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Please consider the following integrality theorem for differentiable manifolds due to K H Mayer:

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I am trying to prove this theorem using Heterotic Super-symmetric Quantum Mechanics described by a Lagrangian density with the form

$$L={\phi}^{T}Q\phi+{\theta}^{T}P\theta$$

where $\phi$ describes bosonic degrees of freedom with an effective propagator denoted $Q$ and $\theta$ describes fermionic degrees of freedom with an effective propagator denoted $P$. The Witten index for this heterotic Susy QM is given by:

$${\it index}=\int \!\!\!\int \!{{\rm e}^{-{\phi}^{T}Q\phi-{\theta}^{T}P \theta}}{d\theta}\,{d\phi}={\it integer} $$

Computing the path integrals we obtain:

$$\int \!{{\rm e}^{-{\theta}^{T}P\theta}}{d\theta}=\sqrt {{\it Det} \left( P \right) }=\sqrt {\prod _{i=1}^{s} \left( 4\,\prod _{n=0}^{ \infty } \left( 1+{\frac {{y_{{i}}}^{2}}{ \left( 2\,n+1 \right) ^{2}{ \pi }^{2}}} \right) ^{2} \right) }={2}^{s}\prod _{i=1}^{s}\cosh \left( \frac{y_{{i}}}{2} \right) $$

$$\int \!{{\rm e}^{-{\phi}^{T}Q\phi}}{d\phi}={\frac {1}{\sqrt {{\it Det} \left( Q \right) }}}={\frac {1}{\sqrt {\prod _ j \left( \prod _{n=1}^{\infty }(1+{\frac {{x_{{j}}}^{2}}{4{\pi }^{2}{n}^{2}} } )\right) ^{2} }}}=\prod _ j{\frac {\frac{x_{{j}}}{2}}{\sinh \left( \frac{x_{{j}}}{2} \right) }} = \hat{A}(M) $$

Then we have:

$$index=\int \!{{\rm e}^{-{\phi}^{T}Q\phi}}{d\phi}\int \!{{\rm e}^{-{\theta}^{ T}P\theta}}{d\theta}=\int \!{\frac {\sqrt {{\it Det} \left( P \right) }}{\sqrt {{\it Det} \left( Q \right) }}}{dM}=\int \hat{A} \left( M \right) {2}^{s}\prod _{i=1}^{s}\cosh \left( \frac{y_{{i}}}{2} \right) {dM} = integer $$

Then my questions are:

  1. Is this heterotic susy proof correct?.

  2. This Mayer theorem has applications to the problem of anomaly for the fivebrane in 11-dimensional M-Theory?

  3. This Mayer Theorem has applications to the problem of anomaly for the sevenbrane in 12-dimensional F-Theory?

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closed as off-topic by ACuriousMind, Brandon Enright, Kyle Kanos, JamalS, Jim Nov 13 '14 at 15:13

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    $\begingroup$ This question appears to be off-topic because it is a check-my-work question. $\endgroup$ – ACuriousMind Nov 13 '14 at 0:40
  • $\begingroup$ An update was made and new questions were formulated $\endgroup$ – Juan Ospina Nov 13 '14 at 13:48