Classical to Quantum

50 views Aug 31, 2024

50 views • Aug 31, 2024

We proceed to give a novel, elementary approach to the classic problem of integrating a polynomial over the unit circle. Complex numbers play an essential role in our treatment. The circle integration numbers of course figure, but there is a big surprise in the simplification that results from exploiting the complex algebraic structure of the unit circle. This is joint work with Dr Kevin Limanta (UNSW). This understanding gives a big clue as to why complex numbers play such a big role in Quantum Mechanics. The Playlist for this series Classical to Quantum is at https://studio.youtube.com/playlist/P... and is accessible to Members of this Channel. Your support is much appreciated. My research papers can be found at my Research Gate page, at https://www.researchgate.net/profile/... My blog is at http://njwildberger.com/, where I will discuss lots of foundational issues, along with other things. Online courses will be developed at openlearning.com. The first one, already underway is Algebraic Calculus One at https://www.openlearning.com/courses/... If you would like to support these new initiatives for mathematics education and research, you could also consider become a Patreon supporter. Thank you. Here are the Wild Egg Maths Playlists (some available only to Members!) • Algebraic Calculus One • Classical to Quantum (Members only)[ • Algebraic Calculus Two](https://www.youtube.com/playlist?list=PLzdiPTrEWyz5VLVr-0LPPgm4T1mtU_DG-) • Advice for mathematics exploration (M... • Solving Polynomial Equations (Members...[ • The Hexagrammum Mysticum (Members only)](https://www.youtube.com/playlist?list=PLzdiPTrEWyz5j1BJdXBw1MFst_nQAqzZ_) • Algebraic Calculus and dCB curves (Me... • Dynamics on Graphs (Members only) • Playlist • Triangle Geometry • Explorations with q-series (Members o... • Six: An elementary course in Pure Mat... • Maxel Inverses and Orthogonal Polynom...Here are the Insights into Mathematics Playlists: • MathHistory: A course in the History ... • WildTrig: Intro to Rational Trigonometry • Math Foundations • Wild Linear Algebra • Famous Math Problems • Box Arithmetic • Elementary Mathematics (K-6) Explained • Year9Maths • Ancient Mathematics • Wild West Banking • Sociology and Pure Mathematics • Sociology and Pure Physics • Old Babylonian Mathematics • Probability and Statistics: an introd... • Boole's Logic and Circuit Analysis • Universal Hyperbolic Geometry • Differential Geometry • Algebraic Topology • Diffusion Symmetry: A bridge between ...*****************

0:02

Hello everyone, thanks for joining me on this adventure, Classical to Quantum. It's a very

0:08

exciting lecture today because I'm going to show you some joint work with Kevin

0:13

Lamont from UNSW that will transform your understanding of integration on circles,

0:21

but also on more complicated things like spheres and L-groups, and other things.

0:29

It's also remarkable that this lecture is going to shed an interesting light on why

0:36

complex numbers are potentially so important in physics, especially the physics of quantum

0:41

mechanics. What I'm suggesting is that that's really a consequence of their natural importance

0:50

in the underlying mathematics, having to do with harmonic analysis on circles and spheres. So,

0:58

the physics is reflecting, ultimately, perhaps a deeper reality coming from mathematics.

1:08

Okay, so I'm going to go a little bit slowly today. We're going to come up with some really

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incredible things, really, but they're almost embarrassingly simple. So, at the end, you'll

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think, "My God, you know, why doesn't everybody know this?" And it's a pretty interesting question. Perhaps one of the reasons, if I may give myself a little bit of a plug here, is that,

1:30

you know, our orientation is a rational one, and that forces us, in some sense,

1:36

to avoid certain easy fixes: you know, the going to infinity, the invoking of irrational numbers,

1:43

and at every stage just, you know, adopting transcendental functions which don't really evaluate as proper functions. So, we can't do those kinds of cheats, and that naturally

1:53

leads us in what I would like to consider "the high ground," right? That's how we're able to

2:00

discover things that are really just sitting right there waiting for people to recognize. But you have to have the right orientation.

2:09

Another thing that is strongly suggested by this is that the general direction of the algebraic calculus course that I've laid out, that you can join, is really the right direction,

2:19

right? So, what we're doing here is very much in the spirit of the algebraic calculus, and, in fact, I will be talking about the topic here today in more detail in one

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of the algebraic calculus courses, maybe AC2 but maybe AC3—we'll see how it goes.

