Monday, December 30, 2013

The Seventh Roots of Unity

I mentioned the other day that I had “known” for a long time that the three cube roots of two were algebraically indistinguishable. When I say I “knew” it, I mean that I had read somewhere that the Galois Group of x^3-2 was S3, the symmetric group. In other words, all permutations of the roots were allowed. This bothered me: I knew you could swap the complex roots, but how could you swap one of the complex roots with the real root? I wrote about this the other day.

This question led to another question: is it always the case that the roots of an irreducible cubic equation are algebraically indistinguishable, or is it possible for the splitting field to have a more specific structure? Specifically, can you write a cubic equation whose Galois group is C3, the cyclic group?

This turns out to be a pretty interesting question, which I was able to answer in the affirmative. I recently stumbled across some old scribbled notes of mine relating to this problem, and I think I’d like to sketch out what I did for the sake of posterity. So here goes…

You start by considering the seventh roots of unity...yes, those seven points in the complex plain located 51.4 degrees apart. We can throw away the obvious root and realize that the remaining six are the roots of the sixth-degree equation:

x^6 + x^5 + x^4 + x^3 + x^2 + x + 1 = 0.

We can take the first complex root and call it omega, and then the other roots are simply omega raised to the power of 2, 3, 4, 5 and 6. And of course the seventh power brings you back to one.

I said you start with the first complex root, but of course the beauty of algebra is that it doesn't matter which one you pick. None of them are distinguishable from any of the others. Yet despite this indistinguishability, there is an order to the roots which flows from their cyclic relationship. It's all very suttle.

What we're going to do is take combinations of these roots to create the roots of a cubic equation. And we'll see that those roots will inherit some of the cyclic nature of the omegas which begat them. You should be able to guess where we're going from here...we're just going to add together complex conjuates, so the (hoped-for) roots of the cubic are:

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

And I'm claiming that A, B and C are the roots of a cubic equation (with integer coefficients). It's not hard to see why this has to be so. You know that the coefficients of any equation are generated by the elementary symmetric polynomials in the roots:

A + B + C
AB + BC + CA
ABC

How can we caluculate these? It looks like it's going to be a fair bit of work, but it turns out to be a piece of cake. The sum of the roots is the most obvious...the sum of A, B and C is just the same as the sum of the omegas, which is just -1. The other sums and products are only slightly harder...because the products of expressions in powers of omega are themselves just powers of omega, and because of the near-symmetries in the A's B's and C's, they collapse with little difficulty. It's not hard to show that A, B and C satisfy the equation:

x^3 +x^2 -2x -1 = 0

Now all we have to do is show that there is a cyclic relationship between the roots...in other words, if you replace A with B, then you have to replace B with C and C with A. Let's leave that for tomorrow...

Automorphisms of the Splitting Field

I learned something over the last few days that has made a big difference in the way I understand Galois theory. I told you the other day that my friend in West Bengal had shown me how to generate functions of the five roots of the quintic which were themselves solutions of a sixth-degree equation, the so-called "resolvent". (It seems the term "resolvent" is used to describe both the functions of the roots and the resulting equation.) In the course of our follow-up discussions, the question arose as to the Galois Group of x^4-2=0....that is, the allowable permutations of the fourth roots of two.

The Galois Group describes the set of automorphisms of the splitting field of a polynomial, and it is well know that any automorphism must map roots of the polynomial to each other. So people normally describe the automorphisms in terms of where they send the roots. My correspondent made the observation that in the present case, the mappings seemed to correspond to flips and rotations of the four points on the circle whose radius was the fourth root of two...in other words, the dihedral group of four elements, or D4.

What I realized is that while all this is technically correct if we just restrict ourselves to looking at the fourth roots of two, it gives a very deceptive picture of what the field automorphisms really look like. Mathematicians are always overly proud of economy of analysis, so having proven that an automorphisms is fully defined by its action on the roots, they are happy to ignore everything else that is going on. And in this case, there is indeed much more going on.

It is true that the splitting field of the polynomial is generated by the fourth roots of two, which are located 90 degrees apart on the circle in the complex plane of that radius. But this same field also includes two more bigger circles that will be of great interest...a second circle containing the fourth roots of four (including the ordinary square roots of two), and third circle containing the fourth roots of eight.

