WEBVTT 00:00:00.450 --> 00:00:05.575 Watch the blue profile. Changing just one boundary makes it bend, and its 00:00:05.575 --> 00:00:11.150 lowest eigenvalue rises. Why? Let us build the explanation, 00:00:11.150 --> 00:00:14.105 then check the same effect in the laboratory. 00:00:15.492 --> 00:00:19.717 The horizontal coordinate is position along an extra dimension. 00:00:19.717 --> 00:00:24.617 The vertical coordinate is field amplitude, not a particle trajectory. 00:00:24.617 --> 00:00:29.847 With zero slope required at both ends, a constant profile is allowed. 00:00:29.847 --> 00:00:34.137 Its curvature, and its normal eigenvalue, are zero. 00:00:35.492 --> 00:00:39.117 Now turn on the boundary coupling, called a. 00:00:39.117 --> 00:00:43.807 The right edge requires slope plus a times amplitude to equal zero. 00:00:43.807 --> 00:00:48.157 Our nonzero constant has zero slope but nonzero amplitude. 00:00:48.157 --> 00:00:53.222 Watch the residual turn red. This profile no longer satisfies the boundary 00:00:53.222 --> 00:00:54.582 condition. 00:00:55.950 --> 00:00:58.885 We keep zero slope at the left edge. 00:00:58.885 --> 00:01:02.360 The bulk equation then gives a cosine profile. 00:01:02.360 --> 00:01:06.520 Change its frequency, x, and watch the right endpoint. 00:01:06.520 --> 00:01:11.530 Most frequencies still fail. Only the frequency that balances slope against 00:01:11.530 --> 00:01:13.860 boundary amplitude is allowed. 00:01:15.242 --> 00:01:19.917 For a cosine, the endpoint slope is minus x sine x. 00:01:19.917 --> 00:01:22.877 The boundary term is a cosine x. 00:01:22.877 --> 00:01:26.867 At coupling one, these two contributions cancel. 00:01:26.867 --> 00:01:30.202 Rearranging gives x tangent x equals a. 00:01:30.202 --> 00:01:33.992 This equation comes directly from the endpoint balance. 00:01:35.367 --> 00:01:40.502 Increase the coupling. The allowed profile bends more, while the first root 00:01:40.502 --> 00:01:45.237 rises. The small graph tracks the same calculation. 00:01:45.237 --> 00:01:50.082 With interval length pi R, the scaled normal mass is x over pi. 00:01:50.082 --> 00:01:53.117 It approaches one half as the coupling grows. 00:01:54.492 --> 00:02:00.457 Now the actual laboratory. We change the boundary strength from zero to four. 00:02:00.457 --> 00:02:03.837 Watch the paired curve lift above the unpaired curve. 00:02:03.837 --> 00:02:09.642 At Wilson angle zero, its intercept is exactly the normal gap we just derived. 00:02:09.642 --> 00:02:13.397 The highlighted input and the plotted result change together. 00:02:14.783 --> 00:02:20.223 The boundary rule changed. The constant profile became inadmissible. 00:02:20.223 --> 00:02:24.073 The lowest allowed mode now has a positive eigenvalue. 00:02:24.073 --> 00:02:28.858 That is the mechanism behind this gap, in this stated interval problem. 00:02:28.858 --> 00:02:34.453 The laboratory computes the effect; the boundary argument explains why it happens. 00:02:35.825 --> 00:02:39.145 We have seen how a boundary selects a profile. 00:02:39.145 --> 00:02:44.320 Now ask a different question: how many profiles are independent? Start 00:02:44.320 --> 00:02:47.435 with three simple two component functions. 00:02:47.435 --> 00:02:52.175 The first lives in the upper component, the second in the lower component, and the third 00:02:52.175 --> 00:02:56.525 changes with position. Watch their curves being drawn. 00:02:57.908 --> 00:03:01.858 Try subtracting half of the first profile from the third. 00:03:01.858 --> 00:03:05.013 Their upper components cancel at the midpoint. 00:03:05.013 --> 00:03:09.088 But move away from that point: the residual returns. 00:03:09.088 --> 00:03:14.028 Dependence would require one fixed set of coefficients to cancel the entire function, 00:03:14.028 --> 00:03:18.988 at every position. A cancellation at one point is not enough. 00:03:20.367 --> 00:03:26.622 The Gram matrix records an overlap for each pair of profiles, integrated over the interval. 00:03:26.622 --> 00:03:31.917 Watch the entries appear. The first and third profiles overlap: their entry 00:03:31.917 --> 00:03:36.057 is one half. That does not make them dependent. 00:03:36.057 --> 00:03:40.982 The determinant is one twelfth, so this matrix is invertible and all three functions are 00:03:40.982 --> 00:03:42.442 independent. 00:03:43.825 --> 00:03:46.740 Use the same coefficients as before. 