
sage: Sq(7).degree()
7

sage: A11 = SteenrodAlgebra(p=11)
sage: A11.degree_on_basis(((), (1,1)))
260
sage: A11.degree_on_basis(((2,), ()))
241

>>> from sage.all import *
>>> SteenrodAlgebra().degree_on_basis((Integer(0),Integer(0),Integer(1)))
7
>>> Sq(Integer(7)).degree()
7

>>> A11 = SteenrodAlgebra(p=Integer(11))
>>> A11.degree_on_basis(((), (Integer(1),Integer(1))))
260
>>> A11.degree_on_basis(((Integer(2),), ()))
241

dimension()[source]#

The dimension of this algebra as a vector space over $$\GF{p}$$.

If the algebra is infinite, return +Infinity. Otherwise, the profile function must be finite. In this case, at the prime 2, its dimension is $$2^s$$, where $$s$$ is the sum of the entries in the profile function. At odd primes, the dimension is $$p^s * 2^t$$ where $$s$$ is the sum of the $$e$$ component of the profile function and $$t$$ is the number of 2’s in the $$k$$ component of the profile function.

EXAMPLES:

sage: SteenrodAlgebra(p=7).dimension()
+Infinity
sage: SteenrodAlgebra(profile=[3,2,1]).dimension()
64
sage: SteenrodAlgebra(p=3, profile=([1,1], [])).dimension()
9
sage: SteenrodAlgebra(p=5, profile=([1], [2,2])).dimension()
20

>>> from sage.all import *
>>> SteenrodAlgebra(p=Integer(7)).dimension()
+Infinity
>>> SteenrodAlgebra(profile=[Integer(3),Integer(2),Integer(1)]).dimension()
64
>>> SteenrodAlgebra(p=Integer(3), profile=([Integer(1),Integer(1)], [])).dimension()
9
>>> SteenrodAlgebra(p=Integer(5), profile=([Integer(1)], [Integer(2),Integer(2)])).dimension()
20

gen(i=0)[source]#

The ith generator of this algebra.

INPUT:

• i – non-negative integer

OUTPUT: the ith generator of this algebra

For the full Steenrod algebra, the $$i^{th}$$ generator is $$\text{Sq}(2^i)$$ at the prime 2; when $$p$$ is odd, the 0th generator is $$\beta = Q(0)$$, and for $$i>0$$, the $$i^{th}$$ generator is $$P(p^{i-1})$$.

For sub-Hopf algebras of the Steenrod algebra, it is not always clear what a minimal generating set is. The sub-Hopf algebra $$A(n)$$ is minimally generated by the elements $$\text{Sq}^{2^i}$$ for $$0 \leq i \leq n$$ at the prime 2. At odd primes, $$A(n)$$ is minimally generated by $$Q_0$$ along with $$\mathcal{P}^{p^i}$$ for $$0 \leq i \leq n-1$$. So if this algebra is $$A(n)$$, return the appropriate generator.

For other sub-Hopf algebras: they are generated (but not necessarily minimally) by the $$P^s_t$$’s (and $$Q_n$$’s, if $$p$$ is odd) that they contain. So order the $$P^s_t$$’s (and $$Q_n$$’s) in the algebra by degree and return the $$i$$-th one.

EXAMPLES:

sage: A = SteenrodAlgebra(2)
sage: A.gen(4)
Sq(16)
sage: A.gen(200)
Sq(1606938044258990275541962092341162602522202993782792835301376)
Sq^4
sage: SteenrodAlgebra(2, basis='pst').gen(2)
P^2_1
sage: B = SteenrodAlgebra(5)
sage: B.gen(0)
Q_0
sage: B.gen(2)
P(5)

sage: SteenrodAlgebra(profile=[2,1]).gen(1)
Sq(2)
sage: SteenrodAlgebra(profile=[1,2,1]).gen(1)
Sq(0,1)
sage: SteenrodAlgebra(profile=[1,2,1]).gen(5)
Traceback (most recent call last):
...
ValueError: This algebra only has 4 generators, so call gen(i) with 0 <= i < 4

