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Word of the Day: Kapuris 
About Those Gaussian Envelopes 


Posted by Marty Green at 11:53 AM No comments: 
The Casimir Effect: Ramanujan Revisited

The topic of the internet video is the series

Posted by Marty Green at 2:16 PM 4 comments: 
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.
Posted by Marty Green at 9:15 PM 6 comments: The Galois Group of the Resolvent

Okay, I thought I was done but I'm not. We generated a sixth-degree equation from the general quintic. People say "that's why you can't solve the quintic...because the resolvent you generate is of higher order than the equation you started with."

That's not the reason you can't solve the quintic. There are solvable sextic equations...maybe our resolvent is one of them? The question is: what is the Galois Group of the resolvent sextic?

I'm thinking it has to be S5. Because there are absolutely no permutations of the roots of the sextic that aren't generated by permutations of the roots of the quintic. And furthermore...every single permutation of the roots of the quintic affects some of the roots of the sextic. Does that make it a one-to-one correspondence?

Because if it doesn't....if there are two different elements of S5 that give the same permutation of the resolvents...then the order of the Galois group is less than 120...which makes it 60 or less. And unless it's A5, then all bets are off...you may indeed have a solvable sextic. So the problem is to show explicitly just what is the Galois Group of the sextic. I'm thinking it has to be S5. But I'm having trouble constructing it.

The Galois Group in question comes from the reshuffling of the six resolvents functions, so it if obviously at most a subgroup of S6. And it's easy to find S5 as a subgroup of S6...just take one letter and hold it fixed while the others get shuffled around. But that's not what the resolvent functions do. They all move under permutation, and they do so in a way that is somehow...symmetric.

Furthermore, there are 5-cycles in the Galois group...permutations that, when repeated 5 times, take us back to where we started. There is only one way to make a five-cycle is S6...hold one element fixed and let the others move. So without loss of generality, we must have

(1 2 3 4 5)

as a permutation, and also:

(2 3 4 5 6)
(1 3 4 5 6)

....except we can't pick and choose all of them willy-nilly, because there are 120 5-cycles of this kind in S6 and that's just too many! We need exactly 24, because that's how many there are in S5...so we have to pick them carefully. It's not that easy.

So that's what I'm working on: to find (and explicitly describe!) the group of permutations of six letters which is isomorphic to S5, and show that this is the Galois Group of the resolvent sextic.
Posted by Marty Green at 5:50 PM No comments: How to Generate Dummit's Resolvents

I said I was going to give this topic a rest, but it turns out I've got one more thing to say. Last time, in the comments, I said you could generate Dummit's Resolvents for a fifth-degree equation by taking a Lagrange Resolvent to the fifth power, then taking the other three Lagrange Resolvents you get from the first one by permuting the fifth roots of unity, taking the fifth power of those and adding all the resulting 480 terms togethers. You collect all the terms symmetric and throw them away, and what's left is Dummit's Resolvent.

And it almost works...but not quite. Lagrange's Resolvent is linear in the roots, so all the terms you get by taking fifth powers are fifth-order in the roots. But Dummit's Resolvents are made up of terms that are fourth-order in the roots:

AA(BE + CD) + BB(CA + DE).... etc.

And so my method can't work exactly. But it's close. The leftover terms when you follow my system look a lot like Dummit's:

AAA(BE + CD) + BBB(CA + DE)...etc; and also,

A(BBEE + CCDD) + B(CCAA + DDEE)....etc.

Since for the sake of Galois Theory we are only interested in the permutation properties of the resolvents, we don't need to keep the whole term...obviously the terms in A^cubed etc. are going to have permute the same way as Dummit's terms in A-squared. (But differently if you reduced it to linear in A, because you need to distinguish the terms outside the brackets from the ones inside!)

