Quantum Error Correction Via Codes Over GF(4)

• p. 15 Example after Theorem 7

  code C is an (85, 2^4) code

• p. 15 Section 4, example after Theorem 7

  The code C is an (85, 2^4) code

• p. 15 paragraph after Theorem 7

  The code C is an (85, 24) code

Judge's note: The page 15 image shows (85, 2^4). On p.14, Theorem 7 says the associated code of a linear [[n,k,d]] code is an (n, 2^{n-k}) code, and the Theorem 6 proof uses the same convention. With n-k = 85-77 = 8, C has 2^8 = 4^4 codewords. That matches C = H^⊥, the dual of the m=4 GF(4) Hamming code, described on p.20.

Second check: another model agreed

Report a problem with this finding

This finding is

wrong right unclear or overstated already corrected (erratum, later version)

I am an author of this paper Send

• p. 16 Theorem 10 statement

  Let Sm be the classical binary simplex code of length n = 2m −1, dimension m

• p. 17 Theorem 10 conclusion

  This yields a [[2^m, 2^m − m − 2, 3]] quantum code.

Only the authors can say what is right here.

Judge's note: My own working, from the p.16–17 images. For binary u, v the trace inner product of u+ωf(u) and v+ωf(v) reduces to u·f(v) + f(u)·v mod 2. For m ≥ 3 this vanishes because simplex codewords meet in 2^{m-2} positions, which is even. For m = 2, S_2 = {000,110,011,101} and the cyclic shift f is a fixed-point-free automorphism. Take u = 110, giving (1, ω̄, ω), and v = 011, giving (ω, 1, ω̄). All 3 coordinates hold distinct nonzero entries, so the inner product is 1 mod 2 and the code is not self-orthogonal. The only other fixed-point-free choice is f^2 = 1+f, which is equivalent under (9). The theorem would otherwise give [[4,0,3]], but Table III (p.33) has d = 2 at n=4, k=0. Theorem 11 states "For m ≥ 2", while Theorem 10 states no bound. The examples start at m = 3, so the missing hypothesis m ≥ 3 is probably implicit. Only the authors can confirm.

Second check: another model agreed

Report a problem with this finding

This finding is

wrong right unclear or overstated already corrected (erratum, later version)

I am an author of this paper Send

• p. 18 Theorem 11, sketch of proof, odd-m enumerator

  (2^{m+2} − 2^m)x^{n−2^m+2}y^{2^m−2} + (2^m − 1)x^{n−2^m}y^{2^m} (m odd)

• p. 17 Theorem 10 (i) weight enumerator of G_m

  x^{2^m} + 4(2^m − 1)x^{2^{m−2}}y^{3.2^{m−2}} + 3y^{2^m}

• p. 18 Theorem 11, sketch of proof, displayed weight enumerator for m odd

  x^n + (2^{m+2} − 2^m)x^{n−2^m+2}y^{2^m−2} + (2^m − 1)x^{n−2^m}y^{2^m} (m odd)

• p. 18 Theorem 11 sketch of proof

  x^n + (2^(m+2) − 2^m)x^(n−2^m+2) y^(2^m−2) + (2^m − 1)x^(n−2^m) y^(2^m)

Only the authors can say what is right here.

Judge's note: From the images: the p.18 enumerator is exactly as quoted, and Theorem 10(i) on p.17 gives x^{2^m} + 4(2^m−1)x^{2^{m−2}}y^{3·2^{m−2}} + 3y^{2^m}. At m=3, n=8, and the proof takes C_3 to be the [[8,3,3]] code, which is G_3 (unique per p.17 and note f). Theorem 10 gives 28 words of weight 6 and 3 of weight 8, but the printed formula gives 24 and 7. Using Theorem 5, the dual weight-1 coefficient is 2^{-(m+2)} Σ A_w(3n−4w). For odd m, 3n = 2^{m+2} − 8, so the weight-(2^m−2) term vanishes and the weight-2^m term has factor −8. The printed A_{2^m} = 2^m − 1 makes this coefficient negative (−1 at m=3), which is impossible, and it contradicts the closing claim that C_m^⊥ has minimal distance 3. A_{2^m} = 2^{m−1} − 1 gives 0. The existence claim of the theorem is unaffected.

Second check: another model agreed

Report a problem with this finding

This finding is

wrong right unclear or overstated already corrected (erratum, later version)

I am an author of this paper Send

• p. 23 Note after Theorem 15

  ωTr(ωu) + ω̄Tr(ωu) = u

Judge's note: The p.23 image shows Tr(ωu) in both terms, with no bar on the second ω. With Tr(x) = x + x^2 and u = ω, Tr(ω^2) = Tr(ω̄) = 1, so the printed left side is ω + ω̄ = 1, not ω. With Tr(ω̄u) = Tr(1) = 0 in the second term, the left side is ω·1 + ω̄·0 = ω.

Second check: another model agreed

Report a problem with this finding

This finding is

wrong right unclear or overstated already corrected (erratum, later version)

I am an author of this paper Send

• p. 23 Remark after Theorem 15

  ωTr(ωu) + ¯ωTr(ωu) = u

Only the authors can say what is right here.