The funny thing about automorphisms of the field is that if you describe them in geometric terms, like flips and rotations, it turns out these three rings behave very differently. The automorphism which rotates the inner ring clockwise simultaneously rotates the outer ring counterclockwise! And the middle ring gets flipped horizontally and vertically. In other words, the complex plane is not smoothly transformed, but rather it is completely scrambled...as it must be. That is the nature of pure algebra...the proximity between two points has absolutely no significance. How can it if rational points are always mapped to each other, while irrational points are swapped about?

So I worked out all the automorphisms of the fourth roots of two, keeping track of all three "rings" of roots instead of focussing just on the inner ring.  It completely changed my perspective on the nature of automorphisms. For example, I have always known that the cube roots of two are theoretically indistinguishable in Galois theory. Obviously the two complex roots can be swapped with each other...I had no problem with that. But why are you allowed to pick one of those complex roots and swap it with the real root? Doesn't that just seem wrong? And the answer is...not when you simultaneously look at what the cube roots of four  are doing at the same time. It turns out that while the real cube root of two is being swapped with its positive complex conjugate, the cube root of four is being swapped with its negative conjugate. So the overall symmetry of the complex plane is somehow restored.

When I wrote last year about the Galois Group of {x^5-2 = 0}, I laboriously tracked the logic of where the roots had go assuming you started by rotating them counter-clockwise. What I didn't realize was the bigger, much more beautiful picture. There are actually four concentric rings worth looking at...the fifth roots of two, four, eight, and sixteen. The automorphism which rotates the inner ring counter-clockwise simultaneously rotates the next ring counter-clockwise by two notches, and so forth, so the outer ring is actually rotated clockwise. There is a complete symmetry! It gets much more intricate when you start considering the allowable permutations between the fifth roots of unity...when you mix these in you ultimately get what is called the Frobenius Group, a set of twenty automorphisms in all. But all those re-shufflings, which appear so arbitrary and downright cock-eyed when you only look at the fifth roots of two, become so much more aesthetically satisfying when you consider what all four rings are doing simultaneously.

Thursday, December 26, 2013

In Which I Get Help From an Unexpected Source

I learned something new about the fifth-degree equation over the last few days. The other day I told you that the reason you can't solve the quintic has something to do with the fact that there exists no set of functions on five letters which "map to themselves under all permutations". Or something like that. Well, it turns out there is...only there isn't. Here's what I learned.

I had constructed these three functions on four letters (corresponding to the four roots of a quartic equation:)

AB + CD = p

AC + BD = q

AD + BC = r

We can see that no matter how you re-shuffle A, B, C and D, you get back the same three functions p, q and r. I said that the reason you can't solve the quintic was that there is no comparable set of four functions on five letters (A, B, C, D and E) which is similarly preserved on taking permutations of letters.

And that's true. But what I didn't know is that you can construct a set of six functions on those five letters which have the desired property...sort of.

I had almost concluded that there were no such functions, and I posted a question on stackexchange.com here to see if I was right. I wasn't. One Balarka Sen pointed out that there were a number of such functions, including one called Dummit's Resolvent, which seems to have been discovered only twenty years ago by a math professor from the U of Vermont. Dummit's functions look something like this:

AA(BE + CD) + BB(CA + DE) + CC(DB + EA) + DD(EC + AB) + EE(AD + BC)

There are actually six of these functions...you can generate the other five by suitable permutations of the A's, B's and C's, as Sen explains in his answer to my question on stackexchange.

Now, what people say next is a little misleading: they say that these six functions generate a sixth-degree equation, and therefore they're not going to help you solved a fifth-degree. That's not quite true. It might be a sixth-degree equation of a simpler form than the fifth-degree which you started from. It's certainly not a sixth-degree equation of the most general type, because the permutations of the roots cannot be more complicated than the permutations of the five letters which generated them. In other words, the Galois Group of the sixth degree equation cannot be bigger than S5, the permutation group on five elements. But could it be a simpler  Galois Group? Maybe it could...

But here's where group theory actually gives us some guidance. I said that Dummet's functions are similar by analogy to my p, q and r which gave me a third-degree resolvent to the general fourth-degree equation. But in one crucial respect they fail to do what p, q and r do. It's true that unlike my p, q and r, Dummits functions are more numerous than the five roots that generated them. That's a problem, but it's not decisive. The real problem is that there are no (non-trivial) permutations on the five letters which leave all six of Dummit's functions in place.

