Speed up fundamental group computations

[enriqueartal]

The computation of plane curve complements uses mapping class group computations and this operation can be very long if attacked directly. Some changes were applied in #36768, but some computations take forever without the changes in this PR.

Note. A previous version of this PR did not use the right branch

📝 Checklist

• The title is concise and informative.
• The description explains in detail what this PR is about.
• I have linked a relevant issue or discussion.
• I have created tests covering the changes.
• I have updated the documentation and checked the documentation preview.

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[enriqueartal]

What is behind this PR is the following problem. If we have an element g of FreeGroup(n) and a braid b of BraidGroup(n), obtaining g * b (mapping class group action) can be very long if the length of the words is very long. The use of the normal forms and the fact that there are explicit formulas when the braid g is a power of the Delta element, can be used to speed up the computations.
I wonder if this could be done directly on the definition of g * b, but I do not know how to do it. Maybe it is work for another PR.

[tscrim]

I don't fully understand the computation, but I'm willing to accept it on faith and passing doctests. Here's some comments based on the code though.

[tscrim]

Why not factor out elt = prod(B(m) for m in rnf[:-1]) from the for gen in F.gens(): loop? In which case, you would do here gen0 = gen0 * elt.

Actually, is there an easy way to know what rnf[-1] is without computing the entire right normal form? If so, then you could get everything you want by having elt = br * ~rnf[-1]. Although rather than the above product, simply doing elt = br * ~rnf[-1] might be faster.

[enriqueartal]

For the first question you are right. I was copying a former strategy where breaking the braid was crucial to speed the computations, it is not so clear in this case. Nevertheless, to find the relevant power of delta involved in the braid I think one must use a normal form, which is quite a fast operation.

I will add this proposed change and explain the real reasons of these changes; I am not sure if they should be added to the documentation, maybe it is a good idea.

[tscrim]

For the description, just to the PR will likely be sufficient.

[tscrim]

[Edit: GH was stupid and brought over partial version of my previous PR comment...Sorry for the noise.]

[tscrim]

For the description, just to the PR will likely be sufficient.

[enriqueartal]

I try to explain the reasons for these changes. The fundamental groups computed are the quotient of a free group by some relations. For some of them, we need to compute $g\cdot b$ where $g$ is a generator of a free group $\mathbb{F}_n$ and $b$ is a braid in $n$ strands. The action is already implemented in sage.

If the length of the word defining $b$ is too big, the computation of $g\cdot b$ can take forever. In some cases, maybe the braid is too big and there is no solution. But in some cases the following solution works. First,
there are some (positive permutation) braids $b_1,\dots,b_r$ and an integer $m\in\mathbb{Z}$ such that two different words defining the same braid produce the same $(b_1,\dots,b_r,m)$. Moreover $b=b_1\cdot\ldots\cdot b_r\cdot\Delta^m$ where $\Delta$ is a special braid; moreover the word $c:=b_1\cdot\ldots\cdot b_r$ is usually much shorter than $b$. Hence the computation of $h:=g\cdot c$ is much faster.

There are closed formulas for $x_i\cdot\Delta$, when $x_i$ is a generator of $\Delta$. Moreover,

$g\cdot(\Delta)^2=(x_1\cdot\ldots\cdot x_n)\cdot g (x_1\cdot\ldots\cdot x_n)^{-1}.$

If $r=r_1 + 2 r_2$, $r_1=0,1$, then,

$g\cdot b = c\cdot \Delta^r = (c\cdot \Delta^{r_1})\cdot(\Delta^2)^{r_2}$

and the closed formulas are much faster than running along the words defined by the powers of $\Delta$.

[enriqueartal]

Some mistakes produced errors in the text. In particular when the right normal form has only a power of the delta element of the braid group, the product of an empty list gave one instead of a braid. On the other side a simplification of the code concerning cnja and cnjb gave a wrong result. The tests passed locally. Waiting to check if they pass here.

[tscrim]

Thank you for the details. Seems like the tests pass here. So I should finish the review then?

[enriqueartal]

For me it is OK

[tscrim]

Then positive review.