Question about Wall crossing formula

[kdv]

I am looking at the wall-crossing formula for N=2 D=4 supersymmetry The Donaldson-Thomas "invariant" is supposedly given by

$$\Omega = -\frac{1}{2} Tr (-1)^{(2J_3)} (2J_3)^2$$
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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 $(0,\pm 2)$, we have $\Omega =-2$ in this normalization. This works if we set $J_3=1$ in the above formula. So I thought that maybe $J_3$ referred to the spin or helicity.

But then they say that for the dyons with charge $(\mp 1, \pm 2n )$, we have $\Omega= 1$. 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 $J_3=1/2$ I get the answer but I don't understand why this should be correct.

Can someone help?

 

[AndyW]

Hello,

Thank you for your question. Let me first clarify the notation used in the formula. The J_3 in the formula represents the third component of the spin for a given particle. This means that for a W boson with charge (0,\pm 2), the spin can take on two values: J_3=1 or J_3=-1. Plugging in these values into the formula, we get:

\Omega = -\frac{1}{2} Tr (-1)^{(2J_3)} (2J_3)^2 = -\frac{1}{2} Tr (-1)^{2} (2)^2 = -2.

This matches the value given for the W bosons with charge (0,\pm 2).

Now, for the dyons with charge (\mp 1, \pm 2n), the spin can take on four values: J_3=1/2, J_3=-1/2, J_3=3/2, J_3=-3/2. Plugging these values into the formula, we get:

\Omega = -\frac{1}{2} Tr (-1)^{(2J_3)} (2J_3)^2 = -\frac{1}{2} Tr (-1)^{1} (1)^2 + (-1)^{-1} (-1)^2 + (-1)^{3} (3)^2 + (-1)^{-3} (-3)^2 = 1.

This matches the value given for the dyons with charge (\mp 1, \pm 2n). The reason why we consider two states with J_3=1/2 is because the dyons have two possible spin states: J_3=1/2 and J_3=-1/2. By taking the sum of these two states, we get the correct value of \Omega=1.

I hope this helps clarify the use of the formula. Let me know if you have any further questions.