Martin-Seysen / mmgroup

Python implementation of the monster group
38 stars 4 forks source link

computing with Y-presentation for M? #8

Closed dimpase closed 1 year ago

dimpase commented 2 years ago

I wonder whether your package may be used for this. It might be useful for extending work I did in https://doi.org/10.1016/j.aim.2017.03.029

Martin-Seysen commented 2 years ago

I just had a quick glance at your paper. Do you need generators of the Y_443 presentation of the Monster in terms of the generators in this project, (and, may be, also vice versa)? At the moment I have no idea how to compute such generators. I think one has to embed a large portion of the Y_443 presentation into a more accessible subgroup of the Monster.

Here is a quick overview of what I can do in the project:

dimpase commented 2 years ago

Do you need generators of the Y_443 presentation of the Monster in terms of the generators in this project, (and, may be, also vice versa)?

yes. Various subgroups of Y_443 are described on p.233 of Atlas of Finite Groups. What generators are used by the project?

Martin-Seysen commented 2 years ago

I stay very close to Conway's construction of the Monster. The essential information about my generators is given here:

https://mmgroup.readthedocs.io/en/stable/api.html#module-mmgroup.mm_group

A kind of rewrite of Conway's paper explaining the choice of my generators is given here:

https://arxiv.org/abs/2002.10921

Martin-Seysen commented 2 years ago

The follwing paper may be helpful:

https://reader.elsevier.com/reader/sd/pii/S0021869313000410?token=12895C2E00FA8F012A3EF12801CC716D8A74366827C9C20B5DD07642F76C8FEFAD024EC64197D1E57954E0930A5803C2&originRegion=eu-west-1&originCreation=20220926203051

dimpase commented 2 years ago

Thanks, this looks useful.

Martin-Seysen commented 1 year ago

There is now a (yet poorly documented) application 'Y555' in the source distribution that maps the Coxeter group Y555 to the BiMonster. I could check e.g. the spider relation in Y555. I hope I can finish the documentation within this year. Ideas for the implementation are taken from [Nor02] and [Far12], with citations as at the end of the document

https://mmgroup.readthedocs.io/en/latest/api.html

Is there still any interest in porting the project to GAP or SAGE?

dimpase commented 1 year ago

Nice - at the moment I'd rather think how to use your code in research (too much Sage and GAP for me now :-)) But yes, inclusion into Sage should not be too hard.

dimpase commented 1 year ago

You might probably simplify dealing with the projective plane of order 3, usually denoted $PG(2,3)$, by using SymPy permutation groups

dimpase commented 1 year ago

If I have some words in the 26 generators, can your code verify a supplied relation between them - as a relation in the subgroup of $(\mathbb{M}\times\mathbb{M}).2$ they generate?

Martin-Seysen commented 1 year ago

Yes, sure!

For verifying e.g. the famous spider relation you just have to code:

from bimm import BiMM, P3_BiMM
spider = ['a', 'b1', 'c1', 'a', 'b2', 'c2', 'a', 'b3', 'c3'] * 10  # ATLAS notation for the Y_555 generators
# An entry in that list may be anything that is accepted 
# by the constructor of class P3_node in module inc_p3.py.
assert P3_BiMM(spider) == BiMM(1)

Remark: At the moment you have to compile the source distribution, install it, and copy the python scripts from subdirectory applications/Y555 to your working directory. All this is yet experimental. In a future version class BiMM and function P3_BiMM may be part of the distribution.

dimpase commented 1 year ago

As far as I understand, any automorphism $\phi$ of the Y555 diagram lift to an automorphism $\Phi$ of PG(2,3). For $\Phi$, one can take, for each of its orbits on the 26 elements, product of the corresponding generators of Bimonster. I wonder whether these products generate the same subgroup of Bimonster as these that lie on the Y555 diagram. E.g. if $\phi^2=1$ then it fixes one "leg" of Y555, so there are 6 length 1 orbits of $\Phi$ there, and 5 length 2 orbits. As every generator outside of Y555 in PG(2,3) is redundant, it's natural to conjecture that the products over all the orbits are like this, too.

Martin-Seysen commented 1 year ago

There is a natural embedding of $\phi$ and of its lift $\Phi$ into the subgroup $\mathbb{M} \times \mathbb{M}$ of the Bimonster. If $\phi$ is the leg-swapping automorphism (with $\phi^2=1$) then the image of $\phi$ in $\mathbb{M} \times \mathbb{M}$ is a pair of 2A involutions in the diagonal of in $\mathbb{M} \times \mathbb{M}$.

The products of the generators in the orbits of $\phi$ centralize the image of $\phi$ in $\mathbb{M} \times \mathbb{M}$. So your question can probably be answered by a calculation in the Baby Monster (which is the cenralizer of a 2A involution).

At present my computation of the image of $\phi$ contains a few unpleasent technicalities which will disappear after integrating my class BiMM (modelling the Bimonster) and its friends into the main project.

dimpase commented 1 year ago

the image of $\phi$ ... is a pair of 2A involutions...

I don't understand how the image can be a pair. Where is the embedding you mentioned described?

Is it known that the products of the generators in the orbits of $\phi$ generate BM? (or, maybe, something like BMxBM ?)

