The Steenrod algebra¶
AUTHORS:
 John H. Palmieri (20080730): version 0.9: Initial implementation.
 John H. Palmieri (20100630): version 1.0: Implemented subHopf 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 \(2(p1)i\).
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
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
SerreCartan 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 nonnegative 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 nonnegative 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 SerreCartan 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 nonnegative 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 nonnegative 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 \(X^m_k = \text{Sq}^{2^m} \text{Sq}^{2^{m1}} ... \text{Sq}^{2^k}\). 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 \(2^i  t_{mi}\).
\(P^s_t\) 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 \(c_{i,k} = [c_{i+k1,1}, c_{i,k1}]\). Thus \(c_{i,k}\) is a \(k\)fold iterated commutator of the elements \(\mathcal{P}(p^i)\), …, \(\mathcal{P}(p^{i+k1})\). 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 leftmultiply 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’.
SubHopf algebras of the Steenrod algebra
The subHopf 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
The Milnor basis is dual to the monomial basis. Furthermore, any subHopf algebra corresponds to a quotient of this of the form
The list of exponents \((e_1, e_2, ...)\) may be considered a function \(e\) from the positive integers to the extended nonnegative integers (the nonnegative integers and \(\infty\)); this is called the profile function for the subHopf algebra. The profile function must satisfy the condition
 \(e(r) \geq \min( e(ri)  i, e(i))\) 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,
and any subHopf algebra corresponds to a quotient of this of the form
Here the profile function has two pieces, \(e\) as at the prime 2, and \(k\), which maps the nonnegative integers to the set \(\{1, 2\}\). These must satisfy the following conditions:
 \(e(r) \geq \min( e(ri)  i, e(i))\) 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 AdamsMargolis [AM1974], for example, for these results on profile functions.)
This module allows one to construct the Steenrod algebra or any of its subHopf algebras, at any prime. When defining a subHopf 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 # nonnegative exponents work as they should
Sq(1,2,1) + Sq(4,1,1)
sage: z**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)
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)
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
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)
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
It can be viewed in other bases:
sage: a.milnor() # same as a
Sq(1,2,1)
sage: a.change_basis('adem')
Sq^9 Sq^4 Sq^1 + Sq^11 Sq^2 Sq^1 + Sq^13 Sq^1
sage: a.change_basis('adem').change_basis('milnor')
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 \(\sum (2^n1) i_n\), 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
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]
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
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)
Applying the antipode twice returns the original element:
sage: y = Sq(8)*Sq(4)
sage: 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: x.support())
[Sq(2,1), Sq(5)]
sage: sorted(c.support())
[(2, 1), (5,)]
sage: Adem = SteenrodAlgebra(basis='adem')
sage: elt = Adem.Sq(10) + Adem.Sq(9) * Adem.Sq(1)
sage: sorted(elt.monomials(), key=lambda x: 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)
sage: a.leading_coefficient()
2
sage: a.leading_monomial()
P(2)
sage: a.leading_term()
2 P(2)
sage: a.change_basis('adem').monomial_coefficients()
{(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
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
See the documentation for SteenrodAlgebra()
for more details and
examples.

sage.algebras.steenrod.steenrod_algebra.
AA
(n=None, p=2)¶ This returns the Steenrod algebra \(A\) or its subHopf algebra \(A(n)\).
INPUT:
 \(n\)  nonnegative 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 subHopf algebra generated by the elements \(\text{Sq}^i\) for \(i \leq 2^n\). Its profile function is \((n+1, n, n1, ...)\). When \(p\) is odd, \(A(n)\) is the subHopf algebra generated by the elements \(Q_0\) and \(\mathcal{P}^i\) for \(i \leq p^{n1}\). Its profile function is \(e=(n, n1, n2, ...)\) 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) subHopf algebra of mod 2 Steenrod algebra, milnor basis, profile function [3, 2, 1] sage: A(2, p=5) subHopf algebra of mod 5 Steenrod algebra, milnor basis, profile function ([2, 1], [2, 2, 2])
 \(n\)  nonnegative integer, optional (default

