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Physics theoremStatement: Literature-sourcedProof: AI-generatedPipeline-generatedprecheck passverified 2026-11-05 (gpt-6.1-sol)
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Wrapped M2 and M5 tensions match the IIA string and D4 tensions

Physical scope

Conditional leading-tension identities for one M2 or M5 wrapped once on the same flat circle, with negligible backreaction; no full M/IIA equivalence or compactification solution is proved.

Statement

Assume the low-energy circle dictionary R11=gsℓs,ℓp3=gsℓs3,ℓs2=α′, the adopted M2/M5 tensions, and the leading constant-dilaton tensions TF=12πα′,TD4=1(2π)4gsℓs5. For a single M2 or M5 wrapped once on the same flat circle of radius R11>0, the wrapped tensions obey 2πR11TM2=TF,2πR11TM5=TD4. These are conditional parameter identities for the leading tensions. They correspond to double dimensional reduction but do not prove full M/IIA equivalence.

Facts & Assumptions

Given: gs>0, ℓs>0, α′=ℓs2, one flat circle of circumference 2πR11, and the exact planned/current supplier interfaces below.

[F1]

The current circle-dictionary carrier adopts R11=gsℓs, ℓp3=gsℓs3, and ℓs=α′, with a flat-circle sector identification and an explicit non-equivalence caveat (M Theory Type IIA Circle Limit Dictionary).

[F2]

The adopted normalization postulate fixes TM2=(2π)−2ℓp−3 and TM5=(2π)−5ℓp−6 (M2 M5 tension normalizations).

[F3]

The current Dp action states μp=((2π)pα′(p+1)/2)−1 and Tp=μp/gs at constant dilaton; this proof uses only its DBI tension clause, not its Wess–Zumino coupling or its explicit local RR-potential crosswalk (Type II Dp Brane DBI WZ Action).

[F4]

The string-tension definition gives ℓs2=α′ and TF=1/(2πα′) (String Tension Alpha Prime).

[A1]

Both wrapped objects use the leading tensions, wrap the same flat circle exactly once, and have negligible backreaction.

Proof

Given: [F1]–[F4], gs>0, ℓs>0, and [A1].

1.1givenF1F2F4algebra

Using R11=gsℓs, ℓp3=gsℓs3, and [F2], 2πR11TM2=(2πgsℓs)(2π)−2(gsℓs3)−1=(2π)−1ℓs−2=12πα′=TF. The positive gs and ℓs hypotheses justify the cancellations and inverses.

1.2givenF1F2F3algebra

Since ℓp6=(ℓp3)2=gs2ℓs6, the M5 product is 2πR11TM5=(2πgsℓs)(2π)−5(gs2ℓs6)−1=(2π)−4gs−1ℓs−5. By [F3] with p=4 and α′=ℓs2, TD4=(2π)−4gs−1ℓs−5. Hence the second equality holds.

2.1A1F3step 1.1step 1.2∎

The two equalities are therefore established under the stated source-reported dictionary and leading-tension assumptions; no RR Wess–Zumino term or full compactification equivalence was used.

Remarks

The Dp item has a current escalation for its RR convention, but its own statement separately authorizes the DBI tension clause used here; this proof does not use its RR Wess–Zumino conversion. The result remains a conditional leading-tension identity, not a proof of full M/IIA equivalence.

Depends on

Dependency tree · 2 levels

Sources