sage: D = SteenrodAlgebra(profile=lambda n: n)
sage: [D.gen(n) for n in range(5)]
[Sq(1), Sq(0,1), Sq(0,2), Sq(0,0,1), Sq(0,0,2)]
sage: D3 = SteenrodAlgebra(p=3, profile=(lambda n: n, lambda n: 2))
sage: [D3.gen(n) for n in range(9)]
[Q_0, P(1), Q_1, P(0,1), Q_2, P(0,3), P(0,0,1), Q_3, P(0,0,3)]
sage: D3 = SteenrodAlgebra(p=3, profile=(lambda n: n, lambda n: 1 if n<1 else 2))
sage: [D3.gen(n) for n in range(9)]
[P(1), Q_1, P(0,1), Q_2, P(0,3), P(0,0,1), Q_3, P(0,0,3), P(0,0,0,1)]
sage: SteenrodAlgebra(p=5, profile=[[2,1], [2,2,2]], basis='pst').gen(2)
P^1_1

>>> from sage.all import *
>>> A = SteenrodAlgebra(Integer(2))
>>> A.gen(Integer(4))
Sq(16)
>>> A.gen(Integer(200))
Sq(1606938044258990275541962092341162602522202993782792835301376)
Sq^4
>>> SteenrodAlgebra(Integer(2), basis='pst').gen(Integer(2))
P^2_1
>>> B = SteenrodAlgebra(Integer(5))
>>> B.gen(Integer(0))
Q_0
>>> B.gen(Integer(2))
P(5)

>>> SteenrodAlgebra(profile=[Integer(2),Integer(1)]).gen(Integer(1))
Sq(2)
>>> SteenrodAlgebra(profile=[Integer(1),Integer(2),Integer(1)]).gen(Integer(1))
Sq(0,1)
>>> SteenrodAlgebra(profile=[Integer(1),Integer(2),Integer(1)]).gen(Integer(5))
Traceback (most recent call last):
...
ValueError: This algebra only has 4 generators, so call gen(i) with 0 <= i < 4

>>> D = SteenrodAlgebra(profile=lambda n: n)
>>> [D.gen(n) for n in range(Integer(5))]
[Sq(1), Sq(0,1), Sq(0,2), Sq(0,0,1), Sq(0,0,2)]
>>> D3 = SteenrodAlgebra(p=Integer(3), profile=(lambda n: n, lambda n: Integer(2)))
>>> [D3.gen(n) for n in range(Integer(9))]
[Q_0, P(1), Q_1, P(0,1), Q_2, P(0,3), P(0,0,1), Q_3, P(0,0,3)]
>>> D3 = SteenrodAlgebra(p=Integer(3), profile=(lambda n: n, lambda n: Integer(1) if n<Integer(1) else Integer(2)))
>>> [D3.gen(n) for n in range(Integer(9))]
[P(1), Q_1, P(0,1), Q_2, P(0,3), P(0,0,1), Q_3, P(0,0,3), P(0,0,0,1)]
>>> SteenrodAlgebra(p=Integer(5), profile=[[Integer(2),Integer(1)], [Integer(2),Integer(2),Integer(2)]], basis='pst').gen(Integer(2))
P^1_1

gens()[source]#

Family of generators for this algebra.

OUTPUT: family of elements of this algebra

At the prime 2, the Steenrod algebra is generated by the elements $$\text{Sq}^{2^i}$$ for $$i \geq 0$$. At odd primes, it is generated by the elements $$Q_0$$ and $$\mathcal{P}^{p^i}$$ for $$i \geq 0$$. So if this algebra is the entire Steenrod algebra, return an infinite family made up of these elements.

For sub-Hopf algebras of the Steenrod algebra, it is not always clear what a minimal generating set is. The sub-Hopf algebra $$A(n)$$ is minimally generated by the elements $$\text{Sq}^{2^i}$$ for $$0 \leq i \leq n$$ at the prime 2. At odd primes, $$A(n)$$ is minimally generated by $$Q_0$$ along with $$\mathcal{P}^{p^i}$$ for $$0 \leq i \leq n-1$$. So if this algebra is $$A(n)$$, return the appropriate list of generators.

For other sub-Hopf algebras: return a non-minimal generating set: the family of $$P^s_t$$’s and $$Q_n$$’s contained in the algebra.

