TFPT scientific source excerpt Original repository path: experiments/theory-contracts/elliptic-moebius-source-bridge-20261026/EQUATORIAL-LIFTS.txt Original file SHA-256: 434eaab2fa67fbf157967a78f4b3f742cb7731d61c9632cdb06f12773c0ba1dd Inclusive original lines: 191-380 Line numbers below refer to the original source. This bounded excerpt is source evidence; it is not a new proof/checker run. Conversation metadata and personal execution details are omitted where indicated. 000191 | documents/tfpt_research_contracts/sections/013-source.tex:139-250 000192 | SHA256 96f670d76ad94d57c496e386fb61cddfa428795ae302024f93933c4d9b0e6dd7. 000193 | 000194 | This corrects the phrase "Original OS seam is x -> 1/bar x". The unit circle 000195 | is the pointwise fixed Euclidean seam. The OS operation along it is family D. 000196 | 000197 | For the two old dividing forms the base actions are 000198 | 000199 | hQkappa: x -> i bar x, 000200 | hQ^3kappa: x -> -i bar x. (7) 000201 | 000202 | These are exactly the two v519 silver-midpoint axes. They swap all four 000203 | marks and their two base cut points lie between marks. Their elliptic fixed 000204 | ovals contain no ramification mark. This is the right slot for the earlier 000205 | OS and Green-kernel comparison. 000206 | 000207 | 1.5 v521 and v525 do not add the missing curve lift 000208 | --------------------------------------------------- 000209 | 000210 | v521 extends the family-D census through the deformed mark family and proves 000211 | that free RP sees the family but does not select the side. Its exact census 000212 | and verdict are at 000213 | 000214 | verification/v521_seam_bit_rp_blind.py:405-480,973-1013 000215 | SHA256 8e46478bfeb77d2052d31901237a2245a7267248205bf584e5357fe6c328b516, 000216 | 000217 | with paper record 000218 | 000219 | paper-source:584d6e2ab5a25deffb5645c2 000220 | documents/changelog/sections/004-source.tex:1-31 000221 | SHA256 f1b117a014f2b1397182f975c03881140094d6d9e566fc331696c8b1b6a450e5. 000222 | 000223 | v525 proves that mark-decorated free states remain side-blind. Its verdict is 000224 | at 000225 | 000226 | verification/v525_seam_bit_twist_blind.py:1486-1528 000227 | SHA256 9b31d49fafb6567f231787df04eb95a7637afc1019e415fd7f90286249ae942f, 000228 | 000229 | with paper record 000230 | 000231 | paper-source:da2365a9cc38790f5b167324 000232 | documents/changelog/sections/003-source.tex:272-293 000233 | SHA256 13da54d7b7a65f7aa2dcd92620b4277a0bde6fee8ff8c105857e2e2882ab28b3. 000234 | 000235 | Neither module constructs (1), the pullback graph (5), or a graph boundary 000236 | domain. Their negative selector results remain unchanged. 000237 | 000238 | 2. Exact lift and group calculation 000239 | ==================================== 000240 | 000241 | 2.1 The holomorphic mirror lift 000242 | -------------------------------- 000243 | 000244 | For S in (2), 000245 | 000246 | (i y/x^2)^2=(1/x)^4-1 000247 | 000248 | after using y^2=x^4-1. Hence S preserves E. A second application gives 000249 | 000250 | S^2(x,y)=(x,-y)=h(x,y). (8) 000251 | 000252 | Thus S has order four, while its base map x -> 1/x has order two. The two 000253 | possible lifts of that base map are S and hS. 000254 | 000255 | Together with h and Q, 000256 | 000257 | h^2=1, Q^4=1, S^2=h, 000258 | h central, 000259 | SQS^-1=hQ^-1. (9) 000260 | 000261 | The holomorphic marked lift group therefore has sixteen elements 000262 | 000263 | h^a Q^b S^c, a,c in Z2, b in Z4. (10) 000264 | 000265 | 2.2 The equatorial anti-holomorphic lifts 000266 | ------------------------------------------ 000267 | 000268 | For x'=1/bar x, the conjugate curve equation gives 000269 | 000270 | x'^4-1=-(bar y)^2/(bar x)^4. 000271 | 000272 | Therefore y' must equal either sign of i bar y/bar x^2. This proves both 000273 | existence and completeness of (1). Direct composition gives 000274 | 000275 | j_+^2=j_-^2=1, Qj_+Q^-1=j_-. (11) 000276 | 000277 | Their fixed sets can be found without uniformization. Put 000278 | 000279 | x=e^(i theta), y=e^(i theta) w. 000280 | 000281 | Then the curve gives 000282 | 000283 | w^2=2i sin(2 theta). (12) 000284 | 000285 | For j_+, fixedness says w=i bar w, hence sin(2 theta)>=0. The two intervals 000286 | [0,pi/2] and [pi,3pi/2], with the two sheets joined at the branch endpoints, 000287 | give two circles. For j_-, fixedness says w=-i bar w, hence 000288 | sin(2 theta)<=0, and the complementary two intervals give two circles. 