sage.algebras.steenrod.steenrod_algebra.
Sq
(*nums)¶ Milnor element Sq(a,b,c,…).
INPUT:
a, b, c, ...
 nonnegative 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
Entries must be nonnegative 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)¶ The mod \(p\) Steenrod algebra
INPUT:
p
 positive prime integer (optional, default = 2)basis
 string (optional, default = ‘milnor’)profile
 a profile function in form specified below (optional, defaultNone
)truncation_type
 0 or \(\infty\) or ‘auto’ (optional, default ‘auto’)precision
 integer orNone
(optional, defaultNone
)generic
 (optional, default ‘auto’)
OUTPUT: mod \(p\) Steenrod algebra or one of its subHopf 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 graded 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)
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
Different bases for the Steenrod algebra:
There are two standard vector space bases for the mod \(p\) Steenrod algebra: the Milnor basis and the SerreCartan basis. When \(p=2\), there are also several other, less wellknown, bases. See the documentation for this module (type
sage.algebras.steenrod.steenrod_algebra?
) and the functionsteenrod_algebra_basis
for full descriptions of each of the implemented bases.This module implements the following bases at all primes:
 ‘milnor’: Milnor basis.
 ‘serrecartan’ or ‘adem’ or ‘admissible’: SerreCartan 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) sage: adem(y) Sq^2 Sq^1 + Sq^3 sage: adem5 = SteenrodAlgebra(5, 'serrecartan') sage: adem5.P(0,2) P^10 P^2 + 4 P^11 P^1 + P^12
If you add or multiply elements defined using different bases, the lefthand 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)
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))
Algebras defined over different bases are not equal:
sage: 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 sage: SteenrodAlgebra(basis='adem') mod 2 Steenrod algebra, serrecartan basis sage: SteenrodAlgebra(basis='adem').basis_name() 'serrecartan' sage: SteenrodAlgebra(basis='woodz').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
SubHopf algebras of the Steenrod algebra:
These are specified using the argument
profile
, along with, optionally,truncation_type
andprecision
. Theprofile
argument specifies the profile function for this algebra. Any subHopf 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 nonnegative integers, plus \(\infty\), corresponding to the subHopf 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
 \(e(r) \geq \min( e(ri)  i, e(i))\) for all \(0 < i < r\).
This is specified via
profile
, and optionallyprecision
andtruncation_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 oftruncation_type
.  a function from positive integers to nonnegative integers (and
\(\infty\)), e.g.,
lambda n: n+2
. None
orInfinity
 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 thanprecision
, which has default value 100. The function is truncated at this point, and all remaining values are set to the value oftruncation_type
.truncation_type
may be 0, \(\infty\), or ‘auto’. If it’s ‘auto’, then it gets converted to 0 in the first case above (whenprofile
is a list), and otherwise (whenprofile
is a function,None
, orInfinity
) it gets converted to \(\infty\).For example, the subHopf 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(4n, 0), truncation_type=0) sage: 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 subHopf algebra:sage: D2 = SteenrodAlgebra(profile=lambda n: max(4n, 0)) sage: A2 == D2 False sage: D2.is_finite() False sage: E2 = SteenrodAlgebra(profile=lambda n: max(4n, 0), truncation_type=Infinity) sage: 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 subHopf 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 subHopf algebra of mod 2 Steenrod algebra, milnor basis, profile function [1, 2, 3, +Infinity, +Infinity, +Infinity, ...] sage: 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
 \(e(r) \geq \min( e(ri)  i, e(i))\) 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
nonnegative integers (and \(\infty\)), one from the nonnegative
integers to the set \(\{1,2\}\), e.g.,
(lambda n: n+2, lambda n: 1 if n<3 else 2)
. None
orInfinity
 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
andtruncation_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
For subHopf 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 subHopf 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

class
sage.algebras.steenrod.steenrod_algebra.
SteenrodAlgebra_generic
(p=2, basis='milnor', **kwds)¶ Bases:
sage.combinat.free_module.CombinatorialFreeModule
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 sage: sage.algebras.steenrod.steenrod_algebra.SteenrodAlgebra_generic(5, 'adem') mod 5 Steenrod algebra, serrecartan basis