EXAMPLES:

sage: A3 = SteenrodAlgebra(3, 'adem')
sage: A3.gens()
Lazy family (<bound method SteenrodAlgebra_generic.gen of mod 3 Steenrod algebra,
serre-cartan basis>(i))_{i in Non negative integers}
sage: A3.gens()[0]
beta
sage: A3.gens()[1]
P^1
sage: A3.gens()[2]
P^3
sage: SteenrodAlgebra(profile=[3,2,1]).gens()
Family (Sq(1), Sq(2), Sq(4))

>>> from sage.all import *
>>> A3.gens()
Lazy family (<bound method SteenrodAlgebra_generic.gen of mod 3 Steenrod algebra,
serre-cartan basis>(i))_{i in Non negative integers}
>>> A3.gens()[Integer(0)]
beta
>>> A3.gens()[Integer(1)]
P^1
>>> A3.gens()[Integer(2)]
P^3
>>> SteenrodAlgebra(profile=[Integer(3),Integer(2),Integer(1)]).gens()
Family (Sq(1), Sq(2), Sq(4))


In the following case, return a non-minimal generating set. (It is not minimal because $$\text{Sq}(0,0,1)$$ is the commutator of $$\text{Sq}(1)$$ and $$\text{Sq}(0,2)$$.)

sage: SteenrodAlgebra(profile=[1,2,1]).gens()
Family (Sq(1), Sq(0,1), Sq(0,2), Sq(0,0,1))
sage: SteenrodAlgebra(p=5, profile=[[2,1], [2,2,2]]).gens()
Family (Q_0, P(1), P(5))
sage: SteenrodAlgebra(profile=lambda n: n).gens()
Lazy family (<bound method SteenrodAlgebra_generic.gen of sub-Hopf algebra
of mod 2 Steenrod algebra, milnor basis, profile function [1, 2, 3, ...,
98, 99, +Infinity, +Infinity, +Infinity, ...]>(i))_{i in Non negative integers}

>>> from sage.all import *
>>> SteenrodAlgebra(profile=[Integer(1),Integer(2),Integer(1)]).gens()
Family (Sq(1), Sq(0,1), Sq(0,2), Sq(0,0,1))
>>> SteenrodAlgebra(p=Integer(5), profile=[[Integer(2),Integer(1)], [Integer(2),Integer(2),Integer(2)]]).gens()
Family (Q_0, P(1), P(5))
>>> SteenrodAlgebra(profile=lambda n: n).gens()
Lazy family (<bound method SteenrodAlgebra_generic.gen of sub-Hopf algebra
of mod 2 Steenrod algebra, milnor basis, profile function [1, 2, 3, ...,
98, 99, +Infinity, +Infinity, +Infinity, ...]>(i))_{i in Non negative integers}


You may also use algebra_generators instead of gens:

sage: SteenrodAlgebra(p=5, profile=[[2,1], [2,2,2]]).algebra_generators()
Family (Q_0, P(1), P(5))

>>> from sage.all import *
>>> SteenrodAlgebra(p=Integer(5), profile=[[Integer(2),Integer(1)], [Integer(2),Integer(2),Integer(2)]]).algebra_generators()
Family (Q_0, P(1), P(5))

homogeneous_component(n)[source]#

Return the nth homogeneous piece of the Steenrod algebra.

INPUT:

• $$n$$ – integer

OUTPUT:

a vector space spanned by the basis for this algebra in dimension $$n$$

EXAMPLES:

sage: A = SteenrodAlgebra()
sage: A.homogeneous_component(4)
Vector space spanned by (Sq(1,1), Sq(4)) over Finite Field of size 2
sage: SteenrodAlgebra(profile=[2,1,0]).homogeneous_component(4)
Vector space spanned by (Sq(1,1),) over Finite Field of size 2

>>> from sage.all import *
>>> A = SteenrodAlgebra()
>>> A.homogeneous_component(Integer(4))
Vector space spanned by (Sq(1,1), Sq(4)) over Finite Field of size 2
>>> SteenrodAlgebra(profile=[Integer(2),Integer(1),Integer(0)]).homogeneous_component(Integer(4))
Vector space spanned by (Sq(1,1),) over Finite Field of size 2


The notation A[n] may also be used:

sage: A[5]
Vector space spanned by (Sq(2,1), Sq(5)) over Finite Field of size 2
sage: SteenrodAlgebra(basis='wall')[4]
Vector space spanned by (Q^1_0 Q^0_0, Q^2_2) over Finite Field of size 2
sage: SteenrodAlgebra(p=5)[17]
Vector space spanned by (Q_1 P(1), Q_0 P(2)) over Finite Field of size 5

