PHYS08200604018 Shamik Banerjee - Homi Bhabha National ...

90CHAPTER 5. PARTITION FUNCTIONS OF TORSION > 1 DYONS IN HETEROTIC STRIN
Thus we have from (5.0.17), (5.0.26)
j = α − δ, m 1 = −sγ = −γ ′ s/K, n 1 = −β/s = −Kβ ′ /s . (5.0.28)
Since gcd(γ ′ , K) = 1, the second equation in (5.0.28) shows that s must be a multiple of K:
s = K ˜s, ˜s ∈ Z . (5.0.29)
Substituting this into the last equation in (5.0.28) and using (5.0.19) we see that
n 1 = a′ b ′
˜s
⇒ a′ b ′
Since gcd(a ′ , b ′ )=1, we must have a unique decomposition
˜s ∈ Z . (5.0.30)
˜s = L 1 L 2 , L 1 |b ′ , L 2 |a ′ . (5.0.31)
On the other hand since s divides r, it follows from (5.0.29) that ˜s must divide r/K = ¯K.
Thus
L 1 | ¯K, L 2 | ¯K . (5.0.32)
It now follows from (5.0.22) that
L 1 |N 1 , L 2 |N 2 . (5.0.33)
Conversely, given any pair (L 1 , L 2 ) satisfying (5.0.33), it follows from (5.0.22) that L 1 , L 2
will satisfy (5.0.31), (5.0.32). This allows us to find integers m 1 , n 1 , j satisfying (5.0.26) via
eqs.(5.0.27)-(5.0.31).
This shows that the poles of ̂Φ(ˇρ, ˇσ, ˇv) −1 at (5.0.24) can come from the s = KL 1 L 2 terms
in (5.0.4) for L 1 |N 1 and L 2 |N 2 . Our next task is to find the residues at these poles to compute
the change in the index as we cross this wall. For this we define
ā = a ′ /L 2 , ¯D = d ′ L 2 , ¯b = b ′ /L 1 , ¯c = c ′ L 1 , s 0 = KL 1 L 2 . (5.0.34)
It follows from (5.0.31) that ā, ¯b, ¯c, ¯D are all integers. In terms of these variables the location
of the pole given in (5.0.24) can be expressed as
We now define
s −1
0 ¯c ¯Dˇρ + ā¯bs 0ˇσ + (ā ¯D + ¯b¯c)ˇv = 0 . (5.0.35)
ˇρ ′ = ¯D 2 ˇρ + ¯b 2 s 2 0ˇσ + 2¯b ¯Ds 0ˇv, ˇσ ′ = ¯c 2 ˇρ + ā 2 s 2 0ˇσ + 2ā¯cs 0ˇv, ˇv ′ = ¯c ¯Dˇρ + ā¯bs 2 0ˇσ + (ā ¯D + ¯b¯c)s 0ˇv .
(5.0.36)
The change of variables from (ˇρ, s 2 0ˇσ, s 0ˇv) to (ˇρ ′ , ˇσ ′ , ˇv ′ ) is an Sp(2, Z) transformation. Thus
we have
Φ 10 (ˇρ, s 2 0ˇσ, s 0ˇv) = Φ 10 (ˇρ ′ , ˇσ ′ , ˇv ′ ) . (5.0.37)