I'm going to leave it for another day to tell you why I consider this circumstance to be critical. Suffice it to say (for now) that there are indeed permutations of A, B, C and D which leave my p, q, and r in place, and the set of all such permutations forms what they call a normal subgroup of S4. But before we get there, I have to tell you one more thing. In one way or another, I've been working on this problem off and on for most of forty years. It's a great problem and I don't think it gets its due in the undergraduate curriculum....I've talked about that previously. So when I posted the question the other day on stackexchange, asking about the existence of these functions, I wasn't sure what I'd get, and I was happy to get a pretty clear answer from that fellow Sen. Now here's the kicker...it turns out that out Balarka Sen is a thirteen-year-old kid from India.

Thirteen years old????

Sunday, December 15, 2013

Towers of Normal Subgroups

When we left off last week, I told you that I thought my way of understanding the quintic equation was different from what you'll find anywhere else. Today I want to elaborate on that.

When I took third-year algebra many years ago, the proof of the unsolvability of the quintic came about three-quarters of the way through the course, and it was just another corollory of a lemma in a long chain of proofs and lemmas that began with something about a "tower of normal subgroups" that had no physical or pictorial motivation. Later, when I asked the prof why you couldn't solve the quintic, she asked me: which line of the proof did you not understand?

That's not what math is to me. When I "understand" something in math, it's not because I've examined the proof line by line and verified that there are no flaws...it means on some level I actually understand it. I tried to explain this to the prof, but she just waved me off dismissively, as though my idea of understanding was just kid's stuff...that real mathematicians worked with logic and rigor and didn't waste their time with airy-fairy ideas about "understanding" things.

Well, just because I had one prof that didn't "get it", does that make me special? Of course not. But over the years I've encountered this attitude repeatedly. It's part of the culture of university professors to take that kind of attitude, and they're awfully smug about it. But I've never bought into it. That's whenever they say there's no "intuitive" way to understand these things, I'm determined to prove the opposite.

The other day I wrote about functions of the roots which map to each other under permutations of the roots. Those correspond to the "normal subgroups" they talk about in Galois theory...or, more correctly, they correspond to the quotient groups which are generated by those normal subgroups. The problem with the way Galois theory is presented is that they stick exclusively to those abstract concepts and never make them concrete by actually showing you those functions that I laid out for you in my last post. If you doubt me, search the internet and see if you can find anyone else talking about things like AB + CD mapping to AC + BD when you swap B and C. You won't find it. But that's the essence of the Galois group of the fourth degree equation.

To show how different my approach is from the traditional, we just have to compare the way Herstein treats the question. I used to have a copy of Herstein, but over the years it's gone astray. (It was actually Bill Leslie's copy...sorry about that, Bill!) If you read Wikipedia, you'll see that Herstein's text was considered the definitive undergraduate treatment for many years. More importantly, Herstein was noted (according to Wikipedia) for his clarity and knack for making things understandable.

As I said, I don't have it in front of me, but I remember a few things about it that stood out. Like every other authority, he's pretty clear that since the fifth degree is unsolvable, it's a waste of time trying to understand why by working your way up through the third and fourth, which are solvable. Then, to emphasize the pointlessness of trying to find sense in the pattern, he writes out the full (and horrifying) solution of the third degree in terms of the coefficients, concluding by saying that the fourth degree is even uglier.

This is completely opposite to my approach, which is to write out the solutions of the third and fourth not in terms of their coefficients, but implicitly in terms of symmetric functions of the roots. And I do it not to show how complicated and ugly they are, but how simple and beautiful.

From everything I've read, there is no one out there who tries to explain these things the way I do (although I give some credit to one Fiona Brunk for her clear and insightful historical narrative on an older website of hers). Oddly enough Herstein hails from Winnipeg (as I do); and in that connection I ought to mention one more ex-Winnipeger, the old-school Jewish genius Robert Israel whose name will be familiar to frequenters of internet math discussion groups, who graduated from St. John's high school just a few years ahead of me, and whose beautiful sister Susan was mooned over longingly by my older brother when we were all in junior high school together. (Aside to my wonderful sister-in-law: give it a rest, Donna! it was forty years ago!)  In this discussion on stackexchange.com, Robert says he doesn't think there is a "dumbed-down simple explanation".

