Researchers believe that cats have an abstract understanding of numbers, often up to about seven. Some claim that mother cats can count as high as six or seven, though three or four is more likely. In contrast, humans seem able to count much further. George Cantor, for example, explored numbers reaching to infinity—and even beyond. Personally, I find numbers challenging, both in practice and theory. Number theory has always been a painful experience for me.
Still, when there’s a need, there’s always a way. So, in this blog post, we draw a final dot to mark the end of our journey through the ring of integer.
I could not find anything about tetrads in Babylonia, but I found them on math.stackexchange: Diophantine equation a2 + b2 =c2 + d2. The complete solution can be found in the textbook L.J. Mordell, "Diophantine Equations", Academic Press 1969, on p. 15.
Well, it is not explicitly complete there, it is somewhat sketchy, but here it is (I skip the proof).
Proposition 1. Every primitive solution of (1) is of the form
a = (mp+nq)/2,
b = (np-mq)/2,
c = (mp-nq)/2,
d = (mq+np)/2,
where m,n,p,q are integers. Conversely, for any integers m,n,p,q such that a,b,c,d are integers, the formula above provides a solution of a2 + b2 =c2 + d2.
There I quoted a Proposition from Mordell's book, without a proof. But now, with Proposition 1 from the previous post, we have a complete proof of Mordell's statement. Let us discuss this in details.
Suppose a,b,c,d are integers satisfying a2 + b2 = c2 + d2. Let x be a vector in R2,2 with components (a,b,c,d). Then x is a null vector: Q(x) = a2 + b2 - c2 - d2 = 0. The matrix x^, defined as in Part 2, is
x^ = {{c+a,b+d},{b-d,c-a}} (1)
is, automatically, of determinant zero, with integer components. Thus we can apply Proposition 1 from Part 4 to deduce that there are vectors v,w with integer components such that
x^ = vwT. (2)
Let (p,-q) be the components of v, and let (m,n) be the components of w. Then (2) takes the form
{{c+a,b+d},{b-d,c-a} = {{pm,pn},{-qm,-qn}} (3).
From (3) we immediately get
which are exactly the formulas from Proposition 1 in Mordell's text. In order to get the exact correspondence we have set (p,-q) to be the components of v, but that does not really matter, since if q runs through all integers, so does -q.
Ark, thanks for another bright, cheerful and mind-refreshing essay!
ReplyDeleteOf course, mother cats need to be able to distinguish numbers to count their kittens. Most often there are 3-4, but can be up to 7 - just as you mentioned above.
Please mind a typo:
"Let (p,-q) be the components of v, and let (m,n) be the components of u" -->
"Let (p,-q) be the components of v, and let (m,n) be the components of w"
A bit more detailed comment will follow...)
Fixed u. Thanks.
DeleteSo we have learned that spinors are some kind of balanced constructions. Thank you, Ark, for this glimpse behind the painted veil hiding the mysterious spinors.
ReplyDeleteI wonder if this elegant construction applies to real numbers as well as to integers. Who knows which of them are more fundamental, I know people who consider integers as solitons or wave packets of real numbers.
Interesting, when we study spinors, we always encounter some kind of special - zero or null - structures within a larger enveloping one (null cone, null-vector subspace, recall the definition of spinors as the minimal left ideal). Does this mean that spinors express some balanced, invariant entity that revives in a permanent flow? Such patterns would be ontologically what we perceive as really existing objects.
These are all good and deep questions. Recently I was busy with trying to give a physical interpretation to ledt and right actions of SL(2,R) - that is what acts on Weyl spinors. As it happens they do not a physical interpretations. Only when we have a pair of spinors, like v and w, and SL(2,R) acts simultaneously on v and w, only then we get a transformation that we know the name for, like Lorentz boost, rotation, translation, dilatation etc.
DeleteI thought I will write about it in details, but then I changed my p[ans. I decided to move on from our sandbox to real life. So I will start a new series - for adults this time. We will be using all our skills acquired so far to build something more serious than so far.
"...physical interpretation to left and right actions of SL(2,R) - that is what acts on Weyl spinors"
DeleteArk, do you mean that there is no physical interpretation for the case when SL(2,R) acts on spinors from R(1,1)? And this fact is in contrast to the ordinary linear transformations resulting from SL-group action on "usual" vectors from R(2), right?
Thank you for asking! So, you did it! Once you keep asking questions, I will answer them. Thus once again change of plans. Postponing the adult stuff for later. I will go into the physical interpretation of group actions in the next post(s). Thank you!
DeleteMea lingua mea inimica est. Now it's my fault that we'll be sitting in the sandbox for a long time instead of diving headfirst into these challenging adult games :)
DeleteAlthough, I like interpretations - physical, geometrical, etc., and will dwell on them gladly. Let the adult games wait a bit, we're in no hurry. Thank you, Ark, for the comfortable pace of our journey!
Anna, I hope to be able to post Part 1 of the closing section of our Kindergarten adventure later today. It will be pdf, as there will be many matrices. and these are clumsy in html..
