arXiv:hep-th/9304011 v1 Apr 5 1993

Semiclassical Seiberg Bound [7,8]
The semiclassical approach provides a key insight into the Seiberg bound (3.40). Consider
the classical equation (3.42) in the neighborhood of a vertex operator insertion. If
we neglect the cosmological constant term, the solution must behave as in (3.45). To check
if this is self-consistent, we insert (3.45) back into (3.42) and note that the neglected term
behaves as
e γφ ∼
1
|z − z i | 2α iγ . (3.53)
If α i γ > 1, the cosmological constant operator is not integrable at z = 0 and we expect
trouble.
Indeed, the careful considerations leading to the classification of solutions in
sec. 3.2 (following eq. (3.16)) show that there is no solution for α i > 1/γ ∼ Q/2. The
essential point is that too much curvature cannot be localized at a single point.
Here are two examples of semiclassical correlators:
Example 1: Consider the three-point function on the sphere,
〈
e
θ 1 φ/γ (z 1 , ¯z 1 ) e θ 2φ/γ (z 2 , ¯z 2 ) e θ 3φ/γ (z 3 , ¯z 3 ) 〉 , (3.54)
where θ i < 1 are considered to be O(1) as γ → 0 and ∑ i θ i > 2, so s < 0. The classical
solution is known in this case and is Möbius invariant. It follows immediately from (3.47)
and the transformation properties of circles under Möbius transformations that
〈
e
θ 1 φ/γ (z 1 , ¯z 1 ) e θ2φ/γ (z 2 , ¯z 2 ) e θ3φ/γ (z 3 , ¯z 3 ) 〉 C[θ i ]
∼ ∣
∣z ∆ 123
12
z ∆ 132
13
z ∆ 231
23
∣ 2 , (3.55)
where ∆ 123 = ∆ 1 + ∆ 2 − ∆ 3 , etc. The coefficient function C is generically nonzero. Sadly,
this example cannot be extended to higher point functions because the classical solutions
to Liouville theory are not known in explicit form, except in special cases which have the
punctures symmetrically located [43].
Example 2: Consider now the three-point function on the sphere, but with s ≥ 0 so we
must fix the area. Using the Möbius invariance of (3.52) gives
C[α i ] =
〈
e
α 1 φ (z 1 , ¯z 1 ) e α2φ (z 2 , ¯z 2 ) e α3φ (z 3 , ¯z 3 ) 〉 s.c.
∼ C[α i ]
A=1
|z ∆ 123
∫ ∞
∫
dλ
0 λ λ4(α 1+α 2 −α 3 )/γ
1
C d2 w (
|w| 2 + 1 ) 2α 1 /γ
= π 4
Γ(j 1 − j 2 − j 3 ) Γ(j 2 − j 3 − j 1 ) Γ(j 3 − j 1 − j 2 )
Γ(−2j 1 ) Γ(−2j 2 ) Γ(−2j 3 )
12
z ∆ 132
13
z ∆ 231
23
| 2
1
(
|w + λ 2 | 2 + 1 ) 2α 2 /γ
Γ(−j 1 − j 2 − j 3 − 1) ,
(3.56)
37