>>> from sage.all import *
>>> A[Integer(5)]
Vector space spanned by (Sq(2,1), Sq(5)) over Finite Field of size 2
>>> SteenrodAlgebra(basis='wall')[Integer(4)]
Vector space spanned by (Q^1_0 Q^0_0, Q^2_2) over Finite Field of size 2
>>> SteenrodAlgebra(p=Integer(5))[Integer(17)]
Vector space spanned by (Q_1 P(1), Q_0 P(2)) over Finite Field of size 5


Note that A[n] is just a vector space, not a Hopf algebra, so its elements don’t have products, coproducts, or antipodes defined on them. If you want to use operations like this on elements of some A[n], then convert them back to elements of A:

sage: sorted(A[5].basis())
[milnor[(2, 1)], milnor[(5,)]]
sage: a = list(A[5].basis())[1]
sage: a  # not in A, doesn't print like an element of A
milnor[(5,)]
sage: A(a) # in A
Sq(5)
sage: A(a) * A(a)
Sq(7,1)
sage: a * A(a) # only need to convert one factor
Sq(7,1)
sage: a.antipode() # not defined
Traceback (most recent call last):
...
AttributeError: 'CombinatorialFreeModule_with_category.element_class' object has no attribute 'antipode'...
sage: A(a).antipode() # convert to elt of A, then compute antipode
Sq(2,1) + Sq(5)

sage: G = SteenrodAlgebra(p=5, profile=[[2,1], [2,2,2]], basis='pst')

>>> from sage.all import *
>>> sorted(A[Integer(5)].basis())
[milnor[(2, 1)], milnor[(5,)]]
>>> a = list(A[Integer(5)].basis())[Integer(1)]
>>> a  # not in A, doesn't print like an element of A
milnor[(5,)]
>>> A(a) # in A
Sq(5)
>>> A(a) * A(a)
Sq(7,1)
>>> a * A(a) # only need to convert one factor
Sq(7,1)
>>> a.antipode() # not defined
Traceback (most recent call last):
...
AttributeError: 'CombinatorialFreeModule_with_category.element_class' object has no attribute 'antipode'...
>>> A(a).antipode() # convert to elt of A, then compute antipode
Sq(2,1) + Sq(5)

>>> G = SteenrodAlgebra(p=Integer(5), profile=[[Integer(2),Integer(1)], [Integer(2),Integer(2),Integer(2)]], basis='pst')

is_commutative()[source]#

True if self is graded commutative, as determined by the profile function. In particular, a sub-Hopf algebra of the mod 2 Steenrod algebra is commutative if and only if there is an integer $$n>0$$ so that its profile function $$e$$ satisfies

• $$e(i) = 0$$ for $$i < n$$,

• $$e(i) \leq n$$ for $$i \geq n$$.

When $$p$$ is odd, there must be an integer $$n \geq 0$$ so that the profile functions $$e$$ and $$k$$ satisfy

• $$e(i) = 0$$ for $$i < n$$,

• $$e(i) \leq n$$ for $$i \geq n$$.

• $$k(i) = 1$$ for $$i < n$$.

EXAMPLES:

sage: A = SteenrodAlgebra(p=3)
sage: A.is_commutative()
False
sage: SteenrodAlgebra(profile=[2,1]).is_commutative()
False
sage: SteenrodAlgebra(profile=[0,2,2,1]).is_commutative()
True

>>> from sage.all import *
>>> A = SteenrodAlgebra(p=Integer(3))
>>> A.is_commutative()
False
>>> SteenrodAlgebra(profile=[Integer(2),Integer(1)]).is_commutative()
False
>>> SteenrodAlgebra(profile=[Integer(0),Integer(2),Integer(2),Integer(1)]).is_commutative()
True


Note that if the profile function is specified by a function, then by default it has infinite truncation type: the profile function is assumed to be infinite after the 100th term.

sage: SteenrodAlgebra(profile=lambda n: 1).is_commutative()
False
sage: SteenrodAlgebra(profile=lambda n: 1, truncation_type=0).is_commutative()
True

sage: SteenrodAlgebra(p=5, profile=([0,2,2,1], [])).is_commutative()
True
sage: SteenrodAlgebra(p=5, profile=([0,2,2,1], [1,1,2])).is_commutative()
True
sage: SteenrodAlgebra(p=5, profile=([0,2,1], [1,2,2,2])).is_commutative()
False