DeleteDidn't finish yet. But you can start reading. The new stuff is in Ch. 5. of Notes
DeleteThank you, Ark! I will start reading after we finish with Vladimirov's seminar.
DeleteArk, I agree that Ts(a)Ts(b) = Ts(a+b), that Ts^T BTs(a) = B, and that det(Ts(a)) = 1, but I cannot obtain the shift Ts(a)τ (x, t) = τ (x + a, t), since the 2nd and 4th elements of the resulting column vector contain x in explicit form "- a x" and I don't know how to get rid of it.
ReplyDeleteAnna, Are you doing it by hand, or you use a program, like, for instance, Reduce?
DeleteAnyway the second component should be, according to (35)
Delete(1-q(x',t'))/2.
Set there x'=x+a, t'=t, calculate (x+a)^2-t^2, and you shouldl get what you need. Similarly for the fourth component.
Sorry, I forgot that q = x^2 - t^2. Everything will be fine now. I need to stop chasing two rabbits at once and doing Exercises while working in the office, otherwise I will soon be unemployed.
Delete...as soon as I came home I noticed the error.
DeleteWhen tired from looking at the monitor, I do it by hand and sometimes using Wolfram alpha. Probably, will try Reduce also.
No, it's not fine :(
DeleteEven for the case of time translations, which is easier than the space ones, I get extra terms:
T(s) tau(x, t) = {x, (1-q)/2 + s^2/2 + st, t+s, -(1+q)/2+s^2/2 + st}
You see, there is one and the same extra term (s^2/2 + st) at the 2nd and 4th places.
The Lorentz boosts and Dilatations are ok.
On the right hand side you should have q(x',t'), niot q (x,t). And this is exactly what you have.
Delete"Probably, will try Reduce also."
DeleteWhenever you want to use Reduce, you can always ask online AI to propose a code for it. And you can ask me if you have problems. Free online AIs are quite good. They often mess up, that's true, but if you insist, they are usually able to fix the code after several attempts. Wolfram Alpha has unfortunate restrictions on the size.
"On the right hand side you should have q(x',t'), niot q (x,t). And this is exactly what you have"
DeleteRight, thank you! Some piece of my mind thought about such possibility yesterday but i was too lazy to get through. Now it is clear, but nevertheless looks like beautiful magic.
"Wolfram Alpha has unfortunate restrictions on the size"
DeleteYes, unfortunately. Wolfram can multiply two 4x4 matrices but already fails when we need to multiply three such objects.
I would be very grateful if you could provide a link to a resource from where I can download Reduce safely.
Here you have it:
Deletehttps://reduce-algebra.sourceforge.io
If you have any problems, questions, need help -ask, and I will assist you.
DeleteHowever, I should say that Reduce or any other AI would not have helped me understand what was wrong with q(x', t'), although such bottlenecks are the most important. But thanks to AI for saving us time, allowing us to understand more.
DeleteAs I realized recently the main profession of the coming years will be "Prompt Engineering". It will be the art and science of using AI for all kind of purposes. Some good, but, probably, mostly evil.
DeleteExercise 5.5: "Conformal inversion" required a bit of attention and accuracy, but it is done.
DeleteIt is hard to believe in this magic - one minus sign in the identity matrix does the complex and precise job of inversion with respect to the zero cone! This fact indicates that generally conformal inversion is a simpler and more fundamental transformation than, for example, space or time translation.
Ark, thanks a lot for the link to Reduce! I have installed it and though the 1391-page Tutorial looks scaring, even managed to do a few simple steps with it :)
DeleteVery good. This extract from a book is shorter than the big manual. Ger\t it and and let me know.
DeleteYes, I have downloaded the book, thank you very much! Since the author is Andrey Grozin, Novosibirsk Institute for Nuclear Research, there is a chance to find analogues in Russian for my exceptional convenience.
DeleteArk, again I do not understand about embedding... In particular, about the "more geometric picture of it". What is the relation between the null cone of R2,2 and the Grassmann manifold ~ S1xS1 = torus?
ReplyDeleteProbably, the torus is a 2d section of the 3d cone?
And why the image of embedding flat space R1,1 into a curved manifold S1xS1 covers only a part of it? On the contrary, if we try to embed R1 into a circle, it needs to be wrapped around the circle many times making a multiple cover ...
Good question. It requires a longer answer. So I will explain everything in the new note. The first part of the new note I will try to post in the next 2-3 hours.
DeleteIn fact I will not be able to do it today. I need to do some graphics, and it always takes time to get what I have in mind. So tomorrow there will be "Sunday special."
DeleteA bit late to the party, but have played a bit with numbers last night and got a bit more 'general conjecture' or sort of a recipe from Prop. 1 WRT "balanced tetrads".
ReplyDeleteIn short, any two integer spinors or pairs of integers, v=(m,n) and w=(p,q), will produce a primitive solution to Diophantine eq. (1) if (a) both v and w are composed of pairs of coprime and odd integers or if (b) one of them is pair of coprimes, while the gcd of the other pair is 2.