Maybe not, depending on how "dumb" you expect. But I still think that's no excuse to throw up your hands and reject the idea of understanding these things altogether.

Friday, December 6, 2013

Why You Can't Solve the Quintic Equation

Last year I wrote a series of blogposts talking about why you can’t solve a quintic equation. I worked through a lot of preliminaries but never really got around to answering the question. Recently someone on stackexchange.com asked for an intuitive explanation of the unsolvability, and I posted and answer. I think I surprised myself at how well I was able to put everything in a nutshell, and so I’m reposting my answer here.

Let the roots of an equation be A, B, C, etc. We are told that the unsolvability of the general quintic equation is related to the unsolvability of the associated Galois group, the symmetric group on five elements. I think I can tell you what this means on an intuitive level.

For three elements A, B, and C, you can create these two functions:
`` ``
``AAB + BBC + CCA``
``ABB + BCC + CAA``

These functions have the interesting property that no matter how you reshuffle the letters A, B and C, you get back the same functions you started with. You might reverse them (as you would if you just swap A and B) or they might both stay put (as they would if you rotate A to B to C) but either way you get them back.

For four elements, something similar happens with these three functions:

``AB + CD``
``AC + BD``
``AD + BC``
`` ``
No matter how you reshuffle A, B, C and D, you get these three functions back. They might be re-arranged, or they might all stay put, but either way you get them back.

For five elements, there exists no such group of functions. Well, not exactly...there is a pair of huge functions consisting of sixty terms each that works, similar to the ones I drew out for the cubic equation...but that's it. There are no groups of functions with three or especially four elements, which is what you would actually want.

If you try to create functions on five letters with this symmetry property, you'll convince yourself that it's impossible. But how can you prove it's impossible? You probably need a little more group theory for that. I'll have a some more to say about that when we return. For now I want to concentrate on the "intuitive", as opposed to the "rigorous" reason why you can't solve the quintic.

For the third degree equation, I identified these functions:

AAB + BBC + CCA = p
ABB + BCC + CAA = q

A, B and C are the roots of a cubic, but p and q are the roots of a quadratic. You can see that because if you look at pq and (p+q), the elementary symmetric polynomials in p and q, you will see they are symmetric in A, B and C. So they are easily expressible in terms of the coefficients of our original cubic equation. And that's why p and q are the stepping stone which gets us to the roots of the cubic.
Similarly, for the fourth degree, we identified these functions:

AB + CD = p
AC + BD = q
AD + BC = r

You can rewrite the previous paragraph word for word but just take everything up a degree, and it remains true. A, B, C, and D are the roots of a quartic, but p,q and r are the roots of a cubic. You can see they must be because if you look at the elementary symmetric polynomials in p, q and r, you will see they are symmetric in A, B, C and D. So they are easily expressible in terms of the coefficients of our original quartic equation. And that's why they are the stepping stone which gets us to the roots of the quartic.

And the intuititve reason why the fifth degree equation is unsolvable is that there is no analagous set of four functions in A, B, C, D, and E which is preserved under permutations of those five letters. I think Lagrange actually understood this intuitively fifty years before Galois. You probably needed a little more group theory to make it completely rigorous, but that's another question.

In particular think Lagrange would have understood the algebraic tricks whereby you went from, say, A B and C to p and q. It involves taking linear functions which mix A B and C with the cube roots of unity and examining the cube of those functions. Its a reversible process, so you can work backward the other way (by taking cube roots of functions in p and q) to solve the cubic. A very similar trick works for the fourth degree. I think Lagrange was able to show conclusively that the same trick does not work for the fifth degree...that's the "intuitive" proof. The "rigorous" proof would have had to show that in the absence of the obvious tricks (analogous to the 3rd and 4th degree), there was no other possible tricks that you could come up with.

I think I'm going to close by saying that the explanation presented here is very different from what you'll find elsewhere. When we come back, I'll elaborate on this point in some more detail.