>>> from sage.all import *
>>> SteenrodAlgebra(profile=lambda n: Integer(1)).is_commutative()
False
>>> SteenrodAlgebra(profile=lambda n: Integer(1), truncation_type=Integer(0)).is_commutative()
True

>>> SteenrodAlgebra(p=Integer(5), profile=([Integer(0),Integer(2),Integer(2),Integer(1)], [])).is_commutative()
True
>>> SteenrodAlgebra(p=Integer(5), profile=([Integer(0),Integer(2),Integer(2),Integer(1)], [Integer(1),Integer(1),Integer(2)])).is_commutative()
True
>>> SteenrodAlgebra(p=Integer(5), profile=([Integer(0),Integer(2),Integer(1)], [Integer(1),Integer(2),Integer(2),Integer(2)])).is_commutative()
False

is_division_algebra()[source]#

The only way this algebra can be a division algebra is if it is the ground field $$\GF{p}$$.

EXAMPLES:

sage: SteenrodAlgebra(11).is_division_algebra()
False
sage: SteenrodAlgebra(profile=lambda n: 0, truncation_type=0).is_division_algebra()
True

>>> from sage.all import *
>>> SteenrodAlgebra(Integer(11)).is_division_algebra()
False
>>> SteenrodAlgebra(profile=lambda n: Integer(0), truncation_type=Integer(0)).is_division_algebra()
True

is_field(proof=True)[source]#

The only way this algebra can be a field is if it is the ground field $$\GF{p}$$.

EXAMPLES:

sage: SteenrodAlgebra(11).is_field()
False
sage: SteenrodAlgebra(profile=lambda n: 0, truncation_type=0).is_field()
True

>>> from sage.all import *
>>> SteenrodAlgebra(Integer(11)).is_field()
False
>>> SteenrodAlgebra(profile=lambda n: Integer(0), truncation_type=Integer(0)).is_field()
True

is_finite()[source]#

True if this algebra is finite-dimensional.

Therefore true if the profile function is finite, and in particular the truncation_type must be finite.

EXAMPLES:

sage: A = SteenrodAlgebra(p=3)
sage: A.is_finite()
False
sage: SteenrodAlgebra(profile=[3,2,1]).is_finite()
True
sage: SteenrodAlgebra(profile=lambda n: n).is_finite()
False

>>> from sage.all import *
>>> A = SteenrodAlgebra(p=Integer(3))
>>> A.is_finite()
False
>>> SteenrodAlgebra(profile=[Integer(3),Integer(2),Integer(1)]).is_finite()
True
>>> SteenrodAlgebra(profile=lambda n: n).is_finite()
False

is_generic()[source]#

The algebra is generic if it is based on the odd-primary relations, i.e. if its dual is a quotient of

$A_* = \GF{p} [\xi_1, \xi_2, \xi_3, ...] \otimes \Lambda (\tau_0, \tau_1, ...)$

Sage also allows this for $$p=2$$. Only the usual Steenrod algebra at the prime $$2$$ and its sub algebras are non-generic.

EXAMPLES:

sage: SteenrodAlgebra(3).is_generic()
True
sage: SteenrodAlgebra(2).is_generic()
False
sage: SteenrodAlgebra(2, generic=True).is_generic()
True

>>> from sage.all import *
>>> SteenrodAlgebra(Integer(3)).is_generic()
True
>>> SteenrodAlgebra(Integer(2)).is_generic()
False
>>> SteenrodAlgebra(Integer(2), generic=True).is_generic()
True

is_integral_domain(proof=True)[source]#

The only way this algebra can be an integral domain is if it is the ground field $$\GF{p}$$.

EXAMPLES:

sage: SteenrodAlgebra(11).is_integral_domain()
False
sage: SteenrodAlgebra(profile=lambda n: 0, truncation_type=0).is_integral_domain()
True

>>> from sage.all import *
>>> SteenrodAlgebra(Integer(11)).is_integral_domain()
False
>>> SteenrodAlgebra(profile=lambda n: Integer(0), truncation_type=Integer(0)).is_integral_domain()
True

is_noetherian()[source]#

This algebra is Noetherian if and only if it is finite.