It's not even that hard to see why Dummit's resolvents "work", if you think about it. It's an awful lot of work to take the fifth power of all those Lagrange Resolvents and add them together: but if you take a step backwards and look at the big picture, you can see there are only a handful of different forms that the resulting terms can take:

AAAAA + BBBBB + CCCCC....etc; (five terms of this type)

AAAAB + BBBBC....etc; (twenty terms of this type)

AAABB etc. (twenty terms)

AAABC etc. (thirty terms)*

AABBC etc. (thirty terms)*

AABCD etc. (twenty terms)

There are only six different types of terms that you get; and only the fourth and fifth (I've marked them with an asterik) survive the symmetrization. And the reason is kind of funny. The terms in AAAAA etc. are obviously symmetric, so they add up to an integer value which you can throw away. The terms in AAAAB etc....well, when you take the fifth powers, these all have a fifth root of unity stuck to them. (Try it if you don't believe me.) There are twenty terms...five of them have a multiplier of w, five of them w^2, etc. You know the sum of the fifth roots of unity is -1; so when you add all twenty terms together, you get:

-(AAAAB + BBBBC +.....+ CCCCA +...etc)

which is also symmetric, so you throw them away.

And the same thing happens whith the third and sixth terms. But the terms I've marked off are a little different. For them, you get all the terms with with a multiplier of a fifth roots of unity: but you also get real terms. You collect the terms with the fifth roots, and they add up to a negative term...the other terms add up positive. You get:

-(AAABC + AAACD +.... + CCCDA +...) + (AAABE + AAACD....)*

You get all thirty terms, but some of them are negative and some of them are positive....the ones I've marked with an asterik. You know how to symmetrize this! It's just

-(all thirty terms) + 2*(AAABE + AAACD)

You throw away the symmetric terms and what's left are the (modified) Dummits Resolvents.

And it's easy to make a sixth degree equation out of them. Remember, there were 120 possible Lagrange resolvents based on the order you choose the roots of the fifth: but when you take fifth powers, this reduces to 24 different ones because some of them are just rotations of others. Then I grouped them according to ones that are related via legal permutations of the roots of unity....this gave me six groups of four. Each group of four generates, by the reduction I've shown above, a bunch of symmetric term (which you throw away) plus a leftover group of ten terms which have the form of Dummit's Resolvent.

It's not hard to convince yourself that while Dummits Resolvents are not symmetric in A,B,C,D and E, their generalized sums and products certainly are. It's a bit of work but before long you can derive therfore integer values for all the elementary symmetric polynomials in the conjugate Dummit Functions.

And those are the coefficients of the sextic equation which is called the "Resolvent Equation" of your original quintic equation.
Posted by Marty Green at 5:46 AM No comments: 
Let's give Galois a rest

I don't think I'll have that much more to say about Galois theory. But I think it's been a pretty good run. I don't think there's anywhere on the Internet where you'll find a better explanation of why you can't solve the quintic than what we've posted here over the last couple of months. At least, you won't find a more understandable explanation.

It's not that I'm smarter than the other people in the game. It's that I'm playing a different game than they are. In academia, the big thing in math is to be rigorous and abstract. What they value is exactness and economy, and if you ask them why something works that way, they say: which line of the proof do you not understand? I don't think that's what math is about.

The difference is most clear when you consider my emphasis on the question: what are the functions on five letters which map to each other under any permutation of those five letters? (That's the question Balarka answered last month on stackexchange which led to our correspondence in the last several blogposts.) Mathematicians don't ask this question. In doing Galois theory, they ask about the existence of normal subgroups. I would venture to say that the great majority of them don't even realize that the significance of those normal subgroups is precisely that they are associated with my "functions that map to each other under permutation." And by the way...it's not a one-to-one association. I didn't realize it at the time, but the function Balarka gave me, "Dummit's Function", is associated with the trivial subgroup consisting of exactly one element...the identity subgroup. So for S5, the resulting quotient group is just S5 itself. What it means is you won't find Dummit's Functions if you're looking for a normal subgroup to generate them.

What makes me think mathematicians don't look at things this way? I think I'm on pretty solid ground here. I would say that the fact that Dummit's Functions weren't even discovered until 1991 (!) is pretty good evidence that this way of thinking is outside the mainstream.

