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.Thu, 08 Aug 2024 19:39:01 +0200Defining several independent Weyl Character Ringshttps://ask.sagemath.org/question/78627/defining-several-independent-weyl-character-rings/ I have the following problem. I want to define expression which contains elements of several Weyl Character Rings. The problem comes from physics. I have expression for theory with several symmetries, which can be of the same or different groups which involves characters of these groups representations. Let's take a simplest example of such expression
$x = \chi_{R_1}(U)\chi_{R_2}(V)$
where $U\in G_1 $, $V\in G_2 $ are matrices in groups $G_1$ and $G_2$ and $R_{1,2}$ are corresponding representations. Let's say I want to find $x^2$ (or any other power) expanded in terms of irreps. If $G_1 \neq G_2$ I can write the
following code that will work
G1 = WeylCharacterRing(['A',2], base_ring=QQ, style="coroots")
G2 = WeylCharacterRing(['A',1], base_ring=G1, style="coroots")
fund1 = G1(G1.fundamental_weights()[1]);
fund2 = G2(G2.fundamental_weights()[1]);
x = fund1*fund2
x**2
which gives me what I want
(A2(0,1)+A2(2,0))*A1(0) + (A2(0,1)+A2(2,0))*A1(2)
Here for simplicity I took $SU(2)$ and $SU(3)$ groups and fundamental representations for both of them.
Now if $G_1 = G_2$ (for example $A_2$ root system in both cases) it also works but result is not very readable now since I can not understand which representation belongs to one group or another. For example
G1 = WeylCharacterRing(['A',2], base_ring=QQ, style="coroots")
G2 = WeylCharacterRing(['A',2], base_ring=G1, style="coroots")
fund1 = G1(G1.fundamental_weights()[1]);
fund2 = G2(G2.fundamental_weights()[1]);
ad1 = G1.adjoint_representation()
ad2 = G2.adjoint_representation()
x = fund1*ad2
x**2
gives me back
(A2(0,1)+A2(2,0))*A2(0,0) + (A2(0,1)+A2(2,0))*A2(0,3) + (2*A2(0,1)+2*A2(2,0))*A2(1,1) + (A2(0,1)+A2(2,0))*A2(3,0) + (A2(0,1)+A2(2,0))*A2(2,2)
which is hard to read. Is there a simple way to keep track which of the expressions belongs to one group or another?
Anton NedelinThu, 08 Aug 2024 19:39:01 +0200https://ask.sagemath.org/question/78627/A question about Auslander-Reiten Quivershttps://ask.sagemath.org/question/78492/a-question-about-auslander-reiten-quivers/I want to use sagemath to draw some Auslander-Reiten quivers.
But when I use the following command by this site [Auslander-Reiten Quivers]
(https://doc.sagemath.org/html/en/reference/quivers/sage/quivers/ar_quiver.html#sage.quivers.ar_quiver.AuslanderReitenQuiver), like
> DA = DiGraph([[1, 2], [2, 3]])
>AR = DA.auslander_reiten_quiver()
It will say
> AttributeError Traceback (most recent call last)
>/tmp/ipykernel_5300/610482001.py in <module>
>----> 1 AR = DA.auslander_reiten_quiver()
>AttributeError: 'DiGraph' object has no attribute 'auslander_reiten_quiver'
I do not know how to solve this question since I am not familiar with sagemath. My version of sagemath is $9.5$.
What should I do for this question ?
Thanks!fushengSat, 27 Jul 2024 09:54:40 +0200https://ask.sagemath.org/question/78492/Auslander-Reiten Quiver - knittinghttps://ask.sagemath.org/question/75981/auslander-reiten-quiver-knitting/Given a quiver representation
A3 = DiGraph({1 : {2 : ['a']},
2 : {3 : ['b']}})
or
A4 = DiGraph({1 : {2 : ['a']},
2 : {3 : ['b']},
3 : {4 : ['b']}})
do you know a script or an algorithm which can be downloaded or is part of a library to perform knitting for determining the Auslander-Reiten quiver, for simple cases such as above. The above are the Dynkin diagrams of type $\Bbb A_3$ and $\Bbb A_4$.