000289 | Both Real curves are therefore dividing and their quotients are annuli. 000290 | 000291 | 2.3 Full anti-holomorphic lift census 000292 | -------------------------------------- 000293 | 000294 | Conjugation satisfies 000295 | 000296 | kappa Q kappa=Q^-1, kappa S kappa=S^-1. (13) 000297 | 000298 | The anti-holomorphic coset of the full marked lift group has sixteen maps. 000299 | The first eight are the old family-D lifts 000300 | 000301 | d_(a,b)=h^a Q^b kappa. 000302 | 000303 | All eight square to one. Their fixed-component census remains 000304 | 000305 | (1,0,1,0; 1,2,1,2), (14) 000306 | 000307 | with the semicolon separating a=0 and a=1. 000308 | 000309 | The other eight are the family-A lifts 000310 | 000311 | a_(a,b)=h^a Q^b S kappa. 000312 | 000313 | Their exact squares are 000314 | 000315 | a_(a,b)^2=h^b Q^(2b). (15) 000316 | 000317 | Consequently b=0,2 gives four further Real involutions. The two b=0 maps 000318 | are j_+ and j_- and each has two ovals. The two b=2 maps project to 000319 | x -> -1/bar x and are fixed-point free. For b=1,3 the square is hQ^2, 000320 | so the four maps have order four and no fixed points. This proves the count 000321 | 000322 | 12 anti-holomorphic involutions + 4 order-four maps. (16) 000323 | 000324 | It also reproduces v519's family-A square dichotomy on the actual curve: 000325 | mu=+1 is equatorial, mu=-1 is fixed-point free, and mu=+-i is order four. 000326 | 000327 | 3. The full equatorial pullback is a graph 000328 | =========================================== 000329 | 000330 | This is the smallest decisive source test. 000331 | 000332 | Away from a branch mark, a point over |x|=1 satisfies exactly one of the two 000333 | fixed equations in (1), according to the sign of sin(2 theta). At each of 000334 | the four marks y=0, it satisfies both. Therefore (5) holds exactly. 000335 | 000336 | In the square uniformization E=C/(Z+iZ), use O=(0,i), T=(1+i)/2 and the 000337 | normalization already used by the C2 calculation: 000338 | 000339 | Qz=iz, 000340 | hz=-z+T, 000341 | kappaz=-i bar z+T. 000342 | 000343 | For the lift S in (2), S*z on the holomorphic differential is i and the 000344 | corresponding affine map is 000345 | 000346 | Sz=iz+1/2. (17) 000347 | 000348 | Hence 000349 | 000350 | j_+z=bar z+i/2, 000351 | j_-z=-bar z+1/2. (18) 000352 | 000353 | The fixed sets are 000354 | 000355 | Fix(j_+): Im z=1/4 or 3/4, 000356 | Fix(j_-): Re z=1/4 or 3/4. (19) 000357 | 000358 | Every horizontal circle meets every vertical circle once. The four 000359 | intersection points solve hz=z and are exactly the four branch marks. Each 000360 | of the four circles is divided into two edges. Thus 000361 | 000362 | V=4, E=8, connected, b1=E-V+1=5. (20) 000363 | 000364 | The four rectangular complementary regions are disks. Equivalently, 000365 | Euler's formula on the torus gives F=E-V=4. 000366 | 000367 | This makes the source issue visible. A fixed oval of j_+ or j_- is smooth, 000368 | but its projection folds two-to-one over one alternating quarter arc. The 000369 | whole original equatorial seam requires all four ovals and the four branch 000370 | vertices. It is not the boundary of either annulus E/j_+ or E/j_-. 000371 | 000372 | A small radial push-off |x|=r with r<1 or r>1 avoids the branch points. Its 000373 | inverse image consists of two unbranched circles because it encloses zero or 000374 | four branch points, hence has even covering monodromy. The original source 000375 | does not choose such a push-off, its side, or the transport back to the 000376 | marked seam. Using it as the physical carrier would therefore be an extra 000377 | source map, not a consequence of (1). 000378 | 000379 | 4. Spin compatibility and its exact limit 000380 | ==========================================