The point of all this is that the unsolvability of the quintic equation makes (almost!) intuitive sense if you play my game and look for functions of five letters that map to each other under permutation...but it becomes totally opaque if you abstract away all the concrete manifestations of the group action, and restrict yourself to talking about towers of normal subgroups. Even the well-known definition of a normal subgroup makes no intuitive sense...it's a subgroup for which "every left coset is also a right coset". What are we supposed to make of that? Mathematicians do that kind of thing all the time...they define two numbers p and q as being relatively prime if "there exists a and b such that ab-pq=1". That's a definition? It's no so harmful in this instance because we all know intuitively what it means for two numbers to be relatively prime, but in group theory, when you cascade one opaque definition like this on top of another, the whole subject rapidly becomes incomprehensible.

So at the end of the day, I'm pretty satisfied that I've explained what's really going on with the fifth degree equation. There may be a few fine points I haven't nailed down, but I'm not too worried about that. If you want to know why the quintic is unsolvable, I don't think you'll find a better place to start looking for answers than this blog right here.

Having gotten that out of the way, I think I'm going to return to my previous order of business, where I was posting my last-year's articles from the Jewish Post. I have to admit that my physics audience is not necessarily taking to this material with enthusiasm, but it's kind of important for me to get them out there. So let's see what's next...

Posted by Marty Green at 11:45 AM 4 comments: 
Galois Group C5: Constructing a Polynomial

A2 = A1^2 -n

So what about the fifth degree?

Posted by Marty Green at 11:37 AM No comments: 
Excel To The Rescue

We've been having some fun with cubic equations. The other day we figured out that the relations A^2-4=B, when applied transitively three times in a row, generates the roots of a cubic equation. The trick is that you have to find a starting point whereby, after three iterations, you get back to where you've started. You can see what is happening on this graph: You can see that starting with A=1.6, we generate B= -1.4 and C= -2.4. Let's look at the elementary symmetric polynomials in A, B and C to see what our cubic equation should look like:

A + B + C = -2.2
AB + BC + CA = -2.72
ABC = 5.376

Now, if we remember what Balarka did the other day, he generated an octic equation by substituing the recursive relationship into itself three times, factored out the trivial quadratic, and then factored the resulting sextic into two terms, one of which was:

x^3 + 2x^2 -3x +5

This is really close to the equations you get from my symmetric polynomials in A, B and C....considering I just drew those lines on the graph by eyeball (no cheating), I'm a bit amazed how well it worked. (Remember the signs alternate when you sub in the coefficients...I told Balarka he had a sign error yesterday, but I was wrong.)

It was actually quite a surprise when Balarka's octic factored cleanly. We had started with the recursive relationship A^2 - 2 = B....going through the same steps, generating the octic, factoring out the quadratic, and factoring the resulting sextic...well, we knew that one had to factor because there was already a known factor that was generated from combining the seventh roots of unity. But we didn't actually expect the arbitrary relationship A^2-4 to also factor.

So what about A^2-6? Well, I could have tried graphing it, but then I had a better idea...why not program the relationship into Excel? It's a super-easy spreadsheet calculation...you literally need only five or six cells. You just keep iterating your seed value until you get the recursion working. And of course, with instant correction, you can easily work it out to four or five decimal places. It's kind of fun.

I tried it first on A^2-4, and verified that the cycle did in fact exist...in fact, there were (of course) two cycles, corresponding to the two cubics that we factored out of the sextic. But how could we tell that these were the roots of a cubic equation in integer coefficients? That was easy! I just got Excel to display the values of the sums and products of the roots. At the sime time as the cyclic relationship converged back on itself, you could see the sums and products zeroing in on integer values. We had not only found the roots but also the cubic equation which they came from?