2:36

Okay, so what are we doing? Well, we're going to be giving a simplified approach to the fundamental question of integrating polynomial or polynumbers in the rational affine plane over the familiar

2:47

unit circle S1, given by the equation x² + y² = 1. And let's re-emphasize: no irrational numbers

2:56

like π or e, or anything else. No transcendental functions like cosines or the exponential function

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will play a role. What will play a role is the complex structure that's sitting there and

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also some pretty remarkable algebraic identities, whose history I don't quite understand completely,

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but perhaps there's some suggestion that they go back to Kummer. The rotational structure of

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the affine plane, closely associated with the complex structure, will play an important role.

3:29

Now, we've already talked about the circle integration numbers, okay? These are these Ω(mn), where m and n are natural numbers, given by 1 over 4^(m+n), 2m factorial, times 2n factorial,

3:42

over m factorial, n factorial, and (m+n) factorial. And here's a table for some of

3:48

these circle integration numbers where the indexing starts at m=n=0. Notice

3:54

prominently there's a one here in the top left corner—that's the case when m and n are both zero.

4:00

Okay, so let's let P be the algebra over the rational numbers of the bipolynomials. For us,

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those are like polynomial expressions in alpha and beta, and the multiplicative identity

4:14

will be denoted with a bold 1. So, we're going to say that a rational linear functional φ on P,

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which is a linear map into the rational numbers (our underlying field here is the rational numbers, but as we'll see in subsequent videos,

4:36

questions about what happens with other fields are incredibly interesting), but anyway, for now,

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the rational numbers. So, if we have such a linear map into the rational numbers, it will be called a circular integral functional precisely when it satisfies the following three conditions.

4:52

First of all, normalization condition: for the multiplicative identity, the constant polynomial

4:59

1, we should have φ(1) = 1. Secondly, locality: if we have a polynomial P which evaluates to

5:10

zero on the circle, so it's identically zero on the circle (meaning that if you have a point on

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the circle and you evaluate the polynomial at that point, you're going to get zero), in that case, φ(P) should equal zero. An example of that would be α² + β² - 1

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as a polynomial. That's a nonzero polynomial, but if you restrict it to the circle,

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it's identically zero, so we would want φ of that to be zero. And finally,

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we want this functional to be invariant under the rotation action. What does that

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mean? It means that if we have a polynomial in the plane and we rotate that polynomial,

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then the φ evaluated on the original P is the same as the φ evaluated on the rotated P. It's

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rotationally invariant, and we're ultimately going to frame that rotational invariance not in terms

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of rotation matrices but more conveniently in terms of multiplication by unit complex numbers.

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Okay, so what are we aiming at today? We're aiming at—well, here's one of our main theorems,

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so this is with Kevin Lam. There is exactly one circular linear functional F. It's exactly one

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such linear functional which has these three properties, and it is given by the rule that

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if you evaluate F on a basic monomial α^k β^l, then you're getting something nonzero only if

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k and l are themselves both even. Say k is 2m and l is 2n, and in that case, you're getting Ω(mn),

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the number that we defined earlier, the circle integration number Ω(mn), that we

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see right here. Otherwise, if the exponents are not both even, then we're going to get zero.

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So, as we've seen, this is somewhat equivalent to traditional formulas going back to Baker and

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Fand for integrating polynomial functions on the circle, but it's a little bit different because

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our normalization of the underlying measure, if you want to use that language, is different. We're

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normalizing the circle to have a mass of one, essentially—that's this normalization property.

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But otherwise, the content is essentially equivalent to the traditional formulas.

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Okay, so our proof of this—so we want to give it an independent proof that's independent of

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the classical derivations, which are quite complicated. If you look at Fand's paper,

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which is a celebrated paper in the American Mathematical Monthly, you'll see arguments involving notions of multivariable calculus or looking at

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integrals involving exponential functions. The gamma function plays an important role. It's

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not at all elementary. So, we're going to be giving a completely independent proof of that,

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and our direction is that we're going to be emphasizing the complex number structure that's

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sort of sitting there on the affine plane once we have the circle firmly in our gaze.

8:50

So, what we're going to do is think about extending a linear functional, say φ, from the space P of bipolynomials to a complex linear functional F_c on the space

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CP of complex bipolynomials. Okay, so we're essentially extending our field from rational

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numbers to rational complex numbers. We can think about rational complex bipolynomials,

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and then we can think about a linear functional F_c, which takes this space

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CP of complex bipolynomials and maps it to the complex numbers—the rational complex numbers.

