Physics theoremStatement: Literature-sourcedProof: AI-generatedPipeline-generatedprecheck passverified 2026-11-05 (gpt-6.1-sol)

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Type Iia D6 Charge Normalization and Cover Arithmetic

Physical scope

Conditional algebraic normalization and source-specific p=6 cover-charge check only. The localized current equation and the O-minus count are adopted source inputs; no worldvolume action, global tadpole theorem, or other orientifold geometry is derived.

Statement

Under the source inputs of the adopted localized-current postulate and the category convention $2 κ_{0}^{2} = (2 π)^{7} α^{' 4}$ , the equality $2 π \sqrt{α^{'}} = 2 κ_{0}^{2} μ_{6}$ is equivalent to $μ_{6} = (2 π)^{- 6} α^{' - 7 / 2}$ . In the specific covering-space convention stated there, an O6-minus charge is $- 2$ D6 units, the count is $2^{9 - 6} = 8$ , and the total is $- 16$ D6 units.

Theorem references

Type Iia D6 O6 Localized F2 Current and Cover Charge

Facts & Assumptions

Given: The source coefficient, charge ratio, and fixed-plane count adopted in the D6/O6 postulate, together with $α^{'} > 0$ and $2 κ_{0}^{2} = (2 π)^{7} α^{' 4}$ .

[A1]

Type Iia D6 O6 Localized F2 Current and Cover Charge adopts $2 π \sqrt{α'} J_{3}$ as the localized source coefficient, the $p = 6$ O-minus charge ratio, and the source-specific $2^{9 - p}$ covering-space count. These are source inputs; the proof below checks only the stated coefficient conversion and $p = 6$ arithmetic.

[A2]

Supergravity Conventions And Normalizations (current batch-56 carrier and current ready Step 1 record inspected; its stated hypotheses match this use): $2 κ_{0}^{2} = (2 π)^{7} α^{' 4}$ .

[F1]

String Tension Alpha Prime: $α^{'} > 0$ has dimension length squared.

[F2]

Graña, §4.2, Eq. (4.18): the source coefficient is $2 π \sqrt{α^{'}} J_{3}$ .

[F3]

Polchinski–Chaudhuri–Johnson, Lecture III §3.3, Eq. (3.25): $μ_{p}^{'} / μ_{p} = - 2^{p - 5}$ for the stated O-minus choice, with $2^{9 - p}$ fixed planes in the cited cover convention.

Proof

1.1givenA1A2F1F2algebra

Since $α^{'} > 0$ , [A2] gives $2 κ_{0}^{2} = (2 π)^{7} α^{' 4} > 0$ . Divide the scalar equality $2 κ_{0}^{2} μ_{6} = 2 π \sqrt{α^{'}}$ by this nonzero number to obtain $μ_{6} = (2 π)^{- 6} α^{' - 7 / 2}$ . Conversely, multiplying this value by $(2 π)^{7} α^{' 4}$ gives $2 π \sqrt{α^{'}}$ . Thus both directions hold even when $J_{3} = 0$ ; no current is cancelled. By [F1], the dimension is length $^{- 7}$ .

1.2givenF3algebra

Substituting $p = 6$ in [F3] gives $μ_{6}^{'} / μ_{6} = - 2^{6 - 5} = - 2$ . Thus each O6-minus plane in this adopted convention carries $- 2$ D6 units.

2.1givenF3step 1.2algebra∎

The cited count at $p = 6$ is $2^{9 - 6} = 2^{3} = 8$ . Adding eight identical charges from step 1.2 gives $8 (- 2) = - 16$ units in that covering-space convention.

Remarks

This verifies only coefficient algebra and the cited finite charge sum. It does not integrate the sourced Bianchi identity over a global compact cycle or infer a charge count in another quotient.

Type Iia D6 Charge Normalization and Cover Arithmetic

Statement

Theorem references

Facts & Assumptions

Proof

Remarks

Depends on

Dependency tree · 2 levels

Sources