EXAMPLES:

sage: SteenrodAlgebra(3).is_noetherian()
False
sage: SteenrodAlgebra(profile=[1,2,1]).is_noetherian()
True
sage: SteenrodAlgebra(profile=lambda n: n+2).is_noetherian()
False

>>> from sage.all import *
>>> SteenrodAlgebra(Integer(3)).is_noetherian()
False
>>> SteenrodAlgebra(profile=[Integer(1),Integer(2),Integer(1)]).is_noetherian()
True
>>> SteenrodAlgebra(profile=lambda n: n+Integer(2)).is_noetherian()
False

milnor()[source]#

Convert an element of this algebra to the Milnor basis

INPUT:

• x – an element of this algebra

OUTPUT: x converted to the Milnor basis

ALGORITHM: use the method _milnor_on_basis and linearity.

EXAMPLES:

sage: Adem = SteenrodAlgebra(basis='adem')
Sq(0,1) + Sq(3)

>>> from sage.all import *
Sq(0,1) + Sq(3)

ngens()[source]#

Number of generators of self.

OUTPUT: number or Infinity

The Steenrod algebra is infinitely generated. A sub-Hopf algebra may be finitely or infinitely generated; in general, it is not clear what a minimal generating set is, nor the cardinality of that set. So: if the algebra is infinite-dimensional, this returns Infinity. If the algebra is finite-dimensional and is equal to one of the sub-Hopf algebras $$A(n)$$, then their minimal generating set is known, and this returns the cardinality of that set. Otherwise, any sub-Hopf algebra is (not necessarily minimally) generated by the $$P^s_t$$’s that it contains (along with the $$Q_n$$’s it contains, at odd primes), so this returns the number of $$P^s_t$$’s and $$Q_n$$’s in the algebra.

EXAMPLES:

sage: A = SteenrodAlgebra(3)
sage: A.ngens()
+Infinity
sage: SteenrodAlgebra(profile=lambda n: n).ngens()
+Infinity
sage: SteenrodAlgebra(profile=[3,2,1]).ngens() # A(2)
3
sage: SteenrodAlgebra(profile=[3,2,1], basis='pst').ngens()
3
sage: SteenrodAlgebra(p=3, profile=[[3,2,1], [2,2,2,2]]).ngens()  # A(3) at p=3
4
sage: SteenrodAlgebra(profile=[1,2,1,1]).ngens()
5

>>> from sage.all import *
>>> A = SteenrodAlgebra(Integer(3))
>>> A.ngens()
+Infinity
>>> SteenrodAlgebra(profile=lambda n: n).ngens()
+Infinity
>>> SteenrodAlgebra(profile=[Integer(3),Integer(2),Integer(1)]).ngens() # A(2)
3
>>> SteenrodAlgebra(profile=[Integer(3),Integer(2),Integer(1)], basis='pst').ngens()
3
>>> SteenrodAlgebra(p=Integer(3), profile=[[Integer(3),Integer(2),Integer(1)], [Integer(2),Integer(2),Integer(2),Integer(2)]]).ngens()  # A(3) at p=3
4
>>> SteenrodAlgebra(profile=[Integer(1),Integer(2),Integer(1),Integer(1)]).ngens()
5

one_basis()[source]#

The index of the element 1 in the basis for the Steenrod algebra.

EXAMPLES:

sage: SteenrodAlgebra(p=2).one_basis()
()
sage: SteenrodAlgebra(p=7).one_basis()
((), ())

>>> from sage.all import *
>>> SteenrodAlgebra(p=Integer(2)).one_basis()
()
>>> SteenrodAlgebra(p=Integer(7)).one_basis()
((), ())

order()[source]#

The order of this algebra.

This is computed by computing its vector space dimension $$d$$ and then returning $$p^d$$.