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And we're going to introduce this complex basis for this space of complex polynomials, given by

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powers U^k and V^l, where U = α + iβ and V = α - iβ, in terms of the original polynomials α and β,

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right? So, what we're doing is we're enlarging our field of coefficients to allow i. That

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allows us to create these linear expressions, calling them U and V, and then they are sort of

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alternate candidates for a basis, say, for the linear space of polynomials. And then,

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taking powers of them, we get alternate bases for higher-dimensional spaces of bipolynomials.

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So, that's sort of the context, and in that context, here is our main theorem

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that we're going to talk about today. It's a very, very important result, I think, with implications and generalizations in many directions.

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Okay, so what is it? There is exactly one complex circular linear functional F_c. So,

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circular linear functional satisfies those three properties: the normalization, the localization,

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and the invariance. But when we extend it to the complex bipolynomials, it's given by quite

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a different rule. The rule is that F_c, acting on one of these basis elements U^k V^l, is either one

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or zero. It's one if k equals l, and it's zero otherwise. So, arguably, this is possibly the

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simplest thing that it actually could be—I mean, it's hard to think of a sensible possible formula

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which is simpler than this one. And that's really worth, you know, going on a long walk and thinking

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about, if you're a mathematician—this fundamental operation of integrating a polynomial on a circle

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is really captured by this incredibly simple fundamental formula that becomes visible once we

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go to the complex domain and we extend our coefficients to complex numbers.

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So, in our view, this should be one of the most widely known and recognized theorems in all of

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integration theory. And, as I mentioned before, in the fullness of time, I will be talking about this in the algebraic calculus course because there's quite a lot to say about it. And,

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you know, it's good to have lots of examples and work problems and so on and so forth. So, there are lots of extensions of this, and in the following lecture,

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you'll see how the previous theorem follows from this more fundamental one. Okay, so we're going to concentrate today on this theorem, but we'll give

12:37

some illustrative examples so you'll be able to see why the previous theorem—the one up here with the Ω's—is naturally connected with this one.

12:50

Okay, so that's what we're going to do today, and it's very important, okay? If you're a student of mathematics, you want to watch this very carefully. Or

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if you're a student of physics, you want to watch it very carefully because, believe me, it has a lot of consequences for physics too, and for quantum mechanics, absolutely.

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Okay, so just to make sure that we all can follow along here, I'm going to say some elementary things just to make sure that our notation is consistent and

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we're sort of on the same page. So, I'm going to illustrate some elementary kind of things,

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so bear with me a little bit if I'm being a little bit pedantic here.

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The default for us is to work over the field of rational numbers, although, as I've said before, the question of how and if this extends to other fields is a very interesting one, and Kevin

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is going to be telling us a lot of interesting things about that. P, as we've said before, is the

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space of rational bipolynomials in α and β, for example, this thing here, okay? It’s got rational

13:50

coefficients: two-fifths minus three α plus 47 αβ minus β³—that’s a bipolynomial in α and β with

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rational coefficients. And P_n denotes the space of those bipolynomials which are homogeneous of

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degree n, for example, this thing here is degree 3: 3 α³ - 5 α²β + 2 β³, okay? That’s in P₃,

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while P(n) denotes the space of such bipolynomials of degree less than or equal to n, so including

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the lower-degree ones. So, this one here, for example, is in P₃: 2 - 3 α - 12 α²β + 7/3 β³,

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so it has some things of degree 3 but also some things of degree one and degree zero.

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Now, this space P_n (the homogeneous ones) is of dimension n+1 over the rational numbers,

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and the basis is just the various powers αⁿ, αⁿ⁻¹β, and so on, decreasing the

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powers of α and increasing the powers of β until you get to βⁿ. There are n+1 of these. So, then the space P(n) is, well, just what you get by stacking these things together. So,

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it’s dimension 1 (that’s the constants), 2 (that’s the linear ones), 3 (that’s the quadratic ones),

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and so on, all the way up to n+1. And that sum is (n+1)(n+2)/2. And we may exhibit a basis for

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this space visually, concretely, by stacking basis elements of P_k for k=0, 1 up to n. So,

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at the top here, there’s a basis for the constant polynomials, then the next row α and β,

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the basis for the linear ones, then the next row α², αβ, β²—those are the quadratic ones—α³, α²β,

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αβ², β³ are the basis for the cubic ones, and so on with some obvious extension.

