On the Refined Index of BPS states

[kdv]

I am trying to reproduce a very simple result that involved evaluating the refined index for a vector supermultiplet but ran into a snag. I hope someone will be able to clear that up easily.

Apparently for a half vector multiplet (meaning that we do not include the CPT conjugated states), the result is (see 0912.1346, equations 3.1 to 3.3):

$$ Tr_{H'} (-y)^{2J_3'} = - y - y^{-1} $$

Here the primes indicate that we must remove from the state a half hypermultiplet, in other words
$$ H = H' \otimes \biggl( [\frac{1}{2}] + 2 [0] \biggr) $$

Now, unless I am mistaken, a half hypermultiplet has two states of helicity 1/2 and two of helicity zero.
And a half vector multiplet has one h=1, two h=1/2 and 1 h=0.

My questions are:

1) First they say that all multiplets contain at least a half hypermultiplet...but I don't see this in the vector multiplet since it has fewer h=0 states than a hypermultiplet.

2) How do they get the trace to be $$ -y-y^{-1}??$$ Clearly they consider only the values $$J_3'=\pm 1/2$$ but why ? My guess was that by J_3' they mean half the helicity, and if I consider the two helicities of a spin 1 particle, I would then get their result , but this leads to several questions:

I) Why would J_3' mean half the helicity?

II) Why would we consider the helicities + 1 and -1 if we are talking about half a vector multiplet? We should not be including the CPT conjugate, I thought?!

III) How does one get rid of the spin 1/2 and spin zero contributions, exactly? It seems to work for the spin 1/2 (vector and hypermultiplets have the same number of h=1/2 components) but it does not work for the h=0 states.

Thanks in advance

 

[kdv]

I am trying to reproduce a very simple result that involved evaluating the refined index for a vector supermultiplet but ran into a snag. I hope someone will be able to clear that up easily.

Apparently for a half vector multiplet (meaning that we do not include the CPT conjugated states), the result is (see 0912.1346, equations 3.1 to 3.3):

$$ Tr_{H'} (-y)^{2J_3'} = - y - y^{-1} $$

Here the primes indicate that we must remove from the state a half hypermultiplet, in other words
$$ H = H' \otimes \biggl( [\frac{1}{2}] + 2 [0] \biggr) $$

Now, unless I am mistaken, a half hypermultiplet has two states of helicity 1/2 and two of helicity zero.
And a half vector multiplet has one h=1, two h=1/2 and 1 h=0.

My questions are:

1) First they say that all multiplets contain at least a half hypermultiplet...but I don't see this in the vector multiplet since it has fewer h=0 states than a hypermultiplet.

2) How do they get the trace to be $$ -y-y^{-1}??$$ Clearly they consider only the values $$J_3'=\pm 1/2$$ but why ? My guess was that by J_3' they mean half the helicity, and if I consider the two helicities of a spin 1 particle, I would then get their result , but this leads to several questions:

I) Why would J_3' mean half the helicity?

II) Why would we consider the helicities + 1 and -1 if we are talking about half a vector multiplet? We should not be including the CPT conjugate, I thought?!

III) How does one get rid of the spin 1/2 and spin zero contributions, exactly? It seems to work for the spin 1/2 (vector and hypermultiplets have the same number of h=1/2 components) but it does not work for the h=0 states.

Thanks in advance

I have my questions. To get the vector multiplet, we must take H' to be [1/2]. Indeed,
$$ [\frac{1}{2}] \otimes \biggl( [\frac{1}{2}] + 2 [0] \biggr) = [1] + [0] + 2 [\frac{1}{2}] $$ which is the particle content of a vector supermultiplet. It's that simple.