#i give up trying to stst up

[woerm-town]

Goodnightt

[thousandmaths]

Dynkin Diagrams and a Proof Sketch

This is the third and final post (1 2 3) in our three-part sequence on classifying finite Coxeter groups. However, don’t despair! Curious readers can find out more about finite Coxeter groups by picking up the main sequence (1 2 3 4 5 6 7 8 9) at Post 6 :)

In the last post, we dove into the history behind Coxeter groups to understand— if not explain— about half of the weirdness in the classification. This post is devoted to the other half :)

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Cleaning up the Classification

At the end of the last post, we mentioned that Dynkin’s new idea of the “simple root” permitted a cleaner way to classify semisimple Lie algebras. It didn’t take long at all for people to apply this idea to the Coxeter systems. The result is a much more memorable rewriting of the classification theorem:

Theorem (Coxeter-Dynkin). Every finite irreducible Coxeter system corresponds to exactly one of the following diagrams.

(source (for picture))

I’ve explained how these Coxeter-Dynkin diagrams* correspond to reflection groups previously, and their relationship to Coxeter groups is very similar.

[ * I will call these Dynkin diagrams in this sequence, although this term is sometimes reserved exclusively for the related Lie theoretic objects. ]

Each dot in the diagram represents a generator; so in particular the number of dots is the rank of the system. As we noted in the last post, most $m_{s,t}$ are $2$, and most of the $m_{s,t}$ which are not $2$ are $3$. So the diagram hides all of those “boring” numbers, and only lists the unusual ones.

whenever there is no line between two dots, that means $m_{s,t}=2$.

whenever there is an unlabeled line between two dots, that means $m_{s,t}=3$.

whenever there is a labeled line between two dots, then $m_{s,t}$ is equal to the label.

You can try your hand at converting one of the Dynkin diagrams above into its $m_{s,t}$ information, perhaps for $B_4$, and see that you really do get the same thing as in the longform description in the first post.

So this really is just a “translation”: we haven’t removed any of the inherent weirdness in the result; even down to preserving the historical idiosyncrasies. But it’s certainly a lot more memorable this way! That’s why I said this was a “psychologically useful” simplification at the end of the last post. And, as it turns out, the graph-theoretic nature of the Dynkin diagram shows up in the proof in a way which is a bit awkward to translate back to the original setting! 

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What’s The Plan? A Sketch of a Proof

We can prove the classification of finite irreducible Coxeter groups by following this outline (which is not mine). Note that, with the exception of Step 1, each of these steps requires quite a bit of effort to prove.

There must exist $m_{s,t}$ for every pair of generators $s$ and $t$. Remember that for general Coxeter groups we don’t have to do this, but if we don’t, then all of the words $st, stst, ststst, stststst, ststststst,$ and so on, are all non-equivalent.

If there is a loop in the diagram, we show that there are infinitely many non-equivalent words. So we cannot have loops. Also, if the diagram is not connected, the Coxeter system is reducible. (In graph-theoretic language, the Dynkin diagram is therefore a tree.)

We note that all regular polytopes give rise to a Dynkin diagram, and we use the classification of regular polytopes. This accounts for Types A and B; also Types F, G, H, and I. After that, we show that any other valid Dynkin diagram must have a point with three lines connecting to it.

Show that if there is both such a point and any labeled edge (i.e. $m_{s,t}>3$ for any generators) then we can find infinitely many non-equivalent words. So this can’t happen, and hence outside of Types ABFGHI, all edges must be unlabelled.

Show that there is only one such point (as in Step 3); the remainder of the proof then focuses on this point.

If $a,b,$ and $c$ are the lengths of the “paths” that start at this point (including the point itself), show that $\frac1a+\frac1b+\frac1c > 1$. This excludes every conceivable diagram that isn’t on the list we gave above.

Finally, we need an explicit construction to show that Types D and E are in fact finite.

We can now answer the questions we had earlier about why the patterns in the exceptional types EFGH (and I) seem to just... stop.

Step 3 explains why the Types F and H patterns do not continue:

$F_4$ corresponds to the $24$-cell, an object unique to four-dimensional geometry,

$H_2, H_3, H_4$ correspond to the pentagon, dodecahedron, and the 120-cell, which do not have analogues in dimensions five or higher.

and Step 6 explains why the Type E pattern does not continue.

In the language of the proof, the Type E pattern is that $a=2$ and $b=3$.

But basic algebra says that $\frac12+\frac13+\frac1c > 1$ is only true when $c<6$, and so the length of the long leg can be at most $5$.

Steps 3 and 4 together also explain why $m_{s,t}$ is so commonly $2$ or $3$: if the diagram’s shape is interesting at all, then it can’t be anything else, and if it’s just a straight line then you have very strong restrictions on the labels.

Of course, you can certainly ask for more details. The formula used in Step 6 comes from evaluating a certain determinant, and it’s probably not very productive to see that derivation laid out in detail. But I would like to make an extended comment on the geometric theorem used in Step 3.