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Okay, and now what we want to do is we want to take this picture and complexify it—we want to

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bring in the complex numbers. So, let’s talk about some basic notation for complex numbers. For us,

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C denotes the field of rational complex numbers (so, no real numbers floating around, no irrationalities, for example, three-fifths plus four-fifths i—the coefficients have to be rational

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numbers). And then we identify the affine plane, usually with coordinates like this: (x,y), which

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is naturally associated with the complex number x + iy. The complex conjugate of z = x + iy is z* =

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x - iy, and the quadrance of z is the product of z with z*, which is the rational number x² + y².

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Then we have this basic identity: the quadrance of z * w = the quadrance of z * the quadrance of w,

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and that shows that quadrance respects the multiplicative structure of complex numbers, and, in particular, the unit complex numbers become a group—it’s closed under multiplication.

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Okay, so now we want to extend our polynomials to just have complex number coefficients. So,

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we’ll let CP be the space of rational complex bipolynomials in α and β, so polynomial in α and

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β with complex number coefficients, like this one here. CP_n denotes the space of those things which

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are homogeneous of degree n, so this one here is degree 3: something α³, something α²β, something

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β³, and those somethings are all complex numbers. And then, as before, CP(n) denotes the space of

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these rational complex bipolynomials of degree less than or equal to n, for example, this thing here has a constant term, a linear term, quadratic term, and a cubic term, so it’s in CP₃. These are

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all linear spaces now over the rational complex numbers, which means you can multiply these things

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not just by ordinary rational numbers but by complex numbers and you stay inside the spaces.

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Okay, so the big advantage of doing all this is that we can introduce a new basis—a new basis for

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these complex bipolynomials, starting just with this linear story here. So, we introduce U—maybe

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I’ll make the definition okay—U is α + iβ, and V = α - iβ. We’ve seen these before, okay? So,

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these are complex bipolynomials, they’re of degree one, but they form an alternate kind of basis. So,

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instead of having α’s and β’s, we have U’s and V’s. And there’s an important relation

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that connects the U’s and V’s to the α’s and β’s: U * V = α² + β². And so, in particular,

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if we’re on the circle, if we’re looking at points on the circle given by α² + β² = 1,

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then UV will be equal to one, identically. So, that’s another way of sort of thinking about what the circle is, in these sort of complex coordinates.

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Okay, great. So, then for a natural number n, the complex polynomials Uⁿ, Uⁿ⁻¹V, up to Vⁿ, form an alternate basis to

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the familiar basis for the space CP_n over C. In other words, the space of degree n homogeneous

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bipolynomials in the complex situation can have now this alternate basis in terms of powers

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of U and V, so the total degree is n instead of powers of α and β with the total degree being n.

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And here’s a very important point: we will see that harmonic analysis on the circle S1 is much

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simplified by moving to these complex bases for the space of bipolynomials.

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So, for example, a basis for the space CP(n) is just as before, obtained by stacking basis elements of the CP_k for k = 0 up to n. For example,

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when n = 3, here is CP₃, a basis for that, consisting of the constant guy 1 (probably

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we should make that bold to be consistent with what we were doing before). Now it’s bold. Then

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the linear ones U and V, the quadratic ones U², UV, V², the cubic ones U³, U²V, UV², V³.

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The polynomials αⁿ and βⁿ have a natural multiplicative property, okay? So,

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if you have a point, say A, with coordinates (x,y) in the plane, and a rational number λ,

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then if you look at this polynomial αⁿ and you evaluate it on the point λA (so the point with

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coordinates λx, λy), well, what does αⁿ do? It takes the x-coordinate and raises

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it to the n-th power, so we’re getting λxⁿ, and just the way numbers work, that’s λⁿxⁿ,

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and xⁿ is αⁿ(A). Similarly, the y-coordinate βⁿ(λA) will be λⁿβⁿ(A). So, geometrically,

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these are transformations corresponding to dilations, central dilations, in the plane.

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Okay, so in the complex case, we have something similar going on, but it turns out to be sort of

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more decisive because they’re more general geometrically. So, I just want to say some

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elementary things about evaluation of polynomials on affine points regarded as complex numbers.