So what about A^2 - 6? (For some reason it seemed reasonable to stick to even values.) This one was different. I found the following triplets with cyclic relationships:

(1.7425, -2.9636, 2.7827)
(2.0547, -1.7782, -2.8380)

but checking the elementary symmetric combinations of these triplets, it was clear that they were not the roots of cubic equations with integer coefficients:

(1.56, -8.56, -14.37)
(-2.56, -4.44, 10.37)

(These triplets came up as integers when we worked with A^2-4). But were the coefficients of the two cubic equations at least something like the roots of a quadratic? Indeed they were. Examining the pair of triplets listed above, you can see by taking sums and products that the coefficients are related as follows:

The x^2 coefficients are given by the relationship x^2 + x - 4 = 0
The x coefficients are given by the relationship x^2 + 13x +38 = 0
The constant coefficients are given by x^2 +4x - 149 = 0

And that's about all I've got to say for now...

Posted by Marty Green at 8:36 AM 4 comments: 
Cubic Equations With Galois Group C3

The other day I identified a cubic equation whose Galois Group, I claimed, was C3. Here is the equation:

I promised to show you that the roots of this equation had a cyclic relationship. Let's remember how we got them. We started with the seventh roots of unity, and generated this cubic equation by letting its roots be

A = w^1 + w^6
B = w^2 + w^5
C = w^3 + w^4

It's very easy to see from these relationships that A^2 = B + 2, and similar equations relate B to C and C back to A. Those are the cyclic relationships.

You can also notice that the splitting field of the original equation is just the Q(A), or the rationals extended by appending A. Or B, or C...you just have to append one of the roots, and the resulting field contains all three. That's different from a typical cubic where the three extensions of Q are quite disjoint. It's obvious if you have two imaginary roots and one real root, the field you get by adjoing the real root is not going to include the comples roots. So no simple equation can link one of the roots to the next, as we obtain in the present case. In short, for a cubic to have a Galois group which is cyclic, it pretty much means that all three roots live in the same simple extension, and hence functions of each other.

Does this give us a recipe for constructing cyclic cubics? In the present case, we had the relationship A^2-2=B. What if we say we're going to look for a field where, for example, A^2-4=B? Will we generate another cubic equation with a cyclic relation among the roots?

I'm not totally sure this will work. A funny thing happens if we try...but let's first try with the case where we know it has to work...with A^2 - 2 = B. If the roots have a cyclic relationship, then we ought to also have:

B^2 - 2 = C; and,

C^2 - 2 = A

But this brings us back full circle...we can substitute one equation into another, and eleminate B and C, giving us:

A^8 - 8A^6 + 20A^4 - 16A^2 -A + 2 = 0

It's not a cubic at all...it's an eight-degree equation. But wait...maybe that makes sense. After all, one "trivial" solution to our cyclic relationship between A, B and C is simply that

A^2-2=A (because A = B = C)

So this should theoretically factor out of our eight-degree equation...and indeed it does, leaving us the sixth-degree equation:

A^6 + A^5 - 5A^4 - 3A^3 + 7A^2 + A - 1 = 0

Still not the cubic we're looking for...or is it? Maybe the cubic we already found is actually buried inside this sixth-degree equation. And in fact, we find that it is...on synthetic division, we find that it does indeed go factor through, leaving the residue:

A^3 - 3A + 1 = 0

And there it is...a whole new cubic, generating a different set of A's, B's and C's than the ones we already had, but with the same cyclic relationship amongst themselves. It's not obvious to me that these new ABC's are especially connected to the seventh roots of unity, but I stand to be corrected. For now, all I can say is...there they are. (I actually found them by approximation....it's not to hard to verify that the numbers (-1.9, 0.35, 1.8) satisfy not only the cubic equation written above, but also the cyclic relationship wherewith they were generated.

So can you start with any similar cyclic relationship and generate a pair of cubic equations with cyclic roots? Sadly, it does not appear so easy. I tried starting with A^2 - 4 = B, generated an eight-degree equation, factored out the quadratic giving me the sixth-degree, but then....I couldn't see any way to factor the sixth degree into two cubics.

And that's about as far as I'm likely to get down this road in the immediate future. Unless anyone out there has any better ideas...

Posted by Marty Green at 2:49 PM 3 comments: