Prove exact Pauli paid-history minimum for arbitrary targets: X/Y/Z squares need 2 returns, not 3 (no new κ)

research/pauli-history-arbitrary-targets/PROOF.md

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+# Exact paid-history minimum for arbitrary targets in the full-Pauli model
+
+This contribution was produced by Nick Petalas. Apache-2.0.
+
+## 0. Model and source
+
+The model is the one in JoyboyBrian's
+[PR #5482](https://github.com/CrocSwap/integer-mult-bounds/pull/5482)
+(`research/pauli-twisted-history-threshold/PROOF.md`, pinned in
+SOURCES.json). Let K=Q(i), X=[[0,1],[1,0]], Z=diag(1,-1), and fix a positive
+width f. Put V=(K^2)^tensor f, P=X^tensor f, C=Z^tensor f and
+R=span_K{I,P,C,PC}. A field is a row (A_1,...,A_N) in End_K(V)^N, meaning
+sum_v A_v w_v for N independent whole-field inputs w_v. Ordinary operations
+are left multiplication by elements of R, K-linear row combinations, COPY
+and CLEAR. Each paid call returns one additional row. The initial history
+is I0=R^N. After returns t_1,...,t_m every available field lies in
+
+    M = I0 + R t_1 + ... + R t_m,                          (0)
+
+because ordinary operations preserve this left R-module and each return
+adjoins one row. #5482 proves (0) and the predecessor facts used below:
+P^2=C^2=I, PC=(-1)^f CP, and dim R=4. We cite these. We do not recheck them.
+
+The new question is the predecessor's own open boundary: it "does not
+classify arbitrary Pauli-nonordinary matrices." Here the requested outputs
+are H_j e_(v_j) for q distinct owners v_j and **arbitrary** H_j in End_K(V),
+up to ordinary background. Equivalently H_j e_(v_j) lies in M.
+
+The **history minimum** m*(H_1,...,H_q) is the least m for which returns
+t_1,...,t_m exist with every H_j e_(v_j) in M. Every #5482 program satisfies
+(0), so m* is a lower bound on its paid returns. The returns achieving m*
+below are explicit rows. Whether one native paid operator applied to one
+current field can produce them is a separate question, which this package
+does not decide (Section 5).
+
+## 1. Structure of R
+
+**Even f.** P and C commute, so R is commutative, with four orthogonal
+idempotents
+
+    E_(e,h) = (I + eP)(I + hC)/4,   e,h in {+1,-1},
+
+summing to I. They satisfy E_chi r = chi(r) E_chi for the character chi of R
+with chi(P)=e and chi(C)=h. So E_chi R = K E_chi. Each E_chi has trace
+2^(f-2), because P, C and PC are traceless for f>=1.
+
+**Odd f.** PC=-CP. Put e11=(I+P)/2, e22=(I-P)/2, e12=(I+P)C/2 and
+e21=(I-P)C/2. Since C(I-P)=(I+P)C, these satisfy the matrix-unit relations
+e_ab e_cd = delta_bc e_ad. Thus R is isomorphic to M_2(K). Every left
+R-module is semisimple and isomorphic to S^d for the two-dimensional simple
+module S. As a left module R is isomorphic to S^2. On V, R acts as
+S^(2^(f-1)), so every nonzero element of R has rank at least 2^(f-1).
+
+## 2. Theorem 1: the exact minimum for arbitrary targets
+
+For a target H define its **nonordinary invariant**:
+
+* even f: the profile Pi(H) = {chi : E_chi H is not in K E_chi};
+* odd f: the excess s(H) = (dim_K(R + R H) - 4)/2, an integer in {0,1,2}.
+
+**Theorem 1.** For distinct owners and arbitrary H_j,
+
+    even f:  m* = max_chi #{ j : chi in Pi(H_j) },
+    odd f:   m* = ceil( (s(H_1) + ... + s(H_q)) / 2 ).
+
+*Lower bound, even f.* Left multiplication by E_chi is R-linear because R
+is commutative. Apply it to (0):
+
+    E_chi M = (+)_v K E_chi e_v + sum_k K E_chi t_k,
+
+so dim(E_chi M / E_chi I0) <= m. The vectors E_chi H_j e_(v_j) lie in
+E_chi M. Modulo E_chi I0, the j-th one is nonzero exactly when chi is in
+Pi(H_j). Distinct owners occupy distinct coordinates, so these vectors are
+independent. Hence m >= #{j : chi in Pi(H_j)} for every chi.
+
+*Lower bound, odd f.* M is a left R-module and each R t_k is a quotient of
+R, so dim(M / I0) <= 4m. The module W = I0 + sum_j R H_j e_(v_j) lies in M.
+By distinct coordinates, dim(W / I0) = sum_j dim((R + R H_j)/R) =
+2 sum_j s(H_j). So 4m >= 2 sum_j s(H_j).
+
+*Attainment, even f.* For each chi list the owners with chi in Pi(H_j) as
+j_(chi,1), j_(chi,2), .... For k = 1,...,mu, with mu the stated maximum, put
+
+    t_k = sum_chi E_chi H_(j_(chi,k)) e_(v_(j_(chi,k)))
+
+over the characters whose list has a k-th entry. Orthogonality gives
+E_chi t_k = E_chi H_j e_(v_j) for the k-th owner j of chi. Then
+
+    H_j e_(v_j) = sum_chi E_chi H_j e_(v_j).
+
+Each term is either E_chi t_k for some k or, when chi is not in Pi(H_j), an
+element c E_chi e_(v_j) of I0. So every target lies in M.
+
+*Attainment, odd f.* The quotient W / I0 is isomorphic to S^d with
+d = sum_j s(H_j). The space e11 (W / I0) has dimension d. Choose lifts
+y_1,...,y_d in e11 W, independent modulo I0. For each pair put
+t = y_a + e21 y_b. Then e11 t = y_a and e12 t = e11 y_b = y_b, so R t contains
+both. An unpaired y is used alone. These ceil(d/2) rows generate W modulo I0.
+
+**Consequences.** At even width the per-owner count q is forced exactly when
+one character is nonordinary for every owner. At odd width it is forced
+exactly when at most one owner has excess one and none has excess zero.
+Neither holds for arbitrary nonordinary targets. The tensor-power
+corollaries below give explicit families where they fail.
+
+The #5482 family (aI+bX)^tensor f, with ab(a^2-b^2) != 0, has excess 2 at
+odd f>=3 by its equation (2). At even f its profile contains (+1,+1) by its
+Section 3. So Theorem 1 gives m* = q there, consistent with that result.
+
+## 3. Corollary 2: every tensor square
+
+Let f=2 and H=M tensor M for M in M_2(K). For a 2x2 matrix B, let psi_B be
+the row vector with psi_B[(x,y)] = B[x,y]. A direct expansion gives
+
+    psi_B^T (M tensor N) = psi_(M^T B N)^T.
+
+The rows psi_I, psi_X, psi_Z and psi_J, with J=XZ, are left eigenvectors of
+(P,C) for the characters (+1,+1), (+1,-1), (-1,+1) and (-1,-1). Each E_chi
+has rank one, so E_chi H is in K E_chi exactly when psi_B^T H is a multiple
+of psi_B^T. Hence
+
+    Pi(M tensor M) = { B in {I,X,Z} : M^T B M is not in K B },     (1)
+
+and the singlet character never occurs, since M^T J M = det(M) J.
+
+**Lemma.** If M^T B M is in K B for two distinct B in {I,X,Z}, then M is in
+K{I,X,Z,XZ}.
+
+*Proof.* Write M=[[a,b],[c,d]]. Then
+
+    M^T M     = [[a^2+c^2, ab+cd], [ab+cd, b^2+d^2]],
+    M^T X M   = [[2ac, ad+bc], [ad+bc, 2bd]],
+    M^T Z M   = [[a^2-c^2, ab-cd], [ab-cd, b^2-d^2]].
+
+*{I,Z}:* ab+cd = ab-cd = 0, a^2+c^2 = b^2+d^2 and a^2-c^2 = d^2-b^2.
+Then ab=cd=0, b^2=c^2 and a^2=d^2. If b=0 then c=0 and M=a diag(1,+-1).
+Otherwise a=d=0 and M=b[[0,1],[+-1,0]], which lies in K{X,XZ}.
+
+*{I,X}:* ac=bd=0, ab+cd=0 and a^2+c^2=b^2+d^2. If a=0 then cd=0. If also
+c=0, then b^2+d^2=0 and bd=0 force M=0; otherwise d=0 and c^2=b^2. If a!=0,
+then c=0, b=0 and a^2=d^2. Every case lies in K{I,X,Z,XZ}.
+
+*{X,Z}:* ac=bd=0, ab=cd and a^2-c^2+b^2-d^2=0. If a=0 then cd=0. With c=0,
+b^2=d^2 and bd=0 force M=0; with d=0, b^2=c^2. If a!=0, then c=0, b=0 and
+a^2=d^2. Again M lies in K{I,X,Z,XZ}. QED.
+
+Conversely, a Pauli matrix sigma satisfies sigma^T B sigma = +-B for every
+Pauli B, so a projectively Pauli M has empty profile.
+
+**Corollary 2.** If no M_j lies in K{I,X,Z,XZ}, then every profile has two
+or three of the characters I, X and Z, and
+
+    m*( M_1^tensor 2, ..., M_q^tensor 2 ) >= ceil(2q/3).
+
+*Proof.* The profiles have total size at least 2q over three characters.
+The bound is attained. 2I+X, 2I+Y and 2I+Z have profiles {I,X}, {X,Z} and
+{I,Z}. Take owners of these three types in numbers a, b and c with
+a+b+c=q, as equal as possible. The largest character count is
+q - min(a,b,c) = ceil(2q/3). For q=3 this gives two returns for three
+owners, whereas the per-owner count is three.
+
+## 4. Corollary 3: every rank-one target at width at least three
+
+Let M = u w^T with u,w nonzero. Put U = u^tensor f, so H = U (w^tensor f)^T
+and every element of R H has rank at most one.
+
+**Tensor-power lemma.** If v_1,...,v_n in K^2 are pairwise nonproportional
+and f >= n-1, then v_1^tensor f,...,v_n^tensor f are linearly independent.
+*Proof.* Suppose sum_i c_i v_i^tensor f = 0. Pair it with g_1 tensor ... tensor g_f,
+where the g_t are linear forms. Choose forms vanishing at each v_k with
+k != i, and pad with forms nonzero at v_i. The result is c_i times a
+nonzero number, so c_i = 0.
+
+Call u a **Pauli eigenvector** if u is an eigenvector of X, Z or XZ. If u is
+not one, then u, Xu, Zu and XZu are pairwise nonproportional. For instance,
+Xu proportional to Zu would make u an eigenvector of ZX, which is
+proportional to XZ. By the lemma, RU = span{U, PU, CU, PCU} has dimension 4
+for f>=3. If u is a Pauli eigenvector sigma u = lambda u, then sigma^tensor f
+fixes the line of U. Then RU = span{U, tau U} for a second Pauli tau, of
+dimension 2 by the lemma.
+
+**Odd f>=3.** Nonzero elements of R have rank at least 2^(f-1) >= 4 > 1, so
+R intersect RH = {0} and dim(R+RH) = 4 + dim RU. Therefore
+
+    s(u w^T tensor power) = 1 if u is a Pauli eigenvector, else 2.
+
+So q rank-one owners whose u_j are Pauli eigenvectors need exactly
+ceil(q/2) returns in the history minimum. For example, two owners with
+targets E11^tensor f need one.
+
+**Even f>=4.** E_chi has rank 2^(f-2) >= 4, and E_chi H = (E_chi U) w^T has
+rank at most one. Hence E_chi H lies in K E_chi only when E_chi U = 0, and
+Pi(H) = {chi : E_chi U != 0}. If u is not a Pauli eigenvector, then
+E_chi U = (1/4) sum_r chi(r) r U is a nonzero combination of four
+independent vectors, so all four characters occur. Otherwise exactly two
+occur:
+
+| u | profile |
+|---|---|
+| X-eigenvector (1,+-1) | {(+1,+1), (+1,-1)} |
+| Z-eigenvector e_1 or e_2 | {(+1,+1), (-1,+1)} |
+| XZ-eigenvector (1,+-i), f = 0 mod 4 | {(+1,+1), (-1,-1)} |
+| XZ-eigenvector (1,+-i), f = 2 mod 4 | {(+1,-1), (-1,+1)} |
+
+*Proof for the table.* For an X-eigenvector PU = U, which kills e=-1.
+(I+hC)U is nonzero because U and CU are independent. The Z case is
+symmetric. For u=(1,+-i), PC U = lambda^f U with lambda^f = (-1)^(f/2), so
+only characters with eh = (-1)^(f/2) can survive. For such a character a
+direct expansion gives (I+eP)(I+hC)U = 2U + (h + e(+-i)^f) CU, which is
+nonzero.
+
+So when f = 0 mod 4, all Pauli-eigenvector rank-one profiles share (+1,+1),
+and q such owners need q returns. When f = 2 mod 4, the three types give
+the same ceil(2q/3) sharing as Corollary 2.
+
+## 5. Scope and limitations
+
+* m* is the exact minimum of the one-row-per-return history count of
+  #5482. The lower bound applies to every #5482 program. Attainment uses
+  explicit returned rows, and Theorem 1 does not show that a native paid
+  operator applied to one current field produces them. #5482's own upper
+  bound has the same kind of hypothesis: each requested H_j must be an
+  available paid primitive.
+* Corollaries 2 and 3 cover every tensor square and every rank-one tensor
+  power at f>=3. Invertible non-Pauli M at f>=3 is **not** classified here.
+  The verifier finds full invariants (excess 2, or all four characters) for
+  the tested invertible cases at f=3,4,5. Those are finite observations, not
+  a proof.
+* No paid-call bound for a physical caller, no recursive profile and no κ
+  change follow. Nonlinear programs, approximation, multiple outputs per
+  call and changed geometry are excluded, as in #5482.
+
+## Credit
+
+JoyboyBrian's PR #5482 supplies the model, the history-module inclusion (0),
+the algebra facts it proves, and the q-call theorem for (aI+bX)^tensor f.
+That package credits Chris R / Clear1nsight, the PR5425/PR5446 history
+authors and Ali Romman's PR5450, and those credits carry over. Theorem 1,
+the module decomposition, the width-two classification, the rank-one
+classification and the sharing constructions are new here. The algebra
+used is classical: matrix units, semisimple modules and tensor-power
+independence.

research/pauli-history-arbitrary-targets/README.md

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+# Exact full-Pauli paid-history minimum for arbitrary targets
+
+This contribution was produced by Nick Petalas. Apache-2.0.
+
+JoyboyBrian's [PR #5482](https://github.com/CrocSwap/integer-mult-bounds/pull/5482)
+proves that, in its phase-complete Pauli history model, q owners with
+targets `(aI+bX)^tensor f` need q paid returns at every width f>=2. It
+states that it "does not classify arbitrary Pauli-nonordinary matrices."
+This package answers that question for the model's history count.
+
+**Theorem 1 (scoped proof).** For arbitrary targets H_j on distinct owners,
+the least number of returned rows is exactly
+
+* even f: the largest number of owners whose target is nonordinary in one
+  common character of the commutative algebra `R=span{I,P,C,PC}`;
+* odd f: `ceil(sum_j s_j / 2)`, where `R` is isomorphic to `M_2(K)` and
+  `s_j = (dim(R+R H_j)-4)/2` is in {0,1,2}.
+
+The lower bound holds for every program in the #5482 model. Explicit
+returned rows attain it.
+
+So the per-owner count q is not a floor for general nonordinary targets:
+
+* **Every tensor square (f=2).** The profile of `M tensor M` is the set of
+  forms B in {I,X,Z} with `M^T B M` not proportional to B. For M not a
+  scalar multiple of a Pauli, it has two or three members. The minimum is
+  at least `ceil(2q/3)`, and this is sharp. Owners with targets
+  `(2I+X)^tensor 2`, `(2I+Y)^tensor 2` and `(2I+Z)^tensor 2` need **two**
+  returns, not three.
+* **Every rank-one tensor power (f>=3).** For `M=u w^T`, the odd-width excess
+  is 1 if u is an eigenvector of X, Z or XZ, and 2 otherwise. Two such
+  owners, for example `E11^tensor 3` twice, share **one** return. At even
+  widths the profile has four characters, or two for Pauli eigenvectors;
+  [PROOF.md](PROOF.md) tabulates them by f mod 4. These profiles force q
+  when f = 0 mod 4 and allow `ceil(2q/3)` when f = 2 mod 4.
+
+At even width the count q is forced exactly when one character is shared
+by every owner. At odd width it is forced exactly when at most one owner
+has excess 1 and none has excess 0. The #5482 family satisfies both
+conditions, consistent with that result.
+
+No κ, physical paid-call count or selected-frontier change is claimed.
+
+## Reproduce
+
+From the repository root, with CPython 3.12 (standard library only):
+
+```sh
+python3.12 -B scripts/research_package.py --package research/pauli-history-arbitrary-targets
+python3.12 -W error -B research/pauli-history-arbitrary-targets/verify.py
+python3.12 -O -W error -B research/pauli-history-arbitrary-targets/verify.py
+```
+
+Each verifier run takes about two seconds and prints a JSON receipt with
+`PASS_PAULI_HISTORY_ARBITRARY_TARGETS` and SHA-256 hashes of the package
+files. It checks the cited proof's hash and uses exact Q(i) arithmetic:
+
+* The algebra identities at widths 1 to 5.
+* Eleven target instances at widths 2 to 5. For each, the formula value,
+  explicit attaining returns, exact membership of every target in the
+  history module, and failure after dropping one return.
+* All 625 width-two matrices with entries in {0,±1,±i}. The character
+  profile equals the form profile (1), the singlet never occurs, and
+  non-Pauli profiles have two or three characters. Size counts are
+  17, 192 and 416 for sizes 0, 2 and 3.
+* 48 vectors u at widths 3 to 9: dim RU is 2 exactly for the 30 Pauli
+  eigenvectors.
+* The even-width rank-one profile table at widths 4, 6 and 8.
+* Five rejection controls.
+
+The finite checks corroborate the written all-width proof. They do not
+replace it.
+
+## Limitations
+
+The minimum counts returned rows in the #5482 history model. Its lower bound
+applies to every program in that model. The attaining rows are explicit, but
+this package does not show that a native operator applied to one current
+field produces them. Invertible non-Pauli tensor powers at widths at least
+three are not classified. The verifier observes full invariants there only
+for finitely many tested matrices.
+
+## Open next questions
+
+1. **Invertible targets at f>=3.** Does every invertible `M` outside
+   `K{I,X,Z,XZ}` have excess 2 at odd f>=3 and all four characters at even
+   f>=4? If so, the q-call count of #5482 extends to all invertible
+   nonordinary tensor powers. First step: the condition `r M^tensor f in R`
+   is a linear relation among the eight tensor powers of `σM` and `σ`;
+   handle the low widths 3 to 6, where the tensor-power lemma alone is too
+   weak.
+2. **Operator realizability of sharing.** Can two paid calls, each a fixed
+   linear operator applied to one current field, deliver the three width-two
+   X/Y/Z-type targets? A positive answer turns the history sharing into a
+   program. A negative answer gives a stronger operator-call floor than
+   Theorem 1.
+3. **Physical caller.** A cached physical boundary must still bind its actual
+   observation module and complete paid bill before either count is charged.
+
+## Credit
+
+JoyboyBrian (PR #5482) supplies the model, the history-module inclusion and
+the q-call theorem for `(aI+bX)^tensor f`. That package's credits carry
+over: Chris R / Clear1nsight, the PR5425/PR5446 history authors and Ali
+Romman's PR5450. Theorem 1, the two classifications and the sharing
+constructions are new here. The algebra is classical.

research/pauli-history-arbitrary-targets/SOURCES.json

@@ -0,0 +1,24 @@
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