Judge's note: The page 23 image shows plain ω inside both traces, with the bar only on the second coefficient. My own working: ω+ω̄=1, so the printed left side equals Tr(ωu), which is binary. Take u=ω: Tr(ω²)=Tr(ω̄)=ω̄+ω=1, so the printed left side is 1, not ω. With the corrected form ωTr(ωu)+ω̄Tr(ω̄u): Tr(ω̄ω)=Tr(1)=0, which gives ω·1+ω̄·0=ω=u. For u=1: ω·1+ω̄·1=1. So the identity is false as printed, while the claim that C′ determines C still holds with Tr(ω̄u) in the second term. That is consistent with C′ being defined from the pair Tr(ωu)|Tr(ω̄u).

Second check: another model agreed

Report a problem with this finding

This finding is

wrong right unclear or overstated already corrected (erratum, later version)

I am an author of this paper Send

• p. 26 Section 6, paragraph on length 24

  There exists a (24, 2^24) d = 8 classical code over GF(2), the Golay code

• p. 26 Section 6, last paragraph

  There exists a (24, 224) d = 8

• p. 26

  There exists a (24, 224) d = 8 classical code over GF(2), the Golay code

Judge's note: The page 26 image shows (24, 2^24) in the Golay sentence. The next sentence gives (24, 4^12) for GF(4) and (24, 2^24) for the additive code. A binary code with 2^24 words of length 24 is all of GF(2)^24, so d = 8 is impossible as printed. This follows from the paper's own definitions, and the parallel codes in the sentence point to exponent 12.

Second check: another model agreed

Report a problem with this finding

This finding is

wrong right unclear or overstated already corrected (erratum, later version)

I am an author of this paper Send

• p. 29 Section 7, example (iii)

  [[19, 3, 3]], [[22, 14, 4]] and [[25, 0, 10]]

• p. 34 Table III cont., n=19, k=13

  2^β

• p. 33 Table III, n=19, k=3

  5 − 6

• p. 34 Table III cont., row n=19, column k=13

  2β

Judge's note: I checked the page 29 crop and Table III on pages 33–34. Note β says these are special upper bounds from Section 7. The β entries are (7,0), (13,0), (15,4), (15,7), (16,8), (18,12), (19,13), (22,14) and (25,0), nine in all. The eliminated sets in Section 7 (i)–(iii) are [[13,0,6]], [[18,12,3]], [[7,0,4]], [[15,4,5]], [[15,7,4]], [[16,8,4]], [[19,3,3]], [[22,14,4]] and [[25,0,10]], also nine. The two lists match except that the text has (19,3) where the table has (19,13). The table entry at (19,3) is 5–6, so a [[19,3,3]] code exists. Section 9 (xi) says "all but ten" (β or γ), which is consistent with nine β entries plus one γ. The reviewer reading of [[19,8,5]] does not fit: that entry is the γ entry, not a Section 7 elimination.

Second check: another model agreed

Report a problem with this finding

This finding is

wrong right unclear or overstated already corrected (erratum, later version)

I am an author of this paper Send

• p. 41 Reference [7]

  On Clifford collineation, transform and similarity groups I, J. Australian Math. Soc., 2 (1961), 80–96.

• p. 41 Reference [6]

  groups I, J. Australian Math. Soc., 2 (1961), 60–79.

• p. 41

  groups I, J. Australian Math. Soc., 2 (1961), 80–96.

Judge's note: Page 41 crop. The two entries have identical titles but different page ranges, so they are distinct papers in a numbered series. [71] is "...similarity groups IV" by Wall, and the text cites [6], [7] and [71] together.

Second check: another model agreed

Report a problem with this finding

This finding is

wrong right unclear or overstated already corrected (erratum, later version)

I am an author of this paper Send

• p. 42 Reference [14]

  Bull. London Math. Soc., 118, pp. 97–122 (1986)

Only the authors can say what is right here.

Judge's note: Rests on outside knowledge: that the 1986 volume of Bull. London Math. Soc. is volume 18. The page 42 crop confirms that the printed volume is 118 in bold.

Second check: another model agreed

Report a problem with this finding

This finding is

wrong right unclear or overstated already corrected (erratum, later version)

I am an author of this paper Send

• p. 20 Note after Theorem 13 / Eq. (10)

  g†(x) ≡ g(x−1) (mod xn −κ)

Only the authors can say what is right here.

Judge's note: My working, from the p.20 image of (10) and the note. Modulo x^n − κ, x^{−j} = κ̄ x^{n−j}. Conjugating the coefficients and then substituting gives ḡ0 + κ̄ Σ ḡ_{n−j} x^j = κ̄·g†(x). Substituting first and then conjugating gives ḡ0 + κ Σ ḡ_{n−j} x^j, which is not equal to g† either, since g0 ≠ 0. Check with the paper's own m=3 example: κ=ω, n=21, g = x^3+x^2+x+ω, so g† = 1 + x^18 + x^19 + x^20, while ḡ(x^{-1}) ≡ ω̄·g† ≠ g†. Only divisibility matters in Theorem 13 and in (11), so the stated results stand; the side remark is off by a unit.

Second check: another model agreed

Report a problem with this finding

This finding is

wrong right unclear or overstated already corrected (erratum, later version)

I am an author of this paper Send