A nice 'addition' stemming from this 'conjecture' is that if v=w (up to ordering), we get Pythagorean triplets.
For example, pair v=w=(1,3) gives triplet (3,4,5), while pair v=w=(1,2) would give 1/2×(3,4,5), so we need to multiply one of them by the factor 2, which 'works' also in general case for different v and w.
Examples of tetrads are then, for v=2×(1,2):
- w=(1,3) -> (1,7|5,5),
- w=(1,4) -> (2,9|6,7),
- w=(1,5) -> (3,11|7,9),
- w=(1,6) -> (4,13|8,11), ...,
and for v=(1,3):
- w=2×(1,4) -> (1,13|7,11),
- w=(1,5) -> (1,8|4,7),
- w=2×(1,6) -> (3,19|9,17),
- w=(1,7) -> (2,11|5,10), and so on.
FWIW.
P.S.
DeleteAs can be seen from the lists at the end of the previous comment, we also have a nice progression 'rule' for obtaining 'new' primitive solutions (a,b|c,d) for "balanced tetrads" once we established the base members of the sequences: in first case we just add (1,2) and in second case (1,3) (or (2,6)) to both sides of our 'base' primitive solutions to get the next member of that particular branch of solutions.
So, in principle, in concert of "initiation" of Pythagorean Brotherhood or Fellowship, as it included also female members and not only males, in respect of knowning "balanced tetrads", it seems they were not very demanding bunch of mystics. ;)
Last night played a bit with 2d geometry and got all those sequences in the comment above from a pair of initial points equidistant from the origin, by applying simple "geometric transformations" that preserve the distance, like translations and reflections or swapping the coordinates.
DeleteIs that what those left and right action matrices in fact do, i.e. their geometric interpretation?
For example, take the pair of points (1,0|0,1) which "make" the line with slope -1/1, translate them perpendicularly to (2,1|1,2), reflect one of them over vertical axis to (1,2|-1,2) and now have new line with slope -3/1. By perpendicular translation of this line or pair of points we get the "odd" sequence branch of primitive solutions above for the spinor (1,3):
(1,2|-1,2)->(3,4|0,5)->(4,7|1,8)->(5,10|2,11)->...
If we take the third element with x2=1 and repeat the "procedure", we get the "odd" sequence branch for spinor (1,5), from which in the same manner we get "odd" sequence branch for spinor (1,7), and so on.
If we take the Pythagorean triple element (3,4|0,5), swap coordinates of the "left" point pair to get (4,3|0,5), we now have "line" with slope -4/2=-2/1, which by perpendicular translation gives the solutions branch for spinor 2×(1,2) above:
(4,3|0,5)->(5,5|1,7)->(6,7|2,9)->(7,9|3,11)->...
From the second element of this sequence, by reflecting 1->-1 and again translating the new line in perpendicular direction we get the gcd=2 branch for spinor (1,3), while from the third element by reflecting 2->-2 and again translating in perpendicular direction, we get solutions for spinor 2×(1,4).
And so we can proceed building new "tetrads" ad infinitum.
Oh well, it seems that some catching up is long overdue, all the way from Lie sphere geometry.
DeleteTo help me get in tune a bit faster, can you answer a few questions for a jump-start:
- Why did you choose torus with arbitrary radii r1 and r2 for PN instead of for example unit sphere or "two-sided" unit circle, i.e. ring, in which case a particular tetrad combinations would give positions of unit isosceles triangle legs, i.e. each tetrad combination would be represented by an arc on the sphere or the ring? Why not just stay on the x-y flat plane, without any scaling?
- As each "side" of the tetrads has 8 possible combinations of coordinates, or 4 if we include only positive coordinates > 0, that give the same "distance" from the origin, do you consider each of 64 or 16 possible combinations for one basic tetrad as "different" points on PN? If the "sides" in the tetrad are swapped, is that new set of points?
There might be some more questions as I go through the material and notes, do please bear with me. :$
Thanks.
Correction:
DeleteThere would be only two combinations if we limit coordinates > 0, (a,b) and (b,a), that give the same distance, which would then give just 4 possible combinations for a particular tetrad, not 16 as said in previous comment.
Perhaps you should write all these discoveries down and ask a couple of different AI's for comments and evaluation? I am often doing it myself and I am finding AI useful and helpful for this kind of considerations.
DeleteOK, thanks for suggestion, will do.
DeleteI kind of see "how" you do things, but the understanding "why" you do things presented here the way you do them, still escapes my grasp. Maybe AI might be of some help there too.
Since those were no discoveries, but maybe just some noticed patterns that could perhaps have eased the visualization or simple planar geometric interpretation, there was no need for AI. And being stupid machine that it is, there was also no use for it.
DeleteWent through your last or recent notes and got most of the answers to my questions, or at least those more "important" ones. It seems that in our shared sandbox, you have already started to build some fancy structures, while I was still just making dots and mainly straight lines in that same sand somewhere at the footing of your towers with cats in them and donut shaped clouds above them.
So, more or less ready for what follows from your blog kitchen.
The construction of towers barely started, so you are not so much behind.
Delete