Gravity and Strings

The harmonic operator on R 3 × S 1 649
In this form it is evident that we have a function with the right periodicity and one can
also immediately check that it is a solution of the Laplace equation. On the other hand, near
the singularity x3 →x3(0), z → z(0), orinthe equivalent limit ℓ → 0inwhich the periodicity
of the fourth coordinate is irrelevant, H R 3 ×S 1 becomes exactly H R 4 plus subdominant
terms.
Now we want to expand in Fourier series the periodic harmonic function H R 3 ×S 1:
The Fourier modes are
H R 3 ×S 1 =
HR3 ×S1 ,n(x3 −x3(0))e
n∈Z
inz
ℓ . (G.6)
H R 3 ×S 1 ,n(x3 −x3(0)) = δn,0 + h/(2ℓ)
|x3 −x3(0)| e− |n|(|x 3 −x 3(0) |−inz (0) )
ℓ . (G.7)
If we consider only the zero mode, we find
H R 3 ×S 1 ,0 = 1 + h/(2ℓ)
which is a harmonic function on R 3 , satisfying
|x3 −x3(0)|
, (G.8)
(3)H R 3 ×S 1 ,0 =− 2πh
ℓ δ(3) (x3 −x3(0)). (G.9)
Using
1
(3)
|x3 −x3(0)| =−4πδ(3) (x3 −x3(0)), (G.10)
it is easy to see that the higher (KK) modes satisfy the massive three-dimensional Laplace
equation
(3) − |n|2
ℓ 2
H R 3 ×S 1 ,n =− 2πh
ℓ δ(3) (x3 −x3(0))e inz (0)
ℓ . (G.11)