ASKSAGE: Sage Q&A Forum - RSS feedhttps://ask.sagemath.org/questions/Q&A Forum for SageenCopyright Sage, 2010. Some rights reserved under creative commons license.Mon, 25 Sep 2023 10:09:52 +0200Differentials on quotient CDGAshttps://ask.sagemath.org/question/73586/differentials-on-quotient-cdgas/The following code throws a `ValueError("The given dictionary does not determine a valid differential")`:
A.<x, y, z> = GradedCommutativeAlgebra(QQ, degrees=(2, 3, 4))
AQ = A.quotient(A.ideal(x^3))
E = AQ.cdg_algebra({y: x^2, z: x*y})
It seems the cause is a check `if not self(self(i)).is_zero()`. Since the error disappears on omitting the `z = x*y` differential, I presume it's evaluating $d(xy) = x^3$ to be something non-zero, maybe because it is being evaluated in `A` and not `AQ`. Is this a bug or am I supposed to define this CDGA in a different way?
ronnoMon, 25 Sep 2023 10:09:52 +0200https://ask.sagemath.org/question/73586/submodules or maybe sub-algebras of CDGAhttps://ask.sagemath.org/question/73175/submodules-or-maybe-sub-algebras-of-cdga/ Let us try to start with a minimum working examle. Start with a graded commutative polynomial algebra, say `$A<wa1,wb1,wb2,wc1,wc2,ya,yb, degrees = ((1,0),(1,0),(2,0),(1,0),(2,0),(0,1),(0,1))` with differential
`d=A.(wa1:a^2, wb1:0, wb2:wb1*yb^2, wc1:0, wc2: wc1*(ya^2 +yb^2)`.
The differential is of total degree 1, as required, and cohomology computations work fine. The bi-degree of the differential is actually (-1,2), so total degree 1. Furthermore the differential just multiplies by even powers of `ya`and `yb`
Thus the algebra `A`, and its cohomology separates into two submodules, `Aeven` where the exponents of both `ya` and `yb` are even, and `Aodd`, where at least one of the exponents is odd.
How can I get Sage to create two subcomplexes, and be able to compute their cohomology. As a start, maybe just `Aeven`, which is a sub-algebra.
I am pretty new to Sage, and Python, so any help would be appreciated.
vinceWed, 06 Sep 2023 05:02:04 +0200https://ask.sagemath.org/question/73175/Exterior Powers of A Cohomologyhttps://ask.sagemath.org/question/69933/exterior-powers-of-a-cohomology/Hi guys, I am working on computing basis of the exterior powers of a CDGA's cohomology. The following code is what I currently have:
L = lie_algebras.Heisenberg(QQ, 2)
A = GradedCommutativeAlgebra(QQ, names=L.basis())
A.inject_variables()
Output: Defining p1, p2, q1, q2, z
B = A.cdg_algebra({p1: 0, p2: 0, q1: 0, q2: 0, z: p1*q1 + p2*q2})
C = B.cohomology(1)
C.basis()
Output: Finite family {[p1]: B[[p1]], [p2]: B[[p2]], [q1]: B[[q1]], [q2]: B[[q2]]}
First, I tried to manipulate the basis to possibly calculate all the possible wedge products but I found out that **type(C)** is **sage.combinat.free_module.CombinatorialFreeModule_with_category**; thus, I cannot do wedges. Then, I tried to compute the dual exterior powers of C using preexisted method. I imported [*ExtPowerDualFreeModule*](https://doc.sagemath.org/html/en/reference/combinat/sage/combinat/free_module.html#sage.combinat.free_module.CombinatorialFreeModule.Element) and run `E = ExtPowerDualFreeModule(C, 2)` but did not succeed. Hence, is there a way to compute the dual exterior powers and see their bases? If not, can the free module C be converted to another object so that I can perform wedges on its basis? It is very important for me to be able to see the basis of the exterior powers of the cohomology.<br>
I apologize for my lack of mathematical knowledge. If I am missing any information, please let me know. I really appreciate your help, thank you!
Hieu NguyenTue, 11 Jul 2023 06:57:42 +0200https://ask.sagemath.org/question/69933/Is there any graded Hopf algebra functionality?https://ask.sagemath.org/question/52298/is-there-any-graded-hopf-algebra-functionality/In the SAGE Reference Manual, there's a brief section on graded Hopf algebras: https://doc.sagemath.org/html/en/reference/categories/sage/categories/graded_hopf_algebras.html
Can one actually define graded Hopf algebras and do computations in them? If not, what is this doing there?davidac897Thu, 02 Jul 2020 09:25:26 +0200https://ask.sagemath.org/question/52298/How to get the graded part of a graded ring?https://ask.sagemath.org/question/47623/how-to-get-the-graded-part-of-a-graded-ring/I have a graded quotient of a graded polynomial ring, say something like
P = PolynomialRing(QQ, , 'x,y,z', order=TermOrder('wdegrevlex',(1,2,3)))
I = P.ideal(x*y^2 + x^5, z*y + x^3*y)
Q = P.quotient(I)
I would like to get the vector space over QQ consisting on vectors of degree, say 9, in Q.
heluaniTue, 27 Aug 2019 15:34:06 +0200https://ask.sagemath.org/question/47623/How to get graded component of graded ringhttps://ask.sagemath.org/question/47622/how-to-get-graded-component-of-graded-ring/I have a graded quotient of a graded polynomial ring, say something like
P = PolynomialRing(QQ, , 'x,y,z', order=TermOrder('wdegrevlex',(1,2,3)))
I = P.ideal(x*y^2 + x^5, z*y + x^3*y)
Q = P.quotient(I)
I would like to get the vector space over QQ consisting on vectors of degree, say 9, in Q.
heluaniTue, 27 Aug 2019 15:32:45 +0200https://ask.sagemath.org/question/47622/What does cohomology_generators actually do?https://ask.sagemath.org/question/46609/what-does-cohomology_generators-actually-do/ I don't understand the output of the function "cohomology_generators" of commutative differential graded algebras. Look at this simple exmaple:
A.<x,y,z,t> = GradedCommutativeAlgebra(QQ, degrees = (2,2,1,2))
B = A.cdg_algebra({z:x-y})
B.cohomology_generators(5)
{2: [t, y, x]}
It's apparently telling me that t, x and y are all generators of the cohomology algebra in degree 2. But in cohomology, x = y, since dz = x - y!wutututTue, 21 May 2019 16:46:44 +0200https://ask.sagemath.org/question/46609/How do I define a homomorphism of a graded commutative algebra?https://ask.sagemath.org/question/42202/how-do-i-define-a-homomorphism-of-a-graded-commutative-algebra/Why does the following throw a `TypeError: images do not define a valid homomorphism`?
E = GradedCommutativeAlgebra(QQ,'x,y',degrees=(1,1))
E.inject_variables()
f = E.hom([x,y])
I expected it to define $f$ to be the identity homomorphism of $E$. What is the right way to define a homomorphism of $E$? I'm more interested in the one that switches $x$ and $y$ than the identity homomorphism, but this seemed a more obvious version of the question.ronnoSat, 28 Apr 2018 19:14:46 +0200https://ask.sagemath.org/question/42202/