00:03:46.740 --> 00:03:50.060 The combined profile is u minus one half. 00:03:50.060 --> 00:03:56.190 Square its amplitude and integrate: the shaded area is one twelfth, not zero. 00:03:56.190 --> 00:04:01.230 This is exactly c star G c. In general it equals the squared norm of the 00:04:01.230 --> 00:04:06.405 combined profile. A zero direction of G is therefore precisely a dependence 00:04:06.405 --> 00:04:07.745 relation. 00:04:09.117 --> 00:04:13.727 Now return to the three exponential profiles used in the laboratory. 00:04:13.727 --> 00:04:18.172 Each has two complex components and depends on two coordinates. 00:04:18.172 --> 00:04:22.027 The surface shows the first component of the middle profile. 00:04:22.027 --> 00:04:24.922 At mass parameter zero it is flat. 00:04:24.922 --> 00:04:29.107 As M increases, it gains a slope in the z direction. 00:04:29.107 --> 00:04:31.817 This new spatial dependence is the key. 00:04:33.200 --> 00:04:37.955 Suppose a fixed linear combination of these three profiles vanishes. 00:04:37.955 --> 00:04:40.570 Differentiate with respect to z. 00:04:40.570 --> 00:04:46.580 Only the middle profile changes, so for nonzero M its coefficient b must vanish. 00:04:46.580 --> 00:04:51.080 The two remaining spinor directions then force a and c to vanish. 00:04:51.080 --> 00:04:56.135 At M zero, that derivative argument disappears and the three constant profiles span 00:04:56.135 --> 00:04:58.150 only two directions. 00:04:59.533 --> 00:05:04.148 Change the profile mass from zero to one in the actual laboratory. 00:05:04.148 --> 00:05:08.523 At zero, one Gram eigenvalue is numerically zero. 00:05:08.523 --> 00:05:11.428 At one, all three are positive. 00:05:11.428 --> 00:05:16.913 Their sizes depend on normalization, and the numerical rank uses a tolerance. 00:05:16.913 --> 00:05:22.198 The independent rational interval certificate also proves positive definiteness at M one, 00:05:22.198 --> 00:05:26.898 with the smallest eigenvalue greater than zero point one one zero five. 00:05:28.283 --> 00:05:31.318 There is an important limit to this conclusion. 00:05:31.318 --> 00:05:37.468 Distinct exponential rates can produce four independent functions, or any finite number. 00:05:37.468 --> 00:05:42.813 Along a suitable line, their independence follows from a Vandermonde determinant. 00:05:42.813 --> 00:05:45.988 For the four displayed rates it is seventy two. 00:05:45.988 --> 00:05:50.998 A physical family count needs the actual operator domain and boundary conditions, not 00:05:50.998 --> 00:05:53.443 just three chosen profiles. 00:05:54.825 --> 00:05:59.550 Next, representation theory becomes an exact finite sum. 00:05:59.550 --> 00:06:04.475 For the positive half spin representation of Spin twenty, each weight is a list of ten 00:06:04.475 --> 00:06:09.770 signs, divided by two. Their product must be positive. 00:06:09.770 --> 00:06:13.535 Choose nine signs freely; the tenth is fixed. 00:06:13.535 --> 00:06:18.895 Watch the allowed weights fill the grid: two to the ninth gives five hundred and twelve. 00:06:20.283 --> 00:06:25.453 At a Cartan probe, each weight contributes a number q, obtained by taking its dot 00:06:25.453 --> 00:06:30.498 product with the probe. For this example the insertion is the product of the first two 00:06:30.498 --> 00:06:35.028 signs. We sum the insertion times q squared. 00:06:35.028 --> 00:06:39.858 Set the first probe coordinates to one and two, with the rest zero. 00:06:39.858 --> 00:06:44.013 Four equal groups of weights give a total of five hundred and twelve. 00:06:45.367 --> 00:06:49.157 Expand the squared dot product before choosing a probe. 00:06:49.157 --> 00:06:53.347 There are fifty five quadratic monomials in ten variables. 00:06:53.347 --> 00:06:57.897 The checker sums each coefficient using exact rational arithmetic. 00:06:57.897 --> 00:07:03.087 Watch the cancellations: only the cross term h one times h two survives, 00:07:03.087 --> 00:07:06.447 with coefficient two hundred and fifty six. 00:07:06.447 --> 00:07:09.792 We have checked the complete quadratic polynomial. 00:07:11.158 --> 00:07:14.378 Now move the probe across that polynomial. 00:07:14.378 --> 00:07:17.858 On either coordinate axis the value is zero. 00:07:17.858 --> 00:07:21.968 Move to one, two and it becomes five hundred and twelve. 00:07:21.968 --> 00:07:28.443 The heat map makes the distinction visible: one zero evaluation cannot establish an identity. 00:07:28.443 --> 00:07:33.088 A nonzero evaluation can refute an identically zero polynomial. 00:07:34.450 --> 00:07:38.950 The laboratory keeps its displayed ten dimensional Cartan probe fixed while we change 00:07:38.950 --> 00:07:43.485 the insertion. Watch the exact trace table change. 