class
Element
¶ Bases:
sage.modules.with_basis.indexed_element.IndexedFreeModuleElement
Class for elements of the Steenrod algebra. Since the Steenrod algebra class is based on
CombinatorialFreeModule
, this is based onIndexedFreeModuleElement
. It has new methods reflecting its role, likedegree()
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: x.support()) [Sq(2,1), Sq(5)] sage: 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.
additive_order
()¶ 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 sage: z.additive_order() 2 sage: (Sq(3) + Sq(3)).additive_order() 1

basis_name
()¶ The basis name associated to self.
EXAMPLES:
sage: a = SteenrodAlgebra().Sq(3,2,1) sage: a.basis_name() 'milnor' sage: a.change_basis('adem').basis_name() 'serrecartan' sage: a.change_basis('wood____y').basis_name() 'woody' sage: b = SteenrodAlgebra(p=7).basis(36)[0] sage: b.basis_name() 'milnor' sage: a.change_basis('adem').basis_name() 'serrecartan'

change_basis
(basis='milnor')¶ 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.
 ‘serrecartan’, ‘serre_cartan’, ‘sc’, ‘adem’, ‘admissible’ for the SerreCartan 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('serrecartan') 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

coproduct
(algorithm='milnor')¶ The coproduct of this element.
INPUT:
algorithm
–None
or a string, either ‘milnor’ or ‘serrecartan’ (or anything which will be converted to one of these by the functionget_basis_name
). IfNone
, default to ‘serrecartan’ if current basis is ‘serrecartan’; 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 sage: c = a.change_basis('adem'); c.coproduct(algorithm='milnor') 1 # Sq^2 + Sq^1 # Sq^1 + Sq^2 # 1 sage: c = a.change_basis('adem'); c.coproduct(algorithm='adem') 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 sage: a.coproduct(algorithm='adem') 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)]
The
support
of an element does not include the coefficients, so at odd primes it may be better to usemonomial_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)]

degree
()¶ 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, callSteenrodAlgebra_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

excess
()¶ Excess of element.
OUTPUT:
excess
 nonnegative integerThe 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: 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

is_decomposable
()¶ 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 = SteenrodAlgebra(p=3, basis='adem') 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]

is_homogeneous
()¶ 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

is_nilpotent
()¶ 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

is_unit
()¶ 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

may_weight
()¶ May’s ‘weight’ of element.
OUTPUT:
weight
 nonnegative integerIf 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

milnor
()¶ 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') sage: a = Adem.basis(4)[1]; a Sq^3 Sq^1 sage: a.milnor() Sq(1,1)

prime
()¶ The prime associated to self.
EXAMPLES:
sage: a = SteenrodAlgebra().Sq(3,2,1) sage: a.prime() 2 sage: a.change_basis('adem').prime() 2 sage: b = SteenrodAlgebra(p=7).basis(36)[0] sage: b.prime() 7 sage: SteenrodAlgebra(p=3, basis='adem').one().prime() 3

wall_height
()¶ Wall’s ‘height’ of element.
OUTPUT: list of nonnegative integers
The height of an element of the mod 2 Steenrod algebra is a list of nonnegative 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 rightlexicographically, 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 \(j \leq k \leq i+j1\).
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]


P
(*nums)¶ The element \(P(a, b, c, ...)\)
INPUT:
a, b, c, ...
 nonnegative 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 nonnegative integers sage: SteenrodAlgebra(basis='serrecartan').P(0,1) Sq^2 Sq^1 + Sq^3 sage: SteenrodAlgebra(generic=True).P(2,0,1) P(2,0,1)

Q
(*nums)¶ The element \(Q_{n0} Q_{n1} ...\) , given by specifying the subscripts.
INPUT:
n0, n1, ...
 nonnegative 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