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So, given a complex number z = x + iy, the polynomial U, or the polynomial U evaluated at z,

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is what? Well, U(z) is U(x + iy), and what is U? We said U is α + iβ of this, so what we should do

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is we should apply α of this, in which case we just get x, and we should add iβ of this. Now,

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β of this is picking out the y-coordinate, so that’s iy. So, we get x + iy, which is just z. So,

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as a polynomial, okay, that you’re evaluating on the plane, U(z) = z itself. And so, if we take a

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power, like Uⁿ, and apply it to z, well, then we’re going to get zⁿ. Similarly with the V:

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V is α - β, so we do the same thing, the only difference is there’s a minus sign appearing here, and we get z*, the conjugate. So, that’s telling us that Vⁿ(z) = z*ⁿ.

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So, it follows that for complex numbers w and z, if we take Uⁿ of the product wz,

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okay? So, well, I guess we don’t really need this thing here, that’s just a repetition,

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so it’s telling us that Uⁿ(wz) = w * zⁿ, and the way complex numbers work,

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that’s wⁿ * zⁿ, and zⁿ is Uⁿ(z). So, this is how the polynomial Uⁿ is transformed when

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you compose it with multiplication by w. Okay, it essentially brings out a factor of wⁿ. Similarly,

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if we do the same thing with Vⁿ as a polynomial on wz, well, then we have to take wz* to the

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power of n. The way complex numbers work, that’s wⁿ * zⁿ, so we get w*ⁿ * Vⁿ(z).

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What’s happened here now is that the w has come out but with a star,

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and then there’s a w*ⁿ coming out.

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Okay, so now suppose F is a rational linear function on the space P of rational numbers.

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We may then extend it to a complex rational linear functional, and we’ll introduce slightly different

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notation just to try to keep these things a little bit separate. So, F is a linear function on the

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space P, and then we’re going to extend that to a complex rational linear function F_c on CP. Well,

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how? Just by defining F_c(iP) to be F(P)i for any polynomial P. So, if F,

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say, of a power combination αⁿ βⁿ is, say, aₙₘ, that’s some rational number—we have such things

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for all natural numbers n and m—then we may evaluate F_c on powers of U and V as follows:

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F_c acting on U = α + βi. By linearity, that’s F(α) + F(β)i, right? The way we define the

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complex extension is we just do the ordinary F on this thing and the ordinary F on the β,

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and then we introduce an i out front here, just as we said up here, okay? So, the F(α),

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that’s a₁₀, and the F(β), that’s a₀₁, and there’s an i here. Similarly, when we evaluate F_c at V,

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we’re going to get F(α) - F(β)i, which is this combination. Similarly, if we just go

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down one more, just to practice a little bit: if we look at the complex function F_c of U²,

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so what is U²? Well, we’ve seen what U² is. U² is α + iβ all squared,

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so that’s α² + β² + 2iαβ. When we apply F_c to that, well, we get F(α²), F(β²), and 2F(αβ)i. So,

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we’re using the linearity, allowing us to go from the complex functional to the underlying original

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functional just on polynomials. And then, in terms of the a’s, we can express that this way.

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Okay, I’ll let you check that in a similar way, U F_c(UV) turns out to be this, and F_c(V²) is this,

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so this is just giving us some practice with this idea of working with the extension of a

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linear functional from an original space to a complexified space. If you’ve done some

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mathematics, this will be quite familiar, but if you haven’t done maybe some linear algebra, it

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might be a little bit unfamiliar, so I just want to give you a little bit of practice with that.

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Okay, but these two bases that we have now, the α’s and β’s and the U’s and V’s,

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they’re pretty well equivalent. I mean, both of them are good. So, we can also go the other way around. If we know the values, say, of the complex functional F_c on U^k V^l, those products,

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and let’s say that’s bₖₗ, then we can recover the values of F on α and β by first inverting

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U = α + iβ and V = α - iβ to get α = ½(U + V) (you add these two things and divide by 2,

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you get α), and taking β to be ½i(U - V) (this minus this and then divide by 2i),

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that’s how we get α and β as a combination of U’s and V’s. So, if we want to figure out what F(α)

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is and we only know the F_c’s on the powers of U and V, well, we have to sort of replace the α’s in

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terms of U’s and V’s. So, there it is, and then we use the linearity. We get ½F_c(U) + ½F_c(V),

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so we get ½(b₁₀) + ½(b₀₁). Similarly, F(β). And just to go to a quadratic case: so, let’s say we

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wanted to calculate F(α²). We’d have to take this expression and square it. So, we get U² + UV + V²,

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and then we apply F_c to that linearly, and we get ¼(b₂₀) + ½(b₁₁) + ¼(b₀₂), and so on.