Knitting is described in A. Schiffler, Quiver Representations, Springer, 2014, p. 70 f. I realize there is a way to calculate the Auslander-Reiten-Translate (https://doc.sagemath.org/html/en/reference/quivers/sage/quivers/representation.html).Tintin1Tue, 13 Feb 2024 22:02:29 +0100https://ask.sagemath.org/question/75981/character table of normalizerhttps://ask.sagemath.org/question/73641/character-table-of-normalizer/Consider the character table of the cyclic permutation group $\mathbb{Z}_4$
Z4 = CyclicPermutationGroup(4)
Z4.character_table()
gives
[ 1 1 1 1]
[ 1 -1 1 -1]
[ 1 zeta4 -1 -zeta4]
[ 1 -zeta4 -1 zeta4]
if $e$ is the identity and $r$ is a rotation then the columns correspond to transformations $e$, $r$, $r^2$ and $r^3$, which of course is intuitive.
$\mathbb{Z}_4$ can for example be obtained from the centralizer of G((1,2,3,4)) with respect to the dihedral group $D_4$
G = DihedralGroup(4)
n = G.centralizer(G((1,2,3,4)))
ctable = n.character_table()
yields the same table as above with the second and fourth columns swapped.
[ 1 1 1 1]
[ 1 -1 1 -1]
[ 1 -zeta4 -1 zeta4]
[ 1 zeta4 -1 -zeta4]
`n.list()` yields `[(), (1,3)(2,4), (1,4,3,2), (1,2,3,4)]` or $[e,r^2,r^3,r]$ while the columns above are ordered as $[e,r^3,r^2,r]$, indicating that the columns are not given standard ordering nor are they given ordering with respect to the normalizer list. Generally, what is the convention for ordering?ayodanThu, 28 Sep 2023 08:37:46 +0200https://ask.sagemath.org/question/73641/Getting Tietze Representation in Original Grouphttps://ask.sagemath.org/question/69590/getting-tietze-representation-in-original-group/ I've got a free group with relators:
sage: H.<a,b,c,d>=FreeGroup();H
Free Group on generators {a, b, c, d}
sage: Y=[[1,2],[3,4]]
sage: G=H/[Y[i] for i in range(2)];G
Finitely presented group < a, b, c, d | a*b, c*d >
sage: I=G.simplification_isomorphism();I
Generic morphism:
From: Finitely presented group < a, b, c, d | a*b, c*d >
To: Finitely presented group < a, c | >
Defn: a |--> a
b |--> a^-1
c |--> c
d |--> c^-1
Cool. Now I want to get the Tietze representation. First, under the isomorphism:
sage: G([4])
d
sage: I(G(([4])))
c^-1
sage: I(G([4])).Tietze()
(-2,)
Sure, since the index of the element is the second in the list under the isomorphism. But, I want to be able to grab the element under the original free group - that is, I want (-3). Something that I think should work:
sage: x=I(G([4]));x
c^-1
sage: H(x).Tietze()
(-2,)
Right? "What is the Tietze representation of the word 'x' in the group 'H'"....but no, still under the isomorphism. Even worse, it looks like it's like, using the Tietze representaton to do something that looks very wrong:
sage: H(I(G([4])))
b^-1
That's very confusing, but at least clearly not what I want. I can do this if I know the word ahead of time:
sage: y=c^-1
sage: y.Tietze()
(-3,)
But the point of the code I'm trying to write is to determine these words under the isomorphism. Anyone know how to do something like my second code block, but with the response (-3)?
thethinkerThu, 29 Jun 2023 00:29:52 +0200https://ask.sagemath.org/question/69590/Dimension of irreducible representations of a grouphttps://ask.sagemath.org/question/67484/dimension-of-irreducible-representations-of-a-group/ It is possible to obtain the dimensions of all irreducible representations of a group by extracting the first column from character table. This process works for small groups, but is very inefficient for larger groups as we need to call the entire character table. Is there any way to obtain the dimensions of the irreducible without using the character table?Diego99Tue, 11 Apr 2023 11:23:39 +0200https://ask.sagemath.org/question/67484/Constructing group representationshttps://ask.sagemath.org/question/46701/constructing-group-representations/I've got a vector space $V= \mathbb{F}_p^n$ and I want to construct rings such as the symmetric algebra $\mathrm{Sym}^*(V)$, the divided power algebra generated by $V$, and so forth. I can do this, but the thing I'm not sure how to do is to build these objects along with the induced action of $\mathrm{GL}_n(\mathbb{F}_p)$ by algebra homomorphisms.
This brings two more general questions:
1. (Likely simple) If I've got a finite group $G$, how do I build a vector space $V$ with an action of $G$? In particular, it would be nice if I could build $V$ with named generators $v_1, v_2, \ldots$.