Q_exp
(*nums)¶ 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

algebra_generators
()¶ 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 subHopf algebras of the Steenrod algebra, it is not always clear what a minimal generating set is. The subHopf 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 \(0 \leq i \leq n1\). So if this algebra is \(A(n)\), return the appropriate list of generators.
For other subHopf algebras: return a nonminimal 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_with_category.gen of mod 3 Steenrod algebra, serrecartan 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))
In the following case, return a nonminimal 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_mod_two_with_category.gen of subHopf 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 ofgens
:sage: SteenrodAlgebra(p=5, profile=[[2,1], [2,2,2]]).algebra_generators() Family (Q_0, P(1), P(5))

an_element
()¶ 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) sage: SteenrodAlgebra(basis='adem').an_element() 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

antipode_on_basis
(t)¶ 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 SerreCartan 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 \(n*2(p1)\), multiplied by \((1)^n\), and the antipode of \(\beta = Q_0\) is \(Q_0\). So convert to the SerreCartan basis, as in the \(p=2\) case.
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) sage: Adem = SteenrodAlgebra(basis='adem') sage: Adem.Sq(4).antipode() 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)

basis
(d=None)¶ Returns basis for self, either the whole basis or the basis in degree \(d\).
INPUT:
 \(d\) – integer or
None
, optional (defaultNone
)
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)) sage: SteenrodAlgebra(2, 'adem').basis(12) 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
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)]
 \(d\) – integer or

basis_name
()¶ The basis name associated to self.
EXAMPLES:
sage: SteenrodAlgebra(p=2, profile=[1,1]).basis_name() 'milnor' sage: SteenrodAlgebra(basis='serrecartan').basis_name() 'serrecartan' sage: SteenrodAlgebra(basis='adem').basis_name() 'serrecartan'

coproduct
(x, algorithm='milnor')¶ Return the coproduct of an element
x
of this algebra.INPUT:
x
– element of selfalgorithm
–None
or a string, either ‘milnor’ or ‘serrecartan’ (or anything which will be converted to one of these by the functionget_basis_name
. IfNone
, default to ‘serrecartan’ if current basis is ‘serrecartan’; otherwise use ‘milnor’.
This calls
coproduct_on_basis()
on the summands ofx
and extends linearly.EXAMPLES:
sage: SteenrodAlgebra().Sq(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='serrecartan') 1 # P(4) + P(1) # P(3) + P(2) # P(2) + P(3) # P(1) + P(4) # 1 sage: SteenrodAlgebra(p=3, basis='serrecartan').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

coproduct_on_basis
(t, algorithm=None)¶ The coproduct of a basis element of this algebra
INPUT:
t
– tuple, the index of a basis element of selfalgorithm
–None
or a string, either ‘milnor’ or ‘serrecartan’ (or anything which will be converted to one of these by the functionget_basis_name
. IfNone
, default to ‘milnor’ unless current basis is ‘serrecartan’, in which case use ‘serrecartan’.
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 SerreCartan basis from this: the coproduct on each \(P^n\) is \(\sum P^i \otimes P^{ni}\) and at odd primes \(\beta\) is primitive. Since the coproduct is an algebra map, one can then compute the coproduct on any SerreCartan basis element.
Which of these methods is used is controlled by whether
algorithm
is ‘milnor’ or ‘serrecartan’.OUTPUT: the coproduct of the corresponding basis element, as an element of
self
tensorself
.EXAMPLES:
sage: A = SteenrodAlgebra() sage: A.coproduct_on_basis((3,)) 1 # Sq(3) + Sq(1) # Sq(2) + Sq(2) # Sq(1) + Sq(3) # 1

counit_on_basis
(t)¶ 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

degree_on_basis
(t)¶ 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

dimension
()¶ 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

gen
(i=0)¶ The ith generator of this algebra.
INPUT:
i
 nonnegative 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 \(P(p^{i1})\).
For subHopf algebras of the Steenrod algebra, it is not always clear what a minimal generating set is. The subHopf 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 \(0 \leq i \leq n1\). So if this algebra is \(A(n)\), return the appropriate generator.
For other subHopf 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) sage: SteenrodAlgebra(2, basis='adem').gen(2) 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

