I am looking at the wall-crossing formula for N=2 D=4 supersymmetry The Donaldson-Thomas "invariant" is supposedly given by(adsbygoogle = window.adsbygoogle || []).push({});

[tex] \Omega = -\frac{1}{2} Tr (-1)^{(2J_3)} (2J_3)^2 [/tex]

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I don't understand how to use this formula. (By the way, the normalization is probably not the same for everyone so you may have seen something slightly different). Then my reference says that for the W bosons with charge [itex] (0,\pm 2)[/itex], we have [itex] \Omega =-2 [/itex] in this normalization. This works if we set [itex]J_3=1[/itex] in the above formula. So I thought that maybe [itex]J_3[/itex] referred to the spin or helicity.

But then they say that for the dyons with charge [itex] (\mp 1, \pm 2n )[/itex], we have [itex]\Omega= 1 [/itex]. Now I don't see how to get this from the formula. I mean, I see that if I consider that there are two states with [itex] J_3=1/2[/itex] I get the answer but I don't understand why this should be correct.

Can someone help?

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# A Question about Wall crossing formula

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