Saturday, January 18, 2014

It's gotta be S5

I think I've got it.

The problem was to show for the general fifth-degree equation that the Galois Group of the resolvent sextic was just S5, the same as the Galois group of the quintic. I already talked about how you could generate the coefficients of the sextic by calculating the elementary symmetric polynomials of Dummits Functions. Which brought us to the question of: what is the Galois Group>

What I did is I wrote out all six Dummit functions in A,B,C,D and E (in a kind of shorthand, actually).
The "standard form" of Dummit's function is

1.  AA(BE + CD) + BB(CA + DE) + CC(DB + EA)....

which is obviously based on the rotation ABCDE. Starting from this one, there are exactly five other "rotations" which work, and these are all of them:

1.  ABCDE  (= ACEBD)
2.  ACBED  (= ABDCE)
3.  AECBD        etc.

The terms in brackets are included to show that the alternate forms generate the same Dummit Function.

Then I took all possible 5-cycles on A,B,C,D and E and examined what there effect was on the Dummit Functions. (EDIT: Let me explain this a little better. First I'll shift A to B, B to C, etc. Obviously the first one is just going to map to a rotation of itself, which means for the sake of the associated Dummit Function, it goes nowhere. But the second one, ACBED, gets transformed to BDCAE...which is the same as No. 6 (or a reverse image, which comes to the same thing. Likewise you can plot where they all go. The first "orbit" is (26435). Then I tries a different five-cycle...actually, I chose a cycle of the 2nd function, because I wanted it to map to itself. The other five mapped to each other in a second five-cycle. I've listed all six five-cycles below:


And then of course you get the images of these under self-compounding...24 5-cycles altogether. (Just like S5.) The point is they didn't mushroom out into every possible 5-cycle in S6. So let's see what happens when you start multiplying them together? At first I started getting  3-cycles...actually, direct products of 3-cycles, like these:


And then I started getting new 5-cycles. But it was easy to verify that each new one was already within the orbits of my original 6. So far so good.

I was starting to think I was getting only even permutations...that would be interesting. Could the Galois Group of the resolvent be A5? I had already 24 5-cycles, and a whole bunch of 3-cycles, so unless I had some odd permutations, it looked like I might be heading there. So I went back to Dummit's Functions and just tried swapping A and B. And I got:


That's an odd permutation, so it can't be A5. I've got 5-cycles, 3-cycles, and odd permutations...if it's not S5, then God help me. 

I know there are guys that would do a lot cleaner job of this than me, but there it is.


Balarka said...

Uh, I haven't checked through all of your calculations (I am pretty ill at this moment, having a hard time with a severe sort of cold) but I believe galois group of the resolvent sextic isn't S5.

I feel this is obvious from the fact that S5 is generated by 5-cycles whereas the galois group of the root of the sextics generate a 6-cycle.

In short, I am more or less pretty sure that the galois group in general has the form PGL(2, 5). An I also believe that such sextics wont help you to solve the quintic, unless of course is reducible and that occurs whenever the quintic has solvable galois group(!).

Marty Green said...

Well, that's what makes it so picturesque. How do you get the group S5 from the allowable permutations of the six roots of the sextic?

I think the easiest way to prove its S5 is to find a one-to-one correspondence between permutations of the roots of the Quintic and permutations of the Dummits Functions. I've been hunting as best I can, and every permutation of this one seems to give me a permutation of that one, without duplicates (not that I've found so far at least.)

And I'm thinking there CAN'T be a six-cycle in the Galois Group of the resolvent: because if every permutation of Dummit's Functions has to come from a permutation of ABCD and E, so how are you going to get a six-cycle out of that?

Other than that, I'm sorry to hear you've been sick. Please get well; I notice (from the fact that we're 12 hours apart) that you've been staying up way past your bedtime to argue with me ;^) and I want to assure you that I'm not going away, so again, please get well. Also, I noticed you flagged my Casimir Effect article in your mymathforum discussion, which got me about a dozen hits. Much appreciated...

Balarka said...

Ah, I get you point. Yes, indeed my analysis does no hold at this point -- man, that was stupid! ;D.

I have been too ill (had a mild fever last night, etc.) draw any plausible argument to show that it isn't S5, but I think I can find reference. See where it agrees with my statement that it has galois group $\text{PGL}(2, 5)$. In fact, it even says that there are no sextic with Galois group S5 though I haven't been able to understand there arguments (for reasons I have given above).

As for the hits, I think there's another reason. I have also mentioned your article on MSE chat.

Marty Green said...

Oddly enough, we may both be right. I don't like to argue by quoting authorities, but this Wikipedia article suggests that PGL(2,5) may be isomorphic to S5:

It's also interesting because the article refers to a non-trivial embedding of S5 in S6, not just one that fixes an arbitrary element. I think that's the same thing I said.

I did look up your hobbes reference, but I have to admit I'm not really capable of reading math at that level. (And I don't have a 102 degree fever to excuse myself either.) Also, I rewrote my Casimir analysis and posted it as a follow-up to this.

Get well soon :^)

Balarka said...

I learned something from you today. Indeed, I see that PGL(2, 5) is isomorphic to S5 which is usually called "exceptional isomorphism", since they arise from two completely different presentations, but still describe the same thing. So, yes, thank you very much for the reference and we are, indeed, describing the same thing and are both right.

Thinking clearly now, I have been a fool. I have read about this thing before happening with the octic resolvent of a septic instead of a sextic resolvent of quintic, so there were no reason to think this should not happen here. As an huge excuse, I want to present that here.

Consider a septic with galois group L(3, 2), which can be visualized as the permutations fixing collinearity in Fano graph (or Fano plane).

Consider a cube now with any two opposite sides marked (1, 2, 3, 4) and (8, 7, 6, 5) in clockwise order and [18] and [27] being two edges of the cube. (I would have provided a picture of what I am saying if I knew how to include that.)

If one applied D2 point symmetry on such a cuboid, then seemingly we bring back L(3, 2).

So, a big DOH and a DERP.

I see you rewrote Casimir effect page. Nicely done!

Marty Green said...

When you say "I have been a fool" you are perhaps being a little hard on yourself. But that's OK.

Glad you liked my Casimir rewrite. I'm still hoping to get you interested in some physics. But I have to admit that I still think Group Theory is the pinnacle of human acheivement. It's so full of mind-boggling cosmic truths that once you start learning about them, you just have to keep digging deeper and deeper. But that road surely leads to madness...(just kidding)...(no, not really kidding)....