EXAMPLES:

sage: SteenrodAlgebra(p=7).order()
+Infinity
sage: SteenrodAlgebra(profile=[2,1]).dimension()
8
sage: SteenrodAlgebra(profile=[2,1]).order()
256
sage: SteenrodAlgebra(p=3, profile=([1], [])).dimension()
3
sage: SteenrodAlgebra(p=3, profile=([1], [])).order()
27
sage: SteenrodAlgebra(p=5, profile=([], [2, 2])).dimension()
4
sage: SteenrodAlgebra(p=5, profile=([], [2, 2])).order() == 5**4
True

>>> from sage.all import *
>>> SteenrodAlgebra(p=Integer(7)).order()
+Infinity
>>> SteenrodAlgebra(profile=[Integer(2),Integer(1)]).dimension()
8
>>> SteenrodAlgebra(profile=[Integer(2),Integer(1)]).order()
256
>>> SteenrodAlgebra(p=Integer(3), profile=([Integer(1)], [])).dimension()
3
>>> SteenrodAlgebra(p=Integer(3), profile=([Integer(1)], [])).order()
27
>>> SteenrodAlgebra(p=Integer(5), profile=([], [Integer(2), Integer(2)])).dimension()
4
>>> SteenrodAlgebra(p=Integer(5), profile=([], [Integer(2), Integer(2)])).order() == Integer(5)**Integer(4)
True

prime()[source]#

The prime associated to self.

EXAMPLES:

sage: SteenrodAlgebra(p=2, profile=[1,1]).prime()
2
sage: SteenrodAlgebra(p=7).prime()
7

>>> from sage.all import *
>>> SteenrodAlgebra(p=Integer(2), profile=[Integer(1),Integer(1)]).prime()
2
>>> SteenrodAlgebra(p=Integer(7)).prime()
7

product_on_basis(t1, t2)[source]#

The product of two basis elements of this algebra

INPUT:

• t1, t2 – tuples, the indices of two basis elements of self

OUTPUT:

the product of the two corresponding basis elements, as an element of self

ALGORITHM: If the two elements are represented in the Milnor basis, use Milnor multiplication as implemented in sage.algebras.steenrod.steenrod_algebra_mult. If the two elements are represented in the Serre-Cartan basis, then multiply them using Adem relations (also implemented in sage.algebras.steenrod.steenrod_algebra_mult). This provides a good way of checking work – multiply Milnor elements, then convert them to Adem elements and multiply those, and see if the answers correspond.

If the two elements are represented in some other basis, then convert them both to the Milnor basis and multiply.

EXAMPLES:

sage: Milnor = SteenrodAlgebra()
sage: Milnor.product_on_basis((2,), (2,))
Sq(1,1)
Sq^3 Sq^1

>>> from sage.all import *
>>> Milnor = SteenrodAlgebra()
>>> Milnor.product_on_basis((Integer(2),), (Integer(2),))
Sq(1,1)
Sq^3 Sq^1


When multiplying elements from different bases, the left-hand factor determines the form of the output:

sage: Adem.Sq(2) * Milnor.Sq(2)
Sq^3 Sq^1
Sq(1,1)

>>> from sage.all import *
Sq^3 Sq^1
Sq(1,1)

profile(i, component=0)[source]#

Profile function for this algebra.

INPUT:

• $$i$$ – integer

• component – either 0 or 1, optional (default 0)

OUTPUT: integer or $$\infty$$

See the documentation for sage.algebras.steenrod.steenrod_algebra and SteenrodAlgebra() for information on profile functions.

This applies the profile function to the integer $$i$$. Thus when $$p=2$$, $$i$$ must be a positive integer. When $$p$$ is odd, there are two profile functions, $$e$$ and $$k$$ (in the notation of the aforementioned documentation), corresponding, respectively to component=0 and component=1. So when $$p$$ is odd and component is 0, $$i$$ must be positive, while when component is 1, $$i$$ must be non-negative.

EXAMPLES:

sage: SteenrodAlgebra().profile(3)
+Infinity
sage: SteenrodAlgebra(profile=[3,2,1]).profile(1)
3
sage: SteenrodAlgebra(profile=[3,2,1]).profile(2)
2

>>> from sage.all import *
>>> SteenrodAlgebra().profile(Integer(3))
+Infinity
>>> SteenrodAlgebra(profile=[Integer(3),Integer(2),Integer(1)]).profile(Integer(1))
3
>>> SteenrodAlgebra(profile=[Integer(3),Integer(2),Integer(1)]).profile(Integer(2))
2


When the profile is specified by a list, the default behavior is to return zero values outside the range of the list. This can be overridden if the algebra is created with an infinite truncation_type:

sage: SteenrodAlgebra(profile=[3,2,1]).profile(9)
0
sage: SteenrodAlgebra(profile=[3,2,1], truncation_type=Infinity).profile(9)
+Infinity

sage: B = SteenrodAlgebra(p=3, profile=(lambda n: n, lambda n: 1))
sage: B.profile(3)
3
sage: B.profile(3, component=1)
1

sage: EA = SteenrodAlgebra(generic=True, profile=(lambda n: n, lambda n: 1))
sage: EA.profile(4)
4
sage: EA.profile(2, component=1)
1