Thursday, December 5, 2013

Answers to Jewish Pop Quiz

For the last couple of months I've been distracted from the physics, so I've been filling time by republishing articles of mine that have previously appeared in the print edition of Winnipeg's Jewish Post. I've still got some more stacked up in the archives, and I do like having them up on the internet, but in the meantime I've written up some interesting stuff on the Fifth Degree Equation, a topic which you might remember from a year and a half ago. I think I'm going to put my new stuff up next week, so we'll be interrupting the Jewish theme at least temporarily. Until then, here's the last (for now) of my Jewish series.

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Jewish Pop Quiz

Last week I left you with seven questions on Jewish culture that would have been child’s play for any kheder yingel in a typical shtetl of Old Russia. Here are the answers, as promised:

1.         What is the deal with the shor she-nagakh es ha-porah?
The case of “the ox who gores the neighbors cow” was an endless torment to the young Talmudist, who was expected to be able to explain the differing degrees of liability falling on the owner of the offending animal, depending on whether the beast was know to have a previous history of erratic behavior. I wasn’t there myself, but I’m guessing that more hours were spent arguing this question than any other single point of the law.

2.         What is the opposite of hayadayim yedey eysav?
“The hands are the hands of Esau, but the voice is the voice of Jacob”: ha-kol kol yaakov. These two catch-phrases came to express the dichotomy between violence and reason. During the Second Intifada, when two Israelis were lynched after straying into Ramallah, a chilling poster was widely circulated showing the famous picture of an Arab raising his bloody hands from a second-story window, over the caption: ha-yadayim yedey oslo.

3.         What is the controversy over the beytzah she-noldah be-yom tov?
They say that the monks of the Middle Ages argued endlessly over how many angels can dance on the head of a pin, but it turns out we could do them one better. The question about what you are allowed to do with an egg that was laid (or was it hatched?) on the Sabbath was taken up by the maskilim as an archtypical example of rabbinical nitpicking. I actually can’t remember if the problem was the laying or the hatching; I just remember that when I asked Rabbi Weizmann, he answered (in all seriousness): “what difference does it make when it was _______?”

4.         What is the difference between a shtut milguf  and a shtut milbar?
The Talmudic tractate Tagrei Lod (The Merchants of Lydda) was named in honor of that city known for its mercernary business practises. One particular sore point for the Talmudist was knowing the difference between a markup or commission taken internally or externally, which couldn’t have been easy to calculate in Ancient Palestine prior to the invention of decimal notation.

5.         What is the solution to the problem of shnayim ukhzin be-talis?
Shnayim ukhzin be-talis…two men are arguing over a tallis; zeh omer ani matzati…this one says “I found it first”…the law says: yekhlukoh…it must be divided. No, not cut in half with a scissors like Solomon and the baby…just sold and the proceeds divided.

6.         Where would you apply the doctrine of kol dalim gevar?
In Talmudic Law, the adjudicator is permitted, in certain circumstances, to consider the vehemence with which the litigant argues his case in assessing the relative credibility of the parties. I guess as a legal principle this falls somewhere between “might makes right” and “the squeaky wheel gets the grease.”

And lastly, a bit of a trick question:

7.         Who was Baba Kama?
This question came up in the blood libel trial of Mendel Beyliss in 1912. Not everyone knows that Beyliss was defended by a “dream team” to rival the one later put together for OJ Simpson. These were mostly Christian lawyers, progressive-minded Russians who considered the accusations an embarrasment for their country. It happened that one of the chief witnesses for the prosecution was Catholic priest Justin Pranaitas who claimed to be an expert in the Talmud. Now, as every Jewish child knew, some of the books of the Talmud have names that sound like Ukrainian grandmothers: Baba Kama, Baba Metzia, Baba Bitra, etc. So on cross-examination, one of the Jewish lawyers on the dream team asked the Pranaitas if he could tell the Court who Baba Kama was. It is said that when the priest answered in all seriousness that he “did not know that person”, the mirth and hilarity among the Jewish spectators in the gallery knew no bounds.

As a final word, it ought to be remembered that Beyliss was ultimately acquitted by an all-Russian jury which included (if Wikipedia is to be believed) no less than seven members of the Black Hundreds! This not-insignificant detail was omitted from Bernard Melamed’s loose finctionalization “The Fixer” (1968). People who want to know the real story ought to ignore Melamed and hunt down a copy of Maurice Samuel’s “Blood Libel” (1966).