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Geometry Again: Classifying Regular Polytopes

A polytope is a high-dimensional generalization of polygons and polyhedra. Regularity is a bit trickier to define, but it comes from the intuitive idea of “looks the same everywhere”:

A polygon is regular if all of its sides have the same length and all of the edges make the same angle at each vertex.

A polyhedron is regular if all of its faces are regular polygons, all of its edges are the same length, and the same number of faces meet at each vertex.

In two dimensions, we know there are infinitely many polygons. But Greek mathematicians famously proved that there are only five regular polyhedra. Their proof goes like this: 

If you want the polygons to “close up”, then they had better meet at an angle which is less than 360º. If they meet at exactly 360º, then they tile the plane, so the shape never closes. And if they meet at more, then they actually get further away from each other; that’s even worse!

You have to have at least three polygons meet at each vertex, otherwise the shape collapses to be two-dimensional.

Now do some arithmetic:

Triangles have angle 60º, and to get $60n<360$ you can put at most 5 of those at each vertex.

Squares have angle 90º, and to get $90n<360$ you can put at most 3 at each vertex.

Pentagons have angle 108º, and to get $108n<360$ you can put at most 3 at each vertex.

Hexagons have angle 120º, so you can’t even put 3 at each vertex.

As you add more sides, the angle always increases, so you can’t use anything with more sides than a pentagon.

This gives you five possible options for how vertices look: either 3, 4, or 5 triangles, or 3 squares, or 3 pentagons.

Finally, check that all of these are actually possible.

(source)

The proof in higher dimensions is more technical, but the argument has the same spirit. In 4D, we get analogues of all five polyhedra, plus the 24-cell, which, as I suggested above, is a rather weird shape that doesn’t translate well to our 3D intuition. 

(source; the 24-cell is the yellow one)

You would think that as dimensions increase far beyond our ordinary experience, these sorts of unintuitive creatures would start popping up all over the place. But amazingly, the equations in dimension 5 and above suddenly become extremely restrictive. In the end, the only things that survive in high dimension are the analogues of the cube, the tetrahedron, and the octahedron.

These geometric restrictions that ultimately descend all the way down to the level of Dynkin diagrams. Therefore, the reason that the exceptional FGH patterns only exist for small rank is because the exceptional polytopes only exist in small dimension.

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[evexe-n]

Depraved Anon: The avid Sparkle and Glittershipper in me MUST know -- do you ship Kisara with anyone?

Late reply curetesy of me not checking my messages yesterday woops

(Also assume I always talk from a fic-reading/writing standpoint, cute fanart needs basically no context for me to like it haha)

Why did this get so long….

Anyway, tough question. Basically, it’s the same for Kisara as with most other characters: if you give me a believable enough scenario, I’ll buy any ship. The only ship with her that I remember liking since way back is the obvious Set/Kisara, and even then it has to be written a specific way, namely, pls give our girl some proper character bc that always felt a bit lacking in canon. I love her, but I need a deeper understanding of motivations etc in order to care in fic-verse.

Gotta admit, I haven’t seen much people ship her with Anzu before, but that could be cute. I mean I have no idea how they’d meet but there are ways around that and you can make up whatever anyway (thanks dso.d) …Anzu would probably be the first to try and become friends because lord knows she (and Kisara too probably) hangs out with way too many guys, who just don’t understand some things she likes. A slow-burn slice of life type fic would work best for these two I guess, damn, that’d actually…..be adorable, wow.

With Mana, I mean, I assume you’re talking about aged-up Mana? Because I doubt 13yo Mana would grow any romantical interest in someone beyond those crushes you get on random rly attractive ppl… That said, 13yo Mana having a one-sided crush on this one really pretty white haired lady she’s seen around the palace would also make for some A+ Cute.

Again, I could see a mutual thing working in an aged-up scenario though, or possibly in ‘oh hey everyone came back from the dead’ scenario, because age is an iffy concept there for the ones that survived Zorc (read: Set, Mana,…) and would have grown up, died, end then just ended up in their younger bodies in the afterlife, but still mentally be more adult.

This reply accidentally turned into me trying to work out how I’d ship those two ships so I guess that’s some form of reply on its own, but to summarize:

tl;dr - I don’t actively ship Kisara with anyone, but depending on how you present it to me, I can ship her with a lot of ppl. From the two mentioned, I think I (surprisingly enough) prefer Sparkleshipping, but they’re both ships that I can easily see working if you give it some proper build-up ♡

+ a somewhat PS: I think I made a joke once about Stepping Stones ‘verse going ‘Can you imagine Kaiba’s reaction if Joey got a crush on Kisara or something?’ and purely for crack!fic value alone I found this worth mentioning. I mean StSt!Kisara is aroace so I doubt I’ll do a proper ship with her there, but hey, it’s my fic I can make up crack scenario’s if I want.

Thanks for dropping by again old buddy old depraved pal✨✨