00:07:43.485 --> 00:07:48.750 For the first insertion, the quadratic moment is five hundred and twelve, consistent 00:07:48.750 --> 00:07:53.595 with our identity. Other insertions produce different moments. 00:07:53.595 --> 00:07:58.425 These are unweighted representation traces, before the normalization and inflow needed 00:07:58.425 --> 00:08:01.140 for physical anomaly coefficients. 00:08:02.492 --> 00:08:06.687 The finite weight sum answers a precise algebraic question. 00:08:06.687 --> 00:08:11.957 Turning it into an anomaly calculation requires the appropriate projectors, normalization 00:08:11.957 --> 00:08:16.287 and descent, together with the contributions of other sectors. 00:08:16.287 --> 00:08:19.562 Global charge data and inflow also matter. 00:08:19.562 --> 00:08:23.302 The moving marker stops at what we have actually computed. 00:08:23.302 --> 00:08:27.587 A trace certificate does not certify the complete quantum theory. 00:08:28.950 --> 00:08:33.825 Our final example is a restricted six dimensional consistency condition. 00:08:33.825 --> 00:08:37.810 Let N count the signed excess of complex Weyl doublets. 00:08:37.810 --> 00:08:43.440 In the stated standalone SU two tensor setting, N must be divisible by six. 00:08:43.440 --> 00:08:48.520 With identical family assignments, three quark colours and an opposite lepton chirality 00:08:48.520 --> 00:08:52.195 contribute three minus one: two per family. 00:08:52.195 --> 00:08:55.815 Watch each family advance the residue clock by two. 00:08:57.200 --> 00:09:01.440 After three such families, the clock returns to zero. 00:09:01.440 --> 00:09:05.070 Six and nine families also pass this condition. 00:09:05.070 --> 00:09:10.475 If the lepton chirality matches instead, each identical family contributes four; 00:09:10.475 --> 00:09:14.130 again, multiples of three return to zero. 00:09:14.130 --> 00:09:17.280 The congruence restricts the possibilities. 00:09:17.280 --> 00:09:20.590 It does not uniquely select three generations. 00:09:21.950 --> 00:09:26.015 Now allow two families with different chirality assignments. 00:09:26.015 --> 00:09:29.575 The first contributes two, the second four. 00:09:29.575 --> 00:09:34.765 Together they make six, so this residue test passes with two families. 00:09:34.765 --> 00:09:39.355 This is a counterexample to dropping the identical assignment assumption. 00:09:39.355 --> 00:09:43.930 It is not a construction of a fully consistent two family physical model. 00:09:45.283 --> 00:09:48.803 Counting complex doublets gives modulus six. 00:09:48.803 --> 00:09:53.278 Counting in half doublet units doubles N and uses modulus twelve. 00:09:53.278 --> 00:09:57.413 Watch both clocks together: they pass at the same time. 00:09:57.413 --> 00:10:00.658 The units changed, not the condition. 00:10:00.658 --> 00:10:06.078 The physical argument also requires integral tensor quantization; factorizing an anomaly 00:10:06.078 --> 00:10:10.338 polynomial alone does not supply all that global information. 00:10:11.700 --> 00:10:16.655 Now change the family number in the laboratory, keeping the lepton chirality opposite 00:10:16.655 --> 00:10:22.740 to the quarks. One and two fail this restricted condition; three passes. 00:10:22.740 --> 00:10:26.360 Four and five fail; six passes again. 00:10:26.360 --> 00:10:29.745 The displayed residue explains every change. 00:10:29.745 --> 00:10:35.055 Keep the identical assignment and standalone theory assumptions attached to this result. 00:10:36.408 --> 00:10:39.353 We have followed four different questions. 00:10:39.353 --> 00:10:42.448 Boundary conditions selected a spectrum. 00:10:42.448 --> 00:10:46.163 Integrated overlaps tested function independence. 00:10:46.163 --> 00:10:48.958 Weight sums determined a polynomial. 00:10:48.958 --> 00:10:52.843 A residue condition constrained identical assignments. 00:10:52.843 --> 00:10:56.868 Each calculation has a reason, inputs and limits. 00:10:56.868 --> 00:11:01.933 Passing these separate diagnostics does not by itself determine a Higgs vacuum, neutrino 00:11:01.933 --> 00:11:05.588 masses or a complete Standard Model construction. 00:11:06.950 --> 00:11:11.950 The companion page links the laboratory, ordinary proofs, exact certificates and their 00:11:11.950 --> 00:11:17.300 checker. It also links the primary papers behind the assumptions. 00:11:17.300 --> 00:11:20.800 No proof assistant formalization is claimed. 00:11:20.800 --> 00:11:26.170 You can change the variables, reproduce the calculations, and challenge the missing physical 00:11:26.170 --> 00:11:31.830 assumptions. The purpose is to make the path between physics and mathematics visible, 00:11:31.830 --> 00:11:33.255 and checkable.