29:16

Okay, so now we’re in a position to think about the proof of this claim that there is essentially one circular integration functional. So,

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suppose that F is a circular integration functional, so that it’s normalized, localized, and an invariant linear functional on the original space P of rational bipolynomials.

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Let’s denote by F_c its complex extension. Let’s determine the values F_c(U^k V^l)—maybe

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we could call them cₖₗ for natural numbers k and l—using just these three properties.

29:59

Okay, so the first one, the normalization condition, tells us that F_c(1), which is the case when k and l are both zero,

30:09

that has to be 1. So, in terms of the array, that normalization condition is

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telling us what the functional applied to this element has to be—it has to be one.

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The second condition, the localization condition, what’s that telling us? It implies that since UV,

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restricted to S1, is just the value 1, okay, that φ_c takes on the same values on all the basis

30:39

elements in any given column of the array because if you look at any column of this array, say, this

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one here, the top one is V, the next one is UV², the next one will be U²V³. Each one is obtained

30:52

from the top one by multiplying by UV, but UV, when we restrict it to the circle, is identically

31:00

1. That means that all of these bipolynomials, when we evaluate them on the circle, are going

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to have exactly the same values. Their values on the circle are going to be the same, and that

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means that this functional, because it has this localization condition, will have the same value

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on all of those elements. So, whatever F_c does to this column, it does the same thing to every

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element in this column. And in particular, if we look at the central column, we see that all of

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these entries under here, they have to be sent to 1 under this circular integration functional F_c.

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So, to figure out what F_c is, it remains only to figure out what F_c is doing to the top elements

31:57

in these non-central columns, namely the U’s: the U², the U³, and so on, and also the V’s:

32:04

the V’s, the V², the V³. Once we know what F_c does to those, then we’ll know what F_c does to anything because what it does to the elements in the column is always the same.

32:17

Okay, so let’s look at one of these. Okay, so we need to determine, say,

32:23

the values φ_c(U^k) or F_c(V^l), corresponding to the top entries for each of these non-central

32:30

columns. So, let’s look at φ_c(U^k), where k is some natural number k > 0,

32:36

so we’re not talking about 1, we’re talking about one of these guys here, okay?

32:41

Now, recall from what we were talking about previously when we were talking about what

32:46

happens to one of these powers if you compose it with multiplication

32:52

by w. Remember that U^k(wz) = w^k * U^k(z), okay? We’ve already talked about this. Okay,

33:06

this is what happens when we compose this functional U^k with multiplication by w.

33:16

Now, in the special case when w is itself on the unit circle, the action of multiplication by w

33:24

is a rotation of the complex plane, okay? But the invariance property says that that cannot

33:35

affect the value of the functional φ, and hence also not the complex extension. That is, that

33:43

rotated bipolynomial is going to have the same φ value or F_c value as the original one did. Well,

33:55

how is F_c going to act on something like this? Well, as far as this thing is concerned, the w^k is just a complex number, so it pulls out—it’s a linear functional. So,

34:08

what we end up getting is that F_c(U^k) = w^k * F_c(U^k) for any complex number w.

34:28

That’s the function’s value on the original polynomial U^k, and then we’ve rotated it

34:34

by composing it with multiplication by w. We’ve seen that the effect is to get this thing here,

34:39

and F of that has to be this. But w is on the unit circle, so that if w^k is not equal to 1,

34:55

this w^k will also be on the unit circle because this is a group. But if it’s not equal to 1,

35:02

then this formula here that we have is telling us that something equals itself multiplied by

35:09

something which is not equal to 1. Okay, so if this thing here is not equal to 1,

35:16

then the only way this can happen is if this F_c(U^k) is actually zero.

35:24

The only number that’s unchanged when you multiply it by something nonzero is zero.

35:32

So, is there a w for any natural number k? Is there a w for which

35:41

we can be sure that w^k is not equal to 1? That’s what we need—that’s what we need to go from this equation to the conclusion that F_c(U^k) = 0.