2. If I've got a vector space $V$ with an action of a group $G$, and I want to define an algebra $\mathcal{F}(V)$ which is functorial in $V$, then how do I port the action of $G$ over?
Alternatively, it might be computationally a lot more efficient to simply construct the divided power algebra over $\mathbb{F}_p$ on $n$ generators from scratch, but if I do this then how do I tell sage how $\mathrm{GL}_n(\mathbb{F}_p)$ acts on it?
(The divided power algebra on one generator $y$ is a ring with polynomial generators $y_1, y_2, y_3, \ldots$ subject to the relation $y_iy_j=\binom{i+j}{i}y_{i+j}$. So intuitively, one thinks of $y_n=\frac{y_1^n}{n!}$. Over a field of characteristic $p$, this amounts to being an algebra on $y_1, y_p, y_{p^2}, \ldots$ where the $p$-th power of each generator equals zero.)ksankarWed, 29 May 2019 20:09:55 +0200https://ask.sagemath.org/question/46701/Branching to Levi Subgroups in Sagehttps://ask.sagemath.org/question/45691/branching-to-levi-subgroups-in-sage/In the Sage computer package, there useful exist tools for branching representations of a simple Lie group to a Levi subgroup. See for example the root system $branching Rules \subseteq $ combinatorics in the Sage manual
Explicitly, one is branching to subgroup corresponding to a Dynkin sub-diagram, obtained by removing a single node.
For example, we can branch from $\operatorname{SL}(n)$ to the subgroup $\operatorname{SL}(n-1)$.
However, $\operatorname{SL}(n-1)$ can be considered as "living" in the larger subgroup
$\operatorname{SL}(n-1) \times \operatorname{U}(1)$. This is true for every subgroup coming from a deleted node, i.e. one can always take the product of the subgroup with $\operatorname{U}(1)$, to obtain a larger subgroup.
How does one branch to this subgroup in Sage. For example, it is done in the LieArt program for mathematica: see A3 of the ArXiv version of Lie Art.
Is this also possible in Sage?nadiasusyWed, 06 Mar 2019 22:10:48 +0100https://ask.sagemath.org/question/45691/Using GAP package in Sagehttps://ask.sagemath.org/question/26163/using-gap-package-in-sage/I would like to use [this GAP package](http://www.gap-system.org/Manuals/pkg/wedderga/doc/chap1.html#X7DB566D5785B7DBC) to do some computations of representations, in particular [these commands](http://www.gap-system.org/Manuals/pkg/wedderga/doc/chap4.html). However, I want to use Sage and not GAP directly for various reasons (such as not having to switch back and forth between systems for where the computations come from).
An ideal "answer" to this question would give me
- Instructions for how to install this package in Sage (assuming Sage's GAP is new enough, which I think it is)
- How to get a group ring nicely with or without this package, from a given (Sage) group
- How to use `PrimitiveCentralIdempotentsByCharacterTable` from within Sage with this
- Ideally, how to apply a given representation to one of these idempotents
That's a tall order, probably, but I don't use the group theory stuff in Sage too often, so it would save me a lot of time if someone who "just knows" the syntax was able to help out. Thanks!kcrismanWed, 11 Mar 2015 23:50:53 +0100https://ask.sagemath.org/question/26163/Order of Irreducible Representation Characters wrong in Sage?https://ask.sagemath.org/question/9300/order-of-irreducible-representation-characters-wrong-in-sage/I am using Sage to calculate Irreducible Representations (IRR) of symmetry groups of graphs. Most of the time the order of the IRR characters match the order of the conjugacy class representatives. For example the trivial representation characters are the first row of the character table (1,1,1,1,...,1) and the first conjugacy class is ( ), the identity representation. But in the example below for a star graph (one central vertex, the 4 other vertices on spokes from the center) the trivial representation characters are the *last* row in the character table and the conjugacy class representatives have the identity as the *first* element.
Example (star graph).
ct=G.character_table()
print ct
[ 1 -1 1 1 -1]
[ 3 -1 -1 0 1]
[ 2 0 2 -1 0]
[ 3 1 -1 0 -1]
[ 1 1 1 1 1] <--- trivial rep. is last
cc=G.conjugacy_classes_representatives()
print cc
[(), (1,2), (1,2)(3,4), (1,2,3), (1,2,3,4)]
^ trivial class is first
Is this a bug or known issue for Sage? How can I get around it so I know the orders of the characters and representatives match up properly?
LouChaosThu, 06 Sep 2012 11:04:45 +0200https://ask.sagemath.org/question/9300/