gens
()¶ 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 subHopf algebras of the Steenrod algebra, it is not always clear what a minimal generating set is. The subHopf 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 \(0 \leq i \leq n1\). So if this algebra is \(A(n)\), return the appropriate list of generators.
For other subHopf algebras: return a nonminimal 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_with_category.gen of mod 3 Steenrod algebra, serrecartan 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))
In the following case, return a nonminimal 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_mod_two_with_category.gen of subHopf 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 ofgens
:sage: SteenrodAlgebra(p=5, profile=[[2,1], [2,2,2]]).algebra_generators() Family (Q_0, P(1), P(5))

homogeneous_component
(n)¶ 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
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
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')

is_commutative
()¶ True if
self
is graded commutative, as determined by the profile function. In particular, a subHopf 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
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

is_division_algebra
()¶ 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

is_field
(proof=True)¶ 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

is_finite
()¶ True if this algebra is finitedimensional.
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

is_generic
()¶ The algebra is generic if it is based on the oddprimary 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 nongeneric.
EXAMPLES:
sage: SteenrodAlgebra(3).is_generic() True sage: SteenrodAlgebra(2).is_generic() False sage: SteenrodAlgebra(2,generic=True).is_generic() True

is_integral_domain
(proof=True)¶ 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

is_noetherian
()¶ 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

milnor
()¶ 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') sage: a = Adem.Sq(2) * Adem.Sq(1) sage: Adem.milnor(a) Sq(0,1) + Sq(3)

ngens
()¶ Number of generators of self.
OUTPUT: number or Infinity
The Steenrod algebra is infinitely generated. A subHopf 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 infinitedimensional, this returns Infinity. If the algebra is finitedimensional and is equal to one of the subHopf algebras \(A(n)\), then their minimal generating set is known, and this returns the cardinality of that set. Otherwise, any subHopf 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

one_basis
()¶ 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() ((), ())

order
()¶ 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

prime
()¶ The prime associated to self.
EXAMPLES:
sage: SteenrodAlgebra(p=2, profile=[1,1]).prime() 2 sage: SteenrodAlgebra(p=7).prime() 7

product_on_basis
(t1, t2)¶ 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 SerreCartan basis, then multiply them using Adem relations (also implemented insage.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) sage: Adem = SteenrodAlgebra(basis='adem') sage: Adem.Sq(2) * Adem.Sq(2) # indirect doctest Sq^3 Sq^1
When multiplying elements from different bases, the lefthand factor determines the form of the output:
sage: Adem.Sq(2) * Milnor.Sq(2) Sq^3 Sq^1 sage: Milnor.Sq(2) * Adem.Sq(2) Sq(1,1)

profile
(i, component=0)¶ 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
andSteenrodAlgebra()
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
andcomponent=1
. So when \(p\) is odd andcomponent
is 0, \(i\) must be positive, while whencomponent
is 1, \(i\) must be nonnegative.EXAMPLES:
sage: SteenrodAlgebra().profile(3) +Infinity sage: SteenrodAlgebra(profile=[3,2,1]).profile(1) 3 sage: SteenrodAlgebra(profile=[3,2,1]).profile(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

pst
(s, t)¶ The Margolis element \(P^s_t\).
INPUT:
s
 nonnegative integert
 positive integerp
 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)

top_class
()¶ 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)

class

class
sage.algebras.steenrod.steenrod_algebra.
SteenrodAlgebra_mod_two
(p=2, basis='milnor', **kwds)¶ Bases:
sage.algebras.steenrod.steenrod_algebra.SteenrodAlgebra_generic
The mod 2 Steenrod algebra.
Users should not call this, but use the function
SteenrodAlgebra()
instead. See that function for extensive documentation. (This differs fromSteenrodAlgebra_generic
only in that it has a methodSq()
for defining elements.)
Sq
(*nums)¶ Milnor element \(\text{Sq}(a,b,c,...)\).
INPUT:
a, b, c, ...
 nonnegative 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)
Entries must be nonnegative integers; otherwise, an error results.