35:54

Well, geometrically, what that’s just saying is: can you pick a complex number on the unit

36:01

circle so that its k-th power is not equal to 1? Well, kind of obviously, right? You

36:07

just sort of pick any one of them and take the k-th power and you’re not going to get 1. Oh,

36:13

but what if it’s perhaps some kind of root of unity? Okay, well, that’s kind of an exception, but those are few and far between, relatively speaking. So, to make that more precise,

36:24

what we could do is we could say, “Well, look, a point on the unit circle has this rational parameterization that we know and love: 1 - t² over 1 + t², 2t over 1 + t².” So, here’s the

36:35

number 1, the way I’m looking at it, here’s the unit circle, here’s the number 1. If I choose a

36:40

complex number on the unit circle which is really close to 1, so, in other words, a t-value which is very close to zero but not quite zero, just a little bit, then if we raise that to the k-th

36:50

power—in terms of angles, every time we double the angle and so on, but in terms of rationals,

36:57

what we’re doing is we’re just reflecting. Each time we’re reflecting one in this thing here, we get that. Then we reflect this, in this, so we’re just reflecting along the circle like this,

37:05

right? So, if this thing is sufficiently close to 1, then this sequence of reflections

37:11

doesn’t have enough room to get back to 1. So, that’s another argument that yes, there always is a point w on the unit circle such that w^k is not equal to 1. That’s what

37:22

we require. We require exactly that w to stick into this equation to conclude that φ_c(U^k) = 0.

37:31

Now, I’m making a little bit of a big deal out of this, which is sort of obvious because I’m interested in what happens with other fields also. So,

37:39

with other fields where our geometric intuition is a little bit different (there are only a finite number of points on the unit circle), well, then maybe the argument needs to be thought about more

37:49

carefully at this point. But in any case, over the rational numbers, I think we’re happy. Okay,

37:58

so we conclude that φ_c(U^k) = 0 when k is bigger than zero, and similarly when you look at the V’s,

38:04

it’s the same story. We see that φ_c(V^l) = 0 whenever l is bigger than zero.

38:11

So, the conclusion is that F_c has a remarkably simple form on this array of powers of U and V. It’s zero on all but the entries in the central column headed by 1,

38:22

and on each of these, it takes the value 1. So, the values are 1 along that central column and 0

38:27

everywhere else—almost the simplest thing it possibly could be.

38:34

So, what have we shown? Well, we’ve shown that if φ exists, then it must be unique, that this

38:39

is the only possibility for it. But in fact, we know what the only possibility for φ_c is. It’s

38:47

what we’ve just said. So, to show existence, we just need to check that this particular linear function that I’ve just described with the values we’ve just described is indeed satisfying the

38:56

three conditions. But basically, we’ve already done that. That’s exactly what we’ve shown:

39:06

that we have the normalization condition, we have the localization condition because all

39:12

the entries in each column are all the same, and we have the invariance condition exactly

39:18

because these U^k’s have this property that when you multiply them by a unit complex number,

39:24

you get a multiple of them, and so the linearity is going to prove that.

39:30

So, that establishes our theorem—the theorem that there is a unique circular integral

39:40

functional and it has the values on the complex basis, the very simple values that we’ve shown,

39:46

namely either 1 or 0, depending on whether the powers of U and V are equal or not.

39:55

True, at least an indication of the direction for that, and we’ll do that just by looking at a few

40:00

examples. Because now we know what this unique circular integral functional is on the basis of

40:10

the powers of U and V, so if we want to evaluate it on a polynomial expressed in α’s and β’s,

40:20

the problem is purely one of algebra. We leave geometry and analysis behind; we just say, "Oh,

40:28

okay, here’s a polynomial in α’s and β’s, all we need to do is just to rewrite that

40:34

in terms of U’s and V’s.” Well, right, we get all these terms in U’s and V’s, and then we just look for the ones that have equal powers of U and V,

40:45

and they get assigned a functional value of 1, and all the others get 0’s. Right, so at this stage,

40:54

all that’s really required to establish the result for the α’s and β’s is a bit of algebra to be able

41:01

to convert polynomials in α and β to polynomials in U and V and to pick out certain coefficients.

41:12

But it turns out that that problem is actually quite a lot more subtle than it might appear at first glance. That itself turns out to be a really interesting problem,

41:23

so let me give you a little hint as to that. Okay, so let’s look at some examples. Suppose we wanted to calculate φ(α²), okay? So, what should

41:33

we do? Okay, so what is α? α is U + V over 2. So, if we square that, we get U² + 2UV + V² over 4.

