room topic changed to Discussion for "The duad-syntheme-total construction for higher values of n": Chatjax: math.ucla.edu/~robjohn/math/mathjax.html [combinatorics] [group-theory]

I think my argument for $\Bbb F_2^3$ generalizes to all $n=2^k$. So it seems odd $n$ are at the opposite extreme in terms of tractibility, with n=3 being the first case and n=5 the second.

@arctictern I wouldn't be surprised if you're right about the $2^k$, although my group theory is so rusty (and wasn't great to begin with) I'll have to really think about it and look stuff up (for me; I'm sure you'll do just fine getting it all sorted out).

I can't imagine the action not being transitive though, to be honest. And in either case, $t(5) \ge 10!/4$, which is just staggeringly huge, proportionally. I'd like to try and eek out $t(6)$, still, although it'll take quite a while (11*12 times as long as $n=5$, which took two to three hours, unless I figure out a way to make it more efficient, but I have no idea how at the moment). I'm curious what'll happen when $n$ is neither odd, nor a power of 2, if it'll be somewhere between.

The total for n=4 (associated to the regular action of F2^3 on itself) has stabilizer S=Aff(2,3), which seems to act 2-transitively on the total. The pairs of synthemes give rise to pairs of 4-cycles. Contrast this with n=3, in which the stabilizer S6 acts 2-transitively again (actually 6-transitively) and pairs of synthemes determine a 6-cycle.

If we think of synthemes as subgraphs of the complete graph on 2n vertices, it'd be interesting to see what kind of subgraphs you can by unioning two synthemes in a given total. Maybe different "isomorphism classes" of totals can give rise to different pair unions.

This is also a way to investigate if S8 does indeed act transitively: see if there is any synthematic total containing the synthemes (12)(34)(56)(78) and (23)(45)(67)(81). Two synthemes' union must either be a pair of 4-cycle graphs or an 8-cycle graph; the first case is covered by the F2^3 argument I think, but it's not obvious by hand why there isn't any total with a pair of synthemes forming an 8-cycle graph.

That is an interesting approach, and I briefly thought about graphs, but differently: I'd placed all the duads in a circle, so that a total is a decomposition of the (n-choose-2)-gon into (2n-1)-gons. But this is pretty clear-cut, I think, so not very informative; I'll definitely be looking into your approach.

I've been seeing the need to do something different for the programming, since as it stands, I have to generate a total by hand, to start the process. It wasn't bad for n=3,4 and annoying for n=5, but I'm really dragging my feet for n=6 and have been trying to find a reasonable way to generate even a single total by calling a function. If I can do that, I'll definitely be able to see if your potential 8-cycle pair of synthemes can be expanded to a total (assuming you don't beat me to it).

Either way, n=6 is the last n-value that's remotely feasible by computer search, without significant technique advances. It'll require half a billion permutations to be checked, but that's a drop in the bucket compared to looking at all 11-subsets of synthemes, or looking at partitions of duads with the right "shape"

@pjs36 I was indeed able to extend {(12)(34)(56)(78),(23)(45)(67)(81)} to a total for n=4, and it's not in the orbit of the other total (coming from reg rep of F2^3). After a nap I'll try drawing some stuff to post.