**Next message:**Vladyslav Shtabovenko: "Re: Imaginary parts and Schouten identity"**Previous message:**Vladyslav Shtabovenko: "Re: Imaginary parts and Schouten identity"**Maybe in reply to:**Nikita Belyaev: "Imaginary parts and Schouten identity"**Next in thread:**Vladyslav Shtabovenko: "Re: Imaginary parts and Schouten identity"**Reply:**Vladyslav Shtabovenko: "Re: Imaginary parts and Schouten identity"**Messages sorted by:**[ date ] [ thread ] [ subject ] [ author ]**Mail actions:**[ respond to this message ] [ mail a new topic ]

Hi Vladyslav ,

*>Could you may be provide me a reference for the occurence of your
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*>formula (obviously I'm not familiar with it)? I want to understand how the general proof goes and what are the steps that might cause problems with computer codes.
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*>
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*>Does this relation hold only in 4 or also in D-dimensions (if yes, in
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*>what scheme, naive or t'Hooft Veltman?). If it is a purely 4D equality then I suppose that the proof might involve SPVAT (scalar,pseudoscalar,vector,axial vector, tensor) decomposition of A and B. This might be a good starting point to investigate things.
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The proof of this formula is much simpler. All you have to do is to use the well-known formula for three Dirac matrices:

GA[a].GA[b].GA[c]=MT[a, b].GA[c]+MT[b, c].GA[a]-MT[a, c].GA[b]-iGA[5].GA[d].Eps[d, a, b, c];

Just substitute it into both traces, multiply them, get 16 terms and simplify all the things. You will get required answer which don't know anything about the structure of A and B. So it is valid for any A and B.

According to D-dimensions - I'm not sure here. Maybe we can generalize this formula to D dimensions, but we did not tried to do that.

Best Regards,

Nikita Belyaev.

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