41:45

So, F(α²) will be ¼ φ_c evaluated on this thing. Okay, but by linearity, it’s evaluating on U²,

41:55

evaluating on UV, evaluating on V². Okay, what do we know about φ_c? Well, we know that it

42:01

kills anything that does not have equal numbers of U’s and V’s in the exponent. This has unequal

42:08

numbers—it’s killed. This one here has unequal numbers—it’s killed. This one here has equal numbers—it’s sent to 1. So, we’re just getting ½, and that is indeed Ω₁₁, because this is a 2 * 1.

42:23

Here’s an example of where we ought to get something 0 when one or two of these exponents

42:31

are odd. So, let’s say α²β³. Okay, so what’s α²β³? Well, here’s α: α is U + V over 2, and

42:37

β is U - V over 2i. So, we have to square this one, we have to cube this one. Okay,

42:46

the 2’s and i’s come out front, we get i over 32, and then we have (U + V)²(U - V)³. Okay, and that

42:56

is an expression in U’s and V’s. Let’s have a look at it: U⁵, U⁴V, U³V², U²V³, UV⁴, V⁵. What are we

43:10

supposed to do? We’re supposed to look to see for terms where the U’s and V exponents are equal.

43:18

None of them are equal. In fact, none of them can be equal because the total degree of this thing

43:24

is 5, okay, and so there can’t be any term with just U’s and V’s appearing with the same power.

43:39

Let’s consider one more example. Let’s look at φ(α⁴β⁶). Okay, so α⁴β⁶, there’s U + V over 2 to

43:47

the 4th, there’s U - V over 2i to the 6th. We can bring the 2’s and 2i’s out front, and we have (U +

43:55

V)⁴(U - V)⁶. Okay, so we have to do some algebra. Okay, we need to expand this thing out. There it

44:03

is. What are we supposed to do? We’re supposed to look through this and find those terms where the

44:11

powers of U and V are equal. Well, you know, here there’s U¹⁰, U⁹V, the powers of U are decreasing,

44:18

the powers of V are increasing. There’s only one place where we get equal powers of U, and that’s

44:25

the central term, the -12 U⁵V⁵. So, when we apply our φ_c, all of these other terms get killed,

44:36

they all get sent to zero, while this U⁵V⁵ gets sent to 1. So, the φ is going to be the constant

44:43

-1 over 210, and we’re going to multiply by -2 for a total of 3 over 256, which is Ω₅₃.

44:51

Let me just convince you that Ω₅₃ is up here. Up here, 5-3. So, this is 0,

44:59

1, 2, and U, and 0, 1, 2, 3. So, we’re looking at this entry here, 3 over 256.

45:10

So, in order to generalize this, to get the general formula, what we need to convince

45:16

ourselves of is that when we’re expanding this thing out in general—so we take α²ⁿ

45:26

and β²ᵐ—we’re getting this thing here. And the point is that when we expand this thing

45:32

out and we look for the coefficient of Uⁿ⁺ᵐVⁿ⁺ᵐ, these are the only terms that

45:43

are going to appear with equal U’s and V’s because this whole thing has degree 2n + 2m,

45:49

half of that is n + m, so that’s the only way of getting equal degrees in such an expression.

45:55

Okay, so we can see that the proof of our original theorem involving the Ω’s comes down to this

46:02

purely algebraic result, which is relatively simple really. It comes down to evaluating a

46:09

product of some power of U + V times some power of U - V, and sort of looking for the coefficient of

46:17

the central term, and that’s supposed to be Ω(mn). So, to complete the proof of our first theorem,

46:26

we need to actually show this. It turns out that even though this is a purely algebraic result, it’s nontrivial, and in fact, it has a pretty interesting story associated with it.

46:42

So, in summary, I’m hoping you can appreciate that if we go back to our theorem—this is what

46:54

we’ve accomplished today—that there’s exactly one complex circular linear functional, φ, and it’s given by this incredibly simple rule. So, all the complexity in the Baker-Folland formulas

47:05

for integration of polynomials is, in some sense, a red herring. It’s just because we’re

47:12

not looking at this in the right way. The right way is obviously to look at it from this complex

47:18

point of view. We are forced to think in terms of complex numbers, and so it’s not surprising—it

47:27

should not be a big surprise—that when it comes to nature and quantum mechanics,

47:34

complex numbers also naturally appear in a very prominent way.

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