>>> from sage.all import *
>>> SteenrodAlgebra(profile=[Integer(3),Integer(2),Integer(1)]).profile(Integer(9))
0
>>> SteenrodAlgebra(profile=[Integer(3),Integer(2),Integer(1)], truncation_type=Infinity).profile(Integer(9))
+Infinity

>>> B = SteenrodAlgebra(p=Integer(3), profile=(lambda n: n, lambda n: Integer(1)))
>>> B.profile(Integer(3))
3
>>> B.profile(Integer(3), component=Integer(1))
1

>>> EA = SteenrodAlgebra(generic=True, profile=(lambda n: n, lambda n: Integer(1)))
>>> EA.profile(Integer(4))
4
>>> EA.profile(Integer(2), component=Integer(1))
1

pst(s, t)[source]#

The Margolis element $$P^s_t$$.

INPUT:

• s – non-negative integer

• t – positive integer

• p – positive prime number

OUTPUT: element of the Steenrod algebra

This returns the Margolis element $$P^s_t$$ of the mod $$p$$ Steenrod algebra: the element equal to $$P(0,0,...,0,p^s)$$, where the $$p^s$$ is in position $$t$$.

EXAMPLES:

sage: A2 = SteenrodAlgebra(2)
sage: A2.pst(3,5)
Sq(0,0,0,0,8)
sage: A2.pst(1,2) == Sq(4)*Sq(2) + Sq(2)*Sq(4)
True
sage: SteenrodAlgebra(5).pst(3,5)
P(0,0,0,0,125)

>>> from sage.all import *
>>> A2 = SteenrodAlgebra(Integer(2))
>>> A2.pst(Integer(3),Integer(5))
Sq(0,0,0,0,8)
>>> A2.pst(Integer(1),Integer(2)) == Sq(Integer(4))*Sq(Integer(2)) + Sq(Integer(2))*Sq(Integer(4))
True
>>> SteenrodAlgebra(Integer(5)).pst(Integer(3),Integer(5))
P(0,0,0,0,125)

top_class()[source]#

Highest dimensional basis element. This is only defined if the algebra is finite.

EXAMPLES:

sage: SteenrodAlgebra(2, profile=(3,2,1)).top_class()
Sq(7,3,1)
sage: SteenrodAlgebra(3, profile=((2,2,1),(1,2,2,2,2))).top_class()
Q_1 Q_2 Q_3 Q_4 P(8,8,2)

>>> from sage.all import *
>>> SteenrodAlgebra(Integer(2), profile=(Integer(3),Integer(2),Integer(1))).top_class()
Sq(7,3,1)
>>> SteenrodAlgebra(Integer(3), profile=((Integer(2),Integer(2),Integer(1)),(Integer(1),Integer(2),Integer(2),Integer(2),Integer(2)))).top_class()
Q_1 Q_2 Q_3 Q_4 P(8,8,2)

class sage.algebras.steenrod.steenrod_algebra.SteenrodAlgebra_mod_two(p=2, basis='milnor', **kwds)[source]#

The mod 2 Steenrod algebra.

Users should not call this, but use the function SteenrodAlgebra() instead. See that function for extensive documentation. (This differs from SteenrodAlgebra_generic only in that it has a method Sq() for defining elements.)

Sq(*nums)[source]#

Milnor element $$\text{Sq}(a,b,c,...)$$.

INPUT:

• a, b, c, … – non-negative integers

OUTPUT: element of the Steenrod algebra

This returns the Milnor basis element $$\text{Sq}(a, b, c, ...)$$.

EXAMPLES:

sage: A = SteenrodAlgebra(2)
sage: A.Sq(5)
Sq(5)
sage: A.Sq(5,0,2)
Sq(5,0,2)

>>> from sage.all import *
>>> A = SteenrodAlgebra(Integer(2))
>>> A.Sq(Integer(5))
Sq(5)
>>> A.Sq(Integer(5),Integer(0),Integer(2))
Sq(5,0,2)


Entries must be non-negative integers; otherwise, an error results.