Martin-Seysen commented 1 year ago

The image of $\phi$ in $\mathbb{M} \times \mathbb{M}$ is $\imath \times \imath$ for some 2A involution $\imath \in \mathbb{M}$. Norton [Nor02] gives a presentation for $\mathbb{M} \times \mathbb{M} \times L_3(3)$, where the generators are derived from the 26 nodes of the projective plane $P3$ and from the automorphism group $L_3(3)$ of $P3$ . The relation in eq. (8) in [Nor02] kills the factor $L_3(3)$; thus embedding the automorphism group $L_3(3)$ into $\mathbb{M} \times \mathbb{M}$.

I dont't know which group is generated by the products of the generators in the orbits of $\phi$. One can explore this group e.g. by calculating orders of many elements in that group. One may also project elements of even length to one of the copies of $\mathbb{M}$ in $\mathbb{M} \times \mathbb{M}$, and try to find standard generatrs of the Baby Monster. But this appears to be a lengthy calculation.

Martin-Seysen commented 1 year ago

I computed 1000 ramdom samples from the intersection $G$ of the subgroup $\mathbb{M} \times \mathbb{M}$ with the subgroup of the Bimonster generated by $a, b_1, c_1, d_1, e_1, f_1, b_2 b_3, c_2 c_3, d_2 d_3, e_2 e_2, f_2 f_3$. The distribution of the orders of the first factor of these elements (in $\mathbb{M} \times \mathbb{M}$) is similar to the expected distibution of the orders of the elements of 2.B, where B is the Baby Monster. See module legswap.py in directory applications.Y555. For the actual computation I have factored $G$ by the image of $\phi$ in $\mathbb{M} \times \mathbb{M}$, which is in the center of $G$.

dimpase commented 1 year ago

Nice! Can you check with the same method that $f_1$ and $f_2f_3$ are redundant, as generators of the group generated by the fixed points of the legswap involution? (which is, as far as I understand, has $G$ as its index 2 subgroup)

I understand that your sampling involved words in $a$, $b_1$,..., $f_1$, $b_2b_3$, ..., $f_2f_3$ - you probably could just filter out words involving $f_k$'s.

dimpase commented 1 year ago

One can certainly try to mimic the $PG(2,3)$ business for the legswap case; a legswap fixes 5 points and 5 lines of $PG(2,3),$ so in the new graph one will have $2\times (5+(13-5)/2)=18$ vertices, and some edges will double. You'll have a subgraph with vertices, in order as on the 1st picture below $f_1,...,b_1,a,b_2 b_3,...,f_2 f_3$:

o-o-o-o-o-o=o-o-o-o-o
  o-o-o-o-o=o-o-o-o

and the conjecture is that you'd generate all the 18 vertices by $e_1,...,b_1,a,b_2 b_3,...,e_2 e_3$ (the 2nd subgraph above). A legswap has a special non-incident vertex-line pair $(p,L)$ in $PG(2,3)$, so that all points on $L$ and all lines through $p$ are fixed - I'm not sure what role, if any, they should play in the new graph.

But before going this route, it would be nice to confirm that the intuition about the redundancy is right, by a computation like I suggested above.

Martin-Seysen commented 1 year ago

The new version of legswap.py confirms the intuition that generators $f_1$ and $f_2 f_3$ are redundant. It computes the distributions of the orders of both factors of the subgroup $G$ of $\mathbb{M} \times \mathbb{M}$ (generated by $a, b_1, c_1, d_1, e_1, b_2 b_3, \ldots, e_2 e_3$) from 1000 samples. After factoring out the center of order $2$ of $2.B$, both distributions resemble the distribution of the orders of the elements of the Baby Monster $B$. The script legswap.py also displays some orders of pairs $g_1, g_2$, where $g = (g_1, g_2) \in G \subset \mathbb{M} \times \mathbb{M}$. This supports the intuition that $G$ is equal to $2.B \times 2.B$, i.e $G$ is not contained in some kind of diagonal of $2.B \times 2.B$.

Martin-Seysen commented 1 year ago

The latest release mmgroup 0.0.12 contains functions for mapping the Coxeter group Y_555 onto the Bimonster in module mmgroup.bimm. For documentation see https://mmgroup.readthedocs.io/en/latest/api.html#module-mmgroup.bimm.readme

So I think we can close this issue now.

dimpase commented 1 year ago

OK, I'll open another issue if I need help with these, or/and to work on more related functionality.

dimpase commented 1 year ago

I've just tested your new release on GitHub's Codespaces, all "not slow" pytest tests pass there.

Martin-Seysen commented 1 year ago

Thank you for testing my package. You are always welcome to ask me qustions about it, eitther via github or directly via e-mail.Von meinem/meiner Galaxy gesendet -------- Ursprüngliche Nachricht --------Von: Dima Pasechnik @.> Datum: 12.01.23 14:15 (GMT+01:00) An: Martin-Seysen/mmgroup @.> Cc: Martin-Seysen @.>, Comment @.> Betreff: Re: [Martin-Seysen/mmgroup] computing with Y-presentation for M? (Issue #8) Closed #8 as completed.

—Reply to this email directly, view it on GitHub, or unsubscribe.You are receiving this because you commented.Message ID: @.***>